Chapter09

Mechanics of Fluids

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200 CHAPTER 9 Compressible Flow 9.1 Bt u ft-l b lb m ft-lb 0.2 4 77 8 32. 2 6012 Bt u slug lbm - R slug -R p c == oo c c R v p = - = - = = 6012 1716 4296 4296 1 778 1 32 2 ft -lb slug- R Btu ft - lb slug lbm o o . Btu 0.171 lbm = o 9.2 c c R c kc c c k R c k R p v p v p p p = + = = + -  = . . . or 1 1 = - c Rk k p /( ). 1 9.3 If s = 0, Eq. 9.1.9 can be written as c n T T R n p p n T T n p p p c R p l l l l 2 1 2 1 2 1 2 1 = = or It follows that, using c c R c c k p v p v = + = and / , T T p p p p R c k p 2 1 2 1 2 1 1 1 = = - / . Using Eq. 9.1.7, T T p p p p p p k k 2 1 2 1 2 1 2 1 1 1 1 2 2 1 1 = = = - - r r r r or / . Finally, this can be written as p p k 2 1 2 1 = r r 9.4 Substitute Eq. 4.5.18 into Eq. 4.5.17 and neglect potential energy change: & & & ~ ~ . Q W m V V p p u u S - = - + - + - 2 2 1 2 2 2 1 1 2 1 2 r r
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201 Enthalpy is defined in Thermodynamics as h u pv u p = + = + ~ ~ / . r Therefore, & & & . Q W m V V h h S - = - + - 2 2 1 2 2 1 2 Assume the fluid is an ideal gas with constant specific heat so that h c T p = . Then ( 29 & & & . Q W m V V c T T S p - = - + - 2 2 1 2 2 1 2 Next, let c c R k c c c R k k p v p v p = + = = - and so that / / ( ). 1 Then, with the ideal gas law T p R = / , r the first law takes the form & & & . Q W m V V k k p p S - = - + - - 2 2 1 2 2 2 1 1 2 1 r r 9.5 Differentiate p c d xy ydx xdy k r - = = + using ( ) : r r r - - - - = k k dp pk d 1 0. Rewrite: dp d k p r r = 9.6 The speed of sound is given by c dp d = / . r For an isothermal process TR p K K = = / , r where is a constant. This can be differentiated: dp Kd RTd = = r r Hence, the speed of sound is c RT = 9.7 Eq. 9.1.4 with & & Q W V c T S p = = + = 0 2 2 is: cons't. V c T V V c T T V V V V c T c T p p p p 2 2 2 2 2 2 2 2 + = + + + = + + + + ( ) ( ) ( ) . 2 2 () 0 22 V VV ∆∆ =+ . . pp c T VVc Th + -∆= ∆ =∆ We neglected ( V) 2 . The velocity of a small wave is V c h c V = = - . . 9.8 For water r r dp d = × 2110 10 6 Pa Since r = 1000 kg /m we see that 3 ,
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202 c dp d = / r = × = 2110 10 1000 1453 6 / m/s 9.9 For water c p dp d = 2245 = × = r r 2110 10 1000 1453 6 m / s. L = × × velocity time = 1453 0.6 = 872 m. 9.10 Since c = 1450 m/s for the small wave, the time increment is t d c = = = 10 1450 0 0069 . seconds 9.11 a) M = = × × = V c 200 1 4 287 288 0 588 . . . b) M 600 / 1. 4 171 6 46 6 0.567. = ××= c) M = × × = 200 1 4 287 223 0 668 / . . d) M / 1. 4 171 6 39 2 0.618. = e) M = × × = 200 1 4 287 238 0 647 / . . 9.12 c kRT d ct = = × × = ∴ = = × = 1 4 287 263 256 256 1 21 309 . . . m /s. m 9.13 a) Assume T = 20 ° C: c kRT = = × × = 1 4 287 293 343 . d c t = = × = 343 2 686 m b) Assume T = 70 ° F: c kRT = = × × = 1 4 1716 530 1130 . fps. d c t = = × = 1130 2 2260 ft. For every second that passes, the lightning flashed about 1000 ft away. Count 5 seconds and it is approximately one mile away. 9.14 c M c V = × × = = = 1 4 287 263 256 1 . sin . m/s. a 1000 si n 0.256 . ta n 0.264 8. 3776 m L L aa = == ∴= t = = 3776 1000 3 776 . . s 1000 m V L
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203 9.15 Use Eq. 9.2.13: a) c V V = = × × = sin . sin a or m/s 1 4 287 288 22 908 o b) c V V = = × × = sin . sin a or fps 1 4 1716 519 22 2980 o 9.16 Eq. 9.2.4: V p c p kRT = - = - = - × × = - r r 0 3 00237 1 4 1716 519 0 113 .
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Chapter09 - CHAPTER 9 Compressible Flow 9.1 c p = 0.24 Btu...

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