Answer to Question #137324 in Mechanical Engineering for Chinny

Question

A reciprocating compressor draws in 500 ft3/min of air whose density is 0.079 lb/ft3 and discharges it with a density of 0.304 lb/ft3. At the suction, p1 = 15 psia; at discharge, p2 = 80 psia. The increase in the specific internal energy is 33.8 Btu/lb and the heat transferred from the air by cooling is 13 Btu/lb. Determine the work on the air in Btu/min and in hp. Neglect change in kinetic energy5 m/s2.

Answer 1

Air sucking with density=0.079 $\frac{lb}{ft^3}$ , discharging density=0.304$\frac{lb}{ft^3}$ ,$P_1=$ 15 psia,$P_2=80 psia$ ,increase in specific internal energy=33.8 Btu/lb and heat transfered=13 Btu/lb

As here in this solution we have to neglect kinetic energy

So,

Apply SFEE concept , as here compressor is open system

$W=\Delta H+ \Delta Q$

Here ,$\Delta H = \Delta U+ (P_2V_2-P_1V_1)$

$\Delta H = \Delta U+ (P_2(\frac{m}{\rho_2})-P_1(\frac{m}{\rho_1}))$

And for unit mass flow rate or in case of specific internal enrgy whole term will be independent of mass

so our Work equation become

$W_{.in}=\Delta U+ (P_2(\frac{1}{\rho_2})-P_1(\frac{1}{\rho_1}))+ \Delta Q$

and mass flow rate is already given as 500 ft³/min=39.5 lb/min

$W_{in}=33.8\times 39.5+ (80(\frac{1}{0.304})-15(\frac{1}{0.709}))\times 39.5+ 13\times 39.5$

$W_{in}=2384 \frac{Btu}{min}$

and in HP

W=56.2 hp