- 1. 1.10 Heat Transfer Aspects of Energy Ibrahim Dincer and Osamah Siddiqui, University of Ontario Institute of Technology, Oshawa, ON, Canada r 2018 Elsevier Inc. All rights reserved. 1.10.1 Introduction 424 1.10.2 Thermodynamics and Laws 424 1.10.2.1 First Law of Thermodynamics 425 1.10.2.1.1 Energy balance 425 1.10.2.1.2 Energy balance of closed systems 425 1.10.2.1.3 Energy and mass balance of control volumes 425 1.10.2.2 Second Law of Thermodynamics 426 1.10.3 Heat Transfer 427 1.10.3.1 Classiﬁcation of Heat Transfer 427 1.10.4 Conduction Heat Transfer 429 1.10.4.1 Heat Conduction Equation 430 1.10.4.2 Steady-State Heat Conduction 431 1.10.4.2.1 Steady-state heat conduction in walls 431 1.10.4.2.2 Steady-state heat conduction in cylinders 432 1.10.4.2.3 Steady-state heat conduction in spheres 432 1.10.4.2.4 Thermal resistance 433 1.10.4.2.4.1 Thermal resistance for a composite wall 434 1.10.4.2.4.2 Thermal resistance for cylinders 434 1.10.4.2.4.3 Thermal resistance for spheres 435 1.10.4.3 Transient Heat Conduction 437 1.10.5 Convection Heat Transfer 442 1.10.5.1 Newton’s Law of Cooling 443 1.10.5.2 The Nusselt Number 445 1.10.5.3 Forced Convection 445 1.10.5.3.1 External ﬂow forced convection 446 1.10.5.3.2 Internal ﬂow forced convection 454 1.10.5.4 Natural Convection 459 1.10.5.5 Boiling Heat Transfer 464 1.10.5.5.1 Pool boiling 464 1.10.5.5.2 Flow boiling 465 1.10.5.6 Condensation Heat Transfer 466 1.10.5.7 Heat Exchangers 466 1.10.6 Radiation Heat Transfer 470 1.10.6.1 The Stefan–Boltzmann Law 471 1.10.7 Case Study 1: Determining the Overall Heat Loss Coefﬁcient for a Flat Plate Solar Collector 473 1.10.8 Case Study 2: Heat Transfer in Parallel-Flow and Counter-Flow Heat Exchangers 475 1.10.9 Future Directions 476 1.10.10 Conclusions 477 References 477 Further Reading 477 Relevant Websites 477 Nomenclature a Acceleration, m/s2 ; absorptivity A Cross-sectional area, m2 ; surface area, m2 Bi Biot number C Experimental constant cp Constant-pressure speciﬁc heat, kJ/kg K cv Constant-volume speciﬁc heat, kJ/kg K c0 Speciﬁc heat ratio D Diameter, m e Speciﬁc energy, J/kg or kJ/kg E Energy, J or kJ Ex Exergy, J or kJ F Force; drag force, N Fo Fourier number g Acceleration due to gravity (¼9.81 m/s2 ) Gr Grashof number h Speciﬁc enthalpy, kJ/kg; heat transfer coefﬁcient, W/m2 1C; head, m H Enthalpy, kJ; overall heat transfer coefﬁcient, W/m2 1C; head, m Comprehensive Energy Systems, Volume 1 doi:10.1016/B978-0-12-809597-3.00109-7 422
- 2. I Electric current, A k Thermal conductivity, W/m 1C L Thickness, m; length, m m Mass, kg; parameter for extended surface; constant _ m Mass ﬂow rate, kg/s n Mole number, kmol Nu Nusselt number P Perimeter, m; Pe Peclet number PE Potential energy, W or kW Pr Prandtl number q Heat rate per unit area, W/m2 ; ﬂow rate per unit width or depth qh Heat generation per unit volume, W/m3 Q Heat transfer, J or kJ _ Q Heat transfer rate, W or kW r Reﬂectivity; radial coordinate; radial distance, m Rt Thermal resistance, 1C/W Ra Rayleigh number Re Reynolds number s Speciﬁc entropy, kJ/kg S Entropy, kJ/K St Stanton number t Time, s; transmissivity T Temperature, 1C or K Ts Absolute temperature of the object surface, K u Speciﬁc internal energy, kJ/kg; velocity in x direction, m/s; variable velocity, m/s U Internal energy, kJ; ﬂow velocity, m/s v Speciﬁc volume, m3 /kg; velocity in y direction, m/s V Volume, m3 ; velocity, m/s Vr Velocity in radial direction, m/s Vx Velocity in x direction, m/s Vy Velocity in y direction, m/s Vz Velocity in z direction, m/s Vy Tangential velocity, m/s _ V Volumetric ﬂow rate, m3 /s W Work, J or kJ _ W Rate of work, W or kW x Quality, kg/kg; Cartesian coordinate; variable Y Characteristic length, m z Height, m Greek letters b Volumetric coefﬁcient of thermal expansion, 1/ K δ Increment; difference D Change in quantity e Surface emissivity, Eddy viscosity Z Efﬁciency y Dimensionless transient temperature; angle f Dimensionless heat transfer rate m Dynamic viscosity, kg/ms; root of the characteristic equation n Kinematic viscosity, m2 /s p Pi number (¼3.1416) r Density, kg/m3 s Stefan–Boltzmann constant, W/m² K4 ; electrical conductivity, 1/ohm m, surface tension S Summation t Shear stress, N/m2 Subscripts and superscripts a Air; medium; surroundings av Average A Fluid A b Black B Fluid B c Cross-sectional, convection, critical cd Conduction cn Condensation cs Control surface cv Control volume des Destroyed dw Dropwise D Diameter e Electrical; end; exit f Fluid; ﬁnal; ﬂow; force; friction fg Vaporization fm Film condition gen Generation hg Heater geometry hs Heat storage H High-temperature ie Internal energy i In in Input inc Inclined j Node l Liquid lat Latent liq Liquid L Low-temperature m Midplane for plane wall; centerline for cylinder mix Mixture n nth value nb Nonblack N nth number o Out out Output p Previous r Radiation s Surface; near surface; saturation; free stream; in direction parallel to streamline sen Sensible sf Surface-ﬂuid sys System t Total Heat Transfer Aspects of Energy 423
- 3. th Thermal tot Total tur Turbulent v Vapor vap Vapor ver Vertical wl Wavy laminar x x direction y y direction z z direction 0 Surroundings; ambient; environment; reference 1 First value; 1st state; initial 1, 2, 3 Points 1.10.1 Introduction The energy of a system constitutes various forms including thermal, chemical, electrical, magnetic, nuclear, kinetic, or potential energy. Macroscopic forms of energy are possessed by a system as a whole compared to a speciﬁed outside reference. Potential and kinetic energies are examples of macroscopic energy forms. Whereas, the forms of energy associated with the molecular structure and level of activity are termed as microscopic forms of energy. The internal energy of a system is deﬁned as the sum of all microscopic energy forms. These include sensible, latent, chemical, and nuclear energy. Sensible energy is related to the kinetic energies of molecules, whereas latent energy is associated with the phase of the system. Chemical energy relates to the atomic bonds of a molecule and nuclear energy, which is associated with the bonds within the atomic nucleus. When a system gains heat, its thermal energy content rises. This is attributed to the increase in molecular activity of the system. The temperature of a given system reﬂects the thermal energy content contained by the system. When two bodies are at different temperatures, a transfer of energy occurs between them until a thermal equilibrium is established. Heat transfer is deﬁned as energy ﬂow between systems or bodies due to the presence of a temperature difference between them. The study of heat transfer involves determining the rate at which energy ﬂows in the form of heat between these bodies having a temperature gradient. The heat transfer rate between any two given bodies is dependent on the existing temperature gradient. Ranging from core industrial processes to everyday household devices, the study of heat transfer is applied in a wide variety of applications and industries. Analysis of heat transfer between systems is based upon the conservation of energy principle, which states energy cannot be created or destroyed; rather it only changes from one form to another. The ﬁrst law of thermodynamics is also an assertion of the principle of conservation of energy. It avers energy as a thermodynamic property. Thermodynamic properties are quantities that either represent the attributes of a complete system or are functions of position that is continuous and does not change rapidly over microscopic distances, except in case of abrupt changes at system boundaries between phases of the given system. Examples of thermodynamic properties include temperature, pressure, volume, viscosity, etc. Thermodynamic aspects deal with quantifying the amount of heat transferred from one body to another, whereas heat transfer studies focus on determining the rate at which the heat is transferred. The objective of this chapter is to present the heat transfer aspects of energy. The underlying thermodynamic principles are discussed ﬁrst, and then followed by heat transfer mechanisms, concepts, and application examples. 1.10.2 Thermodynamics and Laws The term thermodynamics originates from the Greek word “therme” meaning heat and dunamis meaning power. Essentially, it can be deﬁned as the science of energy that includes all characteristics of energy transformations. When two bodies at different temperatures are brought in contact with each other, transfer of energy occurs between them until a thermal equilibrium is established. The Zeroth Law of Thermodynamics asserts that if system A and system B are in thermal equilibrium with a third system C, this signiﬁes systems A and B are in thermal equilibrium with each other. The First Law of Thermodynamics is essentially an avowal of the law of conservation of energy, which simply asserts energy transforms from one form to another, but cannot be created or destroyed. The ﬁrst law is associated with the amount of heat transfer between two systems or bodies; however, the direction in which energy transfer occurs is not speciﬁed. The Second Law of Thermodynamics avers that energy not only has a quantity, but also has an associated quality. The direction in which processes occur is determined by the direction in which the quality of energy decreases. In addition, the second law also determines theoretical performance and efﬁciency limits of energy systems. The Third Law of Thermodynamics asserts that any system would attain its minimum possible energy content at absolute zero temperature. The study of thermodynamics with a macroscopic approach, which does not entail an in depth study of the behavior of microscopic particles, is termed as classical thermodynamics; whereas, the approach of study that takes the micro- scopic molecular activity into consideration is known as statistical thermodynamics. 1.10.2.1 First Law of Thermodynamics As stated above, the ﬁrst law of thermodynamics asserts that energy cannot be created or destroyed; and it can only change from one form to another. This principle allows the analysis of the process of energy transfer between a system and its surroundings by applying an energy balance. 424 Heat Transfer Aspects of Energy
