Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Like this document? Why not share!

- Ch.11 by R G Sanjay Prakash 10802 views
- Chapter 11 thermodynamics solutio... by Muhammad Abdullah... 4067 views
- Thermo 7e sm_chap10-1 by Faiz Shafiq 2885 views
- Thermo 7e sm_chap13-1 by Faiz Shafiq 2751 views
- Thermodynamics: An Engineering Appr... by Mert Bertan 147825 views
- Thermo 7e sm_chap09-1 by Faiz Shafiq 11206 views

No Downloads

Total views

3,008

On SlideShare

0

From Embeds

0

Number of Embeds

1

Shares

0

Downloads

266

Comments

0

Likes

4

No embeds

No notes for slide

- 1. 11-1 Solutions Manual for Thermodynamics: An Engineering Approach Seventh Edition Yunus A. Cengel, Michael A. Boles McGraw-Hill, 2011 Chapter 11 REFRIGERATION CYCLES PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw-Hill. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 2. 11-2 The Reversed Carnot Cycle 11-1C The reversed Carnot cycle serves as a standard against which actual refrigeration cycles can be compared. Also, the COP of the reversed Carnot cycle provides the upper limit for the COP of a refrigeration cycle operating between the specified temperature limits. 11-2C Because the compression process involves the compression of a liquid-vapor mixture which requires a compressor that will handle two phases, and the expansion process involves the expansion of high-moisture content refrigerant. 11-3 A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered. The coefficient of performance, the amount of heat absorbed from the refrigerated space, and the net work input are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) Noting that TH = 40°C = 313 K and TL = Tsat @ 100 kPa = −26.37°C = 246.6 K, the COP of this Carnot refrigerator is determined from COPR,C = 1 1 = = 3.72 TH / TL − 1 (313 K ) / (246.6 K ) − 1 T (b) From the refrigerant tables (Table A-11), h3 = h g @ 40°C = 271.27 kJ/kg h4 = h f @ 40°C = 108.26 kJ/kg Thus, q H = h3 − h4 = 271.27 − 108.26 = 163.0 kJ/kg and 4 QH 3 40°C 100 kPa 1 QL 2 s ⎛ 246.6 K ⎞ q H TH T = ⎯ ⎯→ q L = L q H = ⎜ ⎟ ⎜ 313 K ⎟(163.0 kJ/kg ) = 128.4 kJ/kg q L TL TH ⎠ ⎝ (c) The net work input is determined from wnet = q H − q L = 163.0 − 128.4 = 34.6 kJ/kg PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 3. 11-3 11-4E A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered. The coefficient of performance, the quality at the beginning of the heat-absorption process, and the net work input are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) Noting that TH = Tsat @ 90 psia = 72.78°F = 532.8 R and TL = Tsat @ 30 psia = 15.37°F = 475.4 R. COPR,C = 1 1 = = 8.28 TH / TL − 1 (532.8 R )/ (475.4 R ) − 1 T (b) Process 4-1 is isentropic, and thus ( s1 = s 4 = s f + x 4 s fg ) @ 90 psia = 0.07481 + (0.05)(0.14525) = 0.08207 Btu/lbm ⋅ R ⎛ s1 − s f x1 = ⎜ ⎜ s fg ⎝ ⎞ 0.08207 − 0.03793 ⎟ = = 0.2374 ⎟ 0.18589 ⎠ @ 30 psia 4 1 QH QL 3 2 (c) Remembering that on a T-s diagram the area enclosed represents the net work, and s3 = sg @ 90 psia = 0.22006 Btu/lbm·R, s wnet,in = (TH − TL )(s 3 − s 4 ) = (72.78 − 15.37)(0.22006 − 0.08207) Btu/lbm ⋅ R = 7.92 Btu/lbm PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 4. 11-4 Ideal and Actual Vapor-Compression Refrigeration Cycles 11-5C Yes; the throttling process is an internally irreversible process. 11-6C To make the ideal vapor-compression refrigeration cycle more closely approximate the actual cycle. 11-7C No. Assuming the water is maintained at 10°C in the evaporator, the evaporator pressure will be the saturation pressure corresponding to this pressure, which is 1.2 kPa. It is not practical to design refrigeration or air-conditioning devices that involve such extremely low pressures. 11-8C Allowing a temperature difference of 10°C for effective heat transfer, the condensation temperature of the refrigerant should be 25°C. The saturation pressure corresponding to 25°C is 0.67 MPa. Therefore, the recommended pressure would be 0.7 MPa. 11-9C The area enclosed by the cyclic curve on a T-s diagram represents the net work input for the reversed Carnot cycle, but not so for the ideal vapor-compression refrigeration cycle. This is because the latter cycle involves an irreversible process for which the process path is not known. 11-10C The cycle that involves saturated liquid at 30°C will have a higher COP because, judging from the T-s diagram, it will require a smaller work input for the same refrigeration capacity. 11-11C The minimum temperature that the refrigerant can be cooled to before throttling is the temperature of the sink (the cooling medium) since heat is transferred from the refrigerant to the cooling medium. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 5. 11-5 11-12E A refrigerator operating on the ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The increase in the COP if the throttling process were replaced by an isentropic expansion is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-11E, A-12E, and A-13E), T1 = 20°F ⎫ h1 = h g @ 20° F = 105.98 Btu/lbm ⎬ sat. vapor ⎭ s1 = s g @ 20° F = 0.22341 Btu/lbm ⋅ R P2 = 300 psia ⎫ ⎬ h2 = 125.68 Btu/lbm s 2 = s1 ⎭ T · QH 2 3 300 psia · Win P3 = 300 psia ⎫ h3 = h f @ 300 psia = 66.339 Btu/lbm ⎬ s =s sat. liquid f @ 300 psia = 0.12715 Btu/lbm ⋅ R ⎭ 3 h4 ≅ h3 = 66.339 Btu/lbm ( throttling) T4 = 20°F ⎫ h4 s = 59.80 Btu/lbm (isentropic expansion) ⎬ s 4 = s 3 ⎭ x 4 s = 0.4723 The COP of the refrigerator for the throttling case is COPR = 20°F 4s 4 · QL 1 s qL h -h 105.98 − 66.339 = 1 4 = = 2.012 win h2 -h1 125.68 − 105.98 The COP of the refrigerator for the isentropic expansion case is COPR = h -h qL 105.98 − 59.80 = 1 4s = = 2.344 win h2 -h1 125.68 − 105.98 The increase in the COP by isentropic expansion is 16.5%. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 6. 11-6 11-13 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP and the power requirement are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-11, A-12, and A-13), T1 = 4°C ⎫ h1 = h g @ 4°C = 252.77 kJ/kg ⎬ sat. vapor ⎭ s1 = s g @ 4°C = 0.92927 kJ/kg ⋅ K T P2 = 1 MPa ⎫ ⎬ h2 = 275.29 kJ/kg s 2 = s1 ⎭ P3 = 1 MPa ⎫ ⎬ h = hf sat. liquid ⎭ 3 @ 1 MPa · QH · Win = 107.32 kJ/kg h4 ≅ h3 = 107.32 kJ/kg ( throttling ) The mass flow rate of the refrigerant is & & & ⎯→ m = Q L = m(h1 − h4 ) ⎯ 2 3 1 MPa & QL 400 kJ/s = = 2.750 kg/s h1 − h4 (252.77 − 107.32) kJ/kg 4°C 4s 4 · QL 1 s The power requirement is & & Win = m(h2 − h1 ) = (2.750 kg/s)(275.29 − 252.77) kJ/kg = 61.93 kW The COP of the refrigerator is determined from its definition, COPR = & QL 400 kW = = 6.46 & Win 61.93 kW PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 7. 11-7 11-14 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The rate of heat removal from the refrigerated space, the power input to the compressor, the rate of heat rejection to the environment, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13), P1 = 120 kPa sat. vapor ⎫ h1 = h g @ 120 kPa = 236.97 kJ/kg ⎬s = s g @ 120 kPa = 0.94779 kJ/kg ⋅ K ⎭ 1 P2 = 0.7 MPa ⎫ ⎬ h2 = 273.50 kJ/kg (T2 = 34.95°C ) s 2 = s1 ⎭ P3 = 0.7 MPa ⎫ ⎬ h3 = h f sat. liquid ⎭ @ 0.7 MPa T · QH 2 3 0.7 MPa · Win = 88.82 kJ/kg h4 ≅ h3 = 88.82 kJ/kg (throttling ) Then the rate of heat removal from the refrigerated space and the power input to the compressor are determined from 0.12 4s 4 · QL 1 s & & QL = m(h1 − h4 ) = (0.05 kg/s )(236.97 − 88.82) kJ/kg = 7.41 kW and & & Win = m(h2 − h1 ) = (0.05 kg/s )(273.50 − 236.97 ) kJ/kg = 1.83 kW (b) The rate of heat rejection to the environment is determined from & & & Q H = Q L + Win = 7.41 + 1.83 = 9.23 kW (c) The COP of the refrigerator is determined from its definition, COPR = & QL 7.41 kW = = 4.06 & Win 1.83 kW PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 8. 11-8 11-15 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The rate of heat removal from the refrigerated space, the power input to the compressor, the rate of heat rejection to the environment, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13), P1 = 120 kPa sat. vapor ⎫ h1 = h g @ 120 kPa = 236.97 kJ/kg ⎬s = s g @ 120 kPa = 0.94779 kJ/kg ⋅ K ⎭ 1 P2 = 0.9 MPa ⎫ ⎬ h2 = 278.93 kJ/kg (T2 = 44.45°C ) s 2 = s1 ⎭ P3 = 0.9 MPa ⎫ ⎬ h3 = h f sat. liquid ⎭ @ 0.9 MPa T · QH 2 3 0.9 MPa · Win = 101.61 kJ/kg h4 ≅ h3 = 101.61 kJ/kg (throttling ) Then the rate of heat removal from the refrigerated space and the power input to the compressor are determined from 0.12 MPa 4s 4 · QL 1 s & & QL = m(h1 − h4 ) = (0.05 kg/s )(236.97 − 101.61) kJ/kg = 6.77 kW and & & Win = m(h2 − h1 ) = (0.05 kg/s )(278.93 − 236.97 ) kJ/kg = 2.10 kW (b) The rate of heat rejection to the environment is determined from & & & Q H = Q L + Win = 6.77 + 2.10 = 8.87 kW (c) The COP of the refrigerator is determined from its definition, COPR = & QL 6.77 kW = = 3.23 & Win 2.10 kW PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 9. 11-9 11-16 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The throttling valve in the cycle is replaced by an isentropic turbine. The percentage increase in the COP and in the rate of heat removal from the refrigerated space due to this replacement are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis If the throttling valve in the previous problem is replaced by an isentropic turbine, we would have T s4s = s3 = sf @ 0.7 MPa = 0.33230 kJ/kg·K · QH 2 3 0.7 MPa · Win and the enthalpy at the turbine exit would be ⎛ s3 − s f x4s = ⎜ ⎜ s fg ⎝ ( ⎞ 0.33230 − 0.09275 ⎟ = = 0.2802 ⎟ 0.85503 ⎠ @ 120 kPa h4 s = h f + x 4 s h fg )@ 120 kPa = 22.49 + (0.2802)(214.48) = 82.58 kJ/kg 0.12 MPa 4s 4 · QL 1 s Then, & & Q L = m(h1 − h4 s ) = (0.05 kg/s)(236.97 − 82.58) kJ/kg = 7.72 kW and COPR = & QL 7.72 kW = = 4.23 & Win 1.83 kW & Then the percentage increase in Q and COP becomes & ∆Q L 7.72 − 7.41 & Increase in Q L = = = 4.2% & 7.41 QL Increase in COPR = ∆COPR 4.23 − 4.06 = = 4.2% COPR 4.06 PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 10. 