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### Ch.6

1. 1. 6-1 Chapter 6 THE SECOND LAW OF THERMODYNAMICSThe Second Law of Thermodynamics and Thermal Energy Reservoirs6-1C Water is not a fuel; thus the claim is false.6-2C Transferring 5 kWh of heat to an electric resistance wire and producing 5 kWh of electricity.6-3C An electric resistance heater which consumes 5 kWh of electricity and supplies 6 kWh of heat to aroom.6-4C Transferring 5 kWh of heat to an electric resistance wire and producing 6 kWh of electricity.6-5C No. Heat cannot flow from a low-temperature medium to a higher temperature medium.6-6C A thermal-energy reservoir is a body that can supply or absorb finite quantities of heat isothermally.Some examples are the oceans, the lakes, and the atmosphere.6-7C Yes. Because the temperature of the oven remains constant no matter how much heat is transferredto the potatoes.6-8C The surrounding air in the room that houses the TV set.Heat Engines and Thermal Efficiency6-9C No. Such an engine violates the Kelvin-Planck statement of the second law of thermodynamics.6-10C Heat engines are cyclic devices that receive heat from a source, convert some of it to work, andreject the rest to a sink.6-11C Method (b). With the heating element in the water, heat losses to the surrounding air areminimized, and thus the desired heating can be achieved with less electrical energy input.6-12C No. Because 100% of the work can be converted to heat.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
2. 2. 6-26-13C It is expressed as "No heat engine can exchange heat with a single reservoir, and produce anequivalent amount of work".6-14C (a) No, (b) Yes. According to the second law, no heat engine can have and efficiency of 100%.6-15C No. Such an engine violates the Kelvin-Planck statement of the second law of thermodynamics.6-16C No. The Kelvin-Plank limitation applies only to heat engines; engines that receive heat and convertsome of it to work.6-17 The power output and thermal efficiency of a power plant are given. The rate of heat rejection is to bedetermined, and the result is to be compared to the actual case in practice.Assumptions 1 The plant operates steadily. 2 Heat losses from the working fluid at the pipes and othercomponents are negligible.Analysis The rate of heat supply to the power plant is determined from the thermal efficiency relation, W& 600 MW & QH = net,out = = 1500 MW η th 0.4 FurnaceThe rate of heat transfer to the river water is determined from thefirst law relation for a heat engine, ηth = 40% & & & QL = QH − Wnet,out = 1500 − 600 = 900 MW HE 600 MWIn reality the amount of heat rejected to the river will be lowersince part of the heat will be lost to the surrounding air from the sinkworking fluid as it passes through the pipes and othercomponents.6-18 The heat input and thermal efficiency of a heat engine are given. The work output of the heat engineis to be determined.Assumptions 1 The plant operates steadily. 2 Heat lossesfrom the working fluid at the pipes and other components arenegligible.Analysis Applying the definition of the thermal efficiency tothe heat engine, W net = η th Q H = (0.35)(1.3 kJ ) = 0.455 kJPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
3. 3. 6-36-19E The work output and heat rejection of a heat engine that propels a ship are given. The thermalefficiency of the engine is to be determined.Assumptions 1 The plant operates steadily. 2 Heat lossesfrom the working fluid at the pipes and other components Furnaceare negligible. qHAnalysis Applying the first law to the heat engine gives HE q H = wnet + q L = 500 Btu/lbm + 300 Btu/lbm = 800 Btu/lbm qL wnetSubstituting this result into the definition of the thermal efficiency, sink wnet 500 Btu/lbm η th = = = 0.625 qH 800 Btu/lbm6-20 The work output and heat input of a heat engine are given. The heat rejection is to be determined.Assumptions 1 The plant operates steadily. 2 Heat lossesfrom the working fluid at the pipes and other components Furnaceare negligible. QHAnalysis Applying the first law to the heat engine gives HE Q L = Q H − W net = 500 kJ − 200 kJ = 300 kJ Wnet QL sink6-21 The heat rejection and thermal efficiency of a heat engine are given. The heat input to the engine is tobe determined.Assumptions 1 The plant operates steadily. 2 Heat lossesfrom the working fluid at the pipes and other componentsare negligible. FurnaceAnalysis According to the definition of the thermal qHefficiency as applied to the heat engine, HE wnet =η th q H qL wnet q H − q L =η th q H sinkwhich when rearranged gives qL 1000 kJ/kg qH = = = 1667 kJ/kg 1 − η th 1 − 0.4PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
4. 4. 6-46-22 The power output and fuel consumption rate of a power plant are given. The thermal efficiency is tobe determined.Assumptions The plant operates steadily.Properties The heating value of coal is given to be 30,000 kJ/kg.Analysis The rate of heat supply to this power plant is & Q H = m coal q HV,coal & 60 t/h Furnace = (60,000 kg/h )(30,000 kJ/kg ) = 1.8 × 10 kJ/h 9 = 500 MW coal HE 150 MWThen the thermal efficiency of the plant becomes sink & Wnet,out 150 MW η th = & = = 0.300 = 30.0% QH 500 MW6-23 The power output and fuel consumption rate of a car engine are given. The thermal efficiency of theengine is to be determined.Assumptions The car operates steadily.Properties The heating value of the fuel is given to be 44,000 kJ/kg.Analysis The mass consumption rate of the fuel is m fuel = ( ρV& ) fuel = (0.8 kg/L)(28 L/h) = 22.4 kg/h & Fuel EngineThe rate of heat supply to the car is & 28 L/h 60 kW Q H = m coal q HV,coal & HE = (22.4 kg/h )(44,000 kJ/kg ) = 985,600 kJ/h = 273.78 kW sinkThen the thermal efficiency of the car becomes & Wnet,out 60 kW η th = & = = 0.219 = 21.9% QH 273.78 kWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
