COURSE MATERIAL

II Year B. Tech IV- Semester

CHEMICAL ENGINEERING

PROCESS HEAT TRANSFER

Prepared by :
Dr G Srinivasan
Professor , Department of Chemical Engineering
PUDUCHERRY TECHNOLOGICAL UNIVERSITY
DEPARTMENT OF CHEMICAL ENGINEERING
STEADY STATE HEAT CONDUCTION

Course Contents
2.1 Introduction
2.2 Thermal resistance
2.3 Thermal conductivity of
material
2.4 General heat conduction
equation
2.5 Measurement of thermal
conductivity (Guarded hot
plate method)
2.6 Conduction through a plane
wall
2.7 Conduction through a
composite wall
2.8 Heat flow between surface and
surroundings: cooling and
heating of fluids
2.9 Conduction through a
cylindrical wall
2.10 Conduction through a
multilayer cylindrical wall
2.11 Conduction through a sphere
2.12 Critical thickness of insulation
2.13 Solved Numerical
2.14 References
2.1 Introduction
 The rate of heat conduction in a specified direction is proportional to the
temperature gradient, which is the rate of change in temperature with distance in
that direction. One dimensional steady state heat conduction through homogenous

𝑑𝑡
material is given by Fourier Law of heat conduction:

𝑄 = −𝑘𝐴
𝑑𝑥
𝑄 𝑑𝑡
𝑞 = = − − − − − − − (2.1)
�−𝑘 𝑑
𝑥
Where, �
𝑞 = heat flux, heat conducted per unit time per unit area, 𝑊⁄𝑚2

A = area perpendicular to the direction of heat flow, 𝑚2

Q = rate of heat flow, W

passing, Kelvin K or degree centigrade ℃

dt = temperature difference between the two surfaces across which heat is

The ratio 𝑑𝑡⁄𝑑𝑥 represents the change in temperature per unit thickness, i.e.
dx = thickness of material along the path of heat flow, m

the temperature gradient.
 The negative sign indicates that the heat flow is in the direction of negative
temperature gradient, so heat transfer becomes positive.
 The proportionality factor k is called the heat conductivity or thermal conductivity of
material through which heat is transfer.
 The Fourier law is essentially based on the following assumptions:
1. Steady state heat conduction, i.e. temperature at fixed point does not change
with respect to time.
2. One dimensional heat flow.
3. Material is homogenous and isotropic, i.e. thermal conductivity has a constant
value in all the directions.
4. Constant temperature gradient and a linear temperature profile.
5. No internal heat generation.

𝑑𝑡
 The Fourier law helps to define thermal conductivity of the material.

𝑄 = −𝑘𝐴
𝑑𝑥
Assuming 𝑑𝑥 = 1𝑚; 𝐴 = 𝑚 and 𝑑𝑡 = 1℃, we obtain
2

𝑄=𝑘


 Hence thermal conductivity may be defined as the amount of heat conducted per
unit time across unit area and through unit thickness, when a temperature
difference of unit degree is maintained across the bounding surface.
 Unit of thermal conductivity is given by:

𝑘 = − 𝑄 𝑑𝑥
𝐴 𝑑𝑡
 𝑊 𝑚 𝑊
∴ [𝑘] = 2 =
𝑚 𝑑𝑒𝑔 𝑚 − 𝑑𝑒𝑔
2.2 Thermal Resistance
 In systems, which involve flow of fluid, heat and electricity, the flow quantity is
directly proportional to the driving force and inversely proportional to the flow
resistance.
 In a hydraulic system, the pressure along the path is the driving potential and
roughness of the pipe is the flow resistance.
 The current flow in a conductor is governed by the voltage potential and electrical
resistance of the material.
 Likewise, temperature difference constitutes the driving force for heat conduction
through a medium.

