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

Answer:

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

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

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:

$0 = Q_c -W_c+m((h_1-h_e) + \dfrac{v_i^2-v_e^2}{2}+g(\bar \omega_i - \bar \omega_e)$

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:

$\dfrac{dS}{dt} =\dfrac{Qc}{T}+ m( s_1 -s_2) + \sigma _c$

In a steady state condition $\dfrac{dS}{dt} =0$

Hence;

$0=\dfrac{Qc}{T}+ m( s_1 -s_2) + \sigma _c$

$\sigma _c = m( s_1 -s_2) - \dfrac{Qc}{T}$

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

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

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