Solucionário Çengel...de Calor- 4ª Edição - 177684191 - heat - 4e - sm - chap01, Notas de estudo de Engenharia Química

Baixe Solucionário Çengel...de Calor- 4ª Edição - 177684191 - heat - 4e - sm - chap01 e outras Notas de estudo em PDF para Engenharia Química, somente na Docsity! PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-1 Solutions Manual for Heat and Mass Transfer: Fundamentals & Applications Fourth Edition Yunus A. Cengel & Afshin J. Ghajar McGraw-Hill, 2011 Chapter 1 INTRODUCTION AND BASIC CONCEPTS PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw-Hill. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-2 Thermodynamics and Heat Transfer 1-1C Thermodynamics deals with the amount of heat transfer as a system undergoes a process from one equilibrium state to another. Heat transfer, on the other hand, deals with the rate of heat transfer as well as the temperature distribution within the system at a specified time. 1-2C The description of most scientific problems involves equations that relate the changes in some key variables to each other, and the smaller the increment chosen in the changing variables, the more accurate the description. In the limiting case of infinitesimal changes in variables, we obtain differential equations, which provide precise mathematical formulations for the physical principles and laws by representing the rates of changes as derivatives. As we shall see in later chapters, the differential equations of fluid mechanics are known, but very difficult to solve except for very simple geometries. Computers are extremely helpful in this area. 1-3C (a) The driving force for heat transfer is the temperature difference. (b) The driving force for electric current flow is the electric potential difference (voltage). (a) The driving force for fluid flow is the pressure difference. 1-4C The caloric theory is based on the assumption that heat is a fluid-like substance called the "caloric" which is a massless, colorless, odorless substance. It was abandoned in the middle of the nineteenth century after it was shown that there is no such thing as the caloric. 1-5C The rating problems deal with the determination of the heat transfer rate for an existing system at a specified temperature difference. The sizing problems deal with the determination of the size of a system in order to transfer heat at a specified rate for a specified temperature difference. 1-6C The experimental approach (testing and taking measurements) has the advantage of dealing with the actual physical system, and getting a physical value within the limits of experimental error. However, this approach is expensive, time consuming, and often impractical. The analytical approach (analysis or calculations) has the advantage that it is fast and inexpensive, but the results obtained are subject to the accuracy of the assumptions and idealizations made in the analysis. 1-7C Modeling makes it possible to predict the course of an event before it actually occurs, or to study various aspects of an event mathematically without actually running expensive and time-consuming experiments. When preparing a mathematical model, all the variables that affect the phenomena are identified, reasonable assumptions and approximations are made, and the interdependence of these variables are studied. The relevant physical laws and principles are invoked, and the problem is formulated mathematically. Finally, the problem is solved using an appropriate approach, and the results are interpreted. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-5 1-17 A house is heated from 10°C to 22°C by an electric heater, and some air escapes through the cracks as the heated air in the house expands at constant pressure. The amount of heat transfer to the air and its cost are to be determined. Assumptions 1 Air as an ideal gas with a constant specific heats at room temperature. 2 The volume occupied by the furniture and other belongings is negligible. 3 The pressure in the house remains constant at all times. 4 Heat loss from the house to the outdoors is negligible during heating. 5 The air leaks out at 22°C. Properties The specific heat of air at room temperature is cp = 1.007 kJ/kg⋅°C. 22°C 10°C AIR Analysis The volume and mass of the air in the house are 32 m 600m) )(3m 200(ght)space)(heifloor ( ===V kg 9.747 K) 273.15+K)(10/kgmkPa 287.0( )m kPa)(600 3.101( 3 3 = ⋅⋅ == RT Pm V Noting that the pressure in the house remains constant during heating, the amount of heat that must be transferred to the air in the house as it is heated from 10 to 22°C is determined to be kJ 9038=°−°⋅=−= C)10C)(22kJ/kg kg)(1.007 9.747()( 12 TTmcQ p Noting that 1 kWh = 3600 kJ, the cost of this electrical energy at a unit cost of $0.075/kWh is $0.19== 5/kWh)kWh)($0.07 3600/9038(energy) ofcost used)(Unit(Energy =CostEnegy Therefore, it will cost the homeowner about 19 cents to raise the temperature in his house from 10 to 22°C. 1-18 A 800 W iron is left on the ironing board with its base exposed to the air. The amount of heat the iron dissipates in 2 h, the heat flux on the surface of the iron base, and the cost of the electricity are to be determined. Assumptions Heat transfer from the surface is uniform. Iron 800 W Analysis (a) The amount of heat the iron dissipates during a 2-h period is kWh 1.6==∆= h) kW)(2 8.0(tQQ & (b) The heat flux on the surface of the iron base is W680= W)800)(85.0(base =Q& 2 W/m45,300=== 2 base base m 015.0 W680 A Q q & & (c) The cost of electricity consumed during this period is $0.112=× kWh)/($0.07kWh) (1.6=yelectricit ofCost PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-6 1-19 A 15 cm × 20 cm circuit board houses 120 closely spaced 0.12 W logic chips. The amount of heat dissipated in 10 h and the heat flux on the surface of the circuit board are to be determined. Assumptions 1 Heat transfer from the back surface of the board is negligible. 2 Heat transfer from the front surface is uniform. Analysis (a) The amount of heat this circuit board dissipates during a 10-h period is 20 cm 15 cm Chips, 0.12 W Q& W4.14 W)12.0)(120( ==Q& kWh 0.144==∆= h) kW)(10 0144.0(tQQ & (b) The heat flux on the surface of the circuit board is 2m 03.0)m 2.0)(m 15.0( ==sA 2 W/m480=== 2m 03.0 W4.14 s s A Q q & & 1-20 An aluminum ball is to be heated from 80°C to 200°C. The amount of heat that needs to be transferred to the aluminum ball is to be determined. Assumptions The properties of the aluminum ball are constant. Properties The average density and specific heat of aluminum are given to be ρ = 2700 kg/m3 and cp = 0.90 kJ/kg⋅°C. Analysis The amount of energy added to the ball is simply the change in its internal energy, and is determined from Metal ball )( 12transfer TTmcUE p −=∆= where kg 77.4m) 15.0)(kg/m 2700( 66 333 ==== πρπρ Dm V E Substituting, kJ 515=C80)C)(200kJ/kg kg)(0.90 77.4(transfer °−°⋅=E Therefore, 515 kJ of energy (heat or work such as electrical energy) needs to be transferred to the aluminum ball to heat it to 200°C. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-7 1-21 An electrically heated house maintained at 22°C experiences infiltration losses at a rate of 0.7 ACH. The amount of energy loss from the house due to infiltration per day and its cost are to be determined. Assumptions 1 Air as an ideal gas with a constant specific heats at room temperature. 2 The volume occupied by the furniture and other belongings is negligible. 3 The house is maintained at a constant temperature and pressure at all times. 4 The infiltrating air exfiltrates at the indoors temperature of 22°C. Properties The specific heat of air at room temperature is cp = 1.007 kJ/kg⋅°C. 5°C 0.7 ACH 22°C AIR Analysis The volume of the air in the house is 32 m 450m) )(3m 150(ght)space)(heifloor ( ===V Noting that the infiltration rate is 0.7 ACH (air changes per hour) and thus the air in the house is completely replaced by the outdoor air 0.7×24 = 16.8 times per day, the mass flow rate of air through the house due to infiltration is kg/day 8485 K) 273.15+K)(5/kgmkPa 287.0( day)/m 045kPa)(16.8 6.89( )ACH( 3 3 houseair air = ⋅⋅ × = × == o o o o RT P RT P m VV& & Noting that outdoor air enters at 5°C and leaves at 22°C, the energy loss of this house per day is kWh/day 40.4=kJ/day 260,145C)5C)(22kJ/kg. 007kg/day)(1. 