- 4. It is essential to identify the different forms of energy that may be transferred to or from a system. For any given system, energy is transferred to the system or is extracted from the system in the form of heat, work, or mass ﬂow. In addition, energy gain or loss occurs at the system boundary. When heat transfer occurs to a system, there is a rise in the molecular energy. This leads to an increase in the internal energy of the system. Whereas, when heat transfer occurs from the system to a given surrounding, the internal energy of the system decreases as the energy contained within the molecules is decreased. Energy transfer by work can be understood as the transfer of energy to or from a system that does not occur due to a temperature difference between a given system and its surroundings. When work is done by the system, it results in a decrease in the energy of the system and when work is performed on the system, it increases the system’s energy. Mass ﬂow is also a mechanism for transfer of energy. When mass ﬂows into a given system, the energy content of the system is increased due to the amount of energy carried by the mass entering the system. Similarly, when a mass outﬂow occurs, it causes a decrease in the energy of the system. 1.10.2.1.1 Energy balance During any process, the difference between the total amount of energy entering and leaving a system is the equal to the net total energy change of the system. This is known as the energy balance, which can be expressed as: Ei Eo ¼ DEsys ð1Þ where Ei and Eo represent the amount of energy entering and leaving the system, respectively, and DEsys represents the change in the total energy of the system. When applied on a unit time basis, it can be written as: _ Ei _ Eo ¼ dEsys dt ð2Þ Energy balance can be applied to any type of system for any type of process. In order to apply, it requires an identiﬁcation and quantiﬁcation of all types and forms of energy entering or leaving the system. In the analysis of heat transfer processes, only the energy transfer in the form of heat due to the presence of a temperature gradient between bodies is considered. Hence, for such analyses a heat balance is expedient: Qi Qo þ Qgen ¼ DEsys;th ð3Þ where Qi and Qo denote the heat transfer into and out of the system, respectively. Qgen represents the heat generation, and DEsys,th represents the change in the thermal energy content of the system. 1.10.2.1.2 Energy balance of closed systems A system that has a ﬁxed amount of mass with no mass inﬂow or outﬂow across its boundary is known as a closed system. The forms of energy transfer for such systems are either work or heat transfer as no mass enters or leaves such systems. Energy balance, when applied to closed systems, can be expressed as: Ei Eo ¼ DEsys Qi þ Wi Qo Wo ¼ DEsys ð4Þ ðQi QoÞ ðWo WiÞ ¼ DEsys ð5Þ For most cases, the total energy of a system comprises of its internal energy, particularly for systems that are stationary and do not involve a change in kinetic or potential energies. Hence, the energy balance for closed systems that are stationary can be written as: Ei Eo ¼ DU ð6Þ The change in the internal energy of a body can be expressed in terms of the mass, constant volume speciﬁc heat, and the change in temperature as: DU ¼ mcvDT ð7Þ hence, for closed systems that are stationary, the heat balance can be written as: Qnet ¼ mcvDT ð8Þ where Qnet denotes the net transfer of heat into or out of the system. 1.10.2.1.3 Energy and mass balance of control volumes A control volume or open system is a type of system that also includes mass inﬂow or outﬂow. For such systems, energy can be transferred in the form of heat, work, and mass transfer. Turbines, compressors, pumps, and heat exchangers are a few examples of control volumes. For open systems, the principle of conservation of mass is important to consider. It is expressed as the difference between the total mass entering and leaving a control volume during a speciﬁc time interval is equal to the net change of mass within the control volume during that time interval. mi mo ¼ Dmcv ð9Þ Heat Transfer Aspects of Energy 425
- 5. when considered on a per unit time basis, mass balance can be expressed as: _ mi _ mo ¼ dmcv dt ð10Þ where _ mi and _ mo represent the rate of mass inﬂow and outﬂow, respectively. In heat transfer applications, ﬂuid ﬂow within pipes and ducts is encountered. The mass ﬂow rate in such cases can be determined from the ﬂowing passage cross-sectional area, density, and velocity of the ﬂowing ﬂuid. When the ﬂow is approxi- mated as one-dimensional, the ﬂuid properties are assumed to change only in the direction of the ﬂow. In this case, the mass ﬂow rate is expressed as: _ m ¼ rf VAc ð11Þ where rf denotes the density of the ﬂowing ﬂuid, V denotes the average velocity of the ﬂuid in the direction of ﬂow, and Ac represents the cross-sectional area. In addition, the mass ﬂow rate can also be expressed in terms of the volume ﬂow rate and density of the ﬂuid: _ V ¼ VAc ð12Þ _ m ¼ rf _ V ð13Þ where _ V denotes the volume ﬂow rate of the ﬂuid. For steady-ﬂow processes, there is no change in the amount of mass within the control volume with time. Hence, the total mass inﬂow rates are equal to the mass outﬂow rates. X _ mi ¼ X _ mo ð14Þ In many heat transfer applications, single inlet and single outlet cases are encountered. In such cases, for steady ﬂow conditions, the inlet mass ﬂow rate is equal to the outlet mass ﬂow rate: _ mi ¼ _ mo ð15Þ In control volumes, since energy transfer also takes place by mass transfer, it is necessary to determine the total energy of the ﬂuid that enters or leaves a given system. The total energy of a ﬂow per unit mass entering or leaving a system can be expressed as: etotal ¼ flow energy þ internal energy þ kinetic energy þ potential energy etotal ¼ h þ V2 2 þ gz ð16Þ where h represents the enthalpy of the ﬂuid, which is the sum of the internal energy and ﬂow energy of the ﬂuid that is entering or leaving the system. The energy balance for steady ﬂow processes can thus be expressed as following: _ Ei _ Eo ¼ dEsys dt ð17Þ For steady flow processes; dEsys dt ¼ 0 _ Ei _ Eo _ Qi þ _ Wi þ X _ mi h þ V2 2 þ gz ¼ _ Qo þ _ Wo þ X _ mo h þ V2 2 þ gz ð18Þ During processes that do not include energy transfer by work, and in which kinetic and potential energy changes are negligibly small, the energy balance for systems under steady ﬂow conditions can be expressed as: _ Qnet ¼ _ m ho hi ð Þ ¼ _ mcp To Ti ð Þ ð19Þ where h and T represent the enthalpy and temperature of the ﬂowing ﬂuid, respectively. 1.10.2.2 Second Law of Thermodynamics The ﬁrst law of thermodynamics deals with the transformation and conservation of energy. However, when energy transfer occurs between a system and its surroundings, it only takes place in a particular direction. The ﬁrst law is inadequate to describe the direction in which a process will occur. A process can only occur if it satisﬁes both the ﬁrst and second laws of thermodynamics. The second law of thermodynamics utilizes the property of entropy to indicate the direction in which a process can occur. Entropy is a property that measures the amount of molecular disorder in a given system. Real processes can only occur in the direction that obeys the increase of entropy principle. Real processes are subjected to irreversibilities, which vitiate the performance of any given system. Entropy generation indicates the amount of irreversibilities accompanying any given process. The second law of ther- modynamics states that entropy can only be created but not destroyed. For the ﬁrst law, the energy balance was presented; 426 Heat Transfer Aspects of Energy
- 6. however, the second law includes an entropy and exergy balance. Where entropy is denoted by S, the entropy balance for any system can be expressed as follows: Si So þ Sgen ¼ DSsys ð20Þ The ﬁrst law is concerned about the quantity of energy, whereas the second law of thermodynamics asserts energy also has an associated quality. A measure of the maximum useful work potential of a system in a speciﬁc reference environment is known as exergy. Exergy destruction occurs during processes that entail irreversibilities. The amount of exergy destroyed is related to the entropy generation as follows: Exdest ¼ ToSgen ð21Þ where To represents the temperature of the dead state. The exergy balance for a given system can be expressed as: Exi Exo Exdest ¼ DExsys ð22Þ Similar to the energy balance, exergy balance can also be applied to control volumes. When expressed on a rate basis, it can be expressed as: _ Exi _ Exo _ Exdest ¼ dExsys dt ð23Þ Analysis of heat transfer processes is based on the ﬁrst law of thermodynamics and the energy balance. The proceeding sections discuss the types, mechanisms, analyses, and case studies of heat transfer aspects of energy. 1.10.3 Heat Transfer The form of energy transfer that occurs in the presence of a temperature difference between systems is known as heat. Thermo- dynamic analysis focuses on determining the amount of heat transferred between systems, whereas the study of heat transfer involves determining the heat transfer rate between the given systems. 1.10.3.1 Classiﬁcation of Heat Transfer All substances are capable of holding a certain amount of heat. The property that signiﬁes the amount of heat that the substance can hold is its thermal capacity. When heat is applied to a solid, its temperature increases until it reaches the melting point. The melting point of the solid is the highest temperature, which can be reached by the solid phase before the phase change process from solid to liquid starts. The heat that the solid absorbs while raising the temperature to the melting point is sensible heat. The heat that is required to convert the solid to the liquid phase is called latent heat of fusion. Similarly, when heat is applied to a liquid, its temperature increases until it reaches the boiling point. The boiling point of the liquid is the highest temperature that can be reached by the liquid phase at the measured pressure. The heat that the liquid absorbs while raising the temperature to the boiling point is called sensible heat. The heat that is required to convert the liquid to the vapor phase is called latent heat of vaporization. The condition of a substance or system characterized by values of observable macroscopic properties such as temperature and pressure is called the state or phase of the system. Each of the speciﬁc properties of a substance at a given state has a deﬁnite value independent of how that substance reaches the given state. For instance, when enough heat is added or removed from a substance in a given condition, it undergoes a state change process. The temperature of the substance remains constant during the state change process until the process is complete. This can occur from a solid to liquid phase, liquid to vapor phase, or vice versa. Fig. 1 shows a typical diagram of heat addition to ice, which is changed to liquid water after certain amount of heat addition and the liquid phase changes to the vapor phase after more amount of heat addition. As depicted in Fig. 1, the temperature during a phase change process remains constant. A temperature–volume (T–v) diagram shown in Fig. 2 gives a clearer presentation. The line ABCD denotes the constant pressure line showing the states through which water passes as follows: A–B. This represents the process in which water is heated from a given initial temperature to the saturation temperature at the given pressure. At point B, the state is saturated liquid and the quality (x) is zero. Quality denotes the ratio of the mass of vapor to the total mass in a given saturated liquid vapor mixture. B–C. This line represents the vaporization process that occurs at constant temperature. There is only a phase change from saturated liquid to saturated vapor. The quality increases as this process proceeds. When the phase is completely saturated vapor, the quality reaches 100%. C–D. This represents the superheating of the saturated vapor at constant pressure. Only the vapor phase exists. The saturated vapor formed at point C is heated with an increase in temperature forming superheated vapor. E–F–G. This process represents a vaporization process in which the temperature is not constant. It is a constant pressure process. Point F is known as the critical point. At this point, the saturated liquid and saturated vapor phases are identical. Thermodynamic properties at the critical point are referred to as critical thermodynamic properties, for instance, critical pressure, critical temperature, and critical speciﬁc volume. Heat Transfer Aspects of Energy 427
- 7. H–I. This process is a constant pressure heating process in which no phase change process occurs due to the presence of only one phase. This phenomenon occurs at pressures and temperatures higher than the critical pressure and temperature for a given substance. The internal energy content of a system related to the phase of the system is termed as latent energy. The amount of energy absorbed or released by a system during a phase change process is classiﬁed as latent heat. The temperature of a system remains constant during a phase change process, where m denotes the mass of a system and h represents the speciﬁc latent heat of fusion or vaporization, latent heat is expressed as: Qlat ¼ mh ð24Þ The energy required to convert a unit mass of a particular substance from the solid to liquid phase is known as latent heat of fusion; whereas, the energy required to convert a unit mass of liquid of a particular substance from liquid to gaseous state is known as latent heat of vaporization. The internal energy content of a particular system related to the molecular kinetic energy is termed as sensible energy. The energy that is absorbed or released by a system and that leads to a decrease or increase in the temperature is known as sensible heat. In contrast to latent heat, sensible heat involves a temperature change. Sensible heat can be expressed as: Qsen ¼ mcDT ð25Þ where m represents the mass of the system, c denotes the speciﬁc heat of the system, and DT represents the change in temperature. Temperature Volume Saturated-vapor line Saturated-liquid line Liquid water + water vapor Critical point D G I F C B A E H Fig. 2 Temperature–volume diagram for the phase change of water. Temperature Melting point Melting stage Heat removed All water Water + steam Wet steam stage Dry steam (no superheat) Superheated steam Heat added Ice + water Boiling point All ice Fig. 1 The state–change diagram of water. Adapted from Dincer I, Rosen M. Thermal energy storage: systems and applications. 2nd ed. Hoboken, NJ: Wiley; 2011. 428 Heat Transfer Aspects of Energy
