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جامعة تكريت كلية الهندسة قسم الهندسة الميكانيكية 2018 -2017السنة الدراسية

االختصاص: ميكانيك عام المرحلة: الثالثة

جامعة تكريت كلية الهندسة قسم الهندسة الميكانيكية المرحلة الثالثة ) انتقال الحرارة341(هـ مك

انتقـال )، ثنـائي وثالثـي االبعـاد( انتقـال الحـرارة بالتوصـيل المسـتقر، الحرارة بالتوصـيل (احـادي، مسـتقر) انتقال، مقدمة انتقـال ، منظومات الحمل الحر، ىر عالقات الحمل القس، انتقال الحرارة بالحمل (مباديء)، الحرارة بالتوصيل غير المستقر

، المبادالت الحرارية.الحرارة باالشعاع

Tikrit university

Collage of engineering

Mechanical department

Subject: Heat Transfer

Class: 3P

edP Year

Lecturer: Dr. Tadahmun Ahmed Yassen

References:

1. Principle of heat transfer by Krieth. 2. Heat transfer by Holman. 3. Heat Transfer a practical approach by Yunus Single. 4. Fundamentals of heat and mass transfer by Incropera.

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CHAPTER ONE

Introduction

1.1 Heat Transfer Definision

The science of thermodynamics deals with the amount of heat transfer as a system undergoes a process from one equilibrium state to another, and makes no reference to how long the process will take. But in engineering, we are often interested in the rate of heat transfer, which is the topic of the science of heat transfer.

Heat transfer or (heat) is thermal energy in transmit due to a spacial temperature difference.

Whenever there exist a temperatue difference in a medium or between media, heat transfer must occur.

As shown in Figure 1.1, we refer to different types of heat transfer processes as modes. When a temperature gradient exists in a stationary medium, which may be a solid or a fluid, we use the term conduction to refer to the heat transfer that will occur across the medium. In contrast, the term convection refers to heat transfer that will occur between a surface and a moving fluid when they are at different

temperatures. The third mode of heat transfer is termed thermal radiation. All surfaces of finite temperature emit energy in the form of electromagnetic waves. Hence, in the absence of an intervening medium, there is net heat transfer by radiation between two surfaces at different temperatures.

1.2 Physical Origins and Rate Equations

As engineers it is important that we understand the physical mechanisms which underlie the heat transfer modes and that we be able to use the rate equations that quantify the amount of energy being transferred per unit time.

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1.2.1 Conduction

At mention of the word conduction, we should immediately conjure up concepts of atomic and molecular activity, for it is processes at these levels that sustain this mode of heat transfer. Conduction may be viewed as the transfer of energy from the more energetic to the less energetic particles of a substance due to interactions between the particles. The physical mechanism of conduction is most easily explained by considering a gas and using ideas familiar from your thermodynamics background. Consider a gas in which there exists a temperature gradient and assume that there is no bulk, or macroscopic, motion. The gas may occupy the space between two surfaces that are maintained at different temperatures, as shown in Figure 1.2. We associate the temperature at any point with the energy of gas molecules in proximity to the point. This energy is related to the random translational motion, as well as to the internal rotational and vibrational motions, of the molecules.

Conduction can take place in solids, liquids, or gases. In gases and liquids, conduction is due to the collisions and diffusion of the molecules during their random motion. In solids, it is due to the combination of vibrations of the molecules in a lattice and the energy transport by free electrons.

It is possible to quantify heat transfer processes in terms of appropriate rate equations. These equations may be used to compute the amount of energy being transferred per unit time. For heat conduction, the rate equation is known as Fourier’s law. For the one-dimensional plane wall shown in Figure 1.3, having a temperature distribution T(x), the rate equation is

expressed as:

𝑞𝑥𝐴 𝛼−

𝑑𝑇𝑑𝑥

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Where:

𝑞𝑥: is the heat flow in x-direction, w.

𝐴: is the area normal to the flow direction, m P

2P.

𝑑𝑇𝑑𝑥

: temperature gradiant (slope of the temperature curve), ˚C/m.

𝑞�̈�=𝑞𝑥𝐴 =−𝑘

𝑑𝑇𝑑𝑥 1.1

Where:

𝑘 : the proportional costant (thermal conductivity of the material), W/m.˚C.

𝑞�̈� : heat flux, W/mP

2P.

