7D5 :  CompressedAirDriven Turbine  8 pts 

A small cylinder of compressed air stores energy, just like a battery. When you want to recover the energy from the compressed air, release the air through a turbine and vent the air to the surroundings.  
Use the shaft work to generate electricity. Consider a cylinder that contains air at 400 psia and 1000^{o}F. When the air in the cylinder flows out through the turbine, it produces 250 Btu of shaft work by the time the pressure 

in the cylinder reaches 75 psia. The turbine exhausts to ambient pressure, 14.7 psia. Determine the volume of the cylinder in ft^{3}.  
Assume the air behaves as an ideal gas, the turbine and the cylinder are internally reversible, the entire process is adiabatic and changes in kinetic and potential energies are negligible.  
Read :  The key to this process is that it is entirely isentropic. This will let us determine the initial and final properties of the air in the tank, as well as the properties of the turbine exhaust. The best part is that the properties of the turbine exhaust do not change during the process.  
Given:  P_{1}  400  psia  P_{2}  75  psia  
T_{1}  1000  ^{o}F  P_{out}  14.7  psia  
W_{s}  250  Btu  
Find:  V  ?  ft^{3}  ^{}  
Diagram: 



Assumptions:  1   The system is shown in the diagram.  
2   For the system, heat exchange with the surroundings is negligible.  
3   Changes in kinetic and potential energies are negligible.  
4   The process is reversible.  
5   The air behaves as an ideal gas. This is a very questionable assumption at these pressures, but the problem statement instructed us to make it !  
Equations / Data / Solve:  
We want to evaluate the volume of the tank in the absence of irreversibilities.  
We can begin by applying the 1st Law to this system.  

Eqn 1  
We can simplify Eqn 1 because the process is adiabatic and we have assumed that changes in kinetic and potential energies are negligible and because there is no mass flow into the system.  

Eqn 2  
The mass conservation equation for this process is : 

Eqn 3  
Combining Eqns 1 & 2 yields : 

Eqn 4  
Because the entire process is reversible and adiabatic, the process is isentropic. Therefore, S_{out} can be determined and does not change during the process. Because two intensive properties, S_{out} and P_{out}, are constant, we can conclude that the state of the turbine exhaust is constant and, therefore, T_{out} and H_{out} are constant as well. Therefore, Eqn 4 becomes :  

Eqn 5  
The initial and final moles of air in the tank can be determined from the ideal gas EOS:  

Eqn 6 

Eqn 7  
Apply Eqn 7 to both the initial and final states of the tank contents and combine these with Eqn 5 to get:  

Eqn 8  
Now, we can solve Eqn 8 for the unknown volume of the tank : 

Eqn 9  
An alternate way to express Eqn 9: 

Eqn 9a  
The air remaining in the tank undergoes an isentropic expansion from P_{1}, T_{1} to P_{2}, T_{2}.  
At this point, we can solve this problem by either of two methods. We can apply the 2nd Gibbs Equation for ideal gases and the Ideal Gas Entropy Function or we can use the Ideal Gas Relative Pressure, P_{r}.  
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 10  
We can apply Eqn 10 to the process that the air inside the tank undergoes AND to the process that the air undergoes as it passes through the turbine:  

Eqn 11  
We can solve Eqns 10 & 11 for the unknowns S^{o}_{T2} and S^{o}_{Tout} :  

Eqn 12  

Eqn 13  
We can look up S^{o}_{T1} in the Ideal Gas Property Tables and use it with the known pressures in Eqn 13 to determine S^{o}_{T2} and S^{o}_{Tout} :  
T_{1}  1459.67  ^{o}R  R  1.987  Btu/lbmol^{o}R  
MW  28.97  lb_{m}/lbmol  
T (^{o}R)  S^{o} (Btu/lb_{m}^{o}R)  
1450  0.24808  
1459.67  S^{o}_{T1}  Interpolation yields :  S^{o}_{T1}  0.24981  Btu/lb_{m}^{o}R  
1500  0.25705  S^{o}_{T2}  0.13500  Btu/lb_{m}^{o}R  
S^{o}_{Tout}  0.02323  Btu/lb_{m}^{o}R  
Now, we can use S^{o}_{T2} and S^{o}_{Tout} and the Ideal Gas Property Tables to determine both T_{2} and T_{out} and then U_{1}, U_{2} and U_{out} by interpolation :  
T (^{o}R)  U^{o} (Btu/lb_{m})  
1450  167.28  
1459.67  U^{o}_{1}  Interpolation yields :  U^{o}_{1}  169.17  Btu/lb_{m}  
1500  177.07  
T (^{o}R)  U^{o} (Btu/lb_{m})  S^{o} (Btu/lb_{m}^{o}R)  
930  69.166  0.13406  
T_{2}  U^{o}_{2}  0.13500  Interpolation yields :  T_{2}  933.51  ^{o}R  
940  70.985  0.13674  U^{o}_{2}  69.804  Btu/lb_{m}  
And at the turbine outlet :  
T (^{o}R)  H^{o} (Btu/lb_{m})  S^{o} (Btu/lb_{m}^{o}R)  
590  49.533  0.022643  
T_{out}  H^{o}_{out}  0.02323  T_{out}  591.44  ^{o}R  
600  51.934  0.026678  H^{o}_{out}  49.880  Btu/lb_{m}  
We can plug all of the given and determined values back into Eqns 3, 7, 8 & 9 to evaluate n_{1}, n_{2}, Dn, and finally,V :  
R  10.7316  psiaft^{3} / lbmol^{o}R  V  2.979  ft^{3}  
n_{1} =  0.07606  lbmol  m_{1} =  2.204  lb_{m}  
n_{2} =  0.02230  lbmol  m_{2} =  0.646  lb_{m}  
Dn =  0.05376  lbmol  Dm =  1.558  lb_{m}  
Method 2: Use the Ideal Gas Relative Pressure.  
When an ideal gas undergoes an isentropic process : 