5. 5. 6-56-24E The power output and thermal efficiency of a solar pond power plant are given. The rate of solarenergy collection is to be determined.Assumptions The plant operates steadily. SourceAnalysis The rate of solar energy collection or the rateof heat supply to the power plant is determined from the 350 kWthermal efficiency relation to be Solar pond HE & Wnet,out 350 kW ⎛ 1 Btu ⎞⎛ 3600 s ⎞ 4%&QH = = ⎜ ⎟⎜ ⎟ = 2.986 × 107 Btu/h η th 0.04 ⎜ 1.055 kJ ⎟⎜ 1 h ⎟ ⎝ ⎠⎝ ⎠ sink6-25 The United States produces about 51 percent of its electricity from coal at a conversion efficiency ofabout 34 percent. The amount of heat rejected by the coal-fired power plants per year is to be determined.Analysis Noting that the conversion efficiency is 34%, the amount of heat rejected by the coal plants peryear is Wcoal Wcoal η th = = Coal Furnace Qin Qout + Wcoal Wcoal Qout = − Wcoal 1.878×1012 kWh η th HE & Qout ηth = 34% 1.878 × 10 kWh 12 = − 1.878 × 1012 kWh 0.34 sink = 3.646 × 10 12 kWhPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
6. 6. 6-66-26 The projected power needs of the United States is to be met by building inexpensive but inefficientcoal plants or by building expensive but efficient IGCC plants. The price of coal that will enable the IGCCplants to recover their cost difference from fuel savings in 5 years is to be determined.Assumptions 1 Power is generated continuously byeither plant at full capacity. 2 The time value of money(interest, inflation, etc.) is not considered.Properties The heating value of the coal is given to be28×106 kJ/ton.Analysis For a power generation capacity of 150,000MW, the construction costs of coal and IGCC plants andtheir difference are Construction cost coal = (150,000,000 kW)(\$1300/kW) = \$195 × 10 9 Construction cost IGCC = (150,000,000 kW)(\$1500/kW) = \$225 × 10 9 Construction cost difference = \$225 × 10 9 − \$195 × 10 9 = \$30 × 10 9The amount of electricity produced by either plant in 5 years is & We = WΔt = (150,000,000 kW)(5 × 365 × 24 h) = 6.570 × 1012 kWhThe amount of fuel needed to generate a specified amount of power can be determined from We W Qin We η= → Qin = e or m fuel = = Qin η Heating value η (Heating value)Then the amount of coal needed to generate this much electricity by each plant and their difference are We 6.570 × 1012 kWh ⎛ 3600 kJ ⎞ m coal, coal plant = = ⎜ ⎟ = 2.484 × 10 tons 9 η (Heating value) (0.34)(28 ×10 kJ/ton) ⎝ 1 kWh ⎠ 6 We 6.570 ×1012 kWh ⎛ 3600 kJ ⎞ m coal, IGCC plant = = ⎜ ⎟ = 1.877 ×10 tons 9 η (Heating value) (0.45)(28 × 10 6 kJ/ton) ⎝ 1 kWh ⎠ Δm coal = m coal, coal plant − m coal, IGCC plant = 2.484 × 10 9 − 1.877 × 10 9 = 0.607 × 10 9 tonsFor Δmcoal to pay for the construction cost difference of \$30 billion, the price of coal should be Construction cost difference \$30 × 10 9 Unit cost of coal = = = \$49.4/ton Δm coal 0.607 × 10 9 tonsTherefore, the IGCC plant becomes attractive when the price of coal is above \$49.4 per ton.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
7. 7. 6-76-27 EES Problem 6-26 is reconsidered. The price of coal is to be investigated for varying simple paybackperiods, plant construction costs, and operating efficiency.Analysis The problem is solved using EES, and the solution is given below."Knowns:"HeatingValue = 28E+6 [kJ/ton]W_dot = 150E+6 [kW]{PayBackPeriod = 5 [years]eta_coal = 0.34eta_IGCC = 0.45CostPerkW_Coal = 1300 [\$/kW]CostPerkW_IGCC=1500 [\$/kW]}"Analysis:""For a power generation capacity of 150,000 MW, the construction costs of coaland IGCC plants and their difference are"ConstructionCost_coal = W_dot *CostPerkW_CoalConstructionCost_IGCC= W_dot *CostPerkW_IGCCConstructionCost_diff = ConstructionCost_IGCC - ConstructionCost_coal"The amount of electricity produced by either plant in 5 years is "W_ele = W_dot*PayBackPeriod*convert(year,h)"The amount of fuel needed to generate a specified amount of power can be determinedfrom the plant efficiency and the heating value of coal.""Then the amount of coal needed to generate this much electricity by each plant and theirdifference are" "Coal Plant:" eta_coal = W_ele/Q_in_coal Q_in_coal =m_fuel_CoalPlant*HeatingValue*convert(kJ,kWh) "IGCC Plant:" eta_IGCC = W_ele/Q_in_IGCC Q_in_IGCC =m_fuel_IGCCPlant*HeatingValue*convert(kJ,kWh) DELTAm_coal = m_fuel_CoalPlant - m_fuel_IGCCPlant"For to pay for the construction cost difference of \$30 billion, the price of coal should be"UnitCost_coal = ConstructionCost_diff /DELTAm_coal"Therefore, the IGCC plant becomes attractive when the price of coal is above \$49.4 per ton. "SOLUTIONConstructionCost_coal=1.950E+11 [dollars] ConstructionCost_diff=3.000E+10 [dollars]ConstructionCost_IGCC=2.250E+11 [dollars] CostPerkW_Coal=1300 [dollars/kW]CostPerkW_IGCC=1500 [dollars/kW] DELTAm_coal=6.073E+08 [tons]eta_coal=0.34 eta_IGCC=0.45HeatingValue=2.800E+07 [kJ/ton] m_fuel_CoalPlant=2.484E+09 [tons]m_fuel_IGCCPlant=1.877E+09 [tons] PayBackPeriod=5 [years]Q_in_coal=1.932E+13 [kWh] Q_in_IGCC=1.460E+13 [kWh]UnitCost_coal=49.4 [dollars/ton] W_dot=1.500E+08 [kW]W_ele=6.570E+12 [kWh]PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
8. 8. 6-8 250 PaybackPeriod UnitCostcoal [years] [\$/ton] 1 247 200 2 123.5 UnitCostcoal [\$/ton] 3 82.33 4 61.75 150 5 49.4 6 41.17 100 7 35.28 8 30.87 9 27.44 50 10 24.7 0 1 2 3 4 5 6 7 8 9 10 PayBackPeriod [years] 800 ηcoal UnitCostcoal [\$/ton] 700 0.25 19.98 0.2711 24.22 600 UnitCostcoal [\$/ton] 0.2922 29.6 0.3133 36.64 500 0.3344 46.25 400 0.3556 60.17 0.3767 82.09 300 0.3978 121.7 0.4189 215.2 200 0.44 703.2 100 0 0.24 0.28 0.32 0.36 0.4 0.44 η coal 225 CostPerkWIGCC UnitCostcoal [\$/kW] [\$/ton] 1300 0 180 1400 24.7 UnitCostcoal [\$/ton] 1500 49.4 135 1600 74.1 1700 98.8 1800 123.5 90 1900 148.2 2000 172.9 2100 197.6 45 2200 222.3 0 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 CostPerkW IGCC [\$/kW]PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
