Heat transfer chapter one and two

11/2/2017 Heat Transfer 1
HEAT TRANSFER
(MEng 3121)
INTROUCTION
TO
HEAT TRANSFER
Chapter 1
Debre Markos University
Mechanical Engineering
Department
Prepared and presented by:
Tariku Negash
E-mail: thismuch2015@gmail.com
Lecturer at Mechanical Engineering
Department Institute of Technology, Debre
Markos University, Debre Markos, Ethiopia

11/2/2017Heat Transfer
2
 Mass transfer is the net movement of mass from
one location, usually meaning stream, phase,
fraction or component, to another. Or
 From the region of high concentration to the
lower concentration.
Examples: Evaporation of water from a pond to the
atmosphere.
Flow of energy due solely to a temperature difference
 from 2nd Law of Thermodynamics, heat flows
in direction of decreasing temperature
 heat energy can be transported through a solid,
liquid, gas, or vacuum.
1.1 Definition of Heat and Mass Transfer
a. Heat Transfer
b. Mass Transfer

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1.2 MODES OF HEAT TRANSFER
Heat can be transferred in three different modes:
Conduction, Convection, & Radiation.
A. Conduction: The transfer of energy from the more energetic particles
of a substance to the adjacent less energetic ones as a result of
interactions between the particles.
 In solids, it is due to the combination of
vibrations of the molecules in a lattice and the
energy transport by free electrons (i.e. solids
in metallic form).
 In gases and liquids, conduction is due to the
collisions and diffusion of the molecules
during their random motion.

4
The rate of heat conduction through a plane layer is proportional to
the temperature difference across the layer and the heat transfer
area, but is inversely proportional to the thickness of the layer.
Where:-
K = Thermal conductivity, : A measure of the
ability of a material to conduct heat.
dT/dx = Temperature gradient : The slope of the
temperature curve on a T-x diagram.

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Material Thermal conductivity
k (W.m-1.K-1)
Diamond 2450
Cu 385
Al 205
Brick 0.2
Glass 0.8
Body fat 0.2
Water 0.6
Wood 0.2
Styrofoam 0.01
Air 0.024
Thermal conductivity, k
property of the material
kdiamond very high: perfect heat
sink, e.g. for high power laser
diodes
khuman low: core temp relatively
constant (37oC)
kair very low: good insulator
* home insulation
* woolen clothing
* windows double glazing
i.e, Metals – good conductors: electrons
transfer energy from hot to cold

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Example:. Cold air flows above the hot copper plate.
B. Convection: The mode of energy transfer between a solid surface
and the adjacent liquid or gas that is in motion, and it involves the
combined effects of conduction and fluid motion.
Natural Convection:
Fluid motion occurs due to density variations
caused by temperature difference
Types of convection
Forced convection:
Fluid motion caused by an external agency
In the absence of any bulk fluid motion, heat transfer between a solid
surface and the adjacent fluid is by pure conduction.

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Newton’s law of cooling
h convection heat transfer coefficient, W/m2 · °C
As the surface area through which convection heat transfer takes place
Ts the surface temperature
T the temperature of the fluid sufficiently far from the surface
C. Radiation: The energy emitted by matter in the form of electromagnetic
waves (or photons) as a result of the changes in the electronic
configurations of the atoms or molecules.
• Unlike conduction and convection, the transfer of heat by radiation does
not require the presence of an intervening medium.
 In fact, heat transfer by radiation is fastest (at
the speed of light) and it suffers no attenuation
in a vacuum.
 This is how the energy of the sun reaches the
earth.

8
 In heat transfer studies we are interested in thermal radiation, which is
the form of radiation emitted by bodies because of their temperature.
 All bodies at a temperature above absolute zero emit thermal radiation.
Absorption & Stefan-Boltzmann Law
Stefan–Boltzmann law
Where:-
 Surface Area, A
 Stefan-Boltzmann constant, σ = 5.67 x 10-8 W.m-2.K-4
Applications on radiation heat transfer

Think about
 Why are fireplace pokers made of iron and not copper?
• Some animals have hair which is composed of solid tubular strands, while
others have hollow, air-filled tubes. Where would one more likely find the
latter animal: In cold climates, or warm?
• Two different materials at the same temperature have different emissivities.
Which one glows the brightest?
• Steel reinforcement bars add stability to concrete walls. Do they also
enhance the insulating value of concrete?
• Should you lower the blinds and draw the curtains on a hot day?
• When one steps from a shower on a cold morning, why does the tile floor
seem so much colder than the air?
• Place a wooden spoon and a metal spoon in the freezer. Which will cool
faster? After several hours, what would they feel like?
• Why do people become "flushed" when overheated?
• What is thermal energy? What is the difference between thermal energy and
heat?
11/2/2017 Heat Transfer

