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Alexandra [31]
3 years ago
11

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

ter 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.
Physics
1 answer:
iren2701 [21]3 years ago
5 0

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

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

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Vector A has a magnitude of 30 units. Vector B is perpendicular to vector Aand has a magnitude of 40 units. What would the magni
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here we know that

A = magnitude of vector A

B = magnitude of vector B

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|\vec A + \vec B| = \sqrt{30^2 + 40^2 + 2(30)(40)cos90}

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