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ANEK [815]3 years ago

8 0

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Alex17521 [72]

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3 years ago

Read 2 more answers

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 wa

iren2701 [21]

Answer:

A. 6.36 lbm/s

b. $T_2=341\textdegree F$

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:

$\dot m_{in}-\dot m_{out}=\bigtriangleup \dot m_{sys}=0.0\\\\\dot m_{in}=\dot m_{out}\\\\v_1=\frac{RT_1}{P_1}\\\\v_1=\frac{0.3704\times520}{14.7}\\\\\\v_1=13.1026ft^3/lbm\\\\\dot m=\frac{5000/60}{13.1025}\\\\\dot m=6.36\ lbm/s$

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:

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