- 8. When a temperature difference exists between systems, energy in the form of heat ﬂows from the medium at a higher temperature to the medium at a lower temperature. Heat transfer can occur by three different mechanisms: conduction, con- vection, and radiation. Fig. 3 shows a classiﬁcation of heat transfer. A difference of temperature between mediums is essential for heat transfer to occur. Heat is transferred until a thermal equilibrium is established between the high-temperature medium and the low-temperature medium. 1.10.4 Conduction Heat Transfer Conduction heat transfer is an energy transfer mechanism in which the thermal energy is transferred from more energetic to less energetic neighboring particles. Conduction occurs in solids, liquids, and gases. In solids, the energy transfer occurs due to lattice molecular vibrations as well as movement of free electrons. Whereas, in liquids or gases, the molecules are under constant random motion, and energy transfer occurs due to collision as well as the diffusion of molecules. The conduction heat transfer rate is dependent on the medium material and geometry. In addition, the temperature gradient as well as the thickness of medium affects the heat transfer rate. Conduction heat transfer can be classiﬁed into one-dimensional, two-dimensional, or three-dimensional. The classiﬁcation depends on the variation of temperature within the medium. Three-dimensional cases involve a variation of temperature in all three directions within the medium, leading to a transfer of heat in all three directions. In two-dimensional cases, temperature variation and heat transfer takes place in two directions, as in the third direction, the variation may be negligible. Similarly, for one-dimensional heat transfer problems, temperature variation and transfer of heat is considered only in one direction. Depending on the geometry of the medium, Cartesian, cylindrical, or spherical coordinate systems may be utilized. 1.10.4.1 Heat Conduction Equation The heat conduction equation in a Cartesian coordinate system is obtained by applying the energy balance on a differential rectangular element, and it is expressed as: ∂ ∂x k ∂T ∂x þ ∂ ∂y k ∂T ∂y þ ∂ ∂z k ∂T ∂z þ _ qgen ¼ rc ∂T ∂t ð26Þ where k denotes the thermal conductivity, r represents the density, and c represents the speciﬁc heat of the medium. Heat transfer Latent heat transfer Sensible heat transfer Conduction Steady heat conduction Forced convection External forced convection Internal forced convection Natural convection Transient heat conduction Convection Radiation Fig. 3 Classiﬁcation of heat transfer. Heat Transfer Aspects of Energy 429
- 9. Similarly, for the cylindrical coordinate system, an energy balance is applied on a differential volume in cylindrical coordinates to obtain the heat conduction equation in cylindrical coordinates, and it is expressed as: 1 r ∂ ∂r kr ∂T ∂r þ 1 r2 ∂T ∂f k ∂T ∂f þ ∂ ∂z k ∂T ∂z þ _ qgen ¼ rc ∂T ∂t ð27Þ The heat conduction equation for spherical coordinate system is also obtained by applying the energy balance on a differential spherical coordinate volume element. It is expressed as follows: 1 r2 ∂ ∂r kr2 ∂T ∂r þ 1 r2sin2 y ∂ ∂f k ∂T ∂f þ 1 r2siny ∂ ∂y k sin y ∂T ∂y þ _ qgen ¼ rc ∂T ∂t ð28Þ Example 1: A container wall has a thickness of 2 m and a cross-sectional area of 20 m2 (Fig. 4). It is subjected to a uniform heat generation of 1500 W/m3 . If the density of the wall is 2000 kg/m3 , the thermal conductivity is 20 W/mK, and the speciﬁc heat is 6 kJ/kg K. Determine the rate of change of temperature with time within the wall at x¼0.75 m, when the temperature distribution across the wall is given by T x ð Þ ¼ 500 100x 25x2 Solution: Assumptions: The heat transfer occurring through the wall is considered one-dimensional. And the medium is isotropic with constant (uniform) thermal conductivity. Analysis: The temperature distribution within the wall in the x-direction is given, the heat equation for Cartesian coordinates can be used to determine the rate of change of temperature with time at any given location in the x-direction ∂ ∂x k ∂T ∂x þ ∂ ∂y k ∂T ∂y þ ∂ ∂z k ∂T ∂z þ _ qgen ¼ rc ∂T ∂t For one-dimensional cases, the heat conduction equation can be reduced to ∂ ∂x k ∂T ∂x þ _ qgen ¼ rc ∂T ∂t In the case of constant thermal conductivity, the equation above can be written as k rcp ∂2 T ∂x2 þ _ qgen rcp ¼ ∂T ∂t The temperature distribution is known as T x ð Þ ¼ 500 100x 25x2 hence, ∂2 T ∂x2 can be determined from the above equation as 501C/m2 . Thus, substituting this value in the one-dimensional heat conduction equation to obtain ∂T ∂t ¼ k rcp ∂2 T ∂x2 þ _ qgen rcp ¼ 20 W=mK 2000 kg=m36000 J=kg K 501C=m2 þ 1500 W=m3 2000 kg=m36000 J=kg K L = 2m x T(x) Fig. 4 A wall cross-section with a temperature distribution. 430 Heat Transfer Aspects of Energy
- 10. ∂T ∂t ¼ 8:33 105 1C=s þ 1:25 104 1C=s ∂T ∂t ¼ 4:17 105 1C=s Comments: The temperature in the wall increases with time. In addition, as can be observed from the analysis, the rate of change of temperature with time within the wall is independent of the location x. 1.10.4.2 Steady-State Heat Conduction Steady-state refers to unchanging conditions with time. During conduction heat transfer, when the surface temperatures do not vary with time, such cases can be analyzed without requiring to solve differential equations by utilizing the concept of thermal resistance. In electric circuits, current ﬂows between two points as a virtue of a potential difference between them. Similarly, in thermal circuits, a temperature difference resembles the voltage and the rate of heat transfer resembles the electric current. 1.10.4.2.1 Steady-state heat conduction in walls Energy balance when applied to a plane wall can be expressed as: _ Qi _ Qo ¼ dEwall dt ð29Þ For steady-state conditions, no change of energy of the wall occurs with time, thus the energy balance reduces to _ Qi ¼ _ Qo ð30Þ Henceforth, the conduction heat transfer rate into the wall is equal to the rate of heat transfer within the wall, which is equal to the heat transfer rate out of the wall (Fig. 5). The heat transfer rate occurring through a plane wall under steady-state conditions and one-dimensional cases can be expressed by the Fourier’s law as _ Qi ¼ kA dT dx ð31Þ Under steady-state conditions, since the conduction heat transfer rate through the plane wall is constant, and the thermal conductivity does not vary, dT dx is also a constant. Hence, linear temperature distributions within the wall exist. Therefore, the conduction heat transfer rate can be expressed as: _ Q ¼ kA T1 T2 L ð32Þ Example 2: A wall of a small building has a height of 5 m, width of 4 m, and a thickness of 0.5 m. The thermal conductivity of the wall is determined to be 0.7 W/m K. If the temperature of the inner surface of the wall is measured to be 201C and the outer surface temperature is measured to be 11C. Determine the rate at which heat is lost from the building through this wall. Solution: Assumptions: Steady-state conditions persist. The heat transfer is considered one-dimensional. The medium is isotropic with constant (uniform) thermal conductivity. Analysis: For the case of steady-state conditions, conduction heat transfer rate through a wall can be obtained from the equation _ Q ¼ kA T1 T2 L T1 T2 L x Fig. 5 Heat conduction through a wall section. Heat Transfer Aspects of Energy 431
- 11. The cross-sectional area of the wall A can be determined from the given dimensions as A¼5 m 4 m¼20 m2 . In addition, the thickness of the wall L is given as 0.5 m. Hence, the rate of heat loss from the building through the wall is _ Q ¼ 0:7 W=m 1C ð Þ 20 m2 20 1 ð Þ1C 0:5 m ¼ 532 W 1.10.4.2.2 Steady-state heat conduction in cylinders For the case of steady-state and one-dimensional heat conduction in a hollow cylinder of inner radius r1 and outer radius r2 with constant inner and outer surface temperatures of T1 and T2, respectively, with no heat generation within the cylinder, the rate of heat conduction through the cylinder can be expressed by the Fourier’s law (Fig. 6) _ Q ¼ kA dT dr ð33Þ where at the location r for a cylinder of length L, the area A¼2prL. The above equation can be integrated to obtain _ Q ¼ 2pLk T1 T2 ln r2 r1 ð34Þ Example 3: The inner and outer temperatures of a 1-m-long and 1.5-cm-thick cylindrical stainless steel pipe are measured to be 501C and 471C, respectively. The pipe has a thermal conductivity of 15 W/m K. If the inner radius of the pipe is 0.35 m, determine the rate of heat transfer through the pipe. Solution: Assumptions: Steady-state conditions persist. The heat transfer is considered one-dimensional. The medium is assumed to be isotropic with constant thermal conductivity. Analysis: The inner and outer temperatures of the cylindrical pipe are known. The rate of heat transfer through the pipe can be determined from the equation _ Q ¼ 2pLk T1 T2 ln r2 r1 The outer radius r2 of the pipe is r2¼r1 þ 0.015 m¼0.365 m _ Q ¼ 2p 1 m ð Þ 15 W=mK ð Þ 50 47 ½ 1C ln 0:365 0:35 ¼ 6737:7 W 1.10.4.2.3 Steady-state heat conduction in spheres In the case of a hollow sphere with inner radius r1 and outer radius r2, at an inner surface temperature T1 and outer surface temperature T2 as shown in Fig. 7, with no heat generation and constant thermal conductivity, the steady-state rate of heat conduction through the sphere can be expressed in the form of Fourier’s law: _ Q ¼ kA dT dr ð35Þ where the area A corresponds to the area normal to the direction of heat transfer. For the case of sphere, A¼4p2 . The equation above can be integrated to obtain the following expression: _ Q ¼ k 4pr1r2 ð Þ T1 T2 ð Þ r2 r1 ð36Þ r1 r2 T1 T2 k Fig. 6 Hollow cylinder. 432 Heat Transfer Aspects of Energy
- 12. Example 4: A spherical tank has an inner radius of 10.5 cm and an outer radius of 12.8 cm. The temperatures at the inner and outer surfaces of the tank are maintained at 1801C and 1101C, respectively. If the thermal conductivity of the tank is 39 W/m K, determine the heat transfer rate through the tank. Solution: Assumptions: Steady-state conditions persist. The heat transfer is considered one-dimensional. The medium is isotropic with constant thermal conductivity. Analysis: The inner and outer surface temperatures of the tank are known. The rate of heat transfer through the spherical tank can be determined from _ Q ¼ k 4pr1r2 ð Þ T1 T2 ð Þ r2 r1 _ Q ¼ 39 W=mK ð Þ 4p ð Þ 0:128 m ð Þ 0:105 m ð Þ 180 110 ½ 1C 0:128 0:105 ð Þm ¼ 20:05 kW 1.10.4.2.4 Thermal resistance In heat transfer applications, the concept of thermal resistance (Fig. 8) is useful to analyze steady-state problems. The steady-state heat conduction equation for a wall can be rearranged as _ Q ¼ T1 T2 L kA ð37Þ The thermal resistance against the transfer of heat by conduction through the wall is denoted as Rw ¼ L kA ð38Þ Therefore, the rate of heat conduction through the wall is expressed in terms of the thermal resistance as _ Q ¼ T1 T2 Rw ð39Þ This concept is similar to the electrical resistance concept (Fig. 9), where the rate of heat ﬂow _ Q is analogous to the current ﬂow through an electrical resistor. The temperature difference resembles the voltage difference. I ¼ V1 V2 Relectrical ð40Þ r1 r2 T1 T2 Fig. 7 Hollow sphere. V1 V2 Relectrical I Fig. 8 Electrical resistance concept. T1 T2 Rthermal Fig. 9 Thermal resistance concept. Heat Transfer Aspects of Energy 433
