Sol 10 - Chapter 9 • Compressible Flow 655 9.35 Helium at...

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Unformatted text preview: 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 656 Solutions Manual • Fluid Mechanics, Fifth Edition ⎡ A* RTo ⎤ p(t ) = exp ⎢ −0.6847 t ⎥ = e −0.00993t υ p(0) ⎢ ⎥ ⎣ ⎦ until p(t) drops to 192 kPa or t ≈ 47 s or t ≈ 144 s Ans. (b) Ans. (c) At 500 kPa, we obtain 500/800 = exp(–0.00993t), At choking (192 kPa), 192/800 = exp(–0.00993t), 9.37 Make an exact control volume analysis of the blowdown process in Fig. P9.37, assuming an insulated tank with negligible kinetic and potential energy. Assume critical flow at the exit and show that both po and To decrease during blowdown. Set up first-order differential equations for po(t) and To(t) and reduce and solve as far as you can. Fig. P9.37 Solution: For a CV around the tank, write the mass and the energy equations: mass: d (ρoυ ) = −m, or dt p d ⎛ po ⎞ 0.6847A* υ ⎟ = − B o , where B = ⎜ RT ⎠ dt ⎝ o To R energy: ⎞ dQ dW d⎛ p + = 0 = ⎜ o υ c v To ⎟ + mc p To dt dt dt ⎝ RTo ⎠ We may rearrange and combine these to give a single differential equation for To: dTo 0.6847 3 = −CTo /2 , where C = (k − 1)A* R, or υ dt ⎡1 1⎤ Integrate: To (t) = ⎢ + Ct ⎥ ⎢ To (0) 2 ⎥ ⎣ ⎦ −2 o ∫ To3/2 = −C∫ dt dT Ans. With To(t) known, we could go back and solve the mass relation for po(t), but in fact that is not necessary. We simply use the isentropic-flow assumption: po (t) ⎡ To (t) ⎤ = po (0) ⎢ To (0) ⎥ ⎣ ⎦ k/(k −1) 1 ⎡ ⎤ 1/2 = ⎢1 + CTo (0)t ⎥ 2 ⎣ ⎦ −2k/(k −1) ⎛ 0.6847(k − 1)A* R ⎞ ⎜C = ⎟ υ ⎝ ⎠ Ans. Clearly, tank pressure also decreases with time as the tank blows down. Chapter 9 • Compressible Flow 675 9.63 Sea-level standard air is sucked into a vacuum tank through a nozzle, as in Fig. P9.63. A normal shock stands where the nozzle area is 2 cm2, as shown. Estimate (a) the pressure in the tank; and (b) the mass flow. Solution: The flow at the exit section (“3”) Fig. P9.63 is subsonic (after a shock) therefore must equal the tank pressure. Work our way to 1 and 2 at the shock and thence to 3 in the exit: 101350 p01 = 101350 Pa, A1 /A* = 2.0, thus Ma1 ≈ 2.1972, p1 = ≈ 9520 Pa [1 + 0.2(2.2)2 ]3.5 p2 2.8(2.2)2 − 0.4 = = 5.47, ∴ p 2 = 5.47(9520) ≈ 52030 Pa p1 2.4 Also compute A* /A* ≈ 1.59, or A* = 1.59 cm 2 2 1 2 Also compute po2 = 101350/1.59 = 63800 Pa. Finally compute A 3 /A* = 3/1.59 = 1.89, 2 read Ma3 = 0.327, whence p3 = 63800/[1 + 0.2(0.327)2]3.5 ≈ 59200 Pa. Ans. (a). With To = 288 K, the (critical) mass flow = 0.6847poA*/√(RTo) = 0.0241 kg/s. Ans. (b) 9.64 Air in a large tank at 100°C and 150 kPa exhausts to the atmosphere through a converging nozzle with a 5-cm2 throat area. Compute the exit mass flow if the atmospheric pressure is (a) 100 kPa; (b) 60 kPa; and (c) 30 kPa. Solution: Choking occurs when patmos < 0.5283ptank = 79 kPa. Therefore the first case is not choked, the second two cases are. For the first case, with To = 100°C = 373 K, (a) po 150 = = 1.5 = 1 + 0.2Ma 2 e pe 100 ( ) 3.5 , solve Ma e = 0.784, Te = 373 = 332 K 1 + 0.2(0.784)2 and a e = 1.4(287)(332) ≈ 365 m , Ve = 0.784(365) = 286 m/s, s and ρe = pe /RTe = 1.05 kg/m3 , finally: m = 1.05(0.0005)(286) = 0.150 kg/s Ans. (a) Both cases (b) and (c) are choked, with patm ≤ 79 kPa, and the mass flow is maximum and driven by tank conditions To and po: (b, c) m = m max = 0.6847po A* 0.6847(150000)(0.0005) kg = ≈ 0.157 s √ (RTo ) 287(373) Ans. (b, c) Chapter 9 • Compressible Flow 691 We need the pressure ratios to finish the problem, and then calculate po1 from the Mach number Table B.3 : p2 p = 2.1381 , 1 = 10.9435 p* p* p / p* 10.9435 Then p1 = p 2 1 = (100kPa) = 512 kPa p2 / p * 2.1381 Ans.(b) 2 p o1 = p1[1 + 0.2 Ma1 ]3.5 = 512[1 + 0.2(0.1) 2 ]3.5 = 516 kPa Ans.(c) 9.89 Carbon dioxide flows through an insulated pipe 25 m long and 8 cm in diameter. The friction factor is 0.025. At the entrance, p = 300 kPa and T = 400 K. The mass flow is 1.5 kg/s. Estimate the pressure drop by (a) compressible; and (b) incompressible (Sect. 6.6) flow theory. (c) For what pipe length will the exit flow be choked? Solution: For CO2, from Table A.4, take k = 1.30 and R = 189 J/kg⋅K. Tough calculation, no appendix tables for CO2, should probably use EES. Find inlet density, velocity, Mach number: ρ1 = p1 300000 Pa kg = = 3.97 3 RT1 (189 J/kg⋅K)(400 K) m kg kg ⎞ ⎛ π ⎞ ⎛ = ρ1 A1V1 = ⎜ 3.97 3 ⎟ ⎜ ⎟ (0.08 m)2 V1 , Solve for V1 = 75.2 m/s ⎝ s m ⎠ ⎝ 4⎠ m = 1.5 Ma1 = V1 kRT1 = 75.2 m/s (1.3)(189 m 2 /s2 ⋅K)(400 K) = 75.2 m/s = 0.240 313.5m/s Between section 1 (inlet) and section 2 (exit), the change in (f L /D) equals (0.025)(25 m)/ (0.08 m) = 7.813. We have to find the correct exit Mach number from this change: fL*/D = 1 − Ma 2 k + 1 ⎡ (k + 1) Ma 2 ⎤ + ln ⎢ 2⎥ 2k kMa 2 ⎣ 2 + (k − 1) Ma ⎦ For k = 1.3 and Ma1 = 0.240 compute ( fL*/D)1 = 10.190 Then ( fL*/D)2 = 10.190 − 7.813 = 2.377 for what Mach number? Then iterate (or use EES) to the exit value Ma2 = 0.408 Now compute Then p1 /p* = 4.452 and p2 /p* = 2.600 p2 = p1 ( p2 /p*)/( p1 /p*) = (300 kPa)(2.600)/(4.452) = 175 kPa Ans. (a) The desired compressible pressure drop = 300 − 175 = 125 kPa 692 Solutions Manual • Fluid Mechanics, Fifth Edition (b) The incompressible flow theory (Chap. 6) simply predicts that Δpinc 2 ⎛ 3.97 kg/m 3 ⎞ ⎛ fL ρ1 2 m⎞ V1 = (7.813) ⎜ = ⎟ ⎜ 75.2 ⎟ = 88000 Pa = 88 kPa D2 2 s⎠ ⎝ ⎠⎝ Ans. (b) The incompressible estimate is 30% low. Finally, the inlet value of (fL/D) tells us the maximum possible pipe length for choking at the exit: Lmax = fL * ⎛ D ⎞ ⎛ 0.08 m ⎞ ⎟ = 32.6 m ⎜ ⎟ = (10.19) ⎜ D 1⎝ f ⎠ ⎝ 0.025 ⎠ Ans. (c) 9.90 Air, supplied at p0 = 700 kPa and T0 = 330 K, flows through a converging nozzle into a pipe of 2.5-cm diameter which exits to a near vacuum. If f = 0.022 , what will be the mass flow through the pipe if its length is (a) 0 m, (b) 1 m, and (c) 10 m? Solution: (a) With no pipe (L = 0), the mass-flow is simply the isentropic maximum: m = m max po A* 700000(π /4)(0.025)2 kg = 0.6847 = 0.6847 ≈ 0.764 s √ RTo 287(330) Ans. (a) (b) With