Modern Compressible Flow

John R. Carr, JRC Analytical Services, Mechanical Engineer, PE, MSME, BSME, 6/25/2011 [email protected], www.jrcanalyticalservices.com, 615-218-0131 Recent Problems Solved (42 total problems solved from 5/25 to 6/19/2011) Reference - “Modern Compressible Flow” by Anderson, 3rd edition (2003) Note (for problem solving) the following tables in the book were used: Appendix A Table A.1 – Isentropic flow properties Table A.2 – Normal shock properties Table A.3 – One-dimensional flow with heat addition Table A.4 – One-dimensional flow with friction Table A.5 – Prandtl-Meyer function and Mach angle Chapter 1 – Compressible Flow – Some History and Introductory Thoughts 1.2. In the reservoir of a supersonic wind tunnel, the pressure and temperature of air are 10 atm and 320 K, respectively. Calculate the density, the number density, and the molemass ratio (Note: 1 atm = 1.01 x 105 N/m2). Solution: given are the reservoir conditions (stagnation or total conditions) Using P = ρRT, (10 atm)*(1.01 x 105 N/m2)/(1 atm) = ρ*(287 N-m/kg-K)*(320 K) ρ = 11.00 kg/m3 number density n, P = nkT with k = Boltzmann constant = 1.38 x 10-23 J/K (1.01 x 106 N/m2) = n*(1.38 x 10-23 N-m/K)*(320 K) n = 2.287 x 1026 particles/m3 mole-mass ratio η, Pv = ηRuT where Ru = Universal Gas Constant (1.01 x 106 N/m2)*(1/11 m3/kg) = η*(8314 N-m/kg-mol-K)*(320 K) η = 0.03452 kg-mol/kg 1.4. The pressure and temperature ratios across a given portion of a shock wave in air are P2/P1 = 4.5 and T2/T1 = 1.687, where 1 and 2 denote conditions ahead of and behind the shock wave, respectively. Calculate the change in entropy in units of (a) (ft-lb)/(slug-ºF), and (b) J/(kg-K). Solution: Assuming a calorically perfect gas where cp is constant gives the following equation: s2 – s1 = cp ln (T2/T1) – R ln (P2/P1) equation (1.36) (a) for English units R = 1716 ft-lb/slug-ºR, cp = γR/(γ – 1) = (1.4/0.4)*1716 = 6006 ft-lb/slug-ºR s2 – s1 = (6006)*ln (1.687) – (1716)*ln (4.5) = 560 (ft-lb)/(slug-ºR) (b) for SI units R = 287 J/kg-K, cp = γR/(γ – 1) = (1.4/0.4)*287 = 1004.5 J/kg-K s2 – s1 = (1004.5)*ln (1.687) – (287)*ln (4.5) = 93.6 J/kg-K 1.5. Assume that the flow of air through a given duct is isentropic. At one point in the duct, the pressure and temperature are P1 = 1800 lb/ft2 and T1 = 500 ºR, respectively. At a second point, the temperature is 400 ºR. Calculate the pressure and density at this second point. Solution: Isentropic flow relations give the equation P2/P1 = (T2/T1)γ/(γ-1) giving P2 = P1*(T2/T1)γ/(γ-1) = (1800)*(400/500)3.5 = 824 lb/ft2 P = ρRT gives (824) = ρ2*(1716)*(400) ρ2 = 1.201 x 10-3 slug/ft3 Chapter 2 – Integral Forms of the Conservation Equations for Inviscid Flows Chapter 