Modern Compressible Flow

John R. Carr, JRC Analytical Services, Mechanical Engineer, PE, MSME, BSME, 6/25/2011 [email protected] , www.jrcanalyticalservices.com www.jrcanalyticalservices.com,, 615-218-0131 Recent Problems Solved (42 total problems solved from 5/25 to 6/19/2011) rd Reference - “Modern Compressible Flow” by Anderson, 3 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 mole5 2 mass ratio (Note: 1 atm = 1.01 x 10 N/m ). Solution: given are the reservoir conditions (stagnation or total conditions) 5 2 Using P = ρRT, (10 atm)*(1.01 x 10 N/m )/(1 atm) = ρ*(287 N-m/kg-K)*(320 K) 3 ρ = 11.00 kg/m -23 number density n, P = nkT with k = Boltzmann constant = 1.38 x 10 J/K 6 2 -23 (1.01 x 10 N/m ) = n*(1.38 x 10 N-m/K)*(320 K) 26 3 n = 2.287 x 10 particles/m mole-mass ratio η, Pv = ηRuT where R u = Universal Gas Constant 6 2 3 (1.01 x 10 N/m )*(1/11 m  /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(ft-lb)/(sl ug- ºF), and (b) J/(kg-K). Solution: Assuming a calorically perfect gas where c p is constant gives the following equation: s 2 – s1 = cp ln (T2 /T1) – R ln (P 2 /P1) equation (1.36) (a) for English units R = 1716 ft-lb/slug- ºR, cp = γR/(γ R/(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, c p = γR/(γ R/(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 2 duct, the pressure and temperature are P 1 = 1800 lb/ft and T1 = 500 ºR, respectively. At a second point, the temperature is 400 ºR. Calculate the pressure and density at this second point. γ /(γ-1)

Solution: Isentropic flow relations give the equation P 2 /P1 = (T2 /T1) γ /(γ-1) 3.5 2 P2 = P1*(T2 /T1) = (1800)*(400/500) = 824 lb/ft -3 3 P = ρRT gives (824) = ρ2*(1716)*(400) ρ2 = 1.201 x 10 slug/ft

giving

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 M 1 = 3, 3 P1 = 1 atm, and ρ1 = 1.23 kg/m . Calculate the downstream values of P 2, T2, ρ2, M2, u2, Po2, and To2. Solution: Table A.1 gives P o1 /P1 = 36.73, To1 /T1 = 2.8 giving P o1 = 36.73 atm Using P = ρRT 5 2 3 (1 atm)*(1.01 x 10 N/m )/(1 atm) = (1.23 kg/m )*(287 N-m/kg-K)*T 1 gives T1 = 286.1 K, To1 = (2.8)*(286.1) = 801.1 K Table A.2 gives P 2 /P1 = 10.33, ρ2 / ρ1 = 3.857, T2 /T1 = 2.679, Po2 /Po1 = 0.3283, M2 = 0.4752 gives 3 P2 = 10.33 atm, ρ2 = (3.857)*(1.23) = 4.744 kg/m , T2 = (2.679)*(286.1) = 766.5 K, Po2 = (0.3283)*(36.73) = 12.06 atm, T o2 = To1 = 801.1 K 1/2 1/2 a2 = (γRT2) = [(1.4)*(287)*(766.5)] = 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 T o /T = 289.8 So To = 289.8*T = (289.8)*(270) = 78246 K No, I do not believe the calculation of T o 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 P 1 = 1 atm and T 1 = 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) M 1 = 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 T o1 /T1 = 1.8 so To1 = (1.8)*(288) = 518.4 K Table A.3 gives P/P* = 0.3636, T/T* = 0.5289, T o /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 M 1 = 0.2, Table A.1 gives T o1 /T1 = 1.008 so T o1 = (1.008)*(288) = 290.3 K Table A.3 gives P/P* = 2.273, T/T* = 0.2066, T o /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 P 1 = 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 8 during combustion is 4.5 x 10 ft-lb per slug of fuel. Assuming one-dimensional frictionless flow with γ = 1.4 for the fuel-air mixture, calculate M 2, 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 P o1 /P1 = 1.028 and T o1 /T1 = 1.008 Po1 = (1.028)*(10) = 10.28 atm, T o1 = (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 slug  /slug f  a 8 6 q = 4.5 x 10 ft-lb/slug f  x 0.06 slug /slug = 27 x 10 ft-lb/sluga f  a For the air q = c p(To2 – To1) or q/cp = (To2 – To1) 6 To2 = q/cp + To1 = (27 x 10 ft-lb/slug a)/(6006 ft-lb/slug- ºR) + 1008 = 5503.5 ºR Using equation (3.84) which gives 2 2 2 2 To2 /To1 = {(1 + γM1 )/(1 + γM2 )} *(M2 /M1) * 2 2 {[1 + 0.5*(γ – 1)*M2 ]/[1 + 0.5*(γ – 1)M1 ]} with M1 = 0.2, To1 = 1008 ºR, To2 = 5503.5 ºR, and γ = 1.4 gives 2 2 2 2 5503.5/1008 = 5.4598 = {1.056/(1+1.4*M 2 )} *(M2 /0.2) *{(1 + 0.2*M2 )/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 M 2 = 0.76 Table A.1 gives T o2 /T2 = 1.116, Po2 /P2 = 1.466 T2 = To2 /1.116 = 5503.5/1.116 = 4931 ºR Using eqn. (3.78) to get P 2 2 2 2 2 P2 /P1 = (1 + γM1 )/ (1 + γM2 ) = [1 + (1.4)*(0.2) ]/[1 + (1.4)*(0.76) ] = 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 P o1 /P1 = 1.028 and T o1 /T1 = 1.008 Po1 = (1.028)*(10) = 10.28 atm, T o1 = (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 slug  /slug f  a 8 8 q = 4.5 x 10 ft-lb/slug f  x FA slug /slug f  a = (4.5 x 10 )FA ft-lb/slug a (equation 1) For the air q = c p(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, T o /To* = 0.1736 To* = To2 = To /0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/slug a-ºR)*(5806.45 – 1008) ºR = 28819500 ft-lb/slug a 8 Setting equal to equation 1 above gives 28819500 ft-lb/slug a = FA*(4.5 x 10 ) ft-lb/slug a FA = F/A = 0.06404 slug /slug f  a 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 M 2 = 0.5, P2 = 1 atm, and T2 = 270 K. Assuming adiabatic, one-dimensional flow, with a local friction coefficient of 0.005, calculate M 1, P1, and T1 at the entrance to the pipe. 2

