Question

Air is compressed from 14.7 psia and 60°F to a pressure of 150 psia while being cooled at a rate of 10 Btu/lbm by circulating water through the compressor casing. The volume flow rate of the air at the inlet conditions is 5000 ft³/min, and the power input to the compressor is 700 hp. Determine:(a) the mass ow rate of the air and (b) the temperature at the compressor exit.

Answer 1

A. 6.36 lbm/s

b.
`T_2=341\textdegree F`

Step-by-step explanation:

a. Given the following information;

#Compressor inlet:

Air pressure,
`P_1=14.7psia,T_1=60\textdegree F`

#Compressor outlet:

`Air \ Pressure, P_2=150psia\\\\Air \ volume \ flow \ rate, \dot V_1=5000ft^3/min`

#Cooling rate,

`q_(out)=10Btu/lbm, \dot W=700hp`

# From table A-1E

Gas constant of air
`R=0.3704\ psia.ft^3/lbm.R`

Specific enthalpy at
`P_1=520R-h_1=124.27Btu/lbm`

Using the mass balance:

`\dot m_(in)-\dot m_(out)=\bigtriangleup \dot m_(sys)=0.0\\\\\dot m_(in)=\dot m_(out)\\\\v_1=(RT_1)/(P_1)\\\\v_1=(0.3704*520)/(14.7)\\\\\\v_1=13.1026ft^3/lbm\\\\\dot m=(5000/60)/(13.1025)\\\\\dot m=6.36\ lbm/s`

Hence, the mass flow rate of the air is 6.36lbm/s

b The specific enthalpy at the exit is defined as the energy balance on the system:

`\dot m_(in)-\dot m_(out)=\bigtriangleup \dot m_(sys)=0.0\\\\\dot m_(in)=\dot m_(out)\\\\\dot W_(in)+ \dot mh_1=\dot Q_(out)+\dot m_2\\\\h_2=(\dot W_(in)-\dot Q_(out))/(\dot m)+h_1\\\\h_2=(700* 0.7068-10* 6.36)/(6.36)+124.27\\\\h_2=192.06\ Btu/lbm\\\\#at \ h_2=182.06Btu/lbm, T_2=801R=341\textdegree F`

Hence, the temperature at the compressor exit
`T_2=341\textdegree F`