Solutions Manualfor Problems in Appendix A of SOLAR ENGINEERING OF THERMAL PROCESSES Third Edition J. A. Duffie (Deceased) and W. A. Beckman * John Wiley & Sons, Inc., New York (2006) SOLUTIONS MANUAL for SOLAR ENGINEERING OF THERMAL PROCESSES Third edition CONTENTS Introduction iii Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 14 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 1 24 74 96 113 139 161 176 185 191 213 240 242 249 251 265 286 297 306 320 Course Material 326 ii SOLUTIONS MANUAL for SOLAR ENGINEERING OF THERMAL PROCESSES, 3rd Ed. John A. Duffie (deceased) and William A. Beckman This manual includes solutions to the problems in Appendix A of the third edition of Solar Engineering of Thermal Processes, published by John Wiley & Sons, New York (2006). These solutions have been done with the program EES, Engineering Equation Solver, (available from F-Chart Software, on the web at fchart.com). The equations, tables, solutions, and comments have been translated into a word processor and edited for inclusion in the manual. The equation statements and functions are to a degree the result of the programming style used in EES, but they should be no mystery to those who have done any programming. The EES functions (called the SETP library) useful for solving solar energy problems are freely available at fchart.com under EES "add-ons" The nomenclature in this manual is close to that in the book. Where differences have been required or are convenient, obvious variable names have been used, such as Dec for declination but sometimes δ. There are two exceptions in the matter of style. First, EES allows subscripts in variable names but they show as subscripts only in the Formatted Equations Window and are indicated as in the following examples: I_T is substituted for IT, R_b for Rb, I_o for Io, and A_c for Ac, etc. Second, variables that are ratios can not be shown with a slash "/" as it denotes a division operation; the reverse slash is used in variable names that are ratios, such as: FR'\FR for FR'/FR, and Id\I for the diffuse fraction Id/I. The equation, table, and figure numbers are those in the book. Thermophysical property data are usually obtained from the EES built-in functions in but sometimes are from Appendix E. Monthly average meteorological data are from Appendix G or from various web sites. The EES program allows preprogrammed routines that begin with the word FUNCTION, PROCEDURE, SUBPROGRAM or MODULE and are terminated with the word END. Routines that are not part of the SETP library must always appear at the beginning of the iii program. The use of preprogrammed routines is a personal choice; the equations can be typed in the solutions or the routines can be used. In the early solutions the equations are typically typed in but in later problem the preprogrammed routines are used extensively. The following EES code is the SETP routine for the extraterrestrial radiation on a horizontal surface. The names of all SETP routines end with an underline (e.g., H_zero_). Text between squiggly brackets or between double quotes is for information only. However, a special case exists when the left squiggly bracket on the next line following the function name is followed by a dollar sign. This text is used as help. Text between squiggly brackets does not appear in the Formatted Equation Window. Routines such as this often call other routines (e.g., Dec_). Units are particularly important in all of engineering and are included in the SETP functions. Here units are assigned to the two constants f [deg] and g [MJ/m^2]. The units conversion function in EES Convert(UnitsFrom, UnitsTo) is one of EES’s most useful tools and is illustrated here where the units degrees are converted to radians with Convert(deg, rad). Units cannot be assigned to variables in the Equations Window. The units of SunSetHrAng is set to degrees in the EES program but “[deg]” is only a reminder that the units are degrees. Finally, the value returned by the function must be set using the function name. Function H_Zero_(n,Lat) {Equation 1.10.3} {$H_Zero_ Equation 1.10.3: Calculates the day's extraterrestrial radiation H_Zero. Uses Fctn. Dec_ for Declination. All angles are in degrees and n is the day number. H_Zero is in MJ/m2.} Dec = Dec_(n) "[deg]" SunSetHrAng = arccos(-tan(Lat)*tan(Dec)) "[deg]" f=0.9863[deg] g=37.6[MJ/m^2] H_Zero_ = g*(1+.033*cos(f*n))*(cos(Lat)*cos(Dec)*sin(SunSetHrAng)+Convert(deg, rad)*SunSetHrAng*sin(Lat)*sin(Dec)) "[MJ/m^2]" END The solution to Problem 1.10 uses this function and is as follows: "!1.10 Calculation of extraterrestrial radiation on horizontal surface" “Knowns:” month=2; day=8; lat=45[deg]; G_sc=1367[W/m^2] "Table 1.6.1" Dec = 23.45[deg]*sin(.9863*(284+n)) "Eqn 1.6.1" SunSetHrAng = arccos(-tan(Lat)*tan(Dec)) "Eqn 1.6.10" n=nDay_(month, day) "We could use Equation 1.10.3 the day's extraterrestrial radiations: H_o=(24*3600[s]*G_sc)/pi*(1+0.033*cos(360[deg]*n/365)) *(cos(lat)*cos(dec)& *sin(SunSetHrAng)+(pi*SunSetHrAng/180[deg]*Sin(lat)*sin(dec)))*Convert(J, MJ) " iv "But we will use the SETP function:” H_o = H_Zero_(n, Lat)" "To find the monthly average daily radiation we will use the SETP function with the average day in February." n_ave=AveDay_(month) "Table 1.6.1" H_bar_o=H_Zero_(n_ave, Lat) "To find the energy in the hour 10 to 11 use Equation 1.10.4" Time=10:30 "note that this is the midpoint of the hour" HrAng=(Time-12[hr])*15[deg/hr]; HrAng2=HrAng+7.5[deg]; HrAng1=HrAng-7.5[deg] I_o = (12*3600[s]*G_sc)/pi*(1+.033*cos(360[deg]*n/365))*(cos(Lat)*cos(Dec)*(sin(HrAng2)& -sin(HrAng1))+Convert(deg, rad)*(HrAng2-HrAng1)*sin(Lat)*sin(Dec))*Convert(J, MJ) "We could have used the SETP function I_o=I_Zero_(n, Lat, HrAng1, HrAng2) SOLUTION day=8 Dec=-15.52 [deg] G_sc=1,367 [W/m^2] HrAng=-22.5 [deg] HrAng1=-30 [deg] HrAng2=-15 [deg] H_bar_o=17.63 [MJ/m^2] H_o=15.84 [MJ/m^2] I_o=2.214 [MJ/m^2] lat=45 [deg] month=2 n=39 [-] n_ave=47 [-] SunSetHrAng=73.88 [deg] Time=10:30:00 [hr] There are alternative methods of solving some problems. For example, calculation of an hour’s radiation on sloped surfaces can be done with any of the four methods shown in Section 2.15 and 2.16. In some instances the problem statements call for use of specific methods. In others the choice is left open. In a few cases we have included solutions by more than one method. Students may complain that some data is not included in the problem, such as ground reflectance. We leave it up the student to choose reasonable values. Typical values of 0.2 are used unless there is snow cover then 0.7 seems reasonable. Most of these problems have been used as homework or examination problems in courses at the University of Wisconsin. Many of them have arisen from our research at the Solar Energy Laboratory, from practical experience, or from suggestions of others. We appreciate the assistance of the many people who have contributed ideas and suggestions on which these problems and their solutions are based. For those who teach courses in solar energy to engineering students, we have included on the final pages of this manual a typical outline of a one-semester course we teach to our fourth-year undergraduate and first year graduate student classes. Madison, Wisconsin, April1, 2011 William A. Beckman v P01 01 Equations Knowns: σ = sigma# (1) Dsun = 1.39 × 109 [m] (2) Dearth = 1.27 × 107 [m] (3) RSunEarth = 1.495 × 1011 [m] (4) TSun = 5777 [K] (5) Calculate the emitted solar energy 2 Areasun = π · Dsun ; (6) 4 EmitSol = σ · Areasun · TSun (7) Calculate the fraction intercepted by the earth F raction = Areaearth,proj /Area1au Areaearth,proj = π · (8) 2 Dearth 4 (9) 2 Area1au = 4 · π · RSunEarth (10) Estimate solar constant SolarConstant = F raction · EmitSol/Areaearth,proj (11) Solution Area1au = 2.809 × 1023 m2 Dearth = 1.270 × 107 [m] F raction = 4.510 × 10−10 [-] SolarConstant = 1365 W/m2 Areaearth,proj = 1.266 × 1014 m2 Dsun = 1.390 × 109 [m] RSunEarth = 1.495 × 1011 [m] TSun = 5777 [K] 1 Areasun = 6.070 × 1018 m2 EmitSol = 3.833 × 1026 [W] σ = 5.670 × 10−8 W/m2 -K4 P01 02 Equations Start with Solar Constant = 1367 W/m2 from Problem 1.1. SolarConstV enus will be inversely proportional to the square of the distance from sun: SolarConst .722 SolarConst = 1367 W/m2 SolarConstV enus = (1) (2) Solution SolarConst = 1367 W/m2 SolarConstV enus = 2637 W/m2 2 f0to0.216 (1) f0to2.1.0 = 0.941 [-] 3 .725 [-] f0to0.5to2.0 = 0.P01 03 Equations From Table 1.0 is: f0.5 = 0.941 (2) Therefore the fraction between 0.3.0 = f0to2.5 = 0.5 (3) Solution f0.5to2.216 [-] f0to2.0 = 0.5 and 2.0 − F0to0. λ5 = .10 = 2.787 0. The wavelength divisions for the increments are the wavelengths given in the table for F0 -λ of 0.923. λ3 = . λmid.682 0.8 = 1.731 λ10 = 8.561 0.1.113 1.300.2. λmid.364 0. λmid.3 = .25.561.638.654.525.117 (3) (4) So. the range of wavelengths for each of the 10 bands is: duplicate i = 1. The first increment of a tenth is in the wavelength range of 0 to 0. 10 (5) λrange.416. etc. λmid. 0.5 = .000 0.1 = .638 0.008 1. λ0 = 0.923 1.455 0.7 = .118 0.4 = . λ6 = .008.489 0. λ9 = 1.P01 04 Equations This problem is essentially solved in Table 1.113.410 6.416.525 0. etc.599. λmid. λmid.787.682 λmid.412. λmid.i = λi − λi−1 (6) end (7) Solution Arrays Row λi λmid. (1) (2) λmid.6 = .849.599 0.1b. 0. λ8 = 1. 0.236 0. λ2 = .i λrange. λ1 = . The energy midpoints are the wavelengths in the table for 0.416 0.i 0 1 2 3 4 5 6 7 8 9 10 0. 0.3.117 0.489.731 0.849 1.116 0.244 1. Use the ‘Insert/Modify an Array’ in the Edit menu to simplify data input.159 0. and the wavelength at the energy midpoint of the increment is 0.364.093 0.654 8.077 0. λmid.3.364.05.073 0.072 0.9 = 1.346 4 .2 = .244. λ4 = . λ7 = 1.455.300 0.000.412 2.15. P01 05 Equations Dec = 23.0 40.3 43.4 28.4 61.1 (1) CosIncAng = sin (Dec) · sin (Lat) · Cos(Slope) − sin (Dec) · Cos(Lat) · sin (Slope) · Cos(Surf AzAng)+ Cos(Dec) · Cos(Lat) · Cos(Slope) · Cos(HourAngle)+ Cos(Dec) · sin (Lat) · sin (Slope) · Cos(Surf AzAng) · Cos(HourAngle) + Cos(Dec) · sin (Slope) · sin (Surf AzAng) · sin (HourAngle) Eqn 1. n.9 -14.45 [deg] · sin (. 1.9863 · (284 + n)) Eqn.9 -14.9 -14.9 8.9 θ [deg] 64.3 43.9 -14.3 43. HourAngle.6. Slope.6.1 .2 θ = arccos (CosIncAng) (3) All data and results are in the Parametric Table We could have used the SETP function to find θ: θ = arcCos(CosIncAng(Lat.3 43.3 HourAngle [deg] 30 30 30 30 30 (2) Surf AzAng [deg] 0 0 40 0 90 Slope [deg] 0 60 60 90 90 5 Dec [deg] -14. SurfAzAng)) Table 1 Run n 1 2 3 4 5 41 41 41 41 41 Lat [deg] 43. 04089 · sin (2 · B))(1) B = (n − 1) · 360 [deg] 365 Eqn 1.6.5.032077 · sin (B) − .1 to find the day number) (3) The difference between standard and solar time is: .5.3) in Equation 1.2 [min]·(.2 EqT ime = 229.5.014615 · Cos(2 · B) − .3 n = nDay (month.P01 06 Equations Use of Equation of Time (Eqn 1. day) (2) Use either the SETP function or Table 1.000075 + .001868 · Cos(B) − . . . hr . . . Solar − Standard = (4 [min/deg] · (Long Std − Long Loc ) + EqT ime) · . 016666667 min .0. 49 -6.549 -6.55 208.2 (4) The difference between daylight time and standard time is one hour Daylight − Standard = 1 [hr] (5) Note: In part d the time of 10:00 AM is solar time as nothing else is specified.1 Long Loc [deg] 101 116 70.55 32.549 Solar [hr] 14:02:31 13:02:31 12:00:00 10:00:00 Standard [hr] 14:00:00 14:00:00 11:48:33 10:06:33 Daylight [hr] 15:00:00 15:00:00 12:48:33 11:06:33 .1 208.49 -13.5.5 90 Long Std [deg] 105 105 75 90 6 EqT ime [min] -13. Eqn 1. Table 1 Run 1 2 3 4 month [-] 2 2 7 7 day [-] 3 3 31 31 n [-] 34 34 212 212 B [deg] 32. 3 43.5 22. Dec with Eqn 1.4 89.6.52 -2.6.0 113.3 43.3 43.6.42 DayLength [hr] 10:38 12:00 13:21 11:51 8:51 12:00 15:09 11:41 SunSetHrAngle [deg] 79.42 -23.0 66.6.000907*sin(2*B) Eqn 1.1a SunSetHrAngle = arccos (− tan (Lat) · tan (Dec)) DayLength = 2 [hr] · Eqn 1.33 -22.2 89.33 DayLength [hr] 10:40 12:00 13:23 11:52 8:54 12:00 15:13 11:42 SunSetHrAngle [deg] 80.000758*Cos(2*B) +0.5 22.7 87.3 Dec [deg] -22.6. day) (1) Eqn 1.0 114.6.1a month day n 1 2 3 4 5 6 7 8 1 3 7 3 1 3 7 3 1 22 1 16 1 22 1 16 1 81 182 75 1 81 182 75 Lat [deg] 22.00 23.71 0.71 0. For most solar calculations this difference can be ignored.9 90.8 Dec with Eqn 1. 1.3 Dec [deg] -23.10 arccos (− tan (Lat) · tan (Dec)) 15 [deg] n = nDay (month.01 23.3 43.52 -2.6.399912*cos(B)+0.8 90.5 22.7 Lat [deg] 22.1b can be used dec=(0.5 43.0 100.070257*sin(B)-0.1b Run month day n 1 2 3 4 5 6 7 8 1 3 7 3 1 3 7 3 1 22 1 16 1 22 1 16 1 81 182 75 1 81 182 75 7 (3) (4) The average day for February is the 22.01 -0.12 -2.6.1b B=(n-1)*360[deg]/365 The day lengths using the two equations differ by 2 minutes.3 43.5 22.12 -2.5 43.0 66.0 90.006918-0. AveDay=AveDay(month ) Run (2) .00 23.01 23.01 -0.1 or SETP function The more exact declination equation.1 or SETP function.4 90.9863 · (284 + n)) Eqn 1.P01 07 Equations Determine the sunset hour angle andday length for various conditions Dec = 23.5 22.5 22.2 87.3 43. The average day could be found from Table 1.6.45 [deg] · sin (.0 100.11 Table 1. 5 2 2 2 (cos (ZenithAngle)) + (cos (Dec)) · (sin (HourAngle)) θN S = arccos (7) For polar axis. day) could use Table 1.5a: θP olar = arccos (cos (Dec)) (8) Table 1 Run 1 2 time [hr] 9 12 Dec [deg] -6. Lat = 27. Use parametric table for the two times.45 [deg] · sin (.1 33. month = 3.3a: . use Equation 1.765 6.6.8 Calculation of θbeam for tracking surfaces. Day = 5.P01 08 Equations 1.77 θP olar [deg] 6. n = nDay (month.77 θEW [deg] 59.7.7.1 (3) CosZenithAngle = Cos(Lat) · Cos(Dec) · Cos(HourAngle) + sin (Lat) · sin (Dec) 1. use Equation 1.2a: 2 2 θEW = arccos 1 − (cos (Dec)) · (sin (HourAngle)) (6) For horizontal N-S axis.54 0 8 θN S [deg] 25.765 -6.6.9863 · (284 + n)) (1) (2) 1.6.07 33.765 HourAngle [deg] -45 0 ZenithAngle [deg] 55.5 (4) ZenithAngle = arccos (CosZenithAngle) (5) For horizontal E-W axis.7.765 .1 to find n HourAngle = (T ime − 12hr ) · 15 [deg/hr] Dec = 23. use Equation 1. Dec = 23.8.589 1. day) 1.1 (1) Table 1.8. Dec.6.765 -6.45 [deg] · sin (.4958 0.5 -22.5 (4) CosT heta = sin (Dec) · sin (Lat) · Cos(Slope) − sin (Dec) · Cos(Lat) · sin (Slope) · Cos(Surf AzAng) + Cos(Dec)· Cos(Lat) · Cos(Slope) · Cos(HrAng) + Cos(Dec) · sin (Lat) · sin (Slope) · Cos(Surf AzAng) · Cos(HrAng) + Cos(Dec) · sin (Slope) · sin (Surf AzAng) · sin (HrAng) Rb = CosT heta/CosZen (5) 1. Slope. HrAng.2 (6) We could use the SETP function to find CosTheta CosTheta = CosTheta(Lat.7878 0.9863 · (284 + n)) n = nDay (month.2.5 Dec [deg] -6.5906 . Use parametric table for the two conditions.765 CosT heta [-] 0.413 9 T ime [hr] 14:30 10:30 HrAng [deg] 37. SurfAzAngle) Table 1 Run 1 2 Run 1 2 Lat [deg] 43 43 month day n 3 3 5 5 64 64 Slope [deg] 60 45 Surf AzAng [deg] 0 15 Rb [-] 1.8343 CosZen [-] 0.6.6.1 (2) HrAng = (T ime − 12hr ) · 15 [deg/hr] (3) CosZen = Cos(Lat) · Cos(Dec) · Cos(HrAng) + sin (Lat) · sin (Dec) 1.P01 09 Equations Calculate Rb from Equation 1. 1 1.6. day = 8.P01 10 Equations 1.10 (3) From Equation 1.45 [deg] · sin (.1 SunSetHrAng = arccos (− tan (Lat) · tan (Dec)) (1) (2) 1.6.9863 · (284 + n)) Table 1.6. Gsc = 1367 W/m2 .033 · Cos(360 [deg] · n/365)) π . day) month = 2.10.3 the day’s extraterrestrial radiations is: Ho = 24 · 3600 [s] · Gsc · (1 + 0. lat = 45.10 Calculation of extraterrestrial radiation on horizontal surface n = nDay (month. Dec = 23. . . MJ . . SunSetHrAng · Cos(lat) · Cos(dec) · sin (SunSetHrAng) + π · · sin (lat) · sin (dec) · . . 1 × 10·6 (4) 180 [deg] J . We could have used the SETP function HZero = HZero. Lat) To find the monthly average daily radiation we will use the SETP function with the average day in February.6. nave = AveDay (month) Table 1. (n. Io = (7) HrAng2 = HrAng + 7.5deg (8) 12 · 3600 [s] · Gsc · (1 + .5 [deg] .10.033 · Cos(360 [deg] · n/365)) · (Cos(Lat) · Cos(Dec) π . (nave . Lat) H (6) To find the energy in the hour 10 to 11 use Equation 1. HrAng1 = HrAng − 7.4 T ime = 10 : 30 note that this is the midpoint of the hour HrAng = (T ime − 12hr ) · 15 [deg/hr] .1 (5) ¯ o = HZero. . . . . . . rad . . ·6 MJ . . . ·(sin (HrAng2) − sin (HrAng1))+. 0.017453293 ·(HrAng2 − HrAng1)·sin (Lat)·sin (Dec)·. 1 × 10 (9) . deg J . 5 [deg] Ho = 15. HrAng1. Lat.63 MJ/m2 H HrAng = −22.84 MJ/m2 lat = 45 [deg] month = 2 n = 39 [-] day = 8 Dec = −15.52 [deg] HrAng1 = −30 [deg] Io = 2. HrAng2) Solution HrAng2 = −15 [deg] Gsc = 1367 W/m2 ¯ o = 17. We could have used the SETP function Io =IZero.88 [deg] 10 T ime = 10 : 30 : 00 [hr] .214 MJ/m2 nave = 47 [-] SunSetHrAng = 73. (n. So. this makes an easier to understand plot.e. lat = 40 (2) nave = AveDay (month) 284 + nave dec = 23..45 [deg] · sin 360 · 365 (3) (4) θz = arccos (cos (lat) · cos (dec) · Cos (ω) + sin (lat) · sin (dec)) cos (θz ) · sin (lat) − sin (Dec) γs = − arccos sin (θz ) · Cos (lat) T ime = − ω · 60 [min/hr] 15 [deg/hr] (5) (6) (7) Here Time is the number of minutes before solar noon. the Object Altitude Angle varies from 0 to 90.P01 11 Equations 1. for the mean day of each month we need to find the time (i. 11 . We will use the EES Parametric Table to work through the 12 months. An equation for the Object Azimuth Angle(which has a single value) is: ObjAzAng = − (90 [deg] − (arctan (20.2/10) + 15 [deg])) (1) When the Solar Azimuth Angle (γ s ) is less than ObjAzAng the corner of the window is shaded. ω) at which the ObjAzAng is equal to γ s. the hour angle.11 Calculation of shading by a building ELL As the ell is high compared to the window. 44E-4 6. The plot shows the number of minutes before solar noon at which the sun is incident on the east edge of the window as a function of month.05 nave 17 47 75 105 135 162 198 228 258 288 318 344 γs [deg] -11.These equations are diffucult to solve for the hour angle ω unless very good guesses are provided.92 -12.79 23.34 -11.415 18.91 -23. .74E-5 3.825 -10.1 26.34 -11.373 -3.45 2.34 -11.14E-3 2.959 -5.7 53. The times when shading ceases is determined for the time of year from this plot.255 -7.217 -9.621 -3. An alternative method of solution is to define an error as the absolute difference between γ s and the ObjAzAng and minimize this error.7 43.34 -11.711 -5.06 minutes before solar noon 12 error [deg] 9.0 31.9 38.3 50.53E-4 T ime [min] 43 37 31 24 17 14 16 21 28 35 41 44 .34 -11.418 9.34E-5 4.3 59.26E-4 8.599 -18.95 -2.7 63.0 21.34 -11.02E-3 6.87E-3 6.48E-4 2.9 ω [deg] -10.18 13.005 -8.63E-4 2.34 -11.34 -11.34 -11.2 19.08E-4 1.5 17.01E-5 1.68 -9. Results Run month 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 dec [deg] -20. error = abs (γs − ObjAzAng) (8) The Object Azimuth Angle could have been plotted on a solar position plot for Latitude 40.34 -11.34 -11.34 -11.09 21.896 -4.34 θz [deg] 61.352 -7. 4.1) and the Y axis is towards the south. y=P.P01 12 Equations 1. the corner of the overhang is at x=(W+E)/2. E = 0.5. y. the X axis is towards the west (i. The azimuth angle is arcTan(x/y) and the altitude angle is arcTan(z/sqrt(x2 +y2 )) Expressions for the relationships of distances and angles for any point on the overhang are as follows: W = 3. G = 0. A general point on the overhang is x.g..12 Calculation of shading by an overhang Since the system is symmetrical about the N-S line. For example. Both positive and negative values of γ were placed in the parametric table for plotting purposes. y.25.. H = 1. and z around the boarder of the overhang. The Z axis is vertical. The parametric table is saved as an EES LKT 13 . and z=H/2+G.75. it is necessary to calculate only points on one side (e. Use a coordinate system with the origin at the center of the window. γ = arctan (x/y). z. α = arctan z p x2 P = 0.25 γeast = −γ ! (1) (2) (3) + y2 Set up a parametric table for γ and α with various values of x.75. left in Fig 14. west).e. EES reads this lookup table and overlays the values on a solar postion diagram for a latitude of 34 degrees.table. The program. Area above the red line are shaded. The figure shows the shading boundary plotted on a solar position diagram. $SaveTable ’Parametric’ ’shading. SolarPositionDiagram.lkt’ /N Part b: From the diagram we see that all of February (the 4th month from the bottom) is unshaded and all of July (the second month from the top) is shaded. In August (the 4th month from the top) the center is shaded after 8am and before 4pm. The shading diagram ends at plus or minus 90 degrees since the sun is then behind the window. Plot 1 14 . 1a δha = dec (−n) (9) Equation 1.6.P01 13 Equations 1. δ = dec (n) Equation 1.3 [deg] (4) Long = 104.6 [deg] (5) n = AveDay (month) Or from Table 1.13 Calculation of solar information We will use the EES SETP functions for the calculations.6.6.1 (6) a) Day Length ωs = SunsetHrAng (Lat.high accuracy declination 15 (10) . n) Hrday = ωs · 2 15 [deg/hr] Equation 1. Knowns month = 2 (1) day = 16 (2) γ = 0 [deg] (3) Lat = 38.10 (7) Equation 1.6. See Key variables in Solutions window.6.assumed to be at noon.11 (8) b) Declination .1b . 1 = 0. duplicate i = 7. n.1a f0. 18 (18) ωi = 15 · (i − 12) − 7. n. h) 10AM solar time is what local clock time. (n.1.8.64 = 0.64 (24) Solar1. β. Slope ranging between 0 and 90.1 micro meters.6.5 [deg] .64 = ∆f · Gsc# (25) Note that the solar constant.402 (22) f0. Lat) Equation 1. Use center of the hour.5 [deg]) (20) end (21) g) Find solar radiation in band 0. Lat. ωi − 7.i = IZero. Long std = 105 (26) SolarT Ime = 10 (27) .use arrays with ω ranging from -75 to +75. Gsc is a built-in constant.3 (17) f) Io throughout the day .3. We could have used a third parametric table. θ = arccos (cosIncAng (Lat. γ)) Equation 1.1 (15) β = 60 (16) e) What is Ho? Ho = HZero. ω.1 − f0.At sunrise and sunset the equivalent n is ωs nsunrise = n − 15 [deg/hr] · 24 [hr] nsunset = n + (11) ωs 15 [deg/hr] · 24 [hr] (12) δchange = dec (nsunrise ) − dec (nsunset ) (13) c) change in θ throughout the day for various slopes. (Lat. From Table 1. (n.2 (14) d) Rb vs ω for β=60 . Rb = RBeam. β. ωi + 7.5 (19) Io. γ) Equation 1.use a second parametric table ω -75 to +75.1.0. Use a parametric table with ω ranging from -75 to+75.744 (23) ∆f = f0. ω.10.64 to 1.0.1.0. . . hr . . SolarT Ime − StdT ime = (4 [min/deg] · (Long std − Long) + EqnT ime (n)) · . . 0.016666667 min . (28) Solution Variables in Main program 16 . 04501 0.i 2 MJ/m 0.5 -67.377 θ = 71.744 Lat = 38.346 2.7697 1.0.5 -37.64 = 467.7697 0.5 7.61 [deg] Hrday = 10.0.5 67.847 3.3 [deg] n = 47 [-] ωs = 79.5 37.5 52.5 -7.1487 [deg] f0.6 [hr] Long std = 105 [deg] nsunset = 47.1.5 82.53 [deg] StdT ime = 10.847 2.402 Ho = 21.1 = 0.342 [-] γ = 0 [deg] Long = 104.638 2.21 [hr] .64 = 0.1.22 [-] SolarT Ime = 10 [hr] δ = −12.5 W/m2 Arrays Row 7 8 9 10 11 12 13 14 15 16 17 18 Io.6 [deg] nsunrise = 46.β = 60 [deg] ∆f = 0.95 [deg] f0.5 [deg] day = 16 δha = −12.5 22.5 -22.5 theta vx omega Run 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 θ [deg] β [deg] 0 0 0 0 0 0 0 0 0 0 0 15 15 15 15 15 15 15 15 15 15 15 ω [deg] -75 -60 -45 -30 -15 0 15 30 45 60 75 -75 -60 -45 -30 -15 0 15 30 45 60 75 17 δchange = −0.04501 ωi [deg] -82.5 -52.638 0.107 3.107 2.51 MJ/m2 month = 2 ω = 75 [deg] Solar1.346 1.78 [-] Rb = 5. 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 30 30 30 30 30 30 30 30 30 30 30 45 45 45 45 45 45 45 45 45 45 45 60 60 60 60 60 60 60 60 60 60 60 90 90 90 90 90 90 90 90 90 90 90 -75 -60 -45 -30 -15 0 15 30 45 60 75 -75 -60 -45 -30 -15 0 15 30 45 60 75 -75 -60 -45 -30 -15 0 15 30 45 60 75 -75 -60 -45 -30 -15 0 15 30 45 60 75 Rb vs omega Run 1 2 3 4 5 θ [deg] -75 -74 -73 -72 -71 Rb 5.626 4.432 18 .377 4.106 3.724 3. 651 1.154 2.618 1.83 1.7 1.866 1.8 1.015 2.708 1.444 2.772 1.626 1.201 3.372 2.614 1.737 1.604 1.6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 -70 -69 -68 -67 -66 -65 -64 -63 -62 -61 -60 -59 -58 -57 -56 -55 -54 -53 -52 -51 -50 -49 -48 -47 -46 -45 -44 -43 -42 -41 -40 -39 -38 -37 -36 -35 -34 -33 -32 -31 -30 -29 -28 -27 -26 -25 -24 -23 -22 -21 -20 -19 -18 -17 3.113 2.526 2.645 1.075 2.607 1.2 2.669 1.602 19 .63 1.954 1.786 1.663 1.64 1.815 1.727 1.308 2.041 2.76 1.861 2.731 2.635 1.907 1.717 1.657 1.251 2.684 1.676 1.691 1.611 1.98 1.622 1.848 1.621 2.885 1.748 1.929 1.009 1. 645 1.588 1.583 1.58 1.607 1.584 1.587 1.59 1.58 1.58 1.592 1.599 1.579 1.59 1.651 1.60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 -16 -15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 1.58 1.599 1.684 1.594 1.611 1.582 1.63 1.596 1.7 1.592 1.602 1.676 1.58 1.585 1.64 1.584 1.58 1.588 1.635 1.585 1.579 1.614 1.622 1.581 1.579 1.691 1.594 1.618 1.583 1.581 1.669 1.708 20 .626 1.582 1.596 1.657 1.604 1.587 1.663 1. 815 1.015 3.748 1.786 1.83 1.041 2.737 1.526 2.98 2.009 2.731 2.372 2.885 1.727 1.114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 1.308 2.866 1.2 2.929 1.251 2.848 1.954 1.154 2.76 1.377 21 .724 4.106 4.113 2.201 3.8 1.717 1.626 5.432 3.907 1.861 3.621 2.444 2.075 2.772 1. theta vs omega Rb vs omega 22 . Io vs omega 23 . estimates for the hour increments are as shown in the EES LookUp table. ‘Clear day’ ) (2) Cloudy i = Lookup(i. 12 (1) Cleari = Lookup(i. 12) (6) In the units usually used.7 MJ/m2.01 Determination of irradiation from strip-chart records.e. i. 12) (5) CloudyDay = Sum(Cloudy i . The areas under the curves of Figure 2. . ‘Cloudy day’ ) (3) end (4) The sum of each row is found by: ClearDay = Sum(Cleari .1 are estimated by graphical integration. The sum of the column can be found by first converting the LookUp table values in to arrays: duplicate i = 1. the time spans between the vertical lines on the plots. i = 1.5. i = 1.P02 01 Equations 2.. these are 18. Using hour time increments. respectively.0 and 5. . . MJ/m2 . . ClearDayM J = ClearDay · . . 0.0036 W · hr/m2 . . . . MJ/m2 . . CloudyDayM J = CloudyDay · . . 0036 W · hr/m2 .0. Howver. some sort of program would be needed. (7) (8) For such a simple set of calculations this problem is easier done by hand. if this processes were to be repeated for a year.904 .04 CloudyDay = 1640 Arrays Row 1 2 Clear2i W/m 40 190 Cloudy2 i W/m 10 30 24 CloudyDayM J = 5. Solution ClearDay = 5010 ClearDayM J = 18. 3 4 5 6 7 8 9 10 11 12 360 530 640 720 720 650 530 360 200 70 60 120 200 210 210 300 270 120 90 20 25 . 05.1. the wavelength range is 0.1. etc.434. The wavelength divisions for the increments are the wavelengths given in the table for energy bands of 0. 26 . 0.479 and the energy midpoint wavelength is 0.3.6.300 to 0. For the first one-tenth:increment. 0. This problem is essentially solved in Table 2.P02 02 Equations 2.6.25.02 Division of a terrestrial solar spectrum into 10 equal increments using Table 2.2.1. The energy midpoints are the wavelengths in the table for 0.15. etc. 0. 0. 1 Use the Parametric table for the two months.45 · sin (.1 (2) Dec = 23.P02 03 Equations ¯ from average sunshine hour data. b = 0.7.34. from Table 2.6.30. n = AveDay (month) Lat = 43 [deg] (1) EES function for Table 1.6. a = 0.03 Estimation of H ¯ . n 2.1a (3) SunSetHrAng = arccos (− tan (Lat) · tan (Dec)) Eqn 1.6.9863 · (284 + n)) Eqn 1.10 (4) Gsc = Gsc# Built-in EES constant for the solar constant (5) . . 2. . ¯ Zero = . 1 × 10·6 MJ/m . 9863 · n)) · (Cos(Lat) · Cos(Dec) · sin (SunSetHrAng) H . · Gsc · 3600 · 24 · (1 + .033 · Cos(. J/m2 . 12 ¯ = 22.6.5 10.4 [deg] Table 1 Run month n ¯ n 1 2 1 6 4.0 for January and June.4 and 23.0 MJ/m2 H n = 162 b = 0.09 [deg] Lat = 43 [deg] SunSetHrAng = 113.77 MJ/m2 H n ¯ = 10.1 Dec = 23.1 17 162 Dec [deg] -20.01745 · SunSetHrAng · sin (Lat) · sin (Dec)) Eqn 1.3 AvgDayLngth = (2/15) · arccos (− tan (Lat) · tan (Dec)) ¯ H ¯ Zero = a + b · n H/ ¯ /AvgDayLngth (6) Eqn 1. Note: Appendix G gives Hs Solution a = 0.92 23.09 SunSetHrAng [deg] 69.7.0 shown in the table.2 and 22.34 ¯ Zero = 41.77 AvgDayLngth 9.10.4 27 ¯ HZero2 MJ/m 13. compared to estimates of 6.2 (8) ¯ of 6. π +0.37 41.3 Gsc = 1367 W/m2 month = 6 AvgDayLngth = 15.0 .2 22.12 113.22 15.12 ¯ H 2 MJ/m 6.11 (7) Eqn 2. 402 MJ/m2 (1) Io = IZero.3 and diffuse fraction with Equation 2. (n. Lat.1 n = 9.10. Eqn 2. I = 0. H = 4.11.880 Ho = 14.9.1724 H = 4.984 HourAngle2 = 0 [deg] Io = 2. HourAngle2 = 0 [deg] .10. KT.2 (4) Id /I = fd (5) b Calculation of a day’s clearness index with Equation 2. (n.10.day = Hdif f /H (Lat.11.37 MJ/m2 kT = 0.48 MJ/m2 (6) Ho = HZero.1 or Fig 2.2.2 (8) Fd.day = H/Ho EES function or Eqn 2.4 Eqn 2.9. n.1 or Fig 2.3958 Lat = 40 [deg] . HourAngle1 = −15 [deg] . HourAngle1. Lat = 40 [deg] . HourAngle2) (2) kT = I/Io EES function or Eqn 1.402 MJ/m2 KT.10.9.day ) (9) Solution fd = 0.day = 0.331 MJ/m2 n = 9 [-] Fd.2 and daily diffuse fraction from Eqn 2.3 fd = Idif f /I (kT ) (3) EES function or Eqn 2.48 MJ/m2 I = 0. Lat) (7) KT.P02 04 Equations 2.11.9.day = 0.4a Calculation of hourly clearness index with Equation 2.3117 28 HourAngle1 = −15 [deg] Id = 0. 13 MJ/m2 H n = 75 ¯ = 18.9.10. n = AveDay (month) ¯ o = HZero.327 MJ/m2 H month = 3 .12.8.8 [deg] 29 ¯ d = 4.P02 05 Equations 2.2 (5) ¯ d /H ¯ Fdif f = H (6) ¯ =H ¯b + H ¯d H (7) Solution Fdif f = 0. lat) H ¯ T = H/ ¯ H ¯o K month = 3 (1) EES Function or Table 1.9 MJ/m2 H ¯ T = 0. (n.1 ¯ = 18.5 Calculation of monthly beam and diffuse using Equation 2.6.12.12. n.3 (3) Eqn 2. K EES Function or Eqn 2.1 or Fig 2.7234 [-] K ¯ b = 14.57 MJ/m2 H Lat = 42.1 (2) EES Function or Eqn 1.9.1 (4) ¯T Fdif f = Hdif f bar/Hbar Lat.2289 ¯ o = 26. H Lat = 42. Ho = HZero.13.day ) (5) (6) EES function or Eqn 2.13.find beam and diffuse for the day and then for the hour.1 I = H · rT (3) (4) b) beam and diffuse radiation in hour 10 to 11. KT. Surf AzAng = 0 a) total radiation in hour 10 to 11.2 Fdif f. rT = rt.1 or Table 1. Method 1 .1 Eqn 2.10.9.day = H/Ho EES function or Eqn 1.0 MJ/m2 (1) hrAng = −22.10.3 or Fig 1. n. This problem can be done two ways.2 or Figure 2. lat) KT.day = Hdif f /H (Lat. n = 20.2 or Fig 2.10. lat. (2) slope = 60 [deg] . H = 8. Lat = 43 [deg] .2 30 (7) .P02 06 Equations 2.6 Estimate radiation on tilted surface using various methods starting with daily total radiation. n) EES function or Eqn 12.11.5 [deg] . (n. (hrAng.11. HrAng − 7.2 .1 + Id.LJ. Lat. n.4 or Fig2.Fdif f.day = Hd /H (8) H = H d + Hb (9) rd = rd.6.isotropic = Rb · Ib + Id (23) f) Estimation of IT assuming isotropic sky (Liu & Jordan model). HrAng.nosnow = (Ib + AA · Id ) · Rb + Id · (1 − AA) · Fcs + ρg.snow = 0.15.3 I = Ib.5 (20) (21) d) Find radiation on tilted surface if all treated as beam.9.nosnow · I · Fcg (26) IT. Equation 2.13. Lat. HrAng + 7.HD.16.2 or Fig 2.10.snow = Ib · Rb + Id · Fcs + ρg.2 /I (16) I = Id.5 [deg]) (13) kT. Id = Id.find beam and diffuse for the hour Io = IZero. Ib = Ib.1 (19) CosT heta = cosIncAng (Lat.1 Fcs = 1 + Cos(slope) 2 Fcs + Fcg = 1.2 (18) c) find Rb Rb = CosT heta/CosZen Eqn 1.hour = Id.hour = Idif f /I (kT.2.2 + Ib. ρg.4 AA = Ib /Io (28) IT.hour = I/Io (14) Eqn 2.5deg . Slope.10. n) rd = Id.snow · I · Fcg (27) g) Estimation of IT with Hay & Davies model.3 (15) Fdif f.13. (hrAng.2 (10) Eqn2.LJ.nosnow = 0.hour ) EES function or Eqn 2.7 (25) IT. (24) ρg. Equation 2. HrAng) EES function or Eqn 1.6.nosnow = Ib · Rb + Id · Fcs + ρg.1 EES function or Eqn 1. Surf AzAng) CosZen = CosZenAng (Lat.1 (11) I is from part a (12) Method 2 . (n.8.1 /Hd EES function or Eqn 2.2 (17) I is from part a We will use method 2 results since it relies on one less relationship. IT.3 Fdif f.13.nosnow · I · Fcg (29) 31 . n.allBeam = Rb · I (22) e) Estimation of IT assuming diffuse plus ground-reflected radiation are together isotropic IT. 49 MJ/m2 IT.858 MJ/m2 2 I = 1.1463 [-] Surf AzAng = 0 [deg] .3492 f 13 = −0.16.36 MJ/m IT.nosnow = Ib ·Rb +Id ·(1 − F 1)· IT.P erez.9 Id +Ib /cosZen Id (36) (37) 3 + 5.snow · (35) 2 2 Eqn 2.75 MJ/m Ib.39 MJ/m2 kT.5248 MJ/m2 Io = 2.hour = 0.535 · 10−6 · (arccos (cosZen)) 3 1 + 5.snow = 0.1 = 0.3965 f = 0.hour = 0.13 (42) Solution Variables in Main program a = 0.3927 hrAng = −22.2 = 0. cosZen) = (1 + Cos(Slope)) 1 − Cos(Slope) +Id ·F 1·a/b+Id ·F 2·sin (Slope)+I·ρg.snow = 2.10 (38) =2.226.873.16.873 F 2 = 0.snow · I · Fcg (33) i) Estimation of IT by the Perez model.54 MJ/m2 Lat = 43 [deg] ρg.16.93 = 2. f 12 = −0. f 21 = 0.day = 0.75 2 H = 8 MJ/m Hd = 3.7 rT = 0.20 MJ/m2 KT.16.16.nosnow = 2.2 = 0.nosnow = (Ib + AA · Id ) · Rb + Id · (1 − AA) · Fcs · 1 + f · sin (slope/2) + ρg.day = 0.6019 [-] ρg.346 32 slope = 60 [deg] (41) b = 0.4596 MJ/m 2 IT.535 · 10−6 · (arccos (cosZen)) Eqn 2.11 (40) F 1 = max (0.34 MJ/m2 IT.51 MJ/m 2 IT.HDKR.001 Fdif f.LJ.snow = 2.P erez.P erez.392.24.5.7514 f 12 = −0.1061 f 23 = 0.25 Fdif f.HD.3965 ∆ = 0.12 MJ/m IT.16.16.205 MJ/m Ib.16.nosnow = 0.14 IT.HD.362 f 22 = −0.1507 AA = 0.05 MJ/m2 IT.7 p f = Ib /I (31) 3 IT. The coefficients from Table 2.7459 MJ/m2 2 Id. f 22 = −0.snow = Ib ·Rb +Id ·(1 − F 1)· a = max (0.2 rd = 0.262 F 1 = 0.003 MJ/m2 2 IT.HDKR.allBeam = 2.snow = 2.1 = 0.6807 MJ/m2 Id = 0.4353 [-] Hb = 4.snow = (Ib + AA · Id ) · Rb + Id · (1 − AA) · Fcs · 1 + f · sin (slope/2) + ρg.142 MJ/m2 Ib = 0.83 MJ/m IT.P erez. (f 11 + f 12 · ∆ + π · arccos (cosZen) · f 13/180)) F 2 = f 21 + f 22 · ∆ + π · arccos (cosZen) · f 23/180 Eqn 2.24 f 11 = 0.462.3399 CosZen = 0.5248 MJ/m2 2 IT.16.9 b = max (Cos(85).snow = 2.93 CosT heta = 0.5817 n = 20 Rb = 2.nosnow · I · Fcg (32) 3 IT.001 (39) Note that m=1/cosZen from Eqn 1.nosnow = 2.IT.362.1 and Ion =Io /CosZen so that ∆ = Id /Io Eqn 2. Equation 2.392 f 21 = 0.12 Eqn 2.snow = (Ib + AA · Id ) · Rb + Id · (1 − AA) · Fcs + ρg.HD.snow · I · Fcg (30) h) Estimation of IT with the HDKR model.HDKR. f 13 = −0.462 Fcs = 0.HDKR.226 Fcg = 0.5 [deg] 2 Ho = 13. Equation 2.nosnow = 2.nosnow = 2.6807 MJ/m2 Id.LJ.1 are then: f 11 = 0. f 23 = 0.nosnow · (34) 2 2 Eqn 2.isotropic = 2. cosT heta) (1 + Cos(Slope)) 1 − Cos(Slope) +Id ·F 1·a/b+Id ·F 2·sin (Slope)+I·ρg. ground = I · ρg · (15) 1 − Cos(slope) 2 (16) 33 .T = Rb · Ib Id. HourAngle = 7. n = nDay (month. day) HrAng2 = 0. ρg = 0. n) EES function or Eqn 2.2 (11) (12) (13) On the tilted surface using Liu and Jordan method: Ib. and then estimate the beam and diffuse compone s horizontal radiation.6.3 fd = Idif f /I (kt ) (6) (7) EES function or Eqn 2.10.13. (hrAng.3 or Table 1. HrAng2) kt = I/Io EES function or Eqn 1. n.10.7 Calculation of hour’s radiation on a south-facing vertical surface. calculate I o for that hour.1. This provides the basis for estimating radiation on the vertical surface with Equation 2.7 (1) HrAng1 = −15.8. Here we will solve in detail using the Liu and Jordan model.9.5 Eqn 1. Surf AzAng) CosZenith = CosZenAng (Lat.2 EES function or Eqn 1. Lat = 43. month = 12. They all follow the same basic procedure.10. (n. day = 22.1 Eqn 2.5. Slope. Lat.7 EES function or Table 1. starting with daily total.6. We have 4 model choices: Liu and Jordan.80 MJ/m2 . lat. HrAng.1 (4) I = H · rT (5) Io = IZero. rT = rt. H = 8. GrRef = 0. HrAng1.P02 07 Equations 2.2.10. n.15.10. Slope = 90. Hay and Davies.6.1 (8) Id = I · fd (9) I = Ib + Id (10) CosT heta = cosIncAng (Lat.5. Surf AzAng = 0.1 (2) (3) Method A: Find the hour‘s total radiation with Equation 2.1 or Fig 1. HrAng) Rb = CosT heta/CosZenith EES function or Eqn 1. HDKR and Perez.sky = Id · (14) 1 + Cos(slope) 2 Id. HrAng = −7. 5642 MJ/m IT.LJ.992 MJ/M IT.114 2 Io = 1.HDKR.sky + Id. ρg . Method B: Divide the day‘s radiation into beam and diffuse.138 MJ/m2 Lat = 43 [deg] Rb = 2.7 HrAng1 = −15 [deg] Ib. (I.152 MJ/m2 month = 12 rT = 0. HrAng) (18) IT.8092 ρg = 0.HDKR = IT.314 34 fd = 0.The total radiation on the vertical surface is then: IT = Ib.T = 3.812 MJ/m2 kt = 0. slope.2 gives the hour’ s total from the day‘s total.1.LJ = IT. Solution CosT heta = 0. HrAng) (20) Note that the L&J model is very conservative while the H&D and the HDKR models give similar results.ground = 0.346 MJ/m 2 Id. n. Surf AzAng. Lat.5 [deg] I = 1.5 [deg] 2 Ib = 1. H&D and HDKR models: IT.ground (17) We could use the EES functions for L&J. HrAng) (19) IT. Surf AzAng.9113 GrRef = 0.8 MJ/m2 HrAng2 =0 [deg] Id = 0.13.4. (I.266 MJ/m2 IT = 3. using Equation 2.sky = 0. n.HDKR = 4.HD.133 MJ/m IT.13. (I.HD = 4. Surf AzAng.11.1832 . ρg .LJ = 3. ρg . Lat. slope. and the hour’ s beam is then the difference between the hour‘s total and the hour’ s diffuse. Lat.165 HrAng = −7.7 Surf AzAng = 0 [deg] day = 22 HourAngle = 7. n. slope.3939 H = 8. and then get the hour’ s diffuse from the day‘s diffuse using Equation 2.812 MJ/m2 n = 356 Slope = 90 [deg] CosZenith = 0.612 MJ/m2 2 Id.T + Id. Equation 2.HD = IT. n.2 ¯ = 20.2 (6) ¯ d /H ¯ fdif f use = H (7) ¯ =H ¯b + H ¯d H (8) ¯ T.2. (n.LJ = H ¯b · R ¯b + H ¯ d · 1 + Cos(slope) + H ¯ · ρg · 1 − Cos(slope) H 2 2 Eqn 2.12.65 MJ/m2 H ¯ T.1. (1) (2) Eqn 2.KT = H ¯ T. et seq: ¯ T.281 R ¯ b = 14.LJ =H ¯ T. ρg H EES function or Eqn 2. KT bar = .19. Lat = 35.6.68 ρg = 0.19.P02 08 Equations 2. ¯ Lat. slope) R ¯ o = HZero. ρg ) H b) Using the Klein-Theilacker model.412 MJ/m2 H ¯ T = 0.LJ ( H.8 Calculation of daily average radiation on sloped surfaces a) Using the Liu & Jordan model. K EES function or Eqn 2.1 (5) ¯T fdif f use = Hdif f bar/Hbar Lat. ¯ T.19.KT = 23.12.1 [deg] ¯ b = 1.58 MJ/m2 H KT bar = 0. n.1 ¯b = R ¯ BEAM. (n. Lat.1 (9) This result could have been found from the single EES function.4 (10) Solution fdif f use = 0.06 MJ/m2 H EES function or Table 1.1 or Fig 2.2698 ¯ o = 29.3 (3) EES function or Eqn 10. Equation 2.4a. H.1. lat) H Slope = 35.3 (4) Eqn 2.20. ¯ = 20.6781 K n = 75 . ¯ Lat.9. Slope.68.05 MJ/m2 H month = 3 Slope = 35 [deg] 35 ¯ d = 5.LJ = 24. n = AveDay (month) ¯ T = H/ ¯ H ¯o K ρg = .06 MJ/m2 H ¯ T. Equation 2.78 MJ/m2 H Lat = 35.KT. Slope. n.20.1 month = 3. 2.06. a solution based on the Liu and Jordan model would be similar to that of Problem 2. H ¯ 8 = 26. Slopej . ¯ 4 = 25.53 H (3) (4) The EES duplicate statement will be used with arrays to simplify the calculations. H ¯ 11 = 12. H ¯ 5 = 28. Surf AzAng = 0 (1) Monthly average solar radiation is found in Appendix G. H slope4 = 50. slope2 = 20.87. H ¯ 9 = 22. H ¯ 10 = 17. slope5 = 90 ¯ 6 = 30. Lat. H ¯ 7 = 28. The following Arrays were entered using the Edit/Insert or Modify Array menu ¯ 1 = 11. : duplicate i = 1. 12 (5) monthi = i (6) duplicate j = 1. H slope1 = 0.P02 09 Equations 2. ρg = 0. AveDay (i) .54.8a.23.41 H (2) ¯ 12 = 10. The variable month was used for the abcissa.25.82. Hi .30.KT. H slope3 = 35.i. H ¯ 2 = 15.39. This solution is based on the Klein-Theilacker model.9 Monthly variation of radiation on surfaces of various orientations. Lat = 35.1. ¯ 3 = 20.56. ρg end j (8) (9) end i (10) The results are shown in the plot. 5 (7) ¯ ¯ ¯ H T. Arrays 36 . j = HT. 48 28.22 MJ/m 16.93 16.57 21.58 17.218 9.88 17.42 MJ/m 19.2 28.31 20.52 MJ/m 17.93 22.84 17.46 18.8 10.54 15.i.41 28.09 25.12 MJ/m 11.84 26.78 25.4 28.1 19.56 12.43 24.32 MJ/m 18.29 26.25 13.32 23.08 23.3 23.67 15.17 ¯ HT.75 21.i.76 15.29 26.86 23.28 10.25 26.12 19.18 17.Row 1 2 3 4 5 6 7 8 9 10 11 12 ¯ Hi 2 MJ/m 11.2 17.00 22.97 27.5 15.51 19.84 24.68 Plot 1 37 ¯ HT.9 25.23 20.5 ¯ HT.46 9.32 25.94 26.53 slopei [deg] 0 20 35 50 90 ¯ HT.9 22.i.87 10.57 monthi 1 2 3 4 5 6 7 8 9 10 11 12 .i.48 21.04 23.39 17.36 22.2 25.581 11.4 12.41 18.7 30.06 25.16 ¯ HT.75 22.73 21.27 28.04 22.82 30.03 23.27 25.i.30 28. 19. ¯ Lat. Slope. K EES function or Eqn 2.19.976 R .66 MJ/m2 H Lat = 36 [deg] ωs0 = 73.92 [deg] ¯ d = 3. (n.3a Eqn 2. Equation 2.P02 10 Equations 2.10 Estimation of monthly radiation on a vertical surface. See Problem 2.3b Eqn 2.LJ = H ¯b · R ¯b + H ¯ d · 1 + Cos(slope) + H ¯ · ρg · 1 − Cos(slope) H 2 2 Eqn 2.1 or Table 1.88 [deg] Slope = 90 [deg] (10) ¯ b = 4.LJ = 12.1 ωs = arccos (− tan (Lat) · tan (dec)) (8) Eqn 1.2 or Fig 2.302 MJ/m2 H ¯ T.10.KT = H ¯ T.19.1.20.2 (5) ¯ d /H ¯ Fd = H (6) ¯ =H ¯b + H ¯d H (7) dec = dec (n) EES function or Eqn 1. n.10 or EES function (9) ωs0 = min (arccos (− tan (Lat) · tan (Dec)).2. Lat) H ¯ T = H/ ¯ H ¯o K Slope = 90. ¯ T.4 MJ/m2 H M onth = 1 ρg = 0.4 could also be used. arccos (− tan (Lat − Slope) · tan (Dec))) ¯ b = Cos(lat − slope) · Cos(dec) · sin (ωs0 ) + (π/180) · ωs0 · sin (lat − slope) · sin (dec) R Cos(lat) · Cos(dec) · sin (ωs ) + (π/180) · ωs · sin (lat) · sin (dec) ¯ T.88 [deg] ¯ = 8.6.12.6. H.1 The Klein-Theilacker method is simple when the EES functions are used. This solution is based on the Liu & Jordan method.19. n = AveDay (month) ¯ o = HZero.498 MJ/m2 H ¯ T = 0.10.6.2 38 (11) (12) (13) Solution dec = −20. Equation 2. ρg = 0. ρg H Fd = 0.1 (4) ¯T Fd = Hdif f Bar/Hbar Lat.1 (3) Eqn 2. M onth = 1.KT.805 MJ/m2 H ¯ T.3 or Fig 1. n.47 K ωs = 73.10.4213 ¯ o = 17. ¯ T = 0.KT = 11.12.47 K (1) EES function or Table 1.1 (2) EES function or Eqn 1.9.8. Lat = 36.1 MJ/m2 H n = 17 ¯ b = 1. the Klein-Theilacker method. 45 MJ/m2 H ¯ T = 15.9.2. K month = 10 (1) EES function or Table 1. Lat) H ¯ T = H/ ¯ H ¯o K ρg = 0.1) is valid.5 K ρg = 0.1 (4) ¯T Fd = HDIF F BAR/HBAR Lat. slope) R EES function or Eqn 2.1 Eqn 2.9. (n.3 ¯ · ρg · 1 − Cos(slope) ¯T = H ¯b · R ¯b + H ¯ d · 1 + Cos(slope) + H H 2 2 ¯=H ¯ T /H ¯ R (8) Eqn 2.17 MJ/m2 H n = 288 Slope = 58 [deg] (9) ¯ b = 6.1 (2) EES function or Eqn 1.P02 11 Equations 2.452 R .3911 ¯ o = 20. Slope = 58. ¯ T = 0.12.6.10. Equation 2.1 (5) ¯ d /H ¯ Fd = H (6) ¯ =H ¯b + H ¯d H (7) ¯b = R ¯ BEAM.2 39 ¯ d = 4.3 (3) Eqn 1.2. This solution is based on the Liu & Jordan method.087 MJ/m2 H Lat = 43 [deg] ¯ = 1.363 MJ/m2 H ¯ T = 0. Lat = 43.19.19. (n. n. We also had to assume the monthly diffuse fraction (Eqn 1.50.19.9 MJ/m2 H month = 10 ¯ b = 1. K EES function or Eqn 2.816 R ¯ = 10.11 Estimation of October radiation on a sloped surface. Lat. n = AveDay (month) ¯ o = HZero.2 (10) Solution Fd = 0.1 and an assumption of GrRef = 0.19. 19.54.19. Equation 2. Equation 2.813 MJ/m2 H n = 198 40 ¯ d = 8.P02 12 Equations 2. K ¯ = 21. (9) Solution Fd = 0.20.1 Fd = HDIF F BAR/HBAR Lat.96 MJ/m2 H ¯ T.1: n = AveDay (month) EES function or Table 1.4a.3 ¯T Eqn 2.96. H.602 MJ/m2 H ¯ T = 0. and Klein & Theilacker Lat = 43.2 (8) b) Using the Klein-Theilacker model.KT = H ¯ T. n. (n.1 (2) ¯ o = HZero. et seq.KT = 10.4 et seq.6. slope) R (4) EES function or Eqn 2.1 MJ/m2 H month = 7 Slope = 90 ¯ b = 13.2. ¯ Lat.2482 R ¯ = 21. Lat. H month = 7 (1) Using the Liu & Jordan model.10. n.KT.56 MJ/m2 H Lat = 43 ¯ b = 0.12 Calculation of average radiation on sloped surfaces using Liu & Jordan.12. a) Slope = 90.LJ = H ¯b · R ¯b + H ¯ d · 1 + Cos(slope) + H ¯ · ρg · 1 − Cos(slope) H 2 2 (7) Eqn 2. ¯ T = 0. ¯ T. K (3) ¯ d /H ¯ Fd = H (5) ¯ =H ¯b + H ¯d H (6) ¯b = R ¯ BEAM.20.3917 ¯ o = 40.54 K ρg = 0.3 ¯ T. Lat) H EES function or Equation 1.19. (n.2 .36 MJ/m2 H ¯ T. Slope. ρg H EES function or Eqns 2.LJ = 9. ρg = 0. 6.3377 I = 0.1 (2) EES function or Eqn 1.8.01.1a 2 a∗0 = 0.8.4237 − 0.13 Hourly radiation on a horizontal surface for a standard clear day using Hottel’s method.43 [deg] Ib = 0.01858 1/km2 · (2.6. Lat = 46.4332 a1 = 0.169 dec = −23.5 (11) Just below Eqn 2. HrAng + 7.294 · τb τd = Id /I0 (5) rk = k/k ∗ rk = 1.put hrand in Parametric Table for final results.985 MJ/m2 k = 0.5 [deg]) EES function or Eqn 1.5 Use for checking . (n.00595 1/km2 · (6.7318 Day = 23 I0 = 1.03 τd = 0.5 (12) I0 = IZero.5 [deg] Id = 0.1741 CosZenith = 0.432.1b (6) 2 a∗1 = 0.1 (10) Eqn 2.1c (7) 2 k ∗ = 0.5 [km] − A) Eqn 2.5 (4) τb = a0 + a1 · exp (−k/cosZenith) Eqn 2.1 (3) CosZenith = Cos(Lat) · Cos(Dec) · Cos(HrAng) + sin (Lat) · sin (Dec) Eqn 1. r1 = 1.10.4 (13) Ib = I0 · τb (14) I = Ib + Id (15) Solution A = 0.5.3506 r0 = 1.00 (9) Table 2.5 [km] − A) Eqn 2.1436 Table 1 41 a∗0 = 0. τd = 0.8.6. hrAng=-52.01 .008216 1/km2 · (6 [km] − A) Eqn 2.7246 hrAng = −7.740 MJ/m2 Lat = 46.8.8.5deg .P02 13 Equations 2.432 [km] a∗1 = 0.3506 n = 357 τb = 0. day) dec = dec (n) month = 12.2711 + 0. Day = 23 (1) EES function or Table 1. Lat. The solution will use a parametric table.1d (8) r0 = a0 /a∗0 . HrAng − 7.5055 + 0.03.5 [deg] r1 = 1.8. r0 = 1.245 MJ/m2 month = 12 rk = 1 a0 = 0.8.271 − 0. Arrays could have been used.707 MJ/m2 k ∗ = 0. A = 0. n = nDay (month. r1 = a1 /a∗1 . 491 1.192 0.103 0.Run 1 2 3 4 hrAng [deg] -52.5 -7.707 Ib 2 MJ/m 0.073 1.592 0.529 0.245 I 2 MJ/m 0.5 -37.1785 0.4332 0.338 0.1543 0.985 .5 -22.3148 0.822 0.4825 1.740 42 Id 2 MJ/m 0.197 0.230 0.3971 0.093 0.5 τb τd 0.1436 I0 2 MJ/m 0.1931 0.2142 0. Ibn. Hour5 = 12.7 = 1.35.5. day = 7.80.5.i = (cosZenithAnglei ) + (Cos(Dec)) · (sin (HrAng i )) Eqn 1.5 n = nDay (month.5.6.5 HrAng 8 = 52.7. day) 284 + n Dec = 23.5 Hour8 = 15. month = 1 (1) The following arrays were added using the Edit/Insert Modify Array menu item.42 HrAng 1 = −52. Lat = 38.5. (2) (3) Ibn.5.2 = 0. Ibn. Hour1 = 8. Hour2 = 9.70. 8 (10) CosZenithAnglei = Cos(Lat) · Cos(Dec) · Cos(HrAng i ) + sin (Lat) · sin (Dec) Eqn 1. HrAng 7 = 37.6.05.3a (11) Ib.42 [deg] n=7 Lat = 38 [deg] 43 . Ibn. north-south axis tracking surface. Hour4 = 11.1 = 0.5.5.1 (9) duplicate i = 1.19 (4) (5) HrAng 2 = −37. HrAng 4 = −7. HrAng 3 = −22.i · CosT hetaN S. Ibn. Hour3 = 10.5.5 Ibn. HrAng 6 = 22. HrAng 5 = 7. Ibn.5.30.14 Calculation of hourly beam radiation on a horizontal.5.5.8 = 1.4 = 3. Hour7 = 14.i = Ibn.6 = 3. Ibn.5 = 3.66.45 [deg] · sin 360 · 365 (6) (7) (8) Eqn 1. Hour6 = 13.5.3 = 2.P02 14 Equations 2.i (13) end (14) Solution day = 7 month = 1 (12) Dec = −22.5 q 2 2 2 CosT hetaN S. 66 1.5 10.502 0.80 1.5 9.5 HrAng i [deg] -52.50 1.5 -37.42 CosT hetaN S.19 1.i 2 MJ/m 0.66 3.659 0.343 0.5 14.5 37.30 3.35 0.5 11.5 -7.27 0.19 1.08 .5 22.05 3.5 7.563 0.563 0.5 12.70 2.5 15.i 2 MJ/m 0.i CosZenithAnglei 0.46 1.Arrays Row Houri 1 2 3 4 5 6 7 8 8.502 0.438 0.209 44 Ib.80 1.487 0.659 0.762 0.209 0.53 1.343 0.487 0.5 Ibn.5 52.5 13.438 0.762 0.5 -22. Lat) KT.15 Estimation of diffuse and beam components using clearness indexes Lat = 45.1 or Table 1.day = H/Ho EES function or Eqn 1.12.P02 15 Equations 2.2 (17) ¯ d /H ¯ F¯d = H (18) ¯ =H ¯b + H ¯d H (19) 45 .2 (11) fd.10. nave = AveDay (month) ¯ o = HZero.10. nave .9.3 or Figure 1.1 for hourly calculation: n = nDay (month. Lat) H ¯ T = H/ ¯ H ¯o K (14) EES function or Eqn 1. Ho = HZero.9.1 or Figure 2. K (15) (16) EES function or Eqn 2. HrAng1.1 or Table 1.2 (6) Id /I = fd (7) I = Id + Ib (8) b Estimation of beam and diffuse components for a day.1 ¯T F¯d = HDIF F BAR/HBAR Lat. day = 8.10.10. a) HrAng2 = −15 [deg] .hr = I/Io EES function or Eqn 1.hr ) (4) (5) EES function or Eqn 2. month = 2.day ) (9) (10) EES function or Eqn 2. I = 1.9. Lat. n.4 Eqn 2. (n.57 MJ/m2 ¯ = 8.10.10.day = Hd /H (12) H = H d + Hb (13) c) Estimation of average day’s beam and diffuse. HrAng1 = −30 [deg] .1 or Figure 2.10.1 or Figure 2.67 MJ/m2 H (1) (2) Use Equation 2. H = 10.1 Eqn 2.11. HrAng2) kT.10.2 fd. (n. KT.80 MJ/m2 .11.day = Hdif f /H (Lat. day) (3) Io = IZero.10.1 Eqn 2.10. (nave .3 fd = Idif f /I (kT.12.3 or Figure 1. 