Example Problem with Complete Solution

10E-2 :
Brayton Refrigeration Cycle 
8 pts
In a Brayton Refrigeration Cycle, air is compressed adiabatically from 100 kPa and 540 K to 500 kPa.  Air enters the internally reversible turbine at 510 K and expands adiabatically to 100 kPa.  The isentropic efficiency of the compressor is 85 %.
                   
a.) Calculate the net work per unit mass of air flow for the cycle, in kJ/kg.
b.) Determine the coefficient of performance.    
                   
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 85% and the turbine is isentropic.
Given: P1 = P2 100 kPa P4 = P3 500 kPa
T2 540 K T4   510 K
ηs,comp 85%
Find : a.) Wcycle ? kJ/kg
b.) COPR ?
Diagrams:
 
Assumptions : - Each component is analyzed as an open system operating at steady-state.
- The turbine is 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 exchanged with the surroundings.
Equations / Data / Solve :
Part a.) There is no shaft work occurring
in the HEX's, so Wcycle is :		 Eqn 1
We can determine Wcomp and Wturb by applying the 1st Law to each device.
The 1st Law equations for a steady-state, single-inlet, single outlet adiabatic turbine and compressor with  negligible kinetic and potential energy changes are:
   
Eqn 2 Eqn 3
So, in order to evaluate Wcycle, we must first determine the specific enthalpy of all four streams in the cycle.  We can immediately find H2 and H4 in the Ideal Gas Property Table for air because both T2 and T4 are given.
HoT2 333.66 kJ/kg HoT4 302.17 kJ/kg
Next, let's make use of the fact that the turbine is isentropic (S2 = S1) to evaluate H1.
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 4
We can solve Eqn 4 forSoT1: Eqn 5
We can look upSoT4in the Ideal Gas Property Table for air and use it with the known pressures in Eqn 5 to determineSoT1.  We can do this because the HEX's are isobaric.  P1 = P2 and P3 = P4.
R 8.314 J/mol-K MW 29.00 g/mol
T (K) Ho (kJ/kg) So (kJ/kg-K) SoT4 0.54769 kJ/kg-K
320 107.41 0.07070 SoT1 0.086281 kJ/kg-K
T1 HoT1 0.086281 T1 325.04 K
330 117.45 0.10159 HoT1 112.47 kJ/kg
Method 2: Use the Ideal Gas Relative Pressure.
When an ideal gas undergoes
an isentropic process :		 Eqn 6
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 6 For Pr(T1) , as follows : Eqn 7
Look-up Pr(T4)and use it in Eqn 7
 to determine Pr(T1):		 Pr(T4) 6.7424
Pr(T1) 1.3485
We can now determine T5 and H5 by interpolation on the Ideal Gas Property Table for air.
T (K) Pr Ho (kJ/kg)
320 1.2794 107.41
T1 1.3485 HoT1 T1 324.75 K
330 1.4247 117.45 HoT1 112.18 kJ/kg
Since the two methods differ by less than 0.3%, I will use the results from Method 1 in the remaining calculations of this problem.
Next, we need to evaluate H3.  To do this, we need to use the isentropic efficiency of the compressor.
Eqn 8
Solving Eqn 8 for H3 gives us: Eqn 9
So, in order to determine H3, we must first determine H3S, the enthalpy of stream 3 IF the turbine were isentropic.  We can determine T3S 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.
The Gibbs Equation for Ideal Gases in terms of the Ideal Gas Entropy Function is : Eqn 10
We can solve Eqn 10
 for the unknown SoT3S :		 Eqn 11
We can look up SoT2 in the Ideal Gas Property Table for air and use it with the known compression ratio in Eqn 11 to determine SoT3.  We can do this because the HEX's are isobaric.  P1 = P2 and P3 = P3S = P4.
R 8.314 J/mol-K MW 29.00 g/mol
SoT2 0.60768 kJ/kg-K SoT3S 1.06909 kJ/kg-K
Now, we can use SoT3S and the Ideal Gas Property Table for air to determine T3S and then H3S by interpolation :
T (K) Ho (kJ/kg) So (kJ/kg-K)
820 636.25 1.0583
T3S H3S 1.06909 T3S 827.99 K
830 647.33 1.0718 H3S 645.11 kJ/kg
Method 2: Use the Ideal Gas Relative Pressure.
When an ideal gas undergoes
an isentropic process :		 Eqn 12
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 12 For Pr(T3S), as follows : Eqn 13
Look-up Pr(T2) and use it in
Eqn 13 to determine Pr(T3S):		 Pr(T2) 8.3101
Pr(T3S) 41.5505
We can now determine T3S by interpolation on the the Ideal Gas Property Table for air.
Then, we use T3 to determine H3 from the Ideal Gas Property Table for air.
T (K) Pr Ho (kJ/kg)
820 39.954 636.25
T3S 41.5505 H3S T3S 828.34 K
830 41.868 647.33 H3S 645.49 kJ/kg
Since the two methods differ by less than 0.06%, I will use the results from Method 1 in the remaining calculations of this problem.
Next, we use Eqn 9 to evaluate H3 : H3 700.07 kJ/kg
Now, we go back to Eqns 2 & 3 to
evaluate Wturb and Wcomp :		 Wturb 189.70 kJ/kg
Wcomp -366.41 kJ/kg
Finally, we plug values into Eqn 1 to evaluate Wcycle: Wcycle -176.71 kJ/kg
Part b.) We can determine the coefficient of performance from its definition. Eqn 14
We can evaluate QC by applying the 1st Law to HEX #2 because QC
HEX #2 operates at steady-state, has no shaft work interaction and changes in kinetic and potential energies are negligible.  So, the appropriate form of the 1st Law is:
 
Eqn 15
Plugging values into Eqn 16 gives us: Q12 221.19 kJ/kg
Finally, we can plug values back into Eqn 14 : COPR 1.252
Verify : The assumptions made in the solution of this problem cannot be verified with the given information.
Answers :   Wcycle -177 kJ/kg COPR 1.25
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