3 – One-Dimensional Flow 3.4. Consider a normal shock wave in air. The upstream conditions are given by M1 = 3, P1 = 1 atm, and ρ1 = 1.23 kg/m3. Calculate the downstream values of P2, T2, ρ2, M2, u2, Po2, and To2. Solution: Table A.1 gives Po1/P1 = 36.73, To1/T1 = 2.8 giving Po1 = 36.73 atm Using P = ρRT (1 atm)*(1.01 x 105 N/m2)/(1 atm) = (1.23 kg/m3)*(287 N-m/kg-K)*T1 gives T1 = 286.1 K, To1 = (2.8)*(286.1) = 801.1 K Table A.2 gives P2/P1 = 10.33, ρ2/ρ1 = 3.857, T2/T1 = 2.679, Po2/Po1 = 0.3283, M2 = 0.4752 gives P2 = 10.33 atm, ρ2 = (3.857)*(1.23) = 4.744 kg/m3, T2 = (2.679)*(286.1) = 766.5 K, Po2 = (0.3283)*(36.73) = 12.06 atm, To2 = To1 = 801.1 K a2 = (γRT2)1/2 = [(1.4)*(287)*(766.5)]1/2 = 555 m/s M2 = u2/a2, u2 = M2a2 = (0.4752)*(555) = 263.7 m/s 3.7. During the entry of the Apollo space vehicle into the Earth’s atmosphere, the Mach number at a given point on the trajectory was M = 38 and the atmosphere temperature was 270 K. Calculate the temperature at the stagnation point of the vehicle, assuming a calorically perfect gas with γ = 1.4. Do you think this is an accurate calculation? If not, why? If not, is your answer an overestimate or underestimate? Solution: Using Table A.1 for M = 38 gives To/T = 289.8 So To = 289.8*T = (289.8)*(270) = 78246 K No, I do not believe the calculation of To is accurate because the assumption of a calorically perfect gas is only good up to about M = 5. The answer is an overestimate of the actual value, which I believe to be about 11500 K but I will have to wait until I study Hypersonics a little more before I know how to calculate it. 3.8. Consider air entering a heated duct at P1 = 1 atm and T1 = 288 K. Ignore the effect of friction. Calculate the amount of heat per unit mass (in J/kg) necessary to choke the flow at the exit of the duct, as well as the pressure and temperature at the duct exit, for an inlet Mach number of (a) M1 = 2.0, and (b) M1 = 0.2. Solution: The exit flow is choked when it is sonic giving P* and T* conditions. (a) for M1 = 2.0, Table A.1 gives To1/T1 = 1.8 so To1 = (1.8)*(288) = 518.4 K Table A.3 gives P/P* = 0.3636, T/T* = 0.5289, To/To* = 0.7934 P* = P/0.3636 = 2.750 atm, T* = T/0.5289 = 544.5 K, To2 = To* = To1/0.7934 = 653.4 K q = cp*(To2 – To1) = cp*(To* – To1) cp = γR/(γ – 1) = (1.4)*(287)/(0.4) = 1004.5 J/kg-K q = (1004.5)*(653.4 – 518.4) = 135607.5 J/kg (b) for M1 = 0.2, Table A.1 gives To1/T1 = 1.008 so To1 = (1.008)*(288) = 290.3 K Table A.3 gives P/P* = 2.273, T/T* = 0.2066, To/To* = 0.1736 P* = P/2.273 = 0.440 atm, T* = T/0.2066 = 1394 K, To2 = To* = To1/0.1736 = 1672 K q = cp*(To2 – To1) = cp*(To* – To1) q = (1004.5)*(1672 – 290.3) = 1388154 J/kg 3.9. Air enters a combustor of a jet engine at