Simplifying above equation we get:

𝑞𝑥𝐴𝑑𝑥 = −𝑘𝑑𝑇

∫ 𝑞𝑥𝐴𝑑𝑥 = ∫ −𝑘𝑑𝑇𝑇2

𝑇1𝐿0

Where the temperature is 𝑇1 at x=0 and 𝑇2 at x=L

For constant heat flux and thermal conductivity integrating the above equation we get:

𝑞𝑥𝐴𝐿 = −𝑘(𝑇2 − 𝑇1)

𝑞𝑥𝐴

= −𝐾(𝑇2−𝑇1)𝐿 1.2

Example 1.1

The wall of an industrial furnace is constructed from 0.15-m-thick fireclay brick having a thermal conductivity of 1.7 W/m.K. Measurements made during steadystate operation reveal temperatures of 1400 and 1150 K at the inner and outer surfaces, respectively. What is the rate of heat loss through a wall that is 0.5 m by 1.2 m on a side?

Known: Steady-state conditions with prescribed wall thickness, area, thermal conductivity, and surface temperatures.

Find: Wall heat loss.

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Schematic:

Assumptions:

1. Steady-state conditions.

2. One-dimensional conduction through the wall.

3. Constant thermal conductivity.

Analysis: Since heat transfer through the wall is by conduction, the heat flux may be determined from Fourier’s law. Using Equation 1.2, we have.

𝑞�̈� = −𝐾(𝑇2−𝑇1)𝐿 = 1.7 W/m.K * 250 𝐾

0.15 𝑚 = 2833 W/mP

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The heat flux represents the rate of heat transfer through a section of unit area, and it is uniform (invariant) across the surface of the wall. The heat loss through the wall of area is then

𝑞𝑥 = 𝑞�̈�𝐴 = (𝐻𝑊)𝑞�̈� = (0.5𝑚 ∗ 1.2𝑚) ∗ 2833 W/mP

2P=1700 W.

Thermal Conductivity

It is a property of the material. The value of k, is an indicator of how fast heat is conduct through material. For most substances; thermal conductivity varies with temperature.

𝑘(𝑇) = 𝑘°(1 + 𝛽𝑘𝑇) 1.3

Where:

𝛽𝑘 : imperical constant.

𝑘° : thermal conductivity at datum temperature.

∫ 𝑞𝑥𝐴𝑑𝑥 = ∫ −𝑘°(1 + 𝛽𝑘𝑇)𝑑𝑇𝑇2

𝑇1𝐿0

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𝑞𝑥𝐴𝐿 = −𝑘°�(𝑇2 − 𝑇1) + 𝛽𝑘

2 (𝑇22 − 𝑇12)�

1.2.2 Convection Heat Transfer

Convection is a mode of heat transfer associated with fluid motion and it is of two types:

1. Forced convection: when the flow is caused by external means, such as by a fan, a pump, or atmospheric winds. As an example, consider the use of a fan to provide forced convection air cooling of hot electrical components on a stack of printed circuit boards (Figure 1.5a).

2. Free (or natural) convection: the flow is induced by buoyancy forces, which are due to density differences caused by temperature variations in the fluid. An example is the free convection heat transfer that occurs from hot components on a vertical array of circuit boards in air (Figure 1.5b). Air that makes contact with the components experiences an increase in temperature and hence a reduction in density. Since it is now lighter than the surrounding air, buoyancy forces induce a vertical motion for which warm air ascending from the boards is replaced by an inflow of cooler ambient air.

Figure 1.5: a. Forced convection, b. Free convection.

The heat transfer by convection is found to be proportional to the temperature difference.

𝑞𝑐𝐴𝛼(𝑇𝑠 − 𝑇∞) 1.4

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𝑞𝑐 = ℎ𝑐���𝐴(𝑇𝑠 − 𝑇∞)

Where:

𝑇𝑠 : wall surface temperature, ˚C.

𝑇∞ : surrounding temperature, ˚C.

ℎ𝑐��� : proportionality constant (convection heat transfer coefficient, W/m P

2P. ˚C.

Example 1.2:

Air at 20 ˚C blows over a hot plate 50 by 75 cm maintained at 250 ˚C. The convection heat-transfer

coefficient is 25 W/mP

2P ·˚C. Calculate the heat transfer.

Solution

From Newton’s law of cooling

qc = hc���A(Ts − T∞)

= (25)(0.50*0.75)(250−20)

= 2.156 Kw

Example 1.3:

A plane wall of thermal conductivity (k=2.4 W/m P

2P. ˚C) has its side temperature of 100

˚C and 40 ˚C, the wall thickness is 10 cm and exposed to environment at right side where convection heat transfer coefficient is 10 W/mP

2P. ˚C, calculate:

1. The heat flow per unit area. 2. The temperature of the air at right side.

Assume steady state.

