Question

Ethylene enters a reversible, isothermal, steady-flow compressor at 1 bar and 280 K, and exits at 40 bar. Find the required compressor work (kJ/kmol) using the ideal gas equation of state.

Answer 1

Required compressor work W=8560.44 KJ/Kmol

Step-by-step explanation:

Given that

Initial pressure = 1 bar

`P_1=1\ bar`

Final pressure = 40 bar

`P_2=40\ bar`

Process is isothermal and T=280 K

We know that ,work done in isothermal process given as

`W=P_1V_1\ln (P_1)/(P_2)`

given taht gas is ideal so

P V =m R T

`W=mRT\ln (P_1)/(P_2)`

R for ethylene

R=0.296 KJ/kg.K

Now by putting the values

`W=mRT\ln (P_1)/(P_2)`

`W=1* 0.296* 280 \ln (1)/(40)`

W= -305.73 KJ/kg

Negative sign indicates that work done on the system.

Required compressor work W=305.73 KJ/kg

Molar mass of ethylene M= 28 Kg/Kmol

So W= 305.73 x 28 KJ/Kmol

W=8560.44 KJ/Kmol

Required compressor work W=8560.44 KJ/Kmol