11-10 11-17 A refrigerator with refrigerant-134a as the working fluid is considered. The rate of heat removal from the refrigerated space, the power input to the compressor, the isentropic efficiency of the compressor, and the COP of the refrigerator are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the refrigerant tables (Tables A-12 and A-13), P1 = 0.20 MPa ⎫ h1 = 248.80 kJ/kg ⎬ s = 0.95407 kJ/kg ⋅ K T1 = −5°C ⎭ 1 P2 = 1.2 MPa T2 = 70°C T ⎫ ⎬ h2 = 300.61 kJ/kg ⎭ P2 s = 1.2 MPa ⎫ ⎬ h2 s = 287.21 kJ/kg s 2 s = s1 ⎭ P3 = 1.15 MPa ⎫ ⎬ h3 = h f T3 = 44°C ⎭ @ 44°C = 114.28 kJ/kg h4 ≅ h3 = 114.28 kJ/kg (throttling ) 1.15 MPa 2 1.2 MPa 2s 70°C · Win · QH 3 0.21 MPa 4 · QL 1 0.20 MPa -5°C s Then the rate of heat removal from the refrigerated space and the power input to the compressor are determined from & & Q L = m(h1 − h4 ) = (0.07 kg/s )(248.80 − 114.28) kJ/kg = 9.42 kW and & & Win = m(h2 − h1 ) = (0.07 kg/s )(300.61 − 248.80) kJ/kg = 3.63 kW (b) The isentropic efficiency of the compressor is determined from ηC = h2 s − h1 287.21 − 248.80 = 0.741 = 74.1% = h2 − h1 300.61 − 248.80 (c) The COP of the refrigerator is determined from its definition, COPR = & QL 9.42 kW = = 2.60 & Win 3.63 kW PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 11. 11-11 11-18E An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The mass flow rate of the refrigerant and the power requirement are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-11E, A-12E, and A-13E), T1 = 5°F ⎫ h1 = h g @ 50°F = 103.82 Btu/lbm ⎬ sat. vapor ⎭ s1 = s g @ 5° F = 0.22485 Btu/lbm ⋅ R P2 = 180 psia ⎫ ⎬ h2 = 121.99 Btu/lbm s 2 = s1 ⎭ P3 = 180 psia ⎫ ⎬ h3 = h f sat. liquid ⎭ @ 180 psia T · QH 2 3 180 psia · Win = 51.50 Btu/lbm h4 ≅ h3 = 51.50 Btu/lbm ( throttling) 5°F 4s 4 · QL 1 The mass flow rate of the refrigerant is & & ⎯→ & Q L = m(h1 − h4 ) ⎯ m = & QL 45,000 Btu/h = = 860.1 lbm/h h1 − h4 (103.82 − 51.50) Btu/lbm s The power requirement is 1 kW ⎛ ⎞ & & Win = m(h2 − h1 ) = (860.1 lbm/h)(121.99 − 103.82) Btu/lbm⎜ ⎟ = 4.582 kW ⎝ 3412.14 Btu/h ⎠ PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 12. 11-12 11-19E Problem 11-18E is to be repeated if ammonia is used as the refrigerant. Analysis The problem is solved using EES, and the solution is given below. "Given" x[1]=1 T[1]=5 [F] x[3]=0 P[3]=180 [psia] Q_dot_L=45000 [Btu/h] "Analysis" Fluid$='ammonia' "compressor" h[1]=enthalpy(Fluid$, T=T[1], x=x[1]) s[1]=entropy(Fluid$, T=T[1], x=x[1]) s[2]=s[1] P[2]=P[3] h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) "expansion valve" h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) h[4]=h[3] "cycle" m_dot_R=Q_dot_L/(h[1]-h[4]) W_dot_in=m_dot_R*(h[2]-h[1])*Convert(Btu/h, kW) Solution for ammonia COP_R=4.515 Fluid$='ammonia' m_dot_R=95.8 [lbm/h] Q_dot_L=45000 [Btu/h] W_dot_in=2.921 [kW] Solution for R-134a COP_R=2.878 Fluid$='R134a' m_dot_R=860.1 [lbm/h] Q_dot_L=45000 [Btu/h] W_dot_in=4.582 [kW] PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 13. 11-13 11-20 A commercial refrigerator with refrigerant-134a as the working fluid is considered. The quality of the refrigerant at the evaporator inlet, the refrigeration load, the COP of the refrigerator, and the theoretical maximum refrigeration load for the same power input to the compressor are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From refrigerant-134a tables (Tables A-11 through A-13) P1 = 60 kPa ⎫ ⎬ h1 = 230.03 kJ/kg T1 = −34°C ⎭ P2 = 1200 kPa ⎫ ⎬ h2 = 295.16 kJ/kg T2 = 65°C ⎭ Water 18°C 26°C QH 1.2 MPa 65°C 42°C P3 = 1200 kPa ⎫ ⎬ h3 = 111.23 kJ/kg T3 = 42°C ⎭ h4 = h3 = 111.23 kJ/kg P4 = 60 kPa ⎫ ⎬ x 4 = 0.4795 h4 = 111.23 kJ/kg ⎭ Using saturated liquid enthalpy at the given temperature, for water we have (Table A-4) Condenser 3 Qin 2 Expansion valve 4 Win Compressor Evaporator hw1 = h f @ 18°C = 75.47 kJ/kg 60 kPa -34°C 1 QL h w2 = h f @ 26°C = 108.94 kJ/kg (b) The mass flow rate of the refrigerant may be determined from an energy balance on the compressor & & mR (h2 − h3 ) = mw (hw2 − hw1 ) & mR (295.16 − 111.23)kJ/kg = (0.25 kg/s)(108.94 − 75.47)kJ/kg ⎯ mR = 0.0455 kg/s ⎯→ & The waste heat transferred from the refrigerant, the compressor power input, and the refrigeration load are & & Q H = m R (h2 − h3 ) = (0.0455 kg/s)(295.16 − 111.23)kJ/kg = 8.367 kW & & & Win = m R ( h2 − h1 ) − Qin = (0.0455 kg/s)(295.16 − 230.03)kJ/kg − 0.450 kW = 2.513 kW & & & & Q L = Q H − Win − Qin = 8.367 − 2.513 − 0.450 = 5.404 kW (c) The COP of the refrigerator is determined from its definition COP = & Q L 5.404 = = 2.15 & Win 2.513 (d) The reversible COP of the refrigerator for the same temperature limits is COPmax = 1 1 = = 5.063 T H / T L − 1 (18 + 273) /(−30 + 273) − 1 Then, the maximum refrigeration load becomes & & Q L,max = COPmaxWin = (5.063)(2.513 kW) = 12.72 kW 2 T · QH 2 · Win 3 4 · QL 1 s PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 14. 11-14 11-21 A refrigerator with refrigerant-134a as the working fluid is considered. The power input to the compressor, the rate of heat removal from the refrigerated space, and the pressure drop and the rate of heat gain in the line between the evaporator and the compressor are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the refrigerant tables (Tables A-12 and A-13), h1 = 239.50 kJ/kg P1 = 100 kPa ⎫ s = 0.97207 kJ/kg ⋅ K T1 = −20°C ⎬ 1 ⎭ v 1 = 0.19841 m 3 /kg T P2 = 0.8 MPa ⎫ ⎬ h2 s = 284.07 kJ/kg s 2 s = s1 ⎭ P3 = 0.75 MPa ⎫ ⎬ h3 ≅ h f T3 = 26°C ⎭ @ 26 °C 2s · QH 0.8 MPa · Win 3 = 87.83 kJ/kg h4 ≅ h3 = 87.83 kJ/kg (throttling ) T5 = −26°C sat. vapor 2 0.75 MPa 0.10 MPa 4 ⎫ P5 = 0.10173 MPa ⎬ h = 234.68 kJ/kg ⎭ 5 · QL 1 0.10 MPa -20°C -26°C s Then the mass flow rate of the refrigerant and the power input becomes & m= V&1 0.5/60 m 3 /s = = 0.0420 kg/s v 1 0.19841 m 3 /kg & & Win = m(h2 s − h1 ) / η C = (0.0420 kg/s )[(284.07 − 239.50 ) kJ/kg ] / (0.78) = 2.40 kW (b) The rate of heat removal from the refrigerated space is & & QL = m(h5 − h4 ) = (0.0420 kg/s)(234.68 − 87.83) kJ/kg = 6.17 kW (c) The pressure drop and the heat gain in the line between the evaporator and the compressor are ∆P = P5 − P1 = 101.73 − 100 = 1.73 and & & Qgain = m(h1 − h5 ) = (0.0420 kg/s )(239.50 − 234.68) kJ/kg = 0.203 kW PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 15. 11-15 11-22 Problem 11-21 is reconsidered. The effects of the compressor isentropic efficiency and the compressor inlet volume flow rate on the power input and the rate of refrigeration are to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Given" P[1]=100 [kPa] T[1]=-20 [C] V_dot=0.5 [m^3/min] P[2]=800 [kPa] "Eta_C=0.78" P[3]=750 [kPa] T[3]=26 [C] T[5]=-26 [C] x[5]=1 7 6 1 m 3/min 5 Win [kW] "Analysis" Fluid$='R134a' "compressor" h[1]=enthalpy(Fluid$, P=P[1], T=T[1]) s[1]=entropy(Fluid$, P=P[1], T=T[1]) v[1]=volume(Fluid$, P=P[1], T=T[1]) s_s[2]=s[1] h_s[2]=enthalpy(Fluid$, P=P[2], s=s_s[2]) "expansion valve" x[3]=0 "assumed saturated liquid" h[3]=enthalpy(Fluid$, T=T[3], x=x[3]) h[4]=h[3] "evaporator exit" h[5]=enthalpy(Fluid$, T=T[5], x=x[5]) P[5]=pressure(Fluid$, T=T[5], x=x[5]) "cycle" m_dot=V_dot/v[1]*Convert(kg/min, kg/s) W_dot_in=m_dot*(h_s[2]-h[1])/Eta_C Q_dot_L=m_dot*(h[5]-h[4]) DELTAP=P[5]-P[1] Q_dot_gain=m_dot*(h[1]-h[5]) 4 3 0.5 m 3/min 2 1 0 0.6 0.1 m 3/min 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 0.95 1 ηC 14 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 Win [kW] 3.12 2.88 2.674 2.496 2.34 2.202 2.08 1.971 1.872 QL [kW] 6.168 6.168 6.168 6.168 6.168 6.168 6.168 6.168 6.168 12 1 m 3/min 10 QL [kW] ηc 8 0.5 m 3/min 6 4 0.1 m 3/min 2 0 0.6 0.65 0.7 0.75 0.8 0.85 0.9 ηC PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 16. 11-16 11-23 A refrigerator uses refrigerant-134a as the working fluid and operates on the ideal vapor-compression refrigeration cycle except for the compression process. The mass flow rate of the refrigerant, the condenser pressure, and the COP of the refrigerator are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) (b) From the refrigerant-134a tables (Tables A-11 through A-13) P4 = 120 kPa ⎫ ⎬ h4 = 86.83 kJ/kg x 4 = 0.30 ⎭ h3 = h4 h3 = 86.83 kJ/kg ⎫ ⎬ P3 = 671.8 kPa x 3 = 0 (sat. liq.) ⎭ P2 = P3 P2 = 671.8 kPa ⎫ ⎬ h2 = 298.87 kJ/kg ⎭ P1 = P4 = 120 kPa ⎫ ⎬ h1 = 236.97 kJ/kg x1 = 1 (sat. vap.) ⎭ T2 = 60°C . QH 60°C Condenser 3 2 Expansion valve 4 . Win Compressor 1 Evaporator 120 kPa x=0.3 . QL The mass flow rate of the refrigerant is determined from & W in 0.45 kW & m= = = 0.00727 kg/s h2 − h1 (298.87 − 236.97)kJ/kg (c) The refrigeration load and the COP are T · QH 3 · Win & & Q L = m(h1 − h4 ) = (0.0727 kg/s)(236.97 − 86.83)kJ/kg = 1.091 kW COP = & Q L 1.091 kW = = 2.43 & 0.45 kW Win 2 120 kPa 4s 4 · QL 1 s PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 17. 