Analysis of compressor operation

An ideal gas with a constant pressure heat capacity of 30 kJ/kg K is to be compressed from 1 bar and 25°C to 10 bar.

(a) Compute the final temperature of the gas and the work required if the compressor operates isentropically

(b) Compute the final temperature of the gas and the work required if the compressor has an isentropic efficiency of 75%.

Question

Analysis of compressor operation

An ideal gas with a constant pressure heat capacity of 30 kJ/kg K is to be compressed from 1 bar and 25°C to 10 bar.

(a) Compute the final temperature of the gas and the work required if the compressor operates isentropically

(b) Compute the final temperature of the gas and the work required if the compressor has an isentropic efficiency of 75%.

Accepted Answer

(a) For an isentropic process, from Eq. 4.4-3

[latex]\underline{S}\left(T_{2}, P_{2} ight)-\underline{S}\left(T_{1}, P_{1} ight)=C_{ P }^{*} \ln \left(\frac{T_{2}}{T_{1}} ight)-R \ln \left(\frac{P_{2}}{P_{1}} ight)[/latex] (4.4-3)

[latex]C_{ P }^{*} \ln \frac{T_{2}}{T_{1}}=R \ln \left(\frac{P_{2}}{P_{1}} ight)[/latex]

or

[latex]T_{2, S}=T_{1}\left(\frac{P_{2}}{P_{1}} ight)^{R / C_{ p }^{*}}[/latex]

so that [latex]T_{2, S}[/latex] = 564.38 K and

[latex]\begin{aligned}W &=C_{ P }^{*}\left(T_{2, S}-298.15 ight) \&=30(564.38-298.15)=7986.8 kJ / kg\end{aligned}[/latex]

(b) Here [latex]W_{A}=7986.8 / 0.75=106490.0 kJ / kg . ext { Therefore, } T_{2}=\frac{10649.0}{30}+298.15=[/latex] 653.1 K

Comment
In this case more work is required to compress the gas than with a isentropic compressor, and this extra work requirement results in a higher gas exit temperature.

Question 5.3.2: Analysis of compressor operation An ideal gas with a constan...

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