ch12 - = 0.322 (5%) M = 0.464 (10%) 12.45 Δ / = 48.5% (Not...

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Introduction to Fluid Mechanics 7 th Edition Fox, Pritchard, & McDonald Answers to Selected Problems, Chapter 12 12.1 T = const. p decreases ρ decreases (Irreversible adiabatic process) 12.3 Δ s > 0 so it is feasible for a real (irreversible) adiabatic process 12.5 T 2 = 20 o F p 2 = 100 kPa 12.7 Δ s = – 346 kJ/kg·K ( Δ S = – 1729 J/K) Δ u = – 143 kJ/kg ( Δ U = – 717 kJ) Δ h = – 201 kJ/k ( Δ H = – 1004 kJ) 12.9 h = 57.5% 12.11 W = 176 MJ W s = 228 MJ T s (max) = 858 K Q s = – 317 MJ 12.13 m = 36.7 kg/s T 2 = 572 K V 2 = 4.75 m/s W = 23 MW 12.15 Δ t = 828s ( 14 min) 12.17 Δ = 1.70 x 10 –4 kg/m 3 Δ T = 0.017 K Δ V = 0.049 m/s 12.19 Δ t = 198 μ s E v = 12.7 GN/m 2 12.21 x = 19.2 km 12.23 c = 299 m/s V = 987 m/s V / V bullet = 1.41 12.29 c = 340 m/s (sea level) 12.31 V = 1471 mph α = 31.8 o 12.33 V = 642 m/s (2110 ft/s) 12.35 V = 493 m/s Δ t = 0.398 s 12.37 V = 515 m/s t = 6.92 s 12.39 Δ x 1043 – 1064 m 12.41 Density change < 1.21%, so incompressible 12.43 M = 0.142 (1%) M
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Unformatted text preview: = 0.322 (5%) M = 0.464 (10%) 12.45 Δ / = 48.5% (Not incompressible) 12.47 p dyn = 54.3 kPa p = 152 kPa 12.49 p = 546 kPa h – h = 178 kJ/kg T = 466 K 12.51 p – p = 8.67 kPa V = 195 m/s V = 205 m/s Error using Bernoulli = 5.13% 12.55 T = const (isoenergetic) p decreases (irreversible adiabatic) 12.57 V = 890 m/s T = 677 K p = 212 kPa 12.59 T = const = 294 K (20.6o) (isoenergetic) 1 p = 1.01 MPa, 2 p = 189 kPa (irreversible adiabatic) Δ s = 480 J/kg·K Flow accelerates even with friction due to large pressure drop 12.61 1 T = 2 T = 344 K 1 p = 223 kPa 2 p = 145 kPa Δ s = 0.124 kJ/kg·K 12.63 1 T = 2 T = 445 K 1 p = 57.5 kPa 2 p = 46.7 kPa Δ s = 59.6 J/kg·K 12.65 Δ p = 48.2 kPa (inside higher) 12.67 T * = 260 K p * = 24.7 MPa V * = 252 m/s 12.69 T t = 2730 K p t = 25.5 MPa V t = 1030 m/s...
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This note was uploaded on 09/08/2010 for the course AME 331 taught by Professor Zohar during the Spring '08 term at Arizona.

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