2304 HrAng2 = −15 [deg] ¯ d = 3.Solution day = 8 H = 10.1934 Hb = 8.209 MJ/m2 H Ho = 15.208 MJ/m2 KT.4918 K month = 2 46 fd.6818 n = 39 .day = 0.3992 HrAng1 = −30 [deg] ¯ b = 5.hr = 0.089 MJ/m2 Id = 0.8 MJ/m2 ¯ = 8.57 MJ/m2 ¯ T = 0.461 MJ/m2 H I = 1.214 MJ/m2 Lat = 45 [deg] fd = 0.day = 0.84 MJ/m2 Io = 2.67 MJ/m2 H Hd = 2.3617 MJ/m2 kT.7091 nave = 47 F¯d = 0.711 MJ/m2 ¯ o = 17.63 MJ/m2 H Ib = 1. 6. the estimation of the beam-diffuse distribution.T.9.4 (4) Eqn 2. HrAng = −22. day) (3) Io = IZero.1 or EES function (9) CosT heta = Cos(lat − slope) · Cos(dec) · Cos(hrAng) + sin (lat − slope) · sin (dec) CosZenith = Cos(lat) · Cos(dec) · Cos(hrAng) + sin (lat) · sin (dec) Eqn 1.10.T. n = nDay (month.10. Equation 2. HrAng 2 ) kT = I/Io EES function or Eqn 1.P02 16 Equations 2. (n.6.1 or Fig 2.16 Estimation of an hour’s radiation on south facing surfaces with three sky models HrAng 1 = −30.5 or EES function (11) θ = arccos (CosT heta) (12) θz = arccos (CosZenith) (13) Rb = CosT heta/CosZenith Eqn 1. Lat.LJ = Id · beam by Liu and Jordan 1 + Cos(slope) 2 Ig. and some angles are common to all of the models.15.6.3 fd = Idif f /I (kT ) (5) EES function or Eqn 2.7. day = 8 (1) Surf AzAng = 0 (2) The first part. HrAng 1 .T.1 or EES function (14) a) Use of Liu & Jordan model.LJ = I · ρg · 1 − Cos(slope) 2 (15) diffuse by Liu and Jordan ground reflected by Liu and Jordan 47 (16) (17) .LJ = Ib · Rb Id. ρg = 0. month = 2.5. HrAng 2 = −15. Lat = 45. I = 1.7a or EES function(10) Eqn 1.10.57.1 Ib.2 (6) fd = Id /I (7) I = Ib + Id (8) Dec = 23. slope = 60.45 [deg] · sin 360 · 284 + n 365 Eqn 1.8. 12 Eqn 2.LJx = IT.HDKR = (Ib + Id · Ai ) · Rb (22) 1 + Cos(slope) 2 Id.056 (30) Equation 2.16. Lat.LJ + Id.LJ (18) The total radiation of the sloped surface and could have been easily found from the EES function: IT.T.16.16. Slope.P erez = Id · (1 − F1 ) · + F1 · a/b + F2 · sin (Slope) 2 1 − Cos(Slope) 2 = Ib.HDKR (23) (24) (25) The total radiation of the sloped surface and could have been easily found from the EES function: IT.16.823. HrAng) (26) (c Estimation of radiation on sloped surface using the Perez model.535 · 10−6 1/deg3 · θz3 Id = (1 + (5. f 21 = 0.T.16.T.T.T.T.HDKR + Ig.412.P erez + Id.9 (28) Calculate to use to get f11.T.16.11.T. Lat.7 (21) Ib.T. (I. HrAng) (19) b) Estimation of hourly radiation on sloped surface using the HDKR model.LJ = Ib.14 a = max (0.16.7 Ai = Ib /Io p f = Ib /I Eqn 1. Eqn 2. f 11 + f 12 · ∆ + π · θz · F2 = f 21 + f 22 · ∆ + π · θz · f 23 180 [deg] f 13 180 [deg] Eqn 2. etc from Table 2.HDKRx = IT. f 13 = −0.288.9 b = max (Cos(85). Equation 2. CosZenith) (27) Eqn 2.3 (20) Eqn 2.P erez + Ig.HDKR = Ib. Slope. The coefficients from Table 2.T.HDKR + Id.16.P erez (37) Solution Variables in Main program 48 .11 reduces to: ∆ = Id /Io (31) F1 = max 0.P erez = Ib · Rb (1 + Cos(Slope)) Id. f 23 = .T. ρg .10 (29) is 4. Surf AzAng. ρg .16.LJ + Ig.T.T.1 are: f 11 = 1. n. (I.237.132.HDKR = I · ρg · 2 IT. f12.HDKR.T.LJ.T.P erez = I · ρg · (36) IT.13 Ib.P erez (32) (33) (34) (35) Ig.IT. Surf AzAng.16.16.HDKR = Id · (1 − Ai ) · · 1 + f · sin (slope/2) 2 1 − Cos(slope) Ig.16. f 12 = −1. f 22 = −0.535 · 10−6 ) [1/deg3 ] · θz3 ) Eqn 2. n. CosT heta) Eqn 2.1: Id +Ib /CosZenith + 5. T.T.P erez = 2.27 MJ/m2 Id.7 θ = 21.a = 0.LJx = 3.39 MJ/m2 IT.HDKR = 0.HDKR = 2.HDKRx = 3.106 f 13 = −0.27 MJ/m IT.P erez = 3.T.97 MJ/m2 2 Id.T.LJ = 0.P erez = 0.LJ = 3.57 MJ/m2 2 Io = 2.T.10 MJ/m2 Lat = 45 [deg] Rb = 2.5 [deg] 2 Ib = 1.HDKR = 2.15 MJ/m2 Ib.T.T.LJ = 0.21 MJ/m IT.HDKR = 3.39 MJ/m2 Ig.1634 f 12 = −1.HDKR = 3.4403 = 4.T.55 MJ/m2 2 IT.97 MJ/m2 IT.P erez = 0.38 MJ/m2 kT = 0.9291 CosZenith = 0.P erez = 3.T.216 HrAng 2 = −15 [deg] Ib.P erez = 0.5457 day = 8 f = 0.P erez = 2.57 MJ/m2 Ib.T.8773 f 21 = 0.HDKR = 0.9291 ∆ = 0.10 MJ/m2 month = 2 slope = 60 [deg] CosT heta = 0.2304 [-] I = 1.27 MJ/m2 Ib.208 MJ/m Id = 0.LJ = 0.27 MJ/m2 Ib.T.237 f 23 = 0.4403 Dec = −15.88 [deg] Key Variables Ig.412 F1 = 0.39 MJ/m2 a a a a b b b b c c c c 49 b = 0.15 MJ/m 2 Ig.27 MJ/m2 IT.LJ = 0.LJ = 3.27 MJ/m 2 Ig.55 MJ/m2 2 Id.T.T.T.P erez = 0.11 θz = 63.7 [deg] Ai = 0.132 f 22 = −0.4706 HrAng 1 = −30 [deg] Ib.52 [deg] f 11 = 1.27 MJ/m IT.57 MJ/m2 IT.55 MJ/m2 Id.T.27 MJ/m2 Id.HDKR = 0.LJ = 2.7091 [-] ρg = 0.3617 MJ/m 2 Ig.10 MJ/m Ig.LJ = 2.T.T.823 fd = 0.55 MJ/m2 Id.39 MJ/m2 n = 39 Surf AzAng = 0 [deg] .T.288 F2 = 0.056 HrAng = −22.HDKR = 0. 9. the estimation of the beam-diffuse distribution.10.57.3a or EES function (12) θN S = arccos (CosT hetaN S ) (13) Ib. HrAng 2 ) kT = I/Io EES function or Eqn 1.5.17 From an hour’s radiation on horizontal surface.45 [deg] · sin 360 · 284 + n 365 Eqn 1. month = 2. slope = 60. ρg = 0. HrAng 2 = −15. HrAng 1 . and some angles are common to all of the models. day = 8 (1) Surf AzAng = 0 (2) The first part.2304 2 I = 1.10.3617 MJ/m2 month = 2 Surf AzAng = 0 [deg] .7 θz = 63. get beam radiation on tracking surface HrAng 1 = −30.6.1 or EES function CosZenith = Cos(lat) · Cos(dec) · Cos(hrAng) + sin (lat) · sin (dec) (9) Eqn 1. Lat.208 MJ/m2 kT = 0.5 or EES function θz = arccos (CosZenith) (10) (11) .694 MJ/m2 Lat = 45 [deg] slope = 60 [deg] 50 Dec = −15.P02 17 Equations 2.1 or Fig 2.4403 HrAng = −22.N S = Ib · CosT hetaN S (14) Solution CosT hetaN S = 0.57 MJ/m Io = 2.5 [deg] Ib = 1.5 2 2 CosT hetaN S = cosZenith2 + (Cos(Dec)) · (sin (HrAng)) Eqn 1.N S = 0.7091 ρg = 0.95 [deg] cosZenith = 0. I = 1.6.214 MJ/m2 n = 39 θN S = 54.10.88 [deg] day = 8 HrAng 1 = −30 [deg] Ib.3 fd = Idif f /I (kT ) (5) EES function or Eqn 2. n = nDay (month.4 (4) Eqn 2. (n.7.2 (6) fd = Id /I (7) I = Ib + Id (8) Dec = 23. HrAng = −22.5743 fd = 0. Lat = 45. day) (3) Io = IZero.7.52 [deg] HrAng 2 = −15 [deg] Id = 0. 2 = 0.18 Calculation of monthly average beam.i H ¯ b.59.1 = 0.i EES function or Eqn 1.2. ¯ 1 = 6.i ) CALL H Lat.12.8 = 0.i = Hi K Ho. (H[i].7.p.i = H ¯ i · ρg. and total radiation on sloped surface Lat = 43.P02 18 Equations 2. ¯ 5 = 21.i (15) (16) (17) end (18) This problem is easily done by the following call to an EES procecdure.i · 1 + Cos(Slope) H 2 1 − Cos(Slope) ¯ T.2 ρg. diffuse. H ¯ T.i + HT.4 = 0.i Fd.86. ρg. H ρg.12.i + HT. ¯ 3 = 12.10. The arrays were entered using ‘Insert/modify array’ in the Edit menu. ¯ 2 = 9. H H H H H ¯ 7 = 22.g.2.3 (12) ¯i = H ¯ b.LJ.i = H ¯ b. H ¯ 10 = 10.7 (2) (3) (4) (5) This solution uses the Liu and Jordan model duplicate i = 1.i = HDIF F BAR/HBAR Lat.i .2.9. (ni .11 = 0. ρg.5 = 0.i · H 2 ¯ ¯ ¯ ¯ HT.6 = 0.d. H ¯ T.05.19.1 (9) ¯ T.36. ρg. ¯ ¯ T. H ρg. Duplicate i=1. (ni .b.i H (14) ¯ T.g. ground-reflected.2.i = R ¯ BEAM.i . ρg. Note that the EES procedure returns the three components. Lat. ni .i H (13) ¯ T.2. ρg.i = H ¯ i · Fd. 12 (6) ni = AveDay (i) (7) Ho. ρg. ¯ 12 = 5. H ¯ T.b.g.d.1 ¯ d.10 = 0.9 = 0.58.1 or Figure 2. ρg.i · R ¯ b.37.44.7. ¯ 4 = 16. Slope) R (8) (10) (11) EES function or Eqn 2.i = HZero.i : H End 51 . Slope.3 = 0. Slope = 60 (1) Monthly average radiation data are from Appendix G.b. ¯ 6 = 23. n[i].12 = 0.74 H ρg.d.2. ρg.i = HT. H ¯ 8 = 20.i + H ¯ d.i .i = H ¯ d.89.7 = 0.04 H ¯ 11 = 6.4.33.12 n[i]=AveDay(i ) ¯ T.3 Eqn 2. K EES function or Eqn 2. Lat) ¯ ¯ T. ρg. H ¯ 9 = 14.48.4. 566 0.i2 MJ/m 1.48 6.84 10.50 7.7 Ho.85 10.22 11.062 ρg.i 2 MJ/m 13.423 0.i 0.42 41.47 15.25 10.78 39.07 1.42 9.g.08 2.565 3.49 6.44 9.2 0.76 Fd.71 17.043 1.79 9.25 14.45 3.25 6.72 4.551 0.28 8.06 2.15 .2 0.05 21.617 0.18 11.482 0.90 14.58 20.409 ¯ Hd.29 0.84 8.40 3.73 1.03 14.63 5.27 8.39 ¯ b.557 0.52 0.00 52 ¯ HT.Arrays Row ni 1 2 3 4 5 6 7 8 9 10 11 12 17 47 75 105 135 162 198 228 258 288 318 344 Row 1 2 3 4 5 6 7 8 9 10 11 12 ¯ Hi 2 MJ/m 6.4 0.08 12.435 0.74 ¯ Hb.81 6.59 6.39 ¯ T.960 0.86 16.64 1.7 0.73 0.59 7.13 1.80 10.17 4.i2 MJ/m 1.2 0.89 12.844 1.61 12.436 0.59 10.19 5.2 0.i2 MJ/m 10.67 13.26 7.i K [] 0.482 ¯ HT.4 0.58 16.453 0.15 1.79 8.368 0.i 2 MJ/m 13.63 3.19 1.62 4.377 0.91 ¯ HT.25 15.81 26.367 0.2 0.79 14.788 2.46 6.63 3.i 2 MJ/m 3.381 0.92 28.31 8.37 5.501 0.526 0.i [] 0.b.78 40.69 16.214 1.26 8.13 1.62 11.42 6.36 23.03 33.i R [] 2.390 0.77 13.494 0.36 18.711 0.419 0.7 0.33 8.507 0.04 22.2 0.i 2 MJ/m 2.60 8.390 0.58 15.2 0.76 17.98 2.475 0.813 2.56 35.35 ¯ HT.80 1.657 0.38 5.53 16.542 0.459 0.92 2.d.409 0.02 0.33 14.80 20. 17. ρg.i : H ¯ T b.97.P02 19 Equations 2.3 = 0. .i MJ/m2 2.i call H ¯ T. ρg.2 0. H ¯ T.57 H (3) ρg.72 2. south-facing surface Lat = 39. H ¯ i .6 = 0.7 (5) a Use EES Procedures for both the Liu & Jordan (Section 2.LJ.KT.2 17 47 75 105 135 ¯ HT b.45 12. ρg. ρg.8 [deg] Slope = 90 [deg] Arrays 1 2 3 4 5 (9) (10) Solution Row (8) ¯ Hi 2 MJ/m 6.5. ni .i : H ¯ T b. ρg.2.g.4 0.50 .4.i MJ/m2 8.12 ¯ HT. ni .i2 MJ/m 12.10 = 0.i .42 2.i .68 3.LJ. ρg.p.2.2. H ¯ T.LJ.6 9.72 4.99 6.4 = 0.i call H end Lat = 39.5.7 0.17 ¯ HT d.36.KT.77. H ¯ 2 = 9.i .04 2. H ¯ 7 = 23.20) methods.4.) duplicate i = 1.21 53 ¯ HT.8 13. (2) ¯ 12 = 5.i .LJ.i .11 = 0.LJ.LJ.8 H ¯ 11 = 7.2 = 0.KT. H ¯ 8 = 20.12 = 0.7 0.p.g.19 Estimation of monthly average radiation on a vertical. Slope = 90 ¯ 1 = 6. H ρg. H ¯ T d.LJ.24 10.i ni 0. H ¯ 5 = 21. ρg. H ¯ 4 = 17.05 14.63.13 9.0 17. H ¯ T.20 8.8. Slope. Lat.13.53 2. ρg.5 = 0.18.LJ. H ρg.2 ρg.19) and Klein & Theilacker (Section 2. H ¯ 9 = 16.i MJ/m2 1.7. 12 (6) ni = AveDay (i) (7) ¯ T.9 = 0.2. ρg.68. Lat. H ¯ T. H ¯ T d.97 3.32 3.2 (4) ρg.LJ.8 = 0.KT.2 21.2.1 = 0. Slope.2.KT. (The detailed L & J solution is essentially the same as that of Problem 2. ρg.i . H ¯ 10 = 12.2. H ¯ i .68 4. H (1) ¯ 3 = 12.7.59 1. ¯ 6 = 23.g.7 = 0. 34 2.21 3.59 1.2 0.40 1.g.39 2.8 23.17 2.49 4.21 4.65 1.17 7.2 0.41 8.53 12.4 20.60 8.80 3.73 9.32 9.42 2.53 2.21 3.34 4.95 ¯ HT.72 4.39 2.2 0.5 16.29 9.48 9.91 12.i MJ/m2 7.5 12.03 4.83 10.i MJ/m2 2.97 3.30 13.55 6.04 2.KT.45 6.6 7 8 9 10 11 12 23.32 3.95 ¯ H T.48 9.01 7.75 9.91 54 9.82 3.42 2.82 3.17 2.30 13.74 7.35 10.72 2.7 162 198 228 258 288 318 344 ¯ HT d.58 10.i MJ/m2 11.64 1.KT.57 1 2 3 4 5 6 7 8 9 10 11 12 0.38 2.54 1.40 1.38 2.28 5.34 2.05 1.64 1.4 0.05 1.21 1.KT.12 2.12 4.21 1.2 0.02 11.34 4.88 12.52 9.KT.7 5.88 3.6 Row ¯ HT b.i MJ/m2 1.2 0.54 1.45 13.92 9.62 12.1 7.65 1.46 .05 10.57 3. ) View factor algebra Fc.2.0 Reciprosity relationship (7) (8) 1 + Cos(slope) 2 (9) For ψ=90.2 [m] .a = 0.s + Fc. I = 1. In this solution. Slope = 90.1 [m] ρg = 0. n/m=2. m = 4. It is also assumed that the diffuse radiation on the ground beyond the reflector is not significantly affected by the building.s so the radiation on the receiver is: Ir = Ib + Id · Fr. the diffuse radiation on the reflector is assumed to be reduced to half of that on an unobscured horizontal plane.2=0. Id = 0. n = 2.1/4.1 (1) (2) ρr.70 diffuse reflectance for case b (4) Ac = m · n (5) Ar = p · m (6) Equation 2. (These two assumptions tend to offset. Rb = 2. p/m=2.r =Ar *Fr.17.25 (10) The radiation on the reflector is Ar *Ir =Ar * Ib +As *Id *Fs -r.20 Estimation of effect of finite horizontal diffuse reflector on IT Ib = 1.b = 0.g = 1 Ac · Fc. on the assumption that the building of which the window is part is large in extent.1 is for the case of a horizontal reflector.17.1 [m] .28. and is applicable here.31.5 then from Figure 2.s = Sum of view factors from collector to surroundings must be 1.s + Fr. But with reciprosity As *Fs.r = Ar · Fr.see Figure 2.r + Fc.c = 0.c = 1 (11) Sum of view factors from reflector to surroundings must be 1.0 55 (12) .s Fr.1/4.33.85 diffuse reflectance for case a (3) ρr.2a: Fr.17.5. p = 2.c Fc.P02 20 Equations 2. dimensions of window and reflector .2=0.61. a · Fc.a is: IT.82 m2 Fr.a = 0.1 [m] Rb = 2.a = Ib · Rb + Id · Fc.53 MJ/m2 ρg = 0.c = 0.a = 3.1 (13) And for case b with reflector reflectance ρr.b · Fc.33 MJ/m2 n = 2.61 MJ/m IT.g = 0. the radiation on the window for case a with reflector reflectance ρr.b is: IT.85 56 Fc.28 MJ/m2 m = 4.528 p = 2.25 2 I = 1.s = 0.b = 0.r = 0.2 [m] ρr.82 m2 Fr.b = 3.s + Ir · ρr.r + I · ρg · Fc.25 Ib = 1.47 MJ/m2 ρr.25 Ir = 1.s = 0.r + I · ρg · Fc.g (14) Note that the difference is not very large.7 Fc.31 .17. Solution Ac = 8.s + Ir · ρr.5 Id = 0.75 IT.b = Ib · Rb + Id · Fc.1 [m] Slope = 90 [deg] Ar = 8.2 Fc.Finally.g Eqn 2. AM (5) (6) For the PM .P M b) (7) (8) Tracking on horizontal NS axis: For the AM CosZenithAngleAM = Cos(Lat) · Cos(δ) · Cos(HrAng AM ) + sin (Lat) · sin (δ) Eqn 1. day) (3) δ = dec (n) (4) a) Tracking on horizontal EW axis: For the AM .P M = CosT hetaN S P M · Ibn.31 MJ/m2 .5 2 2 2 CosT hetaN S P M = (cosZenithAngleP M ) + (Cos(δ)) · (sin (HrAng P M )) Eqn 1. month = 2.AM = CosT hetaEW AM · Ibn.P M (14) 57 (13) .6.AM = 2. Ibn.b.5 [deg] n = nDay (month.P M = 1.21 Calculation of beam radiation on the aperture of tracking collectors Lat = 40.AM = CosT hetaN S AM · Ibn.3a (12) IT beamN S. (2) hrAng P M = 52.b.P02 21 Equations 2.7.EW.EW. Ibn.5 .5 [deg] .AM (11) (9) For the PM CosZenithAngleP M = Cos(Lat) · Cos(δ) · Cos(HrAng P M ) + sin (Lat) · sin (δ) Eqn 1.7.2a IT. day = 20.67 MJ/m2 (1) hrAng AM = −7.2a IT.5 .3a (10) IT beamN S.7.P M = CosT hetaEW P M · Ibn.5 2 2 CosT hetaEW AM = 1 − (Cos(δ)) · (sin (HrAng AM )) Eqn 1.5 2 2 2 CosT hetaN S AM = (cosZenithAngleAM ) + (Cos(δ)) · (sin (HrAng AM )) Eqn 1.6.5 2 2 CosT hetaEW P M = 1 − (Cos(δ)) · (sin (HrAng P M )) Eqn 1.7. 9918 CosT hetaN S P M = 0.P M = 1.EW.29 MJ/m2 month = 2 .EW.41 MJ/m2 Lat = 40 [deg] 58 CosT hetaN S AM = 0.b.8435 day = 20 hrAng P M = 52.31 MJ/m2 IT beamN S.P M = 1.6292 CosZenithAngleAM = 0.6282 CosZenithAngleP M = 0.45 MJ/m IT.b.AM = 2.67 MJ/m IT.05 MJ/m2 n = 51 CosT hetaEW P M = 0.615 δ = −11.5 [deg] 2 IT beamN S.58 [deg] Ibn.AM = 2.P M = 1.AM = 1.3278 hrAng AM = −7.Solution CosT hetaEW AM = 0.5 [deg] 2 Ibn. 3 = 1.52 (2) φ1 = 0.06: Xc. i = 1. I¯2 = 1.56 (4) For hour pair 9-10 and 2-3 at Xc =1.15.72/0.47 ¯ = 1.22.68.35 (5) N = N umDay (1) (6) The monthly average utilizable energy for these 3 hour pairs is: U E = 2 · N · Sum I¯i · φi .52 MJ/m2 I[1] φ[1] = 0.68. which is specific to Blue Hill For the hour pair 11-12 and 12-1: Xc. 3 (7) Solution ¯ = 1.3.1 = 0.72/1.68 MJ/m2 I[3] φ[3] = 0.9 MJ/m2 .15.56 Xc. I¯3 = 0.63 ¯ = 0.65 (3) From Fig 2.22 Calculation of utilizable energy $Arrays off I¯1 = 1.65 Xc.2 = 0.52.47 For hour pair 10-11 and 1-2 at Xc =0.15 MJ/m2 I[2] φ[2] = 0.1 = 0.06 59 N = 31 U E = 115.2 = 0.18.P02 22 Equations 2.35 Xc.72/1. (1) Use Figure 2.3 at Xc =0.63: Xc. φ3 = 0. φ2 = 0.3 = 0. K ρg = 0. Lat. Surf AzAng = 0.1 (2) (3) Find the monthly average daily diffuse fraction ¯T fd = HDIF F BAR/HBAR Lat. n = AveDay (month) Ic.13.23 Calculation of utilizability and utilizable energy using generalized utilizability charts. Slope.2 1 + Cos(slope) 1 − Cos(slope) I¯T = Ib · Rb + Id · + I · ρg · 2 2 Liu and Jordan model. Figure 2. H Ic. Eqn 2.1 Find the two dimensionless critical radiation levels . n.6. n. (1) Slope = 60 EES function or Table 1. Lat = 45 month = 2. (hrAng. HrAng = 22.P02 23 Equations 2. ¯ T = 0.1 = 536. Lat.1.67.5.6.23. (Lat.2 (6) EES function or Eqn 2. HrAng.15. ¯ = 8. rT = rt.4 (7) ¯d Id = rd · H (8) ¯ I = rT · H (9) I = Ib + Id (10) Find radiation on the tilted surface Rb = RBeam. Rb . K (4) ¯ d /H ¯ = fd H (5) Find radiation in the given hour.49. The major steps are calculation of month’s diffuse fraction. (HrAng. Xc . I¯T .13. n) rd = rd. n) EES function or Eqn 2. and phi.8.2 = 322. Surf AzAng) EES function or Eqn 1. . MJ/m2 0.0036 W·hr/m2 Xc1 = Ic,1 · 1 [hr] · Eqn2.22.3 - note needed conversion factor ¯IT 60 (11) (12) (13) Xc2 MJ/m2 0036 W·hr/m2 .0. 2.0696 · − 0.96 and Xc =0.23.5b (22) (23) (24) Since Xm > 2 we use the following for φ q 2 φ1×10 qn = abs (g) − g 2 + (1 + 2 · g) · (1 − Xc1 /Xm ) Eqn 2.49 K Nday = 28 φ2eqn = 0.23.T was already used.54 ¯ h = 1.23.169 · g= Xm − 1 (2 − Xm ) (21) ¯h Cos(slope) κ ¯T R + 0.23.7 .5a (25) q 2 g 2 + (1 + 2 · g) · (1 − Xc2 /Xm ) Eqn 2.3 .T not a lower case k since Kbar.75 MJ/m2 Id = 0.00 MJ/m2 κ ¯ T = 0.8 to find utilizability δ = dec (n) ¯h = R I¯T ¯ rT · H ¯ T · rT /rd κ ¯T = K (20) Eqn 2.23.22.23 MJ/m Ic.1 = 536 MJ/m2 ¯ T = 0.526 61 HrAng =22.17 and Xc =0.2 · 1 [hr] · ¯IT Eqn2.5 [deg] Ib = 0.note needed conversion factor (14) ¯b To use Figures 2.25.22.5082 n = 47 φ2 = 0.note this is kappabar.7 MJ/m2 Xm = 2.23.22.95 ¯ = 8.48 MJ/m2 H Ic.5a (26) φ2×10 qn = abs (g) − From the above equation (with Xc1 and Xc2 ) are φ1 = 0.5c Eqn 2.94 rT = 0.34 and φ2 = 0.1 we need to find R ¯b = R ¯ BEAM.54.54 ¯ b = 2.47 MJ/m2 month = 2 φ1eqn = 0.31.2 (19) We could use Eqns 2. = Ic.17 R Slope = 60 [deg] Xc1 = 0. the utilizable energies are: Nday = N umDay (month) (17) U E 1 = 2 · Nday · φ1 · I¯T from statement just under Eqn 2.8 Eqn 2.5 .54 (16) Finally.67 MJ/m2 H I¯T = 2. Slope) R EES function or Eqn 2. Reading Figures 2.3 (15) ¯ b = 2.9 2 I = 1.579 [-] g = −2.1 b and c and interpolating for R Reading and interpolating the graphs is difficult.981 · 2 κ ¯ 2T κ ¯ 2T Cos(δ) Eqn 2.2 (18) U E 2 = 2 · Nday · φ2 · I¯T from statement just under Eqn 2.1416 U E 2 = 60.34 Rb = 1.85 + 0.964 [-] fd = 0. φ1 = 0. φ2 = 0.63 R Surf AzAng = 0 [deg] Xc2 = 0.2 = 322 MJ/m2 Lat = 45 [deg] φ1 = 0.23.23. Xm = 1. Lat. Solution δ = −12.5 MJ/m2 .401 [-] ¯ d = 3.31 ρg = 0.19.1365 U E 1 = 34.6 rd = 0.23.58. (n. P02 24 Equations 2.7. . Itc = 124 W/m2 . Lat = −34.2 HourAngleN oon = 0. month = 12.24 Calculation of daily utilizability and month’s utilizable energy ¯ = 28. Slope = 20. H ρg = . . . MJ/m2 . . c = IT c · 1 [hr] · . IT. . 0036 W · hr/m2 .0. (n.9.12. n) EES function or Eqn 2. (HrAng noon .1 or Fig 2. Lat.13. (HrAng noon . Lat) H ¯ T = H/ ¯ H ¯o K EES function or Table 1.n = rd. (n.n = rt.2 or Fig 2.3 (5) Eqn 2.13.1 (14) rd.13.12.1 (6) ¯ Find radiation on collector and R ¯T f¯d = HDIF F BAR/HBAR Lat. Surf AzAng = 180 Southern hemi facing N (1) (2) (3) ¯T Find K n = AveDay (month) ¯ o = HZero.13. Slope.2 (15) 62 . ρg ) H ¯=H ¯ T /H ¯ R (12) Find Rn HrAng noon = 0 [deg] (13) rt.LJ. n.1 (4) EES function or Eqn 1. (H. n) EES function or Eqn 2. Slope) R (10) ¯T = H ¯b · R ¯b + H ¯ d · 1 + Cos(slope) + ρg · H ¯ · 1 − Cos(slope) H 2 2 (11) The following could have been used instead of the previous five equations ¯ T =H ¯ T.4 or Fig 2.6.2 (7) ¯ d /H ¯ = f¯d H (8) ¯ =H ¯b + H ¯d H (9) ¯b = R ¯ BEAM. n. Lat. ¯ Lat.10. Lat. K EES function or Eqn 2. 3355 Hb = 19.1106 H = 28.11. IT c φ¯f unction = Φ Note that ITc is in W/m2 (27) The utilizable energy is then: ¯ T · φ¯ U E = N umDay (month) · H (28) Solution a = −1. Slope.271 · K (23) ¯ T − 3.001 U E = 678 MJ/m2 c = 0.442 H Itc = 124 W/m2 Lat = −34 [deg] φ¯f unction = 0.170 − 0.n = 0. KT ) EES function or Eqn 2. ¯ n.9403 R Rn = 1.650 K n = 344 [-] ¯ = 0. n.2 (21) Find dimensionless critical level ¯c = X IT.n = RBeam.113 Surf AzAng = 180 [deg] b = −0.650 φ¯ = 0. ¯ H.810 ¯ b = 0.6.24.943 − 9.Rb. fd = Hdif f /H (Lat.1 (16) (17) (18) fd = Hd /H (19) H = Hb + Hd rd.2941 HrAng noon = 0 [deg] ¯ d = 8.602 · K ¯ T2 b = −4.1265 X 63 f¯d = 0.99 ¯ T = 0.n = 1.n · Hd 1 + Cos(slope) 1 − Cos(slope) rd.n = 0.n · Hd · Rb.853 · K (24) ¯ T + 2. HrAng noon . (Lat.2 rd.7 MJ/m2 ¯ = 28.382 fd = 0.7 MJ/m2 H ¯ HT = 26. Slope.6 MJ/m2 KT = 0.1 MJ/m2 ¯ o = 44.26 H Hd = 9.123 ¯ c = 0.C ¯ rT. n.306 · K ¯ · X ¯c + C · X ¯2 φ¯ = exp a + b · Rn /R c (25) (26) φ¯could have been found from the known data using the EES function.c = 0. ¯T KT = K EES function or Eqn 1.n · H rt.873 HourAngleN oon = 0 ¯ b = 20.345 + 8.446 MJ/m2 month = 12 ρg = 0.810 Rb.9194 R rt.936 · K ¯ T2 c = −0.1 MJ/m2 H IT.n · Rn · H (22) FInd φ¯ ¯ T + 4. Surf AzAng) ¯ H = H. ρg .n · H 2 2 (20) Eqn 2.006 Slope = 20 [deg] .031 · K ¯ T2 a = 2.7 A day that has the same total radiation as the monthly average.n + · + ρg · Rn = 1 − rt. Lat. 7(12) cosT heta2 = max (0. The angle ψ 1 and ψ 2 for shading planes 1 and 2 are: P ψ1 = arctan G+H (4) ψ2 = arctan (P/G) (5) The slopes for these two planes are then: slope1 = ψ1 + 90 [deg] (6) slope2 = ψ2 + 90 [deg] (7) The areas of the two planes are: A1 = W · G+H Cos(ψ1 ) (8) A2 = W · G Cos(ψ2 ) (9) Use Equation 1.7a to calculate CosTheta to get the angles of incidence of beam radiation on these planes.6.7(13) 64 .5 (1) Surf AzAng = 0 (2) G = 0.6.6. day) dec = dec (n) EES function or Table 1. Cos(Lat − Slope1 ) · Cos(Dec) · Cos(HrAng) + sin (Lat − Slope1 ) · sin (Dec)) Eqn 1.5.6.1 EES function or Eqn 1. n = nDay (month. P = 0. W = 3 [m] .5 [m] .6. Slope2 = 144. Cos(Lat − Slope2 ) · Cos(Dec) · Cos(HrAng) + sin (Lat − Slope2 ) · sin (Dec)) Eqn 1. H = 2 [m] .P02 25 Equations 2. Lat = 38.6.1 (10) (11) cosT heta1 = max (0.25 Calcuation of beam radiation on a window under a long overhang using the shading plane concept Ibn = 800 · 3600 · 10−6 . Month and day will be set in the parametric table. Slope1 = 105. and also to constrain CosTheta to positive values. HrAng = 22.7 [m] (3) Use to set up problem. 2119 0. Surf AzAng. HrAng + 7. Slope1 . Table 1 Run month day 1 2 1 4 17 15 Run kT fd 1 2 0.1954 Eb.LJ. Lat.2 2 MJ/m 0.15.62 0. HrAng − 7.1 − A2 · IT.2 assumed for both Jan and April since we do not know the location IT.359 Id 2 MJ/m 0. HrAng) (15) Ib = Ibn · CosZen (16) Io = IZero.7228 0.window [MJ] 14.2 (25) Note that the total radiation can be large compared to the beam radiation.Beam radiation on window: Eb. ρg = 0. I = Ib + Id (18) fd = Idif f /I (kT ) (19) fd = Id /I (20) KT = I/Io (21) Use the Liu and Jordan model for finding the radiation on the two planes.1 = IT.7359 0. Lat.989 IT. (I. Lat.1966 Ib 2 MJ/m 1.682 2.this is time consuming by hand.09782 0 I 2 MJ/m 1.415 .window = Ibn · (A1 · CosT heta1 − A2 · CosT heta2 ) (14) We need to estimate the total radiation knowing only the bean normal radiation.5862 Ewindow 65 15.6585 0. n. HrAng) (24) Ewindow = A1 · IT. n.LJ. HrAng) (22) EES function or Eqn 2. ρg .1 (23) IT.85 7.146 IT.5 [deg]) (17) The following four eqns have to be solved simultaneously .24 1.327 3.326 2. Slope2 . (I.04 4.5771 cosT heta2 cosT heta1 0.38 Io 2 MJ/m 2.936 0. Surf AzAng. n.3563 0. CosZen = CosZenAng (Lat. (n.1 2 MJ/m 2. ρg .2 = IT.5deg . 3 [deg] .4 (1) (2) . Lon = 104. day = 16. Surf AzAng = 0 [deg] 66 ρg = 0.26a Solar radiation calculations Due to the nature of this problem it will be solved in 2 EES programs.P02 26a Equations 2. First put hourly data into a lookup table Lonstd = 105 [deg] . month = 2. Lat = 38. Rotation = 15 [deg/hr] .8 [deg] . hour = 1 [hr] .hours = 24 [hr] . zero = 0 [deg] (3) n = nDay (month. day) (4) a) For the hour 10-11 AM what is kT ? slope = 60 [deg] (5) . . I = Lookup(‘Data’ . ‘I’ ) · hour · . 5. . 0036 .0. MJ/m2 . . W · hr/m2 . HrAng2) (7) HrAng1 = −30 [deg] (8) Hour angle at 10AM HrAN g2 = −15 [deg] Hour angle at 11AM (9) kT = I/Io (10) b) What is IT ? What are the beam. diffuse and ground-reflected components of IT ? . (6) Io = IZero. HrAng1. Lat. (n. . . MJ/m2 . . . n = Lookup(‘Data’ . ‘I bn’ ) · hour · . 5. Ib. 0.0036 MJ/m2 W · hr/m2 . 1 (21) I = Ib.5 [deg] at the middle of the hour (12) Cos(θ) = CosIncAng (Lat. Slope. Use your estimated value of Ib.n and estimate IT. (Lat.estimated = I · Idif f /I (kT ) EES function or Eqn 2. HrAng3. Surf AzAng) Ib = Ib. Compare to the value obtained in question 2.T + Ig.n to Ib incidence angle for horizontal radiation (13) (14) find Id (15) IT = Ib.estimated (22) Ib.n − Ib.10.T (16) Rb = RBeam. zero. Id.n (24) 67 .T = Id · 1 + Cos(slope) 2 Ig.estimated (23) errorbn = Ib.T = ρg · I · (19) 1 − Cos(slope) 2 (20) c) Estimate beam normal radiation from I and compare the value with that given in the table.estimated · |100 %| Ib.n. Surf AzAng) (17) Ib. HrAng3 = −22.n · Cos(θ) I = Ib + Id (11) convert Ib.estimated · Cos(θ) = Ib. HrAng3. n.estimated + Id. n.T = Ib · Rb (18) Id.n.T + Id. 10 (27) .334.estimated · Rb + Id. d) Plot Io and kT versus time.Using the Liu and Jordan model IT.316 for an ‘error’ of less than 1%.estimated · errortotal = 1 − Cos(slope) 1 + Cos(slope) + ρg · I · 2 2 IT − IT. Do not expect such good agreement in general.estimated = Ib.use arrays. From part c with an estimated diffuse fraction IT is 2. duplicate j = 1.estimated · |100 %| IT (25) (26) IT from part b is 2. Can do with arrays or Parametric table . . ‘I’ ) · hour · . j. Ij = Lookup(‘Data’ . . 0036 .0. MJ/m2 W · hr/m2 (28) hourj = Lookup(‘Data’ , j, ‘hour’ ) (29) HrAng j = (hourj − 12) · Rotation (30) Io,j = IZero, kT,j = n, Lat, HrAng j − 7.5, HrAng j + 7.5 (31) Ij Io,j (32) end (33) $TabStops 3 cm Solution Variables in Main program day = 16 hours = 24 [hr] I = 1.721 MJ/m2 2 Ib,n,estimated = 1.728 MJ/m 2 Id,T = 0.5396 MJ/m IT,estimated = 2.316 MJ/m2 Lonstd = 105 [deg] Rotation = 15 [deg/hr] θ = 55.41 [deg] errorbn = 2.024 [%] HrAng1 =−30 [deg] 2 Ib = 1.001 MJ/m 2 Ib,T = 1.622 MJ/m Ig,T = 0.1721 MJ/m2 kT = 0.604 month = 2 Rb = 1.62 zero = 0 [deg] Key Variables kT = 0.604 2 IT = 2.334 MJ/m 2 Ib,T = 1.622 MJ/m Id,T = 0.5396 MJ/m2 Ig,T = 0.1721 MJ/m2 Ib,n,estimated = 1.728 MJ/m2 IT,estimated = 2.316 MJ/m2 errorbn = 2.024 [%] errortotal = 0.7555 [%] a b b b b c c c c 68 errortotal = 0.7555 [%] HrAng2 = −15 [deg] Ib,estimated = 0.9811 MJ/m2 Id = 0.7195 MJ/m2 Io = 2.847 MJ/m2 Lat = 38.3 [deg] n = 47 slope = 60 [deg] hour = 1 [hr] HrAng3 = −22.5 [deg] Ib,n = 1.764 MJ/m2 Id,estimated = 0.7397 MJ/m2 IT = 2.334 MJ/m2 Lon = 104.8 [deg] ρg = 0.4 Surf AzAng = 0 [deg] Arrays Row 1 2 3 4 5 6 7 8 9 10 Ii 2 MJ/m 0.2592 0.7848 1.462 1.811 1.721 2.293 1.584 1.667 1.188 0.5976 houri HrAng i 7.5 8.5 9.5 10.5 11.5 12.5 13.5 14.5 15.5 16.5 -67.5 -52.5 -37.5 -22.5 -7.5 7.5 22.5 37.5 52.5 67.5 Io,i 2 MJ/m 0.770 1.638 2.346 2.847 3.107 3.107 2.847 2.346 1.638 0.770 kT,i 0.337 0.479 0.623 0.636 0.554 0.738 0.556 0.710 0.725 0.776 Data Row hour 1 2 3 4 5 6 7 8 9 10 7.5 8.5 9.5 10.5 11.5 12.5 13.5 14.5 15.5 16.5 I 2 W/m 72 218 406 503 478 637 440 463 330 166 Ibn 2 W/m 254 377 741 705 490 853 485 739 756 838 Tamb [C] 2.8 5.4 8.0 10.6 12.1 13.5 15.0 14.8 14.6 14.4 Io and kT vs time 69 P02 26b Equations 2.26 (parts e - i) Solar radiation calculations Due to the nature of this problem it will be solved in 2 EES programs. This program is solved pressing F3 or from ‘solve table’ in the calculate menu. The lookup table from the first part is also used here. Lonstd = 105 [deg] ; Lat = 38.3 [deg] ; hours = 24 [hr] ; Lon = 104.8 [deg] ; Rotation = 15 [deg/hr] ; hour = 1 [hr] ; month = 2; day = 16; ρg = 0.4 Surf AzAng = 0 [deg] zero = 0 [deg] (1) (2) (3) n = nDay (month, day) (4) e) For slopes of 30, 45, 60 and 75 degrees, plot IT vs. time. Use the L&J method with Id estimated as in part c. Set up four parametric tables (one for each slope) with hour angle from -67.5 to +67.5. The hourly values of radiation are assumed to be numerically equal to the average for the hour . MJ/m2 . . T ableRun#. Ib. ‘I bn’ ) · hour · .n = Lookup(‘Data’ . . 0036 W · hr/m2 .0. HrAng) (6) Ib = Ib. n.n · CosZen (7) . (5) CosZen = CosZenAng (Lat. . . MJ/m2 . . . ‘I’ ) · hour · . I = Lookup(‘Data’ . T ableRun#. 0036 W · hr/m2 .0. 1 1 + Cos(slope) 1 − Cos(slope) + I · ρg · 2 2 Eqn 2. Slope. Surf AzAng) IT = Ib · Rb + Id · Eqn 1. n.15. g) What are H. HrAng.8. (Lat. (8) I = Ib + Id (9) Finds Id Rb = RBeam.1 (10) (11) f) If the selection of slope is based on maximizing incident radiation. KT and Ho? H is the sum of I over all hours. what slope would you select for this day? What would you think might be the answer to this question if a summer’s day had been used in the calculations? It does not matter very much but 60 degrees is the highest. . . . MJ/m2 . . . ‘I’ ) · . H = Sumlookup(‘Data’ . 0036 W · hr/m2 .0. (12) 70 . 26 MJ/m2 H day = 16 HrAng = 67.LJ .81 1.853 MJ/m2 H HT = 21.017 MJ/m2 Lat = 38.LJ = 1.5 7.19 MJ/m2 ¯ T GrRef.n = 3.1537 hours = 24 [hr] ¯ T dif f. ρg : H ¯ T beam.37 MJ/m2 Ho = 21.5 37.LJ = 17.58 1.622 month = 2 Rb = 2.LJ = 2.60 MJ/m IT = 1.Ho = HZero.26 0. (n.LJ .78 1.39 1.LJ = 17.09 2.26 2.33 MJ/m2 H 2 I = 0.07 MJ/m2 H ¯ T dif f. ‘I T’ ) (15) ¯ T and what are its components for the month of February a slope of 60 degrees? The monthly average i) What is H radiation for February is 13.5 67.4 Surf AzAng = 0 [deg] Key Variables H = 13.19 0.3 [deg] n = 47 Slope = 60 [deg] . H ¯ T.16 71 H = 13.5 -22.60 IT 2 MJ/m 0.1338 MJ/m2 Lon = 104.LJ.4638 MJ/m2 KT = 0.30 MJ/m2. ¯ Lat.5 -52.46 1.43 1. H.26 MJ/m Ib = 0.33 MJ/m2 H ¯ T beam.57 2.19 2.19 MJ/m2 Id = 0.5 [deg] ¯ T GrRef.5 52.LJ .51 MJ/m2 KT = 0.37 MJ/m2 ¯ = 13. H ¯ T dif f.46 MJ/m2 Lonstd = 105 [deg] Rotation = 15 [deg/hr] zero = 0 [deg] g g g h i i i i Slope=30 Run 1 2 3 4 5 6 7 8 9 10 Slope [deg] 30 30 30 30 30 30 30 30 30 30 HrAng [deg] -67.3 MJ/m2 H From Appendix G (16) ¯ T. ¯ = 13.LJ = 1. H ¯ T GrRef.LJ = 2.72 2.LJ call H (17) $TabStops 1 cm Solution Variables in Main program CosZen = 0.5 -7.3 MJ/m2 H 2 ¯ HT.LJ = 21.853 MJ/m2 H ¯ T. Use the Liu and Jordan method.29 1.09 2.622 HT = 21.LJ = 21.07 MJ/m2 H Ho = 21.5 -37.82 1.5 22. HT = Sumparametric(‘Slope=60’ .8 [deg] ρg = 0.26 3. n. Slope.67 1.p. Lat) (13) KT = H/Ho (14) h) Calculate HT and its components for a slope of 60 degrees.5 I 2 MJ/m 0.793 hour = 1 [hr] ¯ T beam.51 MJ/m2 Ib. 5 22.60 IT 2 MJ/m 0.02 1.5 9.44 2.5 -37.5 -22.29 1.4 8.62 2.51 1.37 3.51 1.76 2.78 1.36 2.38 3.79 2.46 1.50 2.5 37.5 7.5 67.67 1.72 2.5 7.67 1.35 HrAng [deg] -67.5 -22.0 Run 1 2 3 4 5 6 7 8 9 10 Slope [deg] 45 45 45 45 45 45 45 45 45 45 Slope=60 Run 1 2 3 4 5 6 7 8 9 10 Slope [deg] 60 60 60 60 60 60 60 60 60 60 Slope=75 Run 1 2 3 4 5 6 7 8 9 10 Data 72 .5 I 2 MJ/m 0.38 2.5 8.39 2.26 0.Slope=45 HrAng [deg] -67.5 I 2 MJ/m 0.5 67.5 -37.5 -52.07 1.5 -52.48 Row hour 1 2 3 7.19 0.72 2.58 1.60 IT 2 MJ/m 0.5 52.26 0.53 2.21 2.20 2.46 1.58 1.5 -7.5 37.60 IT 2 MJ/m 0.10 2.78 1.5 52.21 2.81 1.16 2.46 Slope [deg] 75 75 75 75 75 75 75 75 75 75 HrAng [deg] -67.17 2.48 1.81 1.5 -22.67 2.5 -7.5 -37.25 3.19 0.5 22.81 1.8 5.5 7.5 22.5 67.5 37.72 2.19 0.58 1.26 0.78 1.5 -52.5 I 2 W/m 72 218 406 Ibn 2 W/m 254 377 741 Tamb [C] 2.5 52.5 I 2 MJ/m 0.67 1.29 1.5 -7.00 1.52 2.57 2.46 1.29 1. 5 16.6 12.8 14.5 12.1 13.4 IT vs time for various slopes 73 .5 14.5 13.5 503 478 637 440 463 330 166 705 490 853 485 739 756 838 10.4 5 6 7 8 9 10 10.5 11.6 14.5 15.5 15.0 14. C1# and C2# λ1 = 10. T2 = 400 (2) λ3 = 1.3 = 3.2 = 90 W/m2 -micrometer T [3] = 6000 [K] λ[3] = 1 [micrometer] Eλ. T3 = 6000 (3) duplicate i = 1.b.b.1 Calculation of blackbody emissive power Arrays are used but a Parametric Table could have been used.i = λ5i · exp (4) C1# C2# λi ·Ti −1 (5) end (6) To show all variables in the Solutions Window use: $Arrays off The results are consistent with Figure 3.b.1.4. T1 = 1000 (1) λ2 = 5. We can use the EES built-in constants.1 = 1164 W/m2 -micrometer λ[1] = 10 [micrometer] T [2] = 400 [K] λ[2] = 5 [micrometer] Eλ.b.742 × 107 W/m2 -micrometer a a a b b b c c c 74 . 3 Eλ.P03 01 Equations 3. Solution Key Variables T [1] = 1000 [K] Eλ. 4 (4) f1 = BlBodyF ract (λ1 · T ) EES function or Eqn 3.4 and taking the difference.6.2 Determination of the fraction of blackbody radiation in a wavelength range This can be done by computing the fraction of the energy in the spectrum from zero to the two wavelengths with Equation 3. T = 300 [K] (1) λ1 = 8 [micrometer] (2) λ2 = 14 [micrometer] (3) f2 = BlBodyF ract (λ2 · T ) EES function or Eqn 3.4 (5) f = f2 − f1 (6) Solution f = 0.6.516 λ1 = 8 [micrometer] 75 . Table 3.1402 T = 300 [K] f2 = 0.3757 λ2 = 14 [micrometer] f1 = 0.6.1a could have been used to find the two fractions.P03 02 Equations 3.6. 4 and sum for 10 terms.P03 03 Equations 3. The second method is used for the SETP EES functions. LT2) Solution Variables in Subprogram BBf f = 0.6. Note that a subprogram was not necessary but was requested in the problem statement.3 We could use EES to integrate Eqn 3. LT 2 : f ) Z LT 2 f = (c1#/sigma#) · LT 1 (1) 1 5 LT · (exp (C2#/LT ) − 1) dLT (2) end (3) LT 1 = 1000 (4) LT 2 = 5000 (5) call BBf (LT 1.3 or we could use Eqn 3. LT 2 : f ) (6) For the given conditions the table gives f=0. LT1. The solution could have been: LT1=1000 LT2=5000 f=(c1#/sigma#)*Integral(1/(LT5 *(exp(C2#/LT)-1)).6335 LT = 5000 [micrometer] LT 1 = 1000 [micrometer] 76 LT 2 = 5000 [micrometer] . SubprogramBBf (LT 1.6335. Here we will do the integration using the EES integral function. LT.6334 and the integral gives 0.6. 670 × 10−8 W/m2 -K4 2 = 0.1 Solution 1 = 0.478 W/m2 -K T2 = 300 [K] 77 qrad = 495.4 Calculation of radiant energy transport between large parallel plates: T1 = 500.4 Eqn 3.45.10.45 σ = 5.P03 04 Equations 3. σ = sigma# Eqn 3.8.7 W/m2 .2 T1 = 500 [K] hrad = 2. T2 = 300. qrad = σ · 1 = 0.20. T14 − T24 (1/1 + 1/2 − 1) qrad = hrad · (T1 − T2 ) 2 = 0. 013 · Cos(15 [deg/hr] · T ime) Radiation heat transfer is given by Equation 3. Note the unusual units in this Equation.7 W/m2 -K hRad = 7.5 Calculation of a heat loss coefficient for combined wind and radiation loss for a plate facing up.1. .P03 05 Equations 3.9.711 + 0. Tamb = ConvertTemp(C.1 [K] Uoverall = 32. Equation 3.9.1 [K] T ime = 0 [hr] Tsky = 262. Divide this heat transfer by (TP late .0056 [1/C] · TDewP oint + 0. K.88 hwind = 25 W/m2 ·K Estimate an equivalent sky temperature using the Berdahl & Martin method.2.657 W/m2 -K TDewP oint = 3 [C] 78 hwind = 25 W/m2 -K TP late = 323. 10) TP late = ConvertTemp(C.Tamb ) to get the radiation coefficient based on the ambient temperature. 50) TDewP oint = 3 [C] T ime = 0 = 0. K.7 [K] .25 2 Tsky = Tamb · 0. hRad = sigma# · · 4 TP4 late − Tsky TP late − Tamb Add the convection coefficient to get the total loss coefficient.000073 1/C2 · (TDP ) + 0.88 Tamb = 283. Uoverall = hRad + hwind Solution = 0. T = T¯ .44 · (sin (1.135 2.020 0.794 5.53 79 176. Ra · −1 5830 N us = hconv · l/k Qconv = hconv · (T1 − T2 ) T 1 + T2 T¯ = 2 beta0 = β Air. k = λ air. 75))) 1 − 1708 · Ra · Cos(min (Slope.3 [kPa] The three conditions are done in a Parametric Table.2 Ra = g# · beta0 · (T1 − T2 ) · l3 µ·α The Nusselt Number is found from Eqn 3.8 3. 75)) ! ·A+B where A and B must be constrained to be positive or zero and for slopes greater than 75.7 3. 75)) B = max 0. Table 1 Run 1 2 3 l [m] 0.11.00 149.07 203.333 Cos(min (Slope. T1 = 100.8 · min (Slope. use slope = 75: 1708 A = max 0. T = T¯ .P03 06 Equations 3.6 .050 0. µ = µ air. T = T¯ Cp = cp air.11.4 1. 1 − Ra · Cos(min (Slope. ρ = ρ air.020 Slope [deg] 0 0 45 N us [-] 2. T = T¯. 75)) !! . T2 = 50. The Rayleigh number is found from Eqn 3.6 N us = 1 + 1.422 Ra [-] 19030 297343 19030 hconv Qconv2 W/m2 -K W/m 4. p = Patm . T = T¯ . α= k ρ · Cp Patm = 101.6 Calculation of convection heat transfer in a collector. Solution a = 0.0428 Tground = 288 [K] b = −0.b = 306 [K] .041 = a + b · 338 [K] + c · 338 [K] .036 = a + b · 463 [K] + c · 463 [K] .03428 Tamb = 298 [K] Tsky = 288 [K] αsolar = 0. 2 0. 2 0.1 the solar absorptance is: αsolar = 0. 4 4 4 4 Tplate.0001241 [1/K] hwind = 23 W/m2 -K Tplate.039 = a + b · 803 [K] + c · 803 [K] b For a plate in the atmosphere with surface normal to the solar radiation and a zenith angle of 90 deg an energy balance on the plate (which absorbes solar radiation on one side.a = 592 [K] 80 c = 1.07095 [-] = 0.P03 07 Equations 3.1.a + c · Tplate.b Gsc# · τatm · αsolar = Sigma# · b · τatm = 0.7 Calculation of equilibrium temperatures of plates exposed to solar radiation a For a plate in space with surface normal to the solar radiation an energy balance on the plate (which absorbes solar radiation on one side and emits radiation from both) is: 4 Gsc# · αsolar = 2 · Sigma# · · Tplate. Tamb = 298 [K] .8. hwind = 23 W/m2 ·K .35 b = 0.a From Table 4.b − Tsky + Tplate.7. exchanges radiation from the top side with the sky and from the bottom side with the ground.35 Assume the emittance of copper. from Table 4.a 2 0. Tground = 288 [K] The convection term in the energy balance dominates so it is clear that the exact value of the emittance is not critical. and loses energy by convection from both sides to the atmosphere) is: 2 b = a + b · Tplate.can be approximated by a parabola 2 = a + b · Tplate.7.8 Tplate.b + c · Tplate.b − Tamb ) Tsky = 288 [K] .050 × 10−7 1/K2 τatm = 0.b − Tground + 2 · hwind · (Tplate. assuming that the disks and cover act like parallel plates for radiation exchange: ! 4 4 Tdisk.90.w − Tcover ) (1/disk + 1/cover − 1) 81 . Tdisk.8 Energy balance calculations on a hypothetical instrument for measuring solar radiation. K.P03 08 Equations 3. K. Knowns hconv = 16 W/m2 ·K . Tdisk.95. cover = 0. disk = 0. τcover = 0.w = ConvertTemp(C. 5). 15).b − Tcover ) G · τcover · αb = 2 · Sigma# · (1/disk + 1/cover − 1) G · τcover · αw = 2 · ! 4 4 Tdisk.35 a An energy balance on each disk.b − Tcover + hconv · (Tdisk. αb = 0.88 αw = 0.95.b = ConvertTemp(C.w − Tcover Sigma# · + hconv · (Tdisk. cover. the nonlinear radiation term would have been linearized using the second two equations. finding the U’s..cover.b + Tcover · Udisk.88 hconv = 16 W/m2 -K disk = 0.cover.w ) · τcover d Calculation of incident radiation In the days before EES.w = 2 · Sigma# · c 2 Tdisk.w + Tcover + hconv · (1/disk + 1/cover − 1) An expression for incident radiation Incident solar energy.let EES do it. is expressed in terms of the temperature difference between black and white disks.b = 288.cover.b − Tdisk.b Heat transfer coefficient from disk to cover: 2 2 Udisk.95 W/m2 -K αw = 0.0 W/m2 Tcover = 272. G.2 [K] Udisk.2 [K] Udisk.cover.4 W/m2 -K 82 τcover = 0.w = 39.b *(Tdisk.cover.w *(Tdisk.b − αw /Udisk.b -Tcover ) For the white disk: G1*τ cover *αw = Udisk.cover.2 [K] Tdisk.0 W/m2 Tdisk.b + Tcover + hconv (1/disk + 1/cover − 1) Tdisk. Note that the first two equations are sufficient to solve for G.b = 40. substituting into the equation of part c and then checking the guess of the cover temperature using one of the commented equations of pary c.w (αb /Udisk. Also note that the ambient temperature was not used except maybe to get a reasonable guess of Tcover.w -Tcover ) Solving for G1 G1 = Tdisk.95 G = 753.35 cover = 0.9 [-] . so the new variable G1 is introduced.w + 2 Tcover Tdisk. Solution αb = 0.b = 2 · Sigma# · Tdisk. For the black disk: G1*τ cover *αb = Udisk.cover.w = 278. The solution would then involve guessing the cover temperature.95 G1 = 753. With EES we do not need to worry about the solution technique . kPa K kJ Tl = ConvertTemp(C. 110 [C]). α= β = 1/T¯ Slope = 60 [deg] µ = µ Air. K.9 Convection loss coefficient and convection heat transfer in a collector $UnitSystem SI Mass Deg Tu = ConvertTemp(C. Tl + Tu T¯ = . ∆T = Tu − Tl . T = T¯ . T = T¯ . K. L = 0.020 [m] . 2 k . Cp = cp Air. k = λ Air.P03 09 Equations 3. T = T¯ . 60 [C]). . ρ · Cp · 1000 ρ = ρ Air. T = T¯.3 [kPa] . P = 101. ν = µ/ρ J .. both Nusselt numbers are identical.069 [-] N u# = 2.069 [-] Slope = 60 [deg] SlopeL = 60 [deg] 83 Ra# = 16. 75 [deg]) Slope is constrained to be between 0 and 75 1708 A = max 0.1328 β = 0.2 [K] α = 0.092 W/m2 -K B = 0. Ra# · −1 5830 As expected.2 [K] µ = 0. q = h · ∆T Solution A = 0. L) or we can evaluate the Rayleigh number and then find the Nusselt number. Tl .002792 [1/K] Cp = 1.2 [K] . Tu .6 0 N u# = 1 + 1.8 · SlopeL)) 1 − 1708 · Ra# · Cos(SlopeL) ! ·A+B SlopeL = min (Slope.00002117 [Pa-s] 0 N u# = 2.6 W/m2 T¯ = 358.9854 kg/m3 Tu = 383.333 Cos(SlopeL) B = max 0. 1 − Ra# · Cos(SlopeL) !! .44 · (sin (1.00003006 m2 /s h = 3.009 [kJ/kg-K] k = 0. 957 [-] Tl = 333. Ra# = g# · β · ∆T · L3 ν·α 1. N u# = h · L/k.02988 [W/m-K] L = 0.00002149 m2 /s ρ = 0.02 [m] q = 154.7985 ∆T = 50 [K] ν = 0. kJ We can use the SETP functions to fing the Nusselt Number N u# = N uF latP late (Slope. k .10 Effect of a slat-type honeycomb on the convection heat transfer of Problem 3. 110 [C]). Tl + Tu T¯ = . ∆T = Tu − Tl . K. L = 0. K. T = T¯ .020 [m] .9 $UnitSystem SI Mass Deg Tu = ConvertTemp(C. T = T¯ . k = k Air. 60 [C]). kPa K kJ Tl = ConvertTemp(C.P03 10 Equations 3. 2 β = 1/T¯ Slope = 60 [deg] µ = µ Air. . T = T¯ . P lSpace = 20 [mm] ρ = ρ Air. ρ · Cp · 1000 Cp = cp Air. α= SlatSpace = 10 [mm] . P = 101. T = T¯.3 [kPa] . J .. 2 at an AspectRatio of 0. C2 = 1.02988 [W/m-K] L = 0.9854 kg/m3 Tl = 333.90 [-] ρ = 0.1 N u# = max 1.2 [K] Slope = 60 [deg] 84 Ra# = 16.00002149 m2 /s SlatSpace = 10 [mm] Tu = 383.1 · C1 · C2 · Ra#0.5: C1 = 0. 957 [-] T¯ = 358.00 We need to evaluate the Rayleigh number and then find the Nusselt number.009 [kJ/kg-K] ∆T = 50 [K] k = 0.113 C2 = 1 Cp = 1.12.00002117 [Pa-s] q = 141.839 W/m2 -K ν = 0.5 β = 0.12.02 [m] N u# = 1. 1 N u# = h · L/k q = h · ∆T Solution α = 0. Ra# = g# · β · ∆T · L3 ν·α From Eqn 3.2 [K] .113. kJ ν = µ/ρ AspectRatio = SlatSpace/P lSpace From Figure 3.2 [K] P lSpace = 20 [mm] µ = 0.28 .9 W/m2 h = 2.00003006 m2 /s AspectRatio = 0.002792 [1/K] C1 = 0. 015 m ˙ [kg/s] 0. Tair = 25 [C] .815 5.14.0606 · (Re# · P r# · Dh /Length) .02551 [W/m-K] P = 2. T = Tair ) .558 ν# [-] 5.P03 11 Equations Heat transfer coefficients for air flowing in a rectangular duct .14.024 Re# 1.015 m2 Length = 2 [m] P r# = 0.367 6.6 is 7.9 + .7 1 + .0075 0. µ = µ (air.279 1.015 0. P = 2 · (width + depth) .26 W/m2 -K Solution Af low = 0.03 [m] .7281 depth = 0. properties and definitions Length = 2.019 9. For conservative design use the laminar flow results.289 2. k = k (air.625 [-] width = 1 [m] k = 0.0 [m] .0909 · (Re# · P r# · Dh /Length) · P r#.00001849 [kg/m-s] Re# = 2.02956 [m] m ˙ = 0.718 85 h = 5. T = Tair ) .7 1. P r# = Pr (air.17 N u# = h · Dh /k For situation 3 the Reynolds number is in the transition region. The turblent heat transfer coefficient from Eqn 3.012 0.625 h W/m2 -K 5. From Eqn 3.015 [m] µ = 0.012 0. $UnitSystem SI Mass Deg kPa C kJ Knowns. 558 Dh = 0.024 [kg/s] Tair = 25 [C] Table 1 Run 1 2 3 depth [m] 0. width = 1 [m] .718 W/m2 -K N u# = 6. T = Tair ) Dh = 4 · Af low P m=ρ*A ˙ f low *V and Re#=ρ*V*Dh /µ so that when ρ is eliminated we have Re# = m ˙ · Dh µ · Af low Use a parametric table for the three situations. Af low = width · depth.197 5.2 N u# = 4. Area = 12 m2 .02 [m] V isc = µ (air.16.918 × 10−5 [kg/m-s] ∆p = 29. M assV el.02 [m] Length = 2 [m] T = 40 [C] . T = 40 [C] .09167 kg/s-m2 86 DenAir = 1.5 or use the SETP function for pressure drop.12 Calculation of pressure drop in a packed bed using the Dunkle & Ellul correlation $UnitSystem SI Length = 2.31 [Pa] M assV el = 0.1 [kg/s] V isc = 1. ∆p = P resDrop (Length. Mass Deg kPa K kJ C M assF lowRate = 1.127 kg/m3 P artDiam = 0. T = T ) M assV el = M assF lowRate/Area Equation 3. p = 101. P artDiam.0.P03 12 Equations 3. P artDiam = 0. T ) Solution Area = 12 m2 M assF lowRate = 1. T = T. DenAir = ρ (Air.1 [kg/s] .3) . 74 + 166 · (1 − V F ) · · V F 3/2 V F 3/2 M assV el · P artDiamM cC Solution Area = 12 m2 Length = 2 [m] P artDiam = 0.127 kg/m M assV el = 0. V F = 0. ∆p = P resDrop (Length.918 × 10−5 [kg/m-s] 87 3 DenAir = 1.45 ∆p = Length · M assV el2 DenAir · P artDiamM cC · (1 − V F ) · 1.01321 [m] V F = 0. M assF lowRate = 1.45 [-] ∆p = 55 [Pa] M assF lowRate = 1.4 by McCorquodale et al.5 1. Length = 2. M assV el.01032 [m] V isc = 1.12 and add the pressure drop. T = T. Area = 12 m2 .0. ∆p = 55 [Pa] T = 40 [C] . P artDiam.P03 13 Equations 3.1 [kg/s] P artDiamM cC = 0.13 Estimation of particle size to get a pressure drop of 55 Pa for the circumstances of Problem 3. T = T ) M assV el = M assF lowRate/Area Equation 3.12 $UnitSystem SI Mass Deg kPa K kJ C Note that with EES it is only necessary to delete the particle diameter in problem 3. V isc = µ (air.16.09167 kg/s-m2 T = 40 [C] . p = 101.3) . T ) An alternative solution is to use Equation 3.5 or use the SETP function for pressure drop.16.1 [kg/s] . DenAir = ρ (Air.5 V isc · 4. Tcover = ConvertTemp(C.4 along with the following: β = 1/Tmean Cp = cp (Air. g = g#.14 Calculation of upward heat loss terms for a collector. 10) L = . 31. L) Tmean = Equation 3. Tplate .11.11.14 Tplate + Tcover 2 k = k (Air. K.11. Tsurr = ConvertTemp(C. Tcover .025 [m] Radiation Heat Transfer Qrad = σ · b 4 4 − Tcover Tplate 1/plate + 1/cover − 1 Radiation coefficient hrad = c cover = 0.88.10. a Slope = 50. Qrad Tplate − Tcover Convection Heat Transfer N u# = N uF latP late (Slope. T = Tmean ) 88 . 100). σ = sigma#. plate = 0.P03 14 Equations 3. $UnitSystem SI Mass Deg kPa K kJ Data: Tplate = ConvertTemp(C.5). K. T = Tmean ) N u# = hconv · L/k Qconv = hconv · (Tplate − Tcover ) Instead of using the NuFlatPlate EES function we could have used either Figure 3.1 or Eqn 3. K. T = Tmean .ρ = ρ (Air.3) ν= µ (Air. T = Tmean ) ρ α= k . P = 101. ρ · Cp · . 1000 Ra = d . J . 00002717 m2 /s cover = 0.Surr = 109.1 [-] 2 hrad = 0.002951 [1/K] plate = 0.Surr = cover · σ · Tcover − Tsurr We can determine QconvCover. 368 [-] Slope = 50 [deg] Tplate = 373.807 m/s2 k = 0.9 [K] β = 0. QconvCover.9 W/m2 Ra = 58.Surr = Qconv + Qrad − QradCover.172 2 Qrad = 60.1 [K] 89 Cp = 1.0285 [W/m-K] N u# = 3.Surr Solution α = 0.1 [K] .88 [-] hconv = 3.025 [m] Qconv = 247.26 W/m ρ = 1.041 kg/m3 Tcover = 304.00001953 m /s QconvCover.8797 W/m 2 -K ν = 0.7 [K] Tsurr = 283.670 × 10−8 W/m2 -K4 Tmean = 338.1 W/m2 σ = 5.Surr = 198.Surr since loss is either convection or radiation.7 W/m2 QradCover. kJ g · β · (Tplate − Tcover ) · L3 ν·α Radiation from cover to surroundings 4 4 QradCover.007 [kJ/kg-K] g = 9.616 W/m2 -K L = 0. b = max 5 W/m2 ·K .313 × 106 [-] TSurr = 283.02519 [W/m-K] Length = 10 [M] N u#wind = 887 [-] Tcover = 304. For empirical equations like this it is best to just turn off unit checking for that equation. a k = k (‘air’ . 