11-17 11-24 A vapor-compression refrigeration cycle with refrigerant-22 as the working fluid is considered. The hardware and the T-s diagram for this air conditioner are to be sketched. The heat absorbed by the refrigerant, the work input to the compressor and the heat rejected in the condenser are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) In this normal vapor-compression refrigeration cycle, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. qH 45°C T · QH 2s Condenser 3 2 Expansion valve Compressor 3 45°C · Win win -5°C 4 Evaporator 2 1 4s 4 · 1 QL sat. vap. -5°C qL s (b) The properties as given in the problem statement are h4 = h3 = hf @ 45°C = 101 kJ/kg h1 = hg @ -5°C = 248.1 kJ/kg. The heat absorbed by the refrigerant in the evaporator is q L = h1 − h4 = 248.1 − 101 = 147.1 kJ/kg (c) The COP of the air conditioner is 1W 1W ⎛ ⎞ ⎛ Btu/h ⎞⎛ ⎞ COPR = SEER ⎜ ⎟ = ⎜16 ⎟⎜ ⎟ = 4.689 W ⎠⎝ 3.412 Btu/h ⎠ ⎝ 3.412 Btu/h ⎠ ⎝ The work input to the compressor is COPR = qL qL 147.1 kJ/kg ⎯ win = ⎯→ = = 31.4 kJ/kg win COPR 4.689 The enthalpy at the compressor exit is win = h2 − h1 ⎯ h2 = h1 + win = 248.1 kJ/kg + 31.4 kJ/kg = 279.5 kJ/kg ⎯→ The heat rejected from the refrigerant in the condenser is then q H = h2 − h3 = 279.5 − 101 = 178.5 kJ/kg PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 18. 11-18 11-25 A vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The amount of cooling, the work input, and the COP are to be determined. Also, the same parameters are to be determined if the cycle operated on the ideal vapor-compression refrigeration cycle between the same temperature limits. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The expansion process through the expansion valve is isenthalpic: h4 = h3. Then, 2 T 2s q L = h1 − h4 = 402.49 − 243.19 = 159.3 kJ/kg qH q H = h2 − h3 = 454.00 − 243.19 = 210.8 kJ/kg win 40°C 3 win = h2 − h1 = 454.00 − 402.49 = 51.51 kJ/kg −15°C q 159.3 kJ/kg COP = L = = 3.093 win 51.51 kJ/kg 4 1 qL s (c) Ideal vapor-compression refrigeration cycle solution: q L = h1 − h4 = 399.04 − 249.80 = 149.2 kJ/kg T q H = h2 − h3 = 440.71 − 249.80 = 190.9 kJ/kg win = h2 − h1 = 440.71 − 399.04 = 41.67 kJ/kg COP = q L 149.2 kJ/kg = = 3.582 win 41.67 kJ/kg Discussion In the ideal operation, the refrigeration load decreases by 6.3% and the work input by 19.1% while the COP increases by 15.8%. qH 3 2 40°C win −15°C 4 1 qL s PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 19. 11-19 11-26 A vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The rate of cooling, the power input, and the COP are to be determined. Also, the same parameters are to be determined if the cycle operated on the ideal vapor-compression refrigeration cycle between the same pressure limits. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the refrigerant-134a tables (Tables A-11 through A-13) Tsat@200 kPa = −10.1°C P1 = 200 kPa ⎫h1 = 253.05 kJ/kg ⎬ T1 = −10.1 + 10.1 = 0°C⎭ s1 = 0.9698 kJ/kg ⋅ K 2 T · QH P2 = 1400 kPa ⎫ ⎬ h2 s = 295.90 kJ/kg s1 = s1 ⎭ Tsat@1400 kPa = 52.4°C 3 200 kPa ⎫ ⎬h3 ≅ h f@48°C = 120.39 kJ/kg T3 = 52.4 − 4.4 = 48°C⎭ h4 = h3 = 120.39 kJ/kg 0.88 = · Win 1.4 MPa P3 = 1400 kPa ηC = 2s 1 · QL 4 s h2 s − h1 h2 − h1 295.90 − 253.05 ⎯ h2 = 301.74 kJ/kg ⎯→ h2 − 253.05 & & QL = m(h1 − h4 ) = (0.025 kg/s)(253.05 − 120.39) = 3.317 kW & & QH = m(h2 − h3 ) = (0.025 kg/s)(301.74 − 120.39) = 4.534 kW & & Win = m(h2 − h1 ) = (0.025 kg/s)(301.74 − 253.05) = 1.217 kW COP = & Q L 3.317 kW = = 2.725 & Win 1.217 kW (b) Ideal vapor-compression refrigeration cycle solution From the refrigerant-134a tables (Tables A-11 through A-13) P1 = 200 kPa ⎫h1 = 244.46 kJ/kg ⎬ x1 = 1 ⎭ s1 = 0.9377 kJ/kg ⋅ K P2 = 1400 kPa ⎫ ⎬ h2 = 285.08 kJ/kg s1 = s1 ⎭ P3 = 1400 kPa ⎫ ⎬h3 = 127.22 kJ/kg x3 = 0 ⎭ h4 = h3 = 127.22 kJ/kg T · QH 2 3 1.4 MPa · Win 200 kPa 4s 4 · QL 1 s & & QL = m(h1 − h4 ) = (0.025 kg/s)(244.46 − 127.22) = 3.931kW & & QH = m(h2 − h3 ) = (0.025 kg/s)(285.08 − 127.22) = 3.947 kW & & Win = m(h2 − h1 ) = (0.025 kg/s)(285.08 − 244.46) = 1.016 kW COP = & Q L 3.931 kW = = 2.886 & Win 1.016 kW Discussion The cooling load increases by 18.5% while the COP increases by 5.9% when the cycle operates on the ideal vapor-compression cycle. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 20. 11-20 Second-Law Analysis of Vapor-Compression Refrigeration Cycles 11-27C The second-law efficiency of a refrigerator operating on the vapor-compression refrigeration cycle is defined as && XQ η II,R = = L & W & & X dest, total Wmin =1− & & W W && where X Q is the exergy of the heat transferred from the low-temperature medium and it is expressed as L T & ⎛ && X Q = −Q L ⎜1 − 0 ⎜ T L L ⎝ ⎞ ⎟. ⎟ ⎠ & & X dest, total is the total exergy destruction in the cycle and W is the actual power input to the cycle. The second-law efficiency can also be expressed as the ratio of the actual COP to the Carnot COP: η II,R = COPR COPCarnot 11-28C The second-law efficiency of a heat pump operating on the a vapor-compression refrigeration cycle is defined as η II,HP = & & ExQ H & W = & & Ex dest,total Wmin =1− & & W W Substituting & W= & QH COPHP and T & ⎛ & & ExQ = Q H ⎜1 − 0 ⎜ T H H ⎝ ⎞ ⎟ ⎟ ⎠ into the second-law efficiency equation η II,HP = & & Ex Q & W H T & ⎛ Q H ⎜1 − 0 ⎜ T H ⎝ = & QH COPHP ⎞ ⎟ ⎟ ⎠ T & ⎛ = Q H ⎜1 − 0 ⎜ T H ⎝ ⎞ COPHP COPHP COPHP ⎟ = = ⎟ Q & TH COPCarnot ⎠ H TH − TL since T0.= TL. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 21. 11-21 11-29C In an isentropic compressor, s2 = s1 and h2s = h2. Applying these to the two the efficiency definitions, we obtain η s,Comp = wisen h2 s − h1 h2 − h1 = = = 1 = 100% w h2 − h1 h2 − h1 η II,Comp = wrev h2 − h1 − T0 ( s 2 − s1 ) h2 − h1 = = = 1 = 100% w h2 − h1 h2 − h1 Thus, the isentropic efficiency and the exergy efficiency of an isentropic compressor are both 100%. The exergy efficiency of a compressor is not necessarily equal to its isentropic efficiency. The two definitions are different as shown in the above equations. In the calculation of isentropic efficiency, the exit enthalpy is found at the hypothetical exit state (at the exit pressure and the inlet entropy) while the exergy efficiency involves the actual exit state. The two efficiencies are usually close but different. In the special case of an isentropic compressor, the two efficiencies become equal to each other as proven above. 11-30 A vapor-compression refrigeration system is used to keep a space at a low temperature. The power input, the COP and the second-law efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis The power input is ⎛ 1 kW ⎞ & & & Win = Q H − Q L = 6000 − 3500 = 2500 kJ/h = (2500 kJ/h)⎜ ⎟ = 0.6944 kW ⎝ 3600 kJ/h ⎠ The COP is COPR = & Q L (3500 / 3600) kW = = 1.4 & 0.6944 kW Win The COP of the Carnot cycle operating between the space and the ambient is COPCarnot = TL 250 K = = 5.208 TH − TL (298 − 250) K The second-law efficiency is then η II = COPR 1.4 = = 0.2688 = 26.9% COPCarnot 5.208 PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 22. 11-22 11-31 A refrigerator is used to cool bananas at a specified rate. The rate of heat absorbed from the bananas, the COP, The minimum power input, the second-law efficiency and the exergy destruction are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The rate of heat absorbed from the bananas is & & Q L = mc p (T1 − T2 ) = (1140 kg/h) (3.35 kJ/kg ⋅ °C)( 28 − 12)°C = 61,100 kJ/h The COP is COP = & Q L (61,100 / 3600) kW 16.97 kW = = = 1.97 & 8.6 kW 8.6 kW Win (b) Theminimum power input is equal to the exergy of the heat transferred from the low-temperature medium: & ⎛ T & & ExQ = −Q L ⎜1 − 0 ⎜ T L L ⎝ ⎞ 28 + 273 ⎞ ⎛ ⎟ = −(16.97 kW)⎜1 − ⎟ = 0.463 kW ⎟ 20 + 273 ⎠ ⎝ ⎠ where the dead state temperature is taken as the inlet temperature of the eggplants (T0 = 28°C) and the temperature of the low-temperature medium is taken as the average temperature of bananas T = (12+28)/2 = 20°C. (c) The second-law efficiency of the cycle is η II = & & Ex Q & Win L = 0.463 = 0.0539 = 5.39% 8 .6 The exergy destruction is the difference between the exergy expended (power input) and the exergy recovered (the exergy of the heat transferred from the low-temperature medium): & & & & Exdest = Win − ExQ = 8.6 − 0.463 = 8.14 kW L PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 23. 11-23 11-32 A vapor-compression refrigeration cycle is used to keep a space at a low temperature. The power input, the mass flow rate of water in the condenser, the second-law efficiency, and the exergy destruction are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The power input is ⎛ 1 kW ⎞ (24,000 Btu/h)(⎜ ⎟ & Q ⎝ 3412 Btu/h ⎠ = 7.034 kW = 3.431 kW & Win = L = COP 2.05 2.05 (b) From an energy balance on the cycle, & & & QH = QL + Win = 7.034 + 3.431 = 10.46 kW The mass flow rate of the water is then determined from & & Q H = mc pw ∆Tw ⎯ m = ⎯→ & & QH 10.46 kW = = 0.2086 kg/s c pw ∆Tw (4.18 kJ/kg ⋅ °C)(12°C) (c) The exergy of the heat transferred from the low-temperature medium is T & ⎛ & & ExQ = −Q L ⎜1 − 0 ⎜ T L L ⎝ ⎞ 20 + 273 ⎞ ⎛ ⎟ = −(7.034 kW)⎜1 − ⎟ = 0.5153 kW ⎟ 0 + 273 ⎠ ⎝ ⎠ The second-law efficiency of the cycle is η II = & & ExQ & Win L = 0.5153 = 0.1502 = 15.0% 3.431 The exergy destruction is the difference between the exergy supplied (power input) and the exergy recovered (the exergy of the heat transferred from the low-temperature medium): & & & & Exdest = Win − ExQ = 3.431 − 0.5153 = 2.916 kW L Alternative Solution The exergy efficiency can also be determined as follows: COPR,Carnot = η II = TL 0 + 273 = = 13.65 TH − TL 20 − 0 COP 2.05 = = 0.1502 = 15.0% COPR,Carnot 13.65 The result is identical as expected. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 24. 