Question

Known Answer: W12=-51.0 Btu, constant Cv: Q12= -19.2 Btu, Table A-22E: Q12=-18.8 Btu....... Please explain in detail how to use the table.
3.145 Air is compressed in a piston-cylinder assembly from pi = 10 lbf/in?, T1 = 500°R to a final volume of K2 = 1 ft in a process described by pun = constant, where n= 1.25. The mass of air is 1/2 lbm. Assuming ideal gas behavior and neglecting kinetic and potential energy effects, determine the work and the heat transfer, each in Btu, using (a) constant specific heats evaluated at 500°R, and (b) data from Table A-22E. Compare the results and discuss.

Answer 1

The detailed solution is given below:

Goven- PE 10 dbf lim? T= 500R M = 0.5 lbm = 183 I n = 125 boly tro bic brocess # - [tral take paran > B = 9(0- ne )_0 вч, уг тет => y = mxRxT, – (0.586m) x (0.3704 ) x (sie) psig. f3 lom. R = 9.26ft3 Equation ☺ become B= 10x1920) = 16153 bill

# work donl, Wi-a = pdv = Pz Vz - PV -n (16153 psi Jx (1f4?) – (lo ksi) x (9.26 ft ?) - 1.25 = 161.53— 92.6 -0.25 = -275.72 psia. ft? sa ale 51.02 btl (negative sign Pliates the core is done -51.02 btu Negative sign indicates the work is done on the system) - 161.53 f ., © → Since wrz is independent of specific heat. i. The work done is same for both cases. 1 + - PVM=C --- > 0 10 9026 F # Para= mR Tz (ft3) = TE = Px X V2 mxr- = 872.19 *R (161.53)x(1 70.5)X (0.3704)

# The energy bakancel: O-W = AU [sincl K.E=P.E=0] => Q = wtau (al At T- 500R ZAOF Cv = 0.171 Brullbmo R [from ideal gas Specific heat table Clirz = Wrzt (U2 -Ui] = (-5102)+ (m XCVXOT) = (5102)+(mxCox (-01) = -5łos+ [0.5% (0.181) x ( 872.19-500)] = -51.02 + 31.82 Qiz : -19.Bt4 O2 = winst Uz-u, = W2 + m (Uz-u) 'U = U@872°R = 149.61 bullbm (by interpolation) Now, 1= U@8th U = U@500k = 85.20 btullbm

=> Q12 = (-51.02)+ (0.5)x (149.61 – 85.20) = -18.81 644 Interpolation: U, -Venik = levici (bero-berada) que = 14150 43983-1550) + (150-1750) 147.50+ 1.76 +0.352 = 149.61btu