2 · (Length + W idth) N u#wind = 0. 31. T = Tprop .5 [M] .15.2 /s0. ρ = ρ (‘air’ .10 are not dimensionally correct unless the 8.1 N u#wind = hwind. K.201 kg/m3 V el = 5 [m/s] 90 k = 0.p03 15 Equations 3.9 and 3. Solution hwind. V el = 5.706 W/m2 -K µ = 0.00001829 [kg/m-s] ρ = 1.5) Evaluate properties at the mean air tempeature.4 Eqn 3. 2 µ = µ (‘air’ .7291 [-] Tprop = 293. With no satisfactory alternative.587 W/m2 -K LChar = 4 [m] Re = 1.6 V olume1/3 . T = Tprop ) .15.3) P r = Pr (air.1 is based.15.15.a · Lchar /k Note: The Re is 1.86 · Re1/2 · P r1/3 Re = LChar · V el · ρ/µ Eqn 3.31*106 . T = Tprop ) .a = 5.9 [K] W idth = 2. W idth = 2. P = 101. hwind.10 Note: Equations 3. T = Tprop ) For a free standing colector: LChar = 4 · Length · W idth .6 has units of (W/m2 -K)/(m0. 8. b For a flush-mounted collector on a building V olume = 564.b = 9.15 Estimation of wind convection coefficients $UnitSystem SI Mass Deg kPa K kJ Data: Length = 10 [M] . use it.7 [K] V olume = 564 m3 P r = 0. 10).1 [K] hwind.5 [M] . K. Tprop = Tsurr + Tcover .6 · ! V el.6 ). TSurr = ConvertTemp(C. which is slightly beyond the range of the experiments on which Equation 3.15.0 [m/s] Tcover = ConvertTemp(C. Glycol − Tin. The solution is shown for the specific case of : m ˙ glycol = 5 Solution Ccold = 15. 000 [W/K] Chot = 18.1 × 105 [W/K] .H2O ) Q = Chot · (Tin.37 [-] Tout.Glycol = 65 [C] . Tin.H2O = 49 [C] 91 Cmax = 18.P03 16 Equations 3.75 [kg/s] . Tin. 524 [W] Tout.0 [kg/s] Tin.981 [-] Q = 246.H2O Chot = m ˙ Glycol · Cp. Ccold = m ˙ H2O · Cp.Glycol = 65 [C] U A = 210. 900 [W/K] C ∗ = 0. Chot ) Cmax = max (Ccold .Glycol = 3. U A = 2.H2O − Tin. 713 [W/K] Cp.16 Effectiveness-NTU calculations for a counterflow heat exchanger $UnitSystem SI Mass Deg kPa K kJ Data: m ˙ H2O = 3.69 [C] .Glycol Cmin = min (Ccold . 900 [W/K] Cp. 190 [J/kg-K] m ˙ H2O = 3.Glycol = 51.Glycol = 3780 [J/kg ·K] . Chot ) N T U = U A/Cmin C ∗ = Cmin /Cmax Ef f = 1 − exp (−N T U · (1 − C ∗ )) 1 − C ∗ · exp (−N T U · (1 − C ∗ )) Q = Ef f · Cmin · (Tin.H2O ) Q = Ccold · (Tout.H2O = 64. 713 [W/K] Ef f = 0.H2O = 49 [C] .Glycol − Tout.8313 [-] N T U = 13. 780 [J/kg-K] m ˙ Glycol = 5. Cp.H2O = 4.75 [kg/s] Tin.956419181 [C] Cmin = 15.Glycol ) Run a parametric table with m ˙ glycol reanging from 0 to 20. Tout vs mdotg lycol 92 . a slope = 60 [deg] . qconv.11. g = 0.88. P lateSp = 0. 85) (1) (2) Heat loss from collector We can use the SETP function for Eqn 3.2.c = qrad. K.p.c + qrad. $UnitSystem SI Mass Deg kPa K kJ Data: p = 0.p. Tsky = Tamb . T = 2 (4) (5) The various energy flows through the cover system are: qconv. Tamb = ConvertTemp(C.sky + qwind (10) 93 .c = σ · (6) 4 4 Tplate − Tcover 1/p + 1/g − 1 (7) qwind = hwind · (Tcover − Tamb ) 4 4 qrad.cover · (Tplate − Tcover ) qrad. P lateSp) (3) N u# = hplate. hwind = 7 W/m2 ·K . Tcover .cover · P lateSp/k Tcover + Tplate k = k air. K.P03 17 Equations 3. Tplate .4 to find hplate.cover N u# = N uF latP late (Slope.p.030 [m] . σ = sigma#.c = hplate.p.17 Heat transfer from the hot absorber plate to the ambient.sky = σ · g · Tcover − Tsky (8) (9) The cover temperature is found from an energy balance on the cover. 15) Tplate = ConvertTemp(C. 88 [-] k = 0.4 292.716 2.p.327 3.971 3.3 305. Tplate = ConvertTemp(C.7 294.103 3.4 296.7 301.9 W/m2 slope = 60 [deg] TplateC = 85 [C] Tamb = 288. edge losses two-dimensional losses in the corners.0 U W/m2 -K 2.03 [m] qrad.p.576 94 .sky = 101.595 2.c + qrad.479 3.c (11) b Heat loss coefficient as a function of plate temperature is found by commenting out the equation above that sets the plate temperature to 85 C and setting the plate temperature (in C) in a parametric table.8 307.7 304.c = 83.813 2.2 [K] Tsky = 288.379 3.3 299.02806 [W/m-K] qloss = 236.03 W/m2 hplate.112 2.5 W/m2 σ = 5.0 290. Solution g = 0.p.379 W/m2 -K Table 1 Run 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 TplateC [C] 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105 Tcover [K] 289.0 295.039 3.163 3.cover = 3.425 2.235 [-] qrad.c = 153. K.p.897 2.220 3.429 3.8 298.2 302.2 291.4 309.2 [K] p = 0.528 3.7 312.2 [K] Tcover = 307.274 3. TplateC ) (12) qloss = U · (Tplate − Tamb ) (13) c What energy transfers are we neglecting in this analysis of a collector? back losses through the insulation.5 W/m2 qwind = 134.670 × 10−8 W/m2 -K4 Tplate = 358.026 W/m2 -K P lateSp = 0.The heat loss through the cover is then: qloss = qconv.4 [K] U = 3.3 314.0 310.2 [-] N u# = 3.6 W/m2 hwind = 7 W/m2 -K qconv. U vs Tplate 95 . 1.∞ αsurf = 1 − ρsurf αsurf = surf Energy balance on surface assuming back is insulated. λsurf = BlBodyF ract (λ · Tsurf ) ρsurf = fo.∞ = 0.322E-7 0.690 0.9 Fraction of the sun’s energy between zero and λ fo. Assume that the 6000 K radiation is normal to the surface and that there is no loss other than radiation from the surface. ρsun = fo.surf αsun surf 0.049 0.005337 0.7 . Table 1 Run 1 2 3 λ [micrometer] 1 2 3 fo.lambda fo. λ + (1 − fo.0 625. S = 1367 W/m2 Use a Parametric table to specify the critical wavelength.885 96 0.P04 01 Equations 4.856 0. λ + (1 − fo.9451 0. λ = BlBodyF ract (λ · Tsun ) From the EES SETP library. find a temperature from the energy balance and repeat the process until the guess and calculated temperatures are equal.lambda. ρ0. λsurf ) · ρλ. Tsun = 6000 [K] . λ) · ρλ. λsurf · ρ0. 4 αsun · S = surf · sigma# · Tsurf If we did this by hand we would need to guess a surface temperature.104 0.9808 3.139 Tsurf [K] 638.∞ αsun = 1 − ρsun Fraction of the IR radiation from surface at Tsurf ace fo. λ = 0.1 Calculation of equilibrium temprerature of a surface in space.7378 0. ρλ. find the fraction of the IR radiation in each band.7 667.100 0. λ · ρ0. 3. αλ. Tsurf ace = ConvertTemp(C. 150) αλ.05 λ = 1. 777 [K] .880 fIR = 0.050 Tsun = 5.P04 02 Equations 4.8 [micrometer] 97 = 0.6.9219 αλ.5 = αλ.95.4 for two bands α = αλ.05. K.2 = 0.2 · (1 − fsol ) (4) fraction of IR radiation below λ from Eqn 3.1 = 0.2 Calculation of absorptance and emittance for a surface with a step change in spectral reflectance $UnitSystem SI Mass Deg kPa K kJ Data: Tsun = 5777 [K] .6.95 fsol = 0.8 [micrometer] (2) fraction of solar radiation below λ from Eqn 3.1 · fIR + αλ. (1) λ = 1. fsol = BlBodyF ract (λ · Tsun ) (3) From Eqn 4.4.1 [K] αλ.1 · fsol + αλ.6.2 · (1 − fIR ) (6) Solution α = 0.6.1 = 0.3.000007637 Tsurf ace = 423.2 = 0.1 for two bands and using αλ = λ from Eqn 4. fIR = BlBodyF ract (λ · Tsurf ace ) (5) From Eqn 4. 0.04 ρ6 = 0. r28 = 0.3 Calculation of absorptance and emittance of a surface from spectral reflectance data.04. 1.04.3 ρ1 = 0.04. $UnitSystem SI Mass Deg kPa K kJ a For the extraterrestrial solar spectrum. the wavelength at center points are: 0. The appropriate wavelenghts are: 3. ρ7 = 0. 4. 6. ρ2 = 0. i = 1. 10) 10 c Emittance of the surface at 350 C.508.8. 1.06.926 = 0. r27 = 0. 0.10. 0.080. 0. r29 = 0. Note that the numbers differ little from part a.6.117.08.787.300.04.94 =1− Sum(r2i .87.59.06.04.159 98 Estimated r2[9] and r2[10] . Using 10 equally spaced intervals from Table 1. r110 = 0. r18 = 0. 0. r15 = 0.3.02. 3. Using 10 equally spaced intervals from Table 2. i = 1. 1.94. r24 = 0.923.38.6.15.1b.93. ρ4 = 0.77. 0.1. r14 = 0. 0. r25 = 0. 8.04. ρ8 = 0.732.525. r17 = 0.682. 10) 10 b For the standard terrestrial solar spectrum.91.923 αterrestrial = 0. 0. 12. r12 = 0. i = 1.455.3 r21 = 0. 20.65.p04 03 Equations 4. r22 = 0. ρ9 = 0. ρ5 = 0.599. 0.3: r11 = 0.822.929. Reading reflectance values from Figure 4. 0.90.04 r16 = 0.31 αterrestrial = 1 − Sum(r1i .05. 2.93.87 r26 = 0.04. 9. 5. ρ10 = 0. 7.05.84.113.581. the appropriate wavelenghts are: 0. Need to use radiation tables to generate the intervals.8. r210 = 0. Reading corresponding reflectance values from figure 4.04. 0. ρ3 = 0. Using 10 points from Table 3.72. r23 = 0.04. 10) 10 Solution αextraterrestrial = 0.426.974.23.364.08.04.412.1.653. r19 = 0.1b. 1.8.32 αextraterrestrial = 1 − Sum(ρi . 1. 0. r13 = 0. Reading reflectance values from figure 4.6. K.90.i = BlBodyF ract (λb.88 175 = 1 − b r14 = .81. 10 fb.i = fa.8. Sum(r1i .87. and from the reflectance curve read the reflectances at the characteristic (midpoint) wavelengths.i · Tb ) end 99 r16 = . Note that it was necessary to limit λ between 0 and 100.5 10 fb. r17 = . 175) Tb = ConvertTemp(C.6.p04 04 Equations 4. 10 i − 0.i · Ta ) end Reading reflectances from curve C of Figure 4. r12 = . r13 = .92. i = 1. r15 = . r110 = . $UnitSystem SI Mass Deg kPa K kJ Data Ta = ConvertTemp(C.5 10 fa.63.4 Calculation of emittance for a real surface at two different temperatures.84. duplicate i = 1.32. r18 = .i = i − 0.i = BlBodyF ract (λa.88 . 10) 10 At 30 C Repeat with new temperature. r19 = .66. 30) a At 175 C Use the 10 equal increments( from either Table 3. duplicate i = 1. K.1b or as calculated here).2 are: r11 = . 92 Sum(r2i .87 0.75 0.92 0.89 41.87 estimated 30 = 1 − r25 = .69 0.21 5.88 0. r27 = .45 0.91 0.93 20.46 6.63 0.28 25.25 0.1 [K] Tb = 303.45 0.66 0.87 0.88.84 0.80 r1i [-] 0. i = 1. r23 = .90.135 Ta = 448.07 9.i [-] 0.55 0.9 0.75 0.82 0.2 are: r21 = .22 8. Software exists to automatically digitize curves.9 0.04 12.88 0.88 0. 10) 10 Note that it would have been useful to read the reflectances from the curve in small increments and place the values in a lookup table.84 11.i [micrometer] 6.85 0.82.87 .35 0.45 13. r22 = .05 0.55 0.51 27.56 9.69.i [micrometer] 4.10 100 r2i [-] 0.95 λa. r28 = .54 16.65 0.15 0.91.65 14.81 0.35 0.91 0.88.56 11.87. r210 = .95 λb.47 8.65 0.229 30 = 0.32 0.25 0. Solution 175 = 0.05 0.The corresponding reflectances from curve C of Figure 4.i [-] 0. r26 = .91.8.85 0.15 0.72 17. r24 = .92 0.88 fb.2 [K] Arrays Row 1 2 3 4 5 6 7 8 9 10 fa.47 7. r29 = . 1 = 0.1 (11) ∆f solar.90 (1) λ3 = 100 (2) αλ.6.75.2 = fsolar.1 · ∆f solar.1 = 0.981 − 0.3 − fsolar.2 (12) From Eqn 4.3 (5) T = 350 [K] (6) From Table 1.4 α = αλ.6.1 = 0.2 · (0.2 + αλ.1 = BlBodyF ract (λ1 · T ) (14) 101 1 .1 + αλ. λ.2 = 0.3 · ∆f solar. λ1 = 0.968 + 0.2 = 0.1 (3) αλ. $UnitSystem SI Mass Deg kPa K kJ Data λ.3 = fsolar.351 (7) fsolar.2 − fsolar.2 · ∆f solar.968) (8) fsolar.3 = 1 − λ.351 (10) ∆f solar.5 Calculation of the α/ ratio for a surface in space.6.3 (13) The fraction of the infrared radiation from 0 to λ is found from Table 3.3 = 1 (9) ∆f solar. λ.P04 05 Equations 4.1a or from the EES SETP function.2 = 1 − λ. λ2 = 2.3.1 for solar radiation the fraction in each band is: fsolar.2 (4) αλ.3 = 0.1 = 1 − λ. fIR.20.6. 2 fIR.1 · ∆f IR.351 [-] λ2 = 2.2 = 0.75 fIR.2 Note that almost all of the energy is in band 3.2 = 0.9997 λ1 = 0.1 = λ.2 − fIR.8997 fIR.1 + λ.1 = 9.3 = 0.0294 [-] λ.1 = fIR.2 + λ.6 [micrometer] αλ.4386 ∆fIR.3 = fIR.091 × 10−26 [-] ∆fsolar.1 = 0.1 = 0.2 = fIR.0001006 [-] ∆fsolar.9 fsolar.3 = 1 [-] Ratio = 0.3 (20) Ratio Ratio = α/ (21) Solution α = 0.9996 [-] = 0.1 ∆fsolar.(16) ∆f IR.3 = 0.6.3 = BlBodyF ract (1000 [micrometer] · T ) Note that we extended the data beyond the 100 micrometer limit.1 (17) ∆f IR.1 = 0.2 = 0.3 = 0.3 = 0.2 = BlBodyF ract (λ2 · T ) (15) fIR.2 · ∆f IR.2 = 0.6 [micrometer] T = 350 [K] αλ.091 × 10−26 fsolar.2 = 0.3 · ∆f IR.9706 [-] λ3 = 100 [micrometer] 102 2 αλ.8 ∆fIR.4875 .25 ∆fIR.1 = 0.0001006 fsolar.fIR.2 = 0.3 − fIR. (19) From Eqn 4.351 [-] λ.1 (18) ∆f IR.3 = 0.6196 [-] λ.1 = 9. 874 ρ[7] = 0. 0. i = 1. 0. 1. ρ10 = 0.0244 × 10−7 · θ4 − 1.2: ρ1 = 0.9937 × 10−13 · θ7 (2) end (3) Use 10 equally spaced intervals from Table 2.1.28. midpoints) are: 0. 1. α30 = αn · alpha/alphan (30) (7) α45 = αn · alpha/alphan (45) (8) α60 = αn · alpha/alphan (60) (9) Solution Variables in Main program α30 = 0.6.04 ρ[8] = 0. 10) 10 (6) Use Equation 4. ρ7 = 0.426.11.08.10.300. ρ9 = 0.1 ρ[10] = 0.581. 0.86 [-] ρ[4] = 0.07.22 ρ[2] = 0. ρ2 = 0.6 Calculation of solar absorptance as a function of angle of incidence.04.14.04 α60 = 0.10.653.3026 × 10−5 · θ3 + 9. 0.7734 × 10−10 · θ6 − 6.1 αn = 0.731 × 10−4 · θ2 − 2.508.6. function alpha/alphan (θ) (1) Equation 4.P04 06 Equations 4.1 alpha/alphan = 1 − 1. ρ3 = 0.732. 1.080.822. 0. 45 and 60 degrees.08 ρ[9] = 0.1 to find α at angles of 30. ρ5 = 0.07 α45 = 0. Reading reflectance values at these wavelengths for curve C of Figure 4. ρ4 = 0.929.e.28 ρ[3] = 0.4 αn = 1 − Sum(ρi ..22.19 (5) From Eqn 4.85 [-] ρ[5] = 0.04 (4) ρ6 = 0. 0.8000 × 10−8 · θ5 + 1.11.5879 × 10−3 · θ + 2.8.9745.19 .81 [-] ρ[6] = 0.14 103 ρ[1] = 0. ρ8 = 0. The appropriate wavelenghts (i. 7 Determine collector solar and IR radiation properties. ‘lambda’ = λ) dλ 0. ‘rho’ . T ) C1 = C1#. This problem is best done in 4 parts. T ) · Interpolate(‘Curve C Fig 4 8 2’ . P lanck = λ5 (1) C2 := C2# (2) C1 C2 −1 · exp λ·T (3) end P lanck (4) Subprogramemittance (T : ) (5) T otal = sigma# · T 4 R 100 [micrometer] P lanck (λ.P04 07a Equations 4.01 [micrometer] = T otal (6) 104 (7) . function P lanck (λ. ‘lambda’ . 05288 0.041 αsolar.4844 0.2677 105 6 2 E0.09135 0.2343 0.315 × 107 W/m2 E0.04327 0.041to25 = 0.09615 0. ‘rho’ .041to25 = 1 − E0.01 0.09615 0.01to100 = 0.6367 0.0.0.04808 0.1202 0.01to100 = 7.21 1.6016 0. ‘lambda’ .01 αsolar.03846 0. It isfortuitous that the two integrals give the same value to 3 significant figures.041to25 = 7.41 micrometers is about 14% of the total.041 to 25 Z 25 P lanck (lambda1.5313 0.1531 0.887 λ = 100 [micrometer] αsolar.141 × 106 W/m2 Tsun = 5.04327 0.1346 0.01 to 100 Z 100 E0.9883 1. Tsun )·Interpolate(‘Curve C Fig 4 8 2’ .01to100 = P lanck (λ. The energy below 0.8711 0.07692 0.041to25 /total (14) The energy beyond 25 micrometers is very small.041to25 = 0.0.5547 0. ‘rho’ .0.end emittance a (8) What is the absorptance for solar radiation assuming the sun is a blackbody at 5777? Tsun = 5777 [K] (9) 4 T otal = sigma# · Tsun (10) wavelength range 0.03846 0. 777 [K] .6953 0.0625 0.139 × 10 W/m T otal = 6. Solution Variables in Main program αsolar.7773 0.012 1.9297 0.309 ρ 0.01to100 /total (12) wavelength range 0. Tsun ) · Interpolate(‘Curve C Fig 4 8 2’ . ‘lambda’ .01to100 = 1 − E0.5195 0.41 0. ‘lambda’ = lambda1) dLambda1(13 E0.09135 0.8359 0.887 Lambda1 = 25 [micrometer] Curve CF ig482 Row 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 λ [micrometer] 0. ‘lambda’ = λ) dλ (11) 0. 9135 0.899 0.984 11.8798 0.1187 0.6664 0.911 4.9039 0.1859 0.2483 0.14 12.9086 0.857 6.21 22.18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 1.8846 0.9135 0.1323 0.52 25 100 0.09428 0.2819 0.122 3.399 2.899 0.5522 0.209 2.515 3.0898 0.63 22.8846 0.454 1.786 5.692 1.7659 0.421 5.5 23.079 5.8894 0.09 20.9135 0.7187 0.9086 0.866 8.8702 0.07 23.8317 0.3901 0.55 14.4903 0.596 4.207 4.9 20.9086 0.8846 0.5281 0.91 18.42 19.7 16.9183 0.45 19.783 3.7933 0.925 9.8846 0.5901 0.9039 0.8798 106 .09443 0.68 13.86 21.242 0.8606 0.591 2.94 24.899 0.927 2.629 7 7.75 23.25 21.9183 0. Plot 1 107 . 01 [micrometer] P lanck (λ.887 whereas using the solar distribution of Table 2.588. limit5 = . limit2 = . T ) · Interpolate(‘Curve C Fig 4 8 2’ . (8) Repeat problem a but assume the solar radiation is distributed as in Table 2.7 Determine collector solar and IR radiation properties. T ) C1 = C1#. ‘lambda’ . P lanck = (1) C2 := C2# (2) C1 C2 λ5 · exp λ·T −1 (3) end P lanck (4) Subprogramemittance (T : ) (5) T otal = sigma# · T 4 (6) R 100 [micrometer] = 0. limit7 = .6. limit3 = .545. limit8 = 1.P04 07b Equations 4. ‘rho’ . The variable limit are the bounds of the bands from Table 2. limit9 = 1.871.6. j = 1. Part of this difference can be attributed to the different methods applied for integration.876.1.28.201.777 (12) (13) Sum(ρj .6. ‘lambda’ = λi ) dλi limiti+1 − limiti end αsolar. This problem is best done in 4 parts. Solution Variables in Main program αsolar. ‘lambda’ = λ) dλ (7) T otal end emittance b.876 108 . limit10 = 1. limit4 = . limit1 = . We will use 10 increments.47.b = 0. function P lanck (λ. ‘lambda’ . 10) 10 (14) Part a used a black body sun to estimate the solar absorptance as 0.b = 1 − limit6 = .691.1. 10 R limiti+1 ρi = limiti (9) (10) (11) Interpolate(‘Curve C Fig 4 8 2’ .012. limit11 = 5 duplicate i = 1.1 the solar absorptance is estimated to be 0.617. ‘rho’ . 7 Determine collector solar and IR radiation properties. ‘rho’ .01to100 = 1 − IR0.041to25 /totalIR (15) There is a very large difference between the two IR emittance values (0. T ) · Interpolate(‘Curve C Fig 4 8 2’ .041 IR.01 [micrometer] = T otal (6) end emittance (8) c (7) What is the emittance for long-wave radiation at 325 K? TIR = 325 [K] (9) 4 T otalIR = sigma# · TIR (10) T $ = ‘Curve C Fig 4 8 2’ (11) wavelength range 0. TIR ) · Interpolate(T $. Solution Variables in Main program 109 .P04 07c Equations 4. ‘rho’ .262) since about 14% of the energy is beyond 25 micrometers.01 to 100 Z 100 IR0.01 IR.01to100 /totalIR (13) wavelength range 0. ‘lambda’ = lambda3) dLambda3(14) 0. This problem is best done in 4 parts.041 to 25 Z 25 IR0. ‘lambda’ .0. P lanck = λ5 (1) C2 := C2# (2) C1 C2 · exp λ·T −1 (3) end P lanck (4) Subprogramemittance (T : ) (5) T otal = sigma# · T 4 R 100 [micrometer] P lanck (λ. TIR )·Interpolate(T $.0. T ) C1 = C1#. ‘lambda’ = λ) dλ 0. function P lanck (λ. ‘lambda’ . ‘rho’ . ‘lambda’ .01to100 = P lanck (lambda2.041to25 = P lanck (lambda3. ‘lambda’ = lambda2) dLambda2(12) 0.376 vs 0.041to25 = 1 − IR0. IR.1 W/m T otalIR = 632.0.3755 Lambda2 = 100 [micrometer] TIR = 325 [K] IR.0.5 W/m2 .041to25 = 467.01to100 = 0.01to100 = 395 W/m2 T $ = ‘Curve C Fig 4 8 2’ 2 IR0.2615 Lambda3 = 25 [micrometer] 110 IR0.041to25 = 0. P04 07d Equations 4.0000106300682*T[i]2 + 5. ‘rho’ . ‘lambda’ .6857 300 350 400 450 500 111 .7 Determine collector solar and IR radiation properties. T ) C1 = C1#. duplicate k = 1. ‘lambda’ = λ) dλ T otal end emittance (7) (8) d Plot the emittance versus surface temperature for a range of temperatures from 300 K to 800 K.0. P lanck = (1) C2 := C2# (2) C1 C2 λ5 · exp λ·T −1 (3) end P lanck (4) Subprogramemittance (T : ) (5) T otal = sigma# · T 4 (6) R 100 [micrometer] = 0. A third order does much better epsilon[i]=-0. 11 (9) TK = 250 + k · 50 (10) call Emittance (TK : k ) (11) end (12) A second order curve fit does not fo a very good job.654 0.51218632 + 0.6931 0. Since EES has a limit on the number of integration variables that can be in the main program. an integral function is used. function P lanck (λ.6856 0.5821 0.00637126087*T[i] . This problem is best done in 4 parts.01 [micrometer] P lanck (λ.39253381E-09*T[i]3 Solution Variables in Main program Arrays Row i Ti 1 2 3 4 5 0. T ) · Interpolate(‘Curve C Fig 4 8 2’ . 5383 550 600 650 700 750 800 epsilon vs T 112 .6205 0.5932 0.6 7 8 9 10 11 0.6463 0.6689 0.5655 0. 526. θ1 ) (1) If(θ1<0 )or (θ1>180 ) then (2) r=0 (3) else (4) If(θ1 = 0) then r= n−1 n+1 (5) 2 (6) else (7) θ2 = arcsin rperp = sin (θ1 ) n sin (θ2 − θ1 ) 2 sin (θ2 + θ1 ) 2 (8) (9) 2 rpar = r= tan (θ2 − θ1 ) (10) 2 tan (θ2 + θ1 ) rperp + rpar 2 (11) endif (12) endif (13) end (14) knowns n = 1.P05 01 Equations Calculate the reflectance of one glass surface. ra = r (n.04494 θc = 50 [deg] rc = 0. θd = 70 rc = r (n. θc ) . rb = r (n.06147 θd = 70 [deg] 113 rd = 0. θd ) Solution Variables in Main program n = 1. (15) rd = r (n.04338 θb = 30 [deg] rb = 0.526 θa = 10 [deg] ra = 0. θb ) . function r (n.1758 (16) . θb = 30. θc = 50. θa = 10. θa ) . 526 θa = 10 [deg] Nc = 3 θb = 70 [deg] τa = 0.7862 τb = 0. Nc = 3. θb = 70 τb = τ (n.P05 02 Equations Transmittance of three covers. Nc .5326 114 (17) . θb ) . τa = τ (n. θa = 10. Nc . (16) The results agree with Fig 5. θ1 ) (1) If(θ1 < 0)or (θ1 > 180) then (2) τ =0 (3) else (4) If(θ1 = 0) then rperp = (5) index − 1 index + 1 2 (6) rpar = rperp (7) else (8) θ2 = arcsin rperp = sin (θ1 ) index sin (θ2 − θ1 ) 2 sin (θ2 + θ1 ) 2 (9) (10) 2 rpar = tan (θ2 − θ1 ) (11) 2 tan (θ2 + θ1 ) endif (12) τ = 1/2 · 1 − rpar 1 − rperp + 1 + (2 · Nc − 1) · rpar 1 + (2 · Nc − 1) · rperp endif (13) (14) end (15) knowns n = 1.3 Solution Variables in Main program n = 1. function τ (index. Nc .1.526. θa ) . 526. Ref rind + 1 (1) (2) (3) τref = 1−r 1 + (2 · N cov − 1) · r (4) else (5) rperp = rpara = sin (Ref rAng − IncAng) sin (Ref rAng + IncAng) 2 tan (Ref rAng − IncAng) tan (Ref rAng + IncAng) (6) 2 (7) τperp = 1 − rperp 1 + (2 · N cov − 1) · rperp (8) τpara = 1 − rpara 1 + (2 · N cov − 1) · rpara (9) τref = τperp + τpara 2 (10) endif (11) τabs = exp −N cov · KL Cos(Ref rAng) (12) T ransIncAng = τabs · τref (13) end (14) Knowns K = 20 [1/m] .P05 03 Equations Transmittance of two covers at two angles. function T ransIncAng (N cov. KL. Ref rInd) sin (IncAng) Ref rAng = arcsin Ref rInd If(incAng ≤ 0) then 2 Ref rind − 1 r= . θb . θa = 0 [deg] . θa .002 [mm] . K · L. L = 0. Rindex) . Rindex = 1. θb = 50 [deg] τb = T ransIncAng (N. IncAng. N = 2. τa = T ransIncAng (N. K · L. Rindex) Solution Variables in Main program 115 (15) (16) . 7814 L = 0.7446 N =2 θa = 0 [deg] Rindex = 1.K = 20 [1/m] τa = 0.526 θb = 50 [deg] 116 .002 [mm] τb = 0. 5 ρd = 0.96 ρd = 0. index) (6) αb = αn · Abs/Absn (θb ) (7) T auAlphaP roductb = τb · α b (1 − (1 − αb ) · ρd ) Eqn 5. index) (3) αa = αn · Abs/Absn (θa ) (4) T auAlphaP roducta = τa · αa (1 − (1 − αa ) · ρd ) Eqn 5.1 (5) Part b τb = T ransIncAng (N. θb = 60.750 θa = 25 [deg] αb = 0. θb . KL.P05 04 Equations τ -α product for two covers at two angles.1 (8) Solution αa = 0. KL.96 N =2 τa = 0.037. index = 1.5.7828 p = 0.6936 .22 τb = 0.526 T auAlphaP roducta = 0.22 (2) Part a τa = T ransIncAng (N.8922 KL = 0. θa = 25. αn = 0.634 θb = 60 [deg] 117 αn = 0.9475 index = 1.96.526 (1) From footnote 2 of Section 5. KL = 0.96. θa .037 T auAlphaP roductb = 0.5. Knowns N = 2. p = 0. We will use SETP functions TransIncAng and Absn. b .49 [deg] KL = 0.b = IncAngEf f Dif f (Slopeb ) (7) = T ransIncAng N. index = 1.a = IncAngEf f GrRef (Slopea ) (2) angled. index (8) τd.a = 56.526.b = 59.b = 0. angled.a = 69. IncAngEffDiff and TransIncAng Knowns N = 1.b Solution angled.a = T ransIncAng N.a = T ransIncAng N.b = 0.70 angleg.33 [deg] N =1 τg. index (5) Part a angleg.83 . KL.81 [deg] slopeb = 90 index = 1.a = 0. KL.81 angled.b = T ransIncAng N.a = 0. slopea = 45. angleg.b .81 118 angleg.a .a = IncAngEf f Dif f (Slopea ) (3) τg. KL = 0. index (9) τg. index (4) τd.b = 59.a .037 τd. angleg.45 [deg] slopea = 45 τg. angled. slopeb = 90 (1) Part a angleg.b = IncAngEf f GrRef (Slopeb ) (6) angled. KL.P05 05 Equations Estimate transmittance for diffuse sky and ground reflected radiation We will use the SETP functions IncAngEffGrRef.037.526 τd. KL. 819 τb = 0. K = 10. angle. (13) L = 0.4 K = 10 119 L = 0. Angle = 58.6. K · L. Ref rInd) sin (IncAng) Ref rAng = arcsin Ref rInd (1) (2) If(incAng ≤ 0) then r= (3) Ref rind − 1 Ref rind + 1 2 . indexb = 1. τa = T ransIncAng (N. indexb ) Solution Variables in Main program Angle = 58 indexa = 1. K · L. KL.6 indexb = 1. indexa ) .P05 06 Equations function T ransIncAng (N cov.002. IncAng.40 Knowns τb = T ransIncAng (N. angle.865 (14) (15) . indexa = 1.002 N =1 τa = 0. τref = 1−r 1 + (2 · N cov − 1) · r (4) else (5) rperp = rpara = sin (Ref rAng − IncAng) sin (Ref rAng + IncAng) 2 tan (Ref rAng − IncAng) tan (Ref rAng + IncAng) (6) 2 (7) τperp = 1 − rperp 1 + (2 · N cov − 1) · rperp (8) τpara = 1 − rpara 1 + (2 · N cov − 1) · rpara (9) τref = τperp + τpara 2 (10) endif (11) T ransIncAng = exp −N cov · KL Cos(Ref rAng) · τref (12) end N = 1. IncAng. τref = 1−r 1 + (2 · N cov − 1) · r (4) else (5) rperp = rpara = sin (Ref rAng − IncAng) sin (Ref rAng + IncAng) 2 tan (Ref rAng − IncAng) tan (Ref rAng + IncAng) (6) 2 (7) τperp = 1 − rperp 1 + (2 · N cov − 1) · rperp (8) τpara = 1 − rpara 1 + (2 · N cov − 1) · rpara (9) τref = τperp + τpara 2 (10) endif (11) τabs = exp −N cov · KL Cos(Ref rAng) (12) T ransIncAng = τabs · τref (13) end (14) Knowns L = 0. index = 1.P05 07 Equations Calculate transmittance and compare exact and approximate methods. KL. function T ransIncAng (N cov.526.0025. N1 = 1. 120 N2 = 2 (15) . θ = 55. Ref rInd) sin (IncAng) Ref rAng = arcsin Ref rInd (1) (2) If(incAng ≤ 0) then r= (3) Ref rind − 1 Ref rind + 1 2 . K = 25. 0003372 τP ar = 0. θ.1% Solution Variables in Main program index = 1.6883 L = 0.001 out of 0. index) (21) Part d .1 (24) (25) The two methods of calculation differ by 0. K · L.2.808 or a little more than 0.calculate single cover transmittance by the exact method L τAbs = exp −K · Eqn 5.single surface sin (θ) θ2 = arcsin Index rP erp = rP ar = r= sin (θ2 − θ) sin (θ2 + θ) tan (θ2 − θ) tan (θ2 + θ) (16) 2 (17) 2 (18) rP erp + rP ar 2 (19) Part b using the approximate method τ1 = T ransIncAng (N1 .47 [deg] 121 N2 = 2 τ2 = 0.9286 .686 r = 0.809 θ = 55 [deg] N1 = 1 τ1Exact = 0.1 · 2 1 + rP erp 1 − (rP erp · τAbs ) τP ar = τAbs · 1 − rP ar 1 + rP ar τ1×10 xact = τP ar + τP erp 2 · (23) ! 1 − rP2 ar 1 − (rP ar · τAbs ) (22) 2 Eqn 5.0025 τ1 = 0. index) (20) Part c using the approximate method τ2 = T ransIncAng (N2 .3.Part a .3.526 rP ar = 0. K · L.074 τAbs = 0. θ.808 θ2 = 32.1472 τP erp = 0.928 K = 25 rP erp = 0.2 Cos(θ2 ) ! 1 − rP2 erp 1 − rP erp τP erp = τAbs · Eqn 5. Ref rInd) sin (IncAng) Ref rAng = arcsin Ref rInd (1) (2) If(incAng ≤ 0) then r= (3) Ref rind − 1 Ref rind + 1 2 . ρg = 0.4 MJ/m2 .2 I = 3 MJ/m2 (15) (16) . IncAng. KL = 0.526. 122 α = 0. β = 25 [deg] . IbT = 4. Id = 0. θb = 25 [deg] .95. KL. function T ransIncAng (N cov. τref = 1−r 1 + (2 · N cov − 1) · r (4) else (5) rperp = rpara = sin (Ref rAng − IncAng) sin (Ref rAng + IncAng) 2 tan (Ref rAng − IncAng) tan (Ref rAng + IncAng) (6) 2 (7) τperp = 1 − rperp 1 + (2 · N cov − 1) · rperp (8) τpara = 1 − rpara 1 + (2 · N cov − 1) · rpara (9) τref = τperp + τpara 2 (10) endif (11) τabs = exp −N cov · KL Cos(Ref rAng) (12) T ransIncAng = τabs · τref (13) end (14) Knowns N = 1.P05 08 Equations Estimate absorbed radiation for an inclined collector.037. index = 1.1 MJ/m2 . 01 · α · τd (23) T auAlphag = 1.1 at 45 degrees or Eqn 5. θg . KL.1 MJ/m2 123 . θd .526 N =1 Sd = 0.787 MJ/m2 I = 3 MJ/m2 KL = 0.037 Sb = 3.1 at 45 degrees or Eqn 5.4.01906 MJ/m2 τb = 0.2 β = 25 [deg] Id = 0. KL.2 θd = 57 (18) τd = T ransIncAng (N.8812 θd = 57 [deg] T auAlphab = 0.2.7906 τg = 0.7068 T auAlphag = 0.3014 MJ/m2 Sg = 0.4.4 MJ/m2 S = 3.For the beam radiataion τb = T ransIncAng (N.95 IbT = 4. index) (21) From Equation 5.01 · α · τb (22) T auAlphad = 1.1 Sb = IbT · T auAlphab Sd = Id · T auAlphad · (25) 1 + Cos(β) 2 Sg = ρg · I · T auAlphag · (26) 1 − Cos(β) 2 (27) S = Sb + Sd + Sg (28) Solution Variables in Main program ρg = 0. KL.6781 θb = 25 [deg] α = 0.824 θg = 69 T auAlphad = 0.01 · α · τg (24) From Equation 5.9.8455 τd = 0.467 MJ/m2 index = 1. θb . index) (17) From Figure 5.1 θg = 69 (20) τg = T ransIncAng (N.2 T auAlphab = 1. index) (19) From Figure 5.4.4. 5. and IDIF F . Zenith = 71. τd59 = T ransIncAng (N. index) T auAlphad59 = (5) τd59 T ransIncAng (n.58 MJ/m2 .845 124 .5.2. β = 63.1 is 59 degrees.4 ρd59 = T auAlphad59 − τd59 (7) 5. 1) (6) 5. Io = 1. TauAlphab =1. 59. θb . Absn . I = 1. The equivalent diffuse angle from Figure 5. Index = 1. a θb = 36. ρg = 0. index) (3) αb = αn · Abs/Absn (θb ) (4) Need diffuse reflectance of the cover for use in Equation 5.853 rather than 0. KL. Note that this is NOT the same angle as required for the diffuse radiation in part b.4. Knowns Lat = 48.526. n = 1.3.3.2.6 T auAlphab = τb · αb 1 − (1 − αb ) · ρd59 (8) Using Equation 5.5.4.03 MJ/m2 (1) (2) For beam radiation τb = T ransIncAng (N. KL.013.955 KL = 0.01*τ b *αb =0. 59.1.P05 09 Equations Estimate absorbed radiation for a collector. KL. αn = 0. We will use EES SETP functions TransIncAng. 2.b For the diffuse radiation The equivalent angle for diffuse sky radiation on a collector sloped at an angle of 63 degrees is 57 from Fig 5. the results are S=1. KL. Ai = Ib /Io p f = Ib /I (24) Sbh = T auAlphab · Rb · (Ib + Id · Ai ) 1 + Cos(β) 3 Sdh = T auAlphad · Id · (1 − Ai ) · 1 + f · (sin (β/2)) · 2 1 − Cos(β) Sgh = T auAlphag · I · ρg · 2 Sh = Sbh + Sdh + Sgh (26) (25) Solution 125 (27) (28) (29) .1. This difference results from the choice of sky models and more than offsets ifferences resulting from choosing Equation 5. θd . the diffuse sky model: Sb = T auAlphab · Ib · Rb (20) 1 + Cos(β) 2 1 − Cos(β) Sg = T auAlphag · I · ρg · 2 S = Sb + Sd + Sg Sd = T auAlphad · Id · (21) (22) (23) Using Equation 5.772 rather than 0.99 rather than 1. θg = 64 (13) τg = T ransIncAng (n.01*τ d *αd =0. the HDKR sky model.5.5. θd = 57 (9) τd = T ransIncAng (n.2. index) (14) αg = αn · Abs/Absn (θg ) (15) αg T auAlphag = τg · 1 − (1 − αd ) · ρd59 (16) d The absorbed solar radiation kT = I/Io Id = I · IDIF F/I (kT ) . index) (10) αd = αn · Abs/Absn (θd ) (11) αd T auAlphad = τd · 1 − (1 − αd ) · ρd59 (12) Using Equation 5.9.1. Rb = (17) Ib = I − Id (18) Cos(θb ) Cos(Zenith) (19) Using Equation 5.766 c For the ground reflected radiation The equivalent angle for ground reflected radiation on a collector sloped at an angle of 63 degrees is 64 from Fig 5.4.1.2 to evaluate TauAlpha.4. TauAlphad =1. KL.75.9. θg . 5 [deg] 126 αn = 0.75 MJ/m2 Sd = 0.857 τd59 = 0.652 αg = 0.7939 Zenith = 71.58 MJ/m2 n=1 S = 1.766 τd = 0.848 θd = 57 [deg] T auAlphad59 = 0.12 MJ/m2 ρg = 0.843 τb = 0.07734 MJ/m2 Sbh = 1.013 αd = 0.1891 MJ/m2 T auAlphab = 0.99 MJ/m2 Sdh = 0.9009 β = 63 [deg] Ib = 0.48 MJ/m2 Sg = 0.937 Ai = 0.4 Rb = 2.021 Sh = 1.8642 f = 0.955 I = 1.2 [deg] .80 MJ/m2 Sgh = 0.03 MJ/m2 Io = 1.526 KL = 0.4369 Index = 1.6876 τg = 0.543 Sb = 1.08 MJ/m2 T auAlphad = 0.6903 MJ/m2 kT = 0.899 θb = 36.αb = 0.836 θg = 64 [deg] T auAlphag = 0.3397 MJ/m2 Lat = 48 [deg] ρd59 = 0.8186 Id = 0. The equivalent diffuse angle from Figure 5. KL. n = 1.P05 10 Equations Determine a months absorbed radiation.526. Index) (4) αb = αn · Abs/Absn (θb ) (5) Need ρd of the cover for use in Equation 5.6.4. . slope = 90. αn = 0.93 (1) (From Appendix G) ¯ = 6.44.1 is 59 degrees -& this is generally NOT the same angle as required for the diffuse sky radiation. Knowns Lat = 43.6 (7) (8) αb 1 − (1 − αb ) · ρd59 (9) For the diffuse radiation: θd = 59 (10) The equivalent angle for diffuse sky radiation on a vertical collector is 59 from Fig 5. ρg = 0. We will use EES SETP functions TransIncAng. 59. Index) T auAlphad59 = τd59 T ransIncAng (n.4. τd = τd59 (11) 127 .5.47 K (2) First need to find average for the three components. Index = 1.013. θb .1. H ¯ t = 0. KL = 0.4 Eqn 5. 1) ρd59 = T auAlphad59 − τd59 T auAlphab = τb · (6) Eqn 5. τd59 = T ransIncAng (n.3. KL.3. Abs/Absn and HDIF F BAR /HBAR. KL.1 the same angle as used in some calculations in part a. 59. For beam radiation: θb = 40 (3) τb = T ransIncAng (n. use Erbs’ correlation for winter .869 ¯ bT = 9.47 K ρg = 0. K ¯T H (15) ¯b = H ¯ −H ¯d H (16) ¯ b = 2.i.932 MJ/m2 H ¯ t = 0.37 MJ/m2 αd = 0.10.e. ω s is less than 81.1 the equivalent angle for ground reflected radiation on a vertical collector is exactly the same as for diffuse sky radiation.526 n=1 slope = 90 [deg] S¯g = 1.4 . T auAlphag = T auAlphad (14) Need to break total into beam and diffuse .727 MJ/m2 H ¯ gT = 1.05 MJ/m2 H Lat = 43 [deg] ¯ b = 2.8951 θd = 59 [deg] ¯ = 6.9102 ¯ b = 3. the diffuse sky model.407 MJ/m2 T auAlphab = 0.97 MJ/m2 ¯ T auAlpha = 0.02137 S¯ = 10.4.013 ρd59 = 0.8572 τd59 = 0.7284 θb = 40 [deg] 128 .765 MJ/m2 H ¯ T = 13.1d The absorbed solar radiation components using Equation 5.357 MJ/m2 H KL = 0. ¯b · R ¯b S¯b = T auAlphab · H (18) ¯ d · 1 + Cos(slope) S¯d = T auAlphad · H 2 (19) ¯ · ρg · 1 − Cos(slope) S¯g = T auAlphag · H 2 (20) b) The total radiation S¯ = S¯b + S¯d + S¯g c) (21) ¯T The average transmittance-absorptance product: Need to find H ¯ bT = H ¯b · R ¯b H (22) ¯ dT = H ¯ d · 1 + Cos(slope) H 2 (23) ¯ gT = H ¯ · ρg · 1 − Cos(slope) H 2 (24) ¯T = H ¯ bT + H ¯ dT + H ¯ gT H (25) ¯ ¯ H ¯T T auAlpha = S/ (26) Solution αb = 0.7941 τb = 0.93 ¯ d = 2.8163 T auAlphad = 0.44 MJ/m2 H ¯ dT = 1.62 R (17) a) From Figure 2.713 MJ/m2 H Index = 1.9881 MJ/m2 αn = 0.αd = αn · Abs/Absn (θd ) T auAlphad = τd · (12) αd 1 − (1 − αd ) · ρd59 (13) For the ground reflected radiation: From Fig 5.7284 τd = 0.use 50 ¯d = H ¯ · Hdif f Bar/HBar Lat.62 R S¯d = 0.8358 T auAlphad59 = 0. n.8358 T auAlphag = 0.19.2.6 S¯b = 7. 5 sigma# · Ti4 (6) end (7) (8) Note that at 45 C and 145 C the transmittances are nearly the same Arrays 129 . T ) P lanck = λ5 · exp (1) C1# C2# λ·T −1 (2) end (3) Although only two temperatures were requested. 12 (4) T C i = −5 + (i − 1) · 50 (5) Ti = ConvertTemp(C. K. At 418 K. Therefore.0 and 25. was used to generate the Lookup table. T C i ) R 100 P lanck(λ. duplicate i = 1. ‘tau’ . ‘lambda’ . Figure 5.7. ‘lambda’ = λ) dλ τi = 2.2 only covers the range out to 15 mircometers.3 micrometers. Ti ) · Interpolate1(‘tau vs lambda’ . However. A plot shows the source temperature dependence of the IR transmittance. the tedlar transmittances needs to cover the range from 4 to 33 micrometers. at a minimum. At 318 K.4 micrometers.P05 11 Equations Transmittance of cover with wavelength-dependent properties function P lanck(λ. 12 temperatures are evaluated. The free program. 95% of the blackbody radiation is between 4. including 45 C and 145 C. 95% of the blackbody radiation is between 5.22and 33. The transmittance beyond 15 will be assumed to be 0. Plot Digitizer.8. 018 8.2 768.857 3.08 10.689 0.919 7.09191 0.5 2.475 4.767 10.5997 0.709 0.062 4.6612 0.8479 0.12 8.71 7.509 3.663 0.700 0.645 0.4901 0.707 0.649 0.06 5.2 718.9726 0.Row 1 2 3 4 5 6 7 8 9 10 11 12 Ti [K] 268.804 9.8489 0.293 7.676 0.4272 130 .8554 0.2 418.2529 0.399 3.921 6.05114 0.705 T Ci [C] -5 45 95 145 195 245 295 345 395 445 495 545 tau vs lambda Row 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 λ [micrometer] 2.708 0.235 4.651 0.2 618.6205 0.371 5.2 518.2 468.1746 0.856 4.4769 0.917 8.301 6.4839 0.605 7.8903 0.011 9.2019 0.018 7.2 668.2 368.7638 0.492 9.9725 0.9226 0.185 7.7598 0.2 318.1332 0.9175 0.8475 0.181 9.675 8.2 568.8483 0.9235 0.2 818.751 5.294 8.799 0.7643 0.373 3.670 0.533 7.9169 0.2 τi [-] 0.9103 0.823 2.7368 0.064 3.1207 0.3877 0.36 τ 0.9231 0.3109 0. 789 0.94 12.42 12.88 13.36 13.98 15 100 0.67 12.63 11.543 0.8224 0.8 tau vs lambda tau vs T 131 .3648 0.84 11.84 11.67 14.8231 0.8 0.84 14.781 0.4062 0.7883 0.56 12.8708 0.3165 0.774 0.12 14.1455 0.8024 0.18 11.36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 10.4681 0.5425 0.12 13. ρg = 0. H k¯t = 0. ¯ = 6. 1) ρd59 = T auAlphad59 − τd59 T auAlphab = τb · (6) Eqn 5. τd59 = T ransIncAng (n.6 (7) (8) αb 1 − (1 − αb ) · ρd59 (9) For the diffuse radiation: θd = 59 (10) The equivalent angle for diffuse sky radiation on a vertical collector is 59 from Fig 5. For beam radiation: θb = 40 (3) τb = T ransIncAng (n. KL.4 Eqn 5.1. Index) (4) αb = αn · Abs/Absn (θb ) (5) Need ρd of the cover for use in Equation 5.44.5. θb . 59.4.0125. KL. 59.92 From Appendix G for January (1) (2) First need to find average for the three components. The equivalent diffuse angle from Figure 5. Index) T auAlphad59 = τd59 T ransIncAng (n. slope = 90.10. Knowns Lat = 43.4.7.1 the same angle as used in some calculations in part a. KL. KL = 0.1 is 59 degrees this is generally NOT the same angle as required for the diffuse sky radiation. This is essentially the same as Problem 5.47 n = 2. τd = τd59 (11) 132 .3.P05 12 Equations Energy absorbed in a room for a month ¯ We will use EES SETP functions TransIncAng. abs/absn and HdiffBar/H.526.3. Index = 1. αn = 0. 0125 ρd59 = 0.38 MJ/m2 H Lat = 43 [deg] ¯ b = 2.254 MJ/m2 H k¯t = 0.87 MJ/m2 αn = 0. ω s is less than 81.1 the equivalent angle for ground reflected radiation on a vertical collector is exactly the same as for diffuse sky radiation.47 ρg = 0.7 S¯b = 7.713 MJ/m2 H Index = 1. n.45 MJ/m2 133 ¯ = 6.51 MJ/m2 T auAlphab = 0.4 .9004 ¯ b = 3.62 R (17) a) From Figure 2.62 R S¯d = 0.1d The absorbed solar radiation components using Equation 5.8144 θd = 59 [deg] αd = 0.2. ¯b · R ¯b S¯b = T auAlphab · H (18) ¯ d · 1 + Cos(slope) S¯d = T auAlphad · H 2 (19) ¯ · ρg · 1 − Cos(slope) S¯g = T auAlphag · H 2 (20) b) The total radiation S¯ = S¯b + S¯d + S¯g c) (21) ¯T The average transmittance-absorptance product: Need to find H ¯ bT = H ¯b · R ¯b H (22) ¯ dT = H ¯ d · 1 + Cos(slope) H 2 (23) ¯ gT = H ¯ · ρg · 1 − Cos(slope) H 2 (24) ¯T = H ¯ bT + H ¯ dT + H ¯ gT H (25) ¯ ¯ H ¯T T auAlpha = S/ (26) Solution αb = 0.use Erbs’ correlation for winter .7111 τb = 0.6436 τd = 0.10.44 MJ/m2 H ¯ dT = 1. the diffuse sky model. T auAlphag = T auAlphad (14) Need to break total into beam and diffuse .use 50 ¯d = H ¯ · Hdif f Bar/HBar Lat.7818 τd59 = 0.i.03704 S¯ = 9.92 ¯ d = 2.4.7448 slope = 90 [deg] S¯g = 1.8597 ¯ bT = 9.7448 T auAlphad59 = 0.19.19 MJ/m2 ¯ T auAlpha = 0.e.727 MJ/m2 H ¯ gT = 2.357 MJ/m2 H KL = 0.765 MJ/m2 H ¯ T = 13.736 T auAlphag = 0.αd = αn · Abs/Absn (θd ) T auAlphad = τd · (12) αd 1 − (1 − αd ) · ρd59 (13) For the ground reflected radiation: From Fig 5. K ¯T H (15) ¯b = H ¯ −H ¯d H (16) ¯ b = 2.526 n=2 T auAlphad = 0.6436 θb = 40 [deg] . 6. . Day) slope = Stringval(T ableN ame$) (3) Note the use of the TableName# function to determine the slope (4) The Boulder data starts at 1am on Jan 8 LookU pRow = (Day − 8) · 24 + T ableRun# + 7 Row 1 in LookTable starts at 8am of Day. nday = nDay (M onth.031.which is both small and in the wrong direction. Finally. Ref rInd = 1.7 minutes (i. lat = 40 [deg] . The Boulder-Week data does not indicate whether it is based on clock time or solar time. We will add a time shift of +20 minutes in order to solve this problem.526.13 Daily solar absorbed by a collector. If we plot Ih versus clock time we see that the data is not symmetrical around solar noon on the 10th and 14th. -1.5 [hr] 134 (5) Start at midpoint of hour between 7 and 8am . Looking at Plot 1. one for each slope.68 degrees of hour angle) .97 which is very high. The conclusion is that this data is not very good. Surf AzAng = 0. ρg = 0. the two clear days (See Plot 1). If we assume time is clock time we find that kT exceeds 1. hour = Lookup(‘Boulder Week’ . ‘hour’ ) − 0. We will use the SETP EES functions as much as possible. shif t = 5 [deg] nCov = 1(1) (2) a) we will use 4 parametric tables. LookU pRow. M onth = 1. However. knowns αn = 0. Long = 105 [deg] . Day = 10.e. the peak radiation between 11 and 12 on the 13th has a kT of 0. KL = 0. for Boulder (meridian = standard meridian) solar time minus clock time is -6.(6) . we need to shift the hour angle by about +5 degrees (about +20 minutes) .85..0 at 4:30 pm.P05 13 Equations 5. HrAng. Surf AzAng) (8) θ = arccos (CosT heta) (9) α = αn · abs/absn (θ) (10) . Slope.HrAng = (hour − 12hr ) · 15 [deg/hr] − Shif t here we used the empirically found Shift (7) CosT heta = cosIncAng (Lat. nDay. . . MJ . . Ih = Lookup(‘Boulder Week’ . ‘I’ ) · . LookU pRow. . 001 kJ .0. c) addition of a second cover. 30 135 . lat. nDay.. Ref rInd. ρg . slope.5 [deg]) (12) kT = Ih /I0 (13) Id = Ih · Idif f /I (kT ) (14) IT = IT. Surf AzAng. lat.2 = ST. HrAng) This function uses Eqns 2. (Ih . (Ih . HrAng. lat.6 MJ/m2.2 = 2 (19) ST. HrAng. (Ih . It is not possible to tell if the second cover is appropriate d) Use a higher absorptance black paint absorber surface. αn ) (16) The general shape of ST and IT versus hour angle are similar.black = ST. Ncov. HrAng + 7.3 hours before and after noon) the values differ by about 10% . KL.HDKR. slope2 = 45 (18) ncov.5deg .6 MJ/m2. Ncov . slope2 . ρg .g.6 MJ/m2 and for two covers was 15. HrAng.16. HrAng − 7. slope. Again it is not possible to to tell if this increased absorbed solar radiation more than offsets the increase in IR radiation losses that will occur with the flat black absorber. For this day the values at slopes 60 and 75 degrees are nearly equal.HDKR. nDay. We have added a sum column option under the main menu Options/Preferences. Ref rInd. ρg .2 .10. (nDay. KL. Lat. slope. nDay. αn ) (20) The absorbed radiation for one cover was 17. (Ih . lat. We need information from Chapter 6 to fully answer parts c and d.7 (15) ST = ST. KL. Surf AzAng.96 (21) ST.8 MJ/m2 whereas the selective surface had an absorber radiation value of 17.HDKR. Although the absorbed radiation was reduced the thermal losses will also be reduced. αblack ) (22) The black absorber had an absorbed solar of 19. b) the average transmittance-absorptance. αblack = 0. nDay. Surf AzAng. taave = ST /IT (17) In the range of greatest importance (e. Surf AzAng.1 and 2.HDKR. We added a column in Parametric table 45 for the absorbed solar radiation with two covers. Ncov . Early in the morning and late in the afternoon the values drops considerably. ρg . Ref rInd. (11) I0 = IZero. 5 θ [deg] 77.352 0.467 3.033 0.5 11.008 1.446 0.5 -12.352 0.770 3.618 1.092 2.5 15.657 IT 2 MJ/m 0.786 0.728 0.7 76.0 62.5 2.308 0.5 77.561 0.80 0.85 0.471 0.84 0.1 16.052 1.725 0.401 0.138 3.5 -57.398 2.790 0.033 0.852 3.009 1.419 α 0.83 0.608 1.5 θ [deg] 71.5 47.286 0.799 0.5 2.193 2.5 16.9 34.5 10.105 0.789 0.5 10.598 3.234 1.286 0.297 0.732 0.5 -42.002 taave [-] 0.004 Id MJ/m2 0.205 0.81 0.592 Id MJ/m2 0.259 0.5 62.640 0.5 HrAng [deg] -72.5 8.736 0.003 Ih 2 MJ/m 0.016 2.5 47.147 0.004 Id MJ/m2 0.9 58.191 Ih 2 MJ/m 0.599 0.658 IT 2 MJ/m 0.002 taave [-] 0.790 0.076 2.805 1.799 0.371 3.5 9.012 0.8 49.003 ST 2 MJ/m 0.5 13.70 75 Run 1 2 136 α 0.471 0.183 2.733 1.705 0.011 1.5 69.5 8.775 0.987 3.80 0.746 0.734 0.5 -27.5 30.84 0.963 2.60 0.012 0.5 θ [deg] 63.740 0.872 1.720 0.62 60 Run 1 2 3 4 5 6 7 8 9 10 11 α 0.84 0.5 62.734 0.017 3.5 13.047 1.5 HrAng [deg] -72.5 -12.6 41.5 12.Run 1 2 3 4 5 6 7 8 9 10 11 hour [hr] 7.5 14.5 13.003 Ih 2 MJ/m 0.003 Ih 2 MJ/m 0.640 0.419 1.592 0.7 51.308 0.0 71.5 -27.012 0.003 ST 2 MJ/m 0.640 0.84 0.0 53.6 kT [-] 0.786 0.84 0.5 -57.670 0.864 3.83 0.3 32.008 1.570 1.805 1.698 45 Run 1 2 3 4 5 6 7 8 9 10 11 α 0.8 52.5 16.033 0.166 2.339 0.747 0.0 35.047 1.269 0.689 0.771 3.5 kT [-] 0.870 2.5 10.880 1.775 0.736 0.5 32.1 36.628 0.473 ST 2 MJ/m 0.012 0.006 0.5 16.83 0.409 2.286 0.84 0.551 2.5 62.269 0.640 0.8 3.419 1.5 11.297 0.692 0.5 11.002 taave [-] 0.733 1.310 2.4 39.352 0.5 47.81 0.78 0.5 15.83 0.259 0.821 1.269 0.75 0.2 56.1 82.519 2.191 0.5 -57.012 1.9 64.208 0.592 0.734 0.5 8.736 0.480 2.872 1.028 taave [-] 0.747 0.714 2.259 0.592 0.003 ST 2 MJ/m 0.5 12.5 77.82 0.606 0.5 77.2 24.734 0.471 0.3 44.665 3.251 0.70 0.330 1.745 0.789 0.5 θ [deg] 67.5 14.5 -42.789 0.250 0.644 0.5 8.191 0.400 hour [hr] 7.643 0.5 HrAng [deg] -72.5 17.877 2.539 0.911 0.2 58.938 3.297 0.739 0.191 0.85 0.6 25.5 -12.799 2.747 4.733 1.82 .84 0.5 14.83 0.005 0.9 17.661 0.047 1.004 Id MJ/m2 0.5 17.83 0.400 hour [hr] 7.4 kT [-] 0.033 0.84 0.742 0.400 hour [hr] 7.5 17.662 0.282 3.5 -42.308 0.931 2.46 IT 2 MJ/m 0.5 17.747 0.635 0.732 0.5 -57.208 0.786 0.205 0.625 0.625 0.242 2.610 0.5 2.83 0.512 0.436 1.83 0.6 20.80 0.5 9.1 44.718 0.643 0.347 0.5 17.872 1.3 45.5 -27.147 0.747 0.7 11.251 0.77 0.76 0.73 0.744 0.5 HrAng [deg] -72.147 0.5 15.5 32.5 32.078 0.570 1.208 0.007 0.799 0.643 0.805 1.5 9.5 17.790 0.419 1.5 31.658 IT 2 MJ/m 0.718 0.044 4.205 0.570 1.5 12.4 kT [-] 0.775 0.251 0. 479 1.5 -42.799 0.84 0.2 43.3 4 5 6 7 8 9 10 11 9.5 2.808 4.0 13.3 17.138 2.003 2.004 Plot 1 IT vs Omega 137 0.807 3.734 0.5 77.352 0.84 0.743 0.640 0.251 0.805 1.775 0.433 0.003 1.5 39.734 0.5 17.83 0.083 4.5 10.147 0.7 0.297 0.747 0.570 1.269 0.400 0.032 3.308 0.5 67.178 0.85 0.471 0.741 0.936 3.84 0.744 0.002 0.81 0.5 12.5 13.208 0.83 0.5 47.790 0.737 0.643 0.5 11.286 0.259 0.84 0.951 2.5 32.047 1.5 15.205 0.172 3.786 0.733 1.5 -27.104 2.728 0.789 0.872 1.2 27.3 55.567 0.375 2.5 14.656 .5 17.5 -12.1 31.74 2.982 3.1 20.5 62.5 16.676 0.720 0. ST vs omega taave vs omega 138 . 430 · 1 − 100 Tpl0 (16) If Slope < 70 then (17) slope0 = Slope (18) else (19) slope0 = 70 (20) endif (21) c = 520 · 1 − 0.07866 · N cov) (15) [K] e = 0. Slope.089 m ·K/W · W indCoef f − 0. Tpl )(13) endif (14) 2 f = 1 + 0.’ . Emitt.Klein. Tpl ) (5) Tamb0 = ConvertTemp(C. Tamb ) (6) else (7) T $ = ‘K’ (8) Tpl0 = Tpl (9) Tamb0 = Tamb (10) endif (11) else (12) call Error(‘Units must be set to SI to use the function U TOP klein. (Tpl . Tamb . K.1166m2 −K/W · W indCoef f · Emitt · (1 + 0.P06 01 Equations function UT op. N cov) If Unitsystem(‘SI’ ) then (1) (2) If Unitsystem(‘C’ ) then (3) T $ = ‘C’ (4) Tpl0 = ConvertTemp(C.000051 1/deg2 · slope02 (22) 139 . W indCoef f . K. Your plate temperature is xxxF1 R or F. Tamb. Slope.4.= 1 m2 ·K/W · N cov/x + 1/W indCoef f + y z [W/m2 ·K] (27) (28) end (29) Calculation of top loss coefficient for flat-plate collector Knowns. hw.b = 1 Tpl.w= Tpl0 − Tamb0 N cov + f (23) If w ≤ 0 then w0 = 0. (Tpl.b =4.a = 10 [C] Ut.b = 40 [C] .4. hw. (Tpl.4a Ut.b .a ) (32) From Fig 6.0 b) Selective absorber Ut.a = 10 [C] .Klein.4 is 45.a = 2(30) (31) Non-selective black absorber Ut. Tamb.b .a = 5 W/m2 -K Tamb.a =3. b .a = 2 Tpl. b = 0.a .b = 100 [C] .1.1 else w0 = w e x = (w0 ) · c/Tpl0 (24) (25) 2 2 y = sigma# · (Tpl0 + Tamb0 ) · Tpl 0 + Tamb0 (26) −1 2 · N cov + f − 1 + 0.b = 20 W/m2 ·K .a = 3.b .a .1 Slope = 45 [deg] Ut. Tpl.b = 100 [C] b = 0.b ) (33) FromFig 6.a = 60 [C] . hw.4.b = 40 [C] Ncov.b = UT op.b = 1 Ncov. Slope.a .95 Ncov. Tamb. Ncov.a = 5 W/m2 ·K .b = 20 W/m2 -K Tamb.b = 3.95. hw. Tamb.Klein.133 · Emitt − N cov + z = Emitt + 0.88 −1 UT op.4f Ut. a . Slope = 45 [deg] . a) a = 0. Tpl.1 Solution Variables in Main program a = 0.03 W/m2 -K hw. Ncov.00591 m2 ·K/W · N cov · W indCoef f 0.Klein.a = 60 [C] .a = UT op.78 W/m2 -K 140 hw. Ncov. Note that the slope used in Figure 6. The temperature units are set to C so we need to be careful where K is needed.K = ConvertTemp(C. K. 8.15. glass = 0. P lateSpacing = 0.4/3 T ime = 12 [hr] (2) (3) (4) Sky temp from Equation 3. InsulT hick = 0.25 2 Tsky = Tamb · 0.3 [kPa] Tdp = 20 [C] .K = ConvertTemp(C. Tplate = 100 [C] .88 Slope = 43 [deg] . plate = 0.5.11.P06 02 Equations Calculation of overall loss coefficient for flat-plate collector.6 hwind = max 5. Tsky ) (6) Back loss from Equation 6. Tplate. kIns = 0. K.4 to determine the heat transfer coefficient between absorber and cover.9.025 [m] .9 W indSp.070 [m] .10 Ub = kIns/InsulT hick (7) Qback = Ub · (TP late − Tamb ) (8) Use Equation 3. (1) Patm = 101.K = ConvertTemp(C.2 . HouseV ol = 300 m3 . σ = sigma#. W indSp = 6.0056 · Tdp + 0.11. We could use the SETP EES functions but we will do this in detail. Wind heat transfer coefficient from Equation 3.013 · Cos(15 · time) (5) Tsky.4.000073 · Tdp + 0.6 · HouseV ol0.034 [W/m ·K] . Tplate ) (9) Tcover. Knowns.711 + 0. Tamb = 25 [C] . Tcover ) (10) Tplate + Tcover T¯ = 2 (11) 141 . K. T = T¯ (12) 1 v Air.k = k Air. P = Patm µ Air. T = T¯. T = T¯ ν= ρ . . . J . . ¯ Cp = cp Air, T = T · 1000 kJ ρ= α= (13) (14) (15) k ρ · Cp (16) ∆T = Tplate − Tcover (17) P lateSpacing 3 ConvertTemp C, K, T¯ · ν · α ! 