Question 6.11: Problem: A compressor takes air from the atmosphere, which i...

Problem: A compressor takes air from the atmosphere, which is at 100 kPa and 17 °C, and compresses it to 500 kPa. The compressor is cooled at a rate of 100 kJ / kg of air throughput and the heat rejected to the surrounding air. Irreversibilities in the compressor increase the entropy of the air flowing through it by 0.2 kJ / kg. Calculate the work done and the entropy generated per unit mass of air.

Find: Work done w and entropy generated ∆s_{tot} per kg of air.
Known: Initial temperature T_1 = 17 °C = 290.15 K, initial pressure P_1 = 100 kPa, final pressure P_2 = 500 kPa, temperature of atmosphere T_{atm} = 290.15 K, heat transfer to compressor q = –100 kJ / kg, increase in entropy of air ∆s_{air} = 0.2 kJ / kg.

Assumptions: Air is an ideal gas with constant specific heats.

Governing Equations:
Entropy change (ideal gas, constant specific heats)                            \Delta s = s_2-s_1=c_p \ln \frac{T_2}{T_1}-R \ln \frac{P_2}{P_1}

Steady flow balance                        w+q=h_2-h_1=c_p(T_2-T_1)

Properties: Air has gas constant R = 0.2870 kJ / kgK (Appendix 1), and at 290.15 K (interpolation) has specific heat c_p = 1.0046 kJ / kgK (Appendix 4).

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We can find the final temperature using the increase in entropy of the air

\Delta s_{air}=c_p \ln \frac{T_2}{T_1}-R \ln \frac{P_2}{P_1}=0.2 \ kJ/kgK, \\ 1.0046 \ kJ /kgK \times \ln\left\lgroup\frac{T_2}{290.15 \ K}\right\rgroup – 0.2870 \ kJ/kgK \times \ln\left\lgroup \frac{500 \ kPa}{100 \ kPa}\right\rgroup , \\ T_2=560.750 \ K. 

Then the work per unit mass is

w=c_p(T_2-T_1)-q=1.0046 \ kJ/kgK \times (560.75 \ K- 290.15 \ K)+100 \ kJ/kg=371.845 \ kJ/kg.

The heat lost gained by the atmosphere from the compressor is q_{atm} = –q = 100 kJ / kg, so the increase in entropy of the atmosphere is

\Delta s_{atm}=\frac{q_{atm}}{T_{atm}}=\frac{100 \ kJ/kg}{290.15\ K}=0.344 \ 65 \ kJ/kgK.

The total entropy generated is then

\Delta s_{tot}=\Delta s _{air} + \Delta s_{atm}=0.2 \ kJ/kgK + 0.344 \ 65 \ kJ/kgK=0.544 \ 65 \ kJ/kgK.

Answer: The work done on the air is 371.8 kJ / kg and the total entropy generated is 0.545 kJ / kgK.

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