Answer :
Answer:
a) - 122 kJ/kg
b) 0.117 kJ/kg K
Explanation:
Given that:
at inlet condition:
Pressure [tex]P_1 = 0.1 \ MPa[/tex]
Temperature [tex]T_1 = 20^0 \ C[/tex]
at Outlet condition:
Pressure [tex]P_2 = 0.4 \ MPa[/tex]
Temperature [tex]T_2 = 90^ ^0 } \ C[/tex]
Heat transfer Q = 0
determine a) the power required by the compressor (kJ/kg)
Applying energy equation to the compressor work by the equation:
[tex]W _c = h_1 -h_2[/tex]
from propane of tables at the given conditions of temperature and pressure; we obtain the following enthalpies
[tex]h_1 = 517.6 \ kJ/kg \\ \\ h_2 = 639.6 \ kJ/kg[/tex]
[tex]W_c = 517.6 - 639 .6[/tex]
[tex]W_c = - 122 \ \ kJ/kg[/tex]
Thus, the power required by the compressor is - 122 kJ/kg
b) the change in specific entropy of propane (kJ/kg K) at the outlet and the inlet of the compressor.
Taking entropies from the propane tables at given conditions of pressure and temperature
[tex]s_1 = 2.194 \ kJ/kgK\\[/tex]
[tex]s_2 = 2.311 \ kJ/kgK[/tex]
Thus; the rate of entropy production is :
[tex]s_{prod.} = s_2-s_1 \\ \\ s_{prod.} = (2.311 - 2.194 ) kJ/kgK \\ \\[/tex]
[tex]s_{prod.} = 0.117 \ kJ/kgK[/tex]
Thus; the change in specific entropy of propane (kJ/kg K) at the outlet and the inlet of the compressor is 0.117 kJ/kg K