1.6 sin (1.8 · Slope) 1708 1 − 1708 · · max 0, 1 − Ra · Cos(Slope) Ra · Cos(Slope) Ra = g# · ∆T · (18) a = 1.44 · (19) b = max 0, Cos(Slope) Ra · 5830 ! 1/3 −1 (20) N usselt = 1 + a + b (21) N usselt = hcp,c · P lateSpacing/k (22) Use Equation 6.4.1 for losses from absorber to cover glass. Qtop = hcp,c · (Tplate − Tcover ) + σ · 4 4 Tplate,K − Tcover,K 1/plate + 1/glass − 1 (23) Equation 6.4.2 for losses from glass to surroundings 4 4 Qtop = hwind · (Tcover − Tamb ) + σ · glass · Tcover,K − Tsky,K (24) The definition of an overall loss coefficient is: Qtop + Qback = UL · (Tplate − Tcover ) (25) Solution a = 1.307 ∆T = 60.01 [C] HouseV ol = 300 m3 kIns = 0.034 [W/m-K] Patm = 101.3 [kPa] ρ = 1.028 kg/m3 Tamb = 25 [C] Tdp = 20 [C] Tsky,K = 297.1 [K] α = 0.0000278 m2 /s glass = 0.88 2 hwind = 12.36 W/m 2 -K ν = 0.00001995 m /s 2 Qback = 36.43 W/m −8 σ = 5.670 × 10 W/m2 -K4 T¯ = 70 [C] Tplate = 100 [C] Ub = 0.4857 W/m2 -K 142 b = 0.8231 plate = 0.11 InsulT hick = 0.07 [m] N usselt = 3.13 [-] Qtop = 276.4 W/m2 Slope = 43 [deg] Tcover = 39.99 [C] Tplate,K = 373.2 [K] UL = 5.214 W/m2 -C Cp = 1, 007 [J/kg-K] hcp,c = 3.606 W/m2 -K k = 0.02881 [W/m-K] P lateSpacing = 0.025 [m] Ra = 48, 304 T ime = 12 [hr] Tcover,K = 313.1 [K] Tsky = 23.93 [C] W indSp = 6.5 [m/s] P06 03 Equations We will use the $If directive to solve for parts a and b. Comment out one of the following two equations: P art$ = ‘a’ (1) Part a, Verfication of convective heat transport shown in Figure 6.4.3(a). $UnitSystem SI kPa K kJ Knowns Tplate = 100; Ta = 10; P lateSpacing = 0.025; T K plate = ConvertTemp(C, K, Tplate ); plate = 0.95; Slope = 45; glass = 0.88; Tsky = Ta T K a = ConvertTemp(C, K, Ta ); σ = sigma#; hwind = 10; g = g# (2) T K sky = ConvertTemp(C, K, Tsky )(3) (4) Heat transfer from absorber to cover glass: $IF Part$=’a’ then N u# = N uF latP late (Slope, T K cover , T K plate , P lateSpacing) (5) $Else ρ = ρ (AIR, T = T K ave , P = 101.3) ν= (6) µ (AIR, T = T K ave ) ρ (7) J . . Cp = cp (AIR. T = T K ave ) · . . 1000 kJ . 1 (10) (11) (12) $EndIf 143 . P lateSpacing 3 ν·α C2 = 1. α= (8) k/Cp ρ (9) Ra = g · (1/T K ave ) · (T K plate − T K cover ) · C1 = 0.28 .0 (From Figure 3.1 · C1 · C2 · Ra0.2) N u# = max 1.12.088. 1 [K] Tsky = 10 [C] 144 Qrad. The convection suppression device provided a 7% reduction.Sky = σ · glass · T K 4cover − T K 4sky (18) (19) Energy Balance on cover: Qrad.c · (T K plate − T K cover ) (16) Qrad.807 m/s2 N u# = 2.599 W/m2 -K plate = 0.8 W/m2 hp. Part b.1 W/m2 σ = 5.1 [K] Tplate = 100 [C] T K sky = 283. T = T K ave ) (14) N u# = hp.Cover + Qconv − Qwind − Qrad.P late.1 [K] T K cover = 321.Cover = σ · T K 4plate − T K 4cover 1/plate + 1/glass − 1 (17) Heat transfer from cover to surroundings: Qwind = hwind · (T K cover − T k a ) Qrad. Solution glass = 0.P late. The top loss coefficient.3a.c = 3.Sky = 211.23 [C] T K plate = 373. T K cover ) (22) In part a here are small differences between these calculated values and those in Figure 6.0291 [W/m-K] Qconv = 178.13.3 (which was done using a simple calculator).4 [K] Tcover = 48.025 [m] Qwind = 382.3 W/m2 T K ave = 347.670 × 10−8 W/m2 -K4 g = 9. Ut .Cover = 415.Sky T Kplate − T Ka (21) Tcover = ConvertTemp(K.c · P lateSpacing/k (15) Qconv = hp. C. Effects of adding convection suppression slats to Problem 6.60 W/m2 and in part b is 6.88 hwind = 10 W/m2 -K P lateSpacing = 0.3 [K] Ta = 10 [C] Ut = 6. in part a is 6.441 W/m2 -K P art$ = ‘a’ Qrad.T K ave = T K plate + T K cover 2 (13) k = k (air.4.6 W/m2 .Sky = 0 (20) Definition of top loss coeffficient: Ut = Qwind + Qrad.P late.956 Slope = 45 T K a = 283.95 k = 0. The only thing that changes between parts a and b is the calculation of the Nusselt Number. 807 m/s2 ν = 0.1 [K] T K sky = 283. 008 [J/kg-K] hp.02 kg/m3 T K a = 283.8 [K] Tcover = 45.7 [C] Slope = 45 T K plate = 373.00002823 m2 /s glass = 0.95 k = 0. 094 [-] Qrad.02901 [W/m-K] P art$ = ‘b’ P lateSpacing = 0.1 [K] Tplate = 100 [C] 145 C2 = 1 g = 9.3 W/m2 Cp = 1.7 W/m2 ρ = 1.215 W/m2 -K N u# = 1.908 Ra = 42.088 plate = 0.025 [m] Qwind = 357 W/m2 Qrad.P late.670 × 10−8 W/m2 -K4 T K cover = 318.88 hwind = 10 W/m2 -K C1 = 0.c = 2.Sky = 195 W/m2 σ = 5.Cover = 431.133 W/m2 -K .1 [K] Tsky = 10 [C] T K ave = 346 [K] Ta = 10 [C] Ut = 6.00002024 m2 /s Qconv = 120.P06 03 b Solution α = 0. We will use all three.c1 = 0.3. K.88 (10) qc2. pl = 0.0 .s = qw + qr0 c2. Tsky = Tamb . K.025 [m] .035.P06 04 Equations Calculation of collector top and overall loss coefficients There are three methods to solve this problem: Equation 6. $UnitSystem SI. Tc2 ) (8) T 0 sky = ConvertTemp(C. kJ.025 [m] .95. 110 [C] . K. Tpl ) (6) T 0 c1 = ConvertTemp(C.c2 + qr0 c1.4. P lSpp.c2 (14) 146 l = 0. hwind = 10 W/m2 ·K . or the detailed method of Example 6. Ncov = 2.c2 = 0. Tamb ) (5) T 0 pl = ConvertTemp(C.c2 = qc0 c1.4. c2 = 0.s (11) qw = hwind · (T 0 c2 − T 0 amb ) 4 4 qr0 c2.4. Space = 0. Tamb = 10 [C] . K.4.9. P lSpc1.88. K.025 [m] . T 0 amb = ConvertTemp(C. kinsulation = 0. K Knowns Slope = 45 [deg] . Figure 6. Tsky ) (9) HT from cover 2 to sky c1 = 0. Tc1 ) (7) T 0 c2 = ConvertTemp(C. Tpl =. P = P o#(2) P lateLength = 2 [m] (3) P lateW idth = 1 [m] (4) Convert temperatures to K.s = sigma# · T 0 c2 − T 0 sky · c2 (12) (13) HT from cover 1 to cover 2 qc1. 02787 [W/m-K] pl = 0. Reading Utop From Figure 6.05 [m] P lateW idth = 1 [m]2 qc2.c1 = sigma# · (22) 4 T 0 pl − T 0 c1 1/pl + 1/c1 − 1 (23) qtop = qp.c2 = sigma# · (16) (17) 4 T 0 c1 − T 0 c2 1/c1 + 1/c2 − 1 (18) HT from plate to cover 1 qp.5 W/m2 qp.035 [W/m-K] P lateLength = 2 [m] qc1.c2 = 401.964 W/m2 -K Ncov = 2 P lSpc1.c1 = 305.41 [C] T 0 c1 = 350.c1 = 401.c1 (19) qc0 p.5 W/m 2 qr0 c1.5 W/m2 qtop = 401.c2 = 136 W/m qr0 c2.64 [C] T 0 amb = 283.c2 condc1.c1 = 95.c1 = hc0 p.c1 = qc1.5 W/m qw = 264.c1 T 0 pl + T 0 c1 = k Air.c2 = 0.4 W/m2 Slope = 45 [deg] Tc2 = 36. T = T 0 c1 + T 0 c2 2 4 qr0 c1.c2 = qc2.c1 = 0.c2 = 265.8 [K] Ub = 0.c2 = hc0 c1. Ut =4.7 W/m2 -K c2 = 0.03051 [W/m-K] hc0 c1.c2 = 0.95 kinsulation = 0.c1 · (T 0 pl − T 0 c1 ) (20) hc0 p.qc0 c1.1 [K] T 0 pl = 383.91 W/m2 qr0 p.5 W/m2 Tamb = 10 [C] Tsky = 10 [C] condp.s (26) Utop = qtop 0 − T0 Tpl amb (27) Using the SETP function.c2 · Condc1.4.c1 (21) condp.715 W/m2 -K . T 0 c1 .2 [K] Utop = 4.s = 137.88 hwind = 10 W/m2 -K P = 101.6 W/m2 Space = 0.c1 + qr0 p.4b.c1 (24) qp. we get the same value for the top loss coefficient.s = 401. T 0 pl .3 [kPa] P lSpp.c2 = k Air.c1 = qc0 p.c1 · condp.6 [K] UL = 4.025 [m] qc0 p.c1 /P lSpp. T 0 c2 .025 [m] 2 qc0 c1.1 [K] 147 c1 = 0.0 W/m2K Ub = kinsulation /l (28) UL = Utop + Ub (29) Solution Condc1.c2 = N uF latP late Slope.c2 (25) qc1.c2 · (T 0 c1 − T 0 c2 ) (15) hc0 c1.c2 /P lSpc1.298 W/m2 -K l = 0.1 W/m2 Tc1 = 77.c1 = N uF latP late Slope.015 W/m2 -K T 0 sky = 283.025 [m] Tpl = 110 [C] T 0 c2 = 309. P lSpp. T 0 c1 .c1 = 0.c1 = 2.88 hc0 p. P lSpc1.c2 = 3. T = 2 4 qr0 p. 12 (7) 1/UL F0 = W· 1 UL ·(D+(W −D)·F ) F lowP ar = mdot/Ac · + 1/Cb + 1 π·Di·hf i 6.8677 0.5. hf i = 300.012.8677 148 FR [-] 0.0005.1 237.2539 0.67 mL [-] 0.9791 0.5 Calculation of collector efficiency factor F’ and flow factor FR Knowns Cb = 1 × 109 .100 Cp = 4180 (1) (2) The temperature is not given .020.P06 05 Equations 6. T = T ) r UL m= 6.8104 .020.4a k·δ (4) mL = m · F = (5) W −D 2 tanh (mL) mL (6) 6.8761 F0 0.8762 0. Di = 0.5.5.9656 0.assume: T = 75 [C] (3) k = k (M aterial$.9439 0.9339 0. δ = .18 (8) Cp UL · F 0 (9) F 00 = F lowP ar · (1 − exp (−1/F lowP ar)) (10) FR = F 0 · F 00 (11) Table 1 Run M aterial$ 1 2 3 Copper Aluminum Carbon Steel k [W/m-K] 397.9 58.6606 F [-] 0. W = 0. D = 0. UL = 8.328 0. mdot/Ac = 0. D = 0. kSolder = 20.791 [1/m] W = 0.005.16 (4) W −D 2 tanh (mL) mL (5) 6.010.002 [m] .5. UL = 3.020.5. k = 385.16 [m] del = 0. W = 0.001 T hickSolder = 0.4a k · del mL = m · F = 6.12 (6) 1/UL F0 = W· (3) 1 UL ·(D+(W −D)·F ) + 1/Cb + 1 π·Di·hf i 6.1954 [-] W idthSolder = 0.18: Cb = kSolder · W idthSolder/T hickSolder r UL m= 6.005 [m] 149 Di = 0. hf i = 300.16.5. W idthSolder = 0.001 [m] hf i = 300 W/m2 -C mL = 0.5.P06 06 Equations 6.9875 [-] kSolder= 20 [W/m-C] UL = 3 W/m2 -C D = 0.5.9332 [-] m = 2.6 Calculation of F’ from UL and plate dimensions Knows Di = 0.01 [m] k = 385 [W/m-C] T hickSolder = 0.18 (7) Solution Cb = 50 [W/m-C] F = 0. del = .02 [m] F 0 = 0.002 (1) (2) Use Equation 6. 7 Derivation of F’ for type (b) air heater The energy balance equation is: Qu = Ac · (S − UL · (Ti − Ta )) (1) Useful energy transfer to air by convection is: Qu /Ac = h1 · (T1 − Tf ) + h2 · (T2 − Tf ) (2) With no back losses: h2 · (T2 − Tf ) = hr · (T1 − T2 ) (3) Solve for T2 T2 = hr · T1 + h2 · Tf hr + h2 (4) Substitute (4) in (2) h2 · hr · (T1 − Tf ) Qu /Ac = h1 + h2 + hr (5) Solve for T1 T1 = Tf + Qu /Ac · 1 h1 + ! (6) h2 ·hr (h2 +hr ) Substitute (6) in (1) Qu /Ac = S − UL · Tf − Ta − (Qu /Ac ) · !! 1 h1 + h2 ·hr (h2 +hr ) 150 (7) .P06 07 Equations 6. Solve for Qu /Ac Qu /Ac = 1 1+ UL h ·h h1 + h 2+hrr · (S − UL · (Ti − Ta )) (8) 2 Then Qu /Ac = F 0 · (S − UL · (Ti − Ta )) where F0 = (9) 1 1+ UL h ·h h1 + h 2+hrr (10) 2 151 . 83. . FRta = 0.L = 6. The radiation (in MJ/m2 ) is not a rate but is total energy for the hour.P06 08 Equations 6.and watch units.3 W/m2 ·C .4 MJ/m2 . Ti = 56 [C] . Knowns IT = 3.6 .7.c FRU. Ta = 14 [C] . Qu/A.8 Calculation of collector output from two parameter model Use Equation 6. . . . ·6 MJ . . L · (Ti − Ta ) · hr · . = FRta · IT − FRU. 1 × 10 J . Solution FRta = 0.4 MJ/m2 hr = 3600 [s] . 600 [s] Ti = 56 [C] 152 IT = 3.869 MJ/m2 FRU.83 Qu /Ac = 1.L = 6.3 W/m2 -C Ta = 14 [C] hr = 3. 6 . Equation 6. FR = 0. Ti = 35 [C] .10 Calculation of a day’s collector output Knowns: UL = 5. hr = 1 Others are set in the table.units are MJ.P06 10 Equations 6. Allow only positive values of Qu \Ac. .2 W/m2 ·C .7.92. . . MJ/hr . . 2 .c = max FR · S − UL · (Ti − Ta ) · hr · . 0 MJ/m Qu/A. . 0.0036 W . 54 0.15 MJ/m2 hr = 1 [hr] Ta = −4 [C] Qu /Ac = 0.2 W/m2 -C .15 153 S = 0. 1.70 2.40 1. T ableRun#) Solution FR = 0.02 3.11 2.94 Ta [C] -3 0 4 5 7 9 11 5 1 -4 35 Qu \Ac2 MJ/m 0. SumQ = Sumparametric(‘Table 1’ .00 0.42 3.85 3. ‘Q uA c’ .21 2.00 12.21 1.30 2. An alternative is to use the SumParametric command and sum from row 1 to the current row solving the table.40 0.07 0.54 1.92 SumQ = 12.52 MJ/m2 UL = 5.52 17.00 MJ/m2 Ti = 35 [C] Table 1 Run Hour 1 2 3 4 5 6 7 8 9 10 Sum 7 8 9 10 11 12 13 14 15 16 115 S 2 MJ/m 0. Under the Menu Options/Preferences we have turned on the sum column.90 0. We cannot sum the whole column at one time since many cells have not yet been filled in.01 0. We can find the daily output two ways.90 2.00 1. 3) . kPa Calculate the dimensionless flow parameter φ and F”.75. F 0 = 0. Knowns: F low/Ac = 10 l/s ·m2 . T = 20. UL = 6. $UnitSystem SI. C. J. P = 101.0 W/m2 ·K 1 v (AIR.11 Calculation of FR for an air heater.P06 11 Equations 6. . . m3 . . . mdot/Ac = F lowAc · ρ · . 001 l .0. ρ= Cp = cp (AIR.687 F 0 = 0.75 ρ = 1. 004 [J/kg-K] F 00 = 0. T = 20) φ= mdot/Ac · Cp UL · F 0 F 00 = φ · (1 − exp (− (1/φ))) FR = F 00 · F 0 Solution Cp = 1.01204 kg/s-m2 154 FR = 0.6263 φ = 2.204 kg/m3 .835 UL = 6 W/m2 -K 2 F low/Ac = 10 l/s-m mdot/Ac = 0. Ti = 30 [C] .L *(Ti -Ta ) or: Q00 u = F RU L · (Tp − Ti ) Solution F RU L = 5. Tp = 118 [C] Equation 6.when the pump is off the absorbed energy equals thermal losses: 0=FR *S-FRU. when the pump is on Qu \Ac =Q”u =FRU.P06 12 Equations 6.5 W/m2 ·C .5 W/m2 -C Q00 u = 484 W/m2 Ti = 30 [C] 155 Tp = 118 [C] .L *(Tp -Ta ).7.12 Calculation of useful gain from no-flow temperature Knowns F RU L = 5. Therefore.L *(Tp -Ta )-FRU.6 . 85 Ti = 55 [C] FR = 0. and FR with Equation 6. F 0 = 0.015 kg/s ·m2 .7.015 Calculate the dimensionless flow parameter.015 kg/s-m2 Ta = 15[C] UL = 4 W/m2 -C 156 F 0 = 0.7.5: φ=m ˙ · Cp F 0 · UL F 00 = φ · (1 − exp (−1/φ)) Eqn 6. Knowns: m ˙ = 0.6 Solution Cp = 4. Cp = 4190 [J/kg ·C] .76 [C] .9719 Qu = 592. φ.9 φ = 17.5 FR = F 00 · F 0 Qu = FR · (IT · T auAlphae · Dust − UL · (Ti − Ta )) 1 − FR T¯plate = Ti + Qu · UL · FR Eqn 6. Ta = 15 [C] .985 IT = 1.8747 m ˙ = 0. 000 W/m2 T auAlphae = 0. Dust = 1 − 0.4 1 − F 00 T¯f luid = Ti + Qu · UL · FR Eqn 6.4 W/m2 T¯plate = 76.13 Calculation of mean plate and fluid temperatures. T auAlphae = 0.2 Eqn 6.9.9.46 T¯f luid = 59. UL = 4 W/m2 ·C . Ti = 55 [C] .21 [C] Dust = 0.7.9.P06 13 Equations 6. 190 [J/kg-C] F 00 = 0.85 IT = 1000 W/m2 . P06 14 Equations 6.8.85.9 Ti = 55 [C] 2 ItCrit =191.85 Ta = 15 [C] .015 . F 0 = 0.9.14 Calculation of critical radiation level UL = 4 W/m2 ·C . See Section 6.1 W/m UL = 4 W/m2 -C 157 T auAlphae = 0. The critical radiation level is that at which the Qu is zero is: 0 = ItCrit · T auAlphae · Dust − UL · (Ti − Ta ) Solution Dust = 0. Ti = 55 [C] . Dust = 1 − 0.985 Ta = 15 [C] F 0 = 0. T auAlphae = 0. 4 indicates that UL rises as plate temperature increases. T auAlphae = 0.P06 15 Equations 6. The value of FR does not matter.3 [C] T auAlpha e = 0. 000 W/m2 Tplate = 224.6. Solution Dust = 0.985 Ti = 55 [C] It = 1. with Q = 0.4. Dust = 1 − 0. Qu =FR *(IT *TauAlphae*Dust-UL *(Tplate -Ta )) Qu =0 0 = (IT · T auAlphae · Dust − UL · (Tplate − Ta )) Figure 6.015 Use Equation 6.7. Ta = 15 [C] . Ti = 55 [C] . It = 1000 W/m2 . Thus the UL used is low and the estimate of Tplate is high.85 UL = 4 W/m2 -C 158 Ta = 15 [C] .85. The calculation above is based on the assumption that UL stays at 4 W/m2C.15 Estimation of no-flow plate temperature Knowns: UL = 4 W/m2 ·C . 7314 × 10−4 1/deg2 · θd2 − 2.4.1388 · Slope + 0.0524. We will solve part a using the basic equations and part b using the SETP functions.1 2 Eqn 5.7734 × 10−10 1/deg6 · θd6 − 6.1. Ref Index = 1. slope = 0 θd = 59.526.8000 × 10−8 1/deg5 · θd5 + 1.P06 16 Equations 6.9 Part a Considering all of the radiation to be diffuse then the equivalent surface slope is zero.5879 × 10−3 [1/deg] · θd + 2.3026 × 10−5 1/deg3 · θd3 + 9. αn = 0. KLb = 0.7 [deg] − 0.2 T RAN Sperp = 1 − REF perp 1 + (2 · N cov − 1) · REF perp T RAN Spara = 1 − REF para 1 + (2 · N cov − 1) · REF para 159 .0125.1the angle is 59.9937 × 10−13 1/deg7 · θd7 α = αn · Abs/Absn Find the transmittance N cov = 1 Ref rAng = arcsin REF perp = REF para = sin (θd ) Ref Index sin (Ref rAng − θd ) sin (Ref rAng + θd ) Eqn 5.1.4 2 tan (Ref rAng − θd ) tan (Ref rAng + θd ) Eqn 5. For diffuse radiation from Figure 5. Knowns: N = 1.0244 × 10−7 1/deg4 · θd4 − 1.1.001497 [1/deg] · Slope2 Abs/Absn = 1. KLa = 0.16 Calculation of effective τ *α product for diffuse radiation on a collector.0000 − 1. 714 T RAN Sperp = 0.15 Eqn 5.2 T RAN Sabs = exp −N cov · Cos(Ref rAng) T RAN Sref = τ = T RAN Sabs · T RAN Sref ρd = 0.7212 θd = 59.8379 N =1 REF perp = 0.15 T auAlphae. KLb .9384 α = 0. θd .a = 0.3.001174 slope = 0 [deg] T auAlphaP rod.10.b = 0.a = 0.a Part b Here we will use the SETP EES function.1828 τ = 0.7 [deg] T RAN Sref = 0.2.694 T auAlphaP rod.526 ρd = 0.4 Using the approximation in footnote 2 of Section 5.5.1. αn ) T auAlphae.9 N cov = 1 Ref rAng = 34.10a. KLa .01 · T auAlphaP rod.b Solution Abs/Absn = 0.1 If we use the EES SETP function T auAlphaP rod.931 KLb = 0.9 2 KLa Eqn 5.9977 160 αn = 0. KL is small so absorption is small and we can use Eqn 6.8443 .b = T auAlphaP rod (N cov.b = 0.46 [deg] T auAlphae.7923 T auAlphaP rod. Ref Index. αn ) According to comments near Eqn 6.a = 1.a2 = T auAlphaP rod (N cov.680 T RAN Sabs = 0.0125 REF para = 0. θd .a = τ · α 1 − (1 − α) · ρd Eqn 5.6909 KLa = 0. Ref Index. T auAlphae.02 · T auAlphaP rod.0524 Ref Index = 1.a2 = 0.b = 1.10.10b T auAlphaP rod.5.T RAN Sperp + T RAN Spara Eqn 5.680 T RAN Spara = 0. T auAlphaP rod. P = P ) (13) Dhyd = Dout − Din (14) From Equation 3.bill2 Equations 7. T = Tave . T = Tave .1 Calculation of collector efficiency factor and collector flow factor for a concentrator with cylindrical receive $UnitSystem SI C J.5 W/m2 ·K (8) Properties: The pipe material in not known but the results are not a strong functions of the pipe thermal conductivity. Tave ) (9) µ = µ (SteamIAP W S . P = P ) (12) Cp = cp (SteamIAP W S . kwall = k (‘Carbon steel AISI1010’ .3 161 1 .045 [m] (2) m ˙ = 0.3 [m] (5) Tave = 100 [C] (6) P = 300 [kPa] (7) UL = 7.0168 [kg/s] (3) Length = 3. P = P ) (10) P r# = Pr (SteamIAP W S . T = Tave .14. T = Tave . kPa Knowns Dout = 0. P = P ) (11) k = k (SteamIAP W S .1 [m] (4) Apertrue = 0.054 [m] (1) Din = 0. 13 (24) Solution Apertrue = 0.7 [-] UL = 7. The abscissa is: X = Re# · P r# · Dhyd /Length (19) At X = 3.1 ν is about 3.3 [m] Dhyd = 0.045 [m] FR = 0.this is close to NusseltT found above.9 W/m2 -K Length = 3. We might have used Figure 3.9497 k = 0.08704 hf i = 371.054 [m] F 0 = 0.749 Tave = 100 [C] Y = 18. f ) (18) The flow in the annulus is laminar. providing confidence in the function. P r#.5 W/m2 -K 162 2 .9 .9758 kwall = 60.4 Cp = 4215 [J/kg-K] Dout = 0.10 ARec = π · Dout · Length Y =m ˙ · (21) (22) Cp ARec · UL · F 0 (23) FR = F 0 · Y · (1 − exp (−1/Y )) Eqn 7.0002818 [kg/m-s] N usseltT = 3.14.6792 [W/m-K] LoverD = 344. RelRough : N usseltT . LoverD.928 [-] P r# = 1.897 [-] Re# = 766. from Figure 3.1 [W/m-K] µ = 0.3.3. hf i · Dhyd /k = N usseltH F0 = (1/UL ) + Dout hf i ·Din The system is subject to a constant heat flux. 1/U L + Dout · ln (Dout /Din ) 2·kwall Eqn 7. N usseltH .4 N usseltH = 4.893 (20) ARec = 0.1 [m] m ˙ = 0.0168 [kg/s] P = 300 [kPa] RelRough = 0 X = 3.14.89.009 [m] f = 0.Re# = Dhyd *Vel*Density/µ = m*D ˙ ˙ hyd /(µ*AXsect ) = 4*m/(µ*π*(D out -Din )) Re# = 4 · m ˙ µ · π · (Dout + Din ) (15) LoverD = Length/Dhyd (16) RelRough = 0 (17) smooth tubes call P ipeF lowN (Re#.5259 m2 Din = 0.1 to estimate the convection heat transfer coefficient in the annulus but this figure is for constant wall temperature conditions. 12 (13) N u# = hw · Dcover /k (14) 163 .1 and 7.30 · Re#. cover = 0. Tf luid.3.18. Dcover.090 [m] . $UnitSystem SI K J kPa Knowns from Example 7. Ttube = 473 [K] . P = Patm k = k Air. Ta = 283 [K] .1 V = 5 [m/s] . Cp = 3260 [J/kg ·K] .0537 [kg/s] .3 [kPa] Tcover + Ta T¯ = 2 (7) patm = 101. Dtube. Dtube = 0.4 [W/m ·k] Tsky = 275 [K] (1) (2) (3) Knowns from Example 7. Dcover = 0.008m .assume atmospheric pressue is 101. tube = 0.P07 02 Equations 7.3.010m .060 [m] . ktube = 16 [W/m ·k] .88.2. Redo Examples 7. T = T¯ (10) (11) Re# = ρ · V · Dcover /µ N u# = 0. K.3 [kPa] (8) µ = µ Air.i = Dcover − 0.i = Dtube − 0.3. T = T¯ (9) ρ = ρ Air.18.3. W = 2.in = ConvertTemp(C.2 with emittance of the tube at 0.2 L = 10 [m] . T = T¯. m ˙ = 0.6 (12) Eqn 3. σ = sigma# kglass = 1.5 [m] . 156. Tcover =290 [K] Properties of air .15.9 [C]) hf i = 300 W/m2 ·K S = 430 W/m2 (4) (5) (6) Guess the cover temperatue to develop the governing equations. 885 m2 DimF lowRate = 41.9783 hw = 40.082 [m] tube = 0.i = 289.1 m2 Cp = 3.13.i ) Calculate area of receiver. 260 [J/kg-K] Dtube = 0. Atube = π · Dtube · L Qloss = Atube · σ · (18) T4 − T4 tube cover.4 [K] Ttube = 473 [K] UL = 2.18 hf i = 300 W/m2 -K ktube = 16 [W/m-k] N u# = 148.i − Tcover )) ln (Dcover /Dcover.235 kg/m3 T¯ = 285.827 m2 ARec = 1. At this point it is necessary to update guesses and then comment out the initial guess for cover temperatue.3.0000179 [kg/m-s] m ˙ = 0. ARec : ARec = π · Dtube · L (22) Calculate unshaded area of aperture.i ·hf i ) + 1/U L Dtube 2·ktube Eqn 7.4 [W/m-k] µ = 0.3 [kPa] ρ = 1.9902 F 00 = 0.i cover 1/tube + 1− · Dtube /Dcover.1 [K] V = 5 [m/s] .8 [K] Tsky = 275 [K] Acover = 2. AAper : AAper = (W − Dcover ) · L (23) Calculate the dimensionless flow rate DimFlowRate to use in calculation of F: DimF lowRate = m ˙ · Cp ARec · UL · F 0 F 00 = DimF lowRate · (1 − exp (−1/DimF lowRate)) (24) Eqn 7.885 m2 Dcover.6 [K] Tcover. Qu = AAper · FR · (S − UL · (ARec /AAper ) · (Tf luid.12.i = 0.88 F 0 = 0.in − Ta )) (27) Solution AAper = 24. 523 [W] S = 430 W/m2 σ = 5.0537 [kg/s] Qloss = 813.3 [W] Q 164u = 9.7 [-] Re# = 31.62 W/m2 -K L = 10 [m] patm = 101.13 (25) FR = F 00 · F 0 (26) Find the useful gain from Eqn 7.i ) (17) Radiation from tube to inside cover must equal heat loss from cover.10 (21) · ln (Dtube /Dtube.Convection and radiation heat loss from cover.988 [-] k = 0.271 W/m2 -K Atube = 1.3.in = 430.1 Qloss = UL · Atube · (Ttube − Ta ) (20) Calculate F’: F0 = 1/UL + Dtube (Dtube.05 [m] cover = 0. 039 Ta = 283 [K] Tf luid.09 [m] Dtube.i cover (19) The collector loss coefficient (based upon the tube temperatue and absorber area) is then found from Eqn 7.02459 [W/m-K] kglass = 1. Qloss = ((2 · π · kglass · L) · (Tcover.670 × 10−8 W/m2 -K4 Tcover = 288.i = 0.3 [-] Dcover = 0.06 [m] FR = 0. Acover = π · Dcover · L (15) 4 4 Qloss = Acover · hw · (Tcover − Ta ) + σ · cover · Tcover − Tsky (16) Find temperature inside tube.3. ω.85. FR = 0. ω)) (12) θb = arccos (cosIncAng (Lat. Surf AzAng = 0. (4) αd = 0. Idif f use = 0.89. β = 60. M onth = 1. 60.P07 03 Equations 7. n. Calculation of absorbed radiation αb = 0. KL.92 (3) Tin = 55 [C] . Surf AzAng)) (13) γs = SolAzAng (Lat. (5) τcover. Day) (9) Rbeam = RBeam. day = 26.20 MJ/m2 .d = T ransIncAng (Ncov . Lat = 43. KL.526. n. β. Ibeam = 1. a nr = 0. Ncov = 1 (2) Ta = −5 [C] .95 Assumed to be the same as for beam. (Lat. β. Surf AzAng) (10) δ = Dec (n) (11) θz = arccos (CosZenAng (Lat.11.3 Estimation of the output of a full CPC Knowns Surf Ref l = 0.36 MJ/m2 . Ref rInd) (7) Angles ω = (T ime − 12 [hr]) · 15 [deg/hr] (8) n = nDay (month.0 W/m2 ·C . UL = 9.b = T ransIncAng (Ncov .0370. Ref rInd = 1. n.1 at an angle of 21deg.95 From Figure 4. θc = 15 (1) time = 10 : 30. ω) (14) 165 . n. Ref rInd) (6) τcover. θb . ω. KL = 0. 2 is used to determine the ContFunct. (β + θc ) .7.1 to calculate S: C= 1 sin (θc ) (18) T ransCP C = Surf Ref lnr (19) AbsBeam = ContF unct · Ibeam · Rbeam · τcover. 1. (arctan (tan (θz ) · Cos(γs ))) .d · T ransCP C · T ransCP C · αd (21) AbsGrRef lected = 0 (Since (beta + thetac ) < 90) (22) S = AbsBeam + AbsDif f use + AbsGrRef lected (23) b Q00 u Calculation of collector gain per unit area .7. F: ContF unct1 = If((β − θc ) . 0) (16) ContF unct = ContF unct1 · ContF unct2 (17) Use Equations 7. 1. 0) (15) ContF unct2 = If((arctan (tan (θz ) · Cos(γs ))) . 1.b · αb · T ransCP C (20) AbsDif f use = (Idif f use /C) · τcover.Equation 7. 1. . . MJ/hr · m2 . . . = FR · S − (UL /C) · 1 [hr] · (Tin − Ta ) · . 0036 W/m2 .0. 998 MJ/m2 τcover.235 τcover.45 [deg] Lat = 43 [deg] nr = 0.95 ContF unct = 1 δ = −19.864 [-] day = 26 Ibeam = 1.5 [hr] UL = 9 W/m2 -C αb = 0.03 [deg] Idif f use = 0.5[deg] S = 1.36 MJ/m2 n = 26 Q00 u = 1.375 MJ/m2 Surf AzAng = 0 θb = 21.8821 θz = 65.95 ContF unct1 = 1 FR = 0.85 θc = 15 [deg] Ta = −5 [C] AbsDif f use = 0. (24) Solution AbsBeam = 1.92 KL = 0.b = 0.945 MJ/m2 αd = 0.8653 .05336 MJ/m2 β = 60 [deg] ContF unct2 = 1 γs = −23.d = 0.2 MJ/m2 M onth = 1 ω = −22.037 Ncov = 1 Ref rInd = 1.39 [deg] Tin = 55 [C] 166 AbsGrRef lected = 0 MJ/m2 C = 3.8051 time = 10.89 Rbeam = 2.48 [deg] T ransCP C = 0.526 Surf Ref l = 0. b · τCP C · 0.88 (13) AbsDif f use = (Id /C) · τcover.4 (12) AbsBeam = ContF unct · Ib.d · τCP C · 0.n · Cos(θb ) · τcover.d = 0.4 Absorbed radiation for a truncated CPC array $UnitSystem SI C J kPa Knowns ρsurf = 0.7. 1. Ib. τcover. 1. n. θc = 9 [deg] .13 MJ/m2 .2 is used to determine the ContFunct F ContF unct1 = If((β − θc ) .37 MJ/m2 . Id = 0.85 HrAng = −22. 0) (9) ContF unct2 = If((arctan (tan (θz ) · Cos(γs ))) . C = 6. Lat = 45. HrAng) (5) θz = arccos (CosZenAng (Lat. day) (4) γs = SolAzAng (Lat.P07 04 Equations 7. nr = 0.7. 1. (arctan (tan (θz ) · Cos(γs ))) . 1.85. β = 65 [deg] τcover.1 is used for the absorbed solar. n = nday (month.thetac < 90) S = AbsBeam + AbsDif f use + AbsGrRef lected (16) Solution 167 . (β + θc ) .90. n.b = 0. r τCP C = ρnsurf Eqn 7. HrAng)) (6) Dec = Dec (n) (7) θb = arccos (cos (Lat − β) · cos (Dec) · cos (HrAng) + sin (Lat − β) · sin (Dec)) (8) Equation 7.n = 1.90.0.88 (14) AbsGrRef lected = 0 (15) Since (beta . 0) (10) ContF unct = ContF unct1 · ContF unct2 (11) Equation 7.7.5 [deg] (1) (2) (3) Use the parametric table for the two dates. 85 τCP C = 0.9 τcover.08 [deg] Lat = 45 [deg] ρsurf = 0.04 MJ/m2 θb = 27.76 0.7212 0 AbsDif f use MJ/m2 0.85 [deg] AbsGrRef lected = 0 MJ/m2 ContF unct1 = 0 γs = −29.03985 0.88 [deg] Table 1 Run month day 1 2 1 3 2 15 Dec [deg] -22.85 θz = 51.819 γs [deg] -21.08 AbsBeam MJ/m2 0.04 .03985 168 AbsGrRef lected MJ/m2 0 0 ContF unct nr 1 0 0.37 MJ/m2 nr = 0.n = 1.9 0.819 [deg] Id = 0.b = 0.93 -2. AbsBeam = 0 MJ/m2 C=6 day = 15 Ib.03985 MJ/m2 ContF unct = 0 Dec = −2.8639 β = 65 [deg] ContF unct2 = 1 HrAng = −22.d = 0.9 θc = 9 [deg] AbsDif f use = 0.5 [deg] month = 3 S = 0.13 MJ/m2 n = 74 τcover.89 -29.9 S 2 MJ/m 0. P07 05 Equations 7. Tin = 120 [C] .5 [m] . Lat = 32 [deg] .7.87 (1) hf i = 3000 W/m2 ·C (2) For linear concentrator on N-S horizontal axis. Ref lconc = 0.3a Use the parametric table for time. use Equation 1.022 [m] .5 Performance of a linear concentrator with given focal zone flux distribution Knowns M onth = 3. L = 10. n = nDay (month. T ransAbsorptP rod = 0. Day) (3) Dec = Dec (n) (4) θz = arccos (CosZenAng (Lat. HrAng)) (5) 169 .78. Wreceiver = 0. day = 16.8 [m] . W = 1. n. 020m.14. there is no temperature gradient in the flow direction.87 Tamb = 9 [C] Wreceiver = 0. At times 9:30 and 11:30. respectively. from the figure. the intercept factors are 17/22 = 0.n = 3.1 [deg] Tin = 120 [C] Areceiver = 0. Assume that all of the absorber is wetted by the boiling liquid.4 [MJ] T ransAbsorptP rod = 0. and FR = F’.θ = arccos 2 2 2 (cos (θz )) + (cos (Dec)) · (sin (HrAng)) .5 35.418 [deg] hf i = 3.5 Eqn 1. Areaaper = (W − Wreceiver ) · L (7) AbsRad = Areaaper · Ib.43 24. 1 + UL /hf i FR = F 0 (10) Areceiver = Wreceiver · L 3600 Qu = FR · AbsRad − UL · Areceiver · · (Tin − Tamb ) 106 (11) (12) Solution AbsRad = 25.5.36 [MJ] Dec = −2.75 0.9968 Ib. the image widths are 0.77 170 AbsRad [MJ] 17.75.3.06 3.5 Ib.1: F0 = 1 .42 25.8 [m] IntF ract Qu [MJ] 16.020/cos(θ).77 M onth = 3 θ = 34. Since liquid is being boiled. By analogy with Figure 6.78 W = 1.14 MJ/m2 n = 75 θz = 35.22 m and 0.1 θz [deg] 49.36 .231 FR = 0.14 Tamb [C] 2 9 θ [deg] 27. and are 0.5 -7. when HrAngs are -37. 000 W/m2 -C Lat = 32 [deg] Ref lconc = 0. From the plot.3a (6) The intercept factors IntFact must be determined.12 to estimate Qu.13.6 W/m2 ·C From Figure 7. These broadened images will be centered on the normal incidence images.7. The image width with radiation normal to the aperture is 0.n · Cos(θ) · Ref lconc · IntF ract · T ransAbsorptP rod UL = 9.022 [m] Areaaper = 18.024 m.5 [m] Qu = 24.9968 IntF ract = 0.77 and 18/24 = 0.n 2 MJ/m 2.1 [deg] UL = 9.4 Table 1 Run 1 2 HrAng [deg] -37.5 [deg] L = 10.1 0.6 [W/m2-C] day = 16 HrAng = −7.2 (8) (9) Use Equation 7. First estimate FR.1 34.67 m2 F 0 = 0.5 and -7. 4. 054 Cmax.5 and 7.32 · 105 θs = arcsin 67 · 106 Cmax.1 θs = 0.circ = 24.3694 171 . Calculate θs for Venus at 67*106 miles from the sun.2.circ = 1 (3) 2 (sin (θs )) For circular concentrator use Equation 7.2.5: Solution Cmax.lin = 155.2. Use Equations 7.6 Calculation of maximum concentration ratios for collectors on Venus.2.6.6: Cmax.lin = 1 sin (θs ) (1) (2) For linear concentrator use Equation 7.P07 06 Equations 7. 2PM and 4PM.L · (Ti − Ta ) (5) Table 1 Run 1 2 3 Ti [C] 75.0 Ta [C] 20. the sun is in the plane perpendicular to the tubes. with Slope nearly equal to the Latitude.2. lat = 38 [deg] .7 Use of Incidence Angle Modifier for a CPC.n · IncAngM od · GT − FRU. n.0 60. and 60 at noon.00 1.0 75. (The exact values are calculated using Equation 1.17.0 GT 2 W/m 925 760 385 HrAng [deg] 0 30 60 θ [deg] 1. FRU.) Values of IncAngMod are read from the plot and entered in the table with the data.0 22.13 0.6.66. Since the sun travels 15 degrees/hr in this plane. γ) (4) Since collector performance data are on the basis of the array area. day = 21 γ=0 (1) (2) At the equinox. 30. β = 40 [deg] .91 172 Qu\Area c W/m2 545 503 168 . day) (3) Cos(θ) = cosIncAng (Lat.2 W/m2 ·C .L = 1. Incident angle n = nDay (month.P07 07 Equations 7.0 75. β.0 IncAngM od 1.0 22. HrAng.n = .9. Knowns FRT A. month = 3. the angles of incidence in the perpendicular plane will be approximately 0. Qu/Areac = FRT A.6 30. use Equation 6. F ocLength = 1. Wperf ect = 2 · 4. This results in a value of Y/f*103 = 13 at an intercept factor of 0. (2) HourAngle = 0 [deg] (3) Calculate RimAngle with Equation 7.8 Output of linear parabolic collector for perfect and imperfect concentrators. use Figure 7. L = 10 [m] (1) Ref lconc = .36 [m] . Knowns: Aperture = 1.95 Wimperf ect = 2 · 13 · F ocLength 103 (6) Neglecting shading by receiver: AAperture = L · Aperture (7) ARec = L · Wimperf ect (8) hr = 1 [hr] (9) 173 .11.10.78.9. Dec = 0 [deg] . Tamb = 7 [C] .1 at σ = 0.P07 08 Equations 7. FR = 0.2: RimAngle = arctan 8 · ! F ocLength/Aperture 2 16 · (F ocLength/Aperture) − 1 (4) For the perfect reflector.87.15. Y/f*103 = 4.95. For the intercept factor to be 0.14 MJ/m2 .26 [m] . UL = 14 W/m2 ·C . T ransAbsP rod = .9.n = 2. Tin = 230 [C] .95. InterceptF act = .2 · F ocLength 103 (5) For the imperfect reflector.1b. It is necessary to graphically integrate to get intercept factor vs Y/f*103. Ib. Lat = 35 [deg] . at a rim angle of 30.2. use Figure 7. Calculate Qu : Qu = FR · AAperture · (Ib.n · InterceptF act · T ransAbsP rod · Ref lconc − . . . MJ . . · (A /A ) · (T − T ) UL · hr · . . 0036 Rec Aperture in amb W · hr .0. 26 [m] InterceptF act = 0.95 [-] Qu = 13.n = 2.9 [-] Ib.57 [MJ] Tamb = 7 [C] Wperf ect = 0.36 [m] F ocLength = 1.3276 m2 HourAngle = 0 [deg] L = 10 [m] RimAngle = 30.87 [-] Tin = 230 [C] ARec = 0. (10) Solution Aperture = 1.01058 [m] AAperture = 13.6 m2 FR = 0.03276 [m] .14 MJ/m2 Ref lconc = 0.78 Wimperf ect = 0.2 [deg] UL = 14 W/m2 -C 174 Dec = 0 [deg] hr = 1 [hr] Lat = 35 [deg] T ransAbsP rod = 0. The relationship between the receiver dimension and the pointing error for L = 100 and W = 10 is: RecDimension0 = W + 2 · L · sin (0..9 Geometry of heliostat .27) + W (2) If there is a pointing error of P degrees.27 half-angle in the same way as a dispersion angle. W = 10 [m] (1) RecDimension = 2 · L · sin (0.08 [m] .1000 m and W = 10 m. For L . and if there is no pointing error.42 [m] 175 RecDimension0 = 41.27 (4) Solution L = 1. it will add to the 0.central receiver collectors. plus the plus the largest linear dimension of the heliostat. If the incident beam radiation is approximately notmal to the heliostat. and with the half-angle subtended by the sun of 0. 000 [m] W = 10 [m] P = 0.27 + P ) (3) so P = arcsin RecDimension0 − W 2·L − 0. W m.27 degrees: knowns: L = 1000 [m] . the receiver dimension is the width of a beam from a point on the heliostat to to the receiver L meters away.6205 RecDimension = 19.P07 09 Equations 7. for example. Ta = 20 [C] . This is not realistic. Track temperature changes on a fully mixed storage tank. Cp = 4190 [J/kg ·C] . Tamb = 20 [C] . Loads and inputs are given as independent of tank temperature. solutions start at row 2. Note that in Row 1 we have entered the tank initial condition so. Assume there are no constraints on temperature: $IF ParametricTable = ’Solution A’ . a real system will have inputs and outputs that are tank-temperature dependent. as shown. U As = 5. in Chapter 10.1.P08 01 Equations 8. hr = 1 [hr](1) We will use two parametric tables.56 [W/C] . Knowna: m = 500 [kg] . . . . . . J/hr . . J . . + . . Ts − Ts · m · Cp = (Qu − Load) · . 1000000 − U As · (Ts − Ta ) · hr · . 3600 MJ . W . ‘Tplus s’ ) (2) Used to extract the tank temperature for the next hour. Ts = Tablevalue(‘Solution A’ . Solution B: Assume that temperature is constrained between 20 and 100. (3) $EndIf If these calculations were carried through more hours on a clear day. The constraints can be programmed thus: $If ParametricTable=’Solution B’ . the tank temperature would go well over 100 C. representing a minimum temperature for energy delivery and a maximum temperature for energy storage. Other constraints could be assumed. The temperature cannot exceed 100C unless the tank is pressurized (not common in solar systems). The temperature can not go below 0 C unless the store goes to ice. tablerun# − 1. or both. which requires a different formulation. In practice the temperature of a store must be constrained by the characteristics of the storage medium or the mechanisms for adding or removing energy. . . . . . J/hr . . J . . + . 3600 TsCalc − Ts · m · Cp = (Qu − Load) · . . 1000000 − U A · (T − T ) · hr · s s a . MJ . W . 14. Ts+0 Ts = Tablevalue(‘Solution A’ . tablerun# − 1. Assuming reasonable values for Qul and Load for the following hours results in temperatures that are constrained to 100. + + Ts+0 = If TsCalc . 20. Solution 176 . ‘Tplus s’ ) (4) (5) (6) Used to extract the tank temperature for the next hour. 100. (7) $EndIf Here the temperature did not reach 100. 20. 18. 100 Ts+ = If Ts+0 . 20. TsCalc . The ‘Load deficiencies’ for the four hours when Tplus s is at 20 are 8. Note also that during hours when the tank temperature is constrained the load must be met by other energy supplies. Thus 61 MJ would have to be supplied from another source. and 21 MJ. 100. 2 9.61 Solution B Run T ime 1 2 3 4 5 6 7 8 9 10 11 12 13 0 1 2 3 4 5 6 7 8 9 10 11 12 Tsplus 45 39.4 16.61 hr = 1 [hr] Ta = 20 [C] + TsCalc = 50.8 -8.8 -8.5 [C] U As = 5.03 33.2 9.5 Load [MJ/hr] Qu [MJ/hr] Ts [C] 12 12 11 11 13 14 18 21 20 20 18 16 0 0 0 0 0 0 0 0 21 41 60 75 45.5 1.548 1.5 1.056 -8.257 22.7 22.1 -8.56 [W/C] .5 Tsplus 45 39.61 Solution A Run T ime 1 2 3 4 5 6 7 8 9 10 11 12 13 0 1 2 3 4 5 6 7 8 9 10 11 12 Load [MJ/hr] Qu [MJ/hr] Ts [C] 12 12 11 11 13 14 18 21 20 20 18 16 0 0 0 0 0 0 0 0 21 41 60 75 45.0 2.034 2.3 22.7 22.0 39.75 22.12 27.03 33.42 20 20 20 20 20 20 22.61 Load = 16 [MJ/hr] Tamb = 20 [C] Ts+0 = 50.787 -8.47 50.0 2.47 50.Cp = 4.1 27.3 22.75 22.1 -8.19 9.0 33.0 33.12 27. 190 [J/kg-C] Qu = 75 [MJ/hr] Ts+ = 50.61 177 m = 500 [kg] Ts = 22.0 39.4 16.42 16.1 27. 0.2 Develop equations for a partially stratified.P08 02 Equations 8. m ˙ L . 0. T1 . two node tank. 0. 0.4.1 becomes: F1c = If(To . Fup = 1 − Fdown (2) For node 1 the energy balance is: m1 · T1+ − T1 = (U A/Cp ) · (T 0 a − T1 ) + F1c · m ˙ c · To − m ˙ L · T1 − Fdown · (F1c · m ˙ c−m ˙ L ) · T1 ∆T + Fup · (F1c · m ˙ c−m ˙ L ) · T2 (3) For node 2 the energy balance is: m2 · T2+ − T2 = (U A/Cp) · (T 0 a − T 2) + (1 − F1c ) · m ˙ c · To + Fdown · (F1c · m ˙ c−m ˙ L ) · T1 ∆T − Fup · (F1c · m ˙ c−m ˙ L ) · T2 (4) 178 . 1) (1) Is the flow between nodes up or down? Fdown = If(F1c · m ˙ c. The two node tank is shown in the diagram Window. Equation 8. It can be assumed that water supplied from the mains at Tm will always enter the bottom node. 1). M assV el. $UnitSystem SI C.0607 kg/s-m2 ρ = 1.8 m2 .053 [m/s] Length = 1.3 [kPa] 179 M assV el = 0. kPa Knowns Length = 1.0125 [m] V = 0. P artDiam = . Ta ) (6) This design does not satisfy the minimum pressure drop of 55 Pa.3 Calculation of pressure drop across a packed bed store.8 m2 m ˙ = 0. Patm = 101. Solution Af rontal = 14. T = Ta ) (3) m ˙ = ρ · Af rontal · V (4) M assV el = m/A ˙ f rontal (5) ∆P = P resDrop (Length.3 [kPa] .8984 [kg/s] Ta = 35 [C] ∆P = 29.0125 [m] . p = Patm .145 kg/m3 .98 [Pa] P artDiam = 0.80 [m] .053 [m/s] estimated (1) (2) ρ = ρ (air.8 [m] Patm = 101. P artDiam. V = 0. Ta = 35 [C] Af rontal = 14.P08 03 Equations 8. 4 Estimation of equivalent heat capacity of a phase change storage medium.P08 04 Equations 8. From Eqn 8.9: EquivHeatCap .8. . = mass · . 1 × 10·6 MJ . Solid = 2890.Liquid = 2890 TP haseChange = 46. 890 [J/kg-C] TP haseChange = 46. 300 [kg] Cp.Liquid = 2. J LatentHeat + 18. 890 [J/kg-C] mass = 1.7 [C] .7 [C] 180 EquivHeatCap = 6.3 TP2 haseChange ! · Cp.7.3 · Cp.Solid + 18.1: Cp.Liquid TP2 haseChange ! (1) Data on the paraffin wax are from Table 13. Assumed same as solid LatentHeat = 209000 [J/kg] . 000 [J/kg] . (2) mass = 1300 [kg] (3) Solution Cp.3 · 2 TP haseChange 1− 18.Solid = 2. Cp.037 [MJ/C] LatentHeat = 209. 3 [1/V] (2) IDI = 2.=V.0 [·] (5) Rsc = 3 [ohm ·amp ·hr] Qm Rsd = 0.864 [·] (4) md = 1.035 · Qm (12) Qd = Qm /0.08 [V] (8) Gd = 0.5 (6) [ohm ·amp ·hr] Qm (7) Gc = 0.85 (13) VDI = 1/KDI · ln Voc = abs (I) +1 IDI (14) Esc + Esd 2 (15) H =1−F If I < 0 then (16) Discharge (17) Vcell.P08 05 Equations function Vcell.08 [V] (9) Esc = 2.25 (10) Esd = 2.oc − VDI − Gd · H + I · Rsd · 1 + md · H Qd /QM − H 181 (18) . F. Qm ) (1) Battery Parameters from Eckstein thesis (1990) KDI = 29.10 (11) Qc = −0. (I.5 [amp] (3) mc = 0. 1 and 8.10. Must uncomment one of these two statements Charging$=‘True’ Charging$ = ‘False’ (23) ChargeEf f iciency = 0.=V. 1. one charging and one discharging.2V will charge the battery in 11 hours.10. 182 . The charging part is trial and error.else Charge (19) Vcell. ChargeEf f iciency) eta = 1 when discharging and eta=0. Select a voltage until the charge time equals 11 hours. F. Imax ) (32) η = If(Ibatt .cell = Qm /Ns (29) Vcell = Vcell.95 (24) Imax = 11 [amp] (25) Ns = 6 (26) Np = 1 (27) Qm = 50 [amp ·hr] (28) Qm. Qm. (22) The problem must be run in two parts.cell ) Simple function to evaluate equations 8.2 (30) Vbatt = Ns · Vcell (31) Ibatt = min (Icell · Np .oc + VDI − Gc · H + I · Rsc · 1 + mc · H Qc /QM − H endif (20) (21) end Vcell. 0.95 when charging F = Q/Qm (33) (34) Z T imef inal Ibatt · η dtime Q = Qo + (35) T imeinitial Z T imef inal Ibatt · Vbatt dtime P = (36) T imeinitial $IF Charging$=’True’ Qo = 10 [Amp ·hr] (37) T imeinitial = 0 [hr] (38) T imef inal = 11 (39) Vbatt = 15.2 (40) By trial and error a voltage of 15. 1. (Icell . 6 42.682 8.000 0.5 19 Q [amp-hr] 50.Q.25 [hr] η = 1 [-] Icell = −3.543 10.487 0.573 -4.5 14 14.439 0.341 -4.857 -4.304 0.066 Vbatt [V] 10.333 [Amp-hr] T imef inal = 19.924 8.3 44.348 0.892 0.946 0.430 -5.220 Ibatt [amp] -5.2 26.0 F [-] 1.393 0.25 [hr] Vbatt = −Ibatt · Rload By trial and error the battery lasts 19.876 9.014 -4.6109 Imax = 11 [amp] Q = 10.5 13 13.084 -5.200 [-] Np = 1 Qm = 50 [amp-hr] time = 19.0 39. The power to charge was 640 Wh and the power delivered discharge was 391 Wh.$Else Rload = 2 [ohm] (41) Qo = 50 [Amp ·hr] (42) T imeinitial = 11 [hr] (43) T imef inal = 19.5.535 0.2 13.210 -4.5 15 15.979 [V] .33 [V] Charging$ = ‘False’ Ibatt = −3.787 0.F.25 [hr] Vcell = 1.261 0. V_batt.380 -5.5 17 17.7 17. (44) Resistance load (45) $EndIF $IntegralTable time: 0.0 47.I_batt.425 10.271 -5.150 -5.684 0.989 [amp] P = −391 [W-hr] Qo = 50 [Amp-hr] T imeinitial = 11 [hr] Integral Table Row 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 time [hr] 11 11.146 8.419 8.2 31.654 10.634 0.759 10. The cycle efficiency is: ηcycle = 391/640 (46) Solution Variables in Main program ChargeEf f iciency = 0.95 F = 0.859 10.4 36.350 9.P Find the cycle efficiency.7 29.25 hr.7 24.5 12 12.5 16 16.938 -4.213 -5.0 19.675 -4.027 9.0 [amp-hr] Rload = 2 [Ω] Vbatt = 7.1 11.713 9.462 -4.cell = 8.839 0.168 10.3 22.584 0.539 9.8 34.5 18 18.327 -5.4 15.301 10.769 -4.735 0.989 [amp] Ns = 6 Qm.132 P [W-hr] 0 -29 -58 -86 -113 -140 -166 -192 -217 -241 -264 -286 -308 -328 -347 -366 -383 183 ηcycle = 0. 989 7.5) 184 .200 -3.18 19.25 10.0 0.979 -391 F vs time (Problem 8. $UnitSystem SI C J kPa Knowns U Aloop = 38.1 [W/C] .P09 01 Equations 9.S = 210 [L/day] . (1) D = 0. Tcold = 12 [C] .019 [m] Thot + Tcold 2 (2) (3) Cp = cp (W ater. U seM. U Atank = 11 [W/C] . T = Tave .F = 610 [L/day] . L = 122 [m] .1 Calculation of water heating loads on a commercial building. Tave = U seS. P = 200) . Thot = 38 [C] Troom = 20 [C] . . . 1/Liter . . . T = Tave . ρ = ρ (W ater. P = 200) · . 0.001 1/m3 . F · 5 [day/week] + U seS. (4) (5) a Energy to heat water only: Loadhw = (U seM.S . . . . ·6 MJ . . · 2 [day/week]) · ρ · Cp · (Thot − Tcold ) · . 1 × 10 J . (6) b Add piping and tank losses: Load = Loadhw + Losstank + Losspiping (7) . . . MJ . . Losstank = U Atank · (Thot − Troom ) · 3600 [s/hr] · 24 [hr/day] · 7 [day/week] · . . 1 × 10·6 j . (8) Losspiping . . . . ·6 MJ . . = U Aloop · (Thot − Troom ) · 3600 [s/hr] · 24 [hr/day] · 7 [day/week] · . 1 × 10 j . 185 (9) . plus the heat to bring pipe and water therein back up to 38 C. assume that piping and water therein cools to 20 C.c With night shut-down: If flow loop is shut down for 14 hours per day. The energy loss from piping will then be losses for 10 hours per day. . . . MJ . . (10) LosspipingRed = U Aloop · (Thot − Troom ) · 3600 [s/hr] · 10 [hr/day] · 7 [day/week] · . . 1 × 10·6 J . . . . . ·6 MJ . . ReheatEnergy = W aterM ass · Cp · 7 [1/week] · (Thot − Troom ) · . 1 × 10 J . (11) . . . liter . W aterM ass = π · D2 · L/4 · . . 1000 3 . . F = 610 [L/day] .8 [MJ/week] Tave = 25 [C] Troom = 20 [C] U seM.3 [MJ/week] LosspipingRed = 172.49 [kg] 186 L = 122 [m] LoadRed = 687 [MJ/week] Losstank = 119.8 [MJ/week] Losspiping = 414.9971 [kg/liter] Thot = 38 [C] U Atank = 11 [W/C] W aterM ass = 34.8 [MJ/week] ReheatEnergy = 18.8 [MJ/week] ρ = 0.S = 210 [L/day] D = 0.18 [MJ/week] Tcold = 12 [C] U Aloop = 38.019 [m] Loadhw = 376.1 [W/C] U seS. · ρ m (12) LoadRed = Loadhw + Losstank + LosspipingRed + ReheatEnergy (13) Solution Cp = 4183 [J/kg-K] Load = 910. 2 Degree-day calculation of house heating load. IntGen = 1. Tamb = −3 (1) Use Equation 9. U Ahouse = 335.P09 02 Equations 9.52 [C] Loadrate = 5870 [W] Thouse = 19 [C] Tamb = −3 [C] U Ahouse = 335 [W/C] 187 .3.5 · 103 .3.3 for balance temperature: Tbase = Thouse − IntGen/U Ahouse (2) Use Equation 9. Knowns Thouse = 19.4 for load rate in terms of Tbase : Loadrate = U Ahouse · (Tbase − Tamb ) (3) Solution IntGen = 1500 [W] Tbase = 14. 2 = 224 [W/C] . 19. σyr = 10. k .3. Checking Units shows a large number of problems.0290 · T¯a.1 σ yr = 10.k = Tset − q˙ U Ah. 9.9 (11) hi.3 (9) duplicate i = 1. k = If ∆T i.k − T¯a.i · Ni1. Use the degree-day method of Section 9. Note that from Example 9.3 Calculation of annual degree-days and loads for building of Problem 9.3. 12 (4) Ni = N umDay (i) (5) σm. 3 (8) Tbase. k = σm. k .3 = 335 [W/C] (1) q˙ = 1500 [W] (2) T¯a.3. U Ah.1. k ln (cosh (1. 22..6.P09 03 Equations 9.6 [C] . 12 hi. k = Tbase. 6 [C] .698 · hi. DegDayAi. 2.396 (12) (13) DegDayB i. 15. k = Ni · ∆T i. 1. -5] (3) duplicate i = 1.5 · ∆T i.10 (6) end (7) duplicate k = 1. DegDayB i. 11.i + 0. We could have used three parametric tables. Tset = 19 [C] .2.45 [C] − 0.i · Ni · 1 [1/day] DegDayAi. 14. Knowns U Ah. U Ah.i Eqn 9. k (14) DegDay 0 i. k DegDay i.0664 · σyr Eqn 9.12 = [-8. -5.3.3. This is due to the empirical nature of some of the Equations.i p σm. 20. k .i = 1.k − T¯a. k . We will use two dimensional arrays for this solution.k Eqn 9. k = (10) Tbase. k = max DegDay 0 i.1 = 112 [W/C] . 0 [C ·day] (15) (16) 188 . k )) + + 0. DegDayAi.2041 2 3. . . . MJ . . k · DegDay i. Li. k = U Ah. k · 24 [hr/day] · 3600 [s/hr] · . . 1 × 10·6 J . 3. Eqn 9.6 (17) end (18) . . . GJ . . . Lk = Sum(Lii. ii = 1. k . 12) · . 001 MJ .0. 50.2 [C-Day] 629.2 546. Solution q˙ = 1500 [W] σyr = 10.3.54 3.i [C] 2.9 GJ Note: Decreasing the building UA has two effects on building heating loads.6 -238.6 [C] Tset = 19 [C] Arrays Row 1 2 3 4 5 6 7 8 9 10 11 12 Row 1 2 3 4 5 6 7 8 9 10 11 12 T¯a.299 2.6 σm.4 .01 309.1 [C-Day] 422 297 143 -102 -260 -402 -508 -446 -282 -167 108 329 DegDayAi.5 350.4 189 DegDayB i.606 1.893 1.11 5. 1] [C-Day] 422 297 168 26 5 0 0 0 3 14 138 329 DegDay 0 [i.54 3.096 2.5 350.6 70.3 [C-Day] 698.4 378.i [C] -8 -5 1 9 14 19 22 20 15 11 2 -5 Ni [day] 31 28 31 30 31 30 31 31 30 31 30 31 DegDayAi.1 421. 5.i [W/C] 112 224 335 Li [GJ] 13.2 [C-Day] 630.7 605. respectively.11 5.125 1.1 536.59 -200.1 485.1 [C-Day] 426 302 168 26 5 0 0 0 3 14 138 336 DegDayB i.i [C] 5.6 50.4 179. 13. 12.41 309.8 U Ah.33 109. and the loss per degree-day. and 89.4 484.4 99.04 12.2 165.2 DegDay 0 [i.67 87.4 484.3 [C-Day] 698.6 12.79 134.01 313.8 -14.7 16.5 125.89 40.386 2.2 375.9 -300.8 39.3 -231.709 9.6.8 39.951 26.5 DegDayAi.835 2.3 89.3 354.8 -169.67 87. It reduces the number of degree-days.9 DegDayB i.5 C and annual Energy requirements are.719 1.11 -52.6 419.6 -80.748 1.606 1.6.603 1.516 1. (19) end (20) For the three different values of (UA)h the base temperatures are.3 14.63 3.6 547.8 537. 2] [C-Day] 629.19 -134.1 536.1 605.574 1.041 50. respectively.299 Tbase.4 125.951 26. and 14.3. 624 0.4 hi.1 [C] 13.3 [-] 1.696 1.2 328.041 50.6 419.393 3.05073 0.872 0.526 -1.39 -9.1 536.8 76.3 3.5327 0.71 139 1337 3182 Li.902 26.8287 hi.46 516.52 19.605 1.4331 0.606 1.143 0.61 4.525 190 ∆T i.3273 -0.2 [C-day] 629.04 12.3 17.9555 0.528 1.1 [MJ] 4082 2874 1629 252.52 5.2 179.1 [C-day] 421.696 -2.1 [-] 1.607 10.7 1470 3889 10873 17517 DegDay i.2864 0.352 DegDay i.61 ∆T i.393 -5.5279 0.6 419.11 5.3 [C] 22.643 -0.6251 -0.696 -6.8 39.9999 2.3 26.4 375.8 Li.63 3.2 Li.393 -8.8783 -0.478 -5.3 11.3 [C-day] 698.6 70.522 12.05368 -0.3894 -0.75 4.478 -7.01 309.6 107.54 3.191 0.3448 1.4 261.9977 -0.5 29.51 -0.39 -14.63 3.951 26.3142 0.2 [C] 20.7628 -1.1 44.607 -3.76 14.2999 2.5 350.7 605.1276 0.1743 -0.2 [-] 1.4 484.3186 -0.52 19.709 9.041 50.36 138.Row 1 2 3 4 5 6 7 8 9 10 11 12 Row 1 2 3 4 5 6 7 8 9 10 11 12 DegDay 0 [i.522 0.4 125.3 17.04 12.79 134.304 10.696 -7.05 4.2 179.67 87.7 605.024 0.2 hi.942 -1.695 1.2 546.39 -16. 