11-24 11-33E A vapor-compression refrigeration cycle is used to keep a space at a low temperature. The mass flow rate of R134a, the COP, The exergy destruction in each component and the exergy efficiency of the compressor, the second-law efficiency, and the exergy destruction are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The properties of R-134a are (Tables A-11E through A-13E) P1 = 20 psia ⎫h1 = 102.73 Btu/lbm ⎬ x1 = 1 ⎭ s1 = 0.2257 Btu/lbm ⋅ R T P2 = 140 psia ⎫h2 = 131.36 Btu/lbm ⎬ T2 = 160°F ⎭ s 2 = 0.2444 Btu/lbm ⋅ R · QH 2s 3 140 psia P3 = 140 psia ⎫h3 = 45.30 Btu/lbm ⎬ x3 = 0 ⎭ s 3 = 0.0921 Btu/lbm ⋅ R h4 = h3 = 45.30 Btu/lbm P4 = 20 psia ⎫ ⎬s 4 = 0.1001 Btu/lbm ⋅ R h4 = 45.30 Btu/lbm ⎭ 2 · Win 20 psia 4s 4 · 1 QL s The energy interactions in each component and the mass flow rate of R-134a are win = h2 − h1 = 131.36 − 102.73 = 28.63 Btu/lbm q H = h2 − h3 = 131.36 − 45.30 = 86.06 Btu/lbm q L = q H − win = 86.06 − 28.63 = 57.43 Btu/lbm & m= & Q L ( 45,000 / 3600) Btu/s = = 0.2177 lbm/s qL 57.43 Btu/lbm The COP is COP = q L 57.43 Btu/lbm = = 2.006 win 28.63 Btu/lbm (b) The exergy destruction in each component of the cycle is determined as follows: Compressor: s gen,1− 2 = s 2 − s1 = 0.2444 − 0.2257 = 0.01874 Btu/lbm ⋅ R & & Ex dest,1-2 = mT0 s gen,1-2 = (0.2177 lbm/s)(540 R)(0.01874 Btu/lbm ⋅ R) = 2.203 Btu/s Condenser: s gen,2−3 = s 3 − s 2 + qH 86.06 Btu/lbm = (0.0921 − 0.2444) Btu/lbm ⋅ R + = 0.007073 Btu/lbm ⋅ R TH 540 R & & Ex dest,2-3 = mT0 s gen,2-3 = (0.2177 lbm/s)(540 R)(0.00707 3 Btu/lbm ⋅ R) = 0.8313 Btu/s Expansion valve: s gen,3− 4 = s 4 − s 3 = 0.1001 − 0.0921 = 0.007962 Btu/lbm ⋅ R & & Ex dest,3-4 = mT0 s gen,3-4 = (0.2177 lbm/s)(540 R)(0.00796 2 Btu/lbm ⋅ R) = 0.9359 Btu/s Evaporator: s gen,4−1 = s1 − s 4 − qL 57.43 Btu/lbm = (0.2257 − 0.1001) Btu/lbm ⋅ R − = 0.003400 Btu/lbm ⋅ R TL 470 R & & Ex dest,4 -1 = mT0 s gen,4 -1 = (0.2177 lbm/s)(540 R)(0.00340 0 Btu/lbm ⋅ R) = 0.3996 Btu/s PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 25. 11-25 The power input and the exergy efficiency of the compressor is determined from & & Win = mwin = (0.2177 lbm/s)(28.63 Btu/lbm) = 6.232 Btu/s η II = 1 − & Ex dest,1− 2 2.203 Btu/s =1− = 0.6465 = 64.7% & 6.232 Btu/s Win (c) The exergy of the heat transferred from the low-temperature medium is & ⎛ T & & ExQ = −Q L ⎜1 − 0 ⎜ T L L ⎝ ⎞ ⎛ 540 ⎞ ⎟ = −(45000 / 3600 Btu/s)⎜1 − ⎟ = 1.862 Btu/s ⎟ 470 ⎠ ⎝ ⎠ The second-law efficiency of the cycle is η II = & & Ex Q & Win L = 1.862 Btu/s = 0.2987 = 29.9% 6.232 Btu/s The total exergy destruction in the cycle is the difference between the exergy supplied (power input) and the exergy recovered (the exergy of the heat transferred from the low-temperature medium): & & & & Ex dest,total = Win − ExQ = 6.232 − 1.862 = 4.370 Btu/s L The total exergy destruction can also be determined by adding exergy destructions in each component: & & & & & Ex dest, total = Ex dest,1-2 + Ex dest,2-3 + Ex dest,3-4 + Ex dest,4-1 = 2.203 + 0.8313 + 0.9359 + 0.3996 = 4.370 Btu/s The result is the same as expected. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 26. 11-26 11-34 A vapor-compression refrigeration cycle is used to keep a space at a low temperature. The mass flow rate of R-134a, the COP, The exergy destruction in each component and the exergy efficiency of the compressor, the second-law efficiency, and the exergy destruction are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and T potential energy changes are negligible. 2 · Analysis (a) The properties of R-134a are (Tables A-11 through A-13) QH 2s P2 = 1.2 MPa ⎫h2 ⎬ T2 = 50°C ⎭ s 2 P3 = 1.2 MPa ⎫ h3 ⎬ x3 = 0 ⎭ s3 = 278.27 kJ/kg 3 1.2 MPa = 0.9267 kJ/kg ⋅ K = 117.77 kJ/kg = 0.4244 kJ/kg ⋅ K The rate of heat transferred to the water is the energy change of the water from inlet to exit & & Q = m c (T − T ) = (0.15 kg/s) ( 4.18 kJ/kg ⋅ °C)( 28 − 20)°C = 5.016 kW H w p w, 2 · Win 4s 4 · 1 QL s w,1 The energy decrease of the refrigerant is equal to the energy increase of the water in the condenser. That is, & QH 5.016 kW & & & Q H = m R (h2 − h3 ) ⎯ ⎯→ m R = = = 0.03125 kg/s h2 − h3 (278.27 − 117.77) kJ/kg The refrigeration load is ⎛ 3412 Btu/h ⎞ & & & QL = Q H − Win = 5.016 − 2.2 = 2.816 kW = (2.816 kW)⎜ ⎟ = 9610 Btu/h ⎝ 1 kW ⎠ The COP of the refrigerator is determined from its definition, & Q 2.816 kW COP = L = = 1.28 & 2.2 kW Win (b) The COP of a reversible refrigerator operating between the same temperature limits is COPCarnot = TL −12 + 273 = = 8.156 TH − TL (20 + 273) − (−12 + 273) The minimum power input to the compressor for the same refrigeration load would be & QL 2.816 kW & Win, min = = = 0.3453 kW COPCarnot 8.156 The second-law efficiency of the cycle is & Win,min 0.3453 η II = = = 0.1569 = 15.7% & 2.2 Win The total exergy destruction in the cycle is the difference between the actual and the minimum power inputs: & & & Ex = W −W = 2.2 − 0.3453 = 1.85 kW dest, total in in, min (c) The entropy generation in the condenser is ⎞ ⎟ + m R ( s3 − s 2 ) & ⎟ ⎠ ⎛ 28 + 273 ⎞ = (0.15 kg/s)(4.18 kJ/kg ⋅ °C)ln⎜ ⎟ + (0.03125 kg/s)(0.4004 − 0.9267) kJ/kg ⋅ K) ⎝ 20 + 273 ⎠ = 0.001191 kW/K The exergy destruction in the condenser is & & Ex =T S = (293 K )(0.001191 kW/K ) = 0.349 kW ⎛ T w, 2 & & S gen,cond = m w c p ln⎜ ⎜T ⎝ w,1 dest,cond 0 gen,cond PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 27. 11-27 11-35 An ideal vapor-compression refrigeration cycle is used to keep a space at a low temperature. The cooling load, the COP, the exergy destruction in each component, the total exergy destruction, and the second-law efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The properties of R-134a are (Tables A-11 through A-13) T1 = −10°C ⎫h1 = 244.51 kJ/kg ⎬ x1 = 1 ⎭ s1 = 0.9377 kJ/kg ⋅ K T P2 = Psat@57.9°C = 1600 kPa ⎫ ⎬ h2 = 287.85 kJ/kg s 2 = s1 ⎭ P3 = 1600 kPa ⎫ h3 = 135.93 kJ/kg ⎬ x3 = 0 ⎭ s 3 = 0.4791 kJ/kg ⋅ K · QH h4 = h3 = 135.93 kJ/kg T4 = −10°C ⎫ ⎬s 4 = 0.5251 kJ/kg ⋅ K h4 = 135.93 kJ/kg ⎭ 2 3 57.9°C · Win −10°C 4s 4 · QL 1 s The energy interactions in the components and the COP are q L = h1 − h4 = 244.51 − 135.93 = 108.6 kJ/kg q H = h2 − h3 = 287.85 − 135.93 = 151.9 kJ/kg win = h2 − h1 = 287.85 − 244.51 = 43.33 kJ/kg COP = q L 108.6 kJ/kg = = 2.506 win 43.33 kJ/kg (b) The exergy destruction in each component of the cycle is determined as follows Compressor: sgen,1− 2 = s2 − s1 = 0 Ex dest,1- 2 = T0 sgen,1- 2 = 0 Condenser: s gen,2−3 = s 3 − s 2 + qH 151.9 kJ/kg = (0.4791 − 0.9377) kJ/kg ⋅ K + = 0.05124 kJ/kg ⋅ K TH 298 K Exdest,2-3 = T0 s gen,2-3 = (298 K)(0.05124 kJ/kg ⋅ K) = 15.27 kJ/kg Expansion valve: s gen,3−4 = s 4 − s 3 = 0.5251 − 0.4791 = 0.04595 kJ/kg ⋅ K Exdest,3-4 = T0 s gen,3-4 = (298 K)(0.04595 kJ/kg ⋅ K) = 13.69 kJ/kg Evaporator: s gen,4−1 = s1 − s 4 − qL 108.6 kJ/kg = (0.9377 − 0.5251) kJ/kg ⋅ K − = 0.02201 kJ/kg ⋅ K TL 278 K Exdest,4-1 = T0 s gen,4-1 = (298 K)(0.02201 kJ/kg ⋅ K) = 6.56 kJ/kg The total exergy destruction can be determined by adding exergy destructions in each component: & & & & & Ex dest,total = Ex dest,1-2 + Ex dest,2-3 + Ex dest,3-4 + Ex dest,4-1 = 0 + 15.27 + 13.69 + 6.56 = 35.52 kJ/kg PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 28. 11-28 (c) The exergy of the heat transferred from the low-temperature medium is ⎛ T Ex qL = − q L ⎜1 − 0 ⎜ T L ⎝ ⎞ 298 ⎞ ⎛ ⎟ = −(108.6 kJ/kg)⎜1 − ⎟ = 7.812 kJ/kg ⎟ 278 ⎠ ⎝ ⎠ The second-law efficiency of the cycle is η II = Ex qL win = 7.812 = 0.1803 = 18.0% 43.33 The total exergy destruction in the cycle can also be determined from Exdest,total = win − Ex qL = 43.33 − 7.812 = 35.52 kJ/kg The result is identical as expected. The second-law efficiency of the compressor is determined from η II,Comp = & & & X recovered W m[h2 − h1 − T0 ( s 2 − s1 )] = rev = & & & m(h2 − h1 ) X expended Wact, in since the compression through the compressor is isentropic (s2 = s1), the second-law efficiency is η II,Comp = 1 = 100% The second-law efficiency of the evaporator is determined from η II, Evap = & & X recovered Q L (T0 − TL ) / TL =1− = & & X expended m[h4 − h1 − T0 ( s 4 − s1 )] & X dest,4-1 & & X −X 4 1 where x 4 − x1 = h4 − h1 − T0 ( s 4 − s1 ) = (135.93 − 244.51) kJ/kg − (298 K )(0.5251 − 0.9377) kJ/kg ⋅ K = 14.37 kJ/kg Substituting, η II, Evap = 1 − x dest,4 -1 x 4 − x1 =1− 6.56 kJ/kg = 0.544 = 54.4% 14.37 kJ/kg PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 29. 11-29 11-36 An ideal vapor-compression refrigeration cycle uses ammonia as the refrigerant. The volume flow rate at the compressor inlet, the power input, the COP, the second-law efficiency and the total exergy destruction are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The properties of ammonia are given in problem statement. An energy balance on the cındenser gives q H = h1 − h4 = 1439.3 − 437.4 = 1361 kJ/kg & Q 18 kW & m= H = = 0.01323 kg/s q H 1361 kJ/kg T · QH 3 2 2 MPa · Win The volume flow rate is determined from & & V1 = mv1 = (0.01323 kg/s)(0.5946 m 3 /kg) = 0.007865 m 3 /s = 7.87 L/s (b) The power input and the COP are 200 kPa 4s 4 · QL 1 s & & Win = m(h2 − h1 ) = (0.01323 kg/s)(1798.3 − 1439.3)kJ/kg = 4.75 kW & & QL = m(h1 − h4 ) = (0.01323 kg/s)(1439.3 − 437.4)kJ/kg = 13.25 kW COP = & Q L 13.25 kW = = 2.79 & 4.75 kW Win (c) The exergy of the heat transferred from the low-temperature medium is & ⎛ T & & ExQ = −Q L ⎜1 − 0 ⎜ T L L ⎝ ⎞ ⎛ 300 ⎞ ⎟ = −(13.25 kW)⎜1 − ⎟ = 1.81 kW ⎟ 264 ⎠ ⎝ ⎠ The second-law efficiency of the cycle is η II = & & ExQ & Win L = 1.81 = 0.381 = 38.1% 4.75 The total exergy destruction in the cycle is the difference between the exergy supplied (power input) and the exergy recovered (the exergy of the heat transferred from the low-temperature medium): & & & & Ex dest,total = Win − ExQ = 4.75 − 1.81 = 2.94 kW L PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 30. 