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# Thermo for Bio Engineering BE 351

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This 87 page Class Notes was uploaded by Eda Donnelly on Saturday September 19, 2015. The Class Notes belongs to BE 351 at Michigan State University taught by Christopher Saffron in Fall. Since its upload, it has received 59 views. For similar materials see /class/207471/be-351-michigan-state-university in Biosystem Engineering at Michigan State University.

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Date Created: 09/19/15

Solutions Manual for Thermodynamics An Engineering Approach Seventh Edition Yunus A Cengel Michael A Boles McGraw Hill 2011 Chapter 6 THE SECOND LAW OF THERMODYNAMICS PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGrawHill Companies Inc McGrawHill and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions and if the recipient does not agree to these restrictions the Manual should be promptly returned unopened to McGrawHill This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced displayed or distributed in any form or by any means electronic or otherwise without the prior written permission of McGraw Hill PROPRIETARY MATERIAL 2011 The McGraWHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission The Second Law of Thermodynamics and Thermal Energy Reservoirs 61C Transferring 5 kWh of heat to an electric resistance Wire and producing 5 kWh of electricity 62C An electric resistance heater which consumes 5 kWh of electricity and supplies 6 kWh of heat to a room 63C Transferring 5 kWh of heat to an electric resistance Wire and producing 6 kWh of electricity 64C No Heat cannot flow from a lowtemperature medium to a higher temperature medium 65C A thermalenergy reservoir is a body that can supply or absorb finite quantities of heat isothermally Some examples are the oceans the lakes and the atmosphere 66C Yes Because the temperature of the oven remains constant no matter how much heat is transferred to the potatoes 67C The surrounding air in the room that houses the TV set PROPRIETARY MATERIAL 2011 The McGraWHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission Heat Engines and Thermal Efficiency 68C No Such an engine violates the KelvinPlanck statement of the second law of thermodynamics 69C Heat engines are cyclic devices that receive heat from a source convert some of it to work and reject the rest to a sink 610C No Because 100 of the work can be converted to heat 611C It is expressed as quotNo heat engine can exchange heat with a single reservoir and produce an equivalent amount of wor quot 612C a No b Yes According to the second law no heat engine can have and efficiency of 100 613C No Such an engine violates the KelvinPlanck statement of the second law of thermodynamics 614C No The KelvinPlank limitation applies only to heat engines engines that receive heat and convert some of it to work 615C Method b With the heating element in the water heat losses to the surrounding air are minimized and thus the desired heating can be achieved with less electrical energy input 616E The rate of heat input and thermal efficiency of a heat engine are given The power output of the heat engine is to be determined Assumptions 1 The plant operates steadily 2 Heat losses from the working uid at the pipes and other components are negligible 4 3gtlt10 Btuh 77m 40 Analysis Applying the definition of the thermal efficiency to the heat engine Wm Wnet 77thQH 1hp 04 3 104Btuh w X X 254453th 472 hp PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 617 The power output and thermal efficiency of a heat engine are given The rate of heat input is to be determined Assumptions 1 The plant operates steadily 2 Heat losses from the working uid at the pipes and other components are negligible 7 40 30 hp Analysis Applying the definition of the thermal efficiency to the heat engine Wnet 30hp 07457kJs QH lhp 0 4 1 559 leS 77m 618 The power output and thermal efficiency of a power plant are given The rate of heat rejection is to be determined 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 uid at the pipes and other components are negligible Analysis The rate of heat supply to the power plant is determined from the thermal efficiency relation W QH quot 600 MW 1500 MW 77d1 04 The rate of heat transfer to the river water is determined from the first law relation 7701 40 for a heat engine 600 MW QL QH 7W out 1500 7 600 900 MW In reality the amount of heat rejected to the river will be lower since part of the heat will be lost to the surrounding air from the working uid as it passes through the pipes and other components 619 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 losses from the working uid at the pipes and other components are negligible Q Analysis Applying the first law to the heat engine gives H QL QH 7 W et 700 kJ7 250kJ 450 kJ Wmt Q L PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 65 620 The heat rejection and thermal efficiency of a heat engine are given The heat input to the engine is to be determined Assumptions 1 The plant operates steadily 2 Heat losses from the working uid at the pipes and other components are negligible Analysis According to the definition of the thermal efficiency as applied to the heat engine 11H wnet 77111 1 H wnet LIH QL 77th qL which when rearranged gives qL 7 1000 k qH g 1667 lekg 177h 1704 621 The power output and fuel consumption rate of a power plant are given The thermal efficiency is to be determined Assumptions The plant operates steadily Properties The heating value of coal is given to be 30000 kJkg Analysis The rate of heat supply to this power plant is QH rhcoaquVcoal 60000 kghX30000 kIkg 18x 109 kJh 150 MW 500Mw Then the thermal efficiency of the plant becomes Wm 150 MW QH 500 MW 77m 0300 300 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 66 622 The power output and fuel consumption rate of a car engine are given The thermal efficiency of the engine is to be determined Assumptions The car operates steadily Properties The heating value of the fuel is given to be 44000 kJkg Analysis The mass consumption rate of the fuel is m pt5m 08 kgmxzz m 176 kgm The rate of heat supply to the car is QH mcoaquVcoal 176 kgh44000 kJkg 774400 kJh 2151kW Then the thermal efficiency of the car becomes Wnetout 7 55 kW QH 2151 kW 77m 0256256 623 The United States produces about 51 percent of its electricity from coal at a conversion efficiency of about 34 percent The amount of heat rejected by the coalfired 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 per year is Wcoal Wcoal 77m r Qin Qout Wcoal W QM 77 7ch th 1878x1012 kWh 71878gtlt1012 kWh 7 034 3646gtlt1012 kWh 71878gtlt10 kWh PROPRIETARY MATERIAL 2011 The McGraWHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 67 624 The projected power needs of the United States is to be met by building inexpensive but inefficient coal plants or by building expensive but efficient IGCC plants The price of coal that will enable the IGCC plants to recover their cost difference from fuel savings in 5 years is to be determined Assumptions 1 Power is generated continuously by either plant at full capacity 2 The time value of money interest inflation etc is not f considered Properties The heating value of the coal is given to be 28gtlt106 kJton Analysis For a power generation capacity of 150000 MW the construction costs of coal and IGCC plants and their difference are Construction cost m1 150000000 kW1300kV 195 gtlt109 Construction cost IGCC 150000000 kV1500kV 225 gtlt109 Construction cost difference 225 gtlt109 7 195 gtlt109 30 gtlt109 The amount of electricity produced by either plant in 5 years is W WAI1500000001ltW5gtlt 365gtlt 24 h 6570gtlt1012 kWh The amount of fuel needed to generate a specified amount of power can be determined from W2 gt Q W2 or mfuel Qiquot W2 Qin 77 m Then the amount of coal needed to generate this much electricity by each plant and their difference are 77 Heating value 77Heating value W 6570x1012 kWh 3600M 77Heating value 7 04028 x106 kJton K 1kWh W 6570x1012 kWh 3600M 77Heating value 7 04828 gtlt106 kIton K IkWh Amml mmcoal pm 7 rnmmymcplnt 2112 gtlt109 71760 gtlt109 0352 gtlt109 tons m coal coal plant 2112gtlt109 tons mcoalIGCC plant 1760gtlt109 tons For Amml to pay for the construction cost difference of 30 billion the price of coal should be Construction cost difference 7 30 X 109 Amml 0352gtlt109 tons Unit cost of coal 852Iton Therefore the IGCC plant becomes attractive when the price of coal is above 852 per ton PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5 as canmuchan casts and apexaung ef cxency Anew The pmblem xs sawed Hang EES and m snluhan xs gven belaw Gwen W7dm15E7 kW cm7cu 3 PaybackVeavs5 yv Ana ysws MNEZPaybackVeavs CunveMyn h CunstCus1 uaW a cue 7 7 7 300 CunstCu DFCuns10us 7 GCOCuns10ust7wa W 27W du me 2 e a7cuavHv7cuax 0unvemkwm m 722ta7GCC HV7cuaD Cunv2MkW M 7cuaLcuaLm7cuaU unstCus1DWDELTAm7cua 1EIEIEI PaybackVeavs UnNCus waw y Mun Unncmma Sflnn 3 PthckYezrs Ivrl mam mu m Asmmwmg my M21 yu mmg Wm Feminism neoal U nitCOStcoal 03 2841 031 31 09 032 3409 033 375 034 41 4 035 459 036 51 14 037 5734 038 6478 039 7387 04 8524 041 9985 042 1 1 93 043 1466 044 1875 045 2557 Cost Gcc UnitCostm kW ton 1300 0 1400 4262 1500 8524 1600 1279 1700 1705 1 800 213 1 1900 2557 2000 2983 2100 3409 2200 3836 UnitCostcoa lton UnitCostcoa lton 034 036 038 Tlcoal 04 042 044 046 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 COStIGcc PROPRIETARY MATERIAL 2011 The McGraWHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation Lfyou are a student using this Manual you are using it Without permission 610 626 The projected power needs of the United States is to be met by building inexpensive but inefficient coal plants or by building expensive but efficient IGCC plants The price of coal that will enable the IGCC plants to recover their cost difference from fuel savings in 3 years is to be determined Assumptions 1 Power is generated continuously by either 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 be 28gtlt106 kJton Analysis For a power generation capacity of 150000 MW the construction costs of coal and IGCC plants and their difference are Construction cost m1 150000000 kW1300kV 195 gtlt109 Construction cost IGCC 150000000 kV1500kV 225 gtlt109 Construction cost difference 225 gtlt109 7 195 gtlt109 30 gtlt109 The amount of electricity produced by either plant in 3 years is We WA 150000000 kW3gtlt 365gtlt 24 h 3942gtlt1012 kWh The amount of fuel needed to generate a specified amount of power can be determined from W2 W Qin W2 gt in 2 or mmel Qin 77 Then the amount of coal needed to generate this much electricity by each plant and their difference are 77 Heating value 77Heating value W 3942gtlt1012 kWh 3600kJ 9 mcoal coal plant 6 1267 X 10 tons 77Heat1ng value 04028 gtlt 10 kJton 1 kWh W 3942x1012 kWh 3600M mcoalIGCCplam j 1055 gtlt109 tons 77Heating value 7 04828 x105 kJton K 1kWh Amml quot1dele 7 mmmccpm 1267 gtlt109 71055 gtlt109 0211gtlt109 tons For Amml to pay for the construction cost difference of 30 billion the price of coal should be Construction cost difference 7 30 X 109 mm 0211gtlt109 tons Unit cost of coal 142ton Therefore the IGCC plant becomes attractive when the price of coal is above 142 per ton 627E The power output and thermal efficiency of a solar pond power plant are given The rate of solar energy collection is to be determined Assumptions The plant operates steadily v Analysis The rate of solar energy collection or the rate of heat supply to the power plant is determined from the thermal efficiency relation to be 350 kW Solar pond W o QH 350 kW 1Btu 3600s 2386X1073tuh 4 77m 004 1055 k lh PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 611 628 A coalburning power plant produces 300 MW of power The amount of coal consumed during a oneday period and the rate of air owing 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 28000 kJkg Analysis 11 The rate and the amount of heat inputs to the power plant are QI Wnetout 300MW quot 032 9375 MW 7711 Qin Qinm 9375 MJs24gtlt 3600s 81gtlt107 MI The amount and rate of coal consumed during this period are 7 mm Qan W2893x106 kg qHV 28 MJkg 6 mew mml 2893gtlt10 kg 3348 kgS AI 24x3600s b Noting that the airfuel ratio is 12 the rate of air owing through the furnace is r39naiI AFrhml 12kg airkg fuel3348 kgs 4018 kgls PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission Refrigerators and Heat Pumps 629C The difference between the two devices is one of purpose The purpose of a refrigerator is to remove heat from a cold medium whereas the purpose of a heat pump is to supply heat to a warm me ium 630C The difference between the two devices is one of purpose The purpose of a refrigerator is to remove heat from a refrigerated space whereas the purpose of an airconditioner is remove heat from a living space 631C No Because the refrigerator consumes work to accomplish this task 632C No Because the heat pump consumes work to accomplish this task 633C The coefficient of performance of a refrigerator represents the amount of heat removed from the refrigerated space for each unit of work supplied It can be greater than unity 634C The coefficient of performance of a heat pump represents the amount of heat supplied to the heated space for each unit of work supplied It can be greater than unity 635C No The heat pump captures energy from a cold medium and carries it to a warm medium It does not create it 636C No The refrigerator captures energy from a cold medium and carries it to a warm medium It does not create it 637C No device can transfer heat from a cold