# Example Problem with Complete Solution

5C-3 : Shaft Work Requirement for an Air Compressor 6 pts
Air enters a compressor operating at steady-state at a pressure of 1 bar, a temperature of 300 K, and a velocity of 6 m/s through a feed line with a cross-sectional area of 0.1 m2.  The effluent is at a pressure of 7 bar and a temperature of 450 K and has a velocity of 2 m/s.  Heat is lost from the compressor at a rate of 180 kJ/min.  If the air behaves as an ideal gas, what is 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, cross-sectional 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 : P1 100 kPa Find :
T1 300 K Ws ??? kW
v1 6 m/s
A1 0.1 m2
P2 700 kPa
T2 450 K
v2 2 m/s
Q -180 kJ/min
-3.00 kW
Diagram : Assumptions:
1 - The compressor operates at steady-state.
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 throughoutthis process.
Equations / Data / Solve :
Let's begin by writing the steady-state form of the 1st Law for open systems. Eqn 1
Solve for Q : Eqn 2
We know the inlet and outlet velocities and we can look up 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 realationship 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 :
V1 0.024942 m3/mole
R 8.314 J/mol-K V1 0.860069 m3/kg
MWair 29 g/mole mdot 0.6976 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 !
Ho1 87.41 kJ/kg Ho2 239.78 kJ/kg
Finally, we can plug values back into Eqn 2 to evaluate the shaft work :
Ws -109.28 kW
Verify : Only the ideal gas assumption can be verified.  Use Eqn 5 for both state 1 and state 2.
V1 24.94 L/mole V2 5.345 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 : Ws -109 kW 