Ch13 - Problem 13.1 Given Air extracted from a large tank Find[2 Mass flow rate Solution Basic equations mrate = ρ⋅ V⋅ A h1 V1 2 V2 2 p = h2 2

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Unformatted text preview: Problem 13.1 Given: Air extracted from a large tank Find: [2] Mass flow rate Solution: Basic equations: mrate = ρ⋅ V⋅ A h1 + V1 2 V2 2 p = h2 + 2 2 ρ k = const The mass flow rate is given by T0 = ( 70 + 273) ⋅ K p0 = 101⋅ kPa J cp = 1004⋅ kg⋅ K k = 1.4 mrate = ρ⋅ A⋅ V A= We need the density and velocity at the nozzle. In the tank From the isentropic relation = const p = 25⋅ kPa D = 15⋅ cm Given or available data T⋅ p ( 1−k) k p⎞ ρ = ρ0⋅ ⎛ ⎟ ⎜p ⎝ 0⎠ ρ0 = 1 k R = 286.9⋅ 2 π⋅ D 4 J kg⋅ K 2 A = 0.0177 m p0 ρ0 = 1.026 R ⋅ T0 kg 3 m kg ρ = 0.379 3 m We can apply the energy equation between the tank (stagnation conditions) and the point in the nozzle to find the velocity 2 h0 = h + V 2 V= ( ) 2⋅ h 0 − h = ⎛ p0 ⎞ T = T0⋅ ⎜ ⎟ ⎝p⎠ Fot T we again use insentropic relations ( Then V= 2⋅ cp⋅ T0 − T The mass flow rate is mrate = ρ⋅ A⋅ V Note that the flow is supersonic at this point Hence we must have a converging-diverging nozzle ) V = 476 ( 2⋅ cp⋅ T0 − T ) ( 1−k) k T = 230.167 K T = −43.0⋅ °C m s kg mrate = 3.18 s c= k⋅ R⋅ T 304 c= m s M= V c M = 1.57 Problem 13.3 [2] Given: Steam flow through a nozzle Find: Speed and Mach number; Mass flow rate; Sketch the shape Solution: Basic equations: mrate = ρ⋅ V⋅ A h1 + V1 2 V2 2 = h2 + 2 2 Assumptions: 1) Steady flow 2) Isentropic 3) Uniform flow 4) Superheated steam can be treated as ideal gas T0 = ( 450 + 273) ⋅ K p0 = 6⋅ MPa p = 2⋅ MPa D = 2⋅ cm Given or available data k = 1.30 R = 461.4⋅ J kg⋅ K (Table A.6) From the steam tables (try finding interactive ones on the Web!), at stagnation conditions J s0 = 6720⋅ kg⋅ K Hence at the nozzle section h0 = 3.302 × 10 ⋅ 6J J and s = s0 = 6720⋅ kg⋅ K p = 2 MPa T = 289 °C From these values we find from the steam tables that ( 2⋅ h 0 − h ) Hence the first law becomes V= The mass flow rate is given by mrate = ρ⋅ A⋅ V = Hence mrate = For the Mach number we need c= A⋅ V v k⋅ R ⋅ T V = 781 A⋅ V v kg 6J h = 2.997 × 10 ⋅ m s 2 A= kg 3 v = 0.1225⋅ π⋅ D 4 A = 3.14 × 10 −4 2 m kg mrate = 2.00 s c = 581 The flow is supersonic starting from rest, so must be converging-diverging m s M= V c M = 1.35 m kg Problem 13.4 Given: Air flow in a passage Find: [2] Mach number; Sketch shape Solution: Basic equations: p0 p ⎛ ⎝ = ⎜1 + k − 1 2⎞ ⋅M ⎟ 2 ⎠ k k−1 c= k⋅ R⋅ T m s T1 = ( 10 + 273) ⋅ K p1 = 150⋅ kPa V1 = 120⋅ p2 = 50⋅ kPa k = 1.4 R = 286.9⋅ The speed of sound at state 1 is c1 = m c1 = 337 s Hence V1 M1 = c1 Given or available data k⋅ R⋅ T1 M1 = 0.356 For isentropic flow stagnation pressure is constant. Hence at state 2 Hence Solving for M2 p0 = p1⋅ ⎛ 1 + ⎜ ⎝ M2 = k−1 2⎞ ⋅ M1 ⎟ 2 ⎠ p0 p2 k−1 2⎞ = ⎛1 + ⋅ M2 ⎟ ⎜ 2 ⎝ ⎠ k k−1 k−1 ⎤ ⎡ ⎢ ⎥ k ⎥ 2 ⎢⎛ p0 ⎞ ⋅ ⎢⎜ ⎟ − 1⎥ k − 1 p2 ⎣⎝ ⎠ ⎦ p0 = 164 kPa M2 = 1.42 Hence, as we go from subsonic to supersonic we must have a converging-diverging nozzle k k−1 J kg⋅ K Problem 13.5 [2] Given: Data on flow in a passage Find: Pressure at downstream location Solution: ft·lbf/lbm·oR R= k= T1 = 53.33 1.4 560 p1 = The given or available data is: 30 psi ft/s o R V1 = 1750 M2 = 2.5 c1 = 1160 M1 = 1.51 p 01 = 111 psi p 02 = 111 psi p2 = 6.52 psi Equations and Computations: From T 1 and Eq. 12.18 Then ft/s From M 1 and p 1, and Eq. 13.7a (using built-in function Isenp (M ,k )) For isentropic flow (p 01 = p 02) From M 2 and p 02, and Eq. 13.7a (using built-in function Isenp (M ,k )) Problem 13.6 [3] Given: Data on flow in a nozzle Find: Mass flow rate; Throat area; Mach numbers Solution: J/kg·K R= k= T0 = p1 = 286.9 1.4 523 200 K kPa A= The given or available data is: 1 cm2 p2 = 50 kPa Equations and Computations: We don't know the two Mach numbers. We do know for each that Eq. 13.7a applies: Hence we can write two equations, but have three unknowns ( 1, M 2, and p 0)! M We also know that states 1 and 2 have the same area. Hence we can write Eq. 13.7d twice: We now have four equations for four unknowns (A *, M 1, M 2, and p 0)! We make guesses (using Solver) for M 1 and M 2, and make the errors in computed A * and p 0 zero. M1 = 0.512 from Eq. 13.7a: p0 = 239 and from Eq. 13.7d: A* = 0.759 cm For: M2 = 1.68 kPa p0 = 239 kPa 0.00% 2 A* = 0.759 cm2 0.00% Note that the throat area is the critical area Sum The stagnation density is then obtained from the ideal gas equation ρ0 = 1.59 kg/m3 The density at critical state is obtained from Eq. 13.7a (or 12.22c) ρ* = Errors 1.01 3 kg/m The velocity at critical state can be obtained from Eq. 12.23) V* = 418 m/s m rate = 0.0321 kg/s The mass flow rate is ρ*V *A * 0.00% Problem 13.8 Given: Air flow in a passage Find: [3] Speed and area downstream; Sketch flow passage Solution: Basic equations: T0 T k−1 2 ⋅M 2 = 1+ c= ⎛ 1 + k − 1 ⋅ M2 ⎞ ⎟ A 1⎜ 2 = ⋅⎜ ⎟ k +1 Acrit M⎜ ⎟ 2 ⎝ ⎠ k ⋅ R⋅ T T1 = ( 32 + 460) ⋅ R p1 = 25⋅ psi M1 = 1.75 T2 = ( 225 + 460) ⋅ R Given or available data k = 1.4 Rair = 53.33⋅ 2 D1 = 3⋅ ft Hence A1 = T0 = T1⋅ ⎛ 1 + ⎜ ⎝ π⋅ D 1 T0 = 793 R k −1 2⎞ ⋅ M1 ⎟ 2 ⎠ 4 A1 = 7.07 ft ft⋅ lbf lbm⋅ R 2 T0 = 334 °F For isentropic flow stagnation conditions are constant. Hence M2 = ⎞ 2 ⎛ T0 ⋅⎜ − 1⎟ k − 1 T2 ⎝ We also have c2 = Hence Acrit = ft c2 = 1283 s V2 = M2⋅ c2 From state 1 M2 = 0.889 ⎠ k⋅ Rair⋅ T2 V2 = 1141 A1⋅ M1 ⎛ 1 + k − 1 ⋅M 2 ⎞ ⎜ 1⎟ 2 ⎜ ⎟ k+ 1 ⎜ ⎟ 2 ⎝ ⎠ Hence at state 2 k+ 1 2⋅ ( k−1) k−1 2⎞ ⎛ Acrit ⎜ 1 + 2 ⋅ M2 ⎟ A2 = ⋅⎜ ⎟ k+ 1 M2 ⎜ ⎟ ⎝ 2 ft s Acrit = 5.10 ft k+ 1 2⋅ ( k−1) A2 = 5.15 ft ⎠ Hence, as we go from supersonic to subsonic we must have a converging-diverging diffuser 2 2 k +1 2⋅ ( k −1) Problem 13.10 [2] Given: Data on flow in a passage Find: Flow rate; area and pressure at downstream location; sketch passage shape Solution: R= k= A1 = 286.9 1.4 0.25 T1 = 283 K p1 = 15 kPa V1 = 590 m/s T2 = 410 M2 = The given or available data is: 0.75 J/kg.K m2 Equations and Computations: From T 1 and Eq. 12.18 (12.18) c1 = Then 337 M1 = 1.75 m/s Because the flow decreases isentropically from supersonic to subsonic the passage shape must be convergent-divergent From p 1 and T 1 and the ideal gas equation ρ1 = 0.185 kg/m3 m rate = 27.2 kg/s The mass flow rate is m rate = ρ1A 1V 1 From M 1 and A 1, and Eq. 13.7d (using built-in function IsenA (M ,k )) (13.7d) A* = 0.180 m2 A2 = 0.192 m2 From M 2 and A *, and Eq. 13.7d (using built-in function IsenA (M ,k )) From M 1 and p 1, and Eq. 13.7a (using built-in function Isenp (M ,k )) (13.7a) p 01 = 79.9 kPa p 02 = 79.9 kPa p2 = 55.0 kPa For isentropic flow (p 01 = p 02) From M 2 and p 02, and Eq. 13.7a (using built-in function Isenp (M ,k )) Problem 13.11 [3] Given: Flow in a converging nozzle to a pipe Find: Plot of mass flow rate Solution: The given or available data is R = 287 J/kg·K k = 1.4 T 0 = 293 K p 0 = 101 kPa Dt = 1 cm 2 A t = 0.785 cm Equations and Computations: The critical pressure is given by p * = 53.4 kPa Hence for p = 100 kPa down to this pressure the flow gradually increases; then it is constant c V = M ·c ρ = p /RT 3 (m/s) (m/s) (kg/m ) 343 41 1.19 342 58 1.18 342 71 1.18 341 82 1.17 341 92 1.16 340 101 1.15 337 138 1.11 335 168 1.06 332 195 1.02 329 219 0.971 326 242 0.925 322 264 0.877 318 285 0.828 315 306 0.778 313 313 0.762 313 313 0.762 313 313 0.762 313 313 0.762 313 313 0.762 Flow (kg/s) 0.00383 0.00539 0.00656 0.00753 0.00838 0.0091 0.0120 0.0140 0.0156 0.0167 0.0176 0.0182 0.0186 0.0187 0.0187 0.0187 0.0187 0.0187 0.0187 Using critical conditions, and Eq. 13.9 for mass flow rate: 53.4 1.000 244 313 313 0.762 0.0185 (Note: discrepancy in mass flow rate is due to round-off error) Flow Rate in a Converging Nozzle 0.020 0.018 0.016 Flow Rate (kg/s) p M T (K) (kPa) (Eq. 13.7a) (Eq. 13.7b) 100 0.119 292 99 0.169 291 98 0.208 290 97 0.241 290 96 0.270 289 95 0.297 288 90 0.409 284 85 0.503 279 80 0.587 274 75 0.666 269 70 0.743 264 65 0.819 258 60 0.896 252 55 0.974 246 53.4 1.000 244 53 1.000 244 52 1.000 244 51 1.000 244 50 1.000 244 0.014 0.012 0.010 0.008 0.006 0.004 0.002 0.000 50 60 70 80 p (kPa) 90 100 Problem 13.12 [2] Given: Flow in a converging-diverging nozzle to a pipe Find: Plot of mass flow rate Solution: J/kg·K R= k= T0 = 286.9 1.4 293 p0 = 101 kPa Dt = 1 At = 0.785 cm cm2 p* = The given or available data is 53.4 kPa K De = 2.5 Ae = 4.909 cm cm2 Equations and Computations: The critical pressure is given by This is the minimum throat pressure For the CD nozzle, we can compute the pressure at the exit required for this to happen A* = A e/A * = 0.785 6.25 M e = 0.0931 p e = 100.4 cm2 (= A t) or 3.41 (Eq. 13.7d) or 67.2 kPa (Eq. 13.7a) Hence we conclude flow occurs in regimes iii down to v (Fig. 13.8); the flow is ALWAYS choked! p* M T * (K) c* V * = c * ρ = p /RT (kPa) 13.7a) (Eq. 13.7b) (m/s) (m/s) (kg/m3) 53.4 1.000 244 313 313 0.762 (Note: discrepancy in mass flow rate is due to round-off error) Flow (kg/s) 0.0187 0.0185 (Using Eq. 13.9) Problem 13.13 [3] Given: Data on tank conditions; isentropic flow Find: Plot cross-section area and pressure distributions Solution: R= k= T0 = The given or available data is: 53.33 1.4 500 ft·lbf/lbm·oR o R p0 = 45 psia pe = 14.7 psia m rate = 2.25 lbm/s Equations and Computations: From p 0, p e and Eq. 13.7a (using built-in function IsenMfromp (M,k)) (13.7a) Me = 1.37 Because the exit flow is supersonic, the passage must be a CD nozzle We need a scale for the area. From p 0, T 0, m flow, and Eq. 13.10c (13.10c) Then At = A* = 0.0146 ft2 For each M , and A *, and Eq. 13.7d (using built-in function IsenA (M ,k ) (13.7d) we can compute each area A . From each M , and p 0, and Eq. 13.7a (using built-in function Isenp (M ,k ) we can compute each pressure p . L (ft) A (ft 2) 1.00 1.25 1.50 1.75 2.00 2.50 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00 12.00 13.00 14.00 14.6 16.00 17.00 18.00 19.00 20.00 p (psia) 0.1234 0.0989 0.0826 0.0710 0.0622 0.0501 0.0421 0.0322 0.0264 0.0227 0.0201 0.0183 0.0171 0.0161 0.0155 0.0150 0.0147 0.0146 0.0146 0.0147 0.0149 0.0152 0.0156 0.0161 M 0.069 0.086 0.103 0.120 0.137 0.172 0.206 0.274 0.343 0.412 0.480 0.549 0.618 0.686 0.755 0.823 0.892 0.961 1.000 1.098 1.166 1.235 1.304 1.372 44.9 44.8 44.7 44.5 44.4 44.1 43.7 42.7 41.5 40.0 38.4 36.7 34.8 32.8 30.8 28.8 26.8 24.9 23.8 21.1 19.4 17.7 16.2 14.7 Area Variation in Passage 0.14 0.12 A (ft2) 0.10 0.08 0.06 0.04 0.02 0.00 0 5 10 15 20 L (ft) p (psia) Pressure Variation in Passage 50 45 40 35 30 25 20 15 10 5 0 0 2 4 6 8 10 L (ft) 12 14 16 18 20 Problem 13.14 Given: Air flow in a converging nozzle Find: [2] Mass flow rate Solution: Basic equations: mrate = ρ⋅ V⋅ A Given or available data pb = 35⋅ psi pb p0 = 0.583 T p0 = 60⋅ psi k = 1.4 Since T0 p = ρ⋅ R ⋅ T Rair = 53.33⋅ Mt = and Tt = ft⋅ lbf lbm⋅ R ct = ρt = k−1 ⎤ ⎡ ⎢ ⎥ k ⎥ 2 ⎢⎛ p0 ⎞ ⋅ ⎢⎜ ⎟ − 1⎥ k−1 ⎣⎝ pt ⎠ ⎦ T0 1+ k−1 2 2 ⋅ Mt k⋅ Rair⋅ Tt pt Rair⋅ Tt mrate = ρt⋅ At⋅ Vt k−1 2 ⋅M 2 T0 = ( 200 + 460) ⋅ R is greater than 0.528, the nozzle is not choked and Hence = 1+ At = π2 ⋅D 4t pt = pb Mt = 0.912 Tt = 566⋅ R Tt = 106⋅ °F Vt = ct Vt = 1166⋅ ρt = 5.19 × 10 ft s − 3 slug ⋅ 3 ft slug mrate = 0.528⋅ s lbm mrate = 17.0⋅ s p0 p ⎛ ⎝ = ⎜1 + k − 1 2⎞ ⋅M ⎟ 2 ⎠ Dt = 4⋅ in At = 0.0873⋅ ft 2 k k −1 Problem 13.15 Given: Isentropic air flow in converging nozzle Find: [2] Pressure, speed and Mach number at throat Solution: Basic equations: T0 T p0 k −1 2 ⋅M 2 =1+ p ⎛ ⎝ k − 1 2⎞ ⋅M ⎟ 2 ⎠ = ⎜1 + m s p1 = 350⋅ kPa V1 = 150⋅ k = 1.4 Given or available data R = 286.9⋅ k k −1 M1 = 0.5 pb = 250⋅ kPa J kg⋅ K The flow will be choked if pb/p0 < 0.528 p0 = p1 ⋅ ⎛ 1 + ⎜ ⎝ Hence so p0 pt k −1 2⎞ ⋅ M1 ⎟ 2 ⎠ k −1 2⎞ = ⎛1 + ⋅ Mt ⎟ ⎜ 2 ⎝ ⎠ Mt = k k −1 where Also V1 = M1⋅ c1 = M1⋅ k⋅ R⋅ T1 Then T0 = T1⋅ ⎛ 1 + ⎜ Hence Tt = Then ct = Finally Vt = Mt⋅ ct ⎝ 2 T0 k−1 2 ⋅ Mt 1+ 2 k⋅ R⋅ Tt p0 = 0.602 (Not choked) k k −1 k−1 ⎤ ⎡ ⎢ ⎥ k ⎥ 2 ⎢⎛ p0 ⎞ ⋅ ⎢⎜ ⎟ − 1⎥ k−1 ⎣⎝ pt ⎠ ⎦ k−1 pb p0 = 415 kPa 2⎞ ⋅ M1 ⎟ ⎠ pt = pb pt = 250 kPa Mt = 0.883 or 1 ⎛ V1 ⎞ T1 = ⋅⎜ ⎟ k⋅ R M1 ⎝⎠ 2 T1 = 224 K T0 = 235 K T0 = −37.9 °C Tt = 204 K Tt = −69.6 °C m ct = 286 s Vt = 252 m s T1 = −49.1 °C Problem 13.16 [3] Given: Data on three tanks Find: Mass flow rate; Pressure in second tank Solution: The given or available data is: R= k= At = 286.9 1.4 1 J/kg.K cm2 We need to establish whether each nozzle is choked. There is a large total pressure drop so this is likely. However, BOTH cannot be choked and have the same flow rate. This is because Eq. 13.9a, below (13.9b) indicates that the choked flow rate depends on stagnation temperature (which is constant) but also stagnation pressure, which drops because of turbulent mixing in the middle chamber. Hence BOTH nozzles cannot be choked. We assume the second one only is choked (why?) and verify later. T 01 = p 01 = p e1 = We make a guess at the pressure at the first nozzle exit: NOTE: The value shown is the final answer! It was obtained using Solver ! p 02 = This will also be tank 2 stagnation pressure: p3 = Pressure in tank 3: Temperature and pressure in tank 1: 308 650 527 K kPa kPa 527 65 kPa kPa Equations and Computations: From the p e1 guess and Eq. 13.17a: Then at the first throat (Eq.13.7b): The density at the first throat (Ideal Gas) is: Then c at the first throat (Eq. 12.18) is: Then V at the first throat is: Finally the mass flow rate is: M e1 = T e1 = 0.556 290 ρ e1 = 6.33 341 190 0.120 c e1 = V e1 = m rate = K kg/m3 m/s m/s kg/s