8485( )( outdoorsindoorsairinfilt =°−°= −= TTcmQ p&& At a unit cost of $0.082/kWh, the cost of this electrical energy lost by infiltration is $3.31/day== 0.082/kWh)kWh/day)($ 4.40(energy) ofcost used)(Unit(Energy =CostEnegy 1-22 The filament of a 150 W incandescent lamp is 5 cm long and has a diameter of 0.5 mm. The heat flux on the surface of the filament, the heat flux on the surface of the glass bulb, and the annual electricity cost of the bulb are to be determined. Assumptions Heat transfer from the surface of the filament and the bulb of the lamp is uniform. Analysis (a) The heat transfer surface area and the heat flux on the surface of the filament are 2cm 785.0)cm 5)(cm 05.0( === ππDLAs 26 W/m101.91×==== 2 2 W/cm191 cm 785.0 W150 s s A Q q & & (b) The heat flux on the surface of glass bulb is 222 cm 1.201cm) 8( === ππDAs 2 W/m7500==== 2 2 W/cm75.0 cm 1.201 W150 s s A Q q & & (c) The amount and cost of electrical energy consumed during a one-year period is Q& Lamp 150 W $35.04/yr= =×=∆= kWh)/.08kWh/yr)($0 (438=Cost Annual kWh/yr 438h/yr) 8kW)(365 15.0(nConsumptioy Electricit tQ& PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-10 1-25 Liquid water is to be heated in an electric teapot. The heating time is to be determined. Assumptions 1 Heat loss from the teapot is negligible. 2 Constant properties can be used for both the teapot and the water. Properties The average specific heats are given to be 0.7 kJ/kg·K for the teapot and 4.18 kJ/kg·K for water. Analysis We take the teapot and the water in it as the system, which is a closed system (fixed mass). The energy balance in this case can be expressed as teapotwatersystemin energies etc. potential, kinetic, internal,in Change system mass and work,heat,by nsferenergy traNet UUUE EEE outin ∆+∆=∆= ∆=− 4342143421 Then the amount of energy needed to raise the temperature of water and the teapot from 15°C to 95°C is kJ 3.429 C)15C)(95kJ/kg kg)(0.7 (0.5C)15C)(95kJ/kg kg)(4.18 (1.2 )()( teapotwaterin = °−°⋅+°−°⋅= ∆+∆= TmcTmcE The 1200-W electric heating unit will supply energy at a rate of 1.2 kW or 1.2 kJ per second. Therefore, the time needed for this heater to supply 429.3 kJ of heat is determined from min 6.0=====∆ s 358 kJ/s 2.1 kJ 3.429 nsferenergy tra of Rate nsferredenergy tra Total transfer in E E t & Discussion In reality, it will take more than 6 minutes to accomplish this heating process since some heat loss is inevitable during heating. Also, the specific heat units kJ/kg · °C and kJ/kg · K are equivalent, and can be interchanged. 1-26 It is observed that the air temperature in a room heated by electric baseboard heaters remains constant even though the heater operates continuously when the heat losses from the room amount to 9000 kJ/h. The power rating of the heater is to be determined. Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of - 141°C and 3.77 MPa. 2 The kinetic and potential energy changes are negligible, ∆ke ≅ ∆pe ≅ 0. 3 The temperature of the room remains constant during this process. Analysis We take the room as the system. The energy balance in this case reduces to We AIR outine outine outin QW UQW EEE = =∆=− ∆=− , , energies etc. potential, kinetic, internal,in Change system mass and work,heat,by nsferenergy traNet 0 4342143421 since ∆U = mcv∆T = 0 for isothermal processes of ideal gases. Thus, kW 2.5=⎟ ⎠ ⎞ ⎜ ⎝ ⎛== kJ/h 3600 kW 1kJ/h 0009, outine QW && PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-11 1-27 A room is heated by an electrical resistance heater placed in a short duct in the room in 15 min while the room is losing heat to the outside, and a 300-W fan circulates the air steadily through the heater duct. The power rating of the electric heater and the temperature rise of air in the duct are to be determined. Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of - 141°C and 3.77 MPa. 2 The kinetic and potential energy changes are negligible, ∆ke ≅ ∆pe ≅ 0. 3 Constant specific heats at room temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications. 3 Heat loss from the duct is negligible. 4 The house is air-tight and thus no air is leaking in or out of the room. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). Also, cp = 1.007 kJ/kg·K for air at room temperature (Table A-15) and cv = cp – R = 0.720 kJ/kg·K. Analysis (a) We first take the air in the room as the system. This is a constant volume closed system since no mass crosses the system boundary. The energy balance for the room can be expressed as )()()( 1212outinfan,ine, outinfan,ine, energies etc. potential, kinetic, internal,in Change system mass and work,heat,by nsferenergy traNet TTmcuumtQWW UQWW EEE v outin −≅−=∆−+ ∆=−+ ∆=− &&& 4342143421 5×6×8 m3 We 300 W 200 kJ/min The total mass of air in the room is kg 284.6 )K 288)(K/kgmkPa 0.287( )m 240)(kPa 98( m 240m 865 3 3 1 1 33 = ⋅⋅ == =××= RT P m V V Then the power rating of the electric heater is determined to be kW 4.93=×°−°⋅+−= ∆−−= + s) 60C)/(181525)(CkJ/kg 0.720)(kg 284.6()kJ/s 0.3()kJ/s 200/60( /)( 12infan,outine, tTTWQW mcv&&& (b) The temperature rise that the air experiences each time it passes through the heater is determined by applying the energy balance to the duct, TcmhmWW hmQhmWW EE p outin ∆=∆=+ ≅∆≅∆+=++ = &&&& &&&&& && infan,ine, 2 0 out1infan,ine, 0)peke (since Thus, C6.2°= ⋅ + = + =∆ )KkJ/kg 1.007)(kg/s 50/60( kJ/s)0.34.93(infan,ine, pcm WW T & && PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-12 1-28 The resistance heating element of an electrically heated house is placed in a duct. The air is moved by a fan, and heat is lost through the walls of the duct. The power rating of the electric resistance heater is to be determined. Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of - 141°C and 3.77 MPa. 2 The kinetic and potential energy changes are negligible, ∆ke ≅ ∆pe ≅ 0. 3 Constant specific heats at room temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications. Properties The specific heat of air at room temperature is cp = 1.007 kJ/kg·°C (Table A-15). Analysis We take the heating duct as the system. This is a control volume since mass crosses the system boundary during the process. We observe that this is a steady-flow process since there is no change with time at any point and thus . Also, there is only one inlet and one exit and thus 0 and 0 CVCV =∆=∆ Em mmm &&& == 21 . The energy balance for this steady-flow system can be expressed in the rate form as )( 0)peke (since 0 12infan,outine, 2out1infan,ine, energies etc. potential, kinetic, internal,in change of Rate (steady) 0 system mass and work,heat,by nsferenergy tranet of Rate TTcmWQW hmQhmWW EEEEE p outinoutin −+−= ≅∆≅∆+=++ =→=∆=− &&&& &&&&& && 444 344 21 & 43421 && We 300 W 250 W Substituting, the power rating of the heating element is determined to be kW 2.97= °°⋅−= C)C)(5kJ/kg 7kg/s)(1.00 (0.6+kW) 3.0()kW 25.0(ine,W& PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-15 1-31E Air gains heat as it flows through the duct of an air-conditioning system. The velocity of the air at the duct inlet and the temperature of the air at the exit are to be determined. Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of - 222°F and 548 psia. 2 The kinetic and potential energy changes are negligible, ∆ke ≅ ∆pe ≅ 0. 3 Constant specific heats at room temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications. Properties The gas constant of air is R = 0.3704 psia·ft3/lbm·R (Table A-1E). Also, cp = 0.240 Btu/lbm·R for air at room temperature (Table A-15E). Analysis We take the air-conditioning duct as the system. This is a control volume since mass crosses the system boundary during the process. We observe that this is a steady-flow process since there is no change with time at any point and thus , there is only one inlet and one exit and thus 0 and 0 CVCV =∆=∆ Em mmm &&& == 21 , and heat is lost from the system. The energy balance for this steady-flow system can be expressed in the rate form as )( 0)peke (since 0 12in 21in energies etc. potential, kinetic, internal,in change of Rate (steady) 0 system mass and work,heat,by nsferenergy tranet of Rate TTcmQ hmhmQ EEEEE p outinoutin −= ≅∆≅∆=+ =→=∆=− && &&& && 444 344 21 & 43421 && 450 ft3/min AIR D = 10 in 2 Btu/s (a) The inlet velocity of air through the duct is determined from ft/min825 ft) 12/5( /minft 450 2 3 2 1 1 1 1 ==== ππrA V VV && (b) The mass flow rate of air becomes slbm 5950lbm/min 35.7 /lbmft 12.6 minft 450 /lbmft 6.12 psia 15 )R 510)(R/lbmftpsia 0.3704( 3 3 1 1 3 3 1 1 1 /. / ==== = ⋅⋅ == v V v & &m P RT Then the exit temperature of air is determined to be F64.0°= °⋅ +°=+= )FBtu/lbm 0.240)(lbm/s 0.595( Btu/s 2 F50in12 pcm Q TT & & PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-16 1-32 A classroom is to be air-conditioned using window air-conditioning units. The cooling load is due to people, lights, and heat transfer through the walls and the windows. The number of 5-kW window air conditioning units required is to be determined. Assumptions There are no heat dissipating equipment (such as computers, TVs, or ranges) in the room. Analysis The total cooling load of the room is determined from gainheatpeoplelightscooling QQQQ &&&& ++= Room 50 people 10 bulbs 12,000 kJ/h where Qcool · kW .333kJ/h 12,000 kW 5kJ/h 000,18kJ/h 36005 kW 1W 10010 gainheat people lights == ==×= =×= Q Q Q & & & Substituting, kW .33933.351cooling =++=Q& Thus the number of air-conditioning units required is units2 87.1 kW/unit 5 kW 9.33 ⎯→⎯= PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-17 Heat Transfer Mechanisms 1-33C Diamond is a better heat conductor. 