Baixe Solucionario Fundamentos da Termodinâmica Van Wylen 7ª ed Cap. 7 e outras Notas de estudo em PDF para Engenharia Elétrica, somente na Docsity! SEVENTH EDITION Fundamentals of Thermodynamics BORGNAKKE | SONNTAG SOLUTION MANUAL CHAPTER 7 Borgnakke and Sonntag CONTENT CHAPTER 7 SUBSECTION PROB NO. In-Text concept questions a-g Concept problems 1-14 Heat engines and refrigerators 15-36 Second law and processes 37-43 Carnot cycles and absolute temperature 44-77 Finite ∆T heat transfer 78-91 Ideal gas Carnot cycles 92-95 review problems 96-113 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.b Geothermal underground hot water or steam can be used to generate electric power. Does that violate the second law? No. Since the earth is not uniform we consider the hot water or steam supply as coming from one energy source (the high T) and we must reject heat to a low temperature reservoir as the ocean, a lake or the atmosphere which is another energy reservoir. Iceland uses a significant amount of steam to heat buildings and to generate electricity. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.c A windmill produces power on a shaft taking kinetic energy out of the wind. Is it a heat engine? Is it a perpetual machine? Explain. Since the wind is generated by a complex system driven by solar heat input and radiation out to space it is a kind of heat engine. Within our lifetime it looks like it is perpetual. However with a different time scale the climate will change, the sun will grow to engulf the earth as it burns out of fuel. There is a storage effect and a non-uniform distribution of states in the system that drives this. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.d heat engines and heat pumps (refrigerators) are energy conversion devices altering amounts of energy transfer between Q and W. Which conversion direction (Q → W or W → Q) is limited and which is unlimited according to the second law. The work output of a heat engine is limited (Q to W). You can transform W to Q unlimited (a heat pump that does not work well or you may think about heat generated by friction). 7.e Ice cubes in a glass of liquid water will eventually melt and all the water approach room temperature. Is this a reversible process? Why? There is heat transfer from the warmer ambient to the water as long as there is a temperature difference. Eventually the temperatures approach each other and there is no more heat transfer. This is irreversible, as we cannot make ice-cubes out of the water unless we run a refrigerator and that requires a work from the surroundings, which does not leave the surroundings unchanged. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag Concept Problems Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.1 Two heat engines operate between the same two energy reservoirs and both receives the same QH. One engine is reversible and the other is not. What can you say about the two QL’s? The reversible heat engine can produce more work (has a higher efficiency) than the irreversible heat engine and due to the energy conservation it then gives out a smaller QL compared to the irreversible heat engine. Wrev = QH - QL rev > Wirrev = QH - QL irrev QL rev < QL irrev Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.2 Compare two domestic heat pumps (A and B) running with the same work input. If A is better than B which one heats the house most? The statement that A is better means it has a higher COP and since QH A = COPA W > QH B = COPB W it can thus provide more heat to the house. The higher heat comes from the higher QL it is able to draw in. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.5 Compare two heat engines receiving the same Q, one at 1200 K and the other at 1800 K; they both reject heat at 500 K. Which one is better? The maximum efficiency for the engines are given by the Carnot heat engine efficiency as ηTH = W . net / Q . H = 1 – TL TH Since they have the same low temperature the one with the highest TH will have a higher efficiency and thus presumably better. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.6 A car engine takes atmospheric air in at 20oC, no fuel, and exhausts the air at – 20oC producing work in the process. What do the first and the second laws say about that? Energy Eq.: W = QH − QL = change in energy of air. OK 2nd law: Exchange energy with only one reservoir. NOT OK. This is a violation of the statement of Kelvin-Planck. Remark: You cannot create and maintain your own energy reservoir. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.7 A combination of two refrigerator cycles is shown in Fig. P7.7. Find the overall COP as a function of COP1 and COP2. The overall COP becomes COP = β = Q . L W . tot = Q . L W . 1 W . 1 W . tot = COP1 W . 1 W . tot = COP1 1 1 + W . 2/W . 1 where we used W . tot = W . 1 + W . 2. Use definition of COP2 and energy equation for refrigerator 1 to eliminate Q . M and we have W . 2 = Q . M / COP2 = (W . 1 + Q . L) / COP2 so then W . 2 / W . 1 = (1 + Q . L/W . 1) / COP2 = (1 + COP1) / COP2 Finally substitute into the first equation and rearrange a little to get COP = β = COP1 COP2 COP1 + COP2 + 1 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.10 If the efficiency of a power plant goes up as the low temperature drops, why do power plants not just reject energy at say –40oC? In order to reject heat the ambient must be at the low temperature. Only if we moved the plant to the North Pole would we see such a low T. Remark: You cannot create and maintain your own energy reservoir. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.11 If the efficiency of a power plant goes up as the low temperature drops why not let the heat rejection go to a refrigerator at, say, –10oC instead of ambient 20oC? The refrigerator must pump the heat up to 20oC to reject it to the ambient. The refrigerator must then have a work input that will exactly offset the increased work output of the power plant, if they are both ideal. As we can not build ideal devices the actual refrigerator will require more work than the power plant will produce extra. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.12 A coal-fired power plant operates with a high T of 600oC whereas a jet engine has about 1400 K. Does that mean we should replace all power plants with jet engines? The thermal efficiency is limited by the Carnot heat engine efficiency. That is, the low temperature is also important. Here the power plant has a much lower T in the condenser than the jet engine has in the exhaust flow so the jet engine does not necessarily have a higher efficiency than the power plant. Gas-turbines are used in power plants where they can cover peak power demands needed for shorter time periods and their high temperature exhaust can be used to boil additional water for the steam cycle. WT Q H Q L . W P, in from coal to ambient Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag Heat Engines and Refrigerators Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.15 A gasoline engine produces 20 hp using 35 kW of heat transfer from burning fuel. What is its thermal efficiency and how much power is rejected to the ambient? Conversion Table A.1: 20 hp = 20 × 0.7457 kW = 14.91 kW Efficiency: ηTH = W . out/Q . H = 14.91 35 = 0.43 Energy equation: Q . L = Q . H - W . out = 35 – 14.91 = 20.1 kW Q . H ⇒ Q . L ⇒ W . out ⇒ Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.16 Calculate the thermal efficiency of the steam power plant cycle described in Example 6.9. Solution: From solution to Example 6.9, wnet = wt + wp = 640.7 – 4 = 636.7 kJ/kg qH = qb = 2831 kJ/kg ηTH = wnet/qH = 636.7 2831 = 0.225 WT QH Q L . WP, in Q 1 2 Notice we cannot write wnet = qH − qL as there is an extra heat transfer 1Q . 2 as a loss in the line. This needs to be accounted for in the overall energy equation. