Solutions Manual Water-Resources Engineering 3rd Edition David a. Chin

Solutions Manual Water-Resources Engineering 3rd Edition David A.Chin Download full at: https://testbankdata.com/download/solutions-manual-water-resources-en gineering-3rd-edition-david-chin/ Chapter 3 Design of Water-Distribution Systems 3.1. (a) For geometric growth: dP = k1 P ∫ ∫ dt dP = k1 dt =⇒ ln P = k1 t + C P Which gives P = P0 ek1 t (b) For arithmetic growth: dP = k2 ∫ ∫ dt dP = k2 dt =⇒ P = k2 t + P0 (c) For declining growth: dP = k3 (Psat − P ) ∫ ∫ dt dP = k3 dt =⇒ ln(Psat − P ) = −k3 t − C ′ Psat − P which gives P = Psat − Ce−k3 t ′ where C = e−C . 3.2. (a) By graphical extension P2030 = 100, 000 (b) For arithmetic growth: P = kt + P0 where P1990 − P1980 61000 − 52000 k= = = 900 10 10 Therefore P2030 = 900t + P1990 = 900(40) + 61000 = 97, 000 49 (c) For geometric growth: P = P0 ekt Therefore P1990 = P1970 ek(20) 61000 = 40000e20k k = 0.021 and hence P2030 = P1990 ek(40) = 61000e(0.021)(40) = 141, 298 (d) For declining growth: P = Psat − Ce−kt (1) where Psat = 100,000 and 1970 : t = 0, P = 40000 1990 : t = 20, P = 61000 Substituting 1970 population into Equation 1 gives 40000 = 100000 − Ce−k(0) = 100000 − C which gives C = 60, 000 Substituting 1990 population into Equation 1 gives 61000 = 100000 − 60000e−k(20) which gives k = 0.0215 Equation 1 can now be used to predict the 2030 population (t = 60) as P2030 = 100000 − 60000e−0.0215(60) = 83, 483 (e) Equations 3.11 and 3.12 give the logistic-curve parameters a and b as Psat − P0 100 − 40 a= = = 1.5 P0 40 [ ] [ ] 1 P0 (Psat − P1 ) 1 40(100 − 52) b= ln = ln = −0.0486 ∆t P1 (Psat − P0 ) 10 52(100 − 40) The logistic curve for predicting the population is given by Equation 3.9 as Psat 100, 000 P = = (2) 1 + ae bt 1 + 1.5e−0.0486 t In 2030, t = 60 years and the population given by Equation 3.19 is P = 92, 487 people 50 3.3. From the given data: P0 = 13,000, P1 = 125,000, and P2 = 300,000. The saturation popula- tion, Psat , is given by Equation 3.10 as 2P0 P1 P2 − P12 (P0 + P2 ) 2(13)(125)(300) − (13)2 (13 + 300) Psat = = = −78 thousand people P0 P2 − P12 (13)(300) − (125)2 Since the calculated value of Psat is negative, estimation of the saturation population using Equation 3.10 is not possible . 3.4. average daily demand = 580(100000) = 58, 000, 000 L/d = 0.671 m3 /s maximum daily demand = 1.8(0.671) = 1.21 m3 /s = 1.04 × 108 L/d maximum hourly demand = 3.25(0.671) = 2.18 m3 /s = 1.89 × 108 L/d 3.5. The NFF can be estimated by Equation 3.20 as NFFi = Ci Oi (X + P )i where the construction factor, Ci , is given by √ Ci = 220F Ai For the 5-story building, F = 1.0 (Table 3.3, Class 2 construction), and Ai = 5000 m2 , hence √ Ci = 220(1.0) 5000 = 16000 L/min where Ci has been rounded to the nearest 1000 L/min. The occupancy factor, Oi , is given by Table 3.4 as 0.85 (C-2 Limited Combustible, Office), (X + P )i can be estimated by the median value of 1.4, and hence the needed fire flow, NFF, is given by NFFi = (16000)(0.85)(1.4) = 19000 L/min This flow must be maintained for a duration of 5 hours (Table 3.6), hence the required volume, V , of water is given by V = 19000 × 5 × 60 = 5.7 × 106 L = 5700 m3 3.6. The needed fire flow for any building should not exceed 45,000 L/min for a duration of 10 hours . 