Example Problem with Complete Solution

10E-1 : Ideal Regenerative Brayton Refrigeration Cycle 9 pts
Consider a Brayton refrigeration cycle with a regenerative heat exchanger. Air enters the compressor at 738oR, 20 psia and is compressed isentropically to 60 psia. Compressed air enters the regenerative heat exchanger at 756oR and is cooled to 738oR before entering the turbine. The expansion through the turbine is isentropic. If the volumetric flow rate at the compressor inlet is 1000 ft3/min, calculate...
a.) the refrigeration capacity, in tons.      
b.) the coefficient of performance.      
               
Read: Make all the usual assumptions for the standard Brayton cycle. Use the ideal gas EOS to convert the volumetric flow rate to a mass flow rate. Determine the specific enthalpy for each stream and then use the 1st Law and the definition of the COP to answer the questions.
Given: V1 1000 ft3/min P2 60 psia
T1 738 oR T3 756 oR
P1 20 psia T4 738 oR
Find : a.) Qin ? tons
b.) COPR ?
Diagrams:
Assumptions :
- Each component is analyzed as an open system operating at steady-state.
- The turbine and compressor processes are isentropic.
- There are no pressure drops for flow through the heat exchangers.
- Kinetic and potential energy changes are negligible.
- The working fluid is air modeled as an ideal gas.
- There is no heat transfer from the heat exchanger to it surroundings.
Equations / Data / Solve :
Stream
T
(oR)
P
(psia)
Ho
(Btu/lbm)
So
(Btu/lbm-oR)
1 738.0 20 85.395  
2   60    
3 756 60 89.803  
4 738.0 60 85.395 0.076853
5 540.2 20 37.637 0.00158
6   20 80.986  
Part a.) The refrigeration capacity is how much heat the refrigerator can remove from the cold reservoir.  So, we need to determine Qin (from the diagram above).  We can accomplish this by applying the 1st Law to HEX #1.  The 1st Law for a steady-state, single-inlet, single outlet, HEX with no shaft work and negligible kinetic and potential energy changes is:
Eqn 1
First, let's determine the mass flow rate of the working fluid.  The key is that we know the volumetric flow rate and the T & P at the compressor inlet.  And, remember that we have assumed that the air behaves as an ideal gas.
Eqn 2
Plugging values into Eqn 2 yields:
R 1545 ft lbf /lbm-oR
MW 29.00 lbm/lbmole mdot 73.250 lb/min
Next, let's determine H5.  We can accomplish this because we assumed the turbine is isentropic.  We can either use the Ideal Gas Entropy Function and the 2nd Gibbs Equation or we can use Relative Properties.   Both methods are presented here.
Method 1: Use the Ideal Gas Entropy Function and the 2nd Gibbs Equation.
2nd Gibbs Equation for Ideal Gases in terms of the Ideal Gas Entropy Function:
   
Eqn 3
   
We can solve Eqn 3 SoT5 : Eqn 4
   
We can look up SoT4 in the Ideal Gas Property Table for air and use it with the known pressures in Eqn 4 to determine SoT5.  We can do this because the HEX's are isobaric.  P1 = P5 = P6 and P2 = P3 = P4.
   
R 1.987 Btu/lbmole-oR MW 29.00 lbm/lbmole
   
T (oR) Ho (Btu/lbm) So (Btu/lbmoR)  
730 83.439 0.074192  
738 HoT4 SoT4 HoT4 85.395 Btu/lbm
740 85.884 0.077519 SoT4 0.076853 Btu/lbm-oR
   
T (oR) Ho (Btu/lbm) So (Btu/lbmoR)  
540 37.580 0.001473 SoT5 0.001579 Btu/lbm-oR
T5 HoT5 0.001579 T5 540.24 oR
550 39.963 0.005848 HoT5 37.637 Btu/lbm
             

Method 2: Use the Ideal Gas Relative Pressure.
             
