Chapter 11 TRANSIENT HEAT CONDUCTION
11-1Solutions ManualforIntroduction to Thermodynamics and Heat TransferYunus A. Cengel2nd Edition, 2008Chapter 11TRANSIENT HEAT CONDUCTIONPROPRIETARY AND CONFIDENTIALThis Manual is the proprietary property of The McGraw-Hill Companies, Inc.(“McGraw-Hill”) and protected by copyright and other state and federal laws. Byopening and using this Manual the user agrees to the following restrictions, and if therecipient does not agree to these restrictions, the Manual should be promptly returnedunopened to McGraw-Hill: This Manual is being provided only to authorizedprofessors and instructors for use in preparing for the classes using the affiliatedtextbook. No other use or distribution of this Manual is permitted. This Manualmay not be sold and may not be distributed to or used by any student or other thirdparty. No part of this Manual may be reproduced, displayed or distributed in anyform or by any means, electronic or otherwise, without the prior written permissionof McGraw-Hill.PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
11-2Lumped System Analysis11-1C In heat transfer analysis, some bodies are observed to behave like a "lump" whose entire bodytemperature remains essentially uniform at all times during a heat transfer process. The temperature of suchbodies can be taken to be a function of time only. Heat transfer analysis which utilizes this idealization isknown as the lumped system analysis. It is applicable when the Biot number (the ratio of conductionresistance within the body to convection resistance at the surface of the body) is less than or equal to 0.1.11-2C The lumped system analysis is more likely to be applicable for the body cooled naturally since theBiot number is proportional to the convection heat transfer coefficient, which is proportional to the airvelocity. Therefore, the Biot number is more likely to be less than 0.1 for the case of natural convection.11-3C The lumped system analysis is more likely to be applicable for the body allowed to cool in the airsince the Biot number is proportional to the convection heat transfer coefficient, which is larger in waterthan it is in air because of the larger thermal conductivity of water. Therefore, the Biot number is morelikely to be less than 0.1 for the case of the solid cooled in the air11-4C The temperature drop of the potato during the second minute will be less than 4 C since thetemperature of a body approaches the temperature of the surrounding medium asymptotically, and thus itchanges rapidly at the beginning, but slowly later on.11-5C The temperature rise of the potato during the second minute will be less than 5 C since thetemperature of a body approaches the temperature of the surrounding medium asymptotically, and thus itchanges rapidly at the beginning, but slowly later on.11-6C Biot number represents the ratio of conduction resistance within the body to convection resistance atthe surface of the body. The Biot number is more likely to be larger for poorly conducting solids since suchbodies have larger resistances against heat conduction.11-7C The heat transfer is proportional to the surface area. Two half pieces of the roast have a much largersurface area than the single piece and thus a higher rate of heat transfer. As a result, the two half pieces willcook much faster than the single large piece.11-8C The cylinder will cool faster than the sphere since heat transfer rate is proportional to the surfacearea, and the sphere has the smallest area for a given volume.11-9C The lumped system analysis is more likely to be applicable in air than in water since the convectionheat transfer coefficient and thus the Biot number is much smaller in air.11-10C The lumped system analysis is more likely to be applicable for a golden apple than for an actualapple since the thermal conductivity is much larger and thus the Biot number is much smaller for gold.11-11C The lumped system analysis is more likely to be applicable to slender bodies than the well-roundedbodies since the characteristic length (ratio of volume to surface area) and thus the Biot number is muchsmaller for slender bodies.PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
