5ME02 HEAT TRANSFER SECTION A
UNIT 1
Introduction, heat transfer in engineering, modes of heat transfer, basic laws of heat transfer and their basic equations. Conduction thermal conductivity and thermal diffusivity effect of phase & temperature on thermal conductivity, one dimensional steady state heat conduction through slab, cylinder & sphere-simple and composite. Combined conduction- convection, overall heat transfer coefficient. General heat conduction differential equation. One dimensional steady state conduction with internal heat generation for infinite slab, wire & cylinder. (8 Hrs)
UNIT 2
Insulations, critical radius of insulation, Conduction through extended surfaces, analysis of a uniform C.S. fin, fin efficiency, fin effectiveness, Biot number. Introduction to unsteady state heat conduction, Newton’s law of cooling, lumped heat capacity analysis. (8 Hrs)
UNIT 3
Radiation- general concepts and definitions, black body & grey body concept. Laws of radiation- Kirchoff’s, Plank’s, Stefan- Boltzman’s, Wien’s law. Concept of shape factor, emissivity factor and radiation heat transfer equation. (No numericals). Radiation errors in temperature, measurement, radiation shield. (7 Hrs)
SECTION B
UNIT 4
Forced convection- heat convection, forced and natural convection, boundary layer theory, hydrodynamic & thermal boundary layers, boundary layer thickness. Laminar & turbulent flow over flat plate and through pipes & tubes (only concept, no derivation & analytical treatment). Dimensionless number and their physical significance Reynolds, Prandtl, Nusselt, Grashof number, empirical correlations for forced convection for flow over flat plate, through pipes & tubes & their applications in problem solving. (8 Hrs)
UNIT 5
Free convection- velocity and thermal boundary layers for vertical plate, free convection over vertical cylinder and horizontal plate/cylinder (only concept, no derivation & analytical treatment). Use of empirical correlations in problem solving. Condensation & Boiling - introduction to condensation heat transfer, film & drop condensation. Boiling heat transfer, pool boiling curves. (7 Hrs)
UNIT 6
Heat exchanger - applications, classification, overall heat transfer coefficient, fouling. L.M.T.D. & E.N.T.U. methods, temperature profiles, selection of heat exchangers. Introduction to working of heat pipe with and without wick. (7 Hrs)
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HEAT TRANSFER 3 ______
UNIT 1 INTRODUCTION
In thermodynamics when two systems are brought into contact with each other some kind of wall, energy transfer such as heat and work take place between them. Work is transfer of energy to a principle which is evidence by changes in its position when acted upon by a force. Heat like work is energy in the process of being transferred. Energy is what is stored, and work and heat are two ways of transferring energy across the boundaries of system. The amount of energy transfer as heat can be determined from energy conservation considerations. Two system are said to be in thermal equilibrium with one another if no energy transfer such as heat occur between them in a finite period when they are connected through the diathermal wall. Temperature is property of matter which two bodies are in equilibrium having in common. Hot and cold are the adjectives used to describe high and low values of temperature. The energy transfer by heat will take place from body with higher temperature to the body with lower temperature, if they two are permitted to interact through a diathermal wall (second law of thermodynamics).
The transfer and conversion of energy from one form to another is basic to all heat transfer processes and hence they are governed by the first as well as second law of thermodynamics. This does not and must not mean that the principles governing the transfer of heat transfer that can be derived from, or are more corollaries of, the basic law of thermodynamics.
The major difference between the thermodynamics and heat transfer is that the former deals with the relation between heat and other forms of energy, whereas the latter is concerned with the analysis of the rate of heat transfer. Thermodynamics deals with the system in equilibrium so it cannot be expected to predict quantitatively the rate of change in process which results from non equilibrium states. Temperature is meant for heat transfer to take place.
The knowledge of temperature distribution is essential in heat transfer studies because of the fact that heat flow takes place only wherever there is temperature gradient in a system. The heat flux (q) is defined as the amount of heat transfer per unit area per unit time, can be calculated from the physical law relating the temperature gradient in the heat flow.
HEAT TRANSFER IN ENGINEERING
The study of temperature distribution and heat transfer is of great importance to engineers because of its almost universal occurrence in many branches of science and engineering. The first step in optimal design of heat exchangers such as boiler, heater, refrigerator and radiator is a detailed analysis of heat transfer. This is essential to determine the feasibility and cost of the undertaking, as well as the size of equipment required to transfer specified amount of heat in given time. A through heat transfer analysis is most important for the proper sizing of fuel elements in the nuclear reactor core to prevent burnout. The performance of air craft is also depending on the ease with which structure and engine can be cooled. The design of chemical plant is usually done on the basis of heat transfer and the analogous mass transfer processes. An accurate heat transfer analysis is necessary in the refrigeration and air-conditioning applications to calculate the heat loads, and to determine the thickness of insulation to avoid excess in heat gains or losses. The utilization of solar
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energy which is so abundantly available also requires a thorough knowledge of heat transfer for the proper design of the solar collector and associated equipment. These are only a few examples in to indicate the importance of heat transfer in engineering sciences. It is clear that engineers and scientist must have a thorough knowledge of science of heat transfer to be able to quantitatively analyse the problem involving the transfer of heat.
MECHANISM OF HEAT TRANSFER
Energy transfer as heat takes place by three distinct modes: conduction convection and radiation. The heat conduction is mode of heat transfer accomplish via two mechanism. By molecular interaction whereby the energy exchanges takes place by the kinetic motion or direct impact of molecule. Molecule at relatively high energy level. Molecule at a relatively high energy level imparts energy to adjacent molecule at lower energy levels. This types of energy transfer always exists so long as there is temperature gradient in a system comprising molecules of solid, liquid or gas. The drift of free electron as in the case of metallic solids. The metallic alloys have different concentration of free electrons, and their ability to conduct heat directly proportional to the concentration of free electron in them. The free electron concentration of non metals very low. Hence materials that are good conductor are good conductor of heat too. Pure conduction is found only in solid.
CONVECTION
It is possible only in the presence of fluid medium. When a fluid flows inside a duct or over a solid body and the temperature of the fluid and the solid surfaces are different the heat transfer between the fluid and the solid surface will take place. This is due to the motion of fluid relative to the surface. This type of heat transfer is called convection. The transport of fluid inseparably linked with the movement of fluid itself. If the fluid motion is set up by buoyancy effect resulting in from the density variation caused by the temperature difference in the fluid, the heat transfer is said to be free or natural
HEAT TRANSFER 5 ______convection. On the other hand if the fluid motion is artificially created by means of an external agency like a blower or fan, the heat transfer is termed as forced convection.
As the energy transferred between the solid surface and the fluid at the surface can take place only by conduction, the heat transfer by convection is always accompanied by conduction.
RADIATION
If two bodies are at different temperature are placed in an evacuated adiabatic enclosure so that they are not in contact through a solid or fluid medium, temperature of the two bodies will tend to become equal. The mode of heat transfer by which this equilibrium is achieved is called thermal radiation. Radiation is an electromagnetic wave phenomenon, and no mediums required for its propagation. In fact the energy transfer by radiation is maximum when the two bodies exchanging energy are separated by a perfect vacuum. Thermal radiation depends only on the temperature and on the optical properties of the emitter.
Apart from the identification of different modes of heat transfer it is also important to determine whether a process is steady or unsteady.
STEADY AND UNSTEADY PROCESS
A steady process is a one which the rate of heat transfer does not vary with time. In steady process there can be no change in the internal energy of a system because the rate of energy influx must be equal to the rate of energy efflux, for example the heat transfer from a hot to cold fluid in a heat exchanger.
When the temperature at various points in a system does change with time, the process is called a transient or unsteady process. Unsteady problem are more complex in nature, but are often encountered in processes like soaking of ingots, heat treatment of metal casting etc.
The basic process of conduction, convection and radiation often combined both in nature and in engineering applications. The steam generating tubes of a boiler, for instance, receives the heat from the products of fuel combustion by all the three modes of heat transfer. Therefore, it is not actually possible to isolate entirely one mode for interaction with the other modes. However, for better understanding, one should study these modes separately.
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BASIC LAWS OF HEAT TRANSFER FOURIERS LAW OF HEAT CONDUCTION
The rate of heat flow through a simple homogeneous solid is directly proportional to the area of the section at right angle to the direction of heat flow, and to change of temperature with respect to length of the path of the heat flow.
The rate of heat transfer per unit area is directly proportional to temperature gradient. Mathematically it can be represented by the equation.
푑푇 푄 ∝ 퐴 푑푥
푄 = Heat flow through the body per unit time in W
퐴 =Surface area of heat flow (perpendicular to the direction of flow) 푚2
푑푥 =Thickness of body in the direction of flow (m)
푑푇 =Temperature difference at the faces of block of thickness dx
푑푇 푄 = −푘퐴 푑푥
푘 =Constant of proportionality and is known as thermal conductivity of the body W/mk.
Negative sign shows that temperature decreases along with the direction of increasing the thickness.
The temperature gradient is always negative on the positive direction of x axis therefore the value of Q becomes positive.
Assumptions :
The following are the assumptions on which Fourier’s law is based : 1. Conduction of heat takes place under steady state conditions. 2. The heat flow is unidirectional. 3. The temperatures gradient is constant and the temperature profile is linear. 4. There is no internal heat generation. 5. The bounding surfaces are isothermal in character. 6. The material is homogeneous and isotropic (i.e., the value of thermal conductivity is constant in all directions).
Some essential features of Fourier’s Law :
Following are some essential features of Fourier’s law : 1. It is applicable to all matter (may be solid, liquid or gas). 2. It is based on experimental evidence and cannot be derived from first principle. 3. It is a vector expression indicating that heat flow rate is in the direction of decreasing
HEAT TRANSFER 7 ______temperature and is normal to an isotherm. 4. It helps to define thermal conductivity ‘k’ (transport property) of the medium through which heat is conducted. THERMAL CONDUCTIVITY OF MATERIAL (k) W/Mk
The amount of heat conducted through a body per unit area, and unit thickness in unit time when the differences in temperature between the faces causing the heat flow is unit temperature. It is also defined as the rate of heat transfer per unit area per unit temperature gradient.
Mathematically it can be represented by
푄 푑푥 푘 = × 퐴 푑푇
Thermal conductivity is a physical property of substance and like viscosity, is primarily function of temperature or position and nature of its substances. It varies significantly with pressure only in the case of gases subjected to high pressure. However, for many engineering problems, materials, are often considered to process a constant thermal conductivity. The thermal conductivity for most materials can be determined experimentally by measuring the rate of heat flow and temperature gradient in the given substances.
Thermal conductivities of common substances at ퟐퟎ풐푪
Substances Thermal conductivity (W/mK)
Silver pure 407.5 Copper pure 386.0 Alluminium pure 175.6 Mild steel 37.2 Lead 29.8 Stainless steel 19.3 Wood 0.15 Asbestos, fibre 0.095 Water 0.51 Air 0.022 Table above shows that the pure metals have the highest values of thermal conductivities while gases and vapors have the lowest, insulating materials and inorganic liquid have thermal conductivities that lie in between those of metal and gases.
The thermal conductivity of material depends upon its state and is a function of its pressure, temperature, humidity and structure. For solids the transport of energy by conduction is due to drift of free electron and lattice vibration waves. The thermal conductivity is sum of electronic component, and lattice component. In pure metal electronic component is much larger than lattice component and so they are good conductor of heat and electricity. At elevated temperature the motion of free electron is hampered due to higher lattice vibration.
The thermal conductivity for most pure metals except aluminum and uranium decreases with increasing temperature. In allows electronic component is less than that of pure metals and their
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values of thermal conductivity is less than that of pure metal. In nonmetallic solid electronics component is very low but lattice component is very high. The value of lattice component increases with increasing temperature due to larger interaction between the atom and lattice.
The thermal conductivity of insulating material is very low. This is because many building and insulating material have porous structure with some fluid, mostly air trapped in. since air is a bad conductor of heat so thermal conductivity of air filled porous material is low. The thermal conductivity of liquid and gases is smaller than that of solid because their intermolecular spacing is much larger and so there is less effective transport of energy.
The thermal conductivity of gas increases with increasing temperature and decreasing molecular weight whereas it generally decreases with increasing temperature for non metallic liquid. There are some materials which have very high thermal conductivity and very low temperature. These are called superconductor for example aluminum.
NEWTON’S LAW OF HEAT CONVECTION
For a fluid moving at mean temperature 푇∞ over a surface at temperature푇푠. Newton proposes following heat convection equation.
It is also known as Newton’s law of cooling which states that the rate of heat transfer from hot is directly proportion to the difference of temperature between the two.
푄 = ℎ × 퐴 × (푇푠−푇∞)
푇푠 =Surface temperature in K
푇∞ =Temperature of fluid in K
ℎ =Convection heat transfer coefficient W/푚2K
CONVECTIVE HEAT TRANSFER COEFFICIENT (h) W/풎ퟐ푲
Convection heat transfer is defined as the ratio of heat transfer to product of surface are and temperature difference.
푄 ℎ = 퐴 × (푇푠−푇∞)
Where,
h- convective heat transfer coefficient
Q- rate of heat transfer
A-surface area through heat transfer is taking place
푇푠-surface temperature
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푇∞-temperature of fluid
THERMAL RADIATION
According to Stefan’s Boltzmann law, the radiation energy emitted by a body is proportional to the fourth power of its absolute temperature.
4 푄 = 휎 × 퐴 × 푇1
휎= Stefans boltzman constat with the value of 5.667× 10−8 푊/푚2퐾4.
Where 푇1is the surface temperature in degree Kelvin Consider a black body of surface area 퐴1and at absolute temperature in 푇1 exchanging radiation with another black body at a temperature푇2. The net heat exchange is proportional to difference in푇4.
4 4 푄 = 휎 × 퐴1 × (푇1 − 푇2 )
The real surfaces like polished metal plate do not radiate as much energy as a black body.
The gray nature of real surface can be accounted for by introducing the factor 휖1called as emissivity.
4 4 푄 = 휎 × 휖1 × 퐴1 × (푇1 − 푇2 )
To account for orientation and geometry of two black surfaces exchanging radiation the above equation is modified to
4 4 푄 = 휎 × 휖1 × 퐴1 × 퐹 × (푇1 − 푇2 )
Where F is the view factor and is dependent upon the geometry of two surfaces exchanging radiation.
THERMAL DIFFUSIVITY (α) 풎ퟐ/풔
It is the ratio of thermal conductivity to the product of density and specific heat.
푘 훼 = 휌 퐶푃
The physical significance of thermal diffusivity is that it tells us how fast heat is propagated or it diffuses through a material during change of temperature with time. The larger the thermal diffusivity, the shorter is the time required for the applied heat to penetrate deeper into the solid. This significance of thermal diffusivity will be better understood while dealing with the problems of unsteady heat conduction.
The larger the value of α, the faster will the heat diffuse through the material and its temperature will change with time. This will result either due to a high value of thermal conductivity k or a low value of heat capacity ρ.c. A low value of heat capacity means the less amount of heat entering the element would be absorbed and used to raise its temperature and more would be available for onward transmission. Metals and gases have relatively high value of α and their response to
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temperature changes is quite rapid. The non-metallic solids and liquids respond slowly to temperature changes because of their relatively small value of thermal diffusivity.
Thermal diffusivity is an important characteristic quantity for unsteady conduction situations.
GENERAL DIFFERENTIAL EQUATION OF HEAT CONDUCTION FOR CARTESIAN COORDINATES y
z
Consider a small volume element in Cartesian coordinate having sides dx , dy ,dz as shown in fig the energy balance equation for the little element is obtained from the first law of thermodynamics’ as follows
푛푒푡 ℎ푒푎푡 푐표푛푑푢푐푡푒푑 푖푛푡표 푡ℎ푒 푖푛푡푒푛푎푙 ℎ푒푎푡 {푒푙푒푚푒푛푡 푑푥 푑푦 푑푧 푝푒푟 푢푛푖푡 푡푖푚푒} + { 푔푒푛푒푟푎푡푖표푛 } (푓푖푟푠푡 푡푒푟푚) (푠푒푐표푛푑 푡푒푟푚) 푖푛푐푟푒푎푠푒푑 푖푛 푖푛푡푒푟푛푎푙 푒푛푒푟푔푦 푤표푟푘 푑표푛푒 푏푦 푒푙푒푚푒푛푡 푝푒푟 = { 푝푒푟 푢푛푖푡 푡푖푚푒 } + { 푢푛푖푡 푡푖푚푒 } (푡ℎ푖푟푑 푡푒푟푚) (푓표푢푟푡ℎ 푡푒푟푚)
The work done by element per unit time is very small because the flow work done by solid due to temperature is negligible.
The three terms 1,2 and 3 of the above equation are evaluated as follows
′ ′ ′ ′ Let 푞푥 be the heat flux in x direction at x, face ABCD and 푞푥+푑푥 the heat flux at x+dx face 퐴 퐵 퐶 퐷 . Then the rate of flow of heat into the element in x direction through face ABCD is given by
푄푋 = 푞푥푑푦 푑푧 휕푇 푄 = −푘 푑푦 푑푧 푥 푥 휕푥
푘푥= is the thermal conductivity of material in x direction and
휕푇 =is the temperature gradient in x direction. 휕푥
The rate of heat flow out of the element in x direction through the face x+dx , 퐴′퐵′퐶′퐷′ is
HEAT TRANSFER 11 ______
휕푄푥 푄 = 푄 + 푑푥 푥+푑푥 푥 휕푥 휕푇 휕 휕푇 푄 = −푘 푑푦 푑푧 − {푘 } 푑푥 푑푦 푑푧 푥+푑푥 푥 휕푥 휕푥 푥 휕푥
Then the net rate of heat entering the element in x direction is the difference between the entering and leaving heat flow rate and is given by.
휕 휕푇 푄 − 푄 = (푘 ) 푑푥 푑푦 푑푧 푋 푥+푑푥 휕푥 푥 휕푥
Similarly the net rate of heat entering the element in y direction is the difference between the entering and leaving the heat flow is given by.
휕 휕푇 푄 − 푄 = (푘 ) 푑푥 푑푦 푑푧 푦 푦+푑푦 휕푦 푦 휕푦
Similarly the net rate of heat entering the element in z direction is the difference between the entering and leaving the heat flow is given by.
휕 휕푇 푄 − 푄 = (푘 ) 푑푥 푑푦 푑푧 푧 푧+푑푧 휕푧 푧 휕푧
The net heat conducted into the element dx dy dz per unit time is given by the following
휕 휕푇 휕 휕푇 휕 휕푇 푓푖푟푠푡 푡푒푟푚 = (푘 ) + (푘 ) + (푘 ) 휕푥 푥 휕푥 휕푦 푦 휕푦 휕푧 푧 휕푧
Let 푞̇ be the internal heat generation per unit time and per unit volume (W/푚3) the rate of energy generation in the element is given by following
푠푒푐표푛푑 푡푒푟푚 = 푞̇푑푥 푑푦 푑푧
The change in internal energy for any element over a period of time dt is product of the following
푐ℎ푎푛푔푒 표푓 푖푛푡푒푟푛푎푙 푒푛푒푟푔푦 = 푚푎푠푠 표푓 푒푙푒푚푒푛푡 × 푠푝푒푐푖푓푖푐 ℎ푒푎푡 × 푐ℎ푎푛푔푒 푖푛 푡푒푚푝푒푟푎푡푢푟푒 표푓 푒푙푒푚푒푛푡 푝푒푟 푢푛푖푡 푡푖푚푒
= 흆 × 풅풙 풅풚 풅풛 × 푪풑 × 풅푻
Where ρ and퐶푝 are the density and specific heat of the material of the element.
Then the change in internal energy per unit time ie the third term is give by the following
흏푻 푡ℎ푖푟푑 푡푒푟푚 = 흆 × 풅풙 풅풚 풅풛 × 푪 × 풑 흏풕
T- Temperature t- Time
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Substituting the values of first, second and third term it leads to the general three dimensional equation of heat conduction
휕 휕푇 휕 휕푇 휕 휕푇 휕푇 (푘 ) + (푘 ) + (푘 ) + 푞̇ = 휌 × 퐶 × 휕푥 푥 휕푥 휕푦 푦 휕푦 휕푧 푧 휕푧 푝 휕푡
Since for the most engineering problem the material can be considered to be isentropic for which the value of thermal conductivity is given by the following
푘 = 푘푥 = 푘푦 = 푘푧
Using the constant thermal conductivity the general heat conduction equation becomes
2 2 2 휕 푇 휕 푇 휕 푇 푞̇ 퐶푝 휕푇 1 휕푇 + + + = 휌 = 휕푥2 휕푦2 휕푧2 푘 푘 휕푥 훼 휕푥
Where α is called as thermal diffusivity of the material.
This above equation is the general heat conduction equation for Cartesian coordinates with internal heat generation.
SPECIAL FORMS OF HEAT CONDUCTION EQUATION
1) Uniform thermal conductivity
2 2 2 휕 푇 휕 푇 휕 푇 푞̇ 퐶푝 휕푇 1 휕푇 + + + = 휌 = 휕푥2 휕푦2 휕푧2 푘 푘 휕푡 훼 휕푡
2) Steady state condition with heat generation
휕2푇 휕2푇 휕2푇 푞̇ + + + = 0 휕푥2 휕푦2 휕푧2 푘
3) Unsteady state heat conduction with no heat generation
휕2푇 휕2푇 휕2푇 1 휕푇 + + = 휕푥2 휕푦2 휕푧2 훼 휕푡
4) One dimensional steady state heat conduction with no heat generation
휕2푇 = 0 휕푥2
5) Two dimensional steady state heat conduction with no heat generation
휕2푇 휕2푇 + = 0 휕푥2 휕푦2
6) Three dimensional steady state heat conduction with no heat generation
HEAT TRANSFER 13 ______
휕2푇 휕2푇 휕2푇 + + = 0 휕푥2 휕푦2 휕푧2
HEAT TRANSFER BY CONDUCTION THROUGH PLANE WALL
Consider a plane wall of material of uniform thermal conductivity k which is assumed to be extending to infinity in y and z directions. For this problem, the temperature is only function of x. the walls of room may be considered as a plane if the energy loss through the edges is negligible.
Starting with the general heat conduction equation
2 2 2 휕 푇 휕 푇 휕 푇 푞̇ 퐶푝 휕푇 1 휕푇 + + + = 휌 = 휕푥2 휕푦2 휕푧2 푘 푘 휕푡 훼 휕푡
For this case as we are assuming the steady state
휕푇 = 0 휕푡
Hence for one dimensional flow
휕2푇 휕2푇 = = 0 휕푦2 휕푧2
And with no heat generation
푞̇ = 0 푘
The conduction equation simplifies to
휕2푇 푑2푇 = 0 표푟 = 0 휕푥2 푑푥2
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The above equation is second order differential equation requiring two boundary condition for its solution these two conditions are
푇 = 푇1 푎푡 푥 = 0
푇 = 푇2 푎푡 푥 = 퐿
푑2푇 = 0 푑푥2
Integrating the above equation
푑푇 = 퐶 푑푥 1
Integrating again with respect to x we get
푇 = 퐶1푥 + 퐶2
Where 퐶1, 퐶2 are determine from the boundary condition at
푇 = 푇1 푎푡 푥 = 0
퐶2 = 푇1
At
푇 = 푇2 푎푡 푥 = 퐿
푇2 = 퐶1퐿 + 퐶2
푇2 = 퐶1퐿 + 푇1 푠푖푛푐푒 퐶2 = 푇1
푇2 − 푇1 = 퐶 퐿 1
So the equation for temperature distribution becomes
푇2 − 푇1 푇 = 푥 + 푇 퐿 1
Differentiating with respect to x we get
푑푇 푇2 − 푇1 = 푑푥 퐿
According to Fourier’s law of heat conduction
푑푇 푄 = −푘퐴 푑푥
(푇2 − 푇1) 푄 = −푘퐴 퐿
HEAT TRANSFER 15 ______
(푇1 − 푇2) 푄 = 푘퐴 퐿
This quantity of heat Q must be supplied to the left face of the wall to maintain a temperature difference of (푇2 − 푇1)across it.
The thermal resistance is defined as the ratio of
퐿 푅 = 푡ℎ푒푟푚푎푙 푘퐴
HEAT TRANSFER BY CONDUCTION THROUGH THE COMPOSITE WALL A B C
푇3
Heat flow 푇1 heat flow
A
푇2 푇4
퐿퐴 퐿퐵 퐿퐶
Consider a transmission of heat through a composite wall consisting of number of slabs
퐿퐴퐿퐵퐿퐶 푏푒 푡ℎ푒 푡ℎ푖푐푘푛푒푠푠 표푓 푠푙푎푏 퐴, 퐵, 퐶
퐾퐴퐾퐵퐾퐶 푏푒 푡ℎ푒 푡ℎ푒푟푚푎푙 푐표푛푑푢푐푡푖푣푖푡푦 표푓 푠푙푎푏 퐴, 퐵, 퐶
푇1,푇4 푏푒 푡ℎ푒 푡푒푚푝푒푟푎푡푢푟푒 표푓 푤푎푙푙 푎푡 푡ℎ푒 표푢푡푒푟 푠푢푟푓푎푐푒푠
푇2,푇3 푏푒 푡ℎ푒 푡푒푚푝푒푟푎푡푢푟푒 표푓 푤푎푙푙 푎푡 푡ℎ푒 푖푛푠푖푑푒 푠푢푟푓푎푐푒푠
Heat transferred through slab A is given by the following equation
퐾퐴 × 퐴 × (푇1 − 푇2) 푄 = 퐿퐴
Heat transferred through Slab B is given by the following equation
퐾퐵 × 퐴 × (푇2 − 푇3) 푄 = 퐿퐵
Heat transferred through Slab C is given by the following equation
퐾퐶 × 퐴 × (푇3 − 푇4) 푄 = 퐿퐶
Since quantity of heat transferred per unit time through each layer is same we have from the above equation
16 HEAT TRANSFER ______
푄 × 퐿퐴 푇1 − 푇2 = 퐾퐴 × 퐴
푄 × 퐿퐵 푇2 − 푇3 = 퐾퐵 × 퐴
푄 × 퐿퐶 푇3 − 푇4 = 퐾퐶 × 퐴
Adding all the three temperature differences across the slab A ,B,C
푄 × 퐿퐴 푄 × 퐿퐵 푄 × 퐿퐶 푇1 − 푇4 = + + 퐾퐴 × 퐴 퐾퐵 × 퐴 퐾퐶 × 퐴
Hence heat transfer across slab A,B,C is given by the following equation
퐴 × (푇1 − 푇4) 푄 = 퐿 퐿 퐿 [ 퐴 + 퐵 + 퐶 ] 퐾퐴 퐾퐵 퐾퐶
The above equation is the heat transfer by conduction through the composite wall.
HEAT TRANSFERRED BY CONDUCTION THROUGH A HOLLOW CYLINDER
Consider a hollow cylinder made up of a material having constant thermal conductivity and insulated at the both ends
HEAT TRANSFER 17 ______
푟1&푟2 푏푒 푡ℎ푒 푖푛푛푒푟 푎푛푑 표푢푡푒푟 푟푎푑푖푢푠
푇1& 푇2푏푒 푡ℎ푒 푡푒푚푝푒푟푎푡푢푟푒 푎푡 푖푛푛푒푟 푎푛푑 표푢푡푒푟 푠푢푟푓푎푐푒푠 표푓 푐푦푙푖푛푑푒푟
푘 푏푒 푡ℎ푒 푐표푛푠푡푎푛푡 푡ℎ푒푟푚푎푙 푐표푛푑푢푐푡푖푣푖푡푦 푤푖푡ℎ푖푛 푡ℎ푒 푐표푛푠푡푎푛푡 푡푒푚푝푒푟푎푡푢푟푒 푟푎푛푔푒
Consider and element of radius r and thickness dr for the length of hollow cylinder through which heat is transferred.
We know area of hollow cylinder
퐴 = 2 × 휋 × 푟 × 퐿
Path length = dr
According to Fourier’s law of heat conduction we can write
푑푇 푄 = −푘퐴 푑푟 푑푇 푄 = −푘 × 2 × 휋 × 푟 × 퐿 × 푑푟 푑푟 푄 × = −푘 × 2 × 휋 × 퐿 × 푑푇 푟
Integrating both sides to the above equation we get
푟2 푑푟 푇2 푄 ∫ = −푘 × 2 × 휋 × 퐿 × ∫ 푑푇 푟 푟1 푇1
[ ] 푄 × [푙푛푟2 − 푙푛푟1] = −푘 × 2 × 휋 × 퐿 × 푇2 − 푇1
푘 × 2 × 휋 × 퐿 × [푇1 − 푇2] 푄 = 푟 푙푛 ( 2) 푟1
This is the equation of heat transfer by conduction through hollow cylinder.
HEAT TRANSFER BY CONDUCTION THROUGH THE COMPOSITE CYLINDER
18 HEAT TRANSFER ______
Consider the flow of heat through the composite cylinder as shown in fig
푇ℎ푓 푏푒 푡ℎ푒 푡푒푚푝푒푟푎푡푢푟푒 표푓 ℎ표푡 푓푙푢푖푑 푖푛푠푖푑푒 푡ℎ푒 푐푦푙푖푛푑푒푟
푇푐푓 푏푒 푡ℎ푒 푡푒푚푝푒푟푎푡푢푟푒 표푓 푐표푙푑 푓푙푢푖푑 푖푛푠푖푑푒 푡ℎ푒 푐푦푙푖푛푑푒푟
푘퐴 푏푒 푡ℎ푒 푡ℎ푒푟푚푎푙 푐표푛푑푢푐푡푖푣푖푡푦 표푓 푖푛푠푖푑푒 푙푎푦푒푟 퐴
푘퐵 푏푒 푡ℎ푒 푡ℎ푒푟푚푎푙 푐표푛푑푢푐푡푖푣푖푡푦 표푓 표푢푡푠푖푑푒 푙푎푦푒푟 퐵
푇1, 푇2, 푇3 푏푒 푡ℎ푒 푡푒푚푝푒푟푎푡푢푟푒 푎푡 푝표푖푛푡 1,2,3
퐿 푏푒 푡ℎ푒 푙푒푛푔푡ℎ 표푓 푐표푚푝표푠푖푡푒 푐푦푙푖푛푑푒푟
ℎℎ푓 푏푒 푡ℎ푒 푖푛푠푖푑푒 ℎ푒푎푡 푡푟푎푛푠푓푒푟 푐표푒푓푓푖푐푖푒푛푡
ℎ푐푓 푏푒 푡ℎ푒 표푢푡푠푖푑푒 ℎ푒푎푡 푡푟푎푛푠푓푒푟 푐표푒푓푓푖푐푖푒푛푡
Let the hot fluid is present inside the composite cylinder
The heat transfer by convection takes place from hot fluid present inside to the inside surface of cylinder it is given by the following equation.
푄 = ℎℎ푓 × 2 × 휋 × 푟1 × 퐿 × (푇ℎ푓 − 푇1)
The heat transferred by conduction through A is given by the following equation
푘퐴 × 2 × 휋 × 퐿 × [푇1 − 푇2] 푄 = 푟 푙푛 ( 2) 푟1
The heat transferred by conduction through B is given by the following equation
푘퐵 × 2 × 휋 × 퐿 × [푇2 − 푇3] 푄 = 푟 푙푛 ( 3) 푟2
The heat transfer by convection takes place from the outside surface of element B with the surrounding is give by the following equation
푄 = ℎ푐푓 × 2 × 휋 × 푟3 × 퐿 × (푇3 − 푇푐푓)
Writing all the above equation in terms of temperature to get the final value of heat transfer that is taking place through the composite cylinder
푄 (푇ℎ푓 − 푇1) = ℎℎ푓 × 2 × 휋 × 푟1 × 퐿 푟 푄 × 푙푛 ( 2) 푟1 [푇1 − 푇2] = 푘퐴 × 2 × 휋 × 퐿
HEAT TRANSFER 19 ______
푟 푄 × 푙푛 ( 3) 푟2 [푇2 − 푇3] = 푘퐵 × 2 × 휋 × 퐿 푄 (푇3 − 푇푐푓) = ℎ푐푓 × 2 × 휋 × 푟3 × 퐿
Adding all the temperature differences across the element we get 푟 푟 푄 × 푙푛 ( 2) 푄 × 푙푛 ( 3) 푄 푟1 푟2 푄 푇ℎ푓 − 푇푐푓 = + + + ℎℎ푓 × 2 × 휋 × 푟1 × 퐿 푘퐴 × 2 × 휋 × 퐿 푘퐵 × 2 × 휋 × 퐿 ℎ푐푓 × 2 × 휋 × 푟3 × 퐿
Hence heat transfer across the element can be obtained by the following equation
2 × 휋 × 퐿 × (푇ℎ푓 − 푇푐푓) 푄 = 푟 푟 푙푛 ( 2) 푙푛 ( 3) 1 푟 푟 1 [ + 1 + 2 + ] ℎℎ푓 × 푟1 푘퐴 푘퐵 ℎ푐푓 × 푟3
The above equation is the heat transfer across the composite cylinder considering the convective heat transfer coefficient at inside and on outside surfaces.
HEAT TRANSFER BY CONDUCTION THROUGH HOLLOW SPHERE
Consider a sphere made up of material having constant thermal conductivity as shown in figure.
푟1푏푒 푡ℎ푒 푖푛푛푒푟 푟푎푑푖푢푠 표푓 푠푝ℎ푒푟푒
푟2푏푒 푡ℎ푒 표푢푡푒푟 푟푎푑푖푢푠 표푓 푠푝ℎ푒푟푒
푇1푏푒 푡ℎ푒 푡푒푚푝푒푟푎푡푢푟푒 표푓 푖푛푛푒푟 푠푢푟푓푎푐푒푠
푇2푏푒 푡ℎ푒 푡푒푚푝푒푟푎푡푢푟푒 표푓 표푢푡푒푟 푠푢푟푓푎푐푒푠
20 HEAT TRANSFER ______
푘 푏푒 푡ℎ푒 푐표푛푠푡푎푛푡 푡ℎ푒푟푚푎푙 푐표푛푑푢푐푡푖푣푖푡푦 표푓 푚푎푡푒푟푖푎푙 푎푡 푔푖푣푒푛 푡푒푚푝푒푟푎푡푢푟푒 푟푎푛푔푒
푐표푛푠푖푑푒푟 푠푚푎푙푙 푒푙푒푚푒푛푡 표푓 푡ℎ푖푐푘푛푒푠푠 푑푟 푎푡 푎푛푦 푟푎푑푖푢푠 푟
푙푒푡 퐴 푏푒 푡ℎ푒 푎푟푒푎 푡ℎ표푟푢푔ℎ 푤ℎ푖푐ℎ 푡ℎ푒 ℎ푒푎푡 푡푟푎푛푠푓푒푟 푖푠 푡푎푘푖푛푔 푝푙푎푐푒
퐴 = 4 × 휋 × 푟2
According to the Fourier’s law of heat conduction the heat transfer through the sphere is given by the following equation
푑푇 푄 = −푘퐴 푑푟 푑푇 푄 = −푘 × 4 × 휋 × 푟2 × 푑푟
Rearranging and integrating the above equation we get
푟2 푑푟 푇2 푄 × ∫ = −푘 × 4 × 휋 × ∫ 푑푇 푟2 푟1 푇1 1 1 푄 × [ − ] = −푘 × 4 × 휋 × [푇2 − 푇1] 푟2 푟1
푘 × 4 × 휋 × 푟1푟2 × [푇1 − 푇2] 푄 = 푟2 − 푟1
This is the equation for heat transfer through the hollow sphere
HEAT TRANSFER THROUGH COMPOSITE SPHERE
푇ℎ푓 푏푒 푡ℎ푒 푡푒푚푝푒푟푎푡푢푟푒 표푓 ℎ표푡 푓푙푢푖푑 푖푛푠푖푑푒 푡ℎ푒 푠푝ℎ푒푟푒
HEAT TRANSFER 21 ______
푇푐푓 푏푒 푡ℎ푒 푡푒푚푝푒푟푎푡푢푟푒 표푓 푐표푙푑 푓푙푢푖푑 표푢푡푠푖푑푒 푡ℎ푒 푠푝ℎ푒푟푒
푘퐴 푏푒 푡ℎ푒 푡ℎ푒푟푚푎푙 푐표푛푑푢푐푡푖푣푖푡푦 표푓 푖푛푠푖푑푒 푙푎푦푒푟 퐴
푘퐵 푏푒 푡ℎ푒 푡ℎ푒푟푚푎푙 푐표푛푑푢푐푡푖푣푖푡푦 표푓 표푢푡푠푖푑푒 푙푎푦푒푟 퐵
푇1, 푇2, 푇3 푏푒 푡ℎ푒 푡푒푚푝푒푟푎푡푢푟푒 푎푡 푝표푖푛푡 1,2,3
ℎℎ푓 푏푒 푡ℎ푒 푖푛푠푖푑푒 ℎ푒푎푡 푡푟푎푛푠푓푒푟 푐표푒푓푓푖푐푖푒푛푡
ℎ푐푓 푏푒 푡ℎ푒 표푢푡푠푖푑푒 ℎ푒푎푡 푡푟푎푛푠푓푒푟 푐표푒푓푓푖푐푖푒푛푡
According to the law of convective heat transfer coefficient the heat transfer between the inside hot fluid to inside surface of layer A is given by the following equation
2 푄 = ℎℎ푓 × 4 × 휋 × 푟1 × (푇ℎ푓 − 푇1)
According to the Newton’s law of heat conduction the heat transferred through layer A is given by the following equation
푘퐴 × 4 × 휋 × 푟1푟2(푇1 − 푇2) 푄 = 푟2 − 푟1
According to the Newton’s law of heat conduction the heat transferred through layer B is given by the following equation
푘퐵 × 4 × 휋 × 푟2푟3(푇2 − 푇3) 푄 = 푟3 − 푟2
According to the law of convective heat transfer coefficient the heat transfer between the inside hot fluid to inside surface of layer A is given by the following equation
2 푄 = ℎ퐶푓 × 4 × 휋 × 푟3 × (푇3 − 푇푐푓)
Writing the above equation in terms of the temperature difference the to get the value of heat transfer through the sphere
푄 (푇ℎ푓 − 푇1) = 2 ℎℎ푓 × 4 × 휋 × 푟1
푄 × (푟2 − 푟1) (푇1 − 푇2) = 푘퐴 × 4 × 휋 × 푟1푟2
푄 × (푟3 − 푟2) (푇2 − 푇3) = 푘퐵 × 4 × 휋 × 푟2푟3 푄 (푇3 − 푇푐푓) = 2 ℎ푐푓 × 4 × 휋 × 푟3
Adding all the temperature differences to get the value of heat transfer through the sphere
22 HEAT TRANSFER ______
푄 푄 × (푟2 − 푟1) 푄 × (푟3 − 푟2) 푄 (푇ℎ푓 − 푇푐푓) = 2 + + + 2 ℎℎ푓 × 4 × 휋 × 푟1 푘퐴 × 4 × 휋 × 푟1푟2 푘퐵 × 4 × 휋 × 푟2푟3 ℎ푐푓 × 4 × 휋 × 푟3
4 × 휋 × (푇ℎ푓 − 푇푐푓) 푄 = 1 (푟2 − 푟1) (푟3 − 푟2) 1 [ 2 + + + 2] ℎℎ푓 × 푟1 푘퐴 × 푟1푟2 푘퐵 × 푟2푟3 ℎ퐶푓 × 푟3
This is the equation of heat transfer through composite sphere if the inside and outside heat transfer coefficients are taken into consideration.
COMBINED MECHANISM OF HEAT TRANSFER
It is not unusual to observe that the heat transfer is taking place due to two or perhaps the all there mechanisms. The most frequently encountered instance is one in which a solid wall plane or cylindrical separates two convecting fluids for example the tube of a heat exchanger. As mentioned earlier, the steam generating tubes of a boiler receives heat from the products of combustion by all the three modes of heat transfer.
The overall heat transfer by the combined modes is usually expressed in terms of an overall conductance or overall heat transfer coefficient U defined by the relation
푄 = 푈 × 퐴 × 훥푇
The overall heat transfer coefficient is a quantity such that the rate of heat flow through a configuration is given by taking a product of U, the surface area and the overall temperature difference.
In case of plane wall as shown in fig heated on one side by a hot fluid A and cooled on the other side by a cold fluid B , the heat transfer rate is give by
HEAT TRANSFER 23 ______
The heat transfer by convection taking place in fluid and surface of wall is given by the following equation
푄 = ℎ퐴 × 퐴 × (푇퐴 − 푇1)
The heat transfer by conduction through the wall is given by the following equation
푘 × 퐴 × (푇1 − 푇2) 푄 = 퐿
The heat transfer by convection from the wall surface to the fluid is given by the following equation
푄 = ℎ퐵 × 퐴 × (푇2 − 푇퐵)
Writing all the above equation in terms of temperature difference so as to get the value of total temperature difference between the fluids
푄 푄 × 퐿 푄 푇퐴 − 푇퐵 = + + ℎ퐴 × 퐴 푘 × 퐴 ℎ퐵 × 퐴
(푇퐴 − 푇퐵) 푄 = 1 퐿 1 + + ℎ퐴 × 퐴 푘 × 퐴 ℎ퐵 × 퐴
퐴 × (푇퐴 − 푇퐵) 푄 = 1 퐿 1 + + ℎ퐴 푘 ℎ퐵
푄 = 푈 × 퐴 × 훥푇
1 푈 = 1 퐿 1 + + ℎ퐴 푘 ℎ퐵
Where U is called as the overall heat transfer coefficient.
The overall hat transfer coefficient depends upon the geometry of the separating wall, its thermal properties and the convective coefficients at the two surfaces. The overall heat transfer coefficient is particularly useful in the case of composite walls, such as in the design of structural walls for boilers, refrigerators air-conditioned building. Use of overall heat transfer coefficient is also made in the design of heat exchanger.
RADIAL HEAT CONDUCTION THROUGH THE CYLINDERICAL SYSTEM
24 HEAT TRANSFER ______
Consider a long cylinder of inside radius 푟𝑖 and outside radius 푟표, and the length L. we consider cylinder to be long sot that the end losses are negligible. The inside and outside surfaces are kept at
constant temperature 푇𝑖 and 푇표 respectively. A steam pipe in a room can be taken as an example of hollow cylinder.