a finite length L = 1 m, the flow will choke in the exit plane instead: Ma e = 1.0, fL 0.022(1.0) = = 0.88, read Ma1 (entrance) ≈ 0.525 D 0.025 Then T1 = 330/[1 + 0.2(0.525)2 ] = 313 K, a1 = 1.4(287)(313) ≈ 354 m/s, V1 = Ma1a1 = 186 m/s, p1 = 700/[1 + 0.2(0.525)2 ]3.5 = 580 kPa, ρ1 = p1 /(RT1 ) = 6.46 kg/m 3 Finally, then, kg m = ρ1A1V1 = (6.46)(π /4)(0.025)2 (186) ≈ 0.590 (23% less) s (c) Repeat part (b) for a much longer length, L = 10 m: fL 0.022(10) m m , V1 = 89 , = = 8.8, Ma1 = 0.246, T1 = 326 K, a1 = 362 D 0.025 s s kg kg also, p1 = 671 kPa, ρ1 = 7.17 3 , m = ρ1A1V1 ≈ 0.314 (59% less) Ans. (c) s m Ans. (b) Chapter 9 • Compressible Flow 693 9.91 Air flows steadily from a tank through the pipe in Fig. P9.91. There is a converging nozzle on the end. If the mass flow is 3 kg/s and the flow is choked, estimate (a) the Mach number at section 1; and (b) the pressure in the tank. Fig. P9.91 Solution: For adiabatic flow, T* = constant = To /1.2 = 373/1.2 = 311 K. The flow chokes in the small exit nozzle, D = 5 cm. Then we estimate Ma2 from isentropic theory: A 2 ⎛ 6 cm ⎞ =⎜ ⎟ = 1.44, read Ma 2 (subsonic) ≈ 0.45, for which fL/D|2 ≈ 1.52, A* ⎝ 5 cm ⎠ p 2 /p* ≈ 2.388, p o2 /p* ≈ 1.449, ρ 2 /ρ * ≈ 2.070, T2 /T* = 1.153 or T2 ≈ 359 K o Given m = 3 kg p2 ⎛π ⎞ 2 = ρ2 A 2 V2 = ⎜ ⎟ (0.06) (0.45) 1.4(287)(359), s 287(359) ⎝ 4 ⎠ 2 Solve for p2 ≈ 640 kPa. Then p* = 640/2.388 ≈ 268 kPa At section 1, fL fL fΔL 0.025(9) = |2 + = 1.52 + ≈ 5.27, read Ma1 ≈ 0.30 DD D 0.06 for which p1 /p* ≈ 3.6, or p1 ≈ 3.6(268) ≈ 965 kPa. Ans. (a) Assuming isentropic flow in the inlet nozzle, ptank ≈ 965[1 + 0.2(0.30)2 ]3.5 ≈ 1030 kPa Ans. (b) ________________________________________________________________________ P9.92 Air enters a 5-cm-diameter pipe at 380 kPa, 3.3 kg/m3, and 120 m/s. The friction factor is 0.017. Find the pipe length for which the velocity (a) doubles; (b) triples; and (d) quadruples. Solution: First find the conditions at the entrance, which we will call section 1: 694 Solutions Manual • Fluid Mechanics, Fifth Edition T1 = p1 380000 = = 401 K ; Ma1 = Rρ1 287(3.3) V1 / V * = 0.3245 , V1 kRT1 = 120 1.4(287)(401) = 0.299 Table B.3 or EES : fL / D = 5.353 (a) Since V* is constant, we simply double (V1/V*) and find the new Mach number: (a) Double V2 = 0.6490 ; Table B.3 : Ma 2 = 0.6144 , fL / D | 2 = 0.4367 V* D fL fL 0.05 Pipe length L = ( |1 − |2 ) = (5.353 − 0.4367) = 14.46 m Ans.(a) fD D 0.017 (b) Again, since V* is constant, we simply triple (V1/V*) and find the new Mach number: (b) Triple V2 = 0.9735 ; Table B.3 : Ma 2 = 0.9684 , fL / D | 2 = 0.0013 V* D fL fL 0.05 Pipe length L = ( |1 − |3 ) = (5.353 − 0.0013) = 15.74 m Ans.(b) fD D 0.017 (c) We are already at choking, it is impossible to quadruple the velocity for this flow. Ans.