Solusion:

TR1R=100 ˚C, TR2R=40 ˚C, k=2.4 W/m. ˚C, hRcR=10 W/m P

2P. ˚C.

�̈� = 𝑞𝐴

= −𝐾 ∆𝑇∆𝑥

= −2.4 (40−100)0.1

= 1440 W/mP

2

𝑞𝐴

= ℎ𝑐(𝑇𝑤 − 𝑇∞) = 10(40 − 𝑇∞)

𝑇∞ = 104˚C

Find 𝑇∞ when ℎ𝑐=100 W/mP

2P. ˚C.

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Convection Energy Balance on a Flow Channel

1.2.3 Radiation

Thermal radiation is energy emitted by matter that is at a nonzero temperature. Although we will focus on radiation from solid surfaces, emission may also occur from liquids and gases. Regardless of the form of matter, the emission may be attributed to changes in the electron configurations of the constituent atoms or molecules. The energy of the radiation field is transported by electromagnetic waves (or alternatively, photons). While the transfer of energy by conduction or convection requires the presence of a material medium, radiation does not. In fact, radiation transfer occurs most efficiently in a vacuum.

There is an upper limit to the emissive power, which is prescribed by the Stefan–Boltzmann law

𝐸𝑏 = 𝜎𝑇𝑠4 1.5

𝜎: is the proportional constant Stefan–Boltzmann constant and have a value of 5.67*10P

-8P W/mP

2P.KP

4P.

Most real materials surfaces do not emit radiation ideally. These are called gray surfaces, emit 𝜀 value of black body which called emissivity, and for gray body:

𝐸 = 𝜀𝜎𝑇4 1.6

𝜀 :is a radiative property of the surface termed the emissivity. With values in the range , the range 0 ≤ 𝜀 ≤ 1.

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Radiation in an Enclosure

A simple radiation problem is encountered when we have a heat-transfer surface at temperature 𝑇1 completely enclosed by a much larger surface maintained at 𝑇2. We will show that the net radiant exchange in this case can be calculated with

𝑞𝑟𝐴

= 𝜎𝜀(𝑇14 − 𝑇24) 1.7

In addition, we must take into account the fact that not all the radiation leaving one surface will reach the other surface since electromagnetic radiation travels in straight lines and some will be lost to the surroundings. We therefore introduce two new factors in Equation (1-6) to take into account both situations, so that

𝑞𝑟𝐴

= 𝐹𝜀𝐹𝐺𝜎(𝑇14 − 𝑇24)

Example 1.4:

Two infinite black plates at 800 ˚C and 300 ˚C exchange heat by radiation. Calculate the heat transfer per unit area.

𝑞𝑟𝐴

= 𝜎(𝑇14 − 𝑇24)

=5.67*10P

-8P (1073P

4P-573P

4P)

=69.03 Kw/m P

2P.

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Example 1.5:

Example 1.5:

Example 1.6:

A horizontal steel pipe having a diameter of 5 cm is maintained at a temperature of 50◦C in a large room where the air and wall temperature are at 20◦C. The heat-transfer coefficient for free convection with this geometry and air is h = 6.5 W/mP

2P · ◦C. The

surface emissivity of the steel may be taken as 0.8. Calculate the total heat lost by the pipe per unit length.

Solution:

T1 = 50◦C = 323◦K and T2 = 20◦C = 293◦K,

The total heat loss is the sum of convection and radiation. The surface area is πdL, so the convection loss per unit length is

The pipe is a body surrounded by a large enclosure so the radiation heat transfer can be calculated from Equation

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The general conduction energy equation:

*The general conduction equation

Consider the one-dimensional system shown in Figure 1-6. If the system is in a steady state, i.e., if the temperature does not change with time, then the problem is a simple one, and we need only integrate Equation (1-1) and substitute the appropriate values to solve for the desired quantity. However, if the temperature of the solid is changing with time, or if there are heat sources or sinks within the solid, the situation is more complex. We consider the general case where the temperature may be changing with time and heat sources may be present within the body. For the element of thickness dx, the following energy balance may be made:

Energy conducted in left face + heat generated within element = change in internal energy + energy conducted out right face These energy quantities are given as follows:

Energy in left face=𝑞𝑥 = −𝑘𝐴 𝜕𝑇𝜕𝑥

Energy generated within element=�̇�𝐴𝑑𝑥

Figure 1-6 Elemental volume for one-dimensional heatconduction analysis.

