# Example Problem with Complete Solution

8C-1 : Shaft Work Requirement for Different Compression Systems 7 pts
Determine the specific shaft work required for the air compressor shown below if the compression is : a.) Isothermal
b.) Isentropic
c.) Polytropic with d = 1.24
d.) A 2-stage compressor with intercooling with d = 1.24 is used instead. Calculate the isothermal efficiency of this system. Assume air behaves as an ideal gas with constant heat capacities and all the compressors are internally reversible.
Data: P1 = 90 kPa, T1 = 310 K, P2 = 1.25 MPa

Read : Parts (a) through (c) are direct applications of equations derived for shaft work in polytropic processes.  Part (a) requires us to assume the fluid is an ideal gas with constant heat capacities so that we can assume d = g.  Part (d) is the application of equations for the internally reversible, polytropic compression of of an ideal gas.  The key is to to determine the optimal intermediate pressure, PX, and use it to to determine the shaft work for each compressor.
Given: P1 90 kPa (isothermal)   dA 1
T1 310 K (isentropic)   gB 1.4
P2 1250 kPa dC 1.24
(ideal, 2-stage w/ intercooling)   dD 1.24
Find: For each part of the problem -WS / mdot ??? kJ/kg Compare results.
Part d.) hT ??? %
Diagram: See the problem statement.
Assumptions: 1 - All compressors operate at steady-state.
2 - Air behaves as an ideal gas.
3 - The heat capacities of the air are constant.
4 - The intercooler in part (d) returns the air to the inlet temperature, T1.
5 - All compressors are internally reversible.
Equations / Data / Solve:
The key equation for parts (a), (c) and (d) is the equation for the specific shaft work in steady-state, polytropic processes. Eqn 1
Part a.) When d = 1 for a polytropic process on an ideal gas, the process is isothermal.
In this case, Eqn 1 does not apply.  Instead, we must use : Eqn 2
Plugging values into Eqn 2 yields : -WS / mdot 234.1 kJ/kg
Part b.) Plug values into Eqn 1 : R 8.314 J/mole-K
MW 28.97 g/mole
-WS / mdot 349.0 kJ/kg
Part c.) Plug values into Eqn 1 : -WS / mdot 305.2 kJ/kg
Part d.) The total work per unit mass of flowing fluid (air in this case) in a 2-stage compression process is the sum of the specific work for each compressor.  The resulting equation is just the application of Eqn 1 to each compressor. Eqn 3
Where PX is the intermediate pressure between the two compressors.
The optimal value of the intermediate pressure can be determined using: Eqn 4
Now, we can plug values into Eqns 4 & 3 : PX 335.4 kPa
Compressor #1:   -WS / mdot 133.29 kJ/kg
Compressor #2:   -WS / mdot 133.29 kJ/kg
Total :   -WS / mdot 266.6 kJ/kg
Notice that when the optimal value of PX is used the compression ratio across each compressor is the same. Eqn 5
As a result, the specific shaft work for each compressor is the same as well.
The isothermal efficiency of the 2-stage compressor can be determined from : Eqn 6
Plugging values into Eqn 6 yields : hT 87.81%
Verify: The assumptions made in the solution of this problem cannot be verified with the given information.
Answers : Part a.) -WS / mdot 234 kJ/kg
This is the standard against which the other compressors are compared. The problem is that an isothermal compressor cannot be built.
Part b.) -WS / mdot 349 kJ/kg
This value seems high, but this compressor does not require any heat exchange because it is adiabatic.
Part c.) -WS / mdot 305 kJ/kg
This polytropic compressor must reject some heat to accomplish the compression with less power input than the isentropic compressor.
Part d.) -WS / mdot 267 kJ/kg hT 87.8%
The 2-stage compressor with intercooling reduces the power requirement by about 13% compared to the compressor in part (c). 