3.7. The (low-lift) supply pumps and the water-treatment plant should be designed for a capacity equal to the maximum daily demand (Table 3.7). With a demand factor of 1.8 (Table 3.2), the per capita demand on the maximum day is equal to 1.8 × 600 = 1080 L/day/capita. Since the population served is 200 000 people, the design capacity, Qdesign of the pumps and water-treatment plant is given by Qdesign = 1080 × 200000 = 2.16 × 108 L/d = 2.50 m3 /s 51  The required fire flow, Qfire , is 28000 L/min = 0.467 m3 /s. According to Table 3.6 the fire flow must be able to be sustained for 7 hours . The volume, Vfire , required for the fire flow will be stored in the service reservoir, and is given by Vfire = 0.467 × 7 × 3600 = 11800 m3 The required flow rate in the distribution pipes is equal to the maximum daily plus fire demand or the maximum hourly demand, whichever is greater. maximum daily + fire demand = 2.50 + 0.467 = 2.98 m3 /s 3.25 maximum hourly demand = × 2.50 = 4.51 m3 /s 1.80 where a demand factor of 3.25 has been assumed for the maximum hourly demand. The design capacity of the supply pipes in the distribution system should therefore be taken as 4.51 m3 /s . 3.8. According to Table 3.8, the minimum acceptable pressure under average daily demand con- ditions is 240 kPa . 3.9. From the given data: P = 200,000, per capita demand = 500 (L/d)/person = 5.79 × 10−6 (m3 /s)/person, and fire flow = 30,000 L/min = 0.5 m3 /s. (a) The required flow rate in the main distribution pipeline is the maximum daily demand plus fire demand or maximum hourly demand, whichever is greater. Taking a peaking factor of 1.8 for maximum daily demand and 3.25 for maximum hourly demand gives maximum daily demand + fire demand = 1.8(200000)(5.79 × 10−6 ) + 0.5 = 2.58 m3 /s maximum hourly demand = 3.25(200000)(5.79 × 10−6 ) = 3.76 m3 /s Therefore, the design capacity of the distribution pipeline should be 3.76 m3 /s . For a maximum allowable velocity of 1.5 m/s, the minimum required pipe diameter, D, requires that π 2 Q 3.76 D = = 4 V 1.5 which yields D = 1.78 m The pipe material to be used in ductile iron pipe (DIP) and a cover of 1.5 m would be appropriate. (b) For DIP with ks = 0.26 mm, D = 1.8 m, A = 2.544 m2 , Q = 3.80 m3 /s, V = Q/A = 1.494 m/s, and Re = V D/ν = (1.494)(1.80)/10−6 = 2.69 × 106 , [ ( )]−2 [ ( )]−2 ks 5.74 0.26 5.74 f = −2 log + = −2 log + = 0.0134 3.7D Re0.9 3.7(1800) (2.69 × 106 )0.9 52  and, taking γ = 9.79 kN/m3 , the Darcy-Weisbach equation gives p1 p2 L V2 − =f γ γ D 2g 550 350 L (1.494)2 − = 0.0134 9.79 9.79 1.8 2(9.81) which yields L = 24.1 km For old DIP, take CH = 80 (Table 2.2) and the Hazen-Williams formula gives ( ) p1 p2 L V 1.85 − = hf = 6.82 1.17 γ γ D CH ( ) 550 350 L 1.494 1.85 − = 6.82 1.17 9.79 9.79 1.8 80 which yields L = 9.4 km The Hazen-Williams equation yields a significantly shorter distance. The Darcy-Weisbach equation is preferable since it covers all flow regimes, while the Hazen-Williams equation is restricted to a narrow range of flow conditions. Also, the value of CH might not be comparable to the value of ks used in the Darcy-Weisbach equation. 