11/2/2017 Heat Transfer 10
HEAT TRANSFER
(MEng 3121)
ONE DIMENSIONAL STEADY
STATE HEAT CONDUCTION
Chapter 2
Debre Markos University
Mechanical Engineering
Department
Prepared & presented by:
Tariku N.
Oct .2017

11
2. ONE DIMENSIONAL STEADY STATE CONDUCTION
For example, consider the steady-state
conduction experiment.
 A cylindrical rod of known material is
insulated on its lateral surface, while
its end faces are maintained at
different, with T1>T2.
2.1 The Conduction Rate Equation
 The temperature difference causes conduction heat transfer in the
positive x- direction.
 How qx depends on the following variables: ∆T, the temperature
difference; ∆x, the rod length; and A, the cross-sectional area.
qx = qx (∆T, ∆x, A) using a chain rule differentiate
equation : by taking one constant and two variable

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Proportionality may be converted to an equality by introducing a
coefficient that is a measure of the material behavior. Hence, we write
What will be qx value for metal and a plastic material for
equal values of ∆T, ∆x, and A?
qx would be smaller for the plastic than for the metal.
Why? Proportionality (∝)
Where k, the thermal conductivity (W/m.k)
Therefore, k is an important property of the material.
Evaluating this expression in the limit as ∆x→ 0, 𝑤𝑒 𝑜𝑏𝑡𝑎𝑖𝑛
For the heat rate For the heat flux
i,e, the minus sign is necessary because heat is always transferred in the
direction of decreasing temperature.

13
Fourier's law (𝒒 𝒙
′′
): The direction of heat flow will always be normal
to a surface of constant temperature, called an isothermal surface.
The rate of heat transfer through a unit thickness of
the material per unit area per unit temperature
difference.
It depends on the physical structure of matter, atomic
and molecular, which is related to the state of the
matter.
The thermal conductivity of a material is a measure
of the ability of the material to conduct heat.
2.2.1 Thermal conductivity (K):
Fig 2.1. A simple experimental setup to determine the
thermal conductivity of a material.
2.2 Thermal Conductivity and Diffusivity

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A high value for thermal conductivity
indicates that the material is a good heat
conductor, and a low value indicates that
the material is a poor heat conductor or
insulator.

15
The thermal conductivities of gases such as
air vary by a factor of 104 from those of pure
metals such as copper.
Pure crystals and metals have the highest
thermal conductivities, and gases and
insulating materials the lowest.
The mechanisms of heat conduction in different phases
of a substance.

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2.2.2 Thermal Diffusivity (𝜶)
 In heat transfer analysis, the ratio of the thermal conductivity to the heat
capacity is an important property termed the thermal diffusivity 𝜶, which is
 It measures the ability of a material to conduct thermal energy relative to its
ability to store thermal energy.
 So what does it mean when materials have large and small value of 𝜶?
 For a large 𝛂 will respond quickly to changes in their thermal environment, and
,For a small 𝜶 will respond more sluggishly, taking longer to reach a new
equilibrium condition.
Example: WAX
Proof its SI units and what is the similar property in fluid flow?
m2/s. This value describes how quickly a material reacts to
a change in temperature.
 In order to predict cooling processes or to simulate temperature fields, the thermal
diffusivity must be known; it is a requisite for solving the Fourier Differential Equation for
unsteady heat conduction.

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Example 2.1: The thermal diffusivity 𝜶 is the controlling transport
property for transient conduction. Using appropriate values of k, 𝜌 , and
cp from Appendix A, (Thermophysical Properties Research Center (TPRC) at
Purdue University) show at incropera (Thermophysical Properties of Matter)
calculate for the following materials at the prescribed temperatures:
a) pure aluminum, 300 and 700 K;
b) silicon carbide, 1000 K;
c) paraffin, 300 K

