# Example Problem with Complete Solution

8D-4 : 2nd Law Efficiency and Lost Work in an Air Compressor 6 pts
For the compressor shown below, calculate : a.)  The actual, reversible and lost work in kW
b.) The 2nd Law efficiency
Assume
air behaves as an ideal gas and the temperature of the surroundings is 310 K.

Read : We can calculate the isentropic work requirement of the compressor because S2 = S1 gives us the additional intensive variable value that we need to fix the state of the effluent stream.  Then we can calculate the actual work from the isentropic work using the isentropic efficiency.
We need to know the actual entropy of the effluent in order to determine the reversible work for the turbine.  We can use the actual work and the 1st Law to determine the actual enthalpy of the effluent.  This gives us the second intensive property we need in order to interpolate on the Ideal Gas Property Table for air to evaluate So2. Once we know So2, we can use the 2nd Gibbs Equation to determine S2. Use S2 to evaluate WS,lost. The 2nd Law Efficiency is the ratio of the actual work to the reversible work that we found in part (a).
Given: hs 72% Find: Ws,act ??? kW
P1 105 kPa Ws,rev ??? kW
T1 310 K Ws,lost ??? kW
P2 4250 kPa hii ??? %
mdot 1.4 kg/s
Tsurr 310 K
Diagram: The diagram in the problem statement is adequate.
Assumptions: 1 - The compressor operates at steady-state.
2 - The compressor is adiabatic.
3 - Changes in kinetic and potential energies are negligible.
Equations / Data / Solve:
Part a.) The isentropic efficiency of an Eqn 1
We can solve Eqn 1 for WS,act : Eqn 2
Because we assumed air is an ideal gas and we know T1 , we can look-up H1 in the Ideal Gas Property Table for air.
H1 97.396 kJ/kg
Next, we need to determine T2S so we can look-up H2S in the Ideal Gas Properties Table for air.
We can either use the Ideal Gas Entropy Function or the Relative Pressure Method.
Method 1: Use the Ideal Gas Entropy Function.
The 2nd Gibbs Equation for Ideal Gases in terms of the Ideal Gas Entropy Function is : Eqn 3
We can solve Eqn 3 for the unknown SoT2 : Eqn 4
We can look up SoT1 in the Ideal Gas Property Table for air and use it with the known pressures in Eqn 4 to determine SoT2 .
R 8.314 J/mol-K SoT1 0.038914 kJ/kg-K
MW 28.97 g/mol SoT2S 1.1010 kJ/kg-K
Now, we can use SoT2 and the Ideal Gas Property Table for air to determine T2 and then H2 by interpolation :
T (K) So (kJ/kg-K) Ho (kJ/kg)
840 1.0850 658.42
T2S 1.1010 H2S T2S 852.19 K
860 1.1112 680.67 H2S 671.98 kJ/kg
Now, we can plug values back into Eqn 2 to get : Ws,act -798.04 kJ/kg
The rate at which work is actually done can be determined using : Eqn 5
Ws,act -1117.2 kW
Method 2: Use the Ideal Gas Relative Pressure.
When an ideal gas undergoes an isentropic process : Eqn 6
Where Pr is the Ideal Gas Relative Pressure, which is a function of T only and we can look-up in the Ideal Gas Property Table for air.
We can solve Eqn 6 For Pr(T2S), as follows : Eqn 7
Look-up Pr(T1) and use it in Eqn 7 To determine Pr(T2S) : Pr(T1) 1.145
Pr(T2S) 46.353
We can now determine T2S by interpolation on the the Ideal Gas Property Table for air.
Then, we use T2S to determine H2S from the Ideal Gas Property Table for air.
T (K) Pr Ho (kJ/kg)
840 43.852 658.42
T2S 46.353 H2S T2S 851.95 K
860 48.039 680.67 H2S 671.71 kJ/kg
Now, we can plug values back into Eqn 2 to get : Ws,act -797.66 kJ/kg
Then, use Eqn 5 to determine the rate at which shaft work is actually done :
Ws,act -1116.73 kW
The two key equations for determining
lost and reversible work are : Eqn 8 Eqn 9
Because the process is adiabatic, Eqn 9 simplifies to : Eqn 10
We can determine S1 from the Ideal Gas Property Table for air but we still need to know S2.
Since we know (Ws)act, we can determine H2,act by applying the 1st Law.  The appropriate form of the 1st Law for this adiabatic, open system, operating at steady-state with negligible changes in kinetic and potential energies is : Eqn 11
Solve Eqn 11 for H2 : Eqn 12
Plug numbers into Eqn 12 : Method 1, So : H2 895.43 kJ/kg
Method 2, Pr :   H2 895.06 kJ/kg
We already obtained SoT1 from the Ideal Gas Property Table for air : SoT1 0.038914 kJ/kg
Now, we can interpolate on the Ideal Gas Property Table for air to determine T2 and SoT2 :
T (K) So (kJ/kg-K) Ho (kJ/kg)
1040 1.3257 883.90 Method 1 Method 2
T2 SoT2 H2 T2 1050.1 1049.7 K
1060 1.3475 906.80 SoT2 1.3367 1.3363 kJ/kg-K
Now, we apply Eqn 3 to the
actual turbine to evaluate DS : Eqn 13
Method 1 Method 2
DS 0.23571 0.23535 kJ/kg-K
Now, plug values back into Eqns 10 & 8 to get : Method 1 Method 2
Ws,lost 102.30 102.14 kW
Ws,rev -1015.0 -1015.1 kW
Part b.) The 2nd Law Efficiency of a turbine is defined as: Eqn 14
Method 1 Method 2
Plugging values into Eqn 14 gives us: hii 90.84% 90.90%
Note: We could have determined the reversible work from : Eqn 14
Verify: Check the Ideal Gas Assumption: V1 = 24.55 L/mole V2 = 2.05 L/mole
Air can be considered to be a diatomic gas, but the molar volume at state 2 is not greater than 5 L/mole.  So, it is not accurate to treat the air as an ideal gas.
Answers : Part a.) Ws,act -1117 kW Part b.) hii 90.8%
Ws,lost 102.3 kW
Ws,rev -1015 kW 