(c) 9.93 Air flows adiabatically in a 3-cm-diameter duct with f = 0.015. At the entrance, V = 950 m/s and T = 250 K. How far down the tube will (a) the Mach number be 1.8; and (b) the flow be choked? Solution: (a) Find the entrance Mach number and its value of fL/d: Ma1 = 950 1.4(287)(250) read f = 3.00; Table B.3: read f L1 = 0.5222; at Ma2 = 1.8, D L2 ⎛ L⎞ = 0.2419; Δ ⎜ f ⎟ = 0.5222 − 0.2419 ≈ 0.28, D ⎝ D⎠ ΔL = 0.28(0.03) = 0.56 m Ans. (a) 0.015 (b) To go all the way to choking requires the full change fΔL1/D = 0.5222, or: ΔL choke = (0.5222)(0.03)/(0.015) = 1.04 m Ans. (b) 712 Solutions Manual • Fluid Mechanics, Fifth Edition By varying V and computing (T, ρ, s) from (b, c, d), we plot the Fanno line and add it to the previous Rayleigh line. The composite graph is as follows: The subsonic intersection is state 1, Ma1 ≈ 0.55, T2 ≈ 588 K, and the supersonic intersection is at Ma2 ≈ 2.20, where, for example, T2 ≈ 316 K. Also, s1 > s2. These two points thus correspond to the two sides of a normal shock wave, where “2” is the supersonic upstream and “1” the subsonic downstream condition. We may check these results in Table B-2, where, at Ma ≈ 2.20, the temperature ratio across the shock is 1.857—for our calculations, this ratio is 588 K/316 K ≈ 1.86 (agreement would be perfect if we kept more significant figures). Shock flow satisfies all the four equations of Rayleigh and Fanno flow combined—continuity, momentum, energy, and the equation of state. 9.112 Air enters a duct subsonically at section 1 at 1.2 kg/s. When 650 kW of heat is added, the flow chokes at the exit at p2 = 95 kPa and T2 = 700 K. Assuming frictionless heat addition, estimate (a) the velocity; and (b) the stagnation pressure at section 1. Solution: Since the exit is choked, p2 = p* and T2 = T* and, of course, Ma2 = 1.0. Then 650 kJ = 542 1.2 kg 542000 Also, T* = 1.2T2 = 1.2(700) = 840 K; hence To1 = 840 − ≈ 301 K o 1005 V2 = V* = kRT* = 1.4(287)(700) ≈ 530 m/s, and q = Q/m = Chapter 9 • Compressible Flow 713 301 Then To1 /T* = = 0.358; Table B.4: read Ma1 ≈ 0.306, read V1 /V* ≈ 0.199 o 840 So V1 = 530(0.199) ≈ 105 m/s Hence po1 ≈ 1.196(180) ≈ 215 kPa Ans. (a) Also read po1 /p* ≈ 1.196, where p* = p2 /0.5283 ≈ 180 kPa, o o Ans. (b) 9.113 Air enters a constant-area duct at p1 = 90 kPa, V1 = 520 m/s, and T1 = 558°C. It is then cooled with negligible friction until it exists at p2 = 160 kPa. Estimate (a) V2; (b) T2; and (c) the total amount of cooling in kJ/kg. Solution: We have enough information to estimate the inlet Ma1 and go from there: m 520 p a1 = 1.4(287)(558 + 273) = 578 , ∴ Ma1 = ≈ 0.90, read 1 = 1.1246, s 578 p* 90 p 160 or p* = = 80.0 kPa, whence 2 = ≈ 2.00, read Ma 2 ≈ 0.38, 1.1246 p* 80 read T2 /T* = 0.575, V2 /V* = 0.287, To2 /T* ≈ 0.493 o We have to back off to section 1 to determine the critical (*) values of T, V, To: Ma1 = 0.9, T1 /T* = 1.0245, T* = also, V1 /V* = 0.911, V* = 558 + 273 = 811 K, T2 = 0.575(811) ≈ 466 K 1.0245 Ans. (b) 520 = 571 m/s, so V2 = 0.287(571) ≈ 164 m/s Ans. (a) 0.911 966 = 973 K To1 /T* = 0.9921, where To1 = T1 + V12 /2c p = 966 K, T* = o o 0.9921 Finally, To2 = 0.493(973) = 480 K, q cooling = c p ΔTo = 1.005(966 − 480) ≈ 489 kJ kg Ans. (c) ...
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