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2.3 General Heat Conduction Equation
 Heat transfer problems are also classified as being one-dimensional,
two dimensional, or three-dimensional, depending on the relative
magnitudes of heat transfer rates in different directions and the level of
accuracy desired.
 In the most general case, heat transfer through a medium 3D.
 That is, the temperature varies along all three primary directions within
the medium during the heat transfer process.
 The temperature distribution throughout the medium at a specified
time as well as the heat transfer rate at any location, can be
described by a set of three coordinates such as the
i. Rectangular (or Cartesian) coordinate system: x, y, and z;
ii. Cylindrical coordinate system: 𝐫, 𝝓 𝑎𝑛𝑑 𝒛
iii. Spherical (or polar) coordinate system: 𝐫, 𝝓 𝑎𝑛𝑑 𝜽

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2.3.1 Rectangular (or Cartesian) coordinate system: x, y, and z
 Consider a small rectangular element of length 𝑑x, width d y, and height dz,
and an infinitesimally small (differential) control volume, d y. 𝑑x.d z
Temperature distribution: T(x. y, z)
 In the absence of motion (or with uniform
motion), there are no changes in mechanical
energy and no work being done on the system.
 Only thermal forms of energy need be
considered. Specifically, if there are temperature
gradients, conduction heat transfer will occur
across each of the control surfaces.
 The conduction heat rates perpendicular
to each of the control surfaces at the x-,y-,and
z- coordinate locations are indicated qx, qy,
and qz.

21
The conduction heat rates at the opposite surfaces can
then be expressed as a Taylor series expansion where,
neglecting higher-order terms,
Within the medium there may also be an energy source term associated
with the rate of thermal energy generation. This term is represented as
q: rate at which energy is generated per unit volume (w/m3)
In addition, changes may occur in the amount of the internal thermal
energy stored by the material in the control volume.

22
Based on applying conservation of energy to a differential control
volume through which energy transfer is exclusively by conduction.
The conduction heat rates in an isotropic material may be evaluated from
Fourier's law, Note: Isotropic material : they have the same
properties in all directions. Or, thermal
conductivity of a material to be independent of
direction.
Fibrous or composite, are some anisontropic
materials

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Net transfer of thermal energy into the
control volume (inflow-outflow)
p
T T T T
k k k q c
x x y y z z t

          
                 
•
Thermal energy
generation
Change in thermal
energy storage
General Cartesian coordinates, of the heat diffusion equation.
In the case of constant thermal conductivity, it reduces to
Equation is called Fourier-Biot equation, and it reduces to these forms
under specified conditions:
Additional
conditions

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2.3.2 Cylindrical coordinate system: 𝐫, 𝝓 𝑎𝑛𝑑 𝒛
When the del operator (𝜵) of heat flux is expressed in cylindrical coordinates,
the general form of the heat flux vector and hence of Fourier's law is
Where,

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Differential control volume, 𝒅𝒓 ∗ 𝒓𝒅𝝓 ∗ 𝒅𝒛, for conduction analysis in
cylinderical coordinates(𝒓, 𝝓,𝒛).
After lengthy manipulations, we obtain
3.2.3 Spherical (or polar) coordinate system: 𝐫, 𝝓 𝑎𝑛𝑑 𝜽

26
Differential control volume, 𝒅𝒓 ∗ 𝒓𝒔𝒊𝒏𝜽𝒅𝝓 ∗ 𝒓𝒅𝜽, for conduction
analysis in spherical coordinates (𝒓, 𝝓,𝜽).
Again after lengthy manipulations, we obtain

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2.4 Steady Heat Conduction In Plane Walls
 For one-dimensional conduction in a plane wall, temperature is a function
of the x-coordinate only and heat is transferred exclusively in this
direction.
 There will be no heat transfer in a direction in which
there is no change in temperature. (isothermal).
2.4.1 Heat transfer through a plane wall: Temperature
distribution and its equivalent thermal circuit.
 For steady-state conditions with no distributed source
or sink of energy within the wall, the appropriate form
of the heat equation is
 No heat generation, the heat ﬂux is a constant, independent of x. and k
is constant , the equation may be integrated twice to obtain the general
solution
(1)
(2)

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To obtain the constants of integration, C1 and
C2 , boundary conditions must be introduced.
x = 0 and x = L, in which case
In which, (3)
At, x = 0,
At, x = L, 𝑇𝑠,2 = 𝐶1 𝐿 + 𝐶2 = 𝐶1 𝐿+𝑇𝑠,1
𝑇𝑠,1 = 𝐶2
Therefore equ. (4) states that the temperature varies linearly with x
Substituting into equ (2) 𝑻 𝒙 = (𝑻 𝒔,𝟐−𝑻 𝒔,𝟏)
𝒙
𝑳
+ 𝑻 𝒔,𝟏 (4)
Heat rate (Fourier's law),
Equation
For the Heat Flux
(5)
(6)
Are constant,
independent
of x