1-34C The thermal conductivity of a material is the rate of heat transfer through a unit thickness of the material per unit area and per unit temperature difference. The thermal conductivity of a material is a measure of how fast heat will be conducted in that material. 1-35C The mechanisms of heat transfer are conduction, convection and radiation. Conduction is the transfer of energy from the more energetic particles of a substance to the adjacent less energetic ones as a result of interactions between the particles. Convection is the mode of energy transfer between a solid surface and the adjacent liquid or gas which is in motion, and it involves combined effects of conduction and fluid motion. Radiation is energy emitted by matter in the form of electromagnetic waves (or photons) as a result of the changes in the electronic configurations of the atoms or molecules. 1-36C In solids, conduction is due to the combination of the vibrations of the molecules in a lattice and the energy transport by free electrons. In gases and liquids, it is due to the collisions of the molecules during their random motion. 1-37C The parameters that effect the rate of heat conduction through a windowless wall are the geometry and surface area of wall, its thickness, the material of the wall, and the temperature difference across the wall. 1-38C Conduction is expressed by Fourier's law of conduction as dx dTkAQ −=cond& where dT/dx is the temperature gradient, k is the thermal conductivity, and A is the area which is normal to the direction of heat transfer. Convection is expressed by Newton's law of cooling as where h is the convection heat transfer coefficient, A )(conv ∞−= TThAQ ss& s is the surface area through which convection heat transfer takes place, Ts is the surface temperature and T∞ is the temperature of the fluid sufficiently far from the surface. Radiation is expressed by Stefan-Boltzman law as where )( 4surr 4 rad TTAQ ss −= εσ& ε is the emissivity of surface, As is the surface area, Ts is the surface temperature, Tsurr is the average surrounding surface temperature and is the Stefan-Boltzman constant. 428 K W/m1067.5 ⋅×= −σ 1-39C Convection involves fluid motion, conduction does not. In a solid we can have only conduction. 1-40C No. It is purely by radiation. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-20 1-54 The convection heat transfer coefficient heat transfer between the surface of a pipe carrying superheated vapor and the surrounding is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is not considered. 3 Rate of heat loss from the vapor in the pipe is equal to the heat transfer rate by convection between pipe surface and the surrounding. Properties The specific heat of vapor is given to be 2190 J/kg · °C. Analysis The surface area of the pipe is 2m 571.1)m 10)(m 05.0( === ππDLAs The rate of heat loss from the vapor in the pipe can be determined from W19710 J/s 19710C )30(C)J/kg 2190)(kg/s 3.0( )( outinloss = =°°⋅= −= TTcmQ p&& With the rate of heat loss from the vapor in the pipe assumed equal to the heat transfer rate by convection, the heat transfer coefficient can be determined using the Newton’s law of cooling: )(convloss ∞−== TThAQQ ss&& Rearranging, the heat transfer coefficient is determined to be C W/m157 2 °⋅= °− = − = ∞ C )20100)(m 571.1( W19710 )( 2 loss TTA Q h ss & Discussion By insulating the pipe surface, heat loss from the vapor in the pipe can be reduced. 1-55 An electrical resistor with a uniform temperature of 90 °C is in a room at 20 °C. The heat transfer coefficient by convection is to be determined. Assumptions 1 Steady operating conditions exist. 2 Radiation heat transfer is negligible. 3 No hot spot exists on the resistor. Analysis The total heat transfer area of the resistor is 222 m 01276.0)m 15.0)(m 025.0(4/)m 025.0(2)4/(2 =+=+= ππππ DLDAs The electrical energy converted to thermal energy is transferred by convection: W30)V 6)(A 5(conv === IVQ& From Newton’s law of cooling, the heat transfer by convection is given as )(conv ∞−= TThAQ ss& Rearranging, the heat transfer coefficient is determined to be C W/m33.6 2 °⋅= °− = − = ∞ C )2090)(m 01276.0( W30 )( 2 conv TTA Q h ss & Discussion By comparing the magnitude of the heat transfer coefficient determined here with the values presented in Table 1-5, one can conclude that it is likely that forced convection is taking place rather than free convection. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-21 1-56 The inner and outer surfaces of a brick wall are maintained at specified temperatures. The rate of heat transfer through the wall is to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values. 2 Thermal properties of the wall are constant. 0.3 m Brick wall Properties The thermal conductivity of the wall is given to be k = 0.69 W/m⋅°C. Analysis Under steady conditions, the rate of heat transfer through the wall is 8°C 26°C W1159=°−×°⋅=∆= m 0.3 C)8(26)m 7C)(4W/m (0.69 2cond L TkAQ& 1-57 The inner and outer surfaces of a window glass are maintained at specified temperatures. The amount of heat transfer through the glass in 5 h is to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the glass remain constant at the specified values. 2 Thermal properties of the glass are constant. Glass Properties The thermal conductivity of the glass is given to be k = 0.78 W/m⋅°C. Analysis Under steady conditions, the rate of heat transfer through the glass by conduction is W 4368 m0.005 C3)(10 )m 2C)(2W/m (0.78 2cond = °− ×°⋅= ∆ = L TkAQ& 3°C 10°C Then the amount of heat transfer over a period of 5 h becomes 0.5 cm kJ 78,620=×=∆= s) 3600kJ/s)(5 (4.368cond tQQ & If the thickness of the glass doubled to 1 cm, then the amount of heat transfer will go down by half to 39,310 kJ. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-22 1-58 Prob. 1-57 is reconsidered. The amount of heat loss through the glass as a function of the window glass thickness is to be plotted. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" L=0.005 [m] A=2*2 [m^2] T_1=10 [C] T_2=3 [C] k=0.78 [W/m-C] time=5*3600 [s] "ANALYSIS" Q_dot_cond=k*A*(T_1-T_2)/L Q_cond=Q_dot_cond*time*Convert(J, kJ) 0.002 0.004 0.006 0.008 0.01 0 50000 100000 150000 200000 250000 300000 350000 400000 L [m] Q co nd [ kJ ] L [m] Qcond [kJ] 0.001 393120 0.002 196560 0.003 131040 0.004 98280 0.005 78624 0.006 65520 0.007 56160 0.008 49140 0.009 43680 0.01 39312 PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-25 1-63 The thermal conductivity of a material is to be determined by ensuring one-dimensional heat conduction, and by measuring temperatures when steady operating conditions are reached. Assumptions 1 Steady operating conditions exist since the temperature readings do not change with time. 2 Heat losses through the lateral surfaces of the apparatus are negligible since those surfaces are well-insulated, and thus the entire heat generated by the heater is conducted through the samples. 3 The apparatus possesses thermal symmetry. Analysis For each sample we have Q&Q& A L L C87482 m 01.0m) 1.0m)( 1.0( W102/20 2 °=−=∆ == == T A Q& Then the thermal conductivity of the material becomes C W/m0.625 °⋅= ° = ∆ =⎯→⎯ ∆ = )C8)(m 01.0( m) W)(0.00510( 2TA LQ k L TkAQ & & 1-64 The thermal conductivity of a refrigerator door is to be determined by measuring the surface temperatures and heat flux when steady operating conditions are reached. q& L = 3 cm 7°C Assumptions 1 Steady operating conditions exist when measurements are taken. 2 Heat transfer through the door is one dimensional since the thickness of the door is small relative to other dimensions. Analysis The thermal conductivity of the door material is determined directly from Fourier’s relation to be 15°C C W/m0.120 °⋅= °− = ∆ =⎯→⎯ ∆ = C)715( m) )(0.03 W/m32( 2 T Lqk L Tkq & & PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-26 1-65 The rate of radiation heat transfer between a person and the surrounding surfaces at specified temperatures is to be determined in summer and in winter. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by convection is not considered. 3 The person is completely surrounded by the interior surfaces of the room. 4 The surrounding surfaces are at a uniform temperature. Properties The emissivity of a person is given to be ε = 0.95 Analysis Noting that the person is completely enclosed by the surrounding surfaces, the net rates of radiation heat transfer from the body to the surrounding walls, ceiling, and the floor in both cases are: (a) Summer: Tsurr = 23+273=296 Qrad Tsurr W84.2= ]KK) (296273)+)[(32m )(1.6.K W/m1067.5)(95.0( )( 4442428 4 surr 4 rad −×= −= − TTAQ ssεσ& (b) Winter: Tsurr = 12+273= 285 K W177.2= ]KK) (285273)+)[(32m )(1.6.K W/m1067.5)(95.0( )( 4442428 4 surr 4 rad −×= −= − TTAQ ssεσ& Discussion Note that the radiation heat transfer from the person more than doubles in winter. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-27 1-66 Prob. 1-65 is reconsidered. The rate of radiation heat transfer in winter as a function of the temperature of the inner surface of the room is to be plotted. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" T_infinity=(20+273) [K] T_surr_winter=(12+273) [K] T_surr_summer=(23+273) [K] A=1.6 [m^2] epsilon=0.95 T_s=(32+273) [K] "ANALYSIS" sigma=5.67E-8 [W/m^2-K^4] "Stefan-Boltzman constant" Q_dot_rad_summer=epsilon*sigma*A*(T_s^4-T_surr_summer^4) Q_dot_rad_winter=epsilon*sigma*A*(T_s^4-T_surr_winter^4) Tsurr, winter [K] Qrad, winter [W] 281 208.5 282 200.8 283 193 284 185.1 285 177.2 286 169.2 287 161.1 288 152.9 289 144.6 290 136.2 291 127.8 281 283 285 287 289 291 120 130 140 150 160 170 180 190 200 210 Tsurr,winter [K] Q ra d, w in te r [W ] PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-30 1-70 A spacecraft in space absorbs solar radiation while losing heat to deep space by thermal radiation. The surface temperature of the spacecraft is to be determined when steady conditions are reached. Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values. 2 Thermal properties of the wall are constant. Properties The outer surface of a spacecraft has an emissivity of 0.8 and an absorptivity of 0.3. α = 0.3 ε = 0.8 . Qrad 950 W/m2 Analysis When the heat loss from the outer surface of the spacecraft by radiation equals the solar radiation absorbed, the surface temperature can be determined from ]K) (0)[KW/m 10(5.670.8) W/m950(3.0 )( 444282 4 space 4 solar radabsorbedsolar −⋅×××=×× −= = − sss ss TAA TTAQ QQ εσα & && Canceling the surface area A and solving for Ts gives Ts = 281.5 K 1-71 The heat generated in the circuitry on the surface of a 5-W silicon chip is conducted to the ceramic substrate. The temperature difference across the chip in steady operation is to be determined. Assumptions 1 Steady operating conditions exist. 2 Thermal properties of the chip are constant. Properties The thermal conductivity of the silicon chip is given to be k = 130 W/m⋅°C. Analysis The temperature difference between the front and back surfaces of the chip is 2m 000036.0m) m)(0.006 006.0( ==A C0.53°= °⋅ ==∆ ∆ = )m 6C)(0.00003 W/m130( m) 0005.0 W)(5( 2kA LQT L TkAQ & & 5 W Chip 6 × 6 × 0.5 mm Ceramic substrate Q& PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-31 1-72 An electric resistance heating element is immersed in water initially at 20°C. The time it will take for this heater to raise the water temperature to 80°C as well as the convection heat transfer coefficients at the beginning and at the end of the heating process are to be determined. Assumptions 1 Steady operating conditions exist and thus the rate of heat loss from the wire equals the rate of heat generation in the wire as a result of resistance heating. 2 Thermal properties of water are constant. 3 Heat losses from the water in the tank are negligible. Properties The specific heat of water at room temperature is c = 4.18 kJ/kg⋅°C (Table A-9). Analysis When steady operating conditions are reached, we have . This is also equal to the rate of heat gain by water. Noting that this is the only mechanism of energy transfer, the time it takes to raise the water temperature from 20°C to 80°C is determined to be W800generated == EQ && h 6.53==°−°⋅= − =∆ −=∆ −= s 510,23 J/s 800 C20)C)(80J/kg kg)(4180 (75)( )( )( in 12 12in 12in Q TTmc t TTmctQ TTmcQ & & water 800 W 120°C The surface area of the wire is 2m 0.00628 = m) m)(0.4 005.0(ππ == DLAs The Newton's law of cooling for convection heat transfer is expressed as . Disregarding any heat transfer by radiation and thus assuming all the heat loss from the wire to occur by convection, the convection heat transfer coefficients at the beginning and at the end of the process are determined to be )( ∞−= TThAQ ss& C W/m3185 C W/m1274 2 2 °⋅= °− = − = °⋅= °− = − = ∞ ∞ C)80120)(m (0.00628 W800 )( C)20120)(m (0.00628 W800 )( 2 2 2 2 1 1 TTA Q h TTA Q h ss ss & & Discussion Note that a larger heat transfer coefficient is needed to dissipate heat through a smaller temperature difference for a specified heat transfer rate. 1-73 A hot water pipe at 80°C is losing heat to the surrounding air at 5°C by natural convection with a heat transfer coefficient of 25 W/m2⋅°C. The rate of heat loss from the pipe by convection is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is not considered. 3 The convection heat transfer coefficient is constant and uniform over the surface. L = 10 m D =5 cm 80°C Analysis The heat transfer surface area is As = πDL = π (0.05 m)(10 m) = 1.571 m2 QUnder steady conditions, the rate of heat transfer by convection is Air, 5°C W2945 C5))(80m C)(1.571W/m(25 22conv =°−°⋅=∆= ThAQ s& PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-32 1-74 A hollow spherical iron container is filled with iced water at 0°C. The rate of heat loss from the sphere and the rate at which ice melts in the container are to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values. 2 Heat transfer through the shell is one-dimensional. 3 Thermal properties of the iron shell are constant. 4 The inner surface of the shell is at the same temperature as the iced water, 0°C. Properties The thermal conductivity of iron is k = 80.2 W/m⋅°C (Table A-3). The heat of fusion of water is given to be 333.7 kJ/kg. Analysis This spherical shell can be approximated as a plate of thickness 0.4 cm and area Iced water 0°C 5°C 0.2 cm A = πD2 = π (0.2 m)2 = 0.126 m2 Then the rate of heat transfer through the shell by conduction is kW 25.3==°−°⋅=∆= W 263,25 m 0.002 C0)(5 )m C)(0.126W/m (80.2 2cond L TkAQ& Considering that it takes 333.7 kJ of energy to melt 1 kg of ice at 0°C, the rate at which ice melts in the container can be determined from kg/s 0.0757=== kJ/kg 333.7 kJ/s 25.263 ice ifh Q m & & Discussion We should point out that this result is slightly in error for approximating a curved wall as a plain wall. The error in this case is very small because of the large diameter to thickness ratio. For better accuracy, we could use the inner surface area (D = 19.6 cm) or the mean surface area (D = 19.8 cm) in the calculations. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-35 1-78 Prob. 1-77 is reconsidered. The amount of power the transistor can dissipate safely as a function of the maximum case temperature is to be plotted. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" L=0.004 [m] D=0.006 [m] h=30 [W/m^2-C] T_infinity=55 [C] T_case_max=70 [C] "ANALYSIS" A=pi*D*L+pi*D^2/4 Q_dot=h*A*(T_case_max-T_infinity) Tcase, max [C] Q [W] 60 0.01555 62.5 0.02333 65 0.0311 67.5 0.03888 70 0.04665 72.5 0.05443 75 0.0622 77.5 0.06998 80 0.07775 82.5 0.08553 85 0.09331 87.5 0.1011 90 0.1089 60 65 70 75 80 85 90 0 0.02 0.04 0.06 0.08 0.1 0.12 Tcase,max [C] Q [ W ] PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-36 1-79E A 300-ft long section of a steam pipe passes through an open space at a specified temperature. The rate of heat loss from the steam pipe and the annual cost of this energy lost are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is disregarded. 3 The convection heat transfer coefficient is constant and uniform over the surface. Analysis (a) The rate of heat loss from the steam pipe is D =4 in 280°F 2ft 2.314ft) 300(ft) 12/4( === ππDLAs Btu/h 433,500≅ °−°⋅⋅=−= Btu/h 433,540= F)50280)(ft 2.314(F)ftBtu/h 6()( 22airpipe TThAQ ss& L=300 ft Q Air,50°F (b) The amount of heat loss per year is Btu/yr 10798.3h/yr) 24Btu/h)(365 540,433( 9×=×=∆= tQQ & The amount of gas consumption per year in the furnace that has an efficiency of 86% is therms/yr161,44 Btu 100,000 therm1 86.0 Btu/yr 10798.3LossEnergy Annual 9 =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛× = Then the annual cost of the energy lost becomes $48,576/yr= = therm)/10.1($) therms/yr(44,161= energy) ofcost loss)(Unitenergy Annual(costEnergy 1-80 A 4-m diameter spherical tank filled with liquid nitrogen at 1 atm and -196°C is exposed to convection with ambient air. The rate of evaporation of liquid nitrogen in the tank as a result of the heat transfer from the ambient air is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is disregarded. 3 The convection heat transfer coefficient is constant and uniform over the surface. 