CHEE 3101 Heat Transfer
MINISTRY OF MANPOWER
DIRECTORATE OF TECHNOLOGICAL EDUCATION
HIGHER COLLEGE OF TECHNOLOGY
ENGINEERING DEPARTMENT
MECHANICAL ENGINEERING SECTION
Specialization: Chemical Engineering
Level
:
Advanced Diploma
Course Name :
Heat Transfer
Course Code :
CHEE 3101
Student Name
Student ID No.
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CHEE 3101 Heat Transfer
CHAPTER 1
Fundamentals of Heat Transfer
(Outcomes 1)
1.1 Introduction
The present standard of living is made possible by the energy available in the form
of heat from various sources like fuels. The process by which this energy is
converted for everyday use is studied under thermodynamics, leaving out the rate
at which the energy is transferred. In all applications, the rate at which energy is
transferred as heat plays an important role. The design of all equipment involving
heat transfer requires the estimate of the rate of heat transfer.
There is no need to list the various equipment where heat transfer rate influences
their operation. The driving potential or the force which causes the transfer of
energy as heat is the difference in temperature between systems. Other such
transport processes are the transfer of momentum, mass and electrical energy. In
addition to the temperature difference, physical parameters like geometry, material
properties like conductivity, flow parameters like flow
velocity also influence the rate of heat transfer.
The aim of this text is to introduce the various rate equations and methods of
determination of the rate of heat transfer across system boundaries under different
situations.
1.2 Thermodynamics
•
•
•
•
Energy can be transferred between a system and its surroundings.
A system interacts with its surroundings by exchanging work and heat
Deals with equilibrium states (Ideal conditions).
Does not give information about:
– Rates at which energy is transferred
– Mechanisms through with energy is transferred
– Factors affecting the energy transfer
– Real time problems & applications
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CHEE 3101 Heat Transfer
In this chapter we will learn
 What is heat transfer
 How is heat transferred
 Relevance and importance
1.3 Thermodynamics Vs Heat transfer
Thermodynamics tells about
• how much heat is transferred
• how much work is done
• final state of the system
Heat transfer tells about
• how heat is transferred (different modes of transfer)
• At what rate heat is transferred
• Temperature distribution inside the body
Heat transfer deals about study of thermal energy transfer that is induced by a
temperature difference (or gradient)
1.4 Modes of heat transfer
• Conduction heat transfer: Occurs when a temperature gradient exists
through a solid or a stationary fluid (liquid or gas).
• Convection heat transfer: Occurs within a moving fluid, or between a solid
surface and a moving fluid, when they are at different temperatures.
• Thermal radiation: Heat transfer between two surfaces (that are not in
contact), often in the absence of an intervening medium.
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1.4.1 Conduction
Transfer of energy from the more energetic to less energetic particles of a
substance by contact of particles of matter.
Mechanisms of conduction:
 Lattice Vibration
 Particle collisions
Lattice Vibration:
Particle collision (observed in matters having lot of free electrons)
• Conduction through electron collisions is more effective than conduction
through lattice vibration
• Metals (particle collision) are good conductors of heat (Eg: Copper) as they
have lot of free electrons.
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CHEE 3101 Heat Transfer
• Ceramic materials, inert gases etc., doesn’t exhibit good conducting nature
as they have very less free electrons. These materials are called Insulators.
Conduction expressed by Fourier’s law of heat conduction as written below
Fig. Heat conduction through a large plane wall of thickness Δx and area A
𝐴𝑟𝑒𝑎 𝑋 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒
Rate of heat conduction α
𝑇ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠
𝑇1−𝑇2
ΔT
dT
𝑄̇ conduction = k A
= - kA = - kA
Δx
Δx
dx
Heat flux is the heat transfer per unit time per unit area, and is equal to
𝑄̇
q′̇ =
𝑨
Where,
•
•
dT
dx
is the temperature gradient (oC/m)
k is the thermal conductivity(W/moC or W/moK)
• A is the area which is normal to the direction of heat transfer.(m2)
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1.4.2 Convection
Energy transfer by random molecular motion (as in conduction) plus bulk
(macroscopic) motion of the fluid. Relative motion between the surfaces is
essential for convection.
Typical convective scenario:
1. Hot surface heats the surrounding fluid (at the interface)
2. The interfacial fluid gets heated and becomes warmer.
3. A fluid movement gets created which replaces the warmer fluid (at the
interface) with cooler fluid (in the main stream), which can draw more
heat away from the surface.
Based on the nature of the fluid movement creation, convection is classified as:
o Natural (free) convection: flow induced by buoyancy forces, arising from
density differences arising from temperature variations in the fluid
o Forced convection: flow movement caused by external means like fan,
blower etc.
Natural Convection:
In this case, the fluid movement is created by the warm fluid itself. The density of
fluid (at the interface) decrease as it is heated; thus, hot fluids become lighter than
cool fluids. Warm fluid surrounding a hot object rises, and is replaced by cooler
fluid resulting in the self-induced fluid movement. Eg: Boiling of water in a kettle
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Forced Convection:
In this case, the fluid movement is created by an external agent like a fan or
blower.
Eg: Forced convection is what makes a windy, winter day feel much colder than a
calm day with same temperature. The heat loss from your body is increased due to
the constant replenishment of cold air by the wind.
Natural wind and fans are the two most common sources of forced convection.
Convective effect is predominantly felt in forced convection rather than free
convection
Convection is expressed by Newton’s law of cooling as
𝑄̇ = h As (Ts - T∞)
Where,
• h is the convection heat transfer coefficient (W/m2.oC or W/m2.oK)
• As is the surface area through which convection heat transfer takes place
(m2)
• Ts is the surface temperature (oC)
• T∞ is the temperature of the fluid sufficiently far from the surface (oC)
1.4.3 Radiation
Radiation is the energy emitted by matter in the form of electromagnetic waves (or
photons) as a result of the changes in the electronic configurations of the atoms or
molecules. Unlike conduction and convection, the transfer of energy by radiation
does not require the presence of an intervening medium. In fact, energy transfer by
radiation is fastest (at the speed of light) and it suffers no attenuation in a vacuum.
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CHEE 3101 Heat Transfer
Any body maintain above absolute zero kelvin temperature can emit radiation.
This is how the energy of the sun reaches the earth.
The maximum rate of radiation that can be emitted from a surface at an absolute
temperature Ts (in K or R) is given by Stefan–Boltzmann law as
𝑄̇ = σ As Ts4
where σ = 5.67 X 10-8 W/m2·K4 is the Stefan–Boltzmann constant. The idealized
surface that emits radiation at this maximum rate is called a blackbody, and the
radiation emitted by a blackbody is called blackbody radiation.
Emissive power E is the rate at which energy is released per unit area (W/m2)
(radiation emitted from the surface)
Irradiation G is the rate of incident radiation per unit area (W/m2) of the surface
(radiation absorbed by the surface), originating from its surroundings.
******************
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CHEE 3101 Heat Transfer
CHAPTER 2

(Outcomes: 2, 3, and 4)
Heat Conduction Equation
 Distinguish between steady and unsteady heat transfer.
 Realize the significance of Fourier’s law, thermal conductivity and heat transfer
coefficients in solving heat transfer problems.
 Calculate temperature distribution and heat transfer rates by conduction through
solids, in flat slabs/plates, in pipes, in cylinders- as single body or in series.
2.1
Steady versus Unsteady (Transient) Heat Transfer
Heat transfer problems are often classified as being steady (also called steady state)
or transient (also called unsteady). The term steady implies no change with time at
any point within the medium, while transient implies variation with time or time
dependence. Therefore, the temperature or heat flux remains unchanged with time
during steady heat transfer through a medium at any location, although both
quantities may vary from one location to another (Fig. 2-1). For example, heat
transfer through the walls of a house will be steady when the conditions inside the
house and the outdoors remain constant for several hours. But even in this case, the
temperatures on the inner and outer surfaces of the wall will be different unless the
temperatures inside and outside the house are the same. The cooling of an apple in
a refrigerator, on the other hand, is a transient heat transfer process since the
temperature at any fixed point within the apple will change with time during
cooling. During transient heat transfer, the temperature normally varies with time
as well as position.
As a special Case, the temperature of the medium changes uniformly with time but
not with position. Such heat transfer systems are called lumped systems. A small
metal object such as a thermocouple junction or a thin copper wire, for example,
can be analyzed as a lumped system during a heating or cooling process. Most heat
transfer problems encountered in practice are transient in nature, but they are
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CHEE 3101 Heat Transfer
usually analyzed under some presumed steady conditions since steady processes
are easier to analyze, and they provide the answers to our questions.
Fig.2-1 Steady and Transient Heat Conduction in Plane Wall
2.2.0 Different coordinate systems
The Various distances and angles involved when describing the location of a point
in different coordinate systems as shown below in the figure.
Fig.2-2 Different Coordinates System
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2.2.1 Multidimensional Heat Transfer
A Heat transfer problems are also classified as being:
 one-dimensional
 two dimensional
 three-dimensional
In the most general case, heat transfer through a medium is three-dimensional.
However, some problems can be classified as two- or one-dimensional depending
on the relative magnitudes of heat transfer rates in different directions and the level
of accuracy desired.
One-dimensional if the temperature in the medium varies in one direction only and
thus heat is transferred in one direction, and the variation of temperature and thus
heat transfer in other directions are negligible or zero.
Two-dimensional if the temperature in a medium, in some cases, varies mainly in
two primary directions, and the variation of temperature in the third direction (and
thus heat transfer in that direction) is negligible.
Fig. 2-3
Fig. 2-4
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CHEE 3101 Heat Transfer
The rate of heat conduction through a
medium in a specified direction (say, in the
x-direction) is expressed by Fourier’s law of
heat conduction for one-dimensional heat
conduction as:
(Eqn. 2-1)
Heat is conducted in the direction of
decreasing temperature, and thus the
temperature gradient is negative when heat
is conducted in the positive x -direction.
Fig. 2-5
The heat flux vector at a point P on the surface of figure must be perpendicular to
the surface, and it must point in the direction of decreasing temperature
If n is the normal of the isothermal surface at point P, the rate of heat conduction
at that point can be expressed by Fourier’s law as
(Eqn. 2-2)
Fig. 2-6
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2.3
Heat generation
A medium through which heat is conducted may involve the conversion of
electrical, nuclear, or chemical energy into heat (or thermal) energy. In heat
conduction analysis, such conversion processes are characterized as heat
generation. For example when electric current passes through wire it generate heat,
likewise a large amount of heat is generated in the fuel elements of nuclear reactor,
another source of heat generation in a medium is exothermic chemical reactions
that may occur throughout the medium.
Note that heat generation is a volumetric phenomenon. Therefore, the rate of heat
generation in a medium is usually specified per unit volume and is denoted by 𝑔̇ ,
whose unit is W/m3. The total rate of heat generation in a medium of volume V can
be determined from
𝐺̇ = ∫𝑣 𝑔̇ 𝑑𝑣
(Eqn.2-3)
In the special case of uniform heat generation, as in the case of electric resistance
heating throughout a homogeneous material, the relation in Eqn. 2.3 reduces to
𝐺̇ = 𝑔̇ 𝑉
(Eqn.2-4)
where 𝑔̇ is the constant rate of heat generation per unit volume.
Example: 2.1.
The resistance wire of a 1200-W hair dryer is 80 cm long and has a diameter of
0.3 cm. Determine the rate of heat generation in the wire per unit volume, in
W/cm3, and the heat flux on the outer surface of the wire as a result of this heat
generation.
Solution: Power consumed by the resistance wire of a hair dryer is given. The heat
generation and the heat flux are to be determined.
Assumptions: Heat is generated uniformly in the resistance wire.
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Analysis : A 1200-W hair dryer will convert electrical energy into heat in the wire
at a rate of 1200 W. Therefore, the rate of heat generation in a resistance wire is
equal to the power consumption of a resistance heater. Then the rate of heat
generation in the wire per unit volume is determined by dividing the total rate of
heat generation by the volume of the wire,
Similarly, heat flux on the outer surface of the wire as a result of this heat
generation is determined by dividing the total rate of heat generation by the surface
area of the wire,
2.4 Thermal conductivity
As discussed earlier, the heat conduction is the transmission of energy by
molecular action. Thermal conductivity is the property of a particular substance
and it can be defined as the rate of heat transfer through a substance (i.e., material)
per unit area per unit temperature difference.
Higher the thermal conductivity more easily will be the heat conduction through
the substance and it indicates that the substance is good conductor, and low value
indicates that the substance is poor heat conductor or good insulator.
It can be realized that the thermal conductivity of a substance would be dependent
on the chemical composition, phase (gas, liquid, or solid), crystalline structure (if
solid), temperature, pressure, and its homogeneity. The general notation of thermal
conductivity is ‘k’ and the unit is W/moC or W/moK.
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Table-2.1: Thermal conductivity of various substances [8]
Table-2.2: Thermal conductivity of mercury at three different phases [8]
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The general results of the careful analysis of the Table-2.1 and 2.2 are as follows,