29
2.5 Thermal Resistance
 Just as an electrical resistance is associated with the conduction of
electricity, a thermal resistance may be associated with the conduction of
heat.
 Defining resistance as the ratio of a driving potential to the
corresponding transfer rate,
𝒂. thermal resistance for conduction in a plane wall is
 For electrical conduction in the same system, Ohm's law
provides an electrical resistance of the form
b. thermal resistance for convection heat transfer at a surface. (Newton's
cooling system)
𝑅 𝑒 =
𝑉1 − 𝑉2
𝐼

30
The equivalent thermal circuit for the plane wall with convection
surface conditions
The heat transfer rate may be determined from
separate consideration of each element in the
network.
Since qx is constant throughout the network, it
follows that
In terms of the overall temperature difference,
c. thermal resistance for radiation
and the total thermal resistance,
Rtot the heat transfer rate

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2.6 The Composite Wall (multilayer plane)
The composite walls that involve any number of series and parallel thermal
resistances due to layers of different materials.
2.7 Overall heat transfer coefﬁcient (U)
With composite systems, it is often convenient to work with an overall
heat transfer coefﬁcient U, which is defined by an expression analogous
to Newton's law of cooling. Accordingly,
𝑤ℎ𝑒𝑟𝑒 ∆T: is the overall temperature difference.(7)

32
In general, we may write
2.7.8 Equivalent thermal circuits for a series–parallel composite wall
Draw thermal circuit for block
diagram
Or,
For case (a) it is presumed that surfaces normal to the x-direction are
isothermal, whereas for case (b) it is assumed that surfaces parallel to
the x-direction are adiabatic

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2.8 Thermal Contact Resistance
 In the analysis of heat conduction through multilayer solids, we assumed
“perfect contact” at the interface of two layers, and thus no temperature
drop at the interface.
 This would be the case when the surfaces are perfectly smooth and
they produce a perfect contact at each point.
 In reality, however, even flat surfaces that appear smooth to the eye
turn out to be rather rough when examined under a microscope,
 The temperature change is attributed to what is known
as the thermal contact resistance, Rt,c.
 From the fig, for a unit area of the
interface, the resistance is defined as
(8) So, how Rt,c can be decrease ?
i. Increasing the joint pressure and/or reducing the
roughness of the mating surfaces.
ii. By selecting an interfacial fluid of large thermal conductivity.

34
a. What will be occurred
if Injera baking (mitad)
is produced by mixing
clay to metal powder.
And what would be
thermal contact resistance
b/n them
b. With what material
did u recommend the clay
would mix in order
to increase thermal
Conductivity or (decrease
Rt,c)
c. Do u have any idea that
increase K of the Mitad
by using composite
materials
Think about it

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2.9 Radial Systems : The Cylinder
A common example is the hollow cylinder whose inner and outer surfaces
are exposed to fluids at different temperatures
For steady-state conditions with no
heat generation, the appropriate form
of the heat equation,
For Fourier's law,
where A = 2𝜋rL is the area
normal to the direction of
heat transfer.
From eq 8, the quantity (𝑘𝑟
𝑑𝑇
𝑑𝑟
= 0) is independent of r, then the conduction
heat transfer rate qr eqn 9 is constant in the radial direction. But, for the heat flux
𝑞 𝑟
′′
is dependent on radial direction
(8)
(9)

36
Double integration for eq (8) by assuming the value of k to be
constant
To obtain the constants of integration C1 and C2 , we introduce the following
boundary conditions:
Applying these conditions to the general solution, we then obtain
Solving for C1 and C2 and substituting into the general solution, we then obtain
(10)
Now substitute eqn (10) into eqn (9) (11)
The thermal resistance for radial system: (12)

37
 For an overall heat transfer coefficient.
If U is defined in terms of the
inside area, A = 2𝜋r1L
(13)
(14)
Eqn. (13) and (14) can be yield
Arbitrary overall heat transfer coefficient
(15)