11-30 11-37 Prob. 11-36 is reconsidered. Using EES software, the problem is to be repeated ammonia, R-134a and R-22 is used as a refrigerant and the effects of evaporator and condenser pressures on the COP, the second-law efficiency and the total exergy destruction are to be investigated. Analysis The equations as written in EES are "GIVEN" P_1=200 [kPa] P_2=2000 [kPa] Q_dot_H=18 [kW] T_L=(-9+273) [K] T_H=(27+273) [K] "PROPERTIES" Fluid$='ammonia' x_1=1 x_3=0 h_1=enthalpy(Fluid$, P=P_1, x=x_1) s_1=entropy(Fluid$, P=P_1, x=x_1) v_1=volume(Fluid$, P=P_1, x=x_1) h_2=enthalpy(Fluid$, P=P_2, s=s_1) s_2=s_1 h_3=enthalpy(Fluid$, P=P_2, x=x_3) s_3=entropy(Fluid$, P=P_2, x=x_3) h_4=h_3 s_4=entropy(Fluid$, P=P_1, h=h_4) q_H=h_2-h_3 m_dot=Q_dot_H/q_H Vol_dot_1=m_dot*v_1 Q_dot_L=m_dot*(h_1-h_4) W_dot_in=m_dot*(h_2-h_1) COP=Q_dot_L/W_dot_in Ex_dot_QL=-Q_dot_L*(1-T_H/T_L) eta_II=Ex_dot_QL/W_dot_in Ex_dot_dest=W_dot_in-Ex_dot_QL The solutions in the case of ammonia, R-134a and R-22 are PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 31. 11-31 Now, we investigate the effects of evaporating and condenser pressures on the COP, the second-law efficiency and the total exergy destruction. The results are given by tables and figures. 4.5 0.6 0.55 4 3.5 0.45 0.4 3 η II COP 0.5 0.35 2.5 0.3 2 100 150 200 250 300 350 0.25 400 P 1 [kPa] PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 32. 11-32 4.5 4 Exdest [kW] 3.5 3 2.5 2 1.5 1 100 150 200 250 300 350 400 P1 [kPa] 5 0.7 0.65 4.55 0.6 4.1 0.5 3.65 η II COP 0.55 0.45 3.2 0.4 2.75 1000 1200 1400 1600 1800 0.35 2000 P2 [kPa] 3 Exdest [kW] 2.6 2.2 1.8 1.4 1 1000 1200 1400 1600 1800 2000 P2 [kPa] PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 33. 11-33 Selecting the Right Refrigerant 11-38C The desirable characteristics of a refrigerant are to have an evaporator pressure which is above the atmospheric pressure, and a condenser pressure which corresponds to a saturation temperature above the temperature of the cooling medium. Other desirable characteristics of a refrigerant include being nontoxic, noncorrosive, nonflammable, chemically stable, having a high enthalpy of vaporization (minimizes the mass flow rate) and, of course, being available at low cost. 11-39C The minimum pressure that the refrigerant needs to be compressed to is the saturation pressure of the refrigerant at 30°C, which is 0.771 MPa. At lower pressures, the refrigerant will have to condense at temperatures lower than the temperature of the surroundings, which cannot happen. 11-40C Allowing a temperature difference of 10°C for effective heat transfer, the evaporation temperature of the refrigerant should be -20°C. The saturation pressure corresponding to -20°C is 0.133 MPa. Therefore, the recommended pressure would be 0.12 MPa. 11-41 A refrigerator that operates on the ideal vapor-compression cycle with refrigerant-134a is considered. Reasonable pressures for the evaporator and the condenser are to be selected. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis Allowing a temperature difference of 10°C for effective heat transfer, the evaporation and condensation temperatures of the refrigerant should be -20°C and 35°C, respectively. The saturation pressures corresponding to these temperatures are 0.133 MPa and 0.888 MPa. Therefore, the recommended evaporator and condenser pressures are 0.133 MPa and 0.888 MPa, respectively. 11-42 A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a is considered. Reasonable pressures for the evaporator and the condenser are to be selected. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis Allowing a temperature difference of 10°C for effective heat transfer, the evaporation and condensation temperatures of the refrigerant should be 4°C and 36°C, respectively. The saturation pressures corresponding to these temperatures are 338 kPa and 912 kPa. Therefore, the recommended evaporator and condenser pressures are 338 kPa and 912 kPa, respectively. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 34. 11-34 Heat Pump Systems 11-43C A heat pump system is more cost effective in Miami because of the low heating loads and high cooling loads at that location. 11-44C A water-source heat pump extracts heat from water instead of air. Water-source heat pumps have higher COPs than the air-source systems because the temperature of water is higher than the temperature of air in winter. 11-45E A heat pump operating on the ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP of the heat pump is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-11E, A-12E, and A-13E), T1 = 40°F ⎫ h1 = h g @ 40° F = 108.78 Btu/lbm ⎬ sat. vapor ⎭ s1 = s g @ 40°F = 0.22189 kJ/kg ⋅ K P2 = 100 psia ⎫ ⎬ h2 = 114.98 Btu/lbm s 2 = s1 ⎭ P3 = 100 psia ⎫ ⎬ h3 = h f sat. liquid ⎭ @ 100 psia · QH 2 3 100 psia · Win = 37.869 Btu/lbm h4 ≅ h3 = 37.869 Btu/lbm ( throttling) The COP of the heat pump is determined from its definition, COPHP = T h -h qH 114.98 − 37.869 = 2 3 = = 12.43 win h2 -h1 114.98 − 108.78 40°F 4s 4 · 1 QL s PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 35. 11-35 11-46 A heat pump operating on the ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP and the rate of heat supplied to the evaporator are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-11, A-12, and A-13), P1 = 200 kPa ⎫ h1 = h g @ 200 kPa = 244.46 kJ/kg ⎬ s =s sat. vapor g @ 200 kPa = 0.93773 kJ/kg ⋅ K ⎭ 1 T P2 = 1000 kPa ⎫ ⎬ h2 = 277.98 kJ/kg s 2 = s1 ⎭ P3 = 1000 kPa ⎫ ⎬ h3 = h f sat. liquid ⎭ @ 1000 kPa · QH 3 1.0 MPa = 107.32 kJ/kg h4 ≅ h3 = 107.32 kJ/kg ( throttling) 2 · Win 200 kPa 4s The mass flow rate of the refrigerant is determined from & Win 6 kJ/s & & Win = m(h2 − h1 ) ⎯ m = ⎯→ & = = 0.179 kg/s h2 − h1 (277.98 − 244.46) kJ/kg 4 · QL 1 s Then the rate of heat supplied to the evaporator is & & QL = m(h1 − h4 ) = (0.179 kg/s)(244.46 − 107.32) kJ/kg = 24.5 kW The COP of the heat pump is determined from its definition, COPHP = q H h2 − h3 277.98 − 107.32 = = = 5.09 win h2 − h1 277.98 − 244.46 PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 36. 11-36 11-47 A heat pump operating on the ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The rate of heat transfer to the heated space and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. & Q H Condenser 3 2 Expansion valve Compressor 4 T 1.4 MPa · & Win 1 Evaporator · 3 1.4 MPa W·in 20°C 4s 4 sat. vap. 20°C 2 QH & QL · 1 QL s (b) The properties as given in the problem statement are h4 = h3 = hf @ 1400 kPa = 127.2 kJ/kg h1 = hg @ 20°C = 261.6 kJ/kg. The enthalpy at the compressor exit is win = h2 − h1 ⎯ h2 = h1 + win = 261.6 kJ/kg + 20 kJ/kg = 281.6 kJ/kg ⎯→ The mass flow rate through the cycle is & & Q L = m(h1 − h4 ) ⎯ m = ⎯→ & & QL 2.7 kJ/s = = 0.02009 kg/s h1 − h4 (261.6 − 127.2) kJ/kg The rate of heat transfer to the heated space is & & QH = m(h2 − h3 ) = (0.02009 kg/s)(281.6 − 127.2) kJ/kg = 3.10 kW (c) The COP of the heat pump is COPHP = & & QH Q 3.10 kW = H = = 7.72 & & mwin (0.02009 kg/s)(20 kJ/kg) Win PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 37. 11-37 11-48 A heat pump vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The hardware and the T-s diagram for this heat pump are to be sketched. The power input and the COP are to be determined. Analysis (a) In a normal vapor-compression refrigeration cycle, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. & QH 1.6 MPa T Condenser · QH 2s 3 Expansion valve 2 & Win 2 3 1.6 MPa · Win Compressor 240 kPa 4 1 4s 4 Evaporator sat. vap. 240 kPa · 1 QL s & QL (b) The properties as given in the problem statement are h4 = h3 = hf @ 1600 kPa = 134 kJ/kg h1 = hg @ 240 kPa = 244 kJ/kg. h2s = 285 kJ/kg. From the definition of isentropic efficiency for a compressor, η Comp = h2 s − h1 h − h1 285 − 244 ⎯ h2 = h1 + 2 s ⎯→ = 244 + = 292.2 kJ/kg 0.85 h2 − h1 η Comp Then the work input to the compressor is win = h2 − h1 = 292.2 − 244 = 48.2 kJ/kg The mass flow rate through the cycle is & QH ⎛ 211/60 kJ/s ⎞ (2 ton)⎜ ⎟ & QH ⎠ = 0.04446 kg/s ⎝ 1 ton & = m(h2 − h3 ) ⎯ m = ⎯→ & = (292.2 − 134) kJ/kg h2 − h3 Then the power input to the compressor is & & Win = mwin = (0.04446 kg/s)(48.2 kJ/kg) = 2.14 kW The COP of the heat pump is COPHP ⎛ 211/60 kJ/s ⎞ (2 ton)⎜ ⎟ & QH ⎠ = 3.29 ⎝ 1 ton = = & 2.14 kW W in PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 38. 11-38 11-49 A geothermal heat pump is considered. The degrees of subcooling done on the refrigerant in the condenser, the mass flow rate of the refrigerant, the heating load, the COP of the heat pump, the minimum power input are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the refrigerant-134a tables (Tables A-11 through A-13) T4 = 20°C⎫ P4 = 572.1 kPa ⎬ x 4 = 0.23 ⎭ h4 = 121.24 kJ/kg h3 = h4 Condenser 3 Expansion valve P1 = 572.1 kPa ⎫ h1 = 261.59 kJ/kg ⎬ x1 = 1 (sat. vap.)