The general heat conduction equation in cylindrical coordinate is given by
휕2푇 1 휕푇 1 휕2푇 휕2푇 푞̇ 1 휕푇 + ( ) + + + = 휕푟2 푟 휕푟 푟2 휕휙2 휕푧2 푘 훼 휕푡
Assuming that heat flow in radial direction the above equation under steady state takes the form
푑2푇 1 푑푇 + = 0 푑푟2 푟 푑푟 푑 1 푑푇 ( ) = 0 푑푟 푟 푑푟
Subject to the boundary condition
푇 = 푇𝑖 푎푡 푟 = 푟𝑖
푇 = 푇표 푎푡 푟 = 푟표 푑 1 푑푇 ( ) = 0 푑푟 푟 푑푟
Integrating the above equation twice we get
푇 = 퐶1 ln(푟) + 퐶2
Using the boundary conditions we get
푇𝑖 = 퐶1푙푛 (푟𝑖) + 퐶2
푇표 = 퐶1푙푛 (푟표) + 퐶2
Subtracting the above two equation we get
푇𝑖 − 푇표 = 퐶1푙 푛(푟𝑖) − 퐶1푙푛 (푟표)
푇𝑖 − 푇표 퐶1 = 푙 푛(푟𝑖) − 푙푛 (푟표)
푇𝑖 − 푇표 퐶1 = 푟 푙푛 ( 𝑖 ) 푟표
푇표 − 푇𝑖 푇𝑖푙푛(푟표) − 푇표푙푛(푟𝑖) 퐶2 = 푇𝑖 − 푟 × 푙푛[푟𝑖] = 푟 푙푛 ( 𝑖 ) 푙푛 ( 표) 푟표 푟𝑖
Substituting the values of 퐶1&퐶2 in the following equation we get
HEAT TRANSFER 25 ______
푇 = 퐶1 ln(푟) + 퐶2
푇𝑖 − 푇표 푇𝑖푙푛(푟표) − 푇표푙푛(푟𝑖) 푇 = 푟 ln(푟) + 푟 푙푛 ( 𝑖 ) 푙푛 ( 표) 푟표 푟𝑖 푑푇 푄 = −푘퐴 푎푡 푟 = 푟 푑푟 𝑖 푑푇 푄 = −푘 × 2 × 휋 × 푟 × 퐿 × 𝑖 푑푟
퐶1 푄 = −푘 × 2 × 휋 × 푟𝑖 × 퐿 × 푟𝑖
푇표 − 푇𝑖 푄 = −푘 × 2 × 휋 × 푟𝑖 × 퐿 × 푟표 푟𝑖 × 푙푛 ( ) 푟𝑖
푇𝑖 − 푇표 푄 = 푘 × 2 × 휋 × 퐿 × 푟 푙푛 ( 표) 푟𝑖
The above equation can alternatively be derived by as follows
푑푇 푄 = −푘퐴 푤ℎ푒푟푒 퐴 = 2 × 휋 × 푟 × 퐿 푑푟 푑푇 푄 = −푘 × 2 × 휋 × 푟 × 퐿 × 푑푟 푑푟 푄 × = −푘 × 2 × 휋 × 퐿 × 푑푇 푟
Integrating the above equation we get
푟표 푑푟 푇표 푄 × ∫ = −푘 × 2 × 휋 × 퐿 × ∫ 푑푇 푟 푟푖 푇푖
푟표 푄 × 푙푛 ( ) = −푘 × 2 × 휋 × 퐿 × (푇표 − 푇𝑖) 푟𝑖
푘 × 2 × 휋 × 퐿 × (푇𝑖 − 푇표) 푄 = 푟 푙푛 ( 표) 푟𝑖
GENERAL HEAT CONDUCTION EQUATION FOR CYLINDRICAL COORDINATES
According to Fourier’s law of heat conduction
휕푇 푄 = −푘 퐴 푟 휕푟
26 HEAT TRANSFER ______
휕푇 푄 = −푘 (푟푑휃 푑푧) 푟 휕푟 휕 푄 = 푄 + (푄 )푑푟 푟+푑푟 푟 휕푟 푟
휕 푄 − 푄 = 푄 − (푄 + (푄 )푑푟) 푟 푟+푑푟 푟 푟 휕푟 푟 휕 휕푇 푄 − 푄 = − (−푘 (푟푑휃 푑푧) ) 푑푟 푟 푟+푑푟 휕푟 휕푟
휕2푇 휕푇 푄 − 푄 = 푘 푑푟 푟푑휃 푑푧 + 푘 푑푟 푑휃 푑푧 푟 푟+푑푟 휕푟2 휕푟
HEAT TRANSFER 27 ______
Similarly
휕푇 푄 = −푘 퐴 휃 푟휕휃 휕푇 푄 = −푘 (푑푟 푑푧) 휃 푟휕휃 휕 푄 = 푄 + (푄 )푟휕휃 휃+푑휃 휃 푟휕휃 휃 휕 푄 − 푄 = 푄 − (푄 + (푄 )푟휕휃) 휃 휃+푑휃 휃 휃 푟휕휃 휃 휕 푄 − 푄 = − (푄 )푟휕휃 휃 휃+푑휃 푟휕휃 휃 휕 휕푇 푄 − 푄 = − (−푘 (푑푟 푑푧) ) 푟휕휃 휃 휃+푑휃 푟휕휃 푟휕휃
1 휕2푇 푄 − 푄 = 푘 푑푟 푑푧 푑휃 휃 휃+푑휃 푟 휕휃2
Similarly
휕푇 푄 = −푘 퐴 푧 휕푧 휕푇 푄 = −푘 (푟푑휃 푑푟) 푧 휕푧 휕 푄 = 푄 + (푄 )푑푧 푧+푑푧 푧 휕푧 푧 휕 푄 − 푄 = 푄 − (푄 + (푄 )푑푧) 푧 푧+푑푧 푧 푧 휕푧 푧 휕 휕푇 푄 − 푄 = − (−푘 (푟푑휃 푑푟) ) 푑푧 푧 푧+푑푧 휕푧 휕푧
휕2푇 푄 − 푄 = 푘 푑푟 푟푑휃 푑푧 푧 푧+푑푧 휕푧2
Net amount of heat in all direction
휕2푇 휕푇 1 휕2푇 휕2푇 푘 푑푟 푟푑휃 푑푧 + 푘 푑푟 푑휃 푑푧 + 푘 푑푟 푑푧 푑휃 + 푘 푑푟 푟푑휃 푑푧 휕푟2 휕푟 푟 휕휃2 휕푧2
Internal heat generated per unit volume is denoted by 푞퐺̇
Internal heat generated
푖푛푡푒푟푛푎푙 ℎ푒푎푡 푔푒푛푒푟푎푡푒푑 = 푞퐺̇ × 푣표푙푢푚푒
푖푛푡푒푟푛푎푙 ℎ푒푎푡 푔푒푛푒푟푎푡푒푑 = 푞퐺̇ × 푑푟 푟푑휃 푑푧
28 HEAT TRANSFER ______
Increase in internal energy
푑푇 푖푛푐푟푒푎푠푒 푖푛 푖푛푡푒푟푛푎푙 푒푛푒푟푔푦 = 푚 퐶 푝 푑푡 푑푇 푖푛푐푟푒푎푠푒 푖푛 푖푛푡푒푟푛푎푙 푒푛푒푟푔푦 = 푑푒푛푠푖푡푦 × 푣표푙푢푚푒 × 퐶 푝 푑푡
Writing the energy balance equation for given element
ℎ푒푎푡 푐표푛푑푢푐푡푒푑 푖푛 푎푙푙 푑푖푟푐푡푖표푛 + 푖푛푡푒푟푛푎푙 ℎ푒푎푡 푔푒푛푒푟푎푡푖표푛 = 푖푛푐푟푒푎푠푒 푖푛 푖푛푡푒푟푛푎푙 ℎ푒푎푡 푔푒푛푒푟푎푡푖표푛 + 푤표푟푘 푑표푛푒 표푛 푒푙푒푚푒푛푡
휕2푇 휕푇 1 휕2푇 휕2푇 푘 푑푟 푟푑휃 푑푧 + 푘 푑푟 푑휃 푑푧 + 푘 푑푟 푑푧 푑휃 + 푘 푑푟 푟푑휃 푑푧 + 푞 ̇ × 푑푟 푟푑휃 푑푧 휕푟2 휕푟 푟 휕휃2 휕푧2 퐺 푑푇 푑푇 = 푚 퐶 = 푑푒푛푠푖푡푦 × 푣표푙푢푚푒 × 퐶 푝 푑푡 푝 푑푡
휕2푇 휕푇 1 휕2푇 휕2푇 푘 푑푟 푟푑휃 푑푧 + 푘 푑푟 푑휃 푑푧 + 푘 푑푟 푑푧 푑휃 + 푘 푑푟 푟푑휃 푑푧 + 푞 ̇ × 푑푟 푟푑휃 푑푧 휕푟2 휕푟 푟 휕휃2 휕푧2 퐺 푑푇 = 휌 × 푑푟 푟푑휃 푑푧 × 퐶 푝 푑푡
2 2 2 휕 푇 1 휕푇 1 휕 푇 휕 푇 푞퐺̇ 1 푑푇 + + + + = 휌 × × 퐶 휕푟2 푟 휕푟 푟2 휕휃2 휕푧2 푘 푘 푝 푑푡
CYLINDER WITH THE INTERNAL HEAT GENERATION
Let us consider a solid cylinder of radius R with internal heat generation such as and electric coil in which heat is generated as a result of electric current in wire or a cylindrical nuclear fuel element in which heat is generated by nuclear fission. The one dimensional heat conduction equation in cylindrical coordinates is
1 푑 푑푇 푞̇ (푟 ) + = 0 푟 푑푟 푑푟 푘 푑 푑푇 푞̇ (푟 ) = − × 푟 푑푟 푑푟 푘
Integrating the above equation we get
푑푇 푞̇ 푟 = − × 푟2 + 퐶 푑푟 2푘 1
HEAT TRANSFER 29 ______
푑푇 푞̇ 퐶1 = − × 푟 + 푑푟 2푘 푟
By second integration
푞̇ 푇 = − × 푟2 + 퐶 푙푛(푟) + 퐶 4푘 1 2 푑푇 푎푡 푟 = 0 푡ℎ푒 푣푎푙푢푒 표푓 = 0 푑푟
푑푇 푞̇ 퐶1 푎푡 푟 = 푅 푡ℎ푒 푣푎푙푢푒 표푓 = − × 푅 + 푑푟 2푘 푅
ℎ푒푎푡 푔푒푛푒푟푎푡푒푑 푖푛 푐푦푙푖푛푑푒푟 푟표푑 = 푞̇ × 휋 × 푅2 × 퐿
The heat is conducted to the surface and then convected away
푑푇 푞̇ × 휋 × 푅2 × 퐿 == −푘 × 2 × 휋 × 푅 × 퐿 × ( ) 푎푡 푟 = 푅 푑푟 푑푇 푞̇ × 푅 ( ) = − 푑푟 푎푡 푟=푅 2푘
Putting 퐶1 = 0
푞̇ 푇 = − × 푟2 + 퐶 4푘 2
푎푡 푟 = 푅 푎푛푑 푇 = 푇푤 푞̇ 푇 = − × 푅2 + 퐶 푤 4푘 2 푞̇ 푇 + × 푅2 = 퐶 푤 4푘 2
Substituting the values of 퐶2 in the value of T we get
푞̇ 푞̇ 푇 = − × 푟2 + 푇 + × 푅2 4푘 푤 4푘 푞̇ 푇 = 푇 + × (푅2 − 푟2) 푤 4푘
This is the temperature variation along the wall radius. The maximum temperature occurs at r=0
푞̇ 푇 = 푇 + × (푅2) 푚푎푥 푤 4푘
In dimensionless form the equation of temperature becomes
2 푇 − 푇푤 푟 = 1 − ( ) 푇푚푎푥 − 푇푤 푅
30 HEAT TRANSFER ______
For hollow cylinder with uniformly distributed heat source and specified surface temperature
푇 = 푇1푎푡 푟 = 푟𝑖 푎푛푑 푇 = 푇2푎푡 푟 = 푟표 푞̇ 푇 = − × 푟 2 + 퐶 ln(푟 ) + 퐶 𝑖 4푘 𝑖 1 𝑖 2 푞̇ 푇 = − × 푟 2 + 퐶 ln(푟 ) + 퐶 푂 4푘 푂 1 푂 2
EXAMPLES A stainless steel plate of 2cm thick is maintained at a temperature of 5500퐶 at one face and 500퐶 at the other end the thermal conductivity of stainless steel is 19.1 푊/푚퐾 calculate the heat transfer through the material per unit area.
Solution
5500퐶
500퐶
2m
푘퐴 푸 = (푇 − 푇 ) 퐿 1 2 푄 푘 = (푇 − 푇 ) 퐴 퐿 1 2 푄 19.1 = (550 − 50) = 477.5 푘푊/푚2 퐴 2 × 10−2
One face of a copper plate 3cm thick is maintained at 4000퐶 and the other face is maintained at1000퐶. How much heat is transferred through the plate?
푘퐴 푸 = (푇 − 푇 ) 퐿 1 2 푄 푘 = (푇 − 푇 ) 퐴 퐿 1 2
Air at 200퐶 blows over a hot plate 50 by 75 cm maintained at2500퐶. The convection heat transfer coefficient is25 푊/푚2퐾. Calculate the heat transfer.
푄 = ℎ 퐴 (푇푆 − 푇∞)
푄 = 25 × 0.50 × 0.75 × (250 − 20)
푄 = 2156.25 푊
HEAT TRANSFER 31 ______
A flat plate of length 1m and width 0.5 m is placed in an air stream at 300퐶 blowing parallel to it. The convective heat transfer coefficient is 30푊/푚2퐾. Calculate the heat transfer if the plate is maintained at a temperature of 3000퐶.
푄 = ℎ 퐴 (푇푆 − 푇∞)
푄 = 30 × 1 × 0.5 × (300 − 30)
푄 = 4050 푊
An electric current is passes through a wire 1mm in diameter and 10 cm long. The wire is submerged at atmospheric pressure, and the current is increased until the water boils. For this situation having convective heat transfer coefficient5000 푊/푚2퐾. And the water temperature will be 1000퐶. How much electric power must be supplied to the wire to maintained the surface temperature of 1140퐶.
푄 = ℎ 퐴 (푇푆 − 푇∞)
푄 = 5000 × 훱 × 퐷 × 퐿 (푇푆 − 푇∞)
푄 = 5000 × 훱 × 1 × 10−3 × 10 × 10−2 × (114 − 100)
푄 = 21.99 푊
The inner surface of plane brick wall at 600퐶 and the outer surface is at350퐶. Calculate the rate of heat transfer per unit area of the surface of wall which is 220mm thick. The thermal conductivity of brick was found to be 0.51 푊/푚퐾
푄 푘 = (푇 − 푇 ) 퐴 퐿 1 2 푄 0.51 = (60 − 35) 퐴 220 × 10−3 푄 = 57.9545 푊/푚2 퐴
An immersion of water of surface area 0.1 푚2 and rating of 1Kw is designed to operate fully submerged in water. Estimate the surface temperature of the heater when the water is at 400퐶 and the heat transfer coefficient is300 푊/푚2퐾. If the heater is by mistake used in air at 400퐶 with the convective heat transfer coefficient of 9 푊/푚2퐾. Calculate the surface temperature.
Solution let 푇푆 be the surface temperature of heater
When water is used
푄 = ℎ 퐴 (푇푆 − 푇∞)
32 HEAT TRANSFER ______
1000 = 300 × 0.1 × (푇푆 − 40) 1000 + 40 = 푇 300 × 0.1 푆
0 푇푆 = 73.33 퐶
When used in air
푄 = ℎ 퐴 (푇푆 − 푇∞)
1000 = 9 × 0.1 × (푇푆 − 40) 1000 + 40 = 푇 9 × 0.1 푆
0 푇푆 = 1151 퐶
This surface temperature is extremely high and will result in melting of the heating element. So an immersion water heater should never be operated in air.
A constant temperature difference of 166.70퐶 is maintained across the surfaces of slab of 3.05cm thickness. Calculate the rate of heat transfer per unit area across the slab for each of the following cases
a) Slab is made of copper having 푘 = 380.7 푊/푚퐾 b) Slab is made of alluminium having 푘 = 225 푊/푚퐾 c) Slab is made of carbon steel having 푘 = 17.3 푊/푚퐾 d) Slab is made of brick having 푘 = 0.865 푊/푚퐾 e) Slab is made of asbestos having 푘 = 0.173 푊/푚퐾
Solution
Slab is made of copper having 푘 = 380.7 푊/푚퐾
푄 푘 = (푇 − 푇 ) 퐴 퐿 1 2 푄 380.7 = (166.7) 퐴 3.05 × 10−2 푄 = 2080743.934 푊/푚2 퐴
Slab is made of alluminium having 푘 = 225 푊/푚퐾
푄 푘 = (푇 − 푇 ) 퐴 퐿 1 2 푄 225 = (166.7) 퐴 3.05 × 10−2
HEAT TRANSFER 33 ______
푄 = 1229754.098 푊/푚2 퐴
Slab is made of carbon steel having 푘 = 17.3 푊/푚퐾
푄 푘 = (푇 − 푇 ) 퐴 퐿 1 2 푄 17.3 = (166.7) 퐴 3.05 × 10−2 푄 = 94554.42 푊/푚2 퐴
Slab is made of brick having 푘 = 0.865 푊/푚퐾
푄 푘 = (푇 − 푇 ) 퐴 퐿 1 2 푄 0.865 = (166.7) 퐴 3.05 × 10−2 푄 = 4727.72 푊/푚2 퐴
Slab is made of asbestos having 푘 = 0.173 푊/푚퐾
푄 푘 = (푇 − 푇 ) 퐴 퐿 1 2 푄 0.173 = (166.7) 퐴 3.05 × 10−2 푄 = 945.54 푊/푚2 퐴
Estimate the loss of heat transfer through a red brick wall of length of 5m, height 4m and thickness 0.25m, if the temperature of the wall surfaces are maintained at 1100퐶 푎푛푑 400퐶 respectively the value of thermal conductivity for brick 푘 = 0.70 푊/푚퐾.
푘 퐴 푄 = (푇 − 푇 ) 퐿 1 2 0.70 × 5 × 4 푄 = (110 − 40) 0.25
푄 = 3920 푊
Water at 600퐶 is flowing through the wall is at850퐶. Calculate the heat flux if the surface heat transfer coefficient is581.5 푊/푚2퐾.
푄 = ℎ 퐴 (푇푆 − 푇∞)
34 HEAT TRANSFER ______
푄/퐴 = 581.5 × (85 − 60)
푄 = 14537.5 푊/푚2 퐴
A composite wall consists of 1.5mm of steel sheet and 10 mm of plywood separated by 2cm of glass wool in between. Calculate the rate of heat flow if the temperature on the steel and plywood sides is 250퐶 푎푛푑 −150퐶 respectively. Given that the value of thermal conductivity of steel, plywood and glass be
푘푠푡푒푒푙 = 23.23 푊/푚퐾 푘푝푙푦푤표표푑 = 0.052 푊/푚퐾
푘푔푙푎푠푠푤표표푙 = 0.014 푊/푚퐾
Solution
푇1 − 푇4 푄 = 퐿 퐿 퐿 ( 푠 + 퐺 + 푃 ) 푘푠퐴 푘퐺퐴 푘푃퐴 25 − 15 푄/퐴 = 1.5 × 10−3 2 × 10−2 1 × 10−3 ( + + ) 23.33 0.014 0.052
푄 = 51.83 푊/푚2 퐴
The inner surface of plane brick wall is 600퐶 and outer surface is at 350퐶. Calculate the rate of heat transfer per unit area of surface of wall which has thickness of 220mm thick and thermal conductivity of brick is 0.52 푊/푚퐾.
Solution
푄 0.52 = (60 − 35) 퐴 220 × 10−3 푄 = 59.09 푊/푚2 퐴
HEAT TRANSFER 35 ______
A steam pipe covered with two layers of insulation. The inner layer of thermal conductivity 0.17푊/푚퐾 and is 30 mm thick and the outher layer of thermal conductivity 0.023 푊/푚퐾 and is 50 mm thick. The pipe is made of steel of thermal conductivity58 푊/푚퐾. And has inner diameter and outer diameter of 160 mm and 170 mm respectively. The temperature of saturated steam is 3000퐶 and the ambient air 500퐶. If the inside and outside surface heat transfer coefficient are 30 푊/ 푚2퐾 and 5.8 푊/푚2퐾 respectively calculate the rate of heat loss per unit length of pipe. solution
Given
푖푛푛푒푟 푟푎푑푖푢푠 표푓 푝푖푝푒 푟1 = 80푚푚 표푢푡푒푟 푟푎푑푖푢푠 표푓 푝푖푝푒 푟2 = 85푚푚
푟3 = 115 푚푚 푟4 = 165 푚푚
2훱(푇𝑖 − 푇표) 푄/푙 = 푟 푟 푟 푙푛 ( 2) 푙푛 ( 3) 푙푛 ( 4) 1 푟 푟 푟 1 + 1 + 2 + 3 + ℎ𝑖푟1 푘1 푘2 푘3 ℎ표푟2
2훱(300 − 50) 푄/푙 = 85 115 165 1 푙푛 ( ) 푙푛 ( ) 푙푛 ( ) + 80 + 85 + 115 30 × 0.008 58 0.17 0.023 푄 = 82.94 푊/푚 푙
An aluminum pipe carries steam at 1100퐶. The pipe having a thermal conductivity of 185 푊/푚퐾 and has inner diameter of 10 cm and outer diameter of 12 cm. the pipe located in room where the ambient air temperature is 300퐶 and the convective heat transfer coefficient is 15 푊/푚2퐾. Determine the heat transfer rate per unit length of the pipe.
To reduce the heat loss from the pipe, it is covered with a 5cm thick layer of insulation having thermal conductivity of0.2 푊/푚퐾. Determine the rate of heat loss per unit length and the percentage reduction in heat loss by the insulation neglect the convective resistance of the steam.
Solution
36 HEAT TRANSFER ______
2훱(푇푠 − 푇∞) 푄/푙 = 푟 푙푛 ( 2) 1 푟 + 1 ℎ표푟2 푘1
2훱(110 − 30) 푄/푙 = 0.06 1 푙푛 ( ) + 0.05 15 × 0.06 185 푄 = 452 푊/푚 푙
For insulated pipe
2훱(푇푠 − 푇∞) 푄/푙 = 푟 푟 푙푛 ( 2) 푙푛 ( 3) 1 푟 푟 + 1 + 2 ℎ표푟3 푘1 푘2
2훱(110 − 30) 푄/푙 = 6 11 1 푙푛 ( ) 푙푛 ( ) + 5 + 6 15 × 0.11 185 0.2 푄 = 138푊/푚 푙
Reduction of heat loss from the pipe after providing insulation
452 − 138 푟푎푡푒 표푓 ℎ푒푎푡 푙표푠푠 = = 0.695 452
A plastic pipe of thermal conductivity 0.5 푊/푚퐾 carries a fluid such that the convective heat transfer coefficient is 300 푊/푚2퐾. The average fluid temperature is 1000퐶. The pipe has an inner diameter of 3cm and outer diameter of 4cm.if the heat transfer rate through the pipe per unit length is 500 W/m. calculate the outside surface temperature of the pipe and the overall heat transfer coefficient based on the outside area.
Solution
2훱(푇1 − 푇2) 푄/푙 = 푟 푙푛 ( 2) 1 푟 + 1 ℎ1푟1 푘1
2훱(100 − 푇2) 500 = 2 1 푙푛 ( ) + 1.5 300 × 1.5 × 10−3 0.5
0 푇2 푖푠 푡ℎ푒 표푢푡푠푖푑푒 푠푢푟푓푎푐푒 푡푒푚푝푒푟푎푡푢푟 표푓 푝푖푝푒 = 36.5 퐶
Overall heat transfer based on the outer surface area
HEAT TRANSFER 37 ______
1 푈0퐴0 = 푟 푙푛 ( 2) 1 푟 + 1 ℎ × 푟1 × 푙 × 2휋 2휋 × 푘1 × 푙 1 푈0 = 푟 푟 × 푙푛 ( 2) 푟 2 푟 2 + 1 ℎ × 푟1 푘1 1 푈 = 0 2 2 0.02 × 푙푛 ( ) + 1.5 1.5 × 300 0.5
2 푈0 = 62.69 푊/푚 퐾
A 0.8 m high and 1.5 m wide double pane window consist of two 4mm thick layers of glass having thermal conductivity of 78 푊/푚퐾 separated by a 10 mm wide stagnant air space of thermal conductivity of 0.026 푊/푚퐾. Determine the rate of heat transfer through this window and the temperature of the inside surface, when the room is maintained at 200퐶 and outside air at −100퐶. Take the convective heat transfer coefficient at inside and outside surfaces of the window as 10 푊/푚2퐾 푎푛푑 40 푊/푚2퐾 respectively. Solution
1 1 푅𝑖 = = = 0.0833 퐾/푊 ℎ1퐴 10 × (0.8 × 1.5)
퐿1 0.004 푅1 = = = 0.00427 퐾/푊 푘1퐴 0.78 × 1.2
퐿2 0.01 푅2 = = = 0.3205 퐾/푊 푘2퐴 0.026 × 1.2 1 1 푅0 = = = 0.0833 퐾/푊 ℎ0퐴 40 × (0.8 × 1.5)
푅푡표푡푎푙 = 푅𝑖 + 푅1 + 푅2 + 푅0
푅푡표푡푎푙 = 0.0833 + 0.00427 + 0.3205 + 0.0833 = 0.4332 퐾/푊
38 HEAT TRANSFER ______
The rate of heat transfer =
푇∞1 − 푇∞2 20 − (−10) 푄 = = 푅푡표푡푎푙 0.4332
푄 = 69.2 푊
For inner surface temperature of the windows
푇∞1 − 푇1 푄 = 푅𝑖
푇1 = 푇푎 − 푄 × 푅𝑖
0 푇1 = 20 − 69.2 × 0.0833 = 14.2 퐶
A steam at 3500퐶 flowing in a pipe of thermal conductivity 80 푊/푚퐾 of 5cm internal diameter and 5.6cm outer diameter and is covered with 3cm thick insulation of thermal conductivity 0.05푊/푚퐾. The heat is lost to the surrounding at50퐶by natural convection and radiation, the combine h being 20 푊/푚2퐾. Taking the heat transfer coefficient inside the pipe as 60 푊/푚2퐾. Determine the rate of heat loss per unit length of the pipe and the temperature drop across the pipe and the insulation.
Solution
푟1 = 2.5 푐푚, 푟2 = 2.8 푐푚, 푟3 = 5.8 푐푚
2 퐴1 = 2 × 휋 × 푟1 × 푙 = 2 × 휋 × 0.025 × 1 = 0.157푚
2 퐴3 = 2 × 휋 × 푟3 × 푙 = 2 × 휋 × 0.058 × 1 = 0.364푚
1 1 푅푖 = = = 0.106 퐾/푊 ℎ푖퐴1 60 × 0.157
HEAT TRANSFER 39 ______
푙푛(푟2/푟1) 푙푛(2.8/2.5) 푅1 = = = 0.00023 퐾/푊 2 × 휋 × 푘1 × 푙 2 × 휋 × 80 × 1
푙푛(푟3/푟2) 푙푛(5.8/2.8) 푅2 = = 푅1 = = 2.318 퐾/푊 2 × 휋 × 푘2 × 푙 2 × 휋 × 0.05 × 1
1 1 푅표 = = = 0.137 퐾/푊 ℎ0퐴3 20 × 0.364
푅푡표푡푎푙 = 푅𝑖 + 푅1 + 푅2 + 푅표
푅푡표푡푎푙 = 0.106 + 0.00023 + 2.318 + 0.137
푅푡표푡푎푙 = 2.56123 퐾/푊
푇∞1 − 푇∞2 350 − 5 푄 = = = 134.7푊 푅푡표푡푎푙 2.56123
훥푇푝𝑖푝푒 = 푄 × 푅1
0 훥푇푝𝑖푝푒 = 134.7 × 0.00023 = 0.03 퐶
훥푇𝑖푛푠푢푙푎푡𝑖표푛 = 푄 × 푅2
0 훥푇𝑖푛푠푢푙푎푡𝑖표푛 = 134.7 × 2.318 = 312.2 퐶
A wall of 0.5m thickness is to be constructed from a material which has an average thermal conductivity of 1.4 푊/푚퐾. The wall is to be insulated with a material having an average thermal conductivity of0.35 푊/푚퐾.
So that the heat loss per square meter will not to exceed the 1450 W. assuming that inner and outer surfaces temperature are 12000퐶 and 150퐶 respectively, calculate the thickness of insulation required.
Solution
40 HEAT TRANSFER ______
let the thickness of insulation be 퐿2
푇1 − 푇2 푄 = 퐿 퐿 1 + 2 푘1퐴 푘2퐴
푇1 − 푇2 푄/퐴 = 퐿 퐿 1 + 2 푘1 푘2
1200 − 15 1450 = 0.5 퐿 + 2 1.4 0.35
퐿2 = 0.161푚
A door of cold storage plant is made from two 6mm thick glass sheets separated by a uniform gap of 2mm. the temperature of air inside the room is −200퐶 and the ambient temperature is300퐶. Assuming the heat transfer coefficient between the glass and air to be 23.26 푊/푚2퐾, determine the rate of heat leaking into the room per unit area of the door. Neglect the convection effect in the air gap. Given that the thermal conductivity of glass is 0.75 푊/푚퐾 and thermal conductivity of air is 0.02푊/푚퐾.
Solution
푄 푇푎 − 푇푏 = 퐴 ∑ 푅
푄 푇푎 − 푇푏 = 1 퐿 퐿 퐿 1 퐴 + 1 + 2 + 3 + ℎ푎 푘1 푘2 푘3 ℎ푎
HEAT TRANSFER 41 ______
푄 30 − (−20) = 1 0.006 0.002 0.006 1 퐴 + + + + 23.26 0.75 0.02 0.75 23.26 푄 50 = = 247.5 푊/푚2 퐴 0.202
Determine the heat loss from the insulated steel pipe carrying a hot liquid, to the surrounding per meter length of the pipe, given the following particular.
Inner diameter of pipe is 10 cm Wall thickness 1 cm Thickness of insulation 3 cm Temperature of hot liquid 850퐶 Temperature of surrounding250퐶 Thermal conductivity for steel 58 푊/푚퐾 Thermal conductivity for insulating material 0.2 푊/푚퐾 Inside heat transfer coefficient is 720 푊/푚2 Outside heat transfer coefficient 9 푊/푚2퐾
Solution
2훱(푇푎 − 푇푏) 푄/푙 = 푟 푟 푙푛 ( 2) 푙푛 ( 3) 1 푟 푟 1 + 1 + 2 + ℎ푎푟1 푘1 푘2 ℎ푏푟3
푄 2훱(85 − 25) = = 114.43 푊/푚 푙 6 9 1 푙푛 ( ) 푙푛 ( ) 1 + 5 + 6 + 720 × 0.05 58 0.2 9 × 0.09
A hollow sphere is made up of two material first with thermal conductivity of 70 푊/푚퐾 is having inner diameter of 10 cm and outer diameter of 30cm and the second with thermal conductivity of 15 푊/푚퐾 forms with outer layer with outer diameter of 40cm. the inside and outside surface temperature are 3000퐶 and 300퐶, respectively. Estimate the rate of heat flow through this sphere assuming perfect contact between two materials.
Solution
4 × 훱 × (푇1 − 푇3) 푄 = 푟 − 푟 푟 − 푟 2 1 + 3 2 푘 × 푟1 × 푟2 푘 × 푟3 × 푟2
4 × 훱 × (300 − 30) 푄 = = 11.25푘푊 0.15 − 0.05 0.2 − 0.15 + 70 × 0.05 × 0.15 15 × 0.15 × 0.2
42 HEAT TRANSFER ______
Determine the heat transfer through the composite wall as shown in figure. Take the conductivities of A, B, C, D and E as 50,10,6.67,20, and 30 W/mK respectively and assume the one dimensional heat transfer.
Solution
퐿 0.05 푅 = = = 1 × 10−3 푊/퐾 푎 푘퐴 50 × 1 퐿 0.1 푅 = = = 2 × 10−2 푊/퐾 푏 푘퐴 10 × 0.5 퐿 0.1 푅 = = = 3 × 10−2 푊/퐾 푐 푘퐴 6.67 × 0.5 퐿 0.05 푅 = = = 2.5 × 10−3 푊/퐾 푑 푘퐴 20 × 1 퐿 0.05 푅 = = = 1.67 × 10−3 푊/퐾 푒 푘퐴 30 × 1
The equivalent resistance for B and C
1 1 1 = + 푅푓 푅푏 푅푐
1 1 1 0.833 = −2 + −2 = −2 푅푓 2 × 10 3 × 10 10
−2 푅푓 = 1.2 × 10 푊/퐾
∑ 푅 = 푅푎 + 푅푓 + 푅푑 + 푅푒
∑ 푅 = 1 × 10−3 + 1.2 × 10−2 + 2.5 × 10−3 + 1.67 × 10−3
HEAT TRANSFER 43 ______
∑ 푅 = 17.17 × 10−3 푊/퐾
푇1 − 푇2 푄 = ∑ 푅
800 − 100 푄 = = 4.07 × 104 푘푊 17.17 × 10−3
Heat source with convection A current of 200 A is passed through a stainless steel wire (k=19 W/푚0퐶 3 mm in diameter. The resistivity of steel may be taken as 70 휇훺 푐푚 and the length of the wire is 1m. the wire is submerged in a liquid at 1100퐶 and experienced a convection heat transfer coefficient of 4 푘푊/푚2퐾. Calculate the centre temperature of wire.
Solution
Assumption: steady one dimensional heat transfer with heat generation. Properties remain constant.
Analysis
All the power generated in the wire must be dissipated by convection to the liquid
2 푃 = 퐼 푅 = 푄 = ℎ퐴(푇푤 − 푇∞)
The resistance of the wire can be calculated from
휌퐿 70 × 10−6 × 100 푅 = = = 0.099훺 퐴 휋 0.152
Where 휌 is the resistivity of the wire. The surface area of the wire is 휋푑퐿
2 푃 = 퐼 푅 = 푄 = ℎ퐴(푇푤 − 푇∞)
2 −3 200 × 0.099 = 4000 × 휋 × 3 × 10 (푇푤 − 110) = 3960 푊
0 푇푤 = 215 퐶
The heat generation rate can be calculated as
푃 = 푞̇ × 휋 × 푟2 × 퐿
3960 푞̇ = = 560.2푀푊/푚3 휋(1.5 × 10−3)2 × 1
Finally the center temperature of the wire is calculated from the equation
2 푞̇푟0 푇 = + 푇 0 4푘 푤
5.602 × 108 × (1.5 × 10−3)2 푇 = + 215 = 231.60퐶 0 4 × 19
44 HEAT TRANSFER ______
A cylindrical cement tube of radius 0.05cm and 1cm has a wire embedded into it along its axis. To maintain a steady temperature difference of 1200퐶 between the inner and outer surface the current of 5 A is made to flow in the wire. Make calculation for the amount of heat generated per meter length and the thermal conductivity of current. Take the resistance of wire equal to 0.1 ohm per centimeter of the length.
Solution
푟푒푠푖푠푡푎푛푐푒 = 0.1훺 푝푒푟 푐푚 표푓 푙푒푛푔푡ℎ
= 10 훺 푝푒푟 푚푒푡푒푟 표푓 푡ℎ푒 푙푒푛푔푡ℎ
ℎ푒푎푡 푔푒푛푒푟푎푡푒푑 = 퐼2푅 = 52 × 10 = 250 푊/푚 푙푒푛푔푡ℎ
Under steady state condition the heat generated equal to the heat transfer through the cylindrical element.
2 × 휋 × 퐿 × 푘 × (푇1 − 푇2) 푄 = 푟 푙푛 2 푟1
2 × 휋 × 푘 × (120) 250 = 1.0 푙푛 0.05
푘 = 0.994 푊/푚퐾
A stainless steel tube with inner diameter of 12 mm thickness 0.2mm length 50cm is heated electrically. The entire 15 푘푊 of heat energy generated in the tube is transferred through its outer surface. Work out the intensity of current flow and the temperature drop across tube wall. For tube material take thermal conductivity 18.5 푊/푚푘 take specific resistance 0.85 훺 푚푚2/푚.
Solution
Specific resistance 휌 = 0.85 × 10−6훺 푚2/푚
푟1 = 6푚푚 , 푟2 = 6 + 0.2 = 6.2푚푚
휌 퐿 8.5 × 10−3 × 50 푅 = = = 5.548 훺 푒 퐴 휋(0.622 − 0.62)
2 푃표푤푒푟 푔푒푛푒푟푎푡푒푑 = 퐼 푅푒 = 15000 푤푎푡푡
15000 푖푛푡푒푛푠푖푡푦 표푓 푐푢푟푟푒푛푡 푓푙표푤 = √ = 52 푎푚푝푠 5.548
Under steady state condition, the heat generated equals the heat transfer through the cylindrical tube. Thus
HEAT TRANSFER 45 ______
2 × 휋 × 퐿 × 푘 × (푇1 − 푇2) 푄 = 푟 푙푛 2 푟1
2 × 휋 × 0.5 × 18.5 × (푇1 − 푇2) 15000 = 6.2 푙푛 6
0 (푇1 − 푇2) = 8.467 퐶
To determine the thermal conductivity of hydrogen, a hollow tube with heating wire concentric with the tube is often used. Essentially the gas between the wire and wall is hollow cylinder and the electric current passing through the wire acts as heat source. Determine the thermal conductivity of gas using the following data.
0 푇1 = 푤푖푟푒 푡푒푚푝푒푟푎푡푢푟푒 = 200 퐶
0 푇2 = 푡푢푏푒 푤푎푙푙 푡푒푚푝푒푟푎푡푢푟푒 = 150 퐶
퐼 = 푐푢푟푟푒푛푡 푖푛 푡ℎ푒 푤푖푟푒 = 0.5 퐴
푉 = 푣표푙푡푎푔푒 푑푟표푝 표푣푒푟 0.3 푚 푠푒푐푡푖표푛 표푓 푤푖푟푒 = 3.6 푉
푟2 = 푡푢푏푒 푟푎푑푖푢푠 = 0.125 푐푚
푟1 = 푤푖푟푒 푟푎푑푖푢푠 = 0.0025 푐푚
2 × 휋 × 퐿 × 퐾 × (푇1 − 푇2) 푄 = 푟 푙푛 ( 2) 푟1 푟 푄 × 푙푛 ( 2) 푟 푘 = 1 2 × 휋 × 퐿 × (푇1 − 푇2) 0.0025 3.6 × 0.5 × 푙푛 ( ) 푘 = 0.125 2 × 휋 × 0.3 × (200 − 150)
푘 = 0.075 푊/푚퐾
A refrigerator stands in a room where the air temperature is 210퐶. The surface temperature on the outside of the refrigerator is 160퐶. The sides are 30 mm thick and have an equivalent thermal conductivity of 0.10 푊/푚퐾. The heat transfer coefficient on the outside is 10푊/푚2퐾. assuming the one dimensional conduction through the sides; calculate the net heat flow rate and the inside surface temperature of refrigerator.
46 HEAT TRANSFER ______
Solution
푄 = ℎ × 퐴 × ∆푇
푄 푞 = = ℎ × (푇 − 푇 ) = 10 × (21 − 16) = 50 푊/푚2 퐴 ∞ 푠표
The heat conducted through the wall
∆푇 푄 = 푘 × 퐴 × 퐿
푄 ∆푇 (푇푠표 − 푇푠𝑖) 푞 = = 푘 × = 푘 × 퐴 퐿 퐿
(16 − 푇푠𝑖) 50 = 0.1 × 0.030
0 푇푠𝑖 = 1 퐶
Find the heat flow rate through the composite wall as shown in Fig. Assume one dimensional flow. kA = 150 W/m°C, kB = 30 W/m°C, kC = 65 W/m°C and kD = 50 W/m°C.
Solution:
The thermal circuit for heat flow in the given composite system has been illustrated in Fig.
HEAT TRANSFER 47 ______
48 HEAT TRANSFER ______
UNIT 2 INSULATION
The material which retards the flow of heat with reasonable effectiveness is known as insulation serves for the following two purposes.
It prevents the heat from system to the surrounding It prevents the heat flow from surrounding to the system
Application of insulation
Boilers and steam pipe Air conditioning system Food preserving stores and refrigerator Insulating bricks Preservation of liquid and gases
Factors affecting the thermal conductivity
Some of the important factor which affect thermal conductivity (k) of insulators the value of k should be always low to reduce the rate of heat flow are as follows
Temperature: for most of the insulating material the value of k increases with increases in temperature Density: there is no mathematical relationship between thermal conductivity and density. The common understanding is that high density insulating material will have higher value of k is not always true Direction of heat flow: for most of the insulating material except few like wood the effect of direction of heat flow on the values of k is negligible Moisture: it is always considered to be necessary to prevent increase of moisture in the insulating material during services, it is however difficult to find the effect of moisture on values of k of different insulating material Air pressure: it has been found that the value of k decreases with decrease in pressure Convection in insulator: the value of k increases due to the phenomenon of convection in insulator.
CRITICAL THICKNESS OF INSULATION
The addition of insulation always increases the conductive thermal resistance. But when the total thermal resistance is made of conductive thermal resistance and convective thermal resistance the addition of insulation in some cases may reduces the convective thermal resistance du tot increase in surface area, as in the case of cylinder and sphere, and the total thermal resistance may actually decreases resulting in increased heat flow. It may be shown that thermal resistance actually decreases and then increases in some cases.
HEAT TRANSFER 49 ______
The thickness up to which the heat flow increase and after which the heat flow decreases is termed as critical thickness of insulation. In case of cylinder and sphere ii is called as critical radius.
CRITICAL THICKNESS OF INSULATION FOR CYLINDER
Consider a solid cylinder of radius 푟1 insulated with an insulation of thickness (푟2 − 푟1) as shown in figure.
Let L be the length of cylinder
푇1 be the surface temperature of cylinder
푇푎𝑖푟 be the temperature of air
ℎ표be the heat transfer coefficient at the outer surface of the insulation
푘 be the thermal conductivity of insulating material
The rate o heat transfer from the surface of solid cylinder to the surrounding is given by the following
2 × 훱 × 퐿 × (푇1 − 푇푎𝑖푟) 푄 = 푟 푙푛 ( 2) 푟 1 1 + 푘 ℎ표 × 푟2
푟 푙푛( 2) 푟1 1 From above equation it is evident that as 푟2 increases the factor increases but the factor 푘 ℎ표×푟2 decreases. Thus Q comes maximum when the denominator is minimum. Required condition is 푟 푙푛 ( 2) 푑 푟 1 [ 1 + ] = 0 푑푟2 푘 ℎ표 × 푟2
50 HEAT TRANSFER ______
1 1 1 1 1 × × + (− 2) = 0 푘 푟2/푟1 푟1 ℎ표 푟2 1 1 − = 0 푘 ℎ표 × 푟2
ℎ표 × 푟2 = 푘 푘 = 푟2 ℎ표
The above relation represents the condition for maximum resistance and consequently maximum heat flow rate. The insulation radius at which resistance to heat flow is minimum is called the critical
radius. The critical radius depends on thermal conductivity k and the ℎ표 and is independent of radius푟1.