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2.10 The Sphere
For the differential control volume of the figure, energy conservation requires
For steady-state, one-dimensional conditions with no
heat generation. The appropriate form of Fourier's
law is
that qr=qr+dr .
Where 𝑨 = 𝟒𝝅𝒓 𝟐 is the area normal to the direction of heat transfer.
(16)
qr is constant, independent of r. eqn (16) may be expressed integral form
Assuming constant k, we then obtain
Thermal resistance for sphere
(17)
(18)

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2.11 Critical Thickness of Insulation : sphere and cylinder
 Let us consider a layer of insulation which might
be installed around a circular pipe, (cylinder) as
shown in Figure.
 The inner temperature of the insulation is ﬁxed at
Ti, and the outer surface exposed to a
convection environment 𝑻∞.
 From the thermal network the heat transfer is
 Now let us manipulate this expression to determine the outer radius of
insulation 𝑟0 , which will maximize the heat transfer. The
maximization condition is
which gives the result (19)
For sphere (20)

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Table 2.3 One-dimensional, steady-state solutions to the heat equation with no
generation (no thermal energy generation)
Summary

41
2.12 Conduction With Thermal Energy Generation
a. Plane wall with thermal energy Generation
Consider the Asymmetric plane wall of Figure a, in
which there is uniform energy generation per unit volume
(q is constant ) the surface is maintained at Ts,1 and Ts,2
 For constant thermal conductivity k, the appropriate
form of the heat equation,
 The general solution is
where C1 and C2 are the constants of integration. For the prescribed
boundary conditions,
Fig a

42
The constants may be evaluated and are of the form
in which case the temperature distribution is
The heat flux at any point in the wall may, of course, be determined by
using Equation (21) with Fourier's law. Note, however, that with
generation the heat ﬂux is no longer independent of x.
(21)
Fig b
For symmetric plane wall Fig b, both surfaces are
maintained at a common temperature,
𝑻 𝒔,𝟏 ≡ 𝑻 𝒔,𝟐 ≡ 𝑻 𝒔.
The temperature distribution is then symmetrical
about the mid plane, using above eqn (21)
(22)
In w/ch plane that the maximum temperature exists?

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The maximum temperature exists at the mid plane
in which case the temperature distribution, Equation (22) , may be
expressed as
 Fig c is represented no heat transfer across the surface
(adiabatic surface at mid plane)
Fig c
 Using eqn. (22), the plane walls that are perfectly
insulated on one side (x=0) and maintained at a fixed
temperature 𝑻 𝒔. on the other side (x=L).
Neglecting radiation and substituting the appropriate
rate equations, the energy balance given by Equation
Substituting from Equation 22 to obtain the temperature gradient at x = L,
it follows that Hence 𝑻 𝒔 may be computed from
knowledge of 𝑇∞. , L, and h.
(23)
(24)

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b. Radial system with thermal energy Generation
Heat generation may occur in a variety of radial geometries.
Consider the long, solid cylinder of figure 2a, which could represent a
current-carrying wire or a fuel element in a nuclear reactor.
For constant thermal conductivity k,
Separating variables and assuming uniform generation,
this expression may be integrated to obtain
Repeating the procedure, the general solution for the temperature distribution
becomes;
To obtain the constants of integration C1 and C2 , we apply boundary conditions:
(25)
(26)

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For solid cylinder the centerline is a line of symmetry (symmetry condn)
 temperature distribution and the temperature gradient must be zero.
 symmetrical boundary conditions (Figure b).
 r = 0 and Equation 25, it is evident that C1=0
Using the surface boundary condition at r = r0 with equation 26, we then
obtain
The temperature distribution is therefore
∴ the heat rate at any radius in the cylinder may, of course, be evaluated
by using Equation (27) with Fourier's law.
(27)
Where T0 is the centerline temperature.
Evaluating Equation 27 at the centerline and dividing the result into Equation
27, we obtain the temperature distribution in non dimensional form,
(28)

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To relate the surface temperature, 𝑻 𝒔, to the temperature of the cold
fluid 𝑻∞, there is two methods
i. surface energy balance
ii. an overall energy balance.
Choosing the second approach, we obtain
Or, (29)
c. Sphere system with thermal energy Generation
𝟏
𝒓 𝟐
𝝏
𝝏𝒓
𝒌𝒓 𝟐
𝝏𝑻
𝝏𝒓
+
𝒒
𝒌
= 0
𝑻 𝒓 = 𝑻 𝒔 +
𝒒𝒓 𝟐
𝟐
𝟔𝒌
𝟏 −
𝒓 𝟐
𝒓 𝟐
𝟐
Eat conduction on sphere (polar) one dimension and steady state with
thermal generation
After some mathematical calculation
(30)