When an ideal gas undergoes an isentropic process :  
   
Eqn 5
   
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 Eqns 5 For Pr(T5) , as follows : Eqn 6
   
Look-up Pr(T4) and use it in Eqn 6 to determine Pr(T5) :  
   
T (oR) Pr Ho (Btu/lbm)  
730 2.9517 83.439 HoT4 85.395 Btu/lbm
738 Pr(T4) HoT4 Pr(T4) 3.0691  
740 3.0985 85.884 Pr(T5) 1.0230  
   
We can now determine T5 and H5 by interpolation on the the Ideal Gas Property Table for air.
T (oR) Pr Ho (Btu/lbm)  
540 1.0217 37.580  
T5 1.0230 HoT5 T5 540.20 oR
550 1.0891 39.963 HoT5 37.628 Btu/lbm
             
Since the two methods differ by only about 0.03%, I will use the results from Method 1 in the remaining calculations of this problem.
Next, we need to evaluate H6.  To do that, we need to apply the 1st Law to the Regenerator.
The 1st Law for a steady-state, multiple-inlet, multiple-outlet, adiabatic HEX with no shaft work and negligible kinetic and potential energy changes is:
Eqn 7
Solve Eqn 7 for H6 : Eqn 8
We already found H4, so we need to find H1 and H3.  We can do this by interpolation in the Ideal Gas Property Table for air because we know both T1 and T3.  Because T1 = T4, H1 = H4 and we already found H4.  So, all we need to work on is H3.
T (oR) Ho (Btu/lbm)
750 88.332 HoT1 85.395 Btu/lbm
756 HoT3
760 90.784 HoT3 89.803 Btu/lbm
Now, we plug values back into Eqn 8 : HoT6 80.986 Btu/lbm
At last, we can plug numbers back into Eqn 1: Qin 3175.3 Btu/min
Converting units to tons of refrigeration yields the answer to part (a) :
1 ton = 200 Btu/min Qin 15.877 tons
Part b.) We can determine the coefficient of performance from its definition.
Eqn 9
Where : Eqn 10
We can determine Wcomp and Wturb by applying the 1st Law to each.
The 1st Law for a steady-state, single-inlet, single outlet adiabatic turbine and compressor with  negligible kinetic and potential energy changes are:
Eqn 11
Eqn 12
We know all of the values we need except H2.  We can determine it because we know the compressor is isentropic.  We can use either Method 1 or 2 described above.
Method 1: Use the Ideal Gas Entropy Function and the 2nd Gibbs Equation.
2nd Gibbs Equation for Ideal Gases in terms of the Ideal Gas Entropy Function:
   
Eqn 13
   
Solving Eqn 13 for SoT2 yields : Eqn 14
   
Because T1 = T4, SoT1 = SoT4.  Then, we can use Eqn 4 to determine SoT2.  We can do this because the HEX's are isobaric.  P1 = P5 = P6 and P2 = P3 = P4.
SoT1 0.076853 Btu/lbm-oR SoT2 0.15213 Btu/lbm-oR
   
T (oR) Ho (Btu/lbm) So (Btu/lbmoR)  
990 147.98 0.14976  
T2 HoT2 0.15213 T2 999.32 oR
1000 150.50 0.15230 HoT2 150.33 Btu/lbm
             
Method 2: Use the Ideal Gas Relative Pressure.
When an ideal gas undergoes an isentropic process :    
   
Eqn 15
   
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 Eqns 5 For Pr(T2) , as follows : Eqn 16
   
Pr(T1) = Pr(T4) because T1 = T4.  Use Pr(T1) in Eqn 16 to determine Pr(T2) :
   
Pr(T1) 3.0691 Pr(T2) 9.2074
   
We can now determine T5 and H5 by interpolation on the the Ideal Gas Property Table for air.
T (oR) Pr Ho (Btu/lbm)  
990 8.8893 147.98  
T2 9.2074 HoT2 T5 999.50 oR
1000 9.2240 150.50   HoT5 150.38 Btu/lbm
Since the two methods differ by only about 0.03%, I will use the results from Method 1 in the remaining calculations of this problem.
At last we return to Eqns 11,12, 10 & 9, in that order:
Wturb 3498.21 Btu/min Wcycle -1258.16 Btu/min
Wcomp -4756.37 Btu/min COPR 2.524
Verify : The assumptions made in the solution of this problem cannot be verified with the given information.
Answers: a.) Qin 15.9 tons
b.) COPR 2.52