Consider air as an ideal gas with Cp=1000 J/(kg.K), R=287 J/(kg.K) and γ≈1.4. It is compressed from 1 bar, 300K to 10 bar using a compressor. The compressor can be considered adiabatic but not reversible. The flow rate is 2 kg/s, and the entropy production rate is 200 W/K.

Question

1.What is the change in specific entropy of the gas (units of J/(kg.K)) as it moves from inlet to exit?

2.If the process was reversible apart from being adiabatic, what would have been the final Temperature (in kelvin) of the gas at the exit?

3.What is the actual temperature of the gas at the exit (in kelvin)?

4.What is the power consumed by the compressor (in kW)? Do not enter the sign.

5.What is the isentropic efficiency of the compressor? Enter the value as a fraction.

Answer 1

1) from 2nd law:

S_gen/m = 200/2 = 100 J/kg-K
s_gen = 0.10 kJ/kg-K

s_gen=s2-s1 = .10kJ/kg-K

look in table for ideal gas for air to find s1

=> s1 = 6.86926 kJ/kg-k

s_gen + s1 = s2
=> 0.10 + 6.86926 = s_gen

3) actual temperature of gas at exit:

apply change in entropy law for ideal gas:
formula can be found in book:

s2-s1=C_p*ln(T2/T1) - R*ln(P2/P1)

all values are given except T2, solve for T2 (be careful about units)

2) if reversible + adiabatic:

0=C_p*ln(T2/T1) - R*ln(P2/P1)

find T2

4) Apply first law w/out Q, potential and kinetic energy

dQ-dW=mC_p*deltaT

W = (2kg/s) * (0.10J/kg-K)*(T2-300)

T2 here is same as T2 found in number 3

5) isentropic efficiency of compressor:

n= W_isen / W_actual

Answer 2

Donho, can you please send the final answer
day after the submission date