It may be noted that the second derivative of the denominator is evaluated; it will come out to be
positive. This would verify that heat flow rate will be maximum when푟2 = 푟푐.
푟 푙푛( 2) 푟1 1 The value of is the conduction thermal resistance which increases with increase in 푟2and . 푘 ℎ표×푟2 Is convective thermal resistance which decreases with increasing푟2. At 푟2 = 푟푐
The rate of increase of conductive resistance of insulation is equal to the rate o decrease of convective thermal resistance thus giving a minimum value for the sum of thermal resistance.
For cylindrical bodies 푟1 < 푟푐 the heat transfer increases by adding insulation till 푟2 = 푟푐. If insulation thickness further increased the rate of heat loss will decreases from this peak value
CRITICAL THICKNESS OF INSULATION FOR SPHERE
Critical thickness of insulation for sphere.
(푇1 − 푇푎𝑖푟) 푄 = 푟2 − 푟1 1 + 2 4 × 훱 × 푘 × 푟2 × 푟1 4 × 훱 × 푟2 × ℎ표
HEAT TRANSFER 51 ______
Adopting the same procedure as that of cylinder we have
푑 푟2 − 푟1 1 [ + 2 ] = 0 푑푟2 4 × 훱 × 푘 × 푟2 × 푟1 4 × 훱 × 푟2 × ℎ표 1 2 2 − 3 = 0 푘 × 푟2 푟2 × ℎ표
3 2 푟2 × ℎ표 = 2 × 푘 × 푟2 2 × 푘 푟2 = 푟푐 = ℎ표
CONDUCTION THROUGH EXTENDED SURFACES
The heat conducted through the solids, wall or boundaries has to be continuously dissipated to the surrounding or environment to maintain the system in a steady state condition. In many engineering application large quantities of heat have to be dissipated from small areas. Heat transfer by convection between a surface and the fluid surrounding it can be increases by attaching to the surface thin strips of metals called as fins. The fin increases the effective area of the surface thereby increasing the heat transfer by convection. The fins are also referred to as extended surfaces. Fins are manufactured in different geometries depending upon the practical application, some of which are as shown below in figure.
The ribs attached along the length of the tube are called as longitudinal fins. The concentric annular discs around tube are termed as circumferential fins.
The fins may be of uniform or variable crossection. They have many different practical applications for example cooling of electronic component cooling of motorcycle engine , compressors electric motor transformer, refrigerator, high efficiency boiler superheater tubes, etc. solid gas turbine blades often acts as a fin, conduction heat down their length to cool a disc.
Use of the fin theory is also made in estimating error in temperature measurement by thermometer or thermocouple.
The problem of determination of heat flow through fin required the knowledge of temperature distribution through it. This can be obtained by regarding the fin as a metallic plate connected at its base to a heated wall and transferring heat to a fluid by convection. The heat flow through the fin is by conduction. Thus the temperature distribution in a fin will depends upon the properties of both the fin material and the surrounding fluid.
A steam pipe of 10 cm inner diameter and 11 cm outer diameter is covered with a n insulating substance of thermal conductivity of 1 푊/푚퐾. The steam temperature and the ambient temperature are 2000퐶 and 200퐶, respectively. If the convective heat transfer coefficient between 2 the insulation surface and air is 8 푊/푚 퐾, find the critical radius of insulation . for this value of 푟0 calculate the heat loss per meter of pipe and the outer surface temperature neglect resistance of the pipe material.
52 HEAT TRANSFER ______
Solution
The crtitcal radius of insulation for cylinder is given by
푘 1 푟 = = = 0.125푚 = 12.5푐푚 푐 ℎ 8
2 × 훱 × 퐿 × (푇𝑖 − 푇푎𝑖푟) 푄 = 푟 푙푛 ( 2) 푟 1 1 + 푘 ℎ표 × 푟2
2 × 훱 × (푇𝑖 − 푇푎𝑖푟) 푄/퐿 = 푟 푙푛 ( 2) 푟 1 1 + 푘 ℎ표 × 푟2
In this case 푟2 = 푟푐
2 × 훱 × (100 − 20) 푄/퐿 = 12.5 × 10−2 푙푛 ( ) 5 × 10−2 1 + 1 8 × 12.5 × 10−2 푄 = 620 푊/푚 퐿
The value of 푟2 푟푎푡ℎ푒푟 푡ℎ푎푛 푟푐 shall yield a smaller value of 푄/퐿
The outer surface temperature can be determined by relation
푄 푇표 − 푇푎𝑖푟 = 퐿 푅0 푄 ( ) 푅 = 푇 − 푇 퐿 0 표 푎𝑖푟 푄 푇 = ( ) 푅 + 푇 표 퐿 0 푎𝑖푟 1 푅0 = 2훱 × ℎ표 × 푟2 1 푇 = 620 × + 20 = 118.720퐶 표 12.5 × 10−2 × 8 × 2훱 1 푅0 = 2훱 × ℎ표 × 푟2
An electrical conductor of copper with a diameter of 1mm is covered with a plastic insulation of thickness 1mm. the temperature of the surrounding is 200퐶. Find the maximum current carried by the conductor in that no part of the plastic is above 800퐶. The following data can be made use of the thermal conductivity of copper is 400 푊/푚퐾 and the thermal conductivity of plastic is 0.5 푊/푚퐾
HEAT TRANSFER 53 ______the convective heat transfer coefficient is 8 푊/푚2퐾 the specific electrical resistance of copper is 3 × 10−8훺m. Discuss the effect of increase or decrease of insulation on the current carrying capacity of the conductor.
Solution
푠푝푒푐푖푓푖푐 푟푒푠푖푠푡푎푛푐푒 푡ℎ푒 푒푙푒푐푡푟푖푐푎푙 푟푒푠푖푡푎푛푐푒 푝푒푟 푚푒푡푒푟 표푓 푙푒푛푔푡ℎ = 푐푟표푠푠푒푐푡푖푛푎푙 푎푟푒푎
3 × 10−8 3 × 10−8 푅 = = = 3.8197훺/푚 훱푟2 훱 × (0.05 × 10−3)2
푄 = 퐼2푅
푄 = 퐼23.8197
The convection resistance of the film insulation per meter length is 푟 푙푛 ( 2) 푟1 1 푅푡ℎ푒 = + 2 × 훱 × 푘 2 × 훱 × ℎ표 × 푟2
1.5 × 10−3 푙푛 ( ) 0.5 × 10−3 1 푅 = + = 13.60 푡ℎ푒 2 × 훱 × 400 2 × 훱 × 8 × 1.5 × 10−3
푇𝑖 − 푇푠푢푟푟 80 − 20 푄 = = = 4.41 푊 푅푡ℎ푒 13.60
푄 = 퐼23.819 = 4.41
퐼 = 1.074 퐴
The maximum safe current limit is 1.074 A for the plastic temperature not to exceed 800퐶
The critical radius is
푘 0.5 푟 = = = 0.062푚 푐 ℎ 8
As the critical value of insulation is much greater than that provided in this problem, the current carrying capacity of the conductor can be raised considerably in increasing the radius of plastic covering upto 6.2 cm. this may however, lead to the problem of having too high temperature at the cable center if the temperature inside the plastic coating has to be kept within the given limit.
EFFICIENCY AND EFFECTIVENESS OF FINS
The efficiency of fin is defined as the ratio of actual heat transferred by fin to the maximum heat transferable by fin if the entire fin area were at a base temperature.
푄푓𝑖푛 휂푓𝑖푛 = 푄푚푎푥
54 HEAT TRANSFER ______
A curious but interesting question is whether the addition of in to a surface always results in increased heat transfer. There may be situation where adding fins may actually decrease the heat transfer from a given surface. This may be explained as follows. Compare the heat transfer rate from a surface with fin to that which would be obtained without fin. This ration is called effectiveness.
푄푤𝑖푡ℎ 푓𝑖푛 퐸 = 푄푤𝑖푡ℎ표푢푡 푓𝑖푛
The effectiveness of a fin must be greater than or equal to unity to justify their addition to surface dissipating heat to the surrounding. In practice, however, finining is hardly justified unless h is less than 0.25. In other words finning is only justified only when h is small. If the value of h is large as experienced in boiling, condensation or high velocity liquids, the fins may actually produced a reduction in heat transfer. It is also apparent that Finning will be more effective with materials of large thermal conductivities.
TEMPERATURE DISTRIBUTION ALONG THE FIN
FINS TYPES
Straight fins of (a) uniform and (b) non-uniform cross sections; (c) annular fin, and (d) pin fin of non-uniform cross section.
HEAT TRANSFER 55 ______
LUMPED HEAT CAPACITY ANALYSIS
Heat transfer in heating or cooling of a body is dependent on both the internal and surface resistance 퐿/푘퐴 and 1/ℎ퐴. The simplest unsteady state problem is a one in which the internal resistance is negligible, that is, the convective resistance at the surface boundary is very large when compared to the internal resistance due to conduction. In other words, the solid has an infinite thermal conductivity so that there is no variation of temperature inside the solid and temperature is a function of time only. This situation cannot exist in reality, because all solids have a finite thermal conductivity, and there will always be a temperature gradient inside the solid whenever heat is added or remove. However, for solids of large thermal conductivity with surface areas that are large in proportion to their volume like thin metallic wires and plates, the internal resistance can be assumed to be small or negligible in comparison with the convective resistance at the surface. Problem such as heat treatment of metal by quenching time response of thermocouple and thermometer etc can be analyzed by this idealization of negligible internal resistance.
The process in which the internal resistance is ignored being negligible in comparison with its surface resistances called Newtonian heating or cooling process.
In Newtonian heating or cooling process the temperature throughout the solid is considered to be uniform at a given time. Such an analysis is also called lumped heat capacity analysis because whole solid, whose energy at any time is function of its temperature and total heat capacity, is considered as one lump.
Let us now consider a solid of area A, whose initial temperature is 푇표throughout and which is suddenly placed in a new environment at a constant temperature 푇∞ as shown in fig the lumped heat capacity of the solid is 훒cV.
Where 훒- density of solid in 푘푔/푚3 c- specific heat of solid in 퐽/푘푔 퐾
V- volume of solid in 푚3
The convective heat transfer coefficient between the solid and surrounding is h in푊/푚2퐾. At any instant of time t. the convective heat loss from the body is equal to the decreases in internal energy of solid. Thus
56 HEAT TRANSFER ______
풅푻 푄 = −ℎ퐴(푇 − 푇 ) = 흆풄푽 ∞ 풅풕
Rewriting the above equation we get
푑푇 −ℎ퐴 = 푑푡 (푇 − 푇∞) 흆풄푽
Integrating the above equation we get
−ℎ퐴 푙푛(푇 − 푇 ) = 푡 + 퐶 ∞ 흆풄푽 1
The constant of integrating can be found out using the initial boundary condition
푇 = 푇0 푎푡 푡 = 0 푔푖푣푒푠
푙푛(푇0 − 푇∞) = 퐶1
Putting the value we get
−ℎ퐴 푙푛(푇 − 푇 ) = 푡 + 푙푛(푇 − 푇 ) ∞ 흆풄푽 0 ∞
(푇 − 푇∞) −ℎ퐴 푙푛 ( ) = 푡 (푇0 − 푇∞) 흆풄푽
ℎ퐴 푇 − 푇∞ (− 푡) = 푒 흆풄푽 푇0 − 푇∞
Above equation gives the temperature distribution for a solid initially at a temperature of 푇0, which is place in a convective enviornemnt at temperature 푇∞
Figure b shows the analogous electrical network for a lumped heat capacity of the system in which
퐶푡ℎ = 흆풄푽 represents the thermal capacity of the system. The 퐶푡ℎcan be easily obtained by noting the similarity between thermal and electrical equation as give by
푇ℎ푒푟푚푎푙 푄 = 흆풄푽 푻 = 퐶푡ℎ푇
푒푙푒푐푡푟푖푐푎푙 푆 = 퐶. 푉
Where S is the capacitor charge, V is the voltage and C is the capacitance of the condenser. In the thermal system energy is stored whereas in the electrical system we store charge. The flow of electric charge in electrical system is called current and flow of energy in the thermal system is known as heat.
BIOT NUMBER 푩풊
Biot number is a non dimensional parameter it is used to test the validity of lumped heat capacity approach
HEAT TRANSFER 57 ______
푖푛푡푒푟푛푎푙 푐표푛푑푢푐푡푖푣푒 푟푒푠푖푠푡푎푛푐푒 퐵 = 𝑖 푐표푛푣푒푐푡푖푣푒 푡ℎ푒푟푚푎푙 푟푒푠푖푠푡푎푛푐푒
퐿푐/푘퐴 퐵 = 𝑖 1/ℎ퐴
ℎ퐿푐 퐵 = 𝑖 푘
Where 퐿푐 is called the characteristic length and is equal to the volume of body divided by its surface area. The characteristic length for some common shapes can be calculated as follows.
퐴 × 2퐿 푝푙푎푛푒 푤푎푙푙 표푓 푡ℎ푖푐푘푛푒푠푠 2퐿 퐿 = = 퐿 푐 2 × 퐴
훱푅2퐿 푅 푙표푛푔 푐푦푙푖푛푑푒푟 표푓 푟푎푑푖푢푠 푅 퐿 = = 푐 2훱푅퐿 2 4 훱푅3 푅 푠푝ℎ푒푟푒 표푓 푟푎푑푖푢푠 푅 퐿 = 3 = 푐 4훱푅2 3
퐿3 퐿 퐶푢푏푒 표푓 푠푖푑푒 퐿 퐿 = = 푐 6퐿2 6
It has been observed that for simple shapes such as plates, cylinder, sphere and cube the lumped heat capacity approach can be used to advantages of having Biot number < 0.1
NUMERICALS A 40 × 40 cm slab of copper 5mm thick at uniform temperature of 2500퐶 suddenly has its surface temperature lowered at 300퐶. Find the time at which the slab temperature becomes 900퐶 given that the value of density is 9000 푘푔/푚3, the specific heat 0.38 푘퐽/푘푔 퐾, thermal conductivity 370 푊/ 푚퐾 and convective heat transfer coefficient is 90푊/푚2퐾.
Solution
퐴푟푒푎 , 퐴 = 2 × 0.40 × 0.40 = 0.32 푚2
푉표푙푢푚푒 , 푉 = 0.05 × 0.40 × 0.40 = 8 × 10−4 푚3
푉 퐴 × 2퐿 8 × 10−4 푚3 푐ℎ푎푟푎푐푒푡푟푖푠푡푖푐 푙푒푛푔푡ℎ 퐿 = = = 퐿 = = 2.5 × 10−3푚 퐶 퐴 2퐴 0.32 푚2
−3 ℎ퐿푐 90 × 2.5 × 10 퐵 = = = 6.1 × 10−4 𝑖 푘 370
푠푖푛푐푒 푡ℎ푒 푣푎푙푢푒 표푓 푏푖표푡 푛푢푚푏푒푟 푖푠 푙푒푠푠 푡ℎ푎푛 0.1 푖푡 푠푎푡푖푠푓푦 푡ℎ푒 푐표푛푑푖푡푖표푛 푓표푟 퐿퐻퐶퐴
ℎ퐴 푇 − 푇∞ (− 푡) = 푒 휌푐푉 푇0 − 푇∞ ℎ퐴 90 × 0.32 = = 0.0105 휌푐푉 9000 × 380 × 8 × 10−4
58 HEAT TRANSFER ______
We know from given data
푇0 푖푠 푡ℎ푒 푠푢푟푓푎푐푒 푡푒푚푝푒푟푎푡푢푟푒
푇 푖푠 푡ℎ푒 푡푒푚푝푒푟푎푡푢푟푒 푤ℎ푖푐ℎ 푖푠 푡표 푏푒 푎푡푡푎푖푛 푎푓푡푒푟 푎 푝푒푟푖표푑 표푓 푡푖푚푒 푡
푇∞ 푖푠 푡ℎ푒 푡푒푚푝푒푟푎푡푢푟푒 표푓 푓푙푢푖푑
ℎ퐴 푇 − 푇∞ (− 푡) = 푒 휌푐푉 푇0 − 푇∞ 90 − 30 = 푒(− 0.015푡) 250 − 30 60 = 푒(− 0.015푡) 220
3.67 = 푒( 0.015푡)
Taking the log of both sides we get
ln (3.67) = ln 푒( 0.015푡)
푡 = 123.83 푠푒푐표푛푑푠
A stainless steel rod of outer diameter 1cm and 1m long originally at a temperature of 3200퐶 is suddenly immersed in a liquid at 1200퐶 for which the convective heat transfer coefficient is 100푊/푚2퐾. Determine the required time for the rod to reach temperature of 2000퐶. For stainless steel take value of density is 7800푘푔/푚3, the specific heat 460 퐽/푘푔 퐾, thermal conductivity 40 푊/푚퐾.
Solution
퐴푟푒푎 , 퐴 = 훱 × 퐷 × 퐿 = 훱 × 0.01 × 1 = 0.0314 푚2
훱 푉표푙푢푚푒 , 푉 = × 0.012 × 1 = 7.8538 × 10−5 푚3 4 푉 퐷 0.01 푐ℎ푎푟푎푐푒푡푟푖푠푡푖푐 푙푒푛푔푡ℎ 퐿 = = = = 2.5 × 10−3푚 퐶 퐴 4 4
−3 ℎ퐿푐 100 × 2.5 × 10 퐵 = = = 6.25 × 10−3 𝑖 푘 40
푠푖푛푐푒 푡ℎ푒 푣푎푙푢푒 표푓 푏푖표푡 푛푢푚푏푒푟 푖푠 푙푒푠푠 푡ℎ푎푛 0.1 푖푡 푠푎푡푖푠푓푦 푡ℎ푒 푐표푛푑푖푡푖표푛 푓표푟 퐿퐻퐶퐴
ℎ퐴 푇 − 푇∞ (− 푡) = 푒 휌푐푉 푇0 − 푇∞
푇0 푖푠 푡ℎ푒 푠푢푟푓푎푐푒 푡푒푚푝푒푟푎푡푢푟푒
푇 푖푠 푡ℎ푒 푡푒푚푝푒푟푎푡푢푟푒 푤ℎ푖푐ℎ 푖푠 푡표 푏푒 푎푡푡푎푖푛 푎푓푡푒푟 푎 푝푒푟푖표푑 표푓 푡푖푚푒 푡
HEAT TRANSFER 59 ______
푇∞ 푖푠 푡ℎ푒 푡푒푚푝푒푟푎푡푢푟푒 표푓 푓푙푢푖푑 ℎ퐴 100 × 0.0314 = = 0.0111425 휌푐푉 7800 × 460 × 7.854 × 10−5
ℎ퐴 푇 − 푇∞ (− 푡) = 푒 휌푐푉 푇0 − 푇∞ 200 − 120 = 푒(− 0.0111425푡) 320 − 120 80 = 푒(− 0.0111425푡) 200 200 푒( 0.0111425푡) = 80
푒( 0.0111425푡) = 2.5
Taking the log of both sides we get
ln (푒( 0.0111425푡)) = ln (2.5)
푡 = 82.18 푠푒푐표푛푑푠
An alluminium sphere weighing 5.5kg and initially at a temperature of 2900퐶 is suddenly immersed in a fluid at temperature of 150퐶. The convective heat transfer coefficient is 58 푊/푚2퐾. Estimate the time required to cool the alluminium to 950퐶, using the lumped heat capacity method of anlaysis. Given that the value of density is 2700 푘푔/푚3, the specific heat 0.9 푘퐽/푘푔 퐾, thermal conductivity 205 푊/푚퐾
Solution
Given mass=5.5 kg
푚푎푠푠 = 푑푒푛푠푖푡푦 × 푣표푙푢푚푒
4 푣표푙푢푚푒 표푓 푎 푠푝ℎ푒푟푒 = × 훱 × 푅3 3 4 푚푎푠푠 = 푑푒푛푠푖푡푦 × × 훱 × 푅3 3 4 5.5 = 2700 × × 훱 × 푅3 3
푅3 = 4.8630 × 10−4
푅 = 0.07863 푚
푉 푅 0.07863 푐ℎ푎푟푎푐푒푡푟푖푠푡푖푐 푙푒푛푔푡ℎ 푓표푟 푠푝ℎ푒푟푒 퐿 = = = = 0.0262 푚 퐶 퐴 3 3
60 HEAT TRANSFER ______
ℎ퐿푐 58 × 0.0262 퐵 = = = 7.41268 × 10−3 𝑖 푘 205
푠푖푛푐푒 푡ℎ푒 푣푎푙푢푒 표푓 푏푖표푡 푛푢푚푏푒푟 푖푠 푙푒푠푠 푡ℎ푎푛 0.1 푖푡 푠푎푡푖푠푓푦 푡ℎ푒 푐표푛푑푖푡푖표푛 푓표푟 퐿퐻퐶퐴
ℎ퐴 푇 − 푇∞ (− 푡) = 푒 휌푐푉 푇0 − 푇∞
푇0 푖푠 푡ℎ푒 푠푢푟푓푎푐푒 푡푒푚푝푒푟푎푡푢푟푒
푇 푖푠 푡ℎ푒 푡푒푚푝푒푟푎푡푢푟푒 푤ℎ푖푐ℎ 푖푠 푡표 푏푒 푎푡푡푎푖푛 푎푓푡푒푟 푎 푝푒푟푖표푑 표푓 푡푖푚푒 푡
푇∞ 푖푠 푡ℎ푒 푡푒푚푝푒푟푎푡푢푟푒 표푓 푓푙푢푖푑 ℎ퐴 ℎ × 3 58 × 3 = = = 9.1 × 10−4 휌푐푉 휌푐 푅 2700 × 900 × 0.07863
ℎ퐴 푇 − 푇∞ (− 푡) = 푒 휌푐푉 푇0 − 푇∞
95 − 15 −4 = 푒(− 9.1×10 푡) 290 − 15
80 −4 = 푒(− 9.1×10 푡) 275
−4 275 푒(9.1×10 푡) = 80
Taking log of both sides we get
−4 ln (푒(9.1×10 푡)) = ln (3.43)
푡 = 1357 푠푒푐표푛푑푠
A steel ball 5cm in diameter and initially at uniform temperature of 4500퐶 is suddenly placed in a controlled environment in which the temperature is maintained at 1000퐶. The convection heat transfer coefficient is 10 푊/푚2퐾. Calculate the time required for the ball to attain the temperature of 1500퐶. Given that the value of density is 7800 푘푔/푚3, the specific heat 0.46 푘퐽/푘푔 퐾, thermal conductivity 35 푊/푚퐾
Solution
4 4 푣표푙푢푚푒 표푓 푎 푠푡푒푒푙 푏푎푙푙 = × 훱 × 푅3 = × 훱 × (2.5 × 10−2)3 = 6.544 × 10−5 푚3 3 3
푎푟푒푎 표푓 푎 푠푡푒푒푙 푏푎푙푙 = 4 × 훱 × 푅2 = 4 × 훱 × (2.5 × 10−2)2 = 7.8539 × 10−3푚2
푉 푅 2.5 × 10−2 푐ℎ푎푟푎푐푒푡푟푖푠푡푖푐 푙푒푛푔푡ℎ 푓표푟 푠푝ℎ푒푟푒 퐿 = = = = 8.33 × 10−3 푚 퐶 퐴 3 3
HEAT TRANSFER 61 ______
−3 ℎ퐿푐 10 × 8.33 × 10 퐵 = = = 2.38 × 10−3 𝑖 푘 35
푠푖푛푐푒 푡ℎ푒 푣푎푙푢푒 표푓 푏푖표푡 푛푢푚푏푒푟 푖푠 푙푒푠푠 푡ℎ푎푛 0.1 푖푡 푠푎푡푖푠푓푦 푡ℎ푒 푐표푛푑푖푡푖표푛 푓표푟 퐿퐻퐶퐴
ℎ퐴 푇 − 푇∞ (− 푡) = 푒 휌푐푉 푇0 − 푇∞
푇0 푖푠 푡ℎ푒 푠푢푟푓푎푐푒 푡푒푚푝푒푟푎푡푢푟푒
푇 푖푠 푡ℎ푒 푡푒푚푝푒푟푎푡푢푟푒 푤ℎ푖푐ℎ 푖푠 푡표 푏푒 푎푡푡푎푖푛 푎푓푡푒푟 푎 푝푒푟푖표푑 표푓 푡푖푚푒 푡
푇∞ 푖푠 푡ℎ푒 푡푒푚푝푒푟푎푡푢푟푒 표푓 푓푙푢푖푑
ℎ퐴 10 × 4 × 훱 × (2.5 × 10−2)2 = 4 휌푐푉 7800 × 460 × × 훱 × (2.5 × 10−2)3 3 ℎ퐴 = 3.344 × 10−4 휌푐푉
ℎ퐴 푇 − 푇∞ (− 푡) = 푒 휌푐푉 푇0 − 푇∞
150 − 100 −4 = 푒(− 3.344×10 푡) 450 − 100
50 −4 = 푒(− 3.344×10 푡) 350
−4 350 푒( 3.344×10 푡) = 50
Taking log of both sides we get
−4 350 ln (푒( 3.344×10 푡)) = ln ( ) 50
푡 = 5819 푠푒푐표푛푑푠
The average heat transfer coefficient for flow of 1000퐶 air over a flat plate is measured by observing the temperature given that the characteristic length is 0.015 m and thick copper slab exposed to 1000퐶 of air. In one test run the initial temperature of the plate was found to be 2100퐶 and in 5 minutes the temperature decreased by 400퐶. Calculate the convective heat transfer coefficient for this case given that density of 9000 푘푔/푚3 the value of specific heat is 380 퐽/푘푔 퐾 the thermal conductivity 370 푊/푚퐾
Solution
ℎ퐴 푇 − 푇∞ (− 푡) = 푒 휌푐푉 푇0 − 푇∞
62 HEAT TRANSFER ______
푇0 푖푠 푡ℎ푒 푠푢푟푓푎푐푒 푡푒푚푝푒푟푎푡푢푟푒
푇 푖푠 푡ℎ푒 푡푒푚푝푒푟푎푡푢푟푒 푤ℎ푖푐ℎ 푖푠 푡표 푏푒 푎푡푡푎푖푛 푎푓푡푒푟 푎 푝푒푟푖표푑 표푓 푡푖푚푒 푡
푇∞ 푖푠 푡ℎ푒 푡푒푚푝푒푟푎푡푢푟푒 표푓 푓푙푢푖푑
ℎ퐴 170 − 100 (− 푡) = 푒 휌푐푉 210 − 100
ℎ퐴 70 (− 푡) = 푒 휌푐푉 110
ℎ퐴 ( 푡) 110 푒 휌푐푉 = 70
ℎ퐴 ( 푡) 푒 휌푐푉 = 1.5714
Taking log of both sides
ℎ퐴 푡 = ln(1.5714) = 0.4519 휌푐푉
0.4519 × 휌푐푉 ℎ = 퐴 푡 0.4519 × 9000 × 380 × 푉 ℎ = 퐴 × 300 푉 푐ℎ푎푟푎푐푒푡푟푖푠푡푖푐 푙푒푛푔푡ℎ 퐿 = = 0.015푚 퐶 퐴 0.4519 × 9000 × 380 × 0.015 ℎ = = 77.27 푊/푚2퐾 300
RELATIONSHIP BETWEEN THE AREA AND PERIMETER
HEAT TRANSFER 63 ______
Surface area of fins through which heat transfer takes place is
퐴푠푢푟푓푎푐푒 = (푊 × 퐿) + (푊 × 퐿) + (푡 × 퐿) + (푡 × 퐿) + (푊 × 푡)
퐴푠푢푟푓푎푐푒 = 2(푊 × 퐿) + 2(푡 × 퐿) + (푊 × 푡)
퐴푠푢푟푓푎푐푒 = 2퐿(푊 + 푡) + (푊 × 푡)
2(푊 + 푡) = 푝푒푟푖푚푒푡푒푟
(푊 × 푡) = 퐴푟푒푎 표푓 ℎ푒푎푡 푡푟푎푛푠푓푒푟 푡ℎ표푟푢푔ℎ 푓푖푛 푡푖푝
Since the heat transfer decreases with the length the temperature distribution is zero at end hence heat transfer by tip of rectangular fin is negligibly small and hence area of fin tip through which heat transfer takes place is neglected.
퐴푠푢푟푓푎푐푒 = 푃푒푟푖푚푒푡푒푟 × 퐿푒푛푔푡ℎ
ANALYSIS OF FINS OF UNIFORM CROSSECTION
Consider a fin of uniform crossection. The principal heat conduction along the x axis whereas the convection loss takes place from the upper and under surfaces of the fin. Assuming the base temperature 푇0, of the fin to be uniform and the thermal conditions at the tip of the fin to be small compared to its length, steady state condition will prevail. Effectively, one dimensional heat conduction will exists and the temperature distribution will be a function of x coordinates the properties of the fins and the convective heat transfer coefficient.
To determine the governing differential equation for the fin, shows in fig. consider the heat flow to and from an element crossection dx thick and distances x from the base. We can write a heat balance for the element as instead of
푄푥 = 푄푥+푑푥 + 푄푐표푛푣푒푐푡𝑖표푛 푑푇 푄 = −푘퐴 푡ℎ푒 ℎ푒푎푡 푐표푛푑푢푐푡푒푑 푖푛 푎푡 푥 푥 푑푥
64 HEAT TRANSFER ______
푑푇 푑 푑푇 푄 = −푘퐴 − (푘퐴 ) 푑푥 ℎ푒푎푡 푐표푛푑푢푐푡푒푑 표푢푡 표푓 푥 + 푑푥 푥+푑푥 푑푥 푑푥 푑푥
푄푐표푛푣푒푐푡𝑖표푛 = ℎ(푃푑푥)(푇 − 푇∞) ℎ푒푎푡 푐표푛푣푒푐푡푒푑 푡ℎ푟표푢푔ℎ 푙푒푛푔푡ℎ 푑푥
Where P is the perimeter and (푃푑푥) is the area for convection and A is the crossectional area of the fin thus energy balance equation gives
푄푥 = 푄푥+푑푥 + 푄푐표푛푣푒푐푡𝑖표푛 푑푇 푑푇 푑 푑푇 −푘퐴 = −푘퐴 − (푘퐴 ) 푑푥 + ℎ(푃푑푥)(푇 − 푇 ) 푑푥 푑푥 푑푥 푑푥 ∞ 푑 푑푇 − (푘퐴 ) 푑푥 + ℎ(푃푑푥)(푇 − 푇 ) = 0 푑푥 푑푥 ∞
푑2푇 푘퐴 − ℎ(푃)(푇 − 푇 ) = 0 푑푥2 ∞
푑2푇 ℎ푃 − (푇 − 푇 ) = 0 푑푥2 푘퐴 ∞
푑2푇 ℎ푃 − 푚2(푇 − 푇 ) = 0 푤ℎ푒푟푒 푚2 = 푑푥2 ∞ 푘퐴
Defining the excess temperature θ as
(푇 − 푇∞) = θ
푑2θ − 푚2θ = 0 푑푥2
The general solution of the above differential equation is given by the following
−m1x m2x θ = C1e + C2e
if 푚1 = 푚2 = 푚To solve the above equation two boundary condition are required to determine the values of C1 푎푛푑 C2. Let us now consider three differential cases of fines each yielding a set of two boundary condition.
−mx mx θ = C1e + C2e
LONG FINS
If the fin is sufficiently long the temperature at its end equals to that of the surrounding fluid. A small rod of 3mm diameter and 30 mm long will approach to this condition. The temperature
at the fin base is 푇0. So the mathematical formulation of this type of the fin becomes
푑2θ − 푚2θ = 0 푖푛 푥 ≥ 0 푑푥2
θ = (푇0 − 푇∞) = θ0 at x = 0
HEAT TRANSFER 65 ______
θ → 0 푎푠 푥 → ∞
Applying the second boundary condition to the solution
−mx mx θ = C1e + C2e 푤푒 푔푒푡 퐶2 = 0
Then the application of first boundary condition gives
−mx C1 = θ0 푎푛푑 푡ℎ푒 푠표푙푢푡푖표푛 푏푒푐표푚푒푠 θ = θ0e
(푇 − 푇∞) = e−mx (푇0 − 푇∞)
After knowing the temperature distribution the heat flow through the fin is obtained either by integrating the heat lost by convection over the entire fin surface by the following relation
∞
푄 = ∫ ℎ 푃(푇 − 푇∞) 푑푥 0
Or by evaluating the conductive heat flow at the fin base according to the relation
푑푇 푄 = −푘퐴 ( ) 푑푥 푎푡 푥=0
Using
∞
푄 = ∫ ℎ 푃(푇 − 푇∞) 푑푥 0
Putting the value of (푇 − 푇∞) we get
∞ −mx 푄 = ∫ ℎ 푃(푇0 − 푇∞)e 푑푥 0
∞ −mx 푄 = ∫ ℎ 푃θ0 e 푑푥 0 1 ℎ푃 푄 = ℎ 푃θ e−mx 푏푢푡 푡ℎ푒 푣푎푙푢푒 표푓푚2 = 0 m 푘퐴
푄 = √ℎ푃푘퐴 θ0
푄 = √ℎ푃푘퐴 휃0 푤ℎ푒푟푒 휃0 = (푇0 − 푇∞)
FIN WITH INSULATED END
It may seem undesirable to use a fin with and insulated tip since such insulation will decreases that transfer from the fin. However, in practice the heat transfer area at the fin tip is
66 HEAT TRANSFER ______
generally small in comparison with the lateral area of the fin for heat transfer. Under these conditions the heat flow at the tip is negligible and this condition can be mathematically expresses as
푑휃 ( ) = 0 푡ℎ푒 푚푎푡ℎ푒푚푎푡푖푐푎푙 푓표푟푚푢푙푎푡푖표푛 표푓 푡ℎ푖푠 푝푟표푏푙푒푚 푏푒푐표푚푒푠 푑푥 푎푡 푥=퐿
푑2θ − 푚2θ = 0 푖푛 0 ≤ 푥 ≤ 퐿 푑푥2
휃 = (푇0 − 푇∞) = 휃0 푎푡 푥 = 0 푑휃 ( ) = 0 푑푥 푎푡 푥=퐿
To solve this situation we start with general solution
−푚푥 푚푥 휃 = 퐶1푒 + 퐶2푒
Applying the first boundary condition gives
휃0 = 퐶1 + 퐶2
Applying the second boundary condition we get
푑휃 = −푚퐶 푒−푚푥 + 푚퐶 푒푚푥 푑푥 1 2
−푚퐿 푚퐿 0 = −푚퐶1푒 + 푚퐶2푒
2푚퐿 퐶1 = 퐶2푒
2mL θ0 = C1 + C2 푝푢푡푡푖푛푔 푣푎푙푢푒 표푓 C1 = C2e
2푚퐿 휃0 = 퐶2푒 + 퐶2
휃0 휃0 퐶 = 푎푛푑 퐶 = 푒2푚퐿 2 푒2푚퐿 + 1 1 푒2푚퐿 + 1
휃0 휃0 퐶 = 푎푛푑 퐶 = 2 푒2푚퐿 + 1 1 푒−2푚퐿 + 1
−푚푥 푚푥 휃 = 퐶1푒 + 퐶2푒 푝푢푡푡푖푛푔 푡ℎ푒 푣푎푙푢푒푠 표푓 퐶2 푎푛푑 퐶1
휃0 휃0 휃 = ( ) 푒−푚푥 + ( ) 푒푚푥 푒−2푚퐿 + 1 푒2푚퐿 + 1
−푚푥 푚푥 휃0 푒 휃0 푒 휃 = ( ) + ( ) 푒−2푚퐿 + 1 푒2푚퐿 + 1
푒−푚푥 푒푚푥 휃 = 휃 [( ) + ( )] 0 푒−2푚퐿 + 1 푒2푚퐿 + 1
HEAT TRANSFER 67 ______
The above equation can be written in more compact form by multiplying the numerator and denominator of the first term of R.H.S by 푒푚퐿 and those of the second term by 푒−푚퐿
푒−푚푥 푒푚퐿 푒푚푥 푒−푚퐿 θ = θ [( ) + ( ) ] 0 푒−2푚퐿 + 1 푒푚퐿 푒2푚퐿 + 1 푒−푚퐿
푒푚(퐿−푥) 푒−푚(퐿−푥) θ = θ [( ) + ( )] 0 푒−푚퐿 + 푒푚퐿 푒푚퐿 + 푒−푚퐿
푒푚(퐿−푥) + 푒−푚(퐿−푥) θ = θ [ ] 0 푒−푚퐿 + 푒푚퐿
푐표푠ℎ 푚(퐿 − 푥) 푒훽 + 푒−훽 휃 = 휃 푠푖푛푐푒 cosh 훽 = 0 푐표푠ℎ 푚퐿 2
(푇 − 푇∞) 푐표푠ℎ[푚(퐿 − 푥)] = (푇0 − 푇∞) 푐표푠ℎ(푚퐿)
The heat flow is given by
푑휃 푄 = −푘퐴 ( ) 푑푥 푎푡 푥=0
푑 푐표푠ℎ[푚(퐿 − 푥)] 푄 = −푘퐴 휃 ( ) 푎푡 푥 = 0 0 푑푥 푐표푠ℎ(푚퐿)
[– 푚 푠푖푛ℎ[푚(퐿 − 푥)]] 푄 = −푘퐴 휃 0 푐표푠ℎ(푚퐿)
ℎ푃 푄 = 푘퐴푚휃 tanh(푚퐿) 푏푢푡 푚2 = 0 푘퐴
푄 = √ℎ푃푘퐴 휃0 tanh(푚퐿)
FINS WITH CONVECTION OFF THE END
This is the case of finite fin where the heat conducted to the end is convected away to the surrounding fluid. The mathematical formulation of the problem is
푑2θ − 푚2θ = 0 푖푛 0 ≤ 푥 ≤ 퐿 푑푥2
θ = (푇0 − 푇∞) = θ0 at x = 0 푑휃 −푘퐴 = ℎ퐴휃 푎푡 푥 = 퐿 푑푥 푑휃 −푘퐴 − ℎ퐴휃 = 0 푎푡 푥 = 퐿 푑푥
68 HEAT TRANSFER ______
푑휃 푘 + ℎ휃 = 0 푎푡 푥 = 퐿 푑푥
Applying the condition to the general solution
−mx mx θ = C1e + C2e 푤푒 푔푒푡 푎푡 푥 = 0 푡ℎ푒 푒푞푢푎푡푖표푛 푏푒푐표푚푒푠
θ0 = C1 + C2 푑휃 푘 + ℎ휃 = 0 푤푒 푔푒푡 푎푡 푥 = 퐿 푡ℎ푒 푒푞푢푎푡푖표푛 푏푒푐표푚푒푠 푑푥 푑 푘 (C e−mx + C emx) = −ℎ(C e−mx + C emx ) 푝푢푡푡푖푛푔 푡ℎ푒 푣푎푙푢푒 푥 = 퐿 푑푥 1 2 1 2 ℎ 푚[−퐶 푒−푚퐿 + 퐶 푒푚퐿] = − (퐶 푒−푚퐿 + 퐶 푒푚퐿 ) 1 2 푘 1 2
This yield 퐶1 푎푛푑 퐶2 푎푠
θ0 − C1 = C2 푑휃 −푘퐴 = ℎ퐴휃 푑푥 푑휃 −푘 = ℎ휃 푑푥 푑휃 ℎ − = 휃 푑푥 푘 푑 ℎ − (C e−mx + C emx) = (C e−mx + C emx) 푑푥 1 2 푘 1 2 ℎ m(−C e−mx + C emx) = − (C e−mx + C emx) 1 2 푘 1 2 ℎ (−C e−mx + C emx) = − (C e−mx + C emx) 1 2 m푘 1 2
푎푡 푥 = 퐿 푝푢푡푡푖푛푔 푡ℎ푒 푣푎푙푢푒 푖푛 푎푏표푣푒 푒푞푢푎푡푖표푛 푤푒 푔푒푡
ℎ (−C e−mL + C emL) = − (C e−mL + C emL) 1 2 m푘 1 2 ℎ ℎ −C e−mL + C emL = − C e−mL − C emL 1 2 m푘 1 m푘 2 ℎ ℎ C emL + C emL = C e−mL − C e−mL 2 m푘 2 1 m푘 1 ℎ ℎ C emL (1 + ) = C e−mL (1 − ) 2 m푘 1 m푘
HEAT TRANSFER 69 ______
mL ℎ C2e (1 + ) C = m푘 1 ℎ e−mL (1 − ) m푘
2mL ℎ C2e (1 + ) C = m푘 1 ℎ (1 − ) m푘
but , θ0 − C1 = C2 putting the value of C1 in this equation we get
2mL ℎ C2e (1 + ) θ − m푘 = C 0 ℎ 2 (1 − ) m푘
2mL ℎ C2e (1 + ) θ = C + m푘 0 2 ℎ (1 − ) m푘 ℎ e2mL (1 + ) θ = C (1 + m푘 ) 0 2 ℎ (1 − ) m푘 ℎ ℎ (1 − ) + e2mL (1 + ) θ = C ( m푘 m푘 ) 0 2 ℎ (1 − ) m푘 ℎ θ0 (1 − ) C = m푘 2 ℎ ℎ (1 − ) + e2mL (1 + ) m푘 m푘
2mL ℎ C2e (1 + ) C = m푘 푝푢푡푡푖푛푔 푡ℎ푒 푣푎푙푢푒 표푓 C 푓푟표푚 푎푏표푣푒 푒푞푢푎푡푖표푛 푤푒 푔푒푡 1 ℎ 2 (1 − ) m푘
2mL ℎ ℎ e (1 + ) θ0 (1 − ) C = m푘 × m푘 1 ℎ ℎ ℎ (1 − ) (1 − ) + e2mL (1 + ) m푘 m푘 m푘
2mL ℎ e (1 + ) θ0 C = m푘 × 1 ℎ ℎ 1 (1 − ) + e2mL (1 + ) m푘 m푘
2mL ℎ θ0e (1 + ) C = m푘 1 ℎ ℎ (1 − ) + e2mL (1 + ) m푘 m푘
−mx mx θ = C1e + C2e putting the values of C1 and C2 we get
70 HEAT TRANSFER ______
2mL ℎ ℎ θ0e (1 + ) θ0 (1 − ) θ = ( m푘 ) e−mx + ( m푘 ) emx ℎ ℎ ℎ ℎ (1 − ) + e2mL (1 + ) (1 − ) + e2mL (1 + ) m푘 m푘 m푘 m푘
Dividing by emL to the numerator and denominator we get
mL ℎ ℎ −mL θ0e (1 + ) θ0 (1 − ) e θ = ( m푘 ) e−mx + ( m푘 ) emx ℎ ℎ ℎ ℎ e−mL (1 − ) + emL (1 + ) e−mL (1 − ) + emL (1 + ) m푘 m푘 m푘 m푘 ℎ ℎ emLe−mx (1 + ) + e−mLemx (1 − ) θ = θ ( m푘 m푘 ) 0 ℎ ℎ e−mL (1 − ) + emL (1 + ) m푘 m푘 ℎ ℎ em(L−x) (1 + ) + e−m(L−x) (1 − ) θ = θ ( m푘 m푘 ) 0 ℎ ℎ e−mL (1 − ) + emL (1 + ) m푘 m푘 ℎ ℎ 푒푚(퐿−푥) + 푒푚(퐿−푥) + 푒−푚(퐿−푥) − 푒−푚(퐿−푥) θ = 휃 ( 푚푘 푚푘 ) 0 ℎ ℎ 푒−푚퐿 − 푒−푚퐿 + 푒푚퐿 + 푒푚퐿 푚푘 푚푘 ℎ ℎ 푒푚(퐿−푥) + 푒−푚(퐿−푥) + 푒푚(퐿−푥) − 푒−푚(퐿−푥) 휃 = 휃 ( 푚푘 푚푘 ) 0 ℎ ℎ 푒−푚퐿 + 푒푚퐿 − 푒−푚퐿 + 푒푚퐿 푚푘 푚푘 ℎ 푒푚(퐿−푥) + 푒−푚(퐿−푥) + (푒푚(퐿−푥) − 푒−푚(퐿−푥)) 휃 = 휃 ( 푚푘 ) 0 ℎ 푒−푚퐿 + 푒푚퐿 + (푒푚퐿 − 푒−푚퐿) 푚푘 ℎ 푐표푠ℎ 푚(퐿 − 푥) + 푆푖푛ℎ 푚(퐿 − 푥) 휃 = 휃 [ 푚푘 ] 0 ℎ 푐표푠ℎ(푚퐿) + 푠푖푛ℎ(푚퐿) 푚푘 푑휃 푄 = −푘퐴 ( ) 푑푥 푎푡 푥=0 ℎ 푑 푐표푠ℎ 푚(퐿 − 푥) + 푆푖푛ℎ 푚(퐿 − 푥) 푄 = −푘퐴 ( 휃 [ 푚푘 ]) 푎푡 푥 = 0 0 ℎ 푑푥 푐표푠ℎ(푚퐿) + 푠푖푛ℎ(푚퐿) 푚푘 ℎ −푚 sinh 푚퐿 − 푚 cosh(푚퐿) 푄 = −푘퐴 휃 [ 푚푘 ] 0 ℎ 푐표푠ℎ(푚퐿) + 푠푖푛ℎ(푚퐿) 푚푘 ℎ sinh 푚퐿 + cosh(푚퐿) 푄 = 푘퐴 휃 푚 [ 푚푘 ] 0 ℎ 푐표푠ℎ(푚퐿) + 푠푖푛ℎ(푚퐿) 푚푘
HEAT TRANSFER 71 ______
ℎ푃 푚2 = 푝푢푡푡푖푛푔 푡ℎ푒 푣푎푙푢푒 표푓 푚 푖푛 푎푏표푣푒 푒푞푢푎푡푖표푛 푤푒 푔푒푡 푘퐴 ℎ ℎ푃 sinh 푚퐿 + cosh(푚퐿) 푄 = 푘퐴 휃 √ [ 푚푘 ] 0 ℎ 푘퐴 푐표푠ℎ(푚퐿) + 푠푖푛ℎ(푚퐿) 푚푘 ℎ sinh 푚퐿 + cosh(푚퐿) 푄 = √ℎ푃푘퐴 휃 [ 푚푘 ] 0 ℎ 푐표푠ℎ(푚퐿) + 푠푖푛ℎ(푚퐿) 푚푘 1 ℎ sinh 푚퐿 + cosh(푚퐿) 푚푘 푄 = √ℎ푃푘퐴 휃 [ ] 0 ℎ 1 1 + 푠푖푛ℎ(푚퐿) 푚푘 푐표푠ℎ(푚퐿)
ℎ tanh 푚퐿 + 푄 = √ℎ푃푘퐴 휃 [ 푚푘 ] 0 ℎ 1 + tanh 푚퐿 푚푘
NUMERICALS One end of very long uniform rod is connected to a wall at 1400퐶, while the other end protrudes into a room whose air temperature is 150퐶. the rod is 3mm in diameter and the heat transfer coefficient between the rod surface and environment is 300푊/푚2퐾. Estimate the total heat dissipated by the rod taking its thermal conductivity as 150 푊/푚퐾 .
