compressor, air enters 100 kPa and 17C and exits compressor 600 kPa and 167C. Ma

Question

compressor, air enters 100 kPa and 17C and exits compressor 600 kPa and 167C. Mass flow rate of air is 2 kg/min & power required to run the compressor is 5 kW. The temp of the surrounding is 17C. The compressor exchanges heat with the surrounding at a rate of Q(dot on top) . Calculate:
(1)the rate of entropy change of air?
(2)the rate of entropy change of the surrounding?
(3)the rate of entropy generation during the process?
The ideal gas constant is R(bar)=314. kJ/kmol x K and molecular weight,air is 29 kg/kmol.

Explanation / Answer

T1 = 17 C = 17+273 = 290 K

T2 = 167 C = 167+ 273 = 440 K

assuming specific heat of air to be constant = cp = 1.005 Kj/Kg.K

W + Q = m x Cp x (T2 - T1)

5 + Q = (2/60) x 1.005 x (440 - 290)

Q = 0.025 KW

1) entropy change of air = S1 = Q/T = m.Cp . dT/T

on integrating from T1 to T2

S1 = (2/60) x 1.005 x ln T2/T1 = (2/60) x 1.005 x ln(440/290)

   = 0.01396 Kj/Kg.K = 13.96 J/Kg.K

2) entropy change of surrounding = S2

   = - Q/T = - 0.025 / (17+273) = - 0.0862 J/Kg.K (- sign as heat flows from the surro.)

3) the rate of entropy generation during the process = S1 + s2

   = 13.96 -  0.0862 = 13.87 j/Kg.K

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