Question

3. A steady state, steady flow compressor draws in 236 liters per second of air whose density is

1.268 kgm/m3 and discharges it with a density of 4.88 kgm/m3. At suction P1= 103.39 kPaa; at

discharge, P2 = 551.428 kPaa. The increase of specific internal energy is 78.45 kJ/kgm and the heat

from the air by cooling is 30.17 kJ/kgm. Neglecting the change in potential and kinetic energies,

determine the work in kJ/min.

Answer 1

We know that

$U=um_1=u\rho _1V$

$Q=qm_2=q\rho_2 V$

Now

$a=\frac{A}{60}$

$a=\frac{U+Q}{60}=\frac{(u\rho_1+q\rho_2)V}{60}$

$a=984KJ/min$