1. http://www.elsolucionario.blogspot.comLIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS DE FORMA CLARA VISITANOS PARA DESARGALOS GRATIS.

2. 1-1Chapter 1 INTRODUCTION AND BASIC CONCEPTS 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 (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. 1-3C 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. 1-4C 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. 1-5C 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. 1-6C 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. 1-7C The right choice between a crude and complex model is usually the simplest model which yields adequate results. Preparing very accurate but complex models is not necessarily a better choice since such models are not much use to an analyst if they are very difficult and time consuming to solve. At the minimum, the model should reflect the essential features of the physical problem it represents.PROPRIETARY MATERIAL. 2007 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. 3. 1-2Heat and Other Forms of Energy& 1-8C The rate of heat transfer per unit surface area is called heat flux q . It is related to the rate of heat & transfer by Q = q&dA . A1-9C Energy can be transferred by heat, work, and mass. An energy transfer is heat transfer when its driving force is temperature difference. 1-10C Thermal energy is the sensible and latent forms of internal energy, and it is referred to as heat in daily life. 1-11C For the constant pressure case. This is because the heat transfer to an ideal gas is mcpT at constant pressure and mcvT at constant volume, and cp is always greater than cv.1-12 A cylindrical resistor on a circuit board dissipates 0.8 W of power. The amount of heat dissipated in 24 h, the heat flux, and the fraction of heat dissipated from the top and bottom surfaces are to be determined. Assumptions Heat is transferred uniformly from all surfaces. Analysis (a) The amount of heat this resistor dissipates during a 24-hour period is& Q = Qt = (0.8 W)(24 h) = 19.2 Wh = 69.1 kJ (since 1 Wh = 3600 Ws = 3.6 kJ) (b) The heat flux on the surface of the resistor isAs = 2 & qs =D 2 4+ DL = 2 (0.4 cm) 2 4+ (0.4 cm)(2 cm) = 0.251 + 2.513 = 2.764 cm 2& Q Resistor 0.8 W& Q 0.80 W = = 0.289 W/cm 2 As 2.764 cm 2(c) Assuming the heat transfer coefficient to be uniform, heat transfer is proportional to the surface area. Then the fraction of heat dissipated from the top and bottom surfaces of the resistor becomesQ top base Q total=Atop base Atotal=0.251 = 0.091 or (9.1%) 2.764Discussion Heat transfer from the top and bottom surfaces is small relative to that transferred from the side surface.PROPRIETARY MATERIAL. 2007 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. 4. 1-31-13E A logic chip in a computer dissipates 3 W of power. The amount heat dissipated in 8 h and the heat flux on the surface of the chip are to be determined. Assumptions Heat transfer from the surface is uniform. Analysis (a) The amount of heat the chip dissipates during an 8-hour period isLogic chip& Q = 3W& Q = Qt = (3 W)(8 h ) = 24 Wh = 0.024 kWh (b) The heat flux on the surface of the chip is & Q 3W & = 37.5 W/in 2 q= = A 0.08 in 21-14 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 areAs = DL = (0.05 cm)(5 cm) = 0.785 cm 2 & Q 150 W & = = 191 W/cm 2 = 1.91 10 6 W/m 2 qs = As 0.785 cm 2& Q Lamp 150 W(b) The heat flux on the surface of glass bulb isAs = D 2 = (8 cm) 2 = 201.1 cm 2 & qs =& Q 150 W = = 0.75 W/cm 2 = 7500 W/m 2 As 201.1 cm 2(c) The amount and cost of electrical energy consumed during a one-year period is & Electricity Consumptio n = Qt = (0.15 kW)(365 8 h/yr) = 438 kWh/yr Annual Cost = (438 kWh/yr)($0.08 / kWh) = $35.04/yrPROPRIETARY MATERIAL. 2007 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. 5. 1-41-15 A 1200 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. Analysis (a) The amount of heat the iron dissipates during a 2-h period isIron 1200 W& Q = Qt = (1.2 kW)(2 h) = 2.4 kWh (b) The heat flux on the surface of the iron base is& Qbase = (0.85)(1200 W) = 1020 W & q=& Qbase 1020 W = = 68,000 W/m 2 Abase 0.015 m 2(c) The cost of electricity consumed during this period is Cost of electricity = (2.4 kWh) ($0.07 / kWh) = $0.171-16 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 isChips, 0.12 W& Q = (120)(0.12 W) = 14.4 W& Q& Q = Qt = (0.0144 kW)(10 h) = 0.144 kWh (b) The heat flux on the surface of the circuit board is15 cmAs = (0.15 m)(0.2 m) = 0.03 m 2 & qs =& Q 14.4 W = = 480 W/m 2 As 0.03 m 220 cmPROPRIETARY MATERIAL. 2007 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. 6. 1-51-17 An aluminum ball is to be heated from 80C to 200C. 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 = 2700 kg/m3 and cp = 0.90 kJ/kgC.Metal ballAnalysis The amount of energy added to the ball is simply the change in its internal energy, and is determined fromE transfer = U = mc p (T2 T1 ) whereE m = V = 6D 3 = 6( 2700 kg/m 3 )(0.15 m) 3 = 4.77 kgSubstituting, E transfer = (4.77 kg)(0.90 kJ/kg C)(200 80)C = 515 kJTherefore, 515 kJ of energy (heat or work such as electrical energy) needs to be transferred to the aluminum ball to heat it to 200C.1-18 The body temperature of a man rises from 37C to 39C during strenuous exercise. The resulting increase in the thermal energy content of the body is to be determined. Assumptions The body temperature changes uniformly. Properties The average specific heat of the human body is given to be 3.6 kJ/kgC. Analysis The change in the sensible internal energy content of the body as a result of the body temperature rising 2C during strenuous exercise isU = mcpT = (80 kg)(3.6 kJ/kgC)(2C) = 576 kJPROPRIETARY MATERIAL. 2007 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. 7. 1-61-19 An electrically heated house maintained at 22C 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 22C. Properties The specific heat of air at room temperature is cp = 1.007 kJ/kgC. Analysis The volume of the air in the house isV = (floor space)(height) = (200 m 2 )(3 m) = 600 m 3 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 0.724 = 16.8 times per day, the mass flow rate of air through the house due to infiltration is& m airP V& P (ACH V house ) = o air = o RTo RTo =0.7 ACH(89.6 kPa)(16.8 600 m 3 / day) (0.287 kPa m 3 /kg K)(5 + 273.15 K)22C AIR5C = 11,314 kg/dayNoting that outdoor air enters at 5C and leaves at 22C, the energy loss of this house per day is & & Q = m c (T T ) infiltairpindoorsoutdoors= (11,314 kg/day)(1.007 kJ/kg.C)(22 5)C = 193,681 kJ/day = 53.8 kWh/dayAt a unit cost of $0.082/kWh, the cost of this electrical energy lost by infiltration is Enegy Cost = (Energy used)(Unit cost of energy) = (53.8 kWh/day)($0.082/kWh) = $4.41/dayPROPRIETARY MATERIAL. 2007 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. 8. 1-71-20 A house is heated from 10C to 22C 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 22C. Properties The specific heat of air at room temperature is cp = 1.007 kJ/kgC. Analysis The volume and mass of the air in the house areV = (floor space)(height) = (200 m 2 )(3 m) = 600 m 3 m=(101.3 kPa)(600 m 3 ) PV = = 747.9 kg RT (0.287 kPa m 3 /kg K)(10 + 273.15 K)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 22C is determined to be22C 10C AIRQ = mc p (T2 T1 ) = (747.9 kg)(1.007 kJ/kg C)(22 10)C = 9038 kJ Noting that 1 kWh = 3600 kJ, the cost of this electrical energy at a unit cost of $0.075/kWh is Enegy Cost = (Energy used)(Unit cost of energy) = (9038 / 3600 kWh)($0.075/kWh) = $0.19Therefore, it will cost the homeowner about 19 cents to raise the temperature in his house from 10 to 22C.1-21E A water heater is initially filled with water at 45F. The amount of energy that needs to be transferred to the water to raise its temperature to 120F is to be determined. Assumptions 1 Water is an incompressible substance with constant specific heats at room temperature. 