P1 = 10 atm, T1 = 1000 ºR, and M1 = 0.2. Fuel is injected and burned, with a fuel-air ratio (by mass) of 0.06. The heat released during combustion is 4.5 x 108 ft-lb per slug of fuel. Assuming one-dimensional frictionless flow with γ = 1.4 for the fuel-air mixture, calculate M2, P2, and T2 at the exit of the combustor. Solution: air inlet state 1 - P1 = 10 atm, T1 = 1000 ºR, and M1 = 0.2 Table A.1 gives Po1/P1 = 1.028 and To1/T1 = 1.008 Po1 = (1.028)*(10) = 10.28 atm, To1 = (1.008)*(1000) = 1008 ºR R = 1716 ft-lb/slug-ºR, cp = γR/(γ -1) = (1.4)*(1716)/(0.4) = 6006 ft-lb/slug-ºR fuel-air ratio (by mass) F/A = 0.06 slugf/sluga q = 4.5 x 108 ft-lb/slugf x 0.06 slugf/sluga = 27 x 106 ft-lb/sluga For the air q = cp(To2 – To1) or q/cp = (To2 – To1) To2 = q/cp + To1 = (27 x 106 ft-lb/sluga)/(6006 ft-lb/slug-ºR) + 1008 = 5503.5 ºR Using equation (3.84) which gives To2/To1 = {(1 + γM12)/(1 + γM22)}2*(M2/M1)2* {[1 + 0.5*(γ – 1)*M22]/[1 + 0.5*(γ – 1)M12]} with M1 = 0.2, To1 = 1008 ºR, To2 = 5503.5 ºR, and γ = 1.4 gives 5503.5/1008 = 5.4598 = {1.056/(1+1.4*M22)}2*(M2/0.2)2*{(1 + 0.2*M22)/1.008} LHS RHS Guess M2 until RHS = LHS M2 RHS 0.5 3.984 0.6 5.235 0.72 5.313 0.76 5.448 close enough, so M2 = 0.76 Table A.1 gives To2/T2 = 1.116, Po2/P2 = 1.466 T2 = To2/1.116 = 5503.5/1.116 = 4931 ºR Using eqn. (3.78) to get P2 P2/P1 = (1 + γM12)/ (1 + γM22) = [1 + (1.4)*(0.2)2]/[1 + (1.4)*(0.76)2] = 0.5839 P2 = (0.5839)*(10) = 5.839 atm 3.10. For the inlet conditions of Prob. 3.9, calculate the maximum fuel-air ratio beyond which the flow will be choked at the exit. Solution: air inlet state 1 - P1 = 10 atm, T1 = 1000 ºR, and M1 = 0.2 Table A.1 gives Po1/P1 = 1.028 and To1/T1 = 1.008 Po1 = (1.028)*(10) = 10.28 atm, To1 = (1.008)*(1000) = 1008 ºR R = 1716 ft-lb/slug-ºR cp = 6006 ft-lb/slug-ºR fuel-air ratio (by mass) F/A = unknown = FA slugf/sluga q = 4.5 x 108 ft-lb/slugf x FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga (equation 1) For the air q = cp(To2 – To1) Exit flow – state 2 – choked flow is assumed For M1 = 0.2 Table A.3 gives P/P* = 2.273, T/T* = 0.2066, To/To* = 0.1736 To* = To2 = To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/sluga Setting equal to equation 1 above gives 28819500 ft-lb/sluga = FA*(4.5 x 108) ft-lb/sluga FA = F/A = 0.06404 slugf/sluga or less to prevent choked flow at the exit 3.12. Air is flowing through a pipe of 0.02-m inside diameter and 40-m length. The conditions at the exit of the pipe are M2 = 0.5, P2 = 1 atm, and T2 = 270 K. Assuming adiabatic, one-dimensional flow, with a local friction coefficient of 0.005, calculate M1, P1, and T1 at the entrance to the pipe. Solution: τw = 0.5ρu2f (friction coefficient f is assumed constant for