A compressor, operating at steadystate, increases the pressure of an air
stream from 1 bar to 10 bar while losing 4.2 kW of heat to the surroundings. 
At the compressor inlet, the air
is at 25^{o}C and has a velocity of 14 m/s. At the compressor outlet, the air is at 350^{o}C and has a velocity of 2.4 m/s. 
If the compressor inlet has a crosssectional area of 500 cm^{2} and the air behaves as an ideal gas, determine the power requirement of the compressor in kW. 











Read : 
The keys here are the 1st Law, the Ideal Gas EOS and the Ideal Gas Property Tables. Since we know the velocity, temperature and pressure of both the feed and effluent, we can determine the change in the specific enthalpy (using the Ideal Gas Properties
Table for air) and the specific kinetic energy. The problem is the mass flow rate. Use the Ideal
Gas EOS to determine the specific volume. Then, use the relationship between velocity, crosssectional area for flow, specific volume and mass flow rate to determine the mass flow rate. After that, plug all the values back into
the 1st Law and solve
for the shaft work. 










Given: 
P_{1} 
1 
bar 



P_{2} 
10 
bar 


100 
kPa 




1000 
kPa 

T_{1} 
25 
°C 



T_{2} 
350 
°C 


298.15 
K 




623.15 
K 

v_{1} 
14 
m/s 



v_{2} 
2.4 
m/s 

A_{1} 
500 
cm^{2} 



Q 
4.20 
kW 


0.050 
m^{2} 
















Find: 
W_{s} 
??? 
kW 
















Diagram: 











Assumptions: 
1  
The compressor operates at steadystate. 


2  
The change in the potential energy of the fluid from
the inlet to the outlet is negligible. 


3  
The air behaves as an ideal gas throughout this process. 










Equations
/ Data / Solve: 


















Let's begin by writing
the steadystate form of the 1st Law for open systems. 








Eqn 1 

Solve for W_{S} : 



Eqn 2 











We know the inlet and outlet velocities and we can lookup the inlet and outlet specific enthalpies in the Ideal Gas Properties Table. So, the only remaining
obstacle to evaluating the shaft work using Eqn 2 is the mass flow rate. 











The following
relationship will let us evaluate the mass flow rate : 


















Eqn 3 











Next, we must use the Ideal Gas Equation of State to
determine the specific volume of the air feed. 


















Eqn 4 











Solve for V : 





Eqn 5 











Convert molar volume to specific volume : 



Eqn 6 











Plugging values into Eqns 5 & 6 yields : 












V_{1} 
0.02479 
m^{3}/mole 

R 
8.314 
J/molK 



V_{1} 
0.85565 
m^{3}/kg 

MW_{air} 
28.97 
g/mole 



m_{dot} 
0.8181 
kg/s 











Next we need to look
up the specific enthalpy of air at the inlet and outlet temperature in the Ideal Gas Properties Table for air. Remember that the enthalpy of an ideal gas does NOT depend on the pressure ! 











At 25^{o}C or 298.15 K, no interpolation is required : 
H^{o}_{1} 
85.565 
kJ/kg 











At 200^{o}C or 473.15 K, interpolation is required : 















T (K) 
H^{o}
(kJ/kg) 








620 
418.55 








623.15 
H^{o}_{2} 








630 
429.25 




H^{o}_{2} 
421.92 
kJ/kg 











Finally, we can plug
values back into Eqn 2 to
evaluate the shaft work
: 
W_{s} 
279.29 
kW 










Verify: 
Only the ideal gas assumption can be
verified. Use Eqn 5 for both state 1 and state 2. 











V_{1} 
24.79 
L/mole 



V_{2} 
5.181 
L/mole 











Because air is made up of diatomic gases, the test for the
applicability of the Ideal Gas EOS is whether the molar volume > 5
L/mole. 











This considition is
satisfied at both the inlet and outlet conditions, so using the Ideal Gas EOS and the Ideal Gas Properties Tables will
yield results accurate to at least 2 significant figures. 










Answers : 
W_{s} 
279 
kW 