Change in internal energy=𝜌𝑐𝐴 𝜕𝑇𝜕𝑡𝑑𝑥

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Energy out right face = 𝑞𝑥+𝑑𝑥 = −𝑘𝐴 𝜕𝑇𝜕𝑥�𝑥+𝑑𝑥

= −𝐴 �𝑘 𝜕𝑇𝜕𝑥

+ 𝜕𝜕𝑥�𝑘 𝜕𝑇

𝜕𝑥� 𝑑𝑥�

−𝑘𝐴 𝜕𝑇𝜕𝑥

+ �̇�𝐴𝑑𝑥 = 𝜌𝑐𝐴 𝜕𝑇𝜕𝑡𝑑𝑥 − 𝐴 �𝑘 𝜕𝑇

𝜕𝑥+ 𝜕

𝜕𝑥�𝑘 𝜕𝑇

𝜕𝑥� 𝑑𝑥�

or

𝜕𝜕𝑥�𝑘 𝜕𝑇

𝜕𝑥� + �̇� = 𝜌𝑐 𝜕𝑇

𝜕𝑡 1.8

This is the one-dimensional heat-conduction equation. To treat more than one-dimensional heat flow, we need consider only the heat conducted in and out of a unit volume in all three coordinate directions, as shown in Figure 1-7a. The energy balance yields

𝑞𝑥 + 𝑞𝑦 + 𝑞𝑧 + 𝑞𝑔𝑒𝑛 = 𝑞𝑥+𝑑𝑥 + 𝑞𝑦+𝑑𝑦 + 𝑞𝑧+𝑑𝑧 + 𝑑𝐸𝑑𝑡

And the energy quantities are given by

𝑞𝑥 = −𝑘𝑑𝑦𝑑𝑧 𝜕𝑇𝜕𝑥

𝑞𝑥+𝑑𝑥 = −�𝑘 𝜕𝑇𝜕𝑥

+ 𝜕𝜕𝑥�𝑘 𝜕𝑇

𝜕𝑥� 𝑑𝑥� 𝑑𝑦𝑑𝑧

𝑞𝑦 = −𝑘𝑑𝑥𝑑𝑧 𝜕𝑇𝜕𝑦

𝑞𝑦+𝑑𝑦 = − �𝑘 𝜕𝑇𝜕𝑦

+ 𝜕𝜕𝑦�𝑘 𝜕𝑇

𝜕𝑦� 𝑑𝑦� 𝑑𝑥𝑑𝑧

𝑞𝑧 = −𝑘𝑑𝑥𝑑𝑦 𝜕𝑇𝜕𝑧

𝑞𝑧+𝑑𝑧 = − �𝑘 𝜕𝑇𝜕𝑧

+ 𝜕𝜕𝑧�𝑘 𝜕𝑇

𝜕𝑧� 𝑑𝑧� 𝑑𝑥𝑑𝑦

𝑞𝑔𝑒𝑛 = �̇�𝑑𝑥𝑑𝑦𝑑𝑧

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𝑑𝐸𝑑𝑡

= 𝜌𝑐𝑑𝑥𝑑𝑦𝑑𝑧 𝜕𝑇𝜕𝑡

So that the general three-dimensional heat-conduction equation is

𝜕𝜕𝑥�𝑘 𝜕𝑇

𝜕𝑥� + 𝜕

𝜕𝑦�𝑘 𝜕𝑇

𝜕𝑦� + 𝜕

𝜕𝑧�𝑘 𝜕𝑇

𝜕𝑧� + �̇� = 𝜌𝑐 𝜕𝑇

𝜕𝑡 1.9

Figure 1.7 Elemental volume for three-dimensional heat-conduction analysis: (a) cartesian coordinates; (b) cylindrical coordinates; (c) spherical coordinates.

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For constant thermal conductivity, Equation (1.9) is written

𝜕2𝑇𝜕𝑥2

+ 𝜕2𝑇𝜕𝑦2

+ 𝜕2𝑇𝜕𝑧2

+ �̇�𝑘

= 1𝛼𝜕𝑇𝜕𝑡

1.9a

Where:

𝛼 = 𝑘𝜌𝑐

:is the thermal difusivity of material, m P

2P/s.

The larger the value of α, the faster heat will diffuse through the material. This may be seen by examining the quantities that make up α. A high value of α could result either from a high value of thermal conductivity, which would indicate a rapid energy-transfer rate, or from a low value of the thermal heat capacity ρc. A low value of the heat capacity would mean that less of the energy moving through the material would be absorbed and used to raise the temperature ofthe material; thus more energy would be available for further transfer. Thermal diffusivity α has units of square meters per second.