3] [C-Day] 698.149 -0.4 375.6 70.52 13.52 .5224 -4.478 -0.3 [MJ] 20209 15822 12133 5197 2027 365.304 -1.61 10.8976 1.8624 -1.2 [MJ] 12181 9377 6782 2435 757 108.422 0.8 297 168.886 -0.709 9.79 134.4776 3.091 1.2 546.393 -13.1 1684 5982 10381 DegDay i.1033 0.3 ∆T i.696 -9. Ls = 5. M ass = 400 [kg] .1 Energy balance on a system for 1 hour. in MJ: . U AT ank = 0. Ac = 5 m2 . F RU L = 4.P10 01 Equations 10. F RT A = 0.76.1.20. Ta = 12 [C] . Cp = 4190 [J/kg ·C] .0/5. for fully mixed Tank.81 [W/C] .3 in finite difference form.80 W/m2 ·C ∆T ime = 1 [hr] (2) T 0 a = 19 [C] (3) Equation 10. Knowns Ts = 43 [C] . IT = 18. . . J . . . M ass · Cp · (Tplus.s − Ts ) = (Qu − Ls − LossT ank ) · . 1000000 MJ . . . . MJ/hr . . . Qu = Ac · F RT A · IT − F RU L · (Ts − Ta ) · 1 [hr] · . 0036 J/s .0. LossT ank (4) (5) . . . MJ/hr . . . = U AT ank · (Ts − T a ) · ∆T ime · . 0.0036 J/s . 07 [MJ] Tplus.8 W/m2 -C Qu = 11 [MJ] U AT ank = 0. 0 (6) Solution 2 Ac = 5 m IT = 3.s = 46.42 [C] ∆T ime = 1 [hr] Ls = 5.6 MJ/m2 Ta = 12 [C] Cp = 4190 [J/kg-C] LossT ank = 0.81 [W/C] .2 [MJ] Ts = 43 [C] 191 (1) F RT A = 0.76 M ass = 400 [kg] T 0 a = 19 [C] F RU L = 4. 78.: . in MJ in the parametric table.1 [W/C] . T 0 a = 20 [C] . FRT auAlpha = 0. U AT ank = 11.5 Knowns Ac = 30 m2 .3.1.L = 7.P10 02 Equations 10.62 W/m2 ·C .5 1 1. ∆T ime = 1 [hr] .5 2 2. Cp = 4190 [J/kg ·c] (1) (2) Use finite difference form of Equation 10.2 Energy balance on fully mixed Tank over 24 hours $Tabstops 0. FRU. Note that the Solve Table must start at run 2. M ass = 1500 [kg] hr = 1 [hr] . . J . (Qu − Ls − LossT ank ) · . 1000000 MJ (Ts+ − Ts ) M ass · Cp · = (3) ∆T ime hr . . . MJ . . Ac · FRT auAlpha · IT − FRU. Qu = max 0.L · hr · (Ts − Ta ) · . . 0.0036 (4) W · hr . . . . MJ . . 0 . LossT ank = U AT ank · hr · (Ts − T a ) · . 0.0036 (5) J · h/s . ‘Tplus s’ ) (6) $DoLast 192 . Ts = Tablevalue(tablerun# − 1. ‘Q u’ . 25) (7) Qloss.day = Sumparametric(‘Table 1’ . 2.day = Sumparametric(‘Table 1’ . ‘Loss tank’ . ‘Tplus s’ ) (10) Tf inal = Tablevalue(25. 25) (9) Tinitial = Tablevalue(1. ‘Tplus s’ ) (11) . 2. 2.Qu. 25) (8) Loadday = Sumparametric(‘Table 1’ . ‘L s’ . . . J . . day − Loadday ) · . M ass · Cp · (Tf inal − Tinitial ) = (Qu.day − Qloss. . 1000000 + EBE MJ . 9 40.1 28.8 0.4 53.1 0.0 0.0 42.41 2.8 51.9 31.0 0.09 1.1 3.78 3.1 67.77 1.5 0.3 0.53 0 0 0 0 0 0 0 -4 -4 -5 -3 -1 0 0 1 0 2 4 10 10 8 8 6 4 4 3 5 6 6 7 6 45.8 48.2 0.3 0.5 40. (12) $EndDoLast Table 1 Run 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 T IM E [HR] Ls [MJ] IT 2 MJ/m Ta [C] Ts [C] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 12 12 11 11 13 14 18 21 20 20 18 16 14 14 13 18 22 24 18 20 15 11 10 9 0 0 0 0 0 0 0 0 1.9 32.5 43.0 0.9 0.4 0.2 34.0 0.1 0.0 0.7 26.9 39.0 0.2 47.7 46.0 1.3 1.1 34.0 3.4 73.6 37.7 46.1 34.1 46.0 37.0 1.1 28.7 0.4 29.6 29.9 40.9 0.5 37.69 3.4 53.8 0.1 51.7 32.3 1.7 24.0 0.7 1.6 0.9 31.7 26.2 47.0 0.5 0.5 37.6 37.5 40.9 23.5 43.0 0.7 193 Qu [MJ] LossT ank [MJ] 0.2 29.8 51.87 3.3 0.6 0.4 .75 2.0 0.1 51.7 24.7 32.9 32.6 29.2 1.0 0.8 0.3 23.3 23.7 0.2 34.9 39.0 0.82 1.0 0.0 37.2 Tsplus [C] 45.0 42.0 0.0 0. P10 03 194 . 4 Calculate collector heat exchanger factor.coll = 3350 [J/kg-C] F 0 R/F R = 0.9494 C˙ coll = 2345 [J/s-C] Cp.tank = 4190 [J/kg-C] m ˙ coll = 0. C˙ coll = m ˙ coll · Cp.75 W/m2 -C .coll (3) −1 ˙ Ccoll − 1 · · C˙ min F 0 R/F R = 1 + Ac · F RU L/C˙ coll Eqn 10.7 [kg/s] .6 m ˙ tank = 0.7 [kg/s] 195 C˙ tank = 2933 [J/s-C] F RU L = 3. m ˙ coll = 0. m ˙ tank = 0.3: C˙ min = min C˙ tank . F RU L = 3. Cp. Knowns = 0. C˙ coll (5) Solution Ac = 50 m2 Cp.7 [kg/s] (1) Cp.2.tank = 4190 [J/kg ·C] (2) C˙ tank = m ˙ tank · Cp.7 [kg/s] (4) C˙ min = 2345 [J/s-C] = 0.P10 04 Equations 10. Ac = 50 m2 .75 W/m2 ·C .60.tank .coll = 3350 [J/kg ·C] . 65 W/m2 -C L = 20 [m] .018 [m] U L0 /U L = 1. insul = 0. F Rta = 0.6 (or 10. F RU L = 4.05 [W/m ·C] (3) Ao = π · Do · L.65 W/m2 ·C . (4) ta0 /ta = 1 1 + Upipe · Ao /C˙ coll U L0 /U L = Upipe = Ai = π · Di · L Eqn 10.1) is then modified as follows: Qu = Ac *(FRta*ta’ \ta *GT .P10 05 Equations 10.5 Pipe loss factors modifying Qu equation.277 Ao = 3. Knowns C˙ coll = 295 [W/C] .018 [m] Ac = 5.2 m2 Di = 0.019 [m] hconv = 25 W/m2 -C ta0 /ta = 0. L = 20 [m] .Ta )) Solution Ac = 5.194 m2 Do = 0.FRUL*UL’ \UL*(Ti .72.456 m2 F Rta = 0.019 [m] .055 [m] Insul = 0.9816 Ai = 1.72 kinsul = 0.2 m2 (1) (2) kinsul = 0.9 (5) i +Ao 1 − Upipe · Ai /C˙ coll + Upipe · AA c ·F RU L 1 + Upipe · Ao /C˙ coll Eqn 10.3.7.601 W/m2 -C 196 C˙ coll = 295 [W/C] F RU L = 4.3. Do = Di + 2 · Insul. Di = 0.10 (6) 1 Do · ln (Do /Di ) (2·kinsul ) (7) + 1/hconv Equation 6.05 [W/m-C] Upipe = 1. hconv = 25 W/m2 ·C .1. m ˙ air = 0. F 0 RU L = 4.78. F 0 Rta = 0. C˙ c (5) Qmax (6) . Ac = 80 m2 . M ass = 5000 [kg] . ∆T ime = 1 [hr] . Tinitial = 47 [C] U AT ank = 140 [kJ/hr ·C] .45 W/m2 ·C .air C˙ min = min C˙ h .water = 4190 [J/kg ·C] (3) For a perfect heat exchanger the fluid with the minimum capacity will go through the maximum temperature rise.water (4) C˙ c = m ˙ air · Cp. Cp.P10 06 Equations 10.6 Air heater simulation with fully mixed liquid storage tank Knowns: TsupplyAir = 35 [C] . Tdelivery = 55 [C] . C˙ h = m ˙ water · Cp.air = 1013 [J/kg ·C] .056 [kg/s] .056 [kg/s] . (1) hr = 1 [hr] (2) Cp. m ˙ water = 0. . . MJ/hr . . ˙ . = Cmin · (Ts − TsupplyAir ) · hr · . 0036 W .0. 197 (7) . The required energy to heat the air from 35 to 55 C is: . . . MJ/hr . . Qreqd = C˙ min · (Tdelivery − TsupplyAir ) · hr · . . 0.0036 J/s . Qreqd − Qmax ) (10) Note that this control would be implemented by reducing the water mass flow rate. Qmax ) (9) The auxiliary energy is then: Aux = max (0. (8) The energy removed from storage is then: Ls = min (Qreqd .2) is: .9. The tank energy balance (Equation 10. . . s − Ts ) MJ . (Tplus. . M ass · Cp.water · = (Qu − Ls − LossTank ) · hr · . . 1 × 10·6 ∆T ime J . ‘T plus s’ ) . Ts = Tablevalue(‘Table 1’ . tablerun# − 1. . . MJ/hr . . Qu = max 0. Ac · F 0 Rta · IT − F 0 RU L · hr · . . 0036 · (T − T ) s a W .0. . . . MJ . . LossT ank = U AT ank · hr · . . 001 · (Ts − Ta ) kJ .0. ‘Aux’ . ‘Loss tank’ . ‘L s’ . ‘Q u’ . 2. T ableRun#) (15) AuxiliaryEnergy required to meet the load EAux = Sumparametric(‘Table 1’ . (11) (12) (13) (14) Energy removed from store to supply load Estore = Sumparametric(‘Table 1’ . T ableRun#) (17) Energy lost from store to surroundings Eloss = Sumparametric(‘Table 1’ . T ableRun#) (16) Energy supplied to store from solar Esolar = Sumparametric(‘Table 1’ . 2. 2. 2. . T ableRun#) (18) Energy gain (or loss) of tank over 12 hours. . . MJ . . water · (Tplus. Estored = M ass · Cp.s − Tinitial ) · . . 1 × 10·6 J . Check = Qreqd · 12 − Estore − Eaux (21) Solution 198 . (19) Store energy balance to determine Energy Balance Error: EBE = Esolar − Eloss − Estored − Estore (20) Energy balance on air stream. 08 4.65 5.9 80.08 4.776 × 10−15 [MJ] Cp.95 6.71 47 46.68 7.77 85.85 49.7 70.55 81.60 5.5 0 0 20.2 46.08 [MJ] Qmax = 9.71 1.38 54.99 4.45 W/m2 -C Ls = 4.air = 1013 [J/kg-C] EAux = 8.00 0.68 9.41 5.37 2.8 76.1 80.00 0.77 [MJ] m ˙ water = 0.28 60.00 0.056 [kg/s] Qu = 0 [MJ] Tplus.4 180.6 46.s [C] 47.97 [MJ] hr = 1 [hr] M ass = 5000 [kg] Qreqd = 4.879 0 821.51 62.8 80.1 823.95 9.24 46.044 [MJ] Estored = 694.6 C˙ c = 56.45 2.08 4.09 9.97 1.8 743.66 1.6 81.62 46.69 3.78 LossT ank = 9.00 0.00 MJ/m2 m ˙ air = 0.9 49.056 [kg/s] Ta = 11 [C] Ts = 80.15 0.6 [J/s-C] ∆T ime = 1 [hr] Esolar = 821.9 80.87 3.084 [MJ] Tinitial = 47 [C] U AT ank = 140 [kJ/hr-C] Check = 1.79 1.53 0.73 [J/s-C] Cp.79 76.69 116.89 [MJ] F 0 Rta = 0.7 70.30 5.41 2.63 1.00 1.82 1.94 3.89 199 Tplus.1 [MJ] F 0 RU L = 4.71 6.75 2.08 40.30 2.60 9.Aux = 0.73 [J/s-C] EBE = 7.8 [C] .2 [MJ] IT = 0.4 54.353 [MJ] Tdelivery = 55 [C] TsupplyAir = 35 [C] Ac = 80 m2 C˙ min = 56.00 8.13 80.4 140 89.42 2.00 0.326 × 10−7 [MJ] Estore = 40.08 4.04 5.00 0.00 22.08 4.water = 4190 [J/kg-C] Eloss = 85.21 5.10 0.09 1.00 [MJ] C˙ h = 234.67 25.s = 80.77 1.0 46.1 [C] Table 1 Run 1 2 3 4 5 6 7 8 9 10 11 12 13 Sum Ta [C] 7 8 9 9 9 12 15 14 12 12 12 11 130 IT MJ/m2 Ts [C] Qu [MJ] Ls [MJ] Aux [MJ] LossT ank [MJ] 0.78 3.1 181.5 62.1 2. C˙ c (6) (7) Calculating the new maximum heat transfer in the HX is the only change from Problem 10.056 [kg/s] .6 with mixed liquid tank and a real load HX. F 0 Rta = 0. m ˙ air = 0. C˙ h = m ˙ water · Cp.air = 1013 [J/kg ·C] . Tdelivery = 55 [C] . Tinitial = 47 [C] U AT ank = 140 [kJ/hr ·C] .7 Air heating simulation of Problem 10. Knowns: TsupplyAir = 35 [C] .58 (3) (4) For a perfect heat exchanger the fluid with the minimum capacity will go through the maximum temperature rise.056 [kg/s] . F 0 RU L = 4.6 . (1) hr = 1 [hr] (2) Cp.45 W/m2 ·C .air C˙ min = min C˙ h .P10 07 Equations 10. Ac = 80 m2 .water (5) C˙ c = m ˙ air · Cp. Cp.water = 4190 [J/kg ·C] = 0. m ˙ water = 0.78. M ass = 5000 [kg] . ∆T ime = 1 [hr] . . . MJ/hr . . ˙ . Qmax = · Cmin · (Ts − TsupplyAir ) · hr · . 0036 W .0. The required energy to heat the air from 35 to 55 C is: . . . MJ/hr . . ˙ . Qreqd = Cmin · (Tdelivery − TsupplyAir ) · hr · . 0.0036 J/s . 9. The tank energy balance (Equation 10.2) is: . Qmax ) (10) The auxiliary energy is then: Aux = max (0. Qreqd − Qmax ) (11) Note that this control would be implemented by reducing the water mass flow rate. (8) (9) The energy removed from storage is then: Ls = min (Qreqd . . . . s − Ts ) ·6 MJ . (Tplus. . water · · hr · . M ass · Cp. 1 × 10 = (Qu − Ls − LossTank ) ∆T ime J . 200 (12) . Ts = Tablevalue(‘Table 1’ . tablerun# − 1. ‘T plus s’ ) . . . MJ/hr . . · (T − T ) Qu = max 0. Ac · F 0 Rta · IT − F 0 RU L · hr · . . 0.0036 s a W . . . . MJ . . . LossT ank = U AT ank · hr · . 0.001 · (Ts − Ta ) kJ . 2. T ableRun#) (19) Energy gain (or loss) of tank over 12 hours. (13) (14) (15) Energy removed from store to supply load Estore = Sumparametric(‘Table 1’ . T ableRun#) (16) AuxiliaryEnergy required to meet the load EAux = Sumparametric(‘Table 1’ . 2. 2. ‘Q u’ . Estored . T ableRun#) (18) Energy lost from store to surroundings Eloss = Sumparametric(‘Table 1’ . ‘Aux’ . 2. ‘L s’ . ‘Loss tank’ . T ableRun#) (17) Energy supplied to store from solar Esolar = Sumparametric(‘Table 1’ . . . . ·6 MJ . . = M ass · Cp.s − Tinitial ) · .water · (Tplus. 1 × 10 J . 58 increased the Aux from 8.03 [C] Ac = 80 m2 C˙ min = 56. (20) Store energy balance to determine Energy Balance Error: EBE = Esolar − Eloss − Estored − Estore (21) Energy balance on air stream.73 [W/C] Cp.6 [W/C] ∆T ime = 1 [hr] Eloss = 86.45 W/m2 -C Ls = 4.21 [MJ] F 0 Rta = 0.084 [MJ] Tinitial = 47 [C] U AT ank = 140 [kJ/hr-C] 201 C˙ c = 56.97 to 33.9 [MJ] IT = 0 MJ/m2 m ˙ air = 0. These are significant chages but general conclusions about the impact over a full year should not be drawn from this one-day simulation.334 × 10−7 [MJ] Esolar = 818.00 [MJ] Tplus.80 [MJ] m ˙ water = 0.08 [MJ] Qmax = 5.65 [MJ] Estored = 698.65 MJ and reduced the energy supplied from the store from 40.452 [MJ] Tdelivery = 55 [C] TsupplyAir = 35 [C] Check = −3.water = 4190 [J/kg-C] EAux = 15.air = 1013 [J/kg-C] = 0.36 [C] .056 [kg/s] Ta = 11 [C] Ts = 81. Check = Qreqd · 12 − Estore − Eaux (22) Reducing the heat exchanger effectiveness from 1 to 0.36 MJ.553 × 10−15 [MJ] Cp. In an annual simulation the expected percentage increase in auxiliary energy is expected to be much smaller.73 [W/C] EBE = 7. Solution Aux = 0.36 [MJ] hr = 1 [hr] M ass = 5000 [kg] Qreqd = 4.78 LossT ank = 9.00 [MJ] C˙ h = 234.056 [kg/s] Qu = 0.5 [MJ] F 0 RU L = 4.58 Estore = 33.s = 80.04 to 15. 66 2.98 71.00 139.99 49.41 5.34 3.36 2.09 1.00 0.03 80.66 46.70 2.08 4.18 80.73 62.16 60.71 47.71 9.11 .82 1.36 1.57 0.23 5.37 81.42 1.66 2.21 202 Tplus.56 54.87 3.03 745.00 0.34 1.18 80.08 33.34 46.08 4.51 115.98 71.61 89.38 1.10 180.00 20.00 15.08 4.66 46.52 1.74 2.60 5.37 81.88 181.00 0.81 81.00 0.00 46.65 5.41 2.42 1.00 818.75 0.80 86.75 2.13 9.Table 1 Run 1 2 3 4 5 6 7 8 9 10 11 12 13 Sum Ta [C] 7 8 9 9 9 12 15 14 12 12 12 11 130 IT MJ/m2 Ts [C] Qu [MJ] Ls [MJ] Aux [MJ] LossT ank [MJ] 0 0 1.72 2.75 0.53 0 22.09 77.32 5.99 49.36 826.09 77.81 81.08 4.78 3.34 46.31 25.00 0.63 9.99 9.31 4.69 3.00 46.38 6.98 6.68 5.77 1.73 62.56 54.72 7.s [C] 47.77 0. M ass = 1000 [kg] . Ac = 20 m2 .P10 08 Equations 10. F RU L = 5 W/m2 ·C . C˙ coll 203 = 0. Tinitial = 35 [C] . Cp = 4190 [J/kg ·C] .3. Equation 10. Tsupply = 11 [C] .8 Simulation of water heater with variable load. F Rta = 0. C˙ coll = 1020 [W/C] Troom = 20 [C] .2. Knowns ∆T ime = 1 [hr] . Use F’R \FR to account for the collector-store heat exchanger.73. U AT ank = 6. C˙ min = min C˙ T ank .0 [W/C] . C˙ tank = 1120 [W/C] .6 hr = 1 [hr] (1) (2) (3) (4) . 2. −1 ˙ Ccoll − 1 · · C˙ min F 0 R/F R = 1 + Ac · F RU L/C˙ coll (5) F 0 Rta = F Rta · F 0 R/F R (6) F 0 RU L = F RU L · F 0 R/F R (7) Use a form of Equation 10. in MJ: .9. . + . Ts − Ts MJ . . = (Qu − Ls − LossTank ) M ass · Cp · · hr · . . 1 × 10·6 ∆T ime J . . . . MJ/hr . . 0 0 . Qu = max 0 [MJ] . Ac · F Rta · IT − F RU L · . 0.0036 · (T − T ) · hr s a W . . . . MJ . . · (Ts − Tsupply ) · hr Ls = m ˙ load · Cp · . . 1 × 10·6 J . . . . MJ/hr . . . LossT ank = U AT ank · . 0036 · (Ts − Troom ) · hr W .0. tableRun#) (14) Eload = Sumparametric(‘Table 1’ . Eu = Sumparametric(‘Table 1’ . ‘Loss tank’ . tableRun#) (13) Eloss = Sumparametric(‘Table 1’ . 2. tableRun#) (15) Tf inal = Ts+ (16) . 2. ‘Tplus s’ ) (12) Find total quantities. ‘Q u’ . 2. (8) (9) (10) (11) Ts = Tablevalue(tablerun# − 1. ‘L s’ . . . . ·6 MJ . . ∆U = M ass · Cp · (Tf inal − Tinitial ) · . 1 × 10 J . 73 F 0 R/F R = 0.9387 [-] Ls = 19.41 [MJ] F 0 RU L = 4.08 [C] Tsupply = 11 [C] C˙ min = 1020 [W/C] ∆U = 42.6852 IT = 2.85 [MJ] F RU L = 5 W/m2 -C hr = 1 [hr] M ass = 1000 [kg] Tf inal = 45.75 MJ/m2 m ˙ load = 150 [kg/hr] Tinitial = 35 [C] Ts+ = 45.99 [MJ] Ta = 20 [C] Ts = 42.08 [C] Table 1 204 C˙ tank = 1120 [W/C] EBE = 8.98 [MJ] Troom = 20 [C] U AT ank = 6 [W/C] .22 [MJ] Eloss = 1.81 [C] C˙ coll = 1020 [W/C] ∆T ime = 1 [hr] Eload = 47.693 W/m2 -C LossT ank = 0.6 F Rta = 0.337 [MJ] F 0 Rta = 0. (17) Tank Energy Balance Error EBE = Eu − Eloss − Eload − ∆U (18) Solution Ac = 20 m2 Cp = 4190 [J/kg-C] = 0.462 × 10−8 [MJ] Eu = 91.49 [MJ] Qu = 29. 28 0.32 32.78 42.75 3.24 0.08 12.00 31.28 29.81 205 Tsplus [C] 35.81 45.32 0.08 .32 32.14 42.49 15.77 0.00 31.00 19.75 150 150 0 150 0.98 0.78 42.45 2.09 1.Run 1 2 3 4 5 Ta [C] IT 2 MJ/m m ˙ load [kg/hr] Qu [MJ] LossT ank [MJ] Ls [MJ] Ts [C] 14 17 18 20 0.00 19.99 35. 8 but here we wanted answers based on knowledge of solar system performance. c) Tplus s will be lower. d) Tplus s will be higher. b) Tplus s will be lower. as FR0 will be higher and Qu will be higher. since there would be increased thermal losses from the collector and tank.9 Effects of changes in the conditions of Problem 10. a) Tplus s will be higher.8 Each change can be tested using the program of Problem 10. resulting in lower Qu.P10 09 Equations 10. 206 . as the temperature of operation of the whole system would be higher. The increase would be less than 3 C. as UL will be higher resulting in higher collector losses and lower Qu. as FR and FR0 will be lower. Ttest = 30 [C] . $UnitSystem C J kPa Knowns F avtan = 0. it is left to the end.EG . Work from the collector back to the tank. v˙ test = 0. Conc = 50.0246 [l/s] . call BRIN EP ROP 2 F I$.050 [l/s] . making a series of corrections on FRta and FRUL. Dpipe = 0. thick insulation = 0.64 m2 . Nparallel = 3.040 Ac = 1. Ncoll = 6 (1) F I$ = ‘EG’ (2) assumed (3) so the flow through each collector is: (4) kinsulation = 0. = 0. v˙ coll = 0.75. Conc.watch units in BRINEPROP.10 Develop Qu equation for collector subsystem.165.P10 10 Equations 10.75/0.043 [W/m ·C] (5) Fluid properties . Pf luid = 200 [kPa] . [l/s] Nparallel Ta = 20 [C] . Ttest Cp.45 Lpipe = 12 [m] . v˙ tank = 0.009 [m] . Nseries = 2. F avU L = 0.012 [m] . The bo correction can be made any time. k . . P rEG : F reezingP tEG . µ . . ρEG . . 1000 J . P = pf luid ) ! (6) (7) 207 . T = Ttest . EG EG kJ ρ = ρ (water. P = Pf luid ) (9) 1.µ = µ (water. P = Pf luid ) (8) Cp = cp (water. T = Ttest . . Section 6. T = Ttest . Convert from basis of Fav to FR .19. . . m3 . . m ˙ test = v˙ test · ρ · . . 0.001 l . Equation 6. Apply flow rate correction. (10) −1 Ac · F avU L F Rtan.1: .1 = F avtan · 1 + 2·m ˙ test · Cp Ac · F avU L F RU L1 = F avU L · 1 + 2·m ˙ test · Cp (11) −1 (12) 2.20. . . m3 . . . m ˙ use = v˙ coll · ρEG · . 0.001 l . 1 · r (13) (14) (15) (16) 3.20.8 (18) F RU L3 = F RU L2 · (1 − K/2) Eqn 10.20. 0 m ˙ use ·Cp.5.5.9 m ˙ use · Nparallel · Cp.EG F RU L4 = F RU L3 · 1− U Ai U Ai +U Ao (m ˙ use ·Nparallel ·Cp.4 Ac m ˙ test · Cp F RU L2 = F RU L1 · r. F Rtan.EG Eqn 10. Two arrays in series: K= Ac · F RU L2 m ˙ use · Cp. Pipe heat loss .EG ) + Ncoll ·Ac ·F RU L3 U Ai 1 + (m˙ use ·Nparallel ·Cp.2 · (1 − K/2) Eqn 10.3.EG Ac ·F U L · 1 − exp − Ac m ˙ use ·Cp.9 (19) 4. −1 U Ai F Rtan.Section 10.EG r= Eqn 6.3 = F Rtan. So the total flow rate and total collector area are used here.2 F RU L1 m ˙ test · Cp Ac 0 F UL = − · ln 1 − F RU L1 · Equation 6.3 · 1 + Eqn 10.5.3.4 = F Rtan.4 (17) F Rtan.3 Dinsul = Dpipe + 2 · thick insulation U Ao = 2 · π · kinsulation · (20) Lpipe ln (Dinsul /Dpipe ) (21) U Ao = U Ai (22) The pipe heat loss is for all 6 collectors.2 = F Rtan.10 (23) (24) .EG ) 208 Eqn 10. EG . Collector-heat exchanger correction C˙ coll = Nparallel · m ˙ use · Cp.5. . . m3 . . C˙ tank = v˙ tank · ρ · Cp · . . 0.001 l . 15*(1/cos(θ)-1) Eqn 6.2. Incidence angle modifier and Qu equation Kta = 1 .5 = F Rtan.5 (31) . C˙ min = min C˙ coll .10 a = Ncoll · Ac · F Rtan.17.0.3 (28) F Rtan. C˙ tank (25) (26) (27) −1 ˙ Ncoll · Ac · F RU L4 Ccoll − 1 F R0 /F R = 1 + · C˙ coll · C˙ min Eqn 10.4 · F R0 /F R (29) F RU L5 = F RU L4 · F R0 /F R (30) 6. . . MJ/hr . . . b = Ncoll · Ac · F RU L5 · . 0.0036 W . 4 = 0.15(1/cos(θ)-1)) .043 [W/m-C] m ˙ test = 0.03 [C] F Rtan.01414 [kg/s] P rEG = 23.2 [J/s-C] Dpipe = 0.002831 [Pa-s] Nparallel = 3 [-] r = 0.728 kinsulation = 0.EG = 3349 [J/kg-K] F avtan = 0.1138*(Ti .498 m2 C˙ coll = 142 [J/s-C] Cp.98 ρEG = 1060 kg/m3 U Ai = 3.64 m2 C˙ min = 142 [J/s-C] Dinsul = 0.0.7237 [-] F Rtan.1458 µ = 0.009 [m] U Ao = 3.012 [m] F I$ = ‘EG’ F Rtan.7 kg/m3 Ttest = 30 [C] v˙ tank = 0.4571 F RU L4 = 4.545 W/m2 -C F Rtan.45 F reezingP tEG = −36.0.75 [-] F Rtan.6436 [-] F RU L2 = 4.03 [m] F avU L = 4.9593 Ta = 20 [C] v˙ coll = 0.3 = 0.01333 [l/s] 209 b = 0.2 = 0.6942 F RU L1 = 4.628 [-] F RU L3 = 3.1138 [MJ/C-hr] C˙ tank = 208.5 = 0.547 W/m2 -C Lpipe = 12 [m] m ˙ use = 0.Ta ) Solution a = 4.207 W/m2 -C F R0 /F R = 0.212 W/m2 -C kEG = 0.386 W/m2 -C F RU L5 = 3.413 W/m2 -C K = 0.901 W/m2 -C F 0 U L = 4.02449 [kg/s] Nseries = 2 [-] ρ = 995.498*IT *(1 .538 [W/C] Ac = 1.1 = 0.0246 [l/s] .538 [W/C] v˙ test = 0. (32) The final equation for Qu in MJ/hr (assuming IT is in MJ/m2 -hr) for the complete assembly is: Qu = 4.3953 [W/m-K] µEG = 0.0007977 [kg/m-s] Ncoll = 6 [-] Pf luid = 200 [kPa] thick insulation = 0.05 [l/s] Conc = 50 Cp = 4183 [J/kg-K] = 0. Ac2 = 1.011 kg/s ·m2 · (Ac1 + Ac2 ) .83. then the angle of incidence of beam radiation is 5 deg on the unshaded collector.88 + 2.6.e. Use as a basis of calculation a unit width of row.P10 11 Equations 10. F RU L = 4. IT 2 = 0. Cp = 4190 [J/kg ·C] . Ti = 40 [C] . bo = −0. Knowns Ac1 = 0.88 MJ/m2 ·hr .4 . 2. the mean angle of incidence is then about 59 deg. that the sun is directly south of the arrays. Refer to the diagram on p. m ˙ = 0.95.10 m2 area. Assume that the combination of diffuse and reflected radiation is isotropic and at the same level as the diffuse on the unshaded collector. 39 for dimensions. i. i.88 m2 ..11 Performance of partially shaded collectors Assume that the solar azimuth angle is 0. IT 1 = 0.14 θ2 = 5 [deg] Ta = 14 [C] (1) (2) (3) Eqn 10.e. θ1 = 59 [deg] .07 W/m2 ·C . F Rtan = 0.22 m2 .. . . MJ/hr . . . Qu = Ac1 · F Rta1 · (1 − K) · IT 1 + Ac2 · F Rta2 · IT 2 − F RU L · (Ac1 · (1 − K) + Ac2 ) · (Ti − Ta ) · . 0.0036 (4) W . 7206 IT 1 = 0.8296 IT 2 = 3.88 m2 F Rta1 = 0.623 [MJ/hr] Ti = 40 [C] Ac2 = 1.0231 [kg/s] Ta = 14 [C] .5 (5) F Rta1 = F Rtan · 1 + bo · 1 −1 Cos(θ1 ) 1 F Rta2 = F Rtan · 1 + bo · −1 Cos(θ2 ) (6) (7) Solution Ac1 = 0.22 m2 F Rta2 = 0.83 MJ/m2 -hr θ1 = 59 [deg] bo = −0.14 F Rtan = 0.07 W/m2 -C m ˙ = 0.0513 θ2 = 5 [deg] 210 Cp = 4190 [J/kg-C] F RU L = 4.88 MJ/m2 -hr Qu = 3.6. K = Ac2 · F RU L m ˙ · Cp Eqn 10.83 K = 0. Rb2 = CosT heta2/CosZen.10 [kg/s] .5 [deg] . (2) bo = −0. Slope2 = 70 [deg] . Cp = 4190 [J/kg ·C] .2 kW. Slope1 = 20 [deg] .it is a very sunny day with kT =0. Ti = 40 [C] .2 (8) CosT heta2 = sin (Dec) · sin (Lat) · Cos(Slope2) − sin (Dec) · Cos(Lat) · sin (Slope2) · Cos(γ) + Cos(Dec) · Cos(Lat) · Cos(Slope2) · Cos(ω) + Cos(Dec) · sin (Lat) · sin (Slope2) · Cos(γ) · Cos(ω) + Cos(Dec) · sin (Slope2) · sin (γ) · sin (ω) Rb1 = Costheta1/CosZen.6.14 G = 580 W/m2 (3) (4) N = nDay (month. Using the isotropic sky model to calculate radiation on the collectors: Go = Gsc# · (1 + .83.1. Ac2 = 50 m2 . This leads to Qu = 42. month = 3. F Rtan = 0.2.12 Output of collectors with two parts at different orientations Knows Lat = 40 [deg] . Ac1 = 50 m2 .033 · Cos(360 [deg] · n/365)) · CosZen 211 (12) .5 (7) CosT heta1 = sin (Dec) · sin (Lat) · Cos(Slope1) − sin (Dec) · Cos(Lat) · sin (Slope1) · Cos(γ) + Cos(Dec) · Cos(Lat) · Cos(Slope1) · Cos(ω) + Cos(Dec) · sin (Lat) · sin (Slope1) · Cos(γ) · Cos(ω) + Cos(Dec) · sin (Slope1) · sin (γ) · sin (ω) Eqn 1.958 and Go = 604 W/m2.2 T heta1 = arccos (CosT heta1) T heta2 = arccos (CosT heta2) (9) (10) (11) It could be assumed that most radiation is beam . so the diffuse fraction will be 0.P10 12 Equations 10. m ˙ = 1.07 W/m2 ·C .10. GrRef = 0. day = 16.6. day) (5) dec = Dec (N ) (6) CosZen = Cos(Lat) · Cos(Dec) · Cos(ω) + sin (Lat) · sin (Dec) Eqn 1. Eqn 1. ω = 52.165 by Equation 2. γ = 0 [deg] (1) Ta = 14 [C] F RU L = 4. Then GT 1 = G*Rb1 and GT 2 = G*Rb2.6. 6.78 [deg] 212 CosT heta1 = 0.04415 m ˙ = 1.5478 dec = −2.3 W/m2 GT 2 = 707 W/m2 month = 3 Qu = 40673 [W] Slope2 = 70 [deg] Ti = 40 [C] Ac2 = 50 m2 CosZen = 0.7 W/m2 K = 0.4388 F Rta1 =0.83 GrRef = 0.07 -C W/m Gb = 484.27 T heta1 = 56.5571 day = 16 F Rtan = 0.4 (20) .1 [kg/s] Rb1 = 1.9582 [-] N = 75 Rb2 = 1.835 (14) GT 1 = Gb · Rb1 + Gd · 1 + Cos(Slope1) 1 − Cos(Slope1) + G · GrRef · 2 2 (15) GT 2 = Gb · Rb2 + Gd · 1 + Cos(Slope2) 1 − Cos(Slope2) + G · GrRef · 2 2 (16) K = Ac2 · F RU L m ˙ · Cp Eqn 10.6. Gb = 580 · 0.Apply the concept of kT to irradiance rather than irradiation: kT = G/Go (13) Gd = 580 · 0.2 W/m2 Lat = 40 [deg] ω = 52.418 [deg] 2 F RU L = 4.7376 2 G = 580 W/m Gd = 95.14 [deg] bo = −0.248 T heta2 = 56.7341 γ = 0 [deg] Go = 605.5 [deg] Slope1 = 20 [deg] Ta = 14 [C] Eqn 10.3 W/m2 kT = 0.165.2 GT 1 = 711.14 Cp = 4190 [J/kg-C] F Rta2 = 0.5 F Rta1 = F Rtan · 1 + bo · F Rta2 = F Rtan · 1 + bo · (17) 1 −1 Cos(T heta1) 1 −1 Cos(T heta2) (18) (19) Qu = Ac1 · F Rta1 · (1 − K) · GT 1 + Ac2 · F Rta2 · GT 2 − F RU L · (Ac1 · (1 − K) + Ac2 ) · (Ti − Ta ) Solution Ac1 = 50 m2 CosT heta2 = 0. N = 10.4.12 [-] N = 10 [-] 213 .1 PW1 = PW2 = A N (2) N (3) (1 + d1) A (1 + d2) Solution A = 5700 [$] P W 1 = 2640 [$] d1 = 0.08. Knowns A = 5700 [$] .P11 01 Equations 11. d1 = .1 Calculation of present worth.08 [-] P W 2 = 1835 [$] d2 = 0.12 (1) From Equation 11. d2 = . 085 (1) NL ! Eqn 11.P11 02 Equations 11.5. 1 PWF = · d−i NL = 15. 1+i 1+d P eriodP ay = M/P W F d = 0.2 (3) Solution d = 0.41 [$] i=0 P W F = 8. 1− i = 0.304 M = 20000 [$] 214 NL = 15 .2 Calculation of uniform loan payment Knowns M = 20000 [$] .1 (2) Eqn 11.5.085 P eriodP ay = 2408. 095 [-] i = 0. PWF = i = .3 Calculation of Present Worth Factor PWF Knowns d = .078 [-] N = 18 [-] P W F = 14.5. 1 · d−i N = 18 1− 1+i 1+d (1) N ! Eqn 11.095.44 [-] 215 .P11 03 Equations 11.1 (2) Solution d = 0.078. 5.4 Calculation of present worth of a series of costs. Knowns A = 1000.03 [-] P W 1 = 10603 [$] d2 = 0.05. d1 = .598 [-] d1 = 0.6 [-] .05 [-] P W F 1 = 10.1 with each discount rate: N ! 1 1+i PWF1 = · 1− d1 − i 1 + d1 (2) PW1 = A · PWF1 1 PWF2 = · d2 − i (3) 1− 1+i 1 + d2 N ! (4) PW2 = A · PWF2 (5) Solution A = 1000 [$] N = 10 [-] P W F 2 = 8.07.07 [-] P W 2 = 8598 [$] 216 i = 0.03. d2 = .P11 04 Equations 11. N = 10 (1) Use Equation 11. i = . We have chosen to use arrays. We can now use the PWF. 6 (9) Y rj = 3 · j (10) Costj = costj−1 · (1 + i3) (11) PWj = Costj (12) j (1 + d3) end (13) P W 2 = Sum(P W k .g. i = 0. Note that the rates are not 3 times the yearly rates. i3. We need to determine the 3-year interest rate from yearly interest (and discount) rates. Cost0 = Glazing cost · Ac (8) duplicate j = 1. Ac = 75 m2 .5 Present worth of a series of triennial payments. Knowns: Glazing cost = 15 $/m2 . The PWF factors of Chapter 11 can be used for any time period. 6) (14) Solution 217 . Nyr = 20 d = 0. month.P11 05 Equations 11.07. e.. k = 1. year or multiple years. The 3-year interest and discount rates are: 3 (3) (1 + d) = (1 + d3) 3 (4) N3 = 6 (5) (1 + i) = (1 + i3) there are 6 three-year periods in the 20 years.10 (1) (2) Two solutions are shown. d3) (7) The second method could use either the parametric table or arrays. 3 Cost3 = Glazing cost · Ac · (1 + i) (6) P W 1 = Cost3 · P W F (N 3. 42 Y ri PWi [$] 3 6 9 12 15 18 1035.44 953.00 1378.87 [$] d = 0.05 683. Ac = 75 m2 Glazing cost = 15 $/m2 Nyr = 20 Cost3 = 1378.1 i3 = 0.07 P W 1 = 5099.91 3802.90 218 d3 = 0.225 P W 2 = 5099.15 807.17 1688.72 3103.01 877.32 2068.32 743.331 N3 = 6 .27 2533.17 [$] i = 0.87 [$] Arrays Row 0 1 2 3 4 5 6 Costi [$] 1125. 544 M onthP ay = 41. d = 0. d) (6) P W F 3 = P W F (N. Knowns: N = 8. 0.5.69 [$] P W int = 1051.3.2 (4) c) The present worth of interest payments of part a.1 (3) b) The monthly payment calculation is the same except that i = 0.105/12 and N = 8*12 = 96. 0. m) AnnualP ay = Loan/P W F1 (2) Eqn 11.239 Loan = 2700 [$] P W F 2 = 5.35 [$] N =8 d = 0. P W F 1 = P W F (N.5.P11 06 Equations 11. is given by Equation 11.095. m/12) Eqn 11.10 [$] .433 219 m = 0. This equation contains 3 PWFs. M onthP ay = Loan P W F (12 · N.105 P W F 3 = 7.2.6 Loan payments and present worth of payments.105. Note that Nmin =N P W int = Loan · (P W F 2 /P W F 1 + P W F 3 · (m − 1/P W F 1 )) (5) P W F 2 = P W F (N.5. m = 0. 0. The PWF is calculated with i = 0 and d = mortgage interest rate. m. Loan = 2700 [$] (1) a) Annual payment to lender is given by Equation 11. d) (7) Solution AnnualP ay = 515.095 P W F 1 = 5.5. 05 [-] Nc = 7 [-] Cb = 984. dd = 0.7 Present worths of single and series payments.08 [-] ie = 0.05. de = 0.1 [$] id = 0. ic = 0. da = 0.05 [-] P W F e = 1432 [$] Ad = 175 [$] dd = 0. PWa = Ca Eqn 11. Na = 7.05 [-] Ne = 10 Ac = 700 [$] da = 0.1(5) Solution Ab = 700. id = 0.00 [$] Cc = 938.P11 07 Equations 11. ie = 0.4 (2) c An inflating cost.71 [$] 220 Ae = 175 [$] de = 0.00 [$] ib = 0. Ne = 10.08. P W d = Ad · (1 + id ) Eqn 11.05.05 [-] Nd = 10 [-] .97 [$] ic = 0. Nb = 7. starting with known cost now: Ab = 700 [$] . with cost at end of first period known: Nd −1 Ad = 175.2 (3) d Present worth of one of a series of inflating costs.08. Nb Cb = Ab · (1 + ib ) Eqn 11. ib = 0.08 [-] Nb = 7 [-] P W d = 125.1 Na (1 + da ) (1) b An inflating cost.05. Nc −1 Cc = Ac · (1 + ic ) 11. Nd = 10.08 [-] Na = 7 [-] P W a = 2625. a Present worth of a single cost in the future: Ca = 4500 [$] .4. 1 · P W F e = Ae · de − ie 1− 1 + ie 1 + de Ne !! Eqn 11.08.4. Nc = 7.75 [$] Ca = 4500.4. starting with known cost at end of first period: Ac = 700.3 Nd (1 + dd ) (4) e Present worth of the series of inflating costs: Ae = 175.5.4.05. P11 08 Equations 11. An increase in the effective income tax rate at any time in the analysis when interest on a mortgage is being paid. tax payments would go up. i. until period NL . (At the same time.8 Effects of increasing income tax rates in late years of an economic analysis. as the additional savings would occur late in the period of the analysis when the present worths of the savings are small.) The dollar impact would not be great. 221 .e. would result in increased income tax savings and thus in increased solar savings. 64.A . year. Load = 124. 0. F = 0. NL + 1. 6. 4) (7) P rincipalBal = max (0.9 Life cycle savings . 0. i) year−1 F uelSav = CF · Load · F · (1 + iF ) .A = (10) SolarSav year (1 + d) (11) b. DownP ayment = 1000. 10 year analysis and no resale value M ortgage = Investment − DownP ayment interest = OldP rincipal · i AnnualP ay = If year. 0. CF = 12.09. 0 Mortgage payments for NL years CF assumed to be value at end of 1st year (5) (6) OldP rincipal = Tablevalue(‘Table 1’ . d = 0. Knowns i = 0. OldP rincipal − AnnualP ay + interest) (8) T axSav = interest · T axBracket (9) SolarSav = F uelSav + T axSav − AnnualP ay P W Sav.10.07. iF = 0. T axBracket = 0. 0) (12) 222 . Investment = 6000.42 (1) (2) a. P W Sav. (3) (4) M ortgage P W F (NL .by tabular method.B = If(year. NL = 5. 5 year analysis and no resale value P W Sav.P11 09 Equations 11. e.80 13157.73 -104. 5.42 CF = 12 F uelSav = 1750.33 1335.80 Table 1 Run year AnnualP ay interest P rincipalBal F uelSav T axSav SolarSav P W Sav.00 0.92 119.98 1090.82 -13.00 0.00 175. the sums of the last three columns.82 -13.99 1318.00 4181.A P W Sav. 2000.65 0.01 3280.71 -70.91 0.01 228.73 -104.18 1529.00 -71.00 0.56 2466.63 1248.21 -20.00 -143.68 1429.00 0.80 iF = 0.00 0.00 0.68 1429.00 -143.22 1636. 5 year analysis with resale of $2000 P W Resale = If(year.21 796.15 1199.00 P W Sav.00 0.26 1750.08 0.00 1594.00 0.00 0.64 Investment = 6000 NL = 5 P W Sav.99 1318.1 Load = 124 P rincipalBal = 0.99 1318.18 1529.B + P W Resale (14) The LCS is the sum of the annual LCS.00 0. i.00 T axSav = 0.00 0.80 7232.00 669.26 1750.00 0..40 -90.73 -104.C = P W Sav.87 -156.67 -1000. These sums are: a: LCS = $2467.00 0.62 -1000.20 -39.31 1166.38 739.56 T axBracket = 0.99 0.00 0.94 500. c: LCS = $-72 Solution AnnualP ay = 0.96 -1000.00 0.B = 0.00 0.00 0. b: LCS = -1372.334 × 10−14 P W Sav.00 0.00 0.C 1 2 3 4 5 6 7 8 9 10 11 Sum 0 1 2 3 4 5 6 7 8 9 10 55 1318.71 -70.00 0.00 15949.91 -56.09 i = 0.A = 739.00 d = 0.00 F = 0.67 -124.C = 0.22 1636.13 2289.20 -39.00 -143.00 0.36 0.81 767.07 OldP rincipal = 6.00 0.71 -70.60 137. 0.94 5000.77 96.00 6594.21 0.10 328.c.00 0.99 1318.00 -1371.B P W Sav.81 223 .82 1286.00 0.37 952. 0) year (1 + d) (13) P W Sav.20 -39.00 year = 10 DownP ayment = 1000 interest = 0.32 1018.00 M ortgage = 5000 P W Resale = 0 SolarSav = 1750.68 210.42 781.46 753.14 50.00 0.00 418.30 1335. t = 0.2 and 11. ND . ND = 1 [yr] i=0 CF = 12. t. 5 [yr] . P2 method. tbar = 0. but with Ne = 5. d. 0. i. C = 0. 10 year analysis and no resale value F uelSav. 0) (10) P 2B = P 2 (m. NL = 5 [yr] .8. C) (11) LCS B = CF · Load · F · P 1B − P 2B · Investment (12) c. Rv. M s. d.10. Ne = 10 [yr] . Resale.3 are used in all three parts of the problem Knowns: m = 0.A = CF · Load · F (6) P 1A = P 1 (Ne . 5 year analysis with no resale value This is the same calculation as in a.64 (3) DownP ayment = Down · Investment (4) M ortgage = Investment − DownP ayment (5) a.9 using the P1. 5 year analysis with resale of $2000 This is the same as b. Down = 1/6. Down. but with resale term added in to P2.)(1) (2) iF = 0. Investment = 6000 [$] .P11 10 Equations 11. Ne . i. 5 [yr] . d. iF . V. M s. 0. tbar. M s. tbar. C) (8) LCS A = F uelSav.07.10 Redo Problem 11.8. 0) (14) P 2C = P 2 (m. 0) (7) P 2A = P 2 (m. Down. ND . V. P 1B = P 1 (5 [yr] . d. d. V. d = 0.1 b. tbar. The functions for Equation 11. NL .A · P 1A − P 2A · Investment (9) Eqn 11. M s = 0. 0.09.8. NL . V = 0. F = 0. t. Rv = 0. C) (15) LCS C = CF · Load · F · P 1B − P 2C · Investment (16) Solution 224 . (Can not be zero. Rv. iF . Resale = 1/3 (13) P 1C = P 1 (5 [yr] . d.42. t. NL . i. iF . Down. ND . Load = 124. 9305 t=0 CF = 12 [$/GJ] F = 0.9305 Rv = 0 .1667 i=0 LCS B = −1371.A = 952.7138 tbar = 0.1 Ne = 10 [yr] P 1C = 4.42 d = 0.64 iF = 0.3333 V =0 225 Down = 0.453 P 2B = 0.C=0 DownP ayment = 1000 [$] Investment = 6000 [$] LCS C = −71.67 [$] M ortgage = 5000 [$] NL = 5 [yr] P 2A = 0.422 P 2C = 0.81 [$] Ms = 0 P 1A = 8.09 F uelSav.07 Load = 124 [GJ] ND = 1 [yr] P 1B = 4.3 [$] LCS A = 2466.96 [$] m = 0.422 Resale = 0. B.A = F uelSav · P artialP 1.64 (3) DownP ayment = Down · Investment (4) M ortgage = Investment − DownP ayment (5) Parts a.i.A. Ne = 10 [yr] . 10 year analysis and no resale value ∆iF = 0. 5 year analysis with no resale value. 0) (14) P artialP 1.B 1+iF (15) d − iF 226 .F.A = P 1 · CF · Load · ∆F Eqn 11.B . Rv = 0. NL = 5 [yr] . Calculations are similar to part a. iF .B · 1+iF 1+d Ne.wrt.i.i. iF . C = 0. d = 0. 0.F. and c refer to the three parts of the previous problem.i.9.5 and 11.42. Load = 124. t = 0.P robable.1 (9) ∆LCS. 0) (6) (7) From Equations 11.F.A = ∆2LCS. a. 0.F = P 1 − Ne · 1+iF 1+d Ne 1+iF (8) d − iF ∆LCS.wrt.F = P 1B − Ne. F = 0. Ne. b.9.A.F.M ax.A Eqn 11.wrt.i.11 Estimation of uncertainties in LCS in results of Problem 11.09.F · ∆iF Eqn 11. tbar = 0.07.A = abs (∆LCS.9.A ) (12) (11) b.9. V = 0.A + ∆2LCS.02. Down = 1/6.11 q ∆LCS. P 1 = P 1 (Ne .B = 5 [yr] (13) P 1B = P 1 (Ne.10.13: P artialP 1. M s = 0.03. F uelSav = CF · Load · F ∆F = 0.9/11.)(1) (2) iF = 0.10 Knowns: m = 0.i.9. ND = 1 [yr] i=0 CF = 12.A ) + abs (∆LCS. Investment = 6000 [$] .3 (10) ∆LCS. d.P11 11 Equations 11.F. d. (Can not be zero.F. M ax.F · ∆iF (16) ∆LCS.34 tbar = 0.i.F.A = 251. and P2 does not enter into the partial derivatives of LCS wrt iF or F.52 [$] F uelSav = 952.70 [$] Down = 0.64 iF = 0.F.i.F.07 Ms = 0 NL = 5 [yr] P artialP 1.B = P 1B · CF · Load · ∆F q ∆LCS.F.112 V =0 .M ax.i.∆LCS.wrt. 5 year analysis with resale of $2000 Answers are the same as part b since adding resale value only changes P2.F.F.60 [$] ∆LCS.i.P robable.89 [$] F = 0.i. Solution C=0 ∆iF = 0.B = 266.B ) + abs (∆LCS.P robable.B = 5 [yr] P artialP 1.i.1 Ne = 10 [yr] P 1B = 4.A = 1012.A.P robable.B = 363.09 ∆LCS.422 t=0 227 d = 0.wrt.55 [$] ∆LCS.F.36 [$] DownP ayment = 1000 [$] Investment = 6000 [$] M ortgage = 5000 [$] Ne.15 [$] ∆LCS.B = 231.B (17) ∆LCS.i.F.B.32 [$] Load = 124 [GJ] ND = 1 [yr] P 1 = 8.B = F uelSav · P artialP 1.B = 131.A = 1232.i.76 [$] ∆LCS.F.B + ∆2LCS.B = ∆2LCS.453 Rv = 0 CF = 12 [$/GJ] ∆LCS.F = 34.1667 i=0 m = 0.B ) (19) (18) c.A = 981.F = 8.02 ∆LCS.B = abs (∆LCS.03 ∆LCS.B.F.wrt.M ax.42 ∆F = 0. 10 Knowns DownP ayment = 1000.6 (3) Solution CF = 12 [$/GJ] Investment = 6000 [$] N pDiscounted = 7.07. Fuel costs discounted iF −d ln 1 + Investment · F ·Load·C F N pDiscounted = 1+iF ln 1+d d = 0. a.07 iF = 0.40 . F = 0.7.64 N p = 5.7. Investment = 6000. DownP ayment = 1000 [$] Load = 124 [GJ] 228 F = 0. Fuel costs not discounted iF ln 1 + Investment · F ·Load·C F Np = ln (1 + iF ) Load = 124.09 iF = 0.12 Calculation of payback times for Problem 11.3 b.64. Eqn 11.P11 12 Equations 11.9/11.09(1) (2) Eqn 11.27 CF = 12. d = 0. 5 year analysis with no resale value P 2B = P 2 (m. d. NL . iF = 0.A = P 2A · Ca P 1A · CF · Load Eqn 11.A = 43 InvestmentA = Ac. Knowns: m = 0. Down = 1/6.1 (9) DownP aymentA = Down · InvestmentA (10) M ortgage = InvestmentA − DownP aymentA (11) LCS A = F uelSav · P 1A − P 2A · InvestmentA Eqn 11. d.8. d. iF .07. 0) (4) F uelSav = CF · Load · F (5) P 2A = P 2 (m. Rv. Rv = 0 Load = 124. 0. t¯. C) (13) P 1B = P 1 (5 [yr] . Down. 5 [yr] . M s. Ne . t = 0. C = 0.10.5.B = P 2B · Ca P 1B · CF · Load (15) 229 . M s. ND . C) (6) Area and F determined from figure with dF/dAc as given by Equation 11. 0) (14) dF/dAc. Ne = 10 [yr] .A · Ca + Ce (8) Eqn 11. t¯.5 (7) From figure with dF/dAc equal to 0.42 (3) a. FA and Ac. NL = 5 [yr] .A are: F = 0.60.10. V = 0. V. V. Ce = 120.8.13 For the system of Problem 11. NL . d.00710. ND = 1 [yr] . Ca = 96. given an F vs Ac curve. iF .1 (12) b. optimize Ac. t. It is necessary to run EES program twice.P11 13 Equations 11. i. 10 year analysis and no resale value P 1A = P 1 (Ne . ND . dF/dAc. Ac. once to find dF/dAc and then find LCS with correct F and Ac.8. Rv.09. t¯ = 0. Can not be zero(1) (2) CF = 12. M s = 0.1. i. 0. t. d = 0.9/11. i = 0. Down. 46 InvestmentC = 2808 [$] LCS C = 1022 [$] Ms = 0 P 1A = 8.01041 1/m2 F uelSav = 892.C = 0. d. 5 [yr] .010.B = 20 m2 Ce = 120 [$] dF /dAc.09 Down = 0.C equal to 0.3333 V =0 C=0 d = 0.B are: FB = 0. ND . C) (20) P 1C = P 1 (5 [yr] .8 [$] InvestmentA = 4248 [$] LCS A = 3594 [$] m = 0. 0.7138 t¯ = 0. d. Ac.B = 0. Down.C = P 2C · Ca P 1C · CF · Load (22) From figure with dF/dAc.42 230 Ac.453 P 2B = 0.38 InvestmentB = 2040 [$] LCS B = 602.422 P 2C = 0.38.1 Ne = 10 [yr] P 1C = 4. 0) (21) dF/dAc.A = 0.46. t. FB and Ac. Ac.C + Ce (24) LCS C = CF · Load · FC · P 1C − P 2C · InvestmentC (25) Solution Ac.From figure with dF/dAc.6 i=0 iF = 0. t¯.9305 t=0 Ac.9305 Rv = 0 .C = 28 m2 CF = 12 [$/GJ] dF /dAc. F and Ac are: FC = 0.2 [$] M ortgage = 3540 [$] NL = 5 [yr] P 2A = 0.B = 20 (16) InvestmentB = Ca · Ac. i.C = 28 (23) InvestmentC = Ca · Ac. V.A = 43 m2 Ca = 96 $/m2 dF /dAc. NL .007102 1/m2 DownP aymentA = 708 [$] FC = 0.422 Resale = 0.014.01358 1/m2 F = 0.B + Ce (17) LCS B = CF · Load · FB · P 1B − P 2B · InvestmentB (18) c. M s.07 Load = 124 [GJ] ND = 1 [yr] P 1B = 4. 5 year analysis with resale of $2000 Resale = 1/3 (19) P 2C = P 2 (m.B equal to 0.1667 FB = 0. iF . Resale. 7 iF = 0. M s.40 load = 3000 [GJ] (1) (2) (3) DownP ayment = Down · Investment (4) M ortgage = Investment − DownP ayment (5) Calculate PW of fuel savings: P 1 = P 1 (Ne .10.01389 [-] P 1 = 13. Ne . ND . Investment = 18000. t¯r = 0. V = 0. F = 0. iF . t = 0.4 CF = 8.06. NL .09.Sav − P W System Eqn 11.14 Calculation of LCS Knowns: m = 0.64 Rv = 0 231 Down = 0. Ne = 15.1 M ortgage = 14400 [$] NL = 10 [yr] P W System = 17846 [$] V =0 d = 0.8. C = 0.1 F uelSav = 18480 [$] LCS = 234154 [$] M s = 0.2 i = 0.9914 t=0 . t. C) (9) P W System = P 2 · Investment (10) LCS = P W F uel. Rv = 0. V. Rv. Any number but zero iF = 0.20.1 (11) Solution C=0 DownP ayment = 3600 [$] Investment = 18000 [$] m = 0. Down. [$] .09 Ne = 15 [yr] P W F uel.Sav = F uelSav · P 1 (8) Calculate PW of system nvestment costs: P 2 = P 2 (m. ND = 1 CF = 8.06 load = 3000 [GJ] ND = 1 [yr] P 2 = 0. 0) (6) F uelSav = CF · Load · F (7) P W F uel.7.8 [$/GJ] F = 0. d. t¯r. 0. i. d. i = 0.8. M s = 250 d = 0. NL = 10.10. investment Down = 0.Sav = 252000 [$] t¯r = 0.P11 14 Equations 11. A = 15 [yr] (6) P 1A = P 1 (Ne. V = 1.Sav.A . C = 0. iF . Rv. M s.A . CF = 10 (1) (2) (3) DownP ayment = Down · Investment (4) M ortgage = Investment − DownP ayment (5) a. t¯. C) (9) P W System. d. NL . Knowns m = 0. Any number but zero i = 0. ND .A = CF · solar · P 1A (8) P 2A = P 2 (m.06. t¯ = 0.40.09.B = 20 [yr] (12) 232 . 0) (7) P W F uel. d = 0.02.15 Calculation of LCS and ROI. 15 year economic analysis Ne. t = 0.A − P W System. Down = 0.A = P 2A · Investment (10) LCS A = P W F uel. solar = 110 [GJ] ND = 1 [yr] Rv = 0. Investment = 11000. Down. iF = 0.80. 20 year economic analysis Ne. NL = 10 [yr] . V.20. d. i. t.Sav.14. 0.P11 15 Equations 11. M s = 0.A (11) b.15. Ne. C = 6065 [$] ROI = 0.C .B = 17630 [$] P W System.A = 15 [yr] P 1A = 12.2 iF = 0.Sav. Down. M s.B . 0) (19) P W F uel.02 Down = 0. t. M s.B = P 2B · Investment (16) LCS B = P W F uel. i. 0.C (23) Thus the Return on Investment ROI.C = CF · solar · P 1C (20) P 2C = P 2 (m.P 1B = P 1 (Ne.Sav.4%.B (17) c.06 LCS B = 9057 [$] Ms = 0 Ne. it would be necessary to solve the problem for a range of discount rates and choose the one for which LCS is zero. Rv.5514 P W System. Rv. as used in part b. d. Without EES or a similar program. If the time of the economic analysis had been 20 years.Sav.C = P 2C · Investment (22) LCS C = P W F uel.C . ND . t¯.15%. V.B = CF · solar · P 1B (14) P 2B = P 2 (m. then the ROI would be 0.B − P W System. ND . NL . Ne.514 P W F uel. iF . ROI.B .3115 or 31.C − P W System.B = 8572 [$] solar = 110 [GJ] 233 d = 0.14 m = 0.Sav.7167 P W F uel.8 V =1 CF = 10 [$/GJ] i = 0.2944 t¯ = 0.A = 13507 [$] P W System. i. t¯.03 P 2C = 0.4 .C = 6065 [$] t = 0.Sav.C = 15 [yr] . V.Sav.09 Ne. Down. Solution C=0 DownP ayment = 2200 [$] LCS A = 5622 [$] M ortgage = 8800 [$] Ne. t. d. LCS C = 0 [$] (18) P 1C = P 1 (Ne.B = 20 [yr] P 1B = 16. Ne.Sav.28 P 2B = 0.7793 P W F uel.A = 7884 [$] Rv = 0. ROI. 0) (13) P W F uel. NL . Ne.15 Investment = 11000 [$] LCS C = 0 [$] ND = 1 [yr] NL = 10 [yr] P 2A = 0. C) (21) P W System. part c. 0.C = 15 [yr] P 1C = 5. iF . is 29. C) (15) P W System. Find ROI for a 15 year economic analysis With EES this is a very simple calculation since it is only necessary to set LCS to zero and let EES find the discount rate which will be the ROI. 07..Sav .16 Comparison of the PW of fuel savings and the cost of a system. iF = 0. Celect = 0.Investment) positive? . (δ=PWF uel. d = 0. load = 19.P11 16 Equations 11.12.e.60 (1) (2) The question is: is ∆ i. Knowns: Investment = 1550 [$] . Ne = 25 [yr] F = 0.045 [$/kwh] . . . $/GJ . . . CF = Celect · . 777778 $/kwh .277. 07 iF = 0.12 P W F uel.Sav = CF · load · F · P 1 (5) ∆ = P W F uel. (3) P 1 = P 1 (Ne . Solution Celect = 0.Sav = 6077 [$] 234 ∆ = 4527 [$] load = 19 [GJ] .6 Ne = 25 [yr] CF = 12.5 [$/GJ] Investment = 1550 [$] P 1 = 42. d. 0) (4) P W F uel. 0.Sav − Investment (6) Since ∆ is positive.045 [$/kwh] F = 0. iF . the criterion is met.64 [-] d = 0. P11 17 235 . 90.25. (5) NL = 20. ND = 0.Equations Find economic optimum solar system.39.2.052 (3) dis = 0. η = 0.8 CF 1 = 11 (6) (7) (8) [$/GJ] η (9) Thermal data Load = (EDHW .020.045 (4) t¯ = 0.050. Down = 0. P rT ax = 0. C=0 inf = . Ms = 0. V al = 0. Ne = 20.058.0001.035. m = 0. Rv = 0. Economic data Collector cost per unit area set in parametric table as either 240[$/m2 ] or 150[$/m2 ] CE = 550 [$] (1) Invest = Ac · CA + CE (2) inf F = 0. . . . ·6 GJ . . + EHouse ) · . 1 × 10 kJ . Rv .Load. Val. ROI. Load. inf F . t¯. Down. LCS Method LCS = LCS (P 1. C) P2’ = P2(m . t¯. 0= LCS(P 10 . P rT ax.7.P2’.CF 1 . CF 1 . Ms .0645 × 107 [kJ] (11) EHouse = 1. inf.e . t¯. SolF ract. Ne . NL . The equations are the same as in part a. C) (15) b. P 2.42 × 108 [kJ] (12) a. EES can solve for the discount rate (i.6 (16) c. ND . Return on Investment Method Return on Investment is the discount rate at which LCS = 0.e. Ne .SolFract. but P1. P2 have new values and the LCS is 0. Ms . NL . PrTax. C) (14) P 2 = P 2 (m. C) 236 . ROI) with LCS set to zero for each of the collector areas. ND . ROI.Invest) P1’ = P1(N. dis. t¯. infF . V al. dis. (10) EDHW = 2.. inf . Invest) (13) P 1 = P 1 (Ne . Down. Payback Time Method inf F −dis ln Invest · SolF ract·Load·C + 1 F1 NP = 1+infF ln 1+dis Eqn 11. Rv . 860 -10.063 0.950 18.2 F uelSav 1 = 1.46 [$] dis = 0.23 -1.7 [$] SolF ract = 0.420 × 108 [kJ] inf F = 0. ROI.891 Rv = 0.25 CA=240 Run 1 2 3 4 5 6 7 Ac2 m 0.7 [$] P rT ax = 0.EES has a problem solving these three equations without very good guesses. Cash Flow (PITI) Method This method compares the solar system out-of-pocket first year expenses .091 CA=150 Run 1 0.008 0.02 LCS = 2. t¯.550 LCS [$] NP -433 4 -1.6 [GJ] Ne = 20 [yr] P 1 = 20. m) First year principle and interest (18) F uelSav 1 = SolF ract · Load · CF 1 First year fuel savings (19) P ropT ax = V al · P rT ax · Invest First year property taxes (20) IncT ax = t¯ · ((1 − Down) · m · Invest + P ropT ax) Ins = Invest · Ms First year income tax savings First year insturance (22) CashF low = F uelSav 1 + IncT ax − P &If irst − P ropT ax − Ins First year cash flow Solution Ac = 75. C) . interest.535 C=0 CF 1 = 13. Down. and insurance (PITI) to the first year fuel savings.43 CA 2 $/m 150 Invest [$] LCS [$] CashF low [$] -550.41 ROI = 0.705 -6. 0.052 Load = 162. taxes.001 (21) NP 237 CashF low [$] ROI (23) .535 CA 2 $/m 240 240 240 240 240 240 240 Invest [$] 550 1.000 0. Rv EES has trouble at zero area (row 1) but all of these values are obvious so run Tables from 2 to 7. Ne . CF 1 .045 EHouse = 1.00 P ropT ax = 371.035 t¯ = 0.350 14.72 -489.0 60.for principle. P &If irst = (1 − Down) · Invest P W F (NL .75 [$/GJ] EDHW = 2. V al. Ms .0 SolF ract Ac2 m 0.91 11. inf . 107 [$] ND = 0.293 0.39 CashF low = −216. ND .0 30. P rT ax.750 4.0 5. Doing a little algebra to eliminate P1’ and P2’ results in: 0 = LCS (P 1 (Ne .5 [$] Invest = 11.09 14.0 15.280 -3. 800 [$] Ms = 0.064 × 107 [kJ] inf = 0.25 15.389 0.081 V al = 0.058 NL = 20 [yr] P 2 = 1. Load. t¯.654 12. d.024.060 -0.0 75. ROI.26 -272.0001 P &If irst = 809.0 SolF ract 0.469 0.65 -112.045 0.05 NP = 10.00 -68.172 0.8 IncT ax = 358.027 -0.69 ROI -0. P 2 (m.0 45.08 -741.91 13.750 11.010 -0.0 m2 CE = 550 [$] η = 0. NL . 196 [$] Ins = 590 [$] m = 0.49 10. SolF ract.150 7.9 CA = 150 $/m2 Down = 0. inf F . 389 0.2 3 4 5 6 7 5.14 -94.800 5.65 -216.550 11.44 7.225 0.0 60.58 -4.050 7.349 2.063 0.109 0.194 0.293 0.300 2.824 3.0 45.152 0.0 0.87 8.107 9.0 15.39 50.300 9.115 0.77 49.952 3.46 0.0 30.800 417 2.00 LCS Yeas to payback 238 -14.25 10.0 75.556 3.535 150 150 150 150 150 150 1.081 .55 9.172 0.469 0.37 7. PITI ROI 239 . and essentially all collected energy will have to be stored. 240 . the store will be cooled more quickly and will be at a lower 24 hour average temperature. But. assuming a fully mixed tank.P12 01 Equations 12. The collector will run a bit hotter because there is no energy delivery during collection hours. if energy is removed from store in the early evening. thus reducing store losses. The effect of concentrating the loads to early evening hours will be determined by two opposing factors. The net effect should be to increase the solar energy delivered by a small amount.1 Effect of load timing. 