Question

Refrigerant 134a at p1 = 30 lbf/in2, T1 = 40oF enters a compressor operating at steady state with a mass flow rate of 200 lb/h and exits as saturated vapor at p2 = 160 lbf/in2. Heat transfer occurs from the compressor to its surroundings, which are at T0 = 40oF. Changes in kinetic and potential energy can be ignored. The power input to the compressor is 2 hp. Determine the heat transfer rate for the compressor, in Btu/hr, and the entropy production rate for the compressor, in Btu/hr·oR.

Answer 1

a)
`\mathbf{Q_c = -3730.8684 \ Btu/hr}`

b)
`\mathbf{\sigma _c = 4.3067 \ Btu/hr ^0R}`

Step-by-step explanation:

From the properties of Super-heated Refrigerant 134a Vapor at
`T_1 = 40^0 F`,
`P_1 = 30 \ lbf/in^2` ; we obtain the following properties for specific enthalpy and specific entropy.

So; specific enthalpy
`h_1 = 109.12 \ Btu/lb`

specific entropy
`s_1 = 0.2315 \ Btu/lb.^0R`

Also; from the properties of saturated Refrigerant 134 a vapor (liquid - vapor). pressure table at
`P_2 = 160 \ lbf/in^2` ; we obtain the following properties:

`h_2 = 115.91 \ Btu/lb\\\\ s_2 = 0.2157 \ Btu/lb.^0R`

Given that the power input to the compressor is 2 hp;

Then converting to Btu/hr ;we known that since 1 hp = 2544.4342 Btu/hr

2 hp = 2 × 2544.4342 Btu/hr

2 hp = 5088.8684 Btu/hr

The steady state energy for a compressor can be expressed by the formula:

`0 = Q_c -W_c+m((h_1-h_e) + (v_i^2-v_e^2)/(2)+g(\bar \omega_i - \bar \omega_e)`

By neglecting kinetic and potential energy effects; we have:

`0 = Q_c -W_c+m(h_1-h_2) \\ \\ Q_c = -W_c+m(h_2-h_1)`

`Q_c = -5088.8684 \ Btu/hr +200 \ lb/hr( 115.91 -109.12) Btu/lb \\ \\`

`\mathbf{Q_c = -3730.8684 \ Btu/hr}`

b) To determine the entropy generation; we employ the formula:

`(dS)/(dt) =(Qc)/(T)+ m( s_1 -s_2) + \sigma _c`

In a steady state condition
`(dS)/(dt) =0`

Hence;

`0=(Qc)/(T)+ m( s_1 -s_2) + \sigma _c`

`\sigma _c = m( s_1 -s_2) - (Qc)/(T)`

`\sigma _c = [200 \ lb/hr (0.2157 -0.2315) \ Btu/lb .^0R - ((-3730.8684 \ Btu/hr))/((40^0 + 459.67^0)^0R)]`

`\sigma _c = [(-3.16 ) \ Btu/hr .^0R + (7.4667 ) Btu/hr ^0R}]`

`\mathbf{\sigma _c = 4.3067 \ Btu/hr ^0R}`