9. 9. 6-96-28 The projected power needs of the United States is to be met by building inexpensive but inefficientcoal plants or by building expensive but efficient IGCC plants. The price of coal that will enable the IGCCplants to recover their cost difference from fuel savings in 3 years is to be determined.Assumptions 1 Power is generated continuously by eitherplant at full capacity. 2 The time value of money (interest,inflation, etc.) is not considered.Properties The heating value of the coal is given to be28×106 kJ/ton.Analysis For a power generation capacity of 150,000 MW,the construction costs of coal and IGCC plants and theirdifference are Construction cost coal = (150,000,000 kW)(\$1300/kW) = \$195 × 10 9 Construction cost IGCC = (150,000,000 kW)(\$1500/kW) = \$225 × 10 9 Construction cost difference = \$225 × 10 9 − \$195 × 10 9 = \$30 × 10 9The amount of electricity produced by either plant in 3 years is & We = WΔt = (150,000,000 kW)(3 × 365 × 24 h) = 3.942 × 1012 kWhThe amount of fuel needed to generate a specified amount of power can be determined from We W Qin We η= → Qin = e or m fuel = = Qin η Heating value η (Heating value)Then the amount of coal needed to generate this much electricity by each plant and their difference are We 3.942 × 1012 kWh ⎛ 3600 kJ ⎞ m coal, coal plant = = ⎜ ⎟ = 1.491×10 tons 9 η (Heating value) (0.34)(28 ×10 6 kJ/ton) ⎝ 1 kWh ⎠ We 3.942 ×1012 kWh ⎛ 3600 kJ ⎞ m coal, IGCC plant = = ⎜ ⎟ = 1.126 ×10 tons 9 η (Heating value) (0.45)(28 ×10 6 kJ/ton) ⎝ 1 kWh ⎠ Δm coal = m coal, coal plant − m coal, IGCC plant = 1.491× 10 9 − 1.126 × 10 9 = 0.365 × 10 9 tonsFor Δmcoal to pay for the construction cost difference of \$30 billion, the price of coal should be Construction cost difference \$30 × 10 9 Unit cost of coal = = = \$82.2/ton Δm coal 0.365 × 10 9 tonsTherefore, the IGCC plant becomes attractive when the price of coal is above \$82.2 per ton.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
10. 10. 6-106-29 A coal-burning power plant produces 300 MW of power. The amount of coal consumed during a one-day period and the rate of air flowing through the furnace are to be determined.Assumptions 1 The power plant operates steadily. 2 The kinetic and potential energy changes are zero.Properties The heating value of the coal is given to be 28,000 kJ/kg.Analysis (a) The rate and the amount of heat inputs to the power plant are & Wnet,out 300 MW & Qin = = = 937.5 MW η th 0.32 & Qin = Qin Δt = (937.5 MJ/s) (24 × 3600 s) = 8.1 × 10 7 MJThe amount and rate of coal consumed during this period are Qin 8.1× 10 7 MW m coal = = = 2.893 × 10 6 kg q HV 28 MJ/kg m coal 2.893 × 10 6 kg m coal = & = = 33.48 kg/s Δt 24 × 3600 s(b) Noting that the air-fuel ratio is 12, the rate of air flowing through the furnace is mair = (AF)mcoal = (12 kg air/kg fuel)(33.48 kg/s) = 401.8 kg/s & &PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
11. 11. 6-11Refrigerators and Heat Pumps6-30C The difference between the two devices is one of purpose. The purpose of a refrigerator is toremove heat from a cold medium whereas the purpose of a heat pump is to supply heat to a warm medium.6-31C The difference between the two devices is one of purpose. The purpose of a refrigerator is toremove heat from a refrigerated space whereas the purpose of an air-conditioner is remove heat from aliving space.6-32C No. Because the refrigerator consumes work to accomplish this task.6-33C No. Because the heat pump consumes work to accomplish this task.6-34C The coefficient of performance of a refrigerator represents the amount of heat removed from therefrigerated space for each unit of work supplied. It can be greater than unity.6-35C The coefficient of performance of a heat pump represents the amount of heat supplied to the heatedspace for each unit of work supplied. It can be greater than unity.6-36C No. The heat pump captures energy from a cold medium and carries it to a warm medium. It doesnot create it.6-37C No. The refrigerator captures energy from a cold medium and carries it to a warm medium. It doesnot create it.6-38C No device can transfer heat from a cold medium to a warm medium without requiring a heat orwork input from the surroundings.6-39C The violation of one statement leads to the violation of the other one, as shown in Sec. 6-4, andthus we conclude that the two statements are equivalent.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
12. 12. 6-126-40 The COP and the refrigeration rate of a refrigerator are given. The power consumption and the rate ofheat rejection are to be determined.Assumptions The refrigerator operates steadily.Analysis (a) Using the definition of the coefficient of performance, thepower input to the refrigerator is determined to be Kitchen air & QL 60 kJ/min & Wnet,in = = = 50 kJ/min = 0.83 kW COP=1.2 COPR 1.2 R(b) The heat transfer rate to the kitchen air is determined from the & QLenergy balance, cool space & & & QH = QL + Wnet,in = 60 + 50 = 110 kJ/min6-41E The heat absorption, the heat rejection, and the power input of a commercial heat pump are given.The COP of the heat pump is to be determined.Assumptions The heat pump operates steadily. ReservoirAnalysis Applying the definition of the heat pump coefficient of & QHperformance to this heat pump gives & HP 2 hp QH 15,090 Btu/h ⎛ 1 hp ⎞ COPHP = = ⎜ ⎟ = 2.97 & W& 2 hp ⎝ 2544.5 Btu/h ⎠ QL net,in Reservoir6-42 The COP and the power input of a residential heat pump are given. The rate of heating effect is to bedetermined.Assumptions The heat pump operates steadily.Analysis Applying the definition of the heat pump coefficient ofperformance to this heat pump gives & & Q H = COPHPW net,in = (1.6)(2 kW ) = 3.2 kW = 3.2 kJ/sPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
13. 13. 6-136-43 The cooling effect and the COP of a refrigerator are given. The power input to the refrigerator is to bedetermined.Assumptions The refrigerator operates steadily. ReservoirAnalysis Rearranging the definition of the refrigerator coefficient & QH COP=1.35of performance and applying the result to this refrigerator gives R & Wnet,in & QL 10,000 kJ/h ⎛ 1 h ⎞ & W net,in = = ⎜ ⎟ = 2.06 kW & QL COPR 1.35 ⎝ 3600 s ⎠ Reservoir6-44 The cooling effect and the COP of a refrigerator are given. The power input to the refrigerator is to bedetermined.Assumptions The refrigerator operates steadily. ReservoirAnalysis Rearranging the definition of the refrigerator & QH COP=1.3coefficient of performance and applying the result to thisrefrigerator gives R & Wnet,in & QL 5 kW & QL & W net,in = = = 3.85 kW COPR 1.3 Reservoir6-45 The COP and the work input of a heat pump are given. The heat transferred to and from this heatpump are to be determined.Assumptions The heat pump operates steadily. ReservoirAnalysis Applying the definition of the heat pump coefficient of QHperformance, Q H = COPHPW net,in = (1.7)(50 kJ) = 85 kJ HP 50 kJ QLAdapting the first law to this heat pump produces Q L = Q H − W net,in = 85 kJ − 50 kJ = 35 kJ ReservoirPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