3.10. The required storage is the sum of three components: (1) volume to supply the demand in excess of the maximum daily demand; (2) fire storage; and (3) emergency storage. The volume to supply the peak demand can be taken as 25% of the maximum daily demand volume. Problem 3.7 gives the maximum daily flow rate as 2.50 m3 /s, hence the storage volume to supply the peak demand fluctuations over a day, Vpeak , is given by Vpeak = (0.25)(2.50 × 86400) = 54000 m3 The required fire flow, Qf , is calculated in Problem 3.7 as 0.467 m3 /s (= 28,029 L/min) and, according to Table 3.6, this fire flow must be maintained for at least 7 hours. The volume to supply the fire demand, Vfire , is therefore given by Vfire = (0.467 × 3600)(7) = 11768 m3 The emergency storage, Vemer , can be taken as the average daily demand, in which case Vemer = 200000 × 600 = 1.2 × 108 L = 120000 m3 The required volume, V , of the service reservoir is therefore given by V = Vpeak + Vfire + Vemer = 54000 + 11768 + 120000 = 185768 m3 The service reservoir should be designed to store about 185 800 m3 of water. 53 3.11. (a) For line AB: population served = 50000 + 20000 = 70000 people average demand = (70000)(0.6) = 42000 m3 /d = 0.486 m3 /s maximum daily demand = (0.486)(1.8) = 0.875 m3 /s maximum hourly demand = (0.486)(3.25) = 1.58 m3 /s fire demand = 15000 + 10000 = 25000 L/min = 0.417 m3 /s maximum daily + fire demand = 0.875 + 0.417 = 1.29 m3 /s design flow = max(1.29, 1.58) = 1.58 m3 /s For line BC: population served = 20000 people average demand = (20000)(0.6) = 12000 m3 /d = 0.139 m3 /s maximum daily demand = (0.139)(1.8) = 0.250 m3 /s maximum hourly demand = (0.139)(3.25) = 0.452 m3 /s fire demand = 10000 L/min = 0.167 m3 /s maximum daily + fire demand = 0.250 + 0.167 = 0.417 m3 /s design flow = max(0.417, 0.452) = 0.452 m3 /s (b) For the steel transmission pipe (k = 1 mm, D = 1200 mm): π π A = D2 = 1.22 = 1.131 m2 4 4 1.58 VAB = = 1.40 m/s 1.131 VAB D (1.40)(1.2) ReAB = = = 1.68 × 106 ν ( 10−6 ) [ ] 1 k/D 5.74 1/1200 5.74 √ = −2 log + = −2 log + fAB 3.7 Re0.9 AB 3.7 (1.68 × 106 )0.9 fAB = 0.0191 0.452 VBC = = 0.400 m/s 1.131 VBC D (0.400)(1.2) ReBC = = = 4.80 × 105 ν ( 10−6 ) [ ] 1 k/D 5.74 1/1200 5.74 √ = −2 log + = −2 log + fBC 3.7 Re0.9 BC 3.7 (4.80 × 105 )0.9 fBC = 0.0196 Applying the energy equation between A and B, neglecting minor losses, gives 2 V2 LAB VAB pB 0 − fAB + hp = + AB + zB D 2g γ 2g 5000 1.40 2 550 1.402 0 − (0.0191) + hp = + + 20 1.2 2(9.81) 9.79 2(9.81) 54  which gives hp = 84.2 m. The required specific speed, ns , (where ω = 1200 rpm = 125.7 rad/s) is therefore given by 1 1 ωQ 2 (125.7)(1.58) 2 ns = 3 = 3 = 1.03 (ghp ) 4 (9.81 × 84.2) 4 This is a centrifugal pump . Applying the energy equation between B and C, neglecting minor losses, gives 2 V2 LBC VBC pC 0 − fBC + hp = + BC + zC D 2g γ 2g 7000 0.400 2 480 0.4002 0 − (0.0196) + hp = + + 20 1.2 2(9.81) 9.79 2(9.81) which gives hp = 69.97 m. The required specific speed, ns , is therefore given by 1 1 ωQ 2 (125.7)(0.452) 2 ns = 3 = 3 = 0.63 (ghp ) 4 (9.81 × 69.97) 4 This is a centrifugal pump . (c) For the storage reservoir, taking the daily service storage