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.19 A coal fired power plant has an efficiency of 35% and produces net 500 MW of electricity. Coal releases 25 000 kJ/kg as it burns so how much coal is used per hour? From the definition of the thermal efficiency and the energy release by the combustion called heating value HV we get W . = η Q . H = η· m . ·HV then m . = W . η × HV = 500 MW 0.35 × 25000 kJ/kg = 500 × 1000 kJ/s 0.35 × 25000 kJ/kg = 57.14 kg/s = 205 714 kg/h Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.20 Assume we have a refrigerator operating at steady state using 500 W of electric power with a COP of 2.5. What is the net effect on the kitchen air? Take a C.V. around the whole kitchen. The only energy term that crosses the control surface is the work input W . apart from energy exchanged with the kitchen surroundings. That is the kitchen is being heated with a rate of W . . Remark: The two heat transfer rates are both internal to the kitchen. Q . H goes into the kitchen air and Q . L actually leaks from the kitchen into the refrigerated space, which is the reason we need to drive it out again. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.21 A room is heated with a 1500 W electric heater. How much power can be saved if a heat pump with a COP of 2.0 is used instead? Assume the heat pump has to deliver 1500 W as the Q . H. Heat pump: β′ = Q . H/W . IN W . IN = Q . H/β′ = 1500 2 = 750 W So the heat pump requires an input of 750 W thus saving the difference W . saved = 1500 W – 750 W = 750 W H Q W L Q T L HP Room incb Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.24 A window air-conditioner unit is placed on a laboratory bench and tested in cooling mode using 750 W of electric power with a COP of 1.75. What is the cooling power capacity and what is the net effect on the laboratory? Definition of COP: β = Q . L / W . Cooling capacity: Q . L = β W . = 1.75 × 750 = 1313 W For steady state operation the Q . L comes from the laboratory and Q . H goes to the laboratory giving a net to the lab of W . = Q . H - Q . L = 750 W, that is heating it. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.25 A water cooler for drinking water should cool 25 L/h water from 18oC to 10oC using a small refrigeration unit with a COP of 2.5. Find the rate of cooling required and the power input to the unit. The mass flow rate is m . = ρV . = 25 × 10-3 0.001002 1 3600 kg/s = 6.93 g/s Energy equation for heat exchanger Q . L = m . (h1 − h2) = m . CP (T1 − T2) Q W L T H H Q REF 1 2 cb = 6.93 × 10-3 × 4.18 × (18 – 10) = 0.2318 kW β = COP = Q . L / W . W . = Q . L / β = 0.2318 / 2.5 = 0.093 kW Comment: The unit does not operate continuously. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.26 A farmer runs a heat pump with a 2 kW motor. It should keep a chicken hatchery at 30oC, which loses energy at a rate of 10 kW to the colder ambient Tamb. What is the minimum coefficient of performance that will be acceptable for the heat pump? Solution: Power input: W . = 2 kW Energy Eq. for hatchery: Q . H = Q . Loss = 10 kW Definition of COP: β = COP = Q . H W . = 10 2 = 5 QleakQ QHL W = 2 kW HP cb Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.29 A water cooler for drinking water should cool 25 L/h water from 18oC to 10oC while the water reservoir also gains 60 W from heat transfer. Assume a small refrigeration unit with a COP of 2.5 does the cooling. Find the total rate of cooling required and the power input to the unit. The mass flow rate is m . = ρV . = 25 × 10-3 0.001002 1 3600 kg/s = 6.93 g/s Energy equation for heat exchanger Q . L = m . (h1 − h2) + Q . H TR = m . CP (T1 − T2) + Q . H TR Q W L T H H Q REF 1 2 cb H.TR Q = 6.93 × 10-3 × 4.18 × (18 – 10) kW + 60 W = 291.8 W β = COP = Q . L / W . W . = Q . L / β = 291.8 / 2.5 = 116.7 W Comment: The unit does not operate continuously. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.30 A car engine delivers 25 hp to the driveshaft with a thermal efficiency of 30%. The fuel has a heating value of 40 000 kJ/kg. Find the rate of fuel consumption and the combined power rejected through the radiator and exhaust. Solution: Heating value (HV): Q . H = m . ·HV From the definition of the thermal efficiency W . = η Q . H = η· m . ·HV m . = W . η·HV = 25 × 0.7355 0.3 × 40 000 = 0.00153 kg/s = 1.53 g/s Conversion of power from hp to kW in Table A.1. Q . L = Q . H - W . = (W . /η −W . ) = ( 1 η −1 )W . = ( 1 0.3 – 1) 25 × 0.7355 = 42.9 kW Exhaust flow Air intake filter Shaft Fan power Fuel line cb Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.31 R-410a enters the evaporator (the cold heat exchanger) in an A/C unit