11-311-12 Relations are to be obtained for the characteristic lengths of a large plane wall of thickness 2L, avery long cylinder of radius ro and a sphere of radius ro.Analysis Relations for the characteristiclengths of a large plane wall of thickness 2L, avery long cylinder of radius ro and a sphere ofradius ro areLc , wall Lc ,cylinder Lc , sphere VAsurfaceVAsurfaceVAsurface 2 LA L2A πro2 h ro 2πro h 2 4πro3 / 34πro 2 ro32ro2ro2L11-13 A relation for the time period for a lumped system to reach the average temperature (Ti T ) / 2 isto be obtained.Analysis The relation for time period for a lumped system to reach the average temperature (Ti T ) / 2can be determined asT (t ) T e bt Ti T Ti T T 2 e btTi T Ti T 1 e bt e bt2(Ti T )2 bt ln 2 t T Tiln 2 0.693 bbPROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
11-411-14 The temperature of a gas stream is to be measured by a thermocouple. The time it takes to register 99percent of the initial ΔT is to be determined.Assumptions 1 The junction is spherical in shape with a diameter of D 0.0012 m. 2 The thermalproperties of the junction are constant. 3 The heat transfer coefficient is constant and uniform over theentire surface. 4 Radiation effects are negligible. 5 The Biot number is Bi 0.1 so that the lumped systemanalysis is applicable (this assumption will be verified).Properties The properties of the junction are given to be k 35 W/m. C , ρ 8500 kg/m 3 , andc p 320 J/kg. C .Analysis The characteristic length of the junction and the Biot number areLc Bi VAsurface πD 3 / 6 D 0.0012 m 0.0002 m66πD 2hLc (90 W/m 2 . C)(0.0002 m) 0.00051 0.1k(35 W/m. C)Since Bi 0.1 , the lumped system analysis is applicable.Then the time period for the thermocouple to read 99% of theinitial temperature difference is determined fromGash, T T (t ) T 0.01Ti T b JunctionDT(t)hAh90 W/m 2 . C 0.1654 s -1ρc pV ρc p Lc (8500 kg/m 3 )(320 J/kg. C)(0.0002 m)-1T (t ) T e bt 0.01 e ( 0.1654 s )t t 27.8 sTi T PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
11-511-15E A number of brass balls are to be quenched in a water bath at a specified rate. The temperature ofthe balls after quenching and the rate at which heat needs to be removed from the water in order to keep itstemperature constant are to be determined.Assumptions 1 The balls are spherical in shape with a radius of ro 1 in. 2 The thermal properties of theballs are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 TheBiot number is Bi 0.1 so that the lumped system analysis is applicable (this assumption will be verified).Properties The thermal conductivity, density, and specific heat of the brass balls are given to be k 64.1Btu/h.ft. F, ρ 532 lbm/ft3, and cp 0.092 Btu/lbm. F.Analysis (a) The characteristic length and theBiot number for the brass balls areLc Bi VAs Brass balls, 250 FπD 3 / 6 D 2 / 12 ft 0.02778 ft66πD 2Water bath, 120 FhLc (42 Btu/h.ft 2 . F)(0.02778 ft ) 0.01820 0.1k(64.1 Btu/h.ft. F)The lumped system analysis is applicable since Bi 0.1. Then the temperature of the balls after quenchingbecomesb hAsh42 Btu/h.ft 2 . F 30.9 h -1 0.00858 s -13ρc pV ρc p Lc (532 lbm/ft )(0.092 Btu/lbm. F)(0.02778 ft)-1T (t ) T T (t ) 120 e bt e (0.00858 s )(120 s) T (t ) 166 FTi T 250 120(b) The total amount of heat transfer from a ball during a 2-minute period ism ρV ρπD 3 (532 lbm/ft 3 )π (2 / 12 ft) 3 1.290 lbm66Q mc p [Ti T (t )] (1.29 lbm)(0.092 Btu/lbm. F)(250 166) F 9.97 BtuThen the rate of heat transfer from the balls to the water becomesQ& total n& ball Q ball (120 balls/min) (9.97 Btu ) 1196 Btu/minTherefore, heat must be removed from the water at a rate of 1196 Btu/min in order to keep its temperatureconstant at 120 F.PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