- 13. 1.10.4.2.4.1 Thermal resistance for a composite wall Composite walls comprise of a number of layers of different materials with different properties and thicknesses (Fig. 10). Such cases include a number of series or parallel thermal resistances arising from the different layers of materials. The rate of heat transfer is associated with the temperature difference and the thermal resistance of each layer. This can be expressed as Q ¼ TA T1 1 h1A ¼ T1 T2 L1 k1A ¼ Tn TB 1 hnA ð41Þ Hence, the one-dimensional heat transfer rate for such systems can be written as Q ¼ TA TB SRt ð42Þ And, SRt ¼ Rtt ¼ 1 HA, where H represents the overall heat transfer coefﬁcient. Therefore, the overall heat transfer coefﬁcient can be expressed as H ¼ 1 Rt;tA ¼ 1 1 h1 þ L1 k1 þ ⋯ þ 1 hn ð43Þ 1.10.4.2.4.2 Thermal resistance for cylinders The thermal resistance concept can also be applied to a cylinder (Fig. 11). Consider a cylinder having an internal radius r1 and an external radius r2. The inner and outer surfaces of the cylinder are subjected to ﬂuids at different temperatures. In case of steady-state conditions with no heat generation, the governing heat conduction equation is 1 r d dr kr dt dr ¼ 0 ð44Þ The rate of conduction heat transfer across the cylindrical surface can be expressed based on the Fourier’s law as Q ¼ kA dT dr ¼ k 2prL ð Þ dT dr ð45Þ Integrating the above equation twice under appropriate boundary conditions and assuming constant thermal conductivity, the following equation can be obtained: Q ¼ k 2pL ð Þ T1 T2 ð Þ ln r1 r2 ¼ T1 T2 Rt ð46Þ Considering the case of a hollow composite cylinder, if the interfacial contact resistances are neglected, the rate of heat transfer can be determined as Q ¼ T1 Tn Rt;t ¼ HA T1 Tn ð Þ ð47Þ where Rt;t ¼ 1 2pr1Lh1 þ ln r2 r1 2pLk1 þ ln r3 r2 2pLk2 þ 1 2prnLhn ð48Þ TA T1 T3 T4 Tz Tn Ln kn k3 k2 k1 LA L1 L2 Rn R3 R2 R1 RA Rz Q Q Tn−1 T2 Fig. 10 Thermal resistances in series in a composite wall. 434 Heat Transfer Aspects of Energy
- 14. 1.10.4.2.4.3 Thermal resistance for spheres Consider conduction heat transfer in a hollow sphere that has an internal radius r1 and external radius r2, with inside and outside temperatures of T1 and T2, and constant thermal conductivity with no heat generation. The rate of heat conduction can be expressed in the form of the Fourier’s law: Q ¼ kA dT dr ¼ k 4pr2 dT dr ð49Þ where the area, A¼4pr2 , denotes the area normal to the direction of heat transfer. Integrating the above equation, the following expression can be obtained Q ¼ k 4p ð Þ T1 T2 ð Þ 1 r2 1 r1 ¼ k 4pr1r2 ð Þ T1 T2 ð Þ r2 r1 ¼ T1 T2 Rt ð50Þ For the case of a composite hollow sphere, neglecting the interfacial contact resistances, the rate of heat transfer becomes Q ¼ T1 Tn Rt;t ¼ HA T1 Tn ð Þ ð51Þ where H denotes the overall heat transfer coefﬁcient. And Rt,t can be expressed as Rt;t ¼ 1 4pr2 1 h1 þ r2 r1 4pr1r2k1 þ r3 r2 4pr2r3k2 þ ⋯ þ 1 4pr2 2 h2 ð52Þ Example 5: A double-pane window shown in Fig. 12 has a height of 1 m and a width of 2 m. The glass layers have a thickness of 5 mm and a thermal conductivity of 0.80 W/mK. A 15-mm stagnant air space with a thermal conductivity of 0.035 W/mK separates the two glass layers. If the inner temperature of the window (T1) is measured to be 201C and the outer temperature (T4) is measured to be 51C, determine the rate of heat loss from the room through this window and the temperature T2. Solution: Assumptions: Steady-state conditions persist. Heat transfer is one-dimensional. The thermal conductivity of the material remains constant and the material is isotropic. r2 r3 rn−1 rn T2 T3 Tn−1 Tn L T1 …... …. Ra R1 R2 R3 Rn Q Q r1 Fig. 11 Thermal resistances in a composite hollow cylinder. Heat Transfer Aspects of Energy 435
- 15. Analysis: For cases involving steady-state heat transfer, the thermal resistance concept can be utilized. For the window given in the example, the rate of conduction heat transfer is constant through the window. Three thermal resistances in series exist in the given window (Fig. 13). R1 ¼ R3 ¼ Lglass kglassA R1 ¼ R3 ¼ 0:005 m 0:8 W=mK ð Þ 2 m2 ð Þ ¼ 0:0031251C=W R2 ¼ Lair kairA ¼ 0:015 m 0:035 W=mK ð Þ 2 m2 ð Þ ¼ 0:21431C=W T1 T2 T3 T4 Fig. 12 Heat loss and temperature distribution through a double-pane window. T1 T4 R1 R2 R3 T1 T2 T3 T4 Fig. 13 Thermal resistances and temperature distribution in a double-pane window. 436 Heat Transfer Aspects of Energy
- 16. As the thermal resistances are in series, the total resistance can be obtained as follows: Rtotal ¼ R1 þ R2 þ R3 Rtotal ¼ 0:003125 þ 0:2143 þ 0:003125 ¼ 0:2211C=W The steady-state heat transfer can then be determined as _ Q ¼ T1 T4 Rtotal ¼ 20 5 ð Þ ½ 1C 0:2211C=W ¼ 113:1 W Under steady operating conditions, the rate of heat transfer through the glass layer can be expressed as _ Q ¼ T1 T2 R1 Thus, the temperature T2 can be determined from the above equation: T2 ¼ T1 _ QR1 ¼ 201C 113:1 W ð Þ 0:003125 1C=W ð Þ ð Þ ¼ 19:61C 1.10.4.3 Transient Heat Conduction The previous chapter was focused on steady-state heat transfer, where the temperatures do not change with time. However, when the heat transfer rate between a given solid object and its surroundings changes with time, it is known as unsteady or transient heat transfer. In such cases, the temperature of the body at any point as well as the heat content varies with time and distance. Time- dependent transient heat transfer problems are encountered in various engineering and energy applications. Exact analysis of transient heat transfer of a solid object during heating or cooling is essential to enhance the processing conditions as well as to save energy, hence leading to high-quality products. Biot number is a dimensionless form of boundary condition that is essential for a transient heat transfer analysis. It represents the ratio of the resistance to conduction heat transfer within the body to convection heat transfer resistance at the surface and can be expressed as Bi ¼ hY k , where h denotes the convective heat transfer coefﬁcient, Y denotes the characteristic length, and k represents the thermal conductivity of the object. For transient heat transfer analysis, three important criteria are considered, which are the cases with Bio0.1, cases with 0.1rBir100, and Bi4100. The case where Bio0.1 is known as the lumped heat capacitance system. Thin or small objects, which have a high thermal conductivity, experience negligible internal conduction resistance within the body and large convective heat transfer resistance at the surface of the body across the ﬂuid boundary layer. For cases involving Bio0.1, negligible temperature gradient exists within the body. For such cases, the temperature of the body as well as the heat transfer rate at a given time can be obtained by applying a heat balance on the object. The case of 0.1rBir100 is referred to as the convection boundary condition or a boundary condition of the third kind. In such cases, there exist ﬁnite external as well as internal resistances to heat transfer to or from a body subjected to heating or cooling, respectively. For such cases, as the external and internal resistances are comparable, the general boundary condition cannot be simpliﬁed. Such problems require a distributed system approach. The series solutions for the dimensionless transient temperature y ¼ TTa TiTa and heat transfer rates f ¼ Q Qi ¼ rcpV TiTma ð Þ rcpV TiTa ð Þ ¼ TiTma TiTa for different geometries are listed in Tables 1 and 2. The case of Bi4100 is referred to as the boundary condition of the ﬁrst kind. Such cases have negligible external resistance and high internal resistance to heat transfer. Since the external resistance is negligible, the heat transfer coefﬁcients are signiﬁcantly high. The series solutions for dimensionless transient temperatures and heat transfer rates for different geometries are also listed in Tables 1 and 2. Example 6: A cylindrical carrot slice, 0.008 m in diameter and 0.1 m in length, at an initial temperature of 271C is cooled to 41C in a forced-air cooling system at a medium temperature of 21C and a ﬂow velocity of 5 m/s. The speciﬁc heat, thermal conductivity, thermal diffusivity, heat transfer coefﬁcient, and density are cp¼3880 J/kg1C, k¼0.69 W/m1C, a¼0.133 106 m2 /s, h¼31.2 W/m2 1C, r¼1298 kg/m3 , respectively. Assuming a uniform temperature variation with time within the product and constant thermal and physical properties, calculate the cooling time required for the product to reach 41C. Solution: From the given data, the Biot number is calculated as Bi ¼ hR=2 ð Þ=k ¼ 31:2 ð Þ 0:004=2 ð Þ=0:69 ¼ 0:09 Hence, the lumped capacitance system assumption is justiﬁable. Furthermore, the dimensionless temperature is y ¼ f fi ¼ T Ta ð Þ= Ti Ta ð Þ ¼ 4 2 ð Þ= 27 2 ð Þ ¼ 0:08 The Fourier number can be calculated as Fo ¼ ln1=y ð Þ=Bi ¼ ln12:5=0:09 ¼ 138:88 Thus, the cooling time is t ¼ FoLc=a ¼ 138:88 0:004 2 2 =0:133 106 ¼ 4176:8 s ¼ 1:16 h Heat Transfer Aspects of Energy 437
- 17. Example 7: An individual dried ﬁg was formed by hand as an inﬁnite slab and refrigerated until the center temperature of 21.61C reached 221C in a freezer cabinet at Ta ¼ 221C. During this refrigeration process, the center temperature distribution of the sample was measured at 30-s intervals, and the measurement was repeated three times. Some physical and thermal properties are L¼0.01 m, l¼0.05 m, X¼0.04 m, k¼0.2367 W/m 1C, a¼9.88 108 m2 /s, and h¼9.045 W/m2 1C. Further information on the experiments and analysis technique, along with the relevant data can be found in Ref. [1]. Here, we will compute the center temperature distribution of this slab product and compare it with the experimental data. Solution: From the data we have, the Biot number is calculated as Bi ¼ hl k ¼ 9:045 0:005=0:2367 ¼ 0:191 here, the Biot number is between 0.1rBir100, leading to a convection boundary condition problem. By knowing the Biot number, the values of m1 and A1 can be found as m1 ¼0.422 and A1 ¼1.03. The theoretical dimensionless center temperature is calculated from the following equation, considering Fo40.2: yt ¼ A1B1 ¼ 2Bi Bi2 þ m2 1 0:5 =m1 Bi2 þ Bi þ m2 1 h i exp m2 1Fo ¼ 1:03 exp 0:4222 Fo where Fo¼at/l2 ¼9.88 108 t/0.0052 . The results are shown in Fig. 14. Example 8: An individual cylindrical eggplant was cooled until the center temperature of 221C reached 71C in the water pool of a hydro cooling unit at a temperature of 11C. During the cooling process, the center temperature distribution of the product was measured at 30-s intervals. Some physical and thermal properties of the product are: R¼0.0225 m, J¼0.142 m, k¼0.6064 W/m 1C, a¼1.438 107 m2 /s, and h¼44.15 W/m2 1C. Further information on the experiments and analysis technique, along with the relevant data, can be found in Ref. [2]. Here, we will determine the theoretical center temperature distribution of the cylindrical eggplant and compare it with the experimental data. Table 1 Dimensionless transient temperatures Equations or series solutions For Bio0.1 y¼eBiFo For 0.1rBir100 Inﬁnite slab y ¼ P 1 n ¼ 1 AnBnCn An ¼ 1 ð Þnþ1 2Bi Bi2 þ m2 n 0:5 =mn Bi2 þ Bi þ m2 n Bn ¼ exp m2 nFo Cn ¼ cosmnG Inﬁnite cylinder y ¼ P 1 n ¼ 1 AnBnCn An ¼ 2Bi= m2 n þ Bi2 J0 mnR ð Þ Bn ¼ exp m2 nFo Cn ¼ J0 mnG ð Þ Sphere y ¼ P 1 n ¼ 1 AnBnCn An ¼ 1 ð Þnþ1 2Bi m2 n þ Bi 1 ð Þ2 h i0:5 = Bi2 Bi þ m2 n Bn ¼ exp m2 nFo Cn ¼ sinmnG=mnG For Bi4100 Inﬁnite slab y ¼ P 1 n ¼ 1 AnBnCn An ¼ 1 ð Þnþ1 2=mn ð Þ Bn ¼ exp m2 nFo Cn ¼ cosmnG mn ¼ 2n 1 ð Þp=2 Inﬁnite cylinder y ¼ P 1 n ¼ 1 AnBnCn An ¼ 2=mnJ1 mn ð Þ Bn ¼ exp m2 nFo Cn ¼ J0 mnG ð Þ J0 mnG ð Þ ¼ 1 m2 n=22 þ m4 n=22 42 m6 n=22 42 62 Sphere y ¼ P 1 n ¼ 1 AnBnCn An ¼ 2 1 ð Þnþ1 Bn ¼ exp m2 nFo Cn ¼ sinmnG=mnG mn ¼ np 438 Heat Transfer Aspects of Energy