First Nozzle! For the presumed choked flow at the second nozzle we use Eq. 13.9a, with T 01 = T 02 and p 02: m rate = 0.120 kg/s For the guess value for p e1 we compute the error between the two flow rates: Δm rate = 0.000 Use Solver to vary the guess value for p e1 to make this error zero! Note that this could also be done manually. kg/s Second Nozzle! Problem 13.17 [2] Problem 13.19 [2] Given: Data on converging nozzle; isentropic flow Find: Pressure and Mach number; throat area; mass flow rate Solution: R= k= A1 = 286.9 1.4 0.05 J/kg.K T1 = 276.3 K V1 = 200 m/s p atm = The given or available data is: 101 kPa m2 Equations and Computations: From T 1 and Eq. 12.18 (12.18) c1 = Then 333 M1 = 0.60 m/s To find the pressure, we first need the stagnation pressure. If the flow is just choked pe = p atm = p* = 101 kPa From p e = p * and Eq. 12.22a (12.22a) p0 = 191 kPa From M 1 and p 0, and Eq. 13.7a (using built-in function Isenp (M ,k ) (13.7a) Then p1 = 150 kPa The mass flow rate is m rate = ρ1A 1V 1 Hence, we need ρ1 from the ideal gas equation. ρ1 = 1.89 kg/m3 m rate = 18.9 kg/s The mass flow rate m rate is then The throat area A t = A * because the flow is choked. From M 1 and A 1, and Eq. 13.7d (using built-in function IsenA (M ,k ) (13.7d) A* = Hence 0.0421 m2 At = 0.0421 m2 Problem 13.20 [2] Problem 13.21 [2] Problem 13.23 [2] Given: Temperature in and mass flow rate from a tank Find: Tank pressure; pressure, temperature and speed at exit Solution: R= k= T0 = 286.9 1.4 273 At = The given or available data is: 0.001 2 m rate = J/kg.K K m2 kg/s Equations and Computations: Because p b = 0 Hence the flow is choked! pe = p* Hence Te = T* From T 0, and Eq. 12.22b (12.22b) T* = Te = 228 228 -45.5 Also Me = Hence Ve = K K o C 1 V* = From T e and Eq. 12.18 ce (12.18) ce = Then 302 m/s Ve = 302 m/s To find the exit pressure we use the ideal gas equation after first finding the exit density. The mass flow rate is m rate = ρeA eV e ρe = Hence 6.62 kg/m3 432 kPa From the ideal gas equation p e = ρeRT e pe = From p e = p * and Eq. 12.22a (12.22a) p0 = 817 kPa We can check our results: From p 0, T 0, A t, and Eq. 13.9a (13.9a) Then m choked = m choked = 2.00 m rate kg/s Correct! Problem 13.24 [2] Given: Isentropic air flow into a tank Find: Initial mass flow rate; Ts process; explain nonlinear mass flow rate Solution: Basic equations: T0 T = 1+ p0 k−1 2 ⋅M 2 p Then p0 = 101⋅ kPa pb = p0 − 10⋅ kPa k = 1.4 Given or available data R = 286.9⋅ A= π2 ⋅D 4 Avena = 65⋅ %⋅ A pb The flow will be choked if pb/p0 < 0.528 Hence p0 p0 k − 1 2⎞ = ⎛1 + ⋅M ⎟ ⎜ pvena ⎝ 2 ⎠ = 0.901 Mvena = Then Tvena = Then cvena = and Vvena = Mvena⋅ cvena Also ρvena = k − 1 2⎞ ⋅M ⎟ 2 ⎠ mrate = ρ⋅ A⋅ V pb = 91⋅ kPa T0 = ( 20 + 273) ⋅ K D = 5⋅ mm 2 Avena = 12.8⋅ mm (Not choked) k k −1 k−1 ⎤ ⎡ ⎢ ⎥ k ⎥ 2 ⎢⎛ p0 ⎞ ⋅ ⎢⎜ − 1⎥ ⎟ k − 1 pvena ⎣⎝ ⎠ ⎦ so Finally J kg⋅ K ⎛ ⎝ = ⎜1 + k k−1 T0 k−1 2 ⋅ Mvena 1+ 2 k⋅ R⋅ Tvena pvena R⋅ Tvena mrate = ρvena⋅ Avena⋅ Vvena where pvena = pb pvena = 91⋅ kPa Mvena = 0.389 Tvena = 284 K Tvena = 11.3⋅ °C m cvena = 338 s Vvena = 131 m ρvena = 1.12 kg s 3 m − 3 kg mrate = 1.87 × 10 s The Ts diagram will be a vertical line (T decreases and s = const). After entering the tank there will be turbulent mixing (s increases) and th comes to rest (T increases). The mass flow rate versus time will look like the curved part of Fig. 13.6b; it is nonlinear because V AND ρ va Problem 13.25 Given: Spherical cavity with valve Find: [3] Time to reach desired pressure; Entropy change Solution: Basic equations: T0 T = 1+ k−1 2 ⋅M 2 p0 p ⎛ ⎝ = ⎜1 + k − 1 2⎞ ⋅M ⎟ 2 ⎠ p = ρ⋅ R ⋅ T Then the inlet area is p0 = 101⋅ kPa Tatm = ( 20 + 273) ⋅ K pf = 45⋅ kPa Given or available data c= Tf = Tatm At = π2 ⋅d 4 k⋅ R⋅ T ρf = 2 ρf = 0.535 k ⎛2⎞ We have choked flow so mrate = At⋅ p0⋅ ⋅⎜ ⎟ R ⋅ T0 ⎝ k + 1 ⎠ Δt = 3 d = 1⋅ mm R = 286.9⋅ pb = pf pb so D = 50⋅ cm J kg⋅ K and tank volume is V = p0 k+ 1 2⋅ ( k−1) J cp = 1004⋅ kg⋅ K π3 ⋅D 3 3 V = 0.131 m = 0.446 (Choked) and final mass is M = ρf ⋅ V M = 0.0701 kg m Since the mass flow rate is constant (flow is always choked) Hence kg k ⎛2⎞ mchoked = At⋅ p0⋅ ⋅⎜ ⎟ R ⋅ T0 ⎝ k + 1 ⎠ T0 = Tatm k = 1.4 At = 0.785 mm pf R ⋅ Tf ⎛ T2 ⎞ ⎛ p2 ⎞ Δs = cp⋅ ln ⎜ ⎟ − R⋅ ln ⎜ ⎟ ⎝ T1 ⎠ ⎝ p1 ⎠ mrate = ρ⋅ A⋅ V The flow will be choked if pb/p0 < 0.528; the MAXIMUM back pressure is The final density is k k−1 M k+ 1 2⋅ ( k−1) Δt = or M mrate − 4 kg mrate = 1.873 × 10 Δt = 374 s mrate M = mrate⋅ Δt s Δt = 6.23 min The air in the tank will be cold when the valve is closed. Because ρ =M/V is constant, p = ρRT = const x T, so as the temperature rises to ambient, the pressure will rise too. ⎛ T2 ⎞ ⎛ p2 ⎞ For the entropy change during the charging process is given by Δs = cp⋅ ln ⎜ ⎟ − R⋅ ln ⎜ ⎟ where T1 = Tatm ⎝ T1 ⎠ ⎝ p1 ⎠ and p1 = p0 p2 = pf Hence ⎛ T2 ⎞ ⎛ p2 ⎞ Δs = cp⋅ ln ⎜ ⎟ − R⋅ ln ⎜ ⎟ ⎝ T1 ⎠ ⎝ p1 ⎠ Δs = 232 T2 = Tatm J kg⋅ K Problem 13.26 [3] Problem 13.27 [3] Problem 13.28 [3] Problem 13.29 [3] Given: Air-driven rocket in space Find: Tank pressure; pressure, temperature and speed at exit; initial acceleration Solution: R= k= T0 = 286.9 1.4 398 At = M= m rate = 25 25 0.05 Because p b = 0 Hence the flow is choked! pe = p* Hence Te = T* The given or available data is: J/kg.K K mm2 kg kg/s Equations and Computations: From T 0, and Eq. 12.22b (12.22b) T* = 332 Te = 332 K 58.7 o Also Me = Hence Ve = K C 1 V* = From T e and Eq. 12.18 ce (12.18) ce = Then 365 m/s Ve = 365 m/s To find the exit pressure we use the ideal gas equation after first finding the exit density. The mass flow rate is m rate = ρeA eV e ρe = 0.0548 pe = Hence 5.21 kg/m3 From the ideal gas equation p e = ρeRT e kPa From p e = p * and Eq. 12.22a (12.22a) p0 = 9.87 kPa We can check our results: From p 0, T 0, A t, and Eq. 13.9a (13.9a) m choked = Then m choked = 0.050 m rate kg/s Correct! The initial acceleration is given by: (4.33) which simplifies to: pe At − Max = mrateV ax = or: 1.25 ax = m/s2 m rate V + p e At M Problem 13.30 Given: Gas cylinder with broken valve Find: [3] Mass flow rate; acceleration of cylinder Solution: Basic equations: T0 T = 1+ k−1 2 ⋅M 2 p0 p ⎛ ⎝ = ⎜1 + k − 1 2⎞ ⋅M ⎟ 2 ⎠ k k−1 p = ρ⋅ R ⋅ T c= k⋅ R⋅ T mrate = ρ⋅ A⋅ V (4.33) Given or available data patm = 101⋅ kPa p0 = 20⋅ MPa d = 10⋅ mm pb = patm T0 Ve = ce The exit pressure is pe = Ve = 313 p0 ⎛1 + k − 1 ⎞ ⎜ ⎟ 2⎠ ⎝ Then k k−1 mrate = ρe⋅ Ae⋅ Ve The momentum equation (Eq. 4.33) simplifies to Hence ax = so Te = 244 K ⎛1 + k − 1 ⎞ ⎜ ⎟ 2⎠ ⎝ The exit speed is Ae = so the nozzle area is The flow will be choked if pb/p0 < 0.528: The exit temperature is Te = T0 = ( 20 + 273) ⋅ K p0 = 5.05 × 10 Ae = 78.5⋅ mm −3 J kg⋅ K R = 286.9⋅ 2 MCV = 65⋅ kg (Choked: Critical conditions) Te = −29⋅ °C ce = and exit density is ρe = k⋅ R⋅ Te m s pe = 10.6⋅ MPa kg mrate = 3.71 s (pe − patm)⋅ Ae − MCV⋅ ax = −Ve⋅ mrate (pe − patm)⋅ Ae + Ve⋅ mrate MCV pb π2 ⋅d 4 k = 1.4 m ax = 30.5 2 s The process is isentropic, followed by nonisentropic expansion to atmospheric pressure pe R ⋅ Te ρe = 151 kg 3 m Problem 13.32 Given: Spherical air tank Find: [4] Air temperature after 30s; estimate throat area Solution: Basic equations: T0 = 1+ T k−1 2 ⋅M 2 p ρ k ⌠ → ⎯→ ⎯ ⎮ ∂⌠ ⎮ ρ dVCV + ⎮ ρ⋅ V dACS = 0 ∂t ⎮ ⌡ ⌡ = const (4.12) Assumptions: 1) Large tank (stagnation conditions) 2) isentropic 3) uniform flow patm = 101⋅ kPa p1 = 2.75⋅ MPa T1 = 450⋅ K D = 2⋅ m ΔM = 30⋅ kg Given or available data Δt = 30⋅ s k = 1.4 R = 286.9⋅ pb = patm The flow will be choked if pb/p1 < 0.528: so pb p1 = 0.037 V= π3 ⋅D 6 3 V = 4.19⋅ m J kg⋅ K (Initially choked: Critical conditions) We need to see if the flow is still choked after 30s ρ1 = The initial (State 1) density and mass are For an isentropic process ρ The final temperature is k = const T2 = p2 ρ2⋅ R To estimate the throat area we use ρ1 = 21.3 R ⋅ T1 so kg 3 M1 = ρ1⋅ V ⎛ ρ2 ⎞ p2 = p1⋅ ⎜ ⎟ ⎝ ρ1 ⎠ M2 = 59.2 kg k p2 = 1.55⋅ MPa T2 = 382 K ρ2 = pb p2 M2 V ρ2 = 14.1 or = 0.0652 At = The average stagnation pressure is p0ave = T1 + T2 2 p1 + p2 2 3 (Still choked) ΔM Δt⋅ ρtave⋅ Vtave where we use average values of density and speed at the throat. T0ave = kg m T2 = 109⋅ °C ΔM = mtave = ρtave⋅ At⋅ Vtave Δt The average stagnation temperature is M1 = 89.2 kg m M2 = M1 − ΔM The final (State 2) mass and density are then p p1 T0ave = 416 K p0ave = 2.15⋅ MPa Hence the average temperature and pressure (critical) at the throat are T0ave Ttave = Hence Finally ⎛1 + k − 1 ⎞ ⎜ ⎟ 2⎠ ⎝ Vtave = At = k⋅ R⋅ Ttave ΔM Δt⋅ ρtave⋅ Vtave Ttave = 347 K and p0ave ptave = ⎛1 + k − 1 ⎞ ⎜ ⎟ 2⎠ ⎝ Vtave = 373 m s At = 2.35 × 10 ρtave = −4 2 m ptave R⋅ Ttave At = 235⋅ mm 2 This corresponds to a diameter Dt = 4⋅ At π Dt = 0.0173 m Dt = 17.3⋅ mm The process is isentropic, followed by nonisentropic expansion to atmospheric pressure k k−1 ptave = 1.14⋅ MPa ρtave = 11.4 kg 3 m Problem 13.33 Given: Ideal gas flow in a converging nozzle Find: [4] Exit area and speed Solution: T0 Basic equations: T Given or available data = 1+ p0 k−1 2 ⋅M 2 p1 = 35⋅ psi p ρ1 = 0.1⋅ lbm ft Check for choking: c1 = Hence 3 V1 M1 = c1 ⎝ k−1 2⎞ ⋅ M1 ⎟ 2 ⎠ ⎛ k + 1⎞ ⎜ ⎟ ⎝2⎠ M2 = From M1 we find Finally from continuity p2 = 25⋅ psi c1 = k⋅ p1 ft c1 = 1424 s ρ1 k = 1.25 p0 = 37.8 psi pcrit = 21.0 psi Hence p2 > pcrit, so NOT choked k−1 ⎤ ⎡ ⎢ ⎥ k ⎥ 2 ⎢⎛ p0 ⎞ ⋅ ⎢⎜ ⎟ − 1⎥ k − 1 p2 ⎣⎝ ⎠ ⎦ ⎛ 1 + k − 1 ⋅M 2 ⎞ ⎜ 1⎟ 2 ⎜ ⎟ k+ 1 ⎜ ⎟ 2 ⎝ ⎠ For isentropic flow 2 A1 = 1⋅ ft k k−1 k k−1 M1⋅ A1 Acrit = ft s ⎛ 1 + k − 1 ⋅ M2 ⎞ ⎟ A 1⎜ 2 = ⋅⎜ ⎟ k+ 1 Acrit M⎜ ⎟ 2 ⎝ ⎠ M1 = 0.351 p0 The critical pressure is then pcrit = Then we have V1 = 500⋅ k − 1 2⎞ ⋅M ⎟ 2 ⎠ k⋅ R⋅ T1 or, replacing R using the ideal gas equation p0 = p1⋅ ⎛ 1 + ⎜ Then ⎛ ⎝ = ⎜1 + k k−1 k+ 1 2⋅ ( k−1) k p⋅ ρ = const ρ⋅ A⋅ V = const k+ 1 2⋅ ( k−1) so so M2 = 0.830 Acrit = 0.557 ft 2 k−1 2⎞ ⎛ Acrit ⎜ 1 + 2 ⋅ M2 ⎟ A2 = ⋅⎜ ⎟ k+ 1 M2 ⎜ ⎟ ⎝ ⎛ p1 ⎞ ρ2 = ρ1⋅ ⎜ ⎟ ⎝ p2 ⎠ 2 1 k A1⋅ ρ1 V2 = V1⋅ A2⋅ ρ2 ρ2 = 0.131 lbm ft V2 = 667 ft s 3 ⎠ k+ 1 2⋅ ( k−1) A2 = 0.573 ft 2 Problem 13.34 [4] Part 1/3 Problem 13.34 [4] Part 2/3 Problem 13.34 [4] Part 3/3 Problem 13.35 [4] Problem 13.36 Given: CD nozzle attached to large tank Find: [2] Flow rate Solution: Basic equations: T0 T = 1+ k−1 2 ⋅M 2 p0 p ⎛ ⎝ k − 1 2⎞ ⋅M ⎟ 2 ⎠ = ⎜1 + p0 = 150⋅ kPa T0 = ( 35 + 273) ⋅ K k = 1.4 Given or available data R = 286.9⋅ For isentropic flow Me = Then Te = Also ce = ρe = Finally J kg⋅ K k−1 ⎤ ⎡ ⎢ ⎥ k ⎥ 2 ⎢⎛ p0 ⎞ ⋅ ⎢⎜ ⎟ − 1⎥ k − 1 pe ⎣⎝ ⎠ ⎦ mrate = ρe⋅ Ve⋅ Ae pe = 101⋅ kPa Ae = π2 ⋅D 4 D = 2.75⋅ cm 2 Ae = 5.94 cm Te = 275 K m ce = 332 s ρe = 1.28 kg 3 m kg mrate = 0.195 s Te = 1.94 °C Ve = Me⋅ ce ⎛ 1 + k − 1 ⋅ M 2⎞ ⎜ e⎟ 2 ⎝ ⎠ pe R ⋅ Te mrate = ρ⋅ V⋅ A Me = 0.773 T0 k⋅ R⋅ Te k k−1 Ve = 257 m s Problem 13.37 [2] Given: Design condition in a converging-diverging nozzle Find: Tank pressure; flow rate; throat area Solution: R= k= T0 = 53.33 1.4 560 Ae = pb = Me = 1 14.7 2 pe = 14.7 ft.lbf/lbm.oR pb pe = The given or available data is: o R in2 psia Equations and Computations: At design condition psia From M e and p e, and Eq. 13.7a (using built-in function Isenp (M ,k ) (13.7a) p0 = 115 psia From M e and A e, and Eq. 13.7d (using built-in function IsenA (M ,k ) (13.7d) A* = Hence 0.593 in2 At = 0.593 in2 From p 0, T 0, A t, and Eq. 13.10a (13.10a) m choked = 1.53 lb/s Problem 13.38 [4] Part 1/2 Problem 13.38 [4] Part 2/2 Problem 13.39 [2] Problem 13.43 [3] Part 1/2 Problem 13.43 [3] Part 2/2 Problem 13.44 Given: Rocket motor on test stand Find: [3] Mass flow rate; thrust force Solution: Basic equations: T0 T = 1+ k−1 2 ⋅M 2 p0 p (patm − pe)⋅ Ae + Rx = mrate⋅ Ve Given or available data pe = 75⋅ kPa patm = 101⋅ kPa d = 25⋅ cm From the pressures The exit speed is Then T0 ⎛ 1 + k − 1 ⋅ M 2⎞ ⎜ e⎟ 2 ⎝ ⎠ Ve = Me⋅ ce p0 = 4⋅ MPa ( ) c= k⋅ R⋅ T mrate = ρ⋅ A⋅ V T0 = 3250⋅ K Ae = π2 ⋅d 4 k = 1.25 R = 300⋅ J kg⋅ K 2 Ae = 491⋅ cm Me = 3.12 Te = 1467 K Ve = 2313 mrate = ρe⋅ Ae⋅ Ve p = ρ⋅ R ⋅ T Momentum for pressure pe and velocity Ve at exit; Rx is the reaction for so the nozzle exit area is The momentum equation (Eq. 4.33) simplifies to Hence k − 1 2⎞ ⋅M ⎟ 2 ⎠ k−1 ⎤ ⎡ ⎢ ⎥ k ⎥ 2 ⎢⎛ p0 ⎞ ⋅ ⎢⎜ ⎟ − 1⎥ k − 1 pe ⎣⎝ ⎠ ⎦ Me = The exit temperature is Te = ⎛ ⎝ = ⎜1 + k k−1 m s ce = k⋅ R⋅ Te and kg mrate = 19.3 s (pe − patm)⋅ Ae − MCV⋅ ax = −Ve⋅ mrate Rx = pe − patm ⋅ Ae + Ve⋅ mrate Rx = 43.5⋅ kN m ce = 742 s ρe = pe R ⋅ Te ρe = 0.170⋅ kg 3 m Problem 13.47 [3] Problem 13.48 [4] Given: Compressed CO2 in a cartridge expanding through a nozzle Find: Throat pressure; Mass flow rate; Thrust; Thrust increase with diverging section; Exit area Solution: Basic equations: Assumptions: 1) Isentropic flow 2) Stagnation in cartridge 3) Ideal gas 4) Uniform flow J kg⋅ K k = 1.29 R = 188.9⋅ p0 = 35⋅ MPa Given or available data: T0 = ( 20 + 273) ⋅ K p0 From isentropic relations pcrit = ⎛1 + k − 1 ⎞ ⎜ ⎟ 2⎠ ⎝ Since pb << pcrit, then pt = pcrit Throat is critical so k k −1 patm = 101⋅ kPa dt = 0.5⋅ mm pcrit = 19.2 MPa mrate = ρt⋅ Vt⋅ At Tt = Vt = pt = 19.2 MPa T0 k−1 1+ 2 k⋅ R⋅ Tt Tt = 256 K Vt = 250 m s 2 At = ρt = π⋅ dt 4 pt R ⋅ Tt mrate = ρt⋅ Vt⋅ At At = 1.963 × 10 ρt = 396 −7 2 kg 3 m kg mrate = 0.0194 s m Rx − ptgage⋅ At = mrate⋅ Vt For 1D flow with no body force the momentum equation reduces to Rx = mrate⋅ Vt + ptgage⋅ At ptgage = pt − patm Rx = 8.60 N When a diverging section is added the nozzle can exit to atmospheric pressuree = patm p k−1 ⎤⎤ ⎡ ⎡ ⎢ ⎢ ⎥⎥ k ⎢ 2 ⎢⎛ p0 ⎞ ⎥⎥ Me = ⎢ ⋅ ⎢⎜ ⎟ − 1⎥⎥ ⎣ k − 1 ⎣⎝ pe ⎠ ⎦⎦ Hence