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Unformatted text preview: cen58933_ch01.qxd 9/10/2002 8:29 AM Page 1 CHAPTER B A S I C S O F H E AT T R A N S F E R he science of thermodynamics deals with the amount of heat transfer as a system undergoes a process from one equilibrium state to another, and makes no reference to how long the process will take. But in engineering, we are often interested in the rate of heat transfer, which is the topic of the science of heat transfer. We start this chapter with a review of the fundamental concepts of thermodynamics that form the framework for heat transfer. We first present the relation of heat to other forms of energy and review the first law of thermodynamics. We then present the three basic mechanisms of heat transfer, which are conduction, convection, and radiation, and discuss thermal conductivity. 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 heat transfer between a solid surface and the adjacent liquid or gas that is in motion, and it involves the combined effects of conduction and fluid motion. Radiation is the 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. We close this chapter with a discussion of simultaneous heat transfer. T 1 CONTENTS 1–1 Thermodynamics and Heat Transfer 2 1–2 Engineering Heat Transfer 4 1–3 Heat and Other Forms of Energy 6 1–4 The First Law of Thermodynamics 11 1–5 Heat Transfer Mechanisms 17 1–6 Conduction 17 1–7 Convection 25 1–8 Radiation 27 1–9 Simultaneous Heat Transfer Mechanism 30 1–10 Problem-Solving Technique 35 Topic of Special Interest: Thermal Comfort 40 1 cen58933_ch01.qxd 9/10/2002 8:29 AM Page 2 2 HEAT TRANSFER 1–1 Thermos bottle Hot coffee Insulation FIGURE 1–1 We are normally interested in how long it takes for the hot coffee in a thermos to cool to a certain temperature, which cannot be determined from a thermodynamic analysis alone. Hot coffee 70°C Cool environment 20°C Heat FIGURE 1–2 Heat flows in the direction of decreasing temperature. I THERMODYNAMICS AND HEAT TRANSFER We all know from experience that a cold canned drink left in a room warms up and a warm canned drink left in a refrigerator cools down. This is accomplished by the transfer of energy from the warm medium to the cold one. The energy transfer is always from the higher temperature medium to the lower temperature one, and the energy transfer stops when the two mediums reach the same temperature. You will recall from thermodynamics that energy exists in various forms. In this text we are primarily interested in heat, which is the form of energy that can be transferred from one system to another as a result of temperature difference. The science that deals with the determination of the rates of such energy transfers is heat transfer. You may be wondering why we need to undertake a detailed study on heat transfer. After all, we can determine the amount of heat transfer for any system undergoing any process using a thermodynamic analysis alone. The reason is that thermodynamics is concerned with the amount of heat transfer as a system undergoes a process from one equilibrium state to another, and it gives no indication about how long the process will take. A thermodynamic analysis simply tells us how much heat must be transferred to realize a specified change of state to satisfy the conservation of energy principle. In practice we are more concerned about the rate of heat transfer (heat transfer per unit time) than we are with the amount of it. For example, we can determine the amount of heat transferred from a thermos bottle as the hot coffee inside cools from 90°C to 80°C by a thermodynamic analysis alone. But a typical user or designer of a thermos is primarily interested in how long it will be before the hot coffee inside cools to 80°C, and a thermodynamic analysis cannot answer this question. Determining the rates of heat transfer to or from a system and thus the times of cooling or heating, as well as the variation of the temperature, is the subject of heat transfer (Fig. 1–1). Thermodynamics deals with equilibrium states and changes from one equilibrium state to another. Heat transfer, on the other hand, deals with systems that lack thermal equilibrium, and thus it is a nonequilibrium phenomenon. Therefore, the study of heat transfer cannot be based on the principles of thermodynamics alone. However, the laws of thermodynamics lay the framework for the science of heat transfer. The first law requires that the rate of energy transfer into a system be equal to the rate of increase of the energy of that system. The second law requires that heat be transferred in the direction of decreasing temperature (Fig. 1–2). This is like a car parked on an inclined road that must go downhill in the direction of decreasing elevation when its brakes are released. It is also analogous to the electric current flowing in the direction of decreasing voltage or the fluid flowing in the direction of decreasing total pressure. The basic requirement for heat transfer is the presence of a temperature difference. There can be no net heat transfer between two mediums that are at the same temperature. The temperature difference is the driving force for heat transfer, just as the voltage difference is the driving force for electric current flow and pressure difference is the driving force for fluid flow. The rate of heat transfer in a certain direction depends on the magnitude of the temperature gradient (the temperature difference per unit length or the rate of change of cen58933_ch01.qxd 9/10/2002 8:29 AM Page 3 3 CHAPTER 1 temperature) in that direction. The larger the temperature gradient, the higher the rate of heat transfer. Application Areas of Heat Transfer Heat transfer is commonly encountered in engineering systems and other aspects of life, and one does not need to go very far to see some application areas of heat transfer. In fact, one does not need to go anywhere. The human body is constantly rejecting heat to its surroundings, and human comfort is closely tied to the rate of this heat rejection. We try to control this heat transfer rate by adjusting our clothing to the environmental conditions. Many ordinary household appliances are designed, in whole or in part, by using the principles of heat transfer. Some examples include the electric or gas range, the heating and air-conditioning system, the refrigerator and freezer, the water heater, the iron, and even the computer, the TV, and the VCR. Of course, energy-efficient homes are designed on the basis of minimizing heat loss in winter and heat gain in summer. Heat transfer plays a major role in the design of many other devices, such as car radiators, solar collectors, various components of power plants, and even spacecraft. The optimal insulation thickness in the walls and roofs of the houses, on hot water or steam pipes, or on water heaters is again determined on the basis of a heat transfer analysis with economic consideration (Fig. 1–3). Historical Background Heat has always been perceived to be something that produces in us a sensation of warmth, and one would think that the nature of heat is one of the first things understood by mankind. But it was only in the middle of the nineteenth The human body Air-conditioning systems Circuit boards Water in Water out Car radiators Power plants Refrigeration systems FIGURE 1–3 Some application areas of heat transfer. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 4 4 HEAT TRANSFER Contact surface Hot body Cold body Caloric FIGURE 1–4 In the early nineteenth century, heat was thought to be an invisible fluid called the caloric that flowed from warmer bodies to the cooler ones. century that we had a true physical understanding of the nature of heat, thanks to the development at that time of the kinetic theory, which treats molecules as tiny balls that are in motion and thus possess kinetic energy. Heat is then defined as the energy associated with the random motion of atoms and molecules. Although it was suggested in the eighteenth and early nineteenth centuries that heat is the manifestation of motion at the molecular level (called the live force), the prevailing view of heat until the middle of the nineteenth century was based on the caloric theory proposed by the French chemist Antoine Lavoisier (1743–1794) in 1789. The caloric theory asserts that heat is a fluidlike substance called the caloric that is a massless, colorless, odorless, and tasteless substance that can be poured from one body into another (Fig. 1–4). When caloric was added to a body, its temperature increased; and when caloric was removed from a body, its temperature decreased. When a body could not contain any more caloric, much the same way as when a glass of water could not dissolve any more salt or sugar, the body was said to be saturated with caloric. This interpretation gave rise to the terms saturated liquid and saturated vapor that are still in use today. The caloric theory came under attack soon after its introduction. It maintained that heat is a substance that could not be created or destroyed. Yet it was known that heat can be generated indefinitely by rubbing one’s hands together or rubbing two pieces of wood together. In 1798, the American Benjamin Thompson (Count Rumford) (1753–1814) showed in his papers that heat can be generated continuously through friction. The validity of the caloric theory was also challenged by several others. But it was the careful experiments of the Englishman James P. Joule (1818–1889) published in 1843 that finally convinced the skeptics that heat was not a substance after all, and thus put the caloric theory to rest. Although the caloric theory was totally abandoned in the middle of the nineteenth century, it contributed greatly to the development of thermodynamics and heat transfer. 1–2 I ENGINEERING HEAT TRANSFER Heat transfer equipment such as heat exchangers, boilers, condensers, radiators, heaters, furnaces, refrigerators, and solar collectors are designed primarily on the basis of heat transfer analysis. The heat transfer problems encountered in practice can be considered in two groups: (1) rating and (2) sizing problems. 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. A heat transfer process or equipment can be studied either experimentally (testing and taking measurements) or analytically (by analysis or calculations). The experimental approach has the advantage that we deal with the actual physical system, and the desired quantity is determined by measurement, within the limits of experimental error. However, this approach is expensive, time-consuming, and often impractical. Besides, the system we are analyzing may not even exist. For example, the size of a heating system of a building must usually be determined before the building is actually built on the basis of the dimensions and specifications given. The analytical approach (including numerical approach) has the advantage that it is fast and cen58933_ch01.qxd 9/10/2002 8:29 AM Page 5 5 CHAPTER 1 inexpensive, but the results obtained are subject to the accuracy of the assumptions and idealizations made in the analysis. In heat transfer studies, often a good compromise is reached by reducing the choices to just a few by analysis, and then verifying the findings experimentally. Modeling in Heat Transfer The descriptions of most scientific problems involve expressions that relate the changes in some key variables to each other. Usually the smaller the increment chosen in the changing variables, the more general and accurate the description. In the limiting case of infinitesimal or differential changes in variables, we obtain differential equations that provide precise mathematical formulations for the physical principles and laws by representing the rates of changes as derivatives. Therefore, differential equations are used to investigate a wide variety of problems in sciences and engineering, including heat transfer. However, most heat transfer problems encountered in practice can be solved without resorting to differential equations and the complications associated with them. The study of physical phenomena involves two important steps. In the first step, all the variables that affect the phenomena are identified, reasonable assumptions and approximations are made, and the interdependence of these variables is studied. The relevant physical laws and principles are invoked, and the problem is formulated mathematically. The equation itself is very instructive as it shows the degree of dependence of some variables on others, and the relative importance of various terms. In the second step, the problem is solved using an appropriate approach, and the results are interpreted. Many processes that seem to occur in nature randomly and without any order are, in fact, being governed by some visible or not-so-visible physical laws. Whether we notice them or not, these laws are there, governing consistently and predictably what seem to be ordinary events. Most of these laws are well defined and well understood by scientists. This 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 timeconsuming experiments. This is where the power of analysis lies. Very accurate results to meaningful practical problems can be obtained with relatively little effort by using a suitable and realistic mathematical model. The preparation of such models requires an adequate knowledge of the natural phenomena involved and the relevant laws, as well as a sound judgment. An unrealistic model will obviously give inaccurate and thus unacceptable results. An analyst working on an engineering problem often finds himself or herself in a position to make a choice between a very accurate but complex model, and a simple but not-so-accurate model. The right choice depends on the situation at hand. The right choice is usually the simplest model that yields adequate results. For example, the process of baking potatoes or roasting a round chunk of beef in an oven can be studied analytically in a simple way by modeling the potato or the roast as a spherical solid ball that has the properties of water (Fig. 1–5). The model is quite simple, but the results obtained are sufficiently accurate for most practical purposes. As another example, when we analyze the heat losses from a building in order to select the right size for a heater, we determine the heat losses under anticipated worst conditions and select a furnace that will provide sufficient heat to make up for those losses. Oven Potato Actual 175°C Water Ideal FIGURE 1–5 Modeling is a powerful engineering tool that provides great insight and simplicity at the expense of some accuracy. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 6 6 HEAT TRANSFER Often we tend to choose a larger furnace in anticipation of some future expansion, or just to provide a factor of safety. A very simple analysis will be adequate in this case. When selecting heat transfer equipment, it is important to consider the actual operating conditions. For example, when purchasing a heat exchanger that will handle hard water, we must consider that some calcium deposits will form on the heat transfer surfaces over time, causing fouling and thus a gradual decline in performance. The heat exchanger must be selected on the basis of operation under these adverse conditions instead of under new conditions. Preparing very accurate but complex models is usually not so difficult. But such models are not much use to an analyst if they are very difficult and timeconsuming to solve. At the minimum, the model should reflect the essential features of the physical problem it represents. There are many significant realworld problems that can be analyzed with a simple model. But it should always be kept in mind that the results obtained from an analysis are as accurate as the assumptions made in simplifying the problem. Therefore, the solution obtained should not be applied to situations for which the original assumptions do not hold. A solution that is not quite consistent with the observed nature of the problem indicates that the mathematical model used is too crude. In that case, a more realistic model should be prepared by eliminating one or more of the questionable assumptions. This will result in a more complex problem that, of course, is more difficult to solve. Thus any solution to a problem should be interpreted within the context of its formulation. 1–3 I HEAT AND OTHER FORMS OF ENERGY Energy can exist in numerous forms such as thermal, mechanical, kinetic, potential, electrical, magnetic, chemical, and nuclear, and their sum constitutes the total energy E (or e on a unit mass basis) of a system. The forms of energy related to the molecular structure of a system and the degree of the molecular activity are referred to as the microscopic energy. The sum of all microscopic forms of energy is called the internal energy of a system, and is denoted by U (or u on a unit mass basis). The international unit of energy is joule (J) or kilojoule (1 kJ 1000 J). In the English system, the unit of energy is the British thermal unit (Btu), which is defined as the energy needed to raise the temperature of 1 lbm of water at 60°F by 1°F. The magnitudes of kJ and Btu are almost identical (1 Btu 1.055056 kJ). Another well-known unit of energy is the calorie (1 cal 4.1868 J), which is defined as the energy needed to raise the temperature of 1 gram of water at 14.5°C by 1°C. Internal energy may be viewed as the sum of the kinetic and potential energies of the molecules. The portion of the internal energy of a system associated with the kinetic energy of the molecules is called sensible energy or sensible heat. The average velocity and the degree of activity of the molecules are proportional to the temperature. Thus, at higher temperatures the molecules will possess higher kinetic energy, and as a result, the system will have a higher internal energy. The internal energy is also associated with the intermolecular forces between the molecules of a system. These are the forces that bind the molecules cen58933_ch01.qxd 9/10/2002 8:29 AM Page 7 7 CHAPTER 1 to each other, and, as one would expect, they are strongest in solids and weakest in gases. If sufficient energy is added to the molecules of a solid or liquid, they will overcome these molecular forces and simply break away, turning the system to a gas. This is a phase change process and because of this added energy, a system in the gas phase is at a higher internal energy level than it is in the solid or the liquid phase. The internal energy associated with the phase of a system is called latent energy or latent heat. The changes mentioned above can occur without a change in the chemical composition of a system. Most heat transfer problems fall into this category, and one does not need to pay any attention to the forces binding the atoms in a molecule together. The internal energy associated with the atomic bonds in a molecule is called chemical (or bond) energy, whereas the internal energy associated with the bonds within the nucleus of the atom itself is called nuclear energy. The chemical and nuclear energies are absorbed or released during chemical or nuclear reactions, respectively. In the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and Pv. For the sake of simplicity and convenience, this combination is defined as enthalpy h. That is, h u Pv where the term Pv represents the flow energy of the fluid (also called the flow work), which is the energy needed to push a fluid and to maintain flow. In the energy analysis of flowing fluids, it is convenient to treat the flow energy as part of the energy of the fluid and to represent the microscopic energy of a fluid stream by enthalpy h (Fig. 1–6). Flowing fluid Stationary fluid Energy = h Energy = u FIGURE 1–6 The internal energy u represents the microscopic energy of a nonflowing fluid, whereas enthalpy h represents the microscopic energy of a flowing fluid. Specific Heats of Gases, Liquids, and Solids You may recall that an ideal gas is defined as a gas that obeys the relation Pv RT or P RT (1-1) where P is the absolute pressure, v is the specific volume, T is the absolute temperature, is the density, and R is the gas constant. It has been experimentally observed that the ideal gas relation given above closely approximates the P-v-T behavior of real gases at low densities. At low pressures and high temperatures, the density of a gas decreases and the gas behaves like an ideal gas. In the range of practical interest, many familiar gases such as air, nitrogen, oxygen, hydrogen, helium, argon, neon, and krypton and even heavier gases such as carbon dioxide can be treated as ideal gases with negligible error (often less than one percent). Dense gases such as water vapor in steam power plants and refrigerant vapor in refrigerators, however, should not always be treated as ideal gases since they usually exist at a state near saturation. You may also recall that specific heat is defined as the energy required to raise the temperature of a unit mass of a substance by one degree (Fig. 1–7). In general, this energy depends on how the process is executed. In thermodynamics, we are interested in two kinds of specific heats: specific heat at constant volume Cv and specific heat at constant pressure Cp. The specific heat at constant volume Cv can be viewed as the energy required to raise the temperature of a unit mass of a substance by one degree as the volume is held constant. The energy required to do the same as the pressure is held constant is the specific heat at constant pressure Cp. The specific heat at constant m = 1 kg ∆T = 1°C Specific heat = 5 kJ/kg·°C 5 kJ FIGURE 1–7 Specific heat is the energy required to raise the temperature of a unit mass of a substance by one degree in a specified way. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 8 8 THERMODYNAMICS Air m = 1 kg 300 → 301 K Air m = 1 kg 1000 → 1001 K 0.718 kJ 0.855 kJ FIGURE 1–8 The specific heat of a substance changes with temperature. pressure Cp is greater than Cv because at constant pressure the system is allowed to expand and the energy for this expansion work must also be supplied to the system. For ideal gases, these two specific heats are related to each other by Cp Cv R. A common unit for specific heats is kJ/kg · °C or kJ/kg · K. Notice that these two units are identical since ∆T(°C) ∆T(K), and 1°C change in temperature is equivalent to a change of 1 K. Also, 1 kJ/kg · °C 1 J/g · °C 1 kJ/kg · K 1 J/g · K The specific heats of a substance, in general, depend on two independent properties such as temperature and pressure. For an ideal gas, however, they depend on temperature only (Fig. 1–8). At low pressures all real gases approach ideal gas behavior, and therefore their specific heats depend on temperature only. The differential changes in the internal energy u and enthalpy h of an ideal gas can be expressed in terms of the specific heats as du Cv dT and dh Cp dT (1-2) The finite changes in the internal energy and enthalpy of an ideal gas during a process can be expressed approximately by using specific heat values at the average temperature as u Cv, ave T and h Cp, ave T (J/g) (1-3) or U IRON 25°C C = Cv = Cp = 0.45 kJ/kg·°C FIGURE 1–9 The Cv and Cp values of incompressible substances are identical and are denoted by C. mCv, ave T and H mCp, ave T (J) (1-4) where m is the mass of the system. A substance whose specific volume (or density) does not change with temperature or pressure is called an incompressible substance. The specific volumes of solids and liquids essentially remain constant during a process, and thus they can be approximated as incompressible substances without sacrificing much in accuracy. The constant-volume and constant-pressure specific heats are identical for incompressible substances (Fig. 1–9). Therefore, for solids and liquids the subscripts on Cv and Cp can be dropped and both specific heats can be represented by a single symbol, C. That is, Cp Cv C. This result could also be deduced from the physical definitions of constant-volume and constantpressure specific heats. Specific heats of several common gases, liquids, and solids are given in the Appendix. The specific heats of incompressible substances depend on temperature only. Therefore, the change in the internal energy of solids and liquids can be expressed as U mCave T (J) (1-5) cen58933_ch01.qxd 9/10/2002 8:29 AM Page 9 9 CHAPTER 1 where Cave is the average specific heat evaluated at the average temperature. Note that the internal energy change of the systems that remain in a single phase (liquid, solid, or gas) during the process can be determined very easily using average specific heats. Energy Transfer Energy can be transferred to or from a given mass by two mechanisms: heat Q and work W. An energy interaction is heat transfer if its driving force is a temperature difference. Otherwise, it is work. A rising piston, a rotating shaft, and an electrical wire crossing the system boundaries are all associated with work interactions. Work done per unit time is called power, and is denoted · by W. The unit of power is W or hp (1 hp 746 W). Car engines and hydraulic, steam, and gas turbines produce work; compressors, pumps, and mixers consume work. Notice that the energy of a system decreases as it does work, and increases as work is done on it. In daily life, we frequently refer to the sensible and latent forms of internal energy as heat, and we talk about the heat content of bodies (Fig. 1–10). In thermodynamics, however, those forms of energy are usually referred to as thermal energy to prevent any confusion with heat transfer. The term heat and the associated phrases such as heat flow, heat addition, heat rejection, heat absorption, heat gain, heat loss, heat storage, heat generation, electrical heating, latent heat, body heat, and heat source are in common use today, and the attempt to replace heat in these phrases by thermal energy had only limited success. These phrases are deeply rooted in our vocabulary and they are used by both the ordinary people and scientists without causing any misunderstanding. For example, the phrase body heat is understood to mean the thermal energy content of a body. Likewise, heat flow is understood to mean the transfer of thermal energy, not the flow of a fluid-like substance called heat, although the latter incorrect interpretation, based on the caloric theory, is the origin of this phrase. Also, the transfer of heat into a system is frequently referred to as heat addition and the transfer of heat out of a system as heat rejection. Keeping in line with current practice, we will refer to the thermal energy as heat and the transfer of thermal energy as heat transfer. The amount of heat transferred during the process is denoted by Q. The amount of heat transferred · per unit time is called heat transfer rate, and is denoted by Q . The overdot · stands for the time derivative, or “per unit time.” The heat transfer rate Q has the unit J/s, which is equivalent to W. · When the rate of heat transfer Q is available, then the total amount of heat transfer Q during a time interval t can be determined from t Q · Q dt (J) (1-6) 0 · provided that the variation of Q with time is known. For the special case of · Q constant, the equation above reduces to Q · Qt (J) (1-7) Vapor 80°C Liquid 80°C Heat transfer 25°C FIGURE 1–10 The sensible and latent forms of internal energy can be transferred as a result of a temperature difference, and they are referred to as heat or thermal energy. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 10 10 HEAT TRANSFER . The rate of heat transfer per unit area normal to the direction of heat transfer is called heat flux, and the average heat flux is expressed as (Fig. 1–11) Q = 24 W = const. 3m A = 6 m2 q· · Q A (W/m2) (1-8) where A is the heat transfer area. The unit of heat flux in English units is Btu/h · ft2. Note that heat flux may vary with time as well as position on a surface. 2m . . Q 24 W q = — = –—— = 4 W/m2 A 6 m2 FIGURE 1–11 Heat flux is heat transfer per unit time and per unit area, and is equal · · to q· Q /A when Q is uniform over the area A. EXAMPLE 1–1 Heating of a Copper Ball A 10-cm diameter copper ball is to be heated from 100°C to an average temperature of 150°C in 30 minutes (Fig. 1–12). Taking the average density and specific heat of copper in this temperature range to be 8950 kg/m3 and Cp 0.395 kJ/kg · °C, respectively, determine (a) the total amount of heat transfer to the copper ball, (b) the average rate of heat transfer to the ball, and (c) the average heat flux. T2 = 150°C SOLUTION The copper ball is to be heated from 100°C to 150°C. The total heat transfer, the average rate of heat transfer, and the average heat flux are to be determined. T1 = 100°C A = πD 2 Q FIGURE 1–12 Schematic for Example 1–1. Assumptions Constant properties can be used for copper at the average temperature. Properties The average density and specific heat of copper are given to be 8950 kg/m3 and Cp 0.395 kJ/kg · °C. Analysis (a) The amount of heat transferred to the copper ball is simply the change in its internal energy, and is determined from Energy transfer to the system Q U mCave (T2 T1) Energy increase of the system where m V 6 D3 6 (8950 kg/m3)(0.1 m)3 4.69 kg Substituting, Q (4.69 kg)(0.395 kJ/kg · °C)(150 100)°C 92.6 kJ Therefore, 92.6 kJ of heat needs to be transferred to the copper ball to heat it from 100°C to 150°C. (b) The rate of heat transfer normally changes during a process with time. However, we can determine the average rate of heat transfer by dividing the total amount of heat transfer by the time interval. Therefore, · Q ave Q t 92.6 kJ 1800 s 0.0514 kJ/s 51.4 W cen58933_ch01.qxd 9/10/2002 8:29 AM Page 11 11 CHAPTER 1 (c) Heat flux is defined as the heat transfer per unit time per unit area, or the rate of heat transfer per unit area. Therefore, the average heat flux in this case is · q ave · Qave A · Qave D2 51.4 W (0.1 m)2 1636 W/m2 Discussion Note that heat flux may vary with location on a surface. The value calculated above is the average heat flux over the entire surface of the ball. 1–4 I THE FIRST LAW OF THERMODYNAMICS The first law of thermodynamics, also known as the conservation of energy principle, states that energy can neither be created nor destroyed; it can only change forms. Therefore, every bit of energy must be accounted for during a process. The conservation of energy principle (or the energy balance) for any system undergoing any process may be expressed as follows: The net change (increase or decrease) in the total energy of the system during a process is equal to the difference between the total energy entering and the total energy leaving the system during that process. That is, Total energy entering the system Total energy leaving the system Change in the total energy of the system (1-9) Noting that energy can be transferred to or from a system by heat, work, and mass flow, and that the total energy of a simple compressible system consists of internal, kinetic, and potential energies, the energy balance for any system undergoing any process can be expressed as E Eout 1in 424 3 Esystem 123 Net energy transfer by heat, work, and mass (J) (1-10) Change in internal, kinetic, potential, etc., energies (W) (1-11) or, in the rate form, as · · E Eout 1in 424 3 dE /dt 1 system 3 424 Rate of net energy transfer by heat, work, and mass Rate of change in internal kinetic, potential, etc., energies Energy is a property, and the value of a property does not change unless the state of the system changes. Therefore, the energy change of a system is zero ( Esystem 0) if the state of the system does not change during the process, that is, the process is steady. The energy balance in this case reduces to (Fig. 1–13) · Ein 123 · Eout 123 Rate of net energy transfer in by heat, work, and mass Steady, rate form: Rate of net energy transfer out by heat, work, and mass (1-12) In the absence of significant electric, magnetic, motion, gravity, and surface tension effects (i.e., for stationary simple compressible systems), the change · Ein · Eout Heat Work Heat Steady system Mass Work Mass · · Ein = Eout FIGURE 1–13 In steady operation, the rate of energy transfer to a system is equal to the rate of energy transfer from the system. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 12 12 HEAT TRANSFER in the total energy of a system during a process is simply the change in its inUsystem. ternal energy. That is, Esystem In heat transfer analysis, we are usually interested only in the forms of energy that can be transferred as a result of a temperature difference, that is, heat or thermal energy. In such cases it is convenient to write a heat balance and to treat the conversion of nuclear, chemical, and electrical energies into thermal energy as heat generation. The energy balance in that case can be expressed as Q Qout 1in 424 3 Egen 123 E 1 424 3 4thermal, system 4 Net heat transfer Heat generation (J) Change in thermal energy of the system (1-13) Energy Balance for Closed Systems (Fixed Mass) Specific heat = Cv Mass = m Initial temp = T1 Final temp = T2 Q = mCv (T1 – T2 ) FIGURE 1–14 In the absence of any work interactions, the change in the energy content of a closed system is equal to the net heat transfer. A closed system consists of a fixed mass. The total energy E for most systems encountered in practice consists of the internal energy U. This is especially the case for stationary systems since they don’t involve any changes in their velocity or elevation during a process. The energy balance relation in that case reduces to Stationary closed system: Ein Eout U mCv T (J) (1-14) where we expressed the internal energy change in terms of mass m, the specific heat at constant volume Cv, and the temperature change T of the system. When the system involves heat transfer only and no work interactions across its boundary, the energy balance relation further reduces to (Fig. 1–14) Stationary closed system, no work: Q mCv T (J) (1-15) where Q is the net amount of heat transfer to or from the system. This is the form of the energy balance relation we will use most often when dealing with a fixed mass. Energy Balance for Steady-Flow Systems A large number of engineering devices such as water heaters and car radiators involve mass flow in and out of a system, and are modeled as control volumes. Most control volumes are analyzed under steady operating conditions. The term steady means no change with time at a specified location. The opposite of steady is unsteady or transient. Also, the term uniform implies no change with position throughout a surface or region at a specified time. These meanings are consistent with their everyday usage (steady girlfriend, uniform distribution, etc.). The total energy content of a control volume during a steady-flow process remains constant (ECV constant). That is, the change in the total energy of the control volume during such a process is zero ( ECV 0). Thus the amount of energy entering a control volume in all forms (heat, work, mass transfer) for a steady-flow process must be equal to the amount of energy leaving it. The amount of mass flowing through a cross section of a flow device per · unit time is called the mass flow rate, and is denoted by m . A fluid may flow in and out of a control volume through pipes or ducts. The mass flow rate of a fluid flowing in a pipe or duct is proportional to the cross-sectional area Ac of cen58933_ch01.qxd 9/10/2002 8:29 AM Page 13 13 CHAPTER 1 the pipe or duct, the density , and the velocity of the fluid. The mass flow · rate through a differential area dAc can be expressed as δm n dAc where n is the velocity component normal to dAc. The mass flow rate through the entire cross-sectional area is obtained by integration over Ac. The flow of a fluid through a pipe or duct can often be approximated to be one-dimensional. That is, the properties can be assumed to vary in one direction only (the direction of flow). As a result, all properties are assumed to be uniform at any cross section normal to the flow direction, and the properties are assumed to have bulk average values over the entire cross section. Under the one-dimensional flow approximation, the mass flow rate of a fluid flowing in a pipe or duct can be expressed as (Fig. 1–15) · m Ac (kg/s) (1-16) where is the fluid density, is the average fluid velocity in the flow direction, and Ac is the cross-sectional area of the pipe or duct. The volume of a fluid flowing through a pipe or duct per unit time is called · the volume flow rate V, and is expressed as · V · m Ac (m3/s) · mh · m Cp T (kJ/s) (1-18) · where Q is the rate of net heat transfer into or out of the control volume. This is the form of the energy balance relation that we will use most often for steady-flow systems. Surface Energy Balance As mentioned in the chapter opener, heat is transferred by the mechanisms of conduction, convection, and radiation, and heat often changes vehicles as it is transferred from one medium to another. For example, the heat conducted to the outer surface of the wall of a house in winter is convected away by the cold outdoor air while being radiated to the cold surroundings. In such cases, it may be necessary to keep track of the energy interactions at the surface, and this is done by applying the conservation of energy principle to the surface. A surface contains no volume or mass, and thus no energy. Thereore, a surface can be viewed as a fictitious system whose energy content remains constant during a process (just like a steady-state or steady-flow system). Then the energy balance for a surface can be expressed as Surface energy balance: · Ein · Eout · m = ρ Ac FIGURE 1–15 The mass flow rate of a fluid at a cross section is equal to the product of the fluid density, average fluid velocity, and the cross-sectional area. (1-17) Note that the mass flow rate of a fluid through a pipe or duct remains constant during steady flow. This is not the case for the volume flow rate, however, unless the density of the fluid remains constant. For a steady-flow system with one inlet and one exit, the rate of mass flow into the control volume must be equal to the rate of mass flow out of it. That · · · is, m in m out m. When the changes in kinetic and potential energies are negligible, which is usually the case, and there is no work interaction, the energy balance for such a steady-flow system reduces to (Fig. 1–16) · Q Ac = π D 2/4 for a circular pipe (1-19) Control volume · m T1 · m T2 · · Etransfer = mCp(T2 – T1) FIGURE 1–16 Under steady conditions, the net rate of energy transfer to a fluid in a control volume is equal to the rate of increase in the energy of the fluid stream flowing through the control volume. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 14 14 HEAT TRANSFER WALL Control surface This relation is valid for both steady and transient conditions, and the surface energy balance does not involve heat generation since a surface does not have a volume. The energy balance for the outer surface of the wall in Fig. 1–17, for example, can be expressed as radiation · Q1 . Q3 conduction . Q1 . Q2 convection FIGURE 1–17 Energy interactions at the outer wall surface of a house. · Q2 · Q3 (1-20) · · where Q 1 is conduction through the wall to the surface, Q 2 is convection from · the surface to the outdoor air, and Q 3 is net radiation from the surface to the surroundings. When the directions of interactions are not known, all energy interactions can be assumed to be towards the surface, and the surface energy balance can · be expressed as E in 0. Note that the interactions in opposite direction will end up having negative values, and balance this equation. EXAMPLE 1–2 Heating of Water in an Electric Teapot 1.2 kg of liquid water initially at 15°C is to be heated to 95°C in a teapot equipped with a 1200-W electric heating element inside (Fig. 1–18). The teapot is 0.5 kg and has an average specific heat of 0.7 kJ/kg · °C. Taking the specific heat of water to be 4.18 kJ/kg · °C and disregarding any heat loss from the teapot, determine how long it will take for the water to be heated. Electric heating element Water 15°C 1200 W FIGURE 1–18 Schematic for Example 1–2. SOLUTION Liquid water is to be heated in an electric teapot. The heating time is to be determined. Assumptions 1 Heat loss from the teapot is negligible. 2 Constant properties can be used for both the teapot and the water. Properties The average specific heats are given to be 0.7 kJ/kg · °C for the teapot and 4.18 kJ/kg · °C for water. Analysis We take the teapot and the water in it as the system, which is a closed system (fixed mass). The energy balance in this case can be expressed as Ein Eout Ein Esystem Usystem Uwater Uteapot Then the amount of energy needed to raise the temperature of water and the teapot from 15°C to 95°C is Ein (mC T )water (mC T )teapot (1.2 kg)(4.18 kJ/kg · °C)(95 (95 15)°C 429.3 kJ 15)°C (0.5 kg)(0.7 kJ/kg · °C) The 1200-W electric heating unit will supply energy at a rate of 1.2 kW or 1.2 kJ per second. Therefore, the time needed for this heater to supply 429.3 kJ of heat is determined from t Total energy transferred Rate of energy transfer Ein · E transfer 429.3 kJ 1.2 kJ/s 358 s 6.0 min cen58933_ch01.qxd 9/10/2002 8:29 AM Page 15 15 CHAPTER 1 Discussion In reality, it will take more than 6 minutes to accomplish this heating process since some heat loss is inevitable during heating. EXAMPLE 1–3 Heat Loss from Heating Ducts in a Basement A 5-m-long section of an air heating system of a house passes through an unheated space in the basement (Fig. 1–19). The cross section of the rectangular duct of the heating system is 20 cm 25 cm. Hot air enters the duct at 100 kPa and 60°C at an average velocity of 5 m/s. The temperature of the air in the duct drops to 54°C as a result of heat loss to the cool space in the basement. Determine the rate of heat loss from the air in the duct to the basement under steady conditions. Also, determine the cost of this heat loss per hour if the house is heated by a natural gas furnace that has an efficiency of 80 percent, and the cost of the natural gas in that area is $0.60/therm (1 therm 100,000 Btu 105,500 kJ). SOLUTION The temperature of the air in the heating duct of a house drops as a result of heat loss to the cool space in the basement. The rate of heat loss from the hot air and its cost are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air can be treated as an ideal gas with constant properties at room temperature. Properties The constant pressure specific heat of air at the average temperature of (54 60)/2 57°C is 1.007 kJ/kg · °C (Table A-15). Analysis We take the basement section of the heating system as our system, which is a steady-flow system. The rate of heat loss from the air in the duct can be determined from · Q · where m is the mass flow rate and air at the inlet conditions is P RT · m Cp T T is the temperature drop. The density of 100 kPa (0.287 kPa · m3/kg · K)(60 1.046 kg/m3 273)K The cross-sectional area of the duct is Ac (0.20 m)(0.25 m) 0.05 m2 Then the mass flow rate of air through the duct and the rate of heat loss become · m Ac (1.046 kg/m3)(5 m/s)(0.05 m2) 0.2615 kg/s and · Q loss · m Cp(Tin Tout) (0.2615 kg/s)(1.007 kJ/kg · °C)(60 1.580 kJ/s 54)°C 5m 20 cm Hot air 100 kPa 60°C 5 m/s 25 cm 54°C · Qloss FIGURE 1–19 Schematic for Example 1–3. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 16 16 HEAT TRANSFER or 5688 kJ/h. The cost of this heat loss to the home owner is Cost of heat loss (Rate of heat loss)(Unit cost of energy input) Furnace efficiency (5688 kJ/h)($0.60/therm) 1 therm 0.80 105,500 kJ $0.040/h Discussion The heat loss from the heating ducts in the basement is costing the home owner 4 cents per hour. Assuming the heater operates 2000 hours during a heating season, the annual cost of this heat loss adds up to $80. Most of this money can be saved by insulating the heating ducts in the unheated areas. EXAMPLE 1–4 P = 12.2 psia atm 50°F 9 ft 70°F 40 ft 50 ft FIGURE 1–20 Schematic for Example 1–4. Electric Heating of a House at High Elevation Consider a house that has a floor space of 2000 ft2 and an average height of 9 ft at 5000 ft elevation where the standard atmospheric pressure is 12.2 psia (Fig. 1–20). Initially the house is at a uniform temperature of 50°F. Now the electric heater is turned on, and the heater runs until the air temperature in the house rises to an average value of 70°F. Determine the amount of energy transferred to the air assuming (a) the house is air-tight and thus no air escapes during the heating process and (b) some air escapes through the cracks as the heated air in the house expands at constant pressure. Also determine the cost of this heat for each case if the cost of electricity in that area is $0.075/kWh. SOLUTION The air in the house is heated from 50°F to 70°F by an electric heater. The amount and cost of the energy transferred to the air are to be determined for constant-volume and constant-pressure cases. Assumptions 1 Air can be treated as an ideal gas with constant properties at room temperature. 2 Heat loss from the house during heating is negligible. 