the length of the pipe) Table A.4 at M2 = 0.5 gives T2/T* = 1.143, P2/P* = 2.138, ρ2/ρ* = 1.871, Po2/Po* = 1.340, 4fL2*/D = 1.069, known is L = 40 m = L1* - L2* or L1* = L + L2* 4fL1*/D = 4fL/D + 4fL2*/D = 4*(0.005)*(40)/(0.02) + 1.069 = 40 + 1.069 = 41.069 From Table A.4 – linear interpolation 4fL1*/D M1 T1/T* P1/P* 32.51 0.14 1.195 7.809 x2 x3 41.069 x1 45.41 0.12 1.197 9.116 8.559/12.9 = (x1 – 0.14)/0.02 - (x2 – 1.195)/0.002 = (x3 – 7.809)/(9.116 – 7.809) x1 = M1 = 0.127, x2 = T1/T* = 1.196, x3 = P1/P* = 8.676 T1 = (T1/T*)*(T*/T2)*T2 = (1.196)*(1/1.143)*(270) = 282.5 K P1 = (P1/P*)*(P*/P2)*P2 = (8.676)*(1/2.138)*(1) = 4.058 atm Chapter 4 – Oblique Shock and Expansion Waves 4.1. Consider an oblique shock wave with a wave angle equal to 35º. Upstream of the wave, P1 = 2000 lb/ft2, T1 = 520 ºR, and V1 = 3355 ft/s. Calculate P2, T2, V2 and the flow deflection angle θ. Solution: upstream - P1 = 2000 lb/ft2, T1 = 520 ºR, V1 = 3355 ft/s a1 = (γRT1)1/2 = [(1.4)*(1716)*(520)]1/2 = 1117.7 ft/s M1 = V1/a1 = 3355/1117.7 = 3.002 Using the θ-β-M plot gives θ = 17.5º (flow deflection angle) Mn1 = M1sin β = (3.002)*sin 35º = 1.722 Using Table A.2 (for M = 1.720) gives P2/P1 = 3.285, T2/T1 = 1.473, Mn2 = 0.6355 P2 = (3.285)*(2000) = 6570 lb/ft2, T2 = (1.473)*(520) = 766 ºR M2 = Mn2/sin (β – θ) = 0.6355/sin (35 – 17.5) = 2.113 a2 = (γRT2)1/2 = [(1.4)*(1716)*(766)]1/2 = 1356.5 ft/s M2 = V2/a2 gives V2 = M2a2 = (2.113)*(1356.5) = 2867 ft/s 4.7. An incident shock wave with a wave angle = 30º impinges on a straight wall. If the upstream flow properties are M1 = 2.8, P1 = 1 atm, and T1 = 300 K, calculate the pressure, temperature, Mach number, and total pressure downstream of the reflected shock. Solution: upstream – state 1 - M1 = 2.8, P1 = 1 atm, T1 = 300 K, β1 = 30º θ-β-M relation gives θ = 11º, Mn1 = M1sin β1 = (2.8)*(sin 30º) = 1.4 From Table A.1 for M1 = 2.8, Po1/P1 = 27.14, To1/T1 = 2.568 Po1 = 27.14 atm, To1 = (2.568)*(300) = 770.4 K From Table A.2 for Mn1 = 1.4, P2/P1 = 2.120, T2/T1 = 1.255, Mn2 = 0.7397 P2 = 2.120 atm, T2 = (1.255)*(300) = 376.5 K M2 = (Mn2)/[sin (β1 – θ)] = 0.7397/sin(30 – 11) = 2.272 For reflected shock M2 = 2.272, θ = 11º, the θ-β-M relation gives β2 = 36º Mn2 = M2sin β2 = (2.272)*sin 36º = 1.335 From Table A.2 (at M = 1.34) P3/P2 = 1.928, T3/T2 = 1.216, P03/P02 = 0.9718, Mn3 = 0.7664 P3 = 1.928*P2 = (1.928)*(2.120) = 4.087 atm T3 = 1.216*T2 = (1.216)*(376.5) = 457.8 K Back to M2 = 2.272, Table A.1 gives Po2/P2 = 11.56, Po2 = (11.56)*(2.120) = 24.5 atm Thus Po3 = 0.9718*Po2 = (0.9718)*(24.5) = 23.8 atm M3 = Mn3/sin (β2 – θ) = 0.7664/sin (36 – 11) = 1.813 4.12. Consider a supersonic flow with an upstream Mach number of 4 and pressure of 1 atm. This flow is