at -20oC, x = 28% and leaves at -20oC, x = 1. The COP of the refrigerator is 1.5 and the mass flow rate is 0.003 kg/s. Find the net work input to the cycle. Energy equation for heat exchanger Q . L = m . (h2 − h1) = m . [hg − (hf + x1 hfg)] = m . [hfg − x1 hfg] = m . (1 – x1)hfg Q L 1 2 cb = 0.003 kg/s × 0.72 × 243.65 kJ/kg = 0.5263 kW β = COP = Q . L / W . W . = Q . L / β = 0.5263 / 1.5 = 0.35 kW Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.34 A large stationary diesel engine produces 15 MW with a thermal efficiency of 40%. The exhaust gas, which we assume is air, flows out at 800 K and the intake air is 290 K. How large a mass flow rate is that, assuming this is the only way we reject heat? Can the exhaust flow energy be used? Heat engine: Q . H = W . out/ηTH = 15 0.4 = 37.5 MW Energy equation: Q . L = Q . H - W . out = 37.5 – 15 = 22.5 kW Exhaust flow: Q . L = m . air(h800 - h290) m . air = Q . L h800 - h290 = 22.5 × 1000 822.2 - 290.43 = 42.3 kg/s The flow of hot gases can be used to heat a building or it can be used to heat water in a steam power plant since that operates at lower temperatures. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.35 In a steam power plant 1 MW is added in the boiler, 0.58 MW is taken out in the condenser and the pump work is 0.02 MW. Find the plant thermal efficiency. If everything could be reversed find the coefficient of performance as a refrigerator. Solution: WT Q H Q L . W P, in CV. Total plant: Energy Eq.: Q . H + W . P,in = W . T + Q . L W . T = 1 + 0.02 – 0.58 = 0.44 MW ηTH = W . T – W . P,in Q . H = 440 – 20 1000 = 0.42 β = Q . L W . T – W . P,in = 580 440 – 20 = 1.38 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.36 Calculate the amount of work input a refrigerator needs to make ice cubes out of a tray of 0.25 kg liquid water at 10oC. Assume the refrigerator has β = 3.5 and a motor-compressor of 750 W. How much time does it take if this is the only cooling load? C.V. Water in tray. We neglect tray mass. Energy Eq.: m(u2 − u1) = 1Q2 − 1W2 Process : P = constant = Po 1W2 = ∫ P dV = Pom(v2 − v1) 1Q2 = m(u2 − u1) + 1W2 = m(h2 − h1) Tbl. B.1.1 : h1 = 41.99 kJ/kg, Tbl. B.1.5 : h2 = - 333.6 kJ/kg 1Q2 = 0.25(-333.4 – 41.99 ) = - 93.848 kJ Consider now refrigerator β = QL/W W = QL/β = - 1Q2/ β = 93.848/3.5 = 26.81 kJ For the motor to transfer that amount of energy the time is found as W = ∫ W . dt = W . ∆t ∆t = W/W . = (26.81 × 1000)/750 = 35.75 s Comment: We neglected a baseload of the refrigerator so not all the 750 W are available to make ice, also our coefficient of performance is very optimistic and finally the heat transfer is a transient process. All this means that it will take much more time to make ice-cubes. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.38 Assume a cyclic machine that exchanges 6 kW with a 250oC reservoir and has a. Q . L = 0 kW, W . = 6 kW b. Q . L = 6 kW, W . = 0 kW and Q . L is exchanged with a 30 oC ambient. What can you say about the processes in the two cases a and b if the machine is a heat engine? Repeat the question for the case of a heat pump. Solution: Heat engine a. Since Q . L = 0 impossible Kelvin – Planck b. Possible, irreversible, ηeng = 0 Ηeat pump a. Possible, irreversible (like an electric heater) b. Impossible, β → ∞, Clausius Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.39 Discuss the factors that would make the power plant cycle described in Problem 6.103 an irreversible cycle. Solution: General discussion, but here are a few of the most significant factors. 1. Combustion process that generates the hot source of energy. 2. Heat transfer over finite temperature difference in boiler. 3. Flow resistance and friction in turbine results in less work out. 4. Flow friction and heat loss to/from ambient in all pipes. 5. Heat transfer over finite temperature difference in condenser. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.40 Discuss the factors that would make the heat pump described in Problem 6.108 an irreversible cycle. Solution: General discussion but here are a few of the most significant factors. 1. Unwanted heat transfer in the compressor. 