Solution
Since the rod is very long
푄 = √ℎ푃푘퐴 θ0
푄 = √ℎ푃푘퐴 휃0 푤ℎ푒푟푒 휃0 = (푇0 − 푇∞)
푃 = 훱퐷 = 훱 × 3 × 10−3 = 9.424 × 10−3푚
훱 훱 퐴 = × 퐷2 = × (3 × 10−3)2 = 7.068 × 10−6푚2 4 4
0 푇0 푖푠 푤푎푙푙 푡푒푚푝푒푟푎푡푢푟푒 = 140 퐶
0 푇∞ 푡푒푚푝푒푟푎푡푢푟푒 표푓 푡ℎ푒 푎푚푏푖푒푛푡 푎푖푟 = 15 퐶
푄 = √ℎ푃푘퐴 (푇0 − 푇∞)
푄 = √3000 × 9.424 × 10−3 × 150 × 7.068 × 10−6 (140 − 15)
푄 = 6.843 푊
In an experiment to determine the thermal conductivity of long solid 2.5cm in diameter its base is placed in a furnace with a large portion of it projecting into the room air at 220퐶. After steady state
72 HEAT TRANSFER ______
condition prevail, the temperatures at two points 10cm apart, are found to be 1100퐶 and 850퐶 respectively. The convective heat transfer coefficient between the rod surface and the surrounding air is 28.4푊/푚2퐾. Determine the thermal conductivity of rod material.
Solution
Since the bar is long it may be treated as an infinitely long rod, for which we can use the following equation
(푇 − 푇∞) = 푒−푚푥 (푇0 − 푇∞)
From the given data the temperature at two points 10 cm apart are 1100퐶 and 850퐶
0 푛표푤 푎푡 푥 = 0 푇0 = 110 퐶
푛표푤 푎푡 푥 = 10푐푚 푇 = 850퐶
0 푇∞ 푖푠 푡ℎ푒 푎푚푏푖푒푛푡 푡푒푚푝푒푟푎푡푢푟푒 표푓 푡ℎ푒 푎푖푟 = 22 퐶 85 − 22 = 푒−푚×0.10 110 − 22 63 = 푒−푚×0.10 88 88 푒푚×0.10 = 63
푚 × 0.10 = 0.33420
푚 = 3.342
ℎ푃 푏푢푡 푚 = √ 푘퐴
ℎ푃 11.1691 = 푘퐴
푃 = 훱퐷 = 훱 × 2.5 × 10−2 = 7.85 × 10−2푚
훱 훱 퐴 = × 퐷2 = × (2.5 × 10−2)2 = 4.91 × 10−4푚2 4 4 ℎ푃 11.1691 = 푘퐴
28.4 × 7.85 × 10−2 11.1691 = 훱 푘 × × (2.5 × 10−2)2 4
푘 = 406.63 푊/푚퐾
HEAT TRANSFER 73 ______
A turbine blade of 6cm long and having crossectional area of 4.65 푐푚2 and perimeter of 12 cm is made of stainless steel of thermal conductivity of 23.3 푊/푚퐾 the temperature at the root is 5000퐶 the blade is exposed to hot gas at 8700퐶. The heat transfer coefficient between the blade surface and gas is 442 푊/푚2퐾. Determine the temperature distribution and rate of heat transfer at the root of the blade. Assume the blade to be insulated.
Solution This is the case of the fin with insulated end for which we can use the following equation for temperature distribution
(푇 − 푇∞) 푐표푠ℎ[푚(퐿 − 푥)] = (푇0 − 푇∞) 푐표푠ℎ(푚퐿)
퐴 = 4.65 푐푚2 = 4.65 × 10−4푚2
퐿 = 6푐푚 = 6 × 10−2푚
퐿 = 12푐푚 = 12 × 10−2푚
푘 = 23.3 푊/푚퐾
ℎ = 442 푊/푚2퐾
0 푇0 = 푡푒푚푝푒푟푎푡푢푟푒 푎푡 푡ℎ푒 푟표표푡 = 500 퐶
0 푇∞ = 푡푒푚푝푒푟푎푡푢푟푒 표푓 푓푙푢푖푑 = 870 퐶
ℎ푃 442 × 12 × 10−2 푏푢푡 푚 = √ = √ = 69.96 푘퐴 4.64 × 10−4 × 23.3
−2 (푇 − 푇∞) 푐표푠ℎ[69.96(6 × 10 − 푥)] = (500 − 870) 푐표푠ℎ(69.96 × 6 × 10−2)
푐표푠ℎ[69.96(6 × 10−2 − 푥)] (푇 − 푇 ) = (500 − 870) ∞ 푐표푠ℎ(69.96 × 6 × 10−2)
−2 (푇 − 푇∞) = −11.12 푐표푠ℎ[69.96(6 × 10 − 푥)]
푄 = √ℎ푃푘퐴 휃0 tanh(푚퐿)
푄 = √442 × 6 × 10−2 × 23.33 × 4.64 × 10−4 (−370) tanh(69.96 × 6 × 10−2) = −280.3 푊
It is required to heat oil to 3000퐶 for the purpose of frying. Along ladle is used in frying pan. The section of the ladle is 5 mm by 18 mm. The surrounding air at 300퐶 . The thermal conductivity of material is 205 푊/푚푘 if the temperature at a distance 380 mm from the oil should not exceed400퐶. Determine the convective heat transfer coefficient.
Solution
74 HEAT TRANSFER ______
푇 − 푇∞ = 푒−푚푥 푇0 − 푇∞
A long handle of the ladle is as shown in figure
0 푇0 = 300 퐶
2 퐴퐶 = 5 푚푚 × 18 푚푚 = 90 푚푚
푃 = 2(푤 + 푡) = 2(18 + 5) = 46 푚푚
0 푇∞ = 30 퐶
푘 = 205 푊/푚퐾
푥 = 380 푚푚
푇 = 400퐶
The long handle of the ladle may be treated as long fin the temperature distribution is given by the following
푇 − 푇∞ = 푒−푚푥 푇0 − 푇∞ 40 − 30 = 푒−푚×0.380 300 − 30 10 = 푒−푚×0.380 270
푚 = 8.673
ℎ푃 푚 = √ 푘퐴퐶
ℎ = 30.17 푊/푚2퐾
HEAT TRANSFER 75 ______
UNIT 3 THERMAL RADIATION
All the substances with body temperature above the absolute zero level continuously emit energy in the form of radiation. The emitted radiant energy is in direct proportional to the temperature of the substance. It may either be considered as being transported by electromagnetic waves in accordance with Maxwell classical electromagnetic theory.
In contrast to the heat transfer by conduction and convection wherein an intervening medium is required for the transport of energy from a higher temperature to a lower temperature, no medium is need to be present between two location for radiant interchange to take place. The radiative energy will pass perfectly through vacuum.
Radiation is the only significant mode of heat transfer where no medium is present. At high absolute temperatures level the contribution of radiation to heat transfer is significant. The solar energy incident upon the earth is also governed by the law of radiation.
The term radiation is applied to that kind of process which transmits energy by means of electromagnetic wave. The amount of radiation mainly depends on the nature temperature and type of surface of the body when radiation are falling on a body three things are happened part of energy absorbed, part of energy reflected, and part of energy transmitted.
‘Radiation’ heat transfer is defined as “the transfer of energy across a system boundary by means of an electromagnetic mechanism which is caused solely by a temperature difference.” Whereas the heat transfer by conduction and convection takes place only in the presence of medium, radiation heat transfer does not require a medium. Radiation exchange, in fact, occurs most effectively in vacuum. Further, the rate of heat transfer by conduction and convection varies as the temperature difference to the first power, whereas the radiant heat exchange between two bodies depends on the difference between their temperature to the ‘fourth power’. Both the amount of radiation and the quality of radiation depend upon temperature. The dissipation from the filament of a vacuum tube or the heat leakage through the evacuated walls of a thermos flask are some familiar examples of heat transfer by radiation.
The contribution of radiation to heat transfer is very significant at high absolute temperature levels such as those prevailing in furnaces, combustion chambers, nuclear explosions and in space applications. The solar energy incident upon the earth is also governed by the laws of radiation.
The energy which a radiating surface releases is not continuous but is in the form of successive and separate (discrete) packet or quanta of energy called photons. The photons are propagated through space as rays ; the movement of swarm of photons is described as electromagnetic waves. The photons travel (with speed equal to that of light) in straight paths with unchanged frequency ; when they approach the receiving surface, there occurs reconversion of wave motion into thermal energy which is partly absorbed, reflected or transmitted through the receiving surface (the magnitude of each fraction depends, upon the nature of the surface that receives the thermal radiation).
76 HEAT TRANSFER ______
Let 푄 푏푒 푡ℎ푒 푡표푡푎푙 푎푚표푢푛푡 표푓 푟푎푑푖푎푛푡 푒푛푒푟푔푦 푖푛푐푖푑푒푛푡 표푛 푡ℎ푒 푠푢푟푓푎푐푒 표푓 푏표푑푦
푄푎푏푒 푡ℎ푒 푎푚표푢푛푡 표푓 푟푎푑푖푎푛푡 푒푛푒푟푔푦 푎푏푠표푟푏푒푑 푏푦 푡ℎ푒 푏표푑푦
푄푟푏푒 푡ℎ푒 푎푚표푢푛푡 표푓 푟푎푑푖푎푛푡 푒푛푒푟푔푦 푟푒푓푙푒푐푡푒푑 푏푦 푡ℎ푒 푏표푑푦
푄푡푏푒 푡ℎ푒 푎푚표푢푛푡 표푓 푟푎푑푖푎푛푡 푒푛푒푟푔푦 푡푟푎푛푠푚푖푡푡푒푑 푏푦 푡ℎ푒 푏표푑푦
Then according to the principal of conservation of energy the total sum must be equal to the incident radiation.
푟푎푖푑푎푛푡 푒푛푒푟푔푦 푖푛푐푖푑푒푛푡 = 푟푎푑푖푎푛푡 푒푛푒푟푔푦 푎푏푠표푟푏푒푑 + 푟푎푑푖푎푛푡 푒푛푒푟푔푦 푟푒푓푙푒푐푡푒푑 + 푟푎푑푖푎푛푡 푒푛푒푟푔푦 푡푟푎푛푠푚푖푡푡푒푑
푄 = 푄푎 + 푄푟 + 푄푡
푄푎 푄푟 푄푡 1 = + + 푄 푄 푄
1 = 훼 + 흆 + 흉
훼 푖푠 푐푎푙푙푒푑 푡ℎ푒 푎푏푠표푟푝푡푖푣푖푡푦 표푓 푡ℎ푒 푠푢푟푓푎푐푒
휌 푖푠푐푎푙푙푒푑 푡ℎ푒 푟푒푓푙푒푐푡푣푖푡푦 표푓 푡ℎ푒 푠푢푟푓푎푐푒
휏 푖푠 푐푎푙푙푒푑 푡ℎ푒 푡푟푎푛푚푖푡푡푖푣푖푡푦 표푓 푠푢푟푓푎푐푒
The bodies may be black, white, transparent and opaque body.
Black body:
A black body is a one in which absorbs the all the incident radiation falling on it for black body훼 = 1, 휌 = 0 푎푛푑 휏 = 0.
Concept of a Black Body
HEAT TRANSFER 77 ______
A black body is an object that absorbs all the radiant energy reaching its surface (for a black body α = 1, ρ = 0, τ = 0). No actual body is perfectly black; the concept of a black body is an idealization with which the radiation characteristics of real bodies can be conveniently compared.
A black body has the following properties:
(i) It absorbs all the incident radiation falling on it and does not transmit or reflect regardless of wavelength and direction.
(ii) It emits maximum amount of thermal radiations at all wavelengths at any specified temperature.
(iii) It is a diffuse emitter (i.e., the radiation emitted by a black body is independent of direction).
Consider a hollow enclosure with a very small hole for the passage of incident radiation as shown in Fig. Incident radiant energy passes through the small opening; some of this energy is absorbed by the inside surface and some is reflected. However, most of this energy is absorbed on a second incidence. Again, a small fraction is reflected. After a number of such reflections the amount unabsorbed is exceedingly small and very little of the original incident energy is reflected back out of the opening. A small hole leading into a cavity (Hohlraum) thus acts very nearly as a black body because all the radiant energy entering through it gets absorbed.
White body
A white body is a one in which it reflects all the incident radiation falling on it. For white body 훼 = 0, 휌 = 1 푎푛푑 휏 = 0
Transparent body
A transparent body is a one in which it transmit all the incident radiation falling on it. Thus for transparent body 훼 = 0, 휌 = 0 푎푛푑 휏 = 1
Opaque body
Opaque body is a one in which it does not transmit the incident radiation falling on it thus for opaque body 훼 + 휌 = 1 푎푛푑 휏 = 0
78 HEAT TRANSFER ______
Solids generally do not transmit unless the material is of very thin section. Metals absorb radiation within a fraction of a micrometre, and insulators within a fraction of millimetre. Glasses and liquids are, therefore, generally considered as opaque.
Gray body
If the radiative properties, α, ρ τ of a body are assumed to be uniform over the entire wavelength spectrum, then such a body is called gray body. A gray body is also defined as one whose absorptivity of a surface does not vary with temperature and wavelength of the incident radiation
Emissive power (E)
It is defined as the total radiant energy emitted by the body at certain temperature per unit time per unit surface area at all wavelength.
Monochromatic emissive power 푬흀
It is defined as the radiant energy emitted by the body per unit time per unit surface area at a particular wavelength and temperature.
The radiation energy emitted by black body per unit time and per unit surface area is given by the following
4 퐸푏 = 휎 × 푇
휎 = 5.67 × 10−8 푊/푚2퐾4
휎 푖푠 푐푎푙푙푒푑 푡ℎ푒 푠푡푒푓푎푛푠 푏표푙푡푧푚푎푛 푐표푛푠푡푎푛푡
푇 푖푠 푐푎푙푙푒푑 푡ℎ푒 푎푏푠표푙푢푡푒 푡푒푚푝푒푟푡푢푟푒 표푓 푡ℎ푒 푠푢푟푓푎푐푒 푖푛 푘푒푙푣푖푛
Stefan’s Boltzmann law
Stefan’s Boltzmann law states that the emissive power of a black body is directly proportional to the fourth power of its absolute temperature.
Emissivity of surface 훜
Emissivity of surface is defined as the ratio of radiation emitted by the surface to the radiation emitted by a black body at the same temperature. It is denoted by (훜) its value varies between 0 and 1. Emissivity is measure of how closely a surface approximates to a black body. For which 훜=1. The emissivity of the surface varies with the temperature of the surface as well as the wavelength and direction of emitted radiation.
KIRCHOFFS LAW
The law states that at any temperature the ratio of total emissive power E to the total absorptivity α is a constant for all substances which are in thermal equilibrium with their environment.
HEAT TRANSFER 79 ______
Let us consider a large radiating body of surface area A which encloses a small body (1) of surface area A1 (as shown in Fig.). Let the energy fall on the unit surface of the body at the rate Eb. Of this energy, generally, a fraction α, will be absorbed by the small body. Thus this energy absorbed by the small body (1) is α1 A1 Eb, in which α1 is the absorptivity of the body. When thermal equilibrium is attained, the energy absorbed by the body (1) must be equal to the energy emitted, say, E1 per unit surface.
A small body of surface area 퐴1 is placed in a hollow evacuated space kept at a constant uniform temperature T after some time the body will attain at steady state the same temperature as that of interior of the space and thereafter will radiate as much energy as it receives.
Let I be the radiant energy falling upon the body per unit time per unit surface area
퐸1푏푒 푡ℎ푒 푒푚푖푠푠푖푣푒 푝표푤푒푟 표푓 푡ℎ푒 푏표푑푦
훼1푏푒 푡ℎ푒 푎푏푠표푟푝푡푖푣푖푡푦 표푓 푡ℎ푒 푏표푑푦
At steady state
푒푛푒푟푔푦 푎푏푠표푟푏푒푑 = 푒푛푒푟푔푦 푒푚푖푡푡푒푑
퐼퐴1훼1 = 퐸1퐴1
퐸1 퐼 = 훼1
If a body is replaced by second body of the same shape and surface area but of different material at exactly the same location at steady state
퐼퐴2훼2 = 퐸2퐴2
퐸2 퐼 = 훼2
Similarly if the second body is replaced at the same location by a black body of same shape and surface area
퐼퐴1훼푏 = 퐸푏퐴1
80 HEAT TRANSFER ______
퐸푏 퐼 = 훼푏
Since I is same on each body we can express I as
퐸1 퐸2 퐸푏 퐼 = = = = 퐸푏 푠푖푛푐푒 훼푏 = 1 푓표푟 푏푙푎푐푘 푏표푑푦 훼1 훼2 훼푏
퐸1 퐸2 퐸푏 = = 훼1 훼2 1
퐸1 퐸2 = = 퐸푏 훼1 훼2
퐸1 훼1 = 퐸푏
퐸2 훼2 = 퐸푏
Therefore for any body
퐸 훼 = 퐸푏
But the ratio of emissive power of body to the emissive power of a black body is called the emissivity.
퐸 훼 = = 흐 퐸푏
훼 = 흐
This is Kirchhoff’s law which states that the emissivity of the surface of body is equal to its absorptivity when the body is in thermal equilibrium with its surrounding.
A black body in addition to being perfect absorber is also a perfect emitter of radiant energy
Kirchhoff’s law holds good for monochromatic radiation also for which
퐸휆1 퐸휆2 퐸푏휆 = = 푠푖푛푐푒 훼푏휆 = 1 훼휆1 훼휆2 훼푏휆
퐸휆1 퐸휆2 퐸푏휆 = = 훼휆1 훼휆2 1
훼휆 = 흐휆
Hence monochromatic emissivity of a body is equal to the monochromatic absorptivity at the same wavelength.
PLANCKS LAW
HEAT TRANSFER 81 ______
Planck’s law is the basic law for thermal radiation. Planck’s deduced by his quantum theory expression for 퐸푏휆 as a function of temperature and wavelength as given below
−5 퐶1휆 퐸 = 푏휆 퐶2 푒휆푇 − 1
Where
−16 2 퐶1 = 3.74 × 10 푊 푚
−2 퐶2 = 1.438 × 10 푚퐾
휆 푖푠 푡ℎ푒 푤푎푣푒푙푒푛푔푡ℎ 푖푛 푚 푎푛푑 푇 푖푠 푡ℎ푒 푎푏푠표푙푢푡푒 푡푒푚푝푒푟푎푡푢푟푒 푖푛 퐾
WIEN’S LAW
푖푓 휆푇 ≪ 퐶2
Or
퐶2 ≫ 1 휆푇
Then the unity present in the equation mention below can be neglected
−5 퐶1휆 퐸 = 푏휆 퐶2 푒휆푇 − 1
And the relationship becomes
−5 퐶1휆 퐸 = 푏휆 퐶2 푒휆푇
The above equation is known as wiens law which is found to be valid for short wavelength.
WIEN’S DISPLACEMENT LAW
In 1893 Wien established a relationship between the temperature of a black body and the wavelength at which the maximum value of monochromatic emissive power occurs. A peak monochromatic emissive power occurs at a particular wavelength. Wien’s displacement law states that the product of λmax and T is constant, i.e.,
Planck’s equation is given by the following
−5 퐶1휆 퐸 = 푏휆 퐶2 푒휆푇 − 1
The maximum value of emissive power 퐸푏 can be obtained by differentiating the Planck’s equation with respect to 휆 and equating it to zero
82 HEAT TRANSFER ______
−5 푑퐸푏휆 푑 퐶1휆 = [ ] 푑휆 푑휆 퐶2 푒휆푇 − 1
퐶 −1 푑퐸푏휆 푑 2 = [퐶 휆−5 × (푒휆푇 − 1) ] 푑휆 푑휆 1
퐶 −2 퐶 퐶 −1 2 퐶2 2 2 0 = 퐶 휆−5 × (−1) (푒휆푇 − 1) × × 푒휆푇 + (푒휆푇 − 1) × 퐶 휆−6 × (−5) 1 휆2푇 1
퐶2 푒휆푇 5휆푇 = 퐶2 퐶 (푒휆푇 − 1) 2
퐶2 (푒휆푇 − 1) 퐶2 퐶2 = 푝푢푡푖푛푔 = 푥 푤푒 푔푒푡 5휆푇 퐶2 휆푇 푒휆푇 푥 푒푥 − 1 = = 1 − 푒−푥 5 푒푥 푥 − 1 + 푒−푥 = 0 푠표푙푢푡푖표푛 표푓 푡ℎ푖푠 푒푞푢푎푡푖표푛 푔푖푣푒푠 5
퐶2 푥 = = 4.965 휆푚푎푥푇
−2 퐶2 1.438 × 10 푚퐾 휆 푇 = = = 2.898 × 10−3푚퐾 푚푎푥 4.965 4.965
−3 휆푚푎푥푇 = 2.898 × 10 푚퐾
This is known as wiens displacement law
STEFANS BOLTZMAN LAW
The law states that the emissive power of a black body is directly proportional to the fourth power of its absolute temperature.
Stefan’s Boltzmann law can be derived from planks law
The total emissive power of a black body is given by
∞ ∞ −5 퐶1휆 퐸 = ∫ 퐸 푑휆 = ∫ 푑휆 푏 푏휆 퐶2 0 0 푒휆푇 − 1
푠푢푏푠푡푖푡푢푡푖푛푔 휆−1 = 푥 표푟 휆−2푑휆 = 푑푥, 표푟 푑휆 = −1/푥2 푑푥 푎푛푑
푤ℎ푒푛 휆 = ∞, 푥 = 0 푎푛푑 휆 = 0, 푥 = ∞
HEAT TRANSFER 83 ______
∞ 5 퐶1푥 1 퐸푏 = ∫ × (− ) 푑푥 퐶2푥 푥2 0 푒 푇 − 1
∞ 3 −1 푥 퐶2푥 퐸 = 퐶 ∫ (푒 푇 − 1) × 푑푥 푏 1 1 0
∞ −퐶 푥 −2퐶 푥 −3퐶 푥 3 2 2 2 퐸푏 = 퐶1 ∫ 푥 × (푒 푇 + 푒 푇 + 푒 푇 + ⋯ … … … ) × 푑푥 0
∞ −퐶 𝑖푥 3 2 퐸푏 = 퐶1 ∫ 푥 푒 푇 푑푥 푤ℎ푒푟푒 푖 = 1,2,3 0
∞ 푛! 푏푢푡 ∫ 푥푛 푒−푎푥푑푥 = 푎푛+1 0
4 3! 6퐶1푇 퐸푏 = 퐶1 = 퐶 푖 3+1 퐶 4푖4 ( 2 ) 2 푇
4 6퐶1푇 1 1 1 퐸푏 = 4 ( 4 + 4 + 4 + ⋯ … . . ) 퐶2 1 2 3
4 4 6퐶1푇 훱 퐸푏 = 4 × 퐶2 90
4 퐸푏 = 휎푇
휎 푖푠 푡ℎ푒 푠푡푒푓푎푛푠 푏표푙푡푧푚푎푛 푐표푛푠푡푎푛푡 푎푛푑 푖푡푠 푣푎푙푢푒 푖푠 푔푖푣푒푛 푏푦
4 −16 4 6퐶1 훱 6 × 3.74 × 10 훱 −8 2 4 휎 = 4 × = −2 4 = 5.67 × 10 푊/푚 퐾 퐶2 90 (1.438 × 10 ) 90
INTENSITY OF RADIATION AND LAMBERT COSINES LAW
When a surface element emits radiation, all of it will be intercepted by a hemispherical surface placed over the element. The intensity of radiation (I) is defined as the rate of energy leaving a surface in a given direction per unit solid angle per unit area of the emitting surface normal to the mean direction in space. A solid angle is defined as a portion of the space inside a sphere enclosed by a conical surface with the vertex of the cone at the centre of the sphere. It is measured by the ratio of the spherical surface enclosed by the cone to the square of the radius of the sphere
Lambert cosine law
The law states that the total emissive power Eθ from a radiating plane surface in any direction is directly proportional to the cosine of the angle of emission. The angle of emission θ is the angle subtended by the normal to the radiating surface and the direction vector of emission of the
84 HEAT TRANSFER ______
receiving surface. If En be the total emissive power of the radiating surface in the direction of its normal, then
Eθ = En cos θ
The above equation is true only for diffuse radiation surface. The radiation emanating from a point on a surface is termed diffused if the intensity, I is constant. This law is also known as Lambert’s law of diffuse radiation.
Subtended plane and solid angle
The plane angle 훼 is defined by a region by the rays of a circle and is measured as the ratio of element of arc of length l on the circle to the radius 훼 = 푙/푟
The solid angle 휔 is defined by a region by the rays of a sphere and is measured as
퐴푛 퐴 퐶표푠휃 휔 = = 푟2 푟2
퐴푛 푖푠 푡ℎ푒 푝푟표푗푒푐푡푖표푛 표푓 푡ℎ푒 푖푛푐푖푑푒푛푡 푠푢푟푓푎푐푒 푛표푟푚푎푙 푡표 푙푖푛푒 표푓 푝푟표푝표푔푎푡푖표푛
퐴 푖푠 푡ℎ푒 푎푟푒푎 표푓 푖푛푐푖푑푒푛푡 푠푢푟푓푎푐푒
휃 푎푛푔푙푒 푏푒푡푤푒푒푛 푡ℎ푒 푛표푟푚푎푙 푡표 푡ℎ푒 푖푛푐푖푑푒푛푡 푠푢푟푓푎푐푒 푙푖푛푒 표푓 푝푟표푝표푔푎푡푖표푛
푟 푖푠 푡ℎ푒 푙푒푛푔푡ℎ 표푓 푡ℎ푒 푙푖푛푒 표푓 푝푟표푝푎푡푖표푛 푏푒푡푤푒푒푛푡 푒ℎ 푟푎푑푖푎푡푖표푛 푎푛푑 푡ℎ푒 푖푛푐푖푑푒푛푡 푠푢푓푎푐푒
INTENSITY OF RADIATION
Consider a small black surface dA arbitararily located at a point in the space under consideration and emitting radiation in different direction a black body radiation collector through which the radiation pass is located at an angular position characterized by zenith angle 휃 towards the surface normal and azimuth angle ∅ of a sphereical coordinate system further the collector subtends a solid angle 푑휔 when viewed from a point on the emitter.
The intensity of radiation 퐼 is the energy emitted of all wavelengths in particular direction per unit surface area and through a unit solid angle. The area is the projected area of the surface on a plane perpendicular to the direction of radiation.
HEAT TRANSFER 85 ______
The collector or incidence surface measures a variation of the emitted radiation depending upon its angular position. Maximum amount of radiation is measured by collector when it is at the position normal to emitter and the intensity in a direction 휃 from normal to the emitter follows the Lambert cosines law
The intensity of radiation in direction 휃 from the normal to a black emitter is proportional to cosines of angle 휃
퐼푛 푑푒푛표푡푒푠 푛표푟푚푎푙 푖푛푡푒푛푠푖푡푦
퐼휃 푟푒푝푟푒푠푒푛푡푠 푡ℎ푒 푖푛푡푒푠푖푡푦 푎푡 푎푛푔푙푒 휃 푓푟표푚 푡ℎ푒 푛표푟푚푎푙
퐼휃 = 퐼푛 cos 휃
Apparently the energy radiated out decreases with increase in the value of 휃 and becomes zero at 휃 = 900.
When the collector is oriented at an angle 휃1 from the normal to the emitter then the radiation striking and being absorbed by the collector can be expressed as
( ) 푑퐸푏 휃1 = 퐼휃1푑휔1푑퐴
( ) 푑퐸푏 휃1 = 퐼푛 cos 휃1 푑휔1푑퐴
푑휔1 푖푠 푡ℎ푒 푠표푙푖푑 푎푛푔푙푒 푠푢푏푡푒푛푑푒푑 푏푦 푐표푙푙푒푐푡표푟 푎푡 푡ℎ푒 푠푢푟푓푎푐푒 표푓 푒푚푖푡푡푒푟
The collector could be located at different angular position and still maintain the same radial distance from the emitter. Let it subtend a solid angle 푑휔2 at the emitter surface when located in a direction 휃2 from the normal. Then the rate of flow of energy through it will be
( ) 푑퐸푏 휃2 = 퐼휃2푑휔2푑퐴
( ) 푑퐸푏 휃2 = 퐼푛 cos 휃2 푑휔2푑퐴
It follows from the equation that for any surface located at angle 휃 from the normal subtending solid angle 푑휔 at the emitter 푑퐴.
(푑퐸푏)휃 = 퐼푛 cos 휃 푑휔 푑퐴
86 HEAT TRANSFER ______
RELATION BETWEEN THE NORMAL INTENSITY AND EMISSIVE POWER
To establish the relation between the normal intensity and the emissive power we relate the differential solid angle 푑휔 to the zenith and azimuth angles by noting that for spherical surface.
Area of collector
= 푟(푑휃) × (푟푠푖푛휃 푑∅)
= 푟2 푠푖푛휃 푑휃 푑∅
HEAT TRANSFER 87 ______
Solid angle 푑휔
푟2 푠푖푛휃 푑휃 푑∅ = 푟2
= 푠푖푛휃 푑휃 푑∅
Then the radiation leaving the emitter and striking the collector is
푑퐸푏 = 퐼푛 cos 휃 푠푖푛휃 푑휃 푑∅ 푑퐴
The total energy 퐸푏 radiated by the emitter and passing through a spherical region can be worked out by integrating over the limits 휋 휃 = 0 푡표 휃 = 2
∅ = 0 푡표 ∅ = 2휋
Thus ,
휋/2 2휋 ∫ 푑퐸푏 = 퐼푛푑퐴 ∫ sin 휃 cos 휃 푑휃 ∫ 푑∅ 0 0 1 퐸 = 퐼 푑퐴 × (2휋) = 퐼 휋푑퐴 푏 푛 2 푛
But the total emissive power of the emitter with area 푑퐴 and temperature T is also given by
4 퐸푏 = 휎푏푇 푑퐴
4 4 휎푏푇 푑퐴 휎푏푇 퐼 = = 푛 휋푑퐴 휋
Thus for a unit surface the intensity of normal radiation 퐼푛 is 1/휋 times the emissive power 퐸푏.
RADIANT HEAT EXCHANGE BETWEEN TWO BLACK BODIES NON ABSORBING MEDIUM
The radiant heat exchange between two bodies depends upon the following
The views the surfaces have of each other, ie how they see each other.
88 HEAT TRANSFER ______
The emitting and absorbing characteristics and The medium that intervenes the two bodies.
Let us assume that the two bodies are black and the medium in nonparticipating in the energy
exchange. Let us consider the area elements 푑퐴1 and 푑퐴2 on the two surfaces. The distance between them is r and the angles made by the normals to the two area elements with the line
joining them are 휙1, 휙2respectively. The projected area of 푑퐴1 in the direction of radiation is 푑퐴1 푐표푠휙1.
Energy leaving 푑퐴1and intercepted by 푑퐴2 is given by
푑퐴2 cos 휙2 푑푄 = 퐼 푑퐴 cos 휙 1−2 1 1 1 푟2
The solid angle subtended by element 푑퐴2 at the center of 푑퐴1 is
푑퐴2 cos 휙2
푟2
Since 퐼1 is independent of direction of 휙1 the energy emitted per unit area 푑퐴1 per unit solid angle is proportional to the cosine of the angle 휙1. This is known as lamberts cosine law.
4 4 휎푇1 푑퐴2 cos 휙2 휎푇1 푑푄 = 푑퐴 cos 휙 푤ℎ푒푟푒 퐼 = 1−2 훱 1 1 푟2 훱
The total radiation leaving 퐴1 and absorbed by 퐴2 is given by
4 휎푇1 cos 휙1 cos 휙2 푄 = ∬ 푑퐴 푑퐴 1−2 훱 푟2 1 2
4 푄1−2 = 퐴1퐹12휎푇1
1 cos 휙1 cos 휙2 퐴 퐹 = ∬ 푑퐴 푑퐴 1 12 훱 푟2 1 2
Similarly energy leaving 푑퐴2 and intercepted by 푑퐴1 is given by
푑퐴1 cos 휙1 푑푄 = 퐼 푑퐴 cos 휙 2−1 2 2 2 푟2
4 4 휎푇2 푑퐴1 cos 휙1 휎푇2 푑푄 = 푑퐴 cos 휙 푤ℎ푒푟푒 퐼 = 2−1 훱 2 2 푟2 2 훱
The total radiation leaving 퐴2 and absorbed by 퐴1 is given by
4 휎푇2 cos 휙1 cos 휙2 푄 = ∬ 푑퐴 푑퐴 2−1 훱 푟2 1 2
4 푄2−1 = 퐴2퐹21휎푇2
1 cos 휙1 cos 휙2 퐴 퐹 = ∬ 푑퐴 푑퐴 2 21 훱 푟2 1 2
HEAT TRANSFER 89 ______
Therefore the net energy exchange between 퐴1 and 퐴2 is obtained by the following
(푄12)푛푒푡 = 푄1−2 − 푄2−1
4 4 (푄12)푛푒푡 = 퐴1퐹12휎(푇1 − 푇2 )
1 cos 휙1 cos 휙2 퐴 퐹 = 퐴 퐹 = ∬ 푑퐴 푑퐴 1 12 2 21 훱 푟2 1 2
Here 퐹12 is the shape factor of 퐴1 with respect to 퐴2 this is the fraction of energy leaving 퐴1 that strikes 퐴2 and is absorbed because the surface is black
Here 퐹21 is the shape factor of 퐴2 with respect to 퐴1 this is the fraction of energy leaving 퐴2 that strikes 퐴1 and is absorbed because the surface is black
퐹12 and 퐹21values depends on how two surfaces are exposed to each other to see each other these are called view factor.
RADIATION ERROR IN TEMPERATURE MEASUREMENT
If the temperature of high temperature gas stream is measured by the insertion of thermometer or thermocouple, the effects of the radiant exchange between the pipe walls and the temperature sensing element introduce considerable errors.
If 푇푔 be the temperature to be measured and 푇푐 is the measured temperature, at steady state, the heat transfer by convection from gas to the thermocouple is equal to the heat transfer by radiation from thermocouple to the wall.
푄 = ℎ퐴푐(푇푔 − 푇푐)
4 4 = 휎퐴푐퐹푐푤(푇푐 − 푇푤 )
Where 퐹푐푤 = 휖푐 the emissivity of thermocouple which is very small compared to the enclosing wall, and 퐴푐 is the area of thermocouple. Here (푇푔 − 푇푐) is called the thermocouple error.
When the couple is shielded the error is considerably reduced. At steady state
90 HEAT TRANSFER ______
The heat transfer by convection from gas to couple is equal to that by radiation from couple to shield and The heat transfer by convection from gas to shield and that by radiation from couple to shield are equal to heat transfer by radiation from shield to wall
4 4 ℎ퐴푐(푇푔 − 푇푐) = 휎퐴푐퐹푐푠(푇푐 − 푇푠 )
4 4 4 4 ℎ2퐴푠(푇푔 − 푇푠) + 휎퐴푐퐹푐푠(푇푐 − 푇푠 ) = 휎퐴푠퐹푠푤(푇푠 − 푇푤 )
1 퐹 = 푤ℎ푒푟푒 퐹 = 휖 푐푠 1 퐴 1 푠푤 푠 + 푐 [ − 1] 휖푐 퐴푠 휖푠
푇푠 푖푠 푡ℎ푒 푒푞푢푖푙푖푏푟푖푢푚 푡푒푚푝푒푟푎푡푢푟푒 표푓 푡ℎ푒 푠ℎ푖푒푙푑푠
퐴푠 푖푠 푡ℎ푒 푠ℎ푖푒푙푑 푠푢푟푓푎푐푒 푎푟푒푎 푎푛푑휖푠 푖푠 푡ℎ푒 푒푚푖푠푠푖푣푖푡푦 표푓 푠ℎ푖푒푙푑
Since heat is transferred by convection from gas to both sides of the shield, the total surface area of
the shield is 2퐴푠as provided in the second equation of first term.
RADIANT HEAT EXCHANGE BETWEEN NON BLACK BODIES
The calculation of radiation heat exchange between black surfaces is relatively easy because the energy incident upon a black surface is totally absorbed by it. The radiation heat exchange between two black bodies can be easily calculated once the shape factor between them has been determined. The radiation exchange between real surfaces is much more complex since not all the incident energy is absorbed a part is reflected back and part may be reflected out of the system entirely. The reflection can be neglected if the absorptivity of the surface is larger than 0.9. but he problem becomes complicated if it involves surface of low absorptivity because the radiant energy can be reflected back and forth between the surface several times. A correct analysis should be take into account the effect of these multiple reflection.
RADIANT HEAT EXCHANGE BETWEEN SMALL GRAY BODIES
Consider two gray bodies 1 and 2 having emissivity’s 휖1 & 휖2 or absorptivites 훼1 & 훼2 . they are said to be small if their size is very small as compared to the distance between them. The radiation emitted by 1 is partly absorbed by 2. The portion of radiation unabsorbed and thus reflected on the first incidence is considered to be lost in space that is nothing return again to surface 1. The same can be said of surface 2 as well.
The energy emitted by body 1
4 = 퐴1휖1휎푇1
The energy incident on body 2
4 = 퐹12퐴1휖1휎푇1
HEAT TRANSFER 91 ______
The energy absorbed by body 2
4 = 훼2퐹12퐴1휖1휎푇1
According to Kirchhoff’s law
훼2 = 휖2
The energy transfer from 1 to 2
4 푄1 = 휖2퐹12퐴1휖1휎푇1
Similarly energy transfer from 2 to 1
4 푄2 = 휖2퐹21퐴2휖1휎푇2
The net radiant heat exchange between two bodies
4 4 푄12 = 휖2퐹퐴휖1휎(푇1 − 푇2 )
퐴퐹 = 퐴1퐹12 = 퐴2퐹21
RADIANT HEAT EXCHANGE BETWEEN LARGE PARALLEL GRAY PLANES
Consider two large parallel gray surfaces as shown in figure small distance apart and exchanging radiation. All the radiation emitted by one plane must reach and be intercepted by other plane. The shape factor is unity. And the effective areas of the surface are same.