- 18. Solution: From the data we have, the Biot number can be calculated as Bi ¼ hR k ¼ 44:15 0:0225=0:6064 ¼ 1:64 Table 2 Transient heat transfer rates Equations or series solutions For Bio0.1 Qt ¼ Qi 1 exp BiFo ð Þ ½ Qi ¼ rcpV Ti Ta ð Þ f ¼ Q Qi ¼ TiTma TiTa For 0.1rBir100 Inﬁnite slab f ¼ P 1 n ¼ 1 AnBn An ¼ 2Bi2 =m2 n Bi2 þ Bi þ m2 n Bn ¼ 1 exp m2 nFo Tma is the temperature at G¼0.57l Inﬁnite cylinder f ¼ P 1 n ¼ 1 AnBn An ¼ 4Bi2 =m2 n Bi2 þ m2 n Bn ¼ 1 exp m2 nFo Tma is the temperature at G¼0.7R Sphere f ¼ P 1 n ¼ 1 AnBn An ¼ 6Bi2 =m2 n Bi2 þ m2 n Bi Bn ¼ 1 exp m2 nFo Tma is the temperature at G¼0.77R For Bi4100 Inﬁnite slab f ¼ P 1 n ¼ 1 AnBn An ¼ 2=m2 n ¼ 8= 2n 1 ð Þ2 p2 Bn ¼ 1 exp m2 nFo Inﬁnite cylinder f ¼ P 1 n ¼ 1 AnBn An ¼ 4=m2 n Bn ¼ 1 exp m2 nFo Sphere f ¼ P 1 n ¼ 1 AnBn An ¼ 6=m2 n ¼ 6=n2 p2 Bn ¼ 1 exp m2 nFo 1.0 0.8 0.6 0.4 0.2 Dimensionless center temperature 0.0 0 5 10 Fourier number 15 20 Computed Experimental (1) 25 30 Fig. 14 Measured and predicted temperature distribution for an individual sample. Adapted from Dincer I. Analytical modelling of heat transfer from a single slab in freezing. Int J Energy Res 1995;19(3):227–33. Heat Transfer Aspects of Energy 439
- 19. Here, the Biot number is between 0.1rBir100, leading to a convection boundary condition problem. By knowing the Biot number as 1.64, the values of mn and An for n¼1 6, which are considered sufﬁcient terms, are taken as m1¼1.497, m2 ¼4.219, m3 ¼7.242, m4 ¼10.332, m5 ¼13.445 and m6 ¼16.571 and A1 ¼1.297, A2 ¼ 0.427, A3¼0.203, A4 ¼ 0.119, A5 ¼0.0823, and A6 ¼ 0.0589. The theoretical dimensionless center temperature is calculated from the following equation, considering Fo40.2: yt ¼ X 1 n ¼ 1 AnBn ¼ X 6 n ¼ 1 Anexp m2 nFo ¼ 1:297 exp 1:4972 Fo 0:427 exp 4:2192 Fo þ 0:203 exp 7:2422 Fo 0:119 exp 10:3322 Fo þ 0:0823 exp 13:4452 Fo 0:0589 exp 16:5712 Fo where Fo¼at/R2 ¼1.438 107 t/0.02252 . The computed results are shown in Fig. 15. The experimental results can also be obtained from Ref. [2]. Example 9: An individual spherical pear was cooled until the center temperature of 221C reached 21C in the water pool of a hydro cooling unit at a temperature of 11C. During the cooling process, the center temperature distribution of the product was measured at 30-s intervals. Some physical and thermal properties of the product are: R¼0.03 m, k¼0.5527 W/m 1C, a¼1.378 107 m2 /s, and h¼160.56 W/m2 1C. Further information on the experiments, and analysis technique, along with the relevant data, can be found in Ref. [2]. Here, we will determine the theoretical center temperature distribution of the spherical pear and compare it with the experimental data. Solution: From the data we have, the Biot number can be calculated as Bi ¼ hR k ¼ 160:56 0:030=0:5572 ¼ 8:64 Here, the Biot number is between 0.1rBir100, leading to a convection boundary condition problem. By knowing the Biot number as 8.64, the values of mn and An for n¼1–6, which are considered sufﬁcient terms, are taken as m1 ¼2.79, m2 ¼5.646, m3 ¼8.581, m4 ¼11.578, m5 ¼14.618 and m6 ¼17.618 and A1 ¼1.903, A2 ¼ 1.657, A3¼ 1.42, A4 ¼ 1.196, A5 ¼1.018, and A6 ¼ 0.878. The theoretical dimensionless center temperature is calculated from the following equation, considering Fo40.2: yt ¼ X 1 n ¼ 1 AnBn ¼ X 6 n ¼ 1 Anexp m2 nFo ¼ 1:903 exp 2:792 Fo 1:675 exp 5:6462 Fo þ 1:42exp 8:5812 Fo 1:196 exp 11:5782 Fo þ 1:018 exp 14:6182 Fo 0:878 exp 17:6862 Fo where Fo¼at/R2 ¼1.378 107 t/0.032 . The computed results are shown in Fig. 16, the experimental results can also be obtained from Ref. [2]. Example 10: Steel billets are used extensively in the manufacturing of various products for numerous industries. The heat treatment process of steel billets involves heating the billets to a temperature of 8001C and immersing them in a water bath until they are cooled to a speciﬁed temperature. The billets have a diameter of 6.5 cm and a length of 100 cm. The thermal conductivity 1.0 0.8 0.6 0.4 Dimensionless center temperature 0.2 0.0 0.1 0.2 Fourier number 0.3 0.4 0.6 0.5 0.7 Fig. 15 Theoretical dimensionless temperature proﬁles for the center of an individual eggplant. 440 Heat Transfer Aspects of Energy
- 20. of the billets is 60.2 W/mK, speciﬁc heat is 490 J/kg K, and density is 8050 kg/m3 . If the water bath is maintained at a constant temperature of 351C, and the convection heat transfer coefﬁcient is 325 W/m2 K, determine the temperature of an immersed billet after 6 min. In addition, determine the amount of heat transfer from the steel billet to the water bath during this time interval. Solution: Assumptions: The convection heat transfer coefﬁcient remains constant. The medium is isotropic and has a constant thermal conductivity. Analysis: As this is a transient analysis, we need to determine if it is appropriate to use the lumped system method. If the Biot number is less than or equal to 0.1, the lumped system method can be considered appropriate. The Biot number can be obtained from the equation Bi ¼ hY k where the characteristic length Y for a cylindrical billet can be obtained from the equation Y ¼ Volume Surface area ¼ pr2 L 2prL ¼ r 2 ¼ 0:0325 2 ¼ 0:01625 m Hence, the Biot number is obtained as Bi ¼ hY k ¼ 325 W=m2 K ð Þ 0:01625 m ð Þ 60:2 W=mK ð Þ ¼ 0:09 Since the Biot number is less than 0.1, the lumped system method can be utilized for this case T t ð ÞT1 TinitialT1 ¼ eat The above equation can be rearranged to obtain T t ð Þ ¼ Tinitial T1 ð Þeat þ T1 where the time constant a can be obtained from the equation a ¼ hLA rVcP ¼ hL rcPY ¼ 325 W=m2 K ð Þ 8050 kg=m3 ð Þ 490 J=kg K ð Þ 0:01625 m ð Þ ¼ 0:00507 s1 At t¼6 min¼360 s, the temperature of the billet can be obtained as T 360 ð Þ ¼ Tinitial T1 ð Þeat þ T1 ¼ 800 35 ½ 1C e 0:00507s1 ð Þ 360s ð Þ þ 351C T 360 ð Þ ¼ 158:31C The amount of heat transferred from the steel billet to the water bath during this time interval can be obtained from Q ¼ mcp Tinitial T 360 ð Þ ½ 1.0 0.8 0.6 0.4 0.2 Dimensionless center temperature 0.0 0.0 0.1 0.2 Fourier number 0.3 0.4 Fig. 16 Theoretical dimensionless temperature proﬁles for the center of an individual sphere. Adapted from Dincer I, Transient temperature distributions in spherical and cylindrical food products subjected to hydrocooling. Int J Energy Res 1994;18(8):741–9. Heat Transfer Aspects of Energy 441
- 21. where the mass of the billet can be obtained from the density and volume m ¼ rV ¼ 8050 kg=m3 p 0:03252 m2 1 m ð Þ ¼ 26:7 kg Hence, Q ¼ 26:7 kg ð Þ 490 J=kg K ð Þ 800 158:3 ð Þ1C 8:4 MJ. Example 11: A long steel shaft at 3001C cools in an environment at 351C. The heat transfer coefﬁcient is calculated to be 60 W/m2 K. The thermal conductivity of the shaft is 15 W/mK, density is 8080 kg/m3 , speciﬁc heat is 490 J/kg K. If the shaft has a diameter of 25 cm, determine the temperature at its center and the amount of heat lost per unit length of the shaft after 1 h. Solution: Assumptions: As the shaft is speciﬁed to be long and contains a thermal symmetry about its centerline, heat transfer is considered one-dimensional. The material is isotropic with constant thermal conductivity. Analysis: In case of one-dimensional transient heat conduction, the temperature at the center of the shaft can be approximated from the ﬁrst term of the series solution if the Fourier number is greater than 0.2. The Fourier number can be calculated as: t ¼ at r2 o ¼ kt rcpr2 0 ¼ 15 W=mK ð Þ 3600 s ð Þ 8080 kg=m3 ð Þ 490 J=kg K ð Þ 0:1252 m2 ð Þ ¼ 0:8740:2 Since the Fourier number is greater than 0.2, one term approximation can be used. Bi ¼ hr0 k ¼ 60 W=m2 K ð Þ 0:125 m ð Þ 15 W=m K ð Þ ¼ 0:5 y0 ¼ T T1 Tinitial T1 ¼ 2 l1 J1 l1 ð Þ J2 0 l1 ð Þ þ J2 1 l1 ð Þ el2 1t J0 l1r r0 where l1 is the root of the eigenfunction l1 j1 l1 ð Þ j0l1 ¼ Bi ¼ 0:5 l1 ¼ 0:9408; J1 l1 ð Þ ¼ 0:4195; J0 l1 ð Þ ¼ 0:7902 Hence; y0 ¼ T0 T1 Tinitial T1 ¼ 2 0:9408 ð Þ 0:4195 0:79022 ð Þ þ 0:41952 ð Þ e0:94082 0:87 ð Þ 0:7902 ð Þ ¼ 0:405 Therefore; T0 ¼ Tinitial T1 ð Þ 0:405 ð Þ þ T1 ¼ 300 35 ð Þ 0:405 ð Þ þ 35 ¼ 142:331C The ratio of the amount of heat lost to the maximum amount of heat the shaft can lose is Q Qmaximum ¼ 1 2y0 J1 l1 ð Þ l1 Hence; Q ¼ Qmaximum 1 2y0 J1 l1 ð Þ l1 ¼ Qmaximum 1 2 0:405 ð Þ 0:4195 ð Þ 0:9408 Q ¼ 0:639 Qmaximum ¼ 0:639rVcP T1 Tinitial ð Þ ¼ 0:639 396:62 kg ð Þ 490 J=kg K ð Þ 35 300 ½ 1C ¼ 32:9 MJ Thus, 32.9 MJ of energy is lost per unit length of the shaft during this process. 1.10.5 Convection Heat Transfer Convection mode of heat transfer takes place within a ﬂuid when one portion of the ﬂuid is mixed with another. Heat transfer by convection can be classiﬁed according to the nature of the ﬂow. Flows that are caused by external means such as a fan, pump, or atmospheric wind are classiﬁed as forced convection. However, when the ﬂow is induced due to buoyancy forces in the ﬂuid arising from density variations, which are caused by temperature differences within the ﬂuid, it is classiﬁed as natural or free convection. For instance, when a hot food item is exposed to the atmosphere, natural convection takes place; on the other hand, for a food product placed in a cold store, forced-convection heat transfer occurs between the air ﬂow and a food item, which is subjected to this ﬂow. When transfer of heat takes place through solid objects, the mode of heat transfer is conduction alone; however, the transfer of heat from a solid surface to a liquid or gas occurs partly due to conduction and partly due to convection. When an appreciable movement of the gas or liquid exists, the conduction heat transfer becomes negligibly small compared with the heat transfer by convection in the gas or liquid. However, a thin boundary layer of ﬂuid always exists on the surface, and conduction heat transfer occurs through this thin ﬁlm. When convection heat transfer occurs within a ﬂuid, it is by combined effects of conduction and bulk ﬂuid motion. Generally, the transferred heat is the sensible heat of the ﬂuid. In cases where a phase change between the liquid and vapor states is taking place, the convection processes also include latent heat exchange. 442 Heat Transfer Aspects of Energy
- 22. 1.10.5.1 Newton’s Law of Cooling The heat transfer that occurs between a solid surface and a ﬂuid is proportional to the surface area and the temperature difference between the ﬂuid and the solid surface; this is known as Newton’s law of cooling. This represents a speciﬁc nature of convection heat transfer and is expressed as _ Q ¼ hA Ts Tf ð Þ ð53Þ where h represents the convection heat transfer coefﬁcient. The heat transfer coefﬁcient includes all effects inﬂuencing convection heat transfer and it depends on the boundary layer conditions, which are dependent on factors such as the nature of the ﬂuid ﬂow, thermal properties, surface geometry, and physical properties. The above equation does not take into account the heat transfer due to radiation. Radiation heat transfer is discussed later. In many cases, the heat transfer due to radiation is negligibly small as compared to conduction or convection heat transfer between a surface and a ﬂuid. However, in heat transfer problems, which involve in high surface temperatures and natural convection, radiation heat transfer is similar in magnitude to the natural convection heat transfer. Consider the wall in Fig. 17, heat transfer occurs from the higher temperature ﬂuid A to the lower temperature ﬂuid B through a wall that has a thickness L. The temperature in ﬂuid A drops to a temperature Ts1 within the wall region. Generally, the bulk ﬂuid temperature is nearly constant, apart from the thin ﬁlms DA or DB near the surface of the wall. The rate of heat transfer per unit surface area from the higher temperature ﬂuid A to the wall and from the wall to the lower temperature ﬂuid B are: q ¼ hA TA Ts1 ð Þ ð54Þ q ¼ hB Ts2 TB ð Þ ð55Þ In addition, the conduction heat transfer in the thin ﬁlms can be expressed as q ¼ kA DA TA Ts1 ð Þ ð56Þ q ¼ kB DB Ts2 TB ð Þ ð57Þ Equating Eqs. (54–57), the convection heat transfer coefﬁcients can be obtained as hA ¼ kA DA and hB ¼ kB DB Hence, the rate of heat transfer per unit surface area in the wall can be obtained as q ¼ k L Ts1 Ts2 ð Þ ð58Þ In case of steady-state heat transfer, Eq. (54) is equal to Eq. (55) and thus to Eq. (58). q ¼ hA TA Ts1 ð Þ ¼ hB Ts2 TB ð Þ ¼ k L Ts1 Ts2 ð Þ ð59Þ L ΔB ΔA TB TA TS2 TS1 Fig. 17 A wall cross-section subject to convection heat transfer from both sides. Heat Transfer Aspects of Energy 443