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Table2.4One-Dimensional,Steady-StateSolutionstothe
HeatEquationforPlane,cylindrical,andSphericalWalls
withUniformGenerationandAsymmetricalSurface
Conditions
(30.1)
(30.2)
(30.3)
(30.4)
(30.5)
(30.6)
(30.7)
(30.8)
(30.9)

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2.13 Heat Transfer from Extended Surfaces (Fins)
 Until now, we have considered heat transfer from the
boundaries of a solid to be in the same direction as
heat transfer by conduction in the solid.
 In contrast, for an extended surface, the direction of
heat transfer from the boundaries is perpendicular to
the principal direction of heat transfer in the solid.
 Consider a strut that connects two walls at different temperatures and
across which there is fluid flow.
 With T1 >T2 temperature gradients in the x-direction sustain heat
transfer by conduction in the strut.
 However, with T1 >T2 >T∞ there is concurrent heat transfer by
convection to the fluid, causing qx, and hence the magnitude of the
Temperature gradient, dT/dx, to decrease with increasing x.

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 If T 𝒔 fixed plane wall of fig a , there are
two ways in which the heat transfer rate
may be increased.
 The convection coefficient h could be increased, by
However, there are many situations for which increasing h to the maximum
possible value is either insufficient to obtain the desired heat transfer rate or
the associated costs are prohibitive (blower or pump power requirements).
i. increasing the fluid velocity, and/or the
ii. increasing fluid temperature T∞ could be reduced.
iii. by increasing the surface area across which the convection occurs on fig.
b, this may be done by employing n 𝒔 that extend from the wall into
the surrounding fluid is called fin.
a b

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Application of Fins

51
2.13 The Fin Equation
Some typical fin configurations:
 For fig a it may be of uniform cross-sectional area (A=t*w), or
Fig b its cross-sectional area may vary with the distance x from the wall.
 Fig (c) an annular fin is one that is circumferentially attached to a
cylinder, and its cross section varies with radius from the wall of the
cylinder. 𝐴 𝑎 = 2𝜋𝑟
 Fig (d) a pin fin, or spine, is an extended surface of circular cross
section. Pin fins may also be of uniform or non uniform cross section.
In any
A straight fin is any extended
surface that is attached to a
plane wall.

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It may depend on space, weight, manufacturing, and cost considerations,
as well as on the extent to which the fins reduce the surface convection
coefficient and increase the pressure drop associated with flow over the
fins.
Selection of a particular fin configuration:
i. A general form of the energy equation for a fin
Applying the conservation of energy
From Fourier's law,
Ac: the fin cross-sectional area which is vary with x
Since the conduction heat rate at x + dx may be expressed as

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The convection heat transfer rate may be expressed as
dAs: is the surface area of the differential element.
Or
Its solution for appropriate boundary conditions provides the temperature
distribution.
ii. Fins of Uniform Cross-Sectional Area
Each fin is attached to a base surface of
temperature T(0) = Tb and extends into a
fluid of temperature 𝑻∞
Where; P: the fin perimeter
Ac: constant the fin cross-sectional area
As : surface area = Px
(31)
Fig. Straight fins of uniform cross
section. (a) Rectangular fin. (b) Pin fin.

54
Equation (31) reduces to
To simplify the form of this equation, we transform the dependent variable
by defining an excess temperature 𝜽 as
𝜽 𝒙 = 𝑻 𝒙 − 𝑻∞
(32)
(33)
where, since 𝑻∞ is a constant, d𝜃/𝑑𝑥 = dT/dx. Substituting Equation (33)
into Equation (32) , we then obtain
Where, 𝑚2 =
ℎ𝑃
𝐾𝐴 𝑐
Eqn. (34) is a linear, homogeneous, second-order differential
equation with constant coefficients. Its general solution is of
the form
(34)
Evaluate the constants C1 and C2 by specifying boundary conditions,
(35)
at the base of the fin (x = 0), (36)

55
2nd condition, specified at the fin tip (x = L), may correspond to one
of four different physical situations.
Case A
Considers convection heat transfer from the fin tip.
Applying an energy balance to a control surface about this tip
Or
That is, the rate at which energy is transferred to the fluid
by convection from the tip must equal the rate at which
energy reaches the tip by conduction through the fin.
(37)
Substitute eqn. 35 in to eqn 36 and 37
And,
(37.1)
(37.2)