Radiation emitted by 퐴1
4 = 휖1휎푇1
Radiation absorbed by 퐴2
4 = 훼2휖1휎푇1
Radiation reflected by 퐴2
4 = 휌2휖1휎푇1
Radiation absorbed by 퐴1
4 = 훼1휌2휖1휎푇1
92 HEAT TRANSFER ______
Radiation reflected by 퐴1
4 = 휌1휌2휖1휎푇1
Radiation absorbed by 퐴2
4 = 훼2휌1휌2휖1휎푇1
According to Kirchhoff’s law
훼2 = 휖2
4 = 휖2휌1휌2휖1휎푇1
So we can prove that energy absorbed on the third incidence by 퐴2
2 4 = (휌1휌2) 휖1휖2 휎푇1
The same equation is applied to surface 2 the net heat exchange for area A is given as
4 4 2 푄 = 휖1휖2 휎(푇1 − 푇2 ){1 + 휌1휌2 + (휌1휌2) +, , , , , , , , , , , , , , , }
2 −1 {1 + 휌1휌2 + (휌1휌2) +, , , , , , , , , , , , , , , } = (1 − 휌1휌2)
4 4 −1 푄12 = 휖1휖2 휎(푇1 − 푇2 )(1 − 휌1휌2)
4 4 휖1휖2 휎(푇1 − 푇2 ) 푄12 = (1 − 휌1휌2)
휌1 = 1 − 휖1
휌2 = 1 − 휖2
4 4 휖1휖2 휎(푇1 − 푇2 ) 푄12 = (1 − (1 − 휖1)(1 − 휖2))
4 4 휎(푇1 − 푇2 ) 푄 = 12 1 1 + − 1 휖1 휖2
RADIATION SHIELDS
HEAT TRANSFER 93 ______
Radiation heat transfer between two surfaces may be reduced either by using the materials which are highly reflective or by introducing radiation shields between them.
Fig shows two infinite parallel gray planes interchanging radiative energy between them without a radiation shield. For case a without a radiation shield.
4 4 푄12 휎(푇1 − 푇2 ) = 1 1 퐴 + − 1 휖1 휖2
For case (b) with radiation shields between the surfaces at equilibrium
4 4 4 4 푄13 푄32 휎(푇1 − 푇3 ) 휎(푇3 − 푇2 ) = = = 1 1 1 1 퐴 퐴 + − 1 + − 1 휖1 휖3 휖3 휖2
푇3 푖푠 푡ℎ푒 푒푞푢푖푙푖푏푟푖푢푚 푡푒푚푝푒푟푎푡푢푟푒 표푓 푡ℎ푒 푠ℎ푖푒푙푑
푇3 푖푠 푘푛표푤푛 푡ℎ푒 ℎ푒푎푡 푡푟푎푛푠푓푒푟 푟푎푡푒 푐푎푛 푏푒 푒푎푠푖푙푦 푐푎푙푐푢푙푎푡푒푑 .
If two parallel plates are of equal emissivity 훜
4 4 푄12 휎(푇1 − 푇2 ) = 푡ℎ푒푛 휖 = 휖 = 흐 1 1 1 2 퐴 + − 1 휖1 휖2
4 4 푄12 휎(푇1 − 푇2 ) = 1 1 퐴 + − 1 흐 흐
4 4 푄12 휎(푇1 − 푇2 ) = 2 퐴 − 1 흐
If the third plane placed between them is also has the same emissivity at equilibrium
4 4 4 4 푄13 휎(푇1 − 푇3 ) 푄32 휎(푇3 − 푇2 ) = = = 1 1 1 1 퐴 + − 1 퐴 + − 1 휖1 휖3 휖3 휖2
4 4 4 4 푄13 휎(푇1 − 푇3 ) 푄32 휎(푇3 − 푇2 ) = = = 2 2 퐴 − 1 퐴 − 1 흐 흐
At thermal equilibrium
94 HEAT TRANSFER ______
푄13 푄32 = 퐴 퐴
Therefore
4 4 4 4 푇1 − 푇3 = 푇3 − 푇2
4 4 4 2푇3 = 푇1 + 푇2
4 4 푇1 + 푇2 푇 4 = 3 2
4 4 4 푇1 + 푇2 휎 (푇1 − ) 푄13 푄32 2 = = 2 퐴 퐴 − 1 흐
4 4 푄13 푄32 1 (푇1 − 푇2 ) = = × 2 퐴 퐴 2 − 1 흐
푄12 1 푄12 ( ) = ( ) 퐴 푤𝑖푡ℎ 푠ℎ𝑖푒푙푑 2 퐴 푤𝑖푡ℎ표푢푡 푠ℎ𝑖푒푙푑
By the use of one radiation shield the net radiant heat transfer is reduced by 50%. The position of shield so long as it does not touch either of the planes does not alter its effectiveness.
If N shield are placed between two planes 1 and 2 , there would be (2N+2) surface resistance, two for each shield and one for each heat transfer surface, and (N+1) space resistance. The total resistance would thus be
1 − 휖 푅 ( 푤푖푡ℎ 푁 푠ℎ푖푒푙푑푠) = (2푁 + 2) + (푁 + 1)(1) 휖 2 = (푁 + 1) ( − 1) 휖
The resistance when no shield is present is
2 푅 ( 푤푖푡ℎ 푛표 푠ℎ푖푒푙푑푠) = ( − 1) 휖
The resistance with the shield in place is (푁 + 1) times as large as when the shields are absent. Thus
푄 1 푄 ( ) = ( ) 퐴 푤𝑖푡ℎ 푁 푠ℎ𝑖푒푙푑 (푁 + 1) 퐴 푤𝑖푡ℎ표푢푡 푠ℎ𝑖푒푙푑
NUMERICAL A black body emits radiation at 2000 K. calculate the monochromatic emissive power at 1 훍m wavelength also calculate the wavelength at which the emission is maximum and calculate the maximum emissive power.
Solution
HEAT TRANSFER 95 ______
From Planck’s law
−5 퐶1휆 퐸 = 푏휆 퐶2 푒휆푇 − 1
3.74 × 10−16(1 × 10−6)−5 퐸 = 푏휆 1.438×10−2 푒 10−6×2000 − 1
11 3 퐸푏휆 = 2.79 × 10 푊/푚
From Wiens displacement law
−3 휆푚푎푥푇 = 퐶표푛푠푡푎푛푡 = 2.898 × 10 푚퐾
2.898 × 10−3 휆 = = 1.449 × 10−6푚 푚푎푥 2000
−5 퐶1휆푚푎푥 (퐸 ) = 푏휆 푚푎푥 퐶2 푒휆푚푎푥푇 − 1
3.74 × 10−16(1.449 × 10−6)−5 (퐸푏휆)푚푎푥 = 0.01433 푒1.449×10−6×2000 − 1
11 3 (퐸푏휆)푚푎푥 = 4.11 × 10 푊/푚
Assuming the sun to be a black body having a surface temperature of 5800 K calculate the total emissive power and the wavelength at which the maximum spectral intensity occurs.
Solution
4 −8 4 퐸푏 = 휎 × 푇 = 5.67 × 10 × (5800)
7 2 퐸푏 = 6.42 × 10 푊/푚
−3 휆푚푎푥푇 = 퐶표푛푠푡푎푛푡 = 2.898 × 10 푚퐾
2.898 × 10−3 휆 = = 5 × 10−7푚 푚푎푥 5800
It is found that the intensity of radiation emitted by the sun is maximum at a wavelength of 0.5훍m assuming the sun to be black body estimate its surface temperature and emissive power.
Solution
According to Wiens displacement law
−3 휆푚푎푥푇 = 퐶표푛푠푡푎푛푡 = 2.898 × 10 푚퐾
96 HEAT TRANSFER ______
−6 휆푚푎푥 0.5 × 10 푇 = = = 5780퐾 퐶표푛푠푡푎푛푡 2.898 × 10−3
According to Stefan’s Boltzmann law
4 −8 4 퐸푏 = 휎 × 푇 = 5.67 × 10 × (5780)
7 2 퐸푏 = 6.33 × 10 푊/푚
A gray surface is maintained at temperatures of 8270퐶. If the maximum spectral emissive power at that temperature is 1.37 × 1010 푊/ 푚3 determine the emissivity of the body and wavelength corresponding to the maximum spectral intensity of radiation.
Solution
퐸푏휆푚푎푥 = 푐표푛푠푡푎푛푡 = 1.307 × 10−5 푊/푚2퐾4 푇5
−5 5 퐸푏휆푚푎푥 = 1.307 × 10 × 1100
10 3 퐸푏휆푚푎푥 = 2.1 × 10 푊/푚
10 퐸휆푚푎푥 1.37 × 10 휖 = = 10 = 0.65 퐸푏휆푚푎푥 2.1 × 10
According to Wiens law
−3 휆푚푎푥푇 = 푐표푛푠푡푎푛푡 = 2.898 × 10 푚퐾
2.898 × 10−3 휆 = 푚푎푥 푇
2.898 × 10−3 휆 = = 2.67흁풎 푚푎푥 1100
HEAT TRANSFER 97 ______
SECTION B UNIT 4 FORCED CONVECTION
Flow of fluid caused by pump, a fan or by the atmospheric winds. These mechanical devices provide a definite circuit for the circulating current and that speeds up the heat transfer rate. Examples of forced convection are cooling of internal combustion engine, air conditioning installation and nuclear reactors, condenser tubes and other heat exchanger equipment.
PRINCIPLE OF HEAT CONVECTION
Convection is the mode of heat transfer, in which energy is transferred between a surface and moving fluid. The convection heat transfer comprises of two mechanism. First is the transfer of energy due to random molecular motion and second is the energy transfer by bulk or macroscopic motion of the fluid. The molecules of the fluid are moving collectively or as aggregates and thus carry energy from a high temperature region to a low temperature region. Hence the heat transfer rate increases in presence of temperature gradient. The convection heat transfer is due to superposition of energy transfer by random motion of the molecules and by the bulk motion of the fluid.
BOUNDARY LAYER CONCEPT
The concept of boundary layer proposed by Prandtl forms the starting point for the simplification of equation of motion and energy. It has been successfully used in many practical problems. In this concept the flow field over a body is divided into two regions
The thin region near the body, called the boundary layer, where the velocity and temperature gradients are large and The region outside the boundary layer where velocity and temperature gradients are very nearly equal to the free stream value.
The thickness of the boundary layer has been arbitrarily defined as the distance from the surface at which the local velocity or temperature reaches to 99 percent of the external velocity or temperature. In general both the velocity boundary layer and thermal boundary layer will exist simultaneously.
VELOCITY BOUNDARY LAYER OR HYDRODYNAMIC BOUNDARY LAYER
98 HEAT TRANSFER ______
Consider a flow over a flat plate as shown in figure. The velocity here in front of the leading edge of the plate is uniform. Due to the no slip condition to be satisfied at the surface of the plate, the velocity is reduced to zero relative to the surface. This results in the retardation of the fluid particles in the adjoining fluid layers until at a distance 푦 = 훿 from the surface called as boundary layer thickness this effect becomes negligible. The deceleration of the fluid particle in the boundary layer is associated with shear stress 훕.
The effect of shear forces originating at the surface extends into the body of the fluids but with increasing distance, y, from the surface, the x velocity component of the fluid, u increases until
it approaches the free stream velocity, 푢∞. The effect of viscosity penetrates further into the free stream resulting in the growth of boundary layer downstream. The boundary layer thickness is
defined as the value of y for which 푢 = 0.99푢∞.
Because of large velocity gradient across the flow frictional shearing stress in the boundary layer are quite large even for the fluids with low viscosity. On the other had, these stress are very small outside the boundary layer. Thus the flow field can be conveniently divided into two parts the thin boundary layer region near the surface in which friction must be considered and an outer flow or potential flow region outside the boundary layer where frictional effect can be neglected and theory of perfect fluids can be used.
HEAT TRANSFER 99 ______
For flow over any surface, there will always exist a velocity boundary layer and hence surface friction. For external flows the velocity boundary layer provides the basis for determining the skin friction coefficient.
휏푠 퐶 = 푓 1 휌푢 2 2 ∞
This is very useful parameter for the calculation of surface frictional drag. The surface shear stress may also be determined by knowing the velocity gradient at surface.
풅풖 휏 = 흁 푠 풅풚
It is most essential to distinguish between the laminar and turbulent boundary layers. Initially, the boundary layer development is laminar as shown in fig for the flow over a flat plat. Depending upon the flow field the fluid properties at some critical distance from the leading edge small disturbances in the flow begin to get amplified, a transition process takes place and the flow becomes turbulent.
In laminar boundary layer, the fluid motion is highly ordered whereas the motion in the turbulent boundary layer is highly irregular with the fluid moving to and from in all direction. Due to fluid mixing resulting from this macroscopic motion, the boundary layer is thicker and the velocity profile in turbulent boundary layer is flatter than that in laminar flow.
The critical distance 푥푐 beyond which the flow cannot retain its laminar character is usually specified in terms of critical Reynolds number 푅푒푥,푐. Depending upon the surface and turbulence level of free stream.
휌푥푐푢∞ 푅푒 = = 5 × 105 푖푠 푛표푟푚푎푙푙푦 푎푠푠푢푚푒푑 푓표푟 푏표푢푛푑푎푟푦 푙푎푦푒푟 푐푎푙푐푢푙푎푡푖표푛 푥,푐 휇
In the turbulent boundary layer, as shown in figure there distinct regimes exists. A laminar sub layer, existing next to the wall has a nearly linear velocity profile. The convective transport in this layer is mainly molecular. In the buffer layer adjacent to sub layer, the turbulent mixing and diffusion effects are comparable. Then there is the turbulent core with large scale turbulence.
THERMAL BOUNDARY LAYER
A thermal boundary layer will develop if the surface temperature and free stream temperature are different. Fig shows the flow of fluid over a flat surface with surface temperature푇푠. At the leading edge the temperature profile is uniform with(푇 = 푇∞). The fluid particle coming into contact with the surface exchange thermal energy with those in the neighboring layer an thermal gradient is set up. With the increasing distance y from the surface the fluid temperature approaches the free stream temperature푇∞. The effects of the heat transfer penetrate further into the free stream resulting in the growth of thermal boundary layer thickness which is defined as the value of y 휃 for which the value of = 0.99 휃0
100 HEAT TRANSFER ______
푇푠 − 푇 휃 휃(푥, 푦) = = = 0.99 푇푠 − 푇∞ 휃0
Since at the surface there is no motion, the energy transfer takes place only by conduction, that is at y=0 the surface heat flux is given by the Fourier’s law.
Analogous to the concept of velocity boundary layer one can visualize the development of
thermal boundary layer with temperature varying from 푇푠 푡표 푇∞ in the boundary layer thickness 훿푡. Let us consider the flow of fluid at uniform temperature 푇∞ flows along a flat plate maintained at constant temperature of 푇푠
HEAT TRANSFER 101 ______
휃 Therefore at each location x along the plate one finds the a location 푦 = 훿푡 in a fluid where = 휃0 휃 0.99. The locus of such point where = 0.99 is called the thermal boundary layer. 휃0
FORCED CONVECTION INSIDE THE TUBES OR DUCTS
Let us consider a fluid entering a circular tube at uniform velocity. The fluid particles in the layer in contact with the surface at the tube will come to complete stop. The layer will cause the fluid particle in the adjacent layer to slow down gradually as a result of friction. To make up for this velocity reduction, the velocity of the fluid at the mid section of the tube will have to increase to keep the mass flow rate through the tube constant. As a result the velocity boundary layer develops along the tube. The thickness of the boundary layer increases in the flow direction until the boundary layer reaches the tube centre and thus fills the entire tube. The region from the tube inlet to the point at which the boundary layer merges at the centerline is called the hydrodynamic entry region, and the length of the this region is called the hydrodynamic entry length. The region beyond this entry region where the velocity profile is fully developed and remains unchanged is called the hydrodynamically developed. Region the velocity in this region is parabolic for laminar flow and some what flatter in turbulent flow.
Now let us consider that a fluid at a uniform temperature enters a circular tube with its wall at different temperature. The fluid particles in the layer in contact with the surface of the tube will assume the tube surface or wall temperature푇푤. This will initiates the convection heat transfer in the tube, and the development of the thermal boundary layer along the tube as shown in figure. The thickness of this boundary layer also increases in the flow direction until the boundary layer reaches the tube centre and thus fills the entire tube. The region of flow over which the thermal boundary layer develops and reaches the tube center is called the thermal entry region, and the length of this region is called the thermal entry length. The region beyond the thermal entry region in which the temperature profile remains unchanged is called thermally developed region.
The region in which the flow is both hydro dynamically and thermally developed is called the fully developed region.
102 HEAT TRANSFER ______
DIMENSIONAL ANALYSIS
Dimensional analysis is the method by which we deduce information about a phenomenon from the single premise that the phenomenon can be described by dimensionally correct equation among certain variables. With a little effort a partial solution of nearly any problem is obtained. On the other hand, complete solution is not obtained nor is the phenomenon revealed by dimensional reasoning alone. The major advantage of dimensional analysis is to reduce the number of independent variables in a problem with the help of dimensional analysis we can combine the variables of the problem into dimensionless group such as Reynolds number, Prandtl number, etc. and the experimental data can be very conveniently presented in terms of these number. Dimensional analysis does not give precise relationship among these non dimensional groups these relations or equations are obtained b correlating experimental data. Thus although dimensional analysis is very powerful tool yet without experimental data it is of no value. The prerequisite to the application of dimensional analysis is the knowledge of the physical variable which influences the given phenomenon. However, once these variables are known dimensional analysis to most problems can be carried out mechanically by set procedure.
HEAT TRANSFER 103 ______
QUANTITY SYMBOL UNIT DIMENSION Mass M Kg 푀1퐿0푡0푇0 Length l,x M 푀0퐿1푡0푇0 Time T Second 푀0퐿0푡1푇0 Velocity V m/s 푀0퐿1푡−1푇0 Acceleration A 푚/푠2 푀0퐿1푡−2푇0 Force F N = kg 푚/푠2 푀1퐿1푡−2푇0 Work W kg 푚2/푠2 푀1퐿2푡−2푇0 Energy or heat E,Q kg 푚2/푠2 푀1퐿2푡−2푇0 Power P kg 푚2/푠3 푀1퐿2푡−3푇0 Density 훒 푘푔/푚3 푀1퐿−3푡0푇0 Pressure P 푁/푚2 푀1퐿−1푡−2푇0 Viscosity 훍 푁푠/푚2 푀1퐿−1푡−1푇0 Kinematic viscosity 훎 푚2/푠 푀0퐿2푡−1푇0 0 2 −2 −1 Specific heat 퐶푝 퐽/푘푔퐾 푀 퐿 푡 푇 Thermal conductivity K 푊/푚퐾 푀1퐿1푡−3푇−1 Thermal diffusivity Α 푚2/푠 푀0퐿2푡−1푇0 Heat transfer coefficient H 푊/푚2퐾 푀1퐿0푡−3푇−1 Coefficient of thermal expansion 훽 1/퐾 푀0퐿0푡0푇−1
There are four primary dimension namely mass (M), length (L), time (t), and temperature (T). The dimensional dimensions of all other physical variable of any phenomenon can be obtained in terms of these four basic dimensions.
DIMENSIONAL ANALYSIS APPLIED TO FORCED CONVECTION
Let us now consider the case of fluid flowing across the heated tube. The various variable parameter pertinent to the problem along with their symbol and dimension are given in the below table.
QUANTITY SYMBOL UNIT DIMENSION Diameter D m 푀0퐿1푡0푇0 Fluid density 훒 푘푔/푚3 푀1퐿−3푡0푇0 Fluid velocity U m/s 푀0퐿1푡−1푇0 Fluid viscosity 훍 푁푠/푚2 푀1퐿−1푡−1푇0 0 2 −2 −1 Specific heat 퐶푝 퐽/푘푔퐾 푀 퐿 푡 푇 Thermal conductivity k 푊/푚퐾 푀1퐿1푡−3푇−1 Heat transfer coefficient h 푊/푚2퐾 푀1퐿0푡−3푇−1
According to the Buckingham 훱 theorem.
푛푢푚푏푒푟 표푓 푣푎푟푖푎푏푙푒 푖푛푣표푙푣푒푑 = 푛 = 7
푛푢푚푏푒푟 표푓 푓푢푛푑푎푚푒푛푡푎푙 푑푖푚푒푛푠푖표푛 = 푚 = 4
푎푐푐표푟푑푖푛푔 푡표 푏푢푐푘푖푛푔ℎ푎푚 Π 푡ℎ푒표푟푒푚 푡ℎ푒 푛푢푚푏푒푟 표푓 훱 푡푒푟푚 = 푛 − 푚 = 7 − 4 = 3
푎1 푏1 푐1 푑1 Π1 = 퐷 ρ 휇 푘 U
104 HEAT TRANSFER ______
푎2 푏2 푐2 푑2 Π2 = 퐷 ρ 휇 푘 퐶푝
푎3 푏3 푐3 푑3 Π3 = 퐷 ρ 휇 푘 h
Considering Π1 term and solving for obtaining the values of 푎1, 푏1, 푐1, 푑1 we get
[푀0퐿0푡0푇0] = [푀0퐿1푡0푇0]푎1 × [푀1퐿−3푡0푇0]푏1[푀1퐿−1푡−1푇0]푐1 × [푀1퐿1푡−3푇−1]푑1[푀0퐿1푡−1푇0]
0 = 0 + 푏1 + 푐1 + 푑1 + 0
0 = 푎1 − 3푏1 − 푐1 + 푑1 + 1
0 = 0 + 0 − 푐1 − 3푑1 − 1
0 = 0 + 0 + 0 − 푑1
On solving the above four equation we get
푑1 = 0
푐1 = −1
푏1 = 1
푎1 = 1
Putting all the above values in the first term of Π1 we get
1 1 −1 0 Π1 = 퐷 ρ 휇 푘 U 퐷휌푈 Π = 1 휇
휌푈퐷 Π = = 푅푒푦푛표푙푑푠 푛푢푚푏푒푟 1 휇
Considering Π2 term and solving for obtaining the values of 푎2, 푏2, 푐2, 푑2 we get
[푀0퐿0푡0푇0] = [푀0퐿1푡0푇0]푎2 × [푀1퐿−3푡0푇0]푏2[푀1퐿−1푡−1푇0]푐2 × [푀1퐿1푡−3푇−1]푑2[푀0퐿2푡−2푇−1]
0 = 0 + 푏2 + 푐2 + 푑2 + 0
0 = 푎2 − 3푏2 − 푐2 + 푑2 + 2
0 = 0 + 0 − 푐2 − 3푑2 − 2
0 = 0 + 0 + 0 − 푑2 − 1
On solving the above equation we get
푑2 = −1
푐2 = 1
푏2 = 0
HEAT TRANSFER 105 ______
푎2 = 0
Putting all the above values in the first term of Π2 we get
푎2 푏2 푐2 푑2 Π2 = 퐷 ρ 휇 푘 퐶푝
0 0 1 −1 Π2 = 퐷 ρ 휇 푘 퐶푝
휇퐶푝 Π = = 푃푟푎푛푑푡푙 푛푢푚푏푒푟 2 푘
Considering Π3 term and solving for obtaining the values of 푎3, 푏3, 푐3, 푑3 we get
[푀0퐿0푡0푇0] = [푀0퐿1푡0푇0]푎3 × [푀1퐿−3푡0푇0]푏3[푀1퐿−1푡−1푇0]푐3 × [푀1퐿1푡−3푇−1]푑3[푀1퐿0푡−3푇−1]
0 = 0 + 푏3 + 푐3 + 푑3 + 1
0 = 푎3 − 3푏3 − 푐3 + 푑3 + 0
0 = 0 + 0 − 푐3 − 3푑3 − 3
0 = 0 + 0 + 0 − 푑3 − 1
On solving the above equation we get
푑3 = −1
푐3 = 0
푏3 = 0
푎3 = 1
Putting all the above values in the term of Π3 we get
푎3 푏3 푐3 푑3 Π3 = 퐷 ρ 휇 푘 h
1 0 0 −1 Π3 = 퐷 ρ 휇 푘 h ℎ 퐷 Π = = 푁푢푠푠푒푙푡 푛푢푚푏푒푟 3 푘
퐹(Π1, Π2, Π3) 푐푎푛 푏푒 푒푥푝푟푒푠푠푒푠 푎푠
푁푢 = 휙(푅푒, 푃푟)
푁푢 푠푡푎푛푡표푛 푛푢푚푏푒푟 푆푡 = 푅푒. 푃푟
Dimensional analysis has thus shown us a way to reduce the seven significant variables of forced convection to three dimensionless parameter. We must now have experimental data in order to determine the functional relationship among these parameter.
DIMENSIONAL ANLYSIS APPLIED TO FREE CONVECTION
106 HEAT TRANSFER ______
Let us now consider the case of natural convection form a vertical plane wall to an adjacent fluid as shown in figure. The convection hat transfer coefficient, h would depends on the same variables as in forced convection excepting the velocity which is not caused by external agency. Instead the fluid circulation is caused by a buoyant force which comes into play because the density variation due to temperature difference.
In free convection the variable velocity is replace by temperature difference acceleration due to gravity and coefficient of thermal expansion.
QUANTITY SYMBOL UNIT DIMENSION Fluid density 훒 푘푔/푚3 푀1퐿−3푡0푇0 Fluid viscosity 훍 푁푠/푚2 푀1퐿−1푡−1푇0 0 2 −2 −1 Specific heat 퐶푝 퐽/푘푔퐾 푀 퐿 푡 푇 Fluid thermal conductivity K 푊/푚퐾 푀1퐿1푡−3푇−1 Coefficient of thermal expansion 훽 1/퐾 푀0퐿0푡0푇−1 Gravitational acceleration G 푚/푠2 푀0퐿1푡−2푇0 Temperature difference Δ푇 K 푀0퐿0푡0푇1 Specific length L M 푀0퐿1푡0푇0 Heat transfer coefficient H 푊/푚2퐾 푀1퐿0푡−3푇−1
Out of these nine variables normally β and g are taken together as single dimension as 푀0퐿1푡−2푇−1
푛푢푚푏푒푟 표푓 푣푎푟푖푎푏푙푒푠 = 푛 = 8
푛푢푚푏푒푟 표푓 푓푢푛푑푎푚푒푛푡푎푙 푑푖푚푒푛푠푖표푛 = 푚 = 4
푛푢푚푏푒푟 표푓 훱 푡푒푟푚 푖푛푣표푙푣푒푑 = 푛 − 푚 = 8 − 4 = 4
푒푎푐ℎ 훱 푡푒푟푚 푤푖푙푙 푐표푛푡푎푖푛 = 푚 + 1 푣푎푟푖푎푏푒푠
So according to Buckingham 훱 theorem
푎1 푏1 푐1 푑1 훱1 = 퐿 휌 휇 푘 Δ푇
HEAT TRANSFER 107 ______
푎2 푏2 푐2 푑2 훱2 = 퐿 휌 휇 푘 βg
푎3 푏3 푐3 푑3 훱3 = 퐿 휌 휇 푘 퐶푝
푎4 푏4 푐4 푑4 훱4 = 퐿 휌 휇 푘 ℎ
Following the same procedure outline in the last section we get the following result
퐿2휌2푘Δ푇 훱 = 1 휇3
퐿 휇 훽 푔 훱 = 2 푘
휇 퐶푝 훱 = = 푃푟 = 푝푟푎푛푑푡푙 푛푢푚푏푒푟 3 푘 ℎ 퐿 훱 = = 푁 = 푛푢푠푠푒푙푡 푛푢푚푏푒푟 4 푘 푢
Several experiment and analytical studies shows that the first two groups always appear together as a single dimensionless group. The parameter so formed is called the Grashof number.퐺푟
퐿2휌2푘Δ푇 퐿 휇 훽 푔 퐺푟 = 훱 훱 = × 1 2 휇3 푘
푔 훽 Δ푇 퐿3휌2 푔 훽 Δ푇 퐿3 퐺푟 = = 휇2 휈2
The free convection heat transfer data may be expressed in non dimensional form as
푁푢 = 휙(퐺푟 , 푃푟)
In natural convection the force is produced by buoyant effect resulting from a temperature difference these effects are included in Grashof number. This parameter replaces the Reynolds number in case of forced convection. Grashof number in free convection is as important as Reynolds number in forced convection. Also as seen above Stanton number has no significance in free convection.
SR. Velocity profile Boundary Boundary condtion 휹 푪풇 NO condition at y=0 at 풚 = 휹 1 푢 푦 푢 = 0 푢 = 푢 3.46푥 1.115 = ∞ 푢∞ 휹 √푅푒푥 √푅푒퐿 2 푢 푦 푦 2 푢 = 0 휕푢 5.48푥 1.46 = 2 ( ) − ( ) 푢 = 푢∞, = 0 푢∞ 휹 휹 휕푦 √푅푒푥 √푅푒퐿 3 푢 3 푦 1 푦 3 휕2푢 휕푢 4.64푥 1.292 = ( ) − ( ) 푢 = 푢 , = 0 푢 = 0, 2 = 0 ∞ 푢∞ 2 휹 2 휹 휕푦 휕푦 √푅푒푥 √푅푒퐿 4 푢 훱 푦 푢 = 0 푢 = 푢 4.795푥 1.31 = sin ( ) ∞ 푢 2 휹 ∞ √푅푒푥 √푅푒퐿 5 Blasius exact solution 5푥 1.328
√푅푒푥 √푅푒퐿
108 HEAT TRANSFER ______
DIMENSIONLESS NUMBER AND THEIR PHYSICAL SIGNIFICANCE
Reynolds Number:
It is the ration of inertia force to the viscous force.
휌푣퐷 = 푅푒푦푛표푙푑푠 푛푢푚푏푒푟 휇
Reynolds number is the indicative of relative importance of the inertial and viscous effect in a fluid motion. At low Reynolds number, the viscous effects dominates and fluid motion is laminar. At high Reynolds number, the inertial effects lead to turbulent flow and the associated turbulence level dominates the momentum and energy flux.
Reynolds number constitutes an important criterion of kinematics and dynamic similarities in forced convection heat transfer. Velocity within the given fields would be similar in magnitude and direction and turbulence pattern when their Reynolds numbers are same.
Grashof Number:
It indicates the relative strength of the buoyant to viscous forces. From its mathematical formulation
푔 훽 Δ푇 퐿3휌2 푔 훽 Δ푇 퐿3 퐺푟 = = 휇2 휈2
Obliviously the Grashof number represents the ratio of the product of buoyant and inertia forces to the square of the viscous forces. Grashof number has role in free convection similar to that played by Reynolds number in forced convection. Free convection is usually suppressed at sufficiently small Grashof number begins at some critical value depending on the arrangement and then becomes more and more effective as Grashof number increases.
Prandtl Number:
It is the indicative of relative ability of the fluid to diffuse momentum and internal energy by molecular mechanism. From its mathematical formulation.
휇 퐶푝 = 푃푟 = 푝푟푎푛푑푡푙 푛푢푚푏푒푟 푘 푘푖푛푒푚푎푡푖푐 푣푖푠푐표푠푖푡푦 휈 푃푟 = = 푡ℎ푒푟푚푎푙 푑푖푓푓푢푠푖푣푖푡푦 훼
Apparently the Prandtl number is the ratio of the kinematic viscosity to thermal diffusivity of the fluid. The kinematic viscosity indicates the momentum transport by molecular friction and thermal diffusivity represents the heat energy transport through the conduction. Obviously Prandtl number provides a measure of the relative effectiveness of momentum and energy transport by diffusion. For highly viscous oils Prandtl number is quite large and that indicates rapid diffusion of momentum by viscous action compared to the diffusion of energy. Prandtl number for gases is near
HEAT TRANSFER 109 ______unity and accordingly the momentum and energy transfer by diffusion are comparable. In contrast the liquid have Prandtl number from 0.003 to 0.01 and that indicates more rapid diffusion of energy compared to that momentum diffusion rate.
The Prandtl number is connecting link between the velocity field and the temperature field, and its value strongly influence the relative growth of the velocity and thermal boundary layers. For oils thermal boundary layer thickness is less than the velocity boundary layer thickness. For gases the thermal boundary layer thickness is equal to velocity boundary layer thickness. For liquid metal the thermal boundary layer thickness is equal to the velocity boundary layer thickness.
Nusselt Number:
Nusselt number establishes the relationship between the convective film coefficient h and the thermal conductivity of the fluid k and the significant length parameter l of the physical system.
ℎ 푙 = 푁푢푠푠푒푙푡 푛푢푚푏푒푟 푘
An energy balance at the surface of a heated plate stipulate that the energy transport by conduction must equal the convective heat transfer into the fluid flowing past the plate.
The Nusselt number is a convenient measure of the convective heat transfer coefficient. For given value of Nusselt number, the convective heat transfer coefficient is directly proportional to thermal conductivity of the fluid and inversely proportional to the significant length parameter.
Stanton number:
It is the ratio of heat transfer coefficient to the flow of heat per unit temperature rise due to velocity of the fluid. The Stanton number can be expressed in terms of the dimensionless number as
푁푢 푠푡푎푛푡표푛 푛푢푚푏푒푟 푆푡 = 푅푒. 푃푟
It should be noted that Stanton number can be used only in correlating the forced convection data. This becomes obvious when we observe the velocity V contained in the expression for Stanton number
Peclet Number:
It is the ratio of heat flow rate by convection to flow rate by conduction under a unit temperature gradient and through thickness l. The Peclet number is the function of Reynolds number and Prandtl number
푃푒 = 푅푒 × 푃푟
BULK MEAN TEMPERATURE AND MEAN FILM TEMPERATURE
The physical properties of a fluid are temperature dependant. Undoubtedly the accuracy of the results obtained by using the theoretical relation and the dimensionless empirical correlation would depends upon the temperature chosen for the evaluation of these properties. No uniform
110 HEAT TRANSFER ______
procedure has been attained in the selection of this reference temperature. However, It is customary to evaluate the fluid properties either on the basis of bulk mean temperature or mean film temperature.
The bulk mean temperature denotes the equilibrium temperature that would result if the fluid at a crossection was thoroughly mixed in an adiabatic container. For turbulent flow of fluids in ducts, these temperatures near the duct axis. In heat exchanger, the fluid flowing through the tubes passage. The bulk temperature is then taken to be the arithmetic mean of the temperature at inlet to and at exit from the heat exchanger tube.
푇𝑖 + 푇푒 푏푢푙푘 푚푒푎푛 푡푒푚푝푒푟푎푡푢푟푒 = 2
The mean film temperature is the arithmetic mean of the surface temperature of a solid and the undisturbed temperature of the fluid which flow past.
푇푠 + 푇∞ 푚푒푎푛 푓푖푙푚 푡푒푚푝푒푟푎푡푢푟푒 = 2
NUMERICAL Air at 250퐶 and at atmospheric pressure flows over a flat plate at 3푚/푠 if the plate is 1m wide and 0 the wall is maintained at 75 퐶, calculate the following at location 푥 = 1푚 and 푥 = 푥푐 from the leading edge of the plate find the following
a) Hydrodynamic and thermal boundary layer thickness b) Local and average friction coefficient c) Local and average heat transfer coefficient d) The total rate of heat transfer and e) Total drag force due to friction
The properties of air at bulk mean temperature of 500퐶 are density 1.093푘푔/푚3 specific heat 1.005 푘퐽/푘푔 퐾 and kinematic viscosity 17.95 × 10−6푚2/푠 and thermal conductivity is 0.0282 푊/ 푚퐾.
Solution:
푢∞ 푥 3 × 1 푚/푠 × 푚 푅푒 = = = = 푑푖푚푒푛푠푖표푛푙푒푠푠 푥 휈 17.95 × 10−6 푚2/푠
5 푅푒푥 = 167130.9192 푤ℎ푖푐ℎ 푖푠 풍풆풇풔풔 푡ℎ푎푛 5 × 10
a) Hydrodynamic and thermal boundary layer thickness
5푥 5 × 1 훿 = = = 0.0122푚 √푅푒푥 √167130.9192
훿 0.0122 훿 = = = 0.0137푚 푡 푃푟1/3 0.71/3
b) Local and average friction coefficient
HEAT TRANSFER 111 ______
0.664 0.664 −3 퐶푓푥 = = = 1.624 × 10 √푅푒푥 √167130.9192
1.328 1.328 −3 퐶푓 = = = 3.248 × 10 √푅푒퐿 √167130.9192 c) Local and average heat transfer coefficient
푘 ℎ = 0.332 (푅푒 )1/2(푃푟)1/3 푥 푥 푥 0.0282 ℎ = 0.332 (167130.9192)1/2(0.699)1/3 = 3.396 푊/푚2퐾 푥 1 푘 ℎ = 0.664 (푅푒 )1/2(푃푟)1/3 푥 푥 0.0282 ℎ = 0.664 (167130.9192)1/2(0.699)1/3 = 6.548 푊/푚2퐾 1
d) Total rate of heat transfer
푄 = ℎ 퐴 (푇푤 − 푇∞)
푄 = 6.548 × 1 × 1 × (75 − 25) = 327.4 푊
e) Total drag force due to friction
퐹퐷 = 휏푤 × 퐴 1 휏 = 퐶 × 휌 × 푢 2 × 푤 푓푥 ∞ 2 1 휏 = × 1.624 × 10−3 × 1.093 × 32 = 7.98764 × 10−3 푁/푚2 푤 2
퐹퐷 = 휏푤 × 퐴
Air at 200퐶 flows over a plate of 60푐푚 × 30푐푚 with the velocity of 20푚/푠 the critical Reynolds number is 5 × 105. Calculate the rate of heat transfer from the plate assuming the flow to be parallel to the side of 60푐푚. The plate temperature is maintained at 1000퐶
The properties of air at bulk mean temperature of 600퐶 are density1.06푘푔/푚3 specific heat 1.005 푘퐽/푘푔 퐾 and kinematic viscosity 18.97 × 10−6푚2/푠 and thermal conductivity is 0.0291 푊/ 푚퐾. solution
푢∞푥푐 20 × 푥푐 푅푒 = , 5 × 105 = = 0.47425푚 푥푐 휈 18.97 × 10−6
a) Hydrodynamic and thermal boundary layer thickness
112 HEAT TRANSFER ______
4.64 푥 4.64 × 0.47425 훿 = = = 3.112 × 10−3 5 √푅푒푥 √5 × 10
0.975 훿 0.975 × 3.112 × 10−3 훿 = = = 3.41726 × 10−3 푡 푃푟1/3 0.71/3
1/2 1/3 푁푢 = 0.332(푅푒푥) (푃푟)
5 1/2 1/3 푁푢 = 0.332(5 × 10 ) (0.7) = 208.44
ℎ푐 푥푐 푁 = 푢 푘
푁푢푘 208.44 × 0.0291 2 ℎ푐 = = = 12.78 푊/푚 퐾 푥푐 0.47425
The average heat transfer coefficient is
ℎ = 2ℎ푐
ℎ = 25.57 푊/푚2퐾
푄 = ℎ 퐴 (푇푤 − 푇∞)
푄 = 25.57 × 0.60 × 0.30 × (100 − 20)
Air at 200퐶 and at atmospheric pressure flows at a velocity of 4.5 푚/푠 past a flat plate with a sharp leading edge. The entire plate surface is maintained at a temperature of 600퐶. Assuming that transition occurs at a critical Reynolds number of 5 × 105, find the distance from the leading edge at which the boundary layer changer from laminar to turbulent. At the location calculate the following.
a) Thickness of hydrodynamic boundary layer. b) Thickness of thermal boundary layer c) Local and average convective heat transfer coefficients d) Heat transfer rate from both sides per unit width of the plate e) Mass entrainment in the boundary layer f) Skin friction coefficient
Assume the cubic velocity profile and approximate method. The themophysical properties of air at mean temperature of 400퐶 are density1.128푘푔/푚3 푃푟 = 0.7 and kinematic viscosity 16.97 × 10−6푚2/푠 and thermal conductivity is 0.02755 푊/푚퐾.
푢∞푥푐 푅푒 = 푥푐 휈
5 −6 푅푒푥푐 × 휈 5 × 10 × 16.96 × 10 푥푐 = = = 1.88푚 푢∞ 4.5
Thickness of hydrodynamic boundary layer
HEAT TRANSFER 113 ______
4.64 푥 4.64 × 1.88 훿 = = = 0.01234푚 5 √푅푒푥 √5 × 10
0.975 훿 0.975 훿 = = = 0.01355푚 푡 푃푟1/3 0.71/3
1/2 1/3 푁푢푥푐 = 0.332(푅푒푥푐) (푃푟)
ℎ푐 푥푐 = 0.332(5 × 105)1/2(0.7)1/3 푘
5 1/2 1/3 푘 ℎ푐 = 0.332(5 × 10 ) (0.7) × 푥푐 0.02755 ℎ = 0.332(5 × 105)1/2(0.7)1/3 × 푐 1.88
2 ℎ푐 = 3.05 푊/푚 퐾
Average heat transfer coefficient
2 ℎ = 2 × ℎ푐 = 6.10 푊/푚 퐾
The heat transfer from both sides per unit width of the plate is
푄 = ℎ (2퐴) (60 − 20)
푄 = 6.10 (2 × 1.88 × 1) (60 − 20)
푄 = 917.4 푊
Maximum entrainment in the boundary layer
5 푚̇ = × 휌 × 푢 × (훿 − 훿 ) 8 2 1 5 푚̇ = × 1.128 × 4.5 × (0.01234 − 0) = 0.039 푘푔/푠 8
Skin friction coefficient
0.664 0.664 퐶 = = = 9.136 × 10−4 푓푥 5 √푅푒푥 √5 × 10
A rectangular plat is 120 cm long in the direction of flow and 200 cm wide. The plate is maintained at 800퐶 when passed in nitrogen that has a velocity of 2.5 m/s and at a temperature of 00퐶 determine the average heat transfer coefficient and thotal heat transfer from plat. The properties of nitrogen at bulk mean temperature of 400퐶 are density1.142푘푔/푚3 퐶푝 = 1.04푘퐽/푘푔 퐾 and kinematic viscosity 15.63 × 10−6푚2/푠 and thermal conductivity is 0.0262 푊/푚퐾.