2 No water flows in or out of the tank during heating. Properties The density and specific heat of water are given to be 62 lbm/ft3 and 1.0 Btu/lbmF. Analysis The mass of water in the tank is 1 ft 3 = 497.3 lbm m = V = (62 lbm/ft )(60 gal) 7.48 gal 120F3Then, the amount of heat that must be transferred to the water in the tank as it is heated from 45 to1120F is determined to be45F WaterQ = mc p (T2 T1 ) = (497.3 lbm)(1.0 Btu/lbm F)(120 45)F = 37,300 BtuPROPRIETARY MATERIAL. 2007 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. 9. 1-8Energy Balance 1-22C Warmer. Because energy is added to the room air in the form of electrical work. 1-23C Warmer. If we take the room that contains the refrigerator as our system, we will see that electrical work is supplied to this room to run the refrigerator, which is eventually dissipated to the room as waste heat.1-24 Two identical cars have a head-on collusion on a road, and come to a complete rest after the crash. The average temperature rise of the remains of the cars immediately after the crash is to be determined. Assumptions 1 No heat is transferred from the cars. 2 All the kinetic energy of cars is converted to thermal energy. Properties The average specific heat of the cars is given to be 0.45 kJ/kgC. Analysis We take both cars as the system. This is a closed system since it involves a fixed amount of mass (no mass transfer). Under the stated assumptions, the energy balance on the system can be expressed asE in E out 1 24 4 3=Net energy transfer by heat, work, and massE system 1 24 4 3Change in internal, kinetic, potential, etc. energies0 = U cars + KE cars 0 = (mc p T ) cars + [m(0 V 2 ) / 2] cars That is, the decrease in the kinetic energy of the cars must be equal to the increase in their internal energy. Solving for the velocity and substituting the given quantities, the temperature rise of the cars becomes T =mV 2 / 2 V 2 / 2 (90,000 / 3600 m/s) 2 / 2 1 kJ/kg = = = 0.69C 0.45 kJ/kg.C mc p cp 1000 m 2 /s 2 PROPRIETARY MATERIAL. 2007 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. 10. 1-91-25 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 & & & & Q =Q +Q +Q coolingwhere & Qlightslightspeopleheat gain= 10 100 W = 1 kW& Q people = 40 360kJ/h = 14,400 kJ/h = 4kW15,000 kJ/h& Q heat gain = 15,000 kJ/h = 4.17 kWSubstituting,Room 40 people 10 bulbsQcool& Qcooling = 1 + 4 + 4.17 = 9.17 kWThus the number of air-conditioning units required is 9.17 kW = 1.83 2 units 5 kW/unit1-26 A resistance heater is to raise the air temperature in the room from 7 to 25C within 15 min. The required power rating of the 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 -141C 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. 4 Heat losses from the room are negligible. Properties The gas constant of air is R = 0.287 kPam3/kgK (Table A-1). Also, cp = 1.007 kJ/kgK for air at room temperature (Table A-15). Analysis We observe that the pressure in the room remains constant during this process. Therefore, some air will leak out as the air expands. However, we can take the air to be a closed system by considering the air in the room to have undergone a constant pressure expansion process. The energy balance for this steady-flow system can be expressed as E in E out 1 24 4 3Net energy transfer by heat, work, and mass=E system 1 24 4 3Change in internal, kinetic, potential, etc. energiesWe,in Wb = U456 m3 7CWe,in = H = m(h2 h1 ) mc p (T2 T1 )or& We,in t = mc p , avg (T2 T1 )The mass of air isWeAIRV = 4 5 6 = 120m 3 m=P1V (100kPa)(120m 3 ) = = 149.3kg RT1 (0.287 kPa m 3 /kg K)(280K)Using cp value at room temperature, the power rating of the heater becomes & W = (149.3 kg)(1.007 kJ/kg C)(25 7) o C/(15 60 s) = 3.01 kW e,inPROPRIETARY MATERIAL. 2007 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. 11. 1-101-27 A room is heated by the radiator, and the warm air is distributed by a fan. Heat is lost from the room. The time it takes for the air temperature to rise to 20C 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 -141C 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. 4 The local atmospheric pressure is 100 kPa. Properties The gas constant of air is R = 0.287 kPam3/kg.K (Table A-1). Also, cp = 1.007 kJ/kgK and cv = 0.720 kJ/kgK for air at room temperature (Table A-15). Analysis We take the air in the room as the system. This is a closed system since no mass crosses the system boundary during the process. We observe that the pressure in the room remains constant during this process. Therefore, some air will leak out as the air expands. However we can take the air to be a closed system by considering the air in the room to have undergone a constant pressure process. The energy balance for this system can be expressed as E in E out 1 24 4 3=Net energy transfer by heat, work, and massE system 1 24 4 35,000 kJ/hChange in internal, kinetic, potential, etc. energiesROOMQin + We,in Wb Qout = U & & & (Qin + We,in Qout )t = H = m(h2 h1 ) mc p (T2 T1 )The mass of air isV = 4 5 7 = 140m4m5m7m Steam3PV (100 kPa)(140 m 3 ) m= 1 = = 172.4 kg RT1 (0.287 kPa m 3 /kg K)(283 K)Wpw12,500 kJ/hUsing the cp value at room temperature,[(12,500 5000)/3600 kJ/s + 0.1 kJ/s]t = (172.4 kg)(1.007 kJ/kg C)(20 10)CIt yields t = 795 s = 13.3 minPROPRIETARY MATERIAL. 2007 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. 12. 1-111-28 A student living in a room turns his 150-W fan on in the morning. The temperature in the room when she comes back 10 h later 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 -141C 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. 4 All the doors and windows are tightly closed, and heat transfer through the walls and the windows is disregarded. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). Also, cp = 1.007 kJ/kgK for air at room temperature (Table A-15) and cv = cp R = 0.720 kJ/kgK. Analysis We take the room as the system. This is a closed system since the doors and the windows are said to be tightly closed, and thus no mass crosses the system boundary during the process. The energy balance for this system can be expressed asE in E out 1 24 4 3=Net energy transfer by heat, work, and massE system 1 24 4 3(insulated)Change in internal, kinetic, potential, etc. energiesROOMWe,in = U We,in = m(u 2 u1 ) mc v (T2 T1 )4m6m6mThe mass of air isV = 4 6 6 = 144 m 3 m=P1V (100 kPa)(144 m 3 ) = = 174.2 kg RT1 (0.287 kPa m 3 /kg K)(288 K)The electrical work done by the fan is & W e = W e t = (0.15 kJ/s)(10 3600 s) = 5400 kJ WeSubstituting and using cv value at room temperature, 5400 kJ = (174.2 kg)(0.720 kJ/kgC)(T2 - 15)C T2 = 58.1C1-29 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 7000 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 -141C 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 toE in E out 1 24 4 3Net energy transfer by heat, work, and mass=E system 1 24 4 3Change in internal, kinetic, potential, etc. energiesWe,in Qout = U = 0AIRWe,in = Qout since U = mcvT = 0 for isothermal processes of ideal gases. Thus, & & We,in = Qout 1kW = 7000kJ/h 3600kJ/h = 1.94 kW WePROPRIETARY MATERIAL. 