14. 14. 6-146-46E The COP and the refrigeration rate of an ice machine are given. The power consumption is to bedetermined.Assumptions The ice machine operates steadily. OutdoorsAnalysis The cooling load of this ice machine is COP = 2.4 Q L = mq L = (28 lbm/h )(169 Btu/lbm) = 4732 Btu/h & & RUsing the definition of the coefficient of performance, the & QLpower input to the ice machine system is determined to be & water Ice ice & QL 4732 Btu/h ⎛ 1 hp ⎞ Wnet,in = = ⎜ 2545 Btu/h ⎟ = 0.775 hp ⎜ Machine ⎟ 55°F 25°F COPR 2.4 ⎝ ⎠6-47 The COP and the power consumption of a refrigerator are given. The time it will take to cool 5watermelons is to be determined.Assumptions 1 The refrigerator operates steadily. 2 The heat gain of the refrigerator through its walls,door, etc. is negligible. 3 The watermelons are the only items in the refrigerator to be cooled.Properties The specific heat of watermelons is given to be c = 4.2 kJ/kg.°C.Analysis The total amount of heat that needs to be removed from the watermelons is QL = (mcΔT )watermelons = 5 × (10 kg )(4.2 kJ/kg ⋅ °C )(20 − 8)o C = 2520 kJThe rate at which this refrigerator removes heat is ( ) Kitchen air QL = (COPR ) Wnet,in = (2.5)(0.45 kW ) = 1.125 kW & &That is, this refrigerator can remove 1.125 kJ of heat per second. 450 W RThus the time required to remove 2520 kJ of heat is COP = 2.5 QL 2520 kJ Δt = = = 2240 s = 37.3 min cool space & QL 1.125 kJ/sThis answer is optimistic since the refrigerated space will gain some heat during this process from thesurrounding air, which will increase the work load. Thus, in reality, it will take longer to cool thewatermelons.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
15. 15. 6-156-48 [Also solved by EES on enclosed CD] An air conditioner with a known COP cools a house to desiredtemperature in 15 min. The power consumption of the air conditioner is to be determined.Assumptions 1 The air conditioner operates steadily. 2 The house is well-sealed so that no air leaks in orout during cooling. 3 Air is an ideal gas with constant specific heats at room temperature.Properties The constant volume specific heat of air is given to be cv = 0.72 kJ/kg.°C.Analysis Since the house is well-sealed (constant volume), the total amount of heat that needs to beremoved from the house is QL = (mcv ΔT )House = (800 kg )(0.72 kJ/kg ⋅ °C )(32 − 20)°C = 6912 kJ OutsideThis heat is removed in 15 minutes. Thus the average rate of heatremoval from the house is & QH & Q 6912 kJ COP = 2.5 QL = L = = 7.68 kW Δt 15 × 60 s ACUsing the definition of the coefficient of performance, the power inputto the air-conditioner is determined to be 32→20°C & QL 7.68 kW House & Wnet,in = = = 3.07 kW COPR 2.5PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
16. 16. 6-166-49 EES Problem 6-48 is reconsidered. The rate of power drawn by the air conditioner required to coolthe house as a function for air conditioner EER ratings in the range 9 to 16 is to be investigated.Representative costs of air conditioning units in the EER rating range are to be included.Analysis The problem is solved using EES, and the results are tabulated and plotted below."Input Data"T_1=32 [C]T_2=20 [C]C_v = 0.72 [kJ/kg-C]m_house=800 [kg]DELTAtime=20 [min]COP=EER/3.412"Assuming no work done on the house and no heat energy added to the housein the time period with no change in KE and PE, the first law applied to the house is:"E_dot_in - E_dot_out = DELTAE_dotE_dot_in = 0E_dot_out = Q_dot_LDELTAE_dot = m_house*DELTAu_house/DELTAtimeDELTAu_house = C_v*(T_2-T_1)"Using the definition of the coefficient of performance of the A/C:"W_dot_in = Q_dot_L/COP "kJ/min"*convert(kJ/min,kW) "kW"Q_dot_H= W_dot_in*convert(KW,kJ/min) + Q_dot_L "kJ/min" EER Win 2.2 [Btu/kWh] [kW] 9 2.184 10 1.965 2 11 1.787 Win [kW] 12 1.638 1.8 13 1.512 14 1.404 15 1.31 1.6 16 1.228 1.4 1.2 9 10 11 12 13 14 15 16 EER [Btu/kWh]PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
17. 17. 6-176-50 A house is heated by resistance heaters, and the amount of electricity consumed during a wintermonth is given. The amount of money that would be saved if this house were heated by a heat pump with aknown COP is to be determined.Assumptions The heat pump operates steadily.Analysis The amount of heat the resistance heaters supply to the house is equal to he amount of electricitythey consume. Therefore, to achieve the same heating effect, the house must be supplied with 1200 kWh ofenergy. A heat pump that supplied this much heat will consume electrical power in the amount of & QH 1200 kWh & Wnet,in = = = 500 kWh COPHP 2.4which represent a savings of 1200 – 500 = 700 kWh. Thus the homeowner would have saved (700 kWh)(0.085 \$/kWh) = \$59.506-51E The COP and the heating effect of a heat pump are given. The power input to the heat pump is to bedetermined.Assumptions The heat pump operates steadily. ReservoirAnalysis Applying the definition of the coefficient of performance, & QH COP=1.4 & QH 100,000 Btu/h ⎛ 1 hp ⎞ & W net,in = = ⎜ ⎟ = 28.1 hp HP & Wnet,in COPHP 1.4 ⎝ 2544.5 Btu/h ⎠ & QL Reservoir6-52 The cooling effect and the rate of heat rejection of a refrigerator are given. The COP of therefrigerator is to be determined.Assumptions The refrigerator operates steadily.Analysis Applying the first law to the refrigerator gives & & & W net,in = Q H − Q L = 22,000 − 15,000 = 7000 kJ/hApplying the definition of the coefficient of performance, & QL 15,000 kJ/h COPR = = = 2.14 W& 7000 net,inPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
18. 18. 6-186-53 The cooling effect and the power consumption of an air conditioner are given. The rate of heatrejection from this air conditioner is to be determined.Assumptions The air conditioner operates steadily. ReservoirAnalysis Applying the first law to the air conditioner gives & QH & & & Q H = W net,in + Q L = 0.75 + 1 = 1.75 kW AC & Wnet,in & QL Reservoir6-54 The rate of heat loss, the rate of internal heat gain, and the COP of a heat pump are given. The powerinput to the heat pump is to be determined.Assumptions The heat pump operates steadily.Analysis The heating load of this heat pump system is the difference 60,000between the heat lost to the outdoors and the heat generated in the kJ/h Househouse from the people, lights, and appliances, & QH & QH = 60,000 − 4,000 = 56,000 kJ / h HPUsing the definition of COP, the power input to the heat pump is COP = 2.5determined to be & QH 56,000 kJ/h ⎛ 1 kW ⎞ Outside & Wnet,in = = ⎜ 3600 kJ/h ⎟ = 6.22 kW ⎜ ⎟ COPHP 2.5 ⎝ ⎠PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