2. Pressure loss (back flow leak) in compressor 3. Heat transfer and pressure drop in line 1 => 2. 4. Pressure drop in all lines. 5. Throttle process 3 => 4. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.43 The water in a shallow pond heats up during the day and cools down during the night. Heat transfer by radiation, conduction and convection with the ambient thus cycles the water temperature. Is such a cyclic process reversible or irreversible? Solution: All the heat transfer takes place over a finite ∆T and thus all the heat transfer processes are irreversible. Conduction and convection have ∆T in the water, which is internally irreversible and ∆T outside the water which is externally irreversible. The radiation is absorbed or given out at the water temperature thus internally (for absorption) and externally (for emission) irreversible. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag Carnot Cycles and Absolute Temperature Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.44 Calculate the thermal efficiency of a Carnot cycle heat engine operating between reservoirs at 300oC and 45oC. Compare the result to that of Problem 7.16. Solution: ηTH = Wnet / QH = 1 – TL TH = 1 – 45 + 273 300 + 273 = 0.445 (Carnot) η7.16 = 0.225 (efficiency about ½ of the Carnot) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.47 Consider the setup with two stacked (temperature wise) heat engines as in Fig. P7.4. Let TH = 900 K, TM = 600 K and TL = 300 K. Find the two heat engine efficiencies and the combined overall efficiency assuming Carnot cycles. The individual efficiencies η1 = 1 – TM TH = 1 – 600 900 = 0.333 η2 = 1 – TL TM = 1 – 300 600 = 0.5 The overall efficiency ηTH = W . net / Q . H = (W . 1 + W . 2) / Q . H = η1 + W . 2 / Q . H For the second heat engine and the energy Eq. for the first heat engine W . 2 = η2 Q . M = η2 (1 – η1) Q . H so the final result is ηTH = η1 + η2 (1 – η1) = 0.333 + 0.5(1 – 0.333) = 0.667 Comment: It matches a single heat engine ηTH = 1 – TL TH = 1 – 300 900 = 2 3 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.48 At a few places where the air is very cold in the winter, like –30oC it is possible to find a temperature of 13oC down below ground. What efficiency will a heat engine have operating between these two thermal reservoirs? Solution: ηTH = 1 – TL TH The ground becomes the hot source and the atmosphere becomes the cold side of the heat engine ηTH= 1 – 273 – 30 273 + 13 = 1 – 243 286 = 0.15 This is low because of the modest temperature difference. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.49 Find the maximum coefficient of performance for the refrigerator in your kitchen, assuming it runs in a Carnot cycle. Solution: The refrigerator coefficient of performance is β = QL/W = QL/(QH - QL) = TL/(TH - TL) Assuming TL ~ 0°C, TH ~ 35°C, β ≤ 273.15 35 - 0 = 7.8 Actual working fluid temperatures must be such that TL < Trefrigerator and TH > Troom A refrigerator does not operate in a Carnot cycle. The actual vapor compression cycle is examined in Chapter 11. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.52 A large heat pump should upgrade 5 MW of heat at 85°C to be delivered as heat at 150°C. What is the minimum amount of work (power) input that will drive this? For the minimum work we assume a Carnot heat pump and Q . L = 5 MW. βHP = Q . H W . in = TH TH - TL = 273.15 + 150 150 - 85 = 6.51 βREF = βHP - 1 = Q . L W . in = 5.51 Now we can solve for the work W . in = Q . L/βREF = 5/5.51 = 0.907 MW Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.53 An air-conditioner provides 1 kg/s of air at 15°C cooled from outside atmospheric air at 35°C. Estimate the amount of power needed to operate the air-conditioner. Clearly state all assumptions made. Solution: Consider the cooling of air which needs a heat transfer as Q . air = m . ∆h ≅ m . Cp∆T = 1 kg/s × 1.004 kJ/kg K × 20 K = 20 kW Assume Carnot cycle refrigerator β = Q . L W . = Q . L / (Q . H - Q . L ) ≅ TL TH - TL = 273 + 15 35 - 15 = 14.4 W . = Q . L / β = 20.0 14.4 = 1.39 kW This estimate is the theoretical maximum performance. To do the required heat transfer TL ≅ 5°C and TH = 45°C are more likely; secondly β < βcarnot HQ W L