Fig. 2.1 Concept of thermal resistance

 From Fourier’s law
𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙
ℎ𝑒𝑎𝑡 𝑓𝑙𝑜𝑤 𝑄 =
(𝑑𝑡) 𝑡ℎ𝑒𝑟𝑚𝑎𝑙
𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 (𝑑𝑥⁄𝑘𝐴)
 Thermal resistance, 𝑅𝑡 = (𝑑𝑥⁄𝑘𝐴), is expressed in the unit 𝑑𝑒𝑔⁄𝑊.
 The reciprocal of thermal resistance is called thermal conductance and it represents
the amount of heat conducted through a solid wall of area A and thickness dx when
a temperature difference of unit degree is maintained across the bounding surfaces.

2.3 Thermal Conductivity of Materials

 Thermal conductivity is a property of the material and it depends upon the material
structure, moisture content and density of the material, and operating conditions of
pressure and temperature.
 Following remarks apply to the thermal conductivity and its variation for different
materials and under different conditions:
 In material thermal conductivity is due to two effects: the lattice vibrational waves
and flow of free electrons.
  In metals the molecules are closely packed; molecular activity is rather small and so
thermal conductivity is mainly due to flow of free electrons.
 In fluids, the free electron movement is negligibly small so conductivity mainly
depends upon the frequency of interactions between the lattice atoms.
 Thermal conductivity is highest in the purest form of a metal. Alloying of metals and
presence of other impurities reduce the conductivity of the metal.

Thermal conductivity of pure copper is 385 W⁄m − deg and that of

nickel is 93W⁄m − deg.


conductivity of only 24 W⁄m − deg.

 Monel metal, an alloy of 30% nickel and 70% copper, has thermal

 Mechanical forming (i.e. forging, drawing and bending) or heat treatment of metal
cause considerable variation in thermal conductivity. Conductivity of hardened steel
is lower than that of annealed steel.
 At elevated temperatures, thermal vibration of the lattice becomes higher and that
retards the motion of free electrons. So, thermal conductivity of metal decreases
with increases of temperature except the aluminium and uranium.

range of 130 ℃ to 370 ℃.

 Thermal conductivity of aluminium remains almost constant within the temperature

 For uranium, heat conduction depends mainly upon the vibrational movement of
atoms. With increase of temperature vibrational movement increase so, conductivity
also increase.
 According to kinetic theory of, conductivity of gases is directly proportional to the
density of the gas, mean molecular speed and mean free path. With increase of
temperature molecular speed increases, so conductivity of gas increases.
Conductivity of gas is independent of pressure except in extreme cases as, for
example, when conditions approach that of a perfect vacuum.
 Molecular conditions associated with the liquid state are more difficult to describe,
and physical mechanisms for explaining the thermal conductivity are not well
understood. The thermal conductivity of nonmetallic liquids generally decreases with
increasing temperature. The water, glycerine and engine oil are notable exceptions.
The thermal conductivity of liquids is usually insensitive to pressure except near the
critical point.
 Thermal conductivity is only very weakly dependent on pressure for solids and for
liquids a well, and essentially dependent of pressure for gases at pressure near
standard atmospheric.
 Non-metallic solids do not conduct heat as efficiently as metals.
 The ratio of the thermal and electrical conductivities is same for all metals at the
same temperature; and that the ratio is directly proportional to the absolute
temperature of the metal.

2.5 Measurement of Thermal Conductivity (Guarded Hot Plate

Method)
 Construction

Main heater 𝐻𝑚 placed at the centre of the unit. It is maintained at a fixed

 The essential elements of the experimental set-up as shown in figure 2.5 are:


Guarded heater 𝐻g which surrounds the main heater on its ends. The guarded
temperature by electrical energy which can be metered.

heater is supplied electrical energy enough to keep its temperature same as that of
main heater.

Fig. 2.5 Elements of guarded hot plate method

1. Steady State Heat Conduction Heat Transfer (-)

 Function of the guarded heater is to ensure unidirectional heat flow and eliminates

Test specimens 𝑆1 and 𝑆2 which are placed on both sides of the heater.
the distortion caused by edge losses.