11-611-16E A number of aluminum balls are to be quenched in a water bath at a specified rate. The temperatureof balls after quenching and the rate at which heat needs to be removed from the water in order to keep itstemperature constant are to be determined.Assumptions 1 The balls are spherical in shape with a radius of ro 1 in. 2 The thermal properties of theballs are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 TheBiot number is Bi 0.1 so that the lumped system analysis is applicable (this assumption will be verified).Properties The thermal conductivity, density, and specific heat of the aluminum balls are k 137Btu/h.ft. F, ρ 168 lbm/ft3, and cp 0.216 Btu/lbm. F (Table A-24E).Analysis (a) The characteristic length and theBiot number for the aluminum balls areLc Bi VA πD 3 / 6 D 2 / 12 ft 0.02778 ft66πD 2Aluminum balls,250 FWater bath, 120 FhLc (42 Btu/h.ft 2 . F)(0.02778 ft ) 0.00852 0.1(137 Btu/h.ft. F)kThe lumped system analysis is applicable since Bi 0.1. Then the temperature of the balls after quenchingbecomesb hAs42 Btu/h.ft 2 . Fh 41.66 h -1 0.01157 s -13ρc pV ρc p Lc (168 lbm/ft )(0.216 Btu/lbm. F)(0.02778 ft)-1T (t ) T T (t ) 120 e bt e ( 0.01157 s )(120 s) T (t ) 152 F250 120Ti T (b) The total amount of heat transfer from a ball during a 2-minute period ism ρV ρπD 3 (168 lbm/ft 3 )π (2 / 12 ft) 3 0.4072 lbm66Q mc p [Ti T (t )] (0.4072 lbm)(0.216 Btu/lbm. F)(250 152) F 8.62 BtuThen the rate of heat transfer from the balls to the water becomesQ& total n& ball Q ball (120 balls/min) (8.62 Btu ) 1034 Btu/minTherefore, heat must be removed from the water at a rate of 1034 Btu/min in order to keep its temperatureconstant at 120 F.PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
11-711-17 Milk in a thin-walled glass container is to be warmed up by placing it into a large pan filled with hotwater. The warming time of the milk is to be determined.Assumptions 1 The glass container is cylindrical in shape with aradius of r0 3 cm. 2 The thermal properties of the milk are takento be the same as those of water. 3 Thermal properties of the milkare constant at room temperature. 4 The heat transfer coefficient isconstant and uniform over the entire surface. 5 The Biot number inthis case is large (much larger than 0.1). However, the lumpedsystem analysis is still applicable since the milk is stirredconstantly, so that its temperature remains uniform at all times.Water60 CMilk3 CProperties The thermal conductivity, density, and specific heat ofthe milk at 20 C are k 0.598 W/m. C, ρ 998 kg/m3, and cp 4.182 kJ/kg. C (Table A-15).Analysis The characteristic length and Biot number for the glass of milk areLc Bi VAs πro2 L2πro L 2πro2 π (0.03 m) 2 (0.07 m) 0.01050 m2π (0.03 m)(0.07 m) 2π (0.03 m) 2hLc (120 W/m 2 . C)(0.0105 m) 2.107 0.1k(0.598 W/m. C)For the reason explained above we can use the lumped system analysis to determine how long it will takefor the milk to warm up to 38 C:b hAs120 W/m 2 . Ch 0.002738 s -1ρc pV ρc p Lc (998 kg/m 3 )(4182 J/kg. C)(0.0105 m)-1T (t ) T 38 60 e bt e ( 0.002738 s )t t 348 s 5.8 minTi T 3 60Therefore, it will take about 6 minutes to warm the milk from 3 to 38 C.PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
11-811-18 A thin-walled glass containing milk is placed into a large pan filled with hot water to warm up themilk. The warming time of the milk is to be determined.Assumptions 1 The glass container is cylindrical in shape with aradius of r0 3 cm. 2 The thermal properties of the milk are takenWaterto be the same as those of water. 3 Thermal properties of the milk60 Care constant at room temperature. 4 The heat transfer coefficient isconstant and uniform over the entire surface. 5 The Biot number inthis case is large (much larger than 0.1). However, the lumpedMilksystem analysis is still applicable since the milk is stirred3 Cconstantly, so that its temperature remains uniform at all times.Properties The thermal conductivity, density, and specific heat ofthe milk at 20 C are k 0.598 W/m. C, ρ 998 kg/m3, and cp 4.182 kJ/kg. C (Table A-15).Analysis The characteristic length and Biot number for the glass of milk areLc Bi VAs πro2 L2πro L 2πro2 π (0.03 m) 2 (0.07 m) 0.01050 m2π (0.03 m)(0.07 m) 2π (0.03 m) 2hLc (240 W/m 2 . C)(0.0105 m) 4.21 0.1(0.598 W/m. C)kFor the reason explained above we can use the lumped system analysis to determine how long it will takefor the milk to warm up to 38 C:hAs240 W/m 2 . Ch 0.005477 s -1ρc pV ρc p Lc (998 kg/m 3 )(4182 J/kg. C)(0.0105 m)b -1T (t ) T 38 60 e bt e (0.005477 s )t t 174 s 2.9 min3 60Ti T Therefore, it will take about 3 minutes to warm the milk from 3 to 38 C.11-19 A long copper rod is cooled to a specified temperature. The cooling time is to be determined.Assumptions 1 The thermal properties of the geometry are constant. 