Q REF 35 C 15 C cb o o Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.54 A cyclic machine, shown in Fig. P7.54, receives 325 kJ from a 1000 K energy reservoir. It rejects 125 kJ to a 400 K energy reservoir and the cycle produces 200 kJ of work as output. Is this cycle reversible, irreversible, or impossible? Solution: ηCarnot = 1 − TL TH = 1 − 400 1000 = 0.6 ηeng = W QH = 200 325 = 0.615 > ηCarnot This is impossible. H Q = 325 kJ W = 200 kJ L Q = 125 kJ T = 1000 K H HE cb T = 400 K L Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7-57 The lowest temperature that has been achieved is about 1 × 10−6 K. To achieve this an additional stage of cooling is required beyond that described in the previous problem, namely nuclear cooling. This process is similar to magnetic cooling, but it involves the magnetic moment associated with the nucleus rather than that associated with certain ions in the paramagnetic salt. Suppose that 10 µJ is to be removed from a specimen at an average temperature of 10−5 K (ten micro- joules is about the potential energy loss of a pin dropping 3 mm). Find the work input to a Carnot heat pump and its coefficient of performance to do this assuming the ambient is at 300 K. Solution: QL = 10 µJ = 10×10 -6 J at TL = 10 -5 K ⇒ QH = QL × TH TL = 10×10-6 × 300 10-5 = 300 J Win = QH - QL = 300 - 10×10 -6 ≅ 300 J β = QL Win = 10×10-6 300 = 3.33×10 -8 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.58 An inventor has developed a refrigeration unit that maintains the cold space at −10°C, while operating in a 25°C room. A coefficient of performance of 8.5 is claimed. How do you evaluate this? Solution: βCarnot = QL Win = TL TH - TL = 263.15 25 - (-10) = 7.52 8.5 > βCarnot ⇒ impossible claim H Q W L Q T = -10C L T = 25C H REF Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.59 Calculate the amount of work input a refrigerator needs to make ice cubes out of a tray of 0.25 kg liquid water at 10oC. Assume the refrigerator works in a Carnot cycle between –8oC and 35oC with a motor-compressor of 750 W. How much time does it take if this is the only cooling load? Solution: C.V. Water in tray. We neglect tray mass. Energy Eq.: m(u2 − u1) = 1Q2 − 1W2 Process : P = constant + Po 1W2 = ∫ P dV = Pom(v2 − v1) 1Q2 = m(u2 − u1) + 1W2 = m(h2 − h1) Tbl. B.1.1 : h1 = 41.99 kJ/kg, Tbl. B.1.5 : h2 = - 333.6 kJ/kg 1Q2 = 0.25(-333.4 – 41.99 ) = - 93.848 kJ Consider now refrigerator β = QL W = QL QH - QL = TL TH - TL = 273 - 8 35 - (-8) = 6.16 W = QL β = - 1Q2 β = 93.848 6.16 = 15.24 kJ For the motor to transfer that amount of energy the time is found as W = ∫ W . dt = W . ∆t ∆t = W W . = 15.24 ×1000 750 = 20.3 s Comment: We neglected a baseload of the refrigerator so not all the 750 W are available to make ice, also our coefficient of performance is very optimistic and finally the heat transfer is a transient process. All this means that it will take much more time to make ice-cubes. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.62 We propose to heat a house in the winter with a heat pump. The house is to be maintained at 20°C at all times. When the ambient temperature outside drops to −10°C, the rate at which heat is lost from the house is estimated to be 25 kW. What is the minimum electrical power required to drive the heat pump? Solution: Minimum power if we assume a Carnot cycle Q . H = Q . leak = 25 kW QleakQ QHL W HP β′ = Q . H W . IN = TH TH-TL = 293.2 30 = 9.773 ⇒ W . IN = 25 9.773 = 2.56 kW Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.63 A certain solar-energy collector produces a maximum temperature of 100°C. The energy is used in a cyclic heat engine that operates in a 10°C environment. What is the maximum thermal efficiency? What is it, if the collector is redesigned to focus the incoming light to produce a maximum temperature of 300°C? Solution: For TH = 100°C = 373.2 K & TL = 283.2 K ηth max = TH - TL TH = 90 373.2 = 0.241 For TH = 300°C = 573.2 K & TL = 283.2 K ηth max = TH - TL TH = 290 573.2 = 0.506 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.64 Helium has the lowest normal boiling point of any of the elements at 4.2 K. At this temperature the enthalpy