2007 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. 13. 1-121-30 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 -141C 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/kgK for air at room temperature (Table A-15) and cv = cp R = 0.720 kJ/kgK. 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 E in E out 1 24 4 3=Net energy transfer by heat, work, and massE system 1 24 4 3Change in internal, kinetic, potential, etc. energiesW e,in + W fan,in Qout = U & & +W (We,in & fan,in Qout )t = m(u 2 u1 ) mc v (T2 T1 )200 kJ/min 568 m3The total mass of air in the room is WeV = 5 6 8 m 3 = 240 m 3 m=P1V (98 kPa )(240 m 3 ) = = 284.6 kg RT1 (0.287 kPa m 3 /kg K )(288 K )300 WThen the power rating of the electric heater is determined to be & & & = Q W W + mc (T T ) / t e,inoutvfan,in21= (200/60 kJ/s) (0.3 kJ/s) + (284.6 kg)(0.720 kJ/kg C)(25 15C)/(18 60 s) = 4.93 kW(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, & & E in = E out & & + mh = Q& & We,in + Wfan,in & & W +W e,in1fan,inout0& + mh2 (since ke pe 0)& & = mh = mc p TThus, T =& & W e,in + W fan,in & mc p=(4.93 + 0.3)kJ/s = 6.2C (50/60 kg/s )(1.007 kJ/kg K )PROPRIETARY MATERIAL. 2007 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. 14. 1-131-31 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 -141C 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/kgC (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 m CV = 0 and E CV = 0 . Also, there is only one inlet and one exit and thus & & & m1 = m2 = m . The energy balance for this steady-flow system can be expressed in the rate form as& & E in E out 1 24 4 3=Rate of net energy transfer by heat, work, and mass& E system 0 (steady) 1442443 4& & = 0 E in = E outRate of change in internal, kinetic, potential, etc. energies& & & & & We,in + Wfan,in + mh1 = Qout + mh2 (since ke pe 0) & & & & We,in = Qout Wfan,in + mc p (T2 T1 ) Substituting, the power rating of the heating element is determined to be & = (0.25 kW ) (0.3 kW) + (0.6 kg/s)(1.007 kJ/kg C)(5C) W e,in250 W We 300 W= 2.97 kWPROPRIETARY MATERIAL. 2007 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. 15. 1-141-32 Air is moved through the resistance heaters in a 1200-W hair dryer by a fan. The volume flow rate of air at the inlet and the velocity 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 -141C 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. 4 The power consumed by the fan and the heat losses through the walls of the hair dryer are negligible. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). Also, cp = 1.007 kJ/kgK for air at room temperature (Table A-15). Analysis (a) We take the hair dryer 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 m CV = 0 and E CV = 0 , and there is only one inlet and one exit and thus & & & m1 = m2 = m . The energy balance for this steady-flow system can be expressed in the rate form as & & E in E out 1 24 4 3=Rate of net energy transfer by heat, work, and mass& & We,in + Wfan,in0& E system 0 (steady) 1442443 4& & = 0 E in = E outRate of change in internal, kinetic, potential, etc. energies& & & + mh1 = Qout 0 + mh2 (since ke pe 0) & & We,in = mc p (T2 T1 )Thus, & m= =P1 = 100 kPa T1 = 22CT2 = 47C A2 = 60 cm2& W e,inc p (T2 T1 ) 1.2 kJ/s = 0.04767 kg/s (1.007 kJ/kg C)( 47 22)C We = 1200 WThen,v1 =RT1 (0.287 kPa m 3 /kg K )(295 K ) = = 0.8467 m 3 /kg 100 kPa P1& V&1 = mv 1 = (0.04767 kg/s )(0.8467 m 3 /kg ) = 0.0404 m 3 /s(b) The exit velocity of air is determined from the conservation of mass equation,v2 = & m=RT2 (0.287 kPa m 3 /kg K )(320 K ) = = 0.9184 m 3 /kg P2 100 kPa 1v2A2V 2 V 2 =& mv 2 (0.04767 kg/s )(0.9184 m 3 /kg ) = = 7.30 m/s A2 60 10 4 m 2PROPRIETARY MATERIAL. 2007 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. 16. 1-151-33 The ducts of an air heating system pass through an unheated area, resulting in a temperature drop of the air in the duct. The rate of heat loss from the air to the cold environment 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 -141C 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/kgC (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 m CV = 0 and E CV = 0 . Also, there is only one inlet and one exit and thus & & & m1 = m2 = m . The energy balance for this steady-flow system can be expressed in the rate form as& & E in E out 1 24 4 3Rate of net energy transfer by heat, work, and mass=& E system 0 (steady) 1442443 4& & = 0 E in = E outRate of change in internal, kinetic, potential, etc. energies& & & mh1 = Qout + mh2 (since ke pe 0) & & Qout = mc p (T1 T2 )90 kg/min AIRSubstituting, Q& & Q out = mc p T = (90 kg/min )(1.007 kJ/kg C )(3C ) = 272 kJ/minPROPRIETARY MATERIAL. 2007 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. 17. 1-161-34E 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 -222F 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 psiaft3/lbmR (Table A-1E). Also, cp = 0.240 Btu/lbmR 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 m CV = 0 and E CV = 0 , there is only one inlet and one exit and thus & & & m1 = m2 = m , and heat is lost from the system. The energy balance for this steady-flow system can be expressed in the rate form as & & E in E out 1 24 4 3& E system 0 (steady) 1442443 4=Rate of net energy transfer by heat, work, and mass& & = 0 E in = E outRate of change in internal, kinetic, potential, etc. energies& & & Qin + mh1 = mh2 (since ke pe 0) & & Qin = mc p (T2 T1 )450 ft3/min(a) The inlet velocity of air through the duct is determined from V1 =V&1 A1=V&1 r 2=450 ft 3 /min (5 / 12 ft) 2AIRD = 10 in2 Btu/s= 825 ft/min(b) The mass flow rate of air becomesRT1 (0.3704 psia ft 3 /lbm R )(510 R ) = = 12.6 ft 3 /lbm P1 15 psia V& 450 ft 3 /min & = 35.7 lbm/min = 0.595 lbm/s m= 1 = v 1 12.6 ft 3 /lbmv1 =Then the exit temperature of air is determined to be T2 = T1 +& Qin 2 Btu/s = 50F + = 64.0F & mc p (0.595 lbm/s)(0.240 Btu/lbm F)PROPRIETARY MATERIAL. 2007 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. 18. 1-171-35 Water is heated in an insulated tube by an electric resistance heater. The mass flow rate of water through the heater is to be determined. Assumptions 1 Water is an incompressible substance with a constant specific heat. 2 The kinetic and potential energy changes are negligible, ke pe 0. 3 Heat loss from the insulated tube is negligible. Properties The specific heat of water at room temperature is cp = 4.18 kJ/kgC. Analysis We take the tube 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 m CV = 0 and E CV = 0 , there is only one inlet and one exit and thus m1 = m2 = m , and the tube is insulated. The energy balance for this steady-flow system can be expressed in the rate form as & & E in E out 1 24 4 3Rate of net energy transfer by heat, work, and mass& E system 0 (steady) 1442443 4=& & = 0 E in = E outRate of change in internal, kinetic, potential, etc. energies& & & We,in + mh1 = mh2 (since ke pe 0) & & We,in = mc p (T2 T1 )Thus,WATER 15C70C7 kW & m=& W e,in c p (T2 T1 )=7 kJ/s = 0.0304 kg/s (4.18 kJ/kg C)(70 15)CPROPRIETARY MATERIAL. 2007 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. 19. 1-18Heat Transfer Mechanisms 1-36C The house with the lower rate of heat transfer through the walls will be more energy efficient. Heat conduction is proportional to thermal conductivity (which is 0.72 W/m.C for brick and 0.17 W/m.C for wood, Table 1-1) and inversely proportional to thickness. The wood house is more energy efficient since the wood wall is twice as thick but it has about one-fourth the conductivity of brick wall. 1-37C 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. 