medium to a warm medium without requiring a heat or work input from the surroundings 638C The violation of one statement leads to the violation of the other one as shown in Sec 64 and thus we conclude that the two statements are equivalent PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 613 639E The COP and the power input of a residential heat pump are given The rate of heating effect is to be determined Reservoir Analysis Applying the definition of the heat pump coefficient of performance to this heat pump gives QH Assumptions The heat pump operates steadily 5 hp COP 24 Reservoir w 30530 Btuh QH COPHPWnetJn 24x5 hp 1 hp 640 The cooling effect and the rate of heat rejection of an air conditioner are given The COP is to be determined Reservoir W netJn Assumptions The air conditioner operates steadily Analysis Applying the first law to the air conditioner gives Wnem QH 7 39L 2572 05kW Applying the definition of the coefficient of performance 4 Reservoir COPR QL 20kW W 05kW netJn 641 The power input and the COP of a refrigerator are given The cooling effect of the refrigerator is to be determined Reservoir Analysis Rearranging the definition of the refrigerator coefficient of 7 performance and applying the result to this refrigerator gives COP71394 QL COPRWHM 1 43 kW 42 kW Assumptions The refrigerator operates steadily W netJn Reservoir PROPRIETARY MATERIAL 2011 The McGraWHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 614 642 A refrigerator is used to keep a food department at a specified temperature The heat gain to the food department and the heat rejection in the condenser are given The power input and the COP are to be determined Assumptions The refrigerator operates steadily Analysis The power input is determined from Wan Q39H 7 Q1 4800 7 3300 1500 kJh 1500 kIh lk W 0417 kW 3600 kJh The COP is cop QL M 22 W 1500 kJh in 643 The COP and the refrigeration rate of a refrigerator are given The power consumption and the rate of heat rejection are to be determined Assumptions The refrigerator operates steadily Analysis a Using the definition of the coefficient of performance the power input to the refrigerator is determined to be 50 kJmin 083 kW W 7 QL 7 60kJmin netm c013R 12 b The heat transfer rate to the kitchen air is determined from the energy balance QH QL Wmin 60 50 110 kJmin 644E 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 Analysis Applying the definition of the heat pump coefficient of H performance to this heat pump gives Assumptions The heat pump operates steadily 2 hp COPHP IQH 15090Btuh lhp 297 Q W 2 hp 25445 Btuh netJn Reservoir PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 615 645 The cooling effect and the COP of a refrigerator are given The power input to the refrigerator is to be determined Reservoir Analysis Rearranging the definition of the refrigerator coefficient of c 013 6 performance and applying the result to this refrigerator gives H 39 Assumptions The refrigerator operates steadily W netJn QL 725000kJh 1h W 434 kW 3600s Wquot c013R 160 Reservoir 646 The COP and the power consumption of a refrigerator are given The time it will take to cool 5 watermelons 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 42 kJkg C Analysis The total amount of heat that needs to be removed from the watermelons is QL chTmmelons 5 X 10 kg42 kJkg c20 7 8 c 2520 k The rate at which this refrigerator removes heat is QL COPRWnem 25045 kw 1125 kW That is this refrigerator can remove 1125 k of heat per second Thus the time required to remove 2520 kl of heat is 7 Q L 2520 kl 224o s373 min QL 1125 kJs At This answer is optimistic since the refrigerated space will gain some heat during this process from the surrounding air which will increase the work load Thus in reality it will take longer to cool the watermelons PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission M7 am Mr canauamnmbe summed Avmnpimxl caalmg 3 A 15 Anxdeal gas wah cmstantspecx c heats almam umpmm Pmpim39m The cmstantwlume specx chutufmns gventabecvr 72 kkg c hnuse 5 HM 8m kg 72 kJkg c35 rznrczz m u This mm xemavedmZU mmuus mum average rate afhzatxemawl rm m hause 15 g mm H 7 7 2 k 5 A Z x s 4 W Uangthz de muan athe cae ment ufperfmmance m pawn puma m m candnumex 15 detammedm be 171kW W SQ ABkW Wquot COPR 28 mam mu m Asmmwmg my M21 yu mmg Wm Feminism in an n s m the BER mung rang are in be mduded Anew The pmblem 15 sawed uuug EES dud m results are ubmmd dud platted belaw Smce d SWEH sea ed we quoteat We nuusu as a c used sys12m cunstamvu ume u demvmme We 1am uMeavan512Hequwedtu mm 012 nuusu App yme vs aw c used system un a me baswsm We nuusu an Data 11 5 c T 2 u c U u 72 kdkdc mjuuse kg DELTA EER ERj 412 Assummg nu Wuvk dunu un mu nuusu and nu heat enevgy added u We nuusu m 012 mu unud de nu unundu m KE and PEJnu ms aw apphed u 012 nuusu s E d m duLu P E m DELTAEidm OiduLL hnuse DELTAujuuseDELTAHNE ya j Usmg the dunnmun unnu cuemme tulpevluvmance unnu AC Widuu 07th LCOP Wmm unvenkamw ij kW n LltW mm OiduLL kJmm Widmjn cunv 2 F Pz h n mu m Madam5mg mu M21 yu musing Wm durum 6 18 649 A refrigerator is used to cool bananas to a specified temperature The power input is given The rate of cooling and the COP are to be determined Assumptions The refrigerator operates steadily Properties The specific heat of banana is 335 kJkgOC Analysis The rate of cooling is determined from QL mop T1 7 T2 215 60 kgmin335 kJkg C24 713 C 132 lemin The COP is COPQL1326Okw W 14kW in 157 650 A refrigerator is used to cool water to a specified temperature The power input is given The flow rate of water and the COP of the refrigerator are to be determined Assumptions The refrigerator operates steadily Properties The specific heat of water is 418 kJkgOC and its density is l kgL Analysis The rate of cooling is determined from QL QH 7 Win 57060 kw7 265kW 685 kW The mass flow rate of water is 685 kW L 009104kgs op T1 7 T2 418 kJkg C23 7 5 c QL hcpT1 T2 quotI39 The volume flow rate is a 546 Lmin p 1kgL lm1n The COP is COP QL 258 W 265 kW m PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 619 651 The rate of heat loss the rate of internal heat gain and the COP of a heat pump are given The power input 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 between the heat lost to the outdoors and the heat generated in the house from the people lights and appliances QH 600007 4000 56000 kJh Using the definition of COP the power input to the heat pump is determined to be quotMquot COPHP 25 W QH 7 56000 kJh 1kW 3600 1an 622 kW 652E The COP and the refrigeration rate of an ice machine are given The power consumption is to be determined Assumptions The ice machine operates steadily u oors Analysis The cooling load of this ice machine is QL rth 28 1bmh169 Btulbm 4732 Btuh Using the definition of the coefficient of performance the power input to the ice machine system is determined to be W 7 QL 7 4732 Btuh lhp netm COPR 24 2545 Btuh 0775 hp 653E An office that is being cooled adequately by a 12000 Btuh window airconditioner is converted to a computer room The number of additional airconditioners that need to be installed is to be determined Assumptions 1 The computers are operated by 7 adult men 2 The computers consume 40 percent of their rated power at any given time Properties The average rate of heat generation from a person seated in a roomoffice is 100 W given Analysis The amount of heat dissipated by the computers is equal to the amount of electrical energy they consume Therefore Q39computers Ratedpower gtlt Usage factor 84 kW04 336 kW Q39people No of people gtlt Q39person 7 X 100 VJ 700 W Q39total Qcompmm Q39people 3360 700 4060 w 13853 Btuh 3 since 1 W 3412 Btuh Then noting that each available air conditioner provides 7000 BtUh 7000 Btuh cooling the number of airconditioners needed becomes Computer 1 Cooling load 13853 Btuh room No of air Cooling capacityof AJC 7000Btuh 198 m 2 Air conditioners PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 654 A decision is to be made between a cheaper but inefficient airconditioner and an expensive but efficient air conditioner for a building The better buy is to be determined 39 The two air quot 39 are in all aspects other than the initial cost and the efficiency Analysis The unit that will cost less during its lifetime is a better buy The total cost of a system during its lifetime the initial operation maintenance etc can be determined by performing a life cycle cost analysis A simpler alternative is to determine the simple payback period The energy and cost savings of the more efficient air conditioner in this case is Energy savings Annual energy usage of A7 Annual energy usage of B Annual cooling loadl COPA 71 COPB 120000kWhyear1327150 7 kW Air Cond A 713500 year COP 32 Cost savings Energy savingsUnit cost of energy 13500 kWhyear010kWh 1350year The installation cost difference between the two airconditioners is Air Cond B Cost difference Cost ofB 7 cost ofA 7000 7 5500 1500 COP 50 Therefore the more efficient airconditioner B will pay for the 1500 cost differential in this case in about 1 year Discussion A cost conscious consumer will have no difficulty in deciding that the more expensive but more efficient air conditioner B is clearly the better buy in this case since air conditioners last at least 15 years But the decision would not be so easy if the unit cost of electricity at that location was much less than 010kWh or if the annual airconditioning load of the house was much less than 120000 kWh 655 A house is heated by resistance heaters and the amount of electricity consumed during a winter month is given The amount of money that would be saved if this house were heated by a heat pump with a known 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 electricity they consume Therefore to achieve the same heating effect the house must be supplied with 1200 kWh of energy A heat pump that supplied this much heat will consume electrical power in the amount of W QH 1200 kWh 500 kWh COPHP 24 which represent a savings of 1200 7 500 700 kWh Thus the homeowner would have saved 700 kWh0085 kWh 5950 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 656 Refrigerant134a flows through the condenser of a residential heat pump unit For a given compressor power consumption the COP of the heat pump and the rate of heat absorbed from the outside air are to be determined Assumptions 1 The heat pump operates steadily 2 The kinetic and potential energy changes are zero 800 kPa QH 800 kPa x0 Properties The enthalpies of R134a at the condenser inlet and exit are P1 800 kPa hl 27122 kJkg T1 35 C P2 800kPa h2 9547 kJkg x2 0 Analysis 11 An energy balance on the condenser gives the heat rejected in the condenser QH r39nh1 7712 0018 kgs27l2279547 kJkg 3164kW QL The COP of the heat pump is cop QHM 254 12 kW 1 The rate of heat absorbed from the outside air Q39L QH 7W1quot 3164712196kW 657 A commercial refrigerator with R 134a as the working uid is considered The evaporator inlet and exit states are specified The mass flow rate of the refrigerant and the rate of heat rejected are to be determined Assumptions 1 The refrigerator operates steadily 2 The kinetic and potential energy changes are zero QH Condenser V Expansion A valve Properties The properties of R 134a at the evaporator inlet and exit states are Tables All through A13 P1 100 kPa x1 02 P2 100 kPa T2 726 C h1 6071 kIkg h2 23474 kJkg Anal sis a The refri eration load is y g 100 kPa QL COPWn 1 20600kW 072 kW QL The mass flow rate of the refrigerant is determined from 072 kW QL 000414 kgls 722 7 771 23474 7 6071 kJkg m b The rate of heat rejected from the refrigerator is QH QL Win 072060132kW PROPRIETARY MATERIAL 2011 The McGraWHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission PerpetualMotion Machines 658C This device creates energy and thus it is aPMIVH 659C This device creates energy and thus it is aPMIVH PROPRIETARY MATERIAL 2011 The McGraWHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 622 Reversible and Irreversible Processes 660C Adiabatic stirring processes are irreversible because the energy stored within the system can not be spontaneously released in a manor to cause the mass of the system to turn the paddle wheel in the opposite direction to do work on the surroundings 661C The chemical reactions of combustion processes of a natural gas and air mixture will generate carbon dioxide water and other compounds and will release heat energy to a lower temperature surroundings It is unlikely that the surroundings will return this energy to the reacting system and the products of combustion react spontaneously to reproduce the natural gas and air mixture 662C No Because it involves heat transfer through a finite temperature difference 663C This process is irreversible As the block slides down the plane two things happen a the potential energy 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 eg perhaps in a spring When we restore the system to its original condition we must a restore the potential energy by lifting the block back to its original elevation and b cool the block and plane back to their original temperatures The potential energy may be restored by returning the energy that was stored during the original process as the block decreased its elevation and releasedpotential energy The portion of the surroundings in which this energy had been stored would then return to its original condition as the elevation of the block is restored to its original condition In order to cool the block and plane to their original temperatures we have to remove heat from the block and plane When this heat is transferred to the surroundings something