Solution: τw = 0.5ρu f (friction coefficient f is assumed constant for the length of the pipe) Table A.4 at M2 = 0.5 gives T 2 /T* = 1.143, P2 /P* = 2.138, ρ2 / ρ* = 1.871, Po2 /Po* = 1.340, 4fL2*/D = 1.069, known is L = 40 m = L 1* - L2* or L1* = L + L2* 4fL1*/D = 4fL/D + 4fL 2*/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 41.069 x1 x2 x3 45.41 0.12 1.197 9.116 8.559/12.9 = (x 1 – 0.14)/0.02 - (x 2 – 1.195)/0.002 = (x 3 – 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 35 º. Upstream of the 2 wave, P1 = 2000 lb/ft , T1 = 520 ºR, and V1 = 3355 ft/s. Calculate P2, T2, V2 and the flow deflection angle θ.

2

Solution: upstream - P 1 = 2000 lb/ft , T1 = 520 ºR, V1 = 3355 ft/s 1/2 1/2 a1 = (γRT1) = [(1.4)*(1716)*(520)] = 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 2 P2 = (3.285)*(2000) = 6570 lb/ft , T2 = (1.473)*(520) = 766 ºR M2 = Mn2 /sin (β – θ) = 0.6355/sin (35 – 17.5) = 2.113 1/2 1/2 a2 = (γRT2) = [(1.4)*(1716)*(766)] = 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 = 30 º impinges on a straight wall. If the upstream flow properties are M 1 = 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 M 1 = 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 M n1 = 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 M 2 = 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 P o2 /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 – M 1 = 4, P1 = 1 atm, expansion corner θ2 = 15º Table A.5 gives v 1 = 65.78º, µ1 = 14.48º gives v 2 = θ2 + v1 = 15 + 65.78 = 80.78º Table A.5 gives M 2 = 5.400 2 2 γ /(γ-1) P1 /P2 = {[1 + 0.5*(γ – 1)*M2 ]/[1 + 0.5*(γ – 1)*M1 ]} 2 2 3.5 = {[1 + 0.2*5.4 ]/[1 + 0.2*4 ]} = 5.490 P2 = P1 /5.490 = 0.182 atm Oblique shock wave – M 2 = 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 P 3 /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 2 properties are M 1 = 3 and P1 = 2116 lb/ft . A Pitot tube is inserted in the flow downstream of the corner. Calculate the value of the pressure measured by the Pitot tube. 2

Solution: upstream M 1 = 3 and P1 = 2116 lb/ft . Table A.1 gives Po1 /P1 = 36.73 2 P01 = (36.73)*(2116) = 77720.64 lb/ft Oblique shock with M 1 = 3, θ =20º. θ-β-M relation gives β = 36.5º Mn1 = M1sin β = (3)*sin 36.5 = 1.784 Table A-2 gives P 2 /P1 = 3.530, Po2 /P01 = 0.8215, Mn2 = 0.6210 2 Po2 = (0.8215)*(77720.68) = 63847.5 lb/ft 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 2 M2 = 2.187, Po2 = 63847.5 lb/ft Table A.2 gives P o3 /Po2 = 0.3733, M3 = 0.4847 2 Po3 = (0.3733)*(63847.5) = 23834 lb/ft (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 A e /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 M e = 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 2 the Pitot tube is 0.6172 atm. The throat area is 0.3 m . Calculate: (a) Exit Mach number Me, (b) Exit area A e, (c) Exit pressure and temperature P e and Te, and (d) mass flow through the nozzle. Solution: (a) there will be a normal shock wave in front of the Pitot tube, P o1 = Po = 10 atm, Po2 = 0.6172 atm (at the Pitot tube), Po2 /Po1 = 0.06172, Table A.2 gives M 1 = Me = 5.00