1-38C 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. 1-39C 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. 1-40C 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. dT & where dT/dx is the 1-41C Conduction is expressed by Fourier's law of conduction as Qcond = kA dx 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 Qconv = hAs (Ts T ) where h is the convection heat transfer coefficient, As 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. 4 & Radiation is expressed by Stefan-Boltzman law as Qrad = As (Ts4 Tsurr ) where is the emissivity of surface, As is the surface area, Ts is the surface temperature, Tsurr is the average surrounding surface temperature and = 5.67 10 8 W/m 2 K 4 is the Stefan-Boltzman constant.1-42C Convection involves fluid motion, conduction does not. In a solid we can have only conduction. 1-43C No. It is purely by radiation. 1-44C In forced convection the fluid is forced to move by external means such as a fan, pump, or the wind. The fluid motion in natural convection is due to buoyancy effects only.PROPRIETARY MATERIAL. 2007 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. 20. 1-191-45C Emissivity is the ratio of the radiation emitted by a surface to the radiation emitted by a blackbody at the same temperature. Absorptivity is the fraction of radiation incident on a surface that is absorbed by the surface. The Kirchhoff's law of radiation states that the emissivity and the absorptivity of a surface are equal at the same temperature and wavelength. 1-46C A blackbody is an idealized body which emits the maximum amount of radiation at a given temperature and which absorbs all the radiation incident on it. Real bodies emit and absorb less radiation than a blackbody at the same temperature. 1-47C No. Such a definition will imply that doubling the thickness will double the heat transfer rate. The equivalent but more correct unit of thermal conductivity is Wm/m2C that indicates product of heat transfer rate and thickness per unit surface area per unit temperature difference. 1-48C In a typical house, heat loss through the wall with glass window will be larger since the glass is much thinner than a wall, and its thermal conductivity is higher than the average conductivity of a wall. 1-49C Diamond is a better heat conductor. 1-50C The rate of heat transfer through both walls can be expressed asT T T T & Q wood = k wood A 1 2 = (0.16 W/m C) A 1 2 = 1.6 A(T1 T2 ) L wood 0.1 m T T2 T T2 & Q brick = k brick A 1 = (0.72 W/m C) A 1 = 2.88 A(T1 T2 ) Lbrick 0.25 m where thermal conductivities are obtained from Table A-5. Therefore, heat transfer through the brick wall will be larger despite its higher thickness. 1-51C The thermal conductivity of gases is proportional to the square root of absolute temperature. The thermal conductivity of most liquids, however, decreases with increasing temperature, with water being a notable exception. 1-52C Superinsulations are obtained by using layers of highly reflective sheets separated by glass fibers in an evacuated space. Radiation heat transfer between two surfaces is inversely proportional to the number of sheets used and thus heat loss by radiation will be very low by using this highly reflective sheets. At the same time, evacuating the space between the layers forms a vacuum under 0.000001 atm pressure which minimize conduction or convection through the air space between the layers. 1-53C Most ordinary insulations are obtained by mixing fibers, powders, or flakes of insulating materials with air. Heat transfer through such insulations is by conduction through the solid material, and conduction or convection through the air space as well as radiation. Such systems are characterized by apparent thermal conductivity instead of the ordinary thermal conductivity in order to incorporate these convection and radiation effects. 1-54C The thermal conductivity of an alloy of two metals will most likely be less than the thermal conductivities of both metals.PROPRIETARY MATERIAL. 2007 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. 21. 1-201-55 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.Brick wallProperties The thermal conductivity of the wall is given to be k = 0.69 W/mC. Analysis Under steady conditions, the rate of heat transfer through the wall is0.3 m 5C20C(20 5)C T & = (0.69 W/m C)(4 7 m 2 ) = 966 W Qcond = kA 0.3 m L1-56 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. Properties The thermal conductivity of the glass is given to be k = 0.78 W/mC.GlassAnalysis Under steady conditions, the rate of heat transfer through the glass by conduction is (10 3)C T & Qcond = kA = (0.78 W/m C)(2 2 m 2 ) = 4368 W L 0.005mThen the amount of heat transfer over a period of 5 h becomes& Q = Qcond t = (4.368 kJ/s)(5 3600 s) = 78,620 kJ 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.10C3C 0.5 cmPROPRIETARY MATERIAL. 2007 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. 22. 1-211-57 EES Prob. 1-56 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)L [m] 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01Qcond [kJ] 393120 196560 131040 98280 78624 65520 56160 49140 43680 39312400000 350000 300000Q cond [kJ]250000 200000 150000 100000 50000 0 0.0020.0040.0060.0080.01L [m ]PROPRIETARY MATERIAL. 2007 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. 23. 1-221-58 Heat is transferred steadily to boiling water in the pan through its bottom. The inner surface temperature of the bottom of the pan is given. The temperature of the outer surface is to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the pan remain constant at the specified values. 2 Thermal properties of the aluminum pan are constant. Properties The thermal conductivity of the aluminum is given to be k = 237 W/mC. Analysis The heat transfer area isA = r2 = (0.075 m)2 = 0.0177 m2 Under steady conditions, the rate of heat transfer through the bottom of the pan by conduction is T T T & = kA 2 1 Q = kA L LSubstituting, 800 W = (237 W/m C)(0.0177 m 2 )T2 105C 0.004 m105Cwhich gives800 WT2 = 105.76C0.4 cm1-59E The inner and outer surface temperatures of the wall of an electrically heated home during a winter night are measured. The rate of heat loss through the wall that night and its cost are to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values during the entire night. 2 Thermal properties of the wall are constant. Properties The thermal conductivity of the brick wall is given to be k = 0.42 Btu/hftF. Analysis (a) Noting that the heat transfer through the wall is by conduction and the surface area of the wall is A = 20 ft 10 ft = 200 ft 2 , the steady rate of heat transfer through the wall can be determined from T T2 (62 25)F & = (0.42 Btu/h.ft.F)(200 ft 2 ) = 3108 Btu/h Q = kA 1 1 ft Lor 0.911 kW since 1 kW = 3412 Btu/h.Brick Wall(b) The amount of heat lost during an 8 hour period and its cost are & Q = Qt = (0.911 kW)(8 h) = 7.288 kWh Cost = (Amount of energy)(Unit cost of energy) = (7.288 kWh)($0.07/kWh) = $0.51Q1 ft62F25FTherefore, the cost of the heat loss through the wall to the home owner that night is $0.51.PROPRIETARY MATERIAL. 2007 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. 24. 1-231-60 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 wellinsulated, and thus the entire heat generated by the heater is conducted through the samples. 