the Mach number at exit is Te = ce = T0 1+ k−1 2 ⋅ Me 2 1 2 Me = 4.334 Te = 78.7 K m ce = 138 s k⋅ R⋅ Te Ve = Me⋅ ce Ve = 600 m s The mass flow rate is unchanged (choked flow) From the momentum equation Rx = mrate⋅ Ve The percentage increase in thrust is 11.67⋅ N − 8.60⋅ N = 35.7 % 8.60⋅ N The exit area is obtained from mrate = ρe⋅ Ve⋅ Ae Ae = T mrate ρe⋅ Ve and ρe = pe T0 pt Tt Conv. Nozzle CD Nozzle Te s ρe = 6.79 R ⋅ Te Ae = 4.77 × 10 p0 pb Rx = 11.67 N kg 3 m −6 2 m Ae = 4.77 mm 2 Problem 13.49 [3] Given: CO2 cartridge and convergent nozzle Find: Tank pressure to develop thrust of 15 N Solution: J/kg·K R= k= T0 = 188.9 1.29 293 pb = 101 kPa Dt = 0.5 mm At = The given or available data is: 0.196 mm2 K Equations and Computations: The momentum equation gives R x = m flowV e Hence, we need m flow and V e pe = pb pe = For isentropic flow 101 kPa If we knew p 0 we could use it and p e, and Eq. 13.7a, to find M e. Once M e is known, the other exit conditions can be found. Make a guess for p 0, and eventually use Goal Seek (see below). p0 = 44.6 MPa From p 0 and p e, and Eq. 13.7a (using built-in function IsenMfromp (M ,k ) (13.7a) Me = 4.5 From M e and T 0 and Eq. 13.7b (using built-in function IsenT (M ,k ) (13.7b) Te = 74.5 From T e and Eq. 12.18 K (12.18) ce = m/s Ve = Then 134.8 606 m/s The mass flow rate is obtained from p 0, T 0, A t, and Eq. 13.10a (13.10a) m choked = 0.0248 kg/s Finally, the momentum equation gives R x = m flowV e = 15.0 We need to set R x to 15 N. To do this use Goal Seek to vary p 0 to obtain the result! N Problem 13.50 Given: Air flow in an insulated duct Find: [2] Mass flow rate; Range of choked exit pressures Solution: T0 Basic equations: T = 1+ k−1 2 ⋅M 2 c= ⎛ 1 + k − 1 ⋅ M2 ⎞ ⎟ A 1⎜ 2 = ⋅⎜ ⎟ k+ 1 Acrit M⎜ ⎟ 2 ⎝ ⎠ k⋅ R⋅ T T0 = ( 80 + 460) ⋅ R p0 = 14.7⋅ psi k = 1.4 Given or available data Rair = 53.33⋅ p1 = 13⋅ psi ft⋅ lbf lbm⋅ R A= π⋅ D 4 k+ 1 2⋅ ( k−1) D = 1⋅ in 2 2 A = 0.785⋅ in Assuming isentropic flow, stagnation conditions are constant. Hence M1 = k−1 ⎤ ⎡ ⎢ ⎥ k ⎥ 2 ⎢⎛ p0 ⎞ ⋅ ⎢⎜ ⎟ − 1⎥ k − 1 p1 ⎣⎝ ⎠ ⎦ c1 = Also ρ1 = M1 = 0.423 m c1 = 341 s k⋅ Rair⋅ T1 p1 ρ1 = 0.0673⋅ Rair⋅ T1 mrate = ρ1⋅ V1⋅ A When flow is choked hence 1 M2 = We also have c2 = From continuity ρ1⋅ V1 = ρ2⋅ V2 k⋅ Rair⋅ T2 1+ k−1 2 2 ⋅ M1 T1 = 521⋅ R T1 = 61.7⋅ °F m s V1 = 144 T2 = 450⋅ R T2 = −9.7⋅ °F V2 = c2 V2 = 1040⋅ lbm 3 lbm mrate = 0.174⋅ s T0 T2 = k−1 1+ 2 ft c2 = 1040⋅ s V1 ρ2 = ρ1⋅ V2 p2 = ρ2⋅ Rair⋅ T2 T0 V1 = M1⋅ c1 ft Hence Hence T1 = ρ2 = 0.0306⋅ ft s lbm ft 3 p2 = 5.11⋅ psi The flow will therefore choke for any back pressure (pressure at the exit) less than or equal to this pressure (From Fanno line function p1 pcrit = 2.545 at M1 = 0.423 so pcrit = p1 2.545 pcrit = 5.11 psi Check!) Problem 13.51 [4] Given: Air flow from converging nozzle into pipe Find: Plot Ts diagram and pressure and speed curves Solution: ft·lbf/lbm·oR R= k= cp = 53.33 1.4 0.2399 T0 = 187 710 p0 = 25 psi pe = 24 psi Me = 0.242 Using built-in function IsenT (M ,k ) Te = 702 Using p e, M e, and function Fannop (M ,k ) p* = 5.34 Using T e, M e, and function FannoT (M ,k ) T* = 592 The given or available data is: Equations and Computations: From p 0 and p e, and Eq. 13.7a (using built-in function IsenMfromp (M ,k )) o Btu/lbm· R o ft·lbf/lbm· R R o o R psi o R We can now use Fanno-line relations to compute values for a range of Mach numbers: M T /T * 0.242 0.25 0.26 0.27 0.28 0.29 0.3 0.31 0.32 0.33 0.34 0.35 0.36 0.37 0.38 0.39 0.4 0.41 0.42 0.43 0.44 0.45 1.186 1.185 1.184 1.183 1.181 1.180 1.179 1.177 1.176 1.174 1.173 1.171 1.170 1.168 1.166 1.165 1.163 1.161 1.159 1.157 1.155 1.153 o T ( R) 702 701 701 700 699 698 720 697 697 700 696 695 680 694 660 T (o693 R) 692 640 691 690 620 689 600 688 687 580 686 0 685 684 682 c (ft/s) 1299 1298 1298 1297 1296 1296 1295 1294 1293 1292 1292 1291 1290 1289 1288 1287 1286 1285 1284 1283 1282 1281 Δs o (ft·lbf/lbm· R) Eq. (12.11b) 315 4.50 24.0 0.00 325 4.35 23.2 1.57 337 4.19 22.3 3.50 350 Ts Curve (Fanno) 4.03 21.5 5.35 363 3.88 20.7 7.11 376 3.75 20.0 8.80 388 3.62 19.3 10.43 401 3.50 18.7 11.98 414 3.39 18.1 13.48 427 3.28 17.5 14.92 439 3.19 17.0 16.30 452 3.09 16.5 17.63 464 3.00 16.0 18.91 477 2.92 15.6 20.14 489 2.84 15.2 21.33 502 2.77 14.8 22.48 514 2.70 14.4 23.58 527 2.63 14.0 24.65 2.56 13.7 30 25.68 10539 20 40 552 2.50 13.4 26.67 . o s 564 2.44 (ft lbf/lbm R) 13.0 27.63 576 2.39 12.7 28.55 V (ft/s) p /p * p (psi) 50 0.46 0.47 0.48 0.49 0.5 0.51 0.52 0.53 0.54 0.55 0.56 0.57 0.58 0.59 0.6 0.61 0.62 0.63 0.64 0.65 0.66 0.67 0.68 0.69 0.7 0.71 0.72 0.73 0.74 0.75 0.76 0.77 0.78 0.79 0.8 0.81 0.82 0.83 0.84 0.85 0.86 0.87 0.88 0.89 0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1 1.151 1.149 1.147 1.145 1.143 1.141 1.138 1.136 1.134 1.132 1.129 1.127 1.124 1.122 1.119 1.117 1.114 1.112 1.109 1.107 1.104 1.101 1.098 1.096 1.093 1.090 1.087 1.084 1.082 1.079 1.076 1.073 1.070 1.067 1.064 1.061 1.058 1.055 1.052 1.048 1.045 1.042 1.039 1.036 1.033 1.029 1.026 1.023 1.020 1.017 1.013 1.010 1.007 1.003 1.000 681 1280 589 2.33 12.4 29.44 680 1279 601 2.28 12.2 30.31 679 1277 613 2.23 11.9 31.14 677 1276 625 Velocity V 2.18 Versus M11.7 (Fanno) 31.94 676 1275 638 2.14 11.4 32.72 675 1274 650 2.09 11.2 33.46 1400 674 1273 662 2.05 11.0 34.19 672 1200 1271 674 2.01 10.7 34.88 671 1270 686 1.97 10.5 35.56 669 1000 1269 698 1.93 10.3 36.21 668 1267 710 1.90 10.1 36.83 800 667 1266 722 1.86 9.9 37.44 V (ft/s) 665 733 1.83 9.8 38.02 600 1265 664 1263 745 1.80 9.6 38.58 400 1262 662 757 1.76 9.4 39.12 661 1260 769 1.73 9.2 39.64 200 659 1259 781 1.70 9.1 40.14 658 792 1.67 8.9 40.62 0 1258 656 1256 804 1.65 8.8 41.09 0.2 0.3 0.4 0.5 0.6 0.7 0.8 655 1255 815 1.62 8.6 41.53 M8.5 653 1253 827 1.59 41.96 652 1252 839 1.57 8.4 42.37 650 1250 850 1.54 8.2 42.77 648 1248 861 1.52 8.1 43.15 647 1247 873 1.49 8.0 43.51 645 1245 884 Pressure p 1.47 Versus M7.8Fanno)43.85 ( 643 1244 895 1.45 7.7 44.18 642 1242 907 1.43 7.6 44.50 30 640 1240 918 1.41 7.5 44.80 638 1239 929 1.38 7.4 45.09 25 636 1237 940 1.36 7.3 45.36 635 20 1235 951 1.35 7.2 45.62 633 1234 962 1.33 7.1 45.86 631 15 1232 973 1.31 7.0 46.10 p (psi) 629 1230 984 1.29 6.9 46.31 628 10 1228 995 1.27 6.8 46.52 626 1227 1006 1.25 6.7 46.71 624 1225 1017 1.24 6.6 46.90 5 622 1223 1027 1.22 6.5 47.07 620 1221 1038 1.20 6.4 47.22 0 619 1219 1049 1.19 6.3 0.2 0.3 0.4 0.5 0.6 0.7 47.37 0.8 617 1218 1059 1.17 6.3 47.50 M 6.2 615 1216 1070 1.16 47.63 613 1214 1080 1.14 6.1 47.74 611 1212 1091 1.13 6.0 47.84 609 1210 1101 1.11 6.0 47.94 607 1208 1112 1.10 5.9 48.02 605 1206 1122 1.09 5.8 48.09 603 1204 1132 1.07 5.7 48.15 601 1202 1142 1.06 5.7 48.20 600 1201 1153 1.05 5.6 48.24 598 1199 1163 1.04 5.5 48.27 596 1197 1173 1.02 5.5 48.30 594 1195 1183 1.01 5.4 48.31 592 1193 1193 1.00 5.3 48.31 0.9 1.0 0.9 1.0 Problem 13.52 [4] Given: Air flow from converging-diverging nozzle into pipe Find: Plot Ts diagram and pressure and speed curves Solution: R= k= cp = 53.33 1.4 0.2399 T0 = p0 = pe = 187 710 25 2.5 Me = Using built-in function IsenT (M ,k ) Te = 368 Using p e, M e, and function Fannop (M ,k ) p* = 6.84 Using T e, M e, and function FannoT (M ,k ) T* = 592 ft·lbf/lbm·oR 2.16 The given or available data is: Equations and Computations: From p 0 and p e, and Eq. 13.7a (using built-in function IsenMfromp (M ,k )) Btu/lbm·oR ft·lbf/lbm·oR R o psi psi o R psi o R We can now use Fanno-line relations to compute values for a range of Mach numbers: M T /T * T (oR) 2.157 2 1.99 1.98 1.97 1.96 1.95 1.94 1.93 1.92 1.91 1.9 1.89 1.88 1.87 1.86 1.85 1.84 1.83 1.82 1.81 1.8 1.79 1.78 1.77 1.76 1.75 1.74 1.73 1.72 1.71 1.7 0.622 0.667 0.670 0.673 0.676 0.679 0.682 0.685 0.688 0.691 0.694 0.697 0.700 0.703 0.706 0.709 0.712 0.716 0.719 0.722 0.725 0.728 0.731 0.735 0.738 0.741 0.744 0.747 0.751 0.754 0.757 0.760 368 394 396 398 400 402 403 405 407 409 410 o 412 T ( R) 414 416 418 420 421 423 425 427 429 431 433 435 436 438 440 442 444 446 448 450 c (ft/s) 650 600 550 500 450 400 350 300 V (ft/s) 940 974 976 978 980 982 985 987 989 991 993 996 998 1000 1002 1004 1007 1009 0 1011 5 1013 1015 1018 1020 1022 1024 1027 1029 1031 1033 1036 1038 1040 2028 1948 1942 1937 1931 1926 1920 1914 1909 1903 1897 1892 1886 1880 1874 1868 1862 1856 1850 10 1844 1838 1832 1826 1819 1813 1807 1801 1794 1788 1781 1775 1768 Δs (ft·lbf/lbm·oR) Eq. (12.11b) 0.37 2.5 0.00 0.41 2.8 7.18 0.41 2.8 7.63 0.41 2.8 8.07 Curve (Fanno) 0.42 2.9 8.51 0.42 2.9 8.95 0.42 2.9 9.38 0.43 2.9 9.82 0.43 2.9 10.25 0.43 3.0 10.68 0.44 3.0 11.11 0.44 3.0 11.54 0.44 3.0 11.96 0.45 3.1 12.38 0.45 3.1 12.80 0.45 3.1 13.22 0.46 3.1 13.64 0.46 3.1 14.05 0.46 15 20 3.2 25 14.46 30 0.47 3.2 o 14.87 . s 0.47 (ft lbf/lbm R) 3.2 15.28 0.47 3.2 15.68 0.48 3.3 16.08 0.48 3.3 16.48 0.49 3.3 16.88 0.49 3.3 17.27 0.49 3.4 17.66 0.50 3.4 18.05 0.50 3.4 18.44 0.50 3.5 18.82 0.51 3.5 19.20 0.51 3.5 19.58 p /p * Ts p (psi) 35 40 1.69 1.68 1.67 1.66 1.65 1.64 1.63 1.62 1.61 1.6 1.59 1.58 1.57 1.56 1.55 1.54 1.53 1.52 1.51 1.5 1.49 1.48 1.47 1.46 1.45 1.44 1.43 1.42 1.41 1.4 1.39 1.38 1.37 1.36 1.35 1.34 1.33 1.32 1.31 1.3 1.29 1.28 1.27 1.26 1.25 1.24 1.23 1.22 1.21 1.2 1.19 1.18 1.17 1.16 1.15 1.14 1.13 1.12 1.11 1.1 1.09 1.08 1.07 1.06 1.05 1.04 1.03 1.02 1.01 1 0.764 0.767 0.770 0.774 0.777 0.780 0.784 0.787 0.790 0.794 0.797 0.800 0.804 0.807 0.811 0.814 0.817 0.821 0.824 0.828 0.831 0.834 0.838 0.841 0.845 0.848 0.852 0.855 0.859 0.862 0.866 0.869 0.872 0.876 0.879 0.883 0.886 0.890 0.893 0.897 0.900 0.904 0.907 0.911 0.914 0.918 0.921 0.925 0.928 0.932 0.935 0.939 0.942 0.946 0.949 0.952 0.956 0.959 0.963 0.966 0.970 0.973 0.976 0.980 0.983 0.987 0.990 0.993 0.997 1.000 452 1042 454 1045 456 1047 458 1049 460 1051 462 1054 464 1056 466 1058 468 1060 470 1063 472 1065 2500 474 1067 476 1069 478 2000 1072 480 1074 482 1500 1076 484 1078 V (ft/s) 486 1080 1000 488 1083 490 1085 500 1087 492 494 1089 496 0 1092 498 1094 2.0 500 1096 502 1098 504 1101 506 1103 508 1105 510 1107 512 1110 514 1112 516 1114 8 518 1116 520 7 1118 522 1121 6 524 1123 527 5 1125 529 4 1127 p (psi) 531 1129 533 3 1132 535 2 1134 537 1136 539 1 1138 541 0 1140 543 2.0 1143 545 1145 547 1147 549 1149 551 1151 553 1153 555 1155 557 1158 559 1160 561 1162 564 1164 566 1166 568 1168 570 1170 572 1172 574 1174 576 1176 578 1179 580 1181 582 1183 584 1185 586 1187 588 1189 590 1191 592 1193 1761 0.52 3.5 19.95 1755 0.52 3.6 20.32 1748 0.53 3.6 20.69 1741 0.53 3.6 21.06 1735 0.53 3.7 21.42 1728 0.54 3.7 21.78 1721 0.54 3.7 22.14 1714 0.55 3.7 22.49 1707 Velocity V 0.55 Versus M 3.8 (Fanno) 22.84 1700 0.56 3.8 23.18 1693 0.56 3.8 23.52 1686 0.57 3.9 23.86 1679 0.57 3.9 24.20 1672 0.58 3.9 24.53 1664 0.58 4.0 24.86 1657 0.59 4.0 25.18 1650 0.59 4.0 25.50 1642 0.60 4.1 25.82 1635 0.60 4.1 26.13 1627 0.61 4.1 26.44 1620 0.61 4.2 26.75 1612 0.62 4.2 27.05 1605 0.62 4.3 27.34 1597 0.63 4.3 27.63 1.8 1.6 1.4 1589 0.63 4.3 27.92 M4.4 1582 0.64 28.21 1574 0.65 4.4 28.48 1566 0.65 4.5 28.76 1558 0.66 4.5 29.03 1550 0.66 4.5 29.29 1542 Pressure p 0.67 Versus M4.6Fanno)29.55 ( 1534 0.68 4.6 29.81 1526 0.68 4.7 30.06 1518 0.69 4.7 30.31 1510 0.69 4.8 30.55 1502 0.70 4.8 30.78 1493 0.71 4.8 31.01 1485 0.71 4.9 31.24 1477 0.72 4.9 31.46 1468 0.73 5.0 31.67 1460 0.74 5.0 31.88 1451 0.74 5.1 32.09 1443 0.75 5.1 32.28 1434 0.76 5.2 32.48 1426 0.76 5.2 32.66 1417 0.77 5.3 1.4 32.84 1.8 1.6 1408 0.78 5.3 33.01 M 5.4 1399 0.79 33.18 1390 0.80 5.4 33.34 1381 0.80 5.5 33.50 1372 0.81 5.6 33.65 1363 0.82 5.6 33.79 1354 0.83 5.7 33.93 1345 0.84 5.7 34.05 1336 0.85 5.8 34.18 1327 0.86 5.9 34.29 1318 0.87 5.9 34.40 1308 0.87 6.0 34.50 1299 0.88 6.0 34.59 1290 0.89 6.1 34.68 1280 0.90 6.2 34.76 1271 0.91 6.2 34.83 1261 0.92 6.3 34.89 1251 0.93 6.4 34.95 1242 0.94 6.5 34.99 1232 0.96 6.5 35.03 1222 0.97 6.6 35.06 1212 0.98 6.7 35.08 1203 0.99 6.8 35.10 1193 1.00 6.8 35.10 1.2 1.2 1.0 1.0 Problem 13.53 [2] Problem 13.54 Given: Air flow in a converging nozzle and insulated duct Find: [3] Pressure at end of duct; Entropy increase Solution: T0 Basic equations: T = 1+ k−1 2 ⋅M 2 p0 ⎛ ⎝ p = ⎜1 + k − 1 2⎞ ⋅M ⎟ 2 ⎠ k k−1 ⎛ T2 ⎞ ⎛ p2 ⎞ Δs = cp⋅ ln ⎜ ⎟ − Rair⋅ ln ⎜ ⎟ ⎝ T1 ⎠ ⎝ p1 ⎠ T0 = ( 250 + 460) ⋅ R p0 = 145⋅ psi p1 = 125⋅ psi k = 1.4 Given or available data Btu cp = 0.2399⋅ lbm⋅ R Rair = 53.33⋅ c= k⋅ R⋅ T T2 = ( 150 + 460) ⋅ R ft⋅ lbf lbm⋅ R Assuming isentropic flow in the nozzle k−1 ⎤ ⎡ ⎢ ⎥ k ⎥ 2 ⎢⎛ p0 ⎞ ⋅ ⎢⎜ ⎟ − 1⎥ k − 1 p1 ⎣⎝ ⎠ ⎦ M1 = M1 = 0.465 T1 = T0 1+ k−1 2 ⋅ M1 2 ⎡⎛ T0 ⎞ ⎤ ⋅ ⎢⎜ ⎟ − 1⎥ k − 1 T2 In the duct T0 (a measure of total energy) is constant, so M2 = ⎣⎝ ⎠ 2 T1 = 681⋅ R T1 = 221⋅ °F M2 = 0.905 ⎦ k⋅ Rair⋅ T1 ft c1 = 1279⋅ s V1 = M1⋅ c1 V1 = 595⋅ k⋅ Rair⋅ T2 ft c2 = 1211⋅ s V2 = M2⋅ c2 V2 = 1096⋅ ft s V1 ρ2 = ρ1⋅ V2 ρ2 = 0.269⋅ lbm ρ1 = Also p1 Rair⋅ T1 ρ1 = 0.4960⋅ p2 = ρ2⋅ Rair⋅ T2 p2 = 60.8⋅ psi (Note: Using Fanno line relations, at M1 = 0.465 so Finally T1 Tcrit p1 pcrit Then T2 Tcrit = 1.031 so M2 = 0.907 p2 pcrit ft s lbm ft mrate = ρ1⋅ V1⋅ A = ρ2⋅ V2⋅ A Hence Then c1 = c2 = At each location 3 ft ⎛ T2 ⎞ ⎛ p2 ⎞ Δs = cp⋅ ln ⎜ ⎟ − Rair⋅ ln ⎜ ⎟ ⎝ T1 ⎠ ⎝ p1 ⎠ T1 = 1.150 Tcrit = = 2.306 pcrit = = 1.119 p2 = 1.119⋅ pcrit 1.150 p1 2.3060 3 Δs = 0.0231⋅ Btu lbm⋅ R Tcrit = 329 K pcrit = 54.2⋅ psi p2 = 60.7⋅ psi Check!) Problem 13.57 [3] Given: Air flow in a CD nozzle and insulated duct Find: Temperature at end of duct; Force on duct; Entropy increase Solution: ( T0 ) Basic equations: Fs = p1 ⋅ A − p2 ⋅ A + Rx = mrate⋅ V2 − V1 Given or available data T1 = ( 100 + 460) ⋅ R p1 = 18.5⋅ psi k = 1.4 k −1 2 ⋅M 2 Btu cp = 0.2399⋅ lbm⋅ R T =1+ M1 = 2 ⎛ T2 ⎞ ⎛ p2 ⎞ Δs = cp⋅ ln ⎜ ⎟ − Rair⋅ ln ⎜ ⎟ ⎝ T1 ⎠ ⎝ p1 ⎠ 2 M2 = 1 Rair = 53.33⋅ A = 1⋅ in ft⋅ lbf lbm⋅ R Assuming isentropic flow in the nozzle k −1 2 ⋅ M1 1+ T0 T2 2 ⋅ = T1 T0 k −1 2 1+ ⋅ M2 2 Also c1 = ρ1 = 1+ k⋅ Rair⋅ T1 V1 = M1⋅ c1 p1 Rair⋅ T1 mrate = ρ1⋅ V1⋅ A ρ1 = 0.0892⋅ V1 = 2320⋅ lbm ft 3 ) 1+ ft s 2 k −1 2 ⋅ M1 2 ⋅ M2 2 T2 = 840⋅ R V2 = 1421⋅ ft s V1 ρ 2 = ρ 1⋅ V2 ρ2 = 0.146⋅ lbm so p2 = ρ2⋅ Rair⋅ T2 ( ) Rx = p2 − p1 ⋅ A + mrate⋅ V2 − V1 Rx = −13.3⋅ lbf Finally ⎛ T2 ⎞ ⎛ p2 ⎞ Δs = cp⋅ ln ⎜ ⎟ − Rair⋅ ln ⎜ ⎟ ⎝ T1 ⎠ ⎝ p1 ⎠ Δs = 0.0359⋅ M1 = 2 T1 Tcrit p1 pcrit = = T1 T2 p1 p2 = 0.6667 = 0.4083 ft p2 = 45.3⋅ psi Hence (Note: Using Fanno line relations, at T2 = 380⋅ °F k ⋅ Rair⋅ T2 V2 = M2⋅ c2 c2 = mrate = ρ1⋅ V1⋅ A = ρ2⋅ V2⋅ A2 lbm mrate = 1.44⋅ s ( T2 = T1⋅ so k −1 (Force is to the right) Btu lbm⋅ R T2 = p2 = p1 0.4083 T1 0.667 p2 = 45.3⋅ psi T2 = 840⋅ R Check!) 