푢∞푥푐 푅푒 = 푥푐 휈
114 HEAT TRANSFER ______
5 −6 푅푒푥푐 × 휈 5 × 10 × 15.63 × 10 푥푐 = = = 3.126푚 푢∞ 2.5
Since the plate length is 120 cm in the flow direction, laminar flow persist in the entire length of the plate for which
1/2 1/3 푁푢 = 0.664푅푒 푃푟
Reynolds number for laminar flow
푢∞퐿 2.5 × 1.2 푅푒 = = = 191.938 휈 15.63 × 10−6
−6 휇 퐶푝 1.04 × 15.63 × 10 × 1.142 푃푟 = = = 0.708 푘 0.0262 × 10−3
1/2 1/3 푁푢 = 0.664 × (191.938) × (0.708) ℎ퐿 = 258.9 푘 푘 0.0262 ℎ = 258.9 = 258.9 × = 5.653 푊/푚2퐾 퐿 1.2
The rate of heat transfer from the plate
푄 = ℎ (퐿 푏)(푇푤 − 푇∞)
푄 = 5.653 (1.2 × 2)(80 − 0) = 1.085 푘푊
Calculate the rate of heat loss from a human body which may be considered as a vertical cylinder 30 cm in diameter and 175 cm high while standing in a 30 km/hr wind at 150퐶. The surface temperature of human body is 350퐶. The properties of air at bulk mean temperature of 250퐶 Are kinematic viscosity 15.33 × 10−6 푚2/푠, thermal conductivity 0.026 푊/푚퐾, Prandtl number 0.7
Solution:
Wind velocity=30 km/hr
30 × 1000 = = 8.33 푚/푠 3600 휌 × 푉 × 푑 푉 × 푑 푅 = = 푒 휇 휈
푉 × 푑 8.33 × 0.3 = = 163014 휈 15.33 × 10−6 ℎ푑 푁 = = 0.664 × (푅 )1/2 × (푃 )1/3 푢 푘 푒 푟
= 0.664 × (163014)1/2 × (0.7)1/3
HEAT TRANSFER 115 ______
ℎ푑 푁 = = 238.33 푢 푘 푘 ℎ = 238.33 × 푑 0.0263 ℎ = 238.33 × 0.3
ℎ = 20.89 푊/푚2퐾
Convective heat loss
푄 = ℎ × 퐴 × ∆푇
푄 = ℎ × 휋 ×× 푑 × 푙 × ∆푇
푄 = 20.89 × 휋 × 0.3 × 1.75 × (35 − 15)
푄 = 688.86 푊푎푡푡
Air moving at 0.3 m/s blows over the top of chest type freezer. The top of the freezer measures 0.9m by 1.5m and is poorly insulated so that the surface remains at 100퐶. If the temperature of the air is 300퐶 make calculation for the maximum heat transfer by forced convection from the top of the freezer. The properties of air at bulk mean temperature of 200퐶 are the kinematic viscosity15.06 × 10−6 푚2/푠, the thermal conductivity 0.0259 푊/푚퐾, Prandtl number 0.703
Solution
The maximum heat transfer occurs when the air flows is in the direction of shorter dimension
휌 × 푉 × 푙 푉 × 푙 푅 = = 푒 휇 휈
푅푒 = 17928 ℎ푙 푁 = = 0.664 × (푅 )1/2 × (푃 )1/3 푢 푘 푒 푟 ℎ푙 푁 = = 0.664 × (17928)1/2 × (0.703)1/3 푢 푘 ℎ푙 푁 = = 79.14 푢 푘 푘 ℎ = 79.14 × 푙 0.0259 ℎ = 79.14 × 0.9
ℎ = 2.28 푊/푚2퐾
116 HEAT TRANSFER ______
푄 = ℎ × 퐴 × ∆푇
푄 = 2.28 × 0.9 × 1.5 × (10 − 30)
푄 = 2.28 × 0.9 × 1.5 × (10 − 30)
푄 = −61.56 푊푎푡푡
Negative sign shows that heat is transfer towards the freezer.
The oil pan of engine approximates a flat plate 0.3m wide by 0.45 m long and protrudes below the framework of the automobile. The engine oil is at 950퐶 and the ambient air is 350퐶 if the automobile runs at 36 푘푚/ℎ푟 make the calculation for rate of heat transfer from the oil pan surface. Assume the negligible resistance to conduction through the oil pan. The properties of air at bulk mean temperature of 650퐶 are kinematic viscosity 18.46 × 10−6 푚2/푠 the thermal conductivity 0.0293 푊/푚퐾 Prandtl number 0.695
Solution
36 × 1000 = = 10 푚/푠 3600 휌 × 푉 × 푙 푉 × 푙 푅 = = 푒 휇 휈
10 × 0.45 = 18.46 × 10−6
5 푅푒 = 2.438 × 10 ℎ푙 푁 = = 0.664 × (푅 )1/2 × (푃 )1/3 푢 푘 푒 푟 ℎ푙 푁 = = 0.664 × (2.438 × 105)1/2 × (0.695)1/3 푢 푘 ℎ푙 푁 = = 290.81 푢 푘 푘 ℎ = 290.81 × 푙 0.0293 ℎ = 290.81 × 0.45
ℎ = 18.93 푊/푚2퐾
푄 = ℎ × 퐴 × ∆푇
푄 = 18.93 × 0.3 × 0.45 × (95 − 35)
푄 = 170.37 푊푎푡푡
HEAT TRANSFER 117 ______
A heat treat steel plate measures 3푚 × 1푚 and is initially at 300퐶 it is cooled by blowing air parallel to 1m edge at 9 푘푚/ℎ푟. If the air is at 100퐶, calculate the convective heat transfer from both sides of the plate the properties of air at bulk mean temperature of 200퐶 are the thermophysical properties of air are 15.06 × 10−6 푚2/푠 the thermal conductivity 0.0259 푊/푚퐾 the Prandtl number 0.703
Solution
9 × 1000 = = 2.5 푚/푠 3600 휌 × 푉 × 푙 푉 × 푙 2.5 × 1 푅 = = = 푒 휇 휈 15.06 × 10−6
푅푒 = 166003 ℎ푙 푁 = = 0.664 × (푅 )1/2 × (푃 )1/3 푢 푘 푒 푟 ℎ푙 푁 = = 0.664 × (166003)1/2 × (0.703)1/3 푢 푘 ℎ푙 푁 = = 240.83 푢 푘 푘 ℎ = 240.83 × 푙 0.0259 ℎ = 240.83 × 1
ℎ = 6.237 푊/푚2퐾
The convective heat transfer from both sides of the plate
푄 = ℎ × 2퐴 × ∆푇
푄 = 6.237 × 2 × 3 × 1 × (30 − 10)
푄 = 747.6 푊푎푡푡
Air flow through a 10 cm internal diameter tube at the rate of 75 푘푔/ℎ푟 . measurement indicate that at a particular point in the tube the pressure and temperature of air are 1.5 bar and 325 K respectively while the tube wall is at temperature of 375 K. make the calculation for the heat transfer rate from one meter length in the region of this point. The general non dimensional correlation for turbulent flow in the tube is
0.8 0.4 푁푢 = 0.023 × (푅푒) × (푃푟)
Where the fluid properties at bulk mean temperature of 325 K are dynamic viscosity 1.967 × 10−5푘푔/푚푠 the thermal conductivity 0.02792 푊/푚퐾 Prandtl number 0.713
118 HEAT TRANSFER ______
Solution
The properties may be assumed to be independent of pressure to an excellent approximation
휌 × 푉 × 푑 휌 × 푚푎푠푠 푓푙표푤 푟푎푡푒 × 푑 푅 = = 푒 휇 휌 × 푎푟푒푎 × 휇
푚푎푠푠 푓푙표푤 푟푎푡푒 × 푑 푅 = 푒 푎푟푒푎 × 휇
푚푎푠푠 푓푙표푤 푟푎푡푒 × 푑 푅 = 푒 휋 × 푑2 × 휇 4 4 × 푚푎푠푠 푓푙표푤 푟푎푡푒 푅 = 푒 휋 × 푑 × 휇
4 × (75⁄3600) 푅 = = 13492 푒 휋 × 0.1 × 1 × 1.967 × 10−5
This is well in excess of critical Reynolds number for flow in tubes the flow is turbulent and the given correlation applies
0.8 0.4 푁푢 = 0.023 × (푅푒) × (푃푟)
0.8 0.4 푁푢 = 0.023 × (13492) × (0.713) ℎ푑 푁 = = 40.46 푢 푘 푘 ℎ = 40.46 × 푑 0.02792 ℎ = 40.46 × = 11.296 푊/푚2퐾 0.1
푄 = ℎ × 퐴 × ∆푇
푄 = ℎ × 휋 × 푑 × 푙 × ∆푇
푄 = 11.296 × 휋 × 0.1 × 1 × (375 − 275)
푄 = 177.35 푊푎푡푡
What factor affects the value of convection coefficient for water flowing inside a circular tube within condenser shell water flows through one hundred thin walled circular tubes having diameter of 22.5 m and length of 5m. The mass flow rate of water is 65 kg/s. and its inlet and outlet temperature are known to be 220퐶 and 280퐶 respectively. Predict the average convective coefficient associated with water flow. Given that the properties of air at bulk mean temperature of 250퐶 the density 996.65 푘푔/푚3 the dynamic viscosity 903.01 × 10−6 푘푔/푚푠 the value of specific heat 4.1776 푘퐽/ 푘푔 퐾 thermal conductivity 2.1893 푘퐽/푚2ℎ푟 퐾
HEAT TRANSFER 119 ______
Solution
휇 × 퐶푝 푃 = 푟 푘
903.01 × 10−6 × 3600 × 4.1776 푃 = 푟 2.1893
푃푟 = 6.2
Mass flow of water through each tube
65 푚 = = 0.65 푘푔/푠 100 휌 × 푉 × 푑 휌 × 푚푎푠푠 푓푙표푤 푟푎푡푒 × 푑 푅 = = 푒 휇 휌 × 푎푟푒푎 × 휇
푚푎푠푠 푓푙표푤 푟푎푡푒 × 푑 푅 = 푒 푎푟푒푎 × 휇
푚푎푠푠 푓푙표푤 푟푎푡푒 × 푑 푅 = 푒 휋 × 푑2 × 휇 4 4 × 푚푎푠푠 푓푙표푤 푟푎푡푒 푅 = 푒 휋 × 푑 × 휇
4 × 0.65 푅 = 푒 휋 × 0.0225 × 903.06 × 10−6
푅푒 = 40749.78
This is well in excess of critical Reynolds number for flow through the tubes 2500 the flow is turbulent and the following condition applies
0.8 0.4 푁푢 = 0.023 × (푅푒) × (푃푟)
0.8 0.4 푁푢 = 0.023 × (40749.78) × (6.2) ℎ푑 푁 = = 232.69 푢 푘 푘 ℎ = 232.69 × 푑 2.1893 ℎ = 232.69 × = 22641 푘퐽/푚2ℎ푟 퐾 0.0225
The crank case of an automobile is approximated as 0.6 m long aand 0.2 m wide, and 0.1 m deep. Assuming that the surface temperature of the crank case is 350 K , estimate the rate of heat flow from the crank case to the atmosphere at 276 K at a road speed of 30 m/s. assume that the vibration of the engine and chasis induced the transmission from laminar to turbulent flow so near to the
120 HEAT TRANSFER ______
leading edge that for practical purpose the boundary layer is turbulent over the entire surface. Ignore radiation and for the front and rear surface use the same heat transfer coefficient as for the bottom and side. Use the following correlation
0.8 1/3 푁푢 = 0.036 푅푒 푃푟
The properties of air at bulk mean temperature are 휌 = 1.092 푘푔/푚3 휇 = 19.123 × 10−6 푁푠/푚2
푘 = 0.0265 푊/푚푘 푃푟 = 0.71
Solution
A crank case of an automobile
휌 × 푉 × 퐿 푅 = 푒 휇
1.092 × 30 × 0.6 푅 = = 1.03 × 106 푒 19.123 × 10−6
From the given relation average Nusselt number is
0.8 1/3 푁푢 = 0.036 푅푒 푃푟
6 0.8 1/3 푁푢 = 0.036 (1.03 × 10 ) (0.71) = 2075
The average convective heat transfer coefficient is
ℎ푙 푁 = = 2075 푢 푘
푁푢푘 2075 × 0.0265 ℎ = = = 91.6 푊/푚2퐾 푙 0.6
The surface area for the crank case
퐴 = (2 × 0.6 × 0.2) + (2 × 0.2 × 0.1) = 0.28 푚2
The rate of heat transfer
푄 = ℎ × 퐴 × ∆푇
푄 = 91.6 × 0.28 × (350 − 275) = 1898 푊푎푡푡
Air at 270퐶 and at 1 atm flows over heated plate with a velocity of 2 m/s. the plate is at uniform temperature of 600퐶. Calculate the heat transfer rate from the first 0.2 m of the plate and for the first 0.4m length of the plate. The properties of air at bulk mean temperature of 43.50퐶 are kinematic viscosity 17.36 × 10−6푚2/푠 thermal conducitivity 0.02749 푊/푚퐾 Prandtl number 0.7 and specific heat 1.006 퐽/푘푔퐾
Solution
HEAT TRANSFER 121 ______
휌 × 푉 × 퐿 푉 × 퐿 푅 = = 푒 휇 푣
The Reynolds number at distance 퐿 = 0.2푚
2 × 0.2 푅 = = 23041 푒 17.36 × 10−6
The rate of heat transfer at distance 퐿 = 0.2 푚
1/2 1/3 푁푢 = 0.332 × 푅푒 × 푃푟
1/2 1/3 푁푢 = 0.664 × (23041) × (0.7) ℎ퐿 푁 = = 0.664 × (23041)1/2 × (0.7)1/3 푢 푘 0.02749 ℎ = 0.664 × (23041)1/2 × (0.7)1/3 × 0.2
ℎ = 12.3 푊/푚2퐾
푄 = ℎ × 퐴 × ∆푇
푄 = 12.3 × 0.2 × (60 − 27) = 81.18 푊/푚 푥
The Reynolds number at distance 퐿 = 0.4푚
2 × 0.4 푅 = = 46082 푒 17.36 × 10−6
The rate of heat transfer at distance 퐿 = 0.4 푚
1/2 1/3 푁푢 = 0.332 × 푅푒 × 푃푟
1/2 1/3 푁푢 = 0.664 × (46082) × (0.7) ℎ퐿 푁 = = 0.664 × (46082)1/2 × (0.7)1/3 푢 푘 0.02749 ℎ = 0.664 × (46082)1/2 × (0.7)1/3 × 0.4
ℎ = 8.7 푊/푚2퐾
푄 = ℎ × 퐴 × ∆푇
푄 = 8.7 × 0.4 × (60 − 27) = 114.8 푊/푚 푥
Air at 100퐶 and at a pressure of 100 kPa is flowing over a plate at velocity of 3 m/s. if the plate is 30 cm wide and at a temperature of 600퐶 calculate the following quantities at a distance of 0.3 m,
122 HEAT TRANSFER ______
boundary layer thickness, local friction coefficient, local shearing stress, total drag force, thermal boundary layer thickness, local and average convective heat transfer coefficient, heat transfer from the plate. The properties of air at bulk mean temperature of 350퐶 are density 1.1373 푘푔/푚3 dynamic viscosity 19 × 10−6 푘푔/푚푠 thermal conductivity 0.0272 푊/푚퐾 , Prandtl number 0.7, specific heat 1.006 푘퐽/푘푔퐾.
휌 × 푉 × 푥 1.1373 × 3 × 0.3 푅 = = = 53872 푒 휇 19 × 10−6
The boundary layer thickness
5푥 5 × 0.3 훿 = = = 6.46 푚푚 √푅푒 √53872
Local friction coefficient
0.664 0.664 −3 퐶푓푥 = = = 2.783 × 10 √푅푒 √53872
Local shear stress
퐶푓푥 휏 = × 휌 × 푉2 푥 2
2.873 × 10−3 휏 = × 1.1373 × 32 = 0.0142 푁/푚2 푥 2
Total drag force
퐹 = 푎푣푒푟푎푔푒 푠ℎ푒푎푟 푠푡푟푒푠푠 × 푠ℎ푒푎푟 푎푟푒푎
푎푣푒푟푎푔푒 푠ℎ푒푎푟 푠푡푟푒푠푠 = 2 × 푙표푐푎푙 푠ℎ푒푎푟 푠푡푟푒푠푠
푎푣푒푟푎푔푒 푠ℎ푒푎푟 푠푡푟푒푠푠 = 2 × 0.0142 = 0.0284 푁/푚2
푎푟푒푎 = 푤 × 퐿 = 0.3 × 0.3 = 0.09 푚2
퐹 = 0.0284 × 0.09 = 2.564 × 10−3푁
Thermal boundary layer thickness
훿 6.46 훿푡ℎ = 1/3 = 1/3 = 7.091푚푚 푃푟 (0.7)
Local convective heat transfer coefficient
HEAT TRANSFER 123 ______
1/2 1/3 푁푢 = 0.664 × 푅푒 × 푃푟 ℎ푥 푁 = = 0.664 × (53872)1/2 × (0.7)1/3 푢 푘 0.0272 ℎ = 0.664 × (53872)1/2 × (0.7)1/3 × = 12.4 푊/푚2퐾 0.3
푄 = ℎ × 퐴 × ∆푇
푄 = 12.4 × 0.3 × 0.3 × (60 − 10) = 55.8 푊푎푡푡
Engine oil at 250퐶 is forced over a 30 cm by 20 cm plate at a velocity of 1.5 m/s. the flow is parallel to the side of 30 cm, which is heated to a uniform temperature of 550퐶. Calculate the rate of heat transfer from the plate to oil. Take the following physical properties of oil at 400퐶. The density of oil 876 푘푔/푚3 , kinematic viscosity 24 × 10−5 푚2/푠 , thermal conductivity 0.144 푊/푚퐾 , Prandtl number 2870. Use the following correlation
1/2 1/3 0.3387 × 푅푒 × 푃푟 푁푢 = 1/4 0.0468 2/3 {1 + ( ) } 푃푟
Solution
The Reynolds number
푉 × 푥 1.5 × 0.3 푅 = = = 1875 푒 푣 24 × 10−5
The Nusselt number is
0.3387 × (1875)1/2 × (2870)1/3 푁푢 = 1/4 = 208.4 0.0468 2/3 1 + ( ) { 2870 }
ℎ푙 푁 = = 208.4 푢 푘 푘 0.144 ℎ = × 208.4 = × 208.4 = 100.026 푊/푚2퐾 푙 0.3 average heat transfer coefficient
ℎ = 200.05 푊/푚2퐾
The rate of heat transfer
푄 = ℎ × 퐴 × ∆푇
푄 = 200.05 × 0.3 × 0.2 × (55 − 25) = 360.09 푊푎푡푡
124 HEAT TRANSFER ______
Air at a velocity of 3 m/s and at 200퐶 flows over a flat plate along its length. The length, width and thickness of the plate are 100 cm, 50 cm and 2 cm respectively. The top surface of the plate is maintained at 1000퐶. Calculate the heat loss by the plate and the temperature at the bottom surface to the plate for steady state condition. Thermal conductivity of the plate is taken as 23 푊/푚퐾 the properties of air at bulk mean temperature of 600퐶 are 휌 = 1.06 푘푔/푚3, kinematic viscosity 18.96 × 10−6푚2/푠, thermal conductivity 푘 = 0.02894 푊/푚퐾, Prandtl number 0.696
,퐶푝 = 1.005 푘퐽/푘푔퐾.
Solution
휌 × 푉 × 퐿 푉 × 퐿 3 × 1 푅 = = = = 1.58 × 105 푒 휇 푣 18.97 × 10−6
Hence the flow is laminar
1/2 1/3 푁푢 = 0.664 × 푅푒 × 푃푟 ℎ퐿 푁 = = 0.664 × (1.58 × 105)1/2 × (0.696)1/3 푢 푘 푘 ℎ = 0.664 × (1.58 × 105)1/2 × (0.696)1/3 × 퐿 0.02894 ℎ = 0.664 × (1.58 × 105)1/2 × (0.696)1/3 × 1
ℎ = 6.67 푊/푚2퐾
The rate of heat transfer
푄 = ℎ × 퐴 × ∆푇
푄 = 6.67 × 1 × 0.5 × (100 − 20) = 270.7 푊/푚
The temperature at bottom surface of the plate is obtained as
∆푇 푄 = 푘 × 퐴 × 푥
HEAT TRANSFER 125 ______
(푇푏 − 푇푠) 푄 = 푘 × 퐴 × 푥
(푇푏 − 100) 270.7 = 23 × 1 × 0.5 × 0.02
0 푇푏 = 100.47 퐶
A flat plat 1m wide and 1.5 m long is maintained at 900퐶 in air with a free stream temperature of 100퐶 flowing along the 1.5 m side of plate determine the velocity of air required to have a rate of energy dissipation of 3.75 kW. Use the following correlation
1/2 1/3 푁푢 = 0.664 × 푅푒 × 푃푟 푓표푟 푙푎푚푖푛푎푟 푓푙표푤
0.8 1/3 푁푢 = [0.036 × 푅푒 − 836] × 푃푟 푓표푟 푡푢푟푏푢푙푒푛푡 푓푙표푤
The properties of air at bulk mean temperature are found to be 휌 = 1.0877 푘푔/푚3,휇 = 2.029 × 10−5 푘푔/푚푠, thermal conductivity 0.028 푊/푚퐾, Prandtl number 0.703, specific heat 1.007 푘퐽/ 푘푔 퐾
Solution
The rate of heat transfer by convection is given by
푄 = ℎ × 퐴 × ∆푇
For both sides of the plate
푄 = ℎ × 2 × 퐴 × ∆푇
3.75 × 103 = ℎ × 2 × 1.5 × 1 × (90 − 10)
ℎ = 15.625 푊/푚2퐾
ℎ퐿 15.625 × 1.5 푁 = = = 837.05 푢 푘 0.028
Assuming the laminar flow along the plate we get
1/2 1/3 푁푢 = 0.664 × 푅푒 × 푃푟
1/2 1/3 837.05 = 0.664 × 푅푒 × (0.703)
6 푅푒 = 2.01 × 10
126 HEAT TRANSFER ______
Since Reynolds number is greater than 5 × 105 and hence the assumption we have made earlier is wrong the fluid flow is turbulent so using the relation for the turbulent flow we get
0.8 1/3 푁푢 = [0.036 × 푅푒 − 836] × 푃푟 푓표푟 푡푢푟푏푢푙푒푛푡 푓푙표푤
0.8 1/3 837.05 = [0.036 × 푅푒 − 836] × (0.703)
5 푅푒 = 7.36 × 10
Since Reynolds number is greater than 5 × 105 and hence fluid flow is turbulent
휌 × 푉 × 퐿 푅 = 푒 휇
2.029 × 10−5 × 7.36 × 105 푉 = = 9.15 푚/푠 1.0877 × 1.5
The local atmospheric pressure at Mahabaleshwar hill station in Maharashtra 1610 m above the sea level is 83.4 kPa. Air at this pressure and 200퐶 flows with the velocity of 8 푚/푠 over a 1.5 푚 × 6 푚 flat plate whose temperature is 1340퐶 determine the rate of heat transfer from the plate if the air flows parallel to 6m long side and 1.5 m long side. The properties of air at bulk mean temperature of 770퐶 are thermal conductivity 0.030 푊/푚퐾, dynamic viscosity 2.075 × 10−5 푘푔/푚푠, Prandtl number 0.697. use the following correlation
0.8 1/3 푁푢 = [0.037 × 푅푒 − 871] × 푃푟
Solution
푃 83.4 휌 = = = 0.991 푘푔/푚3 푅푇 0.287(20 + 273)
When air flows parallel to 6 m side
휌 × 푉 × 퐿 0.991 × 8 × 6 푅 = = = 2292434 푒 휇 2.075 × 10−5
Since Reynolds number is greater than 5 × 105 and hence fluid flow is turbulent
0.8 1/3 푁푢 = [0.037 × 푅푒 − 871] × 푃푟
ℎ퐿 푁 = = [0.037 × (2292434)0.8 − 871] × (0.697)1/3 푢 푘 0.030 ℎ = [0.037 × (2292434)0.8 − 871] × (0.697)1/3 × = 16.25 푊/푚2퐾 6
The heat transfer from the plate is
푄 = ℎ × 푤 × 퐿 × ∆푇
푄 = 16.25 × 1.5 × 6 × (134 − 20) = 16672 푤푎푡푡
HEAT TRANSFER 127 ______
When air flows parallel to 1.5 m side
휌 × 푉 × 퐿 0.991 × 8 × 1.5 푅 = = = 5.73 × 105 푒 휇 2.075 × 10−5
Since Reynolds number is greater than 5 × 105 and hence fluid flow is turbulent
0.8 1/3 푁푢 = [0.037 × 푅푒 − 871] × 푃푟
ℎ퐿 푁 = = [0.037 × (5.73 × 105)0.8 − 871] × (0.697)1/3 푢 푘 0.030 ℎ = [0.037 × (5.73 × 105)0.8 − 871] × (0.697)1/3 × = 11.07 푊/푚2퐾 1.5
The heat transfer from the plate is
푄 = ℎ × 푤 × 퐿 × ∆푇
푄 = 11.07 × 1.5 × 6 × (134 − 20) = 11358 푤푎푡푡
An air stream at 00퐶 is flowing along a heated plate at 900퐶 at a speed of 75 푚/푠 the plate is 45 cm long and 60 cm wide. Assume that the transition of the boundary layer takes place at critical Reynolds number of 5 × 105 calculate the average value of the friction coefficient and heat transfer coefficient for the full length of the plate. Also calculate the heat dissipation from the plate. The properties of air at bulk mean temperature of 450퐶 are
Density 1.113 푘푔/푚3 the specific heat 1.007 푘퐽/푘푔 퐾, the dynamic viscosity 1.928 × 10−5 푘푔/푚푠,
Thermal conductivity 0.0276 푊/푚퐾, Prandtl number 0.693.
0.8 1/3 푁푢 = [0.037 × 푅푒 − 871] × 푃푟
Solution
휌 × 푉 × 퐿 1.113 × 75 × 0.45 푅 = = = 1.95 × 106 푒 휇 1.928 × 10−5
Since Reynolds number is greater than 5 × 105 and hence fluid flow is turbulent
−1/5 1742 퐶푓푥 = 0.074 × 푅푒 − 푅푒
128 HEAT TRANSFER ______
1742 퐶 = 0.074 × (1.95 × 106)−1/5 − = 3.19 × 10−3 푓푥 1.95 × 106
6 0.8 1/3 푁푢 = [0.037 × (1.95 × 10 ) − 871] × (0.693) ℎ퐿 푁 = = 2754.4 푢 푘 2754.4 × 0.0276 ℎ = = 168.94 푊/푚2퐾 0.45
The heat transfer from both sides of the plate
푄 = ℎ × 2 × 퐴 × ∆푇
푄 = ℎ × 2 × 푤 × 퐿 × ∆푇
푄 = 168.94 × 2 × 0.6 × 0.45 × (90 − 0)
푄 = 8210.34 푤푎푡푡
A 10 cm diameter steam pipe is exposes to atmospheric air at 40퐶. The outer surface of the pipe is at 1100퐶 and the air is flowing across the pipe at a velocity of 8 푚/푠 . Determine the rate of heat loss from the pipe per unit of its length. The properties of air at bulk mean temperature of 570퐶 are thermal conductivity 0.0283 푊/푚퐾, kinematic viscosity 1.86 × 10−5푚2/푠, Prandtl number 0.708. Use the following correlation
1/2 1/3 5/8 4/5 0.62 푅푒 × 푃푟 푅푒 푁푢 = 0.3 + 1/4 × [1 + ( ) ] 0.4 2/3 28200 [1 + ( ) ] 푃푟
Solution
푉 × 퐷 8 × 0.10 푅 = = = 43011 푒 푣 1.86 × 10−5
1/2 1/3 5/8 4/5 0.62 푅푒 × 푃푟 푅푒 푁푢 = 0.3 + 1/4 × [1 + ( ) ] 0.4 2/3 28200 [1 + ( ) ] 푃푟
4/5 0.62 (43011)1/2 × (0.708)1/3 43011 5/8 푁푢 = 0.3 + 1/4 × [1 + ( ) ] 0.4 2/3 28200 [1 + ( ) ] (0.708)
푁푢 = 196.34
HEAT TRANSFER 129 ______
ℎ퐷 푁 = = 196.34 푢 푘 196.34 × 0.0283 ℎ = = 55.56 푊/푚2퐾 0.1
푄 = ℎ × 퐴 × ∆푇
푄 = ℎ × 휋 × 퐷 × 퐿 × ∆푇
푄 = 55.56 × 휋 × 0.1 × 1 × (110 − 4) = 1850.35 푊푎푡푡
Air at 270퐶 flowing across the tube with a velocity of 25 푚/푠. The tube could be either square tube with side of 5 cm or a circular tube with a diameter of 5 cm. compare the rate of heat transfer in each case of the if the tube surface is at 1270퐶 use the following correlation
0.805 1/3 푁푢 = 0.027 × 푅푒 × 푃푟 푓표푟 푡ℎ푒 푐푦푙푖푛푑푒푟
0.675 1/3 푁푢 = 0.102 × 푅푒 × 푃푟 푓표푟 푡ℎ푒 푠푞푢푎푟푒 푡푢푏푒
The properties of air at bulk mean temperature of 770퐶 are density 0.955 푘푔/푚3 , thermal conductivity 0.03 푊/푚퐾, kinematic viscosity 20.92 × 10−6푚2/푠, Prandtl number 0.7, specific heat
퐶푝 = 1.009 푘퐽/푘푔퐾.
Solution
Considering cylindrical tube
푉 × 퐷 25 × 0.05 푅 = = = 5.975 × 104 푒 푣 20.92 × 10−6
0.805 1/3 푁푢 = 0.027 × 푅푒 × 푃푟 ℎ퐷 푁 = = 0.027 × 푅 0.805 × 푃 1/3 푢 푘 푒 푟 0.03 ℎ = 0.027 × (5.975 × 104)0.805 × (0.7)1/3 × 0.05
The rate of heat transfer from the cylinder surface per meter of the length
푄 = ℎ × 휋 × 퐷 × ∆푇 퐿
130 HEAT TRANSFER ______
푄 = 100.6 × 휋 × 0.05 × (127 − 27) = 1581푊/푚 퐿
Considering square tube
푉 × 퐿 25 × 0.05 푅 = = = 5.975 × 104 푒 푣 20.92 × 10−6
0.675 1/3 푁푢 = 0.102 × 푅푒 × 푃푟
4 0.675 1/3 푁푢 = 0.102 × (5.975 × 10 ) × (0.7) ℎ퐿 푁 = = 0.102 × 푅 0.805 × 푃 1/3 푢 푘 푒 푟 0.03 ℎ = 0.102 × (5.975 × 104)0.805 × (0.7)1/3 × = 91.02 푊/푚2퐾 0.05
The rate of heat transfer from the square tube per meter of the length
푄 = ℎ × 퐴 × ∆푇 퐿 푄 = 91.02 × 4 × 0.05 × (127 − 27) = 1820 푊/푚 퐿
Air stream at 270퐶 moving at 0.3 m/s across 100 watt incandescent bulb glowing at at 1270퐶 if the bulb is approximated by a 60 m diameter sphere, estimate the heat transfer rate and the percentage of power lost due to convection use the following correlation
0.6 푁푢 = 0.37 푅푒
The properties of air at bulk mean temperature of at 770퐶 are kinematic viscosity 2.09 × 10−5푚2/푠 and the thermal conductivity 0.03 푊/푚퐾
Solution
푉 × 퐷 0.3 × 0.06 푅 = = = 865.3 푒 푣 2.092 × 10−5
HEAT TRANSFER 131 ______
0.6 푁푢 = 0.37 푅푒
0.6 푁푢 = 0.37 (865.3) = 21.4 ℎ × 퐷 ℎ × 0.06 푁 = = = 21.4 푢 푘 0.03
ℎ = 10.7 푊/푚2퐾
The heat transfer rate
푄 = ℎ × 퐴 × (푇푠 − 푇∞)
푄 = 10.7 × 4 × 휋 × 0.062 × (127 − 27)
푄 = 12.10 푊
The percentage heat lost by forced convection
푄 12.10 = × 100 = × 100 = 12.1% 푃 100
Water at 200퐶 flows through a small tube 1mm in diameter at a uniform speed of 0.2 m/s the flow is fully developed at a point beyond which a constant heat flux of 6000 푊/푚2 is imposed. How much farther down the tube will the water reaches to 740퐶 as its hottest point. The properties of air at bulk mean temperature of 470퐶 are density 989 푘푔/푚3, dynamic viscosity 577 × 10−6 푘푔/ 푚푠, thermal conductivity 0.640 푊/푚퐾, Prandtl number 3.77
Solution
푄 = 푚 ̇× 퐶푝 × (푇표 − 푇𝑖)
푞 × 퐴푠푢푟푓푎푐푒 = 푚 ̇× 퐶푝 × (푇표 − 푇𝑖)
푞 × 퐴푠푢푟푓푎푐푒 푇표 = 푇𝑖 + 푚 ̇× 퐶푝
푚̇ = 푑푒푛푠푖푡푦 × 푐푟표푠푠푒푐푡푖표푛 푎푟푒푎 × 푣푒푙표푐푖푡푦 휋 푚̇ = 989 × × 퐷2 × 0.2 = 1.55 × 10−4푘푔/푠 4
퐴푠푢푟푓푎푐푒 = 휋 × 퐷 × 푙 = 휋 × 0.001 × 푙
132 HEAT TRANSFER ______
6000 × 휋 × 0.001 × 푙 70 = 20 + 1.55 × 10−4 × 4180
푙 = 1.86 푚
In a particular solar collector energy is collected by placing a tube at the focal line of a parabolic collector and passing fluid through the tube. The arrangement results in a uniform heat flux of 2000 푊/푚2 along the axis of the tube of diameter 60 mm. determine the length of tube required to heat water from 200퐶 to 800퐶, which flows at the rate of 0.001 푘푔/푠, and the surface temperature at the outlet of tube. The properties of water at bulk mean temperature are found to be 휇 = 352 × 10−6 푁푠/푚2, specific heat 4187 퐽/푘푔퐾, Thermal conductivity 0.67 푊/푚퐾, Prandtl
number 2.2 use the following correlation 푁푢 = 4.36
solution
푒푛푒푟푔푦 푏푎푙푎푛푐푒 푓표푟 푠표푙푎푟 푐표푙푙푒푐푡표푟
푡ℎ푒 푟푎푡푒 표푓 푠표푙푎푟 푒푛푒푟푔푦 푐표푙푙푒푐푡푖표푛 = 푟푎푡푒 표푓 푖푛푐푟푒푎푠푒 표푓 푒푛푡ℎ푎푙푝푦 표푓 푤푎푡푒푟
푞 × 푎푟푒푎 표푓 푐표푙푙푒푐푡표푟 = 푚̇ × 퐶푝 × (푇표 − 푇𝑖)
푞 × 휋 × 푑 × 푙 = 푚̇ × 퐶푝 × (푇표 − 푇𝑖)
2000 × 휋 × 0.06 × 푙 = 0.01 × 4187 × (80 − 20)
푙 = 6.66 푚
휌 × 푉 × 푑 휌 × 푚푎푠푠 푓푙표푤 푟푎푡푒 × 푑 푅 = = 푒 휇 휌 × 푎푟푒푎 × 휇
푚푎푠푠 푓푙표푤 푟푎푡푒 × 푑 푅 = 푒 푎푟푒푎 × 휇
푚푎푠푠 푓푙표푤 푟푎푡푒 × 푑 푅 = 푒 휋 × 푑2 × 휇 4 4 × 푚푎푠푠 푓푙표푤 푟푎푡푒 푅 = 푒 휋 × 푑 × 휇
4 × 0.01 푅 = = 602.8 푒 휋 × 0.06 × 352 × 10−6
The Reynolds number is less than 2500 hence the flow is laminar
푁푢 = 4.36 ℎ × 푑 푁 = = 4.36 푢 푘 0.67 ℎ = 4.36 × = 48.68 푊/푚2퐾 0.06
HEAT TRANSFER 133 ______
푄 = ℎ × 퐴 × (푇푠 − 푇푤)
푞 = ℎ × (푇푠 − 푇푤)
2000 = 48.68 × (푇푠 − 80)
0 푇푠 = 121.08 퐶
134 HEAT TRANSFER ______
UNIT 5 FREE CONVECTION
Circulation of bulk fluid motion is caused by changes in fluid density resulting from temperature gradients between the solid surface and the main mass of the fluid. The stagnant layer of fluid in the immediate vicinity of the hot body gets thermal energy for conduction. The energy thus transferred serves to increase the temperature and internal energy of fluid particle. Because of temperature rise, these particles become less dense and hence lighter than the surrounding fluid particle. The lighter fluid particles move upward to a region of low temperature where they mix with and transfer a part of their energy to cold particles. Simultaneously the cool heavier particle descends downwards to fill the space vacated by the warm fluid particle. The circulation pattern, upward movement of cool fluid, is called the convection current. These currents are set up naturally due to gravity alone and are responsible for heat convection.
When the fluid is heated, the density gradients are developed, and results in buoyancy force, which induces free convection. Such situation is referred to as free or natural convection. The
HEAT TRANSFER 135 ______buoyancy effect is developed due to the presence of fluid density gradients and body force. In free convection the fluid motion is setup by buoyancy force are much smaller than those associated with forced convection. Therefore the heat transfer rate in natural convection is also smaller. There are many situations, where the heat is transferred from a heater to heat a room, heat transfer from pipes, transmission lines, electric transformer, electric motor and electronic equipment are some typical examples.
Empirical correlation for free convection in enclosure
푚 푘푒푓푓푒푐푡𝑖푣푒 퐿 = 퐶(퐺푟 푃푟)푛 ( ) 푘 훿
The value of C , n ,m are calculated from the tables
NOTE 푚 푁푢 = 퐶(퐺푟 푃푟)
C=0.13 for turbulent flow m=1/3 if 퐺푟 푃푟 > 109
C=0.53 for laminar flow m=1/4 퐺푟 푃푟 < 109
EXAMPLE A vertical plate 5m high and 1.5 m wide has one of its surfaces insulated. The other surface is maintained at a uniform temperature of 400K is exposed to quiescent atmospheric air at 300K. Calculate the total rate of heat loss from the plate show that the Grashof number =8.317 × 1011 and use the following equation
136 HEAT TRANSFER ______
0.387푅푎1/6 푁 1/2 = 0.85 + 푓표푟 10−1 < 푅푎 < 1012 푢 [1 + (0.492/푃푟)9/16]8/27
The properties of air at bulk mean temperature of 350 K are 훎= 20.75 × 10−6푚2/푠 푃푟 = 0.697 and the thermal conductivity 0.03 푊/푚퐾
HEAT TRANSFER ACROSS THE VERTICAL AIR GAP Air at atmospheric pressure contain between two 0.5 m square vertical plate separated by a distance of 15mm. the temperatures of the plate are 1000퐶 and 400퐶 respectively. Calculate the free convection heat transfer across the air space. Also calculate the radiation heat transfer across the air space if force surface have 훜=0.2. Given that the properties of air at ambient temperature of 700퐶 are density =1.029 푘푔/푚3 coefficient of thermal expansion 1/343 per K. dynamic viscosity 2.043 × 10−5푘푔/푚푠 the thermal conductivity 0.0295 푊/푚퐾 Prandtl number is 0.7. The value of C=0.197 the value of n=1/4 the value of m= - 1/9.
Solution
Properties remains constant. Steady state condition exists. Plates are isothermal. Air properties are to be estimate at bulk mean temperature.
100 + 40 푏푢푙푘 푚푒푎푛 푡푒푚푝푒푟푎푡푢푟푒 푇 = = 700퐶 = 343퐾 푏 2 1 1 훽 = = = 2.915 × 10−3 퐾−1 푇푏 343 퐾
The Grashof and Prandtl number product is now calculated as,
푔 훽Δ푇 훿3 퐺푟 푃푟 = 푃푟 휈2
9.8 × 2.915 × 10−3 × (100 − 40) × (15 × 10−3)3 × (1.029)2 퐺푟 푃푟 = × 0.7 (2.043 × 10−5)2
퐺푟 푃푟 = 1.027 × 104
Following equation can be used for finding the value of effective thermal conductivity
푚 푘푒 퐿 = 퐶(퐺푟. 푃푟)푛 ( ) 푘 훿
−1/9 푘푒 0.5 = 0.197 × (1.027 × 10−4)1/4 × ( ) = 1.343 푘 0.015
The rate of heat transfer through the air gap having effective thermal conductivity of 1.343 k
푄 푘푒 ( ) = × Δ푇 퐴 푐표푛푣푒푐푡𝑖표푛 훿
1.343 × 0.0295 × 0.5 × 0.5 × (100 − 40) 푄 = = 39.62 푊 0.015
HEAT TRANSFER 137 ______
The radiation heat transfer can be calculated by the following equation
4 4 푄 휎 × (푇1 − 푇2 ) ( ) = 1 1 퐴 푟푎푑𝑖푎푡𝑖표푛 + − 1 휖1 휖2
Considering the emissivity of the plate to be the same
휖1 = 휖2 = 0.2
푄 5.67 × 10−8 × (3734 − 3134) ( ) = = 61.47 푊/푚2 1 1 퐴 푟푎푑𝑖푎푡𝑖표푛 + − 1 0.2 0.2
푄 = 61.47 × 0.5 × 0.5 = 15.37푊
HEAT TRANSFER ACROSS THE HORIZANTAL AIR GAP Two horizontal plates 20 cm on a side are separated by a distance of 1cm with air at 1 atmosphere in the space. The temperatures of the two plates are 1000퐶 for the lower and 400퐶 for the upper plate. Calculate the heat transfer across the air space. Given that the properties of air at bulk mean temperature of 700퐶 are density =1.029 푘푔/푚3 coefficient of thermal expansion 1/343 per K. dynamic viscosity 2.043 × 10−5푘푔/푚푠 the thermal conductivity 0.0295 푊/푚퐾 Prandtl number is 0.7. Take value of C=0.059 n=0.4,m=0
Solution
Properties remains constant. Steady state condition exists. Horizontal plates are isothermal. Air properties are to be evaluated at bulk mean temperature.