Equation (1.9a) may be transformed into either cylindrical or spherical coordinates by standard calculus techniques. The results are as follows:

Cylindrical coordinates: see figure 1.9b

1𝑟𝜕𝜕𝑟�𝑘𝑟 𝜕𝑇

𝜕𝑟� + 1

𝑟2𝜕𝜕∅�𝑘 𝜕𝑇

𝜕∅� + 𝜕

𝜕𝑧�𝑘 𝜕𝑇

𝜕𝑧�+ �̇� = 𝜌𝑐 𝜕𝑇

𝜕𝑡 1.9b

For constant thermal conductivity,

𝜕2𝑇𝜕𝑟2

+ 1𝑟𝜕𝑇𝜕𝑟

+ 1𝑟2

𝜕2𝑇𝜕∅2

+ 𝜕2𝑇𝜕𝑧2

+ �̇�𝑘

= 1𝛼𝜕𝑇𝜕𝑡

Spherical coordinates: see figure 1.9c

1𝑟2

𝜕𝜕𝑟�𝑘𝑟2 𝜕𝑇

𝜕𝑟�+ 1

𝑟2𝑠𝑖𝑛2𝜃𝜕𝜕∅�𝑘 𝜕𝑇

𝜕∅� + 1

𝑟2 sin𝜃 𝜕𝜕𝜃�𝑘 𝑠𝑖𝑛 𝜃 𝜕𝑇

𝜕𝜃� + �̇� = 𝜌𝑐 𝜕𝑇

𝜕𝑡 1.9c

For constant thermal conductivity,

1𝑟𝜕2

𝜕𝑟2(𝑟𝑇) + 1

𝑟2𝑠𝑖𝑛2𝜃𝜕2𝑇𝜕∅2

+ 1𝑟2 sin𝜃

𝜕𝜕𝜃�sin𝜃 𝜕𝑇

𝜕𝜃� + �̇�

𝑘= 1

𝛼𝜕𝑇𝜕𝑡

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Many practical problems involve only special cases of the general equations listed above. As a guide to the developments in future chapters, it is worthwhile to show the reduced form of the general equations for several cases of practical interest.

Steady-state one-dimensional heat flow (no heat generation):

𝒅𝟐𝑻𝒅𝒙𝟐

= 𝟎 1.10

Steady-state one-dimensional heat flow in cylindrical coordinates (no heat generation):

Steady-state one-dimensional heat flow with heat sources:

Two-dimensional steady-state conduction without heat sources:

Example: A 1 m thickness wall made from a material (cp=4000J/kg.k, ρ=1600 kg/m P

3P,

k=40 W/m.k). Temperature distribution is given by 𝑇(𝑥) = 𝑎 + 𝑏𝑥 + 𝑐𝑥2, a=900˚C, b=-300 ˚C/m, c=-50 ˚C/m P

2P, A=10 m P

2P, �̇�=1000 W/m P

3P.

Determine:

1. Rate of heat transfer entering and leaving the wall. 2. Rate of storing energy within the wall. 3. Rate of change of temperature at x=0.25 m and x=0.5 m.

Solution:

Known: Temperature distribution T(x) at an instant of time t in a one-dimensional wall with uniform heat generation.

1. Recall that once the temperature distribution is known for a medium, it is a simple matter to determine the conduction heat transfer rate at any point in the medium, or at its surfaces, by using Fourier’s law. Hence the desired heat rates may be determined by using the prescribed temperature distribution with Equation

1.11

1.12

1.13

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2.1. Accordingly,

1. Recall that once the temperature distribution is known for a medium, it is a simple matter to determine the conduction heat transfer rate at any point in the medium, or at its surfaces, by using Fourier’s law.

2.apply energy balance on the wall:

3.The time rate of change of the temperature at any point in the medium may be determined from the heat equation,

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𝜕2𝑇𝜕𝑥2

+ 𝜕2𝑇𝜕𝑦2

+ 𝜕2𝑇𝜕𝑧2

+ �̇�𝑘

= 1𝛼𝜕𝑇𝜕𝑡

𝜕𝑇𝜕𝑡

= 𝑘𝜌𝑐𝑝

𝜕2𝑇𝜕𝑥2

+ �̇�𝜌𝑐𝑝

𝜕2𝑇𝜕𝑥2

= 𝜕𝜕𝑥�𝜕𝑇𝜕𝑥�

= 𝜕𝜕𝑥

(𝑏 + 2𝑐𝑥) = 2𝑐 = 2(−50°𝐶/𝑚2) = −100°𝐶/𝑚2