Sol 9 - Chapter 9 • Compressible Flow 653 9.32 The large...

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Unformatted text preview: Chapter 9 • Compressible Flow 653 9.32 The large compressed-air tank in Fig. P9.32 exhausts from a nozzle at an exit velocity of 235 m/s. The mercury manometer reads h = 30 cm. Assuming isentropic flow, compute the pressure (a) in the tank and (b) in the atmosphere. (c) What is the exit Mach number? Solution: The tank temperature = To = 30°C = 303 K. Then the exit jet temperature is Fig. P9.32 Te = To − Ve2 (235)2 235 = 303 − = 276 K, ∴ Ma e = ≈ 0.706 Ans. (c) 2c p 2(1005) 1.4(287)(276) 3.5 p tank Then = 1 + 0.2Ma 2 = 1.395 and p tank − pe = (ρmercury − ρtank )gh e pe ( ) Guess ρtank ≈ 1.6 kg/m3 , ∴ po − pe ≈ (13550 − 1.6)(9.81)(0.30) ≈ 39900 Pa Solve the above two simultaneously for pe ≈ 101 kPa and p tank ≈ 140.8 kPa Ans. (a, b) 9.33 Air flows isentropically from a reservoir, where p = 300 kPa and T = 500 K, to section 1 in a duct, where A1 = 0.2 m2 and V1 = 550 m/s. Compute (a) Ma1; (b) T1; (c) p1; (d) m; and (e) A*. Is the flow choked? Solution: Use the energy equation to calculate T1 and then get the Mach number: T1 = To − 2 V1 (550)2 = 500 − = 350 K 2c p 2(1005) Ans. (b) 550 ≈ 1.47 375 Then a1 = 1.4(287)(350) = 375 m/s, Ma1 = V1/a1 = Ans. (a) The flow must be choked in order to produce supersonic flow in the duct. Answer. 2 p1 = po 1 + 0.2 Ma1 ( ) 3.5 = 300 /[1 + 0.2(1.47)2 ]3.5 ≈ 86 kPa Ans. (c) Ans. (d) ρ1 = p1 86000 kg kg = ≈ 0.854 3 , ∴ m = ρ AV = (0.854)(0.2)(550) ≈ 94 RT1 287(350) s m Chapter 9 • Compressible Flow 655 9.35 Helium, at To = 400 K, enters a nozzle isentropically. At section 1, where A1 = 0.1 m2, a pitot-static arrangement (see Fig. P9.25) measures stagnation pressure of 150 kPa and static pressure of 123 kPa. Estimate (a) Ma1; (b) mass flow; (c) T1; and (d) A*. Solution: For helium, from Table A.4, take k = 1.66 and R = 2077 J/kg⋅K. (a) The local pressure ratio is given, hence we can estimate the Mach number: po 150 ⎡ 1.66 − 1 2⎤ = = ⎢1 + Ma1 ⎥ 2 p1 123 ⎣ ⎦ 1.66 /(1.66 −1) , solve for Ma 1 ≈ 0.50 Ans. (a) Use this Mach number to estimate local temperature, density, velocity, and mass flow: T1 = To 400 = ≈ 370 K 2 1 + (k − 1) Ma1 /2 1 + 0.33(0.50)2 p1 123000 kg = ≈ 0.160 3 RT1 2077(370) m Ans. (c) ρ1 = m s kg Finally, m = ρ1 A1V1 = (0.160)(0.1)(565) ≈ 9.03 s Finally, A* can be computed from Eq. (9.44), using k = 1.66: V1 = Ma1a1 = 0.50[1.66(2077)(370)]1/2 ≈ 565 Ans. (b) 2 A1 1 ⎡ 1 + 0.33 Ma1 ⎤ = ⎢ ⎥ A* Ma1 ⎣ (1.66 + 1)/2 ⎦ (1/2)(2.66)/(0.66) ≈ 1.323, A* ≈ 0.0756 m 2 Ans. (d) 9.36 An air tank of volume 1.5 m3 is at 800 kPa and 20°C when it begins exhausting through a converging nozzle to sea-level conditions. The throat area is 0.75 cm2. Estimate (a) the initial mass flow; (b) the time to blow down to 500 kPa; and (c) the time when the nozzle ceases being choked. Solution: For sea level, pambient = 101.35 kPa < 0.528ptank, hence the flow is choked until the tank pressure drops to pambient/0.528 = 192 kPa. (a) We obtain minitial = mmax = 0.6847 po A* RTo = 0.6847 800000(0.75E−4 m 2 ) 287(293) = 0.142 kg s Ans. (a) (b) For a control volume surrounding the tank, a mass balance gives p A* d υ dpo = − m = −0.6847 o , separate the variables: (ρoυ ) = dt RTo dt RTo 670 Solutions Manual • Fluid Mechanics, Fifth Edition Across the shock, at Ma1 = 1.985, At A 3 = 30 cm 2 , Finally, po2 = A* 2 = 1.374, A* = 1.374(12) ≈ 16.5 cm 2 2 A* 1 Ans. (c) A3 30 = = 1.82, whence subsonic Ma 3 ≈ 0.34 A* 16.5 2 Ans. (d) Ans. (b) p o1 