56
Solving C1 and C2 it may be shown, after some manipulation, that
(38)
By using below hyperbolic eqn hint Solving

57
Fin heat transfer rate qf
The amount of heat transferred from the entire fin ca be calculate with two
alternative ways, both of which involve use of the temperature
distribution.
1st The simpler procedure, and the one that we will use, involves
applying Fourier's law at the fin base. That is
Hence, knowing the temperature distribution, 𝜽 𝒙 , qf may be evaluated,
giving
(40)
2nd by using conservation of energy principles
Which is the rate at which heat is transferred by convection from the
fin must equal the rate at which it is conducted through the base of the
fin.
(39)

58
(41)
where Af is the total, including the tip, ﬁn
surface area.
Or, by substituting Eqn (38) in to Eqn
(41) we can get Eqn (40)
Case B: Assume when the fin tip is insulated (adiabatic)
(35)
Substituting from Equation 35 and dividing by m, we then obtain
Where,
(42)
(43)
like Eqn (35), Solving C1 and C2 by using eqn (38.1) and (43) and
substitute into eqn (35) it may be shown, after some manipulation,
(44)
Using this temperature distribution with Equation (39) , the fin heat
transfer rate is then (45)

59
Case C: Temperature is prescribed at the fin tip
That is, the second boundary condition is 𝜽 𝑳 = 𝜽 𝑳, and the resulting
expressions are of the form
(46)
(47)
The very long ﬁn n, is an interesting extension of
these results. In particular, as L→ ∞, 𝜽 𝑳 → 𝟎 and
it is easily verified that,
Case D: For L→ ∞
(48)
(49)

60
Temperature distribution and heat loss for ﬁns of uniform cross section
Summary
See A table of hyperbolic functions at Appendix B1 in your text book (Fundamentals
of Heat and Mass Transfer)

61
2.14 Fin Performance
 Recall that fins are used to increase the heat transfer from a surface by
increasing the effective surface area.
 However, the fin itself represents a conduction resistance to heat
transfer from the original surface.
 For this reason, there is no assurance that the heat transfer rate will
be increased through the use of fins.
 Therefore, we should evaluate the fin (n) effectiveness, 𝜺 𝒇 which is,
𝜺 𝒇 =
𝒒 𝒇
𝒉𝑨 𝒄,𝒃 𝜽 𝒃
𝜺 𝒇 =
heat transfer rate with ﬁn
heat transferexist without the ﬁn
=
𝒒 𝒇
𝒒 𝒏𝒐𝒇
(50) Where, 𝑨 𝒄,𝒃 is the fin cross-sectional
area at the base.
By substituting eqn (49) which is at the case D
for 𝒒 𝒇 in to eqn (50) then,
𝜺 𝒇 =
𝒌𝑷
𝒉𝑨 𝒄,𝒃
(51)

62
i. k should be as high as possible, (copper, aluminum, iron).
Aluminum is preferred: low cost and weight, resistance to corrosion.
𝒊𝒊.
𝑷
𝑨 𝒄,𝒃
should be as high as possible. (Thin plate fins and slender pin fins)
iii. Most effective in applications where h is low. (Use of fins justified if when
the medium is gas and heat transfer is by natural convection).
𝜺 𝒇 =
𝒌𝑷
𝒉𝑨 𝒄,𝒃
Increasing methods of 𝜺 𝒇 from the formula
If
𝜺 𝒇 =
𝒒 𝒇
𝒒 𝒏𝒐𝒇
Therefore, in any rational design the value of should be as large as possible,
𝜺 𝒇 ≥ 𝟐

63
2.15 Measurement of Fin performance .
𝑹 𝒕,𝒇 =
𝜽 𝒃
𝒒 𝒇
𝒘𝒉𝒆𝒓𝒆, 𝜽 𝒃 𝒊𝒔 𝒕𝒉𝒆 𝒅𝒓𝒊𝒗𝒊𝒏𝒈 𝒑𝒐𝒕𝒆𝒏𝒕𝒊𝒂𝒍 = the d/c b/n the base
and the fluid temperature
The thermal resistance due to convection at the exposed base,
𝑹 𝒕,𝒃 =
𝟏
𝒉𝑨 𝒄,𝒃
and substituting from Equation (50) ,it follows that
𝒉𝑨 𝒄,𝒃 =
𝟏
𝑹 𝒕,𝒃
𝜺 𝒇 =
𝑹 𝒕,𝒃
𝑹 𝒕,𝒇
 Hence the fin effectiveness may be interpreted as a ratio of thermal
resistances, and to increase 𝜺 𝒇 it is necessary to reduce the
conduction/convection resistance of the fin.
 If the fin is to enhance heat transfer, its resistance must not exceed that of
the exposed base.
A. Fin performance in terms of a thermal resistance
(52)
(53)
(54)