2.P12 02 Equations 12. but will not double it as Ts and thus Ti in Equation 10. as UL in Equation 10. b. higher Ts in the top of the tank will lead to higher solar energy delivery. Collector losses are very much greater. On the load side. But. The net effect should be to somewhat increase solar energy delivery.2 will be substantially increases. 241 . d.9. Ts will be brought down more quickly by the concentrated load and the 24 hour average Ts will be lower.1 a. Doubling the collector area will increase Qu. The total effect will be to increase Qu and solar energy delivery. Applying the load in the evening hours will mean increased Ts and thus Ti because all energy to load must be stored. c. This means reduced Qu (possibly to zero for the four hours of the problem).2 will increase. reduced Ts and reduced solar energy delivery to load. Increased Ts will lead to greater solar energy delivery to load.2.2 Effects of design changes on performance of the system of Example 10. Stratification of the tank will lead to lower temperatures in the bottom of the tank and lower Ti. 5. Slope. 5. TauAlpha/TauAlphan = 0. K (5) ¯ dif f = H ¯ · Hbardif f /Hbar H (6) SunSetHrAngle0 = min (arccos (− tan (Lat) · tan (δ)).beam = 0. ρg H (10) ¯b = H ¯ −H ¯ dif f H (11) ¯b · R ¯ beam · f¯i · T auAlphabeam + H ¯ dif f · T auAlphadif f · Frs + H ¯ · ρg · T auAlphaGrRef /2 AbsRad = Ar · H Eqn 14. θef f. ρg = 0. n.6. The basic equation to be used is 14. Frs . TauAlpha/TauAlphan = 0.4. H ¯ T = 0.1 [deg] Ar = 6. (n.0 m2 . arccos (− tan (Lat − Slope) · tan (δ))) (7) SunSetHrAngle = arccos (− tan (Lat) · tan (δ)) (8) ¯ beam = R ¯ BEAM. Slope) R (9) ¯T = H ¯ T.P14 01 Equations 14.LJ.1 is 0.dif f and θef f -GrRef both = 59. ¯b. H ¯ d .57 (15) T auAlphadif f = T auAlphan · 0. 5. at this angle.1.82. For this wall and glazing: τn = 0.82 (16) T auAlphaGrRef = 0.2(12) From Fig 5. αn = 0. From Fig. ¯ = 25.6. Knowns M onth = 4. H.71.1 Month’s absorbed radiation on vertical wall with overhang. H ¯ b. ¯ Lat. Lat = 35. R diffuse and ground-reflected components.57.6.2. The normal transmittace (τ n ) for this glazing system from Figure 5. from Fig.3. Lat.6.1.57 (17) 242 1 .78.89 (1) Assumed n = AveDay (month) (2) (3) δ = dec (n) (4) ¯T Hbardif f /Hbar = Hdif f BarHBar Lat.78 (13) T auAlphan = τn · αn (14) T auAlphabeam = T auAlphan · 0.10.71. n.1: θef f. at this angle. and the effective τ -α products for the beam. K Slope = 90 [deg] . from Fig.2. We need to find fibar .3. 4.2 MJ/m2 H Lat = 35.25. Using the charts in Appendix I for w = 1 and w = 4 at Latitudes 35 and 45: At Lat = 35.52: At Lat = 45.4 is 0.4.71 K ρg = 0.415 [deg] ¯ = 25.102 MJ/m2 H M onth = 4 Slope = 90 [deg] T auAlphadif f = 0. so at g = 0. so at g = 0.5692 Ar = 6 m2 Hbardif f /Hbar = 0.89 Frs = 0.2 is 0.3957 τn = 0.51 and at g = 0. f¯i is 0.We do not have f¯i for infinite extensions.78 αn = 0.3883 R T auAlphabeam = 0.3 MJ/m2 H ¯ T = 0.5. f¯i = 0. f¯i is 0. so at g = 0. at e = 0.63. so at g = 0. w = 1: f¯i at g = 0.89.59 ¯ b = 19. f¯i is 0.4 is 0.4.35 [deg] T auAlphan = 0.2 is 0.56 f¯i at g = 0.6 to interpolate for w = 1.26 [MJ] f¯i = 0. w = 4: f¯i at g = 0.4 is 0.76 At Lat = 45.6942 .4 is 0.71.44.76 From Table 14. w = 4: f¯i at g = 0.44 The following are calculated above: Frs = 0.2 is 0.43 and at g = 0.90.3. f¯i is 0.1 [deg] ¯ beam = 0.25.75 Thus at w = 1.25.2 SunSetHrAngle0 = 76.05 MJ/m2 H n = 105 SunSetHrAngle = 96. w = 1: At Lat = 35.5: f¯i = 0.57 243 2 δ = 9.70 and at g = 0.2412 ¯ T = 13.48 Using Equation 14.25.3: Frs = 0.72 and at g = 0.44 ¯ dif f = 6. so will use e = 0.59 (18) Solution AbsRad = 28.69 [deg] T auAlphaGrRef = 0. f¯i = 0.2 is 0. GrRef 00 H (13) ¯ T. ”: ¯ T 00 = H ¯ T. n.dif f 00 = H ¯ dif f /2 H (15) ¯ T. Slope = 90 [deg] (1) Frs = 0.beam + H ¯ T.beam = H ¯ beam · R ¯ beam · f¯i H (8) ¯ T.P14 02 Equations 14.9 [deg] .dif f + H ¯ T. Designate the radiation terms for no overhang with double prime.GrRef 00 = H ¯ · GrRef /2 H (16) Parametric Table: Table 1 244 1 .43 (2) dec = Dec (month) (3) n = AveDay (month) (4) ¯ beam = R ¯ BEAM.beam00 + H ¯ T.1. K (5) (6) ¯T: Rewrite Equation 14.2 Calculation of monthly radiation on a direct gain window.dif f 00 + H ¯ T.GrRef = H ¯ · GrRef /2 H (10) Calculate monthly diffuse and beam: ¯ dif f = H ¯ · Hbardif f /Hbar H (11) ¯ beam = H ¯ −H ¯ dif f H (12) Get f¯i from Appendix I (d) and Frs from Table 14.4.GrRef H (7) ¯ T. Knowns Lat = 44.4 to give the three components of H ¯T = H ¯ T. with and without overhangs. (n.beam00 = H ¯ beam · R ¯ beam H (14) ¯ T. Slope) R ¯T Hbardif f /Hbar = Hdif f barHbar Lat.dif f = H ¯ dif f · Frs H (9) ¯ T.4. Lat. 48 3.86 16.36 23.05 1.05 21.48 6.37 10.2 0.72 Plot 1 245 2 ¯ HT.85 1.46 1.14 2.2 0.58 2.21 ¯ HT.98 1.57 12.24 0.59 0.75 1.Run M onth ¯T K 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 0.73 12.07 7.dif f200 MJ/m 1.34 1.75 .89 12.26 2.33 0.2 0.23 13.79 3.71 3.47 Run 1 2 3 4 5 6 7 8 9 10 11 12 ¯ H T.17 3.20 2.46 0.15 1.27 2.59 3.05 15.33 14.55 0.30 1.91 8.60 3.52 1.6 2.91 7.81 ¯ T.26 2.14 2.4 0.42 11.48 0.93 1.33 8.76 8.54 0.36 9.22 10.07 6.89 1.76 11.29 1.80 3.17 1.dif f2 MJ/m 1.98 1.04 ¯ HT 00 2 MJ/m 14.92 1.55 0.12 4.11 10.85 5.29 1.65 10.69 0.44 13.30 2.27 1.45 7.18 4.04 22.GrRef 00 H MJ/m2 1.84 11.beam MJ/m2 10.58 f¯i GrRef ¯ beam R 1 1 0.4 0.41 4.13 0.74 9.21 10.84 2.26 ¯ HT.46 0.93 1.61 2.40 10.04 5.48 0.2 0.47 11.94 8.55 0.4 0.99 1 1 0.05 1.88 1.03 1.86 15.03 1.77 4.47 0.47 0.57 10.90 7.94 0.28 0.88 0.56 0.46 1.73 11.GrRef MJ/m2 1.78 0.56 3.37 5.04 8.65 10.61 2.49 0.beam200 MJ/m 10.25 3.63 2.39 3.88 13.72 ¯ H 2 MJ/m 6.42 0.2 0.74 ¯ HT 2 MJ/m 13.88 3.27 1.52 0.59 10.2 0.32 3.6 0.68 4.44 9.24 11.2 0.58 20.99 2.81 ¯ HT.76 11.2 0.30 2.36 1.36 1. 1.1. Interpolating between g=0 and g=0. Knowns Lat = 43 [deg] . τ¯dif f = 0.93.2 W/m2 ·C .72 (14) 246 1 . Slope) R (10) ¯ i from Appendix I (d): Getf At Lat=45 and w=4. the room absorptance is 0. Troom = 20 [C] Gap = 0.25 [m] . K T¯a = −7. and at w=25.49.5.82 The interpolation is thus linear for Lat=43 and gives: f¯i = 0.625 [m] .2 (4) n = AveDay (month) (5) dec = dec (n) (6) Calculate monthly beam and diffuse: ¯ dif ¯f /H = Hdif f barHbar Lat. p = P rojection/Height.f¯i =0. GrRef = 0. Slope = 90 [deg] . w = W idth/Height.70 (13) τ¯beam = 0. H ¯ T = 0.f¯i =0.1.96. and at w=25.96 and the ground reflected and diffuse sky transmittances are about 0.5 gives: Frs = 0.70. (n.25 at e=0 and p=0.25 [m] . Lat. Uwindow = 3. W idth = 15 [m] .5.f¯i =0.38 (12) From Example 14. ¯ = 9. K ¯T H (7) ¯ dif f = H ¯ ·H ¯ dif ¯f /H H (8) ¯ beam = H ¯ −H ¯ dif f H (9) ¯ beam = R ¯ BEAM.P14 03 Equations 14.3 Gain (or loss) from a direct-gain south facing window in Example 14.70. e = Extension/Height.83.4. Thus: αr = 0. n.93 At Lat=35 and w=4.91 (11) Frs from Table 14. Height = 1.12. P rojection = 0.f¯i =0. τ¯gr = 0. (1) Extension = 0 [m](2) g = Gap/Height (3) For Feb: month = 2. 3 ¯ beam · R ¯ beam · f¯i · τ¯beam + H ¯ dif f · Frs · τ¯dif f + H ¯ · GrRef · τ¯gr /2 Absorbedroom = Height · W idth · αr · H .From Eqn 14.5. . . MJ . . 5.4 Lossroom = Height · W idth · Uwindow · 24 [hr] · Troom − T¯a · . Eqn 14. . 0.0036 W · hr . (15) (16) The net gain is then: N etGainroom = Absorbedroom − Lossroom (17) The net is nearly zero.401 H n = 47 Slope = 90 [deg] Troom = 20 [C] .2 ¯ dif f /H ¯ = 0.25 [m] ¯ T = 0.12 MJ/m2 H Lat = 43 [deg] p = 0.2 ¯ beam = 5.95 [deg] g = 0. as the monthly daytime gains nearly balance the day-long losses.49 K N etGainroom = 4 [MJ] τ¯beam = 0.625 [m] τ¯gr = 0.463 MJ/m2 H Lossroom = 140 [MJ] P rojection = 0.25 [m] ¯ dif f = 3.38 ¯ = 9.2 W/m2 -C αr = 0.8 [MJ] f¯i = 0.782 R ¯ Ta = −7 [C] Extension = 0 [m] GrRef = 0.657 MJ/m2 H month = 2 ¯ beam = 1.7 w = 12 dec = −12.5 τ¯dif f = 0.7 W idth = 15 [m] 247 2 e=0 Gap = 0.91 Height = 1.72 Uwindow = 3.96 Frs = 0. Solution Absorbedroom = 143. Troom = 20.8 [MJ] ¯L = (1 − fN ) · Uwindow + fN · U Lossroom (2) Uwindow 1 + Rn · Uwindow Eqn 14.5.6.5 (3) .2.4 Calculation of effect of adding night insulation to the window of Problem 14.3 Knowns RN = 1.25. W idth = 15. T¯a = −7. Height = 1.3. the energy absorbed from the room is 144MJ/day: Absorbedroom = 143.P14 04 Equations 14.5 (1) From Problem 14. Uwindow = 3. fN = 0. . . MJ . . ¯ ¯ . = Height · W idth · UL · 24 [hr] · Troom − Ta · . 0.0036 W · hr . 25 [m] Troom = 20 [C] 248 1 Lossroom = 81.38 [MJ] Uwindow = 3.5 T¯a = −7 [C] Height = 1.5.6 C-m2 /W W idth = 15 [m] fN = 0.861 W/m2 -C U N etGainroom = 62.8 [MJ] RN = 1.42 [MJ] ¯L = 1. Eqn 14.4 (4) The net gain is then: N etGainroom = Absorbedroom − Lossroom (5) Solution Absorbedroom = 143.2 W/m2 -C . Equation 18. hw = 10 W/m2 ·K Radiation from basin to cover.1333 converts from Pa to mm Hg. and the condensate and cover are at a single uniform temperature Twg.1 Calculation of energy flows in a basin type still. $UnitSystem K Pa kJ Make the following assumptions: the vapor pressure of solution in the basin is the same as that of water at the same temperature. (2) Tsky Ta = ConvertTemp(C.4. Equation 18. the area of cover is the same as area of the basin.P18 01 Equations 18.6: (The 0. Knowns: Twb = ConvertTemp(C. 20). K. 24) = ConvertTemp(C. 60).9 · σ · Twb − Twg (3) Convection from basin to cover. (1) σ = sigma#.4.) . K. K.3: 4 4 qrad = 0. . . mmHg . . Pwb = P (ST EAM . x = 0) · . T = Twb . . 0.007500615 Pa . . . . mmHg . . . Pwg = P (ST EAM . x = 0) · . T = Twg . 007500615 Pa .0. qrad + qconv + qevap = qwind + qradSky (12) With EES it is as easy to solve for the cover temperature to see if it is the same as that in the problem statement.4.15 × 10 · hc0 · (Pwb − Pwg ) · hf g · 1000 (8) (9) Wind convection: qwind = hw · (Twg − Ta ) (10) Radiation from cover to sky at 20 C. it is necessary that the energy balance on the cover. x = 1) − h (ST EAM . x = 0) −7 qevap = 9. T = Twg . 1/3 Pwb − Pwg hc0 = 0. be satisfied.88 · σ · Twg − Tsky (11) To check the cover temperature. Twg ) (13) 249 1 .9. Equation 18.2.9: hf g = h (ST EAM . Equation 18. T = Twg .4. C.1: 4 4 qradSky = 0.884 · (Twb − Twg ) + Twb · 2016 [mmhg] − Pwb qconv = hc0 · (Twb − Twg ) (4) (5) (6) (7) Heat transfer from basin to cover by evaporation. T C cover = ConvertTemp(K. Equation 3. 2 [K] 2 hw = 10 W/m -K 2 qevap = 321.2 [K] hf g = 2.5 [mmHg] qradSky = 171 W/m2 Ta = 297.5 [mmHg] qrad = 76.2 [K] 250 2 Pwb = 149.83 W/m2 T C cover = 49.Solution hc0 = 2.2 W/m2 Tsky = 293.22 W/m2 qwind = 253.1 W/m σ = 5.454 W/m2 -K Pwg = 89.32 [C] Twg = 322. 384 [kJ/kg] qconv = 26.5 [K] .670 × 10−8 W/m2 -K4 Twb = 333. β = 40 [deg] . The solution shown here was done with EES. This is a simulation problem. T 0 a = 21.526 (2) UL. tan = 0. Many of the functions used in this solution are from problem 6. ρg = 0.1 Design of a water heating system to provide 75% of a week’s loads by solar energy. Lat = 40 [deg] . LoadF low = 45 · 4/14.tank = 1. TRNSYS or other simulation program could have been used.95. Store/Ac = 60 kg/m2 .0125 (4) .77. 251 1 αn = 0. Surf AzAng = 0 [deg] . Knowns: Tmin = 60 [C] . hr = 1 [hr] .05 W/m2 ·C (3) KL = 0. m ˙ c = 50 kg/m2 ·hr .2. H/Dtank = 3. with the basic approach to compute the solar F for a range of collector areas and select that which gives the desired fraction.19. UL = 4 W/m2 ·C . Tmains = 15 [C] .9 (1) Ref rInd = 1. F 0 = 0.P19 01 Equations 19. Ncov = 1. 1.9 m2 (6) The initial tank temperature must be entered into the first row of the Parametric Table and the solution must start in row 2. ‘Day’ ) (7) hour = Lookup(‘Boulder week’ . This system is probably over designed since it supplies a large F in the middle of winter. It is necessary to guess both Ac and Tplus tank and see if the solar F is 0.9 m2 and initial Ttank =69. ρ = 1000 kg/m3 . (finding the ‘steady periodic’ solution removes the effects of initial tank temperature. which for a simulation as short as a week can have a significant effect on system performance. T ableRun#. Day = Lookup(‘Boulder week’ . T ableRun#.75 and if the final tank temperature is equal to the initial tank temperature. This is a double trial and error solution. Cp = 4190 [J/kg ·C] (5) the last iteration Ac = 7.) The results are Ac =7. It would have excess capacity in the summer. ‘hour’ ) . . . MJ . . . T ableRun#. ‘I’ ) · . I = Lookup(‘Boulder week’ . 0.001 kj . The measured horizontal data is suspect since accurate measurements are difficult to make near sunrise and sunset. β. β. KL. Lat. Collector useful output: . Equation 2. Surf AzAng. T ableRun#.LJ. assuming isotropic sky. Lat. Surf AzAng. Day. (I. αn ) (13) Warnings are generated for the 17th hour on January 10. (I.9. Ref rInd.5 [deg] (10) Need to use the middle of the hour.15. ρg . ‘T a’ ) HrAng = 15 [deg/hr] · (hour − 12 [hr]) − 7. (11) Estimation of incident and absorbed solar radiation. day. HrAng) (12) ST = ST.1: IT = IT. ρg . HrAng.LJ. Ncov .1 and 5. (8) (9) Ta = Lookup(‘Boulder week’ . . . m ˙ c · Cp . . ·4 1/s . 77778 × 10 · . x= 2. 0 UL · F 1/hr . 5 .7. (14) FR = F 0 · x · (1 − exp (−1/x)) Eqn 6. . Qu = max 0 [MJ] . Ac · FR · ST − UL · (Ttank − Ta ) · hr · . . 0036 (15) .0. MJ . . W · hr . p = 200) (18) Height = H/Dtank · Dia V olT ank = π · Height · Atank = 2 · π · (19) Dia2 4 (20) Dia2 + π · Dia · Height 4 (21) 252 2 .6 (16) For the tank: M assT ank = Ac · Store/Ac V olT ank = (17) M assT ank ρ (water. Eqn 6. T = 55.7. . . . MJ . . mC tank = Cp · M assT ank · . . 1 × 10·6 J . QtankLoss (22) . . . MJ . . . tank · Atank · hr · (Ttank − T a ) · . = UL. 0.0036 W · hr . 0 (23) The load: On = If(hour. 21. 1. 0. 7. 1. 0) Sets load flow to between 7am and 9pm. 0. 1) · If(hour. . . . . ·6 MJ . . M axLoad = Cp · LoadF low · (Tmin − Tmains ) · hr · . 1 × 10 J . . . . MJ . . Cp · LoadF low · (Ttank − Tmains ) · hr · . QLoad = On · min M axLoad. . 1 × 10·6 J . 02 0.09 0. The Parametric Table shows the first 12 and the last 12 hours of the week.8 64.2 (28) SumQLoad = Tablevalue(T ABLERU N # − 1.84 2.00 0.4 68.42 14.78 176.8 68.41 0.331 0.00 0.15 8.35 173.8 0.00 0.3 0 3.837 0.1 1.08 166.69 0.00 0.733 1.1 2.55 0.00 0.70 12.53 -0.017 0.7 61.758 -3.00 0.1 68.000 0.02 0.000 6.7 67.01 0.42 2.00 0.1 67.67 0.00 2.00 0.00 0.27 9.00 0.2 3.00 0.2 -2.62 0.00 0.57 2.00 -0.00 0.62 0.68 0.13 0.636 0.9 76.00 0. Parametric Table: Table 1 Run 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 0.00 0.73 0.71 0.2 6.8 72.15 0.65 0.000 0.75 0.00 2.42 2.00 0.2 -2.7 7.8 -2.004 0.00 0.70 0.134 0.55 3.78 0.331 0.20 0.00 0.73 12.42 4.2 -1.00 0.93 171.00 0.42 2.00 0.00 0. ‘Tplus tank’ ) + Ttank = Ttank + Qu − QtankLoss − QLoad mCtank (27) Eqn 10.67 0.8 -2.58 0.42 2.000 0.000 0.06 0.00 2.68 0.00 -0.76 0.42 2.00 0.57 0.69 0.00 0.6 77.60 0.00 0.12 14.7 58.46 4.13 0.000 0.66 164.42 2.2 SumQLoad [MJ] 0.72 0.4 73.42 2.42 2.00 0.00 0.1.85 7.8 -2.03 0.68 0.42 2.04 0.096 0.23 2.968 1.01 0.000 0.at end of an hour’s time step: Ttank = Tablevalue(‘Table 1’ .2 161.61 0.7 7.00 0.9.69 0.00 0.00 2. (24) (25) (26) Calculation of new tank temperature . ‘SumQ load’ ) + QLoad F = (29) SumQLoad 7 [day] · 14 [1/day] · M axLoad (30) The results are Ac =7.00 0.71 2.3 62.56 plus Ttank [C] 69.00 0.00 0.42 2.9 m2 and initial Ttank =69.25 0.00 0.2 59.70 0.7 6.2 2.00 14 14 14 14 14 14 14 13 14 15 16 17 18 19 1.22 0.74 Day hour [hr] I 2 MJ/m Ta [C] IT 2 MJ/m ST 2 MJ/m QLoad [MJ] Qu [MJ] QtankLoss [MJ] 8 8 8 8 8 8 8 8 8 8 8 2 3 4 5 6 7 8 9 10 11 12 0. T ableRun# − 1.12 1.76 69.00 0.09 0.4 73.05 0.3 -2.81 0.51 168.0 75.00 0.31 0.00 0.80 0.00 0.00 253 3 F [-] .42 2.00 0.4 65.45 0. 70 0.00 0.21 22 23 24 25 14 14 14 14 14 20 21 22 23 24 0.63 178.00 0.7 0.2 69.00 0.00 0.71 0.00 0.42 0.9 -3.75 .9 -6.00 0.00 0.75 0.9 -3.74 0.00 0.00 0.000 0.71 0.63 178.75 0.00 0.00 0.1 -6.00 254 4 2.000 3.63 178.000 0.63 0.00 0.00 0.6 70.00 0.00 0.000 0.75 0.000 0.00 0.63 178.70 70.5 69.8 69.00 0.1 178.75 0.00 0. This is a simulation problem. Lat = 40 [deg] . with the basic approach to compute the solar F for a range of collector areas and select that which gives the desired fraction. β = 40 [deg] . m ˙ c = 50 kg/m2 ·hr .P19 01a Equations 19.9 (1) Ref rInd = 1. T 0 a = 21.526 (2) UL.05 W/m2 ·C (3) KL = 0. Tmains = 15 [C] . TRNSYS or other simulation program could have been used. Knowns: Tmin = 60 [C] . ρg = 0.0125 (4) . F 0 = 0. H/Dtank = 3.tank = 1. Store/Ac = 60 kg/m2 . 255 1 αn = 0. hr = 1 [hr] . tan = 0. UL = 4 W/m2 ·C .77. LoadF low = 45 · 4/14.95. Surf AzAng = 0 [deg] .19.1 Design of a water heating system to provide 75% of a week’s loads by solar energy.2. Many of the functions used in this solution are from problem 6. The solution shown here was done with EES. Ncov = 1. ) The results are Ac =7.1. This is a double trial and error solution.75 and if the final tank temperature is equal to the initial tank temperature. It is necessary to guess both Ac and Tplus tank and see if the solar F is 0. T ableRun#. This system is probably over designed since it supplies a large F in the middle of winter. T ableRun#. Cp = 4190 [J/kg ·C] (5) the last iteration Ac = 7. ‘Day’ ) (7) hour = Lookup(‘Boulder week’ .9 m2 (6) The initial tank temperature must be entered into the first row of the Parametric Table and the solution must start in row 2.9 m2 and initial Ttank =69. which for a simulation as short as a week can have a significant effect on system performance. (finding the ‘steady periodic’ solution removes the effects of initial tank temperature. Day = Lookup(‘Boulder week’ . It would have excess capacity in the summer. ‘hour’ ) . ρ = 1000 kg/m3 . . . MJ . . I = Lookup(‘Boulder week’ . ‘I’ ) · . T ableRun#. . 0.001 kj . day. (I. HrAng. HrAng) (12) ST = ST.1 and 5. T ableRun#. β. (11) Estimation of incident and absorbed solar radiation.1: IT = IT. αn ) (13) Warnings are generated for the last hour of sunlight on January 10. KL. ρg . Accurate measurements are difficult to make near sunrise and sunset.15. Ref rInd. ‘T a’ ) HrAng = 15 [deg/hr] · (hour − 12 [hr]) − 7. Lat. Surf AzAng. The measured horizontal data is suspect since the hourly value is almost equal to that of the preceding hour.LJ.9. Ncov . Collector useful output: . Equation 2. Lat. ρg .LJ. assuming isotropic sky.5 [deg] (10) Need to use the middle of the hour. (I. (8) (9) Ta = Lookup(‘Boulder week’ . Day. β. Surf AzAng. . . m ˙ c · Cp . . ·4 1/s . x= · . 2.77778 × 10 0 UL · F 1/hr . 7.5 . (14) FR = F 0 · x · (1 − exp (−1/x)) Eqn 6. . Qu = max 0 [MJ] . Ac · FR · ST − UL · (Ttank − Ta ) · hr · . . 0036 (15) .0. MJ . . W · hr . Eqn 6.6 (16) For the tank: M assT ank = Ac · Store/Ac V olT ank = (17) M assT ank ρ (water. T = 55. p = 200) (18) Height = H/Dtank · Dia V olT ank = π · Height · (19) Dia2 4 (20) 256 2 .7. Dia2 + π · Dia · Height 4 . . . . ·6 MJ . . = Cp · M assT ank · . 1 × 10 J . Atank = 2 · π · (21) mC tank (22) . . QtankLoss = UL.tank · Atank · hr · (Ttank − T 0 a ) · . . 0.0036 . MJ . . W · hr . 21. 7. 0) Sets load flow to between 7am and 9pm. 1. 1) · If(hour. (23) The load: On = If(hour. 0. 1. 0. . . . . ·6 MJ . . M axLoad = Cp · LoadF low · (Tmin − Tmains ) · hr · . 1 × 10 J . . . . . ·6 MJ . . Cp · LoadF low · (Ttank − Tmains ) · hr · . QLoad = On · min M axLoad. 1 × 10 J . 1 68.9 m2 and initial Ttank =69. (24) (25) (26) Calculation of new tank temperature .2 ST = 0.8 SumQLoad [MJ] 0.00 0.86 [kg/hr] m ˙ c = 50 kg/m2 -hr QtankLoss = 0.7 [C] T 0 a = 21 [C] V olT ank = 0.424 [MJ] On = 0 [-] Ref rInd = 1.75 [-] hour = 24[hr] I = 0.49 [C] UL.00 [MJ] ρ = 1. 190 [J/kg-C] F 0 = 0.5888 [m] Height = 1.95 hr = 1 [hr] IT = 0.05 W/m2 -C Ac = 7.69 257 3 plus Ttank [C] 69.63 [MJ] Ta = −6.1.5 [deg] KL = 0.0125 M axLoad = 2.1 [C] x = 15.000 -3.00 [MJ] Store/Ac = 60 kg/m2 tan = 0.70 [MJ] ρg = 0.9196 HrAng = 172.00 0.00 .2 (28) SumQLoad = Tablevalue(T ABLERU N # − 1.526 SumQLoad = 178. ‘Tplus tank’ ) + Ttank = Ttank + Qu − QtankLoss − QLoad mCtank (27) Eqn 10. 000 kg/m3 Surf AzAng = 0 [deg] Tmains = 15 [C] + Ttank = 69.00 MJ/m2 M assT ank = 474 [kg] Ncov = 1 Qu = 0. Solution αn = 0.tank = 1.9 m2 Day = 14 FR = 0.00 MJ/m2 Tmin = 60 [C] UL = 4 W/m2 -C Parametric Table: Table 1 Run Day hour [hr] I 2 MJ/m Ta [C] IT 2 MJ/m ST 2 MJ/m QLoad [MJ] Qu [MJ] QtankLoss [MJ] 1 2 8 2 0.00 F [-] -0.00 0.811 m2 Dia = 0.31 β = 40 [deg] F = 0.at end of an hour’s time step: Ttank = Tablevalue(‘Table 1’ .00 0.3 0.00 0.986 [MJ/C] QLoad = 0.9.77 Ttank = 69.000 MJ/m2 LoadF low = 12.9 Cp = 4.766 [m] H/Dtank = 3 Lat = 40 [deg] mC tank = 1. T ableRun# − 1. ‘SumQ load’ ) + QLoad F = (29) SumQLoad 7 [day] · 14 [1/day] · M axLoad (30) The results are Ac =7.4809 m3 Atank = 3. 37 1.49 1.3 -8.8 42.00 0.134 0.1 48.83 26.9 0.000 0.00 0.000 0.00 0.22 0.26 0.8 -3.28 0.00 0.09 0.41 0.00 0.8 2.046 0.8 50.65 1.00 0.27 32.155 0.30 2.85 7.3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 1 2 3 4 5 6 7 8 0.00 0.23 0.000 0.000 0.2 57.31 0.00 0.32 0.8 44.27 32.01 0.8 -8.73 0.19 30.00 0.42 0.000 0.00 0.72 0.23 0.11 0.9 38.3 2.00 0.000 0.27 1.40 0.00 0.796 0.000 0.43 0.22 0.3 62.27 32.00 0.67 0.76 1.69 0.4 68.9 -3.26 54.000 0.000 0.45 0.41 0.8 -2.04 0.3 -3.42 2.02 0.66 0.23 0.00 0.67 0.00 0.00 2.68 0.00 0.00 0.27 32.6 56.000 0.00 0.000 0.26 54.00 0.636 0.54 0.00 0.00 0.00 0.00 0.26 54.7 -1.8 -2.23 0.78 5.000 0.000 0.25 0.19 0.51 0.05 28.00 0.46 1.14 0.5 39.00 0.00 0.20 0.00 0.00 0.000 0.43 0.00 0.00 0.2 51.24 68.41 1.31 0.23 0.9 49.01 0.00 0.00 0.00 0.00 0.00 0.42 2.00 0.98 0.00 0.26 54.27 9.8 3.28 1.1 38.06 0.6 -6.00 1.00 0.98 0.3 -2.017 0.00 0.8 -2.53 0.6 -1.27 0.54 0.00 0.07 2.00 0.13 0.00 0.8 0.7 38.00 0.6 -2.00 0.00 0.62 0.55 0.08 0.00 0.00 0.00 0.8 38.033 -2.8 53.7 67.23 0.60 0.2 -2.8 42.000 0.27 32.00 0.86 37.34 0.6 43.37 0.26 0.00 0.33 0.00 0.00 0.9 -3.167 0.04 1.9 -3.00 0.00 0.00 0.00 0.27 34.23 0.09 0.00 0.14 0.52 56.5 36.10 0.00 0.65 0.22 40.55 1.25 1.000 0.00 0.68 0.31 0.2 54.3 37.00 0.2 -2.3 -3.27 32.26 54.00 0.6 -0.00 0.00 0.11 0.009 0.14 0.42 0.1 2.000 0.15 2.6 42.00 0.2 2.00 0.26 55.25 0.02 258 4 0.00 0.00 2.00 0.00 0.57 1.33 2.6 -8.000 0.55 1.2 39.27 32.3 40.7 43.05 0.10 35.00 0.00 0.9 -3.25 0.9 47.00 0.5 49.00 0.49 1.01 51.00 0.07 0.31 0.3 49.17 0.53 23.5 38.00 0.000 0.00 0.1 43.14 0.40 0.14 0.12 0.47 0.60 1.86 19.00 0.22 0.39 0.2 -1.27 32.7 45.00 0.641 0.14 0.8 64.185 1.26 54.00 0.33 0.8 -3.22 2.00 0.00 0.26 54.00 0.00 0.26 54.26 0.43 0.23 0.20 0.00 0.000 0.00 0.18 0.00 -0.00 0.42 2.27 32.89 54.00 0.27 0.45 0.00 0.36 1.16 0.14 0.2 -2.52 0.71 1.1 57.6 39.29 0.38 2.2 41.000 0.3 -8.23 0.1 1.00 0.00 0.9 -3.00 0.00 3.00 0.00 0.00 0.3 50.42 1.12 14.9 -10.27 32.00 0.6 -0.8 -2.5 52.33 0.389 0.05 0.54 1.23 0.00 0.00 0.27 0.7 49.1 -1.2 -1.13 0.21 0.00 0.00 0.000 0.4 65.97 2.00 0.50 1.000 0.00 0.19 0.42 4.00 0.2 59.000 0.32 0.00 0.41 2.04 0.000 0.3 39.4 39.40 46.00 0.00 0.0 50.1 49.35 0.48 52.17 0.00 0.4 38.6 44.105 1.72 0.53 16.000 0.3 -10 -8.2 -2.00 0.58 0.000 0.41 0.2 0.4 -10 -8.06 0.7 -7.402 1.16 21.000 0.00 0.26 0.00 0.23 0.00 0.00 0.000 0.26 0.14 0.27 32.00 0.24 .00 0.004 0.00 1.000 0.00 0.81 42.2 -2.75 1.134 0.57 39.70 12.7 58.04 0.23 0.37 0.00 0.3 -9.42 0.8 3.14 0.83 0.331 0.3 -5.55 1.51 50.343 0.10 0.00 0.7 -2.7 61.00 0.00 0.2 -2.00 0.00 0.36 0.33 0.56 0.00 0.00 0.96 48.09 0.00 0.26 54.03 0.00 0.00 0.00 0.25 0.00 0.00 0.14 0.14 0.3 -0.252 0.55 1.14 0.00 0.00 0.21 0.000 0.07 0.00 0.85 45.18 0.000 0.000 0.2 -2.00 0.00 0.23 0.8 46.16 0.00 0.5 50.26 54.7 59.1 67.49 0.55 0.00 0.071 0.42 2.3 -3.00 0.53 0.00 0.00 -0.15 0.758 1.00 0.01 0.00 1.62 0.14 0.00 0.30 2.00 0.36 43.02 0.83 1.42 2.00 0. 42 2.41 14.00 0.1 6.00 1.25 0.00 0.6 -1.247 0.04 60.9 62.000 0.36 0.00 0.00 0.004 0.00 0.29 0.30 1.61 80.90 2.01 0.08 0.3 65.00 0.08 0.61 80.7 66.61 80.00 0.22 0.00 0.44 0.00 0.45 0.45 0.00 0.00 0.32 0.1 2.017 0.00 0.45 0.39 99.004 0.94 3.60 0.26 0.00 0.09 0.47 0.56 0.92 73.45 0.68 1.28 1.26 1.57 0.50 0.54 97.6 -6.31 0.89 63.03 0.39 0.52 0.42 3.6 0 0.00 0.22 1.17 0.5 59.05 111.36 0.42 0.07 0.09 0.733 1.43 0.000 0.27 0.000 0.00 0.42 2.00 0.9 40.61 80.092 0.00 0.46 0.0 41.00 0.775 0.23 0.00 0.61 80.31 0.477 -7.8 57.32 107.872 1.13 0.17 0.00 116.8 -5.2 64.00 0.42 2.9 -5 -5.352 0.2 -0.30 0.80 107.52 0.10 0.00 0.419 1.00 0.58 0.00 0.8 54.00 0.00 0.000 0.06 0.6 -2.56 0.61 104.2 42.32 107.2 43.40 0.61 80.41 1.26 2.6 -16.1 59.7 -16.34 0.00 0.1 38.00 0.00 2.25 0.00 0.9 39.38 0.58 0.2 46.000 0.34 0.45 0.34 75.2 41.00 0.1 35.8 39.71 14.34 0.000 0.00 0.000 0.65 0.00 0.00 0.000 0.00 0.64 2.00 0.00 0.62 1.61 113.11 4.39 0.7 44.33 0.24 87.6 -5 -3.00 0.32 107.7 -12.29 0.48 0.4 -8.7 41.00 0.22 0.00 0.6 -5.39 1.30 0.21 36.5 41.7 0.09 0.7 57.00 0.06 1.2 -2.00 0.32 0.45 0.77 78.25 0.51 10.000 0.6 56.00 0.45 0.32 259 5 1.570 1.163 0.00 0.9 66.00 0.42 2.35 0.00 0.042 0.00 0.7 -16.45 0.00 0.00 0.00 0.92 1.46 0.5 34.6 63.331 0.7 6.8 -15.35 112.00 0.45 0.7 49.6 36.06 0.04 0.00 0.37 0.4 -3.43 0.00 0.0 58.83 115.61 80.28 2.00 0.1 -7.00 0.4 -11.91 59.57 91.42 0.34 0.6 0.46 0.87 13.28 2.00 0.00 0.000 0.38 0.00 0.55 0.000 0.32 107.52 0.00 0.3 -11.34 0.00 0.35 2.57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 12 12 12 12 12 12 12 12 12 12 12 12 12 12 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0.19 80.000 0.02 0.39 0.41 0.61 80.00 0.1 3.0 60.00 0.00 0.33 0.05 0.32 107.00 0.57 0.00 0.42 2.22 0.00 0.44 89.3 0.34 0.28 0.00 0.54 0.71 110.74 1.30 0.44 0.66 0.00 0.45 0.1 61.45 0.34 0.57 1.042 0.6 1.00 0.32 61.61 0.99 1.4 60.138 0.00 0.000 0.00 0.00 0.00 0.00 0.29 3.04 0.00 0.00 0.00 0.61 80.23 105.00 0.50 1.34 0.2 45.00 0.00 0.81 66.7 -2.2 -1.5 48.000 0.58 1.00 0.6 61.3 37.19 100.00 0.00 0.32 107.61 95.55 0.6 -5.8 -9.32 107.65 0.18 1.34 1.61 82.34 0.30 0.48 0.00 0.9 50.8 -13.13 0.00 0.000 0.8 36.29 0.00 0.24 0.8 41.205 0.74 2.5 53.00 0.004 0.1 -11.32 107.04 2.4 6.00 0.61 2.81 2.15 0.29 0.00 0.24 0.25 0.000 0.3 61.32 108.02 0.047 1.00 0.57 1.00 0.49 0.00 0.000 0.8 59.45 0.389 0.00 0.00 0.4 41.096 0.00 0.000 0.000 0.000 0.00 0.9 -4.24 0.00 0.00 0.3 59.000 0.30 0.03 0.26 0.4 -5 -5.3 50.00 0.29 0.56 0.61 80.49 .00 6.11 0.2 4.01 0.00 0.80 1.48 2.00 0.6 0.8 -4.00 0.2 -5 -2.180 0.92 2.9 -9.1 42.23 0.6 -6.00 0.3 47.000 0.43 0.17 1.30 0.49 70.2 55.000 0.93 102.34 0.13 2.00 0.000 0.72 2.00 0.071 0.2 51.63 0.06 0.47 0.30 0.07 68.000 0.00 0.97 85.14 1.42 0.14 117.34 0.00 0.12 0.00 0.7 -7.63 93.45 0.00 0.00 0.00 0.34 0.34 2.00 0.805 1.14 1.2 -2.00 0.00 0.00 0.1 38.23 0.00 0.20 2.32 107.7 60.20 1.00 0.00 0.41 0.6 -5.1 -17.32 107.00 0.35 0.00 0.26 0.15 0.86 1.27 0.1 40.1 44.48 0.36 0.1 -5.2 -17.00 0.3 42.47 12.00 0.00 0.000 0.58 0.00 0.28 0.138 0.000 0.00 0.084 0. 452 1.68 0.52 0.02 0.00 0.42 2.58 0.42 2.7 59.00 0.00 0.79 2.01 139.000 0.15 0.31 122.91 152.61 2.00 0.1 -10.299 1.00 0.00 0.00 0.6 -1.38 4.52 0.000 0.6 -8.000 0.53 0.00 0.05 2.15 123.52 0.7 1.00 0.00 0.96 14.28 146.46 1.00 0.12 0.000 0.51 168.50 0.13 0.53 0.52 0.8 -8.3 8.66 146.18 0.5 32.00 0.16 2.00 0.000 0.000 0.1 59.00 0.1 57.7 1.14 0.00 0.49 0.000 0.51 0.24 0.00 0.14 14.00 0.14 0.76 0.66 146.31 119.66 146.00 0.00 0.64 0.27 2.8 31.00 0.42 2.62 0.35 127.00 0.000 0.26 0.2 2.53 0.39 0.000 0.3 -1.5 37.23 2.02 0.00 0.15 123.57 0.08 166.46 0.4 47.00 0.3 0 3.8 -7.8 52.00 0.00 0.1 31.1 56.62 0.00 0.52 0.66 146.48 0.00 0.8 -11.00 0.3 5.00 0.62 0.6 118.16 3.8 72.66 164.57 2.70 0.703 0.7 7.314 0.59 126.09 2.461 0.11 2.12 0.5 63.73 0.02 1.97 0.52 0.15 0.62 2.00 260 6 1.000 0.52 0.51 0.23 161.00 0.00 0.14 0.00 0.000 0.00 0.00 0.66 146.00 0.2 53.9 -1.15 123.7 7.52 0.46 4.78 176.59 0.14 0.000 0.4 73.20 0.00 0.00 0.38 0.14 0.50 0.00 0.42 2.82 150.8 -0.16 0.59 0.00 0.8 -5 -5 -1.00 0.62 0.65 0.31 2.2 70.3 31.2 31.00 0.809 2.00 0.7 4.62 0.00 0.00 0.94 0.13 128.63 0.80 125.00 2.62 0.44 0.000 0.4 -7.69 0.038 0.1 57.096 0.00 0.00 1.00 0.3 55.15 123.64 0.00 0.95 2.9 30.93 154.2 60.24 123.53 0.000 0.7 30.00 0.88 156.00 0.8 63.1 29.00 0.66 146.6 77.42 2.6 29.84 0.49 0.00 0.9 0.81 0.00 0.9 -10.57 0.00 0.00 0.00 0.837 0.99 124.00 0.86 144.42 2.00 0.34 121.1 31.00 0.8 30.00 0.15 123.39 3.00 0.62 0.111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 12 12 12 12 12 12 12 12 12 12 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 15 16 17 18 19 20 21 22 23 24 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 0.7 6.00 6.6 -10 -11.84 2.000 0.63 0.017 0.62 0.52 0.8 55.00 0.00 0.1 0 -2.53 0.00 0.07 1.00 0.9 6.00 0.51 0.000 0.55 3.52 0.62 0.07 0.00 0.15 123.00 0.00 0.03 1.884 1.000 0.52 0.50 0.00 0.00 0.15 123.8 6.03 0.43 141.78 1.89 14.000 0.9 32.7 3.59 2.1 -0.7 2.00 0.81 0.66 146.55 0.66 146.5 69.1 53.76 0.6 -9.67 0.8 30.52 0.000 0.66 15.2 3.73 2.00 0.6 30.1 29.331 0.01 0.6 30.00 0.33 130.76 0.70 3.4 73.00 0.000 0.6 -3.78 0.000 0.32 0.14 0.17 134.13 0.21 1.66 0.6 55.6 -12.00 0.04 2.000 0.12 0.00 0.00 0.608 1.00 0.926 1.87 3.4 56.00 0.15 123.42 2.968 1.7 -4.72 2.00 0.51 0.36 2.70 20.14 0.6 7.733 1.00 0.8 -2.6 -1.34 120.000 0.74 34.79 0.05 0.00 0.7 -10.110 1.66 146.00 0.00 0.53 16.71 0.00 0.34 2.62 0.00 0.48 3.61 0.54 0.00 0.15 0.00 0.74 0.35 173.54 0.00 0.724 1.2 33.42 2.45 0.70 0.6 56.91 0.7 62.15 123.9 76.00 0.2 6.42 0.25 0.72 0.14 0.00 0.93 171.06 10.55 0.66 148.6 -10.15 123.5 51.8 1.00 0.15 0.00 0.532 0.00 0.52 0.00 0.033 0.4 -7.00 0.2 -10.7 -0.00 0.00 0.73 12.4 31.75 .04 0.000 0.19 0.0 75.00 0.000 0.340 0.51 0.1 54.00 0.8 3.52 0.50 0.00 0.60 0.00 0.000 2.42 2.004 0.9 56.7 -2.4 57.2 8.52 0.68 0.66 2.51 0.66 146.00 0.16 0.57 0.00 0.42 2.23 3.42 2.7 57.00 0.00 0.58 137.20 178.15 123.00 0.000 0.12 1.03 0.000 0.71 0.4 1.24 0.000 0.09 0.05 11.00 0.93 159.43 0.71 2.0 30.17 0.00 0.15 8.00 0.42 2.00 0.96 3.00 0.00 0.53 0.059 0.00 0.00 0.74 1.47 0.1 55.00 0.62 0.38 0.56 0.56 0.00 0.750 1.49 0.62 0.000 0.00 0.9 63.80 0.61 0.15 1.06 132.00 0.61 0. 1 178.00 0.00 0.00 0.00 0.63 0.75 .63 178.63 178.000 0.71 0.9 -3.75 0.00 0.00 0.75 0.000 0.5 69.70 70.00 0.00 0.71 0.00 0.00 0.000 -3.1 -6.63 178.00 0.00 0.165 166 167 168 14 14 14 14 21 22 23 24 0.00 261 7 0.75 0.00 0.00 0.000 0.8 69.00 0.70 0.7 0.2 69.9 -6. 2 Add a collector-store HX to problem 19. Day = Lookup(‘Boulder week’ .77. Knowns: Tmin = 60 [C] . ‘hour’ ) .05 W/m2 ·C (3) KL = 0.9 (1) Ref rInd = 1. Store/Ac = 60 kg/m2 . Surf AzAng = 0 [deg] . UL = 4 W/m2 ·C .526 (2) UL. which for a simulation as short as a week can have a significant effect on system performance.95. ‘Day’ ) (8) hour = Lookup(‘Boulder week’ .1. ‘Tplus tank’ ) (7) The initial tank temperature must be entered into the first row of the Parametric Table and the solution must start in row 2.) The results are Ac =7.tank = 1. β = 40 [deg] . tan = 0. Lat = 40 [deg] .2. T 0 a = 21.9 m2 and initial Ttank =69. T ableRun#.P19 02 Equations 19. It is necessary to guess both Ac and Tplus tank and see if the solar F is 0. Ncov = 1. hr = 1 [hr] . m ˙ c = 50 kg/m2 ·hr . LoadF low = 45 · 4/14. This is a double trial and error solution. H/Dtank = 3. ρg = 0. F 0 = 0.1. αn = 0. (finding the ‘steady periodic’ solution removes the effects of initial tank temperature. T ableRun#.0125 Cp = 4190 [J/kg ·C] (4) (5) the last iteration Ac = 8.8 m2 (6) Tinitial = Tablevalue(1. ρ = 1000 kg/m3 . Tmains = 15 [C] . This system is probably over designed since it supplies a large F in the middle of winter. It would have excess capacity in the summer.75 and if the final tank temperature is equal to the initial tank temperature. . . MJ . . I = Lookup(‘Boulder week’ . ‘I’ ) · . T ableRun#. . 0.001 kj . (9) Ta = Lookup(‘Boulder week’ . HrAng) (13) ST = ST. T ableRun#. day. Equation 2. (I. Day. Surf AzAng. ρg . ρg .1: IT = IT. β. KL. Surf AzAng. αn ) (14) 262 1 . β.LJ. Ref rInd.1 and 5. Lat. HrAng.5 [deg] (10) (11) Need to use the middle of the hour. (I.9. (12) Estimation of incident and absorbed solar radiation. Ncov .15. ‘T a’ ) HrAng = 15 [deg/hr] · (hour − 12 [hr]) − 7. assuming isotropic sky.LJ. Lat. Warnings are generated for the 17th hour on January 10. Collector useful output: . The measured horizontal data is suspect since accurate measurements are difficult to make near sunrise and sunset. . . m ˙ c · Cp . . ·4 1/s . 77778 × 10 · x= . 2. 0 UL · F 1/hr . FR = F 0 · x · (1 − exp (−1/x)) (15) Eqn 6.7.5 (16) The following four equations were added and the fifth changed from problem 19.1 . . . 1/s . . ˙ = Ac · m AC ˙ c · . . 2.77778 × 10·4 1/hr . AC. AC. F RU L.35 (19) ˙ cp . HX F 0 R = FR · F R0 /F R Ac . ˙ cp . . (17) F RU L = FR · UL (18) HX = 0. . Ac · F R · ST − UL · (Ttank − Ta ) · hr · . Qu = max 0 [MJ] . . 0.0036 0 (20) . MJ . . W · hr . Eqn 6. p = 200) (23) Height = H/Dtank · Dia V olT ank = π · Height · (24) Dia2 4 (25) Dia2 + π · Dia · Height 4 .6 (21) For the tank: M assT ank = Ac · Store/Ac V olT ank = (22) M assT ank ρ (water. T = 55.7. . . MJ . . = Cp · M assT ank · . . 1 × 10·6 J . Atank = 2 · π · (26) mC tank (27) QtankLoss . . . MJ . . . = UL.tank · Atank · hr · (Ttank − T a ) · . 0036 W · hr .0. . 0 (28) The load: On = If(hour. 1. 0. 0. 1) · If(hour. 1. 0) Sets load flow to between 7am and 9pm. 21. 7. . . . ·6 MJ . . M axLoad = Cp · LoadF low · (Tmin − Tmains ) · hr · . 1 × 10 J . . . . MJ . . Cp · LoadF low · (Ttank − Tmains ) · hr · . QLoad = On · min M axLoad. . 1 × 10·6 J . at end of an hour’s time step: Ttank = Tablevalue(‘Table 1’ . (29) (30) (31) Calculation of new tank temperature . T ableRun# − 1. ‘Tplus tank’ ) 263 2 (32) . 86 [kg/hr] m ˙ c = 50 kg/m2 -hr QtankLoss = 0.8.78 Ac =7.0125 M axLoad = 2.831 [m] H/Dtank = 3 Lat = 40 [deg] mC tank = 2.5356 m3 αn = 0.00 MJ/m2 Tmains = 15 [C] + Ttank = 66.9 m2 and initial Ttank =69.tank = 1.5 [deg] KL = 0.00 [MJ] ρ = 1.71 [MJ] ρg = 0.8 m2 Day = 14 F RU L = 3.17 [C] UL.212 [MJ/C] QLoad = 0.+ Ttank = Ttank + Qu − QtankLoss − QLoad mCtank Eqn 10.31 . 000 kg/m3 Surf AzAng = 0 [deg] Tinitial = 66.096 m2 Dia = 0.7 m2 and initial Ttank =66.526 SumQLoad = 177.000 MJ/m2 LoadF low = 12.00 MJ/m2 M assT ank = 528 [kg] Ncov = 1 Qu = 0. a) epsilonHX =0.9 Cp = 4.2 ST = 0.77 Tmin = 60 [C] UL = 4 W/m2 -C Ac = 8.05 W/m2 -C 264 3 Atank = 4.95 [-] hour = 24[hr] I = 0.823 HrAng = 172.9196 hr = 1 [hr] IT = 0. 190 [J/kg-C] F = 0.00 [MJ] Store/Ac = 60 kg/m2 tan = 0.2 (33) SumQLoad = Tablevalue(T ABLERU N # − 1.1.9. When the HX is large the impact is small but for the small HX there is an 11% incresas in collector area.35 FR = 0.8 [C] x = 15.75 [-] F 0 R = 0.424 [MJ] On = 0 [-] Ref rInd = 1.6103 [m] F 0 = 0.8 [C] T 0 a = 21 [C] V olT ank = 0.7 [C] Ttank = 67. ‘SumQ load’ ) + QLoad F = (34) SumQLoad 7 [day] · 14 [1/day] · M axLoad (35) The results are a) epsilonHX =0.59 [MJ] Ta = −6.679 W/m2 -C Height = 1. Solution ˙ = 0.1222 [kg/s] AC β = 40 [deg] HX = 0.35 Ac =8. ¯ T = 10.3. so no Y correction is needed.1 Calculation of a months performance of a liquid-based heating system with f-Chart.80.96 Vstore = 1650 [l] (1) (2) (3) Load calculation with space heating estimated with Equation 9.4: The store size is 1650/22 = 75 liter/m2. H ¯ ta/ta n = .06 · 109 DegDays = 828 [C ·day] .79. so no X correction is required. Ac = 22. Days = 31 [day] . F 0 RU L = 3. F 0 Rtan = 0.P20 01 Equations 20. T¯a = −8.7 · 106 .2.6: Lhtg = U Ahouse · DegDays · 24 [hr/day] · 3600 [s/hr] (4) Load = Lhtg + Lhw (5) b Calculation of X and Y by Equations 20.3 and 20.2.00. The load heat exchanger parameter is 2. . a Lhw = 2. Knowns: U Ahouse = 145 [W/C] . . . s . . 86400 day . 0215 · Y 3 265 1 (8) .1 (or Figure 20.1): fc = 1.3.029 · Y − .3. X = F 0 RU L · 100 [C] − T¯a · Days · Ac · (6) Load ¯ ¯ Y = F 0 Rtan · ta/ta n · HT · Days · Ac /Load c (7) Calculation of solar fraction with Equation 20.245 · Y 2 + .065 · X − .0018 · X 2 + . 3.945 [-] Ye = 0.245 · Ye2 + .0215 · Ye3 (15) Solution Ac = 22 m2 fd = 0. Equation 20.5 (12) Y e/Y = .65 · exp (−.d Recalculation of X and fraction for reduced store size: Correction for non-standard storage.3: LHX parameter = 0.5 Vstore = 1. Equation 20.4451 [-] Days = 31 [day] fe = 0.065 · Xd − .2918 F 0 RU L = 3.0018 · Xd2 + .029 · Y − .3.243 × 1010 [J] ¯ ta/ta n = 0.0018 · Xd2 + .79 [-] Load = 1.96 X = 1.037 × 1010 [J] T¯a = −8 [C] Xd = 2.3927 266 2 fc = 0.0215 · Y 3 (11) e Recalculation of Y and fraction for reduced load heat exchanger size: Correction for non-standard load heat exchanger.065 · Xd − .029 · Ye − .245 · Y 2 + .060 × 109 [J] U Ahouse = 145 [W/C] Y = 0.2269 LHX parameter = 0.5 Storeactual = 37.2707 ¯ T = 1.313 .070 × 107 J/m2 -day H Lhw = 2.2: Storeactual = 75/2 (9) −.139/LHX parameter ) (13) Ye = Y · Y e/Y (14) fe = 1. 650 [l] Y e/Y = 0.8822 DegDays = 828 [C-day] F 0 Rtan = 0.8 W/m2 -c Lhtg = 1.25 Xd = X · (Storeactual /75) (10) fd = 1.39 + . (1) (2) (3) ¯ T = 15 MJ/m2 H Calculation of month’s loads by degree-day method. Ac = 100 m2 F 0 RU L = 7.95. Daysmonth = 30. Knowns: U Ahouse = 350 [W/C] . DD = 553 [C ·day] . ¯ ta/ta n = 0.0 W/m2 ·C Storeactual = 180 l/m2 F 0 Rtan = 0.P20 02 Equations 20.78.3.2 Liquid solar system f-Chart calculations for one month. T¯ = −2 [C] . Equation 9.6: . . . s . . Load = U Ahouse · DD · . . 86400 day . with Equation 20.25 Storeactual Xc/X = 75 [l/m2 ] (8) 267 1 .3: ∆t = Daysmonth · 24 [hr] · 3600 [s/hr] (6) X = F 0 RU L · 100 [C] − T¯ · ∆t · Ac /Load (7) Correction for non-standard storage. assuming standard storage capacity.2: −.2.3. Equation 20. (4) (5) Calculation of X. 2.4: .a) value of X Xa = X · Xc/X (9) b) value of Y using Equation 20. . . J . . ¯ . ¯ 1000000 · Daysmonth · Ac /Load Yb = F 0 Rtan · ta/ta · H · n T . MJ . 0018 · (Xa /2) + .029 · Yb − .0215 · (Yb /2) Solution Ac = 100 m2 fc = 0.0215 · Yb3 (11) d) Solar fraction with 1/2 the area both X and Y are reduced by 1/2.0018 · Xa2 + .78 Storeactual = 180 l/m2 X = 11. 2 2 3 fd = 1.029 · Yb /2 − .8034 (12) .55 Load = 1.95 Xc/X = 0.672 × 1010 [J] U Ahouse = 350 [W/C] Yb = 1.245 · (Yb /2) + .07 [-] 268 2 ∆t = 2.065 · Xa − .065 · Xa /2 − .994 DD = 553 [C-day] F 0 Rtan = 0.592 × 106 [s] F 0 RU L = 7 W/m2 -C ¯ ta/ta n = 0.245 · Yb2 + . (10) c) Solar fraction for Xa and Yb fc = 1.81 ¯ T = 15 MJ/m2 H T¯ = −2 [C] Xa = 8.891 Daysmonth = 30 [-] fd = 0. .96.19. H T¯a = −8 [C] . DHW = 2.data from App G and Example 2.15/31 Tref = 100 [C] For the month of January .3 a. ta/tan = 0. F ractAlow = 10.P20 03 Equations 20. F RU L = 4. Knowns: U A = 400 [W/C] . F Rtan = 0. Slope = 50 [deg] . Calculation of area required for an air system to meet 50% of January load.1: ¯ T = 11. Store/Ac = 0.6.9 MJ/m2 ·day . month = 1. DD = 828 [C ·day] (1) (2) 821 (3) nDays = N umDay (M onth) (4) ∆t = 3600 [s/hr] · 24 [hr/day] · nDays (5) Determine the Load. X and Y.25.0 W/m2 ·C . . . MJ/day . . . Load = DD · U A · . 3.0864 + DHW · nDays Eqn 9.0.6 J/s . . . . ·6 MJ . × 10 . 1 j . 4.50 and let EES solve for Ac : f=0. X = F RU L · Tref − T¯a · ∆t · Ac · Eqn 20.00187 · X 2 − 0.1 represents the air f-Chart.4 Y = F Rtan · ta/ta · H n (6) (7) (8) Equation 20.1.159 · Y 2 + 0. f = 1.50 f = 0.3 Load ¯ T · Ac · nDays/Load Eqn 20.065 · X − 0.2. and must be used in the range of values shown in Figure 20.4.0095 · Y 3 (9) Set f=0.040 · Y − 0.2.5 (10) Solution 269 1 . Ac = 113.25 U A = 400 [W/C] ∆t = 2.06935 [MJ/day] F RU L = 4 W/m2 -C nDays = 31 [day] T¯a = −8 [C] Y = 0.678 × 106 [s] F Rtan = 0.8449 [-] .601 [-] 270 2 DHW = 0.9 MJ/m2 -day H Slope = 50 [deg] Tref = 100 [C] DD = 828 [C-day] F ractAlow = 10 Load = 28. 618 [MJ] Store/Ac = 0.6 month = 1 ta/tan = 0.8 m2 f = 0.5 ¯ T = 11.96 X = 4. 19.1: ¯ T = 13. H T¯a = −8 [C] . . F Rtan = 0. F ractAlow = 10.6.25.3 a. DD = 821 [C ·day] (3) nDays = N umDay (M onth) (4) ∆t = 3600 [s/hr] · 24 [hr/day] · nDays (5) Determine the Load. month = 1. Knowns: U A = 400 [W/C] . F RU L = 4.7 MJ/m2 ·day . DHW = 2150 Tref = 100 [C] [MJ] 31 [day] (1) (2) For the month of January .0 W/m2 ·C . Calculation of area required for an air system to meet 50% of a January load. X and Y.P20 03a Equations 20. Slope = 50 [deg] . Store/Ac = 0.data from App G and Example 2. ta/tan = 0.96. . . MJ/day . . House = DD · U A · . . 0.0864 J/s . (6) HW = DHW · nDays . . . MJ/day . . Load = DD · U A · . . 3.6 J/s .0.0864 + DHW · nDays Eqn 9. . . . ·6 MJ . × 10 . 1 j . 4.4 Y = F Rtan · ta/ta · H n (7) (8) (9) (10) Equation 20.4.0095 · Y 3 (11) Set f=0.3 Load ¯ T · Ac · nDays/Load Eqn 20.00187 · X 2 − 0.1.159 · Y 2 + 0.2.50 and let EES solve for Ac : f = 0.040 · Y − 0.065 · X − 0. and must be used in the range of values shown in Figure 20.2. X = F RU L · Tref − T¯a · ∆t · Ac · Eqn 20. f = 1.5 (12) 271 1 .1 represents the air f-Chart. 96 X = 3.678 × 106 [s] F Rtan = 0.35 [MJ/day] F RU L = 4 W/m2 -C Load = 30.7 MJ/m2 -day H Slope = 50 [deg] Tref = 100 [C] 272 2 DHW = 69. 150 [MJ] nDays = 31 [day] T¯a = −8 [C] Y = 0.Solution Ac = 98.36 m2 f = 0.7883 [-] ∆t = 2. 524 [MJ] Store/Ac = 0. 374 [MJ] month = 1 ta/tan = 0.6 ¯ T = 13.5 House = 28.728 [-] DD = 821 [C-day] F ractAlow = 10 HW = 2.25 U A = 400 [W/C] . P20 03b Equations 20. -5. 1. 15. 537.2.3 b. 17.. 19. 16. Slope = 50 [deg] .8. . F RU L = 4.0 W/m2 ·C . 14.5. 15.6. 699] (5) T¯a.1 monthi = i (8) ∆ti = 3600 [s/hr] · 24 [hr/day] · nDaysi (9) nDaysi = N umDay (M onthi ) (10) Determine the Load. 20.7. 22..19.6.36 m2 (1) U A = 400 [W/C] . F ractAlow = 10.12 = [13. 2. $Arrays On Knowns: Ac = 98. 14. 656. DHW = 2150 [MJ] 31 [day] (2) Tref = 100 [C] (3) ¯ T. 232. F Rtan = 0. 17. 9.96.6. 15.8.36 m2. 136.7] H (4) DD1. Determine the annual solar fraction for a collector area of 98. X and Y. 15.2. ta/tan = 0. -5] (6) duplicate i = 1. 479. 11.5.25.12 = [-8. 280. Store/Ac = 0. 12. 25. 1.1. 30.7. 103. 16.. 12 (7) Weather data from Appendix G and Example 2. 11.12 = [821. 16.1.4. . . MJ/day . . . Housei = DDi · U A · . 0.0864 J/s . (11) 273 1 . HW i = DHW · nDaysi . . . MJ/day . . 3. + DHW · nDaysi Eqn 9.6 Loadi = DDi · U A · . . 0.0864 J/s . . . . ·6 MJ . × 10 . 1 j . F Rtan · ta/tan · H Yi = min 1.699 Slope = 50 [deg] U A = 400 [W/C] F ractAlow = 10 Store/Ac = 0.91 0.i · ∆ti · Ac · Loadi Xi ¯ T.25 Arrays Table: Main Row monthi 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 nDaysi [day] 31 28 31 30 31 30 31 31 30 31 30 31 DDi [C-day] 821 656 537 280 136 30 1 25 103 232 479 699 ¯ T.850 3. With the limits shown the solar fraction never exceeds 0.91 0.95 0.00 15.677 4.613 20.1 represents the air f-Chart.678E6 2.065 · Xi − 0.36 4.61 2.71 0.592E6 2.678E6 274 2 Housei [MJ] 28.89 13.040 · Yi − 0. fi = 1.592E6 2.91 but this high solar fraction is for months with very small loads so the annual solar fractions is not significantly impacted.678E6 2.4 12.6 16.00 15.700 1.709 11.91 0.185 3.554 24.150 2.34 1.117 2.95 0.91 0.678E6 2. i = 1.73 4.640 10.559 9.00187 · Xi2 − 0.2.04 7.7 ∆ti [s] 2.8 17.5 · 16 Loadi (12) (13) (14) Eqn 20.06 5.36 m2 F Rtan = 0.85 fi [-] 0.4.7 16.60 0.5 15. Eqn 20.678E6 2.61 2.592E6 2.22 5.2.678E6 2.96 DHW = 69.168 18.7 17.1. 12) (18) Solution Ac = 98.678E6 2.61 2.150 2.081 2.2 11.018 16.635 26.0095 · Yi3 (16) end (17) The annual solar fraction is the load weighted average monthly solar fraction.i [C] -8 -5 1 9 14 19 22 20 15 11 2 -5 Xi [-] 3. F RU L · Tref − T¯a.2.04 0.i H MJ/m2 -day 13.i · Ac · nDaysi + 1.81 0. F = Sum(fi · Loadi .8 14.150 2.92 0.374 22. i = 1.671 18.159 · Yi2 + 0.037 35 864 3. and must be used in the range of values shown in Figure 20.150 Loadi [MJ] 30.00 15.00 9.06 1.21 Yi [-] 0.014 5.757 6.419E6 2.942 2.4.79 1.2 15.6 15.50 0.23 15.44 2.150 1.592E6 2.52 .4 (15) Equation 20.94 2.524 24.5 16.35 [MJ/day] F RU L = 4 W/m2 -C Tref = 100 [C] F = 0. 12) Sum(Loadi .6 ta/tan = 0.157 HW i [MJ] 2.150 2.150 2.11 1.307 T¯a.081 2.3 Xi = min 15.081 2.560 8.61 2.081 2. 25.33.5. azim = 0 [deg] (1) F RU L = 4. 656.15 m3 /m2 Tref = 100 [C] good quality glass U Ah = 300 [W/C] (3) (4) (5) Weather data from Appendix G and Problem 2. These two quantities are then used to calcualte the monthly average τ *α product. Slope = 60 [deg] .89. H ¯ 3 = 12. 22.2.1. Ac = 40 m2 .2 (14) 275 1 .1 = 0. Day = 1 [day] (12) τn = 0. 136.7.. 20.2 = 0.8 = 0.3.0125.18 DD1.10 = 0. ¯ 12 = 5.83 (13) Figure 5.P20 04 Equations 20.36.12 = [821.04 H ¯ 11 = 6. 479. 537. 30. αn = . H ¯ 4 = 16. ρg. ¯ 6 = 23. -5] (7) ¯ 1 = 6.6 = 0. StorageV olStd = 0. -5. 232. 2. $UnitSystem SI C J kg KPa Knowns: Lat = 43 [deg] . 699] (6) T¯a.58.25 m3 /m2 .86. StorageV ol = 0.44.76.2 but other assumptions could be made. nCov = 2.05.9 = 0. 15. ρg.12 = [-8.12 = 0.7 = 0.48. ρg.11 = 0.10. H ¯ 8 = 20.2.4..4 = 0. H ¯ 10 = 10.. 19. 103.4. H ¯ 7 = 22.7. This solution uses the Liu and Jordan mode Eqn 5. ρg. 11. ρg.01 · τn · αn Eqn 5.1 tan = 1. H ¯ 2 = 9.74 H ρg.5 = 0. H ¯ 5 = 21. H ρg. V olF lowAirStd = 10 l/m2 ·s . ρg. 1.2.526.2.85 W/m2 ·C (2) V olF lowAir = 12 l/m2 ·s .2. 14.59.2. H ¯ 9 = 14. ρg. H ρg.2 ρg.3 = 0.7 (8) (9) (10) (11) It is first necessary to estimate monthly average incident and absorbed radiation on sloped surfaces.9. Ref rInd = 1. KL = 0. ρg. F Rtan = 0. 1. 280.37. 9.4 Calculation of annual performance for an air system. 5 hrs from noon(18) S¯T. KL.i (19) (20) (21) end (22) F r0 /F r = 1 No collector operating corrections. Slope.LJ.5 · 15)+sin (lat − slope)·sin (δi ) ¯ i .i . Ref rInd.i = S¯T.7a at 2. H ¯ i . ρg.LJ. Slope.3 · (StorageV ol/StorageV olStd) duplicate i = 1. 12 (15) ni = AveDay (i) (16) δi = dec (ni ) Cos θ¯b. Lat. 0. ni . N cov.i = H ¯ T.duplicate i = 1. ρg. αn S¯T.28 Xc/X = (V olF lowAir/V olF lowAirStd) −0. ni .6. (23) System corrections on storage size and air flow.i tabar/tan. H ¯ T.i = Cos(lat − slope)·Cos(δi )·Cos(2.i · tan = ¯ HT. θ¯b. 12 (25) N domi = N umDay (i) . Lat.i H (17) Eqn 1.i . . . s . . . ∆ti = Day · N domi · . 86400 day . . . . s . . . Loadi = U Ah · DDi · . 86400 day . ∆ti . Loadi . Tref . F R0 F R. Xi = Xc/X · XX F rU l. T¯a.i . Ac . (26) (27) (28) X and Y are not needed as they are also calculated internally inside the function for solar fraction. . . J . . 0 ¯ . N domi .i . tabartan. HT. Loadi Yi = Y Y F rtan . F R /F R.i · . . Ac . 1000000 MJ . F rtan . Tref . Fi = SolF ract0 A F rU l. . ∆ti . Ac . Loadi . T¯a. tabartan.i .i . F R0 /F R. . . J . . ¯ . HT.i · . N dom . V olF lowAir i MJ . StorageV ol.1000000 . 0125 Slope = 60 [deg] tan = 0.526 τn = 0. end Fannual = (24) (29) (30) (31) (32) Sum(Fi · Loadi . i = 1. 