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(b) Table A.1 for M e = 5 gives Ae /A* = 25 giving exit area of A e = 25A* = 7.5 m (c) exit pressure P e and temperature T e – using isentropic equations instead of tables 2 -γ /(γ – 1) 2 -3.5 -3 Pe /Po = [1 + (γ – 1)*Me  /2] = (1 +0.2*5 ) = 1.89 x 10 -3 Pe = (1.89 x 10 )*(10) = 0.0189 atm 2 -1 2 -1 Te /To = [1 + (γ – 1)*Me  /2] = (1 +0.2*5 ) = 0.1667 Te = (0.1667)*(500) = 83.33 K (d) mass flow rate mdot = ρAu, using the exit plane 1/2 1/2 ae = (γRTe) = [(1.4)*(287)*(83.33)] = 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 3 (0.0189)*(101325) = ρe*(287)*(83.33), ρe = 0.0801 kg/m 3 2 mdot = ρeAeue = (0.0801 kg/m )*(7.5 m )*(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 -2 x 10 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 – T o = 4000 K, Po = 15 atm To determine exit Mach number M e use isentropic relation from Eqn. (3.30) 2 γ /(γ – 1) P0 /Pe = [1 + (γ -1)*Me  /2] or 2 (γ – 1)/ γ 0.22/1.22 Me = [2/(γ -1)]*[(P 0 /Pe) – 1] = (2/0.22)*[(15/0.0174) – 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 P 1 = 0.01 atm and T 1 = 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: 2 2 2 2 2 1/2 MR /(MR – 1) = MS /(MS – 1)*[1 + 2*(γ – 1)*(MS – 1)*(γ + 1/MS )/(γ + 1) ] = where MS = W/a1 (incident shock wave Mach number) MR = (WR + up)/a2 (reflected shock wave Mach number relative to laboratory) For pressure ratio P 2 /P1 = 1050, Table A.2 gives M S= 30, with γ = 1.4 substitution gives 2 2 2 2 2 1/2 MR /(MR – 1) = [30/(30 – 1)]*[1 + (0.8/2.4 )*(30 -1)*(1.4 + 1/30 )] = 0.4426 2 2 1/2 0.4426*M R – MR – 0.4426 = 0 (quadratic eqn – x 1,2 = [-b +- (b – 4ac) ]/2a 2 1/2 MR = {1 +- [(-1) – 4*(0.4426)*(-0.4426)] }/(2*0.4426) MR = 2.638 (positive root) = (W R + up)/a2 1/2 1/2 a1 = (γRT1) = [(1.4)*(287)*(300)] = 347.2 m/s using Eq. (7.16)

1/2

up = (a1 / γ)*[(P 2 /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)] = 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 1/2 1/2 a2 = (γRT2) = [(1.4)*(287)*(52792)] = 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 P 2 /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 P 2 /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 P 2 = 4.04 atm (b) Using Eq. (7.23) 2 2 2 2 2 1/2 MR /(MR – 1) = MS /(MS – 1)*[1 + 2*(γ – 1)*(MS – 1)*(γ + 1/MS )/(γ + 1) ] 2 2 2 2 2 1/2 MR /(MR – 1) = [5.9/(5.9 – 1)]*[1 + (0.8/2.4 )*(5.9 -1)*(1.4 + 1/5.9 )] = 0.4845 2 0.4845M R – MR – 0.4845 = 0 2 1/2 MR = {1 +- [(-1) – 4*(0.4845)*(-0.4845)] }/(2*0.4845) Taking positive root gives M R = 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, T 5 = (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 C p 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: 2 2 1/2 Cp = Cpo /(1 - M∞ ) = -0.9/(1 – 0.6 ) = -1.125 (b) Laitone’s correction is given by the following: 2 1/2 2 2 2 1/2 Cp = Cpo /{(1 - M∞ ) + [M∞ *(1 + (γ -1)* M∞  /2)/(2*(1 - M∞ ) )]*Cpo} 2 1/2 2 2 2 1/2 = -0.9/{(1 – 0.6 ) + [0.6 *(1 + 0.2*0.6 )/2*(1 – 0.6 ) ]*(-0.9)} = -1.54 (c) The Karman-Tsien rule is given by the following: 2 1/2 2 2 1/2 Cp = Cpo /{(1 - M∞ ) + [M∞  /(1 + (1 - M∞ ) )]*Cpo /2 2 1/2 2 2 1/2 = -0.9/{(1 – 0.6 ) + [0.6  /(1 + (1 – 0.6 ) )]*(-0.9/2) = - 1.27