3 Problem 13.59 [4] Problem 13.61 [4] Problem 13.62 [2] Problem 13.63 Given: Air flow in a converging nozzle and insulated duct Find: [2] Length of pipe Solution: Basic equations: Fanno-line flow equations, and friction factor Given or available data T0 = ( 250 + 460) ⋅ R p0 = 145⋅ psi p1 = 125⋅ psi T2 = ( 150 + 460) ⋅ R D = 2⋅ in k = 1.4 Btu cp = 0.2399⋅ lbm⋅ R Rair = 53.33⋅ From isentropic relations k−1 ⎤⎤ ⎡ ⎡ ⎢ ⎢ ⎥⎥ k ⎢ 2 ⎢⎛ p0 ⎞ ⎥⎥ M1 = ⎢ ⋅ ⎢⎜ ⎟ − 1⎥⎥ ⎣ k − 1 ⎣⎝ p1 ⎠ ⎦⎦ T0 T1 = 1+ k−1 2 ⋅ M1 2 1 2 so M1 = 0.465 T1 = T0 p1 pcrit = Also, for T2 Tcrit p1 2.3044 = 1.031 T1 = 221⋅ °F = 1.150 Tcrit = ⎡ ⎤ fave⋅ Lmax1 1 − M1 k + 1 ⎢ ( k + 1) ⋅ M1 ⎥ = + ⋅ ln ⎢ = 1.3923 2 Dh 2⋅ k ⎛ k − 1 ⋅ M 2⎞ ⎥ k⋅ M1 2⋅ ⎜ 1 + ⎢ 1 ⎟⎥ 2 ⎣⎝ ⎠⎦ k+ 1 ⎛ ⎞ ⎟ 2 1⎜ = = ⋅⎜ ⎟ pcrit p2 M1 k−1 2 ⎜ 1+ ⋅ M1 ⎟ 2 ⎝ ⎠ p1 T1 = 681⋅ R ⎛ 1 + k − 1 ⋅ M 2⎞ ⎜ 1⎟ 2 ⎝ ⎠ 2 Then for Fanno-line flow ft⋅ lbf lbm⋅ R 2 1 2 T1 = 2.3044 Tcrit pcrit = 54.2⋅ psi T2 Tcrit 1+ = 1+ k−1 2 ⋅ M1 2 Tcrit = 592⋅ R k+ 1 2 = k+ 1 2 k−1 2 ⋅ M2 2 leads to M2 = 2 k−1 T1 1.150 Tcrit = 132⋅ °F ⎛ k + 1 Tcrit ⋅⎜ ⎝2 Then ρ1 = T2 ⎞ − 1⎟ ⎠ M2 = 0.906 ⎡ ⎤ fave⋅ Lmax2 1 − M2 k + 1 ⎢ ( k + 1) ⋅ M2 ⎥ = + ⋅ ln = 0.01271 ⎢ ⎛ k−1 2 2⋅ k Dh 2⎞ ⎥ k⋅ M2 ⋅ M2 ⎟ ⎥ 2⋅ ⎜ 1 + ⎢ 2 ⎣⎝ ⎠⎦ Also ⋅ 2 p1 Rair⋅ T1 2 ρ1 = 0.496 lbm ft 3 V1 = M1⋅ k⋅ Rair⋅ T1 V1 = 595 ft s For air at T1 = 221 °F, from Table A.9 (approximately) μ = 4.48 × 10 − 7 lbf ⋅ s ⋅ 2 Re1 = and Re1 = 3.41 × 10 ft For commercial steel pipe (Table 8.1) Hence at this Reynolds number and roughness (Eq. 8.37) Combining results e −4 = 9 × 10 D e = 0.00015⋅ ft These calculations are a LOT easier using the Excel Add-ins! μ f = 0.01924 2 ⋅ ft fave⋅ Lmax2 fave⋅ Lmax1 ⎞ 12 D⎛ L12 = ⋅ ⎜ − ⋅ ( 1.3923 − 0.01271) ⎟= Dh Dh f .01924 ⎝ ρ1⋅ V1⋅ D so ⎠ L12 = 12.0⋅ ft 6 Problem 13.65 [3] Part 1/2 Problem 13.65 [3] Part 2/2 Problem 13.67 Example 13.7 Example 13.7 [3] Problem 13.68 [3] Problem 13.69 [4] Part 1/2 Problem 13.69 [4] Part 2/2 Problem 13.70 Given: Air flow through a CD nozzle and tube. Find: [2] Average friction factor; Pressure drop in tube Solution: Assumptions: 1) Isentropic flow in nozzle 2) Adiabatic flow in tube 3) Ideal gas 4) Uniform flow J kg⋅ K k = 1.40 R = 286.9⋅ p0 = 1.35⋅ MPa Given or available data: T0 = 550⋅ K k −1 ⎤⎤ ⎡ ⎡ ⎢ ⎢ ⎥⎥ k ⎢ 2 ⎢⎛ p0 ⎞ ⎥⎥ From isentropic relations M1 = ⎢ ⋅ ⎢⎜ ⎟ − 1⎥⎥ ⎣ k − 1 ⎣⎝ p1 ⎠ ⎦⎦ p1 = 15⋅ kPa where State 1 is the nozzle exit D = 2.5⋅ cm L = 1.5⋅ m 1 2 M1 = 3.617 Then for Fanno-line flow (for choking at the exit) ⎡ ⎤ fave⋅ Lmax 1 − M1 k + 1 ⎢ ( k + 1) ⋅ M1 ⎥ = + ⋅ ln ⎢ = 0.599 2 Dh 2⋅ k k−1 2⎞ ⎥ ⎛1 + k⋅ M1 ⋅ M1 ⎟ ⎥ ⎢ 2⋅ ⎜ 2 ⎣⎝ ⎠⎦ 2 ⎡ Hence 2 ⎡ 2 2 ⎤⎤ k + 1 ⎢ ( k + 1) ⋅ M1 ⎥⎥ D ⎢ 1 − M1 fave = ⋅ + ⋅ ln ⎢ ⎢ ⎛ k−1 2 2⋅ k L k⋅ M 2⎞ ⎥⎥ ⋅ M1 ⎟ ⎥⎥ ⎢ ⎢ 2⋅ ⎜ 1 + 1 2 ⎣ ⎣⎝ ⎠ ⎦⎦ ⎛ k+ 1 ⎞ ⎟ 2 1⎜ = = ⋅⎜ ⎟ pcrit p2 M1 k−1 2 ⎜ 1+ ⋅ M1 ⎟ 2 ⎝ ⎠ p1 p2 = p1 p1 1⎤ ⎡ ⎢ 2⎥ k+ 1 ⎢ ⎞⎥ ⎛ ⎟⎥ 2 ⎢1 ⎜ ⋅⎜ ⎟ ⎢M k−1 2⎥ ⎢ 1 ⎜ 1 + 2 ⋅ M1 ⎟ ⎥ ⎣ ⎝ ⎠⎦ Δp = p1 − p2 These calculations are a LOT easier using the Excel Add-ins! fave = 0.0100 1 2 = 0.159 p2 = 94.2 kPa Δp = −79.2 kPa Problem 13.71 Given: Air flow in a CD nozzle and insulated duct Find: [3] Duct length; Plot of M and p Solution: Basic equations: Fanno-line flow equations, and friction factor Given or available data T1 = ( 100 + 460) ⋅ R p1 = 18.5⋅ psi M1 = 2 Btu cp = 0.2399⋅ lbm⋅ R k = 1.4 2 M2 = 1 A = 1⋅ in Rair = 53.33⋅ ft⋅ lbf lbm⋅ R Then for Fanno-line flow at M1 = 2 1 2 k+ 1 ⎞ ⎛ ⎜ ⎟ 2 1 = = ⋅⎜ ⎟ = 0.4082 pcrit p2 M1 k−1 2 ⎜ 1+ ⋅ M1 ⎟ 2 ⎝ ⎠ p1 so p1 pcrit = p1 ⎡ ⎤ fave⋅ Lmax1 1 − M1 k + 1 ⎢ ( k + 1) ⋅ M1 ⎥ = + ⋅ ln ⎢ = 0.3050 2 Dh 2⋅ k k−1 2⎞ ⎥ ⎛1 + k⋅ M1 ⋅ M1 ⎟ ⎥ ⎢ 2⋅ ⎜ 2 ⎣⎝ ⎠⎦ 2 2 pcrit = 45.3⋅ psi 0.4082 and at M2 = 1 ⎡ ⎤ fave⋅ Lmax2 1 − M2 k + 1 ⎢ ( k + 1) ⋅ M2 ⎥ = + ⋅ ln ⎢ =0 2 Dh 2⋅ k k−1 2⎞ ⎥ ⎛1 + k⋅ M2 ⋅ M2 ⎟ ⎥ ⎢ 2⋅ ⎜ 2 ⎣⎝ ⎠⎦ Also ρ1 = 2 p1 lbm ρ1 = 0.089⋅ 3 Rair⋅ T1 ft For air at T1 = 100⋅ °F, from Table A.9 2 V1 = M1⋅ k⋅ Rair⋅ T1 μ = 3.96 × 10 − 7 lbf ⋅ s ⋅ 2 V1 = 2320⋅ e = 0.00015⋅ ft Hence at this Reynolds number and roughness (Eq. 8.37) e −3 = 1.595 × 10 D 4⋅ A D = 1.13⋅ in π D= ρ1⋅ V1⋅ D so Re1 = and Re1 = 1.53 × 10 ft For commercial steel pipe (Table 8.1) ft s μ 6 f = .02222 1.13 Combining results ⋅ ft 12 D ⎛ fave⋅ Lmax2 fave⋅ Lmax1 ⎞ L12 = ⋅ ⎜ − = ⋅ ( 0.3050 − 0) ⎟ Dh Dh f ⎝ ⎠ .02222 L12 = 1.29⋅ ft L12 = 15.5⋅ in These calculations are a LOT easier using the Excel Add-ins! The M and p plots are shown in the associated Excel workbook Problem 13.71 (In Excel) Given: Air flow in a CD nozzle and insulated duct Find: [3] Duct length; Plot of M and p Solution: The given or available data is: M 2.00 1.95 1.90 1.85 1.80 1.75 1.70 1.65 1.60 1.55 1.50 1.45 1.40 1.35 1.30 1.25 1.20 1.15 1.10 1.05 1.00 f = 0.0222 p * = 45.3 kPa D= 1.13 in fL max/D ΔfL max/D x (in) p /p * p (psi) 0.305 0.290 0.274 0.258 0.242 0.225 0.208 0.190 0.172 0.154 0.136 0.118 0.100 0.082 0.065 0.049 0.034 0.021 0.010 0.003 0.000 0.000 0.015 0.031 0.047 0.063 0.080 0.097 0.115 0.133 0.151 0.169 0.187 0.205 0.223 0.240 0.256 0.271 0.284 0.295 0.302 0.305 0 0.8 1.6 2.4 3.2 4.1 4.9 5.8 6.7 7.7 8.6 9.5 10.4 11.3 12.2 13.0 13.8 14.5 15.0 15.4 15.5 0.408 0.423 0.439 0.456 0.474 0.493 0.513 0.534 0.557 0.581 0.606 0.634 0.663 0.695 0.728 0.765 0.804 0.847 0.894 0.944 1.000 18.49 19.18 19.90 20.67 21.48 22.33 23.24 24.20 25.22 26.31 27.47 28.71 30.04 31.47 33.00 34.65 36.44 38.37 40.48 42.78 45.30 Fanno Line Flow Curves(M and p ) 45 2.0 1.9 1.8 1.7 1.6 M 1.5 1.4 1.3 1.2 1.1 1.0 40 35 30 p (psi) 25 M Pressure 20 15 0 4 8 x (in) 12 16 Problem *13.73 [2] Given: Isothermal air flow in a duct Find: Downstream Mach number; Direction of heat transfer; Plot of Ts diagram Solution: 2 V1 2 T0 V2 δQ + = h2 + 2 dm h1 + Given or available data T1 = ( 20 + 273) ⋅ K p1 = 350⋅ kPa M1 = 0.1 From continuity mrate = ρ1⋅ V1⋅ A = ρ2⋅ V2⋅ A so ρ1⋅ V1 = ρ2⋅ V2 Also p = ρ⋅ R ⋅ T M= Hence continuity becomes p1 R ⋅ T1 2 T and ⋅ M1⋅ c1 = =1+ k −1 2 ⋅M 2 Basic equations: V c mrate = ρ⋅ V⋅ A p2 = 150⋅ kPa V = M⋅ c or p2 ⋅M ⋅c R ⋅ T2 2 2 Since T1 = T2 c1 = c2 Hence p1 M2 = ⋅M p2 1 From energy ⎛ ⎞⎜ ⎞ V2 ⎟ ⎛ V1 ⎟ δQ ⎜ = ⎜ h2 + − ⎜ h1 + = h02 − h01 = cp⋅ T02 − T01 dm ⎝ 2⎟ ⎝ 2⎟ ⎠ ⎠ M2 = 0.233 2 But at each state p1⋅ M1 = p2⋅ M2 so 2 ( T0 T = 1+ k−1 2 2 ⋅M or T0 = T⋅ ⎛ 1 + ⎜ ⎝ k−1 2 ) 2⎞ ⋅M ⎟ ⎠ p02 Since T = const, but M2 > M1, then T02 > T01, and δQ >0 dm so energy is ADDED to the system T p01 T 01 p1 T02 p2 s Problem *13.74 Given: Isothermal air flow in a pipe Find: [5] Mach number and location at which pressure is 500 kPa Solution: f ⋅ Lmax 2 1 − k⋅ M ) 2 p = ρ⋅ R ⋅ T T1 = ( 15 + 273) ⋅ K p1 = 1.5⋅ MPa V1 = 60⋅ D = 15⋅ cm k = 1.4 R = 286.9⋅ From continuity ρ1⋅ V1 = ρ2⋅ V2 or Since T1 = T2 and V = M⋅ c = M⋅ k⋅ R⋅ T p1 M2 = M1⋅ p2 c1 = m c1 = 340 s V1 M1 = c1 M1 = 0.176 Given or available data Then At M1 = 0.176 At M2 = 0.529 Hence k⋅ R⋅ T1 p1 M2 = M1⋅ p2 f ⋅ Lmax1 D D f ⋅ L12 D 1 − k⋅ M1 2 = 1 − k⋅ M2 2 k⋅ M2 = f ⋅ Lmax2 L12 = 18.2⋅ D D f T1 − 2 + ln ⎛ k⋅ M1 ⎞ = 18.819 ⎝ 2 k⋅ M1 f ⋅ Lmax2 p1 M2 = 0.529 2 = D ⎠ 2 + ln ⎛ k⋅ M2 ⎞ = 0.614 ⎝ f ⋅ Lmax1 D ⎠ = 18.819 − 0.614 = 18.2 L12 = 210 m = ( mrate = ρ⋅ V⋅ A Basic equations: 2 + ln k⋅ M k⋅ M m s ⋅ V1 = f = 0.013 J kg⋅ K p2 T2 ⋅ V2 p2 = 500⋅ kPa Problem *13.75 [2] Problem *13.76 [4] Part 1/2 Problem *13.76 [4] Part 2/2 Problem 13.78 [4] Given: Air flow from converging nozzle into heated pipe Find: Plot Ts diagram and pressure and speed curves Solution: R= k= cp = 53.33 1.4 0.2399 T0 = p0 = pe= 187 710 25 24 Me = Using built-in function IsenT (M ,k ) Te = 702 Using p e, M e, and function Rayp (M ,k ) p* = 10.82 Using T e, M e, and function RayT (M ,k ) T* = 2432 ft·lbf/lbm·oR 0.242 The given or available data is: Equations and Computations: From p 0 and p e, and Eq. 13.7a (using built-in function IsenMfromp (M ,k )) Btu/lbm·oR ft·lbf/lbm·oR R o psi psi o R psi o R We can now use Rayleigh-line relations to compute values for a range of Mach numbers: M T /T * 0.242 0.25 0.26 0.27 0.28 0.29 0.3 0.31 0.32 0.33 0.34 0.35 0.36 0.37 0.38 0.39 0.4 0.41 0.42 0.43 0.44 0.45 0.46 0.289 0.304 0.325 0.346 0.367 0.388 0.409 0.430 0.451 0.472 0.493 0.514 0.535 0.555 0.576 0.595 0.615 0.634 0.653 0.672 0.690 0.708 0.725 T (oR) 702 740 790 841 892 943 3000 994 1046 2500 1097 11492000 1200 1250 T (oR) 1500 1301 13511000 1400 1448 500 1496 0 1543 1589 0 1635 1679 1722 1764 c (ft/s) 1299 1334 1378 1422 1464 1506 1546 1586 1624 1662 1698 1734 1768 1802 1834 1866 1897 1926 1955 1982 2009 2035 2059 V (ft/s) 315 334 358 384Ts 410 437 464 492 520 548 577 607 637 667 697 728 759 790 50 821 852 884 916 947 Δs (ft·lbf/lbm·oR) Eq. (12.11b) 2.22 24.0 0.00 2.21 23.9 10.26 2.19 23.7 22.81 2.18 23.6 34.73 Curve (Rayleigh) 2.16 23.4 46.09 2.15 23.2 56.89 2.13 23.1 67.20 2.12 22.9 77.02 2.10 22.7 86.40 2.08 22.5 95.35 2.07 22.4 103.90 2.05 22.2 112.07 2.03 22.0 119.89 2.01 21.8 127.36 2.00 21.6 134.51 1.98 21.4 141.35 1.96 21.2 147.90 1.94 21.0 154.17 20.8 160.17 100 1.92 150 200 250 1.91 . 20.6 165.92 o s 1.89 (ft lbf/lbm R) 20.4 171.42 1.87 20.2 176.69 1.85 20.0 181.73 p /p * p (psi) 300 0.47 0.48 0.49 0.5 0.51 0.52 0.53 0.54 0.55 0.56 0.57 0.58 0.59 0.6 0.61 0.62 0.63 0.64 0.65 0.66 0.67 0.68 0.69 0.7 0.71 0.72 0.73 0.74 0.75 0.76 0.77 0.78 0.79 0.8 0.81 0.82 0.83 0.84 0.85 0.86 0.87 0.88 0.89 0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1 0.742 0.759 0.775 0.790 0.805 0.820 0.834 0.847 0.860 0.872 0.884 0.896 0.906 0.917 0.927 0.936 0.945 0.953 0.961 0.968 0.975 0.981 0.987 0.993 0.998 1.003 1.007 1.011 1.014 1.017 1.020 1.022 1.024 1.025 1.027 1.028 1.028 1.029 1.029 1.028 1.028 1.027 1.026 1.025 1.023 1.021 1.019 1.017 1.015 1.012 1.009 1.006 1.003 1.000 1805 2083 979 1.83 19.8 186.57 1845 2106 1011 1.81 19.6 191.19 1884 2128 1043 19.4 195.62 Velocity V V1.80 ersus M (Rayleigh) 1922 2149 1075 1.78 19.2 199.86 1958 2170 1107 1.76 19.0 203.92 3000 1993 2189 1138 1.74 18.8 207.80 2027 2208 1170 1.72 18.6 211.52 2500 2060 2225 1202 1.70 18.4 215.08 2091 2242 1233 1.69 18.2 218.48 2000 2122 2258 1265 1.67 18.0 221.73 2150 1500 2274 1296 1.65 17.9 224.84 V (ft/s) 2178 2288 1327 1.63 17.7 227.81 2204 1000 2302 1358 1.61 17.5 230.65 2230 2315 1389 1.60 17.3 233.36 2253 1420 1.58 17.1 235.95 500 2328 2276 2339 1450 1.56 16.9 238.42 2298 1481 1.54 16.7 240.77 0 2350 2318 2361 0.3 1511 1.53 16.5 0.2 0.4 0.5 0.6 0.7 243.010.8 2337 2370 1541 1.51 16.3 245.15 M 2355 2379 1570 1.49 16.1 247.18 2371 2388 1600 1.47 15.9 249.12 2387 2396 1629 1.46 15.8 250.96 2401 2403 1658 1.44 15.6 252.70 2415 2409 1687 1.42 15.4 254.36 2427 2416 1715 1.41 255.93 Pressure p Versus M 15.2 (Rayleigh) 2438 2421 1743 1.39 15.0 257.42 2449 2426 1771 1.37 14.9 258.83 30 2458 2431 1799 1.36 14.7 260.16 2466 2435 1826 1.34 14.5 261.41 25 2474 2439 1853 1.33 14.4 262.59 2480 2442 1880 1.31 14.2 263.71 20 2486 2445 1907 1.30 14.0 264.75 2447 1933 1.28 13.9 265.73 p 2490 15 (psi) 2494 2449 1959 1.27 13.7 266.65 2497 10 2450 1985 1.25 13.5 267.50 2499 2451 2010 1.24 13.4 268.30 2501 5 2452 2035 1.22 13.2 269.04 2502 2452 2060 1.21 13.1 269.73 2502 0 2452 2085 1.19 12.9 270.36 2501 2452 2109 1.18 12.8 0.2 0.3 0.4 0.5 0.6 0.7 270.94 0.8 2500 2451 2133 1.17 12.6 271.47 M 12.5 2498 2450 2156 1.15 271.95 2495 2449 2180 1.14 12.3 272.39 2492 2448 2203 1.12 12.2 272.78 2488 2446 2226 1.11 12.0 273.13 2484 2444 2248 1.10 11.9 273.43 2479 2441 2270 1.09 11.7 273.70 2474 2439 2292 1.07 11.6 273.92 2468 2436 2314 1.06 11.5 274.11 2461 2433 2335 1.05 11.3 274.26 2455 2429 2356 1.04 11.2 274.38 2448 2426 2377 1.02 11.1 274.46 2440 2422 2398 1.01 10.9 274.51 2432 2418 2418 1.00 10.8 274.52 0.9 1.0 0.9 1.0 Problem 13.79 [4] Given: Air flow from converging-diverging nozzle into heated pipe Find: Plot Ts diagram and pressure and speed curves Solution: R= k= cp = 53.33 1.4 0.2399 T0 = p0 = pe = 187 710 25 2.5 Me = Using built-in function IsenT (M ,k ) Te = 368 Using p e, M e, and function Rayp (M ,k ) p* = 7.83 Using T e, M e, and function RayT (M ,k ) T* = 775 ft·lbf/lbm·oR 2.16 The given or available data is: Equations and Computations: From p 0 and p e, and Eq. 13.7a (using built-in function IsenMfromp (M ,k )) Btu/lbm·oR ft·lbf/lbm·oR R o psi psi o R psi o R We can now use Rayleigh-line relations to compute values for a range of Mach numbers: M T /T * T (oR) 2.157 2 1.99 1.98 1.97 1.96 1.95 1.94 1.93 1.92 1.91 1.9 1.89 1.88 1.87 1.86 1.85 1.84 1.83 1.82 1.81 1.8 1.79 1.78 1.77 1.76 1.75 1.74 1.73 1.72 1.71 1.7 0.475 0.529 