10E1 :  Brayton Refrigeration Cycle  8 pts 

A Brayton refrigeration cycle uses air as the working fluid and operates between a high pressure of 800 kPa and a low pressure of 120 kPa. The compressor and turbine inlet temperatures are 540 K and 510 K, respectively.  
The turbine is isentropic and the compressor has an isentropic efficiency of 88%. Calculate W_{net}, in kJ/kg, and the coefficient of performance for the cycle.  
Read :  The key here is that this is a Brayton Cycle. Because air behaves as an ideal gas, we can use the Ideal Gas Property Tables for air. Other key points include the fact that both the compressor and turbine are adiabatic, the compressor has an isentropic efficiency of 88% and the turbine is isentropic.  
Given:  P_{1} = P_{2}  120  kPa  P_{4} = P_{3}  800  kPa  
T_{2}  540  K  T_{4}  510  K  
h_{s,comp}  88%  ^{}  
Find:  a.)  W_{cycle}  ?  kJ/kg  b.)  COP_{R}  ?  
Diagram:  


Assumptions:  1   Each component is analyzed as an open system operating at steadystate.  
2   The turbine is 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 exchanged with the 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  299.12  120  86.535  0.0032422  1.0114  
2  540  120  333.66  0.60768  8.3101  
3  938.62  800  768.78  
3s  892.14  800  716.57  1.1521  55.400  
4  510  800  302.17  0.54769  6.742  
Part a.)  There is no shaft work occurring in the HEX's, so W_{cycle} is : 

Eqn 1  
We can determine W_{comp} and W_{turb} by applying the 1st Law to each device.  
The 1st Law equations for a steadystate, singleinlet, singleoutlet adiabatic turbine and compressor with negligible kinetic and potential energy changes are:  

Eqn 2 

Eqn 3  
So, in order to evaluate W_{cycle}, we must first determine the specific enthalpy of all four streams in the cycle. We can immediately find H_{2} and H_{4} in the Ideal Gas Property Table for air because both T_{2} and T_{4} are given.  
H^{o}_{T2}  333.66  kJ/kg  H^{o}_{T4}  302.17  kJ/kg  
Next, let's make use of the fact that the turbine is isentropic (S_{2} = S_{1}) to evaluate H_{1}.  
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 4  
We can solve Eqn 4 for S^{o}_{T1} : 

Eqn 5  
We can lookup S^{o}_{T4} in the Ideal Gas Property Table for air and use it with the known pressures in Eqn 5 to determine S^{o}_{T1}. We can do this because the HEX's are isobaric. P_{1} = P_{2} and P_{3} = P_{4}.  
R  8.314  J/molK  MW  28.97  g/mol  
T (K)  H^{o} (kJ/kg)  S^{o} (kJ/kg  S^{o}_{T4}  0.54769  kJ/kgK  
298.15  85.565  0.00000  S^{o}_{T1}  0.0032422  kJ/kgK  
T_{1}  H^{o}_{T1}  0.0032422  Interpolation yields :  T_{1}  299.12  K  
300  87.410  0.0061681  H^{o}_{T1}  86.535  kJ/kg  
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_{1}) , as follows : 

Eqn 7  
Lookup P_{r}(T_{4}) and use it in Eqn 7 to determine P_{r}(T_{1}) :  P_{r}(T_{4})  6.7424  
P_{r}(T_{1})  1.0114  
We can now determine T_{5} and H_{5} by interpolation on the Ideal Gas Property Table for air.  
T (K)  P_{r}  H^{o} (kJ/kg)  
298  1.0000  85.565  
T_{1}  1.0114  H^{o}_{T1}  Interpolation yields :  T_{1}  299.12  K  
300  1.0217  87.410  H^{o}_{T1}  86.530  kJ/kg  
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_{3}. To do this, we need to use the isentropic efficiency of the compressor.  

Eqn 8  
Solving Eqn 8 for H_{3} gives us: 

Eqn 9  
So, in order to determine H_{3}, we must first determine H_{3S}, the enthalpy of stream 3 IF the turbine were isentropic. We can determine T_{3S} using either 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 10  
We
can solve Eqn 10 for the unknown S^{o}_{T3S} : 

Eqn 11  
We can look up S^{o}_{T2} in the Ideal Gas Property Table for air and use it with the known compression ratio in Eqn 11 to determine S^{o}_{T3}. We can do this because the HEX's are isobaric. P_{1} = P_{2} and P_{3} = P_{3S} = P_{4}.  
R  8.314  J/molK  MW  28.97  g/mol  
S^{o}_{T2}  0.60768  kJ/kgK  S^{o}_{T3S}  1.1521  kJ/kgK  
Now, we can use S^{o}_{T3S} and the Ideal Gas Property Table for air to determine T_{3S} and then H_{3S} by interpolation : 

T (K)  H^{o} (kJ/kg)  S^{o} (kJ/kg  
880  702.98  1.1369  
T_{3S}  H_{3S}  1.1521  Interpolation yields :  T_{3S}  892.14  K  
900  725.37  1.1620  H_{3S}  716.57  kJ/kg  
Method 2: Use the Ideal Gas Relative Pressure.  
When an ideal gas undergoes an isentropic process : 

Eqn 12  
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 12 For P_{r}(T_{3}), as follows : 

Eqn 13  
Lookup P_{r}(T_{2}) and use it in Eqn 13 to determine P_{r}(T_{3S}):  P_{r}(T_{2})  8.3101  
P_{r}(T_{3S})  55.400  
We can now determine T_{3S} by interpolation on the the Ideal Gas Property Table for air.  
Then, we use T_{3} to determine H_{3} from the Ideal Gas Property Table for air.  
T (K)  P_{r}  H^{o} (kJ/kg)  
880  52.530  702.98  
T_{3S}  55.400  H_{3S}  Interpolation yields :  T_{3S}  891.93  K  
900  57.342  725.37  H_{3S}  716.33  kJ/kg  
Since the two methods differ by less than 0.05%, I will use the results from Method 1 in the remaining calculations of this problem.  
Next, we use Eqn 9 to evaluate H_{3} :  H_{3}  768.78  kJ/kg  
T (K)  H^{o} (kJ/kg)  
920  747.82  
T_{3}  768.78  
940  770.33  Interpolation yields :  T_{3}  938.62  K  
Now, we go back to Eqns 2 & 3 to evaluate W_{turb} and W_{comp} :  W_{turb}  215.64  kJ/kg  
W_{comp}  435.12  kJ/kg  
Finally, we plug values into Eqn 1 to evaluate W_{cycle} :  W_{cycle}  219.48  kJ/kg  
Part b.)  We can determine
the coefficient of performance from its definition. 

Eqn 14  
We can evaluate Q_{C} by applying the 1st Law to HEX #2 because Q_{C} = Q_{12}.  
HEX #2 operates at steadystate, has no shaft work interaction and changes in kinetic and potential energies are negligible. So, the appropriate form of the 1st Law is:  

Eqn 16  
Plugging values into Eqn 16 gives us:  Q_{12}  247.13  kJ/kg  
Finally, we can plug values back into Eqn 14 :  COP_{R}  1.126  
Verify:  The assumptions made in the solution of this problem cannot be verified with the given information.  
Answers :  W_{cycle}  219  kJ/kg  COP_{R}  1.13  