100 + 40 푏푢푙푘 푚푒푎푛 푡푒푚푝푒푟푎푡푢푟푒 푇 = = 700퐶 = 343퐾 푏 2 1 1 훽 = = = 2.915 × 10−3 퐾−1 푇푏 343 퐾
The product of Grashof and Prandtl number is evaluated on the basis of separating distance and is given by the following relation
푔 훽Δ푇 훿3 퐺푟 푃푟 = 푃푟 휈2
푔 훽Δ푇 훿3 퐺푟 = 푤ℎ푒푟푒 훿 푖푠 푡ℎ푒 푎푖푟 푔푎푝 푏푒푡푤푒푒푛 푡푤표 푝푙푎푡푒 휈2
9.8 × 2.915 × 10−3 × (100 − 40) × (0.01)3 × (1.029)2 퐺푟 푃푟 = × 0.7 = 3043 (2.043 × 10−5)2
푚 푘푒 퐿 = 퐶(퐺푟. 푃푟)푛 ( ) 푤ℎ푒푟푒 퐿 푖푠 푡ℎ푒 푙푒푛푔푡ℎ 표푓 푝푙푎푡푒 푡ℎ푒 푣푎푙푢푒 표푓 훿 푖푠 푡ℎ푒 푎푖푟 푔푎푝 푘 훿
0 푘푒 퐿 = 0.059(퐺푟. 푃푟)0.4 ( ) 푘 훿
138 HEAT TRANSFER ______
0 푘푒 0.2 = 0.059(3043)0.4 ( ) = 1.46 푘 0.01
푘푒 = 1.46 푘
푄 푘푒 ( ) = × Δ푇 퐴 푐표푛푣푒푐푡𝑖표푛 훿
푘푒퐴 푄 = × Δ푇 훿 1.46 × 0.0295 × 0.2 × 0.2 푄 = × (100 − 40) = 10.34 푊 0.01
HEAT TRANSFER ACROSS WATER LAYER Two 50 cm horizontal square plate are separated by a distance of 1cm. The lower plate is maintained at a constant temperature of 1000퐶 and the upper plate is maintained at a constant temperature of 800퐶. Water at atmospheric pressure occupies the space between the plates. Calculate the heat lost by lower plate. Given that the properties of water at bulk mean temperature of 900퐶 density 961 푘푔/푚3 kinematic viscosity 0.293 × 10−6 Pr=1.74 thermal conductivity 0.6804 푊/ 푚퐾 푤ℎ푒푟푒 퐶 = 0.13 푛 = 0.3 푎푛푑 푚 = 0
Solution
100 + 80 푏푢푙푘 푚푒푎푛 푡푒푚푝푒푟푎푡푢푟푒 푇 = = 900퐶 = 363퐾 푏 2 1 1 훽 = = = 2.7548 × 10−3 퐾−1 푇푏 363 퐾
푔 훽Δ푇 훿3 퐺푟 = 푤ℎ푒푟푒 훿 푖푠 푡ℎ푒 푎푖푟 푔푎푝 푏푒푡푤푒푒푛 푡푤표 푝푙푎푡푒 휈2
푔 훽Δ푇 훿3 9.8 × 2.7548 × 10−3 × (100 − 80)(0.01)3 퐺푟 = = = 6289424 휈2 (0.293 × 10−6)2
퐺푟. 푃푟 = 6289424 × 1.74 = 10943598.55
푚 푘푒 퐿 = 퐶(퐺푟. 푃푟)푛 ( ) 푤ℎ푒푟푒 퐿 푖푠 푡ℎ푒 푙푒푛푔푡ℎ 표푓 푝푙푎푡푒 푡ℎ푒 푣푎푙푢푒 표푓 훿 푖푠 푡ℎ푒 푎푖푟 푔푎푝 푘 훿
푤ℎ푒푟푒 퐶 = 0.13 푛 = 0.3 푎푛푑 푚 = 0
0 푘푒 퐿 = 0.13(퐺푟. 푃푟)0.3 ( ) 푘 훿
0 푘푒 0.50 = 0.13 × (10943598.55)0.3 ( ) = 16.81 푊/푚퐾 푘 0.01
푘푒 = 16.81 푊/푚퐾 푘
HEAT TRANSFER 139 ______
16.81 × 0.6804 푄 = × 0.50 × 0.50 × (100 − 80) = 5718.762W 0.01
HEAT TRANSFER FROM ISOTHERMAL VERTICAL PLATE A large vertical plate 4m high is maintained at 600퐶 and exposed to atmospheric air at 100퐶. Calculate the heat transfer if the plate is 10m wide. The properties of air at bulk mean temperature of 350퐶 are β=3.25 × 10−3 per K. The kinematic viscosity is 16.5 × 10−6푚2/푠. Pr=0.7 푘 = 0.02685 푊/푚퐾
Use the following correlation below
1/3 푁푢 = 0.10(퐺푟 푃푟)
More complicated relations can be used
0.670푅푎1/4 푁 = 0.68 + 푓표푟 푅푎 < 109 푢 [1 + (0.492/푃푟)9/16]4/9
0.387푅푎1/6 푁 1/2 = 0.825 + 푓표푟 10−1 < 푅푎 < 1012 푢 [1 + (0.492/푃푟)9/16]8/27
Solution
The properties remain constant. At steady state condition exists. Surface is isothermal. Air properties are to be calculated at the bulk mean temperature.
60 + 10 푏푢푙푘 푚푒푎푛 푡푒푚푝푒푟푎푡푢푟푒 푇 = = 350퐶 = 308퐾 푏 2 1 1 훽 = = = 3.2467 × 10−3 퐾−1 푇푏 308 퐾
푔 훽Δ푇 퐿3 퐺푟 푃푟 = 푃푟 휈2
9.8 × 3.2467 × 10−3 × (60 − 10) × 43 퐺푟 푃푟 = × 0.7 (16.5 × 10−6)2
푅푎 = 퐺푟 푃푟 = 2.62 × 1011
0.387푅푎1/6 푁 1/2 = 0.825 + 푓표푟 10−1 < 푅푎 < 1012 푢 [1 + (0.492/푃푟)9/16]8/27
0.387 × (2.62 × 1011)1/6 푁 1/2 = 0.825 + = 26.75 푢 [1 + (0.492/0.7)9/16]8/27
푁푢 = 716
ℎ퐿 푁 = 푢 푘
140 HEAT TRANSFER ______
푁푢 × 푘 716 × 0.02685 ℎ = = = 4.80 푊/푚2퐾 퐿 4
푄 = ℎ × 퐴 × Δ푇
푄 = 4.80 × 4 × 10 × (60 − 10) = 9606 푊
As an alternative we could employ the equation
1/3 푁푢 = 0.10(퐺푟 푃푟)
11 1/3 푁푢 = 0.10(2.62 × 10 ) = 639.9 푊
This gives the value about 10 percent lower than the calculated previously.
HEAT TRANSFER FROM HORIZANTAL TUBE IN WATER A 2cm diameter horizontal heater is maintained at a surface temperature of 380퐶 and submerged in water at 270퐶. Calculate the free convection heat loss per unit length of the heater. The properties of water at bulk mean temperature of 32.50퐶 are density = 995 푘푔/푚3 kinematic viscosity 0.657 × 10−6푚2/푠. Thermal conductivity0.6280 푊/푚퐾 use the following correlation to solve the above problem
1/4 푁푢 = 0.53(퐺푟 푃푟)
Solution
38 + 27 푏푢푙푘 푚푒푎푛 푡푒푚푝푒푟푎푡푢푟푒 푇 = = 32.50퐶 = 305.5퐾 푏 2 1 1 훽 = = = 3.2733 × 10−3 퐾−1 푇푏 305.5퐾
푔 훽Δ푇 푑3 퐺푟 푃푟 = 푃푟 휈2
9.8 × 3.2733 × 10−3 × (38 − 27) × 0.023 퐺푟 푃푟 = × 4.34 = 28382689.67 (0.657 × 10−6)2
퐺푟 푃푟 = 28382689.67
1/4 푁푢 = 0.53(28382689.67) = 38.68
ℎ푑 푁 = 푢 푘
푁푢 × 푘 38.68 × 0.6280 ℎ = = = 1151.752 푊/푚2퐾 푑 0.02
푄 = ℎ × 퐴 × Δ푇
푄/퐿 = 1151.752 × 훱 × 0.02 × (38 − 27) = 796.0334 푊
HEAT TRANSFER 141 ______
HEAT TRANFER FROM FINE WIRE IN AIR A fine wire having a diameter of 0.02mm is maintained at a constant temperature of 540퐶 by an electric current the wire is exposed to air at 1 atm and 00퐶. Calculate the electric power necessary to maintain the wire temperature is the length is 50cm. the properties of air at bulk mean temperature of 270퐶 are kinematic viscosity 15.69 × 10−6푚2/푠. Thermal conductivity0.02624 푊/푚퐾 Pr=0.708 use the following correlation
0.058 푁푢 = 0.675(퐺푟 푃푟)
Solution
54 + 0 푏푢푙푘 푚푒푎푛 푡푒푚푝푒푟푎푡푢푟푒 푇 = = 270퐶 = 300 퐾 푏 2 1 1 훽 = = = 3.3333 × 10−3 퐾−1 푇푏 300 퐾
푔 훽Δ푇 푑3 퐺푟 푃푟 = 푃푟 휈2
9.8 × 3.3333 × 10−3 × (54 − 0) × (0.02 × 10−3)3 퐺푟 푃푟 = × 0.708 = 4.05453 × 10−5 (15.69 × 10−6)2
퐺푟 푃푟 = 4.05453 × 10−5
0.058 푁푢 = 0.675(퐺푟 푃푟)
−5 0.058 푁푢 = 0.675 × (4.05453 × 10 )
푁푢 = 0.375
ℎ푑 푁 = 푢 푘
푁푢 × 푘 0.375 × 0.02624 ℎ = = = 492.6 푊/푚2퐾 푑 0.02 × 10−3
푄 = ℎ × 훱 × 퐷 × 퐿 × Δ푇
푄 = 492.6 × 훱 × 0.02 × 10−3 × 0.50 × (54 − 0) = 0.836 W
HEATED HORIZANTAL PIPE IN AIR A horizontal pipe of 0.3048m in diameter is maintained at temperature of 2500퐶 in a room where the ambient temperature of air is 150퐶. Calculate the free convection heat loss per meter of length.
The properties of air at bulk mean temperature of 132.50퐶 are kinematic viscosity 26.54 × 10−6푚2/푠. Thermal conductivity0.03406 푊/푚퐾 Pr=0.687 use the following correlation
1/4 푁푢 = 0.53(퐺푟 푃푟)
142 HEAT TRANSFER ______
0.387푅푎1/6 푁 1/2 = 0.60 + 푓표푟 10−5 < 퐺푟 푃푟 < 1012 푢 [1 + (0.559/푃푟)9/16]8/27
A simpler equation is also available but it is restricted to laminar range of 10−6 < 퐺푟 푃푟 < 109
0.518 (퐺푟. 푃푟)1/4 푁 = 0.36 + 푢 [1 + (0.559/푃푟)9/16]4/9
Solution
Properties remains constant. Steady state condition exist. Pipe surface is isothermal. Air properties are to be estimated at bulk mean temperature. Considering air as an ideal gas.
250 + 15 푏푢푙푘 푚푒푎푛 푡푒푚푝푒푟푎푡푢푟푒 푇 = = 132.50퐶 = 405.5 퐾 푏 2 1 1 훽 = = = 2.47 × 10−3 퐾−1 푇푏 405.5 퐾
푔 훽Δ푇 푑3 퐺푟 푃푟 = 푃푟 휈2
9.8 × 2.47 × 10−3 × (250 − 15) × 0.30483 퐺푟 푃푟 = × 0.687 = 1.571 × 108 (26.54 × 10−6)2
퐺푟 푃푟 = 1.571 × 108
1/4 푁푢 = 0.53(퐺푟 푃푟)
8 1/4 푁푢 = 0.53 × (1.571 × 10 ) = 59.4
ℎ푑 푁 = 푢 푘
푁푢 × 푘 59.4 × 0.03406 ℎ = = = 6.63 푊/푚2퐾 푑 0.3048
푄 = ℎ × 훱 × 퐷 × 퐿 × Δ푇
푄/퐿 = ℎ × 훱 × 퐷 × Δ푇
푄 = 6.63 × 훱 × 0.3048 × (250 − 15) = 1.49 kW 퐿
An alternative equation can be used to obtain the solution for the given problem
0.387푅푎1/6 푁 1/2 = 0.60 + 푓표푟 10−5 < 퐺푟 푃푟 < 1012 푢 [1 + (0.559/푃푟)9/16]8/27
0.387(1.571 × 108)1/6 푁 1/2 = 0.60 + 푢 [1 + (0.559/0.687)9/16]8/27
푁푢 = 64.7
HEAT TRANSFER 143 ______
The value of Nusselt number is about 8 percent higher
FREE CONVECTION FROM HORIZANTAL PLATES The average heat transfer coefficient is calculated for horizontal plate from the below equation
푚 푁푢 = 퐶(퐺푟 푃푟)
Constant heat flux
The experiments have produced the following correlation for constant heat flux on a horizontal plate. For the heated surface facing upward
1/3 8 푁푢 = 0.13(퐺푟 푃푟) 푓표푟 퐺푟 푃푟 < 2 × 10
And
1/3 8 11 푁푢 = 0.16(퐺푟 푃푟) 푓표푟 2 × 10 < (퐺푟 푃푟) < 10
For heated surface facing downward
1/5 6 11 푁푢 = 0.58(퐺푟 푃푟) 푓표푟 10 < (퐺푟 푃푟) < 10
EXAMPLE A metal plate of 0.609m in height forms the vertical wall of an oven and is at a temperature of 1710퐶. Within the oven is air at temperature of 93.40퐶 and atmospheric pressure. Assuming that 1/4 natural convection condition holds near the plate and that for this case 푁푢 = 0.548(퐺푟. 푃푟) find the mean heat transfer coefficient and the heat taken up by air per second per metre width. For air at 0 −6 −4 132.2 퐶 take = 33.2 × 10 푘푊/푚퐾 휇 = 0.232 × 10 푘푔/푚푠, 퐶푝 = 1.005 푘퐽/푘푔퐾. Assume air as an ideal gas and 푅 = 0.287 푘퐽/푘푔퐾.
Solution
−4 3 휇퐶푝 0.232 × 10 × 1.005 × 10 푃푟 = = = 0.7241 푘 33.2 × 10−6
푃푟 = 0.7241
171 + 93.4 푏푢푙푘 푚푒푎푛 푡푒푚푝푒푟푎푡푢푟푒 푇 = = 132.50퐶 = 405.2 퐾 푏 2 1 1 훽 = = = 2.47 × 10−3 퐾−1 푇푏 405.2 퐾
푔 훽Δ푇 퐿3 퐺푟 = 휈2
푃푉 = 푚푅푇
푃 = 휌푅푇
푃 101.325 휌 = = = 0.8713 푘푔/푚3 푅푇 0.287 × 405.2
144 HEAT TRANSFER ______
휇 0.232 × 10−4 휈 = = = 2.663 × 10−5푚2/푠 휌 0.8713
푔 훽Δ푇 퐿3 퐺푟 = 휈2
9.81 × 2.47 × 10−3 × (171 − 93.4) × 0.06093 퐺푟 = = 5.985 × 108 (2.663 × 10−5)2
퐺푟푃푟 = 5.985 × 108 × 0.7241
8 1/4 푁푢 = 0.548(5.985 × 10 × 0.7241) = 79.07
ℎ퐿 푁 = 푢 푘
−6 푁푢푘 79.07 × 33.2 × 10 ℎ = = = 4.181푊/푚2퐾 퐿 0.0609
푄 = ℎ × 퐴 × Δ푇
푄 = 4.181 × 0.609 × (171 − 93.4) = 197.5 W/m 푏
푏 = 푤푖푑푡ℎ
EXAMPLE The maximum allowable surface temperature of an electrically heated vertical plate of 15 cm high and 10 cm wide is 1400퐶. Estimate the maximum rate of heat dissipation from both sides of the plate in an atmosphere at 200퐶. The radiation heat transfer coefficient is 8.72 푊/푚2퐾. For air at 800퐶 take 훎=21.09 × 10−6푚2/푠 given that Pr=0.682 and thermal conductivity of 0.03 푊/푚퐾. Use 1/4 the following correlation 푁푢 = 0.59(퐺푟. 푃푟)
Solution
140 + 20 푏푢푙푘 푚푒푎푛 푡푒푚푝푒푟푎푡푢푟푒 푇 = = 800퐶 = 353 퐾 푏 2 1 1 훽 = = = 2.8328 × 10−3 퐾−1 푇푏 353 퐾
푔 훽Δ푇 퐿3 퐺푟푃푟 = 푃푟 휈2
9.81 × 2.8328 × 10−3 × (140 − 20) × 0.153 퐺푟푃푟 = × 0.682 = 17510650 (21.09 × 10−6)2
퐺푟푃푟 < 109
The flow is laminar
1/4 푁푢 = 0.59(17510650) = 38.166
HEAT TRANSFER 145 ______
ℎ퐿 푁 = 푢 푘
푁푢푘 38.166 × 0.03 ℎ = = = 7.6332 푊/푚2퐾 퐿 0.15
푄퐶표푛푣푒푐푡𝑖표푛 = 2 × ℎ × 퐴 × Δ푇
푄퐶표푛푣푒푐푡𝑖표푛 = 2 × 7.6332 × 0.15 × 0.10 × (140 − 20) = 27.48 W
푄푟푎푑𝑖푎푡𝑖표푛 = 2 × ℎ푟 × 퐴 × Δ푇
푄푟푎푑𝑖푎푡𝑖표푛 = 2 × 8.72 × 0.15 × 0.10 × (140 − 20) = 31.392W
EXAMPLE 15 cm outer diameter steel pipe lies 2m vertically and 8m horizontally in a large room with an ambient temperature of 300퐶. If the pipe surface is at 2500퐶, calculate the total rate of heat loss from the pipe to the atmosphere. Properties of air at 1400퐶 are density is 0.854푘푔/푚3, specific heat 1.01 푘퐽/푘푔퐾, thermal conductivity 0.0353푊/푚퐾, Pr=0.684 and 훎= 27.8 × 10−6푚2/푠. Use the correlation
1/4 푁푢 = 0.53(퐺푟. 푃푟) 푡ℎ푒 푓푙표푤 푖푠 푙푎푚푖푛푎푟
1 푁푢 = 0.13(퐺푟. 푃푟)3 푡ℎ푒 푓푙표푤 푖푠 푡푢푟푏푢푙푒푛푡
Solution
For vertical part
250 + 30 푏푢푙푘 푚푒푎푛 푡푒푚푝푒푟푎푡푢푟푒 푇 = = 1400퐶 = 413 퐾 푏 2 1 1 훽 = = = 2.4213 × 10−3 퐾−1 푇푏 413 퐾
푔 훽Δ푇 퐿3 퐺푟푃푟 = 푃푟 휈2
9.81 × 2.4213 × 10−3 × (250 − 30) × 23 퐺푟푃푟 = × 0.684 (27.8 × 10−6)2
퐺푟푃푟 = 0.0541 × 1012 × 0.684 = 3.7 × 1010
퐺푟푃푟 > 109
푡ℎ푒 푓푙표푤 푖푠 푡푢푟푏푢푙푒푛푡
1 10 푁푢 = 0.13(3.7 × 10 )3 = 432.7 ℎ퐿 푁 = 푢 푘
146 HEAT TRANSFER ______
푁푢푘 432.7 × 0.0353 ℎ = = = 7.752 푊/푚2퐾 퐿 2
푄 = ℎ × 퐴 × Δ푇
푄 = ℎ × 훱 × 퐷 × 퐿 × Δ푇
푄 = ℎ × 훱 × 0.15 × 2 × 220 = 1570 W
For horizontal part
푔 훽Δ푇 퐿3 퐺푟푃푟 = 푃푟 휈2
9.81 × 2.4213 × 10−3 × (250 − 30) × 0.153 퐺푟푃푟 = × 0.684 (27.8 × 10−6)2
퐺푟푃푟 = 2.287 × 107 × 0.684 = 1.567 × 107
푡ℎ푒 푓푙표푤 푖푠 푙푎푚푖푛푎푟
1/4 푁푢 = 0.53(퐺푟. 푃푟)
7 1/4 푁푢 = 0.53(1.567 × 10 ) = 33.33 ℎ 푑 푁 = 푢 푘
푁푢푘 33.33 × 0.035 ℎ = = = 7.77 푊/푚2퐾 푑 0.15
푄 = ℎ × 퐴 × Δ푇
푄 = 7.77 × 0.15 × 8 × 220 = 6.44 kW
PROBLEM FOR PRACTICE A nuclear reactor with its core constructed of parallel vertical plates 2.2m high and 1.45 m wide has been designed on free convection heating of liquid bismuth. The maximum allowable temperature of bismuth is 3400퐶. The temperature of the wall is 9600퐶 Calculate the maximum possible heat dissipation from both sides of each plate. For the convection coefficient the appropriate correlation is 1/3 0 푁푢 = 0.13(퐺푟. 푃푟) where the properties are evaluated at the mean film temperature of 650 퐶 4 3 for bismuth are density is 10 푘푔/푚 ,=3.12kg/m-h,푐푝 = 150.7 퐽/푘푔퐾, k=13.02푊/푚퐾.
A spherical heater of 20cm diameter at 60표퐶 is immersed in a tank of water at 20표퐶 determine the value of convective heat transfer coefficient . given the properties of water at bulk mean temperature of 40표퐶 are density=992.2 푘푔/푚3, kinematic viscosity 0.659 × 10−6 푚2/푠 prandtl number 4.34, thermal conductivity 0.633 푊/푚퐾, coefficient of thermal expansion 0.41 × 10−3 use the following correlation to determine the convective heat transfer coefficient.
1/4 푁푢 = 2 + 0.43 (퐺푟. 푃푟)
HEAT TRANSFER 147 ______
Solution:
푔 훽Δ푇 퐿3 퐺푟 = 휈2
푔 휌2훽Δ푇 퐿3 퐺푟 = µ2
9.81 × 0.41 × 10−3 × (60 − 20) × 0.203 퐺푟 = (0.659 × 10−6)2
= 2963684803 = 2.96368 × 109
퐺푟푃푟 = 2.96368 × 109 × 4.34 = 1.2862 × 1010
10 1/4 푁푢 = 2 + 0.43 (1.2862 × 10 ) = 146.80 ℎ푑 푁 = = 146.80 푢 푘 ℎ × 0.20 푁 = = 146.80 푢 0.633 ℎ × 0.20 = 146.80 0.633
ℎ = 464.64 푊/푚2퐾
A steam pipe of 50mm diameter and 2.5m long has been placed horizontally and exposed to still air at 250퐶. If the pipe wall temperature is 295표퐶. Determine the rate of heat loss at mean temperature of 160표퐶 the thermophysical properties of air at bulk mean temperature are thermal conductivity 3.74 × 10−2푊/푚퐾 kinematic viscosity 30.09 × 10−6푚2/푠 prandtl number 0.682 the coefficient of thermal expansion 0.00231 per Kelvin.
Solution:
푔 훽Δ푇 퐿3 퐺푟 = 휈2
9.81 × 2.31 × 10−3 × (295 − 25) 0.0503 퐺푟 = (30.09 × 10−6)2
퐺푟 = 844715.3554
퐺푟푃푟 = 844715.3554 × 0.682 = 576095.8724
For laminar flow over a horizontal cylinder the following correlation may be used
1/4 푁푢 = 0.53(퐺푟. 푃푟)
1/4 푁푢 = 0.53 × (576095.8724)
148 HEAT TRANSFER ______
ℎ푑 푁 = = 14.60 푢 푘 ℎ × 0.050 = 14.60 3.64 × 10−2
ℎ = 10.62 푊/푚2퐾
푄 = ℎ × 퐴 × ∆T
푄 = ℎ × 휋 × 푑 × 퐿 × ∆T
푄 = 10.63 × 휋 × 0.05 × 2.5 × (295 − 25) = 1127.07 W
A nuclear reactor with its core constructed of parallel vertical plates of 2.25m high and 1.5m wide has been designed on free convection heating of liquid bismuth metallurgical consideration limits the maximum surface temperature of the plate to 975표퐶 and lowest allowable temperature of bismuth is 325표퐶 estimate the maximum possible heat dissipation from both sides of each plate. The 1/3 appropriate correlation for the convection coefficient is 푁푢 = 0.13 (퐺푟 푃푟) given that the properties of liquid bismuth at bulk mean temperature of 650표퐶 are dynamic viscosity 3.12 푘푔/푚 ℎ푟 Density 10000 푘푔/푚3 specific heat 150.7 퐽/푘푔퐾 thermal conductivity 13.02 푊/푚퐾 coefficient of thermal expansion 0.00108 per K.
Solution
µ 퐶푝 푃푟 = 푘 3.12 × 150.7 푃푟 = = 0.0100 3600 × 13.02
푔 훽Δ푇 퐿3 푔 휌2훽Δ푇 퐿3 퐺푟 = = 휈2 µ2
9.81 × (10000)2 × 0.00108 × (975 − 325) × 2.253 퐺푟 = 3.12 2 ( ) 3600
퐺푟 = 1.0443 × 1016
퐺푟푃푟 = 1.0443 × 1016 × 0.0100 = 1.0443 × 1014
The flow is turbulent use the following correlation
1/3 푁푢 = 0.13 (퐺푟 푃푟)
14 1/3 푁푢 = 0.13 × (1.0443 × 10 ) ℎ푙 = 0.13 × (1.0443 × 1014)1/3 = 6121.99 푘
HEAT TRANSFER 149 ______
ℎ푙 = 6121.99 푘 ℎ × 2.25 = 6121.99 13.02
ℎ = 35425.9616 푊/푚2퐾
푄 = ℎ × 2 × 퐴 × ∆T
푄 = ℎ × 2 × 푙 × 푤 × ∆T
푄 = 35425.9616 × 2 × 2.25 × 1.5 × (975 − 325)
푄 = 155.431 푀푊
A thin wall duct of 0.5m diameter has been laid in an atmosphere of air at 15표퐶 and conveys the gas at 205표퐶 the boundary layer flow is laminar and convective heat transfer is given by
∆푇 0.25 ℎ = 1.37 ( ) 푙 where l is the length of duct in meter how this value of convective heat transfer coefficient compares with the computed from the following nondimensional correlation for the laminar flow natural convection for large vertical cylinder
1/4 푁푢 = 0.57 (퐺푟 푃푟)
Base your calculation on one meter length of duct. Also estimate the heat lost from duct given that the properties of air at bulk mean temperature of 110표퐶 are kinematic viscosity 24.10 × 10−6푚2/푠 thermal conductivity 31.94 푊/푚퐾 prandtl number 0.704
Solution
The convection coefficient of heat transfer is given by
∆푇 0.25 ℎ = 1.37 ( ) 푙
205 − 15 0.25 ℎ = 1.37 ( ) 1
ℎ = 5.086 푊/푚2퐾
1 1 훽 = = = 2.61 × 10−3 푇푏 110 + 273
푔 훽Δ푇 퐿3 퐺푟 = 휈2
9.81 × 2.61 × 10−3 × (205 − 15) × 13 퐺푟 = (24.10 × 10−6)2
150 HEAT TRANSFER ______
퐺푟 = 8375852688
퐺푟 푃푟 = 8375852688 × 0.704 = 5896600293
1/4 푁푢 = 0.57 (퐺푟 푃푟)
1/4 푁푢 = 0.57 × (5896600293) = 157.95 ℎ푑 푁 = = 157.95 푢 푘 ℎ × 1 = 157.95 31.94 × 10−3
ℎ = 5.044923 푊/푚2퐾
푄 = ℎ × 퐴 × ∆T
푄 = ℎ × 휋 × 푑 × 퐿 × ∆T
푄 = 5.044 × 휋 × 0.5 × 1 × (205 − 15) = 1505.66 푊푎푡푡
Estimate the heat transfer from 40 watt bulb at 125표퐶 to 25표퐶 in quiescent air approximate the bulb as a sphere of 50 mm diameter. What percentage of power is lost by free convection the appropriate correlation for convection coefficient is
1/4 푁푢 = 0.60 (퐺푟 푃푟)
Where the different parameter are evaluated at bulk mean temperature of 75표퐶 assume the characteristic length as diameter of the sphere. The properties of air at bulk mean temperature are kinematic viscosity 20.55 × 10−6푚2/푠 thermal conducitivity 0.03 푊/푚퐾 prandtl number 0.693 the value of coefficient of thermal expansion 0.00287 per Kelvin.
Solution
푔 훽Δ푇 퐿3 퐺푟 = 휈2
9.81 × 2.87 × 10−3 × (125 − 25) × 0.0503 퐺푟 = (20.55 × 10−6)2
퐺푟 = 833368.8529
퐺푟푃푟 = 833368.8529 × 0.693 = 577524.6151
1/4 푁푢 = 0.60 (퐺푟 푃푟)
1/4 푁푢 = 0.60 (577524.6151) ℎ푑 푁 = = 16.54 푢 푘
HEAT TRANSFER 151 ______
ℎ × 0.050 = 16.54 0.03
ℎ = 9.9241 푊/푚2퐾
푄 = ℎ × 퐴 × ∆T
푄 = 9.9241 × 4 × 휋 × 0.0252 × (125 − 25)
푄 = 7.79 푊푎푡푡 the percent of heat loss by convection
7.79 = 19.278 % 40
A hot square plate of 40 cm side at 1000퐶 is exposed to atmospheric air at 20표퐶 make the calucation for heat loss from the both the surface of plate if the plate is kept vertical and if the plate is kept horizontal.
The following correlation are suggested for vertical position of plate
1/3 푁푢 = 0.125 (퐺푟 푃푟)
The following correlation is suggested for horizontal position of plate
1/4 푁푢 = 0.72 (퐺푟 푃푟) 푓표푟 푢푝푝푒푟 푠푢푟푓푎푐푒
1/4 푁푢 = 0.35 (퐺푟 푃푟) 푓표푟 푙표푤푒푟 푠푢푟푓푎푐푒 표푓 푡ℎ푒 푝푙푎푡푒
Where the air properties are evaluated at bulk meant temperature of 60표퐶 the thermophysical properties of air are density 1.06 푘푔/푚3 thermal conductivity 0.028 푊/푚퐾 specific heat 1.008 푘퐽/푘푔 kinematic viscosity 18.97 × 10−6 푚2/푠 the coefficient of thermal expansion 0.003 per Kelvin.
Solution
µ × 퐶푝 휌 × 휈 × 퐶푝 푃푟 = = 푘 푘
−6 휌 × 휈 × 퐶푝 1.06 × 18.97 × 10 × 1.008 × 1000 푃푟 = = = 0.724 푘 0.028
For vertical positioning of plate
푔 훽Δ푇 퐿3 퐺푟 = 휈2
9.81 × 0.003 × (100 − 20) × 0.403 퐺푟 = (18.97 × 10−6)2
퐺푟 = 418721789.4
152 HEAT TRANSFER ______
퐺푟푃푟 = 418721789.4 × 0.724 = 303154575.6
The following correlation are suggested for vertical position of plate
1/3 푁푢 = 0.125 × (303154575.6) ℎ푙 푁 = = 83.97 푢 푘 ℎ × 0.40 = 83.97 0.028
ℎ = 5.87 푊/푚2퐾
푄 = ℎ × 2 × 퐴 × ∆T
푄 = 5.87 × 2 × 0.40 × 0.40 × (100 − 20)
푄 = 150.47 푊푎푡푡
When the plate is positioned horizantal
For upper surface
푔 훽Δ푇 퐿3 퐺푟 = 휈2
9.81 × 0.003 × (100 − 20) × 0.403 퐺푟 = (18.97 × 10−6)2
퐺푟 = 418721789.4
퐺푟푃푟 = 418721789.4 × 0.724 = 303154575.6
1/4 푁푢 = 0.72 (퐺푟 푃푟) 푓표푟 푢푝푝푒푟 푠푢푟푓푎푐푒
1/4 푁푢 = 0.72 × (303154575.6)
푁푢 = 95.00 ℎ푙 푁 = = 95.00 푢 푘 ℎ × 0.40 = 95.00 0.028
ℎ = 6.65 푊/푚2퐾
푄푢푝푝푒푟 푠푢푟푓푎푐푒 = ℎ × 퐴 × ∆푇
푄푢푝푝푒푟 푠푢푟푓푎푐푒 = 6.65 × 0.40 × 0.40 × (100 − 20) = 85.12 푊푎푡푡.
For lower surface
HEAT TRANSFER 153 ______
푔 훽Δ푇 퐿3 퐺푟 = 휈2
9.81 × 0.003 × (100 − 20) × 0.403 퐺푟 = (18.97 × 10−6)2
퐺푟 = 418721789.4
퐺푟푃푟 = 418721789.4 × 0.724 = 303154575.6
1/4 푁푢 = 0.35 (퐺푟 푃푟) 푓표푟 푙표푤푒푟 푠푢푟푓푎푐푒 표푓 푡ℎ푒 푝푙푎푡푒
1/4 푁푢 = 0.35 × (303154575.6) = 46.18 ℎ푙 푁 = = 46.18 푢 푘 ℎ × 0.40 = 46.18 0.028
ℎ = 3.23 푊/푚2퐾
푄퐿표푤푒푟 푠푢푟푓푎푐푒 = ℎ × 퐴 × ∆푇
푄푙표푤푒푟 푠푢푟푓푎푐푒 = 3.23 × 0.40 × 0.40 × (100 − 20)
푄푙표푤푒푟 푠푢푟푓푎푐푒 = 41.38 푊푎푡푡.
푄푡표푡푎푙 = 푄푢푝푝푒푟 푠푢푟푓푎푐푒 + 푄푙표푤푒푟 푠푢푟푓푎푐푒
푄푡표푡푎푙 = 85.12 + 46.18 = 126.50 푤푎푡푡.
Calculate the rate of heat loss from a human body which may be considered as vertical cylinder of 30 cm diameter 175 cm high in still air at 15표퐶 the skin temperature is 35표퐶 and emissivity at skin surface is 0.4. neglect the sweating and effect of clothing. Given that the properties of air at bulk mean temperature of 25표퐶 are kinematic viscosity 15.53 × 10−6 푚2/푠 thermal conductivity 0.0263 푊/푚퐾 Prandtl number 0.7
Solution
1 1 훽 = = = 3.3557 × 10−3 푇푏 25 + 273
푔 훽Δ푇 퐿3 퐺푟 = 휈2
9.81 × 0.00335 × (35 − 15) × 1.753 퐺푟 = (15.53 × 10−6)2
퐺푟 = 1.46054 × 1010
퐺푟푃푟 = 1.46054 × 1010 × 0.7 = 1.0222 × 1010
154 HEAT TRANSFER ______
Since the product of grashof and Prandtl is more than 109 hence the flow is turbulent
1/3 푁푢 = 0.13 (퐺푟 푃푟)
10 1/3 푁푢 = 0.13 (1.0222 × 10 )
푁푢 = 282.11 ℎ푙 푁 = = 282.11 푢 푘 ℎ × 1.75 = 282.11 0.0263
ℎ = 4.23 푊/푚2퐾
푄 = ℎ × 퐴 × ∆푇
푄 = ℎ × 휋 × 푑 × 퐿 × ∆푇
푄 = 4.23 × 휋 × 0.30 × 1.75 × (35 − 15)
푄 = 139.53 푊푎푡푡
A thin vertical duct of cylindrical crossection is 0.4 m in diameter. The duct carries a gas at 470K and surrounding air may be considered still at 290K. Determine the heat transfer rate from one meter length of the duct assuming that the boundary layer is laminar. The general non dimensional correlation for laminar flow, natural convection from large vertical cylinder
1/4 푁푢 = 0.56 (퐺푟 푃푟)
The fluid properties are evaluated at bulk mean temperature. The heat transfer coefficient is prescribe by the equation
∆푇 0.25 ℎ = 퐶 ( ) 푙
Where l is the length in meter calculate the value of c which would give same heat transfer rate. The properties of air at bulk mean temperature of gas are density 0.9315 푘푔/푚3 specific heat 1.012 푘퐽/푘푔 퐾 dynamic viscosity 22.106 × 10−6 푘푔/푚 푠 thermal conducitivity 0.03215 푊/푚퐾 coefficient of thermal expansion 0.002631 per Kelvin.
Solution
−6 µ × 퐶푝 22.016 × 10 × 1.012 × 1000 푃푟 = = = 0.693 푘 3.2215 × 10−2
푔 훽Δ푇 퐿3 퐺푟 = 휈2
푔 휌2훽Δ푇 퐿3 퐺푟 = µ2
HEAT TRANSFER 155 ______
9.81 × 0.93152 × 0.002631 × (470 − 290) × 13 퐺푟 = (22.016 × 10−6)2
퐺푟 = 8316704022
퐺푟푃푟 = 8316704022 × 0.693 = 5763475887 correlation for laminar flow, natural convection from large vertical cylinder
1/4 푁푢 = 0.56 (퐺푟 푃푟)
1/4 푁푢 = 0.56 (5763475887)
푁푢 = 154.29 ℎ푙 푁 = = 154.29 푢 푘 ℎ × 1 = 154.29 3.2215 × 10−2
ℎ = 4.97069 푊/푚2퐾
푄 = ℎ × 퐴 × ∆푇
푄 = 4.968 × 휋 × 0.4 × 1 × (470 − 290)
푄 = 1123.16 푊푎푡푡
∆푇 0.25 ℎ = 퐶 ( ) 푙
470 − 290 0.25 4.968 = 퐶 ( ) 1
푐 = 1.3566
BOILING HEAT TRANSFER
Heat transfer to the liquid is a convection process which involves a change of phase from liquid to vapour. In boiling process, the average liquid temperature may remain well below the saturation temperature 푇푠 with the wall temperature 푇푤 above saturation producing local boiling at the wall with subsequent condensation in the colder bulk of the liquid. This is known as subcooled boiling. Boiling in a liquid at saturation temperature is known as saturated or bulk boiling. In some design of evaporator the heating surface is submerged beneath a free surface of liquid. This is known as pool boiling. When the liquid flows through a tube with subcooled or saturated boiling this bounded convection is called forced convection boiling even though circulation may occurs either by density difference or by a pump in the fluid circuit.
156 HEAT TRANSFER ______
BOILING HEAT TRANSFER PHENOMENON
Boiling is a convection process involving the change in phase from liquid to vapour. Boiling may occurs when a liquid is in contact with the surface maintained at a temperature higher than the saturation temperature of the liquid. Heat is transferred from solid surface to the liquid according to the law.
푄 = ℎ퐴 (푇푠 − 푇푠푎푡)
훥푇푒 = (푇푠 − 푇푠푎푡) 푖푠 푘푛표푤푛 푎푠 푒푥푐푒푠푠 푡푒푚푝푒푟푎푡푢푟푒
If heat is added to the liquid form a submerged solid surface, the boiling process is referred to as pool boiling. In this process, the vapour produced may form bubble which grows and subsequently detached themselves from the surface rising to the free surface due to buoyancy effect.
A common example of pool boiling is the boiling of water in kettle on a stove. In contrast to the flow boiling or forced convection boiling occurs in a flowing stream and the boiling surface may itself be a portion of the flow passage. This phenomenon is generally associated with two phase flows through confined passages.
HEAT TRANSFER 157 ______
A necessary condition for occurrence of pool boiling is that the temperature of heating surface exceeds the saturation temperature of the liquid. The type of boiling is determined by the temperature of liquid. If the temperature of the liquid is below the saturation temperature, the process is called subcooled or local boiling. In local boiling the bubbles formed at the surface eventually condense in the liquid. If the liquid is maintained at saturation temperature, the process is called saturated or bulk boiling.
The boiling process depends on the nature of surface, thermo physical properties of fluid and vapour bubble dynamics as a result of the large number of variable involved, general equation describing the boiling process are not available.
There are different regimes of pool boiling in which the heat transfer mechanism differs radically. The temperature distribution in saturated pool boiling with a liquid vapour interface is as shown in figure.
The specific curve has been obtained from an electrically heated platinum wire submerged in water by varying its surface temperature and measuring the surface heat flux.
The six regimes are now described briefly.
Region I called as free convection zone where the excess temperature 훥푇푒 is very small and is equal to=50퐶. Here the liquid near the surface is superheated slightly the convection current circulates the liquid and evaporation takes place at the liquid surface.
Nucleate boiling exists in II and III region. As the excess temperature is 훥푇푒 is increased bubbles begin to form on the surface of the wire at certain localized spot. The bubble condensed in the liquid without reaching the liquid surface. Region II in fact is the beginning of nucleate boiling. As excess temperature is further increased bubbles are formed more rapidly and rise to the surface of liquid resulting in rapid evaporation. This is indicated in
158 HEAT TRANSFER ______
0 region III. Nucleate boiling exists upto 훥푇푒 = 50 퐶. This maximum heat flux is known as critical heat flux, occurs at a point A and is of the order of 1Mw/푚2. The trend of increasing heat flux with increase in excess temperature observed up to region III is reversed in region IV called the film boiling region. This is due to the fact that bubbles now form so rapidly that they blanket the heating surface with vapour film preventing the inflow of fresh liquid from taking their place. Now the heat must be transferred through this vapour film to the liquid to effect any further boiling. Since the thermal conductivity of vapour film is much less than that of the liquid then the value of heat flux must decrease
with increase of 훥푇푒. In region IV the vapour film is stabilized and the heating surface is completely covered by a vapour blanket and the heat flux is lowest as shown in region V.
The surface temperature required to maintained this stable film are high and under these condition a sizeable amount of heat is lost by the surface due to radiation as indicated in region IV. The phenomena of stable film boiling can be observed when a drop of water falls on red hot stove. The drop does not evaporate immediately but dances a few times on the stove. This is due to the formation of stable steam film at the interface between the hot surface and the liquid droplet. From figure it is clear that high heat transfer rate is associated with small values of excess temperature in the nucleate boiling regimes. The equipment used in designing to operate in this region only.
The critical heat flux at A in figure is also called boiling crisis because the boiling process beyond that point is unstable unless of course, point B is reached. The temperature at point B is extremely high and normal above the melting point of the solid. So if the heating of the metallic surface is not limited to point A, the metal may be get damaged or it may even melt. That is why the peak heat flux point is called the burnout point and an accurate knowledge of this point is very important. Our aim should be to operate the equipment close to this value never beyond it.
CONDENSATION HEAT TRANSFER
The process of condensation is reverse of boiling. Whenever a saturated vapour comes in contact with the surface at lower temperature, condensation occurs. There are two modes of condensation. Filmwise in which the condensate wets the surface forming a continuous film which covers the entire surface and dropwise in which the vapour condenses into small liquid droplets of various sizes which falls down the surface in a random fashion. Filmwise condensation occurs on clean uncontaminated surfaces. In this type of condensation the film covering the entire surface grows in thickness as it moves down the surface by gravity. There exists a thermal gradient in the film and so it acts a resistance to heat transfer.
In dropwise condensation a large portion of the area of the plate is directly exposed to the vapour, making heat transfer rate much larger than those in filmwise condensation. Although dropwise condensation would be preferred to filmwise condensation yet it is extremlely difficult to achieve or maintain. This is because most surface becomes wetted after being exposed to condensing vapours over the period of time. Dropwise condensation can be obtained under controlled condition with the help of certain additives to the condensate and various surface coating but its commercial viability has not been proved. For this reason, condensing equipment in use is designed on the basis of filmwise condensation.