3 The volume occupied by the furniture and other things is negligible. Properties The specific heats of air at the average temperature of (50 70)/2 60°F are Cp 0.240 Btu/lbm · °F and Cv Cp R 0.171 Btu/lbm · °F (Tables A-1E and A-15E). Analysis The volume and the mass of the air in the house are V m (Floor area)(Height) (2000 ft2)(9 ft) 18,000 ft3 (12.2 psia)(18,000 ft3) PV 1162 lbm RT (0.3704 psia · ft3/lbm · R)(50 460)R (a) The amount of energy transferred to air at constant volume is simply the change in its internal energy, and is determined from Ein Eout Ein, constant volume Esystem Uair mCv T (1162 lbm)(0.171 Btu/lbm · °F)(70 3974 Btu At a unit cost of $0.075/kWh, the total cost of this energy is 50)°F cen58933_ch01.qxd 9/10/2002 8:29 AM Page 17 17 CHAPTER 1 Cost of energy (Amount of energy)(Unit cost of energy) 1 kWh (3974 Btu)($0.075/kWh) 3412 Btu $0.087 (b) The amount of energy transferred to air at constant pressure is the change in its enthalpy, and is determined from Ein, constant pressure Hair mCp T (1162 lbm)(0.240 Btu/lbm · °F)(70 5578 Btu 50)°F At a unit cost of $0.075/kWh, the total cost of this energy is Cost of energy (Amount of energy)(Unit cost of energy) 1 kWh (5578 Btu)($0.075/kWh) 3412 Btu $0.123 Discussion It will cost about 12 cents to raise the temperature of the air in this house from 50°F to 70°F. The second answer is more realistic since every house has cracks, especially around the doors and windows, and the pressure in the house remains essentially constant during a heating process. Therefore, the second approach is used in practice. This conservative approach somewhat overpredicts the amount of energy used, however, since some of the air will escape through the cracks before it is heated to 70°F. 1–5 I HEAT TRANSFER MECHANISMS In Section 1–1 we defined heat as the form of energy that can be transferred from one system to another as a result of temperature difference. A thermodynamic analysis is concerned with the amount of heat transfer as a system undergoes a process from one equilibrium state to another. The science that deals with the determination of the rates of such energy transfers is the heat transfer. The transfer of energy as heat is always from the higher-temperature medium to the lower-temperature one, and heat transfer stops when the two mediums reach the same temperature. Heat can be transferred in three different modes: conduction, convection, and radiation. All modes of heat transfer require the existence of a temperature difference, and all modes are from the high-temperature medium to a lower-temperature one. Below we give a brief description of each mode. A detailed study of these modes is given in later chapters of this text. 1–6 I CONDUCTION 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. Conduction can take place in solids, liquids, or gases. In gases and liquids, conduction is due to the collisions and diffusion of the cen58933_ch01.qxd 9/10/2002 8:29 AM Page 18 18 HEAT TRANSFER T1 T2 . Q A A ∆x 0 x FIGURE 1–21 Heat conduction through a large plane wall of thickness x and area A. molecules during their random motion. In solids, it is due to the combination of vibrations of the molecules in a lattice and the energy transport by free electrons. A cold canned drink in a warm room, for example, eventually warms up to the room temperature as a result of heat transfer from the room to the drink through the aluminum can by conduction. The rate of heat conduction through a medium depends on the geometry of the medium, its thickness, and the material of the medium, as well as the temperature difference across the medium. We know that wrapping a hot water tank with glass wool (an insulating material) reduces the rate of heat loss from the tank. The thicker the insulation, the smaller the heat loss. We also know that a hot water tank will lose heat at a higher rate when the temperature of the room housing the tank is lowered. Further, the larger the tank, the larger the surface area and thus the rate of heat loss. Consider steady heat conduction through a large plane wall of thickness x L and area A, as shown in Fig. 1–21. The temperature difference across the wall is T T2 T1. Experiments have shown that the rate of heat trans· fer Q through the wall is doubled when the temperature difference T across the wall or the area A normal to the direction of heat transfer is doubled, but is halved when the wall thickness L is doubled. Thus we conclude that the rate of heat conduction through a plane layer is proportional to the temperature difference across the layer and the heat transfer area, but is inversely proportional to the thickness of the layer. That is, Rate of heat conduction (Area)(Temperature difference) Thickness or, · Q cond 30°C kA 20°C . q = 4010 W/m 2 1m (a) Copper (k = 401 W/m·°C) 30°C 20°C . q = 1480 W/m 2 1m (b) Silicon (k = 148 W/m ·°C) FIGURE 1–22 The rate of heat conduction through a solid is directly proportional to its thermal conductivity. T2 T1 kA x T x (W) (1-21) where the constant of proportionality k is the thermal conductivity of the material, which is a measure of the ability of a material to conduct heat (Fig. 1–22). In the limiting case of x → 0, the equation above reduces to the differential form · Q cond kA dT dx (W) (1-22) which is called Fourier’s law of heat conduction after J. Fourier, who expressed it first in his heat transfer text in 1822. Here dT/dx is the temperature gradient, which is the slope of the temperature curve on a T-x diagram (the rate of change of T with x), at location x. The relation above indicates that the rate of heat conduction in a direction is proportional to the temperature gradient in that direction. Heat is conducted in the direction of decreasing temperature, and the temperature gradient becomes negative when temperature decreases with increasing x. The negative sign in Eq. 1–22 ensures that heat transfer in the positive x direction is a positive quantity. The heat transfer area A is always normal to the direction of heat transfer. For heat loss through a 5-m-long, 3-m-high, and 25-cm-thick wall, for example, the heat transfer area is A 15 m2. Note that the thickness of the wall has no effect on A (Fig. 1–23). cen58933_ch01.qxd 9/10/2002 8:29 AM Page 19 19 CHAPTER 1 EXAMPLE 1–5 The Cost of Heat Loss through a Roof The roof of an electrically heated home is 6 m long, 8 m wide, and 0.25 m thick, and is made of a flat layer of concrete whose thermal conductivity is k 0.8 W/m · °C (Fig. 1–24). The temperatures of the inner and the outer surfaces of the roof one night are measured to be 15°C and 4°C, respectively, for a period of 10 hours. Determine (a) the rate of heat loss through the roof that night and (b) the cost of that heat loss to the home owner if the cost of electricity is $0.08/kWh. SOLUTION The inner and outer surfaces of the flat concrete roof of an electrically heated home are maintained at specified temperatures during a night. The heat loss through the roof and its cost that night are to be determined. Assumptions 1 Steady operating conditions exist during the entire night since the surface temperatures of the roof remain constant at the specified values. 2 Constant properties can be used for the roof. Properties The thermal conductivity of the roof is given to be k 0.8 W/m · °C. Analysis (a) Noting that heat transfer through the roof is by conduction and the area of the roof is A 6 m 8 m 48 m2, the steady rate of heat transfer through the roof is determined to be · Q kA T2 T1 L (0.8 W/m · °C)(48 m2) (15 4)°C 0.25 m 1690 W 1.69 kW H A=W×H · Q W L FIGURE 1–23 In heat conduction analysis, A represents the area normal to the direction of heat transfer. Concrete roof 6m 0.25 m 8m 4°C 15°C (b) The amount of heat lost through the roof during a 10-hour period and its cost are determined from Q Cost · Q t (1.69 kW)(10 h) 16.9 kWh (Amount of energy)(Unit cost of energy) (16.9 kWh)($0.08/kWh) $1.35 Discussion The cost to the home owner of the heat loss through the roof that night was $1.35. The total heating bill of the house will be much larger since the heat losses through the walls are not considered in these calculations. Thermal Conductivity We have seen that different materials store heat differently, and we have defined the property specific heat Cp as a measure of a material’s ability to store thermal energy. For example, Cp 4.18 kJ/kg · °C for water and Cp 0.45 kJ/kg · °C for iron at room temperature, which indicates that water can store almost 10 times the energy that iron can per unit mass. Likewise, the thermal conductivity k is a measure of a material’s ability to conduct heat. For example, k 0.608 W/m · °C for water and k 80.2 W/m · °C for iron at room temperature, which indicates that iron conducts heat more than 100 times faster than water can. Thus we say that water is a poor heat conductor relative to iron, although water is an excellent medium to store thermal energy. Equation 1–22 for the rate of conduction heat transfer under steady conditions can also be viewed as the defining equation for thermal conductivity. Thus the thermal conductivity of a material can be defined as the rate of FIGURE 1–24 Schematic for Example 1–5. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 20 20 HEAT TRANSFER TABLE 1–1 The thermal conductivities of some materials at room temperature Material k, W/m · °C* Diamond Silver Copper Gold Aluminum Iron Mercury (l) Glass Brick Water (l) Human skin Wood (oak) Helium (g) Soft rubber Glass fiber Air (g) Urethane, rigid foam 2300 429 401 317 237 80.2 8.54 0.78 0.72 0.613 0.37 0.17 0.152 0.13 0.043 0.026 0.026 *Multiply by 0.5778 to convert to Btu/h · ft · °F. Electric heater Insulation Insulation T1 Sample material k . . T2 Q = We A . L We Insulation . L k = ———— Q A(T1 – T2) FIGURE 1–25 A simple experimental setup to determine the thermal conductivity of a material. heat transfer through a unit thickness of the material per unit area per unit temperature difference. The thermal conductivity of a material is a measure of the ability of the material to conduct heat. A high value for thermal conductivity indicates that the material is a good heat conductor, and a low value indicates that the material is a poor heat conductor or insulator. The thermal conductivities of some common materials at room temperature are given in Table 1–1. The thermal conductivity of pure copper at room temperature is k 401 W/m · °C, which indicates that a 1-m-thick copper wall will conduct heat at a rate of 401 W per m2 area per °C temperature difference across the wall. Note that materials such as copper and silver that are good electric conductors are also good heat conductors, and have high values of thermal conductivity. Materials such as rubber, wood, and styrofoam are poor conductors of heat and have low conductivity values. A layer of material of known thickness and area can be heated from one side by an electric resistance heater of known output. If the outer surfaces of the heater are well insulated, all the heat generated by the resistance heater will be transferred through the material whose conductivity is to be determined. Then measuring the two surface temperatures of the material when steady heat transfer is reached and substituting them into Eq. 1–22 together with other known quantities give the thermal conductivity (Fig. 1–25). The thermal conductivities of materials vary over a wide range, as shown in Fig. 1–26. The thermal conductivities of gases such as air vary by a factor of 104 from those of pure metals such as copper. Note that pure crystals and metals have the highest thermal conductivities, and gases and insulating materials the lowest. Temperature is a measure of the kinetic energies of the particles such as the molecules or atoms of a substance. In a liquid or gas, the kinetic energy of the molecules is due to their random translational motion as well as their vibrational and rotational motions. When two molecules possessing different kinetic energies collide, part of the kinetic energy of the more energetic (higher-temperature) molecule is transferred to the less energetic (lowertemperature) molecule, much the same as when two elastic balls of the same mass at different velocities collide, part of the kinetic energy of the faster ball is transferred to the slower one. The higher the temperature, the faster the molecules move and the higher the number of such collisions, and the better the heat transfer. The kinetic theory of gases predicts and the experiments confirm that the thermal conductivity of gases is proportional to the square root of the absolute temperature T, and inversely proportional to the square root of the molar mass M. Therefore, the thermal conductivity of a gas increases with increasing temperature and decreasing molar mass. So it is not surprising that the thermal conductivity of helium (M 4) is much higher than those of air (M 29) and argon (M 40). The thermal conductivities of gases at 1 atm pressure are listed in Table A-16. However, they can also be used at pressures other than 1 atm, since the thermal conductivity of gases is independent of pressure in a wide range of pressures encountered in practice. The mechanism of heat conduction in a liquid is complicated by the fact that the molecules are more closely spaced, and they exert a stronger intermolecular force field. The thermal conductivities of liquids usually lie between those cen58933_ch01.qxd 9/10/2002 8:29 AM Page 21 21 CHAPTER 1 NONMETALLIC CRYSTALS 1000 PURE METALS k, W/m·°C METAL ALLOYS Silver Copper Aluminum NONMETALLIC alloys Iron SOLIDS 100 Oxides 10 Bronze Steel Nichrome Diamond Graphite Silicon carbide Beryllium oxide Manganese Quartz LIQUIDS Mercury Rock 1 INSULATORS Water Food Fibers Wood Air 0.1 GASES Hydrogen Helium Foams Rubber Oils Carbon dioxide 0.01 of solids and gases. The thermal conductivity of a substance is normally highest in the solid phase and lowest in the gas phase. Unlike gases, the thermal conductivities of most liquids decrease with increasing temperature, with water being a notable exception. Like gases, the conductivity of liquids decreases with increasing molar mass. Liquid metals such as mercury and sodium have high thermal conductivities and are very suitable for use in applications where a high heat transfer rate to a liquid is desired, as in nuclear power plants. In solids, heat conduction is due to two effects: the lattice vibrational waves induced by the vibrational motions of the molecules positioned at relatively fixed positions in a periodic manner called a lattice, and the energy transported via the free flow of electrons in the solid (Fig. 1–27). The thermal conductivity of a solid is obtained by adding the lattice and electronic components. The relatively high thermal conductivities of pure metals are primarily due to the electronic component. The lattice component of thermal conductivity strongly depends on the way the molecules are arranged. For example, diamond, which is a highly ordered crystalline solid, has the highest known thermal conductivity at room temperature. Unlike metals, which are good electrical and heat conductors, crystalline solids such as diamond and semiconductors such as silicon are good heat conductors but poor electrical conductors. As a result, such materials find widespread use in the electronics industry. Despite their higher price, diamond heat sinks are used in the cooling of sensitive electronic components because of the FIGURE 1–26 The range of thermal conductivity of various materials at room temperature. GAS * Molecular collisions * Molecular diffusion LIQUID * Molecular collisions * Molecular diffusion electrons SOLID * Lattice vibrations * Flow of free electrons FIGURE 1–27 The mechanisms of heat conduction in different phases of a substance. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 22 22 HEAT TRANSFER TABLE 1–2 The thermal conductivity of an alloy is usually much lower than the thermal conductivity of either metal of which it is composed Pure metal or alloy k, W/m · °C, at 300 K Copper Nickel Constantan (55% Cu, 45% Ni) 401 91 Copper Aluminum Commercial bronze (90% Cu, 10% Al) 401 237 23 52 TABLE 1–3 Thermal conductivities of materials vary with temperature T, K Copper Aluminum 100 200 300 400 600 800 482 413 401 393 379 366 302 237 237 240 231 218 excellent thermal conductivity of diamond. Silicon oils and gaskets are commonly used in the packaging of electronic components because they provide both good thermal contact and good electrical insulation. Pure metals have high thermal conductivities, and one would think that metal alloys should also have high conductivities. One would expect an alloy made of two metals of thermal conductivities k1 and k2 to have a conductivity k between k1 and k2. But this turns out not to be the case. The thermal conductivity of an alloy of two metals is usually much lower than that of either metal, as shown in Table 1–2. Even small amounts in a pure metal of “foreign” molecules that are good conductors themselves seriously disrupt the flow of heat in that metal. For example, the thermal conductivity of steel containing just 1 percent of chrome is 62 W/m · °C, while the thermal conductivities of iron and chromium are 83 and 95 W/m · °C, respectively. The thermal conductivities of materials vary with temperature (Table 1–3). The variation of thermal conductivity over certain temperature ranges is negligible for some materials, but significant for others, as shown in Fig. 1–28. The thermal conductivities of certain solids exhibit dramatic increases at temperatures near absolute zero, when these solids become superconductors. For example, the conductivity of copper reaches a maximum value of about 20,000 W/m · °C at 20 K, which is about 50 times the conductivity at room temperature. The thermal conductivities and other thermal properties of various materials are given in Tables A-3 to A-16. 10,000 k, W/m·°C Solids Liquids Gases Diamonds 1000 Type IIa Type IIb Type I Silver Copper 100 Gold Aluminum Tungsten Platinum Iron 10 Aluminum oxide Pyroceram glass Clear fused quartz 1 Water 0.1 Helium Carbon tetrachloride FIGURE 1–28 The variation of the thermal conductivity of various solids, liquids, and gases with temperature (from White, Ref. 10). Air Steam Argon 0.01 200 400 600 800 T, K 1000 1200 1400 cen58933_ch01.qxd 9/10/2002 8:29 AM Page 23 23 CHAPTER 1 The temperature dependence of thermal conductivity causes considerable complexity in conduction analysis. Therefore, it is common practice to evaluate the thermal conductivity k at the average temperature and treat it as a constant in calculations. In heat transfer analysis, a material is normally assumed to be isotropic; that is, to have uniform properties in all directions. This assumption is realistic for most materials, except those that exhibit different structural characteristics in different directions, such as laminated composite materials and wood. The thermal conductivity of wood across the grain, for example, is different than that parallel to the grain. Thermal Diffusivity The product Cp, which is frequently encountered in heat transfer analysis, is called the heat capacity of a material. Both the specific heat Cp and the heat capacity Cp represent the heat storage capability of a material. But Cp expresses it per unit mass whereas Cp expresses it per unit volume, as can be noticed from their units J/kg · °C and J/m3 · °C, respectively. Another material property that appears in the transient heat conduction analysis is the thermal diffusivity, which represents how fast heat diffuses through a material and is defined as Heat conducted Heat stored k Cp (m2/s) (1-23) Note that the thermal conductivity k represents how well a material conducts heat, and the heat capacity Cp represents how much energy a material stores per unit volume. Therefore, the thermal diffusivity of a material can be viewed as the ratio of the heat conducted through the material to the heat stored per unit volume. A material that has a high thermal conductivity or a low heat capacity will obviously have a large thermal diffusivity. The larger the thermal diffusivity, the faster the propagation of heat into the medium. A small value of thermal diffusivity means that heat is mostly absorbed by the material and a small amount of heat will be conducted further. The thermal diffusivities of some common materials at 20°C are given in Table 1–4. Note that the thermal diffusivity ranges from 0.14 10 6 m2/s for water to 174 10 6 m2/s for silver, which is a difference of more than a thousand times. Also note that the thermal diffusivities of beef and water are the same. This is not surprising, since meat as well as fresh vegetables and fruits are mostly water, and thus they possess the thermal properties of water. EXAMPLE 1–6 Measuring the Thermal Conductivity of a Material A common way of measuring the thermal conductivity of a material is to sandwich an electric thermofoil heater between two identical samples of the material, as shown in Fig. 1–29. The thickness of the resistance heater, including its cover, which is made of thin silicon rubber, is usually less than 0.5 mm. A circulating fluid such as tap water keeps the exposed ends of the samples at constant temperature. The lateral surfaces of the samples are well insulated to ensure that heat transfer through the samples is one-dimensional. Two thermocouples are embedded into each sample some distance L apart, and a TABLE 1–4 The thermal diffusivities of some materials at room temperature , m2/s* Material Silver Gold Copper Aluminum Iron Mercury (l) Marble Ice Concrete Brick Heavy soil (dry) Glass Glass wool Water (l) Beef Wood (oak) 149 127 113 97.5 22.8 4.7 1.2 1.2 0.75 0.52 0.52 0.34 0.23 0.14 0.14 0.13 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 *Multiply by 10.76 to convert to ft2/s. Cooling fluid Sample Insulation Thermocouple L ∆T1 a Resistance heater Sample a L ∆T1 Cooling fluid FIGURE 1–29 Apparatus to measure the thermal conductivity of a material using two identical samples and a thin resistance heater (Example 1–6). cen58933_ch01.qxd 9/10/2002 8:29 AM Page 24 24 HEAT TRANSFER differential thermometer reads the temperature drop T across this distance along each sample. When steady operating conditions are reached, the total rate of heat transfer through both samples becomes equal to the electric power drawn by the heater, which is determined by multiplying the electric current by the voltage. In a certain experiment, cylindrical samples of diameter 5 cm and length 10 cm are used. The two thermocouples in each sample are placed 3 cm apart. After initial transients, the electric heater is observed to draw 0.4 A at 110 V, and both differential thermometers read a temperature difference of 15°C. Determine the thermal conductivity of the sample. SOLUTION The thermal conductivity of a material is to be determined by ensuring one-dimensional heat conduction, and by measuring temperatures when steady operating conditions are reached. Assumptions 1 Steady operating conditions exist since the temperature readings do not change with time. 2 Heat losses through the lateral surfaces of the apparatus are negligible since those surfaces are well insulated, and thus the entire heat generated by the heater is conducted through the samples. 3 The apparatus possesses thermal symmetry. Analysis The electrical power consumed by the resistance heater and converted to heat is · We (110 V)(0.4 A) VI 44 W The rate of heat flow through each sample is · Q 1 2 · We 1 2 (44 W) 22 W since only half of the heat generated will flow through each sample because of symmetry. Reading the same temperature difference across the same distance in each sample also confirms that the apparatus possesses thermal symmetry. The heat transfer area is the area normal to the direction of heat flow, which is the cross-sectional area of the cylinder in this case: A 1 4 D2 1 4 (0.05 m)2 0.00196 m2 Noting that the temperature drops by 15°C within 3 cm in the direction of heat flow, the thermal conductivity of the sample is determined to be · Q kA T →k L · QL AT (22 W)(0.03 m) (0.00196 m2)(15°C) 22.4 W/m · °C Discussion Perhaps you are wondering if we really need to use two samples in the apparatus, since the measurements on the second sample do not give any additional information. It seems like we can replace the second sample by insulation. Indeed, we do not need the second sample; however, it enables us to verify the temperature measurements on the first sample and provides thermal symmetry, which reduces experimental error. EXAMPLE 1–7 Conversion between SI and English Units An engineer who is working on the heat transfer analysis of a brick building in English units needs the thermal conductivity of brick. But the only value he can cen58933_ch01.qxd 9/10/2002 8:29 AM Page 25 25 CHAPTER 1 find from his handbooks is 0.72 W/m · °C, which is in SI units. To make matters worse, the engineer does not have a direct conversion factor between the two unit systems for thermal conductivity. Can you help him out? SOLUTION The situation this engineer is facing is not unique, and most engineers often find themselves in a similar position. A person must be very careful during unit conversion not to fall into some common pitfalls and to avoid some costly mistakes. Although unit conversion is a simple process, it requires utmost care and careful reasoning. The conversion factors for W and m are straightforward and are given in conversion tables to be 1W 1m 3.41214 Btu/h 3.2808 ft But the conversion of °C into °F is not so simple, and it can be a source of error if one is not careful. Perhaps the first thought that comes to mind is to replace °C by (°F 32)/1.8 since T(°C) [T(°F) 32]/1.8. But this will be wrong since the °C in the unit W/m · °C represents per °C change in temperature. Noting that 1°C change in temperature corresponds to 1.8°F, the proper conversion factor to be used is 1°C 1.8°F Substituting, we get 1 W/m · °C 3.41214 Btu/h (3.2808 ft)(1.8°F) 0.5778 Btu/h · ft · °F k = 0.72 W/m·°C = 0.42 Btu/h·ft·°F which is the desired conversion factor. Therefore, the thermal conductivity of the brick in English units is kbrick 0.72 W/m · °C 0.72 (0.5778 Btu/h · ft · °F) 0.42 Btu/h · ft · °F Discussion Note that the thermal conductivity value of a material in English units is about half that in SI units (Fig. 1–30). Also note that we rounded the result to two significant digits (the same number in the original value) since expressing the result in more significant digits (such as 0.4160 instead of 0.42) would falsely imply a more accurate value than the original one. 1–7 I CONVECTION Convection is the mode of energy transfer between a solid surface and the adjacent liquid or gas that is in motion, and it involves the combined effects of conduction and fluid motion. The faster the fluid motion, the greater the convection heat transfer. In the absence of any bulk fluid motion, heat transfer between a solid surface and the adjacent fluid is by pure conduction. The presence of bulk motion of the fluid enhances the heat transfer between the solid surface and the fluid, but it also complicates the determination of heat transfer rates. FIGURE 1–30 The thermal conductivity value in English units is obtained by multiplying the value in SI units by 0.5778. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 26 26 HEAT TRANSFER Velocity variation of air T T Air flow As Temperature variation of air · Qconv Ts Hot Block FIGURE 1–31 Heat transfer from a hot surface to air by convection. Forced convection Natural convection Air Air hot egg hot egg FIGURE 1–32 Consider the cooling of a hot block by blowing cool air over its top surface (Fig. 1–31). Energy is first transferred to the air layer adjacent to the block by conduction. This energy is then carried away from the surface by convection, that is, by the combined effects of conduction within the air that is due to random motion of air molecules and the bulk or macroscopic motion of the air that removes the heated air near the surface and replaces it by the cooler air. Convection is called forced convection if the fluid is forced to flow over the surface by external means such as a fan, pump, or the wind. In contrast, convection is called natural (or free) convection if the fluid motion is caused by buoyancy forces that are induced by density differences due to the variation of temperature in the fluid (Fig. 1–32). For example, in the absence of a fan, heat transfer from the surface of the hot block in Fig. 1–31 will be by natural convection since any motion in the air in this case will be due to the rise of the warmer (and thus lighter) air near the surface and the fall of the cooler (and thus heavier) air to fill its place. Heat transfer between the block and the surrounding air will be by conduction if the temperature difference between the air and the block is not large enough to overcome the resistance of air to movement and thus to initiate natural convection currents. Heat transfer processes that involve change of phase of a fluid are also considered to be convection because of the fluid motion induced during the process, such as the rise of the vapor bubbles during boiling or the fall of the liquid droplets during condensation. Despite the complexity of convection, the rate of convection heat transfer is observed to be proportional to the temperature difference, and is conveniently expressed by Newton’s law of cooling as The cooling of a boiled egg by forced and natural convection. TABLE 1–5 Typical values of convection heat transfer coefficient Type of convection Free convection of gases Free convection of liquids Forced convection of gases Forced convection of liquids Boiling and condensation h, W/m2 · °C* 2–25 10–1000 25–250 50–20,000 · Q conv hAs (Ts T) (W) (1-24) where h is the convection heat transfer coefficient in W/m2 · °C or Btu/h · ft2 · °F, 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. Note that at the surface, the fluid temperature equals the surface temperature of the solid. The convection heat transfer coefficient h is not a property of the fluid. It is an experimentally determined parameter whose value depends on all the variables influencing convection such as the surface geometry, the nature of fluid motion, the properties of the fluid, and the bulk fluid velocity. Typical values of h are given in Table 1–5. Some people do not consider convection to be a fundamental mechanism of heat transfer since it is essentially heat conduction in the presence of fluid motion. But we still need to give this combined phenomenon a name, unless we are willing to keep referring to it as “conduction with fluid motion.” Thus, it is practical to recognize convection as a separate heat transfer mechanism despite the valid arguments to the contrary. 2500–100,000 *Multiply by 0.176 to convert to Btu/h · ft2 · °F. EXAMPLE 1–8 Measuring Convection Heat Transfer Coefficient A 2-m-long, 0.3-cm-diameter electrical wire extends across a room at 15°C, as shown in Fig. 1–33. Heat is generated in the wire as a result of resistance heating, and the surface temperature of the wire is measured to be 152°C in steady cen58933_ch01.qxd 9/10/2002 8:29 AM Page 27 27 CHAPTER 1 operation. Also, the voltage drop and electric current through the wire are measured to be 60 V and 1.5 A, respectively. Disregarding any heat transfer by radiation, determine the convection heat transfer coefficient for heat transfer between the outer surface of the wire and the air in the room. T• = 15°C 1.5 A 152°C 60 V FIGURE 1–33 SOLUTION 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 When steady operating conditions are reached, the rate of heat loss from the wire will equal the rate of heat generation in the wire as a result of resistance heating. That is, · Q · Egenerated VI (60 V)(1.5 A) 90 W The surface area of the wire is As DL (0.003 m)(2 m) 0.01885 m2 Newton’s law of cooling for convection heat transfer is expressed as · Q conv 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 coefficient is determined to be h · Q conv As(Ts T ) 90 W (0.01885 m2)(152 15)°C 34.9 W/m2 · °C Discussion Note that the simple setup described above can be used to determine the average heat transfer coefficients from a variety of surfaces in air. Also, heat transfer by radiation can be eliminated by keeping the surrounding surfaces at the temperature of the wire. 1–8 I RADIATION Radiation is the 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. Unlike conduction and convection, the transfer of energy by radiation does not require the presence of an intervening medium. In fact, energy transfer by radiation is fastest (at the speed of light) and it suffers no attenuation in a vacuum. This is how the energy of the sun reaches the earth. In heat transfer studies we are interested in thermal radiation, which is the form of radiation emitted by bodies because of their temperature. It differs from other forms of electromagnetic radiation such as x-rays, gamma rays, microwaves, radio waves, and television waves that are not related to temperature. All bodies at a temperature above absolute zero emit thermal radiation. Radiation is a volumetric phenomenon, and all solids, liquids, and gases emit, absorb, or transmit radiation to varying degrees. However, radiation is Schematic for Example 1–8. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 28 28 HEAT TRANSFER · q emit, max = σ Ts4 = 1452 W/ m 2 Ts = 400 K Blackbody (ε = 1) FIGURE 1–34 Blackbody radiation represents the maximum amount of radiation that can be emitted from a surface at a specified temperature. TABLE 1–6 Emissivities of some materials at 300 K Material Emissivity Aluminum foil Anodized aluminum Polished copper Polished gold Polished silver Polished stainless steel Black paint White paint White paper Asphalt pavement Red brick Human skin Wood Soil Water Vegetation 0.07 0.82 0.03 0.03 0.02 0.17 0.98 0.90 0.92–0.97 0.85–0.93 0.93–0.96 0.95 0.82–0.92 0.93–0.96 0.96 0.92–0.96 usually considered to be a surface phenomenon for solids that are opaque to thermal radiation such as metals, wood, and rocks since the radiation emitted by the interior regions of such material can never reach the surface, and the radiation incident on such bodies is usually absorbed within a few microns from the surface. The maximum rate of radiation that can be emitted from a surface at an absolute temperature Ts (in K or R) is given by the Stefan–Boltzmann law as · Q emit, max · · Qref = (1 – α ) Qincident · · Qabs = α Qincident FIGURE 1–35 The absorption of radiation incident on an opaque surface of absorptivity . (W) (1-25) where 5.67 10 8 W/m2 · K4 or 0.1714 10 8 Btu/h · ft2 · R4 is the Stefan–Boltzmann constant. The idealized surface that emits radiation at this maximum rate is called a blackbody, and the radiation emitted by a blackbody is called blackbody radiation (Fig. 1–34). The radiation emitted by all real surfaces is less than the radiation emitted by a blackbody at the same temperature, and is expressed as · Q emit 4 AsT s (W) (1-26) where is the emissivity of the surface. The property emissivity, whose value is in the range 0 1, is a measure of how closely a surface approximates a blackbody for which 1. The emissivities of some surfaces are given in Table 1–6. Another important radiation property of a surface is its absorptivity , which is the fraction of the radiation energy incident on a surface that is absorbed by the surface. Like emissivity, its value is in the range 0 1. A blackbody absorbs the entire radiation incident on it. That is, a blackbody is a perfect absorber ( 1) as it is a perfect emitter. In general, both and of a surface depend on the temperature and the wavelength of the radiation. Kirchhoff’s law of radiation states that the emissivity and the absorptivity of a surface at a given temperature and wavelength are equal. In many practical applications, the surface temperature and the temperature of the source of incident radiation are of the same order of magnitude, and the average absorptivity of a surface is taken to be equal to its average emissivity. The rate at which a surface absorbs radiation is determined from (Fig. 1–35) · Q absorbed · Qincident 4 AsT s · Q incident (W) (1-27) · where Q incident is the rate at which radiation is incident on the surface and is the absorptivity of the surface. For opaque (nontransparent) surfaces, the portion of incident radiation not absorbed by the surface is reflected back. The difference between the rates of radiation emitted by the surface and the radiation absorbed is the net radiation heat transfer. If the rate of radiation absorption is greater than the rate of radiation emission, the surface is said to be gaining energy by radiation. Otherwise, the surface is said to be losing energy by radiation. In general, the determination of the net rate of heat transfer by radiation between two surfaces is a complicated matter since it depends on the properties of the surfaces, their orientation relative to each other, and the interaction of the medium between the surfaces with radiation. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 29 29 CHAPTER 1 When a surface of emissivity and surface area As at an absolute temperature Ts is completely enclosed by a much larger (or black) surface at absolute temperature Tsurr separated by a gas (such as air) that does not intervene with radiation, the net rate of radiation heat transfer between these two surfaces is given by (Fig. 1–36) · Q rad Air As (Ts4 4 Tsurr) (W) hcombined As (Ts T) (W) · Qemitted (1-28) In this special case, the emissivity and the surface area of the surrounding surface do not have any effect on the net radiation heat transfer. Radiation heat transfer to or from a surface surrounded by a gas such as air occurs parallel to conduction (or convection, if there is bulk gas motion) between the surface and the gas. Thus the total heat transfer is determined by adding the contributions of both heat transfer mechanisms. For simplicity and convenience, this is often done by defining a combined heat transfer coefficient hcombined that includes the effects of both convection and radiation. Then the total heat transfer rate to or from a surface by convection and radiation is expressed as · Q total Surrounding surfaces at Tsurr · Qincident ε , As, Ts · Qrad = εσ As (T 4 – T 4 ) s surr FIGURE 1–36 Radiation heat transfer between a surface and the surfaces surrounding it. (1-29) Note that the combined heat transfer coefficient is essentially a convection heat transfer coefficient modified to include the effects of radiation. Radiation is usually significant relative to conduction or natural convection, but negligible relative to forced convection. Thus radiation in forced convection applications is usually disregarded, especially when the surfaces involved have low emissivities and low to moderate temperatures. EXAMPLE 1–9 Radiation Effect on Thermal Comfort It is a common experience to feel “chilly” in winter and “warm” in summer in our homes even when the thermostat setting is kept the same. This is due to the so called “radiation effect” resulting from radiation heat exchange between our bodies and the surrounding surfaces of the walls and the ceiling. Consider a person standing in a room maintained at 22°C at all times. The inner surfaces of the walls, floors, and the ceiling of the house are observed to be at an average temperature of 10°C in winter and 25°C in summer. Determine the rate of radiation heat transfer between this person and the surrounding surfaces if the exposed surface area and the average outer surface temperature of the person are 1.4 m2 and 30°C, respectively (Fig. 1–37). SOLUTION The rates of radiation heat transfer between a person and the surrounding surfaces at specified temperatures are to be determined in summer and 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 0.95 (Table 1–6). Analysis The net rates of radiation heat transfer from the body to the surrounding walls, ceiling, and floor in winter and summer are Room Tsurr 30°C 1.4 m2 · Qrad FIGURE 1–37 Schematic for Example 1–9. cen58933_ch01.qxd 9/10/2002 8:29 AM Page 30 30 HEAT TRANSFER · Q rad, winter As (Ts4 4 Tsurr, winter) (0.95)(5.67 10 8 W/m2 · K4)(1.4 m2) [(30 273)4 (10 273)4] K4 152 W and · Q rad, summer As (Ts4 4 Tsurr, summer) (0.95)(5.67 10 8 W/m2 · K4)(1.4 m2) [(30 273)4 (25 273)4] K4 40.9 W Discussion Note that we must use absolute temperatures in radiation calculations. Also note that the rate of heat loss from the person by radiation is almost four times as large in winter than it is in summer, which explains the “chill” we feel in winter even if the thermostat setting is kept the same. 1–9 T1 OPAQUE SOLID T2 1 mode Conduction T1 GAS T2 Radiation 2 modes Conduction or convection T1 VACUUM Radiation T2 1 mode FIGURE 1–38 Although there are three mechanisms of heat transfer, a medium may involve only two of them simultaneously. I SIMULTANEOUS HEAT TRANSFER MECHANISMS We mentioned that there are three mechanisms of heat transfer, but not all three can exist simultaneously in a medium. For example, heat transfer is only by conduction in opaque solids, but by conduction and radiation in semitransparent solids. Thus, a solid may involve conduction and radiation but not convection. However, a solid may involve heat transfer by convection and/or radiation on its surfaces exposed to a fluid or other surfaces. For example, the outer surfaces of a cold piece of rock will warm up in a warmer environment as a result of heat gain by convection (from the air) and radiation (from the sun or the warmer surrounding surfaces). But the inner parts of the rock will warm up as this heat is transferred to the inner region of the rock by conduction. Heat transfer is by conduction and possibly by radiation in a still fluid (no bulk fluid motion) and by convection and radiation in a flowing fluid. In the absence of radiation, heat transfer through a fluid is either by conduction or convection, depending on the presence of any bulk fluid motion. Convection can be viewed as combined conduction and fluid motion, and conduction in a fluid can be viewed as a special case of convection in the absence of any fluid motion (Fig. 1–38). Thus, when we deal with heat transfer through a fluid, we have either conduction or convection, but not both. Also, gases are practically transparent to radiation, except that some gases are known to absorb radiation strongly at certain wavelengths. Ozone, for example, strongly absorbs ultraviolet radiation. But in most cases, a gas between two solid surfaces does not interfere with radiation and acts effectively as a vacuum. Liquids, on the other hand, are usually strong absorbers of radiation. Finally, heat transfer through a vacuum is by radiation only since conduction or convection requires the presence of a material medium. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 31 31 CHAPTER 1 EXAMPLE 1–10 Room air Heat Loss from a Person 20°C Consider a person standing in a breezy room at 20°C. Determine the total rate of heat transfer from this person if the exposed surface area and the average outer surface temperature of the person are 1.6 m2 and 29°C, respectively, and the convection heat transfer coefficient is 6 W/m2 · °C (Fig. 1–39). · Qconv 29°C · Qrad SOLUTION 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 completely surrounded by the interior surfaces of the room. 3 The surrounding surfaces are at the same temperature as the air in the room. 4 Heat conduction to the floor through the feet is negligible. Properties The emissivity of a person is 0.95 (Table 1–6). Analysis The heat transfer between the person and the air in the room will be by convection (instead of conduction) since it is conceivable that the air in the vicinity of the skin or clothing will warm up and rise as a result of heat transfer from the body, initiating natural convection currents. It appears that the experimentally determined value for the rate of convection heat transfer in this case is 6 W per unit surface area (m2) per unit temperature difference (in K or °C) between the person and the air away from the person. Thus, the rate of convection heat transfer from the person to the air in the room is · Q conv hAs (Ts T) (6 W/m · °C)(1.6 m2)(29 86.4 W 2 20)°C The person will also lose heat by radiation to the surrounding wall surfaces. We take the temperature of the surfaces of the walls, ceiling, and floor to be equal to the air temperature in this case for simplicity, but we recognize that this does not need to be the case. These surfaces may be at a higher or lower temperature than the average temperature of the room air, depending on the outdoor conditions and the structure of the walls. Considering that air does not intervene with radiation and the person is completely enclosed by the surrounding surfaces, the net rate of radiation heat transfer from the person to the surrounding walls, ceiling, and floor is · Q rad As (Ts4 4 Tsurr) (0.95)(5.67 10 8 W/m2 · K4)(1.6 m2) [(29 273)4 (20 273)4] K4 81.7 W Note that we must use absolute temperatures in radiation calculations. Also note that we used the emissivity value for the skin and clothing at room temperature since the emissivity is not expected to change significantly at a slightly higher temperature. Then the rate of total heat transfer from the body is determined by adding these two quantities: · Q total · Q conv · Q rad (86.4 81.7) W 168.1 W · Qcond FIGURE 1–39 Heat transfer from the person described in Example 1–10. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 32 32 HEAT TRANSFER Discussion The heat transfer would be much higher if the person were not dressed since the exposed surface temperature would be higher. Thus, an important function of the clothes is to serve as a barrier against heat transfer. In these calculations, heat transfer through the feet to the floor by conduction, which is usually very small, is neglected. Heat transfer from the skin by perspiration, which is the dominant mode of heat transfer in hot environments, is not considered here. T1 = 300 K · Q T2 = 200 K L = 1 cm ε=1 FIGURE 1–40 EXAMPLE 1–11 Heat Transfer between Two Isothermal Plates Consider steady heat transfer between two large parallel plates at constant temperatures of T1 300 K and T2 200 K that are L 1 cm apart, as shown in Fig. 1–40. Assuming the surfaces to be black (emissivity 1), determine the rate of heat transfer between the plates per unit surface area assuming the gap between the plates is (a) filled with atmospheric air, (b) evacuated, (c) filled with urethane insulation, and (d) filled with superinsulation that has an apparent thermal conductivity of 0.00002 W/m · °C. Schematic for Example 1–11. SOLUTION The total rate of heat transfer between two large parallel plates at specified temperatures is to be determined for four different cases. Assumptions 1 Steady operating conditions exist. 2 There are no natural convection currents in the air between the plates. 3 The surfaces are black and thus 1. Properties The thermal conductivity at the average temperature of 250 K is k 0.0219 W/m · °C for air (Table A-11), 0.026 W/m · °C for urethane insulation (Table A-6), and 0.00002 W/m · °C for the superinsulation. Analysis (a) The rates of conduction and radiation heat transfer between the plates through the air layer are · Q cond kA T2 T1 L (0.0219 W/m · °C)(1 m2) (300 200)°C 0.01 m 219 W and · Q rad 4 A(T 1 (1)(5.67 4 T 2) 10 8 W/m2 · K4)(1 m2)[(300 K)4 (200 K)4] 368 W Therefore, · Q total · Q cond · Q rad 219 368 587 W The heat transfer rate in reality will be higher because of the natural convection currents that are likely to occur in the air space between the plates. (b) When the air space between the plates is evacuated, there will be no conduction or convection, and the only heat transfer between the plates will be by radiation. Therefore, · Q total · Q rad 368 W (c) An opaque solid material placed between two plates blocks direct radiation heat transfer between the plates. Also, the thermal conductivity of an insulating material accounts for the radiation heat transfer that may be occurring through cen58933_ch01.qxd 9/10/2002 8:30 AM Page 33 33 CHAPTER 1 300 K 200 K 300 K 200 K 300 K 200 K 300 K 200 K · Q = 587 W · Q = 368 W · Q = 260 W · Q = 0.2 W 1 cm 1 cm 1 cm 1 cm (a) Air space (b) Vacuum (c) Insulation (d ) Superinsulation FIGURE 1–41 Different ways of reducing heat transfer between two isothermal plates, and their effectiveness. the voids in the insulating material. The rate of heat transfer through the urethane insulation is · Q total · Q cond kA T1 T2 L (300 200)°C 0.01 m (0.026 W/m · °C)(1 m2) 260 W Note that heat transfer through the urethane material is less than the heat transfer through the air determined in (a), although the thermal conductivity of the insulation is higher than that of air. This is because the insulation blocks the radiation whereas air transmits it. (d ) The layers of the superinsulation prevent any direct radiation heat transfer between the plates. However, radiation heat transfer between the sheets of superinsulation does occur, and the apparent thermal conductivity of the superinsulation accounts for this effect. Therefore, · Q total kA T2 T1 L (300 200)°C (0.00002 W/m · °C)(1 m2) 0.01 m 0.2 W 1 which is 1840 of the heat transfer through the vacuum. The results of this example are summarized in Fig. 1–41 to put them into perspective. Discussion This example demonstrates the effectiveness of superinsulations, which are discussed in the next chapter, and explains why they are the insulation of choice in critical applications despite their high cost. EXAMPLE 1–12 Heat Transfer in Conventional and Microwave Ovens The fast and efficient cooking of microwave ovens made them one of the essential appliances in modern kitchens (Fig. 1–42). Discuss the heat transfer mechanisms associated with the cooking of a chicken in microwave and conventional ovens, and explain why cooking in a microwave oven is more efficient. FIGURE 1–42 SOLUTION Food is cooked in a microwave oven by absorbing the electromagnetic radiation energy generated by the microwave tube, called the magnetron. A chicken being cooked in a microwave oven (Example 1–12). cen58933_ch01.qxd 9/10/2002 8:30 AM Page 34 34 HEAT TRANSFER The radiation emitted by the magnetron is not thermal radiation, since its emission is not due to the temperature of the magnetron; rather, it is due to the conversion of electrical energy into electromagnetic radiation at a specified wavelength. The wavelength of the microwave radiation is such that it is reflected by metal surfaces; transmitted by the cookware made of glass, ceramic, or plastic; and absorbed and converted to internal energy by food (especially the water, sugar, and fat) molecules. In a microwave oven, the radiation that strikes the chicken is absorbed by the skin of the chicken and the outer parts. As a result, the temperature of the chicken at and near the skin rises. Heat is then conducted toward the inner parts of the chicken from its outer parts. Of course, some of the heat absorbed by the outer surface of the chicken is lost to the air in the oven by convection. In a conventional oven, the air in the oven is first heated to the desired temperature by the electric or gas heating element. This preheating may take several minutes. The heat is then transferred from the air to the skin of the chicken by natural convection in most ovens or by forced convection in the newer convection ovens that utilize a fan. The air motion in convection ovens increases the convection heat transfer coefficient and thus decreases the cooking time. Heat is then conducted toward the inner parts of the chicken from its outer parts as in microwave ovens. Microwave ovens replace the slow convection heat transfer process in conventional ovens by the instantaneous radiation heat transfer. As a result, microwave ovens transfer energy to the food at full capacity the moment they are turned on, and thus they cook faster while consuming less energy. EXAMPLE 1–13 Heating of a Plate by Solar Energy A thin metal plate is insulated on the back and exposed to solar radiation at the front surface (Fig. 1–43). The exposed surface of the plate has an absorptivity of 0.6 for solar radiation. If solar radiation is incident on the plate at a rate of 700 W/m2 and the surrounding air temperature is 25°C, determine the surface temperature of the plate when the heat loss by convection and radiation equals the solar energy absorbed by the plate. Assume the combined convection and radiation heat transfer coefficient to be 50 W/m2 · °C. 700 W/ m2 α = 0.6 25°C FIGURE 1–43 Schematic for Example 1–13. SOLUTION The back side of the thin metal plate is insulated and the front side is exposed to solar radiation. The surface temperature of the plate is to be determined when it stabilizes. Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the insulated side of the plate is negligible. 3 The heat transfer coefficient remains constant. Properties The solar absorptivity of the plate is given to be 0.6. Analysis The absorptivity of the plate is 0.6, and thus 60 percent of the solar radiation incident on the plate will be absorbed continuously. As a result, the temperature of the plate will rise, and the temperature difference between the plate and the surroundings will increase. This increasing temperature difference will cause the rate of heat loss from the plate to the surroundings to increase. At some point, the rate of heat loss from the plate will equal the rate of solar cen58933_ch01.qxd 9/10/2002 8:30 AM Page 35 35 CHAPTER 1 energy absorbed, and the temperature of the plate will no longer change. The temperature of the plate when steady operation is established is determined from · Egained · Elost or · As q incident, solar hcombined As (Ts T) Solving for Ts and substituting, the plate surface temperature is determined to be Ts T · q incident, solar hcombined 25°C 0.6 (700 W/m2) 50 W/m2 · °C 33.4°C Discussion Note that the heat losses will prevent the plate temperature from rising above 33.4°C. Also, the combined heat transfer coefficient accounts for the effects of both convection and radiation, and thus it is very convenient to use in heat transfer calculations when its value is known with reasonable accuracy. SOLUTION PROBLEM-SOLVING TECHNIQUE The first step in learning any science is to grasp the fundamentals, and to gain a sound knowledge of it. The next step is to master the fundamentals by putting this knowledge to test. This is done by solving significant real-world problems. Solving such problems, especially complicated ones, requires a systematic approach. By using a step-by-step approach, an engineer can reduce the solution of a complicated problem into the solution of a series of simple problems (Fig. 1–44). When solving a problem, we recommend that you use the following steps zealously as applicable. This will help you avoid some of the common pitfalls associated with problem solving. Step 1: Problem Statement In your own words, briefly state the problem, the key information given, and the quantities to be found. This is to make sure that you understand the problem and the objectives before you attempt to solve the problem. Step 2: Schematic Draw a realistic sketch of the physical system involved, and list the relevant information on the figure. The sketch does not have to be something elaborate, but it should resemble the actual system and show the key features. Indicate any energy and mass interactions with the surroundings. Listing the given information on the sketch helps one to see the entire problem at once. Also, check for properties that remain constant during a process (such as temperature during an isothermal process), and indicate them on the sketch. Step 3: Assumptions State any appropriate assumptions made to simplify the problem to make it possible to obtain a solution. Justify the questionable assumptions. Assume reasonable values for missing quantities that are necessary. For example, in the absence of specific data for atmospheric pressure, it can be taken to be AY SY W HARD WAY 1–10 I EA PROBLEM FIGURE 1–44 A step-by-step approach can greatly simplify problem solving. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 36 36 HEAT TRANSFER Given: Air temperature in Denver 1 atm. However, it should be noted in the analysis that the atmospheric pressure decreases with increasing elevation. For example, it drops to 0.83 atm in Denver (elevation 1610 m) (Fig. 1–45). To be found: Density of air Missing information: Atmospheric pressure Assumption #1: Take P = 1 atm (Inappropriate. Ignores effect of altitude. Will cause more than 15% error.) Assumption #2: Take P = 0.83 atm (Appropriate. Ignores only minor effects such as weather.) Step 4: Physical Laws Apply all the relevant basic physical laws and principles (such as the conservation of energy), and reduce them to their simplest form by utilizing the assumptions made. However, the region to which a physical law is applied must be clearly identified first. For example, the heating or cooling of a canned drink is usually analyzed by applying the conservation of energy principle to the entire can. Step 5: Properties FIGURE 1–45 The assumptions made while solving an engineering problem must be reasonable and justifiable. Determine the unknown properties at known states necessary to solve the problem from property relations or tables. List the properties separately, and indicate their source, if applicable. Step 6: Calculations Energy use: $80/yr Substitute the known quantities into the simplified relations and perform the calculations to determine the unknowns. Pay particular attention to the units and unit cancellations, and remember that a dimensional quantity without a unit is meaningless. Also, don’t give a false implication of high accuracy by copying all the digits from the screen of the calculator—round the results to an appropriate number of significant digits. Energy saved by insulation: $200/yr Step 7: Reasoning, Verification, and Discussion IMPOSSIBLE! FIGURE 1–46 The results obtained from an engineering analysis must be checked for reasonableness. Check to make sure that the results obtained are reasonable and intuitive, and verify the validity of the questionable assumptions. Repeat the calculations that resulted in unreasonable values. For example, insulating a water heater that uses $80 worth of natural gas a year cannot result in savings of $200 a year (Fig. 1–46). Also, point out the significance of the results, and discuss their implications. State the conclusions that can be drawn from the results, and any recommendations that can be made from them. Emphasize the limitations under which the results are applicable, and caution against any possible misunderstandings and using the results in situations where the underlying assumptions do not apply. For example, if you determined that wrapping a water heater with a $20 insulation jacket will reduce the energy cost by $30 a year, indicate that the insulation will pay for itself from the energy it saves in less than a year. However, also indicate that the analysis does not consider labor costs, and that this will be the case if you install the insulation yourself. Keep in mind that you present the solutions to your instructors, and any engineering analysis presented to others is a form of communication. Therefore neatness, organization, completeness, and visual appearance are of utmost importance for maximum effectiveness. Besides, neatness also serves as a great checking tool since it is very easy to spot errors and inconsistencies in a neat work. Carelessness and skipping steps to save time often ends up costing more time and unnecessary anxiety. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 37 37 CHAPTER 1 The approach just described is used in the solved example problems without explicitly stating each step, as well as in the Solutions Manual of this text. For some problems, some of the steps may not be applicable or necessary. However, we cannot overemphasize the importance of a logical and orderly approach to problem solving. Most difficulties encountered while solving a problem are not due to a lack of knowledge; rather, they are due to a lack of coordination. You are strongly encouraged to follow these steps in problem solving until you develop your own approach that works best for you. A Remark on Significant Digits In engineering calculations, the information given is not known to more than a certain number of significant digits, usually three digits. Consequently, the results obtained cannot possibly be accurate to more significant digits. Reporting results in more significant digits implies greater accuracy than exists, and it should be avoided. For example, consider a 3.75-L container filled with gasoline whose density is 0.845 kg/L, and try to determine its mass. Probably the first thought that comes to your mind is to multiply the volume and density to obtain 3.16875 kg for the mass, which falsely implies that the mass determined is accurate to six significant digits. In reality, however, the mass cannot be more accurate than three significant digits since both the volume and the density are accurate to three significant digits only. Therefore, the result should be rounded to three significant digits, and the mass should be reported to be 3.17 kg instead of what appears in the screen of the calculator. The result 3.16875 kg would be correct only if the volume and density were given to be 3.75000 L and 0.845000 kg/L, respectively. The value 3.75 L implies that we are fairly confident that the volume is accurate within 0.01 L, and it cannot be 3.74 or 3.76 L. However, the volume can be 3.746, 3.750, 3.753, etc., since they all round to 3.75 L (Fig. 1–47). It is more appropriate to retain all the digits during intermediate calculations, and to do the rounding in the final step since this is what a computer will normally do. When solving problems, we will assume the given information to be accurate to at least three significant digits. Therefore, if the length of a pipe is given to be 40 m, we will assume it to be 40.0 m in order to justify using three significant digits in the final results. You should also keep in mind that all experimentally determined values are subject to measurement errors, and such errors will reflect in the results obtained. For example, if the density of a substance has an uncertainty of 2 percent, then the mass determined using this density value will also have an uncertainty of 2 percent. You should also be aware that we sometimes knowingly introduce small errors in order to avoid the trouble of searching for more accurate data. For example, when dealing with liquid water, we just use the value of 1000 kg/m3 for density, which is the density value of pure water at 0 C. Using this value at 75 C will result in an error of 2.5 percent since the density at this temperature is 975 kg/m3. The minerals and impurities in the water will introduce additional error. This being the case, you should have no reservation in rounding the final results to a reasonable number of significant digits. Besides, having a few percent uncertainty in the results of engineering analysis is usually the norm, not the exception. Given: Volume: Density: V = 3.75 L ρ = 0.845 kg/L (3 significant digits) Also, Find: 3.75 × 0.845 = 3.16875 Mass: m = ρV = 3.16875 kg Rounding to 3 significant digits: m = 3.17 kg FIGURE 1–47 A result with more significant digits than that of given data falsely implies more accuracy. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 38 38 HEAT TRANSFER Engineering Software Packages FIGURE 1–48 An excellent word-processing program does not make a person a good writer; it simply makes a good writer a better and more efficient writer. Perhaps you are wondering why we are about to undertake a painstaking study of the fundamentals of heat transfer. After all, almost all such problems we are likely to encounter in practice can be solved using one of several sophisticated software packages readily available in the market today. These software packages not only give the desired numerical results, but also supply the outputs in colorful graphical form for impressive presentations. It is unthinkable to practice engineering today without using some of these packages. This tremendous computing power available to us at the touch of a button is both a blessing and a curse. It certainly enables engineers to solve problems easily and quickly, but it also opens the door for abuses and misinformation. In the hands of poorly educated people, these software packages are as dangerous as sophisticated powerful weapons in the hands of poorly trained soldiers. Thinking that a person who can use the engineering software packages without proper training on fundamentals can practice engineering is like thinking that a person who can use a wrench can work as a car mechanic. If it were true that the engineering students do not need all these fundamental courses they are taking because practically everything can be done by computers quickly and easily, then it would also be true that the employers would no longer need high-salaried engineers since any person who knows how to use a word-processing program can also learn how to use those software packages. However, the statistics show that the need for engineers is on the rise, not on the decline, despite the availability of these powerful packages. We should always remember that all the computing power and the engineering software packages available today are just tools, and tools have meaning only in the hands of masters. Having the best word-processing program does not make a person a good writer, but it certainly makes the job of a good writer much easier and makes the writer more productive (Fig. 1–48). Hand calculators did not eliminate the need to teach our children how to add or subtract, and the sophisticated medical software packages did not take the place of medical school training. Neither will engineering software packages replace the traditional engineering education. They will simply cause a shift in emphasis in the courses from mathematics to physics. That is, more time will be spent in the classroom discussing the physical aspects of the problems in greater detail, and less time on the mechanics of solution procedures. All these marvelous and powerful tools available today put an extra burden on today’s engineers. They must still have a thorough understanding of the fundamentals, develop a “feel” of the physical phenomena, be able to put the data into proper perspective, and make sound engineering judgments, just like their predecessors. However, they must do it much better, and much faster, using more realistic models because of the powerful tools available today. The engineers in the past had to rely on hand calculations, slide rules, and later hand calculators and computers. Today they rely on software packages. The easy access to such power and the possibility of a simple misunderstanding or misinterpretation causing great damage make it more important today than ever to have a solid training in the fundamentals of engineering. In this text we make an extra effort to put the emphasis on developing an intuitive and physical understanding of natural phenomena instead of on the mathematical details of solution procedures. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 39 39 CHAPTER 1 Engineering Equation Solver (EES) EES is a program that solves systems of linear or nonlinear algebraic or differential equations numerically. It has a large library of built-in thermodynamic property functions as well as mathematical functions, and allows the user to supply additional property data. Unlike some software packages, EES does not solve thermodynamic problems; it only solves the equations supplied by the user. Therefore, the user must understand the problem and formulate it by applying any relevant physical laws and relations. EES saves the user considerable time and effort by simply solving the resulting mathematical equations. This makes it possible to attempt significant engineering problems not suitable for hand calculations, and to conduct parametric studies quickly and conveniently. EES is a very powerful yet intuitive program that is very easy to use, as shown in the examples below. The use and capabilities of EES are explained in Appendix 3. Heat Transfer Tools (HTT) One software package specifically designed to help bridge the gap between the textbook fundamentals and these powerful software packages is Heat Transfer Tools, which may be ordered “bundled” with this text. The software included in that package was developed for instructional use only and thus is applicable only to fundamental problems in heat transfer. While it does not have the power and functionality of the professional, commercial packages, HTT uses research-grade numerical algorithms behind the scenes and modern graphical user interfaces. Each module is custom designed and applicable to a single, fundamental topic in heat transfer to ensure that almost all time at the computer is spent learning heat transfer. Nomenclature and all inputs and outputs are consistent with those used in this and most other textbooks in the field. In addition, with the capability of testing parameters so readily available, one can quickly gain a physical feel for the effects of all the nondimensional numbers that arise in heat transfer. EXAMPLE 1–14 Solving a System of Equations with EES The difference of two numbers is 4, and the sum of the squares of these two numbers is equal to the sum of the numbers plus 20. Determine these two numbers. SOLUTION Relations are given for the difference and the sum of the squares of two numbers. They are to be determined. Analysis We start the EES program by double-clicking on its icon, open a new file, and type the following on the blank screen that appears: x-y=4 x^2+y^2=x+y+20 which is an exact mathematical expression of the problem statement with x and y denoting the unknown numbers. The solution to this system of two cen58933_ch01.qxd 9/10/2002 8:30 AM Page 40 40 HEAT TRANSFER nonlinear equations with two unknowns is obtained by a single click on the “calculator” symbol on the taskbar. It gives x=5 and y=1 Discussion Note that all we did is formulate the problem as we would on paper; EES took care of all the mathematical details of solution. Also note that equations can be linear or nonlinear, and they can be entered in any order with unknowns on either side. Friendly equation solvers such as EES allow the user to concentrate on the physics of the problem without worrying about the mathematical complexities associated with the solution of the resulting system of equations. Throughout the text, problems that are unsuitable for hand calculations and are intended to be solved using EES are indicated by a computer icon. TOPIC OF SPECIAL INTEREST* Thermal Comfort Baby Bird Fox FIGURE 1–49 Most animals come into this world with built-in insulation, but human beings come with a delicate skin. Unlike animals such as a fox or a bear that are born with built-in furs, human beings come into this world with little protection against the harsh environmental conditions (Fig. 1–49). Therefore, we can claim that the search for thermal comfort dates back to the beginning of human history. It is believed that early human beings lived in caves that provided shelter as well as protection from extreme thermal conditions. Probably the first form of heating system used was open fire, followed by fire in dwellings through the use of a chimney to vent out the combustion gases. The concept of central heating dates back to the times of the Romans, who heated homes by utilizing double-floor construction techniques and passing the fire’s fumes through the opening between the two floor layers. The Romans were also the first to use transparent windows made of mica or glass to keep the wind and rain out while letting the light in. Wood and coal were the primary energy sources for heating, and oil and candles were used for lighting. The ruins of south-facing houses indicate that the value of solar heating was recognized early in the history. The term air-conditioning is usually used in a restricted sense to imply cooling, but in its broad sense it means to condition the air to the desired level by heating, cooling, humidifying, dehumidifying, cleaning, and deodorizing. The purpose of the air-conditioning system of a building is to provide complete thermal comfort for its occupants. Therefore, we need to understand the thermal aspects of the human body in order to design an effective air-conditioning system. The building blocks of living organisms are cells, which resemble miniature factories performing various functions necessary for the survival of organisms. The human body contains about 100 trillion cells with an average diameter of 0.01 mm. In a typical cell, thousands of chemical reactions *This section can be skipped without a loss in continuity. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 41 41 CHAPTER 1 occur every second during which some molecules are broken down and energy is released and some new molecules are formed. The high level of chemical activity in the cells that maintain the human body temperature at a temperature of 37.0°C (98.6°F) while performing the necessary bodily functions is called the metabolism. In simple terms, metabolism refers to the burning of foods such as carbohydrates, fat, and protein. The metabolizable energy content of foods is usually expressed by nutritionists in terms of the capitalized Calorie. One Calorie is equivalent to 1 Cal 1 kcal 4.1868 kJ. The rate of metabolism at the resting state is called the basal metabolic rate, which is the rate of metabolism required to keep a body performing the necessary bodily functions such as breathing and blood circulation at zero external activity level. The metabolic rate can also be interpreted as the energy consumption rate for a body. For an average man (30 years old, 70 kg, 1.73 m high, 1.8 m2 surface area), the basal metabolic rate is 84 W. That is, the body is converting chemical energy of the food (or of the body fat if the person had not eaten) into heat at a rate of 84 J/s, which is then dissipated to the surroundings. The metabolic rate increases with the level of activity, and it may exceed 10 times the basal metabolic rate when someone is doing strenuous exercise. That is, two people doing heavy exercising in a room may be supplying more energy to the room than a 1-kW resistance heater (Fig. 1–50). An average man generates heat at a rate of 108 W while reading, writing, typing, or listening to a lecture in a classroom in a seated position. The maximum metabolic rate of an average man is 1250 W at age 20 and 730 at age 70. The corresponding rates for women are about 30 percent lower. Maximum metabolic rates of trained athletes can exceed 2000 W. Metabolic rates during various activities are given in Table 1–7 per unit body surface area. The surface area of a nude body was given by D. DuBois in 1916 as As 0.202m0.425 h0.725 (m2) (1-30) where m is the mass of the body in kg and h is the height in m. Clothing increases the exposed surface area of a person by up to about 50 percent. The metabolic rates given in the table are sufficiently accurate for most purposes, but there is considerable uncertainty at high activity levels. More accurate values can be determined by measuring the rate of respiratory oxygen consumption, which ranges from about 0.25 L/min for an average resting man to more than 2 L/min during extremely heavy work. The entire energy released during metabolism can be assumed to be released as heat (in sensible or latent forms) since the external mechanical work done by the muscles is very small. Besides, the work done during most activities such as walking or riding an exercise bicycle is eventually converted to heat through friction. The comfort of the human body depends primarily on three environmental factors: the temperature, relative humidity, and air motion. The temperature of the environment is the single most important index of comfort. Extensive research is done on human subjects to determine the “thermal comfort zone” and to identify the conditions under which the body feels 1.2 kJ/s 1 kJ/s FIGURE1–50 Two fast-dancing people supply more heat to a room than a 1-kW resistance heater. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 42 42 HEAT TRANSFER TABLE 1–7 Metabolic rates during various activities (from ASHRAE Handbook of Fundamentals, Ref. 1, Chap. 8, Table 4). Metabolic rate* Activity W/m2 Resting: Sleeping Reclining Seated, quiet Standing, relaxed 40 45 60 70 Walking (on the level): 2 mph (0.89 m/s) 3 mph (1.34 m/s) 4 mph (1.79 m/s) 115 150 220 Office Activities: Reading, seated Writing Typing Filing, seated Filing, standing Walking about Lifting/packing 55 60 65 70 80 100 120 Driving/Flying: Car Aircraft, routine Heavy vehicle 60–115 70 185 Miscellaneous Occupational Activities: Cooking Cleaning house Machine work: Light Heavy Handling 50-kg bags Pick and shovel work 95–115 115–140 115–140 235 235 235–280 Miscellaneous Leisure Activities: Dancing, social Calisthenics/exercise Tennis, singles Basketball Wrestling, competitive 140–255 175–235 210–270 290–440 410–505 *Multiply by 1.8 m2 to obtain metabolic rates for an average man. Multiply by 0.3171 to convert to Btu/h · ft2. comfortable in an environment. It has been observed that most normally clothed people resting or doing light work feel comfortable in the operative temperature (roughly, the average temperature of air and surrounding surfaces) range of 23°C to 27°C or 73°C to 80°F (Fig. 1–51). For unclothed people, this range is 29°C to 31°C. Relative humidity also has a considerable effect on comfort since it is a measure of air ’s ability to absorb moisture and thus it affects the amount of heat a body can dissipate by evaporation. High relative humidity slows down heat rejection by evaporation, especially at high temperatures, and low relative humidity speeds it up. The desirable level of relative humidity is the broad range of 30 to 70 percent, with 50 percent being the most desirable level. Most people at these conditions feel neither hot nor cold, and the body does not need to activate any of the defense mechanisms to maintain the normal body temperature (Fig. 1–52). Another factor that has a major effect on thermal comfort is excessive air motion or draft, which causes undesired local cooling of the human body. Draft is identified by many as a most annoying factor in work places, automobiles, and airplanes. Experiencing discomfort by draft is most common among people wearing indoor clothing and doing light sedentary work, and least common among people with high activity levels. The air velocity should be kept below 9 m/min (30 ft/min) in winter and 15 m/min (50 ft/min) in summer to minimize discomfort by draft, especially when the air is cool. A low level of air motion is desirable as it removes the warm, moist air that builds around the body and replaces it with fresh air. Therefore, air motion should be strong enough to remove heat and moisture from the vicinity of the body, but gentle enough to be unnoticed. High speed air motion causes discomfort outdoors as well. For example, an environment at 10°C (50°F) with 48 km/h winds feels as cold as an environment at 7°C (20°F) with 3 km/h winds because of the chilling effect of the air motion (the wind-chill factor). A comfort system should provide uniform conditions throughout the living space to avoid discomfort caused by nonuniformities such as drafts, asymmetric thermal radiation, hot or cold floors, and vertical temperature stratification. Asymmetric thermal radiation is caused by the cold surfaces of large windows, uninsulated walls, or cold products and the warm surfaces of gas or electric radiant heating panels on the walls or ceiling, solar-heated masonry walls or ceilings, and warm machinery. Asymmetric radiation causes discomfort by exposing different sides of the body to surfaces at different temperatures and thus to different heat loss or gain by radiation. A person whose left side is exposed to a cold window, for example, will feel like heat is being drained from that side of his or her body (Fig. 1–53). For thermal comfort, the radiant temperature asymmetry should not exceed 5°C in the vertical direction and 10°C in the horizontal direction. The unpleasant effect of radiation asymmetry can be minimized by properly sizing and installing heating panels, using double-pane windows, and providing generous insulation at the walls and the roof. Direct contact with cold or hot floor surfaces also causes localized discomfort in the feet. The temperature of the floor depends on the way it is constructed (being directly on the ground or on top of a heated room, being made of wood or concrete, the use of insulation, etc.) as well as the floor cen58933_ch01.qxd 9/10/2002 8:30 AM Page 43 43 CHAPTER 1 °C 2.0 Clothing insulation (clo) covering used such as pads, carpets, rugs, and linoleum. A floor temperature of 23 to 25°C is found to be comfortable to most people. The floor asymmetry loses its significance for people with footwear. An effective and economical way of raising the floor temperature is to use radiant heating panels instead of turning the thermostat up. Another nonuniform condition that causes discomfort is temperature stratification in a room that exposes the head and the feet to different temperatures. For thermal comfort, the temperature difference between the head and foot levels should not exceed 3°C. This effect can be minimized by using destratification fans. It should be noted that no thermal environment will please everyone. No matter what we do, some people will express some discomfort. The thermal comfort zone is based on a 90 percent acceptance rate. That is, an environment is deemed comfortable if only 10 percent of the people are dissatisfied with it. Metabolism decreases somewhat with age, but it has no effect on the comfort zone. Research indicates that there is no appreciable difference between the environments preferred by old and young people. Experiments also show that men and women prefer almost the same environment. The metabolism rate of women is somewhat lower, but this is compensated by their slightly lower skin temperature and evaporative loss. Also, there is no significant variation in the comfort zone from one part of the world to another and from winter to summer. Therefore, the same thermal comfort conditions can be used throughout the world in any season. Also, people cannot acclimatize themselves to prefer different comfort conditions. In a cold environment, the rate of heat loss from the body may exceed the rate of metabolic heat generation. Average specific heat of the human body is 3.49 kJ/kg · °C, and thus each 1°C drop in body temperature corresponds to a deficit of 244 kJ in body heat content for an average 70-kg man. A drop of 0.5°C in mean body temperature causes noticeable but acceptable discomfort. A drop of 2.6°C causes extreme discomfort. A sleeping person will wake up when his or her mean body temperature drops by 1.3°C (which normally shows up as a 0.5°C drop in the deep body and 3°C in the skin area). The drop of deep body temperature below 35°C may damage the body temperature regulation mechanism, while a drop below 28°C may be fatal. Sedentary people reported to feel comfortable at a mean skin temperature of 33.3°C, uncomfortably cold at 31°C, shivering cold at 30°C, and extremely cold at 29°C. People doing heavy work reported to feel comfortable at much lower temperatures, which shows that the activity level affects human performance and comfort. The extremities of the body such as hands and feet are most easily affected by cold weather, and their temperature is a better indication of comfort and performance. A hand-skin temperature of 20°C is perceived to be uncomfortably cold, 15°C to be extremely cold, and 5°C to be painfully cold. Useful work can be performed by hands without difficulty as long as the skin temperature of fingers remains above 16°C (ASHRAE Handbook of Fundamentals, Ref. 1, Chapter 8). The first line of defense of the body against excessive heat loss in a cold environment is to reduce the skin temperature and thus the rate of heat loss from the skin by constricting the veins and decreasing the blood flow to the skin. This measure decreases the temperature of the tissues subjacent to the skin, but maintains the inner body temperature. The next preventive 20 25 Sedentary 50% RH ≤ 30 fpm (0.15 m/s) 1.5 30 Heavy clothing 1.0 Winter clothing 0.5 Summer clothing 0 64 68 72 76 80 °F Operative temperature 84 Upper acceptability limit Optimum Lower acceptability limit FIGURE 1–51 The effect of clothing on the environment temperature that feels comfortable (1 clo 0.155 m2 · °C/W 0.880 ft2 · °F · h/Btu) (from ASHRAE Standard 55-1981). 23°C RH = 50% Air motion 5 m/min FIGURE 1–52 A thermally comfortable environment. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 44 44 HEAT TRANSFER Cold window Warm wall Radiation Radiation FIGURE 1–53 Cold surfaces cause excessive heat loss from the body by radiation, and thus discomfort on that side of the body. Brrr! Shivering FIGURE 1–54 The rate of metabolic heat generation may go up by six times the resting level during total body shivering in cold weather. measure is increasing the rate of metabolic heat generation in the body by shivering, unless the person does it voluntarily by increasing his or her level of activity or puts on additional clothing. Shivering begins slowly in small muscle groups and may double the rate of metabolic heat production of the body at its initial stages. In the extreme case of total body shivering, the rate of heat production may reach six times the resting levels (Fig. 1–54). If this measure also proves inadequate, the deep body temperature starts falling. Body parts furthest away from the core such as the hands and feet are at greatest danger for tissue damage. In hot environments, the rate of heat loss from the body may drop below the metabolic heat generation rate. This time the body activates the opposite mechanisms. First the body increases the blood flow and thus heat transport to the skin, causing the temperature of the skin and the subjacent tissues to rise and approach the deep body temperature. Under extreme heat conditions, the heart rate may reach 180 beats per minute in order to maintain adequate blood supply to the brain and the skin. At higher heart rates, the volumetric efficiency of the heart drops because of the very short time between the beats to fill the heart with blood, and the blood supply to the skin and more importantly to the brain drops. This causes the person to faint as a result of heat exhaustion. Dehydration makes the problem worse. A similar thing happens when a person working very hard for a long time stops suddenly. The blood that has flooded the skin has difficulty returning to the heart in this case since the relaxed muscles no longer force the blood back to the heart, and thus there is less blood available for pumping to the brain. The next line of defense is releasing water from sweat glands and resorting to evaporative cooling, unless the person removes some clothing and reduces the activity level (Fig. 1–55). The body can maintain its core temperature at 37°C in this evaporative cooling mode indefinitely, even in environments at higher temperatures (as high as 200°C during military endurance tests), if the person drinks plenty of liquids to replenish his or her water reserves and the ambient air is sufficiently dry to allow the sweat to evaporate instead of rolling down the skin. If this measure proves inadequate, the body will have to start absorbing the metabolic heat and the deep body temperature will rise. A person can tolerate a temperature rise of 1.4°C without major discomfort but may collapse when the temperature rise reaches 2.8°C. People feel sluggish and their efficiency drops considerably when the core body temperature rises above 39°C. A core temperature above 41°C may damage hypothalamic proteins, resulting in cessation cen58933_ch01.qxd 9/10/2002 8:30 AM Page 45 45 CHAPTER 1 of sweating, increased heat production by shivering, and a heat stroke with irreversible and life-threatening damage. Death can occur above 43°C. A surface temperature of 46°C causes pain on the skin. Therefore, direct contact with a metal block at this temperature or above is painful. However, a person can stay in a room at 100°C for up to 30 min without any damage or pain on the skin because of the convective resistance at the skin surface and evaporative cooling. We can even put our hands into an oven at 200°C for a short time without getting burned. Another factor that affects thermal comfort, health, and productivity is ventilation. Fresh outdoor air can be provided to a building naturally by doing nothing, or forcefully by a mechanical ventilation system. In the first case, which is the norm in residential buildings, the necessary ventilation is provided by infiltration through cracks and leaks in the living space and by the opening of the windows and doors. The additional ventilation needed in the bathrooms and kitchens is provided by air vents with dampers or exhaust fans. With this kind of uncontrolled ventilation, however, the fresh air supply will be either too high, wasting energy, or too low, causing poor indoor air quality. But the current practice is not likely to change for residential buildings since there is not a public outcry for energy waste or air quality, and thus it is difficult to justify the cost and complexity of mechanical ventilation systems. Mechanical ventilation systems are part of any heating and air conditioning system in commercial buildings, providing the necessary amount of fresh outdoor air and distributing it uniformly throughout the building. This is not surprising since many rooms in large commercial buildings have no windows and thus rely on mechanical ventilation. Even the rooms with windows are in the same situation since the windows are tightly sealed and cannot be opened in most buildings. It is not a good idea to oversize the ventilation system just to be on the “safe side” since exhausting the heated or cooled indoor air wastes energy. On the other hand, reducing the ventilation rates below the required minimum to conserve energy should also be avoided so that the indoor air quality can be maintained at the required levels. The minimum fresh air ventilation requirements are listed in Table 1–8. The values are based on controlling the CO2 and other contaminants with an adequate margin of safety, which requires each person be supplied with at least 7.5 L/s (15 ft3/min) of fresh air. Another function of the mechanical ventilation system is to clean the air by filtering it as it enters the building. Various types of filters are available for this purpose, depending on the cleanliness requirements and the allowable pressure drop. Evaporation FIGURE 1–55 In hot environments, a body can dissipate a large amount of metabolic heat by sweating since the sweat absorbs the body heat and evaporates. TABLE 1–8 Minimum fresh air requirements in buildings (from ASHRAE Standard 62-1989) Requirement (per person) L/s ft3/min Classrooms, libraries, supermarkets 8 15 Dining rooms, conference rooms, offices 10 20 Hospital rooms 13 25 Application Hotel rooms Smoking lounges 15 30 (per room) (per room) 30 Retail stores 1.0–1.5 (per m2) 60 0.2–0.3 (per ft2) Residential 0.35 air change per buildings hour, but not less than 7.5 L/s (or 15 ft3/min) per person cen58933_ch01.qxd 9/10/2002 8:30 AM Page 46 46 HEAT TRANSFER SUMMARY In this chapter, the basics of heat transfer are introduced and discussed. The science of thermodynamics deals with the amount of heat transfer as a system undergoes a process from one equilibrium state to another, whereas the science of heat transfer deals with the rate of heat transfer, which is the main quantity of interest in the design and evaluation of heat transfer equipment. The sum of all forms of energy of a system is called total energy, and it includes the internal, kinetic, and potential energies. The internal energy represents the molecular energy of a system, and it consists of sensible, latent, chemical, and nuclear forms. The sensible and latent forms of internal energy can be transferred from one medium to another as a result of a temperature difference, and are referred to as heat or thermal energy. Thus, heat transfer is the exchange of the sensible and latent forms of internal energy between two mediums as a result of a temperature difference. The amount of heat transferred · per unit time is called heat transfer rate and is denoted by Q . ·. The rate of heat transfer per unit area is called heat flux, q A system of fixed mass is called a closed system and a system that involves mass transfer across its boundaries is called an open system or control volume. The first law of thermodynamics or the energy balance for any system undergoing any process can be expressed as Eout Ein Esystem When a stationary closed system involves heat transfer only and no work interactions across its boundary, the energy balance relation reduces to Q mCv T where Q is the amount of net heat transfer to or from the sys· tem. When heat is transferred at a constant rate of Q , the amount of heat transfer during a time interval t can be deter· mined from Q Q t. Under steady conditions and in the absence of any work interactions, the conservation of energy relation for a control volume with one inlet and one exit with negligible changes in kinetic and potential energies can be expressed as · · Q m Cp T · · where m Ac is the mass flow rate and Q is the rate of net heat transfer into or out of the control volume. Heat can be transferred in three different modes: 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, and is expressed by Fourier ’s law of heat conduction as · Q cond kA dT dx where k is the thermal conductivity of the material, A is the area normal to the direction of heat transfer, and dT/dx is the temperature gradient. The magnitude of the rate of heat conduction across a plane layer of thickness L is given by · Q cond kA T L where T is the temperature difference across the layer. Convection is the mode of heat transfer between a solid surface and the adjacent liquid or gas that is in motion, and involves the combined effects of conduction and fluid motion. The rate of convection heat transfer is expressed by Newton’s law of cooling as · Q convection hAs (Ts T∞) where h is the convection heat transfer coefficient in W/m2 · °C or Btu/h · ft2 · °F, 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. Radiation is the 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. The maximum rate of radiation that can be emitted from a surface at an absolute temperature Ts is given by the Stefan–Boltzmann · AsTs4 where , 5.67 10 8 W/m2 · K4 law as Q emit, max 8 2 4 or 0.1714 10 Btu/h · ft · R is the Stefan–Boltzmann constant. When a surface of emissivity and surface area As at an absolute temperature Ts is completely enclosed by a much larger (or black) surface at absolute temperature Tsurr separated by a gas (such as air) that does not intervene with radiation, the net rate of radiation heat transfer between these two surfaces is given by · 4 As (Ts4 Tsurr) Q rad In this case, the emissivity and the surface area of the surrounding surface do not have any effect on the net radiation heat transfer. The rate at which a surface absorbs radiation is determined · · · Q incident where Q incident is the rate at which rafrom Q absorbed diation is incident on the surface and is the absorptivity of the surface. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 47 47 CHAPTER 1 REFERENCES AND SUGGESTED READING 1. American Society of Heating, Refrigeration, and AirConditioning Engineers, Handbook of Fundamentals. Atlanta: ASHRAE, 1993. 2. Y. A. Çengel and R. H. Turner. Fundamentals of ThermalFluid Sciences. New York: McGraw-Hill, 2001. 3. Y. A. Çengel and M. A. Boles. Thermodynamics—An Engineering Approach. 4th ed. New York: McGraw-Hill, 2002. 4. J. P. Holman. Heat Transfer. 9th ed. New York: McGrawHill, 2002. 6. F. Kreith and M. S. Bohn. Principles of Heat Transfer. 6th ed. Pacific Grove, CA: Brooks/Cole, 2001. 7. A. F. Mills. Basic Heat and Mass Transfer. 2nd ed. Upper Saddle River, NJ: Prentice-Hall, 1999. 8. M. N. Ozisik. Heat Transfer—A Basic Approach. New York: McGraw-Hill, 1985. 9. Robert J. Ribando. Heat Transfer Tools. New York: McGraw-Hill, 2002. 10. F. M. White. Heat and Mass Transfer. Reading, MA: Addison-Wesley, 1988. 5. F. P. Incropera and D. P. DeWitt. Introduction to Heat Transfer. 4th ed. New York: John Wiley & Sons, 2002. PROBLEMS* 1–1C How does the science of heat transfer differ from the science of thermodynamics? 1–9C What are the mechanisms of energy transfer to a closed system? How is heat transfer distinguished from the other forms of energy transfer? 1–2C What is the driving force for (a) heat transfer, (b) electric current flow, and (c) fluid flow? 1–10C How are heat, internal energy, and thermal energy related to each other? 1–3C What is the caloric theory? When and why was it abandoned? 1–11C An ideal gas is heated from 50°C to 80°C (a) at constant volume and (b) at constant pressure. For which case do you think the energy required will be greater? Why? Thermodynamics and Heat Transfer 1–4C How do rating problems in heat transfer differ from the sizing problems? 1–5C What is the difference between the analytical and experimental approach to heat transfer? Discuss the advantages and disadvantages of each approach. 1–6C What is the importance of modeling in engineering? How are the mathematical models for engineering processes prepared? 1–7C When modeling an engineering process, how is the right choice made between a simple but crude and a complex but accurate model? Is the complex model necessarily a better choice since it is more accurate? Heat and Other Forms of Energy 1–8C What is heat flux? How is it related to the heat transfer rate? *Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and the SI users can ignore them. Problems with a CD-EES icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed CD. Problems with a computer-EES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. 