Answer to Question #288820 in Mechanical Engineering for ros

Question

A steady state, steady flow compressor draws in 500 cfm of air whose density is 0.079 lb/ft^3 and discharges it with a density of 0.304 lb/ft^3. At the suction, p1=15 psia; at discharged p2=80 psia. The increase in specific internal energy is 33.8 Btu/lb, and the heat from the air cooling is 13 Btu/lb.

Answer 1

Solution;

The stead flow energy equation for a compressor is;

$q+h_1=W+h_2$

But;

$h=U+pV$

Given;

Volume flow rate;

$500ft^3/min$

Density at intake;

$0.079lb/ft^3$

Mass flow rate at intake;

$\dot m=500×0.079=39.5lb/min$

Density at discharge;

$0.304lb/ft^3$

Pressure at intake;

$15psia$

Pressure at discharge;

$80psia$

Then ;

$W=h_1-h_2+q$

$W=p_1V_1-p_2V_2-q-\Delta u$

$p_1V_1=\frac{15}{0.079}×\frac{144}{778}=35.144btu/lb$

$p_2V_2=\frac{80}{0.304}×\frac{144}{778}=48.708btu/lb$

Therefore;

$W=35.144-40.708-13-33.8=-60.364Btu/lb$

$W=39.5×60.364=2384.378Btu/min$

Into hp;

$56.25hp$