of evaporation is 83.3 kJ/kmol. A Carnot refrigeration cycle is analyzed for the production of 1 kmol of liquid helium at 4.2 K from saturated vapor at the same temperature. What is the work input to the refrigerator and the coefficient of performance for the cycle with an ambient at 300 K? Solution: For the Carnot cycle the ratio of the heat transfers is the ratio of temperatures QL = n h _ fg = 1 kmol × 83.3 kJ/kmol = 83.3 kJ QH = QL × TH TL = 83.3 × 300 4.2 = 5950 kJ WIN = QH - QL = 5950 - 83.3 = 5886.7 kJ β = QL WIN = 83.3 5886.7 = 0.0142 [ = TL TH - TL ] Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.67 It is proposed to build a 1000-MW electric power plant with steam as the working fluid. The condensers are to be cooled with river water (see Fig. P7.67). The maximum steam temperature is 550°C, and the pressure in the condensers will be 10 kPa. Estimate the temperature rise of the river downstream from the power plant. Solution: W . NET = 10 6 kW, TH = 550°C = 823.3 K PCOND = 10 kPa → TL = TG (P = 10 kPa) = 45.8°C = 319 K ηTH CARNOT = TH - TL TH = 823.2 - 319 823.2 = 0.6125 ⇒ Q . L MIN= 10 6    1 - 0.6125 0.6125 = 0.6327 × 10 6 kW But m . H2O = 60 × 8 × 10/60 0.001 = 80 000 kg/s having an energy flow of Q . L MIN = m . H2O ∆h = m . H2O CP LIQ H2O ∆TH2O MIN ⇒ ∆TH2O MIN = Q . L MIN m . H2O CP LIQ H2O = 0.6327×106 80000 × 4.184 = 1.9°C Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.68 Repeat the previous problem using a more realistic thermal efficiency of 35%. W . NET = 10 6 kW = ηTH ac Q . H, ηTH ac = 0.35 ⇒ Q . L = Q . H - W . NET = W . NET /ηTH ac - W . NET = W . NET(1/ηTH ac – 1) = 106 kW    1 - 0.35 0.35 = 1.857 × 10 6 kW But m . H2O = 60 × 8 × 10/60 0.001 = 80 000 kg/s having an energy flow of Q . L = m . H2O ∆h = m . H2O CP LIQ H2O ∆TH2O ⇒ ∆TH2O = Q . L m . H2O CP LIQ H2O = 1.857 × 106 80 000 × 4.18 = 5.6°C Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.69 A steel bottle V = 0.1 m3 contains R-134a at 20°C, 200 kPa. It is placed in a deep freezer where it is cooled to -20°C. The deep freezer sits in a room with ambient temperature of 20°C and has an inside temperature of -20°C. Find the amount of energy the freezer must remove from the R-134a and the extra amount of work input to the freezer to do the process. Solution: C.V. R-134a out to the -20 °C space. Energy equation: m(u2 − u1) = 1Q2 − 1W2 Process : V = Const => v2 = v1 => 1W2 = 0 Table B.5.2: v1 = 0.11436 m 3/kg, u1 = 395.27 kJ/kg m = V/ v1 = 0.87443 kg State 2: v2 = v1 < vg = 0.14649 Table B.5.1 => 2 phase => x2 = v - vf vfg = 0.11436 - 0.000738 0.14576 = 0.77957 u2 = 173.65 + 0.77957 × 192.85 = 323.99 kJ/kg 1Q2 = m(u2 − u1) = - 62.334 kJ Consider the freezer and assume Carnot cycle β = QL W = QL QH - QL = TL TH - TL = 273 - 20 20 - (-20) = 6.33 Win = QL / β = 62.334 / 6.33 = 9.85 kJ -20 Co R 134a Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.72 Liquid sodium leaves a nuclear reactor at 800°C and is used as the energy source in a steam power plant. The condenser cooling water comes from a cooling tower at 15°C. Determine the maximum thermal efficiency of the power plant. Is it misleading to use the temperatures given to calculate this value? Solution: LIQ Na 800 o C REACTOR ENERGY TO H O 2 COND. COOLING TOWER ENERGY FROM STEAM POWER PLANT TH = 800°C = 1073.2 K, TL = 15°C = 288.2 K ηTH MAX = TH - TL TH = 1073.2 - 288.2 1073.2 = 0.731 It might be misleading to use 800°C as the value for TH, since there is not a supply of energy available at a constant temperature of 800°C (liquid Na is cooled to a lower temperature in the heat exchanger). ⇒ The Na cannot be used to boil H2O at 800°C. Similarly, the H2O leaves the cooling tower and enters the condenser at 15°C, and leaves the condenser at some higher temperature. ⇒ The water does not provide for condensing steam at a constant temperature of 15°C. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.73 A power plant with a thermal efficiency of 40% is located on a river similar to Fig. P7.67. With a total river mass flow rate of 1 × 105 kg/s at 15oC find the maximum power production allowed if the river water should not be heated more than 1 degree. The maximum heating allowed determines the maximum Q . L as Q . L = m . H2O ∆h = m . H2O CP LIQ H2O ∆TH2O = 1 × 105 kg/s × 4.18 kJ/kg-K × 1 K = 418 MW = W . NET(1/ηTH ac – 1) W . NET = Q . L / (1/ηTH ac – 1) = Q . L ηTH ac 1 - ηTH ac = 418 MW × 0.4 1 - 0.4 = 279 MW Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.74 A heat pump is driven by the work output of a heat engine as shown in figure P7.74. If we assume ideal devices find the ratio of the total power Q . L1 + Q . H2 that heats the house to the power from the hot energy source Q . H1 in terms of the temperatures. βHP = Q . H2/W . = Q . H2/(Q . H2- Q . L2) = Troom Troom-Tamb W . = ηHE . Q . H1 = (1- Troom TH ) Q . H1 W . = Q . H2/βHP = Troom Troom-Tamb Q . H2 Q . L1= Q . H1- W . = [1-1 + Troom TH ] Q . H1 Q . H2 + Q . L1 Q . H1 = 1-1 + Troom TH + 1- Troom TH Troom-Tamb Troom = Troom TH + Troom- T 2 room/TH Troom-Tamb = Troom [ 1 TH + 1 - Troom TH Troom - Tamb ] = Troom TH [1 + TH - Troom Troom - Tamb ] = Troom TH [ TH − Tamb Troom − Tamb ] W L1 Q T H H1 Q House Troom HE L2Q H.P. H2 Q T amb Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.77 A large heat pump should upgrade 5 MW of heat at 85°C to be delivered as heat at 150°C. Suppose the actual heat pump has a COP of 2.5 how much power is required to drive the unit. For the same COP how high a high temperature would a Carnot heat pump have assuming the same low T? This is an actual COP for the heat pump as βHP = COP = Q . H/W . in = 2.5 ⇒ Q . L/W . in = 1.5 W . in = Q . L/ 1.5 = 5 / 1.5 = 3.333 MW The Carnot heat pump has a COP given by the temperatures as βHP = Q . H/W . in = TH TH - TL = 2.5 ⇒ TH = 2.5 TH – 2.5 TL ⇒ TH = 2.5 1.5 TL = 5 3 (85 + 273.15) = 597 K Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag Finite ∆T Heat Transfer Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.78 The ocean near Havaii has 20oC near the surface and 5oC at some depth. A power plant based on this temperature difference is being planned. How large an efficiency could it have? If the two heat transfer terms (QH and QL) both require a 2 degree difference to operate what is the maximum efficiency then? Solution: TH = 20°C = 293.2 K; TL = 5°C = 278.2 K ηTH MAX = TH - TL TH = 293.2 - 278.2 293.2 = 0.051 ηTH mod = TH' - TL' TH' = 291.2 - 280.2 291.2 = 0.038 This is a very low efficiency so it has to be done on a very large scale to be economically feasible and then it will have some environmental impact. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.81 A house is heated by an electric heat pump using the outside as the low- temperature reservoir. For several different winter outdoor temperatures, estimate the percent savings in electricity if the house is kept at 20°C instead of 24°C. Assume that the house is losing energy to the outside as in Eq. 7.14. Solution: Heat Pump Q . loss ∝ (TH - TL) Max Perf. Q . H W . IN = TH TH - TL = K(TH - TL) W . IN , W . IN = K(TH - TL) 2 TH A: THA = 24°C = 297.2 K B: THB = 20°C = 293.2 K TL,°C W . INA/K W . INB/K % saving -20 6.514 5.457 16.2 % -10 3.890 3.070 21.1 % 0 1.938 1.364 29.6 % 10 0.659 0.341 48.3 % Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.82 A car engine operates with a thermal efficiency of 35%. Assume the air- conditioner has a coefficient of performance of β = 3 working as a refrigerator cooling the inside using engine shaft work to drive it. How much fuel energy should be spend extra to remove 1 kJ from the inside? Solution: Car engine: W = ηeng Qfuel Air conditioner: β = QL W W = ηeng Qfuel = QL β Qfuel = QL / (ηeng β) = 1 0.35 × 3 = 0.952 kJ W L Q TH H Q T L REF FuelQ H.E. L eng Q FUEL Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 7.83 A refrigerator uses a power input of 2.5 kW to cool a 5°C space with the high temperature in the cycle as 50°C. The QH is pushed to the ambient air at 35°C in a heat exchanger where the transfer coefficient is 50 W/m2K. Find the required minimum heat transfer area. Solution: W . = 2.5 kW = Q . H / βHP Q . H = W . × βHP = 2.5 × [323 / (50 - 5)] = 17.95 kW = h A ∆T A = Q . H h ∆T = 17.95 × 103 50 × 15 = 23.9 m 2 T H Q W T L L Q amb REF = 2.5 kW 35 C 50 C Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.