Thermal conductivity depends on the chemical composition of the
substance.
Thermal conductivity of the liquids is more than the gasses and the metals
have the highest.
Thermal conductivity of the gases and liquids increases with the increase in
temperature.
Thermal conductivity of the metal decreases with the increase in
temperature.
Thermal conductivity is affected by the phase change.
2.5 Thermal diffusivity/ specific heat/ heat capacity
Now we know that the thermal conductivity facilitates the heat to propagate
through the material due to the temperature gradient.
Similarly, specific heat is the capacity of heat stored by a material due to variation
in temperature. Thus the specific heat (unit: kJ/kg·oC) is defined as the amount of
thermal energy required to raise the temperature of a unit amount of material by
1oC.
The product ρCp, which is frequently encountered in heat transfer analysis, is
called the heat capacity of a material. Both the specific heat C p and the heat
capacity ρCp represent the heat storage capability of a material. But Cp expresses
it per unit mass whereas ρCp expresses it per unit volume, as can be noticed from
their units J/kg · °C and J/m3 · °C, respectively.
Another material property that appears in the transient heat conduction analysis is
the thermal diffusivity, which represents how fast heat diffuses through a material
and is defined as
α= Heat conducted / Heat stored = k /ρCp
(Eqn.2.5)
Note that the thermal conductivity k represents how well a material conducts heat,
and the heat capacity ρCp represents how much energy a material stores per unit
volume. Therefore, the thermal diffusivity of a material can be viewed as the ratio
of the heat conducted through the material to the heat stored per unit volume. A
material that has a high thermal conductivity or a low heat capacity will obviously
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have a large thermal diffusivity. The larger the thermal diffusivity, the faster the
propagation of heat into the medium. A small value of thermal diffusivity means
that heat is mostly absorbed by the material and a small amount of heat will be
conducted further.
You may be aware that any flow whether it is electricity flow, fluid flow, or heat
flow needs a driving force. The flow is proportional to the driving force and for
various kinds of flows the driving force is shown in the table 2.3.
Table 2.3. Various flows and their driving forces
Thus the heat flow per unit area per unit time (heat flux,
the following relation,
) can be represented by
Where, proportionality constant k is the thermal conductivity of the material, T is
the temperature and x is the distance in the direction of heat flow. This is known as
Fourier’s law of conduction.
The term steady-state conduction is defined as the condition which prevails in a
heat conducting body when temperatures at fixed points do not change with time.
The term one-dimensional is applied to a heat conduction problem when only one
coordinate is required to describe the distribution of temperature within the body.
2Such a situation hardly exists in real engineering problems. However, by
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considering one-dimensional assumption the real problem is solved fairly upto the
accuracy of practical engineering interest.
2.6 One Dimensional Steady-State Conduction through Constant Area
A simple case of steady-state, one-dimensional heat conduction can be considered
through a flat wall as shown in the fig.2.1.
Fig.2-7: Steady-state conduction through a slab (constant area)
The flat wall of thickness dx is separated by two regions, the one region is at high
temperature (T1 ) and the other one is at temperature T2 . The wall is very large in
comparison of the thickness so that the heat losses from the edges are negligible.
Consider there is no generation or accumulation of the heat in the wall and the
external surfaces of the wall are at isothermal temperatures T1 and T2 . The area of
the surface through which the heat transfer takes place is A. Then the eqn.2.7 can
be written as,
The negative sign shows that the heat flux is from the higher temperature surface
to the lower temperature surface and is the rate of heat transfer through the wall.
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𝑞̇ = −kA
𝑇2 −𝑇1
𝑥2 −𝑥1
(𝑜𝑟)
𝑞̇ =
𝑇1 −𝑇2
𝑥 −𝑥
( 2𝑘𝐴 1 )
(𝑜𝑟)
𝑞̇ =
𝑇1 −𝑇2
(
𝐿
)
𝑘𝐴
(2-11)
Where, L = dx = thickness of the slab
𝑞̇ =
∆𝑇
(2-12)
𝑅𝑡ℎ
Now if we consider a plane wall made up of three different layers of materials
having different thermal conductivities and thicknesses of the layers, the analysis
of the conduction can be done as follows.
Consider the area (A) of the heat conduction (fig.2-8) is constant and at steady state
the rate of heat transfer from layer-1 will be equal to the rate of heat transfer from
layer-2. Similarly, the rate of heat transfer through layer-2 will be equal to the rate
of heat transfer through layer-3. If we know the surface temperatures of the wall
are maintained at T1 and T2 as shown in the fig.2.2, the temperature of the interface
of layer1 and layer 2 is assumed to be at T' and the interface of layer-2 and layer-3
as T".
Fig.2.8: Heat conduction through three different layers
The rate of heat transfer through layer-1 to layer-2 will be,
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and,
The rate of heat transfer through layer 2 to layer 3 will be,
and,
The rate of heat transfer through layer 3 to the other side of the wall,
On adding the above three equations,
2.7 Thermal Resistance Network with Film Contact Resistance
Now consider steady one-dimensional heat flow through a plane wall of thickness
L, area A, and thermal conductivity k that is exposed to convection on both sides to
fluids at temperatures T∞1 and T∞2 with heat transfer coefficients h1 and h2,
respectively, as shown in Fig.. Assuming T∞2 ≤ T∞1, the variation of temperature
will be as shown in the figure. Note that the temperature varies linearly in the wall,
and asymptotically approaches T1 and T2 in the fluids as we move away from the
wall.
Under steady conditions we have
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It is sometimes convenient to express heat transfer through a medium in an
analogous manner to Newton’s law of cooling as
(Eqn.2.18)
Where U is the overall heat transfer coefficient. (unit: W/m2. 0C (or) W/m2. 0K)
2.7.1 The Thermal Resistance Network for Heat Transfer Through A Two-Layer
Plane Wall Subjected to Convection On Both Sides.
The surface temperature of the wall can be determined as described above using
the thermal resistance concept, but by taking the surface at which the temperature
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is to be determined as one of the terminal surfaces. For example, once Q · is
evaluated, the surface temperature T1 can be determined from
(Eqn. 2.19)
2.8 Thermal Resistance (Electrical Analogy):
Physical systems are said to be analogous if that obey the same mathematical
equation. The above relations can be put into the form of Ohm’s law:
V=IRelec
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Using this terminology it is common to speak of a thermal resistance:
A thermal resistance may also be associated with heat transfer by convection at a surface. From
Newton’s law of cooling,
(Eqn. 2.20)
The thermal resistance for convection is then
(Eqn. 2.21)
The unit of the various parameters used above is summarized as follows,
__________________________
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Example 2.2
The two sides of a wall (2 mm thick, with a cross-sectional area of 0.2 m2) are
maintained at 30oC and 90oC. The thermal conductivity of the wall material is 1.28
W/(m·oC). Find out the rate of heat transfer through the wall?
Assumptions
1.Steady-state one-dimensional conduction
2. Thermal conductivity is constant for the temperature range of interest
3. The heat loss through the edge side surface is insignificant
4. The layers are in perfect thermal contact
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Example 2.3
One side of a 1 cm thick stainless steel wall (k 1 = 19 W/moC) is maintained at
180oC and the other side is insulated with a layer of 4 cm fiberglass (k 2 = 0.04
W/moC). The outside of the fiberglass is maintained at 60o C and the heat loss
through then wall is 300 W. Determine the area of the wall.
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The previous discussion showed the resistances of different layers. Now to
understand the concept of equivalent resistance, we will consider the geometry of a
composite as shown in fig.2-9a.
The wall is composed of seven different layers indicated by 1 to 7. The interface
temperatures of the composite are T1 to T5 as shown in the fig.2-9a. The equivalent
electrical circuit of the above composite is shown in the fig 2-9b below,
Fig.2-9. (a) Composite wall, and (b) equivalent electrical circuit
The equivalent resistance of the wall will be,
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where,
Therefore, at steady state the rate of heat transfer through the composite can be
represented by,
Where, R is the equivalent resistance.
-----------------------------------------------------Example: 2.3
Consider a composite wall containing 5-different materials as shown in the fig.210. Calculate the rate of heat flow through the composite from the following data?
Assumptions:
1. Steady-state one-dimensional conduction.
2. Thermal conductivity is constant for the temperature range of interest.
3. The heat loss through the edge side surface is insignificant.
4. The layers are in perfect thermal contact.
5. Area in the direction of heat flow is 1 m2.
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Fig. 2-10 (a) composite, (b) corresponding electrical circuit
The height of the first layer is 4 m (h1 = h2 + h3 ).
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2.9 Thermal contact resistance
In the previous discussion, it was assumed that the different layers of the
composite have perfect contact between any two layers. Therefore, the
temperatures of the layers were taken same at the plane of contact.
However, in reality it rarely happens, and the contacting surfaces are not in perfect
contact or touch as shown in the fig.2-11 (a). It is because as we know that due to
the roughness of the surface, the solid surfaces are not perfectly smooth. Thus
when the solid surfaces are contacted the discrete points of the surfaces are in
contact and the voids are generally filled with the air.
Therefore, the heat transfer across the composite is due to the parallel effect of
conduction at solid contact points and by convection or probably by radiation (for
high temperature) through the entrapped air. Thus an apparent temperature drop
may be assumed to occur between the two solid surfaces as shown in the fig.2-11b.
If TI and TII are the theoretical temperature of the plane interface, then the thermal
contact resistance may be defined as,
Fig.2-11. (a) Contacting surfaces of two solids are not in perfect contact, (b)
temperature drop due to imperfect contact
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where Rc represents the thermal contact resistance.
2.10 Steady-state heat conduction through a variable area
It was observed in the previous discussion that for the given plane wall the area for
heat transfer was constant along the heat flow direction. The plane solid wall was
one of the geometries but if we take some other geometry (tapered plane,
cylindrical body, spherical body etc.) in which the area changes in the direction of
heat flow.
2.10.1 One Dimensional Steady State Heat Conduction in Cylinders and Spheres
Consider a hollow cylinder as shown in the fig.2.9a. The inner and outer radius is
represented by ri and ro , whereas Ti and To (Ti > To ) represent the uniform
temperature of the inner and outer wall, respectively.
Fig. 2-12. (a) Hollow cylinder, (b) equivalent electrical circuit
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Consider a very thin hollow cylinder of thickness dr in the main geometry (fig.212a) at a radial distance r. If ‘dr’ is small enough with respect to r, then the area of
the inner and outer surface of the thin cylinder may be considered to be of same
area. In other words, for very small ‘dr’ with respect to r, the lines of heat flow
may be considered parallel through the differential element in radial outward
direction.
We may ignore the heat flow through the ends if the cylinder is sufficiently large.
We may thus eliminate any dependence of the temperature on the axial coordinate
and for one dimensional steady state heat conduction, the rate of heat transfer for
the thin cylinder,
(Where, A= surface area of cylinder)
Where dT is the temperature difference between the inner and outer surface of the
thin cylinder considered above and k is the thermal conductivity of the cylinder.
On re arranging,
To get the heat flow through the thick wall cylinder, the above equation can be
integrated between the limits,
On solving,
Note: Rcylinder = (ro –ri)/(2𝝅Lk)
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(2.26)
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We can repeat the analysis above for a spherical layer by taking A = 4πr2 and the result
can be expressed as
Note: Rsphere = (ro –ri)/(4𝝅krori)
(2.28)
Now consider steady one-dimensional heat flow through a cylindrical or spherical layer
that is exposed to convection on both sides to fluids at temperatures T1 and T2 with
heat transfer coefficients h1 and h2, respectively, as shown in Fig. The thermal resistance
network in this case consists of one conduction and two convection resistances in series,
just like the one for the plane wall, and the rate of heat transfer under steady conditions
can be expressed as
Where,
For cylindrical layer:
(2.29)
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For spherical layer:
(2.30)
Note that A in the convection resistance relation Rconv = 1/hA is the surface area at which
convection occurs. It is equal to A = 2πrL for a cylindrical surface and A = 4πr 2 for a
spherical surface of radius r. Also note that the thermal resistances are in series, and thus
the total thermal resistance is determined by simply adding the individual resistances, just
like the electrical resistances connected in series.
2.10.2 Multi Layered Cylinders and Spheres
Steady heat transfer through multilayered cylindrical or spherical shells can be handled
just like multilayered plane walls discussed earlier by simply adding an additional
resistance in series for each additional layer. For example, the steady heat transfer rate
through the three-layered composite cylinder of length L shown in Fig. with convection
on both sides can be expressed as
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(2.31)
where A1 = 2πr1L and A4 = 2πr4L. The above Equation can also be used for a threelayered spherical shell by replacing the thermal resistances of cylindrical layers by the
corresponding spherical ones.
The ratio T/R across any layer is equal to Q, which remains constant in one-dimensional
steady conduction.
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2.11 One Dimensional Heat Conduction Equation.
Consider heat conduction through a large plane wall such as the wall of a house,
the glass of a single pane window, the metal plate at the bottom of a pressing iron,
a cast iron steam pipe, a cylindrical nuclear
fuel element, an electrical resistance wire, the
wall of a spherical container, or a spherical
metal ball that is being quenched or tempered.
Heat conduction in these and much other
geometry can be approximated as being onedimensional since heat conduction through
these geometries will be dominant in one
direction and negligible in other directions.
Below we will develop the one dimensional
heat conduction equation in rectangular,
cylindrical, and spherical coordinates.
Heat Conduction Equation in a Large Plane Wall
Consider a thin element of thickness x in a large plane wall, as shown in Figure,
Assume the density of the wall is , the specific heat is C, and the area of the wall
normal to the direction of heat transfer is A. An energy balance on this thin
element during a small time interval t can be expressed as
(Eqn.2.32)
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But the change in the energy content of the element and the rate of heat generation
within the element can be expressed as
By solving, knowing k-thermal conductivity, α-thermal diffusivity=k/ρC, we have
(Eqn.2.33)
(Eqn.2.34)
(Eqn. 2.35)
2.11.1 Boundary Conditions-Initial Conditions – (Plane Wall)
The description of a heat transfer problem in a medium is not complete without a full
description of the thermal conditions at the bounding surfaces of the medium.
Boundary conditions: The mathematical expressions of the thermal conditions at the
boundaries.
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From a mathematical point of view, solving a differential equation is essentially a process
of removing derivatives, or an integration process, and thus the solution of a differential
equation typically involves arbitrary constants
The temperature at any point on the wall at a specified time depends on the condition of
the geometry at the beginning of the heat conduction process.
Such a condition, which is usually specified at time t = 0, is called the initial condition,
which is a mathematical expression for the temperature distribution of the medium
initially.
The
general
solution
of
a
typical To
describe
a
heat
transfer
problem
differential equation involves arbitrary completely, two boundary conditions must be
constants, and thus an infinite number of given for each direction along which heat
transfer is significant.
solutions.
Specified Temperature Boundary Condition
One of the easiest ways to specify the thermal conditions on a surface is to specify
the temperature.
For one-dimensional heat transfer through a plane wall of
thickness L, the specified temperature boundary conditions can be expressed as
T(0, t) = T1
;
T(L, t) = T2, where T1 and T2 are the specified temperatures at
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surfaces at x = 0 and x = L, The specified
temperatures can be constant, which is
the case for steady heat conduction, or
may vary with time.
Specified Heat Flux Boundary Condition
The heat flux in the positive x-direction anywhere in the medium, including the
boundaries, can be expressed by
(Eqn. 2.36)
For a plate of thickness L subjected to heat flux of 50 W/m2 into the medium from
both sides, for example, the specified heat flux boundary conditions can be
expressed as
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Special Case: Insulated Boundary
A well-insulated surface can be modeled as a
surface with a specified heat flux of zero. Then
the boundary condition on a perfectly insulated
surface (at x = 0, for example) can be expressed
as
(Eqn. 2.37)
On an insulated surface, the first derivative of temperature with respect to the
space variable (the temperature gradient) in the direction normal to the insulated
surface is zero.
Another Special Case: Thermal Symmetry
The two surfaces of a large hot plate of thickness L suspended vertically in air is
subjected to the same thermal conditions, and thus the temperature distribution in
one half of the plate is the same as that in the other half.
That is, the heat transfer problem in this plate
possesses thermal symmetry about the center plane
at x = L/2.
Therefore, the center plane can be viewed as an
insulated surface, and the thermal condition at this
plane of symmetry can be expressed as
(Eqn. 2.38)
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which resembles the insulation or zero heat flux boundary
condition.
2.11.2 The solution procedure for solving heat conduction
problems can be summarized as
(1) formulate the problem by obtaining the applicable differential
equation in its simplest form and specifying the boundary conditions,
(2) Obtain the general solution of the differential equation, and
(3) apply the boundary conditions and determine the arbitrary constants in the general
solution
---------------------------------------------------------
Example: 2.4.
Consider a large plane wall of thickness L = 0.2 m, thermal conductivity k = 1.2
W/moK, and surface area A = 15 m2. The two sides of the wall are maintained at
constant temperatures of T1 = 120°C and T2 = 50°C, respectively. Determine (a)
the variation of temperature within the wall and the value of temperature at x = 0.1
m and (b) the rate of heat conduction through the wall under steady conditions.
