| 10E-2 : | Ideal Regenerative Brayton Refrigeration Cycle | 9 pts |
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| Calculate the refrigeration capacity, QC, in tons of refrigeration, and the coefficient of performance of the regenerative Brayton Cycle, shown below. | |||||||||||||||||||
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| 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: | V1 | 1250 | ft3/min | P2 | 70 | psia | |||||||||||||
| T1 | 747 | oR | T3 | 775 | oR | ||||||||||||||
| P1 | 25 | psia | T4 | 747 | oR | ||||||||||||||
| Find: | a.) | Qin | ??? | tons | |||||||||||||||
| b.) | COPR | ??? | |||||||||||||||||
| Diagram: | 
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| Assumptions: | 1 - | Each process is analyzed as an open system operating at steady-state. | |||||||||||||||||
| 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 (oR) |
P (psia) |
Ho (Btu/lbm) |
So (Btu/lbm-oR) |
Pr | ||||||||||||||
| 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 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: | ||||||||||||||||||
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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. | |||||||||||||||||||
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Eqn 2 | ||||||||||||||||||
| Plugging values into Eqn 2 yields: | |||||||||||||||||||
| R | 1545 | ft lbf /lbm-oR | |||||||||||||||||
| MW | 29.00 | lbm/lbmole | mdot | 113.07 | lbm/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. | |||||||||||||||||||
| Apply the 2nd Gibbs Equation for Ideal Gases in terms of the Ideal Gas Entropy Function: | |||||||||||||||||||
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Eqn 3 | ||||||||||||||||||
| We can solve Eqn 3 for SoT5 : | 
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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 | 28.97 | lbm/lbmole | ||||||||||||||
| T (oR) | Ho (Btu/lbm) | So (Btu/lbmoR) | |||||||||||||||||
| 740.00 | 85.884 | 0.077519 | |||||||||||||||||
| 747 | HoT4 | SoT4 | HoT4 | 87.598 | Btu/lbm | ||||||||||||||
| 750 | 88.332 | 0.080805 | SoT4 | 0.079819 | Btu/lbm-oR | ||||||||||||||
| T (oR) | Ho (Btu/lbm) | So (Btu/lbmoR) | |||||||||||||||||
| 550 | 39.963 | 0.005848 | SoT5 | 0.0091996 | Btu/lbm-oR | ||||||||||||||
| T5 | HoT5 | 0.009200 | T5 | 557.79 | oR | ||||||||||||||
| 560 | 42.351 | 0.010149 | HoT5 | 41.824 | Btu/lbm | ||||||||||||||
| Method 2: Use the Ideal Gas Relative Pressure. | |||||||||||||||||||
| When an ideal gas undergoes an isentropic process : | 
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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 Eqn 5 For Pr(T5) , as follows : | 
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Eqn 6 | |||||||||||||||||
| Look-up Pr(T4) and use it in Eqn 6 to determine Pr(T5) : | |||||||||||||||||||
| T (oR) | Pr | Ho (Btu/lbm) | |||||||||||||||||
| 740 | 3.0985 | 85.884 | HoT4 | 87.598 | Btu/lbm | ||||||||||||||
| 747 | Pr(T4) | HoT4 | Pr(T4) | 3.2050 | |||||||||||||||
| 750 | 3.2506 | 88.332 | Pr(T5) | 1.1446 | |||||||||||||||
| We can now determine T5 and H5 by interpolation on the the Ideal Gas Property Table for air. | |||||||||||||||||||
| T (oR) | Pr | Ho (Btu/lbm) | |||||||||||||||||
| 550 | 1.0891 | 39.963 | |||||||||||||||||
| T5 | 1.1446 | HoT5 | T5 | 557.88 | oR | ||||||||||||||
| 560 | 1.1596 | 42.351 | HoT5 | 41.845 | Btu/lbm | ||||||||||||||
| 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 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: | |||||||||||||||||||
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Eqn 7 | ||||||||||||||||||
| Solve Eqn 7 for H6 : | 
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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) | ||||||||||||||||||
| 770 | 93.238 | HoT1 | 87.598 | Btu/lbm | |||||||||||||||
| 775 | HoT3 | ||||||||||||||||||
| 780 | 95.696 | HoT3 | 94.467 | Btu/lbm | |||||||||||||||
| Now, we plug values back into Eqn 8 : | HoT6 | 80.728 | Btu/lbm | ||||||||||||||||
| At last, we can plug numbers back into Eqn 1: | Qin | 4399.1 | Btu/min | ||||||||||||||||
| Converting units to tons of refrigeration yields the answer to part (a) : | 1 ton = | 200 | Btu/min | ||||||||||||||||
| Qin | 21.995 | tons | |||||||||||||||||
| Part b.) | We can determine the coefficient of performance from its definition. | ||||||||||||||||||
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Eqn 9 | ||||||||||||||||||
| Where : | 
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Eqn 10 | |||||||||||||||||
| We can determine Wcomp and Wturb by applying the 1st Law to the compressor and the turbine separately. | |||||||||||||||||||
| The 1st Law for a steady-state, single-inlet, single-outlet adiabatic turbine and compressor with negligible kinetic and potential energy changes are: | |||||||||||||||||||
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Eqn 11 | ||||||||||||||||||
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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. | |||||||||||||||||||
| Apply the 2nd Gibbs Equation for Ideal Gases in terms of the Ideal Gas Entropy Function: | |||||||||||||||||||
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Eqn 13 | ||||||||||||||||||
| Solving Eqn 13 for SoT2 yields : | 
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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 | 0.079819 | Btu/lbm-oR | SoT2 | 0.15044 | Btu/lbm-oR | ||||||||||||||
| T (oR) | Ho (Btu/lbm) | So (Btu/lbmoR) | |||||||||||||||||
| 990 | 147.98 | 0.14976 | |||||||||||||||||
| T2 | HoT2 | 0.15044 | T2 | 992.67 | oR | ||||||||||||||
| 1000 | 150.50 | 0.15230 | HoT2 | 148.65 | Btu/lbm | ||||||||||||||
| Method 2: Use the Ideal Gas Relative Pressure. | |||||||||||||||||||
| When an ideal gas undergoes an isentropic process : | 
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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 Eqn 15 For Pr(T2) , as follows : | 
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Eqn 16 | |||||||||||||||||
| Pr(T1) = Pr(T4) because T1 = T4. Use Pr(T1) in Eqn 16 to determine Pr(T2) : | |||||||||||||||||||
| Pr(T1) | 3.2050 | Pr(T2) | 8.9740 | ||||||||||||||||
| 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 | 147.981 | |||||||||||||||||
| T2 | 8.9740 | HoT2 | T2 | 992.53 | oR | ||||||||||||||
| 1000 | 9.2240 | 150.502 | HoT2 | 148.62 | Btu/lbm | ||||||||||||||
| 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: | |||||||||||||||||||
| Wturb | 5175.83 | Btu/min | Wcycle | -1728.08 | Btu/min | ||||||||||||||
| Wcomp | -6903.91 | Btu/min | COPR | 2.546 | |||||||||||||||
| Verify: | The assumptions made in the solution of this problem cannot be verified with the given information. | ||||||||||||||||||
| Answers : | a.) | Qin | 22.0 | tons | b.) | COPR | 2.55 | ||||||||||||