- 23. Therefore, Eq. (51) leads to the following expression q ¼ TA TB ð Þ 1 hA þ L k þ 1 hB ð60Þ An analogy between Eqs. (46) and (52) leads to the following expression _ Q ¼ H A TA TB ð Þ ð61Þ where H represents the overall heat transfer coefﬁcient and can be expressed as 1 H ¼ 1 hA þ L k þ 1 hB ð63Þ Example 12: Consider a wall that has a height of 1 m, a width of 2 m, a thickness of 0.25 m, and a thermal conductivity of 1.2 W/ mK. The temperature of the room (TA) is measured to be 201C and the outside temperature (TB) is measured to be 51C, if the convection heat transfer coefﬁcients at the inner and outer surfaces of the window are hA ¼12 W/m2 K and hB¼35 W/m2 K, determine the rate of heat loss from the room through this wall. Solution: Assumptions: Steady-state conditions persist. The heat transfer is considered one-dimensional. The medium is isotropic with constant thermal conductivity. Analysis: The rate of heat loss can be determined from the equation _ Q ¼ HA TA TB ð Þ where the overall heat transfer coefﬁcient H can be obtained as 1 H 1 hA þ L k þ 1 hB ¼ 1 12 þ 0:25 1:2 þ 1 35 ¼ 0:32 m2 K=W Thus, H ¼ 1 1:78 ¼ 3:12 W=m2 K. The inside and outside temperatures of the room are known, hence _ Q can be determined as _ Q ¼ HA TA TB ð Þ ¼ 3:12 W=m2 K 2m2 20 ð5Þ ½ 1C _ Q ¼ 156:13 W Thus, 156.13 J of heat is lost every second from the room through this wall. Example 13: The double-pane window shown in Fig. 18, has a height of 1 m and a width of 2 m. The glass layers have a thickness of 5 mm and a thermal conductivity of 0.80 W/mK. A 15-mm stagnant air space with a thermal conductivity of 0.035 W/mK separates the two glass layers. The inner temperature of the room (Ti) is measured to be 231C and the outside temperature (To) is measured to be 101C, if the convection heat transfer coefﬁcient at the inner side of the wall is 12 W/m2 K and at the outer side of the wall is 35 W/m2 K, determine the rate of heat loss from the room through this window. Solution: Assumptions: Steady-state conditions persist. The heat transfer is considered one-dimensional. The medium is assumed to be isotropic with constant thermal conductivity (Fig. 19). Analysis: The rate of heat loss through this window can be determined from the equation _ Q ¼ HA Ti To ð Þ Glass 1 m Air Fig. 18 Double-pane window. 444 Heat Transfer Aspects of Energy
- 24. where H represents the overall heat transfer coefﬁcient. In this case, H can be determined as 1 H ¼ 1 hi þ Lglass kglass þ Lair kair þ Lglass kglass þ 1 ho 1 H ¼ 1 12 þ 0:005 0:8 þ 0:015 0:035 þ 0:005 0:8 þ 1 35 ¼ 0:55 m2 K=W H ¼ 1:81 W=m2 K Since the temperatures inside (Ti) and outside (To) of the room are known, _ Q can be obtained as _ Q ¼ HA Ti To ð Þ ¼ 1:81 W=m2 K ð Þ 2 m2 ð Þ 23 10 ð Þ ½ 1C _ Q ¼ 119:5 W 1.10.5.2 The Nusselt Number In the analysis of convection heat transfer mechanism, the governing equations are commonly nondimensionalized by combining the variables and grouping them into dimensionless numbers; this reduces the total number of variables. The heat transfer coefﬁcient h is nondimensionalized with the Nusselt number. The Nusselt number is deﬁned as Nu ¼ hY kf ð63Þ where Y denotes the characteristic length and k denotes the ﬂuid thermal conductivity. Nusselt number is also known as the dimensionless convection heat transfer coefﬁcient and is named after Wilhelm Nusselt. In the ﬁrst half of the 20th century, Wilhelm Nusselt contributed signiﬁcantly to convective heat transfer. Nusselt number essentially signiﬁes the ratio of the convective heat transfer through a ﬂuid layer of thickness Y, which has a temperature T1 on one side of the ﬂuid and temperature T2 on the other side during the presence of some ﬂuid motion to the conductive heat transfer through the ﬂuid when the ﬂuid layer is stagnant. Nu ¼ _ qconvection _ qconduction ¼ h T1 T2 ð Þ kf T1T2 ð Þ L ¼ hY kf ð64Þ 1.10.5.3 Forced Convection The analysis of forced convection heat transfer deals with the heat transfer taking place between a solid surface and a moving ﬂuid. In order to apply the equation for Newton’s law of cooling as given in Eq. (1), it is required to determine the heat transfer coefﬁcient h. Nusselt–Reynolds correlations can be utilized for this purpose. The deﬁnition of Reynolds number as well as of the Nusselt number is given in Table 3. A few examples of equipment involved in forced convection heat transfer include heat exchangers, forced water and air coolers, forced water and air condensers, and evaporators. Forced convection can occur in various types of cases, such as ﬂow in or across a tube, and ﬂow across a ﬂat plate. These types of cases can be solved mathematically with certain assumptions regarding the boundary conditions. Obtaining exact solutions to T1 Ti To T2 T3 T4 Fig. 19 Heat transfer and temperature distribution through a double-pane window. Heat Transfer Aspects of Energy 445
- 25. such cases can be extremely difﬁcult, particularly for cases that involve in turbulent ﬂows; however, approximate solutions can be obtained by making appropriate assumptions. The ﬁrst step in obtaining a solution of a convective heat transfer problem is to determine whether the boundary layer is turbulent or laminar. The value of the convective heat transfer coefﬁcient h and thus the rate of convective heat transfer are affected by these conditions. Fluid motion within the laminar boundary layer is highly ordered, and streamlines along which the particles move can be identiﬁed. In contrast, in the turbulent boundary layer, ﬂuid motion is highly irregular and it is characterized by the ﬂuctuations in velocity that start to develop in the transitional region; after the transitional boundary layer, a complete turbulent boundary layer exists. These velocity ﬂuctuations increase the transfer of heat, momentum, and species, and thus increase the surface friction and rates of convective heat transfer. In addition, the laminar sublayer is approximately linear, and the transport is primarily domi- nated by diffusion as well as the velocity proﬁle. Furthermore, there exists an adjoining buffer layer in which turbulent mixing and diffusion are comparable. However, transport in the turbulent region is mainly dominated by turbulent mixing. The value of the Reynolds number at which the transition from laminar to turbulent occurs is known as the critical Reynolds number. The critical Reynolds number is dependent on the geometry as well as ﬂow conditions. 1.10.5.3.1 External ﬂow forced convection When a ﬂuid ﬂow is not conﬁned to a speciﬁc channel or passage, and ﬂows unboundedly over any surface such as a pipe, ﬂat plate, cylinder, or sphere, it is classiﬁed as external ﬂow. In order to determine the heat transfer rates for external ﬂow cases, several correlations between the dimensionless parameters Nusselt number, Reynolds number, and Prandtl number are utilized. These correlations were developed based on experimental data. The ﬂuid properties required to obtain these dimensionless parameters are usually taken at the ﬁlm temperature Tfm. The ﬁlm temperature is an average of the ﬂuid free stream and surface temperatures Tfm ¼ TsþTa 2 . The various forced convection heat transfer correlations for external ﬂow for different geometries, with the pertinent parameters listed in Table 4. Example 14: The top cover of a horizontal solar ﬂat plate collector is at a temperature of 501C. The solar collector is located at a location where the wind speed and temperature are determined to be 8.3 m/s and 221C. The length and width of the top cover are 3 m and 1 m, respectively. 1. Determine the rate of convective heat loss from the cover in these conditions. 2. Analyze the effect on the rate of convective heat loss from the cover as the air velocity varies from 5 to 10 m/s while other parameters remain constant. 3. Analyze the effect on the rate of convective heat loss from the cover as the air temperature varies from 5 to 351C while other parameters remain constant. Solution: Schematic (Fig. 20): Assumptions: Steady-state operating conditions persist. Analysis: 1. The rate of heat loss from the top cover of a horizontal solar ﬂat plate collector is to be determined; the ﬂuid in this case is air. The properties of air required to obtain the dimensionless parameters should be evaluated at the ﬁlm temperature Tfm. Tfm ¼ Ts þ Ta 2 ¼ 50 þ 22 ½ 1C 2 ¼ 361C At 361C, density of air r is 1.14 kg/m3 , Prandtl number Pr is 0.71, the kinematic viscosity n is 1.66 105 m2 /s, and the thermal conductivity is 0.0276 W/mK. Table 3 List of important heat transfer dimensionless parameters Name Symbol Deﬁnition Application Biot number Bi hY/k Steady- and unsteady-state conduction Fourier number Fo at/Y2 Unsteady-state conduction Graetz number Gz GY2 cp/k Laminar convection Grashof number Gr gbDTY3 /n2 Natural convection Rayleigh number Ra Gr Pr Natural convection Nusselt number Nu hY/kf Natural or forced convection, boiling, or condensation Peclet number Pe UY/a¼Re Pr Forced convection (for small Pr) Prandtl number Pr cpm/k¼n/a Natural or forced convection, boiling, or condensation Reynolds number Re UY/n Forced convection Stanton number St h/rUcp ¼Nu/Re Pr Forced convection 446 Heat Transfer Aspects of Energy
- 26. The Reynolds number at the plate end can be determined as follows: Re ¼ UY v ¼ 8:3 m=s ð Þ 3m ð Þ ð1:66 105 Þm2=s ¼ 15 105 45 105 Since the Reynolds number is greater than the critical Reynolds number, a suitable correlation to be used for this case is Nu ¼ 0:037Re 4 5 871 Pr 1 3 ¼ 0:037 15 105 4 5 871 0:71 1 3 ¼ 2097 Table 4 Forced convection heat transfer correlations and equations Equation or correlation Correlations for external ﬂow over a ﬂat plate Nu¼0.332Re1/2 Pr1/3 for PrZ0.6 for laminar; local; Tfm Nu¼0.664Re1/2 Pr1/3 for PrZ0.6 for laminar; average; Tfm Nu¼0.565Re1/2 Pr1/2 for Prr0.05 for laminar; local; Tfm Nu¼0.0296Re4/5 Pr1/3 for 0.6rPrr60 for turbulent; local; Tfm, Rer108 Nu¼(0.037Re4/5 871)Pr1/3 for 0.6oPro60 for mixed ﬂow; average; Tfm, Rer108 Correlations for external cross-ﬂow over circular cylinders Nu¼cRen Pr1/3 for PrZ0.7 for average; Tfm; 0.4oReo4 106 where c¼0.989 and n¼0.330 for 0.4oReo4 c¼0.911 and n¼0.385 for 4oReo40 c¼0.683 and n¼0.466 for 40oReo4000 c¼0.193 and n¼0.618 for 4000oReo40,000 c¼0.027 and n¼0.805 for 40,000oReo400,000 Nu¼cRen Prs (Pra/Prs)1/4 for 0.7oPro500 for average; Ta; 1oReo106 where c¼0.750 and n¼0.4 for 1oReo40 c¼0.510 and n¼0.5 for 40oReo1000 c¼0.260 and n¼0.6 for 103 oReo2 105 c¼0.076 and n¼0.7 for 2 105 oReo106 s¼0.37 for Prr10 s¼0.36 for Pr410 Nu¼0.3 þ [(0.62Re1/2 Pr1/3 )/(1 þ (0.4/Pr)2/3 )1/4 ][1 þ (Re/28,200)5/8 ]4/5 for RePr40.2 for average; Tfm Correlations for internal ﬂow in tubes Nu¼4.36 for constant surface heat ﬂux; fully developed; laminar Nu¼3.66 for constant surface temperature; fully developed; laminar Nu¼3.66 þ (0.065(D/L) Re Pr)/(1 þ 0.04[(D/L) Re Pr)]2/3 for constant surface temperature; developing ﬂow; laminar Nu¼0.023Re4/5 Prn for 0.7rPrr160; turbulent; fully developed; ReZ10,000; n¼0.4 for ﬂuid heating; n¼0.3 for ﬂuid cooling Nu¼4.8 þ 0.0156Re0.85 Prs 0.93 for liquid metal ﬂow; constant surface temperature; 104 oReo106 Nu¼6.3 þ 0.0167Re0.85 Prs 0.93 for liquid metal ﬂow; constant surface heat ﬂux; 104 oReo106 Correlations for external cross-ﬂow over spheres Nu/Pr1/3 ¼0.37Re0.6 /Pr1/3 for average; Tfm; 17oReo70,000 Nu¼2 þ (0.4Re1/2 þ 0.06Re2/3 )Pr0.4 (ma/ms)1/4 for 0.71oPro380 for average; Ta; 3.5oReo7.6 104 ; 1o(ma/ms)o3.2. Correlation for falling drop Nu¼2 þ 0.6Re1/2 Pr1/3 [25(x/D)0.7 ] for average; Ta Source: Reproduced from Dincer I. Heat transfer in food cooling applications. Washington, DC: Taylor Francis; 1997. Air Ta = 22°C U = 8.3 m s−1 3m Ts = 50°C Fig. 20 Heat transfer from a ﬂat plate solar collector. Heat Transfer Aspects of Energy 447