in the surroundings has to change its state e g perhaps we warm up some water in the surroundings This change in the surroundings is permanent and cannot be undone Hence the original process is irreversible 664C Because reversible processes can be approached inreality and they form the limiting cases Work producing devices that operate on reversible processes deliver the most work and work consuming devices that operate on reversible processes consume the least work 665C When the compression process is nonquasi equilibrium the molecules before the piston face cannot escape fast enough forming a high pressure region in front of the piston It takes more work to move the piston against this high pressure region 666C When an expansion process is nonquasiequilibrium the molecules before the piston face cannot follow the piston fast enough forming a low pressure region behind the piston The lower pressure that pushes the piston produces less wor PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 624 667C The irreversibilities that occur Within the system boundaries are internal irreversibilities those which occur outside the system boundaries are external irreversibilities 668C A reversible expansion or compression process cannot involve unrestrained expansion or sudden compression and thus it is quasiequilibrium A quasiequilibrium expansion or compression process on the other hand may involve external irreversibilities such as heat transfer through a finite temperature difference and thus is not necessarily reversible PROPRIETARY MATERIAL 2011 The McGraWHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 625 The Carnot Cycle and Carnot39s Principle 669C The four processes that make up the Carnot cycle are isothermal expansion reversible adiabatic expansion isothermal compression and reversible adiabatic compression 670C They are l the thermal efficiency of an irreversible heat engine is lower than the efficiency of a reversible heat engine operating between the same two reservoirs and 2 the thermal efficiency of all the reversible heat engines operating between the same two reservoirs are equal 671C False The second Carnot principle states that no heat engine cycle can have a higher thermal efficiency than the Carnot cycle operating between the same temperature limits 672C Yes The second Carnot principle states that all reversible heat engine cycles operating between the same temperature limits have the same thermal efficiency 673C a No b No They would violate the Carnot principle PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission Carnot Heat Engines 674C No 675C The one that has a source temperature of 600 C This is true because the higher the temperature at which heat is supplied to the working uid of a heat engine the higher the thermal efficiency 676E The source and sink temperatures of a heat engine are given The maximum work per unit heat input to the engine is to be determined 3 Assumptions The heat engine operates steadily Analysis The maximum work per unit of heat that the engine can remove from QH the source is the Carnot efficiency which is determined from W TL 7 510R QL W net 11 0595 QH 77thmax TH 1260 R 677 Two pairs of thermal energy reservoirs are to be compared from a workproduction perspective Assumptions The heat engine operates steadily Analysis For the maximum production of work a heat engine operating between the energy a reservoirs would have to be completely reversible Then for the first pair of reservoirs T 325 K QH 17 L17 0519 mm TH 675 K Q L Wm For the second pair of reservoirs TL 275 K o 17 17 0560 77mm TH 625K The second pair is then capable of producing more work for each unit of heat extracted from the hot reservoir PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission mu w m pawn ampm mm engne is m be detemnned A primx Theheatengnz apexates muddy Carnal ef cxemy whchxs deurmxmdfmm T 292 K 7 5m ax 6D EI mm mm mm TH 477 273 K mun kl Tm the maxxmum pawn Mum uf Lhs but mgquot 15 deurmmzd mm the amman arummu ef cxemyta be 9me kJmm653 kW Wmm quotmgquot u nn snuu mm mam mu m Asmmwmg my M21 yu mmg Wm Feminism m nk temperatures umc 25 C andSEI C are m be pmma Anew The pmblem xs sawed Hang EES and m results are tabulated and planed belaw 7 2 aid at 77 5 c H 55uun klmm Fwst Law apphed m m hea engme 07m 7 aim wimim u Wid LnetiWWiduLnefcunven cm Thema Emmencyr Tempeva may 1 7 1L 273m 273 Dehnmun Mayde game By e aithW7dm m Hum kJmm kW mes mus1 be absu ute WueMI Wm kW SE7 2 543 a 7mm 7 744 E 779 A En7 7 Wna le 831 2 851 n 7855 c EEEEEEEE O am my TH Cl TH Cl mam mu m 557mm 5m M21 yu mmg nwnhm Feminism 680E The sink temperature of a Carnot heat engine the rate of heat rejection and the thermal efficiency 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 thermal efficiency 77m 194 gt 075 17W QH 3200 Btumin QH Q Then the power output of this heat engine can be determined from Wmout mth 0753200 Btumin 2400 Btumin 566 hp 1 For reversible cyclic devices we have Ti 800 Btumin QL 7 TL Thus the temperature of the source THmust be TH TL w 520R2080R QL Kev 800 Btumm 681E 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 11 mm TH 1000R 045 When the first law is applied to the engine above 25445 Btuh 1 hp QH Wnet QL 5 hp 15000 Btuh 27720 Btuh The actual thermal efficiency of the proposed heat engine is then W 5 h 77m quot6 7 p 2544 5 Btuhi 7 0459 QH 27720Btuh lhp Since the thermal efficiency of the proposed heat engine is greater than that of a completely reversible heat engine which uses the same isothermal energy reservoirs the inventor39s claim is invalid PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 630 682 The work output and thermal efficiency of a Carnot heat engine are given The heat supplied to the heat engine the heat rejected and the temperature of heat sink are to be determined Assumptions 1 The heat engine operates steadily 2 Heat losses from the working uid at the pipes and other components are negligible Analysis Applying the definition of the thermal efficiency and an energy QH 77 40 balance to the heat engine the unknown values are determined as follows 500 k W QH itw1250kd QL 77m 04 QL QH 7 W et 1250 7 500 750 kJ T 77mmax1T Lgt04017 STL8838K611OC H TL 1200 273 K 683 The work output and heat rejection of a a Carnot heat engine are given The heat supplied to the heat engine and the source temperature are to be determine Assumptions 1 The heat engine operates steadily 2 Heat losses from the a working uid at the pipes and other components are negligible Analysis Applying the definition of the thermal efficiency and an energy QH balance to the heat engine the unknown values are determined as follows 900 k QH QLWne1509001050kJ QL 150kJ Wnet 900 kJ 0857 77 QH 1050kJ STH 2100K 1827 C T mmax 1Lgt085717w 7 TH TH 684 The thermal efficiency and waste heat rejection of a Carnot heat engine are given The power output and the temperature of the source are to be determined Assumptions 1 The heat engine operates steadily 2 Heat losses from the working uid at the pipes and other components are negligible Analysis Applying the definition of the thermal efficiency and an energy balance to the heat engine the power output and the source temperature are determined as follows 7 179 Lgt07517MgtQH 56kW QH QH Wm mQ39H 07556kW 42 kW gtTH 1152K 879 C T 77th11 75115iK TH TH PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 631 685 A geothermal power plant uses geothermal liquid water at 150 C at a specified rate as the heat source The actual and maximum possible thermal efficiencies and the rate of heat rejected from this power plant are to be determined Assumptions 1 The power plant operates steadily 2 The kinetic and potential energy changes are zero 3 Steam properties are used for geothermal water Properties Using saturated liquid properties Table A4 T 150 C sourced P1ng 63218 kJ ltg xsource 0 T 90 C some 37704 kMltg xsource 0 T 25 C smk kink 10483 kMltg xsink 0 Analysis 11 The rate of heat input to the plant is Q39in mgeo hm1 7 hm2 210 kgs63218 7 377 04 kJkg 53580 kW The actual thermal efficiency is WW 7 8000kW 01493 149 Qin 53580 kW 7701 b The maximum thermal efficiency is the thermal efficiency of a reversible heat engine operating between the source and sink temperatures T mmax 1L 17 02955 295 TH 150 273 K 0 Finally the rate of heat rejection is Q39mt Q39in 7 Vim t 535807 8000 45580 kW 686 The claim that the efficiency of a completely reversible heat engine can be doubled by doubling the temperature 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 reversible engine operates between the same energy reservoirs The thermal efficiency of this completely reversible engine is given by T T 7T 77mer l T L a H QH If we were to double the absolute temperature of the high temperature energy reservoir the new thermal efficiency would be W net TL 7 2TH 7TL TH 7TL Q 17 lt 2 2TH 2TH TH o The thermal efficiency is then not doubled as the temperature of the high temperature reservoir is doubled PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission Carnot Refrigerators and Heat Pumps 687C By increasing TL or by decreasing TH 688C The difference between the temperature limits is typically much higher for a refrigerator than it is for an air conditioner The smaller the difference between the temperature limits a refrigerator operates on the higher is the COP Therefore an airconditioner should have a higher COP 689C The deep freezer should have a lower COP since it operates at a much lower temperature and in a given environment the COP decreases with decreasing refrigeration temperature 690C No At best when everything is reversible the increase in the work produced will be equal to the work consumed by the refrigerator In reality the work consumed by the refrigerator will always be greater than the additional work produced resulting in a decrease in the thermal efficiency of the power plant 691C No At best when everything is reversible the increase in the work produced will be equal to the work consumed by the refrigerator In reality the work consumed by the refrigerator will always be greater than the additional work produced resulting in a decrease in the thermal efficiency of the power plant 692C Bad idea At best when everything is reversible the increase in the work produced will be equal to the work consumed by the heat pump In reality the work consumed by the heat pump will always be greater than the additional work produced resulting in a decrease in the thermal efficiency of the power plant 693 The minimum work per unit of heat transfer from the lowtemperature source for a refrigerator is to be determined Assumptions The refrigerator operates steadily Analysis Application of the first law gives a Wm QH 7Q QH71 QH Q Q Q 9 Wm 39n For the minimum work input this refrigerator would be completely reversible and QL U the thermodynamic definition of temperature would reduce the preceding expression to o W netm THili 303K 710110 QL TL 7273K PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 694 The validity of a claim by an inventor related to the operation of a heat pump is to be evaluated Assumptions The heat pump operates steadily Analysis Applying the definition of the heat pump coefficient of performance copHP M 267 Wmin 75 kW The maximum COP of a heat pump operating between the same temperature limits is l l COP 14 7 m 17 TL TH 17 273 K293 K Since the actual COP is less than the maximum COP the claim is valid 695 The power input and the COP of a Carnot heat pump are given The temperature of the lowtemperature reservoir and the heating load are to be determined Assumptions The heat pump operates steadily Analysis The temperature of the lowtemperature reservoir is T COPHpm H gt87amp gtn 2646K 39 TH 7 TL 2997 TLK The heating load is COPHpm QH 87Q HgtQH 37o kw 39 W 425kW m 696 The refrigerated space and the environment temperatures for a refrigerator and the rate of heat removal 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 reversible manner The coefficient of performance of a reversible refrigerator depends on the temperature limits in the cycle only and is determined from The power input to this refrigerator is determined from the definition of the coefficient of performance of a refrigerator 1 COP Rm THma 25273 K78 273 K71 300 kJmin Wnetinmin COP Rmax 39 L w 3736 kJmin 0623 kW PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 634 697 An inventor claims to have developed a refrigerator The inventor reports temperature and COP measurements The claim is to be evaluated Analysis The highest coefficient of performance a refrigerator can have when removing heat from a cool medium at 12 C to a warmer medium at 25 C is COP Rmax l l COPRJev THmu 25273 K712 273 K71 COP 65 The COP claimed by the inventor is 65 which is below this maximum value thus the claim is reasonable However it is not probable 698E An airconditioning system maintains a house at a specified temperature The rate of heat gain of the house and the rate of internal heat generation are given The maximum power input required is to be determined Assumptions The airconditioner operates steadily Analysis The power input to an airconditioning system will be a minimum when the airconditioner operates in a reversible manner The coefficient of performance of a reversible airconditioner or refrigerator depends on the temperature limits in the cycle only and is determined from 1 r 1767 TH TL 71 100 460 R70 460 R71 COPRYrev The cooling load of this airconditioning system is the sum of the heat gain from the outside and the heat generated within the house QL 800 100 900 Btumin 800 kImin The power input to this refrigerator