Answer to Question #137324 in Mechanical Engineering for Chinny

Question #137324
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.
Expert's answer

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


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

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