Example Problem with Complete Solution

7D-5 : Compressed-Air-Driven 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 1000oF. 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 ft3. 
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: P1 400 psia P2 75 psia
T1 1000 oF Pout 14.7 psia
Ws 250 Btu
Find: V ? ft3
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, Sout can be determined and does not change during the process. Because two intensive properties, Sout and Pout, are constant, we can conclude that the state of the turbine exhaust is constant and, therefore,  Tout and Hout 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 P1, T1 to P2, T2. 
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, Pr. 
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 SoT2 and SoTout :
Eqn 12
Eqn 13
We can look up SoT1 in the Ideal Gas Property Tables and use it with the known pressures in Eqn 13 to determine SoT2 and SoTout :
T1 1459.67 oR R 1.987 Btu/lbmol-oR
MW 28.97 lbm/lbmol
T (oR) So (Btu/lbm-oR)
1450 0.24808
1459.67 SoT1 Interpolation yields : SoT1 0.24981 Btu/lbm-oR
1500 0.25705 SoT2 0.13500 Btu/lbm-oR
SoTout 0.02323 Btu/lbm-oR
Now, we can use SoT2 and SoTout and the Ideal Gas Property Tables to determine both T2 and Tout and then U1, U2 and Uout by interpolation :
T (oR) Uo (Btu/lbm)
1450 167.28
1459.67 Uo1 Interpolation yields : Uo1 169.17 Btu/lbm
1500 177.07
T (oR) Uo (Btu/lbm) So (Btu/lbm-oR)
930 69.166 0.13406
T2 Uo2 0.13500 Interpolation yields : T2 933.51 oR
940 70.985 0.13674 Uo2 69.804 Btu/lbm
And at the turbine outlet :
T (oR) Ho (Btu/lbm) So (Btu/lbm-oR)
590 49.533 0.022643
Tout Hoout 0.02323 Tout 591.44 oR
600 51.934 0.026678 Hoout 49.880 Btu/lbm
We can plug all of the given and determined values back into Eqns 3, 7, 8 & 9 to evaluate n1, n2, Dn, and finally,V :
R 10.7316 psia-ft3 / lbmol-oR V 2.979 ft3
n1 = 0.07606 lbmol m1 = 2.204 lbm
n2 = 0.02230 lbmol m2 = 0.646 lbm
Dn = -0.05376 lbmol Dm = -1.558 lbm
Method 2: Use the Ideal Gas Relative Pressure.
When an ideal gas undergoes an isentropic process :
Eqn 14
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 14 For Pr(T2), as follows :
Eqn 15
Look-up Pr(T1) and use it in Eqn 15 To determine Pr(T2) :
T (oR) Pr
1450 37.310
1459.67 Pr(T1) Pr(T1) 38.318
1500 42.521 Pr(T2) 7.185
Once we know Pr(T2) we can determine T2 by interpolation on the the Ideal Gas Property Table.
We can then use T1 and T2 to determin U1 and U2 from the Ideal Gas Property Tables.
T (oR) Uo (Btu/lbm) T (oR) Pr Uo (Btu/lbm)
1450 167.28 930 7.0696 69.166
1459.67 Uo1 T2 7.185 Uo2
1500 177.07 940 7.3513 70.985
Interpolation yields : Interpolation yields :
Uo1 169.17 Btu/lbm T2 934.08 oR
Uo2 69.909 Btu/lbm
Because the turbine is also an isentropic process, we can determine the relative pressure of the turbine effluent:
Eqn 16 Rearranging:
Eqn 17
Pr(Tout) 1.4082
Now, we can use Pr(Tout) to determine T2 and then Hout using the Ideal Gas Property Tables :
T (oR) Pr Ho (Btu/lbm)
590 1.3914 49.533
Tout 1.4082 Hoout Interpolation yields : Tout 591.99 oR
600 1.4758 51.934 Hoout 50.010 Btu/lbm
We can plug all of the given and determined values back into Eqns 3, 7, 8 & 9 to evaluate n1, n2, Dn, and finally,V :
R 10.7316 psia-ft3 / lbmol-oR V 2.982 ft3
n1 = 0.07614 lbmol m1 = 2.206 lbm
n2 = 0.02231 lbmol m2 = 0.646 lbm
Dn = -0.05383 lbmol Dm = -1.560 lbm
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 psia-ft3 / lbmol-oR
V1 39.16 ft3/lbmol
V2 133.57 ft3/lbmol V3 431.78 ft3/lbmol
The specific volume at state 2 and at the turbine effluent is greater than 80 ft3/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 ft3
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 - Round-off error in the Ideal Gas Property Tables.