1–12 A cylindrical resistor element on a circuit board dissipates 0.6 W of power. The resistor is 1.5 cm long, and has a diameter of 0.4 cm. Assuming heat to be transferred uniformly from all surfaces, determine (a) the amount of heat this resistor dissipates during a 24-hour period, (b) the heat flux, and (c) the fraction of heat dissipated from the top and bottom surfaces. 1–13E A logic chip used in a computer dissipates 3 W of power in an environment at 120°F, and has a heat transfer surface area of 0.08 in2. Assuming the heat transfer from the surface to be uniform, determine (a) the amount of heat this chip dissipates during an eight-hour work day, in kWh, and (b) the heat flux on the surface of the chip, in W/in2. 1–14 Consider a 150-W incandescent lamp. The filament of the lamp is 5 cm long and has a diameter of 0.5 mm. The diameter of the glass bulb of the lamp is 8 cm. Determine the heat flux, in W/m2, (a) on the surface of the filament and (b) on the surface of the glass bulb, and (c) calculate how much it will cost per year to keep that lamp on for eight hours a day every day if the unit cost of electricity is $0.08/kWh. Answers: (a) 1.91 106 W/m2, (b) 7500 W/m2, (c) $35.04/yr 1–15 A 1200-W iron is left on the ironing board with its base exposed to the air. About 90 percent of the heat generated in the iron is dissipated through its base whose surface area is 150 cm2, and the remaining 10 percent through other surfaces. Assuming the heat transfer from the surface to be uniform, cen58933_ch01.qxd 9/10/2002 8:30 AM Page 48 48 HEAT TRANSFER D = 8 cm Filament d = 0.5 mm L = 5 cm FIGURE P1–14 determine (a) the amount of heat the iron dissipates during a 2-hour period, in kWh, (b) the heat flux on the surface of the iron base, in W/m2, and (c) the total cost of the electrical energy consumed during this 2-hour period. Take the unit cost of electricity to be $0.07/kWh. 1–16 A 15-cm 20-cm circuit board houses on its surface 120 closely spaced logic chips, each dissipating 0.12 W. If the heat transfer from the back surface of the board is negligible, determine (a) the amount of heat this circuit board dissipates during a 10-hour period, in kWh, and (b) the heat flux on the surface of the circuit board, in W/m2. Chips 15 cm 1–18 The average specific heat of the human body is 3.6 kJ/kg · °C. If the body temperature of a 70-kg man rises from 37°C to 39°C during strenuous exercise, determine the increase in the thermal energy content of the body as a result of this rise in body temperature. 1–19 Infiltration of cold air into a warm house during winter through the cracks around doors, windows, and other openings is a major source of energy loss since the cold air that enters needs to be heated to the room temperature. The infiltration is often expressed in terms of ACH (air changes per hour). An ACH of 2 indicates that the entire air in the house is replaced twice every hour by the cold air outside. Consider an electrically heated house that has a floor space of 200 m2 and an average height of 3 m at 1000 m elevation, where the standard atmospheric pressure is 89.6 kPa. The house is maintained at a temperature of 22°C, and the infiltration losses are estimated to amount to 0.7 ACH. Assuming the pressure and the temperature in the house remain constant, determine the amount of energy loss from the house due to infiltration for a day during which the average outdoor temperature is 5°C. Also, determine the cost of this energy loss for that day if the unit cost of electricity in that area is $0.082/kWh. Answers: 53.8 kWh/day, $4.41/day 1–20 Consider a house with a floor space of 200 m2 and an average height of 3 m at sea level, where the standard atmospheric pressure is 101.3 kPa. Initially the house is at a uniform temperature of 10°C. Now the electric heater is turned on, and the heater runs until the air temperature in the house rises to an average value of 22°C. Determine how much heat is absorbed by the air assuming some air escapes through the cracks as the heated air in the house expands at constant pressure. Also, determine the cost of this heat if the unit cost of electricity in that area is $0.075/kWh. 1–21E Consider a 60-gallon water heater that is initially filled with water at 45°F. Determine how much energy needs to be transferred to the water to raise its temperature to 140°F. Take the density and specific heat of water to be 62 lbm/ft3 and 1.0 Btu/lbm · °F, respectively. The First Law of Thermodynamics 20 cm FIGURE P1–16 1–17 A 15-cm-diameter aluminum ball is to be heated from 80°C to an average temperature of 200°C. Taking the average density and specific heat of aluminum in this temperature range to be 2700 kg/m3 and Cp 0.90 kJ/kg · °C, respectively, determine the amount of energy that needs to be transAnswer: 515 kJ ferred to the aluminum ball. 1–22C On a hot summer day, a student turns his fan on when he leaves his room in the morning. When he returns in the evening, will his room be warmer or cooler than the neighboring rooms? Why? Assume all the doors and windows are kept closed. 1–23C Consider two identical rooms, one with a refrigerator in it and the other without one. If all the doors and windows are closed, will the room that contains the refrigerator be cooler or warmer than the other room? Why? 1–24C Define mass and volume flow rates. How are they related to each other? cen58933_ch01.qxd 9/10/2002 8:30 AM Page 49 49 CHAPTER 1 1–25 Two 800-kg cars moving at a velocity of 90 km/h have a head-on collision on a road. Both cars come to a complete rest after the crash. Assuming all the kinetic energy of cars is converted to thermal energy, determine the average temperature rise of the remains of the cars immediately after the crash. Take the average specific heat of the cars to be 0.45 kJ/kg · °C. 1–26 A classroom that normally contains 40 people is to be air-conditioned using window air-conditioning units of 5-kW cooling capacity. A person at rest may be assumed to dissipate heat at a rate of 360 kJ/h. There are 10 lightbulbs in the room, each with a rating of 100 W. The rate of heat transfer to the classroom through the walls and the windows is estimated to be 15,000 kJ/h. If the room air is to be maintained at a constant temperature of 21°C, determine the number of window airAnswer: two units conditioning units required. 1–27E A rigid tank contains 20 lbm of air at 50 psia and 80°F. The air is now heated until its pressure is doubled. Determine (a) the volume of the tank and (b) the amount of heat Answers: (a) 80 ft3, (b) 2035 Btu transfer. 1–28 A 1-m3 rigid tank contains hydrogen at 250 kPa and 420 K. The gas is now cooled until its temperature drops to 300 K. Determine (a) the final pressure in the tank and (b) the amount of heat transfer from the tank. 1–29 A 4-m 5-m 6-m room is to be heated by a baseboard resistance heater. It is desired that the resistance heater be able to raise the air temperature in the room from 7°C to 25°C within 15 minutes. Assuming no heat losses from the room and an atmospheric pressure of 100 kPa, determine the required power rating of the resistance heater. Assume constant specific heats at room temperature. Answer: 3.01 kW 1–30 A 4-m 5-m 7-m room is heated by the radiator of a steam heating system. The steam radiator transfers heat at a rate of 10,000 kJ/h and a 100-W fan is used to distribute the warm air in the room. The heat losses from the room are estimated to be at a rate of about 5000 kJ/h. If the initial temperature of the room air is 10°C, determine how long it will take for the air temperature to rise to 20°C. Assume constant specific heats at room temperature. 5000 kJ/h Room 4m×6m×6m Fan FIGURE P1–31 1–31 A student living in a 4-m 6-m 6-m dormitory room turns his 150-W fan on before she leaves her room on a summer day hoping that the room will be cooler when she comes back in the evening. Assuming all the doors and windows are tightly closed and disregarding any heat transfer through the walls and the windows, determine the temperature in the room when she comes back 10 hours later. Use specific heat values at room temperature and assume the room to be at 100 kPa and 15°C in the morning when she leaves. Answer: 58.1°C 1–32E A 10-ft3 tank contains oxygen initially at 14.7 psia and 80°F. A paddle wheel within the tank is rotated until the pressure inside rises to 20 psia. During the process 20 Btu of heat is lost to the surroundings. Neglecting the energy stored in the paddle wheel, determine the work done by the paddle wheel. 1–33 A room is heated by a baseboard resistance heater. When the heat losses from the room on a winter day amount to 7000 kJ/h, it is observed that the air temperature in the room remains constant even though the heater operates continuously. Determine the power rating of the heater, in kW. 1–34 A 50-kg mass of copper at 70°C is dropped into an insulated tank containing 80 kg of water at 25°C. Determine the final equilibrium temperature in the tank. 1–35 A 20-kg mass of iron at 100°C is brought into contact with 20 kg of aluminum at 200°C in an insulated enclosure. Determine the final equilibrium temperature of the combined system. Answer: 168°C 1–36 An unknown mass of iron at 90°C is dropped into an insulated tank that contains 80 L of water at 20°C. At the same Room 4m×5m×7m Steam · Wpw FIGURE P1–30 Water Iron Wpw 10,000 kJ/ h FIGURE P1–36 cen58933_ch01.qxd 9/10/2002 8:30 AM Page 50 50 HEAT TRANSFER time, a paddle wheel driven by a 200-W motor is activated to stir the water. Thermal equilibrium is established after 25 minutes with a final temperature of 27°C. Determine the mass of the iron. Neglect the energy stored in the paddle wheel, and Answer: 72.1 kg take the density of water to be 1000 kg/m3. 1–37E A 90-lbm mass of copper at 160°F and a 50-lbm mass of iron at 200°F are dropped into a tank containing 180 lbm of water at 70°F. If 600 Btu of heat is lost to the surroundings during the process, determine the final equilibrium temperature. 1–38 A 5-m 6-m 8-m room is to be heated by an electrical resistance heater placed in a short duct in the room. Initially, the room is at 15°C, and the local atmospheric pressure is 98 kPa. The room is losing heat steadily to the outside at a rate of 200 kJ/min. A 200-W fan circulates the air steadily through the duct and the electric heater at an average mass flow rate of 50 kg/min. The duct can be assumed to be adiabatic, and there is no air leaking in or out of the room. If it takes 15 minutes for the room air to reach an average temperature of 25°C, find (a) the power rating of the electric heater and (b) the temperature rise that the air experiences each time it passes through the heater. 1–39 A house has an electric heating system that consists of a 300-W fan and an electric resistance heating element placed in a duct. Air flows steadily through the duct at a rate of 0.6 kg/s and experiences a temperature rise of 5°C. The rate of heat loss from the air in the duct is estimated to be 250 W. Determine the power rating of the electric resistance heating element. 1–40 A hair dryer is basically a duct in which a few layers of electric resistors are placed. A small fan pulls the air in and forces it to flow over the resistors where it is heated. Air enters a 1200-W hair dryer at 100 kPa and 22°C, and leaves at 47°C. The cross-sectional area of the hair dryer at the exit is 60 cm2. Neglecting the power consumed by the fan and the heat losses through the walls of the hair dryer, determine (a) the volume flow rate of air at the inlet and (b) the velocity of the air at the Answers: (a) 0.0404 m3/s, (b) 7.30 m/s exit. T2 = 47°C A 2 = 60 cm 2 P1 = 100 kPa T1 = 22°C · We = 1200 W FIGURE P1–40 1–41 The ducts of an air heating system pass through an unheated area. As a result of heat losses, the temperature of the air in the duct drops by 3°C. If the mass flow rate of air is 120 kg/min, determine the rate of heat loss from the air to the cold environment. 1–42E Air enters the duct of an air-conditioning system at 15 psia and 50°F at a volume flow rate of 450 ft3/min. The diameter of the duct is 10 inches and heat is transferred to the air in the duct from the surroundings at a rate of 2 Btu/s. Determine (a) the velocity of the air at the duct inlet and (b) the temperaAnswers: (a) 825 ft/min, (b) 64°F ture of the air at the exit. 1–43 Water is heated in an insulated, constant diameter tube by a 7-kW electric resistance heater. If the water enters the heater steadily at 15°C and leaves at 70°C, determine the mass flow rate of water. 70°C Water 15°C Resistance heater, 7 kW FIGURE P1–43 Heat Transfer Mechanisms 1–44C Define thermal conductivity and explain its significance in heat transfer. 1–45C What are the mechanisms of heat transfer? How are they distinguished from each other? 1–46C What is the physical mechanism of heat conduction in a solid, a liquid, and a gas? 1–47C Consider heat transfer through a windowless wall of a house in a winter day. Discuss the parameters that affect the rate of heat conduction through the wall. 1–48C Write down the expressions for the physical laws that govern each mode of heat transfer, and identify the variables involved in each relation. 1–49C How does heat conduction differ from convection? 1–50C Does any of the energy of the sun reach the earth by conduction or convection? 1–51C How does forced convection differ from natural convection? 1–52C Define emissivity and absorptivity. What is Kirchhoff ’s law of radiation? 1–53C What is a blackbody? How do real bodies differ from blackbodies? 1–54C Judging from its unit W/m · °C, can we define thermal conductivity of a material as the rate of heat transfer through the material per unit thickness per unit temperature difference? Explain. 1–55C Consider heat loss through the two walls of a house on a winter night. The walls are identical, except that one of them has a tightly fit glass window. Through which wall will the house lose more heat? Explain. 1–56C Which is a better heat conductor, diamond or silver? cen58933_ch01.qxd 9/10/2002 8:30 AM Page 51 51 CHAPTER 1 1–57C Consider two walls of a house that are identical except that one is made of 10-cm-thick wood, while the other is made of 25-cm-thick brick. Through which wall will the house lose more heat in winter? 1–58C How do the thermal conductivity of gases and liquids vary with temperature? 105°C 1–59C Why is the thermal conductivity of superinsulation orders of magnitude lower than the thermal conductivity of ordinary insulation? 1–60C Why do we characterize the heat conduction ability of insulators in terms of their apparent thermal conductivity instead of the ordinary thermal conductivity? 1–61C Consider an alloy of two metals whose thermal conductivities are k1 and k2. Will the thermal conductivity of the alloy be less than k1, greater than k2, or between k1 and k2? 1–62 The inner and outer surfaces of a 5-m 6-m brick wall of thickness 30 cm and thermal conductivity 0.69 W/m · °C are maintained at temperatures of 20°C and 5°C, respectively. Determine the rate of heat transfer through the wall, in W. Answer: 1035 W 800 W FIGURE P1–65 1–66E The north wall of an electrically heated home is 20 ft long, 10 ft high, and 1 ft thick, and is made of brick whose thermal conductivity is k 0.42 Btu/h · ft · °F. On a certain winter night, the temperatures of the inner and the outer surfaces of the wall are measured to be at about 62°F and 25°F, respectively, for a period of 8 hours. Determine (a) the rate of heat loss through the wall that night and (b) the cost of that heat loss to the home owner if the cost of electricity is $0.07/kWh. 1–67 In a certain experiment, cylindrical samples of diameter 4 cm and length 7 cm are used (see Fig. 1–29). The two thermocouples in each sample are placed 3 cm apart. After initial transients, the electric heater is observed to draw 0.6 A at 110 V, and both differential thermometers read a temperature difference of 10°C. Determine the thermal conductivity of the Answer: 78.8 W/m · °C sample. Brick wall 20°C 0.4 cm 5°C 30 cm FIGURE P1–62 1–63 The inner and outer surfaces of a 0.5-cm-thick 2-m 2-m window glass in winter are 10°C and 3°C, respectively. If the thermal conductivity of the glass is 0.78 W/m · °C, determine the amount of heat loss, in kJ, through the glass over a period of 5 hours. What would your answer be if the glass were Answers: 78,624 kJ, 39,312 kJ 1 cm thick? Reconsider Problem 1–63. Using EES (or other) software, plot the amount of heat loss through the glass as a function of the window glass thickness in the range of 0.1 cm to 1.0 cm. Discuss the results. 1–68 One way of measuring the thermal conductivity of a material is to sandwich an electric thermofoil heater between two identical rectangular samples of the material and to heavily insulate the four outer edges, as shown in the figure. Thermocouples attached to the inner and outer surfaces of the samples record the temperatures. During an experiment, two 0.5-cm-thick samples 10 cm 10 cm in size are used. When steady operation is reached, the heater is observed to draw 35 W of electric power, and the temperature of each sample is observed to drop from 82°C at the inner surface to 74°C at the outer surface. Determine the thermal conductivity of the material at the average temperature. Samples Insulation Wattmeter 1–64 1–65 An aluminum pan whose thermal conductivity is 237 W/m · °C has a flat bottom with diameter 20 cm and thickness 0.4 cm. Heat is transferred steadily to boiling water in the pan through its bottom at a rate of 800 W. If the inner surface of the bottom of the pan is at 105°C, determine the temperature of the outer surface of the bottom of the pan. Insulation ~ Source 0.5 cm Resistance heater FIGURE P1–68 1–69 Repeat Problem 1–68 for an electric power consumption of 28 W. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 52 52 HEAT TRANSFER 1–70 A heat flux meter attached to the inner surface of a 3-cm-thick refrigerator door indicates a heat flux of 25 W/m2 through the door. Also, the temperatures of the inner and the outer surfaces of the door are measured to be 7°C and 15°C, respectively. Determine the average thermal conductivity of Answer: 0.0938 W/m · °C the refrigerator door. 1–77 A 50-cm-long, 800-W electric resistance heating element with diameter 0.5 cm and surface temperature 120°C is immersed in 60 kg of water initially at 20°C. Determine how long it will take for this heater to raise the water temperature to 80°C. Also, determine the convection heat transfer coefficients at the beginning and at the end of the heating process. 1–71 Consider a person standing in a room maintained at 20°C at all times. The inner surfaces of the walls, floors, and ceiling of the house are observed to be at an average temperature of 12°C in winter and 23°C in summer. Determine the rates of radiation heat transfer between this person and the surrounding surfaces in both summer and winter if the exposed surface area, emissivity, and the average outer surface temperature of the person are 1.6 m2, 0.95, and 32°C, respectively. 1–78 A 5-cm-external-diameter, 10-m-long hot water pipe at 80°C is losing heat to the surrounding air at 5°C by natural convection with a heat transfer coefficient of 25 W/m2 · °C. Determine the rate of heat loss from the pipe by natural conAnswer: 2945 W vection, in W. 1–72 Reconsider Problem 1–71. Using EES (or other) software, plot the rate of radiation heat transfer in winter as a function of the temperature of the inner surface of the room in the range of 8°C to 18°C. Discuss the results. 1–79 A hollow spherical iron container with outer diameter 20 cm and thickness 0.4 cm is filled with iced water at 0°C. If the outer surface temperature is 5°C, determine the approximate rate of heat loss from the sphere, in kW, and the rate at which ice melts in the container. The heat from fusion of water is 333.7 kJ/kg. 5°C 1–73 For heat transfer purposes, a standing man can be modeled as a 30-cm-diameter, 170-cm-long vertical cylinder with both the top and bottom surfaces insulated and with the side surface at an average temperature of 34°C. For a convection heat transfer coefficient of 15 W/m2 · °C, determine the rate of heat loss from this man by convection in an environment at Answer: 336 W 20°C. Iced water 0.4 cm 1–74 Hot air at 80°C is blown over a 2-m 4-m flat surface at 30°C. If the average convection heat transfer coefficient is 55 W/m2 · °C, determine the rate of heat transfer from the air to Answer: 22 kW the plate, in kW. FIGURE P1–79 1–75 Reconsider Problem 1–74. Using EES (or other) software, plot the rate of heat transfer as a function of the heat transfer coefficient in the range of 20 W/m2 · °C to 100 W/m2 · °C. Discuss the results. Reconsider Problem 1–79. Using EES (or other) software, plot the rate at which ice melts as a function of the container thickness in the range of 0.2 cm to 2.0 cm. Discuss the results. 1–76 The heat generated in the circuitry on the surface of a silicon chip (k 130 W/m · °C) is conducted to the ceramic substrate to which it is attached. The chip is 6 mm 6 mm in size and 0.5 mm thick and dissipates 3 W of power. Disregarding any heat transfer through the 0.5-mm-high side surfaces, determine the temperature difference between the front and back surfaces of the chip in steady operation. 1–81E The inner and outer glasses of a 6-ft 6-ft doublepane window are at 60°F and 42°F, respectively. If the 0.25-in. space between the two glasses is filled with still air, determine the rate of heat transfer through the window. Silicon chip 0.5 mm Ceramic substrate FIGURE P1–76 6 6 mm m m 3W 1–80 Answer: 439 Btu/h 1–82 Two surfaces of a 2-cm-thick plate are maintained at 0°C and 80°C, respectively. If it is determined that heat is transferred through the plate at a rate of 500 W/m2, determine its thermal conductivity. 1–83 Four power transistors, each dissipating 15 W, are mounted on a thin vertical aluminum plate 22 cm 22 cm in size. The heat generated by the transistors is to be dissipated by both surfaces of the plate to the surrounding air at 25°C, which is blown over the plate by a fan. The entire plate can be assumed to be nearly isothermal, and the exposed surface area of the transistor can be taken to be equal to its base area. If the average convection heat transfer coefficient is 25 W/m2 · °C, determine the temperature of the aluminum plate. Disregard any radiation effects. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 53 53 CHAPTER 1 1–84 An ice chest whose outer dimensions are 30 cm 40 cm 40 cm is made of 3-cm-thick Styrofoam (k 0.033 W/m · °C). Initially, the chest is filled with 40 kg of ice at 0°C, and the inner surface temperature of the ice chest can be taken to be 0°C at all times. The heat of fusion of ice at 0°C is 333.7 kJ/kg, and the surrounding ambient air is at 30°C. Disregarding any heat transfer from the 40-cm 40-cm base of the ice chest, determine how long it will take for the ice in the chest to melt completely if the outer surfaces of the ice chest are at 8°C. 1–87E A 200-ft-long section of a steam pipe whose outer diameter is 4 inches passes through an open space at 50°F. The average temperature of the outer surface of the pipe is measured to be 280°F, and the average heat transfer coefficient on that surface is determined to be 6 Btu/h · ft2 · °F. Determine (a) the rate of heat loss from the steam pipe and (b) the annual cost of this energy loss if steam is generated in a natural gas furnace having an efficiency of 86 percent, and the price of natural gas is $0.58/therm (1 therm 100,000 Btu). Answers: (a) 289,000 Btu/h, (b) $17,074/yr Answer: 32.7 days Tair = 30°C Ice chest 0°C 3 cm 0°C Styrofoam FIGURE P1–84 1–85 A transistor with a height of 0.4 cm and a diameter of 0.6 cm is mounted on a circuit board. The transistor is cooled by air flowing over it with an average heat transfer coefficient of 30 W/m2 · °C. If the air temperature is 55°C and the transistor case temperature is not to exceed 70°C, determine the amount of power this transistor can dissipate safely. Disregard any heat transfer from the transistor base. 1–88 The boiling temperature of nitrogen at atmospheric pressure at sea level (1 atm) is 196°C. Therefore, nitrogen is commonly used in low temperature scientific studies since the temperature of liquid nitrogen in a tank open to the atmosphere will remain constant at 196°C until the liquid nitrogen in the tank is depleted. Any heat transfer to the tank will result in the evaporation of some liquid nitrogen, which has a heat of vaporization of 198 kJ/kg and a density of 810 kg/m3 at 1 atm. Consider a 4-m-diameter spherical tank initially filled with liquid nitrogen at 1 atm and 196°C. The tank is exposed to 20°C ambient air with a heat transfer coefficient of 25 W/m2 · °C. The temperature of the thin-shelled spherical tank is observed to be almost the same as the temperature of the nitrogen inside. Disregarding any radiation heat exchange, determine the rate of evaporation of the liquid nitrogen in the tank as a result of the heat transfer from the ambient air. N2 vapor Tair = 20°C Air 55°C 1 atm Liquid N2 –196°C Power transistor Ts ≤ 70°C 0.6 cm 0.4 cm FIGURE P1–85 1–86 Reconsider Problem 1–85. Using EES (or other) software, plot the amount of power the transistor can dissipate safely as a function of the maximum case temperature in the range of 60°C to 90°C. Discuss the results. FIGURE P1–88 1–89 Repeat Problem 1–88 for liquid oxygen, which has a boiling temperature of 183°C, a heat of vaporization of 213 kJ/kg, and a density of 1140 kg/m3 at 1 atm pressure. 1–90 Reconsider Problem 1–88. Using EES (or other) software, plot the rate of evaporation of liquid nitrogen as a function of the ambient air temperature in the range of 0°C to 35°C. Discuss the results. 1–91 Consider a person whose exposed surface area is 1.7 m2, emissivity is 0.7, and surface temperature is 32°C. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 54 54 HEAT TRANSFER Determine the rate of heat loss from that person by radiation in a large room having walls at a temperature of (a) 300 K and Answers: (a) 37.4 W, (b) 169.2 W (b) 280 K. 1–92 A 0.3-cm-thick, 12-cm-high, and 18-cm-long circuit board houses 80 closely spaced logic chips on one side, each dissipating 0.06 W. The board is impregnated with copper fillings and has an effective thermal conductivity of 16 W/m · °C. All the heat generated in the chips is conducted across the circuit board and is dissipated from the back side of the board to the ambient air. Determine the temperature difference between Answer: 0.042°C the two sides of the circuit board. 1–93 Consider a sealed 20-cm-high electronic box whose base dimensions are 40 cm 40 cm placed in a vacuum chamber. The emissivity of the outer surface of the box is 0.95. If the electronic components in the box dissipate a total of 100 W of power and the outer surface temperature of the box is not to exceed 55°C, determine the temperature at which the surrounding surfaces must be kept if this box is to be cooled by radiation alone. Assume the heat transfer from the bottom surface of the box to the stand to be negligible. 40 cm 40 cm 100 W ε = 0.95 Ts = 60°C Electronic box 20 cm Stand FIGURE P1–93 1–94 Using the conversion factors between W and Btu/h, m and ft, and K and R, express the Stefan–Boltzmann constant 5.67 10 8 W/m2 · K4 in the English unit Btu/h · ft2 · R4. 1–95 An engineer who is working on the heat transfer analysis of a house in English units needs the convection heat transfer coefficient on the outer surface of the house. But the only value he can find from his handbooks is 20 W/m2 · °C, which is in SI units. The engineer does not have a direct conversion factor between the two unit systems for the convection heat transfer coefficient. Using the conversion factors between W and Btu/h, m and ft, and °C and °F, express the given convection heat transfer coefficient in Btu/h · ft2 · °F. Answer: 3.52 Btu/h · ft2 · °F Simultaneous Heat Transfer Mechanisms 1–96C Can all three modes of heat transfer occur simultaneously (in parallel) in a medium? 1–97C Can a medium involve (a) conduction and convection, (b) conduction and radiation, or (c) convection and radiation simultaneously? Give examples for the “yes” answers. 1–98C The deep human body temperature of a healthy person remains constant at 37°C while the temperature and the humidity of the environment change with time. Discuss the heat transfer mechanisms between the human body and the environment both in summer and winter, and explain how a person can keep cooler in summer and warmer in winter. 1–99C We often turn the fan on in summer to help us cool. Explain how a fan makes us feel cooler in the summer. Also explain why some people use ceiling fans also in winter. 1–100 Consider a person standing in a room at 23°C. Determine the total rate of heat transfer from this person if the exposed surface area and the skin temperature of the person are 1.7 m2 and 32°C, respectively, and the convection heat transfer coefficient is 5 W/m2 · °C. Take the emissivity of the skin and the clothes to be 0.9, and assume the temperature of the inner surfaces of the room to be the same as the air temperature. Answer: 161 W 1–101 Consider steady heat transfer between two large 290 K and parallel plates at constant temperatures of T1 T2 150 K that are L 2 cm apart. Assuming the surfaces to be black (emissivity 1), determine the rate of heat transfer between the plates per unit surface area assuming the gap between the plates is (a) filled with atmospheric air, (b) evacuated, (c) filled with fiberglass insulation, and (d) filled with superinsulation having an apparent thermal conductivity of 0.00015 W/m · °C. 1–102 A 1.4-m-long, 0.2-cm-diameter electrical wire extends across a room that is maintained at 20°C. Heat is generated in the wire as a result of resistance heating, and the surface temperature of the wire is measured to be 240°C in steady operation. Also, the voltage drop and electric current through the wire are measured to be 110 V and 3 A, respectively. Disregarding any heat transfer by radiation, determine the convection heat transfer coefficient for heat transfer between the outer surface of the wire and the air in the room. Answer: 170.5 W/m2 · °C Room 20°C 240°C Electric resistance heater FIGURE P1–102 1–103 Reconsider Problem 1–102. Using EES (or other) software, plot the convection heat transfer coefficient as a function of the wire surface temperature in the range of 100°C to 300°C. Discuss the results. 1–104E A 2-in-diameter spherical ball whose surface is maintained at a temperature of 170°F is suspended in the middle of a room at 70°F. If the convection heat transfer coefficient is 12 Btu/h · ft2 · °F and the emissivity of the surface is 0.8, determine the total rate of heat transfer from the ball. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 55 55 CHAPTER 1 1–105 A 1000-W iron is left on the iron board with its base exposed to the air at 20°C. The convection heat transfer coefficient between the base surface and the surrounding air is 35 W/m2 · °C. If the base has an emissivity of 0.6 and a surface area of 0.02 m2, determine the temperature of Answer: 674°C the base of the iron. Iron 1000 W of the collector is 100°F. The emissivity of the exposed surface of the collector is 0.9. Determine the rate of heat loss from the collector by convection and radiation during a calm day when the ambient air temperature is 70°F and the effective sky temperature for radiation exchange is 50°F. Take the convection heat transfer coefficient on the exposed surface to be 2.5 Btu/h · ft2 · °F. 20°C Tsky = 50°F Solar collector 70°F FIGURE P1–105 1–106 The outer surface of a spacecraft in space has an emissivity of 0.8 and a solar absorptivity of 0.3. If solar radiation is incident on the spacecraft at a rate of 950 W/m2, determine the surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed. 1–107 A 3-m-internal-diameter spherical tank made of 1-cmthick stainless steel is used to store iced water at 0°C. The tank is located outdoors at 25°C. Assuming the entire steel tank to be at 0°C and thus the thermal resistance of the tank to be negligible, determine (a) the rate of heat transfer to the iced water in the tank and (b) the amount of ice at 0°C that melts during a 24-hour period. The heat of fusion of water at atmospheric pressure is hif 333.7 kJ/kg. The emissivity of the outer surface of the tank is 0.6, and the convection heat transfer coefficient on the outer surface can be taken to be 30 W/m2 · °C. Assume the average surrounding surface temperature for radiation exAnswer: 5898 kg change to be 15°C. The roof of a house consists of a 15-cm-thick concrete slab (k 2 W/m · °C) that is 15 m wide and 20 m long. The emissivity of the outer surface of the roof is 0.9, and the convection heat transfer coefficient on that surface is estimated to be 15 W/m2 · °C. The inner surface of the roof is maintained at 15°C. On a clear winter night, the ambient air is reported to be at 10°C while the night sky temperature for radiation heat transfer is 255 K. Considering both radiation and convection heat transfer, determine the outer surface temperature and the rate of heat transfer through the roof. If the house is heated by a furnace burning natural gas with an efficiency of 85 percent, and the unit cost of natural gas is $0.60/therm (1 therm 105,500 kJ of energy content), determine the money lost through the roof that night during a 14-hour period. FIGURE P1–109E Problem Solving Technique and EES 1–110C What is the value of the engineering software packages in (a) engineering education and (b) engineering practice? 1–111 Determine a positive real root of the following equation using EES: 2x3 1–112 10x0.5 3x 3 Solve the following system of two equations with two unknowns using EES: x3 3xy y2 y 7.75 3.5 1–108 1–109E Consider a flat plate solar collector placed horizontally on the flat roof of a house. The collector is 5 ft wide and 15 ft long, and the average temperature of the exposed surface 1–113 Solve the following system of three equations with three unknowns using EES: 2x y z 3x2 2y xy 2z 1–114 5 z 8 2 Solve the following system of three equations with three unknowns using EES: x x2y z 3y0.5 xz xyz 1 2 2 Special Topic: Thermal Comfort 1–115C What is metabolism? What is the range of metabolic rate for an average man? Why are we interested in metabolic cen58933_ch01.qxd 9/10/2002 8:30 AM Page 56 56 HEAT TRANSFER rate of the occupants of a building when we deal with heating and air conditioning? 1–116C Why is the metabolic rate of women, in general, lower than that of men? What is the effect of clothing on the environmental temperature that feels comfortable? 1–117C What is asymmetric thermal radiation? How does it cause thermal discomfort in the occupants of a room? 1–118C How do (a) draft and (b) cold floor surfaces cause discomfort for a room’s occupants? Resistance heater 1–119C What is stratification? Is it likely to occur at places with low or high ceilings? How does it cause thermal discomfort for a room’s occupants? How can stratification be prevented? 1–120C Why is it necessary to ventilate buildings? What is the effect of ventilation on energy consumption for heating in winter and for cooling in summer? Is it a good idea to keep the bathroom fans on all the time? Explain. Review Problems 1–121 2.5 kg of liquid water initially at 18°C is to be heated to 96°C in a teapot equipped with a 1200-W electric heating element inside. The teapot is 0.8 kg and has an average specific heat of 0.6 kJ/kg · °C. Taking the specific heat of water to be 4.18 kJ/kg · °C and disregarding any heat loss from the teapot, determine how long it will take for the water to be heated. 1–122 A 4-m-long section of an air heating system of a house passes through an unheated space in the attic. The inner diameter of the circular duct of the heating system is 20 cm. Hot air enters the duct at 100 kPa and 65°C at an average velocity of 3 m/s. The temperature of the air in the duct drops to 60°C as a result of heat loss to the cool space in the attic. Determine the rate of heat loss from the air in the duct to the attic under steady conditions. Also, determine the cost of this heat loss per hour if the house is heated by a natural gas furnace having an efficiency of 82 percent, and the cost of the natural gas in that area is $0.58/therm (1 therm 105,500 kJ). Answers: 0.488 kJ/s, $0.012/h 4m 65°C 3 m/s Hot air 60°C FIGURE P1–122 1–123 Reconsider Problem 1–122. Using EES (or other) software, plot the cost of the heat loss per hour as a function of the average air velocity in the range of 1 m/s to 10 m/s. Discuss the results. 1–124 Water flows through a shower head steadily at a rate of 10 L/min. An electric resistance heater placed in the water pipe heats the water from 16°C to 43°C. Taking the density of FIGURE P1–124 water to be 1 kg/L, determine the electric power input to the heater, in kW. In an effort to conserve energy, it is proposed to pass the drained warm water at a temperature of 39°C through a heat exchanger to preheat the incoming cold water. If the heat exchanger has an effectiveness of 0.50 (that is, it recovers only half of the energy that can possibly be transferred from the drained water to incoming cold water), determine the electric power input required in this case. If the price of the electric energy is 8.5 ¢/kWh, determine how much money is saved during a 10-minute shower as a result of installing this heat exchanger. Answers: 18.8 kW, 10.8 kW, $0.0113 1–125 It is proposed to have a water heater that consists of an insulated pipe of 5 cm diameter and an electrical resistor inside. Cold water at 15°C enters the heating section steadily at a rate of 18 L/min. If water is to be heated to 50°C, determine (a) the power rating of the resistance heater and (b) the average velocity of the water in the pipe. 1–126 A passive solar house that is losing heat to the outdoors at an average rate of 50,000 kJ/h is maintained at 22°C at all times during a winter night for 10 hours. The house is to be heated by 50 glass containers each containing 20 L of water heated to 80°C during the day by absorbing solar energy. A thermostat-controlled 15-kW back-up electric resistance heater turns on whenever necessary to keep the house at 22°C. (a) How long did the electric heating system run that night? (b) How long would the electric heater have run that night if the house incorporated no solar heating? Answers: (a) 4.77 h, (b) 9.26 h 1–127 It is well known that wind makes the cold air feel much colder as a result of the windchill effect that is due to the increase in the convection heat transfer coefficient with increasing air velocity. The windchill effect is usually expressed in terms of the windchill factor, which is the difference between the actual air temperature and the equivalent calm-air cen58933_ch01.qxd 9/10/2002 8:30 AM Page 57 57 CHAPTER 1 50,000 kJ/h and the surrounding air temperature is 10°C, determine the surface temperature of the plate when the heat loss by convection equals the solar energy absorbed by the plate. Take the convection heat transfer coefficient to be 30 W/m2 · °C, and disregard any heat loss by radiation. 1–129 A 4-m 5-m 6-m room is to be heated by one ton (1000 kg) of liquid water contained in a tank placed in the room. The room is losing heat to the outside at an average rate of 10,000 kJ/h. The room is initially at 20°C and 100 kPa, and is maintained at an average temperature of 20°C at all times. If the hot water is to meet the heating requirements of this room for a 24-hour period, determine the minimum temperature of the water when it is first brought into the room. Assume constant specific heats for both air and water at room temperature. 22°C Water 80°C Answer: 77.4°C FIGURE P1–126 temperature. For example, a windchill factor of 20°C for an actual air temperature of 5°C means that the windy air at 5°C feels as cold as the still air at 15°C. In other words, a person will lose as much heat to air at 5°C with a windchill factor of 20°C as he or she would in calm air at 15°C. For heat transfer purposes, a standing man can be modeled as a 30-cm-diameter, 170-cm-long vertical cylinder with both the top and bottom surfaces insulated and with the side surface at an average temperature of 34°C. For a convection heat transfer coefficient of 15 W/m2 · °C, determine the rate of heat loss from this man by convection in still air at 20°C. What would your answer be if the convection heat transfer coefficient is increased to 50 W/m2 · °C as a result of winds? What is the windAnswers: 336 W, 1120 W, 32.7°C chill factor in this case? 1–130 Consider a 3-m 3-m 3-m cubical furnace whose top and side surfaces closely approximate black surfaces at a temperature of 1200 K. The base surface has an emissivity of 0.7, and is maintained at 800 K. Determine the net rate of radiation heat transfer to the base surface from the top and Answer: 594,400 W side surfaces. 1–131 Consider a refrigerator whose dimensions are 1.8 m 1.2 m 0.8 m and whose walls are 3 cm thick. The refrigerator consumes 600 W of power when operating and has a COP of 2.5. It is observed that the motor of the refrigerator remains on for 5 minutes and then is off for 15 minutes periodically. If the average temperatures at the inner and outer surfaces of the refrigerator are 6°C and 17°C, respectively, determine the average thermal conductivity of the refrigerator walls. Also, determine the annual cost of operating this refrigerator if the unit cost of electricity is $0.08/kWh. 1–128 A thin metal plate is insulated on the back and exposed to solar radiation on the front surface. The exposed surface of the plate has an absorptivity of 0.7 for solar radiation. If solar radiation is incident on the plate at a rate of 700 W/m2 17°C 6°C 700 W/ m2 α = 0.7 FIGURE P1–128 10°C FIGURE P1–131 1–132 A 0.2-L glass of water at 20°C is to be cooled with ice to 5°C. Determine how much ice needs to be added to the water, in grams, if the ice is at 0°C. Also, determine how much water would be needed if the cooling is to be done with cold water at 0°C. The melting temperature and the heat of fusion of ice at atmospheric pressure are 0°C and 333.7 kJ/kg, respectively, and the density of water is 1 kg/L. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 58 58 HEAT TRANSFER Ice, 0°C Water 0.2 L 20°C FIGURE P1–132 0.6 cm and a thermal conductivity of 0.7 W/m · C. Heat is lost from the outer surface of the cover by convection and radiation with a convection heat transfer coefficient of 10 W/m2 · °C and an ambient temperature of 15°C. Determine the fraction of heat lost from the glass cover by radiation. 1–138 The rate of heat loss through a unit surface area of a window per unit temperature difference between the indoors and the outdoors is called the U-factor. The value of the U-factor ranges from about 1.25 W/m2 · °C (or 0.22 Btu/h · ft2 · °F) for low-e coated, argon-filled, quadruple-pane windows to 6.25 W/m2 · °C (or 1.1 Btu/h · ft2 · °F) for a singlepane window with aluminum frames. Determine the range for the rate of heat loss through a 1.2-m 1.8-m window of a house that is maintained at 20°C when the outdoor air temperature is 8°C. 1.2 m 1–133 Reconsider Problem 1–132. Using EES (or other) software, plot the amount of ice that needs to be added to the water as a function of the ice temperature in the range of 24°C to 0°C. Discuss the results. Indoors 20°C 1–134E In order to cool 1 short ton (2000 lbm) of water at 70°F in a tank, a person pours 160 lbm of ice at 25°F into the water. Determine the final equilibrium temperature in the tank. The melting temperature and the heat of fusion of ice at atmospheric pressure are 32°F and 143.5 Btu/lbm, respectively. . Q Answer: 56.3°F 1–135 Engine valves (Cp 440 J/kg · °C and 7840 kg/m3) are to be heated from 40°C to 800°C in 5 minutes in the heat treatment section of a valve manufacturing facility. The valves have a cylindrical stem with a diameter of 8 mm and a length of 10 cm. The valve head and the stem may be assumed to be of equal surface area, with a total mass of 0.0788 kg. For a single valve, determine (a) the amount of heat transfer, (b) the average rate of heat transfer, and (c) the average heat flux, (d) the number of valves that can be heat treated per day if the heating section can hold 25 valves, and it is used 10 hours per day. 1–136 The hot water needs of a household are met by an electric 60-L hot water tank equipped with a 1.6-kW heating element. The tank is initially filled with hot water at 80°C, and the cold water temperature is 20°C. Someone takes a shower by mixing constant flow rates of hot and cold waters. After a showering period of 8 minutes, the average water temperature in the tank is measured to be 60°C. The heater is kept on during the shower and hot water is replaced by cold water. If the cold water is mixed with the hot water stream at a rate of 0.06 kg/s, determine the flow rate of hot water and the average temperature of mixed water used during the shower. 1–137 Consider a flat plate solar collector placed at the roof of a house. The temperatures at the inner and outer surfaces of glass cover are measured to be 28°C and 25°C, respectively. The glass cover has a surface area of 2.2. m2 and a thickness of Outdoors –8°C 1.8 m FIGURE P1–138 1–139 Reconsider Problem 1–138. Using EES (or other) software, plot the rate of heat loss through the window as a function of the U-factor. Discuss the results. Design and Essay Problems 1–140 Write an essay on how microwave ovens work, and explain how they cook much faster than conventional ovens. Discuss whether conventional electric or microwave ovens consume more electricity for the same task. 1–141 Using information from the utility bills for the coldest month last year, estimate the average rate of heat loss from your house for that month. In your analysis, consider the contribution of the internal heat sources such as people, lights, and appliances. Identify the primary sources of heat loss from your house and propose ways of improving the energy efficiency of your house. 1–142 Design a 1200-W electric hair dryer such that the air temperature and velocity in the dryer will not exceed 50°C and 3/ms, respectively. 1–143 Design an electric hot water heater for a family of four in your area. The maximum water temperature in the tank cen58933_ch01.qxd 9/10/2002 8:30 AM Page 59 59 CHAPTER 1 and the power consumption are not to exceed 60°C and 4 kW, respectively. There are two showers in the house, and the flow rate of water through each of the shower heads is about 10 L/min. Each family member takes a 5-minute shower every morning. Explain why a hot water tank is necessary, and determine the proper size of the tank for this family. 1–144 Conduct this experiment to determine the heat transfer coefficient between an incandescent lightbulb and the surrounding air using a 60-W lightbulb. You will need an indoor– outdoor thermometer, which can be purchased for about $10 in a hardware store, and a metal glue. You will also need a piece of string and a ruler to calculate the surface area of the lightbulb. First, measure the air temperature in the room, and then glue the tip of the thermocouple wire of the thermometer to the glass of the lightbulb. Turn the light on and wait until the temperature reading stabilizes. The temperature reading will give the surface temperature of the lightbulb. Assuming 10 percent of the rated power of the bulb is converted to light, calculate the heat transfer coefficient from Newton’s law of cooling. cen58933_ch01.qxd 9/10/2002 8:30 AM Page 60 ...
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This note was uploaded on 01/28/2010 for the course HEAT ENG taught by Professor Ghaz during the Spring '10 term at University of Guelph.

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