0.533 0.536 0.540 0.544 0.548 0.552 0.555 0.559 0.563 0.567 0.571 0.575 0.579 0.584 0.588 0.592 0.596 0.600 0.605 0.609 0.613 0.618 0.622 0.626 0.631 0.635 0.640 0.645 0.649 0.654 368 410 413 416 418 421 424 427 430 433 436 o 440 T ( R) 443 446 449 452 455 459 462 465 468 472 475 479 482 485 489 492 496 499 503 507 c (ft/s) 800 750 700 650 600 550 500 450 400 350 300 940 993 996 1000 1003 1007 1010 1014 1017 1021 1024 1028 1032 1035 1039 1043 1046 1050 0 1054 10 1057 1061 1065 1069 1073 1076 1080 1084 1088 1092 1096 1100 1104 V (ft/s) 2028 1985 1982 1979Ts 1976 1973 1970 1966 1963 1960 1957 1953 1950 1946 1943 1939 1936 1932 1928 20 1925 1921 1917 1913 1909 1905 1901 1897 1893 1889 1885 1880 1876 p /p * 0.32 0.36 0.37 0.37 Curve 0.37 0.38 0.38 0.38 0.39 0.39 0.39 0.40 0.40 0.40 0.41 0.41 0.41 0.42 0.42 30 0.43 s 0.43 0.43 0.44 0.44 0.45 0.45 0.45 0.46 0.46 0.47 0.47 0.48 Δs (ft·lbf/lbm·oR) Eq. (12.11b) 2.5 0.00 2.8 13.30 2.9 14.15 2.9 14.99 (Rayleigh) 2.9 15.84 2.9 16.69 3.0 17.54 3.0 18.39 3.0 19.24 3.0 20.09 3.1 20.93 3.1 21.78 3.1 22.63 3.2 23.48 3.2 24.32 3.2 25.17 3.2 26.01 3.3 26.86 40 3.3 50 27.70 60 3.3 o 28.54 . (ft lbf/lbm R) 3.4 29.38 3.4 30.22 3.4 31.06 3.5 31.90 3.5 32.73 3.5 33.57 3.6 34.40 3.6 35.23 3.6 36.06 3.7 36.89 3.7 37.72 3.7 38.54 p (psi) 70 80 1.69 1.68 1.67 1.66 1.65 1.64 1.63 1.62 1.61 1.6 1.59 1.58 1.57 1.56 1.55 1.54 1.53 1.52 1.51 1.5 1.49 1.48 1.47 1.46 1.45 1.44 1.43 1.42 1.41 1.4 1.39 1.38 1.37 1.36 1.35 1.34 1.33 1.32 1.31 1.3 1.29 1.28 1.27 1.26 1.25 1.24 1.23 1.22 1.21 1.2 1.19 1.18 1.17 1.16 1.15 1.14 1.13 1.12 1.11 1.1 1.09 1.08 1.07 1.06 1.05 1.04 1.03 1.02 1.01 1 0.658 0.663 0.668 0.673 0.677 0.682 0.687 0.692 0.697 0.702 0.707 0.712 0.717 0.722 0.727 0.732 0.737 0.742 0.747 0.753 0.758 0.763 0.768 0.773 0.779 0.784 0.789 0.795 0.800 0.805 0.811 0.816 0.822 0.827 0.832 0.838 0.843 0.848 0.854 0.859 0.865 0.870 0.875 0.881 0.886 0.891 0.896 0.902 0.907 0.912 0.917 0.922 0.927 0.932 0.937 0.942 0.946 0.951 0.956 0.960 0.965 0.969 0.973 0.978 0.982 0.986 0.989 0.993 0.997 1.000 510 1107 514 1111 517 1115 521 1119 525 1123 529 1127 532 1131 536 1135 540 1139 544 1143 548 1147 2500 551 1151 555 1155 559 2000 1159 563 1164 567 1500 1168 571 1172 V (ft/s) 575 1176 1000 579 1180 583 1184 500 1188 587 591 1192 595 0 1196 599 1200 2.0 603 1204 607 1208 612 1213 616 1217 620 1221 624 1225 628 1229 632 1233 636 1237 9 641 1241 645 8 1245 649 7 1249 653 1253 6 657 1257 662 5 1261 p (psi) 666 4 1265 670 3 1269 674 1273 2 678 1277 682 1 1281 686 0 1285 690 2.0 1288 694 1292 699 1296 703 1300 706 1303 710 1307 714 1310 718 1314 722 1318 726 1321 730 1324 733 1328 737 1331 741 1334 744 1337 747 1341 751 1344 754 1347 757 1349 761 1352 764 1355 767 1358 769 1360 772 1362 775 1365 1872 1867 1863 1858 1853 1849 1844 1839 1834 Velocity V 1829 1824 1819 1814 1809 1803 1798 1793 1787 1782 1776 1770 1764 1758 1752 1.8 1746 1740 1734 1728 1721 1715 1708 Pressure p 1701 1695 1688 1681 1674 1667 1659 1652 1645 1637 1629 1622 1614 1606 1.81598 1589 1581 1573 1564 1555 1546 1537 1528 1519 1510 1500 1491 1481 1471 1461 1451 1441 1430 1420 1409 1398 1387 1376 1365 0.48 3.8 39.36 0.48 3.8 40.18 0.49 3.8 41.00 0.49 3.9 41.81 0.50 3.9 42.62 0.50 3.9 43.43 0.51 4.0 44.24 0.51 4.0 45.04 45.84 V0.52 ersus M (4.1 Rayleigh) 0.52 4.1 46.64 0.53 4.1 47.43 0.53 4.2 48.22 0.54 4.2 49.00 0.54 4.3 49.78 0.55 4.3 50.56 0.56 4.3 51.33 0.56 4.4 52.10 0.57 4.4 52.86 0.57 4.5 53.62 0.58 4.5 54.37 0.58 4.6 55.12 0.59 4.6 55.86 0.60 4.7 56.60 0.60 4.7 57.33 1.6 1.4 0.61 4.8 58.05 M4.8 0.61 58.77 0.62 4.9 59.48 0.63 4.9 60.18 0.63 5.0 60.88 0.64 5.0 61.56 0.65 5.1 62.24 Versus M (Rayleigh) 0.65 5.1 62.91 0.66 5.2 63.58 0.67 5.2 64.23 0.68 5.3 64.88 0.68 5.3 65.51 0.69 5.4 66.14 0.70 5.5 66.76 0.71 5.5 67.36 0.71 5.6 67.96 0.72 5.6 68.54 0.73 5.7 69.11 0.74 5.8 69.67 0.74 5.8 70.22 0.75 5.9 70.75 0.76 6.0 1.4 71.27 1.6 0.77 6.0 71.78 M 6.1 0.78 72.27 0.79 6.2 72.75 0.80 6.2 73.21 0.80 6.3 73.65 0.81 6.4 74.08 0.82 6.4 74.50 0.83 6.5 74.89 0.84 6.6 75.27 0.85 6.7 75.63 0.86 6.7 75.96 0.87 6.8 76.28 0.88 6.9 76.58 0.89 7.0 76.86 0.90 7.1 77.11 0.91 7.1 77.34 0.92 7.2 77.55 0.93 7.3 77.73 0.94 7.4 77.88 0.95 7.5 78.01 0.97 7.6 78.12 0.98 7.6 78.19 0.99 7.7 78.24 1.00 7.8 78.25 1.2 1.2 1.0 1.0 Problem 13.80 [2] Problem 13.81 [2] Problem 13.82 Given: Frictionless air flow in a pipe Find: [2] Heat exchange per lb (or kg) at exit, where 500 kPa Solution: Basic equations: mrate = ρ⋅ V⋅ A p = ρ⋅ R ⋅ T Given or available data T1 = ( 15 + 273) ⋅ K ρ1 = ) ( p1 − p2 = ρ1⋅ V1⋅ V2 − V1 (Energy) ) p1 = 1⋅ MPa p1 R = 286.9⋅ c1 = m c1 = 340 s kg 3 k⋅ R⋅ T1 m V1 = 119 From momentum V2 = + V1 From continuity ρ1⋅ V1 = ρ2⋅ V2 m s V1 ρ 2 = ρ 1⋅ V2 ρ2 = 3.09 T2 = 564 K p2 J kg⋅ K m s V2 = 466 V1 = M1⋅ c1 ρ1⋅ V1 J cp = 1004⋅ kg⋅ K (Momentum) p2 = 500⋅ kPa ρ1 = 12.1 R ⋅ T1 p1 − p2 M1 = 0.35 k = 1.4 D = 5⋅ cm At section 1 ( δQ = cp⋅ T02 − T01 dm T2 = 291 °C Hence T2 = and T02 = T2⋅ ⎛ 1 + ⎜ k −1 with T01 = T1⋅ ⎛ 1 + ⎜ k −1 Then kg m 3 δQ Btu kJ = cp⋅ T02 − T01 = 164⋅ = 383⋅ dm lbm kg ρ2⋅ R ⎝ ⎝ ( 2 2⎞ ⋅ M2 ⎟ T02 = 677 K T02 = 403 °C 2⎞ ⋅ M1 ⎟ T01 = 295 K T01 = 21.9 °C ⎠ 2 ⎠ ) (Note: Using Rayleigh line functions, for M1 = 0.35 T0 T0crit so = 0.4389 T0crit = T01 0.4389 T0crit = 672 K close to T2 ... Check!) M2 = 1 Problem 13.83 Given: Frictionless flow of Freon in a tube Find: [2] Heat transfer; Pressure drop NOTE: ρ2 is NOT as stated; see below Solution: Basic equations: mrate = ρ⋅ V⋅ A ( p = ρ⋅ R ⋅ T Q = mrate⋅ h02 − h01 Btu Given or available data h1 = 25⋅ lbm ρ1 = 100⋅ lbm ft π2 ⋅D 4 3 ) 2 ( h0 = h + V 2 p1 − p2 = ρ1⋅ V1⋅ V2 − V1 h2 = 65⋅ Btu lbm ρ2 = 0.850⋅ mrate V1 = ρ1⋅ A ft V1 = 8.03 s ft V2 = 944 s V2 3 lbm mrate = 1.85⋅ s h01 = h1 + 2 mrate V2 = ρ2⋅ A Then A= lbm ft 2 D = 0.65⋅ in ) A = 0.332 in 2 ( ) The heat transfer is Q = mrate⋅ h02 − h01 The pressure drop is Δp = ρ1⋅ V1⋅ V2 − V1 ( V1 h02 = h2 + 2 Q = 107 ) Btu s Δp = 162 psi h01 = 25.0 Btu lbm h02 = 82.8 Btu lbm 2 (74 Btu/s with the wrong ρ2!) (-1 psi with the wrong ρ2!) Problem 13.84 [3] Problem 13.86 [3] Problem 13.87 [3] Given: Frictionless flow of air in a duct Find: Heat transfer without choking flow; change in stagnation pressure Solution: Basic equations: T0 T p0 π2 ⋅D 4 2 A = 78.54 cm ρ1 = From continuity p1 − p2 = k = 1.4 ( 3 m s then ) 2 From continuity ρ1⋅ V1 = or p1 p1 ⋅ M1⋅ c1 = ⋅ M ⋅ k⋅ R⋅ T1 = R ⋅ T1 R ⋅ T1 1 k⋅ R⋅ T1 T2 T02 = T2⋅ ⎛ 1 + ⎜ k−1 ⎝ p02 = p2⋅ ⎛ 1 + ⎜ 2 k−1 ⎝ ( 2 2⎞ ⋅ M2 ⎟ 2⎞ ⋅ M2 ⎟ p02 = 58.8 kPa ) δQ MJ = cp⋅ T02 − T01 = 1.12⋅ dm kg (Using Rayleigh functions, at M1 = 0.215 T01 T0crit = T01 2 2 p 2 2 ⋅ k⋅ R⋅ T⋅ M = k⋅ p⋅ M R⋅ T p2 = 31.1 kPa k p1⋅ M1 ⋅ = ρ2⋅ V2 = R T1 k k−1 ⎠ 2 J kg⋅ K M1 = 0.215 ⎛ 1 + k⋅ M 2 ⎞ 1⎟ p2 = p1⋅ ⎜ ⎜ 1 + k⋅ M 2 ⎟ 2⎠ ⎝ T02 = 1394 K ⎠ V1 M1 = c1 ρ ⋅ V = ρ⋅ c ⋅ M = but ⎛ p2 M2 ⎞ T2 = T1⋅ ⎜ ⋅ ⎟ ⎝ p1 M1 ⎠ p2⋅ M2 T1 Finally m c1 = 331 s kg mrate 2 2 ⋅ V2 − V1 = ρ2⋅ V2 − ρ1⋅ V1 A p1 − p2 = k⋅ p2⋅ M2 − k⋅ p1⋅ M1 Then c1 = V1 = 71.2 2 = R = 286.9⋅ m Hence Hence J cp = 1004⋅ kg⋅ K ρ1 = 0.894 R ⋅ T1 p1⋅ M1 D = 10⋅ cm δQ = cp⋅ T02 − T01 dm mrate V1 = ρ1⋅ A From momentum kg mrate = 0.5⋅ s M2 = 1 ) p1 At state 1 ) mrate = ρ⋅ A⋅ V p1 = 70⋅ kPa ( Given or available data T1 = ( 0 + 273) ⋅ K A= k − 1 2⎞ ⋅M ⎟ 2 ⎠ p = ρ⋅ R ⋅ T p mrate ⋅ V2 − V1 A p1 − p2 = ⎛ ⎝ = ⎜1 + ( k−1 2 ⋅M 2 = 1+ k k−1 k p2⋅ M2 ⋅ R T2 2 T2 = 1161 K T01 = T1⋅ ⎛ 1 + ⎜ ⎝ p01 = p1⋅ ⎛ 1 + ⎜ Δp0 = p02 − p01 T01 = 0.1975 T02 = T02 0.1975 ⎝ T2 = 888 °C k−1 ⎠ 2 k−1 2 2⎞ ⋅ M1 ⎟ 2⎞ ⋅ M1 ⎟ ⎠ T01 = 276 K k k−1 p01 = 72.3 kPa Δp0 = −13.5 kPa T02 = 1395 K and ditto for p02 ...Check!) Problem 13.88 [3] Problem 13.93 [3] Given: Data on flow through gas turbine combustor Find: Maximum heat addition; Outlet conditions; Reduction in stagnation pressure; Plot of process Solution: R= k= cp = T1 = p1 = M1 = The given or available data is: 286.9 1.4 1004 773 1.5 0.5 p02 J/kg·K T02 J/kg·K K MPa p2 T p01 T01 T1 Equations and Computations: From p1 = ρ 1 RT1 ρ1= 6.76 V1 = M 1 kRT1 V1 = 279 p1 kg/m3 From T2 m/s s Using built-in function IsenT (M,k): T 01 /T 1 = 1.05 T 01 = 812 K Using built-in function Isenp (M,k): p 01 /p 1 = 1.19 p 01 = 1.78 MPa M2 = For maximum heat transfer: 1 Using built-in function rayT0 (M,k), rayp0 (M,k), rayT (M,k), rayp (M,k), rayV (M,k): * T 01 /T 0* = T0 = 0.691 1174 K * p 01 /p 0 = T /T * = * p /p = ρ /ρ = * * p0 = T* = 1.114 0.790 * 1.60 MPa ( = p 02) 978 K ( = T 02) ( = p 2) 1.778 p= 0.844 0.444 ρ= 3.01 MPa 3 kg/m -182 kPa * Note that at state 2 we have critical conditions! Hence: From the energy equation: p 012 – p 01 = δQ dm -0.182 MPa = c p (T02 − T01 ) δ Q /dm = 364 kJ/kg ( = T 02) ( = ρ 2) Problem 13.94 [3] Problem 13.95 [3] Problem 13.96 [3] Problem 13.97 [4] Part 1/2 Problem 13.97 [4] Part 2/2 Problem 13.98 [4] Part 1/2 Problem 13.98 [4] Part 2/2 Problem 13.99 Given: Normal shock due to explosion Find: Shock speed; temperature and speed after shock [3] V Shock speed Vs S hift coordinates: (Vs – V) ( Vs) Solution: Shock at rest 2 M1 + k−1 2 Basic equations: 2 M2 = p2 p1 = V = M⋅ c = M⋅ k⋅ R⋅ T ⎛ 2⋅ k ⎞ ⋅ M 2 − 1 ⎜ ⎟1 ⎝ k − 1⎠ ⎛ 1 + k − 1 ⋅ M 2⎞ ⋅ ⎛ k⋅ M 2 − k − 1 ⎞ ⎜ ⎟ 1⎟⎜ 1 2 2⎠ ⎠⎝ =⎝ 2 T1 ⎛ k + 1 ⎞ ⋅M 2 ⎜ ⎟1 ⎝2⎠ T2 2⋅ k 2 k−1 ⋅ M1 − k+1 k+1 k = 1.4 From the pressure ratio M1 = Then we have ⎛ 1 + k − 1 ⋅ M 2⎞ ⋅ ⎛ k⋅ M 2 − k − 1 ⎞ ⎜ ⎟ 1⎟⎜ 1 2⎠ 2 ⎠⎝ T2 = T1⋅ ⎝ 2 ⎛ k + 1 ⎞ ⋅M 2 ⎜ ⎟1 ⎝2⎠ M2 = R = 286.9⋅ J kg⋅ K Given or available data ⎛ k + 1 ⎞ ⋅ ⎛ p2 + k − 1 ⎞ ⎜ ⎟⎜ ⎟ ⎝ 2⋅ k ⎠ ⎝ p1 k + 1 ⎠ p2 = 30⋅ MPa p1 = 101⋅ kPa T1 = ( 20 + 273) ⋅ K T2 = 14790 K T2 = 14517⋅ °C M1 = 16.0 2 2 M1 + k−1 M2 = 0.382 ⎛ 2⋅ k ⎞ ⋅ M 2 − 1 ⎜ ⎟1 ⎝ k − 1⎠ m Then the speed of the shock (Vs = V1) is V1 = M1⋅ k⋅ R⋅ T1 V1 = 5475 After the shock (V2) the speed is V2 = M2⋅ k⋅ R⋅ T2 V2 = 930 m V = Vs − V2 V = 4545 m But we have V2 = Vs − V s Vs = V1 Vs = 5475 s s These results are unrealistic because at the very high post-shock temperatures experienced, the specific heat ratio will NOT be constant! The extremely high initial air velocity and temperature will rapidly decrease as the shock wave expands in a spherical manner and thus weakens. m s Problem 13.100 [3] Given: C-D nozzle with normal shock Find: Mach numbers at the shock and at exit; Stagnation and static pressures before and after the shock Solution: ⎛ 1 + k − 1 ⋅ M2 ⎞ ⎟ A 1⎜ 2 = ⋅⎜ ⎟ k+ 1 Acrit M⎜ ⎟ 2 ⎝ ⎠ Basic equations: Isentropic flow Given or available data 2 2 M1 + k−1 2 M2 = Normal shock k+ 1 2⋅ ( k−1) Rair = 53.33⋅ 2 At = 1.5⋅ in p1 = p02 2⋅ k 2 k−1 ⋅ M1 − k+1 k+1 ⎛ ⎝ = ⎜1 + p p2 ⎛ 2⋅ k ⎞ ⋅ M 2 − 1 ⎜ ⎟1 ⎝ k − 1⎠ k = 1.4 p0 p01 = k − 1 2⎞ ⋅M ⎟ 2 ⎠ k k−1 ⎛ k + 1 ⋅M 2 ⎞ 1⎟ ⎜ 2 ⎜ ⎟ ⎜ 1 + k − 1 ⋅ M12 ⎟ 2 ⎝ ⎠ k k−1 ⎛ 2⋅ k ⋅ M 2 − k − 1 ⎞ ⎜ ⎟ 1 k + 1⎠ ⎝k + 1 ft⋅ lbf lbm⋅ R T0 = ( 175 + 460) ⋅ R (Shock area) 2 As = 2.5⋅ in p01 = 125⋅ psi 1 k−1 Ae = 3.5⋅ in 2 Because we have a normal shock the CD must be accelerating the flow to supersonic so the throat is at critical state. Acrit = At At the shock we have As = 1.667 ⎛ 1 + k − 1 ⋅M 2 ⎞ 1⎟ 1⎜ 2 At this area ratio we can find the Mach number before the shock from the isentropic relation = ⋅⎜ ⎟ k+ 1 Acrit M1 ⎜ ⎟ 2 ⎝ ⎠ Acrit k+ 1 2⋅ ( k−1) As Solving iteratively (or using Excel's Solver, or even better the function isenMsupfromA from the Web site!) M1 = 1.985 The stagnation pressure before the shock was given: p01 = 125 psi The static pressure is then p1 = p01 ⎛ 1 + k − 1 ⋅ M 2⎞ ⎜ 1⎟ 2 ⎝ ⎠ k k−1 p1 = 16.4 psi After the shock we have Also M2 = 2 2 M1 + k−1 M2 = 0.580 ⎛ 2⋅ k ⎞ ⋅ M 2 − 1 ⎜ ⎟1 ⎝ k − 1⎠ ⎛ k + 1 ⋅M 2 ⎞ 1⎟ ⎜ 2 ⎜ ⎟ ⎜ 1 + k − 1 ⋅ M12 ⎟ 2 ⎝ ⎠ p02 = p01⋅ k k−1 ⎛ 2⋅ k ⋅ M 2 − k − 1 ⎞ ⎜ ⎟ 1 k + 1⎠ ⎝k + 1 and p02 = 91.0 psi 1 k−1 2⋅ k 2 k − 1⎞ p2 = p1⋅ ⎛ ⋅ M1 − ⎜ ⎟ k + 1⎠ ⎝k + 1 p2 = 72.4 psi Finally, for the Mach number at the exit, we could find the critical area change across the shock; instead we find the new critical area from isentropic conditions at state 2. − ⎛ 1 + k − 1 ⋅M 2 ⎞ ⎜ 2⎟ 2 Acrit2 = As⋅ M2⋅ ⎜ ⎟ k+ 1 ⎜ ⎟ 2 ⎝ ⎠ At the exit we have Ae Acrit2 k+ 1 2⋅ ( k−1) 2 Acrit2 = 2.06 in = 1.698 ⎛ 1 + k − 1 ⋅M 2 ⎞ e⎟ 1⎜ 2 At this area ratio we can find the Mach number before the shock from the isentropic relation = ⋅⎜ ⎟ k+ 1 Acrit2 Me ⎜ ⎟ 2 ⎝ ⎠ k+ 1 2⋅ ( k−1) Ae Solving iteratively (or using Excel's Solver, or even better the function isenMsubfromA from the Web site!) These calculations are obviously a LOT easier using the Excel functions available on the Web site! Me = 0.369 Problem 13.101 Given: Normal shock near pitot tube Find: [2] Air speed Solution: Basic equations: ( p1 − p2 = ρ1⋅ V1⋅ V2 − V1 ) Given or available data T1 = 285⋅ R Rair = 53.33⋅ k −1 ⎤ ⎡ ⎢ ⎥ k ⎥ 2 ⎢⎛ p02 ⎞ ⋅ ⎢⎜ − 1⎥ ⎟ k −1 ⎣⎝ p2 ⎠ ⎦ At state 2 M2 = From momentum p1 − p2 = ρ2⋅ V2 − ρ1⋅ V1 2 2 2 2 ⎤ 1 ⎡ p2 ⎛ 2 ⋅ ⎢ ⋅ 1 + k⋅ M2 ⎞ − 1⎥ ⎝ ⎠ k p1 ⎣ Also c1 = Then ⎦ k⋅ Rair⋅ T1 V1 = M1⋅ c1 k − 1 2⎞ ⋅M ⎟ 2 ⎠ p02 = 10⋅ psi M2 = 0.574 2 2 p1 2 p 2 2 ⋅ k ⋅ R⋅ T⋅ M = k ⋅ p ⋅ M R⋅ T ρ ⋅ V = ρ⋅ c ⋅ M = or 2 2 p1⋅ ⎛ 1 + k⋅ M1 ⎞ = p2⋅ ⎛ 1 + k⋅ M2 ⎞ ⎝ ⎠ ⎝ ⎠ M1 = 2.01 ft c1 = 827 s ft s V1 = 1822⋅ p2 p2 = 8⋅ psi ft⋅ lbf lbm⋅ R V1 = 1666 Note: With p1 = 1.5 psi we obtain (Using normal shock functions, for ⎛ ⎝ = ⎜1 + but p1 − p2 = k⋅ p2⋅ M2 − k⋅ p1⋅ M1 M1 = p p1 = 1.75⋅ psi k = 1.4 Hence p0 (Momentum) k k−1 = 4.571 we find M1 = 2.02 ft s M2 = 0.573 Check!) Problem 13.103 [2] Problem 13.104 [2] Given: Normal shock Find: Speed and temperature after shock; Entropy change Solution: R= k= cp = The given or available data is: 53.33 1.4 0.2399 