- 27. Once the Nusselt number is determined, the heat transfer coefﬁcient h can be obtained from the equation Nu ¼ hY k h ¼ Nu k Y ¼ 2097 0:0276 W=mK 3 m ¼ 18:92 W=m2 K Hence, the rate of heat loss from the collector cover can be determined as _ Q ¼ hA Ts Tf ¼ 18:92 W=m2 K 3 m 1 m ð Þ 50 22 ½ 1C ¼ 1589 W 2. The effect of changing air velocity U on the rate of convective heat loss from the cover while all other parameters are held constant is shown in Fig. 21. As can be depicted from the graph, at 10 m/s the rate of heat loss increases to 1939 W, while at 5 m/s, it drops to 863.7 W. 3. The effect of varying air temperature Ta on the rate of convective heat loss from the cover while all other parameters are held constant is shown in Fig. 22. As can be depicted from the graph, as the air temperature rises to 351C, the rate of heat loss drops to 832 W, whereas at an air temperature of 51C, the rate of heat loss rises to 2635 W. Example 15: In various energy applications, the excess energy available during a process is stored in a medium for later use. Thermal energy storage involves storing the excess produced energy in the form of thermal energy. Consider a cylindrical thermal energy storage tank of diameter 3 m and a height of 5 m. The efﬁciency of a thermal energy storage decreases as the thermal losses from the tank increase. Assume the temperature of the outer surface of the thermal energy storage tank remains constant at 501C. 1. Determine rate of convective heat loss from the tank if the local wind speed is 2 m/s and temperature is 221C. 2. Analyze the effect of changing air temperature on the rate of heat loss from the tank. 3. If the overall energy efﬁciency of a closed thermal energy storage system is deﬁned as Zoverall ¼ Energy recovered Energy input Determine the overall energy efﬁciency for a day for the conditions given above if the total energy input was 24 GJ and analyze the effect of air temperature on the overall efﬁciency. Solution: Assumptions: Steady-state operating conditions persist. The external surface temperature of the tank remains uniform and constant. 1. The external surface temperature of the tank is assumed to be constant at 501C. The velocity of cross-ﬂow wind and temperature are also known. The properties of air need to be determined at ﬁlm temperature Tfm. The ﬁlm temperature can be evaluated as Tfm ¼ Ts þ Ta ð Þ 2 ¼ 50 þ 22 ½ 1C 2 ¼ 361C 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 U (m s−1) Q loss (W) Fig. 21 Effect of air velocity on convective heat loss. 448 Heat Transfer Aspects of Energy
- 28. At this temperature, density (r) of air is 1.1 kg/m3 , the kinematic viscosity (n) is 1.66 105 m2 /s, Prandtl number (Pr) is 0.71, and the thermal conductivity (k) is 0.027 W/mK. After obtaining the properties, the Reynolds number can be determined from the equation Re ¼ UY v For cases involving external cross ﬂow over a cylinder, the diameter represents the characteristic length Y. Hence, the Reynolds number is calculated as Re ¼ UY v ¼ 2 m=s ð Þ 3 m ð Þ 1:66 105 m2=s ð Þ ¼ 361; 445:8 Based on the Reynolds number, the suitable correlation can be chosen from Table 2: Nu ¼ 0:027Re0:805 Pr 1 3 ¼ 0:027 361; 445:780:805 0:71 1 3 ¼ 717:8 From the Nusselt number, the convective heat transfer coefﬁcient h can be obtained as h ¼ Nu k Y ¼ 717:8 ð Þ 0:027 W=mK ð Þ ð3 mÞ ¼ 6:5 W=m2 K After obtaining the convective heat transfer coefﬁcient, the Newton’s law of cooling can be applied to determine the rate of convective heat loss from the tank: _ Q ¼ hA Ts Ta ð Þ ¼ 6:5 W=m2 K p 3 m ð Þ 5 m ð Þ 50 22 ½ 1C ¼ 8576:5 W 2. The effect of air temperature on the rate of heat loss is shown in Fig. 23. As the air temperature decreases to 101C, the rate of heat loss increases to 18,859 W. However, when the air temperature reaches 351C, then the rate of heat loss decreases to 4507 W. 3. The overall energy efﬁciency of the storage tank can be expressed as Zoverall ¼ Energy recovered Energy input ¼ Qrecovered Qin ¼ 1 QL Qch where QL denotes the energy lost and Qch represents the energy input to the tank during charging. Hence, the overall efﬁciency can be determined as Zoverall ¼ 1 8576 24 3600 J ð Þ 24 109 J ð Þ ¼ 0:97 Hence, the overall efﬁciency of the storage tank is 97% assuming the temperature of the tank and other given conditions remain constant. 5 10 15 20 25 30 35 1000 1250 1500 1750 2000 2250 2500 Ta (°C) Q loss (W) Fig. 22 Effect of ambient temperature on convective heat loss. Heat Transfer Aspects of Energy 449
- 29. The effect of air temperature on the overall efﬁciency is shown in Fig. 24. The efﬁciency drops to 93% at an air temperature of 101C and increases to 98.4% at a temperature of 351C. Example 16: Condensers are used extensively in refrigeration cycles to extract heat from the working ﬂuid. Consider a section of a circular cylindrical condenser pipe that has an external surface temperature of 651C. Air at a velocity of 15 m/s is passes over the pipe. The pipe has a length of 2 m and a diameter of 15 cm. If the temperature of air passing over the pipe is 181C, 1. Determine the rate of heat loss from the surface of the pipe due to convection. 2. Analyze the effect of changing the air velocity from 5 to 35 m/s on the convective heat transfer rate. Solution: Assumptions: Steady-state operating conditions persist. The external surface temperature of the pipe remains uniform and constant. Analysis: 1. The outer surface temperature of the condenser is known. Also, the cross-ﬂow air velocity and temperature are also known. To determine the dimensionless parameters required, properties of air at the ﬁlm temperature Tfm are necessary. The ﬁlm −10 −5 0 5 10 15 20 25 30 35 0 2000 4000 6000 8000 10,000 12,000 14,000 16,000 18,000 20,000 Ta (°C) Q loss (W) Fig. 23 Effect of ambient temperature on heat loss rate. −10 −5 0 5 10 15 20 25 30 35 0.93 0.935 0.94 0.945 0.95 0.955 0.96 0.965 0.97 0.975 0.98 0.985 Ta (°C) overall Fig. 24 Effect of ambient temperature on overall efﬁciency. 450 Heat Transfer Aspects of Energy
- 30. temperature can be evaluated as Tfm ¼ Ts þ Ta ð Þ 2 ¼ 65 þ 18 ½ 1C 2 ¼ 41:51C At this temperature, density (r) of air is 1.1 kg/m3 , the kinematic viscosity (n) is 1.72 105 m2 /s, Prandtl number (Pr) is 0.71, and the thermal conductivity (k) is 0.027 W/mK. The Reynolds number can be determined from Re ¼ UY v where Y represents the characteristic length. For the case of external cross ﬂow over a cylinder, the characteristic length is represented by its diameter. Hence, the Reynolds number is calculated as Re ¼ UY v ¼ 15 m=s ð Þ 0:15 m ð Þ 1:72 10 5 m2=s ð Þ ¼ 130; 814 Based on the Reynolds number, the suitable correlation can be chosen from Table 2 Nu ¼ 0:027Re0:805 Pr 1 3 ¼ 0:027 130; 8140:805 0:71 1 3 ¼ 316:7 After determining the Nusselt number, the convective heat transfer coefﬁcient h can be obtained as h ¼ Nu k Y ¼ 316:7 ð Þ 0:027 W=mK ð Þ ð0:15 mÞ ¼ 57 W=m2 K Once the convective heat transfer coefﬁcient is determined, the Newton’s law of cooling can be applied _ Q ¼ hA Ts Ta ð Þ ¼ 57 W=m2 K p 0:15 m ð Þ 2 m ð Þ 65 18 ½ 1C ¼ 2524:9 W 2. The effect of increasing or decreasing air velocity on the convective heat transfer rate is depicted in Fig. 25. As can be observed, the rate of heat transfer increases considerably as the air velocity increases. At a velocity of 35 m/s, the rate of convective heat transfer is 5083 W. However, at a velocity of 5 m/s, the rate of heat transfer drops to 1061 W. Example 17: It is essential to measure the temperature of the combustion gases produced in the combustion chamber of a power plant. A thermocouple that has a spherical junction of 0.85 mm diameter is inserted in a passage in which the combustion gases ﬂow with a velocity of 3.5 m/s. The thermal conductivity of the junction is 80 W/mK, density is 8575 kg/m3 , and speciﬁc heat is 410 J/kg K. Whereas, the combustion gases are estimated to have a thermal conductivity of 0.075 W/mK, a kinematic viscosity of 39 106 m2 /s, and a Prandtl number of 0.71. 1. When the thermocouple is inserted in the passage, its temperature rises and the temperature difference between the thermo- couple junction and combustion gases decreases. Determine the time required by the thermocouple for the temperature difference to reach 5% of its initial value. 2. Analyze how the response time of the thermocouple changes as the speciﬁc heat of the spherical junction is varied while other parameters remain constant. 3. Analyze the effect of changing velocity of combustion gases and changing thermal conductivity of the junction on the time required by the thermocouple for the temperature difference to reach 5% of its initial value. 5 10 15 20 25 30 35 1600 2400 3200 4000 4800 5600 U (m s−1 ) Q (W) Fig. 25 Effect of air velocity on heat loss rate. Heat Transfer Aspects of Energy 451
- 31. Solution: Assumptions: 1. The combustion gases have a constant temperature. Radiation heat transfer is negligible. Analysis: This case involves changing of temperature with time. If the Biot number is less than 0.1, the lumped system method can be utilized. Bi ¼ hY k where the characteristic length Y for a sphere can be obtained from the equation Y ¼ Volume Surface area ¼ 4 3 pr3 4pr2 ¼ r 3 ¼ 0:000425 3 ¼ 0:00014 m In order to determine the heat transfer coefﬁcient h, the Nusselt number is required. The appropriate correlation for the Nusselt number can be chosen from Table 2. For the case of external cross-ﬂow over spheres, the suitable correlation is Nu ¼ 2 þ 0:4Re 1 2 þ 0:06Re 2 3 Pr0:4 ma ms 1 4 where the Reynolds number can be calculated as Re ¼ UY v ¼ 3:5 m=s ð Þ 0:00085 m ð Þ 39 106 m2=s ð Þ ¼ 76:3 Substituting the Reynolds number in the Nusselt number correlation Nu ¼ 2 þ 0:4 76:3 ð Þ 1 2 þ 0:06 76:3 ð Þ 2 3 0:710:4 ¼ 5:98 Once the Nusselt number is obtained, the heat transfer coefﬁcient can be determined h ¼ Nu k D ¼ 5:98 0:075 W=m K ð Þ 0:00085 m ð Þ ¼ 527:65 W=m2 K The Biot number can then be obtained as Bi ¼ hY k 527:65 W=m2 K ð Þ 0:00014 m ð Þ 80 W=mK ð Þ ¼ 9:2 104 Since the Biot number is less than 0.1, the lumped system method can be utilized. The time required by the thermocouple to reach 5% of the initial temperature difference can be calculated by the lumped system method as t ¼ rVcp hA ln Ti Ta T Ta ¼ rcpr 3h ln Ti Ta 0:05 Ti Ta ð Þ ¼ 8575 kg=m3 ð Þ 410 J=kg K ð Þ 0:000425 m ð Þ 3 ð Þ 527:65 W=m2 K ð Þ ln 20 ð Þ t ¼ 2:83 s Hence, it will take approximately 2.83 s for the thermocouple junction to reach 5% of the initial temperature difference. 2. The effect of changing speciﬁc heat on the response time of the thermocouple is shown in Fig. 26; as can be observed, the response increases to 5.5 s at a speciﬁc heat value of 800 J/kg K and drops to 0.7 s at a speciﬁc heat value of 100 J/kg K. 3. The effect of changing velocity of combustion gases from 2 to 35 m/s on the thermocouple time to reach 5% of initial temperature difference is shown in Fig. 27. As can be depicted from the ﬁgure, the response time of the thermocouple drops signiﬁcantly as the velocity increases. At a velocity of 35 m/s, the response time decreases to 1.1 s; whereas, at a low velocity of 2 m/s, the response time increases to 3.4 s. Example 18: In a heat exchanger, hot water at 951C ﬂows through the tubes of a tube bank. The tubes are arranged as shown in Fig. 28 and have an outer diameter of 1 cm. Air at a mean velocity of 6 m/s enters the tube bank and ﬂows in a normal direction over the tubes. The tube bank contains 20 rows of tubes and 10 tubes in each row. If air enters the tube bank with a temperature of 181C, determine the outlet temperature. Also, determine rate of heat transfer that occurs between the ﬂowing air and tubes per unit length of tubes. In addition, analyze how the air exit temperature and the rate of heat transfer changes with the number of rows of tubes in the tube bank. Solution: Assumptions: Steady operation conditions persist. The temperature of hot water ﬂowing through the tubes is equal to the outer surface temperature of the tubes. Analysis: The initial step in such analysis involves determining the properties of the ﬂuid at the mean temperature, which in this case is air. However, since the exit temperature of air is unknown, an initial mean temperature value of 251C is taken to determine 452 Heat Transfer Aspects of Energy
