Page 1 Chapter 1 Basics of Heat Transfer 1-1 Chapter 1 BASICS OF HEAT TRANSFER 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. Page 2 Chapter 1 Basics of Heat Transfer 1-1 Chapter 1 BASICS OF HEAT TRANSFER 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. Chapter 1 Basics of Heat Transfer 1-2 Heat 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 ? ? A dA q Q ? ? . 1-9C 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 mC p ?T at constant pressure and mC p ?T at constant volume, and C p is always greater than C v. 1-12 A cylindrical resistor on a circuit board dissipates 0.6 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 kJ 51.84 = Wh 14.4 ? ? ? ? h) W)(24 6 . 0 ( t Q Q ? (since 1 Wh = 3600 Ws = 3.6 kJ) (b) The heat flux on the surface of the resistor is 2 2 2 cm 136 . 2 885 . 1 251 . 0 cm) cm)(1.5 4 . 0 ( 4 cm) 4 . 0 ( 2 4 2 ? ? ? ? ? ? ? ? ? ? ? DL D A s 2 W/cm 0.2809 ? ? ? 2 cm 136 . 2 W 60 . 0 s s A Q q ? ? (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 becomes Q Q A A top base total top base total or (11.8%) ? ? ? ? ? 0 251 2136 . . 0.118 Discussion Heat transfer from the top and bottom surfaces is small relative to that transferred from the side surface. Q ? Resistor 0.6 W Page 3 Chapter 1 Basics of Heat Transfer 1-1 Chapter 1 BASICS OF HEAT TRANSFER 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. Chapter 1 Basics of Heat Transfer 1-2 Heat 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 ? ? A dA q Q ? ? . 1-9C 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 mC p ?T at constant pressure and mC p ?T at constant volume, and C p is always greater than C v. 1-12 A cylindrical resistor on a circuit board dissipates 0.6 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 kJ 51.84 = Wh 14.4 ? ? ? ? h) W)(24 6 . 0 ( t Q Q ? (since 1 Wh = 3600 Ws = 3.6 kJ) (b) The heat flux on the surface of the resistor is 2 2 2 cm 136 . 2 885 . 1 251 . 0 cm) cm)(1.5 4 . 0 ( 4 cm) 4 . 0 ( 2 4 2 ? ? ? ? ? ? ? ? ? ? ? DL D A s 2 W/cm 0.2809 ? ? ? 2 cm 136 . 2 W 60 . 0 s s A Q q ? ? (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 becomes Q Q A A top base total top base total or (11.8%) ? ? ? ? ? 0 251 2136 . . 0.118 Discussion Heat transfer from the top and bottom surfaces is small relative to that transferred from the side surface. Q ? Resistor 0.6 W Chapter 1 Basics of Heat Transfer 1-3 1-13E 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 is Q Q t ? ? ? ? ? ( )( ) ? 3 8 24 W h Wh 0.024 kWh (b) The heat flux on the surface of the chip is 2 W/in 37.5 ? ? ? 2 in 08 . 0 W 3 s s A Q q ? ? 1-14 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 are 2 cm 785 . 0 ) cm 5 )( cm 05 . 0 ( ? ? ? ? ?DL A s 2 6 W/m 10 1.91 ? ? ? ? ? 2 2 W/cm 191 cm 785 . 0 W 150 s s A Q q ? ? (b) The heat flux on the surface of glass bulb is 2 2 2 cm 1 . 201 cm) 8 ( ? ? ? ? ?D A s 2 W/m 7500 ? ? ? ? 2 2 W/cm 75 . 0 cm 1 . 201 W 150 s s A Q q ? ? (c) The amount and cost of electrical energy consumed during a one-year period is Electricity Consumption kW)(365 8 h / yr) 438 kWh / yr Annual Cost = (438 kWh / yr)($0.08 kWh) ? ? ? ? ? ? ( . / Q t ? 015 $35.04 / yr 1-15 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 is Q Q t ? ? ? ? ( . ? 12 kW)(2 h) 2.4 kWh (b) The heat flux on the surface of the iron base is ? ( . )( Q base W) = 1080 W ? 0 9 1200 ? ? . q Q A ? ? ? base base 2 W m 1080 0 015 72,000 W / m 2 (c) The cost of electricity consumed during this period is Cost of electricity = (2.4 kWh) ($0.07 kWh) ? ? / $0.17 Logic chip W 3 ? Q ? Iron 1200 W Q ? Lamp 150 W Page 4 Chapter 1 Basics of Heat Transfer 1-1 Chapter 1 BASICS OF HEAT TRANSFER 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. Chapter 1 Basics of Heat Transfer 1-2 Heat 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 ? ? A dA q Q ? ? . 