Assumptions
1 Heat conduction is steady.
2 Heat conduction is one-dimensional since the wall is large relative to its
thickness and the thermal conditions on both sides are uniform.
3 Thermal conductivity is constant.
4 There is no heat generation.
Analysis (a) Taking the direction normal to the surface of the wall to be the xdirection, the differential equation for this problem can be expressed as
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With boundary conditions
T(0) = T1 = 120°C
T(L) = T2 = 50°C
Integrating the differential equation once with respect to x yields
where C1 is an arbitrary constant.
Integrating one more time, we obtain
T(x) = C1x + C2
This is the general solution of the differential equation.
The first boundary condition can be interpreted as in the general solution, re-place
all the x's by zero and T(x) by T1.
T(0) = C1 x 0 + C2  C2 = T1
The second boundary condition can be interpreted as in the general solution,
replace all the x's by L and T(x) by T2. That is,
Substituting the C1 and C2 expressions into the general solution, we obtain
Substituting the given information, the value of the temperature at x = 0.1 m is
determined to be
(b) The rate of heat conduction anywhere in the wall is determined from Fourier's
law to be
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The numerical value of the rate of heat conduction through the wall is deter-mined
by substituting the given values to be
Example: 2.5
Consider a 0.8-m-high and 1.5-m-wide double-pane window consisting of two 4mm-thick layers of glass (k = 0.78 W/m.°C) separated by a 10-mm-wide stagnant
air space (k = 0.026 W/m.°C). Determine the steady rate of heat transfer through
this double-pane window and the temperature of its inner surface for a day during
which the room is maintained at 20°C while the temperature of the outdoors is
10°C. Take the convection heat transfer coefficients on the inner and outer surfaces
of the window to be h1 = 10 W/m2.°C and h2= 40 W/m2.°C, which includes the
effects of radiation.
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2.12 One Dimensional General Heat Conduction Equation in a Long Cylinder
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(Eqn. 2.39)
2.13 Heat Conduction Equation in a Sphere
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(Eqn. 2.40)
Example: 2.6
A 3-m internal diameter spherical tank made of 2-cm-thick stainless steel (k = 15
W/m · °C) is used to store iced water at T1 = 0°C. The tank is located in a room
whose temperature is T2 = 22°C. The walls of the room are also at 22°C. The
outer surface of the tank is black and heat transfer between the outer surface of the
tank and the surroundings is by natural convection and radiation. The convection
heat transfer coefficients at the inner and the outer surfaces of the tank are h 1 = 80
W/m2 °C and h2 = 10 W/m2 °C, respectively. Determine (a) the rate of heat transfer
to the iced water in the tank and (b) the amount of ice at 0°C that melts during a
24-h period.
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Example: 2.7
Steam at T1 =320°C flows in a cast iron pipe (k = 80 W/m.°C) whose inner and
outer diameters are D1 =5 cm and D2 = 5.5 cm, respectively. The pipe is covered
with 3-cm-thick glass wool insulation with k =0.05 W/m.°C. Heat is lost to the
surroundings at T2 = 5°C by natural convection and radiation, with a combined
heat transfer coefficient of h2 = 18 W/m2.°C. Taking the heat transfer coefficient
inside the pipe to be h1 = 60 W/m2 · °C, determine the rate of heat loss from the
steam per unit length of the pipe. Also determine the temperature drops across the
pipe shell and the insulation.
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That is, the temperatures between the inner and the outer surfaces of the pipe differ
by 0.02°C, whereas the temperatures between the inner and the outer surfaces of
the insulation differ by 284°C.
2.14 CRITICAL RADIUS OF INSULATION
Adding more insulation to a wall or to the attic always decreases heat transfer. The
thicker the insulation, the lower the heat transfer rate. This is expected, since the
heat transfer area A is constant, and adding insulation always increases the thermal
resistance of the wall without increasing the convection resistance. Adding
insulation to a cylindrical pipe or a spherical shell, however, is a different matter.
The additional insulation increases the conduction resistance of the insulation layer
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but decreases the convection resistance of the surface because of the increase in the
outer surface area for convection.
Consider a cylindrical pipe of outer radius r1 whose outer surface temperature T1 is
maintained constant (Fig). The pipe is now insulated with a material whose thermal
conductivity is k and outer radius is r2. Heat is lost from the pipe to the surrounding
medium at temperature T, with a convection heat transfer coefficient h. The rate of
heat transfer from the insulated pipe to the surrounding air can be expressed as
The variation of Q with the outer radius of the insulation r2
The value of r2 at which Q reaches a maximum is determined from the requirement
that dQ/dr2 = 0 (zero slope). Performing the differentiation and solving for r2 yields
the critical radius of insulation for a cylindrical body to be
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Note that the critical radius of insulation depends on the thermal conductivity of
the insulation k and the external convection heat transfer coefficient h. The rate of
heat transfer from the cylinder increases with the addition of insulation for r 2  rcr,
reaches a maximum when r2 = rcr, and starts to decrease for r2  rcr. Thus, insulating
the pipe may actually increase the rate of heat transfer from the pipe instead of
decreasing it when r2  rcr.
In a similar manner that the critical radius of insulation for a spherical shell is
The effectiveness of insulation is often given in terms of its R-value, the thermal
resistance of the material per unit surface area, expressed as
Where L is the thickness and k is the thermal conductivity of the material.
Example: 2.7
Calculate the critical radius of insulation for the pipe which is used for transport
hot water, consider the lowest possible value of h=5 W/m2°C for the case of
natural convection of gases. The thermal conductivity of common insulating
materials which is used for insulation is about 0.05 W/m°C.
Given :
h=5 W/m2°C ; k=0.05 W/m °C
The value of the critical radius, rc = kins/h = 0.05/5 = 0.01 m
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Example: 2.8
A 3-mm-diameter and 5-m-long electric wire is tightly wrapped with a 2-mm thick
plastic cover whose thermal conductivity is k = 0.15 W/m.°C. Electrical
measurements indicate that a current of 10 A passes through the wire and there is a
voltage drop of 8 V along the wire. If the insulated wire is exposed to a medium at
T = 30°C with a heat transfer coefficient of h = 12 W/m2.°C, determine the
temperature at the interface of the wire and the plastic cover in steady operation.
Also determine whether doubling the thickness of the plastic cover will increase or
decrease this interface temperature.
In steady operation, the rate of heat transfer becomes equal to the heat generated
within the wire, which is determined to be
The values of the two thermal resistances are determined to be
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CHEE 3101 Heat Transfer
and therefore the total resistance to be
Then the interface temperature can be determined from
Note that we did not involve the electrical wire directly in the thermal resistance
network, since the wire involves heat generation.
To answer the second part of the question, we need to know the critical radius of
insulation of the plastic cover. It is determined to be
This is larger than the radius of the plastic cover. Therefore, increasing the
thickness of the plastic cover will enhance heat transfer until the outer radius of the
cover reaches 12.5 mm. As a result, the rate of heat transfer Q̇ will increase when
the interface temperature T1 is held constant, or T1 will decrease when Q̇ is held
constant, which is the case here.
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Assignment Questions:
Q1. Thermal conductivity of water __________ with rise in temperature
(A)
(B)
(C)
(D)
Increases
remains same
decreases
may increase or decrease depending upon temperature
Q2. Heat transfer takes place according to the statement of,
(A) Zeroth Law of Thermodynamics
(B) First Law of Thermodynamics
(C) Second Law of Thermodynamics
(D) Third Law of Thermodynamics
Q3. For a current carrying metallic ball of 20 mm diameter exposed to air (h=25
W/m2K), maximum heat distribution occurs when the thickness of the insulation (K=0.5
W/mK) is:
(A)10 mm
(B) 20 mm
(C) 30 mm
(D) 40 mm
Q4. For a current carrying metallic rod of 20 mm diameter exposed to air (h=25
W/m2K), maximum heat distribution occurs when the thickness of the insulation (K=0.5
W/mK) is:
(A) 10 mm
(B) 20 mm
(C) 30 mm
(D) 40 mm
Q5. Heat is lost through a brick wall (k=0.72W/moK), which is 4m long, 3m wide, and
25cm thick at a rate of 500W. If the inner surface of the wall is at 22 oC, the temperature
at the mid plane of the wall is
(A) 0oC
(B) 11oC
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Q6. Consider the following:
1. Water
2.Wood
3. Air
4. Mercury
Which of the following options correspond to the order of correct thermal conductivity
at room temperature?
(A)4>1>3>2
(B) 4>1>2>3
C) 1>4>3>2
(D) 1>4>2>3
Q7. It is proposed to coat a 1mm diameter wire with enamel paint (k=0.1 W/mK) to
increase heat transfer with air. If the air side heat transfer coefficient is 100 W/m2K, the
optimum thickness of enamel paint should be
(A) 0.25 mm
(B) 0.5 mm
(C) 1 mm
(D) 2 mm
Q8. A 2-kW electric resistance heater submerged in 30-kg water is turned on and kept
on for 10 min. During the process, 400 kJ of heat is lost from the water. The
temperature rise of water is
(A) 5.6°C
(B) 6.4°C
(C) 13.6°C
(D) 23.3°C
Q11. A furnace is made of a red brick wall of thickness 0.5 m and conductivity 0.7
W/mK. For the same heat loss and the temperature drop, this can be replaced by a layer
of diatomite earth of conductivity of 0.14 W/mK and thickness should be:
(A) 0.5 m
(B) 0.1 m
(C) 0.2 m
(D) 0.4 m
Q9. A gas filled tube has 2 mm inside diameter and 25 cm length. The gas is heated by
an electrical wire of diameter 50 cm located along the axis of the tube. Current and
voltage drop across the heating element are 0.5 amps and 4 volts, respectively. If the
measured wire and inside tube wall temperatures are 175oC and 150oC respectively, find
the thermal conductivity of the gas filling tube.
(A) 0.12 W/moK
(B) 0.15 W/moK
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Q10. Electric current is passed through a copper ball 1 mm in diameter. The ball is
submerged in liquid water at atmospheric pressure, and the current is increased until the
water boils. For this situation h=5000W/m2 oC. How much electric power must be
supplied to the wire to maintain the ball at 114oC?
(A) 0.2199W
(B) 21.99W
(C) 25.64W
(D) 0.2564 W
Q12. Determine the correctness or otherwise of the following Assertion [a] and the
Reason [r].
Assertion [a]: in case of insulating electrical wire the optimum thickness should
be below critical thickness.
Reason [r]: upto critical radius increment of conductive resistance is the
predominant effect than decrement of convective resistance.
(A) Both [a] and [r] are true and [r] is the correct reason for [a].
(B) Both [a] and [r] are true but [r] is NOT the correct reason for [a].
(C) [a] is true but [r] is false.
(D) [a] is false but [r] is true.
Q13. Determine the correctness or otherwise of the following Assertion [a] and the
Reason [r].
Assertion [a]: in case of gases thermal conductivity increases with temperature.
Reason [r]: in fluids conduction is due to Lattice vibrations & motion of free
electrons.
(A)
(B)
(C)
(D)
Both [a] and [r] are true and [r] is the correct reason for [a].
Both [a] and [r] are true but [r] is NOT the correct reason for [a].
[a] is true but [r] is false.
[a] is false but [r] is true.
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Q14. Match the following :
(Group1)
Temperature distribution along plane walls in series:
(Group2)
WALL
MATERIAL
(P) SLAB1
- (T) Copper
(Q) SLAB2
- (U) Silver
(R) SLAB3
-
(V) Aluminium
(S) SLAB4
-
(W) Stainless steel
(A) P-U,Q-V,R-W,S-T
(B) P-T,Q-W,R-V,S-U
(C) P-W,Q-U,R-T,S-V
(D) P-V,Q-T,R-U,S-W
Q15. A furnace wall is of three layers, first layer of insulation brick of 12 cm thickness
of conductivity 0.6 W/mK. The face is exposed to gases at 870°C with a convection
coefficient of 110 W/m2K. This layer is backed by a 10 cm layer of firebrick of
conductivity 0.8 W/mK. There is a contact resistance between the layers of 2.6 × 10 –4 m2
°C/W. The third layer is the plate backing of 10 mm thickness of conductivity 49
W/mK. The contact resistance between the second and third layers is 1.5 × 10–4 m2
°C/W. The plate is exposed to air at 30°C with a convection coefficient of 15 W/m 2K.
Determine the heat flow, the surface temperatures and the overall heat transfer
coefficient.
[Answer: T1 = 850.97°C; T2 = 432.40°C; T3 = 431.86°C; T4= 170.26°C
T5 = 169.95°C; T6 = 169.52°C]
Q16. A composite wall is made up of 3 layers of thicknesses 25 cm, 10 cm and 15 cm
with thermal conductivities of 1.7, kB and 9.5 W/mK. The outside surface is exposed to
air at 20°C with convection coefficient of 15 W/m2K and the inside is exposed to gases
at 1200°C with a convection coefficient of 28 W/m2K and the inside surface is at
1080°C. Determine the unknown thermal conductivity, all surface temperatures,
resistances of each layer and the overall heat transfer coefficient per unit area.
[Answer: kB=1.163 W/moC; U=2.85 W/m2oC; T1 = 585.9°C; T2 = 297°C;
T3 = 243.95°C]
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Q15. A pipe carrying steam at 230°C has an internal diameter of 12 cm and the pipe
thickness is 7.5 mm. The conductivity of the pipe material is 49 W/mK and the
convective heat transfer coefficient on the inside is 85 W/m2K. The pipe is insulated by
two layers of insulation one of 5 cm thickness of conductivity 0.15 W/mK and over it
another 5 cm thickness of conductivity 0.48 W/mK. The outside is exposed to air at
35°C with a convection coefficient of 18 W/m2K. Determine the heat loss for 5 m
length. Also determine the interface temperatures and the overall heat transfer
coefficient based on inside and outside areas.
[Answer: T1 = 222.3°C; T2 = 222.2°C; T3 = 77.04°C; T4 = 48.03°C]
Q16. Derive the unidirectional temperature distribution equation for plane wall, hollow
cylinder and hollow sphere from Laplace equation.
Q17. With neat sketch derive the general two dimensional unsteady state heat
conduction equation for an isotropic slab with internal heat generation. Simplify the
obtained equation to Poisson’s equation, Fourier’s equation and Laplace equation.
Q18. Using the general conduction equation for a Cartesian system, derive the
expressions for the position and value of maximum temperature for plane wall with
internal heat generation having,
a. different surface temperature,
b. same surface temperature,
c. One surface insulated.
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CHAPTER 3 (outcomes 5, 6, 7 and 12)
3.1
PHYSICAL MECHANISM OF CONVECTION
The convection is the mechanism of heat transfer through a fluid in the presence of
bulk fluid motion. Convection is classified as natural (or free) and forced
convection, depending on how the fluid motion is initiated. In forced convection,
the fluid is forced to flow over a surface or in a pipe by external means such as a
pump or a fan. In natural convection, any fluid motion is caused by natural means
such as the buoyancy effect, which manifests itself as the rise of warmer fluid and
the fall of the cooler fluid. Convection is also classified as external and internal,
depending on whether the fluid is forced to flow over a surface or in a channel.
3.2
Internal versus External Flow
A fluid flow is classified as being internal and external, depending on whether the
fluid is forced to flow in a confined channel or over a surface. The flow of an
unbounded fluid over a surface such as a plate, a wire, or a pipe is external flow.
The flow in a pipe or duct is internal flow if the fluid is completely bounded by
solid surfaces. Water flow in a pipe, for example, is internal flow, and air flow over
an exposed pipe during a windy day is external flow
(Fig.). The flow of liquids in a pipe is called openchannel flow if the pipe is partially filled with the
liquid and there is a free surface. The flow of water in
rivers and irrigation ditches are examples for that.
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3.3
Laminar versus Turbulent Flow
The highly ordered fluid motion characterized by smooth
streamlines is called laminar. The flow of high-viscosity fluids such as oils at low
velocities is typically laminar. The highly disordered fluid motion that typically
occurs at high velocities characterized by velocity fluctuations is called turbulent.
The flow of low-viscosity fluids such as air at high velocities is typically turbulent.
The flow regime greatly influences the heat transfer rates and the required power
for pumping.
3.4
3.4.1
Use of Non Dimensional Numbers in Heat Transfer Analysis
Nusselt Number
In convection studies, it is common practice to nondimensionalize the governing
equations and combines the variables, which group together into dimensionless
numbers in order to reduce the number of total variables. It is also common
practice to nondimensionalize the heat transfer coefficient h with the Nusselt
number, defined as
Eq. 3.1
where k is the thermal conductivity of the fluid and Lc is the characteristic length.
Significance: the Nusselt number represents the enhancement of heat transfer
through a fluid layer as a result of convection relative to conduction across the
same fluid layer. The larger the Nusselt number, the more effective the convection.
A Nusselt number of Nu = 1 for a fluid layer represents heat transfer across the
layer by pure conduction.
3.4.2
Prandtl Number
The relative thickness of the velocity and the thermal boundary layers is best
described by the dimensionless parameter Prandtl number, defined as
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Eq. 3.2
The Prandtl numbers of fluids range from less than 0.01 for liquid metals to more
than 100,000 for heavy oils. Note that the Prandtl number is in the order of 10 for
water.
The Prandtl numbers of gases are about 1, which indicates that both momentum
and heat dissipate through the fluid at about the same rate. Heat diffuses very
quickly in liquid metals (Pr
« 1) and very slowly in oils (Pr » 1) relative to
momentum. Consequently the thermal boundary layer is much thicker for liquid
metals and much thinner for oils relative to the velocity boundary layer.
3.4.3
Reynolds Number
The transition from laminar to turbulent flow depends on the surface geometry,
surface roughness, free-stream velocity, surface temperature, and type of fluid,
among other things. The flow regime depends mainly on the ratio of the inertia
forces to viscous forces in the fluid. This ratio is called the Reynolds number,
which is a dimensionless quantity, and is expressed for external flow
Eq. 3.3
where V is the upstream velocity (equivalent to the free-stream velocity u∞ for
a flat plate), Lc is the characteristic length of the geometry, and  = /ρ is the
kinematic viscosity of the fluid. For a flat plate, the characteristic length is the
distance x from the leading edge. Note that kinematic viscosity has the unit m2/s,
which is identical to the unit of thermal diffusivity, and can be viewed as viscous
diffusivity or diffusivity for momentum.
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At large Reynolds numbers, the inertia forces, which are proportional to the
density and the velocity of the fluid, are large relative to the viscous forces, and
results rapid fluctuations of the fluid, called as Turbulent flow.
At small Reynolds numbers, however, the viscous forces are large enough to
overcome the inertia forces and to keep the fluid “in line.” Thus the flow is
laminar. The Reynolds number at which the flow becomes turbulent is called the
critical Reynolds number. (flow over plate Recritical = 5 105).
3.4.4 Schmidt Number (Sc) – defined as the ratio of the molecular diffusivity of
momentum to the molecular diffusivity of mass.
Sc 