20. 20. 6-206-57 Refrigerant-134a flows through the condenser of a residential heat pump unit. For a given compressorpower consumption the COP of the heat pump and the rate of heat absorbed from the outside air are to bedetermined.Assumptions 1 The heat pump operates QHsteadily. 2 The kinetic and potential energy 800 kPa 800 kPachanges are zero. x=0 35°C CondenserProperties The enthalpies of R-134a at thecondenser inlet and exit are Expansion Win P = 800 kPa ⎫ valve ⎬h1 = 271.22 kJ/kg 1 Compressor T1 = 35°C ⎭ P2 = 800 kPa ⎫ ⎬h2 = 95.47 kJ/kg x2 = 0 ⎭ EvaporatorAnalysis (a) An energy balance on the condensergives the heat rejected in the condenser QL & Q H = m(h1 − h2 ) = (0.018 kg/s)(271.22 − 95.47) kJ/kg = 3.164 kW &The COP of the heat pump is Q& 3.164 kW COP = &H = = 2.64 Win 1.2 kW(b) The rate of heat absorbed from the outside air & & & Q L = Q H − Win = 3.164 − 1.2 = 1.96 kWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
21. 21. 6-216-58 A commercial refrigerator with R-134a as the working fluid is considered. The evaporator inlet andexit states are specified. The mass flow rate of the refrigerant and the rate of heat rejected are to bedetermined.Assumptions 1 The refrigerator operates steadily. 2 QHThe kinetic and potential energy changes are zero. CondenserProperties The properties of R-134a at theevaporator inlet and exit states are (Tables A-11through A-13) Expansion Win valve P1 = 120 kPa ⎫ Compressor ⎬h1 = 65.38 kJ/kg x1 = 0.2 ⎭ P2 = 120 kPa ⎫ ⎬h2 = 238.84 kJ/kg Evaporator T2 = −20°C ⎭ 120 kPa 120 kPaAnalysis (a) The refrigeration load is x=0.2 QL -20°C & & Q L = (COP)Win = (1.2)(0.45 kW) = 0.54 kWThe mass flow rate of the refrigerant is determined from & QL 0.54 kW mR = & = = 0.0031 kg/s h2 − h1 (238.84 − 65.38) kJ/kg(b) The rate of heat rejected from the refrigerator is & & & Q H = Q L + Win = 0.54 + 0.45 = 0.99 kWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
22. 22. 6-22Perpetual-Motion Machines6-59C This device creates energy, and thus it is a PMM1.6-60C This device creates energy, and thus it is a PMM1.Reversible and Irreversible Processes6-61C This process is irreversible. As the block slides down the plane, two things happen, (a) the potentialenergy of the block decreases, and (b) the block and plane warm up because of the friction between them.The potential energy that has been released can be stored in some form in the surroundings (e.g., perhapsin a spring). When we restore the system to its original condition, we must (a) restore the potential energyby lifting the block back to its original elevation, and (b) cool the block and plane back to their originaltemperatures. The potential energy may be restored by returning the energy that was stored during the originalprocess as the block decreased its elevation and released potential energy. The portion of the surroundingsin which this energy had been stored would then return to its original condition as the elevation of theblock is restored to its original condition. In order to cool the block and plane to their original temperatures, we have to remove heat fromthe block and plane. When this heat is transferred to the surroundings, something in the surroundings hasto change its state (e.g., perhaps we warm up some water in the surroundings). This change in thesurroundings is permanent and cannot be undone. Hence, the original process is irreversible.6-62C Adiabatic stirring processes are irreversible because the energy stored within the system can not bespontaneously released in a manor to cause the mass of the system to turn the paddle wheel in the oppositedirection to do work on the surroundings.6-63C The chemical reactions of combustion processes of a natural gas and air mixture will generatecarbon dioxide, water, and other compounds and will release heat energy to a lower temperaturesurroundings. It is unlikely that the surroundings will return this energy to the reacting system and theproducts of combustion react spontaneously to reproduce the natural gas and air mixture.6-64C No. Because it involves heat transfer through a finite temperature difference.6-65C Because reversible processes can be approached in reality, and they form the limiting cases. Workproducing devices that operate on reversible processes deliver the most work, and work consuming devicesthat operate on reversible processes consume the least work.6-66C When the compression process is non-quasi equilibrium, the molecules before the piston facecannot escape fast enough, forming a high pressure region in front of the piston. It takes more work tomove the piston against this high pressure region.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
23. 23. 6-236-67C When an expansion process is non-quasiequilibrium, the molecules before the piston face cannotfollow the piston fast enough, forming a low pressure region behind the piston. The lower pressure thatpushes the piston produces less work.6-68C The irreversibilities that occur within the system boundaries are internal irreversibilities; thosewhich occur outside the system boundaries are external irreversibilities.6-69C A reversible expansion or compression process cannot involve unrestrained expansion or suddencompression, and thus it is quasi-equilibrium. A quasi-equilibrium expansion or compression process, onthe other hand, may involve external irreversibilities (such as heat transfer through a finite temperaturedifference), and thus is not necessarily reversible.The Carnot Cycle and Carnots Principle6-70C The four processes that make up the Carnot cycle are isothermal expansion, reversible adiabaticexpansion, isothermal compression, and reversible adiabatic compression.6-71C They are (1) the thermal efficiency of an irreversible heat engine is lower than the efficiency of areversible heat engine operating between the same two reservoirs, and (2) the thermal efficiency of all thereversible heat engines operating between the same two reservoirs are equal.6-72C False. The second Carnot principle states that no heat engine cycle can have a higher thermalefficiency than the Carnot cycle operating between the same temperature limits.6-73C Yes. The second Carnot principle states that all reversible heat engine cycles operating between thesame temperature limits have the same thermal efficiency.6-74C (a) No, (b) No. They would violate the Carnot principle.Carnot Heat Engines6-75C No.6-76C The one that has a source temperature of 600°C. This is true because the higher the temperature atwhich heat is supplied to the working fluid of a heat engine, the higher the thermal efficiency.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
