PROBLEM 3.130

PROBLEM 3.117

3 o 2 As shown in Fig. P3.117, 20 ft of air at T1 = 600 R, 100 lbf/in. undergoes a polytropic expansion to a final pressure of 51.4 lbf/in.2 The process follows pV1.2 = constant. The work is W = 194.34 Btu. Assuming ideal gas behavior for the air, and neglecting kinetic and potential energy effects, determine (a) the mass of air, in lb, and the final temperature, in oR. (b) the heat transfer, in Btu.

KNOWN: Air undergoes a polytropic process in a piston-cylinder assembly. The work is known.

FIND: Determine the mass of air, the final temperature, and the heat transfer.

SCHEMATIC AND GIVEN DATA:

o T1 = 600 R Q 2 p1 = 100 lbf/in. 3 V1 = 20 ft Air 2 P2 = 51.4 lbf/in. W = 194.34 Btu 1.2 pV = constant pv1.2 = constant p 1 100 . T1 ENGINEERING MODEL: 1. The air is a closed system. 2. Volume change is the only work mode. 3. The process is 1.2 2 polytropic, with pV = constant and W = 194.34 Btu. 51.4 . T2 4. Kinetic and potential energy effects can be neglected.

ANALYSIS: (a) The mass is determined using the ideal gas equation of state. v

m = = = 9.00 lb

1.2 To get the final temperature, we use the polytropic process, pV = constant, to evaluate V2 as follows.

3 3 V2 = = (20 ft ) = 34.83 ft

Now

PROBLEM 3.117 (CONTINUED)

o T2 = = = 537 R

Alternative solution for T2

The work for the polytropic process can be evaluated using W = . For the process pV1.2 = constant, and incorporating the ideal gas equation of state, we get

W = =

Solving for T2 and inserting values

o o T2 = + T1 = + (600 R) = 537 R

(b) Applying the energy balance; ΔKE + ΔPE + ΔU = Q – W. With ΔU = m(u2 – u1), we get

Q = m(u2 – u1) + W

From Table A-22E: u(600oR) = 102.34 Btu/lb and u(537oR) = 91.53 Btu/lb. Thus,

Q = (9.00 lb)(91.53 – 102.34)Btu/lb + (194.34 Btu) = 97.05 Btu (in)

PROBLEM 3.33

Two kg of Refrigerant 134A undergoes a polytropic process in a piston-cylinder assembly from an initial state of saturated vapor at 2 bar to a final state of 12 bar, 80oC. Determine the work for the process, in kJ.

KNOWN: Refrigerant 134A undergoes a polytropic process in a piston-cylinder assembly.

FIND: Determine the work.

SCHEMATIC AND GIVEN DATA:

p

R-134A 2 m = 2 kg . 12 bar 80oC pvn = constant

. 2 bar ENGINEERING MODEL: 1. The refrigerant is a 1 closed system. 2. The process is polytropic: pvn = constant. v

ANALYSIS: The work for the polytropic process is determined using Eq. 2.17, with pvn = constant. Following the procedure of part (a) of Ex. 2.1

W = (*)

In order to evaluate this expression, we need to determine the specific volumes and the polytropic exponent, n.

3 State 1: From Table A-11; v1 = vg1 = 0.0993 m /kg o 3 State 2: From Table A-12, at 12 bar, 80 C; v2 = 0.02051 m /kg

The polytropic exponent is found from pvn = constant as follows.

→ → n = ln(p1/p2) / ln(v2/v1)

n = ln(2/12) / ln(0.02051/0.0993) = 1.136

Inserting values in Eq. (*) and converting units, we get

PROBLEM 3.33 (CONTINUED)

W = = -69.88 kJ (in)