 DAB
μ – Dynamic viscosity;


DAB
Eq. 3.4
γ – Kinematic viscosity;
DAB – Diffusion coefficient
3.5 Flow over a flat plate
Heat Transfer from flat surfaces – Flow over a flat plate (External flow)
i) Mean film temperature,
– Plate surface temperature
– Fluid temperature
All the thermo physical properties of the fluid (like density, viscosity, specific heat,
thermal conductivity) should be taken corresponding to mean film temperature
ii) Criteria for flow type
Re < 5 x 105
 Laminar flow
Re > 5 x 105
 Turbulent flow
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3.5.1 For flat plate laminar flow:
1. Local Nusselt number, Nux = 0.332 (Re)0.5 (Pr)0.333
2. Local heat transfer coefficient, hx = (Nux k / L)
3. Average heat transfer coefficient, h = 2hx
4. Heat transfer rate, Q = h A (Tw-Tα)
5. Velocity boundary layer thickness, =
 hx  5 x Re0.5
6. Thermal boundary layer thickness, =  Tx   hx Pr 
0.333
7. Local friction coefficient, Cfx = 0.664 (Re)-0.5
8. Average friction coefficient, C fL  1.328(Re) 0.5
3.5.2 For flat plate Turbulent flow:
1. Local Nusselt number, Nux = 0.0296 (Re)0.8 (Pr)0.333
2. Local heat transfer coefficient, hx = (Nux k / L)
3. Average heat transfer coefficient, h = 1.25 hx
4. Heat transfer rate, Q = h A (Tw-Tα)
5. Velocity boundary layer thickness,
6. Thermal boundary layer thickness,
 hx  0.381 x Re0.2
 Tx   hx
7. Local friction coefficient,
Cfx = 0.592 (Re)-0.2
Cfx = 0.37 (log10 Re)-2.584 for
for
Re > 5 x 105 & Re < 107
Re > 107 & Re < 109
Note: The flow is fully turbulent right from the leading edge of the plate
3.5.3 For flat plate Combined flow
1. Average Nusselt number, Nu = Pr0.333 [0.037(Re)0.8 – 371]
2. Average heat transfer coefficient, h = (Nu k / L)
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CHEE 3101 Heat Transfer
3. Heat transfer rate, Q = h A (Tw-Tα)
4. Average friction coefficient,
C fL  0.074(Re) 0.2  1742(Re) 1.0
Note: The flow is laminar at the start and becomes turbulent as it flows over
the plate
Example 3.1
Air at 20oC, at a pressure of 1 bar is flowing over a flat plate at a velocity of 3 m/s.
If the plate is maintained at 60oC, calculate the heat rate transfer per unit width of
the plate taking the length of the plate along the flow of air is 2 m.
Given:
Fluid Temperature, Tά = 20o C ; Pressure, P = 1 bar;
Plate surface temperature = 60o C;
Velocity, U = 3m/s;
Width, W = 1m;
Length, L = 2m
To find: Heat transfer rate per unit width (Q̇ /W)
Solution:
o
Film Temp, Tf = (Tw+ Tά) /2  Tf = 40 C
o
Properties of air at 40 C (From table);
3
ρ = 1.128 kg/m ;
k = 0.02756 W/moK;
Re = (UL/γ) = 3.5377 x 10
5