4 The temperature of the thin-shelled spherical tank is nearly equal to the temperature of the nitrogen inside. Properties The heat of vaporization and density of liquid nitrogen at 1 atm are given to be 198 kJ/kg and 810 kg/m3, respectively. Analysis The rate of heat transfer to the nitrogen tank is Vapor 222 m 27.50m) 4( === ππDAs Air 20°C Q& 1 atm Liquid N2 W430,271 C)]196(20)[m 27.50(C) W/m25()( 22air = °−−°⋅=−= TThAQ ss& Then the rate of evaporation of liquid nitrogen in the tank is determined to be kg/s 1.37===⎯→⎯= kJ/kg 198 kJ/s 430.271 fg fg h Q mhmQ & &&& PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-37 1-81 A 4-m diameter spherical tank filled with liquid oxygen at 1 atm and -183°C is exposed to convection with ambient air. The rate of evaporation of liquid oxygen in the tank as a result of the heat transfer from the ambient air is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is disregarded. 3 The convection heat transfer coefficient is constant and uniform over the surface. 4 The temperature of the thin-shelled spherical tank is nearly equal to the temperature of the oxygen inside. Properties The heat of vaporization and density of liquid oxygen at 1 atm are given to be 213 kJ/kg and 1140 kg/m3, respectively. Vapor Analysis The rate of heat transfer to the oxygen tank is Air 20°C Q& 1 atm Liquid O2 -183°C 222 m 27.50m) 4( === ππDAs W120,255 C)]183(20)[m 27.50(C). W/m25()( 22air = °−−°=−= TThAQ ss& Then the rate of evaporation of liquid oxygen in the tank is determined to be kg/s 1.20===⎯→⎯= kJ/kg 213 kJ/s 120.255 fg fg h Q mhmQ & &&& PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-40 1-85 A sealed electronic box dissipating a total of 120 W of power is placed in a vacuum chamber. If this box is to be cooled by radiation alone and the outer surface temperature of the box is not to exceed 55°C, the temperature the surrounding surfaces must be kept is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by convection is disregarded. 3 The emissivity of the box is constant and uniform over the exposed surface. 4 Heat transfer from the bottom surface of the box to the stand is negligible. Properties The emissivity of the outer surface of the box is given to be 0.95. Analysis Disregarding the base area, the total heat transfer area of the electronic box is 2m 65.0)m 5.0)(m 2.0(4m) m)(0.5 5.0( =×+=sA The radiation heat transfer from the box can be expressed as [ ]4surr42428 4 surr 4 rad )K 27355()m 65.0)(K W/m1067.5)(95.0( W120 )( T TTAQ ss −+⋅×= −= − εσ& 120 W ε = 0.95 Ts =55°C which gives Tsurr = 300.4 K = 27.4°C. Therefore, the temperature of the surrounding surfaces must be less than 27.4°C. 1-86E Using the conversion factors between W and Btu/h, m and ft, and K and R, the Stefan-Boltzmann constant is to be expressed in the English unit, . 428 K W/m1067.5 ⋅×= −σ 42 RftBtu/h ⋅⋅ Analysis The conversion factors for W, m, and K are given in conversion tables to be R 1.8 =K 1 ft 3.2808 = m 1 Btu/h 3.41214 = W 1 Substituting gives the Stefan-Boltzmann constant in the desired units, 42 RftBtu/h 0.171 ⋅⋅=×⋅= 42 42 R) (1.8ft) 2808.3( Btu/h 3.412145.67=K W/m67.5σ PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-41 1-87E Using the conversion factors between W and Btu/h, m and ft, and °C and °F, the convection coefficient in SI units is to be expressed in Btu/h⋅ft2⋅°F. Analysis The conversion factors for W and m are straightforward, and are given in conversion tables to be ft 3.2808 = m 1 Btu/h 3.41214 = W 1 The proper conversion factor between °C into °F in this case is F1.8=C1 °° since the °C in the unit W/m2⋅°C represents per °C change in temperature, and 1°C change in temperature corresponds to a change of 1.8°F. Substituting, we get FftBtu/h 1761.0 F) (1.8ft) 2808.3( Btu/h 3.41214=C W/m1 2 2 2 °⋅⋅= ° °⋅ which is the desired conversion factor. Therefore, the given convection heat transfer coefficient in English units is FftBtu/h 3.87 2 °⋅⋅=°⋅⋅×°⋅= FftBtu/h 0.176122=C W/m22 22h 1-88 An aircraft flying under icing conditions is considered. The temperature of the wings to prevent ice from forming on them is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer coefficient is constant. Properties The heat of fusion and the density of ice are given to be 333.7 kJ/kg and 920 kg/m3, respectively. Analysis The temperature of the wings to prevent ice from forming on them is determined to be C34.1°= °⋅ +°=+= C W/m150 J/kg) 00m/s)(333,7 )(0.001/60kg/m 920( C0 2 3 icewing h Vh TT if ρ PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-42 1-89 The convection heat transfer coefficient for heat transfer from an electrically heated wire to air is to be determined by measuring temperatures when steady operating conditions are reached and the electric power consumed. Assumptions 1 Steady operating conditions exist since the temperature readings do not change with time. 2 Radiation heat transfer is negligible. Analysis In steady operation, the rate of heat loss from the wire equals the rate of heat generation in the wire as a result of resistance heating. That is, W330= A) V)(3 110(generated === IEQ V&& D =0.2 cm Q 180°C The surface area of the wire is 2m 0.01319 = m) m)(2.1 002.0(ππ == DLAs L = 2.1 m The Newton's law of cooling for convection heat transfer is expressed as Air, 20°C )( ∞−= TThAQ ss& Disregarding any heat transfer by radiation, the convection heat transfer coefficient is determined to be C W/m156 2 °⋅= °− = − = ∞ C)20180)(m (0.01319 W330 )( 21 TTA Qh s & Discussion If the temperature of the surrounding surfaces is equal to the air temperature in the room, the value obtained above actually represents the combined convection and radiation heat transfer coefficient. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-45 1-96 Two large plates at specified temperatures are held parallel to each other. The rate of heat transfer between the plates is to be determined for the cases of still air, evacuation, regular insulation, and super insulation between the plates. Assumptions 1 Steady operating conditions exist since the plate temperatures remain constant. 2 Heat transfer is one- dimensional since the plates are large. 3 The surfaces are black and thus ε = 1. 4 There are no convection currents in the air space between the plates. Properties The thermal conductivities are k = 0.00015 W/m⋅°C for super insulation, k = 0.01979 W/m⋅°C at -50°C (Table A- 15) for air, and k = 0.036 W/m⋅°C for fiberglass insulation (Table A-6). Analysis (a) Disregarding any natural convection currents, the rates of conduction and radiation heat transfer T2T1 [ ] W511=+=+= =−⋅×= −= = − °⋅= − = − 372139 W372)K 150()K 290()m1)(K W/m1067.5(1 )( W139 m 0.02 K )150290( )m C)(1 W/m01979.0( radcondtotal 442428 4 2 4 1rad 2221 cond QQQ TTAQ L TT kAQ s &&& & & εσ 2 cm Q · (b) When the air space between the plates is evacuated, there will be radiation heat transfer only. Therefore, W372== radtotal QQ && (c) In this case there will be conduction heat transfer through the fiberglass insulation only, W252=−⋅= − == m 0.02 K )150290()m C)(1 W/m036.0( 2o21condtotal L TT kAQQ && (d) In the case of superinsulation, the rate of heat transfer will be W1.05=−°⋅= − == m 0.02 K )150290( )m C)(1 W/m00015.0( 221condtotal L TT kAQQ && Discussion Note that superinsulators are very effective in reducing heat transfer between to surfaces. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-46 1-97 The outer surface of a wall is exposed to solar radiation. The effective thermal conductivity of the wall is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat transfer coefficient is constant and uniform over the surface. Properties Both the solar absorptivity and emissivity of the wall surface are given to be 0.8. Analysis The heat transfer through the wall by conduction is equal to net heat transfer to the outer wall surface: [ ] ) W/m150)(8.0( )K 27344()K 27340()K W/m10(0.8)(5.67C)44C)(40W/m (8 m 0.25 C27)-(44 )()( 2 44428-2 solar 4 2 4 surr2 12 solarradconvcond + +−+⋅×+°−°⋅= ° +−+−= − ++= k qTTTTh L TT k qqqq so αεσ &&&& 44ºC αs = ε = 0.8 air, 40°C h . Qrad 150 W/m2 27ºC Solving for k gives C W/m0.961 °⋅=k 1-98E A spherical ball whose surface is maintained at a temperature of 170°F is suspended in the middle of a room at 70°F. The total rate of heat transfer from the ball is to be determined. Assumptions 1 Steady operating conditions exist since the ball surface and the surrounding air and surfaces remain at constant temperatures. 2 The thermal properties of the ball and the convection heat transfer coefficient are constant and uniform. Properties The emissivity of the ball surface is given to be ε = 0.8. Analysis The heat transfer surface area is As = πD2 = π(2/12 ft) 2 = 0.08727 ft2 Under steady conditions, the rates of convection and radiation heat transfer are Btu/h 9.4 ]R) 460+(70R) 460+)[(170RftBtu/h 10)(0.1714ft 70.8(0.0872 )( Btu/h 9.130F70))(170ft F)(0.08727ftBtu/h (15 444282 44 rad 22 conv = −⋅⋅×= −= =°−°⋅⋅=∆= − oss s TTAQ ThAQ εσ& & Air 70°F 170°F D = 2 in Q Therefore, Btu/h 140.3=+=+= 4.99.130radconvtotal QQQ &&& Discussion Note that heat loss by convection is several times that of heat loss by radiation. The radiation heat loss can further be reduced by coating the ball with a low-emissivity material. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-47 1-99 An 800-W iron is left on the iron board with its base exposed to the air at 20°C. The temperature of the base of the iron is to be determined in steady operation. Assumptions 1 Steady operating conditions exist. 2 The thermal properties of the iron base and the convection heat transfer coefficient are constant and uniform. 3 The temperature of the surrounding surfaces is the same as the temperature of the surrounding air. Iron 800 W Properties The emissivity of the base surface is given to be ε = 0.6. Analysis At steady conditions, the 800 W energy supplied to the iron will be dissipated to the surroundings by convection and radiation heat transfer. Therefore, W 800radconvtotal =+= QQQ &&& where K) 293(0.7K) 293()m K)(0.02 W/m(35 22conv −=−⋅=∆= sss TTThAQ& and ]K) (293[100.06804 ]K) (293)[KW/m 10)(5.67m 0.6(0.02)( 448 44428244 rad −×= −⋅×=−= − − s soss T TTTAQ εσ& Substituting, ]K) (293[1006804.0)K 293(7.0 W800 448 −×+−= − ss TT Solving by trial and error gives C601°== K 874sT Discussion We note that the iron will dissipate all the energy it receives by convection and radiation when its surface temperature reaches 874 K. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-50 1-102E A flat plate solar collector is placed horizontally on the roof of a house. The rate of heat loss from the collector by convection and radiation during a calm day are to be determined. Assumptions 1 Steady operating conditions exist. 2 The emissivity and convection heat transfer coefficient are constant and uniform. 3 The exposed surface, ambient, and sky temperatures remain constant. Properties The emissivity of the outer surface of the collector is given to be 0.9. Air, 70°F Tsky = 50°F Solar collector Q&Analysis The exposed surface area of the collector is 2ft 75ft) ft)(15 5( ==sA Noting that the exposed surface temperature of the collector is 100°F, the total rate of heat loss from the collector to the environment by convection and radiation becomes Btu/h 3551 ])R 46050(R) 460100)[(RftBtu/h 10)(0.1714ft 75)(9.0()( Btu/h 5625F)70100)(ft F)(75Btu/h.ft 5.2()( 44428-244 surrrad 22 conv = +−+⋅⋅×=−= =°−°⋅=−= ∞ ss ss TTAQ TThAQ σε& & and Btu/h 9176=+=+= 35515625radconvtotal QQQ &&& PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-51 1-103 Temperature of the stainless steel sheet going through an annealing process inside an electrically heated oven is to be determined. Assumptions 1 Steady operating conditions exist. 2 Temperature of the stainless steel sheet is uniform. 3 Radiation heat transfer between stainless steel sheet and surrounding oven surfaces is between a small object and a large enclosure. Properties The emissivity of the stainless steel sheet is given to be 0.40. Analysis The amount of heat transfer by radiation between the sheet and the surrounding oven surfaces is balanced by the convection heat transfer between the sheet and the ambient air: 0convrad =− qq && 0)()( 44surr =−−− ∞TThTT ssεσ 0K )]273600()[K W/m10(K ])273750)[(K W/m1067.5)(40.0( 2444428 =+−⋅−−+⋅× − ss TT Solving the above equation by EES software (Copy the following line and paste on a blank EES screen to verify solution): 0.40*5.67e-8*((750+273)^4-T_s^4)-10*(T_s-(600+273))=0 The temperature of the stainless steel sheet is C 736 °== K 1009sT Discussion Note that the energy balance equation involving radiation heat transfer used for solving the stainless steel sheet temperature must be used with absolute temperature. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-52 1-104 The upper surface temperature of a silicon wafer undergoing heat treatment in a vacuum chamber by infrared heater is to be determined. Assumptions 1 Steady operating conditions exist. 2 Radiation heat transfer between upper wafer surface and surroundings is between a small object and a large enclosure. 3 One-dimensional conduction in wafer. 4 The silicon wafer has constant properties. 5 No hot spot exists on the wafer. Properties The thermal conductivity of silicon at 1000 K is 31.2 W/m · K (Table A-3). Analysis The heat transfer through the thickness of the wafer by conduction is equal to net heat transfer at the upper wafer surface: radabscond qqq &&& −= )( 4surr 4 ,IR , TTq L TT k us lss,u −−= − εσα & 444 , 428 2 6 K )310)(K W/m1067.5)(70.0( ) W/m200000)(70.0( m) 10725( K )1000( )K W/m2.31( −⋅×− = × − ⋅ − − us s,u T T Copy the following line and paste on a blank EES screen to solve the above equation: 31.2*(T_su-1000)/725e-6=0.70*200000-0.70*5.67e-8*(T_su^4-310^4) Solving by EES software, the upper surface temperature of silicon wafer is K 1002=usT , Discussion Excessive temperature difference across the wafer thickness will cause warping in the silicon wafer. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-55 Special Topic: Thermal Comfort 1-111C The metabolism refers to the burning of foods such as carbohydrates, fat, and protein in order to perform the necessary bodily functions. The metabolic rate for an average man ranges from 108 W while reading, writing, typing, or listening to a lecture in a classroom in a seated position to 1250 W at age 20 (730 at age 70) during strenuous exercise. The corresponding rates for women are about 30 percent lower. Maximum metabolic rates of trained athletes can exceed 2000 W. We are interested in metabolic rate of the occupants of a building when we deal with heating and air conditioning because the metabolic rate represents the rate at which a body generates heat and dissipates it to the room. This body heat contributes to the heating in winter, but it adds to the cooling load of the building in summer. 1-112C The metabolic rate is proportional to the size of the body, and the metabolic rate of women, in general, is lower than that of men because of their smaller size. Clothing serves as insulation, and the thicker the clothing, the lower the environmental temperature that feels comfortable. 1-113C Asymmetric thermal radiation is caused by the cold surfaces of large windows, uninsulated walls, or cold products on one side, and the warm surfaces of gas or electric radiant heating panels on the walls or ceiling, solar heated masonry walls or ceilings on the other. Asymmetric radiation causes discomfort by exposing different sides of the body to surfaces at different temperatures and thus to different rates of heat loss or gain by radiation. A person whose left side is exposed to a cold window, for example, will feel like heat is being drained from that side of his or her body. 1-114C (a) Draft causes undesired local cooling of the human body by exposing parts of the body to high heat transfer coefficients. (b) Direct contact with cold floor surfaces causes localized discomfort in the feet by excessive heat loss by conduction, dropping the temperature of the bottom of the feet to uncomfortable levels. 1-115C Stratification is the formation of vertical still air layers in a room at difference temperatures, with highest temperatures occurring near the ceiling. It is likely to occur at places with high ceilings. It causes discomfort by exposing the head and the feet to different temperatures. This effect can be prevented or minimized by using destratification fans (ceiling fans running in reverse). 1-116C It is necessary to ventilate buildings to provide adequate fresh air and to get rid of excess carbon dioxide, contaminants, odors, and humidity. Ventilation increases the energy consumption for heating in winter by replacing the warm indoors air by the colder outdoors air. Ventilation also increases the energy consumption for cooling in summer by replacing the cold indoors air by the warm outdoors air. It is not a good idea to keep the bathroom fans on all the time since they will waste energy by expelling conditioned air (warm in winter and cool in summer) by the unconditioned outdoor air. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-56 1-117 The windows of a house in Atlanta are of double door type with wood frames and metal spacers. The average rate of heat loss through the windows in winter is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat losses associated with the infiltration of air through the cracks/openings are not considered. Q& Window 22°C 11.3°C Analysis The rate of heat transfer through the window can be determined from )( oiwindowoverallavg window, TTAUQ −=& where Ti and To are the indoor and outdoor air temperatures, respectively, Uoverall is the U- factor (the overall heat transfer coefficient) of the window, and Awindow is the window area. Substituting, W535=°−°⋅= C)3.1122)(m C)(20 W/m50.2( 22avg window,Q& Discussion This is the “average” rate of heat transfer through the window in winter in the absence of any infiltration. 