is determined from the definition of the coefficient of performance of a refrigerator Q39 900 Btumin WWWquot W W 5093 Btumin 12o hp Rm ax PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 699 A heat pump maintains a house at a specified temperature The rate of heat loss of the house and the power 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 reversible manner The coefficient of performance of a reversible heat pump depends on the temperature limits in the cycle only and is determined rom 1 1 110000 kl q 17TL TH 17 2 273 K22273 K COPHPJW 1475 The required power input to this reversible heat pump is determined from the definition of the coefficient of performance to be QH 7110000k 1 1h W 3600 s net in min 207 kW 39 39 COPHP 1475 This heat pump is powerful enough since 5 kW gt 207 kW 6100E The power required by a reversible refrigerator with specified reservoir temperatures is to be determined Assumptions The refrigerator operates steadily Analysis The COP of this reversible refrigerator is 7 TL 7 450R 7 5 Km TH 7 TL 540 R7 450 R 9 Wmquot 15000 Btuh COP Using this result in the coefficient of performance expression yields QL 715000Btuh 1kW W 0879 kW 5 341214Btuh netJn COP Rmax 6101 The power input and heat rejection of a reversed Carnot cycle are given The cooling load and the source temperature are to be determined Assumptions The refrigerator operates steadily Analysis Applying the definition of the refrigerator coefficient of performance QL QH 7 Wmin 2000 7 200 1800 kW Applying the definition of the heat pump coefficient of performance cop QL M R Wmin 200kW The temperature of the heat source is determined from T L 9 T COP L R TH 7 TL 300 7 TL 7 TL 270K 73 C PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 636 6102 The power input and the cooling load of an air conditioner are given The rate of heat rejected in the condenser the COP and the rate of cooling for a reversible operation are to be determined Assumptions The air conditioner operates steadily Analysis a The rate of heat rejected is QH QL Vfin 1055 kJ 3600 kl q 18000Btuh 34k lt l 1311 lt of 11w l 31230 leh b The COP is 18000Btuhj co1gtQ L t 1552 W 34 kW m c The rate of cooling if the air conditioner operated as a Carnot refrigerator for the same power input is TL 295 K 2682 TH 4L 337 221lt COPKev 3412Btuh Q39Lm COvaWinmin 268234kW 11W 311130 Btuh 7 7 6103 The rate of heat removal and the source and sink temperatures are given for a Carnot refrigerator The COP of the refrigerator and the power input are to be determined Assumptions The refrigerator operates steadily Analysis The COP of the Carnot refrigerator is determined from TL 28 8 K COP 1371 Rm TH 7TL 36 715 K The power input is l6000 kJh SW3 1167 kJh 0324 kW COPKmaX 57L 1371 In In The rate of heat rejected is QH QL Wuetyin 16000 kM11 167 kJh 17167 leh PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6104 A heat pump maintains a house at a specified temperature in winter The maximum COPS of the heat pump for different outdoor temperatures are to be determined Analysis The coefficient of performance of a heat pump will be a maximum when the heat pump operates in a reversible manner The coefficient of performance of a reversible heat pump depends on the temperature limits in the cycle only and is determined for all three cases above to be 1 1 COP 7 7 293 H 17 TL TH 1 7 10 273K20 273K 1 l lt COP i i 117 HPreV 1 TL TH 17 7 5 273K 20 273K COPHPM 1 7 1 7 586 o 17 TL TH 17 730 273K20 273K 6105E A heat pump maintains a house at a specified temperature The rate of heat loss of the house is given The minimum power inputs required for different source temperatures are to be determined Assumptions The heat pump operates steadily Analysis 11 The power input to a heat pump will be a minimum when the heat pump operates in a reversible manner If the outdoor air at 25 F is used as the heat source the COP of the heat pump and the required power input are determined to be l COPHPmax COPmrev A 55000 Btuh 17TL TH 1 House 1015 0 1725 460 R78460 R 78 F and 39 55000 Btuh 1 hp Wnemmin QH 213 hp COPHPM 1015 2545 Btuh b If the wellwater at 50 F is used as the heat source the COP of the 25 F or heat pump and the required power input are determined to be 50 F COP COP 1 i 1 i 19 2 HR HP 17 TL TH 17 50 460 R78 460 R 39 and 39 55000 Btuh 1 h Wnemmin Q H p 113 hp COPHRmax 192 2545 Btuh PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 638 6106 A Carnot heat pump consumes 66kW of power when operating and maintains a house at a specified temperature The average rate of heat loss of the house in a particular day is given The actual running time of the heat pump that day the heating cost and the cost if resistance heating is used instead are to be determined Analysis 11 The coefficient of performance of this Carnot heat pump depends on the temperature limits in the cycle only and is determined from 1296 1 1 COP HP 17 TL TH 1e 2 2731lt25 273 K 51000 km The amount of heat the house lost that day is QH QH 1 day 55000 kJh24 h 1320000 kJ Then the required work input to this Carnot heat pump is determined from 66 kW the definition of the coefficient of performance to be QH 1320000 k WnetJn COPHP 1296 101880 kI Thus the length of time the heat pump ran that day is Wmquot 7101880 k At W 66 kJs netjn 15440 s 429 h b The total heating cost that day is Cost W X price Wmin gtlt AtXprice 66 kW429 h0085 kWh 241 0 If resistance heating were used the entire heating load for that day would have to be met by electrical energy Therefore the heating system would consume 1320000 kJ of electricity that would cost 1 kWh 3600 kl New Cost QH gtlt price 1 320000k j0085 kWh 312 PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 639 6107 A Carnot heat engine is used to drive a Carnot refrigerator The maximum rate of heat removal from the refrigerated space and the total rate of heat rejection to the ambient air are to be determined Assumptions The heat engine and the refrigerator operate steadily Analysis a The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot efficiency which is determined from TL 7 300 K 7 17 1 77mmax 77mg TH 1173 K Then the maximum power output of this heat engine is determined from the definition of thermal efficiency to be Wmout an39H 0744800 kJmin 5952 kJmin which is also the power input to the refrigerator Wnetvin The rate of heat removal from the refrigerated space will be a maximum if a Carnot refrigerator is used The COP of the Carnot refrigerator is COP 7 1 7 1 7837 RJeV TH TL71 7 27 273 K75 273 K1 Then the rate of heat removal from the refrigerated space becomes Q39LR COPRJeVXWnem 8375952 kImin 4982 kJmin b The total rate of heat rejection to the ambient air is the sum of the heat rejected by the heat engine QLYHE and the heat discarded by the refrigerator Q39HyR Q39LHE Q39HHE 7 Wmout 800 7 5952 2048 kJmin Q39HR Qle Wmin 4982 5952 55772 kImin Q39ambient Q39LHE Q39HR 2048 55772 5782 kJmin PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6108E A Carnot heat engine is used to drive a Carnot refrigerator The maximum rate of heat removal from the refrigerated space and the total rate of heat rejection to the ambient air are to be determined Assumptions The heat engine and the refrigerator operate steadily Analysis 11 The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot efficiency which is determined from TL 7 540 R 17 1 77mmax 77mc TH 2160 R 075 Then the maximum power output of this heat engine is determined from the definition of thermal efficiency to be Wmout an39H 075700 Btumin 525 Btumin which is also the power input to the refrigerator Wmin The rate of heat removal from the refrigerated space will be a maximum if a Carnot refrigerator is used The COP of the Carnot refrigerator is COP 1 1 80 R TH TL71 80 460 R20460 R71 Then the rate of heat removal from the refrigerated space becomes Q39LR COPRJWXWMUH 80525 Btumin 4200 Btumin b The total rate of heat rejection to the ambient air is the sum of the heat rejected by the heat engine QLvHE and the heat discarded by the refrigerator Q39HyR Q39LHE Q39HHE 7 Wmout 700 7 525 175 Btumin Q39HR Q39le Wmin 4200 525 4725 Btumin Q39ambient Q39LHE Q39HR 175 4725 4900 Btumin PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 641 6109 A heat pump that consumes 4kW of power when operating maintains a house at a specified temperature The house is losing heat in proportion to the temperature difference between the indoors and the outdoors The lowest outdoor temperature for which this heat pump can do the job is to be determined Assumptions The heat pump operates steadily Analysis Denoting the outdoor temperature by TL the heating load of this house can be expressed as QH 3800 kJh K297 7 TL 1056 kWIlt297 7 TL K The coefficient of performance of a Carnot heat pump depends on the temperature limits in the cycle only and can be expressed as 3800 kJhK copHP 4 17 TL TH 17TL297 K or as 4 kW COPHP 1056 kWKX297 7 TL K WWquot 4 kW Equating the two relations above and solving for TL we obtain TL 2635K 95 C PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 642 6110 An airconditioner with R l34a as the working uid is considered The compressor inlet and exit states are specified The actual and maximum COPs and the minimum volume flow rate of the refrigerant at the compressor inlet are to be determined Assumptions 1 The airconditioner operates steadily 2 The kinetic and potential energy changes are zero Properties The properties of R l34a at the compressor inlet and exit states are Tables All through Al3 P1 400 kPah1 25555 kJkg QH x1l U1005120m3kg PPM a h 3006111k MI lt T2 70 c 2 g Wm Analysis 11 The mass flow rate of the refrigerant and the power consumption of the compressor are 3 100Lmi 1m 1mm V1 1000L60s 003255kgs Q 21 005120ng 400 kPa sat vap Win th h2 7 hl 003255kgs300617 25555 kJkg 1467 kW The heat gains to the room must be rejected by the airconditioner That is lmin 0 QL QM Q39equipment 250 kJmin 6 09 kW 5067 kW S Then the actual COP becomes COPQ5067kW Win 1467kW 345 b The COP of a reversible refrigerator operating between the same temperature limits is COP 1 1 max 2114 TH TL 71 372732327371 c The minimum power input to the compressor for the same refrigeration load would be Q 7 5067kW W 02396kW quot39quot COPmax The minimum mass flow rate is 7 0005318 kgs mRvm quot h2 7h 30061725555kJkg Finally the minimum volume flow rate at the compressor inlet is Vmin1 mRmin U1 0005318 kgs005120 ng 00002723 m3s 163 Lmin PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 643 6111 The COP of a completely reversible refrigerator as a function of the temperature of the sink is to be calculated and plotted Assumptions The refrigerator operates steadily Analysis The coefficient of performance for this completely reversible refrigerator is given by TL 7 250K COP 7 RM THiTL TH7250K Using EES we tabulate and plot the variation of COP with the sink temperature as follows cOPRmax n I I I I 300 340 380 420 TH K 6112 An expression for the COP of a completely reversible refrigerator in terms of the thermalenergy reservoir 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 Wnemn QH QL This result may be used to reduce the coefficient of performance a QPRJev QLQL QH WnetJn QH QL QH QL 1 6 Since this refrigerator is completely reversible the thermodynamic Wmin definition of temperature tells us that QL QH TH 0 QL TL When this is substituted into the COP expression the result is COP 1 TL 1 TH TL 71 TH 7T PROPRIETARY MATERIAL 2011 The McGraWHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission Special Topic Household Refrigerators 6113C The energy consumption of a household refrigerator can be reduced by practicing good conservation measures such as 1 opening the refrigerator door the fewest times possible and for the shortest duration possible 2 cooling the hot foods to room temperature first before putting them into the refrigerator 3 cleaning the condenser coils behind the refrigerator 4 checking the door gasket for air leaks 5 avoiding unnecessarily low temperature settings 6 avoiding excessive ice buildup on the interior surfaces of the evaporator 7 using the powersaver switch that controls the heating coils that prevent condensation on the outside surfaces in humid environments and 8 not blocking the air flow passages to and from the condenser coils of the refrigerator 6114C It is important to clean the condenser coils of a household refrigerator a few times a year since the dust that collects on them serves as insulation and slows down heat transfer Also it is important not to block air flow through the condenser coils since heat is rejected through them by natural convection and blocking the air flow will interfere with this heat rejection process A refrigerator cannot work unless it can reject the waste heat 6115C Today s refrigerators are much more efficient than those built in the past as a result of using smaller and higher efficiency motors and compressors better insulation materials larger coil surface areas and better door seals 6116C It is a bad idea to overdesign the refrigeration system of a supermarket so that the entire airconditioning needs of the store can be met by refrigerated air without installing any airconditioning system This is because the refrigerators cool the air to a much lower temperature than needed for air conditioning and thus their efficiency is much lower and their operating cost is much higher 6117C It is a bad idea to meet the entire refrigeratorfreezer requirements of a store by using a large freezer that supplies sufficient cold air at 20 C instead of installing separate refrigerators and freezers This is because the freezers cool the