T 01 = 1250 p1 = 20 M1 = 300.02 V1 = 764 Using built-in function IsenT (M,k): T 01 /T 1 = 0.0685 Btu/lbm·R Btu/lbm·R R o 2.5 ρ1 = ft·lbf/lbm·R 2.25 psi Equations and Computations: From p1 = ρ1 RT1 kg/m3 m/s o R o F o R 728 T1 = o F 143 psi 556 96 Using built-in function NormM2fromM (M,k): M2 = 0.513 Using built-in function NormTfromM (M,k): T 2 /T 1 = 2.14 Using built-in function NormpfromM (M,k): p 2 /p 1 = From V 2 = M 2 kRT 2 From ⎛T Δ s = c p ln ⎜ 2 ⎜T ⎝1 T2 = p2 = 7.13 V2 = 867 Δs = 0.0476 37.1 ft/s ⎛p ⎞ ⎞ ⎟ − R ln ⎜ 2 ⎟ ⎜p ⎟ ⎟ ⎝ 1⎠ ⎠ Btu/lbm·R ft·lbf/lbm·R 1188 Problem 13.106 [2] Given: Normal shock Find: Pressure after shock; Compare to isentropic deceleration Solution: R= k= T 01 = 286.9 1.4 550 p 01 = 650 M1 = The given or available data is: 2.5 J/kg·K K kPa Equations and Computations: Using built-in function Isenp (M,k): p 01 /p 1 = 17.09 Using built-in function NormM2fromM (M,k): M2 = 0.513 Using built-in function NormpfromM (M,k): p 2 /p 1 = 7.13 p1 = 38 kPa p2 = 271 kPa p2 = 543 kPa Using built-in function Isenp (M,k) at M 2: p 02 /p 2 = But for the isentropic case: Hence for isentropic deceleration: 1.20 p 02 = p 01 Problem 13.107 [2] Given: Normal shock Find: Speed and Mach number after shock; Change in stagnation pressure Solution: R= k= T1 = 53.33 1.4 445 p1 = 5 V1 = 2000 mph c1 = 1034 ft/s M1 = The given or available data is: 2.84 ft·lbf/lbm·R o 0.0685 Btu/lbm·R R psi 2933 ft/s 793 ft/s Equations and Computations: From c1 = kRT1 Then Using built-in function NormM2fromM (M,k): M2 = 0.486 Using built-in function NormdfromM (M,k): ρ 2 /ρ 1 = 3.70 Using built-in function Normp0fromM (M,k): p 02 /p 01 = 0.378 Then V2 = ρ1 V1 ρ2 V2 = 541 mph Using built-in function Isenp (M,k) at M 1: p 01 /p 1 = 28.7 From the above ratios and given p 1: p 01 = 143 psi p 02 = 54.2 psi p 01 – p 02 = 89.2 psi Problem 13.108 [2] Given: Normal shock Find: Speed; Change in pressure; Compare to shockless deceleration Solution: R= k= T1 = 53.33 1.4 452.5 p1 = 14.7 psi V1 = 1750 mph c1 = 1043 ft/s M1 = The given or available data is: 2.46 ft·lbf/lbm·R o 0.0685 Btu/lbm·R R 2567 ft/s p2 = 101 psi p2 – p1 = 86.7 psi 781 ft/s p2 = 197 psi p2 – p1 = 182 psi Equations and Computations: From c1 = kRT1 Then Using built-in function NormM2fromM (M,k): M2 = 0.517 Using built-in function NormdfromM (M,k): ρ 2 /ρ 1 = 3.29 Using built-in function NormpfromM (M,k): p 2 /p 1 = 6.90 Then V2 = ρ1 V1 ρ2 V2 = 532 mph Using built-in function Isenp (M,k) at M 1: p 01 /p 1 = 16.1 Using built-in function Isenp (M,k) at M 2: p 02 /p 2 = 1.20 From above ratios and p 1, for isentropic flow (p 0 = const): Problem 13.109 [2] Problem 13.111 [2] Problem 13.112 Given: Normal shock Find: [4] Rankine-Hugoniot relation Solution: 2 Momentum: p1 + ρ1⋅ V1 = p2 + ρ2⋅ V2 Energy: Basic equations: h1 + 2 Mass: Ideal Gas: 1 1 2 2 ⋅ V1 = h2 + ⋅ V2 2 2 ( ) ( ) ρ1 V1 = ρ2⋅ V2 p = ρ⋅ R ⋅ T 2 2 ( )( From the energy equation 2⋅ h2 − h1 = 2⋅ cp⋅ T2 − T1 = V1 − V2 = V1 − V1 ⋅ V1 + V2 From the momentum equation p2 − p1 = ρ1⋅ V1 − ρ2⋅ V2 = ρ1⋅ V1⋅ V1 − V2 Hence V1 − V2 = Using this in Eq 1 2 ( 2 ) ) (1) where we have used the mass equation p2 − p1 ρ1⋅ V1 p2 − p1 p2 − p1 ⎛ V2 ⎞ p2 − p1 ⎛ 2⋅ cp⋅ T2 − T1 = ⋅ V1 + V2 = ⋅ ⎜1 + ⋅ ⎜1 + ⎟= ρ1⋅ V1 ρ1 ρ1 ⎝ V1 ⎠ ⎝ ( ) ( ) ρ1 ⎞ 1⎞ ⎛1 ⎟ = (p2 − p1)⋅ ⎜ + ⎟ ρ2 ⎠ ⎝ ρ1 ρ2 ⎠ where we again used the mass equation Using the idea gas equation ⎛ p2 2⋅ cp⋅ ⎜ ⎝ ρ2⋅ R − p1 ⎞ 1⎞ ⎛1 ⎟ = (p2 − p1)⋅ ⎜ + ⎟ ρ1⋅ R ⎠ ⎝ ρ1 ρ2 ⎠ Dividing by p1 and multiplying by ρ2, and using R = cp - cv, k = cp/cv 2⋅ Collecting terms cp ⎛ p2 ρ2 ⎞ ⎞ ⎛ ρ2 ⎞ k ⎛ p2 ρ2 ⎞ ⎛ p2 ⋅⎜ − ⋅⎜ − ⎟ = 2⋅ ⎟ = ⎜ − 1⎟ ⋅ ⎜ + 1⎟ R p1 ρ1 k − 1 p1 ρ1 ⎝ ⎠ ⎝ ⎠ ⎝ p1 ⎠ ⎝ ρ1 ⎠ p2 ⎛ 2⋅ k ⋅⎜ −1− p1 k − 1 ⎝ p2 p1 For an infinite pressure ratio = ρ2 ⎞ 2⋅ k ρ2 ρ2 ⋅ − −1 k − 1 ρ1 ρ1 ⎛ 2⋅ k −1− ⎜ ⎝k − 1 ( k + 1) − ( k − 1) ⋅ 2⋅ k ρ2 ρ2 ⋅ − −1 ⎟= ρ1 ⎠ k − 1 ρ1 ρ1 ρ2 ρ1 ρ2 ⎞ ⎟ ρ1 ⎠ =0 = ( k + 1) ρ2 ⋅ −1 ( k − 1) ρ1 ( k + 1) ( k − 1) or − ρ2 or p2 p1 ( k + 1) ⋅ = ρ1 ρ1 − ( k − 1) ( k + 1) − ( k − 1) ⋅ ρ1 ρ2 ρ2 = k+1 k−1 ρ2 ρ1 (= 6 for air) Problem 13.113 [3] Problem 13.114 [3] Problem 13.115 [3] Problem 13.117 [3] Problem 13.119 [4] Problem 13.120 [2] Problem 13.121 [2] Problem 13.124 [3] Given: Normal shock in CD nozzle Find: Exit pressure; Throat area; Mass flow rate Solution: R= k= T 01 = p 01 = M1 = 286.9 1.4 550 700 2.75 A1 = 25 cm2 Ae = The given or available data is: 40 cm2 J/kg·K K kPa Equations and Computations (assuming State 1 and 2 before and after the shock): Using built-in function Isenp (M,k): p 01 /p 1 = 25.14 p1 = 28 kPa Using built-in function IsenT (M,k): T 01 /T 1 = 2.51 T1 = 219 K Using built-in function IsenA (M,k): A 1 /A 1* = 3.34 A 1* = A t = 7.49 cm2 p 02 = 284 kPa Me= 0.279 pe= 269 Then from the Ideal Gas equation: ρ1 = c1 = V1 = Also: So: Then the mass flow rate is: 0.4433 297 815 m rate = m rate = kg/m3 m/s m/s ρ 1 V 1A 1 0.904 kg/s For the normal shock: Using built-in function NormM2fromM (M,k): M2 = 0.492 Using built-in function Normp0fromM (M,k) at M 1: p 02 /p 01 = 0.41 For isentropic flow after the shock: Using built-in function IsenA (M,k): A 2 /A 2* = A2 = But: 1.356 A1 A 2* = 18.44 Hence: Using built-in function IsenAMsubfromA (Aratio,k): A e /A 2* = For: 2.17 Using built-in function Isenp (M,k): p 02 /p e = 1.06 cm2 kPa Problem 13.125 [2] Problem 13.128 [3] Problem 13.129 [3] Problem 13.131 [3] Problem 13.132 [4] Problem 13.133 [3] Problem *13.135 [5] Problem *13.136 [2] Problem *13.137 [4] Problem 13.138 [3] Given: Normal shock Find: Approximation for downstream Mach number as upstream one approaches infinity Solution: Basic equations: 2 M2n = 2 2 M1n + k −1 (13.48a) ⎛ 2⋅ k ⎞ ⋅ M 2 − 1 ⎜ ⎟ 1n ⎝ k − 1⎠ M2n = M2⋅ sin ( β − θ) (13.47b) 2 2 M1n + k −1 Combining the two equations M2n M2 = = sin( β − θ) 1+ M2 = ⎛ 2⋅ k ⎞ ⋅ M 2 − 1 ⎜ ⎟ 1n ⎝ k − 1⎠ = sin( β − θ) 2 2 M1n + k −1 ⎡⎛ 2⋅ k ⎞ ⋅ M 2 − 1⎤ ⋅ sin( β − θ) 2 ⎢⎜ ⎟ 1n ⎥ ⎣⎝ k − 1 ⎠ ⎦ 2 ( k − 1) ⋅ M1n 2 ⎡⎛ 2⋅ k ⎞ − 1 ⎥ ⋅ sin( β − θ) 2 ⎤ ⎢⎜ ⎟ ⎝ k − 1 ⎠ M 2⎥ ⎢ 1n ⎦ ⎣ As M1 goes to infinity, so does M1n, so M2 = 1 ⎛ 2⋅ k ⎞ ⋅ sin ( β − θ) 2 ⎜ ⎟ ⎝ k − 1⎠ M2 = k−1 2⋅ k⋅ sin ( β − θ) 2 Problem 13.139 [3] Given: Data on an oblique shock Find: Mach number and pressure downstream; compare to normal shock Solution: R= k= p1 = 286.9 1.4 80 M1 = 2.5 β= The given or available data is: 35 J/kg.K kPa o Equations and Computations: From M 1 and β M 1n = 1.43 M 1t = 2.05 From M1n and p1, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) (13.48d) p2 = 178.6 V t1 = V t2 The tangential velocity is unchanged Hence c t1 M t1 = (T 1 ) 1/2 c t2 M t2 M t1 = (T 2)1/2 M t2 M 2t = (T 1/T 2)1/2 M t1 From M1n, and Eq. 13.48c (using built-in function NormTfromM (M ,k )) T 2/T 1 = Hence 1.28 M 2t = 1.81 kPa Also, from M1n, and Eq. 13.48a (using built-in function NormM2fromM (M ,k )) (13.48a) M 2n = 0.726 The downstream Mach number is then M 2 = (M 2t2 + M 2n2)1/2 M2 = 1.95 Finally, from geometry V 2n = V 2sin(β - θ) Hence θ = β - sin-1(V 2n/V 2) or θ = β - sin-1(M 2n/M 2) θ= 13.2 o 570 kPa For the normal shock: From M1 and p1, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) p2 = Also, from M1, and Eq. 13.48a (using built-in function NormM2fromM (M ,k )) M2 = 0.513 For the minimum β : The smallest value of β is when the shock is a Mach wave (no deflection) β = sin-1(1/M 1) β= 23.6 o Problem 13.140 [3] Given: Oblique shock in flow at M = 3 Find: Minimum and maximum β, plot of pressure rise across shock Solution: The given or available data is: R= k= M1 = 286.9 1.4 3 J/kg.K Equations and Computations: The smallest value of β is when the shock is a Mach wave (no deflection) β = sin-1(1/M 1) β= The largest value is 19.5 o β= 90.0 o The normal component of Mach number is M 1n = M 1sin(β) (13.47a) For each β, p2/p1 is obtained from M1n, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) (13.48d) Computed results: β (o) M 1n p 2/p 1 19.5 20 30 40 50 60 70 75 80 85 90 1.00 1.03 1.50 1.93 2.30 2.60 2.82 2.90 2.95 2.99 3.00 1.00 1.06 2.46 4.17 5.99 7.71 9.11 9.63 10.0 10.3 10.3 Pressure Change across an Oblique Shock 12.5 10.0 p 2/p 1 7.5 5.0 2.5 0.0 0 30 60 β() o 90 Problem 13.141 [3] Given: Velocities and deflection angle of an oblique shock Find: Shock angle β; pressure ratio across shock Solution: R= k= V1 = 286.9 1.4 1250 V2 = The given or available data is: 650 θ= 35 J/kg.K m/s m/s o Equations and Computations: From geometry we can write two equations for tangential velocity: For V 1t V 1t = V 1cos(β) (1) For V 2t V 2t = V 2cos(β - θ) (2) For an oblique shock V 2t = V 1t, so Eqs. 1 and 2 give V 1cos(β) = V 2cos(β - θ) Solving for β (3) β = tan-1((V 1 - V 2cos(θ))/(V 2sin(θ))) β= (Alternatively, solve Eq. 3 using Goal Seek !) 62.5 o For p 2/p 1, we need M 1n for use in Eq. 13.48d (13.48d) We can compute M 1 from θ and β, and Eq. 13.49 (using built-in function Theta (M ,β, k )) (13.49) θ= 35.0 o β= 62.5 o M1 = For 3.19 This value of M 1 was obtained by using Goal Seek : Vary M 1 so that θ becomes the required value. (Alternatively, find M 1 from Eq. 13.49 by explicitly solving for it!) We can now find M 1n from M 1. From M 1 and Eq. 13.47a M 1n = M 1sin(β) Hence M 1n = 2.83 Finally, for p 2/p 1, we use M 1n in Eq. 13.48d (using built-in function NormpfromM (M ,k ) p 2 /p 1 = 9.15 (13.47a) Problem 13.142 Given: Data on an oblique shock Find: Deflection angle θ; shock angle β ; Mach number after shock Solution: The given or available data is: R= k= M1 = T1 = p 2 /p 1 = 286.9 1.4 3.25 283 5 J/kg.K K Equations and Computations: From p 2/p 1, and Eq. 13.48d (using built-in function NormpfromM (M ,k ) and Goal Seek or Solver ) (13.48d) p 2 /p 1 = 5.00 M 1n = For 2.10 From M 1 and M 1n, and Eq 13.47a M 1n = M 1sin(β ) β= 40.4 (13.47a) o From M 1 and β , and Eq. 13.49 (using built-in function Theta (M ,β , k ) (13.49) θ= 23.6 o To find M 2 we need M 2n. From M 1n, and Eq. 13.48a (using built-in function NormM2fromM (M ,k )) (13.48a) M 2n = 0.561 The downstream Mach number is then obtained from from M 2n, θ and β , and Eq. 13.47b M 2n = M 2sin(β - θ) Hence M2 = 1.94 (13.47b) Problem 13.143 [4] Given: Airfoil with included angle of 20o Find: Mach number and speed at which oblique shock forms Solution: R= k= T1 = 286.9 1.4 288 θ= The given or available data is: 10 J/kg.K K o Equations and Computations: From Fig. 13.29 the smallest Mach number for which an oblique shock exists at a deflection θ = 10o is approximately M 1 = 1.4. By trial and error, a more precise answer is (using built-in function Theta (M ,β, k ) M1 = 1.42 β= θ= 67.4 10.00 c1 = V1 = 340 483 A suggested procedure is: 1) Type in a guess value for M 1 2) Type in a guess value for β o o m/s m/s 3) Compute θ from Eq. 13.49 (using built-in function Theta (M ,β, k )) (13.49) 4) Use Solver to maximize θ by varying β o 5) If θ is not 10 , make a new guess for M 1 6) Repeat steps 1 - 5 until θ = 10 o Computed results: β() θ() 67.4 56.7 45.5 39.3 35.0 31.9 27.4 22.2 19.4 17.6 16.4 10.0 10.0 10.0 10.0 10.0 10.0 10.0 10.0 10.0 10.0 10.0 o M1 1.42 1.50 1.75 2.00 2.25 2.50 3.00 4.00 5.00 6.00 7.00 o Error 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% Sum: 0.0% To compute this table: 1) Type the range of M 1 2) Type in guess values for β 3) Compute θ from Eq. 13.49 (using built-in function Theta (M ,β, k ) o Compute the absolute error between each θ and θ = 10 4) 5) Compute the sum of the errors 6) Use Solver to minimize the sum by varying the β values (Note: You may need to interactively type in new β values if Solver generates β values that lead to no θ, or to β values that correspond to a strong rather than weak shock) Oblique Shock Angle as a Function of Aircraft Mach Number 90 75 60 β (o) 45 30 15 0 1 2 3 4 M 5 6 7 Problem 13.144 [4] Given: Airfoil with included angle of 60o Find: Plot of temperature and pressure as functions of angle of attack Solution: R= k= T1 = p1 = V1 = The given or available data is: 286.9 1.4 276.5 75 1200 δ= 60 c1 = 333 M1 = J/kg.K K kPa m/s 3.60 o Equations and Computations: From T 1 Then m/s Computed results: α( ) β() θ ( ) Needed θ() 0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 16.00 18.00 20.00 22.00 24.00 26.00 28.00 30.00 47.1 44.2 41.5 38.9 36.4 34.1 31.9 29.7 27.7 25.7 23.9 22.1 20.5 18.9 17.5 16.1 30.0 28.0 26.0 24.0 22.0 20.0 18.0 16.0 14.0 12.0 10.0 8.0 6.0 4.0 2.0 0.0 30.0 28.0 26.0 24.0 22.0 20.0 18.0 16.0 14.0 12.0 10.0 8.0 6.0 4.0 2.0 0.0 o o o o Sum: Error 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% M 1n p 2 (kPa) T 2 (oC) 2.64 2.51 2.38 2.26 2.14 2.02 1.90 1.79 1.67 1.56 1.46 1.36 1.26 1.17 1.08 1.00 597 539 485 435 388 344 304 267 233 202 174 149 126 107 90 75 357 321 287 255 226 198 172 148 125 104 84 66 49 33 18 3 597 357 Max: To compute this table: 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) Type the range of α Type in guess values for β Compute θNeeded from θ = δ/2 - α Compute θ from Eq. 13.49 (using built-in function Theta (M ,β, k ) Compute the absolute error between each θ and θNeeded Compute the sum of the errors Use Solver to minimize the sum by varying the β values (Note: You may need to interactively type in new β values if Solver generates β values that lead to no θ) For each α, M 1n is obtained from M 1, and Eq. 13.47a For each α, p 2 is obtained from p 1, M 1n, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) For each α, T 2 is obtained from T 1, M 1n, and Eq. 13.48c (using built-in function