Cooling unit plates 𝐶1 and 𝐶2 are provided for circulation of cooling medium. Flow


of cooling medium is maintained to keep the constant surface temperature of
specimen.
 Thermocouples attached to the specimens at the hot and cold faces.
 Desired measurement

𝑑𝑡 𝑘𝐴
 From the Fourier’s law of heat conduction
𝑄 = −𝑘𝐴
(𝑡ℎ − 𝑡𝑐)
= 𝑋
𝑑𝑥
𝑄 𝑋
∴𝑘 = − − − − − − − (2.46)
𝐴 (𝑡ℎ − 𝑡� �
)
 So to measure thermal conductivity k following measurements are required
 Heat flow Q from the main heart through a test specimen; it will be half of the total
electrical input to the main heater

Temperature drop across the specimen (𝑡ℎ − 𝑡𝑐); subscripts h and c refer to the hot
 Thickness of the specimen X

and cold faces respectively
 Area A of heat flow; the area for heat flow is taken to be the area of main heater
plus the area of one-half of air gap between it and the guarded heater
 For the specimen of different thickness, the respective temperature at the hot and
cold faces would be different and then the thermal conductivity is worked out from

𝑄 𝑋 𝑋
the following relation:
𝑘= ) − − − − − − − (2.47)
( 1 +
2

𝐴 (𝑡ℎ1 − 𝑡𝑐1) (𝑡ℎ2 − 𝑡𝑐2)

 Where suffix 1 is for the upper specimen and 2 is for the lower specimen.
 Here Q is the total electrical input to the main heater.

2.6 Conduction Through a Plane Wall:-

 Alternatively, The Fourier rate equation may be used directly to determine the heat

Consider an elementary strip of thickness dx located at a distance x from the

flow rate.

reference plane. Temperature difference across the strip is dt, and temperature


gradient is dt⁄dx.
 Heat transfer through the strip is given by

𝑄 = −𝑘 𝐴 𝑑𝑡
𝑑𝑥
 Fig. 2.6 Heat conduction through plane wall

transfer through the wall. So integrate the equation between the limits, t = t1
 For steady state condition, heat transfer through the strip is equal to the heat

at x = 0 and t = t2 at x = δ, thus
𝑡2

𝑄 ∫ 𝑑𝑥 = −𝑘 𝐴 ∫ 𝑑𝑡
ẟ

0 𝑡1

𝑄𝛿 =𝑘 𝑘 𝐴 (𝑡1 − 𝑡2)
𝐴( 𝑡1 − ); 𝑄 = − − − − − − − (2.53)
𝛿
𝑡2

 To determine the temperature at any distance x from the wall surface, the Fourier

a) 𝑥 = 0 where the temperature is stated to be 𝑡1

rate equation is integrated between the limit:

b) 𝑥 = 𝑥 where the temperature is to be worked out

 Thus,
𝑥 𝑡

𝑄 ∫ 𝑑𝑥 = −𝑘 𝐴 ∫ 𝑑𝑡
0
𝑡1

𝑄 𝑥 = 𝑘 𝐴( 𝑡1 𝑘 𝐴 ( 𝑡1
− 𝑡); 𝑄−=𝑡)
𝑥
 Substituting the value of Q in above equation

𝑘 𝐴 ( 𝑡1 − 𝑡2 𝑘 𝐴 ( 𝑡1 − 𝑡)
=
 𝑡2 − 𝑡1
∴ 𝑡 = 𝑡1 + ( ) 𝑥 − − − − − − − (2.54)
𝛿
 The expression for the heat flow rate can be written as
𝑡1 − 𝑡2 𝑡1 −
𝑄 =
𝛿⁄𝑘 𝐴 𝑡2 − − − − − − − (2.55)
=
𝑅𝑡

 Where Rt = δ⁄k A is the thermal resistance to heat flow. Equivalent thermal circuit
for flow through a plane wall has been included in figure 2.6.
 Let us develop the condition when weight, not space, required for insulation of a
plane wall is the significant criterion.
 For one dimensional steady state heat condition
𝑘 𝐴 (𝑡1 − 𝑡2) 𝑡1 − 𝑡2
𝑄= =
𝛿 𝛿⁄𝑘 𝐴

Thermal resistance of the wall, Rt = δ⁄k A

Weight of the wall, W = ρ A δ


Eliminating the wall thickness δ from expression



𝑊

𝑅𝑡 =
𝜌𝑘𝐴2

𝑊 = (𝜌𝑘)𝑅𝑡𝐴2 − − − − − − − (2.56)

 From the equation when the product (ρk) for a given resistance is smallest, the
weight of the wall would also be so. It means for the lightest insulation for a
specified thermal resistance, product of density times thermal conductivity should
be smallest.