1-9C 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 mC p ?T at constant pressure and mC p ?T at constant volume, and C p is always greater than C v. 1-12 A cylindrical resistor on a circuit board dissipates 0.6 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 kJ 51.84 = Wh 14.4 ? ? ? ? h) W)(24 6 . 0 ( t Q Q ? (since 1 Wh = 3600 Ws = 3.6 kJ) (b) The heat flux on the surface of the resistor is 2 2 2 cm 136 . 2 885 . 1 251 . 0 cm) cm)(1.5 4 . 0 ( 4 cm) 4 . 0 ( 2 4 2 ? ? ? ? ? ? ? ? ? ? ? DL D A s 2 W/cm 0.2809 ? ? ? 2 cm 136 . 2 W 60 . 0 s s A Q q ? ? (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 becomes Q Q A A top base total top base total or (11.8%) ? ? ? ? ? 0 251 2136 . . 0.118 Discussion Heat transfer from the top and bottom surfaces is small relative to that transferred from the side surface. Q ? Resistor 0.6 W Chapter 1 Basics of Heat Transfer 1-3 1-13E 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 is Q Q t ? ? ? ? ? ( )( ) ? 3 8 24 W h Wh 0.024 kWh (b) The heat flux on the surface of the chip is 2 W/in 37.5 ? ? ? 2 in 08 . 0 W 3 s s A Q q ? ? 1-14 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 are 2 cm 785 . 0 ) cm 5 )( cm 05 . 0 ( ? ? ? ? ?DL A s 2 6 W/m 10 1.91 ? ? ? ? ? 2 2 W/cm 191 cm 785 . 0 W 150 s s A Q q ? ? (b) The heat flux on the surface of glass bulb is 2 2 2 cm 1 . 201 cm) 8 ( ? ? ? ? ?D A s 2 W/m 7500 ? ? ? ? 2 2 W/cm 75 . 0 cm 1 . 201 W 150 s s A Q q ? ? (c) The amount and cost of electrical energy consumed during a one-year period is Electricity Consumption kW)(365 8 h / yr) 438 kWh / yr Annual Cost = (438 kWh / yr)($0.08 kWh) ? ? ? ? ? ? ( . / Q t ? 015 $35.04 / yr 1-15 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 is Q Q t ? ? ? ? ( . ? 12 kW)(2 h) 2.4 kWh (b) The heat flux on the surface of the iron base is ? ( . )( Q base W) = 1080 W ? 0 9 1200 ? ? . q Q A ? ? ? base base 2 W m 1080 0 015 72,000 W / m 2 (c) The cost of electricity consumed during this period is Cost of electricity = (2.4 kWh) ($0.07 kWh) ? ? / $0.17 Logic chip W 3 ? Q ? Iron 1200 W Q ? Lamp 150 W Chapter 1 Basics of Heat Transfer 1-4 1-16 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 is W 4 . 14 W) 12 . 0 )( 120 ( ? ? Q ? kWh 0.144 ? ? ? ? h) kW)(10 0144 . 0 ( t Q Q ? (b) The heat flux on the surface of the circuit board is 2 m 03 . 0 ) m 2 . 0 )( m 15 . 0 ( ? ? s A 2 W/m 480 ? ? ? 2 m 03 . 0 W 4 . 14 s s A Q q ? ? 1-17 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 ? = 2,700 kg/m 3 and 90 . 0 ? p C kJ/kg. ?C. Analysis The amount of energy added to the ball is simply the change in its internal energy, and is determined from E U mC T T transfer ? ? ? ? ( ) 2 1 where m V D ? ? ? ? ? ? ? ? 6 6 2700 015 4 77 3 ( )( . . kg / m m) kg 3 3 Substituting, E transfer kg)(0.90 kJ / kg. C)(200-80) C = ? ? ? ( . 4 77 515 kJ 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. 1-18 The body temperature of a man rises from 37°C to 39°C 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/kg.°C. Analysis The change in the sensible internal energy content of the body as a result of the body temperature rising 2 ?C during strenuous exercise is ?U = mC ?T = (70 kg)(3.6 kJ/kg. ?C)(2 ?C) = 504 kJ Chips, 0.12 W 15 cm 20 cm Q ? Metal ball E Page 5 Chapter 1 Basics of Heat Transfer 1-1 Chapter 1 BASICS OF HEAT TRANSFER 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. Chapter 1 Basics of Heat Transfer 1-2 Heat 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 ? ? A dA q Q ? ? . 1-9C 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 mC p ?T at constant pressure and mC p ?T at constant volume, and C p is always greater than C v. 1-12 A cylindrical resistor on a circuit board dissipates 0.6 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 kJ 51.84 = Wh 14.4 ? ? ? ? h) W)(24 6 . 