3 The apparatus possesses thermal symmetry. Analysis The electrical power consumed by the heater and converted to heat isQ& W e = VI = (110 V )(0.6 A ) = 66 WThe rate of heat flow through each sample is& 66 W & W Q= e = = 33 W 2 23 cmThen the thermal conductivity of the sample becomes A=D 2 4= (0.04 m) 2 43 cm = 0.001257 m2& QL (33 W)(0.03 m) T & Q = kA k = = = 78.8 W/m.C L AT (0.001257 m 2 )(10C)1-61 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 wellinsulated, 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 = 25 / 2 = 12.5 W& Q& QA = (0.1 m)(0.1 m) = 0.01 m 2 T = 82 74 = 8C Then the thermal conductivity of the material becomes & (12.5 W)(0.005 m) QL T & k= = = 0.781 W/m.C Q = kA L AT (0.01 m 2 )(8C)LLAPROPRIETARY MATERIAL. 2007 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. 25. 1-241-62 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 wellinsulated, 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 & = 20 / 2 = 10 W QA = (0.1 m)(0.1 m) = 0.01 m 2 T = 82 74 = 8C LThen the thermal conductivity of the material becomes & (10 W)(0.005 m) QL T & k = = = 0.625 W/m C Q = kA L AT (0.01 m 2 )(8C)LA1-63 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. 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 Fouriers relation to be 15C& q=k& qL (25 W/m )(0.03 m) T k = = = 0.09375 W/m C T (15 7)C L 2Door& q7C L = 3 cm1-64 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 = 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 & Q = A (T 4 T 4 ) T radsssurr= (0.95)(5.67 10surr8W/m .K )(1.6 m )[(32 + 273) (296 K) ]K 242444= 84.2 W(b) Winter: Tsurr = 12+273= 285 K 4 & Q rad = As (Ts4 Tsurr )Qrad= (0.95)(5.67 10 8 W/m 2 .K 4 )(1.6 m 2 )[(32 + 273) 4 (285 K) 4 ]K 4 = 177.2 WDiscussion Note that the radiation heat transfer from the person more than doubles in winter.PROPRIETARY MATERIAL. 2007 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. 26. 1-251-65 EES Prob. 1-64 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] 281 282 283 284 285 286 287 288 289 290 291Qrad, winter [W] 208.5 200.8 193 185.1 177.2 169.2 161.1 152.9 144.6 136.2 127.8210 200 190 180Q rad,w inter [W ]170 160 150 140 130 120 281283285287289291T surr,w inter [K]PROPRIETARY MATERIAL. 2007 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. 27. 1-261-66 A person is standing in a room at a specified temperature. The rate of heat transfer between a person and the surrounding air by convection is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is not considered. 3 The environment is at a uniform temperature.Tair QconvAnalysis The heat transfer surface area of the person isAs = DL= (0.3 m)(1.70 m) = 1.602 m2 Under steady conditions, the rate of heat transfer by convection is Room air& Qconv = hAs T = (20 W/m 2 C)(1.602 m 2 )(34 18)C = 513 W1-67 Hot air is blown over a flat surface at a specified temperature. The rate of heat transfer from the air to the plate 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. Analysis Under steady conditions, the rate of heat transfer by convection is80C Air 30C& Qconv = hAs T = (55 W/m 2 C)(2 4 m 2 )(80 30)C = 22,000 WPROPRIETARY MATERIAL. 2007 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. 28. 1-271-68 EES Prob. 1-67 is reconsidered. The rate of heat transfer as a function of the heat transfer coefficient is to be plotted. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" T_infinity=80 [C] A=2*4 [m^2] T_s=30 [C] h=55 [W/m^2-C] "ANALYSIS" Q_dot_conv=h*A*(T_infinity-T_s) h [W/m2.C] 20 30 40 50 60 70 80 90 100Qconv [W] 8000 12000 16000 20000 24000 28000 32000 36000 4000040000 35000 30000Q conv [W ]25000 20000 15000 10000 5000 20304050607080901002h [W /m -C]PROPRIETARY MATERIAL. 2007 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. 29. 1-281-69 The heat generated in the circuitry on the surface of a 3-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/mC. Analysis The temperature difference between the front and back surfaces of the chip isA = (0.006 m)(0.006 m) = 0.000036 m 2& Q 3WCeramic T & substrate Q = kA L &L (3 W)(0.0005 m) Q T = = = 0.32C kA (130 W/m C)(0.000036 m 2 )Chip 6 6 0.5 mm1-70 An electric resistance heating element is immersed in water initially at 20C. The time it will take for this heater to raise the water temperature to 80C 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/kgC (Table A-9). & & Analysis When steady operating conditions are reached, we have Q = E = 800 W . This is also generatedequal 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 20C to 80C is determined to be Qin = mc(T2 T1 ) & Qin t = mc(T2 T1 ) t =mc(T2 T1 ) (75 kg)(4180 J/kg C)(80 20)C = = 23,510 s = 6.53 h & 800 J/s Qwater 800 W120CinThe surface area of the wire is As = DL = (0.005 m)(0.4 m) = 0.00628 m 2& The Newton's law of cooling for convection heat transfer is expressed as Q = hAs (Ts T ) . 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 & Q 800 W = = 1274 W/m 2 C As (Ts T1 ) (0.00628 m 2 )(120 20)C & Q 800 W = = 3185 W/m 2 C h2 = 2 As (Ts T 2 ) (0.00628 m )(120 80)C h1 =Discussion Note that a larger heat transfer coefficient is needed to dissipate heat through a smaller temperature difference for a specified heat transfer rate.PROPRIETARY MATERIAL. 2007 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. 30. 1-291-71 A hot water pipe at 80C is losing heat to the surrounding air at 5C by natural convection with a heat transfer coefficient of 25 W/m2C. 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.80CAnalysis The heat transfer surface area isAs = DL = (0.05 m)(10 m) = 1.571 m2D =5 cmUnder steady conditions, the rate of heat transfer L = 10 m by convection is & Q = hA T = (25W/m 2 C)(1.571 m 2 )(80 5)C = 2945 W convQ Air, 5Cs1-72 A hollow spherical iron container is filled with iced water at 0C. 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, 0C. Properties The thermal conductivity of iron is k = 80.2 W/mC (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 areaA = D2 = (0.2 m)2 = 0.126 m25CThen the rate of heat transfer through the shell by conduction is (5 0)C T & Qcond = kA = (80.2 W/m C)(0.126 m 2 ) = 12,632 W L 0.004 mConsidering that it takes 333.7 kJ of energy to melt 1 kg of ice at 0C, the rate at which ice melts in the container can be determined from & Q 12.632 kJ/s & = = 0.038 kg/s mice = 333.7 kJ/kg hifIced water 0C0.4 cmDiscussion 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 = 19.2 cm) or the mean surface area (D = 19.6 cm) in the calculations.PROPRIETARY MATERIAL. 2007 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. 31. 1-301-73 EES Prob. 1-72 is reconsidered. The rate at which ice melts as a function of the container thickness is to be plotted. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" D=0.2 [m] L=0.4 [cm] T_1=0 [C] T_2=5 [C] "PROPERTIES" h_if=333.7 [kJ/kg] k=k_('Iron', 25) "ANALYSIS" A=pi*D^2 Q_dot_cond=k*A*(T_2-T_1)/(L*Convert(cm, m)) m_dot_ice=(Q_dot_cond*Convert(W, kW))/h_ifL [cm] 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2mice [kg/s] 0.07574 0.03787 0.02525 0.01894 0.01515 0.01262 0.01082 0.009468 0.008416 0.0075740.08 0.07 0.06m ice [kg/s]0.05 0.04 0.03 0.02 0.01 0 0.20.40.60.811.21.41.61.82L [cm ]PROPRIETARY MATERIAL. 2007 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. 32. 