first expanded around an expansion corner with θ = 15º, and then compressed through a compression corner with equal angle θ = 15º so that it is returned to its original upstream direction. Calculate the Mach number and pressure downstream of the compression corner. Solution: upstream – M1 = 4, P1 = 1 atm, expansion corner θ2 = 15º Table A.5 gives v1 = 65.78º, µ1 = 14.48º gives v2 = θ2 + v1 = 15 + 65.78 = 80.78º Table A.5 gives M2 = 5.400 P1/P2 = {[1 + 0.5*(γ – 1)*M22]/[1 + 0.5*(γ – 1)*M12]}γ/(γ-1) = {[1 + 0.2*5.42]/[1 + 0.2*42]}3.5 = 5.490 P2 = P1/5.490 = 0.182 atm Oblique shock wave – M2 = 5.400, θ3 = 15º, θ-β-M relation gives β = 23.5º Mn2 = M2sin β = (5.4)*sin 23.5º = 2.153 Table A.2 gives P3/P2 = 5.336, Mn3 = 0.5540 P3 = (5.226)*(0.182) = 0.952 atm M3 = Mn3/sin (β – θ) = 0.5540/sin (23.5 – 15) = 3.748 4.14. Consider a supersonic flow past a compression corner with θ = 20º. The upstream properties are M1 = 3 and P1 = 2116 lb/ft2. A Pitot tube is inserted in the flow downstream of the corner. Calculate the value of the pressure measured by the Pitot tube. Solution: upstream M1 = 3 and P1 = 2116 lb/ft2. Table A.1 gives Po1/P1 = 36.73 P01 = (36.73)*(2116) = 77720.64 lb/ft2 Oblique shock with M1 = 3, θ =20º. θ-β-M relation gives β = 36.5º Mn1 = M1sin β = (3)*sin 36.5 = 1.784 Table A-2 gives P2/P1 = 3.530, Po2/P01 = 0.8215, Mn2 = 0.6210 Po2 = (0.8215)*(77720.68) = 63847.5 lb/ft2 M2 = Mn2/sin (β – θ) = 0.6210/sin (36.5 -20) = 2.187 There will be a normal shock in front of the Pitot tube M2 = 2.187, Po2 = 63847.5 lb/ft2 Table A.2 gives Po3/Po2 = 0.3733, M3 = 0.4847 Po3 = (0.3733)*(63847.5) = 23834 lb/ft2 (Pitot tube total pressure measurement) Chapter 5 – Quasi-One-Dimensional Flow 5.8 A blunt-nosed aerodynamic model is mounted in the test section of a supersonic wind tunnel. If the tunnel reservoir pressure and temperature are 10 atm and 800 ºR, respectively, and the exit-to-throat area ratio is 25, calculate the pressure and temperature at the nose of the model. Solution: from Table A.1 for Ae/A* = 25, Me = 5.000, Po/Pe = 529.1, To/Te = 6 Pe = Po/529.1 = 10/529.1 = 0.01890 atm Te = To/6 = 800/6 = 133.3 ºR There will be a normal shock in front of the nose of the blunt body Table A.2 for Me = 5 gives P2/Pe = 29, T2/Te = 5.80 P2 = 29*Pe = 29*(0.01890) = 0.548 atm T2 = 5.8*Te = 5.8*(133.3) = 773.3 ºR as the static properties on the nose of the blunt body. 5.10 Consider a supersonic nozzle with a Pitot tube mounted at the exit. The reservoir pressure and temperature are 10 atm and 500 K, respectively. The pressure measured by the Pitot tube is 0.6172 atm. The throat area is 0.3 m2. Calculate: (a) Exit Mach number Me, (b) Exit area Ae, (c) Exit pressure and temperature Pe