24. 24. 6-246-77 Two pairs of thermal energy reservoirs are to be compared from a work-production perspective.Assumptions The heat engine operates steadily.Analysis For the maximum production of work, a heat engine operatingbetween the energy reservoirs would have to be completely reversible. THThen, for the first pair of reservoirs QH TL 325 K HE η th,max = 1 − = 1− = 0.519 TH 675 K QL WnetFor the second pair of reservoirs, TL TL 275 K η th,max = 1 − = 1− = 0.560 TH 625 KThe second pair is then capable of producing more work for each unit of heat extracted from the hotreservoir.6-78 The source and sink temperatures of a power plant are given. The maximum efficiency of the plant isto be determined.Assumptions The plant operates steadily. 1200 KAnalysis The maximum efficiency this plant can have is the QHCarnot efficiency, which is determined from HE T 300 K QL Wnet η th,max = 1− L = 1− = 0.75 TH 1200 K 300 K6-79 The source and sink temperatures of a power plant are given. The maximum efficiency of the plant isto be determined.Assumptions The plant operates steadily. 1000 KAnalysis The maximum efficiency this plant can have is the QHCarnot efficiency, which is determined from HE T 300 K QL Wnet η th,max = 1− L = 1− = 0.70 TH 1000 K 300 RPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
25. 25. 6-256-80 [Also solved by EES on enclosed CD] The source and sink temperatures of a heat engine and the rateof heat supply are given. The maximum possible power output of this engine is to be determined.Assumptions The heat engine operates steadily.Analysis The highest thermal efficiency a heat engine operating between two specified temperature limitscan have is the Carnot efficiency, which is determined from T 298 K 550°C η th, max = η th,C = 1 − L = 1 − = 0.638 or 63.8% TH 823 K 1200 kJ/minThen the maximum power output of this heat engine is determined HEfrom the definition of thermal efficiency to be Wnet,out = η th QH = (0.638)(1200 kJ/min ) = 765.6 kJ/min = 12.8 kW & & 25°CPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
26. 26. 6-266-81 EES Problem 6-80 is reconsidered. The effects of the temperatures of the heat source and the heatsink on the power produced and the cycle thermal efficiency as the source temperature varies from 300°Cto 1000°C and the sink temperature varies from 0°C to 50°C are to be studied. The power produced and thecycle efficiency against the source temperature for sink temperatures of 0°C, 25°C, and 50°C are to beplotted.Analysis The problem is solved using EES, and the results are tabulated and plotted below."Input Data from the Diagram Window"{T_H = 550 [C]T_L = 25 [C]}{Q_dot_H = 1200 [kJ/min]}"First Law applied to the heat engine"Q_dot_H - Q_dot_L- W_dot_net = 0W_dot_net_KW=W_dot_net*convert(kJ/min,kW)"Cycle Thermal Efficiency - Temperatures must be absolute"eta_th = 1 - (T_L + 273)/(T_H + 273)"Definition of cycle efficiency"eta_th=W_dot_net / Q_dot_H 0.8 T = 50 C L ηth TH WnetkW 0.75 = 25 C [C] [kW] = 0C 0.7 0.52 300 10.47 0.59 400 11.89 0.65 0.65 500 12.94 0.69 600 13.75 0.6 η th 0.72 700 14.39 0.55 0.75 800 14.91 Q = 1200 kJ/m in 0.77 900 15.35 0.5 H 0.79 1000 15.71 0.45 0.4 300 400 500 600 700 800 900 1000 T [C] H 20 18 T = 50 C L 16 = 25 C 14 = 0C 12 W net,KW [kW ] 10 8 6 Q = 1200 kJ/m in H 4 2 0 300 400 500 600 700 800 900 1000 T [C] HPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
27. 27. 6-276-82E The sink temperature of a Carnot heat engine, the rate of heat rejection, and the thermalefficiency are given. The power output of the engine and the source temperature are to be determined.Assumptions The Carnot heat engine operates steadily.Analysis (a) The rate of heat input to this heat engine is determined from the definition of thermalefficiency, & QL 800 Btu/min η th = 1 − ⎯ ⎯→ 0.75 = 1 − ⎯ & ⎯→ Q H = 3200 Btu/min & Q & Q H HThen the power output of this heat engine can be determined from TH W net,out = η th Q H = (0.75)(3200 Btu/min ) = 2400 Btu/min = 56.6 hp & &(b) For reversible cyclic devices we have HE ⎛ QH & ⎞ ⎛T ⎞ 800 Btu/min ⎜ ⎟ =⎜ H ⎟ ⎜Q & ⎟ ⎜ ⎟ ⎝ L ⎠ rev ⎝ TL ⎠ 60°FThus the temperature of the source TH must be ⎛Q& ⎞ ⎛ 3200 Btu/min ⎞ TH = ⎜ H ⎜ 800 Btu/min ⎟(520 R ) = 2080 R ⎟ TL = ⎜ ⎜Q& ⎟ ⎟ ⎝ L ⎠ rev ⎝ ⎠6-83E The source and sink temperatures and the power output of a reversible heat engine are given. Therate of heat input to the engine is to be determined.Assumptions The heat engine operates steadily.Analysis The thermal efficiency of this reversible heat engine 1500 Ris determined from & QH TL 500 R η th,max = 1 − = 1− = 0.6667 HE TH 1500 R & QL & WnetA rearrangement of the definition of the thermal efficiency produces 500 R W& & W net 5 hp ⎛ 2544.5 Btu/h ⎞ η th,max = net ⎯ & ⎯→ Q H = = ⎜ ⎟ = 19,080 Btu/h & QH η th,max 0.6667 ⎜ ⎝ 1 hp ⎟ ⎠PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