1-118 Boiling experiments are conducted by heating water at 1 atm pressure with an electric resistance wire, and measuring the power consumed by the wire as well as temperatures. The boiling heat transfer coefficient is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat losses from the water container are negligible. Water 100°C Heater 120°C Analysis The heat transfer area of the heater wire is 2m 004398.0m) m)(0.70 002.0( === ππDLA Noting that 4100 W of electric power is consumed when the heater surface temperature is 120°C, the boiling heat transfer coefficient is determined from Newton’s law of cooling to be )( satTThAQ s −=& C W/m46,600 2 °⋅= °− = − = C)100)(120m (0.004398 W4100 )( 2satTTA Qh s & PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-57 Review Problems 1-119 The power required to maintain the soldering iron tip at 400 °C is to be determined. Assumptions 1 Steady operating conditions exist since the tip surface and the surrounding air temperatures remain constant. 2 The thermal properties of the tip and the convection heat transfer coefficient are constant and uniform. 3 The surrounding surfaces are at the same temperature as the air. Properties The emissivity of the tip is given to be 0.80. Analysis The total heat transfer area of the soldering iron tip is 24 2 2 m 1062.1 )m 02.0)(m 0025.0(4/)m 0025.0( 4/ −×= += += ππ ππ DLDAs The rate of heat transfer by convection is W541 C )20400)(m 1062.1)(C W/m25( )( 242 tipconv . TThAQ s = °−×°⋅= −= − ∞ & The rate of heat transfer by radiation is W451 K ])27320()273400)[(m 1062.1)(K W/m1067.5)(80.0( )( 44424428 4 surr 4 tiprad . TTAQ s = +−+×⋅×= −= −− εσ& Thus, the power required is equal to the total rate of heat transfer from the tip by both convection and radiation: W2.99=+=+= W45.1 W54.1radconvtotal QQQ &&& Discussion If the soldering iron tip is highly polished with an emissivity of 0.05, the power required to maintain the tip at 400 °C will reduce to 1.63 W, or by 45.5%. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-60 1-124 Engine valves are to be heated in a heat treatment section. The amount of heat transfer, the average rate of heat transfer, the average heat flux, and the number of valves that can be heat treated daily are to be determined. Assumptions Constant properties given in the problem can be used. Properties The average specific heat and density of valves are given to be cp = 440 J/kg.°C and ρ = 7840 kg/m3. Analysis (a) The amount of heat transferred to the valve is simply the change in its internal energy, and is determined from kJ 26.35=C40)C)(800kJ/kg kg)(0.440 0788.0( )( 12 °−°⋅= −=∆= TTmcUQ p Engine valve T1 = 40°C T2 = 800°C D = 0.8 cm L = 10 cm (b) The average rate of heat transfer can be determined from W87.8== × = ∆ = kW 0878.0 s 605 kJ 35.26 avg t QQ& (c) The average heat flux is determined from 24 W/m101.75×==== m) m)(0.1 008.0(2 W8.87 π2 avgavg ave πDL Q A Q q s && & (d) The number of valves that can be heat treated daily is valves 3000=×= min 5 valves)min)(25 6010( valvesofNumber 1-125 The glass cover of a flat plate solar collector with specified inner and outer surface temperatures is considered. The fraction of heat lost from the glass cover by radiation is to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the glass remain constant at the specified values. 2 Thermal properties of the glass are constant. Properties The thermal conductivity of the glass is given to be k = 0.7 W/m⋅°C. Analysis Under steady conditions, the rate of heat transfer through the glass by conduction is W 583 m 0.006 C)31(33)m C)(2.5W/m (0.7 2cond = °− °⋅= ∆ = L TkAQ& A = 2.5 m2 31°C 33°C L=0.6 Air, 15°C h=10 W/m2.°C Q& The rate of heat transfer from the glass by convection is W400C)15)(31m C)(2.5W/m (10 22conv =°−°⋅=∆= ThAQ& Under steady conditions, the heat transferred through the cover by conduction should be transferred from the outer surface by convection and radiation. That is, W183400583convcondrad =−=−= QQQ &&& Then the fraction of heat transferred by radiation becomes 0.314=== 583 183 cond rad Q Q f & & (or 31.4%) PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-61 1-126 The range of U-factors for windows are given. The range for the rate of heat loss through the window of a house is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat losses associated with the infiltration of air through the cracks/openings are not considered. 20°C -8°C Window Q&Analysis The rate of heat transfer through the window can be determined from )(windowoverallwindow oi TTAUQ −=& where Ti and To are the indoor and outdoor air temperatures, respectively, Uoverall is the U- factor (the overall heat transfer coefficient) of the window, and Awindow is the window area. Substituting, Maximum heat loss: W378=°−−×°⋅= C)]8(20)[m 8.1C)(1.2 W/m25.6( 22max window,Q& Minimum heat loss: W76=°−−×°⋅= C)]8(20)[m 8.1C)(1.2 W/m25.1( 22min window,Q& Discussion Note that the rate of heat loss through windows of identical size may differ by a factor of 5, depending on how the windows are constructed. 1-127 Prob. 1-126 is reconsidered. The rate of heat loss through the window as a function of the U-factor is to be plotted. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" A=1.2*1.8 [m^2] T_1=20 [C] T_2=-8 [C] U=1.25 [W/m^2-C] 1 2 3 4 5 6 7 50 100 150 200 250 300 350 400 U [W/m2-C] Q w in do w [ W ] "ANALYSIS" Q_dot_window=U*A*(T_1-T_2) U [W/m2.C] Qwindow [W] 1.25 75.6 1.75 105.8 2.25 136.1 2.75 166.3 3.25 196.6 3.75 226.8 4.25 257 4.75 287.3 5.25 317.5 5.75 347.8 6.25 378 PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-62 1-128 An electric heater placed in a room consumes 500 W power when its surfaces are at 120°C. The surface temperature when the heater consumes 700 W is to be determined without and with the consideration of radiation. Assumptions 1 Steady operating conditions exist. 2 The temperature is uniform over the surface. Analysis (a) Neglecting radiation, the convection heat transfer coefficient is determined from ( ) CW/m02 C20120)m 25.0( W500 )( 2 2 °⋅= °− = − = ∞TTA Q h s & The surface temperature when the heater consumes 700 W is C160°= °⋅ +°=+= ∞ )m 25.0(C)W/m02( W700C20 22hA Q TTs & eW& T∞ , h qrad qconv Tw Ts A, ε (b) Considering radiation, the convection heat transfer coefficient is determined from [ ] ( ) CW/m58.12 C20120)m 25.0( K) 283(K) 393()KW/m1067.5)(m 5(0.75)(0.2- W500 )( )( 2 2 444282 4 surr 4 °⋅= °− −⋅× = − −− = − ∞TTA TTAQ h s sσε& Then the surface temperature becomes ( ) [ ] C152.9°== −×+−= −+−= − ∞ K 9.425 K) 283()1067.5)(5(0.75)(0.2)293)(25.0)(58.12(700 )( 448 4 surr 4 s ss ss T TT TTATThAQ σε& Discussion Neglecting radiation changed Ts by more than 7°C, so assumption is not correct in this case. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-65 1-135 A 2-kW electric resistance heater in a room is turned on and kept on for 50 minutes. The amount of energy transferred to the room by the heater is (a) 2 kJ (b) 100 kJ (c) 6000 kJ (d) 7200 kJ (e) 12,000 kJ Answer (c) 6000 kJ Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. We= 2 [kJ/s] time=50*60 [s] We_total=We*time [kJ] "Wrong Solutions:" W1_Etotal=We*time/60 "using minutes instead of s" W2_Etotal=We "ignoring time" 1-136 A hot 16 cm × 16 cm × 16 cm cubical iron block is cooled at an average rate of 80 W. The heat flux is (a) 195 W/m2 (b) 521 W/m2 (c) 3125 W/m2 (d) 7100 W/m2 (e) 19,500 W/m2 Answer (b) 521 W/m2 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. a=0.16 [m] Q_dot=80 [W] A_s=6*a^2 q=Q_dot/A_s "Some Wrong Solutions with Common Mistakes" W1_q=Q_dot/a^2 "Using wrong equation for area" W2_q=Q_dot/a^3 "Using volume instead of area" PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-66 1-137 A 2-kW electric resistance heater submerged in 30-kg water is turned on and kept on for 10 min. During the process, 500 kJ of heat is lost from the water. The temperature rise of water is (a) 5.6°C (b) 9.6°C (c) 13.6°C (d) 23.3°C (e) 42.5°C Answer (a) 5.6°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. C=4.18 [kJ/kg-K] m=30 [kg] Q_loss=500 [kJ] time=10*60 [s] W_e=2 [kJ/s] "Applying energy balance E_in-E_out=dE_system gives" time*W_e-Q_loss = dU_system dU_system=m*C*DELTAT “Some Wrong Solutions with Common Mistakes:” time*W_e = m*C*W1_T "Ignoring heat loss" time*W_e+Q_loss = m*C*W2_T "Adding heat loss instead of subtracting" time*W_e-Q_loss = m*1.0*W3_T "Using specific heat of air or not using specific heat" 1-138 Eggs with a mass of 0.15 kg per egg and a specific heat of 3.32 kJ/kg⋅°C are cooled from 32°C to 10°C at a rate of 200 eggs per minute. The rate of heat removal from the eggs is (a) 7.3 kW (b) 53 kW (c) 17 kW (d) 438 kW (e) 37 kW Answer (e) 37 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. C=3.32 [kJ/kg-K] m_egg=0.15 [kg] T1=32 [C] T2=10 [C] n=200 "eggs/min" m=n*m_egg/60 "kg/s" "Applying energy balance E_in-E_out=dE_system gives" "-E_out = dU_system" Qout=m*C*(T1-T2) "kJ/s" “Some Wrong Solutions with Common Mistakes:” W1_Qout = m*C*T1 "Using T1 only" W2_Qout = m_egg*C*(T1-T2) "Using one egg only" W3_Qout = m*C*T2 "Using T2 only" W4_Qout=m_egg*C*(T1-T2)*60 "Finding kJ/min" PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-67 1-139 Steel balls at 140°C with a specific heat of 0.50 kJ/kg⋅°C are quenched in an oil bath to an average temperature of 85°C at a rate of 35 balls per minute. If the average mass of steel balls is 1.2 kg, the rate of heat transfer from the balls to the oil is (a) 33 kJ/s (b) 1980 kJ/s (c) 49 kJ/s (d) 30 kJ/s (e) 19 kJ/s Answer (e) 19 kJ/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. c=0.50 [kJ/kg-K] m1=1.2 [kg] T1=140 [C] T2=85 [C] n=35 "balls/min" m=n*m1/60 "kg/s" "Applying energy balance E_in-E_out=dE_system gives" "-E_out = dU_system" Qout=m*c*(T1-T2) "kJ/s" “Some Wrong Solutions with Common Mistakes:” W1_Qout = m*c*T1 "Using T1 only" W2_Qout = m1*c*(T1-T2) "Using one egg only" W3_Qout = m*c*T2 "Using T2 only" W4_Qout=m1*c*(T1-T2)*60 "Finding kJ/min" 1-140 A cold bottled drink (m = 2.5 kg, cp = 4200 J/kg⋅°C) at 5°C is left on a table in a room. The average temperature of the drink is observed to rise to 15°C in 30 minutes. The average rate of heat transfer to the drink is (a) 23 W (b) 29 W (c) 58 W (d) 88 W (e) 122 W Answer: (c) 58 W Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. c=4200 [J/kg-K] m=2.5 [kg] T1=5 [C] T2=15 [C] time = 30*60 [s] "Applying energy balance E_in-E_out=dE_system gives" Q=m*c*(T2-T1) Qave=Q/time “Some Wrong Solutions with Common Mistakes:” W1_Qave = m*c*T1/time "Using T1 only" W2_Qave = c*(T2-T1)/time "Not using mass" W3_Qave = m*c*T2/time "Using T2 only" PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-70 1-145 Steady heat conduction occurs through a 0.3-m thick 9 m by 3 m composite wall at a rate of 1.2 kW. If the inner and outer surface temperatures of the wall are 15°C and 7°C, the effective thermal conductivity of the wall is (a) 0.61 W/m⋅°C (b) 0.83 W/m⋅°C (c) 1.7 W/m⋅°C (d) 2.2 W/m⋅°C (e) 5.1 W/m⋅°C Answer (c) 1.7 W/m⋅°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. A=9*3 [m^2] L=0.3 [m] T1=15 [C] T2=7 [C] Q=1200 [W] Q=k*A*(T1-T2)/L "Wrong Solutions:" Q=W1_k*(T1-T2)/L "Not using area" Q=W2_k*2*A*(T1-T2)/L "Using areas of both surfaces" Q=W3_k*A*(T1+T2)/L "Adding temperatures instead of subtracting" Q=W4_k*A*L*(T1-T2) "Multiplying by thickness instead of dividing by it" 1-146 Heat is lost through a brick wall (k = 0.72 W/m·ºC), which is 4 m long, 3 m wide, and 25 cm thick at a rate of 500 W. If the inner surface of the wall is at 22ºC, the temperature at the midplane of the wall is (a) 0ºC (b) 7.5ºC (c) 11.0ºC (d) 14.8ºC (e) 22ºC Answer (d) 14.8ºC Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. k=0.72 [W/m-C] Length=4 [m] Width=3 [m] L=0.25 [m] Q_dot=500 [W] T1=22 [C] A=Length*Width Q_dot=k*A*(T1-T_middle)/(0.5*L) "Some Wrong Solutions with Common Mistakes" Q_dot=k*A*(T1-W1_T_middle)/L "Using L instead of 0.5L" W2_T_middle=T1/2 "Just taking the half of the given temperature" PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-71 1-147 A 10-cm high and 20-cm wide circuit board houses on its surface 100 closely spaced chips, each generating heat at a rate of 0.12 W and transferring it by convection and radiation to the surrounding medium at 40°C. Heat transfer from the back surface of the board is negligible. If the combined convection and radiation heat transfer coefficient on the surface of the board is 22 W/m2⋅°C, the average surface temperature of the chips is (a) 41°C (b) 54°C (c) 67°C (d) 76°C (e) 82°C Answer (c) 67°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. A=0.1*0.2 [m^2] Q= 100*0.12 [W] Tair=40 [C] h=22 [W/m^2-C] Q= h*A*(Ts-Tair) "Wrong Solutions:" Q= h*(W1_Ts-Tair) "Not using area" Q= h*2*A*(W2_Ts-Tair) "Using both sides of surfaces" Q= h*A*(W3_Ts+Tair) "Adding temperatures instead of subtracting" Q/100= h*A*(W4_Ts-Tair) "Considering 1 chip only" 1-148 A 40-cm-long, 0.4-cm-diameter electric resistance wire submerged in water is used to determine the convection heat transfer coefficient in water during boiling at 1 atm pressure. The surface temperature of the wire is measured to be 114°C when a wattmeter indicates the electric power consumption to be 7.6 kW. The heat transfer coefficient is (a) 108 kW/m2⋅°C (b) 13.3 kW/m2⋅°C (c) 68.1 kW/m2⋅°C (d) 0.76 kW/m2⋅°C (e) 256 kW/m2⋅°C Answer (a) 108 kW/m2⋅°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. L=0.4 [m] D=0.004 [m] A=pi*D*L [m^2] We=7.6 [kW] Ts=114 [C] Tf=100 [C] “Boiling temperature of water at 1 atm" We= h*A*(Ts-Tf) "Wrong Solutions:" We= W1_h*(Ts-Tf) "Not using area" We= W2_h*(L*pi*D^2/4)*(Ts-Tf) "Using volume instead of area" We= W3_h*A*Ts "Using Ts instead of temp difference" PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-72 1-149 Over 90 percent of the energy dissipated by an incandescent light bulb is in the form of heat, not light. What is the temperature of a vacuum-enclosed tungsten filament with an exposed surface area of 2.03 cm2 in a 100 W incandescent light bulb? The emissivity of tungsten at the anticipated high temperatures is about 0.35. Note that the light bulb consumes 100 W of electrical energy, and dissipates all of it by radiation. (a) 1870 K (b) 2230 K (c) 2640 K (d) 3120 K (e) 2980 K Answer (b) 2230 K Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. e =0.35 Q=100 [W] A=2.03E-4 [m^2] Q=e*A*sigma#*T^4 1-150 Commercial surface coating processes often use infrared lamps to speed the curing of the coating. A 1-mm-thick, teflon (k = 0.45 W/m⋅K) coating is applied to a 4 m × 4 m surface using this process. Once the coating reaches steady-state, the temperature of its two surfaces are 50oC and 45oC. What is the minimum rate at which power must be supplied to the infrared lamps steadily? (a) 36 kW (b) 40 kW (c) 44 kW (d) 48 kW (e) 52 kW Answer (a) 36 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. k=0.45 [W/m-K] A=16 [m^2] t=0.001 [m] dT=5 [C] Q=k*A*dT/t PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-75 1-153 A 3-m2 black surface at 140°C is losing heat to the surrounding air at 35°C by convection with a heat transfer coefficient of 16 W/m2⋅°C, and by radiation to the surrounding surfaces at 15°C. The total rate of heat loss from the surface is (a) 5105 W (b) 2940 W (c) 3779 W (d) 8819 W (e) 5040 W Answer (d) 8819 W Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. sigma=5.67E-8 [W/m^2-K^4] eps=1 A=3 [m^2] h_conv=16 [W/m^2-C] Ts=140 [C] Tf=35 [C] Tsurr=15 [C] Q_conv=h_conv*A*(Ts-Tf) Q_rad=eps*sigma*A*((Ts+273)^4-(Tsurr+273)^4) Q_total=Q_conv+Q_rad “Some Wrong Solutions with Common Mistakes:” W1_Ql=Q_conv "Ignoring radiation" W2_Q=Q_rad "ignoring convection" W3_Q=Q_conv+eps*sigma*A*(Ts^4-Tsurr^4) "Using C in radiation calculations" W4_Q=Q_total/A "not using area" PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-76 1-154 A person’s head can be approximated as a 25-cm diameter sphere at 35°C with an emissivity of 0.95. Heat is lost from the head to the surrounding air at 25°C by convection with a heat transfer coefficient of 11 W/m2⋅°C, and by radiation to the surrounding surfaces at 10°C. Disregarding the neck, determine the total rate of heat loss from the head. (a) 22 W (b) 27 W (c) 49 W (d) 172 W (e) 249 W Answer: (c) 49 W Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. sigma=5.67E-8 [W/m^2-K^4] eps=0.95 D=0.25 [m] A=pi*D^2 h_conv=11 [W/m^2-C] Ts=35 [C] Tf=25 [C] Tsurr=10 [C] Q_conv=h_conv*A*(Ts-Tf) Q_rad=eps*sigma*A*((Ts+273)^4-(Tsurr+273)^4) Q_total=Q_conv+Q_rad "Wrong Solutions:" W1_Ql=Q_conv "Ignoring radiation" W2_Q=Q_rad "ignoring convection" W3_Q=Q_conv+eps*sigma*A*(Ts^4-Tsurr^4) "Using C in radiation calculations" W4_Q=Q_total/A "not using area" PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 1-77 1-155 A 25-cm-long, 0.4-cm-diameter electric resistance wire is used to determine the convection heat transfer coefficient in air at 25°C experimentally. The surface temperature of the wire is measured to be 230°C when the electric power consumption is 180 W. If the radiation heat loss from the wire is calculated to be 60 W, the convection heat transfer coefficient is (a) 186 W/m2⋅°C (b) 280 W/m2⋅°C (c) 373 W/m2⋅°C (d) 585 W/m2⋅°C (e) 620 W/m2⋅°C Answer (a) 186 W/m2⋅°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. L=0.25 [m] D=0.004 [m] A=pi*D*L We=180 [W] Ts=230 [C] Tf=25 [C] Qrad = 60 We- Qrad = h*A*(Ts-Tf) “Some Wrong Solutions with Common Mistakes:” We- Qrad = W1_h*(Ts-Tf) "Not using area" We- Qrad = W2_h*(L*D)*(Ts-Tf) "Using D*L for area" We+ Qrad = W3_h*A*(Ts-Tf) "Adding Q_rad instead of subtracting" We= W4_h*A*(Ts-Tf) "Disregarding Q_rad"