as 25% of the maximum daily demand volume and noting that 10000 L/min fire flow is to be maintained for 3 h, and 15000 L/min for 4 h, gives service storage = (0.25)(1.8)(42000) = 18900 m3 fire storage = (10)(60)(3) + (15)(60)(4) = 5400 m3 emergency storage = 42000 m3 required storage = 18900 + 5400 + 42000 = 66300 m3 3.12. From the given data: Q = 4.67 L/s, L = 110 m, p1 = 380 kPa, and ∆z = 3 m. The pipe velocity is given by Q 4.67 × 10−3 0.00595 V = = π 2 = A 4 D D2 Hence, for V < 2.4 m/s, √ 0.00595 D> = 0.0498 m = 49.8 mm 2.4 Take D = 50 mm and see if this is adequate for the pressure. For copper, ks = 0.0023 mm, 55  and at 20◦ C, ν = 1.00 × 10−6 m/s2 , 4.67 × 10−3 V = π 2 = 2.38 m/s 4 (0.050) vD (2.38)(0.050) Re = = = 119 × 105 ν 1 × 10−6 0.25 0.25 f=[ ( )]2 = [ ( )]2 = 0.0175 ks log 3.7D + 5.74 0.9 0.0023 log 3.7(50) 5.74 + (1.19×10 5 )0.9 Re L V2 110 2.382 hf = f = 0.0175 = 11.12 m D 2g 0.050 2(9.81) ( ) ( ) p1 380 p2 = γ − ∆z − hf = 9.79 − 3 − 11.12 = 241 kPa γ 9.79 Since p2 > 240 kPa, a 50 mm copper line is (barely) adequate. 3.13. The design calculations for this problem are summarized in Table 3.1. 3.14. From the given data: Qref = 200 L/min = 0.00333 m3 /s, L1 = 20 m, L2 = 5 m, ∆z1 = 2 m, ∆z2 = 3 m, p0 = 380 kPa, and p2 = 240 kPa. For galvanized iron, ks = 0.15 mm = 1.5 × 10−4 . From the supply pipe to the first floor: p0 V02 p1 V12 L1 V12 + + z0 = + + z1 + f1 γ 2g γ 2g D1 2g where π 2 A1 = D = 0.7854D2 4 2Qref 2(0.00333) 0.008487 V1 = = 2 = A1 0.7854D D2 ( ) V1 D 0.008487 D 8487 Re1 = = 2 −6 = ν D 10 D 0.25 0.25 f1 = [ ( )]2 = [ ( )]2 −4 4.054×10−5 0.9 log 1.5×10 + 5.74 0.9 log D + 0.001671D 3.7D ( 8487 D ) Combining the above equations and substituting known quantities yields p1 0.25 7.342 × 10−5 = 36.02 − [ ( )]2 (1) γ 4.054×10−5 0.9 D5 log D + 0.001671D From the first to the second floor, p1 V12 p2 V22 L2 V22 + + z1 = + + z2 + f2 γ 2g γ 2g D2 2g 56  Table 3.1: Head Loss in Standard Fittings in Terms of Equivalent Pipe Lengths Starting Fitting Total Friction Other Elev Terminal Terminal Head Flow Length Diam Velocity Length Length Loss Losses Diff Head Pressure Pipe (m) (L/min) (m) (mm) (m/s) (m) (m) (m) (m) (m) (m) (kPa) AB 38.82 409 17.0 64 2.12 5.12 22.12 1.37 15.12 0 22.33 216 57 BB’ 22.33 144 2.7 51 1.17 4.66 7.36 0.21 0 −1.0 23.12 226 B’C’ 23.12 144 2.5 51 1.17 3.05 5.55 0.16 0 2.5 20.46 200 C’F’ 20.46 45 46.0 25 2.62 1.52 47.52 5.15 0 0 15.31 146 C’D’ 20.46 45 4.0 51 0.37 2.13 6.13 0.02 0 4.0 16.44 161 D’E’ 16.44 45 46.0 25 1.52 1.52 47.52 5.15 0 0 11.29 109 where π 2 A2 = D = 0.7854D2 4 Qref 0.00333 0.004244 V2 = = 2 = A2 0.7854D D2 ( ) V2 D 0.004244 D 4244 Re2 = = 2 −6 = ν D 10 D 0.25 0.25 f1 = [ ( )]2 = [ ( )]2 −4 4.054×10−5 0.9 log 1.5×10 + 5.74 0.9 log D + 0.003118D 3.7D ( 4244 D ) Combining the above equations and substituting known quantities yields p1 0.25 4.590 × 10−6 = 27.51 + [ ( )]2 (2) γ 4.054×10−5 D5 log D + 0.003118D0.9 Solving Equations 1 and 2 for D and taking the next larger available diameter yields D = 0.0508 m = 2 in. as the required pipe diameter. For this diameter, the actual pressure on the second floor (p2 ) is 262 kPa and the pressure on the first floor (p1 ) is 295 kPa . 