472 Q CHAPTER 16
I c = VkRT= V(1.40)(1716)(460 + 100) = 1160 ft/s NM = vie = = 0.474

Since [v„ = 550 ft/s] < [c = 1160 ft/s], p ,/p„ = [1 + (u0/c0)2(A: - l ) / 2 ] * / ( t - l ) ps = 292 psia cpT„ + vlllg = cpTs 7 ; = 5 85 0 R 16.18 p ,/250 = [1 + ( ^ ) 2 ( 1 . 4 0 - l)/2]*
2

•"-»

(6000/32.2)(460 + 100) + 550 /[(2)(32.2)] = (6000/32.2)(7S) or 125 °F

Show that the equation ps=p0 + ( p 0 )(uo/2)[l + (NM)l/4 + • • •] results from the binomial expansion of the e quation pjp0 = [1 + (v0/c)2(k - \)l2]k'^\ I Ps/Po = {1 + (Wc)2[(A: - l )/2]}*' (t - 1 >; substitute k = fw*Jp» Po
L
n

\

2p0
x

/J

E xpanding by the binomial theorem: (a + b) = a" + {na"~ )(b) + [n(n - l )/2!]a" _1 Z> 2 + • • • a nd then simplifying l eads to ps = p0 + ( p 0 fo/2)[l + ( A f i / 4 ) + • • •] Q.E.D. 16.19 C ompute the value of Ä from the values of k a nd cp for air. f 16.20 R = [(k- i)/k](cp) = [(1.40 -1)/1.40](6000) = 1714 ft-lb/(slug-°R)

C ompute the enthalpy change in 5 kg of oxygen when the initial conditions are pt = 130 kPa abs and 7", = 10 °C, a nd the final conditions are p2 = 500 kPa abs and T2 = 95 °C [cp = 0.917 kJ/kg • K].

f
16.21

A2 = Ä i = ( c P ) ( 7 2 - 7 ; )

f / 2 - H 1 = ( 5 )(c p )(7 2 -r 1 ) = ( 5)(0.917)(95-10) = 390kJ

A cylinder containing 2 kg nitrogen at 0.14 MPa abs and 5 °C is compressed isentropically to 0.30 MPa abs. Find t he final temperature and the work required [c„ = 0.741 kJ/kg • K]. I 7 2 = {TiXpJptf-1** = (273 + 5)(0.30/0. i4)0-«o-.vi-«o = 3 4 6 K o r 7 3 <>c F r o m t h e p r i n c i p l e o f conservation o f energy, the work done on the gas must equal its increase in internal energy, since there is no heat transfer in an isentropic process; that is, u2-ux = ( c „ ) ( 7 2 - ľ,) = work per kilogram: Work = (2)(0.741)(73 - 5) = 101 kJ.

16.22

If 3.0 slugs of air are involved in a reversible polytropic process in which the initial conditions px = 12 psia and 7 i = 6 0 °F change to p 2 = 2 0 psia and volume V = 1011 ft3, determine the ( a) f ormula for the process and (A) w ork done on the air. I (a) pt = pJRTi = (12)(144)/(53.3)(32.17)(460 + 60) = 0.00194 slug/ft3. R w as converted to foot-pounds per slug a nd degree Rankine by multiplying by 32.17. Also, p 2 = IÔTI = 0.002967 slug/ft3; pjp" = p 2 / p 2 , n = [In (p 2 /pi)]/[ln (pz/pi)] = [In (i)]/[ln (0.002967/0.00194)] = 1.20; hence p/p12 = c onst describes the polytropic p rocess. ( 6) W ork of expansion is W=\

ŕ2

pdV

T his is the work done by the gas on its surroundings. Since py V" = p2V2 = pV, by substituting into the integral,

if m is the mass of gas. V2 = 1011 ft3 a nd Vi = V^pjpxf = ( 1 0 1 1 ) ( i ) w 2 = 1547 ft3. Then W = [(20)(144)(1011) - (12)(144)(1548)]/(1 - 1.2) = - 1 1 8 4 000 ft-lb. Hence, the work done on the gas is 1 184 000 ft-lb. 16.23 F or Prob. 16.22, find the ( a) a mount of heat transfer and (b) e ntropy change. f ( a) F rom the first law of thermodynamics the heat added minus the work done by the gas must equal the i ncrease in internal energy; that is, Q„ - W = U2 - t/, = cvm(T2 - T,). First T2 = p2/p2R = (20)(144)/(0.002965)(53.3)(32.17) = 566 °R. Then Q„ = - (1184000/778) + (0.17)(32.17)[3(566 - 520)] = —761 Btu and 761 Btu was transferred from the mass of air.