328 = 328 kPa, p3 = ≈ 303 kPa 1.374 [1 + 0.2(0.34)2 ]3.5 9.56 Air from a reservoir at 20°C and 500 kPa flows through a duct and forms a normal shock downstream of a throat of area 10 cm2. By an odd coincidence it is found that the stagnation pressure downstream of this shock exactly equals the throat pressure. What is the area where the shock wave stands? Solution: If a shock forms, the throat is sonic, A* = 10 cm2. Now p* = 0.5283p o1 = 0.5283(500) ≈ 264 kPa = p o2 1 also Then po2 264 = = 0.5283: Table B.2, read Ma1 2.43 p o1 500 2 3.0 [1 + 0.2(2.43) ] So A1 /A* = 1 1.728(2.43) ≈ 2.47, or A1 (at shock) = 2.47(10) ≈ 24.7 cm 2 Ans. 9.57 Air flows from a tank through a nozzle into the standard atmosphere, as in Fig. P9.57. A normal shock stands in the exit of the nozzle, as shown. Estimate (a) the tank pressure; and (b) the mass flow. Solution: The throat must be sonic, and the area ratio at the shock gives the Mach number: 3 Fig. P9.57 2 ⎡ ⎤ 14 ⎣1 + 0.2Ma1 ⎦ , solve Ma ≈ 1.76 upstream of the shock A1 /A* = = 1.4 = 1 10 1.728Ma1 Chapter 9 • Compressible Flow 671 Then p2 /p1 |shock = 2.8(1.76)2 − 0.4 101350 ≈ 3.46, p2 = 1 atm, p1 = ≈ 29289 Pa 2.4 3.46 Thus p tank = po1 = 29289[1 + 0.2(1.76)2 ]3.5 ≈ 159100 Pa m = m max = 0.6847po A* RTo = 0.6847(159100)(0.001) ≈ 0.333 kg s Ans. (b) Ans. (a) Given that To = 100°C = 373 K and a critical throat area of 10 cm2, we obtain 287(373) 9.58 Argon (Table A.4) approaches a normal shock with V1 = 700 m/s, p1 = 125 kPa, and T1 = 350 K. Estimate (a) V2, and (b) p2. (c) What pressure p2 would result if the same velocity change V1 to V2 were accomplished isentropically? Solution: For argon, take k = 1.67 and R = 208 J/kg⋅K. Determine the Mach number upstream of the shock: a1 = kRT1 = 1.67(208)(350) ≈ 349 Then p2 p1 m 700 ; Ma1 = V1 /a1 = ≈ 2.01 s 349 Ans. (b) |shock = 2(1.67)(2.01) 2 − 0.67 1.67 + 1 ≈ 4.79, or p2 = 4.79(125) ≈ 599 kPa and V2 /V1 = 0.67(2.01)2 + 2 m ≈ 0.437, or V2 = 0.437(700) = 306 2 s 2.67(2.01) Ans. (a) 672 Solutions Manual • Fluid Mechanics, Fifth Edition For an “isentropic” calculation, assume the same density ratio across the shock: ρ2 /ρ1 = V1 /V2 = 1 = 2.29; Isentropic: p2 /p1 ≈ ( ρ2 /ρ1 )k , 0.437 or: p2,isentropic ≈ 125(2.29)1.67 ≈ 498 kPa Ans. (c) 9.59 Air, at stagnation conditions of 450 K and 250 kPa, flows through a nozzle. At section 1, where the area = 15 cm2, there is a normal shock wave. If the mass flow is 0.4 kg/s, estimate (a) the Mach number; and (b) the stagnation pressure just downstream of the shock. Solution: If there is a shock wave, then the mass flow is maximum: mmax = 0.4 Then p A* kg 250000 A* , solve A* = 0.000840 m 2 = 0.6847 o = 0.6847 s RTo 287(450) A1 0.0015 = = 1.786 Table B.1: Read Ma1,upstream ≈ 2.067 A* 0.00084 Ans. (a) Finally, from Table B.2, read Ma 1, downstream ≈ 0.566 Also, Table B.2: po 2 = 0.690, p0,downstream = 0.690(250) ≈ 172 kPa po1 Ans. (b) 9.60 When a pitot tube such as Fig. (6.30) is placed in a supersonic flow, a normal shock will stand in front of the probe. Suppose the probe reads p0 = 190 kPa and p = 150 kPa. If the stagnation temperature is 400 K, estimate the (supersonic) Mach number and velocity upstream of the shock. Fig. P9.60 Solution: We can immediately find Ma inside the shock: po2 /p2 = 190 = 1.267 = 1 + 0.2Ma 2 2 150 ( ) 3.5 , solve Ma 2 ≈ 0.591 Ans. 2 Then, across the shock, Ma1 = 0.4(0.591)2 + 2 , solve Ma 1 ≈ 1.92 2.8(0.591)2 − 0.4 Chapter 9 • Compressible Flow 673 T1 = 400 = 230 K, a1 = 1.4(287)(230) ≈ 304 m/s, [1 + 0.2(1.92)2 ] V1 = Ma1a1 = (1.92)(304) ≈ 