# Example Problem with Complete Solution

10E-1 : 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 Wnet, 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: P1 = P2 120 kPa P4 = P3 800 kPa
T2 540 K T4 510 K
hs,comp 88%
Find: a.) Wcycle ? kJ/kg b.) COPR ?
Diagram: Assumptions: 1 - Each component is analyzed as an open system operating at steady-state.
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
(oR)
P
(psia)
Ho
(Btu/lbm)
So
(Btu/lbm-oR)
Pr
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 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.

Apply the 2nd Gibbs Equation for Ideal Gases in terms of the Ideal Gas Entropy Function: Eqn 4

We can solve Eqn 4 for SoT1 : Eqn 5

We can look-up SoT4 in the Ideal Gas Property Table for air and use it with the known pressures in Eqn 5 to determine SoT1.  We can do this because the HEX's are isobaric.  P1 = P2 and P3 = P4.

R 8.314 J/mol-K MW 28.97 g/mol

T (K) Ho (kJ/kg) So (kJ/kg- SoT4 0.54769 kJ/kg-K
298.15 85.565 0.00000 SoT1 0.0032422 kJ/kg-K
T1 HoT1 0.0032422 Interpolation yields : T1 299.12 K
300 87.410 0.0061681 HoT1 86.535 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) 1.0114

We can now determine T5 and H5 by interpolation on the Ideal Gas Property Table for air.

T (K) Pr Ho (kJ/kg)
298 1.0000 85.565
T1 1.0114 HoT1 Interpolation yields : T1 299.12 K
300 1.0217 87.410 HoT1 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 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.

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 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 28.97 g/mol
SoT2 0.60768 kJ/kg-K SoT3S 1.1521 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-
880 702.98 1.1369
T3S H3S 1.1521 Interpolation yields : T3S 892.14 K
900 725.37 1.1620 H3S 716.57 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(T3), as follows : Eqn 13

Look-up Pr(T2) and use it in Eqn 13 to determine Pr(T3S): Pr(T2) 8.3101
Pr(T3S) 55.400

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)
880 52.530 702.98
T3S 55.400 H3S Interpolation yields : T3S 891.93 K
900 57.342 725.37 H3S 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 H3 : H3 768.78 kJ/kg
T (K) Ho (kJ/kg)
920 747.82
T3 768.78
940 770.33 Interpolation yields : T3 938.62 K
Now, we go back to Eqns 2 & 3 to evaluate Wturb and Wcomp : Wturb 215.64 kJ/kg
Wcomp -435.12 kJ/kg
Finally, we plug values into Eqn 1 to evaluate Wcycle : Wcycle -219.48 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 = Q12.
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 16
Plugging values into Eqn 16 gives us: Q12 247.13 kJ/kg
Finally, we can plug values back into Eqn 14 : COPR 1.126
Verify: The assumptions made in the solution of this problem cannot be verified with the given information.
Answers : Wcycle -219 kJ/kg COPR 1.13 