-6
2
γ = 16.96 x 10 m /s;
Re < 5 x 10
5
Pr = 0.699
Laminar flow
For a flat plate laminar flow,
Local Nux = 0.332 (Re)
Nux = 175.27
Local hx = (Nux k / L)
0.5
(Pr)
0.333
2
hx = 2.415 W/m K
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Heat transfer rate, Q̇ = h A (Tw – Tά)
Q̇ = h (LW) (Tw – Tά)
(Q̇ /W) = h L (Tw – Tά)
Average h = 2 hx (for laminar flow)
2
h = 4.83 W/m K
(𝐐̇ /W) = 386.4 W/m
Example 3.2
Air at 20oC, at atmospheric pressure flows over a flat plate of at a velocity of 3
m/s. If the plate is 1m wide and is at 80oC, calculate the following at x=300mm i)
Hydrodynamic boundary layer thickness, ii) Thermal boundary layer thickness, iii)
Local friction coefficient, iv) Average friction coefficient, v) Local heat transfer
coefficient, vi) Average heat transfer coefficient, vii) Heat transfer rate.
Given:
Fluid Temperature, Tά = 20o C ; Distance, x = 0.3 m ; Velocity, U = 3m/s;
Plate surface temperature = 80o C;
Width, W = 1m
To find: i) δhx ii) δTx, iii) Cfx, iv) CfL, iv) hx, v) h, vi) Q̇
Solution:
Film Temp, Tf = (Tw+ Tά) /2
 Tf = 50o C
Properties of air at 50o C (From table);
ρ = 1.093 kg/m3;
k = 0.02826 W/mK;
Rex = (Ux/γ) = 0.501 x 105
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γ = 17.95 x 10-6 m2/s;
 Re < 5 x 105
Pr = 0.698
Laminar flow
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CHEE 3101 Heat Transfer
Example: 3.3
Air at 20oC is flowing over a flat plate of 1 m long, 0.5 m wide at a velocity of 100
m/s. The flow over the whole length of the plate is made turbulent. Calculate the
following i) Thickness of the boundary layer, ii) Mean value of the heat transfer
coefficient
Given:
Fluid Temperature, Tά = 20o C;
Length, L = 1m;
Velocity, U = 100 m/s ;
Width = 0.5 m
The flow is turbulent over the entire length
To find: i) δ, ii) h
Solution:
Film Temp, Tf = (Tw+ Tά) /2  Tf = 20o C
Properties of air at 20o C (from table);
ρ = 1.205 kg/m3;
k = 0.02593 W/mK;
γ = 15.06 x 10-6 m2/s;
Pr = 0.703
Rex = (UL/γ) = 6.64 x 106  Re > 5 x 105 Turbulent flow
No check is required for combined flow as the problem clearly tells that it is fully
turbulent flow
For a flat plate turbulent flow,
Local Nux = 0.0296 (Re)0.8 (Pr)0.333;
where Re = 1.23 x 106
Nux = 7552
Local hx = (Nux k / L)
hx = 195.8 W/m2K
Average h = 1.25 hx (for turbulent flow)
hturbulent = 244.75 W/m2K
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Boundary layer thickness
δ = 0.37 χ (Re)-0.2
δhx = 0.0159 m
Boundary layer thickness = 0.0159 m
Average heat transfer coefficient = 244.75 W/m2K
3.6 Heat Transfer from circular surfaces – (External flow)
Two regions

Boundary layer region near the surface

Inviscid region away from the surface
 Pressure gradient along the cylinder is responsible for the
development of a separated flow region on the backside of the
cylinder
 Separation of flow affects the drag force on a curved surface to great
extent
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CHEE 3101 Heat Transfer
3.6.1 Heat Transfer from circular surfaces – Flow over a cylinder (External
flow)
i) Mean film temperature,
All the thermo physical properties of the fluid (like density, viscosity, specific heat,
thermal conductivity) should be taken corresponding to mean film temperature
Tf 
Tw  T
2
Tw – Plate surface temperature
ii) Reynolds number,
Re 
Tα – Fluid temperature
UD

iii) Nusselt number, Nu = C (Re)m (Pr)0.333
ReD
C
m
0.4 – 4.0
0.989
0.330
4.1 – 40.0
0.911
0.385
40.1 – 4000
0.683
0.466
4001 – 40,000
0.193
0.618
40,001 – 400,000
0.0266
0.805
iv) Average heat transfer coefficient, h = (Nu k /D)
v) Heat transfer rate, Q̇ = h A (Tw- T∞);
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CHEE 3101 Heat Transfer
3.6.2 Heat Transfer from circular surfaces – Flow over a sphere (External
flow)
i) Nusselt number, Nu = 0.37 (Re)0.6
for, 17 < Re < 70,000
ii) Average heat transfer coefficient, h = (Nu k /D)
iii) Heat transfer rate, Q̇ = h A (Tw- T∞);
Where A = 4π r2
Example 3.4
Air at 15oC, 30 km/h flows over a cylinder of 400 mm diameter and 1500 mm
height with surface temperature of 45oC. Calculate the heat loss.
To find: Heat loss (Q̇)
Given:
Fluid Temperature, T∞ = 15o C; Velocity, U=30 km/h=8.33 m/s;
Diameter, D = 0.4 m; Plate surface temperature, Tw = 45o C;
Length L = 1.5 m
Properties of air at 30o C (from table),
ρ = 1.165 kg/m3; k = 0.02675 W/mK; γ = 16 x 10-6 m2/s; Pr = 0.701
Re = (UD/γ) = 2.08 x 105
Average Nu = C (Re)m (Pr)0.333
Re = 2.08 x 105  Corresponding ‘C’ value is 0.0266 and ‘m’ value is 0.805.
(from table)
Nu = 451.3
Average h = (Nu k / D)
h = 30.18 W/m2K
Heat loss, Q̇ = h A (Tw – T∞)
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Q̇ = h (π D L) (Tw – T∞)
𝐐̇ = 1706.6 W
3.7 Heat Transfer from circular surfaces – Flow through a cylinder (Internal
flow)
Retardation of fluid flow near the walls causes Boundary layer development
(Shown by dotted line in the fig)
 Thickness of the BL is limited to radius of the pipe since the flow is
within a confined passage
 BL from the pipe walls meet at the centre of the pipe and the entire
flow region acquires the same pattern of the flow
 Once the BL thickness becomes equal to the radius of the pipe there
will be no further changes in the velocity distribution. This invariant
velocity profile distribution is called fully developed velocity profile
i.e Poiseulle flow
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CHEE 3101 Heat Transfer
i)
Mean film temperature,
Tmo
Tf 
Tmo  Tmi
2
Tmi – Fluid outlet temperature
– Fluid inlet temperature
All the thermo physical properties of the fluid (like density, viscosity, specific heat,
thermal conductivity) should be taken corresponding to mean film temperature
ii) Reynolds number,
UD
Re 