⎭ s1 = 0.9223 kJ/kg P2 = 1400 kPa ⎫ ⎬ h2 = 280.00 kJ/kg s 2 = s1 ⎭ From the steam tables (Table A-4) h w1 = h f @ 50°C = 209.34 kJ/kg . QH 4 Compressor sat. vap. 20°C x=0.23 40°C Water 50°C T · QH Tsat @ 1400 kPa = 52.40°C 2 1.4 MPa P3 = 1400 kPa ⎫ ⎬ T3 = 48.59°C (from EES) h3 = 121.24 kJ/kg ⎭ · Win 3 Then, the degrees of subcooling is . Win 1 Evaporator h w2 = h f @ 40°C = 167.53 kJ/kg The saturation temperature at the condenser pressure of 1400 kPa and the actual temperature at the condenser outlet are 2 1.4 MPa s2 = s1 4s 4 ∆Tsubcool = Tsat − T3 = 52.40 − 48.59 = 3.81°C · QL 1 s (b) The rate of heat absorbed from the geothermal water in the evaporator is & & Q L = m w (hw1 − hw2 ) = (0.065 kg/s)(209.34 − 167.53)kJ/kg = 2.718 kW This heat is absorbed by the refrigerant in the evaporator & mR = & QL 2.718 kW = = 0.01936 kg/s h1 − h4 (261.59 − 121.24)kJ/kg (c) The power input to the compressor, the heating load and the COP are & & & Win = m R (h2 − h1 ) + Qout = (0.01936 kg/s)(280.00 − 261.59)kJ/kg = 0.6564 kW & & Q H = m R (h2 − h3 ) = (0.01936 kg/s)(280.00 − 121.24)kJ/kg = 3.074 kW COP = & QH 3.074 kW = = 4.68 & Win 0.6564 kW (d) The reversible COP of the cycle is COPrev = 1 1 − TL / TH = 1 = 12.92 1 − (25 + 273) /(50 + 273) The corresponding minimum power input is & Win, min = & QH 3.074 kW = = 0.238 kW COPrev 12.92 PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 39. 11-39 11-50 An actual heat pump cycle with R-134a as the refrigerant is considered. The isentropic efficiency of the compressor, the rate of heat supplied to the heated room, the COP of the heat pump, and the COP and the rate of heat supplied to the heated room if this heat pump operated on the ideal vapor-compression cycle between the same pressure limits are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The properties of refrigerant-134a are (Tables A-11 through A-13) P2 = 800 kPa ⎫ ⎬ h2 = 291.76 kJ/kg ⎭ T3 = Tsat@750 kPa = 29.06°C T2 = 55°C . QH 750 kPa P3 = 750 kPa ⎫ ⎬ h3 = 87.91 kJ/kg T3 = (29.06 − 3)°C⎭ h4 = h3 = 87.91 kJ/kg Tsat@200 kPa = −10.09°C P1 = 200 kPa ⎫h1 = 247.87 kJ/kg ⎬ T1 = (−10.09 + 4)°C⎭ s1 = 0.9506 kJ/kg P2 = 800 kPa ⎫ ⎬h2 s = 277.26 s 2 = s1 ⎭ 800 kPa 55°C Condenser 3 2 Expansion valve . Win Compressor 1 4 Evaporator . QL The isentropic efficiency of the compressor is 2 T 277.26 − 247.87 h −h ηC = 2 s 1 = = 0.670 291.76 − 247.87 h2 − h1 2 · Q (b) The rate of heat supplied to the room is · Win 3 & & Q H = m(h2 − h3 ) = (0.018 kg/s)(291.76 − 87.91)kJ/kg = 3.67 kW (c) The power input and the COP are · QL 4 & & Win = m(h2 − h1 ) = (0.018 kg/s)(291.76 − 247.87)kJ/kg = 0.790 kW COP = 1 s & QH 3.67 = = 4.64 & Win 0.790 (d) The ideal vapor-compression cycle analysis of the cycle is as follows: h1 = h g @ 200 kPa = 244.46 kJ/kg s1 = s g @ 200 kPa = 0.9377 kJ/kg.K P2 = 800 kPa ⎫ ⎬ h2 = 273.25 kJ/kg s 2 = s1 ⎭ T · QH 2 3 0.8 MPa · Win h3 = h f @ 800 kPa = 95.47 kJ/kg h4 = h3 h − h3 273.25 − 95.47 COP = 2 = = 6.18 h2 − h1 273.25 − 244.46 0.2 MPa 4s 4 · QL 1 s & & Q H = m(h2 − h3 ) = (0.018 kg/s)(273.25 − 95.47)kJ/kg = 3.20 kW PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 40. 11-40 Innovative Refrigeration Systems 11-51C Performing the refrigeration in stages is called cascade refrigeration. In cascade refrigeration, two or more refrigeration cycles operate in series. Cascade refrigerators are more complex and expensive, but they have higher COP's, they can incorporate two or more different refrigerants, and they can achieve much lower temperatures. 11-52C Cascade refrigeration systems have higher COPs than the ordinary refrigeration systems operating between the same pressure limits. 11-53C The saturation pressure of refrigerant-134a at -32°C is 77 kPa, which is below the atmospheric pressure. In reality a pressure below this value should be used. Therefore, a cascade refrigeration system with a different refrigerant at the bottoming cycle is recommended in this case. 11-54C We would favor the two-stage compression refrigeration system with a flash chamber since it is simpler, cheaper, and has better heat transfer characteristics. 11-55C Yes, by expanding the refrigerant in stages in several throttling devices. 11-56C To take advantage of the cooling effect by throttling from high pressures to low pressures. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 41. 11-41 11-57 A two-stage compression refrigeration system with refrigerant-134a as the working fluid is considered. The fraction of the refrigerant that evaporates as it is throttled to the flash chamber, the rate of heat removed from the refrigerated space, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 The flash chamber is adiabatic. Analysis (a) The enthalpies of the refrigerant at several states are determined from the refrigerant tables (Tables A-11, A12, and A-13) to be h1 = 234.44 kJ/kg, h2 = 262.68 kJ/kg T 1.4 MPa h3 = 255.55 kJ/kg, 4 h5 = 127.22 kJ/kg , h6 = 127.22 kJ/kg h7 = 63.94 kJ/kg, h8 = 63.94 kJ/kg The fraction of the refrigerant that evaporates as it is throttled to the flash chamber is simply the quality at state 6, x6 = h6 − h f h fg 127.22 − 63.94 = = 0.3303 191.62 0.4 MPa 5 2 A 7 6 8 B 0.10 MPa 3 9 · QL 1 s (b) The enthalpy at state 9 is determined from an energy balance on the mixing chamber: & & & E in − E out = ∆E system 0 (steady) =0 & & E in = E out & & ∑m h = ∑m h e e i i (1)h9 = x6 h3 + (1 − x6 )h2 h9 = (0.3303)(255.55) + (1 − 0.3303)(262.68) = 260.33 kJ/kg P9 = 0.4 MPa ⎫ ⎬ s 9 = 0.9437 kJ/kg ⋅ K h9 = 260.33 kJ/kg ⎭ also, P4 = 1.4 MPa ⎫ ⎬ h4 = 287.07 kJ/kg s 4 = s 9 = 0.9437 kJ/kg ⋅ K ⎭ Then the rate of heat removed from the refrigerated space and the compressor work input per unit mass of refrigerant flowing through the condenser are & & m B = (1 − x 6 )m A = (1 − 0.3303)(0.25 kg/s ) = 0.1674 kg/s & & Q L = m B (h1 − h8 ) = (0.1674 kg/s )(234.44 − 63.94 ) kJ/kg = 28.55 kW & & & & & Win = WcompI,in + WcompII,in = m A (h4 − h9 ) + m B (h2 − h1 ) = (0.25 kg/s )(287.07 − 260.33) kJ/kg + (0.1674 kg/s )(262.68 − 234.44 ) kJ/kg = 11.41 kW (c) The coefficient of performance is determined from COPR = & QL & W net,in = 28.55 kW = 2.50 11.41 kW PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 42. 11-42 11-58 A two-stage compression refrigeration system with refrigerant-134a as the working fluid is considered. The fraction of the refrigerant that evaporates as it is throttled to the flash chamber, the rate of heat removed from the refrigerated space, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 The flash chamber is adiabatic. Analysis (a) The enthalpies of the refrigerant at several states are determined from the refrigerant tables (Tables A-11, A12, and A-13) to be h1 = 234.44 kJ/kg, h2 = 271.40 kJ/kg h3 = 262.40 kJ/kg, T h5 = 127.22 kJ/kg, h7 = 81.51 kJ/kg, h8 = 81.51 kJ/kg 1.4 MPa h6 = 127.22 kJ/kg The fraction of the refrigerant that evaporates as it is throttled to the flash chamber is simply the quality at state 6, x6 = h6 − h f h fg = 4 0.6 MPa 5 2 A 7 127.22 − 81.51 = 0.2527 180.90 6 B 8 (b) The enthalpy at state 9 is determined from an energy balance on the mixing chamber: & & & E in − E out = ∆Esystem 0 (steady) 0.10 MPa 3 9 · QL 1 s =0 & & E in = E out & & ∑m h = ∑m h e e i i (1)h9 = x6 h3 + (1 − x6 )h2 h9 = (0.2527 )(262.40) + (1 − 0.2527 )(271.40) = 269.13 kJ/kg P9 = 0.6 MPa ⎫ ⎬ s 9 = 0.9443 kJ/kg ⋅ K h9 = 269.13 kJ/kg ⎭ also, P4 = 1.4 MPa ⎫ ⎬ h4 = 287.28 kJ/kg s 4 = s 9 = 0.9443 kJ/kg ⋅ K ⎭ Then the rate of heat removed from the refrigerated space and the compressor work input per unit mass of refrigerant flowing through the condenser are & & m B = (1 − x 6 )m A = (1 − 0.2527 )(0.25 kg/s ) = 0.1868 kg/s & & Q L = m B (h1 − h8 ) = (0.1868 kg/s )(234.44 − 81.51) kJ/kg = 28.57 kW & & & & & Win = WcompI,in + WcompII,in = m A (h4 − h9 ) + m B (h2 − h1 ) = (0.25 kg/s )(287.28 − 269.13) kJ/kg + (0.1868 kg/s )(271.40 − 234.44 ) kJ/kg = 11.44 kW (c) The coefficient of performance is determined from COPR = & QL & W net,in = 28.57 kW = 2.50 11.44 kW PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 43. 11-43 11-59 Problem 11-57 is reconsidered. The effects of the various refrigerants in EES data bank for compressor efficiencies of 80, 90, and 100 percent is to be investigated. Analysis The problem is solved using EES, and the results are tabulated and plotted below. Fluid$='R134a' "Input Data" P[1]=100 [kPa] P[4] = 1400 [kPa] P[6]=400 [kPa] "Eta_comp =1.0" m_dot_A=0.25 [kg/s] "High Pressure Compressor A" P[9]=P[6] h4s=enthalpy(Fluid$,P=P[4],s=s[9]) "State 4s is the isentropic value of state 4" h[9]+w_compAs=h4s "energy balance on isentropic compressor" w_compA=w_compAs/Eta_comp"definition of compressor isentropic efficiency" h[9]+w_compA=h[4] "energy balance on real compressor-assumed adiabatic" s[4]=entropy(Fluid$,h=h[4],P=P[4]) "properties for state 4" T[4]=temperature(Fluid$,h=h[4],P=P[4]) W_dot_compA=m_dot_A*w_compA "Condenser" P[5]=P[4] "neglect pressure drops across condenser" T[5]=temperature(Fluid$,P=P[5],x=0) "properties for state 5, assumes sat. liq. at cond. exit" h[5]=enthalpy(Fluid$,T=T[5],x=0) "properties for state 5" s[5]=entropy(Fluid$,T=T[5],x=0) h[4]=q_H+h[5] "energy balance on condenser" Q_dot_H = m_dot_A*q_H "Throttle Valve A" h[6]=h[5] "energy balance on throttle - isenthalpic" x6=quality(Fluid$,h=h[6],P=P[6]) "properties for state 6" s[6]=entropy(Fluid$,h=h[6],P=P[6]) T[6]=temperature(Fluid$,h=h[6],P=P[6]) "Flash Chamber" m_dot_B = (1-x6) * m_dot_A P[7] = P[6] h[7]=enthalpy(Fluid$, P=P[7], x=0) s[7]=entropy(Fluid$,h=h[7],P=P[7]) T[7]=temperature(Fluid$,h=h[7],P=P[7]) "Mixing Chamber" x6*m_dot_A*h[3] + m_dot_B*h[2] =(x6* m_dot_A + m_dot_B)*h[9] P[3] = P[6] h[3]=enthalpy(Fluid$, P=P[3], x=1) "properties for state 3" s[3]=entropy(Fluid$,P=P[3],x=1) T[3]=temperature(Fluid$,P=P[3],x=x1) s[9]=entropy(Fluid$,h=h[9],P=P[9]) "properties for state 9" T[9]=temperature(Fluid$,h=h[9],P=P[9]) "Low Pressure Compressor B" x1=1 "assume flow to compressor inlet to be saturated vapor" h[1]=enthalpy(Fluid$,P=P[1],x=x1) "properties for state 1" T[1]=temperature(Fluid$,P=P[1], x=x1) PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 44. 