585 m/s Ans. P9.61 Air flows from a large tank, where T = 376 K and p = 360 kPa, to a design condition where the pressure is 9800 Pa. The mass flow is 0.9 kg/s. However, there is a normal shock in the exit plane just after this condition is reached. Estimate (a) the throat area; and, just downstream of the shock, (b) the Mach number, (c) the temperature, and (d) the pressure. Solution: The low design pressure definitely indicates a supersonic condition: p 9800 2 = = 0.0272 = (1 + 0.2 Ma1 ) −3.5 , solve for po 360000 Choked : m = mmax = 0.6847 po A * RTo = 0.6847 (360000) A * 287(376) Ma1 = 3.00 , Solve A * = 0.0012 m 2 Ans.(a) Find T1 in front of the shock and then use the normal shock conditions: 2 T1 = To /(1 + 0.2 Ma1 ) = (376 K ) /[1 + 0.2(3.0) 2 ] = 134 K Table B.2, Ma1=3.0 : Ma 2 = 0.475 p 2 / p1 = 10.333 , Ans.(b) Ans.(c) Ans.( d ) T2 / T1 = 2.679 , T2 = (2.679)(134 K ) = 360 K p 2 = (10.333)(9800) = 101,300 Pa The problem worked out so that the external pressure was 1 atm. 9.62 An atomic explosion propagates into still air at 14.7 psia and 520°R. The pressure just inside the shock is 5000 psia. Assuming k = 1.4, what are the speed C of the shock and the velocity V just inside the shock? Solution: The pressure ratio tells us the Mach number of the shock motion: 2 5000 2.8Ma1 − 0.4 = 340 = , solve for Ma1 ≈ 17.1 14.7 2.4 ft a1 = 1.4(1717)(520) = 1118 ft/s, ∴ V1 = C = 17.1(1118) ≈ 19100 Ans. (a) s p 2 /p1 = Chapter 9 • Compressible Flow 673 T1 = 400 = 230 K, a1 = 1.4(287)(230) ≈ 304 m/s, [1 + 0.2(1.92)2 ] V1 = Ma1a1 = (1.92)(304) ≈ 585 m/s Ans. P9.61 Air flows from a large tank, where T = 376 K and p = 360 kPa, to a design condition where the pressure is 9800 Pa. The mass flow is 0.9 kg/s. However, there is a normal shock in the exit plane just after this condition is reached. Estimate (a) the throat area; and, just downstream of the shock, (b) the Mach number, (c) the temperature, and (d) the pressure. Solution: The low design pressure definitely indicates a supersonic condition: p 9800 2 = = 0.0272 = (1 + 0.2 Ma1 ) −3.5 , solve for po 360000 Choked : m = mmax = 0.6847 po A * RTo = 0.6847 (360000) A * 287(376) Ma1 = 3.00 , Solve A * = 0.0012 m 2 Ans.(a) Find T1 in front of the shock and then use the normal shock conditions: 2 T1 = To /(1 + 0.2 Ma1 ) = (376 K ) /[1 + 0.2(3.0) 2 ] = 134 K Table B.2, Ma1=3.0 : Ma 2 = 0.475 p 2 / p1 = 10.333 , Ans.(b) Ans.(c) Ans.( d ) T2 / T1 = 2.679 , T2 = (2.679)(134 K ) = 360 K p 2 = (10.333)(9800) = 101,300 Pa The problem worked out so that the external pressure was 1 atm. 9.62 An atomic explosion propagates into still air at 14.7 psia and 520°R. The pressure just inside the shock is 5000 psia. Assuming k = 1.4, what are the speed C of the shock and the velocity V just inside the shock? Solution: The pressure ratio tells us the Mach number of the shock motion: 2 5000 2.8Ma1 − 0.4 = 340 = , solve for Ma1 ≈ 17.1 14.7 2.4 ft a1 = 1.4(1717)(520) = 1118 ft/s, ∴ V1 = C = 17.1(1118) ≈ 19100 Ans. (a) s p 2 /p1 = 674 Solutions Manual • Fluid Mechanics, Fifth Edition We then compute the velocity ratio across the shock and thence the relative motion inside: V2 /V1 = 0.4(17.1)2 + 2 = 0.1695, ∴ V2 = 0.1695(19100) = 3240 ft/s 2.4(17.1)2 Then Vinside = C − V2 = 19100 − 3240 ≈ 15900 ft/s Ans. (b) ...
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This note was uploaded on 05/18/2011 for the course MEEG 332 taught by Professor Staff during the Spring '08 term at University of Delaware.

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