Question

Determine the work input and entropy generation during the compression of steam from 100 kPa to 1 MPa in an adiabatic compressor if the inlet state is saturated vapor and the isentropic efficiency is 85 percent.

Answer 1

Answer:

The value of work input to the compressor $W_{in}$ = 1505.7 $\frac{KJ}{kg}$

The value of entropy generation inside the compressor

$S_{gen} = - 1.064 \frac{KJ}{kg K}$

Explanation:

Initial pressure $P_{1}$ = 100 k pa

Final pressure $P_{2}$ = 1 M pa = 1000 k pa

At initial pressure $P_{1}$ = 100 k pa the steam is saturated vapour. From the steam tables the value of volume at this state is $v$ = 1.673 $\frac{m^{3} }{kg}$

Thus the work input to the compressor $W_{in}$ = $v$ $dP$

⇒ $W_{in}$ = $v$  ( $P_{2}$ - $P_{1}$ )

⇒ $W_{in}$ = 1.673 ( 1000 - 100 )

⇒ $W_{in}$ = 1505.7 $\frac{KJ}{kg}$

The isentropic efficiency is  = 85 %

So the work input to the compressor $W_{in}$ = $\frac{1505.7}{0.85}$

⇒ $W_{in}$ =  1771.42 $\frac{KJ}{kg}$

This is the value of work input to the compressor.

$S_{gen} = - R \ln \frac{P_{2} }{P_{1} }$

$S_{gen} = - 0.462 \ln \frac{1000}{100}$

$S_{gen} = - 1.064 \frac{KJ}{kg K}$

This is the value of entropy generation inside the compressor.