air to a much lower temperature than needed for refrigeration and thus their efficiency is much lower and their operating cost is much higher PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 645 6118 A refrigerator consumes 300 W when running and 74 worth of electricity per year under normal use The fraction of the time the refrigerator will run in a year is to be determined Assumptions The electricity consumed by the light bulb is negligible Analysis The total amount of electricity the refrigerator uses a year is Totalcost of energy 7 74year 7 1057 kWhyear Unit cost of energy 007kWh Total electric energy used W27total The number of hours the refrigerator is on per year is WM 71057kWhyear W 3524 hyear 03 kW Total operating hours AI Noting that there are 365x248760 hours in a year the fraction of the time the refrigerator is on during a year is determined to be Total operating hours 7 3524year Total hours per year 7 8760 hyear Time fraction on 0402 Therefore the refrigerator remained on 402 of the time 6119 The light bulb of a refrigerator is to be replaced by a 25 energy efficient bulb that consumes less than half the electricity It is to be determined if the energy savings of the efficient light bulb justify its cost Assumptions The new light bulb remains on the same number of hours a year Analysis The lighting energy saved a year by the energy efficient bulb is Lighting energy saved Lighting power savedOperating hours 40 718W60 hyear 39 1320Wh132kWh This means 132 kWh less heat is supplied to the refrigerated space by the light bulb which must be removed from the refrigerated space This corresponds to a refrigeration savings of 102 kWh Refrigeration energy saved nghtmg energy saved 139321kWh COP Then the total electrical energy and money saved by the energy efficient light bulb become Total energy saved Lighting Refrigeration energy saved 132 102 234 kWh year Money saved Total energy savedUnit cost of energy 234 kWh year008 kWh 019 year That is the light bulb will save only 19 cents a year in energy costs and it will take 25019 132 years for it to pay for itself from the energy it saves Therefore it is not justified in this case PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 646 6120 A person cooks three times a week and places the food into the refrigerator before cooling it first The amount of money this person Will save a year by cooling the hot foods to room temperature before refrigerating them is to be determined Assumptions 1 The heat stored in the pan itself is negligible 2 The specific heat of the food is constant Properties The specific heat of food is c 390 kIkg C given Analysis The amount of hot food refrigerated per year is mfood 5 lltgpan3 pansweek52 weeksyear 780 kgyear The amount of energy removed from food as it is unnecessarily cooled to room temperature in the refrigerator is Energy removed Qout mfoochT 780 kgyear390 kJkg C95 7 23 C 219024 kJyear 7 Energy removed 7 219024 kJyear 1kWh V 7 COP 7 15 3600 kl Money saved CEnergy savedUnit cost of energy 4056 kWhyear010kVh 406lyear Energy saved E 1 4056 kWhyear Therefore cooling the food to room temperature before putting it into the refrigerator Will save about four dollars a year PROPRIETARY MATERIAL 2011 The McGraWHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 647 6121 The door of a refrigerator is opened 8 times a day and half of the cool air inside is replaced by the warmer room air The cost of the energy wasted per year as a result of opening the refrigerator door is to be determined for the cases of moist and dry air in the room Assumptions 1 The room is maintained at 20 C and 95 kPa at all times 2 Air is an ideal gas with constant specific heats at room temperature 3 The moisture is condensed at an average temperature of 4 C 4 Half of the air volume in the refrigerator is replaced by the warmer kitchen air each time the door is opened Properties The gas constant of air is R 0287 kPam3kgK Table Al The specific heat of air at room temperature is cp 1005 kJkg C Table A2a The heat ofvaporization of water at 4 C is hfg 2492 kJkg Table A4 Analysis The volume of the refrigerated air replaced each time the refrigerator is opened is 03 m3 half of the 06 m3 air volume in the refrigerator Then the total volume of refrigerated air replaced by room air per year is Vaimeplaced 03 m3 8day365 daysyear 876 mSyear The density of air at the refrigerated space conditions of 95 kPa and 4 C and the mass of air replaced per year are P p0 01195kgm3 RT 0287 kPa m 3kgK4 273 K malr plair 1195kgm 3876m 3year 1047 kgyear The amount of moisture condensed and removed by the refrigerator is mmoistm mm moisture removed per kg air 1047 kg airyear0006 kgkg air 628 kgyear The sensible latent and total heat gains of the refrigerated space become aninsensible maircp Troom refrig 1047 kgyear1005 kJkg C20 7 4 C 16836 kJyear aninlatent mmoisturehfg 628 kgyear2492 kJkg 15650 kJyear anintotal aninsensible aninlatent 16gt83615gt650 31486 klyear For a COP of 14 the amount of electrical energy the refrigerator will consume to remove this heat from the refrigerated space and its cost are Electrical energy used total Q 31 32 486 kJyear 1 kWh 14 3600 kJ Cost of energy used total Energy usedUnit cost of energy 645 kWhyear0075kVh 048year 645 kWhyear If the room air is very dry and thus latent heat gain is negligible then the amount of electrical energy the refrigerator will consume to remove the sensible heat from the refrigerated space and its cost become Electrical energy used sensible W W 14 3600 kl Cost of energy used sensible Energy usedUnit cost of energy 334 kWhyear0075kVh 025year 334kWhyear PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission Review Problems 6122 The source and sink temperatures of a heat engine are given The maximum work per unit heat input to the engine is to be determined Assumptions The heat engine operates steadily IQ Analysis The maximum work per unit of heat that the engine can remove from H the source is the Carnot efficiency which is determined from W e 7 TL 290K Wnet 77 1 17 o773 H quotm TH 1280K Q L 6123 The work output and the source and sink temperatures of a Carnot heat engine are given The heat supplied to and rejected from the heat engine are to be determined Assumptions 1 The heat engine operates steadily 2 Heat losses from the working uid at the pipes and other components are negligible Analysis Applying the definition of the thermal efficiency and an energy QH balance to the heat engine the unknown parameters are determined as f 11 500 kJ o ows QL T mvmax1f1 Maul w TH 7 1200 273 K W QH net 500 kJ 640 kJ 77m 0781 QL QH 7W 6407500140kJ 6124E The operating conditions of a heat pump are given The minimum temperature of the source that satisfies the second law of thermodynamics is to be determined Assumptions The heat pump operates steadily Analysis Applying the first law to this heat pump gives 341214Btuh 39 39 7W 32000Btuh718kW Qt QH lkw 25860 Btuh In the reversible case we have TL Q39L TH QH Then the minimum temperature may be determined to be 25860 Btuh 32000 Btuh TL TH Q L 530R 428R QH PROPRIETARY MATERIAL 2011 The McGraWHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 649 6125 A heat pump with a specified COP is to heat a house The rate of heat loss of the house and the power consumption of the heat pump are given The time it will take for the interior temperature to rise from 3 C to 22 C is to be determined Assumptions 1 Air is an ideal gas with constant specific heats at room temperature 2 The house is wellsealed so that no air leaks in or out 3 The COP of the heat pump remains constant during operation Properties The constant volume specific heat of air at room temperature is CV 0718 kkg C Table A2 Analysis The house is losing heat at a rate of Q39Loss 40000 kJh 1111kJs 40900 km 22 C The rate at which this heat pump supplies heat is C7 QH COPmWnem 248 kW 192 kW 3 That is this heat pump can supply heat at a rate of 192 kJs Taking the house as the system a closed system the energy balance can be written as Wm Ein Eout AEsystem p0 ential etc energle Qin Qout AU quot 2 1 Qin Qout quot10002 T1 Qin QoutAI quot10002 T1 Substituting 192 71111kJsAt 2000kg0718kMltg c122 7 3 c Net energy tmsfer Change in internal kinetic by heat work and mass 1 5 Solving for AI itwill take AI 3373 s 0937 h for the temperature in the house to rise to 22 C PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 650 6126E A refrigerator with a watercooled condenser is considered The cooling load and the COP of a refrigerator are given The power input the exit temperature of water and the maximum possible COP of the refrigerator are to be determined Assumptions The refrigerator operates steadily Analysis a The power input is water 58 F 7 Q 724000Btuh1055k1 1h 3974 kW 113m 3600 s W quot COP 177 b The rate of heat rejected in the condenser is QH Q Win 24000 Btuh 3974 kW lBtu 3600s 1055kJ 1h 37560Btuh The exit temperature of the water is QH mcpT2 T1 QH rncp T2T1 37560Btuh 3600s 1h 58 F 2 F 145 lbms 100 Btulbm or c Taking the temperature of hightemperature medium to be the average temperature of water in the condenser 25 460 T L 133 TH 7 TL 0558 6527 25 COPKev 6127 A Carnot heat engine cycle is executed in a closed system with a fixed mass of R 134a The thermal efficiency of the cycle is given The net work output of the engine is to be determined Assumptions All components operate steadily Properties The enthalpy of vaporization of R 134a at 50 C is hfg 15179 kIkg Table All Analysis The enthalpy of vaporization kg at a given T or P represents the amount of heat transfer as 1 kg of a substance is converted from saturated liquid to saturated vapor at that T or P Therefore the amount of heat transfer to R134a during the heat addition process of the cycle is QH mhfg50c 001 kg15179 kJkg 1518 kl R 134a Carnot HE Then the work output of this heat engine becomes W mm mth o151518 kl 0228 kJ n PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 651 6128 A heat pump with a specified COP and power consumption is used to heat a house The time it takes for this heat pump to raise the temperature of a cold house to the desired level is to be determined Assumptions 1 Air is an ideal gas with constant specific heats at room temperature 2 The heat loss of the house during the warpup period is negligible 3 The house is wellsealed so that no air leaks in or out Properties The constant volume specific heat of air at room temperature is cv 0718 kIkg C Analysis Since the house is wellsealed constant volume the total amount of heat that needs to be supplied to the house is QH chAThouse 1500 kg0718 kIkg C227 7 C 16155 k A The rate at which this heat pump supplies heat is House H COPMWMin 285 kW 14 kW That is this heat pump can supply 14 kJ of heat per second Thus the time required to 5 kW supply 16155 k of heat is ArQ w1154s192 min QH 14 kJs 6129 A solar pond power plant operates by absorbing heat from the hot region near the bottom and rejecting waste heat to the cold region near the top The maximum thermal efficiency that the power plant can have is to be determine Analysis The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot efficiency which is determined rom 77mmnmyc1i1io127 or 127 W In reality the temperature of the working fluid must be above 35 C in the condenser and below 80 C in the boiler to allow for any effective heat transfer Therefore the maximum efficiency of the actual heat engine will be lower than the value calculated above PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 652 6130 A Carnot heat engine cycle is executed in a closed system with a fixed mass of steam The net work output of the cycle and the ratio of sink and source temperatures are given The low temperature in the cycle is to be determined Assumptions The engine is said to operate on the Carnot cycle which is totally reversible Analysis The thermal efficiency of the cycle is T 1 CarnotHE 77th17 L17 05 TH 2 Also W W 60k 77th QH 120kI QH 77111 05 QL QH 7W12076060kl and 0025 kg H20 QL 60 kJ 2400kIk h qL m 0025kg g fgTL since the enthalpy of vaporization kg at a given T or P represents the amount of heat transfer as 1 kg of a substance is converted from saturated liquid to saturated vapor at that T or P Therefore TL is the temperature that corresponds to the hfg value of 2400 kJkg and is determined from the steam tables Table A4 to be TL 425 C PROPRIETARY MATERIAL 2011 The McGraWHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission a a m Anew The pmblem xs sawed Hang EES and m results are ubumed and planed belaw TL 0 27m 8 A 252 a 232 a 4 2mg 9 184 5 154 A 12m 8 9 5 Ba 17 g 42 5 Wnamm le mam mu m 55mm my M21 yu mmg Wm Feminism 654 6132 A Carnot refrigeration cycle is executed in a closed system with a fixed mass of R 134a The net work input and the ratio of maximumtominimum temperatures are given The minimum pressure in the cycle is to be determined Assumptions The refrigerator is said to operate on the reversed Carnot cycle Which is totally reversible Analysis The coefficient of performance of the cycle is 1 T TH 12TL COPR 5 THTri 1271 Also COPR Vg L A QL c013R me 5X22 kJ110k in QH QLW11022132kJ and U 7 QH 7 132kJ 7 7 qH 777mil375kJkg ihfgTH since the enthalpy of vaporization kg at a given T or P represents the amount of heat transfer per unit mass as a substance is converted from saturated liquid to saturated vapor at that T or P Therefore T H is the temperature that corresponds to the hfg value of 1375 kJkg and is determined from the R 134a tables to be TH 2 613 C 3343 K Then T T H L 12 g 2786 K 2 56 C Therefore Pmin Psat5 6 c 355 kPa PROPRIETARY MATERIAL 2011 The McGraWHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission a a m fuman uf my wk mpum Anew The pmblem xs sawed Hang EES and m results are ubumed and planed belaw Ana1ys1s The cuemmem Mpe uvmance m m cyc e 1s gwen by mj134a u 913 kg THmTLR ahu 1 2 1H 1 2T4 WJH 221lt ependmg my me vam D1 Wim ad1u51 mg guessva ue c1111 COP 1 J 1THtuTLR ahur 1 n COP R F1151 1aw apphed 1 me yemgevatmn cyc e we ds 07L W7m 07H smauymuw ana1ys1s m m cundensev we ds mjmw mj134a U 031 0 H maww 3174 an h 1 3 r U a1su T 3T 0 h T mjmw g 39 g enthamy Aa TZTiH XZWr enthamym Aa TZTiH FD HTHmTLRa1m1L The rmmmum pvessuve 15mg samenun pvessuve cunespundmg 1 LL mm vessuve Aa iL x cunvenkPa MPa 1L 7L 7273 Pmquot MPz mam mu m Asmmwmg my M21 yu mmg mm1 Feminism 656 6134 Two Carnot heat engines operate in series between specified temperature limits If the thermal efficiencies of both engines are the same the temperature of the intermediate medium between the two engines is to be determined Assumptions The engines are said to operate on the Carnot cycle which is totally o reversib e Analysis The thermal efficiency of the two Carnot heat engines can be expressed as T T l 7 and l 7 L 77m1 TH 77am T Equating 1 i 1 TL TH T Solving for T E T JTHT Jasoo K300 K 735 K 6135E The thermal efficiency of a completely reversible heat engine as a function of the source temperature is to be calculated and plotted Assumptions The heat engine operates steadily Analysis With the specified sink temperature the thermal efficiency of this completely reversible heat engine is TL 7 7 500R 17 7 1 77threv TH TH Using EES we tabulate and plot the variation of thermal efficiency with the source temperature T1threv 500 800 1100 1400 1700 2000 TH PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 657 6136 A Carnot heat engine drives a Carnot refrigerator that removes heat from a cold medium at a specified rate The rate of heat supply to the heat engine and the total rate of heat rejection to the environment are to be determined Analysis 11 The coefficient of performance of the Carnot refrigerator is 1 1 COP 6143 M TH TL 71 300 K258 K71 Then power input to the refrigerator becomes 250 kJmjn 407 kJmin Wnemn Q 7 250 kJmin COP 6143 R39C QH R which is equal to the power output of the heat engine W net0ut The thermal efficiency of the Carnot heat engine is determined from T 300 K 7 17 L17 06667 39 T H K Then the rate of heat input to this heat engine is determined from the definition of thermal efficiency to be Q Wnemu 407 kJmin 61 1lemin Hm 77mm 06667 39 b The total rate of heat rejection to the ambient air is the sum of the heat rejected by the heat engine QLYHE and the heat discarded by the refrigerator Q39HyR Qua QHHE Wnetyom 611 407 204 kJmin Q Q39LR Wnetyin 250 407 2907 kImin and Q39Ambient Q39LHE Q39mR 204 2907 311kJmin PROPRIETARY MATERIAL 2011 The McGraWHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission umpen xes af2753EEI andZZS K Anew The pmmm xs sawed Hang EES and m kesu ts are ubumed and planed belaw aim L 25EIKJmm Tisunr K 1H mum TiLic 15c 1L T7L70273 CueMmem uvpemmance mum Camm vemgevamv 1H 15411 P my 171 Puwev mpuHu me vemgevamv W m m we m L weep R mum sij Widmimj The a mmncy ume m1 mm 15 iLiHE 7 1 etaiHE 1 r TiLiHEKH The ma mm mpm m m m1 mm 15 07d c9131 E 7 W74 my 4131 EetaiHE F1151 1aw apphed 1 me m1 mm and vemgevamv OiduLLiHE iduLHiH rvvidutiuuLHE OiduLHj of ng W7dm7m7R om OM IkJmm WW1 3E 51 2313 E 3m 41 231 4 7 27 13 277 1 25 1 275 1 23 72 273 7 AL 3 E 3 TL Om em 1 c kJmm KJmm 1 2m 31 3 231 3 a r13 28 24 27B 2 715 25 21 275 2 714 22 24 272 2 712 1a 31 2139 3 711 113 43 21313 4 Va 13 SE 2E3 E TH K 45 1m 79 251 3 74 a E13 258 V2 5 314 255 3 u 2 E37 252 E mam mu 13 45mm 55 mm yu mmg mm1 Feminism 440 420 400 380 360 340 05quot kJmin 320 300 280 260 500 600 70 0 900 1000 QH HE 12 8 TLc C 05quot kJmin PROPRIETARY MATERIAL 2011 The McGraWHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation Lfyou are a student using this Manual you are using it Without permission 659 660 6138 Half of the work output of a Carnot heat engine is used to drive a Carnot heat pump that is heating a house The minimum rate of heat supply to the heat engine is to be determined Assumptions Steady operating conditions exist Analysis The coefficient of performance of the Carnot heat pump is l l COP 1475 H 17TLTH 172273K22273K Then power input to the heat pump which is supplying heat to the house at the same rate as the rate of heat loss becomes W QH 62000 kIh 75 4203 kJh COPHPYC 14 which is half the power produced by the heat engine Thus the power output of the heat engine is W net out netJn 24203 kJh 8406 1am To minimize the rate of heat supply we must use a Carnot heat engine whose thermal efficiency is determined from TL 293 K 1 TH 1073K 77mc 1 Then the rate of heat supply to this heat engine is determined from the definition of thermal efficiency to be W QH HE quot v 8406 km 11560 kJh 39 77mm 0727 6139E An extraordinary claim made for the performance of a refrigerator is to be evaluated Assumptions Steady operating conditions exist Analysis The performance of this refrigerator can be evaluated by comparing it with a reversible refrigerator operating between the same temperature limits 1 l COP COP 124 Rmquot Rm TH TL 71 75 460 35 460 71 Discussion This is the highest COP a refrigerator can have when absorbing heat from a cool medium at 35 F and rejecting it to a warmer medium at 75 F Since the COP claimed by the inventor is above this maximum value the claim is false PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 661 6140 A Carnot heat pump cycle is executed in a steady ow system with R 134a owing at a specified rate The net power input and the ratio of the maximumtominimum temperatures are given The ratio of the maximum to minimum pressures is to be determined Analysis The coefficient of performance of the cycle is T COP i i 6 0 HP liTLTH 17112 39 TH12TL and TH QH c013HP gtltWin 605 kW 300 kJs QH 300 kJs l3636kk h T qH m 022 kgs g fgTH L since the enthalpy of vaporization kg at a given T or P represents the amount of heat transfer per unit mass as a substance is converted from V gt saturated liquid to saturated vapor at that T or P Therefore TH is the temperature that corresponds to the hfg value of 13636 kIkg and is determined from the R134a tables to be TH 2 620 C 3351 K and Pmax sat620 c 1763kPa T 3351K TL H 2914K 2183OC 125 Also Pmin sat183 c 542 kPa Then the ratio of the maximum to minimum pressures in the cycle is Pm 71763kPa P 542 kPa mm 325 PROPRIETARY MATERIAL 2011 The McGraWHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 662 6141 Switching to energy efficient lighting reduces the electricity consumed for lighting as well as the cooling load in summer but increases the heating load in winter It is to be determined if switching to efficient lighting will increase or decrease the total energy cost of a building Assumptions The light escaping through the windows is negligible so that the entire lighting energy becomes part of the internal heat generation Analysis 11 Efficient lighting reduces the amount of electrical energy used for lighting yeararound as well as the amount of heat generation in the house since light is eventually converted to heat As a result the electrical energy needed to air condition the house is also reduced Therefore in summer the total cost of energy use of the household definitely decreases b In winter the heating system must make up for the reduction in the heat generation due to reduced energy used for lighting The total cost of energy used in this case will still decrease if the cost of unit heat energy supplied by the heating system is less than the cost of unit energy provided by lighting The cost of 1 kWh heat supplied from lighting is 008 since all the energy consumed by lamps is eventually converted to thermal energy Noting that 1 therm 105500 kJ 293 kWh and the furnace is 80 efficient the cost of 1 kWh heat supplied by the heater is Cost of lkWh heat supplied by furnace Amount of useful energy77fum ce Price 1 therm 1kWh 080 140th e139m293 kWh I 0060 per kWh heat ii which is less than 008 Thus we conclude that switching to energy efficient lighting will reduce the total energy cost of this building both in summer and in winter Discussion To determine the amount of cost savings due to switching to energy efficient lighting consider 10 h of operation of lighting in summer and in winter for 1 kW rated power for lighting Current lighting Lighting cost Energy usedUnit cost 1 kW10 h008kWh 080 Increase in air conditioning cost Heat from lightingCOPunit cost 10 kWh35008kWh 023 Decrease in the heating cost Heat from lightingEffunit cost1008 kWh140293kWh 060 Total cost in summer 080023 10339 Total cost in winter 080060 020 Energy e icient lighting Lighting cost Energy usedUnit cost 025 kV10 h008kWh 020 Increase in air conditioning cost Heat from lightingCOPunit cost 25 kWh35008kWh 006 Decrease in the heating cost Heat from lightingEffunit cost2508 kWh140293kWh 015 Total cost in summer 020006 02639 Total cost in winter 020015 005 Note that during a day with 10 h of operation the total energy cost decreases from 103 to 026 in summer and from 020 to 005 in winter when efficient lighting is used PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 663 6142 A heat pump is used to heat a house The maximum money saved by using the lake water instead of outside air as the heat source is to be determined Assumptions 1 Steady operating conditions exist 2 The kinetic and potential energy changes are zero Analysis When outside air is used as the heat source the cost of energy is calculated considering a reversible heat pump as follows COPmax 1 1 1192 14L TH 17 0 273 25 273 W m QH 1400003600kW 3262 kw 39 COP 1192 Costah 3262 kW100 h0085kWh 2773 Repeating calculations for lake water COPmax 1 1 1987 14L TH 1410 273 25273 W 7 QH 7 1400003600 kW 71957 kw quotquotquot quot COP 1987 39 Costlake 1957 kW100 h0085kWh 1663 Then the money saved becomes Money Saved Cost a 7 Costl ke 2773 1663 1110 PROPRIETARY MATERIAL 2011 The McGraWHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 664 6143 The cargo space of a refrigerated truck is to be cooled from 25 C to an average temperature of 5 C The time it Will take for an 8kW refrigeration system to precool the truck is to be determined Assumptions 1 The ambient conditions remain constant during precooling 2 The doors of the truck are tightly closed so that the infiltration heat gain is negligible 3 The air inside is sufficiently dry so that the latent heat load on the refrigeration system is negligible 4 Air is an ideal gas with constant specific heats Properties The density of air is taken 12 kgm3 and its specific heat at the average temperature of 15 C is cp 10 kkg C Table A2 Analysis The mass of air in the truck is malr palyka 12 kgm312m X 23 m X 35 m 116 kg Truck The amount of heat removed as the air is cooled from 25 to 5 C T1 ZSOC leingvm GMAT 116 kg10 kIkg C25 7 5 c T2 5 C 2320 k X Noting that UA is given to be 80 W C and the average air temperature in the truck Q during precooling is 2552 15 C the average rate of heat gain by transmission is determined to be UAAT 80W C25 715 c 800 w 080kJs Quansmission ve Therefore the time required to cool the truck from 25 to 5 C is determined to be merig Al Qcooliugair thnsmission Al 39 39 2 320k em 322s 54min Qrefrig thnsmission 8 0 8 kIS PROPRIETARY MATERIAL 2011 The McGraWHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 665 5144 A refrigeration system is to cool bread loaves at a rate of 1200 per hour by refrigerated air at 30 C The rate ofheat removal from the breads the required volume ow rate of air and the size of the compressor of the refrigeration system are to be determined Assumptions 1 Steady operating conditions exist 2 The thermal properties of the bread loaves are constant 3 The cooling section is wellinsulated so that heat gain through its walls is negligible Properties The average specific and latent heats of bread are given to be 293 kJkg C and 1093 kJkg respectively The gas constant of air is 0287 kPam3kgK Table Al and the specific heat of air at the average temperature of 30 222 26 C e 250 K is C 10 kJkg C Table A2 Ambirix a Noting that the breads are cooled at a rate of 500 loaves per hour breads can be considered to ow steadily through the cooling section at a mass ow rate of mm 1200 breadsh0350 kgbread 420 kgh 01157 kgs Hill Then the rate ofheat removal from the breads as they are cooled from 30 C to 10 C and frozen becomes Qbread mpAmmd 420 kgh293 kJkg C307 710 C49224 kJh th momma 420 kghX1093kmltg 419061001 and Q39Mal Q39bmd QMquotg 49224 45905 95130 kJh 10m hv the breads is an oiueuu 39 air 39 air39 H A 8 C 39 39 39 mw auu ofair 39 39 39 39 mm A L akjhu 117891 kgh CPAT m 10 kJkg cxs c pi 101393kpa 1453kgm3 RT 0287 kPam3kgK30 2731lt 2m quot393quot 1139891kgh3 8185m3h p3 1453 kgm c For a COP of 12 the size of the compressor ofthe refrigeration system must be W W 79275 km 2202kW 3 COP 12 PROPRIETARY MATERIAL 2011 r Inc 4 01 nrmamtim 666 6145 The drinking water needs of a production facility with 20 employees is to be met by a bobbler type water fountain The size of compressor of the refrigeration system of this water cooler is to be determined Assumptions 1 Steady operating conditions exist 2 Water is an incompressible substance with constant properties at room temperature 3 The cold water requirement is 04 Lh per person Properties The density and specific heat of water at room temperature are p l0 kgL and c 418 kIkg CC Table A 3 Analysis The refrigeration load in this case consists of the heat gain of the reservoir and the cooling of the