58 Chapter 4 Fundamentals of Flow in Open Channels 4.1. V = 1 m/s A = (b + my)y = (5 + 2 × 2)2 = 18 m2 Therefore Q = V A = (1)(18) = 18 m3 /s 4.2. Q = 8 m3 /s, w1 = 4 m, V1 = 1 m/s, and Q = w1 y1 V1 which leads to Q 8 y1 = = = 2m w1 V1 (4)(1) w2 = 5 m, y2 = y1 − 0.5 m = 1.5 m, and Q = w2 y2 V2 which leads to Q 8 V2 = = = 1.07 m/s w2 y2 (5)(1.5) 4.3. The hydraulic radius, R, is defined by A R= P where, for circular pipes, πD2 A= and P = πD 4 Hence πD2 /4 D R= = πD 4 or D = 4R 59 4.4. The shear stress, τ0 , on the perimeter of the channel is given by τ0 = γRS0 (1) From the given data b = 5 m, y = 1.8 m, m = 1.5, and the geometric properties of the channel are A = by + my 2 = 5(1.8) + 1.5(1.8)2 = 13.86 m2 √ √ P = b + 2 1 + m2 y = 5 + 2 1 + 1.52 (1.8) = 11.49 m A 13.86 R= = = 1.21 m P 11.49 From the given data, τ0 = 3.5 N/m2 , and since γ = 9790 N/m2 , Equation 1 gives the maximum allowable slope, S0 , as τ0 3.5 S0 = = = 0.00030 γR (9790)(1.21) For the excavated channel, ks = 3 mm = 0.003 m, and ν = 1.00 × 10−6 m2 /s at 20◦ C. Substituting these data into Equation 4.38 gives the flow rate, Q, as ( ) √ ks 0.625ν Q = −2A 8gRS0 log10 + √ 12R R 32 8gS0 [ ] √ 0.003 0.625(1.00 × 10−6 ) Q = −2(13.86) 8(9.81)(1.21)(0.00030) log10 + √ 12(1.21) (1.21) 32 8(9.81)(0.00030) = 17.2 m3 /s Therefore, for the given flow depth restrictions in the channel, the flow capacity of the channel is 17.2 m3 /s . 4.5. From the given data: b = 8 m, S0 = 0.0001, ks = 2 mm = 0.002 m, and Q = 15 m3 /s. At 20◦ C, µ = 1.00 × 10−6 m2 /s, and, for a rectangular channel, by A = by and R= 2y + b Substituting into Equation 4.38 gives ( ) √ ks 0.625ν Q = −2A 8gRS0 log10 + 3√ 12R R 2 8gS0 √ [ ] 8y 0.002 0.625(1.00 × 10−6 ) 15 = −2(8y) 8(9.81)( )(0.0001) log10 8y + 8y 3 √ 2y + 8 12( 2y+8 ) ( 2y+8 ) 2 8(9.81)(0.0001) which yields y = 2.25 m Therefore, the uniform-flow depth in the channel is 2.25 m . 60 4.6. Hydraulically rough flow conditions occur in open channels when u∗ ks ≥ 70 (1) ν where √ u∗ = gRSf (2) Equation 4.46 can be rearranged and put in the form ( n )6 ks = = 2.84 × 108 n6 (3) 0.039 Substituting Equations 2 and 3 into Equation 1 and noting that ν = 1.00 × 10−6 m2 /s at 20◦ C and g = 9.81 m/s2 yields √ √ 9.81 RSf × 2.84 × 108 n6 ≥ 70 1.00 × 10−6 which simplifies to √ n6 RSf ≥ 7.9 × 10−14 (4) From the given data: b = 5 m, S0 = 0.05% = 0.0005, and by definition: A by 5y R= = = (5) P 2y + b 2y + 5 Equation 4, can be combined with Equation 5 to give the following condition for fully turbulent flow, √( ) 5y (0.013)6 (0.0005) ≥ 7.9 × 10−14 2y + 5 This condition is satisfied when y ≥ 0.683 m . 4.7. The Darcy-Weisbach uniform-flow equation is given by Equation 4.38 as ( ) √ ks 0.625ν Q = −2A 8gRS0 log10 12R R 32 √8gS0 + (1) where the following variables are known: y = 2.20 m S0 = 0.0006 ks = 2 mm = 0.002 m ν = 1.00 × 10−6 m2 /s g = 9.81 m/s2 A = 3.6y + 2y 2 = 3.6(2.20) + 2(2.20)2 = 17.6 m2 3.6y + 2y 2 3.6(2.20) + 2(2.20)2 R= √ = √ = 1.31 m 3.6 + 2 5y 3.6 + 2 5(2.20) 61  Substituting these variables into Equation 1 yields Q = 34.0 m3 /s . Since y = 2.20 m corresponds to A = 17.6 m2 , then V = 34.0/17.6 = 1.93 m/s . The Manning’s equation gives the average velocity, V , as 1 2 12 V = R 3 S0 n Table 4.2 indicates that a mid-range roughness coefficient for concrete is n = 0.015. The average velocity given by the Manning equation is 1 2 1 V = (1.31) 3 (0.0006) 2 = 1.96 m/s 0.015 and the corresponding flow rate, Q, is Q = AV = (17.6)(1.96) = 34.5 m3 /s Hence, in this case, the Darcy-Weisbach and Manning equations give the similar results . 