Question 3.10: Methane is compressed from 1 bar and 290 K to 10 bar. If the...

From the Mollier diagram, shown diagrammatically in Figure 3.5

H_{1} = 4500 cal/mol,
H_{2} = 6200 cal/mol (isentropic path),
Isentropic work = 6200 – 4500
= \underline{\underline{1700 cal/mol } }

For an isentropic efficiency of 0.85:
Actual work done on gas = \frac{1700}{0.85} = \underline{\underline{2000 cal/mol} }
So, actual final enthalpy
H^{\prime}_{2}=H_{1} +2000=\underline{\underline{6500 cal/mol} }

From Mollier diagram, if all the extra work is taken as irreversible work done on the gas, the exit gas temperature = \underline{\underline{480 K} }
Molecular weight methane = 16
Energy required = (mols per hour) × (specific enthalpy change)
= \frac{10,000}{16}\times 2000\times 10^{3}
= 1.25 × 10^{9} cal/h
= 1.25 × 10^{9} × 4.187
= 5.23 × 10^{9} J/h
Power = \frac{5.23 \times 10^{9} }{3600} =\underline{\underline{1.45 MW} }