Eqn 14  
Where P_{r} 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 14 For P_{r}(T_{2}), as follows : 

Eqn 15  
Lookup P_{r}(T_{1}) and use it in Eqn 15 To determine P_{r}(T_{2}) :  
T (^{o}R)  P_{r}  
1450  37.310  
1459.67  P_{r}(T_{1})  P_{r}(T_{1})  38.318  
1500  42.521  P_{r}(T_{2})  7.185  
Once we know P_{r}(T_{2}) we can determine T_{2} by interpolation on the the Ideal Gas Property Table.  
We can then use T_{1} and T_{2} to determin U_{1} and U_{2} from the Ideal Gas Property Tables.  
T (^{o}R)  U^{o} (Btu/lb_{m})  T (^{o}R)  P_{r}  U^{o} (Btu/lb_{m})  
1450  167.28  930  7.0696  69.166  
1459.67  U^{o}_{1}  T_{2}  7.185  U^{o}_{2}  
1500  177.07  940  7.3513  70.985  
Interpolation yields :  ^{}  Interpolation yields :  
U^{o}_{1}  169.17  Btu/lb_{m}  T_{2}  934.08  ^{o}R  
U^{o}_{2}  69.909  Btu/lb_{m}  
Because the turbine is also an isentropic process, we can determine the relative pressure of the turbine effluent:  

Eqn 16  Rearranging: 

Eqn 17  
P_{r}(T_{out})  1.4082  
Now, we can use P_{r}(T_{out}) to determine T_{2} and then H_{out} using the Ideal Gas Property Tables :  
T (^{o}R)  P_{r}  H^{o} (Btu/lb_{m})  
590  1.3914  49.533  
T_{out}  1.4082  H^{o}_{out}  Interpolation yields :  T_{out}  591.99  ^{o}R  
600  1.4758  51.934  H^{o}_{out}  50.010  Btu/lb_{m}  
We can plug all of the given and determined values back into Eqns 3, 7, 8 & 9 to evaluate n_{1}, n_{2}, Dn, and finally,V :  
R  10.7316  psiaft^{3} / lbmol^{o}R  V  2.982  ft^{3}  
n_{1} =  0.07614  lbmol  m_{1} =  2.206  lb_{m}  
n_{2} =  0.02231  lbmol  m_{2} =  0.646  lb_{m}  
Dn =  0.05383  lbmol  Dm =  1.560  lb_{m}  
Verify:  The ideal gas assumption needs to be verified.  
We need to determine
the specific volume at each state and check if: 


Air can be considered a diatomic gas.  
Solving the Ideal Gas EOS for molar volume yields : 


Use :  R  10.7316  psiaft^{3} / lbmol^{o}R  
V_{1}  39.16  ft^{3}/lbmol  
V_{2}  133.57  ft^{3}/lbmol  V_{3}  431.78  ft^{3}/lbmol  
The specific volume at state 2 and at the turbine effluent is greater than 80 ft^{3}/lbmol. Air can be considered to be a diatomic gas, so the ideal gas assumption is valid here. The ideal gas assumption is not valid in state 1 and this makes the solution somewhat questionable, but we were instructed to make the ideal gas assumption in the problem statement.  
Answers :  Method 1  Method 2  
The volume of the tank is:  V  2.979  2.982  ft^{3}  
The difference between the two methods is caused by the following issues (ranked from most important to least important).  
1   Errors associated with linearly interpolating between values of a functions that are not really linear.  
3   Roundoff error in the Ideal Gas Property Tables. 