4.8. The Darcy-Weisbach uniform-flow equation is given by Equation 4.38 as ( ) √ ks 0.625ν Q = −2A 8gRS0 log10 12R R 32 √8gS0 + (1) where the following variables are either known or can be expressed in terms of the uniform-flow depth, y: S0 = 0.0001 ks = 1 mm = 0.001 m Q = 18 m3 /s ν = 1.00 × 10−6 m2 /s g = 9.81 m/s2 A = 5y + 2y 2 5y + 2y 2 R= √ 5 + 2 5y Substituting these variables into Equation 1 and solving for y yields y = 2.19 m . Check u∗ ks /ν and R/ks to determine the state of the flow and the validity of the Manning equation. Taking y = 2.19 m gives R = 1.39 m and √ √ u∗ ks gRS0 ks (9.81)(1.39)(0.0001)(0.001) = = = 37 ν ν 1.00 × 10−6 R 1.39 = = 1390 ks 0.001 Therefore, since 5 ≤ u∗ ks /ν ≤ 70 (i.e., 5 ≤ 37 ≤ 70) then according to Equation 5.19 the flow is in transition . Since u∗ ks /ν < 70 (i.e., 37 < 70) and R/ks > 500 (i.e., 1390 > 500), then the Manning equation is not valid . 62  4.9. Comparing the Manning and Darcy-Weisbach equations √ 1 8g R6 = f n which gives √ 1 1 1 1 1 fR6 f 2 R6 f 2 R6 n= √ =√ = 8g 8(9.81) 8.86 If the friction factor, f , is taken as a constant, the above relation indicates that n will be a function of the depth (since R is a function of the depth). If f ∼ R− 3 , n would be a constant 1 in the above equation. So the answer to the question is no . 4.10. Given: Q = 20 m3 /s, n = 0.015, S0 = 0.01 (a) Manning equation is given by 5 1 2 1 1 An3 12 Q = An Rn3 S02 = S n n P 23 0 n where An = [b + myn ]yn = [2.8 + 2yn ]yn √ √ Pn = b + 2 1 + m2 yn = 2.8 + 2 5yn = 2.8 + 4.472yn Substituting into the Manning equation yields 5 1 [(2.8 + 2yn )yn ] 3 1 20 = 2 (0.01) 2 0.015 (2.8 + 4.472yn ) 3 or 5 [(2.8 + 2yn )yn ] 3 2 = 3.0 (2.8 + 4.472yn ) 3 Solving by trial and error yields yn = 0.91 m (b) Comparing the Manning and Darcy-Weisbach equations gives √ 1 8g R6 = f n which leads to 8gn2 f= 1 R3 63  In this case A = (2.8 + 2y)y = (2.8 + 2 × 0.91)(0.91) = 4.2 m2 P = 2.8 + 4.472(0.91) = 6.87 m A 4.20 R= = = 0.611 m P 6.87 therefore 8(9.81)(0.015)2 f= 1 = 0.0208 (0.611) 3 For fully turbulent flow, where the Manning equation applies, [ ] 1 ks √ = −2 log f 12R [ ] 1 ks √ = −2 log 0.0208 12(0.611) 6.93 = −2 log[0.136ks ] which leads to ks = 0.00249 m = 2.5 mm 4.11. From the given information, 1 n = 0.039d 6 where d is in m. In this case, d = 30 mm = 0.030 m, and a 70% error in d is 0.7(0.030) = 0.021 m. Hence, d = 0.030 m ± 0.021 m. Hence, the “best estimate” of n, denoted by n̄, is given by 1 n̄ = 0.039(0.030) 6 = 0.022 The lower estimate of n, nL , is given by 1 nL = 0.039(0.030 − 0.021) 6 = 0.018 and the upper estimate of n, nU , is given by 1 nU = 0.039(0.030 + 0.021) 6 = 0.024 The maximum percentage error in estimating n is therefore given by 0.022 − 0.018 error = × 100 = 18% 0.022 4.12. According to Equation 4.45, ( )1 6 √1 R n 8g ks 1 = ( ) (1) ks6 2.0 log 12 kRs 64