1-311-74E The inner and outer glasses of a double pane window with a 0.5-in air space are at specified temperatures. The rate of heat transfer through the window is to be determined Assumptions 1 Steady operating conditions exist since the Glass surface temperatures of the glass remain constant at the specified values. 2 Heat transfer through the window is onedimensional. 3 Thermal properties of the air are constant. Properties The thermal conductivity of air at the average Air temperature of (60+48)/2 = 54F is k = 0.01419 Btu/hftF (Table A-15E). & Q Analysis The area of the window and the rate of heat loss through it areA = (4 ft) (4 ft) = 16 m 260F 48FT T2 (60 48)F & Q = kA 1 = (0.01419 Btu/h.ft. F)(16 ft 2 ) = 131 Btu/h L 0.25 / 12 ft1-75 Two surfaces of a flat plate are maintained at specified temperatures, and the rate of heat transfer through the plate is measured. The thermal conductivity of the plate material is to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the plate remain constant at the specified values. 2 Heat Plate transfer through the plate is one-dimensional. 3 Thermal properties of the plate are constant. Q Analysis The thermal conductivity is determined directly from the steady one-dimensional heat conduction relation to be & T T (Q / A) L (500 W/m 2 )(0.02 m) & = = 0.125 W/m C Q = kA 1 2 k = 0C 80C (T T ) (80 0)C L 121-76 Four power transistors are mounted on a thin vertical aluminum plate that is cooled by a fan. The temperature of the aluminum plate is to be determined. Assumptions 1 Steady operating conditions exist. 2 The entire plate is nearly isothermal. 3 Thermal properties of the wall are constant. 4 The exposed surface area of the transistor can be taken to be equal to its base area. 5 Heat transfer by radiation is disregarded. 6 The convection heat transfer coefficient is constant and uniform over the surface. Analysis The total rate of heat dissipation from the aluminum plate and the total heat transfer area are & Q = 4 15 W = 60 WAs = (0.22 m)(0.22 m) = 0.0484 m 2 Disregarding any radiation effects, the temperature of the aluminum plate is determined to be15 W Ts& Q 60 W & = 25C + = 74.6C Q = hAs (Ts T ) Ts = T + 2 hAs (25 W/m C)(0.0484 m 2 )PROPRIETARY MATERIAL. 2007 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. 33. 1-321-77 A styrofoam ice chest is initially filled with 40 kg of ice at 0C. The time it takes for the ice in the chest to melt completely is to be determined. Assumptions 1 Steady operating conditions exist. 2 The inner and outer surface temperatures of the ice chest remain constant at 0C and 8C, respectively, at all times. 3 Thermal properties of the chest are constant. 4 Heat transfer from the base of the ice chest is negligible. Properties The thermal conductivity of the styrofoam is given to be k = 0.033 W/mC. The heat of fusion of ice at 0C is 333.7 kJ/kg. Analysis Disregarding any heat loss through the bottom of the ice chest and using the average thicknesses, the total heat transfer area becomesA = (40 3)(40 3) + 4 (40 3)(30 3) = 5365 cm 2 = 0.5365 m 2 The rate of heat transfer to the ice chest becomes (8 0)C T & Q = kA = (0.033 W/m C)(0.5365 m 2 ) = 4.72 W L 0.03 mThe total amount of heat needed to melt the ice completely isIce chest, 0C& QQ = mhif = (28 kg)(333.7 kJ/kg) = 9344 kJ3 cmThen transferring this much heat to the cooler to melt the ice completely will take t =Q 9344,000 J = = 1.98 10 6 s = 22.9 days & 4.72 J/s Q1-78 A transistor mounted on a circuit board is cooled by air flowing over it. The transistor case temperature is not to exceed 70C when the air temperature is 55C. The amount of power this transistor can dissipate safely is to be determined. Air, Assumptions 1 Steady operating conditions exist. 2 Heat 55C transfer by radiation is disregarded. 3 The convection heat transfer coefficient is constant and uniform over the surface. 4 Heat transfer from the base of the transistor is negligible. Power Analysis Disregarding the base area, the total heat transfer area transistor of the transistor isAs = DL + D 2 / 4 = (0.6 cm)(0.4 cm) + (0.6 cm) 2 / 4 = 1.037 cm 2 = 1.037 10 4 m 2 Then the rate of heat transfer from the power transistor at specified conditions is& Q = hAs (Ts T ) = (30 W/m2 C)(1.03710-4 m2 )(70 55)C = 0.047 W Therefore, the amount of power this transistor can dissipate safely is 0.047 W.PROPRIETARY MATERIAL. 2007 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. 34. 1-331-79 EES Prob. 1-78 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] 60 62.5 65 67.5 70 72.5 75 77.5 80 82.5 85 87.5 90Q [W] 0.01555 0.02333 0.0311 0.03888 0.04665 0.05443 0.0622 0.06998 0.07775 0.08553 0.09331 0.1011 0.10890.120.1Q [W ]0.080.060.040.020 60657075808590T case,m ax [C]PROPRIETARY MATERIAL. 2007 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. 35. 1-341-80E A 200-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 isAs = DL = (4 / 12 ft)(200 ft) = 209.4 ft 2280F D =4 in L=200 ftQAir,50F& Q pipe = hAs (Ts Tair ) = (6 Btu/h ft 2 F)(209.4 ft 2 )(280 50)F = 289,000 Btu/h(b) The amount of heat loss per year is & Q = Qt = (289,000 Btu/h)(365 24 h/yr) = 2.531 10 9 Btu/yr The amount of gas consumption per year in the furnace that has an efficiency of 86% isAnnual Energy Loss =2.531 10 9 Btu/yr 1 therm 100,000 Btu = 29,435 therms/yr 0.86 Then the annual cost of the energy lost becomesEnergy cost = (Annual energy loss)(Unit cost of energy) = (29,435 therms/yr)($1.10 / therm) = $32,380/yr1-81 A 4-m diameter spherical tank filled with liquid nitrogen at 1 atm and -196C 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 thinshelled 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 isVaporAs = D 2 = (4 m) 2 = 50.27 m 2 & Q = hAs (Ts Tair ) = (25 W/m 2 C)(50.27 m 2 )[20 (196)]C = 271,430 WThen the rate of evaporation of liquid nitrogen in the tank is determined to be & Q 271.430 kJ/s & & & m = = = 1.37 kg/s Q = mh fg h fg 198 kJ/kgAir 20C 1 atm& QLiquid N2-196CPROPRIETARY MATERIAL. 2007 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. 36. 1-351-82 A 4-m diameter spherical tank filled with liquid oxygen at 1 atm and -183C 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 thinshelled 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 isAs = D 2 = (4 m) 2 = 50.27 m 2 & Q = hAs (Ts Tair ) = (25 W/m 2 .C)(50.27 m 2 )[20 (183)]CAir 20C= 255,120 WThen the rate of evaporation of liquid oxygen in the tank is determined to be & Q 255.120 kJ/s & & & m = = = 1.20 kg/s Q = mh fg h fg 213 kJ/kg& Q1 atm Liquid O2 -183CPROPRIETARY MATERIAL. 2007 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. 37. 1-361-83 EES Prob. 1-81 is reconsidered. The rate of evaporation of liquid nitrogen as a function of the ambient air temperature is to be plotted. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" D=4 [m] T_s=-196 [C] T_air=20 [C] h=25 [W/m^2-C] "PROPERTIES" h_fg=198 [kJ/kg] "ANALYSIS" A=pi*D^2 Q_dot=h*A*(T_air-T_s) m_dot_evap=(Q_dot*Convert(J/s, kJ/s))/h_fgTair [C]mevap [kg/s]0 2.5 5 7.5 10 12.5 15 17.5 20 22.5 25 27.5 30 32.5 351.244 1.26 1.276 1.292 1.307 1.323 1.339 1.355 1.371 1.387 1.403 1.418 1.434 1.45 1.4661.51.45m evap [kg/s]1.41.351.31.251.2 05101520253035T air [C]PROPRIETARY MATERIAL. 2007 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. 38. 1-371-84 A person with a specified surface temperature is subjected to radiation heat transfer in a room at specified wall temperatures. The rate of radiation heat loss from the person is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by convection is disregarded. 