12) Sum(Loadi .7545 V olF lowAirStd = 10 l/m2 -s azim = 0 [deg] F rU l = 4.9 F rtan = 0.15 m3 /m2 Tref = 100 [C] Xc/X = 1.85 W/m2 -C Lat = 43 [deg] StorageV ol = 0.25 m3 /m2 U Ah = 300 [W/C] Day = 1 [day] Fannual = 0.76 KL = 0.83 V olF lowAir = 12 l/m2 -s . 12) (33) Solution αn = 0. i = 1.227 276 2 Ac = 40 m2 F R0 /F R = 1 N cov = 2 StorageV olStd = 0.54 Ref rInd = 1. i [C] -8 -5 1 9 14 19 22 20 15 11 2 -5 277 3 Loadi [J] 2.68E6 2.59E6 2.58 20.6 -18.59E6 2.18 11.67 3.6 45.59 10.46 0.01 11.1 21.869 3.01E9 1.60 0.734 15.06 10.86 0.69 19.89 2.98 0.47 15.76 17.2 0.74 8.67E9 6.7 37.78E8 2.98 1.2 0.08 11.86 534.2 0.9 -13.70E10 1.64 .33 14.6 36.93 0.661 12.38 ¯ Hi 2 MJ/m 6.00 0.4 39.93 10.4 18.87 1.9 -23.64 9.533 7.87 0.90 0.4 9.00 1.2 -9.80 0.26E9 3.918 78.13E10 1.92 0.25 15.3 53.00 1.71 17.53 16.53E9 7.89 12.2 0.80 0.42E6 2.23 0.67 13.4 0.88 0.7 0.48 6.5 51.69 16.04 22.79 14.Arrays Table: Main 1 2 3 4 5 6 7 8 9 10 11 12 DDi [C-Day] 821 656 537 280 136 30 1 25 103 232 479 699 δi [deg] -20.58 16.88 0.0 Row N domi ρg.2 35.433 4.57 0.59E7 6.i [deg] 35.37 5.86 16.55 64.36 23.94 0.5 2.2 0.0 -2.81E10 Xi [-] 3.7 Row ∆ti [s] 2.93 0.85 11.694 ni 17 47 75 105 135 162 198 228 258 288 318 344 Yi [-] 0.05 21.07 10.7 0.i 2 MJ/m 9.0 47.52 0.7 T¯a.44 9.555 4.59E6 2.12 22.88 0.35 0.2 0.48E8 2.94 ¯ HT.87 4.59E6 2.68E6 2.94 0.68E6 2.4 0.4 54.58 15.39E10 7.8 23.90 0.15 θ¯b.9 41.92 16.24E10 1.25 1.2 13.87 10.7 35.74 tabar\tan.i 1 2 3 4 5 6 7 8 9 10 11 12 31 28 31 30 31 30 31 31 30 31 30 31 0.92 0.68E6 ¯ ST.68E6 2.68E6 2.8 10.i 0.i 2 MJ/m 13.68E6 2.243 9.235 3.2 0.292 Fi [-] 0. H ¯ 3 = 12. H ¯ 8 = 20. (1) αn = .2 but other assumptions could be made.9. The radiation calculations are the same as Problems 2. 14.7 = 0.2. 19.72. 30. H ¯ 9 = 14.2 ρg.5 W/m2 ·C . H ¯ 5 = 21. azim = 0 [deg] LoadDHW = 1.86.74 H ρg.12 = [-8.2.18 and 20. 280.36. V olF lowAirStd = 10 l/m2 ·s .48. KL = 0.3 = 0.. Knowns: Ac = 50 m2 .1 = 0. U Ah = 300 [W/C] .2 = 0. 1. These two quantities are then used to calcualte the monthly average τ *α product.04 H ¯ 11 = 6.2.44. Lat = 43 [deg] . ρg.37. -5.11 = 0.4.89. H ¯ 7 = 22. 2. 537. 656.9 [GJ] V olF lowAir = 12 l/m2 ·s . ρg. 479. 699] (6) T¯a. 15.1.526. Slope = 60 [deg] . Ref rInd = 1. ρg. 103.5 Calculation of annual solar fraction for a combined water and space heating load. ¯ 6 = 23. nCov = 2.4 = 0. ρg. ρg.2.12 = 0. 1. This solution uses the Liu and Jordan mode Eqn 5.10 = 0.6 = 0.05. 22. -5] (7) ¯ 1 = 6.P20 05 Equations 20.58. F 0 Rtan = 0.7. Day = 1 [day] (12) 278 1 .2.. H ¯ 2 = 9. F 0 RU L = 3. H ¯ 4 = 16. ρg.10. 136.5 = 0..4 DD1. 20. 9. ρg. It is first necessary to estimate monthly average radiation on sloped surfaces.9 = 0. H ρg.12 = [821. (2) StorageV ol = 5000 [l] Tref = 100 [C] good quality glass StorageV olStd = Ac · 75 l/m2 (3) (4) (5) Weather data from Appendix G and Problem 20.33.4.0125.59.8 = 0.4. H ¯ 10 = 10. ρg.7 (8) (9) (10) (11) It is first necessary to estimate monthly average incident and absorbed radiation on sloped surfaces. H ρg.7. ¯ 12 = 5. 232. 25.2. 11. Lat.i = H ¯ T. 12 (28) N domi = N umDay (i) .6.i · tan = ¯ HT.2 (14) duplicate i = 1. αn S¯T. θ¯b. H ¯ T.5 · 15)+sin (lat − slope)·sin (δi ) ¯ i .i = Cos(lat − slope)·Cos(δi )·Cos(2.01 · τn · αn (13) Eqn 5. (23) System corrections on storage size and air flow. ni . ρg.i (20) (21) end F r0 /F r = 1 (19) (22) Collector operating corrections contained in F‘Rtan and F‘RUL. Ref rInd. N cov. ni .25 Xc/X = (StorageV ol/StorageV olStd) (24) load heat exchanger: =1 (25) · Cmin /U Ah = 2 (26) StoreCap = StorageV ol/Ac (27) duplicate i = 1.i = S¯T.i .1 tan = 1.83 Figure 5. Slope.5 hrs from noon(18) S¯T.LJ.3. −0. H ¯ i .i .LJ. KL. 12 (15) ni = AveDay (i) (16) δi = dec (ni ) Cos θ¯b.7a at 2.i H (17) Eqn 1.i tabar/tan.τn = 0. Lat. ρg.5. Slope. . . s . . . ∆ti = Day · N domi · . 86400 day . . . . . . . . s . . 9 J . . 1 × 10 + Load · Loadi = U Ah · DDi · . . 86400 DHW . . day GJ . F R0 F R.i . Tref . Loadi . Xi = Xc/X · XX F 0 RU L. X and Y are not needed as they are also calculated internally inside the function for solar fraction. Ac . ∆ti . T¯a. . . . ¯ T.i · . 1000000 J . Ac . H .i . tabartan. F R0 /F R. Loadi Yi = Y Y F 0 Rtan . N domi .. MJ . i . tabartan. Ac . ∆ti . Fi = SolF ract0 L F 0 RU l. . F R0 /F R. Tref . T¯a. F 0 Rtan .i . Loadi . . . J . . ¯ . HT.i · . . StoreCap. U Ah MJ . Cmin . N domi .1000000 . i = 1. 12) (36) Solution 279 2 . end (29) (30) (31) (32) (33) (34) (35) Find the annual solar fraction Sum(Fi · Loadi . 12) Fannual = Sum(Loadi . i = 1. 60 -18. 750 [l] Tref = 100 [C] Xc/X = 0.7 0.2 0.46 2.72 LoadDHW = 1.37 5.18 13.6 36.592E6 2.678E9 1.7545 V olF lowAirStd = 10 l/m2 -s azim = 0 [deg] =1 F 0 RU l = 3.4 0.33 14.0125 Ref rInd = 1.6165 0.000 1.2 35.92 -12.8 8.67 13.4 0.678E6 ¯ Hi 2 MJ/m 6.384 2.67 13.06 10.80 10.526 StoreCap = 100 l/m2 U Ah = 300 [W/C] Cmin = 600 [W/C] Fannual = 0.548E9 4.2 0.000 0.592E6 2.015 5.05 21.90 0.565 3.94 0.000 1.77 17.79 23.89 2.581E10 9.519 0.7 2.24 9.2 0.63 [-] Lat = 43 [deg] Slope = 60 [deg] τn = 0.007 0.0 47.73 4.678E6 2.915 12. 000 [l] tan = 0.2 0.7 37.93 10.i 2 MJ/m 9.926E9 2.93 0.04 22.9306 Ac = 50 m2 F R0 /F R = 1 KL = 0.93 0.89 12.88 0.7 35.88 0.94 .8529 1.01 11.678E6 2.i Xi Yi 17 47 75 105 135 162 198 228 258 288 318 344 31 28 31 30 31 30 31 31 30 31 30 31 0.4 54.678E6 2.000 1.25 15.2 0.425E9 2.913E9 1.5 W/m2 -C N cov = 2 StorageV olStd = 3.76 17.5 51.906 2.58 15.195 6.58 20.592E6 2.678E6 2.577 0.48 6.8243 0.22 -9.69 16.9 [GJ] StorageV ol = 5.678E6 2.032 2.2 0.36 23.158E9 5.15 280 3 ¯ ST.6866 ¯ HT.975 0.000 1.420 0.889 3.αn = 0.09 21.91 -23.3 53.53 16.44 9.570E9 7.652 0.66 12.7 7.79 14.87 0.6 45.849 1.i 2 MJ/m 13.318E10 1.288 0.74 8.7 δi [deg] -20.i [C] -8 -5 1 9 14 19 22 20 15 11 2 -5 ∆ti [s] 2.29 Fi [-] 0.05 T¯a.83 V olF lowAir = 12 l/m2 -s Arrays Table: Main Row 1 2 3 4 5 6 7 8 9 10 11 12 Row 1 2 3 4 5 6 7 8 9 10 11 12 DDi [C-day] 821 656 537 280 136 30 1 25 103 232 479 699 θ¯b.457 tabar\tan.002E10 ni N domi ρg.i 0.18 11.188 2.74 Loadi [J] 2.42 9.2 0.90 0.07 10.87 10.41 18.890E10 1.i [deg] 35.95 -2.86 16.92 0.852 4.71 17.9 Day = 1 [day] F 0 Rtan = 0.513 6.58 16.08 11.4 39.022 1.59 10.592E6 2.419E6 2.47 15.678E6 2.92 0.7 0.9 41.431E10 2.85 11.94 0. Qaux ) (1) Three cases must be considered: (1) At low temperatures the COP is less than 1: use all auxiliary. (Load − Delmax ) · hours) (11) goto 4 (12) 281 1 . $SumRow on Since there is logic required in the heat pump performance.max>Qload : HP runs only part of the time and auxiliary is zero. COP := Delmax /W ork max (2) If(COP > 1) then (3) goto 2 Case 1 Qwork := 0 (4) Qaux := Load · hours (5) Qdel := 0 (6) goto 4 (7) Case 2 2 : IF (Delmax>Load ) then goto 3 (8) Qwork := W ork max · hours (9) Qdel := Delmax · hours (10) Qaux := max (0.6 Calculation using the bin method of energy requirements to heat a building with a heat pump. (2) COP>1 and Qdel. Qwork . procedure HP (Load. hours.max<Qload : use a combination of auxiliary and Qdel. Tbin . (3) Qdel.P20 06 Equations 20. Delmax . a Procedure is used. W ork max : Qdel . Column 1 is the temperature. T ableRun#. ‘T bin’ ) (20) Hours = Lookup(‘Bins’ . Tbin = Lookup(‘Bins’ . The various energy quanities are also in the parametric table.Case 3 3 : F ractionT ime := Load/Delmax (13) Qwork := F ractionT ime · hours · W ork max (14) Qdel := Load · hours (15) Qaux := 0 (16) 4 : end HP (17) The heat pump work and delivered energy as a function of ambient temperature are placed in a lookup table.3 [C] (18) hr = 1 (19) A parametric table is made with the bins and associated hours as knowns. ‘hours’ ) (21) The Load in each bin is: . Tbal = 18. T ableRun#. Knowns: U A = 300 [W/C] . 2 is the work and 3 is the delivered energy. . . kW . . U A · (Tbal − Tbin ) · . Load = max 0. . 0.001 W. Lookuprow(‘HP’ . Qaux ) (26) Qpurchased = Qwork + Qaux (27) The month’s auxiliary energy requirement is 2545 kWh. Qwork .5 GJ Solution Variables in Main program Delmax = 7.2 GJ. Tbin ). 1. 1. ‘Q del’ ) (24) W ork max = Lookup(‘HP’ . the work is 1525 kWh. Delmax . W ork max : Qdel . ‘Work’ ) (25) call HP (Load. Lookuprow(‘HP’ . Tbin . (22) Qload = Load · hours (23) The maximum HP input and output at T=Tbin : Delmax = Lookup(‘HP’ . Tbin ). or 9.09 [kW] Qpurchased = 0 [kWh] U A = 300 [W/C] . or 5.3 [C] hr = 1 Qload = 0 [kW-Hr] Tbin = 8 [C] 282 2 Load = 3.67 [kW] Qaux = 0 [kWh] Qwork = 0 [kWh] W ork max = 3. hours.2 [kW] Hours = 0 [Hr] Qdel = 0 [kWh] Tbal = 18. 70 -16 2.69 6.25 -24 1.5 3.0 Lookup Table: HP Row Tbin [C] W ork [kW] Qdel [kW] 283 3 Qpurchased [kWh] 73 51 139 138 362 335 356 350 302 467 389 385 356 249 98 15 5 0 0 4.70 2 6.29 3.0 92.5 34.89 4.5 8.97 Hours [Hr] 5.8 1.49 4.14 -14 2.0 40.5 4.49 7.09 3.0 2.5 4 11.89 6 7.5 3 0 0 744 Load [kW] 13.26 0 5.67 8 7.7 1.0 110.03 -10 3.37 -4 4.28 8 7.29 6.09 11.0 44.8 2.2 2.5 Qdel [kWh] 0 0 0 0 57 73 103 133 149 302 332 443 578 568 227 36 11 0 0 3.1 2.25 -20 1.69 9.5 1.89 10.3 2.070 Qwork [kWh] 0 0 0 0 54 58 72 84 86 162 167 212 264 249 98 15 5 0 0 1.5 1.9 2.48 -8 3.0 43.5 2.49 10.25 -18 1.6 .5 34 40 44 43 77 76 92 110 103.25 -22 1.81 -2 5.29 12.0 3.0 11.5 46.69 3.09 8.92 -6 4.69 12.Parametric Table: Table 1 Run 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 Sum Tbin Delmax [C] [kW] -26 1.5 12 33.0 0.50 4 6.525 Qload [kW-Hr] 73 51 139 138 365 350 388 400 365 608 554 615 670 568 227 36 11 0 0 5.5 8.5 1.6 1.89 7.09 150.5 12.5 46.9 3.09 5.5 1.0 33.0 103.2 3.0 77.4 2.29 9.58 -12 3.67 -154 77.2 41.558 W ork max [kW] 1.545 Lookup Table: Bins Row 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Tbin [C] -26 -24 -22 -20 -18 -16 -14 -12 -10 -8 -6 -4 -2 0 2 4 6 8 Hours [Hr] 5.013 Qaux [kWh] 73 51 139 138 308 277 284 267 216 306 222 173 92 0 0 0 0 0 0 2.0 76. 9 1.8 3.1 2 3 4 5 6 -100 -20 0 2 20 100 1.7 6.9 3.25 5.5 1.25 1.5 2.5 2.5 10 10 284 4 . 013 [kWh] F W 0 = 0.1921 [-] W ork = 1.3205 [-] Load = 5. 067 [kWh] 285 1 F AT M 0 = 0.2677 [-] FSolar = 0. 558 [kWh] (2) (3) Equation 20. work fraction with solar F W = W ork W ithSolar /Load F AT M 0 = Load = 5558 [kWh] .2744 [-] W ork W ithSolar = 1.3. 545 [kWh] F AU X = 0.30 (1) Data from Problem 20. 781 [kWh] F W = 0. Del = 3013 [kWh] Work fraction for stand alone HP F W = F W 0 · (1 − FSolar ) (5) Non-pruchased fraction for stand alone HP F AU X = (1 − F AT M 0 − F W 0 ) · (1 − FSolar ) F AU X = AuxW ithSolar /Load (4) Find WorkW ith Solar Load − Aux − W ork Load (6) Equation 20. calculate month’s performance.all in kWh: W ork = 1525 [kWh] . 525 [kWh] Del = 3.3 . F W 0 = W ork/Load Aux = 2545 [kWh] .P20 07 Equations 20.6 .7.6. Knowns: FSolar = 0.7. aux fraction with solar (7) Find AuxW ithSolar (8) Solution Aux = 2.2. AuxW ithSolar = 1.7 Add a solar heating system in parallel with the heat pump of Problem 20. 45 [deg] · sin 360 · 365 (5) (6) ωs = (arccos (− tan (lat) · tan (δ))) (7) a = 0.5016 · sin (ωs − 60 [deg]) (8) b = 0. slope = 60 [deg] . F RU L = 4. Surf AzAng = 0 Ac = 40 m2 T¯a = −8 [C] .04: ¯ = 6. (n. ρg = 0.76. month = 1.P21 01 Equations 21.7. H F Rtan = 0. Tin = 20 [C] .1 Calculation of a month’s performance of a heating system using generalized utilizability charts. Knowns: lat = 43. From problem 20.4767 · sin (ωs − 60 [deg]) (9) ¯ T = (a + b · Cos(HrAng)) · K ¯T Lk (10) ¯ o = HZero.6609 − 0.3 [GJ] (1) (2) (3) HrAng = (T ime − 12 [hr]) · 15 [deg/hr] (4) n = AveDay (month) 284 + n δ = 23. Load = 21.85 W/m2 ·C .409 + 0.44 MJ/m2 . Lat) H (11) 286 1 . Lat. n.85 + 0.981 · 2 2 2 ¯ ¯ Cos(δ) Lk Lk T T (19) Xc = I¯T. HrAng.c /I¯T (20) Xm − 1 2 − Xm q 2 2 φ = abs abs (g) − g + (1 + 2 · g) · (1 − Xc /Xm ) g= (21) (22) hr = 1 [hr] I¯T.c = (17) (23) . Lat.169 · ¯T ¯h R Cos(slope) Lk − 0. (HrAng. (HrAng. Surf AzAng) ¯T Hd/Hbar = HDIF F BARHBAR Lat.¯ =K ¯T · H ¯o H (12) Rb = RBeam. n. (Lat. Slope. K (13) rd = rd. n) (15) rt = rt.0696 · − 0. n) ¯T · H ¯ o · (rt − Hd/Hbar · rd ) · Rb + HdHbar · rd · 1 + Cos(Slope) + ρg · rt · 1 − Cos(Slope) I¯T = K 2 2 (16) ¯h = R (14) I¯T ¯ rt · H (18) Xm = 1. . F RU L · Tin − T¯a · hr · . 0036 .0. MJ/hr . J/s . 1. (24) F Rtan · tabar/ta The hourly energy is then: QBar/Ac = F Rtan · tabarta · I¯T · φ (25) The following statement sums the QBar \Ac to get Sum for the day: Sum = Sumparametric(‘Table 1’ . T ableRun#) (26) The fraction by solar is: f = 31 · 2 · Ac · Sum · . ‘QBarA c’ . . 0.001 . GJ . 5853 F RU L = 4.5 [hr] Xm = 3.76 slope = 60 [deg] T ime = 15.3 [GJ] φ = 0.3517 [-] Hd/Hbar = 0.4091 ¯ o = 13.4819 K month = 1 QBar/Ac = 0. MJ (27) Load Problem 20.44 MJ/m2 H ¯ T = 0.32 MJ/m2 rd = 0.c = 0.709 b = 0.4071 Lk ωs = 69.04 estimated the solar fraction for January as 35%.055 tabar/ta = 0.12 [deg] Rb = 3. Solution a = 0.065 Surf AzAng = 0 [deg] Tin = 20 [C] .78 [-] Ac = 40 m2 F Rtan = 0.021 MJ/m2 287 [C] T¯a = −8 2 δ = −20.94 Xc = 0.5 [deg] I¯T.7 rt = 0.88 MJ/m2 ¯ T = 0.4885 f = 0.36 MJ/m2 H lat = 43 [deg] n = 17 ρg = 0.85 W/m2 -C HrAng = 52.585 ¯ = 6.6843 MJ/m2 Load = 21.76 hr = 1 [hr] I¯T = 0.92 [deg] g = −1. Here the estimate is also 35%.477 R Sum = 3.51 ¯ h = 2. 94 0.88 Xc [-] 0.5 15.87 1.51 QBar\A c MJ/m2 1.64 0.165 0.152 0.055 0.28 2.5 rt rd tabar\ta 0.108 0.32 0.Parametric Table: T able 1 Run 1 2 3 4 T ime [hr] 12.147 0.47 0.72 0.62 0.96 0.78 288 3 Rb φ 2.40 2.76 0.12 1.43 0.176 0.94 ¯ IT 2 MJ/m 2.72 3.5 14.96 0.36 0.32 .065 0.5 13.95 0.113 0.69 0.11 0. (ω.2 W/m2 ·C .2 and 1.12.7: hr = 1 [hr] (12) 289 1 .3 Calculation of a month’s collected energy with generalized utilizability method. n) ¯ · rd · Rb + H¯d H ¯o · K ¯ T · rt − H¯d /H ¯ · rd · 1 + Cos(Slope) + ρg · rt · 1 − Cos(Slope) I¯T = H 2 2 (10) (11) Find critical level with Equation 21.2 with K ¯ = 0.9.13. n) (9) rd = rd. (1) (2) Ta = 0 [C] . Lat. Ti = 50 [C] .39 H¯d /H (7) Find hourly radiation on tilted surface from Equation 2. (Lat. F Rta = 0.4: From Figure 2.1. Lat = 40 [deg] . Lat. ω. Knowns: F RU L = 4. n. month = 1.13. ¯ T = 0.06 MJ/m2 H Surf AzAng = 0 [deg] .4 ω = 15 [deg/hr] · Hour Hour is set in the parametric table.P21 03 Equations 21.5 K ¯ o = 15.75. Slope = 55 [deg] . Dec = −20. Slope.23. (ω. It is hours from solar noon.4: Rb = RBeam.5 and ω s < 81. Surf AzAng) (8) rt = rt. ρg = 0.2 Assumed n = AveDay (month) (3) (4) Find rt and rd from Equations 2. ωs = arccos (− tan (Lat) · tan (Dec)) (5) (6) ¯ T =0. Xc = F RU L · (Ti − Ta ) · hr · . . . 0.0036 . MJ/hr . W . 5 22.477 0.00 MJ/m2 rd = 0.95 104.5 3.00 [-] ¯ b = 2.58 0.4 (16) .5 ω [deg] 7.5 K ω = 67.00 290 2 SumQ2u MJ/m 56.19.1. Then find the hourly useful energy from Equation 21.06 MJ/m2 H Lat = 40 [deg] ωs = 71.00 φ [-] 0.77 0.9 [deg] hr = 1 [hr] ¯ T = 0. Slope) R (14) ¯ b .486 [-] Hour = 4.5 67.0148 Surf AzAng = 0 [deg] Parametric Table: Table 1 Run 1 2 3 4 5 hour [hr] 0.06 2.15 2.5 [hr] I¯T = 0.92 0. (n.091 2.5: With critical level and R Qu/Ac = F Rta · I¯T · φ (15) The monthly output is two times sum of the five hourly values times the number of days in the month: SumQu = 2 · N umDay (month) · Sumparametric(‘Table 1’ .2 rt = 0.9 133.87 1.5 4.4 142.23.14 0.11 1.3: Find R ¯b = R ¯ BEAM. 1. (13) FRta · ¯IT ¯ b from Equation 2.486 Qu \Ac2 MJ/m 0.8 142. Lat.87 Slope = 55 [deg] Ti = 50 [C] F RU L = 4.87 I¯T MJ/m2 2.31 [deg] Rb = 9.20 0.01072 Ta = 0 [C] F Rta = 0. T ableRun#) Solution Dec = −20.5 37.5 Rb 2.92 0.43 0.47 0.472 R SumQu = 142.696 1.39 H¯d /H month = 1 φ = 0.5 2.1(c).41 MJ/m2 n = 17 Qu /Ac = 0.5 [deg] ρg = 0.5 52.39 3.41 Xc [-] 0.4 MJ/m2 Xc = 2.2 W/m2 -C ¯ = 0.538 0.45 0.5 1. ‘Q uAc’ .11 9.55 0. find φ from Figure 2.75 ¯ o = 15. Slope) R (9) Find radiation on tilted surface from Equation 2. lat.P21 04 Equations 21. n.1: 1 + Cos(Slope) 1 − Cos(Slope) ¯ ¯ ¯ ¯ HT = Ho · KT · (1 − Hdbar/Hbar) · Rb + HdbarHbar · + ρg · 2 2 ¯ ¯ = HT R ¯o · K ¯T H (10) (11) Find Rb.3: Find R ¯b = R ¯ BEAM.2 and 1. Slope = 55 [deg] . Slope.2 Assumed n = AveDay (month) (3) (4) Find rt and rd at noon from Equations 2. n) (6) rd. (noon. month = 1.19. Ti = 50 [C] .noon = rt. (n. Surf AzAng) (12) From Figure 2. Dec = −20.noon = rd. n) (7) ¯ T =0. (Lat.noon (14) 291 1 . Knowns: F RU L = 4.19.4: From Figure 2. ρg = 0.06 MJ/m2 H Surf AzAng = 0 [deg] .4 ¯T Hd/H = Hdif f H Lat.75. Lat = 40 [deg] . F Rta = 0. ¯ T = 0. n.5 K ¯ o = 15.2 W/m2 ·C .13.4 Calculation of a month’s collected energy with daily utilizability method.2 with K ¯T Hdbar/Hbar = HDIF F BARHBAR Lat. K (13) Dif = rd.noon · Hd/H/rt. (noon. K (8) ¯ b from Equation 2.12.9 [deg] . (1) (2) Ta = 0 [C] .5 and ω s < 81. n.2 with ω s < 81. 0 [deg] . lat.13. Lat.4: noon = 0 [deg] (5) rt.11.noon from Equation 2.noon = RBeam.2: Rb.24. noon + Dif · 1 + Cos(Slope) 1 − Cos(Slope) + ρg · 2 2 (15) Find critical level with Equation 21.Rnoon = (1 − Dif ) · Rb.1.7: hr = 1 [hr] (16) Xc = F RU L · (Ti − Ta ) · hr · . . 0.0036 . MJ . 4 ¯ b = 2.856 R Rnoon = 1.06 MJ/m2 H month = 1 φ = 0.noon = 2.3911 ¯ T = 13.97 MJ/m2 H n = 17 RBar/RN oon = 1.2 rd.2(c) with R/R ¯ noon RBar/RN oon = R/R (18) φ = 0.5293 Hd/H = 0.noon = 0.24.31: Find φ from Figure 2.3: ndom = numDay (month) (20) ¯T · φ SumQ/Ac = ndom · F Rta · H (21) Solution Dec = −20.472 R rt.1622 SumQ/Ac = 129.noon · Rnoon · H ¯ N oon =1.5 K noon = 0 [deg] ρg = 0.75 hr = 1 [hr] Lat = 40 [deg] ndom = 31 ¯ = 1.2 W/m2-C ¯ o = 15.046 Slope = 55 [deg] Ti = 50 [C] Dif = 0.40 (19) The monthly output is found from Equation 21.5398 F Rta = 0.noon = 0.422 Surf AzAng = 0 [deg] 292 2 F RU L = 4.9 MJ/m2 Xc = 0.305 Rb.9 [deg] Hdbar/Hbar = 0.5688 ¯ T = 0.2.1744 Ta = 0 [C] . W·hr (17) ¯o · K ¯T FRta · rt. 1 T¯a = 28 [C] .5 and ω s < 81. Slope = 25 [deg] .2. month = 7. K (10) ¯ b from Equation 2. COP (1) (2) (3) (4) n = AveDay (month) (5) numDay = N umDay (month) (6) Find rt and rd at noon from Equations 2.19.7.13. ρg = 0. n) (8) rd. Load = 5.2 assumed ¯ = 20. Dec = 21. Lat = 25. Lat.13.2 and 1. Storage = 4180.1: ¯=H ¯ T /H ¯ R ¯T = H ¯ · (1 − Hdbar/Hbar) · R ¯ b + HdbarHbar · 1 + Cos(Slope) + ρg · 1 − Cos(Slope) H 2 2 293 1 (12) (13) . n. lat. Tmin = 70 [C] . (noon.63.12.2 with K ¯T Hdbar/Hbar = HDIF F BARHBAR Lat.01 MJ/m2 . Use method of Section 21. F Rtan = 0. COP = 0. (n. n) (9) ¯ T =0.3: Find R ¯b = R ¯ BEAM.3 Knowns: Ac = 25 m2 .4: noon = 0 [deg] (7) rt. ¯ T = 0. Slope) R (11) Find radiation on tilted surface from Equation 2. neglecting storage losses.72.5.noon = rd. H K Surf AzAng = 0 [deg] [GJ] . lat.19.5 Calculation of July solar contribution to operation of a n absorption air conditioner.P21 05 Equations 21.8 [deg] .noon = rt.4: From Figure 2. F RU L = 2. (noon. 1(a).: Find φmax from Figure 2. Slope.6.24. The equivalent beam radiation angle is 50 from Figure 5.11.noon · Hd/H/rt.84 and 0.2: Rb.32.2: hr = 1 [hr] XcM in (23) MJ |0.3.95 and XcM in =0.2 with ω s < 81.92 · (1 − Hdbar/Hbar) · R dif f use = 0.noon (16) 1 + Cos(Slope) 1 − Cos(Slope) + ρg · 2 2 Rnoon = (1 − Dif ) · Rb. n.32 · ρg · (18) 1 + Cos(Slope) 2 (19) 1 − Cos(Slope) 2 (20) ¯ ¯ ta/ta n · R = beam + dif f use + ground (21) ¯ = F Rtan · ta/ta ¯ F Rta n (22) Find minimum critical level with Equation 21.4.2(c) with R/R ¯ N oon = R/R ¯ noon R/R Tmin − T¯a IT c = F RU L · ¯ F Rta (25) (26) ¯ n.1 the ratios of (TauAlpha)/(T auAlpha) for these three angles are 0. (Lat. n. Surf AzAng) (14) From Figure 2. 0 [deg] . ρg .10. (27) Find X’ and Y: ∆t = 3600 [s/hr] · 24 [hr] · N umDay 0 X = Ac · F RU L · 100 [C] · ∆t · .92. Lat.0036 W·hr | F RU L · Tmin − T¯a · hr · ¯ FRta = ¯ rt. ¯b beam = 0.4.noon = RBeam.noon from Equation 2. Slope.noon + Dif · (17) Find average TauAlpha from Equation 5. IT c φ¯max = φ¯ H.1 From Figure 5.26 or from φ¯function.24.noon · Rnoon · H (24) ¯ N oon =0.2: Dif = rd. K (15) Find Rnoon from Equation 2.84 · Hdbar/Hbar · ground = 0.24.10. 0.Find Rb.4 ¯T Hd/H = Hdif f H Lat. The equiva¯ lent diffuse and ground reflected angles are 57 and 76 from Figure 5. . 1 × 10·9 (28) . GJ . J Load . . . 0.001 GJ . 3.3.3 294 2 (29) (30) . MJ ¯ ·H ¯ T · N umDay · Y = Ac · F Rta Load Eqn 21.4 Eqn 21. 2 Rb.noon = 0.85 · f ) − 1) · (1 − exp (−0.6319 F Rta Hdbar/Hbar = 0.74 ground = 0.15 · X )) · 0 !0.3436 F RU L = 2.3.noon = 0.8538 R rt.218 [-] (32) .12 MJ/m2 H Load = 7.9404 Slope = 25 [deg] ¯ ta/ta n = 0.4484 Dif = 0.5: Storage/Ac = Storage · 4.8 W/m2 month = 7 φ¯max = 0.190 [kJ/kg] Ac (31) f = Y · φ¯max − 0. from Equation 21.8 [deg] noon = 0 [deg] ¯ = 0.015 · (exp (3. 180 [kg] T¯a = 28 [C] Y = 1.5 K n = 198 ρg = 0.6458 ¯ b = 0.6 kJ/m2 Tmin = 70 [C] beam = 0.01 MJ/m2 H Lat = 25.1287 Surf AzAng = 0 [deg] XcM in = 0.6083 IT c = 174.9058 R rd.63 W/m2 -C hr = 1 [hr] ¯ T = 0.4291 ¯ T = 18.1185 Storage = 4.noon = 0.002998 ¯ = 20.77 kJ/m2 350 Storage/Ac Solution Ac = 25 m2 ∆t = 2.678 × 106 [s] ¯ = 0.56 F Rtan = 0.Find the solar fraction.7 dif f use = 0.2554 COP = 0.417 [-] 295 3 Dec = 21.286 [GJ] numDay = 31 ¯ N oon = 0.2 [deg] f = 0.8777 X 0 = 2. f.72 Hd/H = 0.9466 R/R Rnoon = 0.9569 Storage/Ac = 700. 6 kJ/m2 Tmin = 70 [C] X 0 = 1.8089 [GJ] T¯a = 28 [C] U Ast = 7 [W/C] Y = 0.4678 ¯ T = 18.904 [-] COP = 0.1 MJ/m2 H Loado = 8.5808 Loadhouse = 5.P21 06b Solution Variables in Main program Ac = 25 m2 ¯ = 0.9584 [-] 296 1 ∆t = 2.947 R/R ¯ Ts = 71.5 month = 7 φ¯max = 0.14 [C] Xc = 0.441 [GJ] φ¯ = 0.7 F RU L = 2.678 × 106 [s] fT L = 0.63 W/m2 -C Load = 9. 180 [kg] Tmax = 72.1 [GJ] P arta $ = ‘False’ ¯ N oon = 0.29 [C] XcM in = 0.2689 f = 0.255 .25 [GJ] numDay = 31 Qst = 0.625 Storage/Ac = 700.606 Storage = 4.632 F Rta ¯ KT = 0. work through the radiation calcuLations.082 − 3. Y.294 · Y )) (12) ACP S = min (P · X · P hiBarc + (1 − P ) · (3. and apply unutilizability ideas to building performance calcuLations.031 · K) · K (2) b = −4. P hiBarc .13 0.10: . function P hiBar (X. P hiBaru ) Equations 22.652 (11) P = (1 − exp (−.943 − (9.306 − 2.271 − 4.5.652 (7) P = (1 − exp (−0.170 − K · (0. In this problem we start with meteorological data from Appendix G.142 · P hiBar) · (1 − exp (−0. P hiBar) Equations 22. R.345 + K · (8.082 − 3.329 · X)) .P22 01 Equations 22.936 · K) (4) P hiBar= exp ((a + b · R) · (X · (1 + c · X))) (5) end φ¯ (6) function DirectGain (X.3. K) Equation 2.853 − 3.24.1 Estimation of a month’s auxiliary energy requirement for a direct-gain building.142 · P hiBaru ) · (1 − exp (−.329 · X))) (9) end DirectGain (10) function ACP S (X.4: (1) a = 2. P · X + (1 − P ) · (3. Y. 1) (13) end ACP S (14) Knowns: 297 1 .294 · Y )) (8) DirectGain= min (1.602 · K) (3) c = 0. N cov = 2. H. Lat = 35 [deg] . Ref rInd. Slope.4 (16) Awindow = 16. KL. Lat) H (24) ¯ T = H/ ¯ H ¯o K (25) Find incident and absorbed radiation on window.10 for the effective beam incidence angle. (n. Uwindow = 3.5 hour approximation of section 5.2 [MJ/C] . αn = . H (26) ¯=H ¯ T /H ¯ R (27) Use the 2. Tbase = 18.3 [C] ρg = 0. (15) (17) ∆T b = 5 [C] (18) Assume the following: KL = 0. n. H.5 [deg] . H T¯a = 6 [C] . ρg .1 m2 mchouse = 20. (20) n = AveDay (M onth) (21) DaysinM onth = N umDay (M onth) (22) dec = Dec (n) (23) ¯ o = HZero. Azimuth) ¯ Lat.Slope = 90 [deg] . Azimuth = 0 [deg] . ρg ¯T = H ¯ T. Slope.526. month = 2 Ref rInd = 1. n. αn S¯T = S¯T. Cos(IncAngEf f Beam) = cosIncAng (Lat.97 (19) All 12 months could be evaluatedby using a parametric table with Month=tablerun#. IncAngEf f Beam. n. 37.LJ.75 MJ/m2 .LJ. ¯ Lat.0125. U Ahouse = 200 [W/C] . ¯T ¯ = S¯T /H ta (28) (29) (30) Utilizability calcuLations: .45 W/m2 ·C . ¯ = 9. DD = 346. Slope. N cov. . . MJ . . . Load = U Ahouse · DD · 24 [hr/day] · . 0.0036 W · hr . Itc = U Ahouse · Tbase − T¯a ¯ Awindow · ta (31) Eqn 22. n) (34) rd. (ωnoon . n) (35) ¯T Hd/H = HDIF F H Lat.3. lat.5 (32) ωnoon = 0 [deg] (33) rt.n = rt. K (36) 298 2 . (ωnoon . n. lat.n = rd. Slope.24.2 Rn = (1 − (rd. n. Azimuth) Eqn 2.Rb.n /rt.n ) · Hd/H) · Rb.n + (rd.n ) · HdH · 1 + Cos(Slope) 1 − Cos(Slope) + ρg · 2 2 ¯ RnORbar = Rn /R ¯ c = Itc · 1 [hr] · X LaZ MJ W·hr .n /rt. (Lat. ωnoon .n = RBeam. . n · Rn · H (40) (41) For a zero thermal capacity building: = Load − 1 − φ¯ · Awindow · S¯T · DaysinM onth FzeroCap = Eqn 2. Eqn 2.3 ¯ rt.24.2(38) (39) .24. . n = 0.7908 ¯ = 0.3.8 Eqn 22.33 MJ/m2 H KL = 0.7279 IncAngEf f Beam = 51. 079 [MJ] n = 47 QD = 1. φ¯ Eqn 22.4851 ¯ = 9.539 Hd/H = 0.9 W/m2 Load = 5.16 [deg] Lat = 35 [deg] LaZ = 3.5974 ¯ o = 23.7176 ta Uwindow = 3.136 R RnORbar = 0.3.95 [deg] Fusef ul.5616 X .4179 K LaDeltaT 5 = 3.0036 ¯ c . 979 [MJ] mchouse = 20. φ¯ (50) LaDeltaT 10 = Load · (1 − FDoubleDeltaT ) (51) Solution Variables in Main program αn = 0.2 [MJ/C] ωnoon = 0 [deg] ρg = 0.4 Rn = 0.n = 0. RnORbar .12 (47) (48) LaDeltaT 5 = Load · (1 − FF initeCap ) (49) c For a building with finite thermal capacity with a temperature swing of 10C: FDoubleDeltaT = DirectGain X.3. 442 [MJ] Rb.07 MJ/m2 H ¯ T = 0.75 MJ/m2 H Itc = 212.solar = ¯ ST · Awindow · DaysInM onth b (44) For a building with finite thermal capacity with a temperature swing of 5C: ¯ T · DaysinM onth · Awindow · ta/Load ¯ X=H QD = Awindow · S¯T · DaysinM onth · φ¯ Y = mchouse · ∆T b · DaysinM onth/QD FF initeCap = DirectGain X.1445 Slope = 90 [deg] Tbase = 18.n = 1. Y.3.7 (45) (46) Eqn 22.1 m2 ∆Tb = 5 [C] FzeroCap = 0.5991 Azimuth = 0 [deg] dec = −12.4026 ¯ = 1.1559 T¯a = 6 [C] X = 0.898 S¯T = 7.109 rt. 839 [MJ] N cov = 2 Ref rInd = 1.3 [C] ¯ c = 0. 2 · Y. K ¯T φ¯ = P hiBar X a (37) Load − LaZ Load Eqn 22.solar= 0.961 Awindow = 16.945 MJ/m2 U Ahouse = 200 [W/C] Y = 1.45 W/m2 -C 299 3 DaysinM onth = 28 FDoubleDeltaT = 0.526 rd.97 DD = 346 [C-day] FF initeCap = 0.3 (42) (43) The fraction of absorbed energy that reduces auxiliary energy demand is: Load − LaZ Fusef ul. 756 [MJ] month = 2 φ¯ = 0.0125 LaDeltaT 10 = 2.3579 ¯ T = 11.9 Eqn 22.0. ¯ T = 0.3 [C] Slope = 55 [deg] .n = 0.174. P hiBaru ) Equations 22.853 − 3.306 − 2. R. Ta = −3 [C] .082 − 3.142 · P hiBaru ) · (1 − exp (−.62.271 − 4.943 − (9.63 MJ/m2 .1 and other data are: ¯ = 6.1 W/m2 ·C . function P hiBar (X.P22 02 Equations 22. DD = 657 [C ·Day] .602 · K) (3) c = 0.81. ta T¯i = 21 [C] .294 · Y )) (8) ACP S = min (P · X · P hiBarc + (1 − P ) · (3.436. 1) (9) end ACP S (10) Data for Jan from Example 22. Lat = 40 [deg] .5.4: (1) a = 2.345 + K · (8.50 MJ/m2 Tbase = 18. ρg = . (12) Azimuth = 0 [deg] . ∆T b = 5 [C] .3 DaysinM onth = 31 (13) (14) (15) .10: (7) .30. S¯ = 9. P hiBarc . hr = 1 [hr] .329 · X)) .24.2 Estimation of a month’s auxiliary energy requirements for an ACPS system. K) Equation 2. rt.936 · K) (4) P hiBar= exp ((a + b · R) · (X · (1 + c · X))) (5) end φ¯ (6) function ACP S (X. αn = 0.846 300 1 (11) ¯ = 0. R U Ah = 175 [W/C] . F RT An = 0. Rn = 1.764.652 P = (1 − exp (−.5. F RU L = 3.93. H ¯ = 1.170 − K · (0. Y. K tan = 0.031 · K) · K (2) b = −4. ¯ = F RT An · ta/ta ¯ F Rta n (16) FR = F RT An /tan (17) ¯ Rn/Rbar = Rn /R (18) Critical radiation levels and utilizabilities: T¯i − Ta 2 IT cc = max F RU L · . 0 W/m ¯ F Rta . . . 0.0036 MJ . w·hr Xcc = IT cc · hr · ¯ rt.2 (19) (20) (21) U Ah · (Tbase − Ta ) + Area · F RU L · T¯i − Ta ¯ Area · F Rta . K Eqn 22.5.n · Rn · H ¯T P hiBarc = P hiBar Xcc . Rn/Rbar . . . 0036 MJ .0. Rn/Rbar .8 (24) (26) Load and dumped energy for January: QD = Area · FR · S¯ · DaysinM onth · P hiBard Eqn 22.5.4 P hiBaru = 1 − P hiBarc + P hiBard Eqn 22.n · Rn · H ¯T P hiBard = P hiBar Xcd .5.5.5 . K IT cd = max ! 2 . 0 W/m Eqn 22. w·hr Xcd = IT cd · hr · ¯ rt. . . MJ . . Load = 24 [hr/day] · U Ah · DD · . . 0036 W · hr .0. Eqn 22. Y. P hiBaru ) . P hiBarc .9 (27) (28) (29) f = ACP S (X.5. . . GJ . . LAux = Load · (1 − f ) · . . 0.001 Eqn 22.11 MJ .5. b With collector area doubled and medium construction The January auxiliary decreased from 7.2 GJ in Example 22.5.2 GJ in Example 22. c With doubled collector area and heavy construction The January auxiliary decreased from 7. Thus these two changes probably would not be cost effective. The percentage change on an annual basis will certainly be even smaller.1 to 7. a (23) (25) X = S¯ · DaysinM onth · Area · FR /Load Y = Capbldg · ∆T b · DaysinM onth/QD (22) (30) (31) For the building with heavy’ construction The January auxiliary decreased from 7. d Advisability of the changes The large increase in building mass and collector area do not make large changes in the January auxiliary energy.2 to 5.0 GJ when the collector area is doubled.1 GJ on doubling the collector area and with heavier construction. The larger building capacity will result in smaller interior temperature swings and thus more comfortable conditions.5. 301 2 .1 GJ when heavy construction is assumed.1 to 6. 3 W/m2 LAux = 5.7329 IT cd = 282.6665 Rn = 1. 771 [MJ] Rn /Rbar = 0.1 Area 2 m 20 40 40 P hiBard P hiBaru 0.33 1.771 Y X 17.436 K P hiBarc = 0.4345 0.737 ¯ = 0.n = 0. 934 [MJ] P hiBaru = 0.3 W/m2 Load = 9.9 62.846 U Ah = 175 [W/C] Y = 3.81 ta Tbase = 18.93 DaysinM onth = 31 ¯ = 0.1 22.1 6.321 0.62 ¯ = 6.5 2.764 R Slope = 55 [deg] Ta = −3 [C] X = 0.3 rt.473 Area = 40 m2 DD = 657 [C-Day] F RT An = 0.8691 0.473 0.1 [GJ] QD = 2.174 tan = 0.5 MJ/m2 T¯i = 21 [C] Xcc = 0.49 .4742 0.6545 ρg = 0.6665 302 3 QD [MJ] 555.3009 [-] Capbldg = 62.321 0.771 2.3 [C] Xcd = 0.0 5.1 [MJ/C] f = 0.8691 Azimuth = 0 [deg] ∆Tb = 5 [C] F RU L = 3.28 0.40 0.321 ¯ = 1.Solution αn = 0.63 MJ/m2 H Lat = 40 [deg] P hiBard = 0.1 f 0.1287 0.5936 F Rta hr = 1 [hr] ¯ T = 0.6777 [-] Parametric Table: Table 1 Run P art$ 1 2 3 a b c Capbldg [MJ/C] 62.281 3.49 FR = 0.1 W/m2-C IT cc = 125.3 S¯ = 9.8691 LAux [GJ] 7.6665 0. Area = 15 [m] · 1. H f¯i = 0. solution to this problem would be improved by using hourly utilizability.936 · K) (4) P hiBar= exp ((a + b · R) · (X · (1 + c · X))) (5) end φ¯ (6) Knowns: Lat = 40 [deg] .2 W/m2 ·C . Frs = 0. K) Equation 2. Ref Index = 1.P22 03 Equations 22. Nglazings = 2.271 − 4.345 + K · (8. Uw = 3. then the and the the monthly average absorbed solar radiation. θ¯b = 57 [deg] .170 − K · (0. function P hiBar (X. With this assumption we can use monthly utilizbility ¯ If hourly shading algorithms were readily available. T¯a = 3 [C] .306 − 2. S. H ¯ d = 6. H (7) ρg = 0. αroom = . Thus the critical level must be: IT c = Uw · T¯i − T¯a (12) ¯ Calculate H nDay = AveDay (month) (13) Dec = Dec (N day) (14) 303 1 .93 MJ/m2 .96.67.87. M onth = 3.24. Slope = 90 [deg] .1 ¯ = 19.2 (8) (9) (10) ¯ b = 13.943 − (9.3 CalcuLation of month’s gain of a perfectly controlled direct gain system. KL = 0. Azimuth = 0 [deg] .602 · K) (3) c = 0.031 · K) · K (2) b = −4.526.42 (11) The monthly utilizability method can be used to estimate the maximum useful gain for the month if we assume that the effects of shading and transmission are constant throughout the month. the solar energy per unit area absorbed inside of the building must be at least equal to Uw *(T¯i -T¯a ) or there will be a net loss.25 [m] θ¯d = 59 [deg] From Example 14. T¯i = 20 [C] .4: (1) a = 2.5. R.853 − 3.0370.26. With a window loss coefficient of Uw . θ¯d .n = RBeam. KL. K (28) (29) Rn = (1 − rd. KL. Slope. KL. Ref Index τdif f = T ransIncAng Nglazings . Lat.n ) · Rb. Ref Index τgrnd = T ransIncAng Nglazings .n · HdH/rt. θ¯b . (HrAng n .n = rd. nDay. θ¯d . HrAng n .n · Hd/H/rt.n = rt.¯b = R ¯ BEAM. Lat) (16) ¯T = H ¯b · R ¯ b · f¯i + H ¯ d · Frs + H ¯ · ρg · 1 − Cos(Slope) H 2 ¯ ¯ ¯ R = HT /H (17) (18) ¯ CalcuLate S¯ and T auAlpha: τBeam = T ransIncAng Nglazings . nDay) (27) rt. (HrAng n .n · 1 − Cos(Slope) 1 + Cos(Slope) + ρg · 2 2 ¯ = Rn /R ¯ Rn/R (30) (31) Utilizability calcuLations: hr = 1 [hr] ¯ c = IT c · hr · X (32) . (Lat. nDay.n + rd. Lat. Ref Index 1 − Cos(Slope) ¯ ¯ ¯ ¯ ¯ ¯ S = αroom · fi · Hb · Rb · τBeam + Hd · τDif f · Frs + H · ρg · τGrnd · 2 ¯ ¯ H ¯T T auAlpha = S/ (19) (20) (21) (22) (23) Noon values on an average day: HrAng n = 0 [deg] (24) ¯ T = H/H ¯ o K (25) Rb. (nDay. Lat. nDay) ¯T Hd/H = Hdif f H Lat. Azimuth) (26) rd. (nDay. Slope) R (15) Ho = HZero. . 0036 MJ W·hr .0. . Eqn 2.Rn \R. ¯ T =H ¯ b *R ¯ b +H ¯ d /2+H*ρ ¯ g *(1-cos(Slope))/2 = 18.Alternate *3600/(rtn *Rn *H*1e6 =0.24. The two methods give nearly the same results.120 ¯ ¯ ¯ ¯ P hialternate =PhiBar( X c .8 W ¯ XBarc =IT c. K ¯T φ¯ = P hiBar X (33) (34) The maximum useful energy gain is given by: QuM ax = Area · S¯ · φ¯ (35) An alternate solution is given below in which calcuLations are based upon radiation incident upon an unshaded vertical window.K T ) = 0.3 except that Sbar is used in place of Rn *HBar ¯ rt. transmittance and room absorptance.2 MJ/m2 H Monthly average shading.n · S ¯ c . Rn/R ¯ . transmittance and absorptance: ¯ ¯ tBar.82 ¯ ¯ T *φ¯*f T auAlpha QuM ax.60 f T auAlpha= S/H ¯ ¯ ¯ IT c =Uw *(T i -T a )/f T auAlpha = 90.alternate =Area*H = 168 MJ/day 304 2 .noshade = 0. The critical level is increased by the monthly average effects of shading. Solution Variables in Main program αroom = 0.1239 X .7164 θ¯d = 59 [deg] ¯ c = 0.18 Rn /R ¯ T auAlpha = 0.7019 T¯a = 3 [C] Area = 18.96 f¯i = 0.44 MJ/m2 Lat = 40 [deg] φ¯ = 0.418 [deg] hr = 1 [hr] ¯ d = 6.8086 ¯ = 0.99 MJ/m2 H ¯ T = 0.956 R ¯ = 1.1336 Slope = 90 [deg] τdif f = 0.037 nDay = 75 Ref Index = 1.1445 τBeam = 0.93 MJ/m2 H Ho = 27.4 W/m2 M onth = 3 QuM ax = 165.7263 K Nglazings = 2 ρg = 0.2 rd.42 ¯ = 19.526 Rb.9137 rt.n = 0.175 ¯ b = 13.67 MJ/m2 H IT c = 54.6838 θ¯b = 57 [deg] Uw = 3.2 W/m2 -C 305 3 Dec = −2.93 MJ/m2 τgrnd = 0.n = 0.n = 0.75 m2 Frs = 0.7019 T¯i = 20 [C] Azimuth = 0 [deg] Hd/H = 0.9468 S¯ = 10.26 MJ/m2 H KL = 0.8024 R Rn = 0.87 HrAng n = 0 [deg] ¯ T = 15.8 [MJ] ¯ b = 0. 6 [A] (1) Io = .2 (7) V = I · Rload (8) Solution a = 23.6 [A] Rload = 20 [Ω] 306 1 .1 (6) Eqn 23.6 [V] (3) Rs = 0.6 [V] Io = 0.2.9 [ohm] (4) Rload = 20 [ohm] (5) Cell Equations for I-V and P-V: V + I · Rs I = IL − Io · exp −1 a P =I ·V Eqn 23.P23 01 Equations 23.9 [V] IL = 13.008 [A] Rs = 0.9 [Ω] I = 7.008 [A] (2) a = 23.496 [A] P = 1124 [W] V = 149.1 Calculation of power from known array conditions and constants Knowns: IL = 13.2. 26 [A] . µV oc = −0. Vmp.6 [A] .ref = 11. Voc. Imp.6 [V] . k = k#.4 [V] . GT. (1) (4) C = 0.716 [V/K] .ref .ref = 13.2 Calculation of maximum power from known array conditions and constants $UnitSystem J K Pa Knowns: Isc.P23 02 Equations 23.ref = 186.0002677 [1/K] .02 [A/K] Ns = 300 (2) Tc. Eg.ref = 1000 W/m2 . µIsc = 0.ref = 131. GT = 800 W/m2 .ref = 298 [K] (3) Tc = 303 [K] q = e#. . . . ·19 J . . = 1.12 [eV] · . 60207 × 10 eV .1. 71 [ohm] (6) aref = 8.ref )) Rsh /Rsh.ref = 14.2.11 (15) 307 1 .ref − Eg /Tc )) Eg = Eg.ref = GT.9 3 Io = Io.ref + µIsc · (Tc − Tc.8 (11) IL = (GT /GT.ref Eqn 23.1022 [V] (7) IL.ref = 82.2.177 × 10−9 [A] (9) Rs.ref /Tc.10b (12) Eqn 23.09 [A] (8) Io.ref · (1 − C · (Tc − Tc.995 [ohm] (10) Temperature and radiation dependent parameters are: a/aref = Tc /Tc.2.ref · (Tc /Tc.ref )) Eqn 23.ref ) · exp ((1/k) · (Eg.2.ref /GT Eqn 23. (5) The following reference parameters were found from the EES program FINDREFPAR Rsh.ref ) · (IL.2.ref = 2.10a (13) (14) Eqn 23.ref = 1. 6 [A] Ns = 300 Rsh = 103.1 (17) Eqn 23.995 [Ω] Vmp.35 [A] Io.0002677 [1/K] GT.02 [A/K] q = 1.691 × 10−9 [A] µIsc = 0.ref = 11.716 [V/K] Rs = 2.ref = 82.ref = 298 [K] C = 0.4 [Ω] Tc.2. Solution a = 8.09 [A] Isc.792 × 10−19 [J] I = 9. Allow V to vary between 0 and 150 V.ref = 186.ref = 131.ref = 1.102 [V] GT = 800 W/m2 IL.ref = 1.ref = 2.12 (16) Cell Equations for I-V and P-V: V + I · Rs V + I · Rs I = IL − Io · exp −1 − a Rsh P =I ·V Eqn 23.168 [A] Io = 2.26 [A] k = 1.2 (18) The easiest way to find the maximum power with EES is to use the Min/Max solution method to maximize P with respect to V.2. 206 [W] Rsh.177 × 10−9 [A] µV oc = −0.Rs = Rs. 000 W/m2 Imp.602 × 10−19 [Coulomb] Rs.381 × 10−23 [J/K] P = 1.ref = 14.ref Eqn 23.ref = 13.995 [Ω] Tc = 303 [K] Voc.5 [V] 308 2 Eg = 1.71 [Ω] V = 131.2.238 [V] Eg.4 [V] .6 [V] aref = 8.ref = 1.794 × 10−19 [J] IL = 11. 2.549.2.2 · Vmax. and 12 ohms.782 × 10−9 . 309 1 . 7.4 Eqn 23. the combination of Ms=2 & Mp=4 is best but Ms=4 and Mp=2 is essentially the same but Ms=3 & Mp=2 is a close second with only 6 cells.2 (11) For Rload=4.3 Calculation of best series/parallel connections for resistive loads of 4.P23 03 Equations 23. and Mp is # strings in parallel: V 2 = Ms · V (3) I2 = M p · I (4) V 2 = I2 · Rload (5) P = I2 · V 2 (6) Imax Vmax + Imax · Rs Vmax + Imax · Rs −1 − = IL − Io · exp a Rsh Eqn 23.1 IL = 4.567. For Rload=12. Io = 4.2 · Rmax (10) Pmax = Imax. I-V for an individual module: V + I · Rs V + I · Rs I = IL − Io · exp −1 − a Rsh a = 0.2 = Imax. Knowns from Example 23.2. Ms=4 and Mp=2 is best.2 = Vmax · M s (9) Vmax.2 = Imax · M p (8) Vmax. For Rload=7.1 (1) (2) I-V for various series and parallel combinations: (Ms is # modules in series in each string.9352. Rs = 0. Ms=4 & Mp=2 is best. Rsh = 52.1 (7) Imax. 9 300.5 188.6 370.6 310 2 .6 226.Parametric Table: Table 1 Run Ms Mp 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 2 3 2 4 1 2 3 2 4 1 2 3 2 4 1 2 2 4 2 1 2 2 4 2 1 2 2 4 2 Rload [Ω] 4 4 4 4 4 7 7 7 7 7 12 12 12 12 12 P [W] 56.7 456.9 225.8 41.6 292.4 321.0 27.0 107.6 166.7 115.1 300. By definition.92 GT N OCT = 800 W/m2 TaN OCT = 20 [C] . (1) (2) a. α = 0.2.3. but probably ranges between 5 and 15%.3: The efficiency is unknown.85 [C] TcN OCT = 47.3. UL = 25 W/m2 ·C . GT = 1100.45 [C] UL = 25 W/m2 -C ηmax = 0.05. 100 W/m2 T cM in = 66.15 (3) T cM in = Ta + (GT · τ · α/UL ) · 1 − ηmax τ ·α Eqn 23. ηnotc = 0. τ = 0.2 (6) τ ·α Solution α = 0. the cell efficiency at NOCT conditions is zero. Calculate the cell temperature at NOCT conditions with Equation 23.92 GT = 1. ηnotc TcN OCT = TaN OCT + (GT N OCT · τ · α/UL ) · 1 − Eqn 23. Knowns Ta = 35.94 TaN OCT = 20 [C] 311 1 ηnotc = 0 T cM ax = 70.P23 04 Equations 23. Calculate cell operating temperature with Equation 23.67 [C] .05 τ = 0.4 Calculation of cell operating temperature and NOCT. This fixes minimum and maximum possible temperatures: ηmin = .3 (4) ηmin T cM ax = Ta + (GT · τ · α/UL ) · 1 − τ ·α (5) b.3.15 GT N OCT = 800 W/m2 Ta = 35 [C] ηmin = 0.3.94. ηmax = 0. Tc. Voc = 15 [V] .0005 [A/K] .596 [ohm] (8) Calculate parameters at operating conditions: a/aref = Tc /Tc.ref = 1.2.P23 05 Equations 23.ref = 0. GT.ref ) · exp .5 Power calculation from data at reference conditions.ref = 1.5 [V] .ref + µIsc · (Tc − Tc.2 [A] .ref = (Tc /Tc.ref )) 3 Io /Io.ref = 298 [K] . Tc = 288 [K] .ref Eqn 23.21 [A] (6) Io. µV oc = −.464 × 10−11 [A] (7) Rs. Imp = 1 [A] Eg. N s = 24.ref = 78.12 [eV] RLoad = 10 [ohm] (1) (2) (3) Use the EES program 23. GT = 750 W/m2 .ref = 1.RefProp to find the reference parameters.601 [V] (5) IL.8 (9) IL = (GT /GT. µIsc = . Rsh.ref = 1000 W/m2 . $Units J K Knowns: Isc = 1.ref ) · (IL.05 [V/K] .78 [ohm] (4) aref = 0. Vmp = 12. . ref ) Eqn 23.2.1.ref = 1 − C · (Tc − Tc.60207 × 10·19 k# Eg /Eg.9 ! . J . ref − Eg /Tc ) Eqn 23. eV (10) · (Eg.0002677 (13) Rsh /Rsh.2.10b Eqn 23.ref = GT.2.10a (11) (12) C = 0.ref /GT (14) 312 1 .ref /Tc. 0005 [A/K] Pmax = 9.2.9038 [A] Io = 2.706 [W] RLoad = 10 [Ω] Rsh.05 [V/K] aref = 0.78 [Ω] V = 8.ref = 0.ref = 1.7467 [A] Isc = 1.2.12 (15) a Find Power with Rload Cell Equations for I-V and P-V: V + I · Rs V + I · Rs −1 − I = IL − Io · exp a Rsh P =I ·V Eqn 23.209 [V] Rmax = 17.1 Imax = IL − Io · exp a Rsh Pmax = Imax · Vmax Eqn 23.2 (17) V = I · RLoad (18) b Find Rmax that maximizes the Power.21 [A] Io.ref = 298 [K] Voc = 15 [V] 313 2 .2 (19) (20) Vmax = Imax · Rmax (21) Solution a = 0. The EES Min/Max option was used to find the value of Rmax that maximizes the power.464 × 10−11 [A] N s = 24 C = 0.2.2.ref = 1.5808 [V] Eg.ref = 78.576 × 10−12 [A] µV oc = −0.1 (16) Eqn 23.00 [V] Rs = 0.601 [V] GT = 750 W/m2 IL. 000 W/m2 Imax = 0.739 [W] Eg = 1. Vmax + Imax · Rs Vmax + Imax · Rs −1 − Eqn 23.2 [A] P = 6.123 [eV] I = 0.12 [eV] IL = 0.8209 [A] Imp = 1 [A] µIsc = 0.0002677 [1/K] GT.596 [Ω] Vmax = 13.ref = 1.2.41 [Ω] Rs.596 [Ω] Tc = 288 [K] Vmp = 12.Rs = Rs.ref = 1.ref Eqn 23.5 [V] Rsh = 105 [Ω] Tc. 40) = ConvertTemp(C. 20).N OCT (4) (5) Determine the array efficiencies at the two given temperatures: η1 = P1 Area · GT.92.1 m2 (1) T1 = ConvertTemp(C.g.N OCT TN OCT = ConvertTemp(C.2 TN OCT − Tref. P2 = 10. Area = 0. Tc − Ta = (GT /GT.3.2. UL is assumed to remain constant. K.9 Use NOCT Conditions for finding cell loss coefficient: ta UL = GT. Data: Measured conditions: GT = 800 W/m2 .2. GT.ref · Ef f Equip (7) Determine the temperature coefficient of power from : η2 − η1 µP mp = T2 − T1 (8) Use Equationn 23. T2 = ConvertTemp(C.16 (10) Use definition of maximum power point efficiency to find power output of system.0 [W] .2 [W] . 30). η − η1 = µP mp · (Tc − T1 ) Eqn 23. condition one) as a reference and use Equation 23. K.N OCT = 800 W/m2 .16 to find array efficiency as a function of cell temperature. P1 = 12. 59. K.ref = 1000 W/m2 .6) (2) Ef f Equip = 0.6 Calculation of power at a new set of conditions from data at measured conditions. 18.3 to find cell temperature at the new operating radiation level and ambient temperature. Ta = ConvertTemp(C. K. K. Tref.noct ) · (1 − η/ta) (9) TN OCT − Tref.ref · Ef f Equip (6) η2 = P2 Area · GT. (3) ta = 0.N OCT · Eqn 23.2). P = Area · GT · Ef f Equip · η (11) Solution 314 1 .P23 06 Equations 23.3. GT.noct Choose one of the measured conditions (e. 2 [K] 315 2 η1 = 0.1156 GT.4 [K] Tref.8 [K] TN OCT = 313.51 [W] T1 = 291.6 [K] Ef f Equip = 0.N OCT = 800 W/m2 P1 = 12.ref = 1. Area = 0. 000 W/m2 P2 = 10 [W] Ta = 303.2 [K] η = 0.1087 µP mp = −0.N OCT = 293.2 [K] UL = 36 W/m2 -K .92 GT = 800 W/m2 P = 8.2 [W] T2 = 332.1326 GT.000589 [1/K] ta = 0.9 [-] Tc = 320.1 m2 η2 = 0. kT . ηref = 0. Eqn 2.169 · ¯ 2 − 0. slope. K. Only the March results are presented.9226 · kT (10) a2 = Rb · (1 + cosB − 2 · Rb ) + ρ · (1 + cosB − 2 · Rb ) · 1 − cosB 2 b2 = . Ac = 4 · 0.1551 + .0544 · ln (kT ) a3 = 1 − cosB − Rb 2 (11) (12) 2 (13) b3 = (. ¯ Rh.104.9. Slope.0696 · 2 ¯2 K K (Cos(Dec)) Xm − 1 2 − Xm q 2 p = abs abs (g) − g 2 + (1 + 2 · g) · (1 − XCrit/Xm) g= φClark = (1 − Step(XCrit − Xm)) · p (2) (3) (4) (5) end (6) function Zi(Rb .3184 · kT ) · kT (14) 2 Zi = (a1 · b1 + a2 · b2 + a3 · b3) · (I0/IT ) (15) end (16) Knowns: TN OCT = 46 [C] .85 + 0.981 · Xm = 1. ρ) Eqn 23. IT .23.5 (1) ¯ Cos(Slope) K Rh − 0.P23 07 Equations 23.1 Without EES and the SETP functions this is a long problem which starts with the month’s average daily radiation and works through the PV array characteristics and output. I0.427 m2 316 1 (17) .1456 + .7.7 A computer program to do Example 23.7. Dec function φClark XCrit. ηmppt = 0. The first sets of calculations give the absorbed radiation hour by hour through a day estimated from monthly average daily radiation.6 (7) cosB = Cos(slope) (8) 2 a1 = Rb2 + ρ · (1 − cosB) · Rb + (ρ · (1 − cosB)) 4 (9) b1 = −.2769 − . Month. ¯ = 13. ¯ Lat. (19) (20) (21) N day = AveDay (month) (22) HrAng = (time − 12 [hr]) · 15 [deg/hr] (23) dec = dec (N day) (24) rt = rt. Lat.( H.ref ) = ta · GN OCT Eqn 23. Rh = I¯T /I (33) ¯ T · rt /rd k¯ = K (34) I¯o = GZero.2. µM P = −.ref = 20 [C] .9 ρg = 0. HrAng. SurfAzAng. TN OCT. ηbatt = 0. Lat.00050 [1/C] . Slope. Lat = 40 [deg] . k.2 ¯T · ta · (1 − ηref ) · Z T − T µ · I a ref M P . Surf AzAng) (29) ¯ I = rt · H (30) ¯T Hd/H = HDIF F BARHBAR Lat. (Lat. N day. (N day. I¯T . (35) UL · (TN OCT − TN OCT. hr = 1 [hr] Surf AzAng = 0 [deg] . Lat) (27) ¯ T = H/H ¯ o K (28) Rb = RBeam. (HrAng. Tref = 25 [C] . N day) (25) rd = rd. N day. K ¯ · (rt − Hd/H · rd ) · Rb + HdH · rd · 1 + Cos(slope) + ρg · rt · 1 − Cos(slope) I¯T = H 2 2 (31) (32) We could have used the following call to find I¯T and eliminated the previous 9 lines but some of the intermediate results are used below. ρg ) I¯T =I¯T. (HrAng. (18) M onth = 3. N day) (26) Ho = HZero. HrAng) ¯ I¯o . HrAng.2 MJ/m2 H GN OCT = 800 W/m2 Assumed but does not matter as it cancels out. Lat.8.Bc = 1200 [W ·h] . (N day. ρg Z = Zi Rb . ta = 0. slope = 35 [deg] . Slope. Slope.3. . η¯array = ηmppt · ηref · 1 + µM P · + . . ηmppt hr · ηmppt · UL · . 0.0036 MJ/hr J/s . (37) . . ¯ ¯ Ei = η¯array · Ac · IT /hr · . . 777778 Xc = .277. ¯ i · . . 0036 L .0. MJ/hr . W . Ac · η¯array · I¯T (36) . W . . MJ/hr . 7 (40) 317 2 .7.7. Eqn 23.4 (38) (39) · hr Eqn 23. 7.3361 ηmppt = 0.8 ¯L.5 [deg] I = 0. ‘E bar ex i’ .226 rt = 0.418 [deg] do = 0. (41) ¯ex.315 − 0.i = 0 [W] E GN OCT = 800 W/m2 HrAng = 82.2 MJ/m2 H H MJ/m2 o = 27.2306 η¯array = 0.00 [-] Z = 2. ¯ L · hr P = 1.i = 7 [W] E ¯L.3.44 I¯o = 0.1847/K Bc (50) x = do + ∆f max p x − x2 − 4 · P · do · ∆f max ∆f s = 2·P (51) fmonth = fo + ∆f s (53) (52) $ShowWindow Solution Solution Variables in Main program Ac = 1.12 (47) ¯ex /L ¯ do = ηbatt · E Bc ∆f max = min 1 − fo . 1.01317 slope = 35 [deg] time = 17.9 (43) Find the Sum row in the Parametric table. 200 [W-h] ∆fmax = 0.7156 Hd/H = 0.3613 MJ/m2 I¯T = 0. Slope. 1.0005 [1/C] φ = 1.i = E ¯i − E ¯ex.24 [-] dec = −2.69 W/m2 -C Xc = 0.094 [-] ηbatt = 0.ref = 20 [C] x = 0.9 Ta = 9 [C] TN OCT = 46 [C] Tref = 25 [C] UL = 27.009238 Surf AzAng = 0 [deg] ta = 0.1219 MJ/m2 k¯ = 0.7.6205 318 3 .000 [--] rd = 0. ‘E bar L i’ . T ableRun#) + 0.734 Rh = 1. 1.i = E ¯i · φ E (42) Eqn 23.15 MJ/m2 ¯ T = 0. T ableRun#) E (44) ¯L = Sumparametric(‘Table 1’ .104 E ¯ ¯L = 429 [W] Ei = 7 [W] E fmonth = 0.9461 fo = 0.9 ¯ex.1059 · (46) (48) (49) ¯ · hr fo · L ¯T − 0.8931 ρg = 0.i E Eqn 23. k.2 Rb = 1.001 is to prevent divide by zero Eqn 23.2844 [-] ∆fs = 0.3374 ¯ = 600 [W] L µM P = −0.5 [hr] TN OCT. ‘L bar i’ . ¯ Rh . ¯ex = Sumparametric(‘Table 1’ .001 L ¯L /L ¯ fo = E The 0. T ableRun#) E (45) ¯ = Sumparametric(‘Table 1’ .481 K Lat = 40 [deg] ¯ i = 0 [W] L M onth = 3 N day = 75 P = 0.8 ¯ex = 252 [W] ηref = 0.4476 hr = 1 [hr] ¯ = 13. dec φ = φClark Xc .708 m2 Bc = 1. 5 11.000 1.13 1.61 1.00 0.090 0.094 0.093 0.094 0.61 1.24 1.089 1.5 15.Parametric Table: Table 1 Run 1 2 3 4 5 6 7 8 9 10 11 12 time [hr] 6.00 0.5 9.60 0.15 0.174 0.220 0.000 ¯ex.000 1.13 1.094 ¯i E [W] 7 27 49 71 89 98 98 89 71 49 27 7 ¯i L [W] 0 0 0 100 100 100 100 100 100 0 0 0 319 4 Xc [-] 0.60 1.5 Ta [C] 4 5 5 6 8 10 11 12 12 11 10 9 ¯ IT 2 MJ/m 0.23 2.219 0.5 13.15 Z [-] 2.i E [W] 7 27 49 6 16 22 22 15 6 49 27 7 ¯L.5 10.00 1.093 0.40 1.5 8.000 0.5 17.53 1.094 0.01 2.49 1.92 2.11 1.41 0.24 η¯array [-] 0.61 2.00 0.00 φ [--] 1.01 1.000 1.11 0.00 0.61 1.093 0.74 1.5 12.49 1.23 2.02 1.92 1.74 1.094 0.53 1.5 7.093 0.093 0.093 0.095 0.i E [W] 0 0 0 65 73 77 77 73 65 0 0 0 .175 0.000 1.5 16.5 14.02 1. 