2 The heat transfer coefficient isconstant and uniform over the entire surface.Properties The properties of copper are k 401 W/m ºC, ρ 8933 kg/m3, and cp 0.385 kJ/kg ºC (TableA-24).Analysis For cylinder, the characteristic length and the Biot number areLc VAsurface (πD 2 / 4) L D 0.02 m 0.005 m44πDLhL(200 W/m 2 . C)(0.005 m)Bi c 0.0025 0.1k(401 W/m. C)D 2 cmTi 100 ºCSince Bi 0.1 , the lumped system analysis is applicable. Then the cooling time is determined fromb hAh200 W/m 2 . C 0.01163 s -1ρc pV ρc p Lc (8933 kg/m 3 )(385 J/kg. C)(0.005 m)-1T (t ) T 25 20 e bt e ( 0.01163 s )t t 238 s 4.0 minTi T 100 20PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
11-911-20 The heating times of a sphere, a cube, and a rectangular prism with similar dimensions are to bedetermined.Assumptions 1 The thermal properties of the geometries are constant. 2 The heat transfer coefficient isconstant and uniform over the entire surface.Properties The properties of silver are given to be k 429 W/m ºC, ρ 10,500 kg/m3, and cp 0.235kJ/kg ºC.Analysis For sphere, the characteristic length and the Biot number areLc Bi V AsurfaceπD 3 / 6 D 0.05 m 0.008333 m66πD 25 cmhLc (12 W/m 2 . C)(0.008333 m) 0.00023 0.1k(429 W/m. C)Airh, T Since Bi 0.1 , the lumped system analysis is applicable. Then the time period for the sphere temperatureto reach to 25ºC is determined fromb hAh12 W/m 2 . C 0.0005836 s -13ρc pV ρc p Lc (10,500 kg/m )(235 J/kg. C)(0.008333 m)-1T (t ) T 25 33 e bt e ( 0.0005836 s )t t 2428 s 40.5 minTi T 0 33Cube:Lc Bi b VAsurfaceL3L 0.05 m 2 0.008333 m666L2hLc (12 W/m . C)(0.008333 m) 0.00023 0.1k(429 W/m. C)5 cm5 cmAirh, T 5 cmhAh12 W/m 2 . C 0.0005836 s -1ρc pV ρc p Lc (10,500 kg/m 3 )(235 J/kg. C)(0.008333 m)-1T (t ) T 25 33 e bt e ( 0.0005836 s )t t 2428 s 40.5 minTi T 0 33Rectangular prism:Lc Bi b VAsurface (0.04 m)(0.05 m)(0.06 m) 0.008108 m2(0.04 m)(0.05 m) 2(0.04 m)(0.06 m) 2(0.05 m)(0.06 m)hLc (12 W/m 2 . C)(0.008108 m) 0.00023 0.1k(429 W/m. C)4 cmhAh ρc pV ρc p Lc12 W/m 2 . C(10,500 kg/m 3 )(235 J/kg. C)(0.008108 m)5 cm 0.0005998 s -1Airh, T 6 cm-1T (t ) T 25 33 e bt e ( 0.0005998 s )t t 2363 s 39.4 minTi T 0 33The heating times are same for the sphere and cube while it is smaller in rectangular prism.PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
11-1011-21E A person shakes a can of drink in a iced water to cool it. The cooling time of the drink is to bedetermined.Assumptions 1 The can containing the drink is cylindrical in shapewith a radius of ro 1.25 in. 2 The thermal properties of the drinkare taken to be the same as those of water. 3 Thermal properties ofthe drinkare constant at room temperature. 4 The heat transfercoefficient is constant and uniform over the entire surface. 5 TheBiot number in this case is large (much larger than 0.1). However,the lumped system analysis is still applicable since the drink isstirred constantly, so that its temperature remains uniform at alltimes.Water32 FDrinkMilk3 C90 FProperties The density and specific heat of water at roomtemperature are ρ 62.22 lbm/ft3, and cp 0.999 Btu/lbm. F(Table A-15E).Analysis Application of lumped system analysis in this case givesLc b VAs πro2 L2πro L 2πro 2 π (1.25 / 12 ft) 2 (5 / 12 ft) 0.04167 ft2π (1.25 / 12 ft)(5/12 ft) 2π (1.25 / 12 ft) 2hAs30 Btu/h.ft 2 . Fh 11.583 h -1 0.00322 s -1ρc pV ρc p Lc (62.22 lbm/ft 3 )(0.999 Btu/lbm. F)(0.04167 ft)-1T (t ) T 40 32 e ( 0.00322 s )t t 615 s e bt 90 32Ti T Therefore, it will take 10 minutes and 15 seconds to cool the canned drink to 40 F.PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
