Example Problem with Complete Solution

8C-1 : Shaft Work Requirement for Different Compression Systems 7 pts
Air is compressed by an internally reversible compressor from 100 kPa and 300 K to a pressure of 900 kPa.  Calculate the specific shaft work requirement for the compressor for each of the following cases:
               
a.) Isentropic compression with γ = 1.4    
b.) Polytropic compression with δ = 1.3    
c.) Isothermal compression        
d.) Internally reversible 2-stage compression with intercooling, δ = 1.3
  Calculate the isothermal efficiency of this 2-stage compressor.
               
Read: Figure 1a
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 100 kPa Find:
T1 300 K For each part of the problem…
P2 900 kPa -WS / mdo ??? kJ/kg
gA 1.4 (isentropic) Compare.
dB 1.3
dC 1 (isothermal)
dD 1.3
(ideal, 2-stage w/ intercooling)
Diagrams:
Parts a - c : Part d.)
Figure 1b
Figure 2
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 (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.
Equation 1
Eqn 1
Part a.) Plug values into Eqn 1 :
R 8.314 J/mole-K
MW 29.0 g/mole -WS / mdo 262.93 kJ/kg
Part b.) Plug values into Eqn 1 : -WS / mdo 246.12 kJ/kg
Part c.) 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 :
Equation 2
Eqn 2
Plugging values into Eqn 2 yields : -WS / mdo 188.98 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.
Equation 3
Eqn 3
Where PX is the intermediate pressure between the two compressors.
The optimal value of the intermediate pressure can be determined using:
Equation 4
Eqn 4
Now, we can plug values into Eqns 4 & 3 : PX 300 kPa
Compressor #1:   -WS / mdo 107.55 kJ/kg
Compressor #2:   -WS / mdo 107.55 kJ/kg
Total :   -WS / mdo 215.09 kJ/kg
Notice that when the optimal value of PX is used the compression ratio across each compressor is the same.
Equation 5 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 :
Equation 6 Eqn 6
Plugging values into Eqn 6 yields : ηT 87.86%
Verify : The assumptions made in the solution of this problem cannot be verified with the given information.
Answers : Part a.) -WS / md 263 kJ/kg
Part b.) -WS / md 246 kJ/kg
Better than isentropic, but not adiabatic !
Part c.) -WS / md 189 kJ/kg
Wow, this is really low !
Part d.) -WS / md 215 kJ/kg   ηT 88%
Not bad at all !