NormTfromM (M ,k )) Pressure on an Airfoil Surface as a Function of Angle of Attack 700 p 2 (kPa) 600 500 400 300 200 100 0 0 5 10 15 20 25 30 25 30 α( ) o Temperature on an Airfoil Surface as a Function of Angle of Attack 400 350 o T 2 ( C) 300 250 200 150 100 50 0 0 5 10 15 α( ) o 20 Problem 13.145 [4] Given: Airfoil with included angle of 60o Find: Angle of attack at which oblique shock becomes detached Solution: R= k= T1 = p1 = V1 = The given or available data is: 286.9 1.4 276.5 75 1200 δ= 60 c1 = 333 M1 = J/kg.K K kPa m/s 3.60 o Equations and Computations: From T 1 Then m/s From Fig. 13.29, at this Mach number the smallest deflection angle for which an oblique shock exists is approximately θ = 35o. By using Solver , a more precise answer is (using built-in function Theta (M ,β, k ) M1 = 3.60 β= θ= 65.8 37.3 o o A suggested procedure is: 1) Type in a guess value for β 2) Compute θ from Eq. 13.49 (using built-in function Theta (M ,β, k )) (13.49) 3) Use Solver to maximize θ by varying β For a deflection angle θ the angle of attack α is α = θ - δ/2 α= 7.31 o Computed results: α (o) β (o) θ (o) Needed θ (o) 0.00 1.00 2.00 3.00 4.00 5.50 5.75 6.00 6.25 6.50 6.75 7.00 7.25 7.31 47.1 48.7 50.4 52.1 54.1 57.4 58.1 58.8 59.5 60.4 61.3 62.5 64.4 65.8 30.0 31.0 32.0 33.0 34.0 35.5 35.8 36.0 36.3 36.5 36.8 37.0 37.3 37.3 30.0 31.0 32.0 33.0 34.0 35.5 35.7 36.0 36.2 36.5 36.7 37.0 37.2 37.3 Sum: Error 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% M 1n 2.64 2.71 2.77 2.84 2.92 3.03 3.06 3.08 3.10 3.13 3.16 3.19 3.25 3.28 0.0% Max: p 2 (kPa) 597 628 660 695 731 793 805 817 831 845 861 881 910 931 T 2 (oC) 357 377 397 418 441 479 486 494 502 511 521 533 551 564 931 564 To compute this table: Type the range of α Type in guess values for β Compute θNeeded from θ = α + δ/2 Compute θ from Eq. 13.49 (using built-in function Theta (M ,β, k ) Compute the absolute error between each θ and θNeeded Compute the sum of the errors Use Solver to minimize the sum by varying the β values (Note: You may need to interactively type in new β values if Solver generates β values that lead to no θ) For each α, M 1n is obtained from M 1, and Eq. 13.47a For each α, p 2 is obtained from p 1, M 1n, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) For each α, T 2 is obtained from T 1, M 1n, and Eq. 13.48c (using built-in function NormTfromM (M ,k )) 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) Pressure on an Airfoil Surface as a Function of Angle of Attack 1000 p 2 (kPa) 900 800 700 600 500 0 2 4 6 8 6 8 o α( ) Temperature on an Airfoil Surface as a Function of Angle of Attack 600 500 o T 2 ( C) 550 450 400 350 300 0 2 4 α (o) Problem 13.146 Given: Data on airfoil flight Find: Lift per unit span Solution: R= k= p1 = 286.9 1.4 70 M1 = 2.75 δ= c= The given or available data is: 7 1.5 J/kg.K kPa o m Equations and Computations: The lift per unit span is L = (p L - p U)c (1) (Note that p L acts on area c /cos(δ), but its normal component is multiplied by cos(δ)) For the upper surface: pU = p1 pU = 70.0 kPa For the lower surface: We need to find M 1n θ= δ The deflection angle is θ= 7 o From M 1 and θ, and Eq. 13.49 (using built-in function Theta (M , β,k )) (13.49) θ= 7.0 o β= 26.7 o M 1n = 1.24 For (Use Goal Seek to vary β so that θ = δ) From M 1 and β From M 1n and p 1, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) (13.48d) p2 = pL = p2 pL = From Eq 1 113 113 kPa L= 64.7 kN/m kPa Problem 13.147 [3] Given: Data on airfoil flight Find: Lift per unit span Solution: R= k= p1 = 286.9 1.4 75 M1 = 2.75 δU = 5 o δL = c= 15 2 o The given or available data is: J/kg.K kPa m Equations and Computations: The lift per unit span is L = (p L - p U)c (1) (Note that each p acts on area c /cos(δ), but its normal component is multiplied by cos(δ)) For the upper surface: We need to find M 1n(U) θU = δU θU = The deflection angle is 5 o From M 1 and θU, and Eq. 13.49 (using built-in function Theta (M , β,k )) (13.49) For θU = 5.00 o βU = 25.1 o (Use Goal Seek to vary βU so that θU = δU) From M 1 and βU M 1n(U) = 1.16 From M 1n(U) and p 1, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) (13.48d) p2 = 106 pU = p2 pU = 106 θL = δL θL = 15 o θL = 15.00 o βL = 34.3 o kPa kPa For the lower surface: We need to find M 1n(L) The deflection angle is From M 1 and θL, and Eq. 13.49 (using built-in function Theta (M , β,k )) For (Use Goal Seek to vary βL so that θL = δL) From M 1 and βL M 1n(L) = 1.55 From M 1n(L) and p 1, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) p2 = pL = p2 pL = From Eq 1 198 198 kPa L= 183 kN/m kPa Problem 13.148 [3] Given: Oblique shock Mach numbers Find: Deflection angle; Pressure after shock Solution: k= p1 = M1 = 1.4 75 4 M2 = The given or available data is: 2.5 β= 33.6 kPa Equations and Computations: We make a guess for β: o From M 1 and β, and Eq. 13.49 (using built-in function Theta (M , β,k )) (13.49) θ= From M 1 and β From M 2, θ, and β 21.0 M 1n = M 2n = 2.211 0.546 o (1) We can also obtain M 2n from Eq. 13.48a (using built-in function normM2fromM (M ,k )) (13.48a) M 2n = 0.546 (2) We need to manually change β so that Eqs. 1 and 2 give the same answer. Alternatively, we can compute the difference between 1 and 2, and use Solver to vary β to make the difference zero Error in M 2n = 0.00% Then p 2 is obtained from Eq. 13.48d (using built-in function normpfromm (M ,k )) (13.48d) p2 = 415 kPa Problem 13.149 [4] Given: Air flow into engine Find: Pressure of air in engine; Compare to normal shock Solution: k= p1 = 1.4 50 M1 = The given or available data is: 3 θ= 7.5 kPa o Equations and Computations: Assuming isentropic flow deflection p 0 = constant p 02 = p 01 For p 01 we use Eq. 13.7a (using built-in function Isenp (M , k )) (13.7a) p 01 = kPa p 02 = For the deflection 1837 1837 kPa θ= 7.5 o From M 1 and Eq. 13.55 (using built-in function Omega (M , k )) (13.55) ω1 = Deflection = Applying Eq. 1 49.8 ω2 - ω1 = ω(M 2) - ω(M 1) o (1) ω2 = ω1 - θ ω2 = 42.3 (Compression!) o From ω2, and Eq. 13.55 (using built-in function Omega (M , k )) ω2 = 42.3 M2 = For 2.64 o (Use Goal Seek to vary M 2 so that ω2 is correct) Hence for p 2 we use Eq. 13.7a (using built-in function Isenp (M , k )) p 2 = p 02/(p 02/p 2) p2 = For the normal shock (2 to 3) 86.8 M2 = 2.64 kPa From M 2 and p 2, and Eq. 13.41d (using built-in function NormpfromM (M ,k )) (13.41d) p3 = 690 kPa For slowing the flow down from M 1 with only a normal shock, using Eq. 13.41d p= 517 kPa Problem 13.150 [3] Given: Air flow in a duct Find: Mach number and pressure at contraction and downstream; Solution: k= M1 = 1.4 2.5 θ= p1 = The given or available data is: 7.5 50 o kPa Equations and Computations: For the first oblique shock (1 to 2) we need to find β from Eq. 13.49 (13.49) We choose β by iterating or by using Goal Seek to target θ (below) to equal the given θ Using built-in function theta (M, β,k ) θ= 7.50 o β= 29.6 o Then M 1n can be found from geometry (Eq. 13.47a) M 1n = 1.233 Then M 2n can be found from Eq. 13.48a) Using built-in function NormM2fromM (M,k ) (13.48a) M 2n = 0.822 Then, from M 2n and geometry (Eq. 13.47b) M2 = 2.19 From M 1n and Eq. 13.48d (using built-in function NormpfromM (M ,k )) (13.48d) p 2/p 1 = 1.61 p2 = 80.40 Pressure ratio We repeat the analysis of states 1 to 2 for states 2 to 3, to analyze the second oblique shock We choose β for M 2 by iterating or by using Goal Seek to target θ (below) to equal the given θ Using built-in function theta (M, β,k ) θ= 7.50 o β= 33.5 o Then M 2n (normal to second shock!) can be found from geometry (Eq. 13.47a) M 2n = 1.209 Then M 3n can be found from Eq. 13.48a) Using built-in function NormM2fromM (M,k ) M 3n = 0.837 Then, from M 3n and geometry (Eq. 13.47b) M3 = 1.91 From M 2n and Eq. 13.48d (using built-in function NormpfromM (M ,k )) p 3/p 2 = 1.54 p3 = 124 Pressure ratio Problem 13.151 [3] NOTE: Angle is 30o not 50o! Given: Air flow in a duct Find: Mach number and pressure at contraction and downstream; Solution: k= M1 = 1.4 2.5 β= p1 = The given or available data is: 30 50 o kPa Equations and Computations: For the first oblique shock (1 to 2) we find θ from Eq. 13.49 (13.49) Using built-in function theta (M, β,k ) θ= 7.99 o Also, M 1n can be found from geometry (Eq. 13.47a) M 1n = 1.250 Then M 2n can be found from Eq. 13.48a) Using built-in function NormM2fromM (M,k ) (13.48a) M 2n = 0.813 Then, from M 2n and geometry (Eq. 13.47b) M2 = 2.17 From M 1n and Eq. 13.48d (using built-in function NormpfromM (M ,k )) (13.48d) p 2/p 1 = 1.66 p2 = 82.8 Pressure ratio We repeat the analysis for states 1 to 2 for 2 to 3, for the second oblique shock We choose β for M 2 by iterating or by using Goal Seek to target θ (below) to equal the previous θ Using built-in function theta (M, β,k ) θ= 7.99 o β= 34.3 o Then M 2n (normal to second shock!) can be found from geometry (Eq. 13.47a) M 2n = 1.22 Then M 3n can be found from Eq. 13.48a) Using built-in function NormM2fromM (M,k ) M 3n = 0.829 Then, from M 3n and geometry (Eq. 13.47b) M3 = 1.87 From M 2n and Eq. 13.48d (using built-in function NormpfromM (M ,k )) p 3/p 2 = 1.58 p3 = 130 Pressure ratio Problem 13.152 [3] Given: Deflection of air flow Find: Pressure changes Solution: R= k= p= M= θ1 = 286.9 1.4 95 1.5 15 θ2 = The given or available data is: 15 J/kg.K kPa o o Equations and Computations: We use Eq. 13.55 (13.55) and Deflection = ωa - ωb = ω(M a) - ω(M b) From M and Eq. 13.55 (using built-in function Omega (M , k )) ω= 11.9 θ1 = ω1 - ω ω1 = θ1 + ω ω1 = 26.9 o For the first deflection: Applying Eq. 1 From ω1, and Eq. 13.55 (using built-in function Omega (M , k )) o (1) ω1 = 26.9 M1 = 2.02 For o (Use Goal Seek to vary M 1 so that ω1 is correct) Hence for p 1 we use Eq. 13.7a (13.7a) The approach is to apply Eq. 13.7a twice, so that (using built-in function Isenp (M , k )) p 1 = p (p 0/p )/(p 0/p 1) p1 = 43.3 kPa For the second deflection: We repeat the analysis of the first deflection Applying Eq. 1 θ2 + θ1 = ω2 - ω ω2 = θ2 + θ1 + ω ω2 = 41.9 o (Note that instead of working from the initial state to state 2 we could have worked from state 1 to state 2 because the entire flow is isentropic) From ω2, and Eq. 13.55 (using built-in function Omega (M , k )) ω2 = 41.9 M2 = For 2.62 o (Use Goal Seek to vary M 2 so that ω2 is correct) Hence for p 2 we use Eq. 13.7a (using built-in function Isenp (M , k )) p 2 = p (p 0/p )/(p 0/p 2) p2 = 16.9 kPa Problem 13.153 [3] Given: Deflection of air flow Find: Mach numbers and pressures Solution R= k= p2 = M2 = 286.9 1.4 10 4 θ1 = 15 o θ2 = The given or available data is: 15 o J/kg.K kPa Equations and Computations: We use Eq. 13.55 (13.55) and Deflection = ωa - ωb = ω(M a) - ω(M b) From M and Eq. 13.55 (using built-in function Omega (M , k )) ω2 = 65.8 o For the second deflection: Applying Eq. 1 ω1 = ω2 - θ2 ω1 = 50.8 o From ω1, and Eq. 13.55 (using built-in function Omega (M , k )) ω1 = 50.8 M1 = For 3.05 o (Use Goal Seek to vary M 1 so that ω1 is correct) (1) Hence for p 1 we use Eq. 13.7a (13.7a) The approach is to apply Eq. 13.7a twice, so that (using built-in function Isenp (M , k )) p 1 = p 2(p 0/p 2)/(p 0/p 1) p1 = 38.1 kPa For the first deflection: We repeat the analysis of the second deflection Applying Eq. 1 θ2 + θ1 = ω2 - ω ω = ω2 - (θ2 + θ1) ω= 35.8 o (Note that instead of working from state 2 to the initial state we could have worked from state 1 to the initial state because the entire flow is isentropic) From ω, and Eq. 13.55 (using built-in function Omega (M , k )) For ω= M= 35.8 2.36 o (Use Goal Seek to vary M so that ω is correct) Hence for p we use Eq. 13.7a (using built-in function Isenp (M , k )) p = p 2(p 0/p 2)/(p 0/p ) p= 110 kPa Problem 13.154 [4] Given: Mach number and deflection angle Find: Static and stagnation pressures due to: oblique shock; compression wave Solution: R= k= p1 = 286.9 1.4 50 M1 = 3.5 θ= 35 o θ= 35 o The given or available data is: J/kg.K kPa Equations and Computations: For the oblique shock: We need to find M 1n The deflection angle is From M 1 and θ, and Eq. 13.49 (using built-in function Theta (M , β,k )) (13.49) θ= 35.0 o β= For 57.2 o (Use Goal Seek to vary β so that θ = 35o) From M 1 and β M 1n = 2.94 From M 1n and p 1, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) (13.48d) p2 = 496 kPa To find M 2 we need M 2n. From M 1n, and Eq. 13.48a (using built-in function NormM2fromM (M ,k )) (13.48a) M 2n = 0.479 The downstream Mach number is then obtained from from M 2n, θ and β, and Eq. 13.47b M 2n = M 2sin(β - θ) M2 = Hence (13.47b) 1.27 For p 02 we use Eq. 12.7a (using built-in function Isenp (M , k )) (13.7a) p 02 = p 2/(p 02/p 2) p 02 = 1316 kPa For the isentropic compression wave: p 0 = constant For isentropic flow p 02 = p 01 p 01 = 3814 kPa p 02 = 3814 kPa For p 01 we use Eq. 13.7a (using built-in function Isenp (M , k )) (Note that for the oblique shock, as required by Eq. 13.48b (13.48b) p 02/p 01 = 0.345 (using built-in function Normp0fromM (M ,k ) p 02/p 01 = 0.345 (using p 02 from the shock and p 01) θ= −θ θ= For the deflection -35.0 (Compression ) o We use Eq. 13.55 (13.55) and Deflection = ω2 - ω1 = ω(M 2) - ω(M 1) From M 1 and Eq. 13.55 (using built-in function Omega (M , k )) ω1 = 58.5 ω2 = ω1 + θ ω2 = 23.5 o ω2 = 23.5 o M2 = 1.90 Applying Eq. 1 o From ω2, and Eq. 13.55 (using built-in function Omega (M , k )) For (Use Goal Seek to vary M 2 so that