Answer:
a) [tex]\mathbf{Q_c = -3730.8684 \ Btu/hr}[/tex]
b) [tex]\mathbf{\sigma _c = 4.3067 \ Btu/hr ^0R}[/tex]
Explanation:
From the properties of Super-heated Refrigerant 134a Vapor at [tex]T_1 = 40^0 F[/tex], [tex]P_1 = 30 \ lbf/in^2[/tex] ; we obtain the following properties for specific enthalpy and specific entropy.
So; specific enthalpy [tex]h_1 = 109.12 \ Btu/lb[/tex]
specific entropy [tex]s_1 = 0.2315 \ Btu/lb.^0R[/tex]
Also; from the properties of saturated Refrigerant 134 a vapor (liquid - vapor). pressure table at [tex]P_2 = 160 \ lbf/in^2[/tex] ; we obtain the following properties:
[tex]h_2 = 115.91 \ Btu/lb\\\\ s_2 = 0.2157 \ Btu/lb.^0R[/tex]
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:
[tex]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)[/tex]
By neglecting kinetic and potential energy effects; we have:
[tex]0 = Q_c -W_c+m(h_1-h_2) \\ \\ Q_c = -W_c+m(h_2-h_1)[/tex]
[tex]Q_c = -5088.8684 \ Btu/hr +200 \ lb/hr( 115.91 -109.12) Btu/lb \\ \\[/tex]
[tex]\mathbf{Q_c = -3730.8684 \ Btu/hr}[/tex]
b) To determine the entropy generation; we employ the formula:
[tex]\dfrac{dS}{dt} =\dfrac{Qc}{T}+ m( s_1 -s_2) + \sigma _c[/tex]
In a steady state condition [tex]\dfrac{dS}{dt} =0[/tex]
Hence;
[tex]0=\dfrac{Qc}{T}+ m( s_1 -s_2) + \sigma _c[/tex]
[tex]\sigma _c = m( s_1 -s_2) - \dfrac{Qc}{T}[/tex]
[tex]\sigma _c = [200 \ lb/hr (0.2157 -0.2315) \ Btu/lb .^0R - \dfrac{(-3730.8684 \ Btu/hr)}{(40^0 + 459.67^0)^0R}][/tex]
[tex]\sigma _c = [(-3.16 ) \ Btu/hr .^0R + (7.4667 ) Btu/hr ^0R}][/tex]
[tex]\mathbf{\sigma _c = 4.3067 \ Btu/hr ^0R}[/tex]
