Operation of an isothermal compressor
Methane at 260 K is to be isothermally compressed from 0.1 MPa to 1.0 MPa.
a) What is the minimum work required, and how much heat must be removed to keep the compression process isothermal?
b) If the compressor is only 75% efficient what is the work required, and how much heat must be removed to keep the compression process isothermal?
a) The steady-state energy and entropy balances for this case are
\begin{aligned}0 &=\hat{H}(T=260 K , P=0.1 MPa )-\hat{H}(T=260 K , P=1 MPa )+W+Q \\&=882 kJ / kg -897 kJ / kg +W+Q\end{aligned}or W = 15 kJ/kg – Q and
\begin{aligned}0 &=\hat{S}(T=260 K , P=0.1 MPa )-\hat{S}(T=260 K , P=1 MPa )+\frac{Q}{T} \\&=(7.1-5.9) kJ / kg K +\frac{Q}{260 K }\end{aligned}so that Q = –312 kJ/kg and then W = 15 + 312 = 327 kJ/kg
[The Aspen process simulator does not provide the option of an isothermal compressor, so there are no results to compare with.]
b) Since the turbine is only 75% efficient, the actual work done by the compressor is 436 kJ/kg. Since the temperature and pressure of the inlet and exit streams are fixed, the enthalpies of the streams remain the same as in part a. So
W = 436 kJ/kg = 15 kJ/kg – Q or Q = 15 – 436 = –421 kJ/kg
Therefore, the extra work that must be added as a result of the turbine efficiency turns into heat that must be removed.