11-44 s[1]=entropy(Fluid$,P=P[1],x=x1) P[2]=P[6] h2s=enthalpy(Fluid$,P=P[2],s=s[1]) " state 2s is isentropic state at comp. exit" h[1]+w_compBs=h2s "energy balance on isentropic compressor" w_compB=w_compBs/Eta_comp"definition of compressor isentropic efficiency" h[1]+w_compB=h[2] "energy balance on real compressor-assumed adiabatic" s[2]=entropy(Fluid$,h=h[2],P=P[2]) "properties for state 2" T[2]=temperature(Fluid$,h=h[2],P=P[2]) W_dot_compB=m_dot_B*w_compB "Throttle Valve B" h[8]=h[7] "energy balance on throttle - isenthalpic" x8=quality(Fluid$,h=h[8],P=P[8]) "properties for state 8" s[8]=entropy(Fluid$,h=h[8],P=P[8]) T[8]=temperature(Fluid$,h=h[8],P=P[8]) "Evaporator" P[8]=P[1] "neglect pressure drop across evaporator" q_L + h[8]=h[1] "energy balance on evaporator" Q_dot_L=m_dot_B*q_L "Cycle Statistics" W_dot_in_total = W_dot_compA + W_dot_compB COP=Q_dot_L/W_dot_in_total "definition of COP" ηcomp 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 QL [kW] 28.55 28.55 28.55 28.55 28.55 28.55 28.55 28.55 28.55 COP 1.438 1.57 1.702 1.835 1.968 2.101 2.234 2.368 2.501 R134a 250 200 T [°C] 150 100 4 5 1400 kPa 50 7 400 kPa 0 100 kPa -50 -100 -0.25 2,3,9 6 1 8 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 s [kJ/kg-K] PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 45. 11-45 2.8 2.6 COP 2.4 2.2 R134a ammonia 2 R22 1.8 1.6 1.4 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 0.9 0.95 1 800 900 η comp 300 250 ammonia QL [kW] 200 150 100 R22 50 0 0.6 R134a 0.65 0.7 0.75 0.8 η comp 0.85 R134a 2.6 2.5 COP 2.4 2.3 2.2 2.1 2 1.9 100 200 300 400 500 600 700 1000 P[6] [kPa] PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 46. 11-46 11-60 A two-stage cascade refrigeration cycle is considered. The mass flow rate of the refrigerant through the upper cycle, the rate of heat removal from the refrigerated space, and the COP of the refrigerator are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The properties are to be obtained from the refrigerant tables (Tables A-11 through A-13): h1 = h g @ 200 kPa = 244.46 kJ/kg . QH s1 = s g @ 200 kPa = 0.9377 kJ/kg.K P2 = 500 kPa ⎫ ⎬ h2 s = 263.30 kJ/kg s 2 = s1 ⎭ h −h η C = 2s 1 h2 − h1 0.80 = 263.30 − 244.46 ⎯ ⎯→ h2 = 268.01 kJ/kg h2 − 244.46 Condenser 7 Expansion valve 8 h3 = h f @ 500 kPa = 73.33 kJ/kg h4 = h3 = 73.33 kJ/kg h5 = h g @ 400 kPa = 255.55 kJ/kg s 5 = s g @ 400 kPa = 0.9269 kJ/kg.K P6 = 1200 kPa ⎫ ⎬ h6 s = 278.33 kJ/kg s6 = s5 ⎭ ηC = 0.80 = 6 . Win Compressor Evaporator 5 Condenser 3 Expansion valve 4 h 6 s − h5 h6 − h5 2 . Win Compressor Evaporator . QL 1 278.33 − 255.55 ⎯ ⎯→ h6 = 284.02 kJ/kg h6 − 255.55 h7 = h f @ 1200 kPa = 117.77 kJ/kg h8 = h7 = 117.77 kJ/kg The mass flow rate of the refrigerant through the upper cycle is determined from an energy balance on the heat exchanger & & m A (h5 − h8 ) = m B (h2 − h3 ) & & m A (255.55 − 117.77)kJ/kg = (0.15 kg/s)(268.01 − 73.33)kJ/kg ⎯ ⎯→ m A = 0.212 kg/s (b) The rate of heat removal from the refrigerated space is & & Q L = m B (h1 − h4 ) = (0.15 kg/s)(244.46 − 73.33)kJ/kg = 25.67 kW (c) The power input and the COP are & & & Win = m A (h6 − h5 ) + m B (h2 − h1 ) = (0.15 kg/s)(284.02 − 255.55)kJ/kg + (0.212 kg/s)(268.01 − 244.46)kJ/kg = 9.566 kW COP = & Q L 25.67 = = 2.68 & Win 9.566 PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 47. 11-47 11-61 A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The cooling rate of the high-temperature evaporator, the power required by the compressor, and the COP of the system are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Condenser 2 T 3 Compressor 1 Expansion valve Expansion valve · QH 3 800 kPa Evaporator 1 4 5 0°C 4 -26.4°C Expansion valve 7 2 6 Evaporator 2 · Win 5 · 7 1 QL s 6 Analysis From the refrigerant tables (Tables A-11, A-12, and A-13), P3 = 800 kPa ⎫ ⎬ h3 = h f sat. liquid ⎭ @ 800 kPa = 95.47 kJ/kg h4 = h6 ≅ h3 = 95.47 kJ/kg ( throttling ) T5 = 0°C ⎫ ⎬ h = h g @ 0°C = 250.45 kJ/kg sat. vapor ⎭ 5 T7 = −26.4°C ⎫ ⎬ h7 = h g @ − 26.4°C = 234.44 kJ/kg sat. vapor ⎭ The mass flow rate through the low-temperature evaporator is found by & QL 8 kJ/s & & & Q L = m 2 ( h7 − h6 ) ⎯ ⎯→ m 2 = = = 0.05757 kg/s h7 − h6 (234.44 − 95.47) kJ/kg The mass flow rate through the warmer evaporator is then & & & m1 = m − m 2 = 0.1 − 0.05757 = 0.04243 kg/s Applying an energy balance to the point in the system where the two evaporator streams are recombined gives & & m h + m 2 h7 (0.04243)(250.45) + (0.05757)(234.44) & & & m1 h5 + m 2 h7 = mh1 ⎯ h1 = 1 5 ⎯→ = = 241.23 kJ/kg & m 0.1 Then, P1 = Psat @ − 26.4°C ≅ 100 kPa ⎫ ⎬ s1 = 0.9789 kJ/kg ⋅ K h1 = 241.23 kJ/kg ⎭ P2 = 800 kPa ⎫ ⎬ h2 = 286.26 kJ/kg s 2 = s1 ⎭ The cooling rate of the high-temperature evaporator is & & Q = m (h − h ) = (0.04243 kg/s)(250.45 − 95.47) kJ/kg = 6.58 kW L 1 5 4 The power input to the compressor is & & W = m(h − h ) = (0.1 kg/s)(286.26 − 241.23) kJ/kg = 4.50 kW in 2 1 The COP of this refrigeration system is determined from its definition, & Q (8 + 6.58) kW COPR = L = = 3.24 & 4.50 kW Win PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 48. 11-48 11-62E A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The power required by the compressor and the COP of the system are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Condenser 2 T 3 Compressor 1 Expansion valve Expansion valve Evaporator 1 Expansion valve 7 2 3 180 psia · Win 5 30 psia 4 4 5 · QH 10 psia 6 · 7 1 QL Evaporator 2 6 s Analysis From the refrigerant tables (Tables A-11E, A-12E, and A-13E), P3 = 180 psia ⎫ ⎬ h3 = h f sat. liquid ⎭ @ 180 psia = 51.50 Btu/lbm h4 = h6 ≅ h3 = 51.50 Btu/lbm ( throttling ) P5 = 30 psia ⎫ ⎬ h = h g @ 30 psia = 105.32 Btu/lbm sat. vapor ⎭ 5 P7 = 10 psia ⎫ ⎬ h = h g @ 10 psia = 98.68 Btu/lbm sat. vapor ⎭ 7 The mass flow rates through the high-temperature and low-temperature evaporators are found by & Q L ,1 9000 Btu/h & & Q L ,1 = m1 (h5 − h4 ) ⎯ m1 = ⎯→ & = = 167.2 lbm/h h5 − h4 (105.32 − 51.50) Btu/lbm & QL,2 24,000 Btu/h & & Q L , 2 = m 2 (h7 − h6 ) ⎯ m 2 = ⎯→ & = = 508.6 lbm/h h7 − h6 (98.68 − 51.50) Btu/lbm Applying an energy balance to the point in the system where the two evaporator streams are recombined gives & & m h + m2 h7 (167.2)(105.32) + (508.6)(98.68) & & & & = 100.33 Btu/lbm m1h5 + m 2 h7 = (m1 + m 2 )h1 ⎯ h1 = 1 5 ⎯→ = & 1 +m 2 & 167.2 + 508.6 m Then, P1 = 10 psia ⎫ ⎬ s = 0.2333 Btu/lbm ⋅ R h1 = 100.33 Btu/lbm ⎭ 1 P2 = 180 psia ⎫ ⎬ h2 = 127.05 Btu/lbm s 2 = s1 ⎭ The power input to the compressor is 1 kW ⎛ ⎞ & & & Win = (m1 + m 2 )(h2 − h1 ) = (167.2 + 508.6) lbm/h(127.05 − 100.33) Btu/lbm⎜ ⎟ = 5.29 kW ⎝ 3412.14 Btu/h ⎠ The COP of this refrigeration system is determined from its definition, & Q (24,000 + 9000) Btu/h ⎛ 1 kW ⎞ COPR = L = ⎜ ⎟ = 1.83 & 5.29 kW W ⎝ 3412.14 Btu/h ⎠ in PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 49. 11-49 11-63E A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The power required by the compressor and the COP of the system are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Condenser 2 3 Compressor 1 Expansion valve Expansion valve Evaporator 1 · QH 2 3 180 psia · Win 5 60 psia 4 5 7 T 4 Expansion valve 10 psia 6 · 7 1 QL Evaporator 2 6 s Analysis From the refrigerant tables (Tables A-11E, A-12E, and A-13E), P3 = 180 psia ⎫ ⎬ h3 = h f sat. liquid ⎭ @ 180 psia = 51.50 Btu/lbm h4 = h6 ≅ h3 = 51.50 Btu/lbm ( throttling ) P5 = 60 psia ⎫ ⎬ h = h g @ 60 psia = 110.11 Btu/lbm sat. vapor ⎭ 5 P7 = 10 psia ⎫ ⎬ h = h g @ 10 psia = 98.68 Btu/lbm sat. vapor ⎭ 7 The mass flow rates through the high-temperature and low-temperature evaporators are found by & Q L ,1 30,000 Btu/h & & Q L,1 = m1 (h5 − h4 ) ⎯ m1 = ⎯→ & = = 511.8 lbm/h h5 − h4 (110.11 − 51.50) Btu/lbm & QL,2 24,000 Btu/h & & Q L , 2 = m 2 (h7 − h6 ) ⎯ m 2 = ⎯→ & = = 508.6 lbm/h h7 − h6 (98.68 − 51.50) Btu/lbm Applying an energy balance to the point in the system where the two evaporator streams are recombined gives & & m h + m2 h7 (511.8)(110.11) + (508.6)(98.68) & & & & = 104.41 Btu/lbm m1h5 + m2 h7 = (m1 + m 2 )h1 ⎯ h1 = 1 5 ⎯→ = & 1 +m 2 & 511.8 + 508.6 m Then, P1 = 10 psia ⎫ ⎬ s = 0.2423 Btu/lbm ⋅ R h1 = 104.41 Btu/lbm ⎭ 1 P2 = 180 psia ⎫ ⎬ h2 = 132.69 Btu/lbm s 2 = s1 ⎭ The power input to the compressor is 1 kW ⎛ ⎞ & & & Win = (m1 + m 2 )(h2 − h1 ) = (511.8 + 508.6) lbm/h(132.69 − 104.41) Btu/lbm⎜ ⎟ = 8.46 kW ⎝ 3412.14 Btu/h ⎠ The COP of this refrigeration system is determined from its definition, & Q (24,000 + 30,000) Btu/h ⎛ 1 kW ⎞ COPR = L = ⎜ ⎟ = 1.87 & 8.46 kW 3412.14 Btu/h ⎠ Win ⎝ PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 50. 11-50 11-64 A two-stage cascade refrigeration system is considered. Each stage operates on the ideal vapor-compression cycle with upper cycle using water and lower cycle using refrigerant-134a as the working fluids. The mass flow rate of R-134a and water in their respective cycles and the overall COP of this system are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 The heat exchanger is adiabatic. Analysis From the water and refrigerant tables (Tables A-4, A-5, A-6, A-11, A-12, and A-13), T1 = 5°C ⎫ h1 = h g @ 5°C = 2510.1 kJ/kg ⎬ sat. vapor ⎭ s1 = s g @ 5°C = 9.0249 kJ/kg ⋅ K P2 = 1.6 MPa ⎫ ⎬ h2 = 5083.4 kJ/kg s 2 = s1 ⎭ P3 = 1.6 MPa ⎫ ⎬ h3 = h f sat. liquid ⎭ @ 1.6 MPa T 2 3 = 858.44 kJ/kg h4 ≅ h3 = 858.44 kJ/kg ( throttling) T5 = −40°C ⎫ h5 = h g @ − 40°C = 225.86 kJ/kg ⎬ sat. vapor ⎭ s 5 = s g @ − 40°C = 0.96866 kJ/kg ⋅ K 7 1.6 MPa 6 5°C 4 400 kPa 1 8 -40°C · QL P6 = 400 kPa ⎫ ⎬ h6 = 267.59 kJ/kg s6 = s5 ⎭ P7 = 400 kPa ⎫ ⎬ h7 = h f sat. liquid ⎭ @ 400 kPa 5 s = 63.94 kJ/kg h8 ≅ h7 = 63.94 kJ/kg ( throttling) The mass flow rate of R-134a is determined from & & & Q L = m R (h5 − h8 ) ⎯ ⎯→ m R = & QL 20 kJ/s = = 0.1235 kg/s h5 − h8 (225.86 − 63.94) kJ/kg An energy balance on the heat exchanger gives the mass flow rate of water & & m R (h6 − h7 ) = m w (h1 − h4 ) & & ⎯ ⎯→ m w = m R h6 − h7 267.59 − 63.94 = (0.1235 kg/s) = 0.01523 kg/s h1 − h4 2510.1 − 858.44 The total power input to the compressors is & & & Win = m R (h6 − h5 ) + m w (h2 − h1 ) = (0.1235 kg/s)(267.59 − 225.86) kJ/kg + (0.01523 kg/s)(5083.4 − 2510.1) kJ/kg = 44.35 kJ/s The COP of this refrigeration system is determined from its definition, COPR = & QL 20 kJ/s = = 0.451 & 44.35 kJ/s Win PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 51. 