Example Problem with Complete Solution

5C-3 : Shaft Work Requirement for an Air Compressor 6 pts
A compressor, operating at steady-state, 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 25oC and has a velocity of 14 m/s. At the compressor outlet, the air is at 350oC and has a velocity of 2.4 m/s.
If the compressor inlet has a cross-sectional area of 500 cm2 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, 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 1 bar P2 10 bar
100 kPa 1000 kPa
T1 25 C T2 350 C
298.15 K 623.15 K
v1 14 m/s v2 2.4 m/s
A1 500 cm2 Q -4.20 kW
0.050 m2
Find: Ws ??? kW
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 throughout this process.
Equations / Data / Solve:
Let's begin by writing the steady-state form of the 1st Law for open systems.
Eqn 1
Solve for WS :
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 :
V1 0.02479 m3/mole
R 8.314 J/mol-K V1 0.85565 m3/kg
MWair 28.97 g/mole mdot 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 25oC or 298.15 K, no interpolation is required : Ho1 85.565 kJ/kg
At 200oC or 473.15 K, interpolation is required :
T (K) Ho (kJ/kg)
620 418.55
623.15 Ho2
630 429.25 Ho2 421.92 kJ/kg
Finally, we can plug values back into Eqn 2 to evaluate the shaft work : Ws -279.29 kW
Verify: Only the ideal gas assumption can be verified. Use Eqn 5 for both state 1 and state 2.
V1 24.79 L/mole V2 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 : Ws -279 kW