7 with k=2 equals P(U)=(2/c)*(U/c)*EXP(-(U/c)2 ) Line before Fig 24.P24 01 P24_01_D.6 ¯ 2 =c2 *Γ ( 1+1/2)=c2 *π/4 U 320 1 . This can be done analytically. The Weibull probablity distribution Eqn 24.jpg Equations Show Weibull distribution with k=2 is identical to Rayleigh distribution.2.2. 0492 321 2 .841 k=2 U = 10 PRay = 0.2.10 2 −1 Γ (1 + 1/k) k−1 PW ei = (k/c) · (U/c) PRay = (π/2) · U ¯ U2 k · exp − (U/c) (4) ¯ 2 · exp − (π/4) · U/U (5) Solution c = 16. k=2 (1) ¯ = c · Γ (1 + 1/k) U (2) ¯2 · σ2 = U (3) Eqn 24.93 σ = 7.2.2. ¯ 2 /π c2 =4*U Substitute c2 into the Weibull probability distribution results in ¯ 2 )*Exp(-(π/4)*(U/U ¯ )2 ) P(U)=(π/2)*(U/U which is Eqn 24.0492 ¯ = 15 U PW ei = 0.Solve for c2 .11. the Rayleigh distribution In the following we show both the Rayleigh and Weibull (with k=2) distributions on the same plot.9 ! Γ (1 + 2/k) Eqn 24. 782 ¯ = 9 [m/s] U 322 1 .P24 02 Equations Find solutions to two simultaneous nonlinear equations.5 [m/s] k = 2. ¯ = 9 [m/s] U (1) σ = 3.11 [m/s] σ = 3.5 [m/s] (2) ¯ = c · Γ (1 + 1/k) U (3) 2 2 σ =u ¯ · Γ (1 + 2/k) 2 Γ (1 + 1/k) ! −1 (4) Solution c = 10. 5 [m/s] U . U p = 101.5) U (4) The power per unit area for the two average velocities is found from Eqn 24.5) P/Amax = ρ¯ · · 2 2 3 cmin Γ (2.5 W/m2 ¯max = 7. $UnitSystem SI kPa C Knowns: ¯min = 7 [m/s] . Then the parameter c can then be found from Eqn 24.5 [m/s] .204 kg/m3 p = 101.3 [kPa] T = 20 [C] 323 1 P/Amax = 242.13.3 [kPa] . U ¯max = 7.9: ¯min = cmin · Γ (1. P = p) (1) (2) For Rayleigh distribution the parameter k is equal to 2. T = T.2 W/m2 ¯min = 7 [m/s] U cmin = 7.P24 03 Equations Determine the wind speed range for a range in power. This analysis gives 394 to 485 W/m2 which is pobably OK in a wind atlas.463 [m/s] P/Amin = 197. Solution cmax = 8.899 [m/s] ρ¯ = 1. T = 20 [C] Standard conditions ρ¯ = ρ (air.2.2.5) U (3) ¯max = cmax · Γ (1. 3 cmax Γ (2.5) P/Amin = ρ¯ · · 2 2 (5) (6) The power range in the Wind Atlas for these two wind speeds is 400 to 500 W/m2. 225 (16) ¯ high =4. σ low =2.89.σ high =2. A Parametric Table is defined From Figure 24.38 − 15.=6936.4 3 2 U U U else (10) (11) h Cp. ρ = 1. U with a range of average wind speeds and standard deviations. A = 300 m2 .61[m/s]. k call F indCK U (17) 324 1 . Find Weibull c and k ¯ . σ : c.3.2602 (m/s) · U Cp.09 + 0.6[m/s]. ¯ . σ : c.=1388 3 (m/s) i (12) U3 endif (13) endif end (9) (14) function (15) Knowns η = 0.20[m/s]:. U ¯ low =3. k SubprogramF indCK U (1) ¯ = c · γ (1 + 1/k) U (2) γ (1 + 2/k) ¯2 · σ2 = U 2 γ (1 + 1/k) ! −1 (3) end CK (4) function Cp.75[m/s]. (U ) (5) If((U < 8) or (U > 22)) then (6) Cp.=0 (7) else (8) If U < 18 then h i h i 2 3 h i (m/s) (m/s) [(m/s)] -1 − 2506 + 309.2.P24 05 Equations Find average power knowing only average wind speed and standard deviation. (U ) · U 3 · p dU P¯ = · 2 8 [m/s] .k−1 p = (k/c) · (U/c) k · exp − (U/c) Weibull probability density function Integrate from 8 to 22 Z 22 [m/s] A·ρ·η Cp. . . kW . . · . . 0.001 W. 00 2.554 1.635 5.68 2.484 1.75 3.04 3.99 3.01 3.5 4.862 2. Parametric Table: Pbar vs Ubar Run 1 2 3 4 5 6 7 8 9 10 11 12 ¯ U [m/s] 3.2 2.125 4.776 2.75 σ [m/s] 2. (18) (19) ¯ for various values of σ that the maximum power occurs when both the average and It is clear from the graph of P¯ vs U standard deviation of the velocity are at a maximu.3 2.3 2.125 4.698 1.125 4.985 P¯ [kW] 1.910 4.4 2.2 2.2 2. The worst year is at a low velocity and low standard deviation.289 1.5 4.359 k [-] 1.077 1.4 2.178 1.5 4.75 3.632 1.75 3.623 5.956 2. The worst year is 1980.5 c [m/s] 3.5 2.50 3.645 5.44 3.55 1.420 1.5 2.64 4.364 3.3 2. Probably the best year is then 1988 or 1990.73 Pbar vs Ubar 325 2 .125 4.31 2.652 5.4 2.363 3.872 4.848 4.893 4.92 2.5 4.362 3. these are important parts of the assignments and should not be neglected. You will be assigned to work with a new partner on each problem set. There will be a design project that will integrate various aspects of the course. Subsequent problem sets (with one or two exceptions) will treat in some detail parts of the simulation: Calculation of solar radiation absorbed by the collector. It Project: will replace the final examination. Calculation of system performance. and which represents a practical solar heating system operating in a specific location) and prepare a brief summary and interpretation of the results. A. Second (or subsequent) Printing. Calculation of system economics. Most of these problems are best done with EES. 326 .Solar Energy A semester course for engineering seniors and graduate students Text: Solar Engineering of Thermal Processes . Most of the problem sets constitute an interrelated series of calculations. Note that some of the problems require qualitative statements or interpretations. 2006. you and your partner will turn in one solution. The problem sets will be evaluated and the grades will contribute to the final grade.Third Edition. You will run a simulation program (which will be provided. will be an introduction to the course and will be a prediction of performance of a solar process. with the results of one set providing information for subsequent sets. if time permits. Supplemental notes will be provided for several assignments. A report will be required and will be due on final examination day. Problem sets are due on the day indicated and must be turned in at the end of class. Calculation of performance of collector and other components in the system. The first problem set. start on the next day’s material. using the EES/SETP functions that are provided. by Duffie and Beckman Assignments: Noted as sections of the text. Problem due dates will be used to discuss the solution and. Problems: There will be 12 weekly problem sets (called A through L) plus a semester project. 7 Set H 10.2 22.24 Chapter 7 Set G Chapter 8 Ch. FR.1-2.1-5. 10.2 5.1-22.21 Set B 2.1-10.13 2.6.14-2. Systems Simulation of a Solar Energy System Solar Radiation: Definitions. Angles.25. Shading.7 Set E 6. 2. Radiation Data & Processing Radiation on Sloped Surfaces Utilizability Concepts.10 3. Absorbed Radiation Flat Plate Collectors.5-6. Utilizability Basic and Life Cycle Economics P1.22-2.1-3.Solar Energy Technology Day 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 Topic Introduction: Collector.3 –5.8-10.11-3.1-6. F'.5 Ch 15. 22. Liquid and Air Heaters Collector Characterizations and Tests. 9. Optimizations.4 6. Heat Transfer Topics Heat Transfer Topics.8-6. Mean Temperatures.8-11. 18 327 . P2 Method. Spectral Radiation Properties Spectral Radiation Properties. Cell Energy Balances More photovoltaics Design Methods: f-Charts. Practical Considerations Concentrating and Evacuated Collectors Energy Storage Solar Process Loads. Heat Capacity Effects.7-4.10 Chapter 23 Set I Chapters 20 & 21 Set J 11. Extraterrestrial Radiation Engineering Equation Solver (EES) Measurements. 5.16-6.22. 3.3 .9-2. 17. System Analysis System Analysis and Simulation Photovoltaics. Transmittance Transmitted Radiation.7 11.13. Uncertainties Wind Power More wind Power Project discussion Review EXAM II Passive Processes Passive and Hybrid Systems Other Applications Closure Assignment (Chapter 10) Chapter 1 Set A 2. PV Notes Set D 6.10 Set K Chapter 24 Set L Ch 14.1-4. F". 4.17. Storage Energy Balances. QU Review EXAM I Critical Radiation.1-11.6 Set C 4. Solar Geometry.15 Set F 6.12. Loss Coefficients F. Your task is to calculate how much energy this system will deliver to meet the building heating and water heating loads of the house for a typical year. The default value of 0.013)” when you choose to “Select solar collector from a list”. and is to supply space and hot water heat for a residence. The collectors are mounted flush on the building (in a manner similar to that in Figure 13.5b and c). The total roof area suitable for collectors is 75 m2.50. For this simulation assume a value of 0.1 is typical of modern collectors.1 and 6. The system has the following major components: The collector is a flat-plate liquid heater with one glass cover. The “negative of the second order loss coefficient” indicates how much the thermal losses deviate from being linear with temperature. WI. These parameters are built into the “Average performing flat-plate (0.8.1.013 W/m2C. A schematic diagram of the system is shown below.8.50 W/m2C. The collector parameter “intercept efficiency” is essentially the optical efficiency of the collector and is 0.1. The “negative of the first order loss coefficient” is essentially an overall thermal loss coefficient from the collector fluid to the ambient and is 3. The collectors are facing south at a slope of 60 degrees from the horizontal. 328 . The “First order IAM from ASHRAE test” is related to the change in optical efficiency as a function of the angle of the beam solar radiation.Problem Set A A solar heating system for a residence is to be located in Madison. 0.2. similar to those shown in Figures 6.3. 3. circulated through a water-to-air load heat exchanger (LHX) and returned to the tank. defined as the ratio of the solar contribution to the load divided by the load.6. including infiltration. The hot water load is 60 liters per person per day for a family of five. The building has an overall loss coefficient. Use plots or diagrams to illustrate your results. The domestic hot water (DHW) subsystem consists of a heating coil (heat exchanger) in the main storage tank. The collector slope can conceivably vary between 30º and 75º. The auxiliary energy maintains the upper portion of the tank to 60C. you can change several of the design parameters of the system. The flow rate of the water is controlled to exactly meet the house heating load. which cannot exceed 75 m2.zip on the course homepage). The default case uses 75 liters/m2. The usual performance figure is the solar fraction. Express the error as a percentage of the delivered solar energy. Details of how systems are controlled and related matters will be dealt with as the course progresses. The Auxiliary heat exchanger is located in the top of the tank and is of sufficient capacity that it can supply all of the domestic hot water and house heating energy needs if necessary. The effectiveness is assumed to be 0. WI for the entire year using the default parameter set. Perform an energy balance on the main solar tank for the entire year. 329 . You are to write a brief report that is intended to inform a group of contractors and architects about the performance of the system and the effects of changes in the design on system performance. Storage unit size normally varies in proportion to the collector area. 75 and 100 liters/ m2 can be assumed. The Problem Using the simulation program provided (CombiSys. 100*Error/(Solar Energy in) = %error.0 W/m2-C that is sized in proportion to the collector area. In addition. Note that the solar contribution to the load is NOT ESol due to losses from the tank. Mains water enters this heat exchanger and is heated to the delivery temperature of 45C. The auxiliary energy supply maintains the upper section of the tank at 60C. The main storage tank is an insulated tank with a loss coefficient of 1. The space heating system can meet the entire space heating load. Sixty degree or higher water is removed from the top of the tank. These include: The collector area. you can investigate storage size only for the 60º slope. (UA) of 350 W/C. simulate the operation of this system in Madison. The mains temperature is assumed to be constant at 10C. The LHX is large enough to meet the entire house load. If solar energy heats the domestic hot water above 45C (as it will in the summer) then a bypass system (not shown) takes mains water and mixes it with the too-hot water to deliver water at 45C. (Energy in – Energy out – Energy Stored = Error) The error is due to numerical tolerances in solving the equations. To reduce the number of runs. Typical values range from 30 to 100 liters/m2.The collector-heat exchanger isolates the anti-freeze solution in the collector loop from the water storage tank loop. F. ratios of 50. 2. Running TRNSYS 1. click on the various plot identifiers at the top of the plot – the individual plots should disappear and reappear. Click and drag the mouse over part of the plot for a blow-up of a region. In the c:\temp\combisys directory there is a program called TRNSED (pronounced Transed). These include the heating load of the building and the characteristics of the collector. The right mouse button will start and stop the simulation. 2. The various system options should be self-explanatory. 3. Double click it to run. The second file CombiSys_defaults. Right click on the file and select to install into the directory c:\temp. With the plot on the screen. Installing: 1. After the simulation is complete select NO to the question “Exit on-line Plotter”. Help is available by ckicking on the question marks.trd and a window describing the system in the problem statement will open. Click “OK” and a dialog box will come up asking you what file you would like to open. 3. Energy rates are integrated to give monthly energy quantities.trd and is provided so you can restore the default settings. There are many other design parameters that for this problem you cannot change or do not need to change.zip from the course homepage and place it in the c:\temp directory. Quantities like solar energy collected. Comments and suggestions: It is suggested that you first simulate the system operation for one week.Estimate how much the home owner can afford to pay for the solar equipment if the auxiliary energy is (a) natural gas and (b) electricity. You MUST install in this directory. You should end up with a new directory called c:\temp\combisys on your computer. Installing TRNSYS on your computer: Information on installing TRNSYS (pronounced Transys) follows. Choose CombiSys.trd is identical to CombiSys. and you can use this weeklong operation to get accustomed to the program and what it does. energy lost from the tank. We will look at the impact of other design parameters during the semester. Download the file CombiSys. auxiliary energy and various temperatures are computed as a function of time. Click near the top or bottom on 330 . You can run the system by either pressing the F8 key or by choosing “Calculate” from the TRNSYS menu. The computing time will be very small. Examining the On-Line plots You may manipulate the on-line plots in a variety of ways. h. you will find an output file. go to menu item Calculations and choose Exit.either the right or left had axis numbers to change the scale. is essential information in determining the economics of the application. COMBISYS. a value will be printed for the completed portion of the last month. 331 . Subsequent sets will treat absorbed radiation. g. you will look at a series of aspects of this heating simulation. and you will have a picture of how to calculate long-term system performance.OUT. A few hours extra on this problem will be well spent. c. There are two tabs at the bottom for looking at either instantaneous values or integrated values. If the simulation is more than one page you may move back and forth in time with the tabs at the bottom. d. Examining the Output Once the simulation has completed and you have returned to TRNSED. f. Use EES and the EES/SETP function (available on the course home page) for your solution. This problem is concerned with calculations on basic aspects of solar radiation. you will find a summary of the performance for the total time of the simulation. Copy the file and paste into EXCEL (or other spreadsheet program) to do your calculations. In this and subsequent sets. When finished. The output is a text file that you can copy and paste into Excel. It is necessary to put in the time to learn to use EES and the EES/SETP functions otherwise future problems will take MUCH longer. the performance of a solar heating system in Madison. and systems calculations. You will need to add up the monthly values to obtain yearly values. PROBLEM SET B In Problem Set A. Annual information of this type. as will be seen later. If the simulation ends within a month. under the “Windows” menu item at the top of the screen. UTank: change in internal energy of the storage tank from the start of each month ESol: the integrated energy transferred across the collector loop heat exchanger EAux: integrated auxiliary energy added to the space to meet heating load requirements ELossTank: the integrated energy loss from the tank (assumed to be in an unheated area of the house) EMains: the integrated energy entering the tank with the water from the mains EDHW: the integrated energy leaving the tank with the DHW EHouse: integrated energy losses from the house EIncSol: the integrated solar radiation incident on the collectors EHorSol: the integrated solar radiation incident on a horizontal surface Values are printed for each month. b. e. The values printed in the output file are as follows: a. Set C will use the concepts in this set to deal with calculation of radiation on sloped surfaces starting with measured data on a horizontal surface. At the end of this series. WI was explored. other component performances. i. collector energy balances. we will have explored the basis of all of the aspects of Set A. If you look at “View Simulation Results”. are given in the table. Add R_b to your table of part 5 and show a plot of Rb vs.1 -5. Ibn and Tamb are the averages over the time from 8am to 9am). 3.3 -3. time_clock and Theta.100 µm? Express the result both as a fraction of the total and as energy rate in W/m2.7 -5..1) in Madison. Note that in EES you can have a final row in a parametric table that contains the sum of the values in the column – see Options\Preferences\Options\Include a sum row in the parametric table. solar time.4 -6.4º the hour-by-hour radiation on a horizontal surface. The time is clock time (not solar time) in the central time zone and represents the integrated values for the previous hour (e. What is the day length? At 10:30 AM solar time. 7.6. February 16. WI. I.8 -4. what is the local clock time? At what time does sunrise occur (both clock time and solar time)? For this day. How much of the extraterrestrial solar radiation is in a wavelength range of 0. Make a plot of theta versus solar time (use the midpoint of the hour). Problem Set C For the same location and date as in Set B (Madison. units and data. 5. Add Go to your table of part 6 and show a plot of Go [W/m2] versus solar time. Tamb. Ibn. Due to measurement “problems” it may be necessary to set the radiation values for hours that include sunrise or sunset to zero. the normal beam radiation. WI: 1. 8. In the upper left-hand corner of the table select Paste Special and then select to paste column names. time hr 1 2 3 4 5 6 7 8 9 10 11 I 2 W/m 0 0 0 0 0 0 0 31 201 390 503 Ibn 2 W/m 0 0 0 0 0 0 0 55 674 813 883 Tamb ºC -3 -3. 6.640 µm to 1.g.1 -3 -2. south facing surfaces. including the column header and units and then copy. 2. In EES open a “Lookup Table” with 4 columns and 24 rows. select the entire table. what is the declination? Is it significantly different at sunrise and sunset? How does the angle of incidence of beam radiation (Theta) vary through the day for a surface sloped 60 degrees to the south? Make a parametric table with time_solar. for the row at time 9 the values of I.1 -3. 4. 9.9 -5. and the ambient temperature. longitude=89. To transfer the following information into EES.For February 16 (the average day for February from table 1.2 332 . What is the extraterrestrial radiation on a horizontal surface for this day in W-hr/m2? Calculate using the appropriate SETP/EES function and also add up the values of Go (times one hour) from part 7. 5 -14. 3. How do the values compare to those obtained in question 2? Do not use the function I_T_LJ_ to do the calculations.12 13 14 15 16 17 18 19 20 21 22 23 24 585 597 415 213 198 69 17 0 0 0 0 0 0 913 917 360 40 287 47 42 0 0 0 0 0 0 -4.01 micrometers by adding a point 333 .4 -5 -5. which is not sufficient for the calculations of this problem. 6. for a slope of 60 degrees to the south and for a ground reflectance of 0. Use the Liu and Jordan method. Io [W/m2] and I [W/m2] as functions of hour (use the midpoint of the clock-time hour). WI heating system has radiation properties of the absorber plate as given in Figure 4. time. 45. The collector is expected to deliver most of its output when the absorber surface temperature is about 325 K. If the selection of slope is based on maximizing incident radiation for the day (as we will see. 5.4071 micrometers to 25 micrometers. Explain why they are not the same. Problem Set D A collector that is to be used on the Madison.2 Curve C as a lookup file'.1 -12. 4. 60 and 75 degrees. Comment on any unusual aspects of the plots.5 -6. curve C. For slopes of 30. Use the L&J method and the function I_T_LJ_ to do the calculations.6 -9.8 -13. How might your answer to this question change if a summer’s day had been used in the calculations? What are H and Ho [MJ/m2] and KT for this location and day? What is H T [MJ/m2] for the month of February for this location using the data in Appendix G and a slope of 60 degrees.2. plot IT [W/m2] vs.8.1 -10. 2. Compare your answers with the value calculated by TRNSYS in set A.4: 1. Using the data in the table. Plot kT .3 -15 For this day. The data in the table ranges from 0. Extend the data down to 0.1 -7. diffuse and ground-reflected components of IT as a function of clock time.6 -11. plot IT [W/m2] and the beam. Use the EES parametric table and the appropriate SETP/EES functions. Use both the Liu and Jordan method and the K-T method.3 -12. this is not necessarily the correct basis). Why can’t we use the SETP/EES procedure I_T_LJ_p_ to do this calculation? Estimate Ib and Id from I and use your estimated value of Ib and Id to estimate IT using the L & J method.8. (See the course homepage for 'Figure 4. what slope would you select for this day? (Note: under Options\Preferences you can have EES sum a parametric table column). air mass 2). 1.4071 to 25 micrometers.4071. Plot S and IT for slopes of 30. Do not use the beam radiation values given in the weather data (the SETP functions for S uses total radiation and estimates the beam and diffuse breakup).96 and an emittance of 0.4. Use the EES integral function and your Planck function to integrate Equation 4.121 per sheet. 60. Use the same day in your calculations as was used in Set C. Extend to 100 micrometers in the same manner. 2.6. Repeat problem 1 but assume the solar radiation is distributed as in Table 2. For a slope of 45 degrees compare the absorbed radiation for this day for the two cover collector with that of the one cover collector.5 to evaluate the solar absorptance. Plot (τα)ave for the four slopes as a function hour angle..5. and 75 degrees as a function of hour angle (four graphs). Plot the emittance versus surface temperature for a range of temperatures from 300 K to 800 K. 45.1 where epsilon and epsilon_lambda are replaced by rho and rho_lambda.at 0. What are the general trends? 2. 1. Use the HDKR method in your analysis so we all use the same algorithms. Can you draw any general conclusions? 3.01 to 100 micrometers versus integrating from 0.85.86.1. Use the EES curve fit option under the menu Plots and find a 2nd order polynomial (in degrees Kelvin) that fits your results. Comment on the differences when you integrate from 0. What is the absorptance for solar radiation assuming the sun is a blackbody at 5777K? Write an EES function called Planck_(lambda.6. This surface is proposed for use in other collectors that will operate at different temperatures.1 (i. What is the emittance for long-wave radiation at 325 K? Comment on the error associated with extrapolating the data. There is also the possibility of using a flat-black absorber surface which has an absorptance at normal incidence of 0. 3.e. What effect would use of this surface have on absorbed radiation for the 45 degree slope? 334 .1). What can you speculate about the utility of second cover? 4. Use the Insert\modify Array command from the Edit menu to insert the wavelength bands from Table 2.6. Find an average rho in each band and use Equation 4. 4. It is proposed that the collector design is to be modified by the addition of a second cover. Problem Set E The surface to be used as the absorber plate in collector has a normal incidence absorptance of 0. T) that will evaluate the blackbody energy at a specified wavelength and temperature (Equation 3. The collector has a single glass cover of low iron content and a KL of 0.01 with the same reflectance value as at 0. ηcoll = 0. 2.000345085[1/K]*T_K + 2.) " "!Data" Ins_back=0. T_bar_plate)) 3. with a water inlet temperature of 35C. FR. 'T'. Interpolate1. T=ConvertTemp(C.07 [m] k_ins=0. The heat transfer coefficient inside the tubes is given as 175 W/m2-K. You will be using this program for Sets G and H so you want to get it working correctly. The plate emittance is assumed to be 0. WI application has the following characteristics: (If you download the Word copy of this problem you can copy the following from Word and paste into EES.0245 [W/m-K] "back insulation thickness" "polyurethane rigid foam" 335 . K.14. The following equation representing the plate emittance as a function of plate temperature was obtained from the solution to Problem set D4: εplate= . You can copy and paste the fragment into EES. F``. You can easily extract information from the lookup table using the linear interpolate function. UL. The collector in the Madison.00443111 + 0. Check to see if it will be necessary to account for this variation in the plate emittance in evaluating annual performance. to extract the viscosity at the mean plate temperature (assumed to be the same as the tube wall temperature) use: mu_glycol_wall=Interpolate1('PropyleneGlycol'. This table contains the properties of 50-50 propylene glycol and water from 244. 4. Use the 'Uncertainty Propagation' under the 'Calculate’ menu to estimate the impact on not knowing the wind coefficient better than plus or minus 5 W/m2-K. The temperatures in the table are in K. For example. Use this value in your solution to parts 1 and 2. print values of: ηcoll . T_bar_bar_fluid and Qu. If your program is working correctly. 'mu'. S. For February 16th in Madison.487. F`.LKT' from the course homepage \Computer Stuff and read it into your EES program with Tables\Open Lookup Table. Your job is to complete the program and answer the following questions: 1. Use the Lookup table 'PropyleneGlycol.61859E-07[1/K^2]*T_K^2 where T_K is the mean plate temperature in K. compute the hourly output of the collector (MJ/hr) and the outlet fluid temperature. To show that your program is working correctly.0.Problem Set F A program fragment is given below that describes a solar collector for use in Madison WI. Calculate the value of the heat transfer coefficient inside the tubes and compare it to the 175 W/ m2-K. WI for the hour 11 to 12.3K. T_plate. The wind heat transfer coefficient has been assumed to be equal to 10 W/m2-K.3K to 394. 5 [m] "length in the flow direction" W=Tubes*Sp_tube "width of collector" A_c=L*W Fluid$='PG' "Propylene Glycol / Water" Conc=50 [%] "Glycol/Water concentration" alpha_n=0. Row.61859E-07[1/K^2]*T_K^2 336 .14 "IR emittance of absorber plate" $Else epsilon_plate=-0. 'I'))*Convert(W-hr/m^2.01[W-hr/m^2]. day) SurfAzAng=0 [deg] GrRef=0. ‘Data’. Row. MJ/m^2) $Ifnot parametrictable T_amb=lookup('Data'.028 [m] "plate to cover spacing" T_in=35 [C] "also check using 60 and 85" slope=45 [deg] Long=89.526 "Glass refractive index" C_b=1e6 [W/m-K] "bond conductance" h_fi=200 [W/m^2-K] “heat transfer coefficient inside the tubes" N_cov=1 "Number of covers" h_w=10 [W/m^2-K] "with a plus or minus 5 W/m^2-K uncertainty" "!Get weather data from lookup table.115 [m] "distance between tube centers" L=2.U_edge=0." Row=12 HrAng=Lookup('Data'.neglect the impact of the headers" Sp_tube=0. 'T_amb') $EndIf $If parametricTable='Fixed epsilon' epsilon_plate=0.4 hour=1 [hr] zero=0 Tubes=8 "number of tubes in parallel . divided evenly among the tubes" KL = 0.025 [l/s] "flow through the collector.1 [deg] Month=2 Day=16 n=nDay_(month.0005 [m] k_plate=385 [W/m^2-K] "copper" Flow=0.000345085[1/K]*T_K + 2.121 "Glass KL product" Refrind=1.881 "solar absorptance of absorber plate – from problem D4" Dia_Tube=0.098 W/m^2-K] "edge loss coefficient" Space=0.4 [Deg] Long_std=90 [deg] Lat=43. 'HrAng') T_sky=T_amb I=max(0.008 [m] "use for both inner and outer diameter" Thick_plate=0. Row.00443111 + 0. lookup('Data'. Assume the tests are carried out in Madison. You should set up a parametric table with each row representing one test point for 337 .g.. Prandtl and Nusselt numbers. “ Problem Set G This problem simulates a collector test.2C) to 100C.$EndIf "Find average plate and fluid temperatures – make guess to get started – later replace by appropriate functions" T_bar_plate= T_bar_Fluid= "Find U_top and U_L" U_top= U_L= "Find F`. Use a range of inlet temperatures from just a little above T_amb (e. Modify your program from set F to generate plots of efficiency versus (T_in-T_amb)/I_T and K_tau_alpha versus (1/cos(theta)-1). collector efficiency factor" F`= "Find F_R and F`` " F_R= F``= "Find absorbed solar using HDKR method" S= "Find useful energy gain" Q_u= "Find the outlet temperature" "Find the collector efficiency" “Find the actual inside tube heat transfer coefficient. The collector slope is such that the incident beam radiation is normal to the collector surface when the surface azimuth angle is zero. All tests are completed at noon. WI on February 16 but that the day is clear with kt equal to 0. Reynolds. In reality it takes time to change the test settings and to let the system come to a new equilibrium condition.1 C. Need to find properties. 12.7 at solar noon and the ambient temperature remains constant at 12. Your program should include the effects of the plate emittance as a function of plate temperature and the effects of the inside-thetubes heat transfer coefficient as a function of flow rate and the mean fluid and mean plate temperatures. Comment on the use of the flow rate correction factor (Section 6.19.4º). WI on February 16. Use your program from Set G to evaluate the impact of flow rate on collector performance. 3. I and the ambient temperature.20) in evaluating the impact of varying flow rate on the collector parameters FR(τα)n and FRUL.7. Plot efficiency versus Delta T over I for the conditions of set F and for double and half of the flow rate. Make a plot showing both eta vs (T_ave . Use your program to investigate the impact of test-day conditions on collector parameters (e.1). Plot collector efficiency versus DELTAT/I and plot the ratio FR(τα)/ FR(τα)n versus (1/Cos(theta)-1). What is the collector slope? 2.1º.g.17.3 for converting data from one presentation method to the other. kT of 0. FRUL and bo. Tamb. (Latitude 43.2C.. the conditions of Set F). 1.2 and 6. In a second parametric table vary the azimuth angle from 0 to some value such that theta goes from 0 degrees to 60 degrees in order to generate the plot of Kta versus (1/Cos(theta)-1). 3.7 and Tamb=25 instead of 12.19. the hour-by-hour radiation on a horizontal surface. Show uncertainty bars on a plot of efficiency vs DELTA T over I for the nominal conditions (i.T_amb)/I and comment on the use of Equations 6.T_amb)/I and eta vs (T_in . 2. Also determine the parameters of equation 6. Estimate the resulting uncertainty in determining the linear collector parameters (FR(τα)n and FRUL) when the uncertainty in all temperature measurements is ±0. are given in the table of problem set 338 . Problem Set H 1. in flow rate is measurements is ±3% and in solar radiation measurements is ±5% 5. There is always an uncertainty in any measured variable. The following two questions were not used but would make an excellent follow-on problem. In Europe the test data is presented in terms of (T_ave-T_amb) rather than (T_in-T_amb) where T_ave = (T_in + T_out)/2.5 instead of 0.the DELTA T over I plot. Longitude=89. 4.e.explain. Problem Set I For Madison. Note that EES can plot uncertainty bars. Do you expect that FR(τα)n found using the two different models will be the same . Use the Curve fit option in the Plot menu to determine FR(τα)n. Determine the “steady-periodic” solution for this one day.e. The tank is sized with the rule-of-thumb of 75 liters per square meter of collector. Consider a solar system with a collector.. Plot the steady-periodic tank temperature as a function of time. 7-8) and 18 (i. electrical use.1 Day=16 Month=2 FRta=0. Qu.e. Qhouse for the 24 hour period all in MJ/hr. 5. 3. 2.30 UA_house=300 T_house=23 G_house=1500 hour=1 DELTAtau=1 [deg] [W/m^2-K] [deg] [deg] [m^2] [l/m^2] “tank volume per unit of collector area" [m] “tank height" [W/m^2-C] “U-value of tank insulation. 17-18) to zero radiation. The losses from the house can be modeled as an overall loss coefficient (UAhouse) times the difference in the house temperature and the ambient temperature." [hr] [hr] 1. 4. Qtank loss and the heating demand of the house..B. The characteristics of the various components are as follows: "!Data" Lat=43.4 A_c=35 b_o=-0. Discuss your results. The time is clock time at the local standard meridian (90º). a fully mixed storage tank and a house heating load.. the tank loss.7 FRUL=4 slope=50 SurfAzAng=0 GrRef=0. That is." [W/C] "loss coefficient of the house" [C] "set temperature of house" [W] "internal gains of house due to people. The storage tank is located outside. find the initial tank temperature at midnight such that when the system is operated for the 24 hour period the tank temperature returns to the initial temperature. Assume that the solar energy and ambient temperature recorded at the end of an hour was constant for that hour. How would you have to change your program if the collector area were 10 m2? 339 . Plot the collector useful gain.14 Vtank\A_c=75 h_tank=2 U_tank=0. Due to sunrise and sunset “problems” set the hours 8 (i. The house has average internal gains due to people and electrical use of 1500W. etc. Problem Set J The plot shows current and voltage for a silicon PV module operating at 1000 W/m2 at two cell temperatures. TableRun#-1. 10 500 Tcell=25 °C 9 400 8 Tcell = 50 °C 6 300 5 4 200 3 2 100 1 0 0 10 20 30 40 50 Voltage [volts] Reading (very carefully) from the plots: V_oc_ref=63. So.6 "[Volt]" 340 60 0 70 Power [W] Current [amps] 7 . ‘T|Plus_tank’) where T|plus_tank is the previously calculated tank temperature (note that EES will format T|Plus_tank as T+tank). the temperature of the tank at the start of an hour is in the row above the one you are currently working on) use: T_tank=TableValue(‘TableName’. The MAX function returns the maximum of two expressions.e. Determine the maximum power point current. IL. Rs. 25 and 50ºC. with the EES expression Z=MAX(0. voltage and power for this module when the ambient temperature is 10ºC and the incident radiation is 900 W/m2.. Z will never go below zero no matter the values of X and Y.914 m2. Use a parametric table with 25 entries where the first entry is the initial condition. Use Euler integration with a one hour time step. The number of cells in series string is 110 and the number of series strings in parallel is 60. The NOCT for this module is 51ºC. a µIsc and µVoc.You may wish to use the EES MAX function. To extract information out of a completed row (i. The module area is 3. You will need to start your “table solution” in row 2 since row 1 should contain the initial conditions. Determine values for Io.2 "[Volt]" V_oc_50=57. X-Y). Life Cycle Savings b. This is sometimes called the PITI (Principle.8 % General inflation rate = 2. d.3) with fuel costs discounted.5 % Effective income tax rate = 39 % Down payment = 0. The four criteria are: a. we will use these results in an economic study of this solar application.3 24.1 As noted in Chapter 11.I_sc_ref=9.5 54.4 V_mp=51. Payback time B (Section 11.1 13. Interest.6 41. In this set. Economic parameters are as follows (with rates in percent per year): Mortgage interest rate = 5. Return on Investment.25 341 . Ac 0 5 7.050 Ratio of assessed valuation to investment = 0.4 "[Amp]" "[Amp]" "[Amp]" "[Volt]" Problem Set K In Problem Set A. c.2 of investment Ratio year 1 miscellaneous costs to investment =0.0 I_sc_50=9. considering principle and interest payments.0 % Period of analysis = 20 years Term of loan = 20 years Property tax rate = 3. payments for taxes and insurance. Taxes and Insurance) method. and F vs Ac curves were generated for three ratios of storage capacity to collector area. there are a number of “figures of merit” that are favored by or used by those designing and evaluating solar energy applications.5 15 30 45 60 75 F 0 9.0 63. In this problem you are to consider LCS and three additional criteria.90 Ratio of resale value to investment = 0. and fuel purchase savings. Use the F vs Ac data from Problem Set A for 75 liters per square meter of collector. Cash Flow.3 I_mp=8. the annual thermal performance of a solar heating system was determined.6 71. The criterion of economic viability that we recommend is Life Cycle Solar Savings. all in the first year. 5 % Area dependent cost = $ 150m2 Fixed cost = $ 550 The loads (from problem set A) are 2..C=0 (A non income producing system) First year fuel cost = $11/GJ (what you pay for fuel as it enters the furnace) Furnace efficiency = 0.2475*108 kJ for the house heating. What might be interesting variables to investigate? Semester Project The following is an example of a system simulation problem given as a semester project. storage tank. this one includes some complications that can be omitted if a shorter project is wanted. The major system components are: collector. you can easily explore the effects of other economic parameters. 1. This requires the integration and application to systems of many of the ideas used in short problems in this appendix. collector heat exchanger. The results of the thermal analysis will 342 . why? 2. a continuous supply of heated air is required at a rate of 1. They are arranged as shown in the diagram. If you have an EES program to do this problem. To do this. load heat exchanger. and auxiliary heat exchanger. Are the results very much different? 3. INDUSTRIAL AIR HEATER SIMULATION The Application For an industrial operation. and allow you to explore the effects of several critical variables. The location is Albuquerque. There are many variations on problems of this type. A liquid solar heating system has been proposed to supply part of the needed energy. The project is open-ended in the sense that the student must decide what variables will be treated and how the results will be presented. The air supply is at a constant temperature of 22 C. The object is to find economic optimum collector areas for the 75 liters/m2 systems (if optima exist) using each the four methods and compare them. at a temperature of at least 65 C.40 kg/s. NM.2 % Discount rate = 4. assuming that the collector unit area costs are $240/m2 instead of $150/m2. Repeat the process.0645*107 kJ for the domestic hot water and 1. Appropriate meteorological data must be available. You are assigned the task of evaluating this system and reporting to your management on the technical and economic feasibility of the system. How do the results using these four criteria compare? Would you expect them to be different? If so. you are to write a simulation program that will allow you to do year-long simulations of the system.8 Energy cost inflation rate = 5. the short-term variations of energy flows and temperatures. You should explore the effects of these design variables on: LA.81 and FRUL = 4. such as piping. It is to leave the heater section at a temperature no lower than 65 C. The heat capacity of air under the conditions of this operation can be taken as 1013 J/kgC. It is recommended that the flow rate on the water-side be fixed at 0. and is then heated in the auxiliary heat exchanger. β = latitude and Ac is variable Collector heat exchanger: The UA of the exchanger and thus also its effectiveness εc can be specified at any level. based on past experience and on equipment that is available. the energy losses from the storage tank. Any assumptions about them should be clearly stared. Storage: A wide range of storage tank sizes is available. the auxiliary energy required Ls. The Thermal Analysis Some thought about this problem will lead you to the following: There are three design variables that are or may be important in determining thermal performance and thus economic feasibility.then be used as the basis for an economic analysis. the temperature of the air leaving the auxiliary heat exchanger 343 . bo = –0.30 kg/s. the tank temperature TAs. assume the following: Collector: FR (τα)n = 0. Use the effectiveness of this exchanger εL to describe its operation. It is suggested that for a time period of your choice (a period which would be important in process design) you consider what happens with time to the following variables: T. if anything. In the proposed process the air enters the heating section at 22 C. is heated first by solar energy. They are the collector area Ac. For purposes of these simulations. You should consider what. the solar energy delivered across the load heat exchanger QL..) Freeze protection is to be provided by use of an antifreeze solution in the collector loop. The auxiliary heat exchanger has the capacity to provide all of the needed energy if necessary.5 W/m2C.12.e. should be done about boiling. ductwork. the collector heat exchanger effectiveness εc. are standard. Load heat exchanger: Cross flow water-to-air heat exchangers are available in a wide range of sizes. the temperature of the air leaving the load heat exchanger TAo. etc. (Thus its cost is common to all systems and does not enter into economic comparisons of systems. controls. The Equipment Some of the component design parameters can be fixed. pumps. Based on past experience it is recommended that the ratio of storage tank capacity to collector area be kept at 100 l/m2. γ = 0. Assume a loss coefficient for the tanks of 5. Assume that the tanks are cylindrical with a ratio of height to diameter of 1. Other Components: The other components. i.015 kg/m2s. and the load heat exchanger effectiveness εL.. It is also of interest to explore process dynamics.45 W/m2C at mɺ /Ac = 0. assumptions made.) As a first suggestion. they are not a substitute for the simulations. This does not need to be done in detail. but should be sufficient to indicate to your management what the economic feasibility of this project might be. and 344 . and thus you can get a rough first estimate of the solar contribution of a system. In particular. You will have to make assumptions in order to write the simulation program. The technical report itself should follow standard technical report format. It should be one or two pages in length and should contain the essential facts of what you did and what conclusions you reached. notes of what might be done about practical problems such as boiling and freezing. Considering it to be a fully-mixed tank leads to the simplest program. the radiation incident on the collector LA. It should include such items as: the basis of the calculations. It is intended for managers who are not familiar with the details of solar processes but who have strong engineering backgrounds. assume the following economic parameters: Cost of collectors = $ 250/m2 Cost of storage = $ 0. The Economic Analysis Based on the thermal analysis. Also note that you can calculate the total minimum amount of energy required to heat the air from 22 to 65 C. you will have to decide how you wish to model the storage tank. an economic analysis is to be prepared. Your Report An executive summary should be included in your report. You will need two meteorological data files. you should cover a range of at least 100 to 400 m2 (but this may not be an adequate range). tables or charts of data.50/liter Fixed costs = $ 20. (These methods provide a first estimate only. analysis of your results. the rate of delivery of auxiliary energy to the load. the rate of solar energy delivery to the tank IT. In order to do this. and has the advantage of producing conservative results. This can be obtained by use of the design methods described in Chapters 20 and 21. You will need some guidance in selecting collector area ranges to explore. you can estimate roughly what the annual efficiency of the collector might be.000 Cost now of delivering auxiliary energy = $ 10/GJ Cost of heat exchangers = $ 5 per unit of W/C Your Procedure Prepare a simulation program in the language of your choice.Ls the rate of solar energy delivery to the load Qu. one a one-day file of hourly data for debugging your program and a second a TMY3 (Typical Meteorological Year) file of hourly data which you will use for the simulations. It should be the kind of report you would prepare for your supervisor who has some familiarity with the technology. 345 .whatever else may be appropriate.