Criteria for flow type
Re < 2300 - Laminar flow
Re > 2300 - Turbulent flow
Laminar Flow:
Nu = 3.66
Turbulent Flow:
Nu = 0.023 (Re)0.8 (Pr)n
(Dittus-Boelter equation)
n = 0.4 – Heating process
n = 0.3 – Cooling process
Equivalent diameter for rectangular section
Dh (or) De =(4A / P) = [4LW/2(L+W)]
Where A – Area in m2, P – Perimeter in m, L – Length in m, W – Width in m.
Equivalent diameter for hollow cylinder (annular spaces)


4  Do2  Di2 
4

Dh (or) De = (4A / P) =
 Do  Di 

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CHEE 3101 Heat Transfer
Where Do – Outer diameter in m, Di – Inner diameter in m
Heat transfer
Q̇ = h A (Tw - Tm) where A = π D L (m2)
Tm – Mean temperature oC, =
Tmi – Inlet temperature oC,
Tmo  Tmi
2
Tmo – Outlet temperature oC.
Tw – Tube wall temperature oC
Q̇ = m Cp(Tmo - Tmi)
Mass flow rate
m = ρ x A x U (kg/s)
where ρ – Density in kg/ m3 , A – Area, (π/4)D2 in m2 , U – Velocity in m/s.
Example 3.5
Water flows inside a tube of 20 mm diameter and 3 m long flows at a velocity of
0.03 m/s. The water gets heated from 40o C to 120o C while passing through the
tube. The tube wall is maintained at constant temperature of 160 o C. Find heat
transfer rate
Given:
Diameter, D = 0.020 m;
Length, L = 3 m; Velocity, U = 0.03 m/s;
Inner temperature of water, Tmi = 40o C; Outer temperature of water, Tmo = 120o C
Wall temperature, Tw = 160o C
To find: Heat transfer rate (Q̇)
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Solution:
Bulk mean Temp, Tm = (Tmi+T mo) / 2 = 80oC
Properties of water at 80oC (from table)
ρ = 974 kg/m3;
γ = 0.364 x 10-6 m2/s;
Pr = 2.22;
k = 0.6687 W/ mK
Re = (UD/γ) = 1648.35
Since Re < 2300,  flow is laminar.
For laminar flow, Nu = 3.66
Average heat transfer coefficient, h = Nu k / D
h =122.39 W/m2K
Heat transfer rate, Q̇= h A (Tw - Tm) = h πDL (Tw-Tm)
𝐐̇ = 1845.29 W
Example 3.6
Water at 50o C enters 50 mm diameter and 4 m long tube with a velocity of 0.8
m/s. The tube wall is maintained at a constant temperature of 90 o C. Determine the
heat transfer coefficient and the total amount of heat transferred if exit water
temperature is 70o C.
Given:
Diameter, D = 0.05 m;
Length, L = 4 m;
Inner temperature, Tmi = 50o C;
Velocity, U = 0.8 m/s
Exit temperature, Tmo = 70o C
Tube wall temperature, Tw = 90o C
To find: i) Heat transfer coefficient ( h ), ii) Heat transfer rate (Q̇)
Solution:
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CHEE 3101 Heat Transfer
Bulk mean Temp,
Tm = (Tmi+T mo) / 2 = 60o C
Properties of water at 60o C ( from table),
ρ = 985 kg/m3;
γ = 0.478 x 10-6 m2/s;
Pr = 3.020; k = 0.6513 W/mK
Re = (UD/γ) = 8.36 x 104  Turbulent flow
So, Nu =0.023(Re)0.8(Pr)n
The process involved is heating, hence n = 0.4
Nu = 310
Heat transfer coefficient, h = Nu k / D = 4039.3 W/m2K
Heat transfer, Q̇ = h A (Tw-Tm)= h πDL (Tw-Tm) = 76,139 W
Example 3.7
Air at 15o C, 35 m/s, flows through a hollow cylinder of 4 cm inner diameter and 6
cm outer diameter and leaves at 45o C. Tube wall is maintained at 60o C. Calculate
the heat transfer coefficient between the air and the inner tube.
Given:
Inlet temperature, Tmi = 15o C ;
Inner Diameter, Di = 0.04 m
Velocity, U = 35 m / s ; Exit temperature, Tmo = 45o C;
Outer Diameter, Do = 0.06 m;
Wall temp, Tw = 60o C
To find: Heat transfer coefficient (h)
Solution:
Mean Temp, Tm = (Tmi+T mo) / 2 = 30o C
Properties of air at 30o C (from table)
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CHEE 3101 Heat Transfer
ρ = 1.165 kg/m3;
γ = 16 x 10-6 m2/s; Pr = 0.701; k = 0.02675 W/mK
Hydraulic or Equivalent diameter
De = 4A/P
= (4(π/4)[Do2 – Di2])/(π )[Do + Di]) = 0.02 m
ReDe = (U De /γ) = 43750
Since Re > 2300, flow is turbulent.
For turbulent flow, general equation, Nu =0.023(Re)0.8(Pr)n
This is heating process, so n=0.4
Nu = 102.9
Heat transfer coefficient, h = Nu k / De
h = 137.7 W/m2K
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CHEE 3101 Heat Transfer
Assignment Questions:
Q1. Air at 25oC, flows over a flat plate maintained at 135oCat a speed of 5 m/s.
The plate is 3 m long and 1.5 m wide. Calculate the local heat transfer coefficient
at x=0.5 m and the heat transferred from the first 0.5 m of the plate.
[ Answer : hx=0.5 = 6.14 W/m2K;
Q̇avg uptox=0.5 = 1013.9 W]
Q2. Air at 30oC, 0.2 m/s flows across a 120 W spherical electric bulb at 130 oC.
Find the heat transfer rate and power lost due to convection if bulb diameter is 70
mm.
[Answer: Q̇= 12.22 W;
% heat lost = 10.18 %]
Q3. Water at 30o C, 20 m/s flows through a straight tube of 60 mm diameter. The
tube surface is maintained at 70o C and outlet temperature of water is 50o C. Find
the heat transfer coefficient from the tube surface to the water, heat transferred and
the tube length.
[ Answer: h = 43726.59 W/m2K;
Q̇= 4.69 x 106 W;
L = 18.96 m]
Q4. Air at 30o C, 6 m/s flows in a rectangular section of size 300 x 800 mm.
Calculate the heat leakage per meter length per unit temperature difference.
[Answer: (Q̇ / L ΔT) = 39.79 W]
Q5. Air at 333 K, 1bar pressure, flow through 12 cm diameter tube. The surface
temperature of the tube is maintained at 400 K and mass flow rate is 75 kg/hr.
Calculate the heat transfer rate for 1.5 m length of the tube.
[Answer: Q̇ = 300.82 W]
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CHEE 3101 Heat Transfer
3.8 Boiling and Condensation
• In the previous chapter, Fluid was considered as a homogenous single phase
system
• In many situations, fluid changes its phase during convective heat transfer.
• Boiling & condensation are such convective heat transfer processes that are
associated with change in phase of liquid.
Boiling
• The change of phase from liquid to vapour state is known as boiling.
Condensation
• The change of phase from vapour to vapour state is known as condensation.
Applications of Phase change heat transfer
• Thermal and nuclear power plant.
• Refrigerating & Air-conditioning systems.
• Process of heating & cooling
• Heating of metal in furnaces
3.8a. Boiling Heat Transfer
• Change of phase from liquid to vapour state
• Possible only when the temperature of heating surface (Tw) exceeds the
saturation temperature of liquid (Tsat) at given pressure.
• According to convection law,
Q = h A ( Tw – Tsat ) = h A (ΔT)
Where, ΔT – is known as excess temperature
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CHEE 3101 Heat Transfer
3.8a.1 Pool Boiling
• If heat is added to a liquid
from a submerged solid
surface, the boiling process is
referred as pool boiling.
• The liquid above the hot
surface is essentially stagnant and its motion (as well as heat transfer) near
the surface is due to
i) Free convection
ii) Mixing induced by bubble growth & detachment
Typical boiling curve for water at 1 atm pressure.
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CHEE 3101 Heat Transfer
Boiling Curve
Interface evaporation (Region I)
• Evaporation process with no bubble formation.
• Excess temperature ΔT is very small (5oC).
• Liquid near the surface is super heated slightly and evaporation takes place
at the liquid surface.
• The liquid above the hot surface is essentially stagnant and its motion (as
well as heat transfer) near the surface is due to
(i) Free convection
(ii) Mixing induced by bubble growth and detachment
Nucleate Boiling (Region II & III)
• Starts at region II.
• As excess temperature is further increased, bubbles are formed more rapidly
and rapid evaporation takes place. This is indicated in region III.
• Nucleate boiling exists upto ΔT = 50oC.
• The maximum flux, known as critical flux occurs at point A
Film boiling (Region IV, V and VI)
• Region IV – vapour film formed is not stable and collapses & reforms
rapidly.
• With further increase in ΔT, the vapour film is stabilized as indicated in
region V.
• The surface temperature required to maintain a stable film is high and under
these conditions a sizable amount of heat is lost by the surface due to
radiation. This is indicated in region VI.
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CHEE 3101 Heat Transfer
 Inference from the boiling curve
Rapid increase in heat transfer rates is associated with small values of the
excess temperature in nucleate boiling region.
3.8a.2 Flow or Forced convection boiling
• Flow boiling or forced convection boiling may occur when a fluid is forced
through a pipe or over a surface which is maintained at a temperature higher
than the saturation temperature of the fluid.
• Occurs in water tube boilers involving forced convection
3.8b. Condensation
The change of phase from vapour to liquid state is known as condensation
Modes of condensation
There are two modes of condensation
3.8b.1 Filmwise condensation: The liquid condensate wets the solid surface,
spreads out and forms a continuous film over the entire surface is known as
filmwise condensation. Film condensation occurs when a vapour is free from
impurities.
3.8b.2 Dropwise condensation: In dropwise condensation, the vapour
condenses into small liquid droplets of various sizes which fall down the
surface in a random fashion. Heat transfer rates in dropwise condensation may
be as much as 10 times higher than in filmwise condensation.
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CHEE 3101 Heat Transfer
CHAPTER 4
(outcomes 8, 9, 10 and 11)
RADIATION HEAT TRANSFER
4.1 Concepts:
• Any surfaces at all temperatures (but above absolute zero temperature) emit
thermal radiation.
• In addition to emitting radiation, the surfaces of a body has the capacity for
absorbing all or part of the radiation emitted by surrounding surfaces and
falling on it.
• A surface emits radiation in all directions encompassed by a hemisphere.
- We shall be concerned only with situations involving radiation exchange
between surfaces, in which the space between the surfaces is a vacuum or is
occupied by a gas which does not participate in the radiation exchange in any
way.
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CHEE 3101 Heat Transfer
4.1.1 Definition
The heat transfer from one body to another without any transmitting medium is
known as radiation
 Electromagnetic wave phenomenon
 Propagates at the speed of light
4.1.2 Emission properties
The rate of emission of radiation by a body depends upon the following factors:
 Wavelength or frequency of radiation
 Temperature of the surface
 Nature of the surface
4.1.3 Emissive power
- The emissive power is defined as the total amount of radiation emitted by a
body per unit time and unit area. It is expressed in W/m2
- Monochromatic emissive power is the energy emitted at a given wavelength per
unit time per unit area in all directions.
4.1.4 Concept of Black body
Black body is an ideal surface having the following properties
 Absorbs all incident radiation, regardless of wavelength and direction
 Emits the maximum energy for a given temperature and wavelength
4.1.5 Concept of Gray body
 If a body absorbs a definite percentage of incident radiation irrespective of
their wavelength, the body is known as gray body.
 The emissive power of a gray body is always less than that of the black
body.
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CHEE 3101 Heat Transfer
4.2 Absorption, Reflection and Transmission
Energy balance yields, Q = Qa + Qr + Qt
Dividing the above equation by Q,
1 = (Qa /Q) + (Qr /Q) + (Qt /Q) (or) 1 = α + ρ + τ
Absorptivity (α) = Radiation absorbed / Incident radiation
Reflectivity (ρ) = Radiation reflected / Incident radiation
Transmissivity (τ) = Radiation transmitted / Incident radiation
Note: Transmissivity and Emissivity are different
4.2.1 Emissivity
It is defined as the ability of the surface of a body to radiate heat.
It is also defined as the ratio of the emissive power of any body (gray body) to
the emissive power of a black body of equal temperature.
Emissivity, ε = E / Eb
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CHEE 3101 Heat Transfer
4.3 Laws of Radiation
4.3.1 Planck’s Distribution Law
The relationship between the monochromatic emissive power of a black body
and wave length of a radiation at a particular temperature is given by the
following expression, by Planck
Ebλ = (c1 λ-5) / (exp(c2/ λT) - 1)
where Ebλ = Monochromatic emissive power W/m2
λ = Wavelength – m
c1 = 0.374 x 10-15 W m2
c2 = 14.4 x 10-3 mK
4.3.2 Wien’s Displacement Law
The Wien’s law gives the relationship between temperature and wavelength
corresponding to the maximum spectral emissive power of the black body at
that temperature
λmax T = 2898 µmK
=>
λmax T = 2.9 x 10-3 mK
[since µ = 10-6 m]
4.3.3 Stefan – Boltzmann Law
The emissive power of a black body is proportional to the fourth power of
absolute temperature
Eb α T4
 Eb = σ T4
where Eb = Emissive power – W/m2
σ = Steffan – Boltzmann constant
= 5.67 x 10-8 W/m2K4
T = Absolute Temperature – K
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4.3.4 Kirchoff’s Law Of Radiation
This law states that the ratio of total emissive power to the absorptivity is
constant for all surfaces which are in thermal equilibrium with the
surroundings. This can be written as
E1 / α1 = E2 / α2 = E3 / α3 ………
It also states that the emissivity of the body is always equal to its absorptivity
when the body remains in thermal equilibrium with its surroundings.
α1 = ε1 ; α2 = ε2 and so on.
Example: 4.1
A black body at 3000 K emits radiation. Calculate the following,
a)
b)
c)
d)
Monochromatic emissive power at 1 µm wave length,
Wave length at which emission is maximum.
Total emissive power,
Calculate the total emissive of the furnace if it is assumed as a real surface
having emissivity equal to 0.85
Given:
Surface temperature, T = 3000 K
To find:
a.
b.
c.
d.
Monochromatic emissive power Ebλ at λ = 1 µ = 1 x 10-6 m
Maximum Wave length, (λmax )
Total emissive power, Eb
Emissive power of real surface at ε = 0.85
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CHEE 3101 Heat Transfer
Solution:
a. Monochromatic emissive power:
From Planck’s distribution law,
Ebλ = (c1 λ-5) / (exp(c2/ λT) - 1)
where c1 = 0.374 x 10-15 W m2
c2 = 14.4 x 10-3 mK
λ = 1 x 10-6 m [Given]
=> Ebλ = 3.10 x 1012 W/m2
b. Maximum Wave length, (λmax ) :
From Wien’s law,
λmax T = 2.9 x 10-3 mK
=> λmax = 0.966 x 10-6 m
c. Total emissive power, Eb :
From Steffan - Boltzmann law,
Eb = σ T4 W/m2
=>
Where σ = Steffan Boltzmann constant
= 5.67 x 10-8 W/m2 K4
Eb = 4.59 x 106 W/m2
d. Total emissive power of a real surface :
(Eb)real = ε σ T4
where
ε – Emissivity = 0.85
(Eb)real = 3.90 x 106 W/m2
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4.4 Radiation – Gray body heat exchange formulae
Heat exchange between 2 large parallel plate is given by:

Q12 =
σ A (T14 – T24)
Where emissivity,
 
1
1
σ

1
1
2
1
= Steffan Boltzmann constant
= 5.67 x 10-8 W/m2 K4
ε1 = Emissivity of surface 1
ε2 = Emissivity of surface 2
T1 = Temperature of surface 1 – in K
T2 = Temperature of surface 2 – in K
Heat exchange between 2 large concentric cylinder (or) sphere is given by :
Q12 =

A1 σ (T14 – T24)
 
where
1

1
A  1

 1 

1

1
A2 

 2

For cylinder, Area, A = 2 π r L
For sphere, Area, A = 4 π r2
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Example: 4.2
Calculate the net radiant interchange per sq.m, for two large planes at a
temperature of 900 K and 400 K respectively. Assume that the emissivity of hot
plane is 0.9 and that of cold plane is 0.7.
Given:
Hot plane temperature, T1 = 900 K
Cold plane temperature, T2 = 400 K
Emissivity of hot plane, ε1 = 0.9
Emissivity of cold plane, ε2 = 0.7
To find;
Net radiant heat exchange per square meter.
Solution:
The heat exchange between two large parallel plate is given by
Q =
where
σ =
=
σ A (T14 – T24)
 (1)
= 1 / ((1/ ε1) + (1/ ε2) – 1) = 0.649
Steffan Boltzmann constant
5.67 x 10-8 W/m2 K4
Equation ( 1 ) => Q / A = 23.20 x 103 W/m2 = 23.20 kW/m2
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CHEE 3101 Heat Transfer
Example: 4.3
Calculate the heat exchange by radiation between the surfaces of two long
cylinders having radii 120 mm and 60 mm respectively. The axis of the cylinders is
parallel to each other. The inner cylinder is maintained at a temperature of 130 o C
and emissivity of 0.6. Outer cylinder is maintained at a temperature of 30 o C and
emissivity of 0.5.
Given:
r1 = 60 mm = 0.060 m
r2 = 120 mm = 0.12 m
T1 = 130 + 273 = 403 K
T2 = 30 + 273 = 303 K
ε1 = 0.6
ε2 = 0.5
To find:
Heat exchange, (Q)
Solution:
The heat exchange between two large concentric cylinder is given by
Q12 =
σ A (T14 – T24)
 ( 1)
= 1 / ((1/ ε1) + (A1/A2)((1/ ε2) – 1))
= 0.46
Equation ( 1 )
=>
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[since A = π D L and L1 / L2 =1]
Q12 = 176.47 W
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CHEE 3101 Heat Transfer
Assignment Questions:
Q1. Estimate the net radiant heat exchange per square meter between two large
plates at a temperature of 550o C and 320o C. Assume that emissivity of hot plate is
0.8 and cold plate is 0.6
[Answer: Q / A = 9880.6 W/m2 = 9.88 kW/m2]
Q2. Liquid oxygen is stored in double walled spherical vessel. Inner wall
temperature is – 160o C and outer wall temperature is 30o C. Inner diameter of
sphere is 20 cm and outer diameter is 32 cm. Calculate the following.
a. Heat transfer if emissivity of spherical surface is 0.05
b. Rate of evaporation of liquid oxygen if its latent heat is 200 kJ/kg.
[Answer: Q12 = - 2.12 W;
m = 1 x 10-5 kg/s]
Q3. A pipe of outside diameter 30 cm having emissivity 0.6 and at a temperature
of 600 K runs centrally in a brick duct of 40 cm side square section having
emissivity 0.8 and at a temperature of 300 K. Calculate the heat exchange per
metre length.
[Answer: Q12 = 3569.2 W]
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CHEE 3101 Heat Transfer
CHAPTER 5 HEAT EXCHANGERS
(Outcomes 13 to 16)
5.1 Definition
A heat exchanger is defined as an equipment which transfers the heat from a hot
fluid to a cold fluid.
5.1.1 Types of heat exchangers:
Criteria for classification of heat exchangers:
(i)
Nature of heat exchange process
(ii)
Relative direction of fluid motion
(iii)
Design and constructional features
(iv)
Physical state of fluids
I. Nature of heat exchange process:
a) Direct contact heat exchangers or Open heat exchangers
Examples: Cooling towers, Direct contact feed heaters
b) Indirect contact heat exchangers.
Examples: IC engines, gas turbines, Air pre heaters, Economizers.
II. Relative direction of fluid motion:
a.
Parallel flow heat exchanger – Fluids move in same direction
b. Counter flow heat exchanger – Fluids move in opposite direction
c. Cross flow heat exchanger – Fluids make right angle to each other
III. Design and constructional features:
a.
Concentric tubes
b.
Shell and tube (most widely used)
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CHEE 3101 Heat Transfer
c.
Multiple shell and tube passes
d.
Compact heat exchangers
IV. Physical state of fluids:
a.
Condensers
b.
Evaporators.
5.1.2 Applications
• Automobiles (Radiators, Air preheater, Exhaust gas heat removal)
• Gas turbines
• Cooling towers
• Industrial Furnaces (Economizers)
# As discussed above, heat exchangers are manufactured in a variety of types,
The simplest being the double-pipe heat exchanger. One fluid in a double-pipe
heat exchanger flows through the smaller pipe while the other fluid flows through
the annular space between the two pipes. Two types of flow arrangement are
possible in a double-pipe heat exchanger: (1) parallel flow, and (2) counter flow.
In a parallel-flow type, both the hot and cold fluids enter the heat exchanger at the
same end and move in the same direction,
whereas in a counter-flow type, the hot and cold fluids enter the heat exchanger at
opposite ends and flow in opposite directions.
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CHEE 3101 Heat Transfer
Temperature variation in a heat exchanger
Parallel flow
Counter flow
5.2 Introduction to LMTD
• Temperature difference between the hot and cold fluids in a heat exchanger
varies from point to point
• In addition various modes of heat transfer are involved
• Hence, based on the concept of appropriate mean temperature difference,
also called logarithmic mean temperature difference, the total heat transfer
rate in the heat exchanger is expressed as
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CHEE 3101 Heat Transfer
Q = UA (∆T)lm
where
U – Overall heat transfer co-efficient, W/m2K
A – Area, m2
(∆T)lm – Logarithmic mean temperature difference.
FOR PARALLE FLOW
(∆T) lm =[(Th1 – Tc1) – (Th2 – Tc2)] / ln ((Th1 – Tc1) / (Th2 – Tc2))
FOR COUNTER FLOW
(∆T) lm = [(Th1 – Tc2) – (Th2 – Tc1)] / ln ((Th1 – Tc2) / (Th2 – Tc1))
Heat lost by hot fluid = Heat gained by cold fluid
Qh = Qc
Q = mhCph(Th1 – Th2) = mcCpc(Tc2 – Tc1)
Where
mh – Mass flow rate of hot fluid
mc – Mass flow rate of cold fluid
Cph – Specific heat of hot fluid
Cpc – Specific heat of cold fluid
Surface area of cylindrical tube
A = π Di L
Where, Di – Inner diameter
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CHEE 3101 Heat Transfer
In case of evaporators / Condensers
Q = m x hfg
Where , hfg – Latent heat (J/kg K)
Example 5.1
In a double pipe heat exchanger, hot fluid with a specific heat of 2300 J/kg enters
at 380oC and leaves at 300oC. Cold fluid enters at 25oC and leaves at 210oC.
Calculate the heat exchanger area required for
(i)
Counter flow
(ii)
Parallel flow.
Take overall heat transfer coefficient as 750 W/m2 K and mass flow rate of hot
fluid is 1 kg/s.
Given:
Mass flow rate of hot fluid, mh = 1 kg/sec
Entry temperature of cold fluid, Tc1 = 25o C
Exit temperature of cold fluid, Tc2 = 210o C
Specific heat of oil (Hot fluid), Cph = 2300 J/kg K
Entry temperature of hot fluid, Th1 = 380o C
Exit temperature of hot fluid, Th2 = 300o C
Overall heat transfer co-efficient, U = 750 W/m2K
To find: Heat exchanger area (A) for (i) Counter flow (ii) Parallel flow
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CHEE 3101 Heat Transfer
Solution:
We know that,
Heat transfer, Q = mcCpc(Tc2 – Tc1) = mh Cph(Th1 – Th2)
=>
Q = mhCph(Th1 – Th2)
[Since Specific heat & flow rate of hot fluid is known]
Q = 184 x 103 W
We know that,
Heat transfer, Q = UA (∆T) lm 
(1)
(i) For counter flow
(∆T) lm = [(Th1 – Tc2) – (Th2 – Tc1)] / ln ((Th1 – Tc2) / (Th2 – Tc1))
= 218.3o C
Substitute (∆T) lm, Q and U values in Equation (1),
(1) 
Q = UA (∆T) lm
A = Q/U (∆T)lm = 1.12 m2
(ii) For parallel flow
(∆T) lm =[(Th1 – Tc1) – (Th2 – Tc2)] / ln ((Th1 – Tc1) / (Th2 – Tc2))
= 191.3o C
Substitute (∆T) lm, Q and U values in Equation (1),
(1) 
Q = UA (∆T) lm
A = Q/U (∆T) lm = 1.27 m2
Area required for Counter flow = 1.12 m2
Area required for Parallel flow = 1.27 m2
Inference:
Counter flow heat exchanger is more effective compared to parallel flow heat
exchanger
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CHEE 3101 Heat Transfer
5.3 Cross Flow Heat Exchangers
In compact heat exchangers, the two fluids move perpendicular to each other, and
such a flow configuration is called cross-flow. The cross-flow is further classified
as unmixed and mixed flow, depending on the flow configuration, as shown in
Figure. In (a) the cross-flow is said to be unmixed since the plate fins force the
fluid to flow through a particular inter fin spacing and prevent it from moving in
the transverse direction (i.e., parallel to the tubes). The cross-flow in (b) is said to
be mixed since the fluid now is free to move in the transverse direction. Both fluids
are unmixed in a car radiator. The presence of mixing in the fluid can have a
significant effect on the heat transfer characteristics of the heat exchanger.
Other common type of heat exchangers in industrial applications is the shell-andtube heat exchangers. Shell-and-tube heat exchangers contain a large number of
tubes (sometimes several hundred) packed in a shell with their axes parallel to that
of the shell. Heat transfer takes place as one fluid flows inside the tubes while the
other fluid flows outside the tubes through the shell. Baffles are commonly placed
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CHEE 3101 Heat Transfer
in the shell to force the shell-side fluid to flow across the shell to enhance heat
transfer and to maintain uniform spacing between the tubes. Despite their
widespread use, shell and- tube heat exchangers are not suitable for use in
automotive and aircraft applications because of their relatively large size and
weight. Note that the tubes in a shell-and-tube heat exchanger open to some large
flow areas called headers at both ends of the shell, where the tube-side fluid
accumulates before entering the tubes and after leaving them.
Fig. The schematic of a shell-and-tube heat exchanger (Shell & tube Condenser).
Shell-and-tube heat exchangers are further classified according to the number of
shell and tube passes involved. Heat exchangers in which all the tubes make one
U-turn in the shell, for example, are called one-shell-pass and two tube-passes heat
exchangers. Likewise, a heat exchanger that involves two passes in the shell and
four passes in the tubes is called a two-shell-passes and four-tube-passes heat
exchanger
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5.4 Fouling of a heat exchanger
Surface of the heat exchangers do not remain clean after it has been in use for
some time.
Surface becomes fouled with scales or deposits which in turns affects the value of
heat transfer coefficient (U).
This fouling effect is taken care by introducing an additional thermal resistance
called the fouling resistance or fouling factor (Rf) .
U outer 
1
r  1
r r  r 
1
 R fo  o ln  o    o  R fi   o 
ho
k  ri   ri 
 ri  hi
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Causes for fouling
1. Presence of inherent salt & scales in the fluids used in the heat exchangers
2. Condensation of inorganic vapor components
3. Usage of flue gas which contains ash & effluents with it
4. Chemical reaction resulting from the fluids & the heat exchanger
components
5. Usage of low melting matter
5.5 The effectiveness–NTU method
 Usually, a heat exchanger can be designed by the LMTD method when the
inlet and outlet temperatures of both fluids are known.
 However, when the inlet and outlet temperatures of the fluids are not known
and if the problem is to determine the inlet or exit temperature of heat
exchanger, effectiveness – NTU method is used.
 The plot between effectiveness and NTU for various values of (Cmin/Cmax) is
available in the chart for different heat exchangers
Number of Transfer units (NTU)
NTU 
Where
UA
Cmin
U = Overall heat transfer coefficient
A = Heat exchanger area
Cmin = minimum heat Capacity
Heat capacity, C = m x Cp
Effectiveness of a heat exchanger
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Ratio of actual heat transfer to the maximum possible heat transfer
Effectiven ess,  
Actual heat transfer
Q