HEAT TRANSFER 159 ______
Calculate the cooling capacity by natural convection in air of heat sink that has four rectangular thin fins size 20 푚푚 × 25 푚푚. The fins may be assumed to have a constant surface temperature 600퐶 in ambient air at 200퐶. Take the fin efficiency as 60 % . the properties of air at bulk mean temperature of 400퐶 are density 1.128 푘푔/푚3 , specific heat 1005 퐽/푘푔퐾, dynamic viscosity 1.91 × 10−5 푘푔/푚푠, thermal conductivity 0.0276 푊/푚퐾, kinematic viscosity 16.96 × 10−6 푚2/푠, Prandtl number 0.70 use the following correlation
2
1/6 0387 × (푅푎) 푁푢 = 0.825 + 8/27 0.492 9/16 {1 + ( ) } [ 푃푟 ]
Solution
Let the characteristic length is 20 mm
1 1 훽 = = 푇푏 313
푔 훽Δ푇 퐿3 퐺푟 = 휈2 1 9.81 × × (60 − 20) × (0.02)3 퐺푟 = 313 = 34868 (16.96 × 10−6)2
푅푎 = 퐺푟 × 푃푟
푅푎 = 34868 × 0.70 = 24280
2
1/6 0387 × (푅푎) 푁푢 = 0.825 + 8/27 0.492 9/16 {1 + ( ) } [ 푃푟 ]
160 HEAT TRANSFER ______
2
0387 × (24280)1/6 푁푢 = 0.825 + 8/27 = 6.60 0.492 9/16 {1 + ( ) } [ 0.7 ]
ℎ × 퐿 푁 = = 6.60 푢 푘 푘 0.0276 ℎ = 6.60 × = 6.60 × = 9.11 푊/푚2퐾 퐿 0.02
푄 = ℎ × 퐴 × (푇푠 − 푇∞) × 푒푓푓푖푐푖푒푛푐푦 표푓 푓푖푛
푄 = ℎ × (4 × (2 × 푙 × 푤)) × (푇푠 − 푇∞) × 푒푓푓푖푐푖푒푛푐푦 표푓 푓푖푛
푄 = 9.11 × (4 × (2 × 0.02 × 0.025)) × (60 − 20) × 0.60
푄 = 0.875 푤푎푡푡
HEAT TRANSFER 161 ______
UNIT 6 HEAT EXCHANGER
A ‘heat exchanger’ may be defined as an equipment which transfers the energy from a hot fluid to a cold fluid, with maximum rate and minimum investment and running costs. In heat exchangers the temperature of each fluid changes as it passes through the exchangers, and hence the temperature of the dividing wall between the fluids also changes along the length of the exchanger.
Examples of heat exchangers:
(i) Intercoolers and preheaters ; (ii) Condensers and boilers in steam plant ; (iii) Condensers and evaporators in refrigeration units ; (iv) Regenerators ; (v) Automobile radiators ; (vi) Oil coolers of heat engine ; (vii) Milk chiller of a pasteurizing plant ; (viii) Several other industrial processes.
Heat exchanger is any device used for affecting the process of heat exchange between two fluids that are at different temperatures. The heat exchanger are useful in many engineering process like those in refrigerating and air conditioning system, power system, food processing system, chemical reactor and space or aeronautical application. A heat exchanger in which two fluids exchange heat by coming into direct contact is called a direct contact heat exchanger. Example of this type is open feed water heater, superheater and jet condenser.
Recuperators are heat exchanger in which the fluids are separate by a wall. The wall may be a simple plane wall or a tube or a complex configuration involving fins baffles and multiple passes of tubes. These units are also called surface heat exchanger, are more commonly used because they can be constructed with a large hat transfer surface in relatively small volume and are suitable for heating cooling, evaporating or condensing application. A periodic flow type of heat exchanger is called a regenerator. In this type of heat exchanger the same space is alternately occupies by the hot and cold gases between which heat is exchanged. Regenerator find their application in preheater for steam power plant, blast furnace oxygen producers etc. the heat exchange process is heat exchanger can be described by the principle of conduction, convection radiation and evaporation or condensation.
In order to meet the widely varying applications, several types of heat exchangers have been developed which are classified on the basis of nature of heat exchange process, relative direction of fluid motion, design and constructional features, and physical state of fluids.
Heat exchangers, on the basis of nature of heat exchange process, are classified as follows :
(i) Direct contact (or open) heat exchangers.
(ii) Indirect contact heat exchangers.
(a) Regenerators. (b) Recuperators.
CLASSIFICATION OF HEAT EXCHANGER
162 HEAT TRANSFER ______
CLASSIFICATION ACCORDING TO THE TRANSFER PROCESS
Direct contact type:
In this type of heat exchanger the tow immiscible fluids at different temperature come in direct contact. For heat exchange between two fluids, one fluid is sprayed through the other cooling tower; jet condenser and scrubber are best examples
Direct transfer type heat exchanger:
In this type of heat exchanger, the cold and hot fluid flow simultaneously through the device and the heat is transferred through the wall separating them. These types of heat exchanger are most commonly used in almost all fields of engineering.
Regenerator:
HEAT TRANSFER 163 ______
These are also called storage type of heat exchanger, in which hot and cold fluid flow alternatively on the same surface. The hot fluid gives heat to the surface and cold fluid extracts heat from it. In many applications a rotating disc type matrix is used, continuous flows of both the hot and cold fluid are maintained. These are preheater for steam power plant, blast furnaces, oxygen producer etc.
Recuperator:
These are also called transfer type heat exchanger. In these heat exchangers, the hot and cold fluids are separated by a plane wall or tube surfaces, hence heat is indirectly transferred from hot fluid to cold fluid by convection and conduction.
Compact heat exchanger:
These are special class of heat exchanger in which the heat transfer area per unit volume is greater than 700푚2/푚3. These heat exchanger have denser arrays of finned tubes or plates, when at least one of the fluid used is gas for example automobile radiator have are density of the order of 1100 푚2/푚3.
BASED ON CONSTRUCTION
Tabular heat exchanger:
These are also called as tube in tube or concentric tube or double pipe heat exchanger. These are widely used in many sizes and different flow arrangement and types.
Finned tube type:
When a high operating pressure or an enhanced heat transfer rate is required, the extended surfaces are used on one side of the heat exchanger. These heat exchangers are used for liquid to gas heat exchange. Fins are always used on gas side. The tube fins are used in gas turbine, automobile, airplanes, heat pumps, refrigeration , electronics, cryogenics air conditioning system, etc.
164 HEAT TRANSFER ______
BASED ON FLOW ARRANGEMENT
TYPES OF HEAT EXCHANGER
The heat exchanger may be classified in several ways. One classification is according to the fluid flow arrangement or the relative direction of hot and cold fluid. The fluid may be separated by plane wall but more commonly be a concentric tube arrangement as shown in figure. If both the fluids are moving in same direction as shown in fig (a). The arrangement is called the parallel flow type. In the counter flow arrangement the fluid moves in parallel but in opposite direction.
In double pipe heat exchanger either the hot or cold fluid occupies the annular space and other fluids moves through the inner pipe. Since both the fluids streams traverse the exchanger only once this arrangement is called single pass heat exchanger.
Another flow configuration is one in which the fluids move at right angles to each other through the heat exchanger. This type of arrangement as illustrated in figure called cross flow type. Cross flow heat exchanger are used in air or gas heating application where a gas is forced across a tube bundle through which another fluids moves. Two type of cross flow arrangement are possible.
HEAT TRANSFER 165 ______
In this gas flow across the finned tube bundle and is thus unmixed as it is confined in a separate channel between the fin. Similarly the other fluid passing through the tube is also unmixed. The heat exchanger is widely used in air conditioning application.
A different type of heat exchanger is as shown in figure (b). Here the tubes are without fins the gas flowing across the tubes is said to be mixed because it can move about freely in the exchanger as it exchanges heat.
For mixed flow condition the temperature varies only in the flow direction because there will be a tendency for the fluid temperature to equalize in the direction normal to the flow due to mixing. When a fluid is unmixed, the temperature variation will be in both the flow and normal direction the overall heat transfer rate for the exchanger will thus depends on the nature of the mixing condition.
When large quantities of heat are to be transferred the heat transfer area requirement of the heat exchanger also becomes large. In single pass heat exchanger this requirement can be met either by increasing the length of the tubes or by decreasing their diameter and increasing the number o tubes at the same time.
Neither of these method is practicable due to limitation of size, the length of the tube cannot increase arbitrarily and large pressure drop would occur with smaller diameter tubes. These difficulties lead us to the concept of multipass arrangement. Fig below shows such arrangement.
The fluid flowing through the tube called as the tube fluid whereas the fluid flowing outside the tube is called shell fluid. Depending upon the heat transfer area requirement we can have multiple tube and or shell passes. Flow condition for shell and tube type heat exchanger are neither parallel flow nor counter flow type. To create turbulence in the shell side fluid and to enhance a
166 HEAT TRANSFER ______
cross flow velocity of this fluid relative ot the tube baffles are provided. This result in higher heat transfer coefficient for the outer tube surface.
Figure below shows the shell and tube exchanger with a bundle of tubes arranged in one shells pass with two tubes passes and baffles.
OVERALL HEAT TRANSFER COEFFICIENT
The thermal design of heat exchanger involves the calculation of the necessary surface area required to transfer heat at a given flow rate and fluid temperatures. The concept of overall heat transfer coefficient U is of great significance in the heat exchanger calculation.
푄 = 푈퐴(훥푇)푚
(훥푇)푚 푖푠 푡ℎ푒 푎푣푒푟푎푔푒 푒푓푓푒푐푡푖푣푒 푡푒푚푝푒푟푎푡푢푟푒 푑푖푓푓푒푟푒푛푐푒 푓표푟 푡ℎ푒 푒푛푡푖푟푒 ℎ푒푎푡 푒푥푐ℎ푎푛푔푒푟
For plane wall
1 푈 = 1 퐿 1 + + ℎ0 푘 ℎ𝑖
For cylindrical wall
1 푈 = 표 1 푟 푟 푟 1 + 표 푙푛 ( 표) + 표 ℎ표 푘 푟𝑖 푟𝑖 ℎ𝑖
1 푈 = 𝑖 1 푟 푟 푟 1 + 𝑖 푙푛 ( 표) + 𝑖 ℎ𝑖 푘 푟𝑖 푟표 ℎ표
HEAT TRANSFER 167 ______
Where i and o represent the inside and outside surface of the wall respectively
Since surface area for heat transfer on both inner and outer surface are not the same , so we have two overall coefficient as defined above. However, for the sake of compatibility.
푈𝑖퐴𝑖 = 푈표퐴표
FOULING FACTORS
The following equations are in fact valid for clean surface only.
1 푈 = 1 퐿 1 + + ℎ0 푘 ℎ𝑖
For cylindrical wall
1 푈 = 표 1 푟 푟 푟 1 + 표 푙푛 ( 표) + 표 ℎ표 푘 푟𝑖 푟𝑖 ℎ𝑖
1 푈 = 𝑖 1 푟 푟 푟 1 + 𝑖 푙푛 ( 표) + 𝑖 ℎ𝑖 푘 푟𝑖 푟표 ℎ표
However it is a well known fact that the surface of heat exchanger does not remain clean after it has been in use for some time. The surface becomes fouled with scaling or deposits which are formed due to impurities in the fluid chemical reaction between the fluids and the wall material, rust formation etc. the effect of these deposits is felt in terms of greatly increased surface resistance affecting the value of overall heat transfer coefficients. This effect is taken care of by introducing additional thermal resistances called the fouling resistance 푅푓. The value of fouling resistance must be determine experimentally by testing the heat exchanger in both clean and dirty condition being defined by the following
1 1 = 푅푓. + 푈푓표푢푙. 푈퐶푙푒푎푛.
LMTD METHOD OF HEAT EXCHANGER ANALYSIS
The thermal analysis of any heat exchange involves variables like inlet and outlet fluid temperatures, the overall heat transfer coefficient, total surface area for heat transfer and total heat transfer rate. Since the hot fluid is transferring part of its energy to the cold fluid, there will be an increased in enthalpy of the cold fluids and corresponding decrease in enthalpy of hot fluid. This may be expressed as
푄 = 푚̇ℎ푐ℎ(푇ℎ,𝑖 − 푇ℎ,표)
푄 = 푚̇푐푐푐(푇푐,표 − 푇푐,𝑖)
푚̇ = 푚푎푠푠 푓푙표푤 푟푎푡푒
168 HEAT TRANSFER ______
푐 = 푐표푛푠푡푎푛푡 표푓 푠푝푒푐푖푓푖푐 ℎ푒푎푡
푡ℎ푒 푠푢푏푠푐푟푖푝푡 푐 푎푛푑 ℎ 푟푒푝푟푒푠푒푛푡 푐표푙푑 푎푛푑 ℎ표푡 푓푙푢푖푑
푠푢푏푠푐푟푖푝푡 푖 푎푛푑 표 푟푒푝푟푒푠푒푛푡푠 푡ℎ푒 푓푙푢푖푑 푖푛푙푒푡 푎푛푑 표푢푡푙푒푡 푐표푛푑푖푡푖표푛
If we denote the temperature difference between the hot and cold fluid as
훥푇 = (푇ℎ − 푇푐)
Since 훥푇 varies with the position in heat exchanger, the actual rate equation for heat transfer will be
푄 = 푈퐴(훥푇푚)
(훥푇푚) 푖푠 푡ℎ푒 푠푢푖푡푎푏푙푒 푚푒푎푛 푡푒푚푝푒푟푎푡푢푟푒 푑푖푓푓푒푟푒푛푐푒 푎푐푟표푠푠 푡ℎ푒 ℎ푒푎푡 푒푥푐ℎ푎푛푔푒푟.
퐿푀푇퐷 푖푠 푟푒푓푒푟푒푑 푡표 푎푠 푙표푔푎푟푖푡ℎ푚푖푐 푚푒푎푛 푡푒푚푝푒푟푎푡푢푟푒 푑푖푓푓푒푟푒푛푐푒 푚푒푡ℎ표푑 표푓 푎푛푎푙푦푠푖푠
LMTD METHOD OF HEAT EXCHANGER ANALYSIS FOR PARALLEL FLOW HEAT EXCHANGER
Let us consider the parallel flow heat exchanger as shown in figure assuming that
The overall heat transfer coefficient is uniform throughout the exchanger. The potential and kinetic energy changes are negligible The specific heat of the fluid are constant The heat exchange takes place only between two fluids. Temperatures of the fluid are constant over given cross-section and may be represented by their bulk mean temperature.
The heat transfer for hot and cold fluid for a differential element of length dx is given b the following
푑푄 = 푈푑퐴(훥푇)
훥푇 = 푇ℎ − 푇푐 푖푠 푡ℎ푒 푙표푐푎푙 푡푒푚푝푒푟푎푡푢푟푒 푑푖푓푓푒푟푒푛푐푒 푏푒푡푤푒푒푛 ℎ표푡 푎푛푑 푐표푙푑 푓푙푢푖푑
푑퐴 = 푤푖푑푡ℎ × 푑푥
HEAT TRANSFER 169 ______
Energy balance over hot and cold fluid for differential element gives
푑푄 = −푚̇ℎ푐ℎ푑푇ℎ = −퐶ℎ푑푇ℎ 푤ℎ푒푟푒 퐶ℎ = 푚̇ℎ푐ℎ
푑푄 = +푚̇푐푐푐푑푇푐 = +퐶푐푑푇푐 푤ℎ푒푟푒 퐶푐 = 푚̇푐푐푐
퐶ℎ 푎푛푑 퐶푐 푎푟푒 푡ℎ푒 ℎ표푡 푎푛푑 푐표푙푑 푓푙푢푖푑 ℎ푒푎푡 푐푎푝푎푐푖푡푦 푟푎푡푒푠.
ℎ푒푎푡 푐푎푝푎푐푖푡푦 푟푎푡푒푠 푎푟푒 푡ℎ푒 푝푟표푑푢푐푡 표푓 푚푎푠푠 푓푙표푤 푟푎푡푒 푎푛푑 푠푝푒푐푖푓푖푐 ℎ푒푎푡
훥푇 = (푇ℎ − 푇푐)
Differentiating the above equation we get
푑(훥푇) = (푑푇ℎ − 푑푇푐)
Putting the value of 푑푇ℎ 푎푛푑 푑푇푐 푤푒 푔푒푡
푑푄 푑푄 푑(훥푇) = − − 퐶ℎ 퐶푐
1 1 푑(훥푇) = −푑푄 ( + ) 퐶ℎ 퐶푐
1 1 푑(훥푇) = −푈푑퐴훥푇 ( + ) 퐶ℎ 퐶푐
Integrating the above equation within the temperature limit 훥푇1 푎푛푑훥푇2 we get
훥푇2 2 푑(훥푇) 1 1 ∫ = −푈 ( + ) ∫ 푑퐴 훥푇 퐶ℎ 퐶푐 훥푇1 1
훥푇2 1 1 푙푛 ( ) = −푈퐴 ( + ) 훥푇 퐶ℎ 퐶푐
By using the following equation and putting the values in the above equation
푄 = 푚̇ℎ푐ℎ(푇ℎ,𝑖 − 푇ℎ,표) 푤ℎ푒푟푒 푚̇ℎ푐ℎ = 퐶ℎ
푄 = 푚̇푐푐푐(푇푐,표 − 푇푐,𝑖) 푤ℎ푒푟푒 푚̇푐푐푐 = 퐶푐
훥푇2 (푇ℎ,𝑖 − 푇ℎ,표) (푇푐,표 − 푇푐,𝑖) 푙푛 ( ) = −푈퐴 ( + ) 훥푇1 푄 푄
훥푇2 −푈퐴 푙푛 ( ) = ((푇ℎ,𝑖 − 푇ℎ,표) + (푇푐,표 − 푇푐,𝑖)) 훥푇1 푄
훥푇2 −푈퐴 푙푛 ( ) = ((푇ℎ,𝑖 − 푇푐,𝑖) + (푇푐,표 − 푇ℎ,표)) 훥푇1 푄
170 HEAT TRANSFER ______
훥푇2 −푈퐴 푙푛 ( ) = ((푇ℎ,𝑖 − 푇푐,𝑖) − (−푇푐,표 + 푇ℎ,표)) 훥푇1 푄
훥푇2 −푈퐴 푙푛 ( ) = (훥푇1 − 훥푇2) 훥푇1 푄
훥푇2 푈퐴 푙푛 ( ) = (훥푇2 − 훥푇1) 훥푇1 푄
푈퐴(훥푇2 − 훥푇1) 푄 = 훥푇 푙푛 ( 2) 훥푇1
Where
훥푇1 = (푇ℎ,𝑖 − 푇푐,𝑖) 푎푛푑 훥푇2 = (−푇푐,표 + 푇ℎ,표)
푄 = 푈퐴훥푇푙푚
(훥푇2 − 훥푇1) 훥푇 = 푙푚 훥푇 푙푛 ( 2) 훥푇1
Or the above equation may also be written as
(훥푇1 − 훥푇2) 훥푇 = 푙푚 훥푇 푙푛 ( 1) 훥푇2
LMTD METHOD OF HEAT EXCHANGER ANALYSIS FOR COUNTER FLOW HEAT EXCHANGER
A counter flow heat exchanger where the fluids moves in parallel but in opposite direction, is shown in figure. The change in temperature difference between the two fluids is greatest at the entrance and of the parallel flow heat exchanger but it may not be so in counter flow arrangement. The analysis of counter flow heat exchanger can be done exactly in the same manner as outlined in the previous section for a parallel flow heat exchanger. In fact the derivation for parallel flow heat exchanger is valid for any heat exchanger. By taking a differential area element for a counter flow exchange and proceeding it can be easily shown and valid in this case too.
푄 = 푈퐴훥푇푙푚
(훥푇2 − 훥푇1) 훥푇 = 푙푚 훥푇 푙푛 ( 2) 훥푇1
훥푇1 = (푇ℎ,𝑖 − 푇푐,표) 푎푛푑 훥푇2 = (−푇푐,𝑖 + 푇ℎ,표)
Comparing the parallel flow and counter flow heat exchanger it is observed that for same terminal temperature of the fluid LMTD for counter flow heat exchanger is more than the LMTD for parallel flow heat exchanger.
HEAT TRANSFER 171 ______
A special case of counter flow heat exchanger exists when the capacity rates are 퐶ℎ 푎푛푑퐶푐 are equal then
(푇ℎ,𝑖 − 푇ℎ,표) = (푇푐,표 − 푇푐,𝑖)
훥푇1 = 훥푇2
In this case the LMTD according to the above equation is of the form 0/0 and so undefined. But it is oblivious that since 훥푇 is constant throughout the exchanger hence
훥푇푙푚 = 훥푇1 = 훥푇2
EFFECTIVENESS NTU METHOD OF HEAT EXCHANGER ANALYSIS
In thermal analysis of various types of heat exchanger by LMTD method an equation derived in earlier section has to be used. This equation is pretty simple and can be used in the design of heat exchanger when all the terminal temperatures are known or are easily determined. The difficulty arises if the temperatures off the fluid leaving the heat exchanger are no known. This type of situation is encountered in selection of heat exchanger or when the exchanger is to be run at off design conditions. Although the outlet temperatures and heat flow rates can still be found with the help of charts yet it would be possible only through tedious trial and error procedure. In such cases, it is preferable to utilize an altogether different method known as effectiveness NTU method.
172 HEAT TRANSFER ______
the effectiveness method is based on the effectiveness of heat exchanger in transferring given amount of heat. To obtained an expression for the rate of heat transfer without involving any of the outlet temperature let us first introduce the term effectiveness as ϵ
푎푐푡푢푎푙 ℎ푒푎푡 푡푟푎푛푠푓푒푟 푒푓푓푒푐푡푖푣푒푛푒푠푠 휖 = 푚푎푥푖푚푢푚 푝표푠푠푖푏푙푒 ℎ푒푎푡 푡푟푎푛푠푓푒푟
푄 휖 = 푄푚푎푥
The actual rate of heat transfer is given by following equation
푄 = 푚̇ℎ푐ℎ(푇ℎ,𝑖 − 푇ℎ,표)
푄 = 푚̇푐푐푐(푇푐,표 − 푇푐,𝑖)
푄푚푎푥 푖푠 푡ℎ푒 푚푎푥푖푚푢푚 푟푎푡푒 표푓 ℎ푒푎푡 푡푟푎푛푠푓푒푟
푄푚푎푥 is the maximum rate of heat transfer that a counter flow heat exchanger of infinite area would transfer with given inlet temperature, flow rates and specific heats. Also we recognize that the maximum possible heat transfer would obtained if one of the fluids was to undergo a temperature change is equal to maximum temperature difference present in the exchanger. We consider two distinct cases to illustrate
1) 퐶ℎ > 퐶푐
For this type of heat exchanger with no external heat loss, the outlet temperature of the cold fluid will equal to inlet temperature of the hot fluid. The temperature distribution in the fluid are shown in fig. the maximum rate of heat transfer is given by
푄푚푎푥 = 퐶푐(푇푐,표 − 푇푐,𝑖)
푇푐,표 = 푇ℎ,𝑖
푄푚푎푥 = 퐶푐(푇ℎ,𝑖 − 푇푐,𝑖) 푎푙푠표
푄 = 퐶푐(푇푐,표 − 푇푐,𝑖)
1) 퐶ℎ < 퐶푐
HEAT TRANSFER 173 ______
In this case the outlet temperature of the hot fluid would equal to inlet temperature of cold fluid as shown in figure.
푄푚푎푥 = 퐶ℎ(푇ℎ,𝑖 − 푇ℎ,표)
푇ℎ,표 = 푇푐,𝑖
푄푚푎푥 = 퐶ℎ(푇ℎ,𝑖 − 푇푐,𝑖)
Also
푄 = 퐶ℎ(푇ℎ,𝑖 − 푇ℎ,표)
Hence comparing above equation we can express
푄푚푎푥 = 퐶푚𝑖푛(푇ℎ,𝑖 − 푇푐,𝑖)
퐶푚𝑖푛 is the smaller of 퐶ℎ and 퐶푐 using the definition of effectiveness it follows that
퐶ℎ(푇ℎ,𝑖 − 푇ℎ,표) 휖 = 퐶푚𝑖푛(푇ℎ,𝑖 − 푇푐,𝑖)
퐶푐(푇푐,표 − 푇푐,𝑖) 휖 = 퐶푚𝑖푛(푇ℎ,𝑖 − 푇푐,𝑖)
Once effectiveness of heat exchanger is known, its actual rate of heat transfer can be determined by
푄 = 휖 퐶푚𝑖푛(푇ℎ,𝑖 − 푇푐,𝑖)
푄 = 휖 퐶푚푎푥
The equation above is very significant because it express the actual rate of heat transfer by heat exchanger in terms of the effectiveness, 퐶푚𝑖푛 and the difference between the inlet temperatures of the two fluids. It does not refer to the outlet fluid temperature and can replace the LMTD analysis effectively.
As will be shown in the following subsection, effectiveness for any heat exchanger can be expresses as
푈퐴 퐶푚𝑖푛 휖 = 휖 ( , ) 퐶푚𝑖푛 퐶푚푎푥
Where
퐶푚𝑖푛 퐶푐 퐶ℎ = 표푟 퐶푚푎푥 퐶ℎ 퐶푐 푈퐴 푖푠 푐푎푙푙푒푑 푡ℎ푒 푛푢푚푏푒푟 표푓 푡푟푎푛푠푓푒푟 푢푛푖푡푠, 푁푇푈 퐶푚𝑖푛 푈퐴 푁푇푈 = 퐶푚𝑖푛
174 HEAT TRANSFER ______
Thus NTU is the dimensionless parameter. It is measure of the heat transfer size of the exchanger. The larger the value of NTU, the closer the heat exchanger reaches its thermodynamic limit of operation.
EFFECTIVENESS FOR PARALLEL FLOW HEAT EXCHANGER Assuming
퐶푚𝑖푛 = 퐶푐
푇푐,표 − 푇푐,𝑖 휖 = 푇ℎ,𝑖 − 푇푐,𝑖
푄 = 푚̇ℎ푐ℎ(푇ℎ,𝑖 − 푇ℎ,표)
푄 = 푚̇푐푐푐(푇푐,표 − 푇푐,𝑖)
From above equation we get
퐶푚𝑖푛 푚̇푐 × 퐶푐 푇ℎ,𝑖 − 푇ℎ,표 = = 퐶푚푎푥 푚ℎ ×̇ 퐶ℎ 푇푐,표 − 푇푐,𝑖
훥푇2 푇ℎ,표 − 푇푐,표 푈퐴 퐶푚𝑖푛 푙푛 ( ) = 푙푛 ( ) = − (1 + ) 훥푇1 푇ℎ,𝑖 − 푇푐,𝑖 퐶푚𝑖푛 퐶푚푎푥
푈퐴 퐶 푇ℎ,표 − 푇푐,표 − (1+ 푚푖푛 ) = 푒 퐶푚푖푛 퐶푚푎푥 푇ℎ,𝑖 − 푇푐,𝑖
The above equation can be rearranged as
푇ℎ,표 − 푇푐,표 푇ℎ,표 − 푇푐,𝑖 + 푇푐,𝑖 − 푇푐,표 = 푇ℎ,𝑖 − 푇푐,𝑖 푇ℎ,𝑖 − 푇푐,𝑖
Substituting the value of 푇ℎ,표
퐶푚𝑖푛 푇ℎ,표 = 푇ℎ,𝑖 − (푇푐,표 − 푇푐,𝑖) 퐶푚푎푥
퐶푚𝑖푛 (푇ℎ,𝑖 − 푇푐,𝑖) − (푇푐,표 − 푇푐,𝑖) + 푇푐,𝑖 − 푇푐,표 푇ℎ,표 − 푇푐,표 퐶 = 푚푎푥 푇ℎ,𝑖 − 푇푐,𝑖 푇ℎ,𝑖 − 푇푐,𝑖
푇ℎ,표 − 푇푐,표 퐶푚𝑖푛 = 1 − 휖 − 휖 푇ℎ,𝑖 − 푇푐,𝑖 퐶푚푎푥
푇ℎ,표 − 푇푐,표 퐶푚𝑖푛 = 1 − 휖 (1 + ) 푇ℎ,𝑖 − 푇푐,𝑖 퐶푚푎푥
푈퐴 퐶 − (1+ 푚푖푛 ) 퐶푚𝑖푛 푒 퐶푚푖푛 퐶푚푎푥 = 1 − 휖 (1 + ) 퐶푚푎푥
HEAT TRANSFER 175 ______
퐶 −푁푇푈(1+ 푚푖푛 ) 퐶푚𝑖푛 푒 퐶푚푎푥 = 1 − 휖 (1 + ) 퐶푚푎푥
퐶 퐶푚𝑖푛 −푁푇푈(1+ 푚푖푛 ) 휖 (1 + ) = 1 − 푒 퐶푚푎푥 퐶푚푎푥
퐶 −푁푇푈(1+ 푚푖푛 ) 1 − 푒 퐶푚푎푥 휖 = 퐶 (1 + 푚𝑖푛 ) 퐶푚푎푥
Note:
Notice that the expression for 휖 contains U,A and the heat capacities only. Also had we started with
퐶푚𝑖푛 = 퐶ℎ we would have obtained the same expression for 휖.
HEAT PIPE
Heat pipe is the smaller and more compact heat exchange, the better the design for transferring energy from one location to another. The heat pipe is a novel device for high thermal conductance i.e. it allows the transfer of large quantities of heat through the small surface areas with very little temperature difference from end to end. The heat pipe is similar to a thermal syphon but has added advantages of being used in any orientation unlike thermal syphon wherein the evaporator must be situated at the lowest point. A heat pipe in its simplest form is as shown in figure. It consist of circular pipe with a layer of wicking material covering the inside surface and hollow core in the centre. A condensable fluid like water is placed in the pipe and the liquid permeates the wicking material by capillary action. When one end of pipe is heated the liquid is vaporized in the wick and the vapours moves to the central core. At the other end of the pipe, heat is removed at the condenser section and the vapour condenser back into the wick. The liquid is replenished in the evaporator section by capillary action. A typical heat pipe of copper using water as a working fluid can transport an axial heat flux of 6500 kW/푚2 at 2000퐶. Heat pipe are now being used in several application in engineering practice of cooling of micro electric circuit, power transistor, cryogenics target in nuclear accelerator, temperature control devices. Waste heat recovery boiler, heating and cooling of engines and space vehicle.
A heat pipe is basically a sealed container, normally in the form of a tube, containing wick lining the inside wall. It is used to transfer heat from the source to sink by means of evaporation and condensation of a fluid in the sealed system. The purpose of the wick is to transport the working fluid in the liquid form from one end to other end by capillary action. The heat pipe can transfer heat much more effectively than a solid conductor of the same cross-section. The sequence of operation is as follows
Heat in Evaporation Vapour flow Condensation Heat out
176 HEAT TRANSFER ______
Liquid return in wick
The heat pipe has no moving parts requires no external energy other than the heat it transmit and is reversible in operation and completely silent and is rugged like any piece of tube or pipe and can withstand a lot of abuse. The heat pipe is sensitive to the effect of gravity and hence its inclination to horizontal is much more effective.
The wick provides an axial pumping it distribute the liquid circumferentially and ensure that in the evaporator section the available surface is wetted so that it can support al the radial heat flux. The wick itself is integral with the walls of the heat pipe in the form of groove and may increases the heat transfer.
APPLICATION OF HEAT PIPE
HEAT TRANSFER 177 ______
NUMERICALS ---HEAT EXCHANGER ANALYSIS Hot oil with a capacity rate of 2500 푊/퐾 flows through a double pipe heat exchanger. It enters at 3600퐶 and leaves at 3000퐶. Cold fluid enters at 300퐶 and leaves at 2000퐶. If the overall heat transfer coefficient is 800푊/푚2퐾 , determine the heat exchanger area required for parallel and counter flow heat exchanger.
Solution
The heat transfer from the oil is
푄 = 푚̇ℎ푐ℎ(푇ℎ,𝑖 − 푇ℎ,표)
푄 = 퐶ℎ(푇ℎ,𝑖 − 푇ℎ,표)
퐶ℎ 푖푠 푡ℎ푒 ℎ푒푎푡 푐푎푝푎푐푖푡푦 푟푎푡푒푠
퐶ℎ = 2500 푊/퐾
0 푇ℎ,𝑖 = 360 퐶
0 푇ℎ,표 = 300 퐶
푄 = 2500(360 − 300) = 150푘푊
For parallel flow heat exchanger the temperature distribution is given by the following equation
(훥푇1 − 훥푇2) 훥푇 = 푙푚 훥푇 푙푛 ( 1) 훥푇2
훥푇1 = (푇ℎ,𝑖 − 푇푐,𝑖) 푎푛푑 훥푇2 = (−푇푐,표 + 푇ℎ,표)
푄 = 푈퐴훥푇푙푚
0 훥푇1 = (푇ℎ,𝑖 − 푇푐,𝑖) = 360 − 30 = 330 퐶
0 훥푇2 = (−푇푐,표 + 푇ℎ,표) = 300 − 200 = 100 퐶
178 HEAT TRANSFER ______
(훥푇1 − 훥푇2) 훥푇 = 푙푚 훥푇 푙푛 ( 1) 훥푇2
(330 − 100) 훥푇 = = 192.640퐶 푙푚 330 푙푛 ( ) 100
The heat exchanger area may be calculated by the following equation
푄 = 푈퐴훥푇푙푚 푄 150000 퐴 = = = 0.973푚2 푈퐴훥푇푙푚 800 × 192.64
For counter flow heat exchanger the temperature distribution equation is same only the value of
훥푇1 and 훥푇2 changes.
훥푇1 = (푇ℎ,𝑖 − 푇푐,표) 푎푛푑 훥푇2 = (−푇푐,𝑖 + 푇ℎ,표)
0 훥푇1 = (푇ℎ,𝑖 − 푇푐,표) = 360 − 200 = 160 퐶
0 훥푇2 = (−푇푐,𝑖 + 푇ℎ,표) = 300 − 30 = 270 퐶
(160 − 270) 훥푇 = = 210.220퐶 푙푚 160 푙푛 ( ) 270 푄 150000 퐴 = = = 0.892푚2 푈퐴훥푇푙푚 800 × 210.22
Thus we see that for the same terminal temperatures of fluids the surface area required for a counter flow arrangement is less than that in parallel flow arrangement
A counter flow concentric tube heat exchanger is used to cool engine oil of specific heat 2130 J/kg K. from 1600퐶 to 600퐶 with water available at 250퐶as the cooling medium. The flow rate of cooling water through the inner tube of 0.5m diameter is 2 kg/s. if the value of overall heat transfer coefficient is 250 푊/푚2퐾, how long the heat exchanger must be to meet the cooling requirement.
Solution
The required heat transfer equation for hot fluid is given by the following
푄 = 푚̇ℎ푐ℎ(푇ℎ,𝑖 − 푇ℎ,표)
푄 = 2 × 2130(160 − 60) = 426푘푊
The required heat transfer equation for cold fluid is given by the following
푄 = 푚̇푐푐푐(푇푐,표 − 푇푐,𝑖)
HEAT TRANSFER 179 ______
푄 푇푐,표 = + 푇푐,𝑖 푚̇푐푐푐 426000 푇 = + 25 = 75.880퐶 푐,표 2 × 4186
The length of the heat exchange can be obtained from the below equation
푄 = 푈퐴훥푇푙푚
퐴 = 훱 × 퐷𝑖 × 퐿
훥푇1 = (푇ℎ,𝑖 − 푇푐,표) 푎푛푑 훥푇2 = (−푇푐,𝑖 + 푇ℎ,표)
0 훥푇1 = (푇ℎ,𝑖 − 푇푐,표) = 160 − 75.88 = 84.12 퐶
0 훥푇2 = (−푇푐,𝑖 + 푇ℎ,표) = 60 − 25 = 35 퐶
(84.12 − 35) 훥푇 = = 56.010퐶 푙푚 84.12 푙푛 ( ) 35
푄 = 푈퐴훥푇푙푚
푄 퐴 = 푈 × 훥푇푙푚 푄 426 × 100 훱 × 퐷𝑖 × 퐿 = = 푈 × 훥푇푙푚 250 × 56.01
426 × 100 퐿 = = 19.37푚 훱 × 0.5 × 250 × 56.01
In double pipe counter flow heat exchanger of 10,000 kg/hr of a oil is having specific heat of 2095 J/kg K is cooled from 800퐶 to 500퐶 by 8000 kg/hr of water entering at 250퐶. Determine the heat exchanger area for an overall heat transfer coefficient of 300푊/푚2퐾 take specific heat of water as 4180 J/kgK.
Solution
The outlet temperature of cold fluid can be found by equating the heat lost by oil and that gained by water
푄 = 푚̇푐푐푐(푇푐,표 − 푇푐,𝑖)
푄 = 푚̇ℎ푐ℎ(푇ℎ,𝑖 − 푇ℎ,표)
푚̇ℎ푐ℎ(푇ℎ,𝑖 − 푇ℎ,표) = 푚̇푐푐푐(푇푐,표 − 푇푐,𝑖)
10000 × 2095 × (80 − 50) = 8000 × 4180(푇푐,표 − 25)
180 HEAT TRANSFER ______
10000 × 2095 × (80 − 50) 푇 = + 25 = 43.80퐶 푐,표 8000 × 4180
For counter flow heat exchanger
0 훥푇1 = (푇ℎ,𝑖 − 푇푐,표) = 80 − 43.8 = 36.2 퐶
0 훥푇2 = (푇ℎ,표 − 푇푐,𝑖) = 50 − 25 = 25 퐶
(훥푇1 − 훥푇2) 푄 = 푈퐴 훥푇 푙푛 ( 1) 훥푇2
푄 = 푚̇ℎ푐ℎ(푇ℎ,𝑖 − 푇ℎ,표)
10000 푄 = × 2095 × (80 − 50) 3600
푄 = 174583.33푘푊
훥푇 푄 × 푙푛 ( 1) 훥푇 퐴 = 2 푈 × (훥푇1 − 훥푇2) 36.2 174583.33 × 푙푛 ( ) 퐴 = 25 800 × (36.2 − 25)
퐴 = 19.23 푚2
BASICS CONCEPT RELATED TO HEAT TRANSFER
Heat is a form of energy can neither be created nor destroyed this makes it possible to consider the conversion of one form of energy to other
Modes of heat transfer
Conduction convection radiations are three modes of heat transfer. The first two mode of heat transfer appears to be more dominant in many practical applications. However the radiation is the most important mode of heat transfer at high temperatures.
Conduction
It is atomic or molecular process. It occurs in the presence of temperature gradient it is the only mode of heat transfer in a solid medium. Fourier law is the basic law for heat conduction.
Convection
The capacity of moving matter to carry heat energy from one region to another region. Convection means transport of heat energy by means of displacement of fluid elements from one point to another, which is at different temperature. Convection may be of two types natural convection or free convection and forced convection, the free or natural convection occurs adverse
HEAT TRANSFER 181 ______density gradient as result of temperature difference, this happens when the temperature of the fluid at a lower level because greater than that at upper level. Forced convection occurs when the motion in the medium is caused by external mechanism such as pump blower etc.
Radiation
A body at a temperature above absolute zero always emits energy in the form of electromagnetic waves. The rate of release of such energy is proportional to the fourth power of the absolute temperature of the body. This phenomenon is called radiation and the governing law is known as Stefan’s Boltzmann law.
LAW GOVERNING THE DIFFERENT MODES OF HEAT TRANSFER Kirchhoff’s law of radiation
It states that the ratio of total emissive power to Absorptivity is constant for all substances which are in thermal equilibrium with the surrounding
퐸1 퐸2 퐸3 = = = 푐표푛푠푡푎푛푡 훼1 훼2 훼3
Fourier’s law of heat conduction
It states that the rate of heat transfer through a flat plat is proportional to the surface area of the plate, the temperature difference between inner and outer surface inversely proportional to the thickness of plate.
푑푇 푄 ∝ 퐴 푑푥
푄 = Heat flow through the body per unit time in W
퐴 =Surface area of heat flow (perpendicular to the direction of flow) 푚2
푑푥 =Thickness of body in the direction of flow (m)
푑푇 =Temperature difference at the faces of block of thickness dx
푑푇 푄 = −푘퐴 푑푥
푘 = Constant of proportionality and is known as thermal conductivity of the body W/mk.
Newton’s law of cooling
When a fluid flows over a solid surface which is at different temperature ie at higher or lower temperature the convection heat transfer takes place.
푄 = ℎ × 퐴 × (푇푠−푇∞)
푇푠 =Surface temperature in K
푇∞ =Temperature of fluid in K
182 HEAT TRANSFER ______
ℎ =Convection heat transfer coefficient W/푚2K
Stefan’s Boltzmann law
It states that the amount of heat energy emitted per unit area per unit time by radiation from an ideal surface is directly proportional to the fourth power of the absolute temperature
푄 ∝ 휎 푇4
휎 푆푡푒푓푎푛푠 푏표푙푡푧푚푎푛푛 푐표푛푠푡푎푛푡
휎 = 5.67 × 10−8 푊/푚2퐾4
Weins displacement law
For a black emissive spectrum the wavelength giving the maximum emissive power at a particular temperature may be given by
3 휆푚푎푥푇 = 2.89 × 10 푚퐾
Planks distribution law
The spectral distribution of the radiation intensity of the black body is given by e…. this relation gives s the magnitude of the emitted energy at each of the wavelength which composes the whole radiation spectra.
−5 퐶1휆 퐸 = 푏휆 퐶2 푒휆푇 − 1
Lambert cosines law
This states that the diffuse surface radiates energy in such manner that the rate of energy radiated in any particular direction is proportion to the cosines of the angle between the direction under consideration and normal to the surface
IMPORTANT DEFINITION
Thermal conductivity
It is defined as the amount of heat that can flow per unit time across a unit crossection area when the temperature gradient is unity. The material having lower thermal conductivity is called insulators. The best conductors are pure metal and poorest one are gases lies in between non metallic solids and liquids.
Black body
A black body is one which absorbs all the incident radiate energy and does not reflect or transmit the energy.
Emissive power
HEAT TRANSFER 183 ______
The emissive power of a black body is defined as the energy emitted by the surface per unit time per unit area and is dependent upon a number of parameters like surface material and roughness
Monochromatic emissive power is defined as the rate of energy radiated per unit area of the surface per unit wavelength. It is measured in W/m3
Irradiation G
It is the total radiation incident upon a surface per unit time per unit area.
Radiosity (J)
It is the term used to indicate the total radiation leaving a surface per unit time per unit area.
Grey body
Grey body is one which absorbs only a definite percentage of incident radiation waves irrespective of their wavelength
Colored body
This represents a surface who’s Absorptivity varies with the wavelength of radiation waves.
Reynolds number
It is the ration of inertia force to the viscous force. It indicates the relative importance of the inertial and viscous effect and the fluid motion is laminar or turbulent. At high Reynolds number the inertia effect leads to turbulent flow the associated turbulent level dominates the momentum and energy flow.
휌푉퐷 푅 = 푒 휇
Prandtl number
This number represents the relative importance momentum and energy transfer by diffusion. For gases the Prandtl number is equal to one and for oils is greater than one
휇퐶푝 푃 = 푟 푘
Grashof number
This is the ratio of product of inertia force and the buoyancy force to the square of the viscous force.
푔훽∆푇퐿3 퐺푟 = 휈2
Nusselt number
184 HEAT TRANSFER ______
This represents the ratio of heat transfer by convection to the heat transfer by conduction
across the fluid of thickness 퐿푐.
ℎ퐿푐 푁 = 푢 푘
Absorptivity
It is the ratio of radiant heat energy absorbed by a body to the total incident radiant heat energy on the surface of the body
Reflectivity
It is the ratio of radiant heat energy reflected by a body to the total incident radiant heat energy on the body surface
Trasnsmissvity
It is the ratio of radiant heat energy transmitted by a body to the total incident heat energy on the surface of the body.