11-51 11-65 A two-stage vapor-compression refrigeration system with refrigerant-134a as the working fluid is considered. The process with the greatest exergy destruction is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From Prob. 11-55 and the water and refrigerant tables (Tables A-4, A-5, A-6, A-11, A-12, and A-13), s1 = s 2 = 9.0249 kJ/kg ⋅ K s 3 = 2.3435 kJ/kg ⋅ K s 4 = 3.0869 kJ/kg ⋅ K s 5 = s 6 = 0.96866 kJ/kg ⋅ K s 7 = 0.24757 kJ/kg ⋅ K s 8 = 0.27423 kJ/kg ⋅ K & m R = 0.1235 kg/s & m w = 0.01523 kg/s q L = h5 − h8 = 161.92 kJ/kg q H = h2 − h3 = 4225.0 kJ/kg T L = −30°C = 243 K T H = 30°C = 303 K T0 = 30°C = 303 K T 3 7 1.6 MPa 2 6 5°C 4 400 kPa 1 8 -40°C · QL 5 s The exergy destruction during a process of a stream from an inlet state to exit state is given by ⎛ q q x dest = T0 s gen = T0 ⎜ s e − s i − in + out ⎜ Tsource Tsink ⎝ ⎞ ⎟ ⎟ ⎠ Application of this equation for each process of the cycle gives ⎛ q ⎞ & & X destroyed, 23 = m w T0 ⎜ s 3 − s 2 + H ⎟ ⎜ TH ⎟ ⎝ ⎠ 4225.0 ⎞ ⎛ = (0.01523)(303 K )⎜ 2.3435 − 9.0249 + ⎟ = 33.52 kJ/s 303 ⎠ ⎝ & & X destroyed, 34 = m w T0 ( s 4 − s 3 ) = (0.01523)(303)(3.0869 − 2.3435) = 3.43 kJ/s & & X destroyed, 78 = m R T0 ( s 8 − s 7 ) = (0.1235)(303)(0.27423 − 0.24757) = 0.996 kJ/s ⎛ q ⎞ & & X destroyed, 85 = m R T0 ⎜ s 5 − s 8 − L ⎟ ⎜ TL ⎟ ⎝ ⎠ 161.92 ⎞ ⎛ = (0.1235)(303)⎜ 0.96866 − 0.27423 − ⎟ = 1.05 kJ/s 243 ⎠ ⎝ & & & X destroyed, heat exch = T0 [m w ( s1 − s 4 ) + m R ( s 7 − s 6 )] = (303)[(0.01523)(9.0249 − 3.0869) + (0.1235)(0.24757 − 0.96866)] = 0.417 kJ/s For isentropic processes, the exergy destruction is zero: & X destroyed, 12 = 0 & X destroyed, 56 = 0 Note that heat is absorbed from a reservoir at -30°C (243 K) and rejected to a reservoir at 30°C (303 K), which is also taken as the dead state temperature. Alternatively, one may use the standard 25°C (298 K) as the dead state temperature, and perform the calculations accordingly. The greatest exergy destruction occurs in the condenser. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 52. 11-52 11-66 A two-stage cascade refrigeration cycle with a flash chamber with refrigerant-134a as the working fluid is considered. The mass flow rate of the refrigerant through the high-pressure compressor, the rate of refrigeration, the COP are to be determined. Also, the rate of refrigeration and the COP are to be determined if this refrigerator operated on a single-stage vapor-compression cycle under similar conditions. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the refrigerant-134a tables (Tables A-11 through A-13) h1 = h g @ −10°C = 244.51 kJ/kg s1 = s g @ −10°C = 0.9377 kJ/kg.K P2 = 450 kPa ⎫ ⎬ h2 s = 261.07 kJ/kg ⎭ s 2 = s1 ηC = 0.86 = h2 s − h1 h2 − h1 261.07 − 244.51 ⎯ h2 = 263.76 kJ/kg ⎯→ h2 − 244.51 T 1.6 MPa 4s 4 5 A 7 6 9 B 3 −10°C 8 h3 = h g @ 450 kPa = 257.53 kJ/kg 2s 2 0.45 MPa 1 s h5 = h f @ 1600 kPa = 135.93 kJ/kg h6 = h5 = 135.93 kJ/kg h7 = h f @ 450 kPa = 68.81 kJ/kg h8 = h7 = 68.81 kJ/kg h6 = 135.93 kJ/kg ⎫ ⎬ x 6 = 0.3557 P6 = 450 kPa ⎭ The mass flow rate of the refrigerant through the high pressure compressor is determined from a mass balance on the flash chamber & m= & m7 0.11 kg/s = = 0.1707 kg/s 1 − x 6 1 − 0.3557 Also, & & & m3 = m − m7 = 0.1707 − 0.11 = 0.06072 kg/s (b) The enthalpy at state 9 is determined from an energy balance on the mixing chamber: & & & mh9 = m7 h2 + m3 h3 (0.1707 kg/s) h9 = (0.11 kg/s)(263.76 kJ/kg) + (0.06072 kg/s)(257.53 kJ/kg) ⎯ h9 = 261.54 kJ/kg ⎯→ Then, P9 = 450 kPa ⎫ ⎬s 9 = 0.9393 kJ/kg h9 = 261.54 kJ/kg ⎭ P4 = 1600 kPa ⎫ ⎬ h4 s = 288.41 kJ/kg s 4 = s9 ⎭ ηC = 0.86 = h4 s − h9 h4 − h9 288.41 − 261.54 ⎯ h4 = 292.78 kJ/kg ⎯→ h4 − 261.54 PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 53. 11-53 The rate of heat removal from the refrigerated space is & & QL = m7 (h1 − h8 ) = (0.11 kg/s)(244.51 − 68.81)kJ/kg = 19.33 kW (c) The power input and the COP are & & & Win = m7 (h2 − h1 ) + m(h4 − h9 ) = (0.11 kg/s)(263.76 − 244.51)kJ/kg + (0.1707 kg/s)(292.78 − 261.54)kJ/kg = 7.45 kW COP = & QL 19.33 = = 2.59 & 7.45 Win (d) If this refrigerator operated on a single-stage cycle between the same pressure limits, we would have h1 = h g @ −10°C = 244.51 kJ/kg s1 = s g @ −10°C = 0.9377 kJ/kg.K P2 = 1600 kPa ⎫ ⎬ h2 s = 287.85 kJ/kg s 2 = s1 ⎭ h −h η C = 2s 1 h2 − h1 2 T · QH 2s · Win 1.6 MPa 3 287.85 − 244.51 ⎯ h2 = 294.90 kJ/kg ⎯→ 0.86 = h2 − 244.51 −10°C 4 · QL 1 s h3 = h f @ 1600 kPa = 135.93 kJ/kg h4 = h3 = 135.93 kJ/kg & & QL = m(h1 − h4 ) = (0.1707 kg/s)(244.51 − 135.93)kJ/kg = 18.54 kW & & Win = m(h2 − h1 ) = (0.1707 kg/s)(294.90 − 244.51)kJ/kg = 8.60 kW COP = & QL 18.54 = = 2.16 & 8.60 Win Discussion The cooling load decreases by 4.1% while the COP decreases by 16.6% when the cycle operates on the singlestage vapor-compression cycle. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 54. 11-54 Gas Refrigeration Cycles 11-67C The ideal gas refrigeration cycle is identical to the Brayton cycle, except it operates in the reversed direction. 11-68C In the ideal gas refrigeration cycle, the heat absorption and the heat rejection processes occur at constant pressure instead of at constant temperature. 11-69C The reversed Stirling cycle is identical to the Stirling cycle, except it operates in the reversed direction. Remembering that the Stirling cycle is a totally reversible cycle, the reversed Stirling cycle is also totally reversible, and thus its COP is COPR, Stirling = 1 TH / TL − 1 11-70C In aircraft cooling, the atmospheric air is compressed by a compressor, cooled by the surrounding air, and expanded in a turbine. The cool air leaving the turbine is then directly routed to the cabin. 11-71C No; because h = h(T) for ideal gases, and the temperature of air will not drop during a throttling (h1 = h2) process. 11-72C By regeneration. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 55. 11-55 11-73 An ideal-gas refrigeration cycle with air as the working fluid is considered. The rate of refrigeration, the net power input, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Analysis (a) We assume both the turbine and the compressor to be isentropic, the turbine inlet temperature to be the temperature of the surroundings, and the compressor inlet temperature to be the temperature of the refrigerated space. From the air table (Table A-17), T1 = 280 K ⎯ ⎯→ h1 = 280.13 kJ/kg Pr1 = 1.0889 T3 = 310 K ⎯ ⎯→ h3 = 310.24 kJ/kg Pr3 = 1.5546 Thus, T 37°C 7°C P ⎛ 160 ⎞ ⎯→ T2 = 431.5 K Pr2 = 2 Pr1 = ⎜ ⎟(1.0889 ) = 4.978 ⎯ P1 ⎝ 35 ⎠ h2 = 432.96 kJ/kg Pr4 = 2 · QH 3 1 · QL 4 s P4 ⎛ 35 ⎞ Pr3 = ⎜ ⎯→ T4 = 200.6 K ⎟(1.5546 ) = 0.3401 ⎯ P3 ⎝ 160 ⎠ h4 = 200.57 kJ/kg Then the rate of refrigeration is & & & QL = m(q L ) = m(h1 − h4 ) = (0.2 kg/s )(280.13 − 200.57) kJ/kg = 15.9 kW (b) The net power input is determined from & & & Wnet, in = Wcomp, in − Wturb, out where & & Wcomp,in = m(h2 − h1 ) = (0.2 kg/s )(432.96 − 280.13) kJ/kg = 30.57 kW & & Wturb,out = m(h3 − h4 ) = (0.2 kg/s )(310.24 − 200.57) kJ/kg = 21.93 kW Thus, & W net,in = 30.57 − 21.93 = 8.64 kW (c) The COP of this ideal gas refrigeration cycle is determined from COPR = & QL & W net,in = 15.9 kW = 1.84 8.64 kW PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
- 56. 11-56 11-74 An ideal-gas refrigeration cycle with air as the working fluid is considered. The rate of refrigeration, the net power input, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Analysis (a) We assume the turbine inlet temperature to be the temperature of the surroundings, and the compressor inlet temperature to be the temperature of the refrigerated space. From the air table (Table A-17), T1 = 280 K ⎯ ⎯→ h1 = 280.13 kJ/kg Pr1 = 1.0889 2 T T3 = 310 K ⎯ ⎯→ h3 = 310.24 kJ/kg Pr3 = 1.5546 37°C Thus, Pr2 = P2 ⎛ 160 ⎞ ⎯→ T2 s = 431.5 K Pr1 = ⎜ ⎟(1.0889 ) = 4.978 ⎯ P1 ⎝ 35 ⎠ h2 s = 432.96 kJ/kg · QH 3 7°C P ⎛ 35 ⎞ ⎯→ T4 s = 200.6 K Pr4 = 4 Pr3 = ⎜ ⎟(1.5546) = 0.3401 ⎯ P3 ⎝ 160 ⎠ h4 s = 200.57 kJ/kg 2 4s 1 · 4 QL s Also, ηT = h3 − h4 ⎯ ⎯→ h4 = h3 − η T (h3 − h4 s ) h3 − h4 s = 310.24 − (0.85)(310.24 − 200.57 ) = 217.02 kJ/kg Then the rate of refrigeration is & & & QL = m(q L ) = m(h1 − h4 ) = (0.2 kg/s )(280.13 − 217.02) kJ/kg = 12.6 kW (b) The net power input is determined from & & & Wnet, in = Wcomp, in − Wturb, out where & & & Wcomp,in = m(h2 − h1 ) = m(h2 s − h1 ) / η C = (0.2 kg/s )[(432.96 − 280.13) kJ/kg ]/ (0.80 ) = 38.21 kW & & W turb,out = m(h3 − h4 ) = (0.2 kg/s )(310.24 − 217.02 ) kJ/kg = 18.64 kW Thus, & W net,in = 38.21 − 18.64 = 19.6 kW (c) The COP of this ideal gas refrigeration cycle is determined from COPR = & QL & W net,in = 12.6 kW = 0.643 19.6 kW PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

No public clipboards found for this slide

Be the first to comment