3 The emissivity of the person is constant and uniform over the exposed surface. Properties The average emissivity of the person is given to be 0.5. 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) Tsurr = 300 K 4 & Qrad = As (Ts4 Tsurr )Tsurr= (0.5)(5.67 108 W/m 2 .K 4 )(1.7 m 2 )[(32 + 273) 4 (300 K)4 ]K 4 = 26.7 W(b) Tsurr = 280 KQrad4 & Q rad = As (Ts4 Tsurr )32C= (0.5)(5.67 10 8 W/m 2 .K 4 )(1.7 m 2 )[(32 + 273) 4 (280 K) 4 ]K 4 = 121 WDiscussion Note that the radiation heat transfer goes up by more than 4 times as the temperature of the surrounding surfaces drops from 300 K to 280 K.1-85 A circuit board houses 80 closely spaced logic chips on one side, each dissipating 0.06 W. All the heat generated in the chips is conducted across the circuit board. The temperature difference between the two sides of the circuit board is to be determined. Assumptions 1 Steady operating conditions exist. 2 Thermal properties of the board are constant. 3 All the heat generated in the chips is conducted across the circuit board. Properties The effective thermal conductivity of the board is given to be k = 16 W/mC. Analysis The total rate of heat dissipated by the chips is& Q = 80 (0.06 W) = 4.8 W& QChipsThen the temperature difference between the front and back surfaces of the board isA = (0.12 m)(0.18 m) = 0.0216 m2& (4.8 W)(0.003 m) QL T & T = = = 0.042C Q = kA L kA (16 W/m C)(0.0216 m 2 ) Discussion Note that the circuit board is nearly isothermal.PROPRIETARY MATERIAL. 2007 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. 39. 1-381-86 A sealed electronic box dissipating a total of 100 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 55C, 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 isAs = (0.4 m)(0.4 m) + 4 (0.2 m)(0.4 m) = 0.48 m 2 The radiation heat transfer from the box can be expressed as 4 & Qrad = As (Ts4 Tsurr )[4 100 W = (0.95)(5.67 10 8 W/m 2 K 4 )(0.48 m 2 ) (55 + 273 K ) 4 Tsurr]100 W = 0.95 Ts =55Cwhich gives Tsurr = 296.3 K = 23.3C. Therefore, the temperature of the surrounding surfaces must be less than 23.3C.1-87E Using the conversion factors between W and Btu/h, m and ft, and K and R, the Stefan-Boltzmann constant = 5.67 10 8 W/m 2 K 4 is to be expressed in the English unit, Btu/h ft 2 R 4 . Analysis The conversion factors for W, m, and K are given in conversion tables to be 1 W = 3.41214 Btu/h 1 m = 3.2808 ft 1 K = 1.8 RSubstituting gives the Stefan-Boltzmann constant in the desired units, = 5.67 W/m 2 K 4 = 5.67 3.41214 Btu/h (3.2808 ft) 2 (1.8 R) 4= 0.171 Btu/h ft 2 R 41-88E 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/hft2F. Analysis The conversion factors for W and m are straightforward, and are given in conversion tables to be 1 W = 3.41214 Btu/h 1 m = 3.2808 ftThe proper conversion factor between C into F in this case is 1C = 1.8Fsince the C in the unit W/m2C represents per C change in temperature, and 1C change in temperature corresponds to a change of 1.8F. Substituting, we get 1 W/m 2 C =3.41214 Btu/h (3.2808 ft) 2 (1.8 F)= 0.1761 Btu/h ft 2 Fwhich is the desired conversion factor. Therefore, the given convection heat transfer coefficient in English units is h = 14 W/m 2 C = 14 0.1761 Btu/h ft 2 F = 2.47 Btu/h ft 2 FPROPRIETARY MATERIAL. 2007 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. 40. 1-391-89 A cylindrical sample of a material is used to determine its thermal conductivity. The temperatures measured along the sample are tabulated. The variation of temperature along the sample is to be plotted and the thermal conductivity of the sample material is to be calculated. Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional (axial direction). Analysis The following table gives the results of the calculations. The plot of temperatures is also given below. A sample calculation for the thermal conductivity is as follows:A=D 2k12 = =4= (0.025 m) 2 4= 0.00049 m 2& Q& QL A(T1 T2 ) (83.45 W)(0.010 m) (0.00049 m 2 )(6.13C)= 277.8 W/m C)Distance from left face, cmTemperature, C0 1 2 3 4 5 6 7 8T1= 89.38 T2= 83.25 T3= 78.28 T4= 74.10 T5= 68.25 T6=63.73 T7= 49.65 T8= 44.40 T9= 40.000Temperature difference (C) T1-T2= 6.13 T2-T3= 4.97 T3-T4= 4.18 T4-T5= 5.85 T5-T6= 4.52 T6-T7= 14.08 T7-T8= 5.25 T8-T9= 4.40 T1-T2= 6.13123456 78x, cmThermal conductivity (W/mC) 277.8 342.7 407.4 291.1 376.8 120.9 324.4 387.1 277.890Temperature [C]8070605040 012345678Distance [cm]Discussion It is observed from the calculations in the table and the plot of temperatures that the temperature reading corresponding to the calculated thermal conductivity of 120.9 is probably not right, and it should be discarded.PROPRIETARY MATERIAL. 2007 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. 41. 1-401-90 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 T wing = Tice +Vhif h= 0C +(920 kg/m 3 )(0.001/60 m/s)(333,700 J/kg) 150 W/m 2 C= 34.1 CSimultaneous Heat Transfer Mechanisms1-91C All three modes of heat transfer can not occur simultaneously in a medium. A medium may involve two of them simultaneously. 1-92C (a) Conduction and convection: No. (b) Conduction and radiation: Yes. Example: A hot surface on the ceiling. (c) Convection and radiation: Yes. Example: Heat transfer from the human body. 1-93C The human body loses heat by convection, radiation, and evaporation in both summer and winter. In summer, we can keep cool by dressing lightly, staying in cooler environments, turning a fan on, avoiding humid places and direct exposure to the sun. In winter, we can keep warm by dressing heavily, staying in a warmer environment, and avoiding drafts. 1-94C The fan increases the air motion around the body and thus the convection heat transfer coefficient, which increases the rate of heat transfer from the body by convection and evaporation. In rooms with high ceilings, ceiling fans are used in winter to force the warm air at the top downward to increase the air temperature at the body level. This is usually done by forcing the air up which hits the ceiling and moves downward in a gently manner to avoid drafts.PROPRIETARY MATERIAL. 2007 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. 42. 1-411-95 The total rate of heat transfer from a person by both convection and radiation to the surrounding air and surfaces at specified temperatures is to be determined. Assumptions 1 Steady operating conditions exist. 2 The person is Tsurr completely surrounded by the interior surfaces of the room. 3 The surrounding surfaces are at the same temperature as the air in the 23C room. 4 Heat conduction to the floor through the feet is negligible. 5 The convection coefficient is constant and uniform over the entire surface of the person. Qrad Properties The emissivity of a person is given to be = 0.9. 32C Analysis The person is completely enclosed by the surrounding =0.9 surfaces, and he or she will lose heat to the surrounding air by Qconv convection and to the surrounding surfaces by radiation. The total rate of heat loss from the person is determined from & Q = A (T 4 T 4 ) = (0.90)(5.67 10 8 W/m 2 .K 4 )(1.7 m 2 )[(32 + 273) 4 (23 + 273) 4 ]K 4 = 84.8 W radsssurr& Qconv = hAs T = (5W/m 2 K)(1.7m 2 )(32 23)C = 76.5W And& & & Q total = Qconv + Qrad = 84.8 + 76.5 = 161.3 WDiscussion Note that heat transfer from the person by evaporation, which is of comparable magnitude, is not considered in this problem.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 = 1. 