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

Answer:

Required compressor work W=8560.44  KJ/Kmol

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

$W=P_1V_1\ln \dfrac{P_1}{P_2}$

given taht gas is ideal so

P V =m R T

$W=mRT\ln \dfrac{P_1}{P_2}$

R for ethylene

R=0.296 KJ/kg.K

Now by putting the values

$W=mRT\ln \dfrac{P_1}{P_2}$

$W=1\times 0.296\times 280 \ln \dfrac{1}{40}$

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