0 ( t Q Q ? (since 1 Wh = 3600 Ws = 3.6 kJ) (b) The heat flux on the surface of the resistor is 2 2 2 cm 136 . 2 885 . 1 251 . 0 cm) cm)(1.5 4 . 0 ( 4 cm) 4 . 0 ( 2 4 2 ? ? ? ? ? ? ? ? ? ? ? DL D A s 2 W/cm 0.2809 ? ? ? 2 cm 136 . 2 W 60 . 0 s s A Q q ? ? (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 becomes Q Q A A top base total top base total or (11.8%) ? ? ? ? ? 0 251 2136 . . 0.118 Discussion Heat transfer from the top and bottom surfaces is small relative to that transferred from the side surface. Q ? Resistor 0.6 W Chapter 1 Basics of Heat Transfer 1-3 1-13E 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 is Q Q t ? ? ? ? ? ( )( ) ? 3 8 24 W h Wh 0.024 kWh (b) The heat flux on the surface of the chip is 2 W/in 37.5 ? ? ? 2 in 08 . 0 W 3 s s A Q q ? ? 1-14 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 are 2 cm 785 . 0 ) cm 5 )( cm 05 . 0 ( ? ? ? ? ?DL A s 2 6 W/m 10 1.91 ? ? ? ? ? 2 2 W/cm 191 cm 785 . 0 W 150 s s A Q q ? ? (b) The heat flux on the surface of glass bulb is 2 2 2 cm 1 . 201 cm) 8 ( ? ? ? ? ?D A s 2 W/m 7500 ? ? ? ? 2 2 W/cm 75 . 0 cm 1 . 201 W 150 s s A Q q ? ? (c) The amount and cost of electrical energy consumed during a one-year period is Electricity Consumption kW)(365 8 h / yr) 438 kWh / yr Annual Cost = (438 kWh / yr)($0.08 kWh) ? ? ? ? ? ? ( . / Q t ? 015 $35.04 / yr 1-15 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 is Q Q t ? ? ? ? ( . ? 12 kW)(2 h) 2.4 kWh (b) The heat flux on the surface of the iron base is ? ( . )( Q base W) = 1080 W ? 0 9 1200 ? ? . q Q A ? ? ? base base 2 W m 1080 0 015 72,000 W / m 2 (c) The cost of electricity consumed during this period is Cost of electricity = (2.4 kWh) ($0.07 kWh) ? ? / $0.17 Logic chip W 3 ? Q ? Iron 1200 W Q ? Lamp 150 W Chapter 1 Basics of Heat Transfer 1-4 1-16 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 is W 4 . 14 W) 12 . 0 )( 120 ( ? ? Q ? kWh 0.144 ? ? ? ? h) kW)(10 0144 . 0 ( t Q Q ? (b) The heat flux on the surface of the circuit board is 2 m 03 . 0 ) m 2 . 0 )( m 15 . 0 ( ? ? s A 2 W/m 480 ? ? ? 2 m 03 . 0 W 4 . 14 s s A Q q ? ? 1-17 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 ? = 2,700 kg/m 3 and 90 . 0 ? p C kJ/kg. ?C. Analysis The amount of energy added to the ball is simply the change in its internal energy, and is determined from E U mC T T transfer ? ? ? ? ( ) 2 1 where m V D ? ? ? ? ? ? ? ? 6 6 2700 015 4 77 3 ( )( . . kg / m m) kg 3 3 Substituting, E transfer kg)(0.90 kJ / kg. C)(200-80) C = ? ? ? ( . 4 77 515 kJ 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. 1-18 The body temperature of a man rises from 37°C to 39°C 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/kg.°C. Analysis The change in the sensible internal energy content of the body as a result of the body temperature rising 2 ?C during strenuous exercise is ?U = mC ?T = (70 kg)(3.6 kJ/kg. ?C)(2 ?C) = 504 kJ Chips, 0.12 W 15 cm 20 cm Q ? Metal ball E Chapter 1 Basics of Heat Transfer 1-5 1-19 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 C p = 1.007 kJ/kg. ?C (Table A-15). Analysis The volume of the air in the house is V ? ? ? ( ( floor space)(height) m )(3 m) m 2 3 200 600 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 314 , 11 K) 273.15 + /kg.K)(5 kPa.m 287 . 0 ( day) / m 600 kPa)(16.8 6 . 89 ( ) ACH ( 3 3 house air air ? ? ? ? ? ? o o o o RT V P RT V P m ? ? Noting that outdoor air enters at 5 ?C and leaves at 22 ?C, the energy loss of this house per day is kWh/day 53.8 = kJ/day 681 , 193 C ) 5 C)(22 kJ/kg. .007 kg/day)(1 314 , 11 ( ) ( outdoors indoors p air infilt ? ? ? ? ? ? ? T T C m Q ? ? At a unit cost of $0.082/kWh, the cost of this electrical energy lost by infiltration is $4.41/day ? ? ) $0.082/kWh kWh/day)( 8 . 53 ( energy) of cost t used)(Uni (Energy = Cost Enegy 5 ?C 0.7 ACH 22 ?C AIRRead More

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