Max possible heat transfer Qmax
Where,
Q = mcCpc(Tc2 – Tc1) = mhCph(Th1 – Th2)
Qmax = Cmin (Th1 – Tc1)
Cmin = minimum heat capacity
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Effectiveness Vs NTU Chart
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Example 5.2
A parallel flow heat exchanger is used to cool 4.2 kg/min of hot liquid of specific
heat 3.5 kJ/kg K at 130oC. A cooling water of specific heat 4.18 kJ/kg K is used for
cooling purpose at a temperature of 15oC. The mass flow rate of cooling water is
17 kg/min. calculate the following.
a. Outlet temperature of hot liquid
b. Outlet temperature of water
c. Effectiveness of heat exchanger.
Take Overall heat transfer co-efficient as 1100 W/m2K. Heat exchanger area is
0.30 m2
Given:
Mass flow rate of hot fluid, mh= 4.2 kg/min = 0.07 kg/s
Specific heat of hot fluid, Cph = 3.5 kJ/kg K = 3.5 x 103 J/kg K
Inlet temperature of hot fluid, Th1 = 130o C
Mass flow rate of cooling water, mc= 17 kg/min = 0.28 kg/s
Specific heat of water, Cpc = 4.18 kJ/kg K = 4.18 x 103 J/kg K
Inlet temperature of cooling water, Tc1 = 15o C
Overall heat transfer co-efficient, U = 1100 W/m2K
Area, A = 0.30 m2
To find:
a. Outlet temperature of liquid (Th2)
b. Outlet temperature of water (Tc2)
c. Effectiveness of heat exchanger (ε).
Solution:
Capacity rate of hot liquid, Ch = mh x Cph = 245 W/K
Capacity rate of water,
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Cc = mc x Cpc = 1170.4 W/K
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 (1)
 (2)
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CHEE 3101 Heat Transfer
From (1) and (2),
Cmin = 245 W/K;
Cmax = 1170.4 W/K
 (3)
=> Cmin / Cmax = 0.209
Number of transfer units, NTU = UA / Cmin
=>
 (4)
NTU = 1.34
To find effectiveness ε, refer data chart for parallel flow heat exchanger
From graph,
Xaxis  NTU =1.34
Curve  Cmin / Cmax = 0.209
Corresponding Yaxis value is 64%
i.e., Effectiveness, ε = 0.64
Maximum possible heat transfer
Qmax = Cmin (Th1 – Tc1) = 28175 W
Actual heat transfer rate
Q = ε x Qmax = 18032 W
We know that,
Heat transfer, Q = mc Cpc (Tc2 – Tc1)
=>
18032 = 1170.4 Tc2 - 17556
Tc2 = 30.40o C
=>
Outlet temperature of cold water, Tc2= 30.40o C
We know that,
Heat transfer, Q = mh Cph (Th1 – Th2)
=>
=>
18032 = 31850 – 245 Th2
Th2 = 56.4o C
Outlet temperature of hot liquid, Th2 = 56.4o C
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CHEE 3101 Heat Transfer
Assignment Questions:
Q1. Water flows at the rate of 65 kg/min through a double pipe, counter flow heat
exchanger. Water is heated from 50oC to 75oC by oil flowing through the tube. The
specific heat of the oil is 1.780 kJ/kg K. The oil enters at 115 oC and leaves at 70o
C. The overall heat transfer co-efficient is 340 W/m2K. Calculate the following
1. Heat exchanger area
2. Rate of heat transfer
[Answer: Q = 113 x 103 W; A = 11.54 m2]
Q2. A counter flow double pipe heat exchanger is used to cool the engine oil from
150oC to 55oC with water, available at 23oC as the cooling medium. The specific
heat of oil is 2125 J/kg K. The flow rate of cooling water through the inner tube of
0.4 m diameter is 2.2 kg/s. The flow rate of oil through the outer tube of 0.75 m
diameter is 2.4 kg/s. If the value of the overall heat transfer co-efficient is 240
W/m2K, how long must the heat exchanger be to meet its cooling requirement?
[Answer: L = 31.9 m]
Q3. Saturated steam at 126oC is condensing on the outer tube surface of a single
pass heat exchanger. The heat exchanger heats 1050 kg/h of water from 20 oC to
95oC. The overall heat transfer co-efficient is 1800 W/m2K. Calculate the
following
1. Area of heat exchanger
2. Rate of condensation of steam.
Take hfg = 2185 kJ/kg
[Answer: Rate of condensation of steam, mh = 0.0416 kg/s; A = 0.828 m2]
Q4. In a counter flow heat exchanger, water is heated from 20 o C to 80o C by oil
with a specific heat of 2.5 kJ/kg-K and mass flow rate of 0.5 kg/s. The oil is cooled
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CHEE 3101 Heat Transfer
from 110o C to 40o C. If the overall heat transfer co-efficient is 1400 W/m2K, find
the following by using NTU method.
a. Mass flow rate of water
b. Effectiveness of heat exchanger
b. Surface area.
[Answer: Mass flow rate of water, mc = 0.348 kg/s
Effectiveness, ε = 0.77
Surface area, A = 3.03 m2]
**************************
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References:
1. Yunus A.Cengel, “Heat Transfer” 2nd Edition Reprint, McGraw Hill Higher
Education, 2002.
2. P S Ghoshdastidar, “Heat Transfer” 2nd Edition, Oxford University Press,
2005.
3. J. P. Holman, “Heat Transfer” 8th International Edition, McGraw Hill
Higher Education INC, 1997.
4. D S Kumar, “Heat and Mass Transfer”, Laxmi Publications, 2000.
5. Ozisik M.N, “Heat Transfer”, McGraw-Hill Book Co., 1994.
6. Nag P.K, “ Heat Transfer”, Tata McGraw-Hill, New Delhi, 2002.
7. C.P Kothandaraman, “Heat and Mass Transfer Data Book”, New Age
international (P) limited, publishers, New Delhi,2012.
8. NPTEL (National Programme on Technology Enhanced Learning)
nptel.ac.in.
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