incoming water The water fountain must be able to provide water at a rate 0 rhmter pVWmer l kgL04Lh person20 persons 80kgh To cool this water from 22 C to 8 C heat must removed from the Water in water at a rate of 22 C Qcooling rhcpTin Tout aquotI 80 kgh418 kJkg C22 8 C lt 468 kJh 130w since 1 W36kJh Then total refrigeration load becomes QrefrigJotal Qcoolmg thnsfer 130 45 175 W 3 lt Noting that the coefficient of performance of the refrigeration Refrig system is 29 the requiredpower input is Water out I a Qrefri 175 W 80c i W 7 g 603 W R s COP 29 Therefore the power rating of the compressor of this refrigeration system must be at least 603 W to meet the cold water requirements of this office PROPRIETARY MATERIAL 2011 The McGraWHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 667 6146E A washing machine uses 33year worth of hot water heated by a gas water heater The amount of hot water an average family uses per week is to be determined Assumptions 1 The electricity consumed by the motor of the washer is negligible 2 Water is an incompressible substance with constant properties at room temperature Properties The density and specific heat of water at room temperature are p 621 lbmft3 and c 100 Btulbm F Table A3E Analysis The amount of electricity used to heat the water and the net amount transferred to water are Total cost of energy 7 33year 7 2727 thermsyear Unit cost of energy 121therm Total energy used gas Total energy transfer to water Ein EfficiencyTotal energy used 058 X 2727 therm syear 100000 Btu 1 1 year 1 1582 therms ear 1582 therms ear y y 1therm 52 weeks 30420 Btuweek Then the mass and the volume of hot water usedper week become 7 30420 Btuweek 10Btulbm F13060 F Ein mcTout 11quot a m m CT out 7 T 4346 lbmweek 1 Vwam 4346lbmwe3ek 70 ft3Week p 6211bmft 74804 gal 1f 3 j 524 galweek t Therefore an average family uses about 52 gallons of hot water per week for washing clothes PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission be summed Armapn39mxl Th2 pnce uf elecmcxtyrzmmns Canaan 2 Water 5 mmcampresable substance wnh can mlpmpemes anaam temperature 3 Txme mu uf m may 0nt m auan xs nut Carma126 Anew The mamt af elecmcny used in heat the wam and the ml amaunttxansfenedta wam are Tam castaf emrgy Tam amagymmxamcag 7 Um cam my 7 sunya 39snnxnkWh 7 ZSkWhyur Tam may mama wan 5 mammaan mama n 95 am may rig gkwlqyeaz Th2 mamt af elecmcny cansumedbythe heatpump anams cast are 2969kWhyeaz z z 299 5 kWhyear Energy mgr in water Em usa e afheat xn 7 g P P COP 57 97yeaz Thznthe maney may yearbythz hzatpump And39he simple payback penadbecame Mamyswed megycast uf elennchutuj 7megycast uf heatpump 25n7719717x n Adniumalmsta auancast My yam Mmeysave sznn sm 2 aback enad 7 9 W P 178Uyur the heat pump In summer and mus am serving as an mxrcandluaner mmaaa man an asmmmm Oh Mann yu musing Maw Feminism 2 am number ufyurs requuedta break even are m be bdmdmd Anew The pmme 15 sawed Hang EES and up results are ubumed and p aned belaw Enevgy supphed by the wam heatenu the wam p21 yea s EiEbcHea ev Oust p21 yeaHu upevate exedymwam heate uv DIVE yea s Cus17E 2mHea ev25E Eayam Enevgy supphed m the Way by Emma beam damu bmeydy puvchased a 95 E7E 22 Heatey eta CusLE ectHeateyUthus1 kWhyeav UMCUSFD ma EakWh Watev mp Enevgy supphed by heat pump heatev Enevgy supphed by exam beam EiHeatPump EbmHeamWWWyear E edvma Wuvk enegy supphed m heat pump Heat added m wamCOP 00H 3 Wjeadpum EiHeatPumpCOP kWhyeav Oust p21 yea m upevate me heat pump 5 CuSLHeatPumpZVViHeatpump Uthu Let vakEyen be We numbev Myeavs m bveak even Veamb bveak eyem HEQ EEHHQ the cusHu bunuw the ewe iaEEIEI m mstaH heat pump me a Cus1DWU cm d0usHNi vkEven CusLH2atPumpCusLE edHea ev 5 Addc am Cus ummw cpsuaeupm eavs iayeav eav E D95 HE E 5 452 m3 3 5 m2 9135 A 759 m a A 551 74 22 A 392 E7 BE A 2E7 EZ 5 A 155 57 93 4 mm 53 98 4 cm 5m 53 3 951 47 5 Flyznunn mu m dsmmxmg Oh M21 yu mmg uwm Rmsm N BrkEven years Cost lyear Heat punp PROPRIETARY MATERIAL 2011 The McGraWHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation Lfyou are a student using this Manual you are using it Without permission 6149 A home owner is to choose between a highefficiency natural gas furnace and a groundsource heat pump The system with the lower energy cost is to be determined Assumptions The two heater are comparable in all aspects other than the cost of energy Analysis The unit cost of each kJ of useful energy supplied to the house by each system is Natural gasfarnace Unit cost of useful energy W 138 gtlt10 kI 097 105500 kJ Heat Pump System Unit cost of useful energy 73gtlt10 6 kJ 35 3600 kl The energy cost of groundsource heat pump system will be lower 6150 The ventilating fans of a house discharge a houseful of warmed air in one hour ACH 1 For an average outdoor temperature of 5 C during the heating season the cost of energy vented out by the fans in 1 h is to be determined Assumptions 1 Steady operating conditions exist 2 The house is maintained at 22 C and 92 kPa at all times 3 The infiltrating air is heated to 22 C before it is vented out 4 Air is an ideal gas with constant specific heats at room temperature 5 The volume occupied by the people furniture etc is negligible Properties The gas constant of air isR 0287 kPam3kgK Table A l The specific heat of air at room temperature is cp 10 kIkg C Table A2a Analysis The density of air at the indoor conditions of 92 kPa and 22 C is 5 C 7 P0 7 92km 1087kgm3 92kPa RTg 0287kPam3kgK22273K Po Bathroom Noting that the interior volume of the house is 200 gtlt 28 560 m3 the fan T T mass flow rate of air vented out becomes mar pvajr 1087 kgm3560m3h 6087 kgh 0169 kgs Noting that the indoor air vented out at 22 C is replaced by infiltrating outdoor air at 5 C this corresponds to energy loss at a rate of Q10ssfan maircp Tindoors Toutdoors 22 c 0169 kgs10 kkg C227 5 C 2874 kJs 2874 kW Then the amount and cost of the heat vented out per hour becomes Fuel energy loss QilossyfmAt fmace 2874 kV1h096 2994 kWh Money loss Fuel energy lossUnit cost of energy 1 therm 2994kWh 120th X39 erm293kWh 0123 Discussion Note that the energy and money loss associated with ventilating fans can be very significant Therefore ventilating fans should be used sparingly PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 672 6151 The ventilating fans of a house discharge a houseful of airconditioned air in one hour ACH 1 For an average outdoor temperature of 28 C during the cooling season the cost of energy vented out by the fans in 1 h is to be determined Assumptions 1 Steady operating conditions exist 2 The house is maintained at 22 C and 92 kPa at all times 3 The infiltrating air is cooled to 22 C before it is vented out 4 Air is an ideal gas with constant specific heats at room temperature 5 The volume occupied by the people furniture etc is negligible 6 Latent heat load is negligible Properties The gas constant of air isR 0287 kPam3kgK Table Al The specific heat of air at room temperature is cp 10 kJkg C Table A2a Analysis The density of air at the indoor conditions of 92 kPa and 22 C is P 7 0 7 92km 1087kgm3 RT 0287kPam3kgK22273K 28 C Pa Noting that the interior volume of the house is 200 X 28 560 m3 the mass flow rate of air vented out becomes majr primr 1087 kgm 3 560 m 3m 6087 kgh 0169 kgs Noting that the indoor air vented out at 22 C is replaced by infiltrating outdoor air at 28 C this corresponds to energy loss at a rate of Qlossfan maircp Toutdoors indoors 0169 kgs10kkg C287 22 C1014ks 1014 kW Then the amount and cost of the electric energy vented out per hour becomes Electric energy loss QlossyfmAtCOP 1014 kV1h23 0441 kWh Money loss Fuel energy lossUnit cost of energy 0441kVh010kh 0044 Discussion Note that the energy and money loss associated with ventilating fans can be very significant Therefore ventilating fans should be used sparingly PROPRIETARY MATERIAL 2011 The McGraWHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 673 6152 A geothermal heat pump with R 134a as the working uid is considered The evaporator inlet and exit states are specified The mass flow rate of the refrigerant the heating load the COP and the minimum power input to the compressor are to be determined Assumptions 1 The heat pump operates steadily 2 The kinetic and potential energy changes are zero 3 Steam properties are used for geothermal water Properties The properties of R 134a and water are QH Steam and R 134a tables Condenser T1 12 C h19655 kJkg x1 015 P1 4433 kPa V Expansion Win P2 P1 4433kPa A valve h2 25727 kIkg x2 1 TW 1 60 C 39 hw125118kJkg xw1 0 7 TW 2 40 C 39 hw 2 16753 kJkg xwv2 0 39 Geo water 60 C 40 C Analysis 11 The rate of heat transferred from the water is the energy change of the water from inlet to exit QL rhw hwy1 7 hm2 0065 kgs25118 716753kJkg 5437 kW The energy increase of the refrigerant is equal to the energy decrease of the water in the evaporator That is 5437kW QL h 7 QL 2 hi mR hr1 2572779655kJkg 00338 kgls b The heating load is QH QL Win 543716 704 kW 0 The COP of the heat pump is determined from its definition 7 704 kW COPQ H 16kW W m 440 d The COP of a reversible heat pump operating between the same temperature limits is 1 1 COP 9 17 TL TH 17 25 273 60 273 max 51 Then the minimum power input to the compressor for the same refrigeration load would be W 7 QH 704kW quot COP 951 max 0740 kW PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission qua suppnea m the heat pump are m be damned Table A3 QampTrTx v 7 TrTx 3 W4 kpkgecxswmec lmzkg 5513 kW 5 The cop uf areva39sxble h the spem e eatpump upemhng between a temperature um 15 LTH 1 1n 1 z733n 273 Than the mnmum puwe mpm mum be QR W m COPm pmmp xrymme a 5mm Ismg ups Mama pm m using u when pnnls an memass uwmte uflakevater emu be dammed wm nn 1 7 c wmm mm Luke wmu pumy Lm mm m H m exchanch 14kg 6 C cop 41129 TLTH 2732727K Thenmemmmumpuwermpulvmuldbe Wm 2 Auunz uukw mum CUP 1429 27 The le ufhal absurbed mm me lake 15 Q Qerm 17 7871244 16 SKkW An enagy balance unthehal Exchanga39 gves me mass uw rate uflake Water m W 9 amp 0751kgS cpAT AIBkJkgTXST weth mm a A Mummg my Mm m m mg mum mm 676 6155 It is to be proven that a refrigerator39s COP cannot exceed that of a completely reversible refrigerator that shares the same thermalenergy reservoirs Assumptions The refrigerator operates steadily Analysis We begin by assuming that the COP of the general refrigerator B is greater than that of the completely reversible refrigeratorA COPE gt COPA When this is the case a rearrangement of the coefficient of performance expression yields WE Q lt Qt WA COPE COP A That is the magnitude of the work required to drive refrigerator B is less than that needed to drive completely reversible refrigeratorA Applying the first law to both refrigerators yields QHB lt QHA since the work supplied to refrigerator B is less than that supplied to refrigeratorA and both have the same cooling effect QL Since A is a completely reversible refrigerator we can reverse it without changing the magnitude of the heat and work transfers This is illustrated in the figure below The heat QL which is rejected by the reversed refrigerator A can now be routed directly to refrigerator B The net effect when this is done is that no heat is exchanged with the TL reservoir The magnitude of the heat supplied to the reversed refrigerator A QHA has been shown to be larger than that rejected by refrigerator B There is then a net heat transfer from the TH reservoir to the combined device in the dashed lines of the figure whose magnitude is given by QHA 7 QHB Similarly there is a net work production by the combined device whose magnitude is given by WA 7 W3 WA W3 4 The combined cyclic device then exchanges heat with a reservoir at a single temperature and produces work which is clearly a violation of the KelvinPlanck statement of the second law Our assumption the COPE gt COPA must then be wrong PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6156 It is to be proven that the COP of all completely reversible refrigerators must be the same when the reservoir temperatures are the same Assumptions The refrigerators operate steadily Analysis We begin by assuming that COPA lt COPE When this is the case a rearrangement of the coefficient of performance expression yields QL gt Qt WE COP A COPE WA That is the magnitude of the work required to drive refrigerator A is greater than that needed to drive refrigerator B Applying the first law to both refrigerators yields QHA gtQHB since the work supplied to refrigeratorA is greater than that supplied to refrigerator B and both have the same cooling effect QL Since A is a completely reversible refrigerator we can reverse it without changing the magnitude of the heat and work transfers This is illustrated in the figure below The heat QL which is rejected by the reversed refrigeratorA can now be routed directly to refrigerator B The net effect when this is done is that no heat is exchanged with the TL reservoir The magnitude of the heat supplied to the reversed refrigeratorA WA WB lt QHA has been shown to be larger than that rejected by refrigerator B There is then a net heat transfer from the TH reservoir to the combined device in the dashed lines of the figure whose magnitude is given by QHA 7 QHB Similarly there is a net work production by the combined device whose magnitude is given by WA 7 W3 The combined cyclic device then exchanges heat with a reservoir at a single temperature and produces work which is clearly a violation of the KelvinPlanck statement of the second law Our assumption the COPA lt COPE must then be wrong If we interchange A and B in the previous argument we would conclude that the COPE cannot be less than