8D4 :  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 S_{2} = S_{1} 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 S^{o}_{2}. Once we know S^{o}_{2}, we can use the 2nd Gibbs Equation to determine S_{2}. Use S_{2} to evaluate W_{S,lost}. The 2nd Law Efficiency is the ratio of the actual work to the reversible work that we found in part (a).  
Given:  h_{s}  72%  Find:  W_{s,act}  ???  kW  
P_{1}  105  kPa  W_{s,rev}  ???  kW  
T_{1}  310  K  W_{s,lost}  ???  kW  
P_{2}  4250  kPa  h_{ii}  ???  %  
m_{dot}  1.4  kg/s  
T_{surr}  310  K  
Diagram:  The diagram in the problem statement is adequate.  
Assumptions:  1   The compressor operates at steadystate.  
2   The compressor is adiabatic.  
3   Changes in kinetic and potential energies are negligible.  
Equations / Data / Solve:  
Part a.)  The isentropic
efficiency of an adiabatic compressor is defined by: 

Eqn 1  
We can solve Eqn 1 for W_{S,act} : 

Eqn 2  
Because we assumed air is an ideal gas and we know T_{1} , we can lookup H_{1} in the Ideal Gas Property Table for air.  
H_{1}  97.396  kJ/kg  
Next, we need to determine T_{2S} so we can lookup H_{2S} 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 S^{o}_{T2} : 

Eqn 4  
We can look up S^{o}_{T1} in the Ideal Gas Property Table for air and use it with the known pressures in Eqn 4 to determine S^{o}_{T2} .  
R  8.314  J/molK  S^{o}_{T1}  0.038914  kJ/kgK  
MW  28.97  g/mol  S^{o}_{T2S}  1.1010  kJ/kgK  
Now, we can use S^{o}_{T2} and the Ideal Gas Property Table for air to determine T_{2} and then H_{2} by interpolation :  
T (K)  S^{o} (kJ/kgK)  H^{o} (kJ/kg)  
840  1.0850  658.42  
T_{2S}  1.1010  H_{2S}  T_{2S}  852.19  K  
860  1.1112  680.67  H_{2S}  671.98  kJ/kg  
Now, we can plug values back into Eqn 2 to get :  W_{s,act}  798.04  kJ/kg  
The rate at which work is actually done can be determined using : 

Eqn 5  
W_{s,act}  1117.2  kW  
Method 2: Use the Ideal Gas Relative Pressure.  
When an ideal gas undergoes an isentropic process : 

Eqn 6  
Where P_{r} is the Ideal Gas Relative Pressure, which is a function of T only and we can lookup in the Ideal Gas Property Table for air.  
We can solve Eqn 6 For P_{r}(T_{2S}), as follows : 

Eqn 7  
Lookup P_{r}(T_{1}) and use it in Eqn 7 To determine P_{r}(T_{2S}) :  P_{r}(T_{1})  1.145  
P_{r}(T_{2S})  46.353  
We can now determine T_{2S} by interpolation on the the Ideal Gas Property Table for air.  
Then, we use T_{2S} to determine H_{2S} from the Ideal Gas Property Table for air.  
T (K)  P_{r}  H^{o} (kJ/kg)  
840  43.852  658.42  
T_{2S}  46.353  H_{2S}  T_{2S}  851.95  K  
860  48.039  680.67  H_{2S}  671.71  kJ/kg  
Now, we can plug values back into Eqn 2 to get :  W_{s,act}  797.66  kJ/kg  
Then, use Eqn 5 to determine the rate at which shaft work is actually done :  
W_{s,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 S_{1} from the Ideal Gas Property Table for air but we still need to know S_{2}.  
Since we know (W_{s})_{act}, we can determine H_{2,act} by applying the 1st Law. The appropriate form of the 1st Law for this adiabatic, open system, operating at steadystate with negligible changes in kinetic and potential energies is :  

Eqn 11  
Solve Eqn 11 for H_{2} : 

Eqn 12  
Plug numbers into Eqn 12 :  Method 1, S^{o} :  H_{2}  895.43  kJ/kg  
Method 2, P_{r} :  H_{2}  895.06  kJ/kg  
We already obtained S^{o}_{T1} from the Ideal Gas Property Table for air :  S^{o}_{T1}  0.038914  kJ/kg  
Now, we can interpolate on the Ideal Gas Property Table for air to determine T_{2} and S^{o}_{T2} :  
T (K)  S^{o} (kJ/kgK)  H^{o} (kJ/kg)  
1040  1.3257  883.90  Method 1  Method 2  
T_{2}  S^{o}_{T2}  H_{2}  T_{2}  1050.1  1049.7  K  
1060  1.3475  906.80  S^{o}_{T2}  1.3367  1.3363  kJ/kgK  
Now, we apply Eqn 3 to
the actual turbine to evaluate DS : 

Eqn 13  
Method 1  Method 2  
DS  0.23571  0.23535  kJ/kgK  
Now, plug values back into Eqns 10 & 8 to get :  Method 1  Method 2  
W_{s,lost}  102.30  102.14  kW  
W_{s,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:  h_{ii}  90.84%  90.90%  
Note: We could have determined the reversible work from : 

Eqn 14  
Verify:  Check the Ideal Gas Assumption: 


V_{1} =  24.55  L/mole  V_{2} =  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.)  W_{s,act}  1117  kW  Part b.)  h_{ii}  90.8%  
W_{s,lost}  102.3  kW  
W_{s,rev}  1015  kW 