10E2 :  Ideal Regenerative Brayton Refrigeration Cycle  9 pts 

Calculate the refrigeration capacity, Q_{C}, in tons of refrigeration, and the coefficient of performance of the regenerative Brayton Cycle, shown below.  


Assume that both the compressor and the turbine are isentropic.  
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:  V_{1}  1250  ft^{3}/min  P_{2}  70  psia  
T_{1}  747  ^{o}R  T_{3}  775  ^{o}R  
P_{1}  25  psia  T_{4}  747  ^{o}R  
Find:  a.)  Q_{in}  ???  tons  
b.)  COP_{R}  ???  
Diagram: 


Assumptions:  1   Each process is analyzed as an open system operating at steadystate.  
2   The turbine and compressor are isentropic.  
3   There are no pressure drops for flow through the heat exchangers.  
4   Kinetic and potential energy changes are negligible.  
5   The working fluid is air modeled as an ideal gas.  
6   There is no heat transfer from the heat exchanger to its surroundings.  
Equations / Data / Solve:  
Stream  T (^{o}R) 
P (psia) 
H^{o} (Btu/lb_{m}) 
S^{o} (Btu/lb_{m}^{o}R) 
P_{r}  
1  747  25  87.598  3.2050  
2  992.7  70  8.9740  
3  775  70  94.467  
4  747  70  87.598  0.079819  3.2050  
5  557.8  25  41.824  0.0091996  1.1446  
6  25  80.728  
Part a.)  The refrigeration capacity is how much heat the refrigerator can remove from the cold reservoir. So, we need to determine Q_{in} (from the diagram above). We can accomplish this by applying the 1st Law to HEX #1. The 1st Law for a steadystate, singleinlet, singleoutlet, 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 lb_{f }/lb_{m}^{o}R  
MW  29.00  lb_{m}/lbmole  m_{dot}  113.07  lb_{m}/min  
Next, let's determine H_{5}. 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.  
Apply the 2nd Gibbs Equation for Ideal Gases in terms of the Ideal Gas Entropy Function:  

Eqn 3  
We can solve Eqn 3 for S^{o}_{T5} : 

Eqn 4  
We can lookup S^{o}_{T4} in the Ideal Gas Property Table for air and use it with the known pressures in Eqn 4 to determine S^{o}_{T5}. We can do this because the HEX's are isobaric. P_{1} = P_{5} = P_{6} and P_{2} = P_{3} = P_{4}.  
R  1.987  Btu/lbmole^{o}R  MW  28.97  lb_{m}/lbmole  
T (^{o}R)  H^{o} (Btu/lb_{m})  S^{o} (Btu/lb_{m}^{o}R)  
740.00  85.884  0.077519  
747  H^{o}_{T4}  S^{o}_{T4}  H^{o}_{T4}  87.598  Btu/lb_{m}  
750  88.332  0.080805  S^{o}_{T4}  0.079819  Btu/lb_{m}^{o}R  
T (^{o}R)  H^{o} (Btu/lb_{m})  S^{o} (Btu/lb_{m}^{o}R)  
550  39.963  0.005848  S^{o}_{T5}  0.0091996  Btu/lb_{m}^{o}R  
T_{5}  H^{o}_{T5}  0.009200  T_{5}  557.79  ^{o}R  
560  42.351  0.010149  H^{o}_{T5}  41.824  Btu/lb_{m}  
Method 2: Use the Ideal Gas Relative Pressure.  
When an ideal gas undergoes an isentropic process : 

Eqn 5  
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 5 For P_{r}(T_{5}) , as follows : 

Eqn 6  
Lookup P_{r}(T_{4}) and use it in Eqn 6 to determine P_{r}(T_{5}) :  
T (^{o}R)  P_{r}  H^{o} (Btu/lb_{m})  
740  3.0985  85.884  H^{o}_{T4}  87.598  Btu/lb_{m}  
747  P_{r}(T_{4})  H^{o}_{T4}  P_{r}(T_{4})  3.2050  
750  3.2506  88.332  P_{r}(T_{5})  1.1446  
We can now determine T_{5} and H_{5} by interpolation on the the Ideal Gas Property Table for air.  
T (^{o}R)  P_{r}  H^{o} (Btu/lb_{m})  
550  1.0891  39.963  
T_{5}  1.1446  H^{o}_{T5}  T_{5}  557.88  ^{o}R  
560  1.1596  42.351  H^{o}_{T5}  41.845  Btu/lb_{m}  
Since the two methods differ by less than 0.01%, I will use the results from Method 1 in the remaining calculations of this problem.  
Next, we need to evaluate H_{6}. To do that, we need to apply the 1st Law to the Regenerator.  
The 1st Law for a steadystate, multipleinlet, multipleoutlet, adiabatic HEX with no shaft work and negligible kinetic and potential energy changes is:  