COPA The only alternative left is that COPA COPE PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6157 An expression for the COP of a completely reversible heat pump in terms of the thermalenergy reservoir temperatures TL and TH is to be derived Assumptions The heat pump operates steadily Analysis Application of the first law to the completely reversible heat pump yields Wnetjn QH Q This result may be used to reduce the coefficient of performance COPHPrev QH QH 1 Wnemn QH QL 1QL QH Since this heat pump is completely reversible the thermodynamic definition of temperature tells us that QL TL Q H TH When this is substituted into the COP expression the result is l T COPHPrev H 14L TH TH 7TL 6158 A Carnot heat engine is operating between specified temperature limits The source temperature that Will double the efficiency is to be determined Analysis Denoting the new source temperature by TH the thermal efficiency of the Carnot heat engine for both cases can be expressed as T t T 77mc 1 L and 77am 1 i 277mg TH TH 0 Substituting Solving for TH T m o TH72TL which is the desired relation PROPRIETARY MATERIAL 2011 The McGraWHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 679 6159 A Carnot cycle is analyzed for the case of temperature differences in the boiler and condenser The ratio of overall temperatures for which the power output will be maximum and an expression for the maximum net power output are to be determined Analysis It is given that QHhAHTH7T TH Therefore W 77mQH 1 TH TH 17 JTH n where we defined r andx as r TLTH and x l 7 THTH For a reversible cycle we also have TQH1 mom mlHrHlirgTH T Q r hAtlTLTL hAtTHlTXTVTLTH but TL i T rl7x TH T TH Substituting into above relation yields 1 7 H x r hALr17x7 TL TH Solving for x 7 r 7 TL TH x rlthAH mm 1 Substitute 2 into 1 r 7 TL TH W METHO rrhAH hAL 1 Taking the partial derivative aa W holding everything else constant and setting it equal to zero gives r which is the desired relation The maximum net power output in this case is determined by substituting 4 into 3 It simplifies to PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 680 Fundamentals of Engineering FE Exam Problems 6160 The label on a washing machine indicates that the washer will use 85 worth of hot water if the water is heated by a 90 efficiency electric heater at an electricity rate of 009kWh 1f the water is heated from 18 C to 45 C the amount of hot water an average family uses per year in metric tons is a 116 tons b 158 tons c 271 tons d 301 tons e 335 tons Answer b 271 tons Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Eff090 C418 quotkJkgCquot T118 quotCquot T245 quotCquot Cost85 quotquot Price009 quotkWhquot EinCostPrice3600 quotkJquot EinmCT2 T1Eff quotkJquot quotSome Wrong Solutions with Common Mistakesquot EinW1mCT2 T1Eff quotMultiplying by Eff instead of dividingquot EinW2mCT2 T1 quotIgnoring efficiencyquot EinW3mT2 T1Eff quotNot using specific heatquot EinW4mCT2T1Eff quotAdding temperaturesquot PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 681 6161 A 24m high 200m2 house is maintained at 22 C by an airconditioning system whose COP is 32 It is estimated that the kitchen bath and other ventilating fans of the house discharge a houseful of conditioned air once every hour If the average outdoor temperature is 32 C the density of air is 120 kgm3 and the unit cost of electricity is 010kWh the amount of money vented out by the fans in 10 hours is a 050 b 160 c 500 d 1100 e 1600 Answer a 050 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values COP32 T1 22 quotCquot T232 quotCquot Price010 quotkWhquot Cp1005 quotkJkgCquot rho120 quotkgmquot3quot V24200 quotmquot3quot mrhoV mtotam10 Ein mtotaCpT2T1COP quotkJquot CostEin3600Price quotSome Wrong Solutions with Common Mistakesquot W1CostPrice3600mtotaCpT2T1COP quotMultiplying by Eff instead of dividingquot W2CostPrice3600mtotaCpT2T1 quotIgnoring efficiencyquot W3CostPrice3600mCpT2T1COP quotUsing m instead of mtotaquot W4CostPrice3600mtotaCpT2T1COP quotAdding temperaturesquot 6162 The drinking water needs of an office are met by cooling tab water in a refrigerated water fountain from 23 C to 6 C at an average rate of 10 kgh 1f the COP of this refrigerator is 31 the required power input to this refrigerator is a 197 W b 612 W c 64 W d 109 W e 403 W Answer c 64 W Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values COP31 Cp418 quotkJkgCquot T1 23 quotCquot T26 quotoquot mdot103600 quotkgsquot QLmdotCpT1T2 quotkW39 WinQL1000COP quotW39 quotSome Wrong Solutions with Common Mistakesquot W1WinmdotCpT1T2 1000COP quotMultiplying by COP instead of dividingquot W2WinmdotCpT1T2 1000 quotNot using COPquot W3WinmdotT1T2 1000COP quotNot using specific heatquot W4Win mdotCpT1T2 1000COP quotAdding temperaturesquot PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 682 6163 A heat pump is absorbing heat from the cold outdoors at 5 C and supplying heat to a house at 25 C at a rate of 18000 kJh Ifthe power consumed by the heat pump is 19 kW the coefficient of performance of the heat pump is a 13 b 26 c 30 d 38 e 139 Answer b 26 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values TL5 quotCquot TH25 quotCquot QH180003600 quotkJsquot Win19 quotkW39 COPQHNVin quotSome Wrong Solutions with Common Mistakesquot W1COPWinQH quotDoing it backwardsquot W2COPTHTHTL quotUsing temperatures in Cquot W3COPTH273THTL quotUsing temperatures in Kquot W4COPTL273THTL quotFinding COP of refrigerator using temperatures in Kquot 6164 A heat engine cycle is executed with steam in the saturation dome The pressure of steam is 1 MPa during heat addition and 04 MPa during heat rejection The highest possible efficiency of this heat engine is a 80 b 156 c 202 d 798 e 100 Answer a 80 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values PH1000 quotkPaquot PL400 quotkPaquot THTEMPERATURESteamAPWSXOPPH TLTEMPERATURESteamIAPWSXOPPL EtaCarnot1TL273TH273 quotSome Wrong Solutions with Common Mistakesquot W IEtaCarnot1PLPH quotUsing pressuresquot W2EtaCarnot1TLTH quotUsing temperatures in Cquot W3EtaCarnotTLT H quotUsing temperatures ratioquot PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 683 6165 A heat engine receives heat from a source at 1000 C and rejects the waste heat to a sink at 50 C lfheat is supplied to this engine at a rate of 100 kJs the maximum power this heat engine can produce is a 254 kW b 554 kW c 746 kW d 950 kW e 1000 kW Answer c 746 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values TH1000 quotCquot TL50 quotCquot Qin100 quotkWquot Eta1TL273TH273 Wout EtaQin quotSome Wrong Solutions with Common Mistakesquot W1Wout1TLTHQin quotUsing temperatures in Cquot W2WoutQin quotSetting work equal to heat inputquot W3WoutQinEta quotDividing by efficiency instead of multiplyingquot W4WoutTL273TH273Qin quotUsing temperature ratioquot 6166 A heat pump cycle is executed with R134a under the saturation dome between the pressure limits of 14 MPa and 016 MPa The maximum coefficient of performance ofthis heat pump is a 11 b 38 c 48 d 53 e 29 Answer c 48 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values PH1400 quotkPaquot PL160 quotkPaquot THTEMPERATURER134ax0PPH quotoquot TLTEMPERATURER134ax0PPL quotoquot COPHPTH273THTL quotSome Wrong Solutions with Common Mistakesquot W1COPPHPHPL quotUsing pressuresquot W2COPTHTHTL quotUsing temperatures in Cquot W3COPTLTHTL quotRefrigeration COP using temperatures in Cquot W4COPTL273THTL quotRefrigeration COP using temperatures in Kquot PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation Lfyou are a student using this Manual you are using it without permission 684 6167 A refrigeration cycle is executed with R 134a under the saturation dome between the pressure limits of 16 MPa and 02 MPa 1f the power consumption of the refrigerator is 3 kW the maximum rate of heat removal from the cooled space of this refrigerator is a 045 kJs 078 kJs c 30 kJs d 116 kJs e 146 kJs Answer d 116 kIs Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values PH1600 quotkPaquot PL200 quotkPaquot Win3 quotkW39 THTEMPERATURER134ax0PPH quotoquot TLTEMPERATURER134aX0PPL quotoquot COPTL273THTL QLWinCOP quotkW39 quotSome Wrong Solutions with Common Mistakesquot W1QLWinTLTHTL quotUsing temperatures in Cquot W2QLWin quotSetting heat removal equal to power inputquot W3QLWinCOP quotDividing by COP instead of multiplyingquot W4QLWinTH273THTL quotUsing COP definition for Heat pumpquot 6168 A heat pump with a COP of 32 is used to heat a perfectly sealed house no air leaks The entire mass within the house air furniture etc is equivalent to 1200 kg of air When running the heat pump consumes electric power at a rate of 5 kW The temperature of the house was 7 C when the heat pump was turned on If heat transfer through the envelope of the house walls roof etc is negligible the length of time the heat pump must run to raise the temperature of the entire contents of the house to 22 C is a 135 min b 431 min c 138 min d 188 min e 808 min Answer a 135 min Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values corgt32 Cv0718 quotkJkgCquot m1200 quotkgquot T1 7 quot0quot T222 quotoquot QHmCvT2T1 Win5 quotkWquot WintimeQHCOP60 quotSome Wrong Solutions with Common Mistakesquot WinW ltime60mCvT2T l COP quotMultiplying by COP instead of dividingquot WinW2time60mCvT2T1 quotIgnoring COPquot WinW3timemCvT2T1 COP quotFinding time in seconds instead of minutesquot WinW4time60mCpT2T l COP quotUsing Cp instead of Cvquot Cp1005 quotkJkgKquot PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 685 6169 A heat engine cycle is executed with steam in the saturation dome between the pressure limits of 7 MPa and 2 MPa If heat is supplied to the heat engine at a rate of 150 kIs the maximum power output of this heat engine is a 81 kW b 197 kW c 386kW d 107 kW e 130 kW Answer b 197 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values PH7000 quotkPaquot PL2000 quotkPaquot Qin150 quotkWquot THTEMPERATURESteamlAPWSXOPPH quotCquot TLTEMPERATURESteamlAPWSxOPPL quotCquot Eta1TL273TH273 Wout EtaQin quotSome Wrong Solutions with Common Mistakesquot W1Wout1TLTHQin quotUsing temperatures in Cquot W2Wout lPLPHQin quotUsing pressuresquot W3WoutQinEta quotDividing by efficiency instead of multiplyingquot W4WoutTL273TH273Qin quotUsing temperature ratioquot 6170 An airconditioning system operating on the reversed Carnot cycle is required to remove heat from the house at a rate of 32 kIs to maintain its temperature constant at 20 C If the temperature of the outdoors is 35 C the power required to operate this airconditioning system is a 058 kW b 320 kW c 156 kW d 226 kW e 164 kW Answer e 164 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values TL20 quotoquot TH35 quotoquot QL32 quotkJsquot COPTL273THTL COPQLNVin quotSome Wrong Solutions with Common Mistakesquot QLW1WinTLTHTL quotUsing temperatures in Cquot QLW2Win quotSetting work equal to heat inputquot QLW3WinCOP quotDividing by COP instead of multiplyingquot QLW4WinTH273THTL quotUsing COP of HPquot PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 686 6171 A refrigerator is removing heat from a cold medium at 3 C at a rate of 7200 kJh and rejecting the waste heat to a medium at 30 C If the coefficient of performance of the refrigerator is 2 the power consumed by the refrigerator is a 01 kW b 05 kW c 10 kW d 20 kW e 50 kW Answer c 10 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values TL3 quotCquot TH30 quotCquot QL72003600 quotkJsquot CO 2 QLWinCOP quotSome Wrong Solutions with Common Mistakesquot QLW1WinTL273THTL quotUsing Carnot COPquot QLW2Win quotSetting work equal to heat inputquot QLW3WinCOP quotDividing by COP instead of multiplyingquot QLW4WinTLTHTL quotUsing Carnot COP using Cquot 6172 Two Carnot heat engines are operating in series such that the heat sink of the first engine serves as the heat source of the second one If the source temperature of the first engine is 1300 K and the sink temperature of the second engine is 300 K and the thermal efficiencies of both engines are the same the temperature of the intermediate reservoir is a 625 K b 800 K c 860 K d 453 K e 758 K Answer a 625 K Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values TH1300 quotKquot TL300 quotKquot quotSetting thermal efficiencies equal to each otherquot 1Tmid l H1TLTmid quotSome Wrong Solutions with Common Mistakesquot W1TmidTLTH2 quotUsing average temperaturequot PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 687 6173 Consider a Carnot refrigerator and a Carnot heat pump operating between the same two thermal energy reservoirs If the COP of the refrigerator is 34 the COP of the heat pump is a 17 b 24 c 34 d 44 e 50 Answer d 44 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values COPR34 COPHPCOPR1 quotSome Wrong Solutions with Common Mistakesquot W1COPCOPR1 quotSubtracting 1 instead of adding 1quot W2COPCOPR quotSetting COPs equal to each otherquot 6174 A typical new household refrigerator consumes about 680 kWh of electricity per year and has a coefficient of performance of 14 The amount of heat removed by this refrigerator from the refrigerated space per year is a 952 MJyr b 1749 MJyr c 2448 MJyr d 3427 MJyr e 4048 MJyr Answer d 3427 MJyr Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Win68036 quotMJquot COPR14 QLWinCOPR quotMJquot quotSome Wrong Solutions with Common Mistakesquot W1QLWinCOPR36 quotNot using the conversion factorquot W2QLWin quotIgnoring COPquot W3QLWinCOPR quotDividing by COP instead of multiplyingquot PROPRIETARY MATERIAL 2011 The McGrawHill Companies Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission

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