ω2 = 23.5o) Hence for p 2 we use Eq. 13.7a (using built-in function Isenp (M , k )) p 2 = p 02/(p 02/p 2) p2 = 572 kPa (1) Problem 13.155 [3] Given: Wedge-shaped airfoil Find: Lift per unit span assuming isentropic flow Solution: The given or available data is: R k p M = = = = δ= c= 286.9 1.4 70 2.75 7 1.5 J/kg.K kPa o m Equations and Computations: The lift per unit span is L = (p L - p U)c (1) (Note that p L acts on area c /cos(δ), but its normal component is multiplied by cos(δ)) For the upper surface: pU = p pU = 70 θ= −δ θ= -7.0 kPa For the lower surface: o We use Eq. 13.55 (13.55) and Deflection = ωL - ω = ω(M L) - ω(M ) (2) From M and Eq. 13.55 (using built-in function Omega (M , k )) ω= 44.7 θ= ωL - ω ωL = θ+ω ωL = 37.7 o ωL = 37.7 o ML = 2.44 Applying Eq. 2 o From ωL, and Eq. 13.55 (using built-in function Omega (M , k )) For (Use Goal Seek to vary M L so that ωL is correct) Hence for p L we use Eq. 13.7a (13.7a) The approach is to apply Eq. 13.7a twice, so that (using built-in function Isenp (M , k )) p L = p (p 0/p )/(p 0/p L) pL = From Eq 1 113 kPa L= 64.7 kN/m Problem 13.156 [4] Given: Mach number and airfoil geometry Find: Lift and drag per unit span Solution: R= k= p1 = 286.9 1.4 50 M1 = The given or available data is: 1.75 α= c= 18 1 J/kg.K kPa o m Equations and Computations: F = (p L - p U)c The net force per unit span is Hence, the lift force per unit span is L = (p L - p U)c cos(α) (1) D = (p L - p U)c sin(α) (2) The drag force per unit span is For the lower surface (oblique shock): We need to find M 1n θ= α θ= The deflection angle is 18 o From M 1 and θ, and Eq. 13.49 (using built-in function Theta (M , β,k )) (13.49) θ= 18.0 o β= For 62.9 o (Use Goal Seek to vary β so that θ is correct) From M 1 and β M 1n = 1.56 From M 1n and p 1, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) (13.48d) p2 = 133.2 pL = p2 pL = 133.2 kPa kPa For the upper surface (isentropic expansion wave): p 0 = constant For isentropic flow p 02 = p 01 For p 01 we use Eq. 13.7a (using built-in function Isenp (M , k )) (13.7a) p 01 = 266 kPa p 02 = 266 kPa θ= α θ= For the deflection 18.0 (Compression ) o We use Eq. 13.55 (13.55) and Deflection = ω2 - ω1 = ω(M 2) - ω(M 1) From M 1 and Eq. 13.55 (using built-in function Omega (M , k )) ω1 = ω2 = ω1 + θ ω2 = Applying Eq. 3 19.3 37.3 o o From ω2, and Eq. 13.55 (using built-in function Omega (M , k )) ω2 = 37.3 M2 = For 2.42 o (Use Goal Seek to vary M 2 so that ω2 is correct) Hence for p 2 we use Eq. 13.7a (using built-in function Isenp (M , k )) p 2 = p 02/(p 02/p 2) p2 = 17.6 pU = p2 pU = 17.6 kPa From Eq. 1 L= 110.0 kN/m From Eq. 2 D= 35.7 kN/m kPa (3) Problem 13.157 Given: Mach number and airfoil geometry Find: Plot of lift and drag and lift/drag versus angle of attack Solution: The given or available data is: k= p1 = 1.4 50 M1 = 1.75 α= c= kPa o 12 1 m Equations and Computations: The net force per unit span is F = (p L - p U)c Hence, the lift force per unit span is L = (p L - p U)c cos(α) (1) The drag force per unit span is D = (p L - p U)c sin(α) (2) For each angle of attack the following needs to be computed: For the lower surface (oblique shock): We need to find M 1n Deflection θ= α From M 1 and θ, and Eq. 13.49 (using built-in function Theta (M , β,k )) (13.49) β find (Use Goal Seek to vary β so that θ is the correct value) From M 1 and β find M 1n From M 1n and p 1, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) (13.48d) find p2 and pL = p2 [4] For the upper surface (isentropic expansion wave): p 0 = constant For isentropic flow p 02 = p 01 For p 01 we use Eq. 13.7a (using built-in function Isenp (M , k )) (13.7a) find p 02 = θ= Deflection 266 kPa α we use Eq. 13.55 (13.55) and Deflection = ω2 - ω1 = ω(M 2) - ω(M 1) (3) From M 1 and Eq. 13.55 (using built-in function Omega (M , k )) ω1 = Applying Eq. 3 19.3 ω2 = find ω1 + θ o (4) From ω2, and Eq. 12.55 (using built-in function Omega (M , k )) From ω2 find M2 (Use Goal Seek to vary M 2 so that ω2 is the correct value) Hence for p 2 we use Eq. 13.7a (using built-in function Isenp (M , k )) p 2 = p 02/(p 02/p 2) pU = p2 Finally, from Eqs. 1 and 2, compute L and D Computed results: o α( ) β (o) θ (o) 0.50 1.00 1.50 2.00 4.00 5.00 10.00 15.00 16.00 16.50 17.00 17.50 18.00 35.3 35.8 36.2 36.7 38.7 39.7 45.5 53.4 55.6 56.8 58.3 60.1 62.9 0.50 1.00 1.50 2.00 4.00 5.00 10.0 15.0 16.0 16.5 17.0 17.5 18.0 Error 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% Sum: 0.0% M 1n p L (kPa) ω2 ( ) ω2 from M 2 ( ) 1.01 1.02 1.03 1.05 1.09 1.12 1.25 1.41 1.44 1.47 1.49 1.52 1.56 51.3 52.7 54.0 55.4 61.4 64.5 82.6 106.9 113.3 116.9 121.0 125.9 133.4 19.8 20.3 20.8 21.3 23.3 24.3 29.3 34.3 35.3 35.8 36.3 36.8 37.3 19.8 20.3 20.8 21.3 23.3 24.3 29.3 34.3 35.3 35.8 36.3 36.8 37.3 o o Error 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% Sum: 0.0% M2 p U (kPa) L (kN/m) D (kN/m) 1.77 1.78 1.80 1.82 1.89 1.92 2.11 2.30 2.34 2.36 2.38 2.40 2.42 48.7 47.4 46.2 45.0 40.4 38.3 28.8 21.3 20.0 19.4 18.8 18.2 17.6 2.61 5.21 7.82 10.4 20.9 26.1 53.0 82.7 89.6 93.5 97.7 102.7 110 0.0227 0.091 0.205 0.364 1.46 2.29 9.35 22.1 25.7 27.7 29.9 32.4 35.8 L/D 115 57.3 38.2 28.6 14.3 11.4 5.67 3.73 3.49 3.38 3.27 3.17 3.08 To compute this table: 1) Type the range of α 2) Type in guess values for β 3) Compute θ from Eq. 13.49 (using built-in function Theta (M ,β, k ) 4) Compute the absolute error between each θ and α 5) Compute the sum of the errors 6) Use Solver to minimize the sum by varying the β values (Note: You may need to interactively type in new β values if Solver generates β values that lead to no θ) 7) For each α, M 1n is obtained from M 1, and Eq. 13.47a 8) For each α, p L is obtained from p 1, M 1n, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) 9) For each α, compute ω2 from Eq. 4 10) For each α, compute ω2 from M 2, and Eq. 13.55 (using built-in function Omega (M ,k )) 11) Compute the absolute error between the two values of ω2 12) Compute the sum of the errors 13) Use Solver to minimize the sum by varying the M 2 values (Note: You may need to interactively type in new M 2 values) if Solver generates β values that lead to no θ) 14) For each α, p U is obtained from p 02, M 2, and Eq. 13.47a (using built-in function Isenp (M , k )) 15) Compute L and D from Eqs. 1 and 2 Lift and Drag of an Airfoil as a Function of Angle of Attack L and D (kN/m) 120 100 80 Lift Drag 60 40 20 0 0 2 4 6 8 10 12 14 16 18 20 α (o) Lift/Drag of an Airfoil as a Function of Angle of Attack 140 120 L/D 100 80 60 40 20 0 0 2 4 6 8 10 α (o) 12 14 16 18 20 Problem 13.158 [4] Given: Mach number and airfoil geometry Find: Drag coefficient Solution: R= k= p1 = M1 = The given or available data is: 286.9 1.4 95 2 α= δ= 0 10 J/kg.K kPa o o Equations and Computations: The drag force is D = (p F - p R)cs tan(δ/2) (1) (s and c are the span and chord) This is obtained from the following analysis Airfoil thickness (frontal area) = 2s (c /2tan(δ/2)) Pressure difference acting on frontal area = (p F - p R) (p F and p R are the pressures on the front and rear surfaces) 2 C D = D /(1/2ρV A ) The drag coefficient is (2) But it can easily be shown that ρV 2 = pkM 2 Hence, from Eqs. 1 and 2 C D = (p F - p R)tan(δ/2)/(1/2pkM 2) (3) For the frontal surfaces (oblique shocks): We need to find M 1n From M 1 and θ, and Eq. 13.49 (using built-in function Theta (M , β,k )) θ= δ/2 θ= The deflection angle is 5 o (13.49) θ= β= 5.0 34.3 M 1n = 1.13 For o o (Use Goal Seek to vary β so that θ = 5o) From M 1 and β From M 1n and p 1, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) (13.48d) p2 = 125.0 pF = p2 pF = 125.0 kPa kPa To find M 2 we need M 2n. From M 1n, and Eq. 13.48a (using built-in function NormM2fromM (M ,k )) (13.48a) M 2n = 0.891 The downstream Mach number is then obtained from from M 2n, θ and β, and Eq. 13.47b M 2n = M 2sin(β - θ) M2 = Hence (13.47b) 1.82 For p 02 we use Eq. 13.7a (using built-in function Isenp (M , k )) (13.7a) p 02 = 742 kPa For the rear surfaces (isentropic expansion waves): Treating as a new problem Here: M 1 is the Mach number after the shock and M 2 is the Mach number after the expansion wave p 01 is the stagnation pressure after the shock and p 02 is the stagnation pressure after the expansion wave M 1 = M 2 (shock) M1 = 1.82 p 01 = p 02 (shock) p 01 = 742 kPa p 0 = constant For isentropic flow p 02 = p 01 p 02 = 742 θ= δ θ= For the deflection 10.0 kPa o We use Eq. 13.55 (13.55) and Deflection = ω2 - ω1 = ω(M 2) - ω(M 1) From M 1 and Eq. 13.55 (using built-in function Omega (M , k )) ω1 = ω2 = ω1 + θ ω2 = Applying Eq. 3 21.3 31.3 o o From ω2, and Eq. 13.55 (using built-in function Omega(M, k)) For ω2 = M2 = 31.3 2.18 o (Use Goal Seek to vary M 2 so that ω2 = 31.3o) Hence for p 2 we use Eq. 13.7a (using built-in function Isenp (M , k )) p 2 = p 02/(p 02/p 2) p2 = pR = p2 pR = Finally, from Eq. 1 71.2 71.2 CD = 0.0177 kPa kPa (3) Problem 13.159 [4] FU 1 FL RU RL Given: Mach number and airfoil geometry Find: Lift and Drag coefficients Solution: R= k= p1 = M1 = The given or available data is: 286.9 1.4 95 2 α= δ= 12 10 J/kg.K kPa o o Equations and Computations: Following the analysis of Example 13.14 the force component perpendicular to the major axis, per area, is F V/sc = 1/2{(p FL + p RL) - (p FU + p RU)} (1) and the force component parallel to the major axis, per area, is F H/sc = 1/2tan(δ/2){(p FU + p FL) - (p RU + p RL)} (2) using the notation of the figure above. (s and c are the span and chord) The lift force per area is F L/sc = (F Vcos(α) - F Hsin(α))/sc (3) The drag force per area is F D/sc = (F Vsin(α) + F Hcos(α))/sc C L = F L/(1/2ρV 2A ) The lift coefficient is (4) (5) But it can be shown that ρV 2 = pkM 2 (6) Hence, combining Eqs. 3, 4, 5 and 6 2 C L = (F V/sc cos(α) - F H/sc sin(α))/(1/2pkM ) (7) Similarly, for the drag coefficient C D = (F V/sc sin(α) + F H/sc cos(α))/(1/2pkM 2) (8) For surface FL (oblique shock): We need to find M 1n θ= α + δ/2 θ= The deflection angle is 17 o From M 1 and θ, and Eq. 13.49 (using built-in function Theta (M , β,k )) (13.49) For θ= β= 17.0 48.2 o o (Use Goal Seek to vary β so that θ = 17o) From M 1 and β M 1n = 1.49 From M 1n and p 1, and Eq. 13.48d (using built-in function NormpfromM (M ,k )) (13.48d) p2 = 230.6 p FL = p2 p FL = 230.6 kPa kPa To find M 2 we need M 2n. From M 1n, and Eq. 13.48a (using built-in function NormM2fromM (M ,k )) (13.48a) M 2n = 0.704 The downstream Mach number is then obtained from from M 2n, θ and β, and Eq. 13.47b M 2n = M 2sin(β - θ) Hence M2 = (13.47b) 1.36 For p 02 we use Eq. 13.7a (using built-in function Isenp (M , k )) (13.7a) p 02 = 693 kPa For surface RL (isentropic expansion wave): Treating as a new problem Here: M 1 is the Mach number after the shock and M 2 is the Mach number after the expansion wave p 01 is the stagnation pressure after the shock and p 02 is the stagnation pressure after the expansion wave M 1 = M 2 (shock) M1 = 1.36 p 01 = p 02 (shock) p 01 = 693 kPa p 0 = constant For isentropic flow p 02 = p 01 p 02 = 693 θ= δ θ= For the deflection 10.0 kPa o We use Eq. 13.55 (13.55) and Deflection = ω2 - ω1 = ω(M 2) - ω(M 1) From M 1 and Eq. 13.55 (using built-in function Omega (M , k )) ω1 = ω2 = ω1 + θ ω2 = Applying Eq. 3 7.8 17.8 o o From ω2, and Eq. 13.55 (using built-in function Omega (M , k )) For ω2 = M2 = 17.8 1.70 o (Use Goal Seek to vary M 2 so that ω2 = 17.8o) Hence for p 2 we use Eq. 13.7a (using built-in function Isenp (M , k )) p 2 = p 02/(p 02/p 2) p2 = 141 p RL = p2 p RL = 141 kPa kPa (3) For surface FU (isentropic expansion wave): M1 = 2.0 p 0 = constant For isentropic flow p 02 = p 01 p 01 = p 02 = 743 743 For p 01 we use Eq. 13.7a (using built-in function Isenp (M , k )) θ= α - δ/2 θ= For the deflection 7.0 kPa o We use Eq. 13.55 and Deflection = ω2 - ω1 = ω(M 2) - ω(M 1) From M 1 and Eq. 13.55 (using built-in function Omega (M , k )) ω1 = ω2 = ω1 + θ ω2 = Applying Eq. 3 26.4 33.4 o o From ω2, and Eq. 13.55 (using built-in function Omega(M, k)) For ω2 = M2 = 33.4 2.27 o (Use Goal Seek to vary M 2 so that ω2 = 33.4o) Hence for p 2 we use Eq. 13.7a (using built-in function Isenp (M , k )) p 2 = p 02/(p 02/p 2) p2 = 62.8 p FU = p2 p FU = 62.8 kPa kPa For surface RU (isentropic expansion wave): Treat as a new problem. Flow is isentropic so we could analyse from region FU to RU but instead analyse from region 1 to region RU. M1 = For isentropic flow 2.0 p 0 = constant p 02 = p 01 (3) p 01 = p 02 = 743 743 θ= α + δ/2 θ= TOTAL deflection 17.0 kPa kPa o We use Eq. 13.55 and Deflection = ω2 - ω1 = ω(M 2) - ω(M 1) From M 1 and Eq. 13.55 (using built-in function Omega (M , k )) ω1 = ω2 = ω1 + θ ω2 = Applying Eq. 3 26.4 43.4 o o From ω2, and Eq. 13.55 (using built-in function Omega(M, k)) ω2 = M2 = For 43.4 2.69 o (Use Goal Seek to vary M 2 so that ω2 = 43.4o) Hence for p 2 we use Eq. 13.7a (using built-in function Isenp (M , k )) p 2 = p 02/(p 02/p 2) p2 = 32.4 p RU = p2 p RU = 32.4 kPa p FL = p RL = p FU = p RU = 230.6 140.5 62.8 32.4 kPa kPa kPa kPa kPa The four pressures are: From Eq 1 F V/sc = 138 kPa From Eq 2 F H/sc = 5.3 kPa From Eq 7 CL = 0.503 From Eq 8 CD = 0.127 (3) ...
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This note was uploaded on 07/10/2011 for the course CHE 144 taught by Professor Tuzla during the Spring '11 term at Lehigh University .

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