Solutions manual for introduction to fluid mechanics 7th edition download

Introduction to Fluid Mechanics 7th Edition Fox, Pritchard, & McDonald Link download full: Answers to Selected Problems, Chapter 13 http://testbankair.com/download/solutions-manual-for-introduction-to-fluid-mechanics-7th-edition/ 13.1 m=3.18 kg/s 13.3 V = 781 m/s M = 1.35 m=3.18 kg/s 13.5 p2 = 6.52 psi 13.7 M2 = 1.20 Supersonic diffuser 13.9 M2 = 1.20 Supersonic diffuser 13.15 pt = 250 kPa Vt = 252 m/s Mt = 0.883 13.17 pt = 166 kPa 13.19 p = 150 kPa M = 0.60 At = 0.0421 m2 m=18.9 kg/s 13.21 At = 1.94 x 10–3 m2 13.23 p0 = 817 kPa pe = 432 kPa Te = 288 K (– 45.5oC) Ve = 302 m/s 13.25 Δt = 374 s (6.23 min) Δs = 232 J/kg·K 13.27 pe = 687 kPa m=0.0921 kg/s arfx = 1.62 m/s2 13.29 p0 = 9.87 kPa (abs) pe = 5.21 kPa (abs) Te = 332 K (58.7oC) Ve = 365 m/s ax = 1.25 m/s2 13.31 Rx = 304 lbf (Tension) 13.33 A2 = 0.573 ft2 V2 = 667 ft/s 13.35 Me = pe = 381 kPa Pressure and flow decrease asymptotically o Tf = 228 K (– 45 C) 13.37 p0 = 115 psia m=1.53 lb/s At = 0.593 in2 13.39 pe = 125 kPa (abs) m=0.401 kg/s 13.41 V1 = 1300 m/s m=87.4 kg/s 13.43 m=3.57 lbm/s Mass flow rate decreases by a factor of 13.45 Rx = 950 N 13.47 pe = 88.3 kPa m=0.499 kg/s Rx = – 1026 N (to left) 13.49 p0 = 44.6 MPa 13.53 M1 = 0.200 m=3.19 x 10–3 kg/s p2 = 47.9 kPa (abs) 13.55 pmin = 18.5 psia Vmax = 1040 ft/s o o 13.57 Te = 840 R (380 F) Rx = 13.3 lbf (to right) Δs = 0.359 Btu/lbm·R o 13.59 p0t = 56.6 psia T2 = 433 R p02 = 27.8 psia m=0.0316 lbm/s 13.61 T2 = 238 K p2 = 26.1 kPa (abs) Δs = 172 J/kg·K 13.63 L = 12.0 ft 13.65 L = 18.8 ft 13.69 T2 = 551oR m=5.33 slug/s 13.71 L = 15.5 ft 13.73 M2 = 0.233 Heat added 13.75 p2 = 198 psia (Isothermal) p2 = 153 psia (Adiabatic) 13.77 Q = 1.84 x 10 ft /day 13.81 δQ/dm = 449 kJ/kg Δs = 0.892 kJ/kg·K 13.83 Note: ρ2 = 0.850 lbm/ft3 Q = 107 Btu/s Δp = 162 psi With wrong ρ2 = 100 lbm/ft : Q = 74 Btu/s Δp = – psi 13.85 δQ/dm = 18 kJ/kg Δs = 0.0532 kJ/kg·K Δp0 = 2.0 kPa 13.87 δQ/dm = 1.12 MJ/kg Δp0 = – 13.5 kPa 13.89 M2 = 0.50 T02 = 1556 K T2 = 1480 K Q=1.86 MJ/s 13.91 δQ/dm = 447 kJ/kg Δs = 0.889 kJ/kg·K Δp0 = 22 kPa 13.93 δQ/dm = 364 kJ/kg Δp0 = – 182 kPa T02 = 1174 K p02 = 1.60 MPa T2 = 978 K p2 = 0.844 MPa ρ2 = 3.01 kg/m3 13.95 M2 = 0.60 T02 = 966 K δQ/dm = 343 kJ/kg (61.6% of max) Q= 4010 kW 13.97 M2 = 1.74 p2 = 4.49 psia 13.99 Vs = 5475 m/s V = 4545 m/s (Constant specific heats unrealistic) 13.101 V = 1666 ft/s 13.103 p1 = 1.28 psia ρ1 = 0.00653 lbm/ft3 V1 = 2260 ft/s T01 = 954oR p01 = 10.0 psia T02 = 954oR p02 = 7.22 psia 13.105 T2 = 520 K p02 = 1.29 MPa (abs) 13.107 M2 = 0.486 V2 = 541 mph (793 ft/s) Δp0 = 89.2 psi 13.109 T01 = 426 K p01 = 207 kPa (abs) p02 = 130 kPa (abs) 13.111 M1 = 2.48 V1 = 2420 ft/s p02 = 29.1 psia p2 = 24.3 psia 13.113 M2d = 0.547 p2d = 512 kPa p02d = 628 kPa A*s = 0.111 m2 13.115 M1 = 2.20 p02 = 178 kPa V1 = 568 m/s (“Isentropic”) 13.117 T0 = 533 K p3 – p2 = 37.4 kPa s4 – s1 = – 30.5 J/kg·K 13.119 V2 = 268 m/s (Relative to wave), = – 276 m/s (Relative to ground) 13.121 Me = 1.452 m=0.808 lbm/s 13.123 Me = 2.94 p0 = 3.39 MPa pb1 = 3.35 MPa pb2 = 1.00 MPa pb3 = 101 kPa 13.125 pb = 301 kPa 13.127 M1 = 1.50 13.129 patm < p0 < 112 kPa (abs); p0 > 743 kPa (abs) 13.131 p3 = 66.6 psia 13.133 pb = 301 kPa 13.137 V2 = 2140 ft/s Δs = 0.0388 Btu/lbm·oR 13.139 M2 = 1.95 p2 = 179 kPa M2 = 0.513 (Normal shock) p2 = 570 kPa (Normal shock) βmin = 23.6o 13.141 β = 62.5o p2/p1 = 9.15 13.143 M1 = 1.42 V1 = 483 m/s β = 67.4o 13.145 α = 7.31o pmax = 931 kPa Tmax = 564oC 13.147 L/w = 183 kN/m 13.149 p = 690 kPa p = 517 kPa (Normal shock only) 13.151 p = 130 kPa (Note: The angle is 30o, NOT 50o; with 50o there is no second shock!) 13.153 M1 = 3.05 p1 = 38.1 kPa M = 2.36 p = 110 kPa 13.155 L/w = 64.7 kN/m 13.159 CL = 0.503 CD = 0.127 ... 533 K p3 – p2 = 37.4 kPa s4 – s1 = – 30.5 J/kg·K 13.119 V2 = 268 m/s (Relative to wave), = – 276 m/s (Relative to ground) 13.121 Me = 1.452 m=0.808 lbm/s 13.123 Me = 2.94 p0 = 3.39 MPa pb1