64
A. Fin performance in terms of a its efficiency (𝜼 𝒇)
𝜼 𝒇 =
Actual heat transfer rate from the fin
Ideal heat transfer rate from the fin if the entire fin were at base temperature
Case B
Case D
Where, 𝑚 =
ℎ𝑃
𝐾𝐴 𝑐
 In the limiting case of zero thermal resistance or
infinite thermal conduction ( k → ∞ ) the
temperature of the fin will be uniform at the base
value of 𝑇𝑏
𝜼 𝒇 =
𝒒 𝒇
𝒒 𝒎𝒂
=
𝒒 𝒇
𝒉𝑨 𝒇 𝜽 𝒃
(55)
(56)
(57)

65
2.16 Corrected fin length (𝑳 𝒄)
For a Case B using equation (45)
Corrected fin length of the form 𝑳 𝒄 = 𝑳 + (𝒕/𝟐)
for a rectangular fin and 𝑳 𝒄 = 𝑳 + (𝑫/𝟒) for a pin
fin.
By using equation (57)
If the width of a rectangular fin is much larger than
its thickness, w >> t, the perimeter may be
approximated as P =2w, and
Where, 𝑚 =
ℎ𝑃
𝐾𝐴 𝑐

66
Figure 2.1 Efficiency of straight fins (rectangular, triangular, and
parabolic profiles).
(58)
Hence, as shown in Figures 2.1 and 2.2, the efficiency of a rectangular fin with tip
convection may be represented as a function of
Where,

67
Figure 2.2 Efficiency of annular fins of rectangular profile.

68
2.17 Expressions for the efficiency and surface area of several common
fin geometries are summarized in Table 2.6.
Table 2.6 Efﬁciency of common ﬁn shapes

69
Table2.6continued

70
2.18 Overall Surface Efﬁciency
In the contrast the fin efficiency 𝜼 𝒇 , which characterizes the performance of
a single fin, the overall surface efﬁciency 𝜼 𝟎 characterizes an array of fins
and the base surface to which they are attached
The total rate of heat transfer by convection from the fins and the prime (un finned)
surface may be expressed as
Where,
If there are N fins in the array, each of
surface area Af , and the area of the prime
h is assumed to be equivalent for the finned and prime surfaces.
𝜼 𝒇 is the efficiency of a
single fin.
surface is designated as At, the total surface area.
(59)
(60)

71
Substituting Equation (59) into (60), it follows that
Equation (59) may be used to infer an expression for the thermal resistance
( 𝑹 𝒕,𝒇) using eqn (52) of a fin array. That is,
𝑹 𝒕,𝒇 =
𝜽 𝒃
𝒒 𝒇
(61)
(62)
Where, 𝑹 𝒕,𝟎s an effective resistance
that accounts for parallel heat flow
paths by conduction/convection in the
fins and by convection from the prime
surface.

72
2.19 Fins attachment Methods
1st Fins are machined as
an integral part of the
wall from which they
extend more
commonly, fig a or
manufactured
separately and are
attached to the wall by
a metallurgical or
adhesive joint.
2nd By a press ﬁt, for
which the fins are
forced into slots
machined on the wall
material. fig b
fig a
fig b
Draw thermal circuits for each array

73
 In such cases (fig b),there is a thermal contact resistance 𝑹 𝒕.𝒄, may
adversely influence overall thermal performance.
 An effective circuit resistance may again be obtained, where,
with the contact resistance,
the corresponding overall surface efficiency is
Where,
In manufacturing, care must be taken to render

74
Reference
This lecture power point adapted from
1. Yunus Cengel, Heat and Mass Transfer A Practical Approach,
3rd edition
2. Jack P. Holman, Heat Transfer, Tenth Edition.
3. Frank P. Incropera, Theodore l. Bergman, Adrienne S.
Lavine, and David P Dewitt, fundamental of Heat and Mass
Transfer, 7th edition
4. Lecture power point of heat transfer by Mehmet Kanoglu
University of Gaziantep

Heat transfer chapter one and two

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