- 32. 100 200 300 400 500 600 700 800 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 t (s) cp (J/kgK) Fig. 26 Effect of speciﬁc heat on response time. 2 5 8 11 14 17 20 23 26 29 32 35 0.5 1.0 1.5 2.0 2.5 3.0 3.5 U (m s−1) t (s) Fig. 27 Effect of combustion gases velocity on thermocouple response time. 3 cm 3 cm Air V = 6 m s−1 T = 18°C Fig. 28 Cooling of water in a tube bank. Heat Transfer Aspects of Energy 453
- 33. the required properties. After obtaining the ﬁnal result, this value will be compared to the obtained value. The density of air at 251C and atmospheric pressure is 1.18 kg/m3 , Prandtl number at 251C is 0.7073 and at a surface temperature is 0.7006, kinematic viscosity 1.55 105 m2 /s, speciﬁc heat is 1006 J/kg K and thermal conductivity is 0.026 W/mK. In cases involving tube banks, Reynolds number is evaluated at the maximum velocity, which occurs while the ﬂuid ﬂows within the tube bank. The maximum velocity can be calculated from the distances between the tubes given in the ﬁgure as Vmaximum ¼ dt dt D V, where dt denotes the distance between the tubes in the transverse direction, V represents the approaching velocity, and D represents the diameter of the tubes. Vmaximum ¼ dt dt D V ¼ 0:03 m 0:03 0:01 ð Þm 6 m=s ð Þ ¼ 9 m=s Re ¼ VmaximumD v ¼ 9 m=s ð Þ 0:01 m ð Þ 1:55 105 m2=s ð Þ ¼ 5806:5 Based on the Reynolds number, the applicable Nusselt number correlation can be utilized: Nu ¼ 0:27Re0:63 Pr0:36 Pr Prs 0:25 ¼ 0:27 5806:50:63 0:70730:36 0:7073 0:7006 0:25 ¼ 56:2 Once the Nusselt number is determined, heat transfer coefﬁcient h can be obtained as h ¼ Nu k D ¼ 56:2 0:026 W=mK ð Þ 0:01 m ¼ 146:1 W=m2 K There are 20 rows with 10 tubes in each row, hence, there are 200 tubes in total. The surface area per unit length of tube subjected to heat transfer can be calculated as A ¼ npDL ¼ 200p 0:01 m ð Þ 1 m ð Þ ¼ 6:28 m2 The mass ﬂow rate of air can be calculated as _ ma ¼ rV 10 0:03 m ð Þ 1 m ð Þ ð Þ ¼ 1:18 kg=m3 6 m=s ð Þ 0:3 m2 ¼ 2:124 kg=s The exit temperature of air can be determined from the equation Te ¼ Ts Ts Ti ð Þexp hA _ macp ¼ 95 95 18 ð Þexp 146:1 W=m2 K ð Þ 6:28 m2 ð Þ 2:124 kg=s ð Þ _ 1006 J=m2 K ð Þ ! Te ¼ 44:91C Hence, the air exits the tube bank of the heat exchanger at a temperature of 44.91C. The rate of heat transfer per unit length of the tubes can be determined from the equation _ Q ¼ _ mcp Te Ti ð Þ ¼ 2:124 kg=s ð Þ 1006 J=kg K ð Þ 44:9 18 ½ 1C ¼ 57:5 103 W ¼ 57:5 kW The mean temperature assumed was 251C, whereas the mean temperature from the obtained results is TiþTe ð Þ 2 ¼ 18þ44:9 2 ¼ 31:451C. When the above calculations are repeated with a mean temperature of 31.451C, a _ Q value of 57.23 kW is obtained. The effect on the air exit temperature and heat transfer rate as the number of rows of tubes changes is shown in Fig. 29. As can be observed from the ﬁgure, the heat transfer rate increases to 88 kW when the 36 rows of tubes are used, while the exit temperature increases to 601C (Figs. 29 and 30). 1.10.5.3.2 Internal ﬂow forced convection When a ﬂuid ﬂow is restricted to ﬂow within a given passage such as pipe or duct, it is classiﬁed as internal ﬂow. Internal ﬂow forced convection cases are encountered in various energy applications. Heat gain or loss by the air ﬂowing in the ducts of air conditioning systems is an example of internal ﬂow forced convection. In addition, heat loss from steam pipes in a power plant is also a typical example. In internal ﬂow cases, as the ﬂuid ﬂow is restricted within a pipe or duct, the boundary layer has a limit unto which it can grow. In order to determine the heat transfer rates, correlations between dimensionless parameters such as Nusselt number, Reynolds number, and Prandtl number are utilized. Pertinent correlations for different types of cases have been listed in Table 2. The correlations can be used to determine the Nusselt number and thus the heat transfer coefﬁcient for a given case. Example 19: A boiler utilizes hot air to generate steam. The hot air ﬂows through a 3-meter-long tube passing through the boiler. Consider a case in which the water in the boiler is being boiled at a temperature of 1051C. Air enters the 10-cm-diameter tube at a temperature of 2801C and ﬂows with an average velocity of 5 m/s. If the temperature of the tube at its outer surface is equal to the temperature of the boiling water, determine 1. The convective heat transfer coefﬁcient of the air ﬂowing in the tubes. 2. Rate of steam generation in the boiler. 454 Heat Transfer Aspects of Energy
- 34. Solution: Assumptions: • Steady-state operation conditions persist. • The outer temperature of the tube is equal to the temperature of the boiling water in the boiler. • Negligible pipe thermal resistance. • Tube inner surface is smooth. Analysis: 1. The exit temperature of air is not known, hence, the properties of air are obtained at an assumed mean temperature of 2001C. After obtaining the exit temperature, the calculations will be repeated with the obtained new mean temperature. The density of air is 0.73 kg/m3 , Prandtl number is 0.69, speciﬁc heat is 1026 J/kg K, thermal conductivity is 0.039 W/mK, and kinematic viscosity is 3.59 105 . The average velocity of air through the tube is known to be 5 m/s, so the Reynolds number can be calculated: Re ¼ VD v ¼ 5 m=s ð Þ 0:1 m ð Þ 3:59 105 m2=s ð Þ 13; 927:6 16 20 24 28 32 36 40 45 50 55 60 Number of rows T e (°C) Fig. 29 Effect of number of rows on air exit temperature. 16 20 24 28 32 36 40 50 60 70 80 90 100 Number of rows Q (kW) Fig. 30 Effect of number of rows on the heat transfer rate. Heat Transfer Aspects of Energy 455
- 35. The Reynolds number is greater than 10,000, hence the ﬂow is turbulent, assuming the ﬂow is fully developed throughout the tube. The appropriate Nusselt number correlation is chosen from Table 2. Nu ¼ 0:023Re0:8 Pr0:3 ¼ 0:023 13; 927:6 ð Þ0:8 0:69 ð Þ0:3 ¼ 42:5 Thus, the convective heat transfer coefﬁcient h is obtained from the equation h ¼ Nu k D ¼ 42:5 0:039 W=m K ð Þ ð0:1 mÞ ¼ 16:6 W=m2 K 2. The temperature of air at the exit of the tube can be obtained as Te ¼ Ts Ts Ti ð Þexp hA _ mcp ¼ 1051C ð Þ 105 280 ½ 1C exp 16:6 W=m2 K ð Þ p 0:1 m ð Þ 3 m ð Þ ð Þ 0:029 kg=s ð Þ 1026 J=kg K ð Þ Te ¼ 210:61C In case of internal ﬂow constant surface temperature, the logarithmic mean temperature difference is calculated as DTlm ¼ Te Ti ð Þ ln TsTe TsTi ¼ 257:3 280 ½ 1C ln 105275:3 105280 ¼ 137:41C For internal ﬂow constant surface temperature cases, the rate of convective heat transfer can be obtained from the equation _ Q ¼ hADTlm ¼ 16:6 W=m2 K p 0:1 m ð Þ 3 m ð Þ ð Þ 137:41C ð Þ ¼ 2083 W Once the rate of heat transfer is obtained, the rate of steam generation in the boiler can be determined _ msteam ¼ _ Q hfg ¼ 2556:3 W 2; 257; 000 J=kg ¼ 0:00092 kg=s The properties of air were determined at an initial assumed mean temperature of 2001C. Evaluating the air properties at the new mean temperature T ¼ 210:6þ280 2 ¼ 245:31C , and repeating the above calculations, we obtain Te ¼2091C; _ Q¼1962 W _ msteam ¼ 0:00087 kg=s. Example 20: During the design of an air conditioning system, the heat loss from a duct while operating in the heating mode of operation is being considered. The duct has a square cross-section of dimensions 0.3 m 0.3 m and has a length of 10 m (Fig. 31). The temperature of air as it enters the duct is measured to be 651C. The temperature of the duct remains nearly constant at 401C. If air ﬂows through the duct at an average velocity of 4 m/s, determine the rate at which heat is lost from the air as it travels through the duct. Solution: Assumptions: Steady-state operation conditions persist. The duct inner surface is smooth. And the temperature of the duct remains constant. Analysis: The exit temperature of air is not known; hence, the bulk ﬂuid mean temperature cannot be determined. The properties of air are determined at the inlet temperature of 651C. After obtaining the outlet temperature, the calculations will be repeated with the obtained mean temperature. At 651C, the density of air is 1.044 kg/m3 , speciﬁc heat is 1008 J/kg K, thermal conductivity is 0.029 W/mK, kinematic viscosity is 1.9 105 m2 /s, and Prandtl number of 0.7. As the duct is noncircular, the hydraulic diameter needs to be determined to evaluate the Reynolds number. The hydraulic diameter Dh is obtained as Dh ¼ 4Acs p ¼ 4 0:3 m 0:3 m ð Þ ð4 0:3 mÞ ¼ 0:3 m Air 65°C Ts = 40°C L = 10 m Fig. 31 Heat loss from a duct. 456 Heat Transfer Aspects of Energy
- 36. The Reynolds number can then be calculated using the hydraulic diameter Re ¼ VDh v ¼ 4 m=s ð Þ 0:3 m ð Þ 1:9 105 m2=s ð Þ ¼ 63157:9 Since the Reynolds number is greater than 10,000, the ﬂow is turbulent. The entry lengths are approximately equal to 10Dh¼3 m. As the length of the duct is much longer than the entry lengths, the ﬂow can be assumed to be fully developed throughout the duct. To obtain the Nusselt number, the appropriate correlation for the case of fully developed turbulent internal ﬂow can be applied Nu ¼ 0:023Re0:8 Pr0:3 ¼ 0:023 63157:90:8 0:70:3 ¼ 143:1 The heat transfer coefﬁcient can be determined from the Nusselt number and the hydraulic diameter h ¼ Nu k Dh 143:1 0:029 W=m K ð Þ ð0:3 mÞ ¼ 13:8 W=m2 K After obtaining the heat transfer coefﬁcient, the exit temperature of air can be determined Te ¼ Ts Ts Ti ð Þexp hA _ mcp ¼ 401C ½ 40 65 ½ 1C exp 13:8 W=m2 K ð Þ 4 0:3 m ð Þ 10 m ð Þ ð Þ 0:38 kg=s ð Þ 1008 J=kg K ð Þ Te ¼ 56:221C As this is a case of constant surface temperature, the logarithmic mean temperature difference is calculated DTlm ¼ Ti Te ð Þ ln TsTe TsTi ¼ 65 56:22 ½ 1C ln 4056:22 4065 ¼ 20:31C The rate of heat loss from the duct can be obtained as _ Q ¼ hADTlm ¼ 13:8 W=m2 K 4 0:3 m 10 m ð Þ 20:31C ð Þ ¼ 3361:7 W The rate of heat loss from the duct is obtained to be 3361.7 W. After repeating the above calculations at the new mean temperature of T ¼ 65þ56:22 2 ¼ 60:611C, the rate of heat loss is obtained as 3391 W. Example 21: A solar thermal power plant utilizes parabolic trough solar collectors. The temperature of the concentrator ﬂuid as it leaves the solar collector is required to be 3901C, while it enters the solar collector at a temperature of 2901C. The heat ﬂux concentrated on the tube by the concentrator is approximated to be 18,500 W/m2 . The concentrator ﬂuid has density of 850 kg/m3 , thermal conductivity of 0.09 W/mK, speciﬁc heat of 3000 J/kg K and a kinematic viscosity of 1.9 107 m2 /s. If the ﬂuid ﬂows through a single tube with a mass ﬂow rate of 3.2 kg/s, 1. Determine the length of the concentrator required if the tube has a diameter of 10 cm. 2. Plot the ﬂuid temperature and surface temperature of the tube as a function of the collector length. 3. Analyze the effect of tube diameter on the required collector length. Solution: Assumptions: Steady operation conditions persist. The tube is thin walled. In addition, the heat ﬂux on the tube surface is constant and uniform. Analysis: 1. The collector tube has a constant surface heat ﬂux of 18,500 W/m2 , the inlet temperature of the concentrator ﬂuid is known, and the required tube length is to be determined to achieve the outlet temperature required. Applying an energy balance on the tube qA ¼ _ mcp To Ti ð Þ q pDL ð Þ ¼ _ mcp To Ti ð Þ L ¼ _ mcp To Ti ð Þ q pD ð Þ ¼ 3:2 kg=s ð Þ 3000 J=kg K ð Þ 390 290 ½ 1C 18; 500 W=m2 ð Þ p 0:1 m ð Þ ð Þ ¼ 165:2 m The rate of heat transfer with this length of collector is _ Q ¼ q pDL ð Þ ¼ 18; 500 W=m2 p 0:1 m ð Þ 165:2 m ð Þ ð Þ ¼ 9; 60; 133 W 2. The surface temperature of the tube can be determined from Newton’s law of cooling q ¼ h Ts Tf ð Þ Heat Transfer Aspects of Energy 457