Eqn 7  
Solve Eqn 7 for H_{6} : 

Eqn 8  
We already found H_{4}, so we need to find H_{1} and H_{3}. We can do this by interpolation in the Ideal Gas Property Table for air because we know both T_{1} and T_{3}. Because T_{1} = T_{4}, H_{1} = H_{4} and we already found H_{4}. So, all we need to work on is H_{3}.  
T (^{o}R)  H^{o} (Btu/lb_{m})  
770  93.238  H^{o}_{T1}  87.598  Btu/lb_{m}  
775  H^{o}_{T3}  
780  95.696  H^{o}_{T3}  94.467  Btu/lb_{m}  
Now, we plug values back into Eqn 8 :  H^{o}_{T6}  80.728  Btu/lb_{m}  
At last, we can plug numbers back into Eqn 1:  Q_{in}  4399.1  Btu/min  
Converting units to tons of refrigeration yields the answer to part (a) :  1 ton =  200  Btu/min  
Q_{in}  21.995  tons  
Part b.)  We can determine the coefficient of performance from its definition.  

Eqn 9  
Where : 

Eqn 10  
We can determine W_{comp} and W_{turb} by applying the 1st Law to the compressor and the turbine separately.  
The 1st Law for a steadystate, singleinlet, singleoutlet 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 H_{2}. 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.  
Apply the 2nd Gibbs Equation for Ideal Gases in terms of the Ideal Gas Entropy Function:  

Eqn 13  
Solving Eqn 13 for S^{o}_{T2} yields : 

Eqn 14  
Because T_{1} = T_{4}, S^{o}_{T1} = S^{o}_{T4}. Then, we can use Eqn 4 to determine S^{o}_{T2}. We can do this because the HEX's are isobaric. P_{1} = P_{5} = P_{6} and P_{2} = P_{3} = P_{4}.  
S^{o}_{T1}  0.079819  Btu/lb_{m}^{o}R  S^{o}_{T2}  0.15044  Btu/lb_{m}^{o}R  
T (^{o}R)  H^{o} (Btu/lb_{m})  S^{o} (Btu/lb_{m}^{o}R)  
990  147.98  0.14976  
T_{2}  H^{o}_{T2}  0.15044  T_{2}  992.67  ^{o}R  
1000  150.50  0.15230  H^{o}_{T2}  148.65  Btu/lb_{m}  
Method 2: Use the Ideal Gas Relative Pressure.  
When an ideal gas undergoes an isentropic process : 

Eqn 15  
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 15 For P_{r}(T_{2}) , as follows : 

Eqn 16  
P_{r}(T_{1}) = P_{r}(T_{4}) because T_{1} = T_{4}. Use P_{r}(T_{1}) in Eqn 16 to determine P_{r}(T_{2}) :  
P_{r}(T_{1})  3.2050  P_{r}(T_{2})  8.9740  
We can now determine T_{5} and H_{5} by interpolation on the the Ideal Gas Property Table for air.  
T (^{o}R)  P_{r}  H^{o} (Btu/lb_{m})  
990  8.8893  147.981  
T_{2}  8.9740  H^{o}_{T2}  T_{2}  992.53  ^{o}R  
1000  9.2240  150.502  H^{o}_{T2}  148.62  Btu/lb_{m}  
Since the two methods differ by less than 0.01%, 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:  
W_{turb}  5175.83  Btu/min  W_{cycle}  1728.08  Btu/min  
W_{comp}  6903.91  Btu/min  COP_{R}  2.546  
Verify:  The assumptions made in the solution of this problem cannot be verified with the given information.  
Answers :  a.)  Q_{in}  22.0  tons  b.)  COP_{R}  2.55 