2.7 Conduction Through a Composite Wall

 A composite wall refers to a wall of a several homogenous layers.
 Wall of furnace, boilers and other heat exchange devices consist of several layers; a
layer for mechanical strength or for high temperature characteristics (fire brick), a
layer of low thermal conductivity material to restrict the flow of heat (insulating
brick) and another layer for structural requirements for good appearance (ordinary
brick).
 Figure 2.7 shows one such composite wall having three layers of different materials

The layers have thickness δ1, δ2, δ3 and their thermal conductivities correspond to
tightly fitted to one another.


The surface temperatures of the wall are t1 and t4 and the temperatures at the
the average temperature conditions.

interfaces are t2 and t3.


 Fig. 2.7 Heat conduction through composite wall
 Under steady state conditions, heat flow does not vary across the wall. It is same for
𝑘1𝐴 𝑘2𝐴 𝑘3𝐴
𝑄 = (𝑡 ) = (𝑡 )= (𝑡
every layer. Thus
) − − − − − − − (2.57)
−𝑡 −𝑡 −𝑡
𝛿1 𝛿2 𝛿3
1 2 2 3 4
3

𝑄 𝛿1 𝑄 𝛿2 𝑄 𝛿3
 Rewriting the above expression in terms of temperature drop across each layer,

𝑡1 − 𝑡2 = 1 ; 𝑡2 − 𝑡3 2 ; 𝑡3 − 𝑡4 =
𝑘 =
𝐴 𝐴 𝑘 𝐴
𝑘 3
 Summation gives the overall temperature difference across the wall
𝑡1 − 𝑡4 = 𝑄 ( 𝛿1 𝛿3
+ 𝛿 + )
𝑘1𝐴 𝑘2 𝑘3𝐴
2

𝐴
(𝑡1 − 𝑡4)
Then

𝑄=
+ 𝑘2 + 𝑘3 𝐴
ẟ1 ẟ2 ẟ3
𝑘1 𝐴
𝐴

𝑄= (𝑡1 − 𝑡4) ( 𝑡1 −
𝑡4) − − − − − − − (2.58)
𝑅𝑡1 + 𝑅𝑡2 + =
𝑅𝑡
𝑅𝑡3
  Where Rt = Rt1 + Rt2 + Rt3, is the total resistance.
 Analysis of the composite wall assumes that there is a perfect contact between
layers and no temperature drop occurs across the interface between materials.

2.8 Heat Flow Between Surface and Surroundings: Cooling and

Heating of Fluids
 When a moving fluid comes into contact with a stationary surface, a thin boundary
layer develops adjacent to the wall and in this layer there is no relative velocity with
respect to surface.
 respect to surface.

Fig. 2.8 Heat conduction through a wall separating two fluids

 In a heat exchange process, this layer is called stagnant film and heat flow in the
layer is covered both by conduction and convection processes. Since thermal
conductivity of fluids is low, the heat flow from the moving fluid of the wall is mainly
due to convection.
 The rate of convective heat transfer between a solid boundary and adjacent fluid is
given by the Newton-Rikhman law:

𝑄 = ℎ 𝐴(𝑡𝑠 − 𝑡𝑓) − − − − − − − (2.59)

Where, tf is the temperature of moving fluid, ts is the temperature of the wall

surface, A is the area exposed to heat transfer and h is the convective co-efficient.


The dimension of h is W⁄m2 − deg.

