Question

A 500 cubic feet per minute of air enters a system whose density is 0.075 lbm/ft3 and discharges it with a density of 0.34 lbm/ft3. The pressure at the suction is 10 psia and at the discharge is 80 psia. The increase in the specific internal energy is 35.8 BTU/lbm. Determine the work in Hp.

Answer 1

$m = \rho V$

$m= 0.075 \times 500 = 37.5$ lb/min

According to the balancing work equation

$w= \frac{-2384.378}{42.4}$

= 56.235 hp