28. 28. 6-286-84E The claim of an inventor about the operation of a heat engine is to be evaluated.Assumptions The heat engine operates steadily.Analysis If this engine were completely reversible, the thermal efficiency would be TL 550 R η th,max = 1 − = 1− = 0.45 TH 1000 R 1000 RWhen the first law is applied to the engine above, & QH & & & ⎛ 2544.5 Btu/h ⎞ HE 5 hp Q H = W net + Q L = (5 hp)⎜ ⎜ ⎟ + 15,000 Btu/h = 27,720 Btu/h ⎟ ⎝ 1 hp ⎠ 15,000 Btu/hThe actual thermal efficiency of the proposed heat engine is then 550 R W& 5 hp ⎛ 2544.5 Btu/h ⎞ η th = net = ⎜ ⎜ ⎟ = 0.459 ⎟ & QH 27,720 Btu/h ⎝ 1 hp ⎠Since the thermal efficiency of the proposed heat engine is greater than that of a completely reversible heatengine which uses the same isothermal energy reservoirs, the inventors claim is invalid.6-85 The claim that the efficiency of a completely reversible heat engine can be doubled by doubling thetemperature of the energy source is to be evaluated.Assumptions The heat engine operates steadily.Analysis The upper limit for the thermal efficiency of any heat engine occurs when a completely reversibleengine operates between the same energy reservoirs. The thermal efficiency of this completely reversibleengine is given by TL TH − TL TH η th,rev = 1 − = TH TH QHIf we were to double the absolute temperature of the high temperature HEenergy reservoir, the new thermal efficiency would be QL Wnet TL 2T − T L T − TL η th,rev = 1 − = H <2 H TL 2T H 2T H THThe thermal efficiency is then not doubled as the temperature of the high temperature reservoir is doubled.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
29. 29. 6-296-86 An inventor claims to have developed a heat engine. The inventor reports temperature, heat transfer,and work output measurements. The claim is to be evaluated.Analysis The highest thermal efficiency a heat engine operating between two specified temperature limitscan have is the Carnot efficiency, which is determined from TL 290 K 500 K η th, max = η th,C = 1 − = 1− = 0.42 or 42% TH 500 K 700 kJThe actual thermal efficiency of the heat engine in question is HE 300 kJ Wnet 300 kJ η th = = = 0.429 or 42.9% QH 700 kJ 290 Kwhich is greater than the maximum possible thermal efficiency. Therefore, this heat engine is a PMM2 andthe claim is false.6-87E An inventor claims to have developed a heat engine. The inventor reports temperature, heat transfer,and work output measurements. The claim is to be evaluated.Analysis The highest thermal efficiency a heat engine operating between two specified temperature limitscan have is the Carnot efficiency, which is determined from TL 540 R η th, max = η th,C = 1 − = 1− = 0.40 or 40% TH 900 R 900 RThe actual thermal efficiency of the heat engine in question is 300 Btu Wnet 160 Btu HE η th = = = 0.533 or 53.3% 160 Btu QH 300 Btuwhich is greater than the maximum possible thermal efficiency. 540 RTherefore, this heat engine is a PMM2 and the claim is false.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
30. 30. 6-306-88 A geothermal power plant uses geothermal liquid water at 160ºC at a specified rate as the heat source.The actual and maximum possible thermal efficiencies and the rate of heat rejected from this power plantare to be determined.Assumptions 1 The power plant operates steadily. 2 The kinetic and potential energy changes are zero. 3Steam properties are used for geothermal water.Properties Using saturated liquid properties, the source and the sink state enthalpies of geothermal waterare (Table A-4) Tsource = 160°C ⎫ ⎬hsource = 675.47 kJ/kg xsource = 0 ⎭ Tsink = 25°C ⎫ ⎬hsink = 104.83 kJ/kg xsink = 0 ⎭Analysis (a) The rate of heat input to the plant may be taken as the enthalpy difference between the sourceand the sink for the power plant & Qin = mgeo (hsource − hsink ) = (440 kg/s)(675.47 − 104.83) kJ/kg = 251,083 kW &The actual thermal efficiency is & W net,out 22 MW η th = = = 0.0876 = 8.8% & Qin 251.083 MW(b) The maximum thermal efficiency is the thermal efficiency of a reversible heat engine operatingbetween the source and sink temperatures TL (25 + 273) K η th, max = 1 − = 1− = 0.312 = 31.2% TH (160 + 273) K(c) Finally, the rate of heat rejection is & & & Qout = Qin − Wnet,out = 251.1 − 22 = 229.1 MWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
31. 31. 6-31Carnot Refrigerators and Heat Pumps6-89C By increasing TL or by decreasing TH. 16-90C It is the COP that a Carnot refrigerator would have, COPR = . TH / TL − 16-91C No. At best (when everything is reversible), the increase in the work produced will be equal to thework consumed by the refrigerator. In reality, the work consumed by the refrigerator will always begreater than the additional work produced, resulting in a decrease in the thermal efficiency of the powerplant.6-92C No. At best (when everything is reversible), the increase in the work produced will be equal to thework consumed by the refrigerator. In reality, the work consumed by the refrigerator will always begreater than the additional work produced, resulting in a decrease in the thermal efficiency of the powerplant.6-93C Bad idea. At best (when everything is reversible), the increase in the work produced will be equalto the work consumed by the heat pump. In reality, the work consumed by the heat pump will always begreater than the additional work produced, resulting in a decrease in the thermal efficiency of the powerplant.6-94 The minimum work per unit of heat transfer from the low-temperature source for a heat pump is to bedetermined.Assumptions The heat pump operates steadily.Analysis Application of the first law gives TH W net,in QH − QL QH QH = = −1 QL QL QL HPFor the minimum work input, this heat pump would be completely Wnet,in QLreversible and the thermodynamic definition of temperature wouldreduce the preceding expression to TL W net,in TH 535 K = −1 = − 1 = 0.163 QL TL 460 KPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
32. 32. 6-326-95E The claim of a thermodynamicist regarding to the thermal efficiency of a heat engine is to beevaluated.Assumptions The plant operates steadily. 1260 RAnalysis The maximum thermal efficiency would be achieved whenthis engine is completely reversible. When this is the case, QH TL 510 R HE η th,max = 1 − = 1− = 0.595 QL Wnet TH 1260 RSince the thermal efficiency of the actual engine is less than this, this 510 Rengine is possible.6-96 An expression for the COP of a completely reversible refrigerator in terms of the thermal-energyreservoir temperatures, TL and TH is to be derived.Assumptions The refrigerator operates steadily.Analysis Application of the first law to the completely reversible refrigerator yields W net,in = Q H − Q LThis result may be used to reduce the coefficient of performance, TH QL QL 1 COPR, rev = = = QH W net,in Q H − Q L Q H / Q L − 1Since this refrigerator is completely reversible, the R Wnet,inthermodynamic definition of temperature tells us that, QL Q H TH = TL QL TLWhen this is substituted into the COP expression, the result is 1 TL COPR, rev = = TH / TL − 1 TH − TLPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