 Heat transfer by convection may be written as

𝑄 = 𝑡𝑠 − 𝑡𝑠 −
𝑡𝑓 = 𝑡𝑓 − − − − − − − (2.60)
1 𝑅𝑡
ℎ𝐴
 Where Rt = 1⁄h A is the convection resistance.
 The heat transfer through a wall separating two moving fluids involves: (i) flow of
heat from the fluid of high temperature to the wall, (ii) heat conduction through the
wall and (iii) transport of heat from the wall to the cold fluid.

𝑘𝐴
 Under steady state conditions, the heat flow can be expressed by the equations:

𝑄 = ℎ𝑎 𝐴(𝑡𝑎 − 𝑡1) = (𝑡1 − 𝑡2) = ℎ𝑏 𝐴(𝑡2 − 𝑡𝑏)

𝛿
Where ha and hb represent the convective film coefficients, k is thermal conductivity
of the solid wall having thickness δ. These expressions can be presented in the form:

 𝑡𝑎 − 𝑡1 = 𝑄 𝑄𝛿 𝑄
ℎ ; 𝑡1 − ; 𝑡2 −
𝑡 ℎ 𝐴
𝐴2 = 𝑡
𝑘𝐴𝑏 =
𝑎 𝑏

1 𝛿 1
 Summation of these gives
𝑡𝑎 − 𝑡𝑏 = 𝑄 ( + + )
ℎ𝑎 𝑘 ℎ𝑏 𝐴
𝐴 𝐴

∴𝑄 = (𝑡𝑎 − 𝑡𝑏)
1 1 − − − − − − − (2.61)
+ +
ẟ
ℎ𝑎𝐴
𝑘 ℎ𝘣𝐴
𝐴

 The denominator (1⁄haA + δ⁄kA + 1⁄hbA) is the sum of thermal resistance

of difference sections through which heat has to flow.
 Heat flow through a composite section is written in the form

𝑄 = 𝑈𝐴(𝑡𝑎 (𝑡𝑎 − 𝑡𝑏)

− ) = −−−−−−−
1⁄
𝑡𝑏 (2.62) 𝑈𝐴
 Where, U is the overall heat transfer coefficient.
 It represents the intensity of heat transfer from one fluid to another through a wall
separating them.
 Numerically it equals the quantity of heat passing through unit area of wall surface in

dimensions of W⁄m2 − deg.

unit time at a temperature difference of unit degree. The coefficient U has

 By comparing the equation

1 1 𝛿 1
= + +
𝑈𝐴 𝐴 𝑘𝐴 = 𝑅𝑡 − − − − − − − (2.63)
ℎ𝑏 𝐴
ℎ𝑎

 So heat transfer coefficient is reciprocal of unit thermal resistance to heat flow.

 The overall heat transfer coefficient depends upon the geometry of the separating
wall, its thermal properties and the convective coefficient at the two surfaces.
 The overall heat transfer coefficient is particularly useful in the case of composite
walls, such as in the design of structural walls for boilers, refrigerators, air-
conditioned buildings, and in the design of heat exchangers.

2.9 Conduction Through a Cylindrical Wall

cylindrical element at radius r.

 In the alternative approach to estimate heat flow, consider an infinitesimally thin

Let thickness of this elementary ring be dr and the change of temperature across it
be dt.


 Then according to Fourier law of heat conduction

𝑄 = −𝑘𝐴 𝑑 𝑑𝑟
𝑡 = −𝑘(2𝜋𝑟𝑙)
𝑑𝑡 𝑑𝑟
𝑑𝑟
𝑄 = 𝑑𝑡
𝑘(2𝜋𝑟𝑙)
 Integrate the equation within the boundary condition

𝑄 𝑟 𝑡2
2 𝑑
2𝜋 = ∫ 𝑑𝑡
∫𝑟
𝑘𝑙
𝑟1 𝑟
𝑡1

𝑄 𝑟
𝑙𝑜𝑔 2 = − 𝑡 )
(𝑡
2𝜋𝑘𝑙 𝑒
𝑟1 1 2

(𝑡1 − 𝑡2) (𝑡1 − 𝑡2)