and Te, and (d) mass flow through the nozzle. Solution: (a) there will be a normal shock wave in front of the Pitot tube, Po1 = Po = 10 atm, Po2 = 0.6172 atm (at the Pitot tube), Po2/Po1 = 0.06172, Table A.2 gives M1 = Me = 5.00 (b) Table A.1 for Me = 5 gives Ae/A* = 25 giving exit area of Ae = 25A* = 7.5 m2 (c) exit pressure Pe and temperature Te – using isentropic equations instead of tables Pe/Po = [1 + (γ – 1)*Me2/2]-γ/(γ – 1) = (1 +0.2*52)-3.5 = 1.89 x 10-3 Pe = (1.89 x 10-3)*(10) = 0.0189 atm Te/To = [1 + (γ – 1)*Me2/2]-1 = (1 +0.2*52)-1 = 0.1667 Te = (0.1667)*(500) = 83.33 K (d) mass flow rate mdot = ρAu, using the exit plane ae = (γRTe)1/2 = [(1.4)*(287)*(83.33)]1/2 = 183.0 m/s Me = ue/ae gives ue = Meae = (5)*(183) = 914.9 m/s Using perfect gas equation of state P = ρRT for exit flow properties (0.0189)*(101325) = ρe*(287)*(83.33), ρe = 0.0801 kg/m3 mdot = ρeAeue = (0.0801 kg/m3)*(7.5 m2)*(914.9 m/s) = 549.5 kg/s 5.16 Consider a rocket engine burning hydrogen and oxygen. The combustor chamber temperature and pressure are 4000 K and 15 atm, respectively. The exit pressure is 1.174 x 10-2 atm. Calculate the Mach number at the exit. Assume that γ = const = 1.22 and that R = 519.6 J/kg-K. Solution: Reservoir/total property conditions – To = 4000 K, Po = 15 atm To determine exit Mach number Me use isentropic relation from Eqn. (3.30) P0/Pe = [1 + (γ -1)*Me2/2]γ/(γ – 1) or Me2 = [2/(γ -1)]*[(P0/Pe)(γ – 1)/γ – 1] = (2/0.22)*[(15/0.0174)0.22/1.22 – 1] = 21.668 Me = 4.655 Chapter 6 – Differential Conservation Equations for Inviscid Flows – no homework problems. Chapter 7 – Unsteady Wave Motion 7.5 Consider an incident normal shock wave that reflects from the end wall of a shock tube. The air in the driven section of shock tube (ahead of the incident wave) is at P1 = 0.01 atm and T1 = 300 K. The pressure ratio across the incident shock is 1050. With the use of Eq. (7.23), calculate (a) the reflected shock wave velocity relative to the tube, and (b) the pressure and temperature behind the reflected shock. Solution: Eq. (7.23) is the following: MR/(MR2 – 1) = MS/(MS2 – 1)*[1 + 2*(γ – 1)*(MS2 – 1)*(γ + 1/MS2)/(γ + 1)2]1/2 = where MS = W/a1 (incident shock wave Mach number) MR = (WR + up)/a2 (reflected shock wave Mach number relative to laboratory) For pressure ratio P2/P1 = 1050, Table A.2 gives MS= 30, with γ = 1.4 substitution gives MR/(MR2 – 1) = [30/(302 – 1)]*[1 + (0.8/2.42)*(302 -1)*(1.4 + 1/302)]1/2 = 0.4426 0.4426*MR2 – MR – 0.4426 = 0 (quadratic eqn – x1,2 = [-b +- (b2 – 4ac)1/2]/2a MR = {1 +- [(-1)2 – 4*(0.4426)*(-0.4426)]1/2}/(2*0.4426) MR = 2.638 (positive root) = (WR + up)/a2 a1 = (γRT1)1/2 = [(1.4)*(287)*(300)]1/2 = 347.2 m/s using Eq. (7.16) up = (a1/γ)*[(P2/P1) – 1]*{[2γ/(γ + 1)]/[(P2/P1) + (γ -1)/(γ + 1)]}1/2 = (347.2/1.4)*(1050 – 1)*[(2.8/2.4)/(1050 + 1/6)]1/2 = 8671 m/s Using Eq. (7.10) T2/T1 = (P2/P1)*{[(γ + 1)/(γ – 1) + P2/P1]/[1 + (γ + 1)(P2/P1)/(γ – 1)]} = (1050)*{[6 + 1050]/[1 + (6*1050)]} = 176 T2 = 176*T1 = 176*300 = 52792 K a2 = (γRT2)1/2 = [(1.4)*(287)*(52792)]1/2 = 4606 m/s (a) MRa2 = WR + up gives WR = MRa2 - up = (2.638)*(4606) – 8671 = 3480 m/s (b) P2 = 10.50 atm, T2 = 52792 K 7.7 Consider a blunt-nosed aerodynamic model mounted inside the driven section of a shock tube. The axis of the model is aligned parallel to the axis of the shock tube, and the nose of the model faces towards the on-coming incident shock wave. The driven gas is air initially at a temperature and pressure of 300 K and 0.1 atm, respectively. After the diaphragm is broken, an incident shock wave with a pressure ratio of P2/P1 = 40.4 propagates into the driven section. (a) Calculate the pressure and temperature at the nose of the model shortly after the incident shock sweeps by the model. (b) Calculate the pressure and temperature at the nose of the model after the reflected shock sweeps by the model. Solution: (a) Using Table A.2 for P2/P1 = 40.4 gives MS = 5.900 Using Eq. (7.10) T2/T1 = (P2/P1)*{[(γ + 1)/(γ – 1) + P2/P1]/[1 + (γ + 1)/(P2/P1)/(γ – 1)]} = (40.4)*[(6 + 40.4)/(1 + 6*40.4)] = 7.072 T2 = 2310 K P2/P1 = 40.4 gives P2 = 4.04 atm (b) Using Eq. (7.23) MR/(MR2 – 1) = MS/(MS2 – 1)*[1 + 2*(γ – 1)*(MS2 – 1)*(γ + 1/MS2)/(γ + 1)2]1/2 MR/(MR2 – 1) = [5.9/(5.92 – 1)]*[1 + (0.8/2.42)*(5.92 -1)*(1.4 + 1/5.92)]1/2 = 0.4845 0.4845MR2 – MR – 0.4845 = 0 MR = {1 +- [(-1)2 – 4*(0.4845)*(-0.4845)]1/2}/(2*0.4845) Taking positive root gives MR = 2.469 and using Table A.2 (with linear interpolation) P5/P2 = 6.952, T5/T2 = 2.108 P5 = (6.952)*(4.04) = 28.09 atm, T5 = (2.108)*(2310) = 4869 K Chapter 8 – General Conservation Equations Revisited: Velocity Potential Equation – no homework problems Chapter 9 – Linearized Flow 9.2 In low-speed flow, the pressure coefficient at a point on an airfoil is -0.90. Calculate the value of Cp at the same point for M∞ = 0.6 by means of (a) The Prandtl-Glauert rule, (b) Laitone’s correction, and (c) The Karman-Tsien rule. Solution: (a) The Prandtl-Glauert rule is given by the following: Cp = Cpo/(1 - M∞2) = -0.9/(1 – 0.62)1/2 = -1.125 (b) Laitone’s correction is given by the following: Cp = Cpo/{(1 - M∞2)1/2 + [M∞2*(1 + (γ -1)* M∞2/2)/(2*(1 - M∞2)1/2)]*Cpo} = -0.9/{(1 – 0.62)1/2 + [0.62*(1 + 0.2*0.62)/2*(1 – 0.62)1/2]*(-0.9)} = -1.54 (c) The Karman-Tsien rule is given by the following: Cp = Cpo/{(1 - M∞2)1/2 + [M∞2/(1 + (1 - M∞2)1/2)]*Cpo/2 = -0.9/{(1 – 0.62)1/2 + [0.62/(1 + (1 – 0.62)1/2)]*(-0.9/2) = - 1.27