4 There are no convection currents in the air space between the plates. Properties The thermal conductivities are k = 0.00015 W/mC for super insulation, k = 0.01979 W/mC at -50C (Table A-15) for air, and k = 0.036 W/mC for fiberglass insulation (Table A-6). Analysis (a) Disregarding any natural convection currents, the rates of conduction and radiation heat transfer T2 T1 T1 T2 2 2 ( 290 150) K & Qcond = kA = (0.01979 W/m C)(1 m ) = 139 W L 0.02 m & Q = A (T 4 T 4 ) rads12= 1(5.67 10 & & =Q +Q8total[]radW/m 2 K 4 )(1m 2 ) (290 K ) 4 (150 K ) 4 = 372 WQ& Q total cond rad = 139 + 372 = 511 W (b) When the air space between the plates is evacuated, there will be radiation heat transfer only. Therefore, & & Q = Q = 372 W2 cm (c) In this case there will be conduction heat transfer through the fiberglass insulation only, T T2 ( 290 150) K & & Q total = Qcond = kA 1 = (0.036 W/m o C)(1 m 2 ) = 252 W 0.02 m L (d) In the case of superinsulation, the rate of heat transfer will be T T2 ( 290 150) K & & Q total = Qcond = kA 1 = (0.00015 W/m C)(1 m 2 ) = 1.05 W 0.02 m L Discussion Note that superinsulators are very effective in reducing heat transfer between to surfaces.PROPRIETARY MATERIAL. 2007 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. 43. 1-421-97 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 0.8.150 W/m227C44CAnalysis The heat transfer through the wall by conduction is equal to net heat transfer to the outer wall surface:s = = 0.8 air, 40C h . Qrad& & & & q cond = q conv + q rad + q solarT2 T1 4 = h(To T2 ) + (Tsurr T24 ) + s q solar L (44 - 27)C = (8 W/m 2 C)(40 44)C + (0.8)(5.67 10 -8 W/m 2 K 4 ) (40 + 273 K ) 4 (44 + 273 K ) 4 k 0.25 m k[]+ (0.8)(150 W/m 2 )Solving for k gives k = 0.961 W/m C1-98 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, & & Q = E generated = VI = (110 V)(3 A) = 330 W240CThe surface area of the wire is D =0.2 cmAs = DL = (0.002 m)(1.4 m) = 0.00880 m 2 The Newton's law of cooling for convection heat transfer is expressed asL = 1.4 mQAir, 20C& Q = hAs (Ts T ) Disregarding any heat transfer by radiation, the convection heat transfer coefficient is determined to be & Q 330 W = = 170.5 W/m 2 C h= 2 As (T1 T ) (0.00880 m )(240 20)C 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. 2007 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. 44. 1-431-99 EES Prob. 1-98 is reconsidered. The convection heat transfer coefficient as a function of the wire surface temperature is to be plotted. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" L=1.4 [m] D=0.002 [m] T_infinity=20 [C] T_s=240 [C] V=110 [Volt] I=3 [Ampere] "ANALYSIS" Q_dot=V*I A=pi*D*L Q_dot=h*A*(T_s-T_infinity)h [W/m2.C] 468.9 375.2 312.6 268 234.5 208.4 187.6 170.5 156.3 144.3 134Ts [C] 100 120 140 160 180 200 220 240 260 280 300500 450 4002h [W /m -C]350 300 250 200 150 100 100140180220260300T s [C]PROPRIETARY MATERIAL. 2007 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. 45. 1-441-100E A spherical ball whose surface is maintained at a temperature of 170F is suspended in the middle of a room at 70F. 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 = 0.8.Air 70F170FAnalysis The heat transfer surface area isD = 2 inAs = D2 = (2/12 ft) 2 = 0.08727 ft2QUnder steady conditions, the rates of convection and radiation heat transfer are & Qconv = hAs T = (15 Btu/h ft 2 F)(0.08727 ft 2 )(170 70)F = 130.9 Btu/h & Q rad = As (Ts4 To4 ) = 0.8(0.08727 ft 2 )(0.1714 10 8 Btu/h ft 2 R 4 )[(170 + 460 R) 4 (70 + 460 R) 4 ] = 9.4 Btu/hTherefore,& & & Qtotal = Qconv + Qrad = 130.9 + 9.4 = 140.3 Btu/hDiscussion 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.1-101 A 1000-W iron is left on the iron board with its base exposed to the air at 20C. 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 1000 WProperties The emissivity of the base surface is given to be = 0.6. Analysis At steady conditions, the 1000 W energy supplied to the iron will be dissipated to the surroundings by convection and radiation heat transfer. Therefore,& & & Q total = Qconv + Qrad = 1000 W where and& Qconv = hAs T = (35 W/m 2 K)(0.02 m 2 )(Ts 293 K) = 0.7(Ts 293 K) & Qrad = As (Ts4 To4 ) = 0.6(0.02 m 2 )(5.67 10 8 W/m 2 K 4 )[Ts4 (293 K) 4 ]= 0.06804 10 8 [Ts4 (293 K) 4 ]Substituting,1000 W = 0.7(Ts 293 K ) + 0.06804 10 8 [Ts4 (293 K) 4 ]Solving by trial and error gives T s = 947 K = 674CDiscussion We note that the iron will dissipate all the energy it receives by convection and radiation when its surface temperature reaches 947 K.PROPRIETARY MATERIAL. 2007 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. 46. 1-451-102 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. 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 from950 W/m2& & Qsolar absorbed = Q rad 4 & Qsolar = As (Ts4 Tspace )0.3 As (950 W/m 2 ) = 0.8 As (5.67 10 8 W/m 2 K 4 )[Ts4 (0 K) 4 ] = 0.3 = 0.8Canceling the surface area A and solving for Ts gives.QradTs = 281.5 K1-103 A spherical tank located outdoors is used to store iced water at 0C. The rate of heat transfer to the iced water in the tank and the amount of ice at 0 C that melts during a 24-h period are 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 tank and the convection heat transfer coefficient is constant and uniform. 3 The average surrounding surface temperature for radiation exchange is 15C. 4 The thermal resistance of the tank is negligible, and the entire steel tank is at 0C. Properties The heat of fusion of water at atmospheric pressure is hif = 333.7 kJ/kg . The emissivity of the outer surface of thetank is 0.75.Air 25CAnalysis (a) The outer surface area of the spherical tank isAs = D 2 = (3.02 m) 2 = 28.65 m 2 Then the rates of heat transfer to the tank by convection and radiation become& Q0C Iced water 0C1 cm& Qconv = hAs (T Ts ) = (30 W/m 2 C)(28.65 m 2 )(25 0)C = 21,488 W 4 & Q rad = As (Tsurr Ts4 ) = (0.75)(28.65 m 2 )(5.67 10 -8 W/m 2 K 4 )[(288 K) 4 (273 K ) 4 ] = 1614 W & & & Q =Q + Q = 21,488 + 1614 = 23,102 W = 23.1 kW totalconvrad(b) The amount of heat transfer during a 24-hour period is& Q = Qt = (23.102 kJ/s)(24 3600 s) = 1,996,000 kJ Then the amount of ice that melts during this period becomes Q = mhif m =Q 1,996,000 kJ = = 5980 kg 333.7 kJ/kg hifDiscussion The amount of ice that melts can be reduced to a small fraction by insulating the tank.PROPRIETARY MATERIAL. 2007 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. 47. 1-461-104 The roof of a house with a gas furnace consists of a 15-cm thick concrete that is losing heat to the outdoors by radiation and convection. The rate of heat transfer through the roof and the money lost through the roof that night during a 14 hour period are to be determined. Assumptions 1 Steady operating conditions exist. 2 The emissivity and thermal conductivity of the roof are constant. Properties The thermal conductivity of the concrete is given to be k = 2 W/mC. The emissivity of the outer surface of the roof is given to be 0.9. Analysis In steady operation, heat transfer from the outer surface of the roof to the surroundings by convection and radiation must be equal to the heat transfer through the roof by conduction. That is, & & & Q=Q =Q roof, condroof to surroundin gs, conv + radThe inner surface temperature of the roof is given to be Ts,in = 15C. Letting Ts,out denote the outer surface temperatures of the roof, the energy balance above can be expressed as Ts,in Ts,out & Q = kA = ho A(Ts,out Tsurr ) + A (Ts,out 4 Tsurr 4 ) L & Tsky = 255 K Q 15C Ts,out & Q = (2 W/m C)(300 m 2 ) 0.15 m = (15 W/m 2 .C)(300 m 2 )(Ts,out 10)C[+ (0.9)(300 m 2 )(5.67 10 8 W/m 2 K 4 ) (Ts,out + 273 K) 4