11-1111-22 An iron whose base plate is made of an aluminum alloy is turned on. The time for the platetemperature to reach 140 C and whether it is realistic to assume the plate temperature to be uniform at alltimes are to be determined.Assumptions 1 85 percent of the heat generated in the resistance wires is transferred to the plate. 2 Thethermal properties of the plate are constant. 3 The heat transfer coefficient is constant and uniform over theentire surface.Properties The density, specific heat, and thermal diffusivity of the aluminum alloy plate are given to be ρ 2770 kg/m3, cp 875 kJ/kg. C, and α 7.3 10-5 m2/s. The thermal conductivity of the plate can bedetermined from k αρcp 177 W/m. C (or it can be read from Table A-24).Analysis The mass of the iron's base plate is3Air22 C2m ρV ρLA (2770 kg/m )(0.005 m)(0.03 m ) 0.4155 kgNoting that only 85 percent of the heat generated is transferred to theplate, the rate of heat transfer to the iron's base plate isQ& 0.85 1000 W 850 WIRON1000 WinThe temperature of the plate, and thus the rate of heat transfer from theplate, changes during the process. Using the average plate temperature,the average rate of heat loss from the plate is determined from 140 22 Q& loss hA(Tplate, ave T ) (12 W/m 2 . C)(0.03 m 2 ) 22 C 21.2 W2 Energy balance on the plate c
Introduction to Thermodynamics and Heat Transfer Yunus A. Cengel 2nd Edition, 2008 Chapter 11 TRANSIENT HEAT CONDUCTION . If you are a student using this Manual, you are using it without permission. 11-3 11-12 Relations are to be obtained for the characteristic lengths of a large plane wall of thickness 2L, a
Boiling water CONDUCTION CONVECTION RADIATION 43. Frying a pancake CONDUCTION CONVECTION RADIATION 44. Heat you feel from a hot stove CONDUCTION CONVECTION RADIATION 45. Moves as a wave CONDUCTION CONVECTION RADIATION 46. Occurs within fluids CONDUCTION CONVECTION RADIATION 47. Sun’s rays reaching Earth CONDUCTION CONVECTION RADIATION 48.
steady state response, that is (6.1) The transient response is present in the short period of time immediately after the system is turned on. If the system is asymptotically stable, the transient response disappears, which theoretically can be recorded as "! (6.2) However, if the system is unstable, the transient response will increase veryFile Size: 306KBPage Count: 29Explore furtherSteady State vs. Transient State in System Design and .resources.pcb.cadence.comTransient and Steady State Response in a Control System .www.electrical4u.comConcept of Transient State and Steady State - Electrical .electricalbaba.comChapter 5 Transient Analysis - CAUcau.ac.krTransient Response First and Second Order System .electricalacademia.comRecommended to you b
Part One: Heir of Ash Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26 Chapter 27 Chapter 28 Chapter 29 Chapter 30 .
more bone conduction communication studies—both external and by ARL — have been conducted to investigate the various characteristics of bone conduction communication systems. Progress has been made in understanding the nature of bone conduction hearing and speech perception, bone conduction psychophysics, and bone conduction technology.
Heat Transfer: Conduction, Convection and Latent Heat In addition to radiation, energy can also be transferred in the form of heat. There are three ways this can happen: When heat simply diffuses through an object, passing from molecule to molecule, that's called conduction As it turns out, air is actually very poor at diffusing heat
TO KILL A MOCKINGBIRD. Contents Dedication Epigraph Part One Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Part Two Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18. Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26
the voltage applied and resistance of the heat source respectively, then using the energy balance, the net heat transfer (Q net) within the enclosure is given by- QQ Q Qnet conduction convection radiation ( 2 net Q V R ) The conduction heat transfer (Q conduction) at the enclosure side walls is determined as- wall conduction wall wall wall .
Tipologi diagnosa keperawatan keluarga 11. Aktual : masalah keperawatan yang memerlukan tindakan yang cepat 22. Resiko/ resiko tinggi : masalah yang belum terjadi tetapi tanda untuk menjadi masalah aktual 33. Potensial : Keadaan sejahtera, keluarga telah mampu memenuhi kebutuhannya dan mempunyai sumber penunjang kesehatan . Contoh : Gangguan pemenuhan keb. Nutrisi pada An.S keluarga Bp.B .
