Example Problem with Complete Solution

8C-3 : Isentropic Efficiency of an Ideal Gas Compressor 7 pts
Consider the adiabatic air compressor shown below.
a.) Calculate the minimum power input required and T2
b.) The outlet temperature from a real, adiabatic compressor that accomplishes the same compression is 520K. Calculate the actual power input and the isentropic efficiency of the real compressor.
Read : Determine So(T2) for an isentropic process and then interpolate to obtain both T2S and H2S. Then, an energy balance will give you (WS)min. Use the isentropic efficiency and (WS)min to determine (WS)act.
Given: m 9.5 kg/s Find: a.) (WS)min ??? kW
P1 110 kPa T2S ??? K
T1 310 K b.) (WS)act ??? kW
P2 550 kPa hS, comp ???
T2, part (b) 520 K
Diagram: The diagram in the problem statement is adequate.
Assumptions: 1 - The compressor operates at steady-state and there is no significant heat transfer.
2 - Kinetic and potential energy changes are negligible.
3 - Air is modeled as an ideal gas.
Equations / Data / Solve:
Part a.) An isentropic compressor requires the minimum power input.
We can determine the isentropic work by applying the 1st Law to an isentropic compressor that takes in the same feed and yields an effluent at the same pressure.
For a steady-state, single-inlet, single outlet system with negligible heat transfer, kinetic and potential energy changes, the 1st Law is:
Eqn 1
The entropy change for this process can determined using the 1st Gibbs Equation in terms of the Ideal Gas Entropy Function.
Eqn 2
We can also apply Eqn 2 to our
hypothetical, isentropic compressor:
Eqn 3
We can solve Eqn 3 for the unknown SoT2S :
Eqn 4
We can evaluate SoT1 using the Ideal Gas Property Tables: So(T1) 0.0061681 kJ/kg-K
We can get HoT1 while we are looking in the Ideal Gas Property Tables because we will need it later when we evaluate Eqn 1.
H1 87.410 kJ/kg
Now, we can plug values into Eqn 4 : R 8.314 kJ/kmol-K
MW 28.97 kg/kmol
So(T2S) 0.46806 kJ/kg-K
Now, we can use SoT2S and the Ideal Gas Property Tables to determine T2S and H2S by interpolation.
T (K) H (kJ/kg) So (kJ/kg-K)
470 260.49 0.46258
T2S H2S 0.46806 T2S 472.50 K
480 270.88 0.48445 H2S 263.09 kJ/kg
Now, we can plug values back into Eqn 1 : (WS)min -1669.0 kW
Part b.) We can determine the actual power input for the compressor by applying the 1st Law to the real compressor, just as we did in Eqn 1 for the isentropic compressor.
Eqn 5
We can evaluate HoT2 using the Ideal Gas Property Tables: T2, part (b) 520 K
H2 312.65 kJ/kg
Now, we can evaluate WS,act using Eqn 5: (WS)act -2139.78 kW
The isentropic efficiency of a compressor is defined by:
Eqn 6
Since we determined the isentropic work in part (a) and the actual work in part (b), we are ready to plug numbers into Eqn 6 and wrap up this problem.
hS, comp 78.00%
Verify: Check the Ideal Gas assumption:
V1 = 23.43 L/mole
V2 = 7.86 L/mole
Since air can be considered to be a diatomic gas and both molar volumes are greater than 5 L/mole, it is accurate to treat the air as an ideal gas.
Answers : Part a.) (WS)min -1670 kW Part b.) (WS)act -2140 kW
T2S 473 K hS, comp 78.0%