33. 33. 6-336-97 The rate of cooling provided by a reversible refrigerator with specified reservoir temperatures is to bedetermined.Assumptions The refrigerator operates steadily. 300 KAnalysis The COP of this reversible refrigerator is & QH TL 250 K COPR, max = = =5 T H − T L 300 K − 250 K R 10 kWRearranging the definition of the refrigerator coefficient of performance & QLgives & & 250 K Q L = COPR, max W net,in = (5)(10 kW) = 50 kW6-98 The refrigerated space and the environment temperatures for a refrigerator and the rate of heatremoval from the refrigerated space are given. The minimum power input required is to be determined.Assumptions The refrigerator operates steadily.Analysis The power input to a refrigerator will be a minimum when the refrigerator operates in a reversiblemanner. The coefficient of performance of a reversible refrigerator depends on the temperature limits in thecycle only, and is determined from COPR, rev = 1 = 1 = 8.03 25°C (TH / TL ) − 1 (25 + 273 K )/ (− 8 + 273 K ) − 1The power input to this refrigerator is determined from the definition of the Rcoefficient of performance of a refrigerator, 300 kJ/min & QL 300 kJ/min & Wnet,in, min = = = 37.36 kJ/min = 0.623 kW COPR, max 8.03 -8°CPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
34. 34. 6-346-99 The refrigerated space temperature, the COP, and the power input of a Carnot refrigerator are given.The rate of heat removal from the refrigerated space and its temperature are to be determined.Assumptions The refrigerator operates steadily.Analysis (a) The rate of heat removal from the refrigerated space is determined from the definition of theCOP of a refrigerator, QL = COPRWnet,in = (4.5)(0.5 kW ) = 2.25 kW = 135 kJ/min & & 25°C(b) The temperature of the refrigerated space TL is determined fromthe coefficient of performance relation for a Carnot refrigerator, 500 W R 1 1 COPR, rev = ⎯ ⎯→ 4.5 = COP = 4.5 (TH / TL ) − 1 (25 + 273 K )/TL − 1 TLIt yields TL = 243.8 K = -29.2°C6-100 An inventor claims to have developed a refrigerator. The inventor reports temperature and COPmeasurements. The claim is to be evaluated.Analysis The highest coefficient of performance a refrigerator can have when removing heat from a coolmedium at -12°C to a warmer medium at 25°C is 1 1 25°C COPR, max = COPR, rev = = = 7.1 (TH / TL ) − 1 (25 + 273 K )/ (− 12 + 273 K ) − 1The COP claimed by the inventor is 6.5, which is below this Rmaximum value, thus the claim is reasonable. However, it is notprobable. COP= 6.5 -12°CPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
35. 35. 6-356-101E An air-conditioning system maintains a house at a specified temperature. The rate of heat gain ofthe house and the rate of internal heat generation are given. The maximum power input required is to bedetermined.Assumptions The air-conditioner operates steadily.Analysis The power input to an air-conditioning system will be a minimum when the air-conditioneroperates in a reversible manner. The coefficient of performance of a reversible air-conditioner (orrefrigerator) depends on the temperature limits in the cycle only, and is determined from 1 1 COPR, rev = = = 26.75 (TH / TL ) − 1 (95 + 460 R )/ (75 + 460 R ) − 1The cooling load of this air-conditioning system is 95°Fthe sum of the heat gain from the outside and the heatgenerated within the house, & A/C Q L = 800 + 100 = 900 Btu/min HouseThe power input to this refrigerator is determined 800 kJ/min 75°Ffrom the definition of the coefficient of performanceof a refrigerator, & QL 900 Btu/min & Wnet,in, min = = = 33.6 Btu/min = 0.79 hp COPR, max 26.756-102 A heat pump maintains a house at a specified temperature. The rate of heat loss of the house isgiven. The minimum power input required is to be determined.Assumptions The heat pump operates steadily.Analysis The power input to a heat pump will be a minimum when the heat pump operates in a reversiblemanner. The COP of a reversible heat pump depends on the temperature limits in the cycle only, and isdetermined from 1 1 80,000 kJ/h COPHP, rev = = = 10.2 1 − (TL / TH ) 1 − (− 5 + 273 K )/ (24 + 273 K ) HouseThe required power input to this reversible heat pump is determined 24°Cfrom the definition of the coefficient of performance to be & QH 80,000 kJ/h ⎛ 1 h ⎞ & ⎜ Wnet,in, min = = ⎜ 3600 s ⎟ = 2.18 kW ⎟ HP COPHP 10.2 ⎝ ⎠which is the minimum power input required. -5°CPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
36. 36. 6-366-103 A heat pump maintains a house at a specified temperature. The rate of heat loss of the house and thepower consumption of the heat pump are given. It is to be determined if this heat pump can do the job.Assumptions The heat pump operates steadily.Analysis The power input to a heat pump will be a minimum when the heat pump operates in a reversiblemanner. The coefficient of performance of a reversible heat pump depends on the temperature limits in thecycle only, and is determined from 1 1COPHP, rev = = = 14.75 1 − (TL / TH ) 1 − (2 + 273 K )/ (22 + 273 K ) 110,000 kJ/h HouseThe required power input to this reversible heat pump is 22°Cdetermined from the definition of the coefficient ofperformance to be & QH 110,000 kJ/h ⎛ 1 h ⎞ HP 5 kW & ⎜ Wnet,in, min = = ⎜ 3600 s ⎟ = 2.07 kW ⎟ COPHP 14.75 ⎝ ⎠This heat pump is powerful enough since 5 kW > 2.07 kW.6-104E The power required by a reversible refrigerator with specified reservoir temperatures is to bedetermined.Assumptions The refrigerator operates steadily. 540 RAnalysis The COP of this reversible refrigerator is TL 450 R COPR, max = = =5 R . Wnet,in T H − T L 540 R − 450 R 15,000 Btu/hUsing this result in the coefficient of performance expression yields 450 R & QL 15,000 Btu/h ⎛ 1 kW ⎞ & W net,in = = ⎜ ⎟ = 0.879 kW COPR, max 5 ⎝ 3412.14 Btu/h ⎠PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.