𝑄 = 2𝜋𝑘𝑙 𝑟 = − − − − − − − (2.68)
𝑙𝑜𝑔 2⁄𝑟 𝑅
1
𝑒 𝑡

𝑟
 For conduction in hollow cylinder, the thermal resistance is given by:
𝑙𝑜𝑔𝑒 21⁄𝑟
𝑅𝑡 = − − − − − − − (2.69)
2𝜋𝑘𝑙
 Special Notes
 Heat conduction through cylindrical tubes is found in power plant, oil refineries and
most process industries.
 The boilers have tubes in them, the condensers contain banks of tubes, the heat
exchangers are tubular and all these units are connected by tubes.
 Surface area of a cylindrical surface changes with radius. Therefore the rate of heat
conduction through a cylindrical surface is usually expressed per unit length rather
than per unit area as done for plane wall.
 Logarithmic Mean Area
 It is advantageous to write the heat flow equation through a cylinder in the same
form as that for heat flow through a plane wall.

Then thickness 𝛿 will be equal to (𝑟2 − 𝑟1) and the area 𝐴 will be an equivalent area
Fig. 2.10 Logarithmic mean area concept

𝐴𝑚. Thus


𝑘𝐴 (𝑡1 − 𝑡2)
𝑄 = (𝑡 − ) = − − − − − − − (2.70)
𝑡 𝑘𝐴
𝛿 1 2 𝑚
(𝑟2 − 𝑟1)

(t1 − t2) (𝑡1 − 𝑡2)

 Comparing equations 3.68 and 3.70

𝑄 = 2πkl = 𝑘𝐴
log r2⁄r1 2 − 𝑟1)
𝑚

(𝑟
e

2π(𝑟2 − 𝑟1)𝑙 𝐴2 − 𝐴1
𝐴𝑚 = − − − − − − − (2.71)
log r2
⁄ log A2⁄A
1
e

r=
e

Where 𝐴1 and 𝐴2 are the inner and outer surface areas of the cylindrical tube.
The equivalent area 𝐴𝑚 is called the logarithmic mean area of the tube. Further


2π(𝑟2 − 𝑟1)𝑙
𝐴𝑚 = 2𝜋𝑟𝑚𝑙 =
log r2⁄r1
e
  Obviously, logarithmic mean radius of the cylindrical tube is:
(𝑟2 − 𝑟1)
𝑟𝑚 =
log r2 r − − − − − − − (2.72)
e
⁄1

2.10 Conduction Through a Multilayer Cylindrical Wall

 Multi-layer cylindrical walls are frequently employed to reduce heat looses from
metallic pipes which handle hot fluids.
 The pipe is generally wrapped in one or more layers of heat insulation.
 For example, steam pipe used for conveying high pressure steam in a steam power
plant may have cylindrical metal wall, a layer of insulation material and then a layer
of protecting plaster.
 The arrangement is called lagging of the pipe system.

Fig. 2.11 Steady state heat conduction through a composite cylindrical wall
 Figure 2.11 shows conduction of heat through a composite cylindrical wall having
three layers of different materials.
 There is a perfect contact between the layers and so an equal interface temperature
for any two neighbouring layers.
 For steady state conduction, the heat flow through each layer is same and it can be

(𝑡1 − 𝑡2)
described by the following set of equations:
𝑄 = 2𝜋𝑘1𝑙 𝑟
log 2⁄𝑟 1
𝑒

(𝑡2 − 𝑡3)
= 2𝜋𝑘2𝑙
log 𝑟3⁄𝑟
2
𝑒
 (𝑡3 − 𝑡4)
= 2𝜋𝑘3𝑙
log 𝑟4⁄𝑟3
𝑒

 These equations help to determine the temperature difference for each layer of the
composite cylinder,
(𝑡 � 𝑙𝑜 𝑟2
−𝑡 )= � 𝑔
1 2
2𝜋𝑘1𝑙 𝑒
𝑟1
(𝑡 𝑄 𝑙𝑜 𝑟3
−𝑡 )=
𝑔
2 3
2𝜋𝑘2𝑙 𝑒
𝑟2
(𝑡 𝑄 𝑙𝑜 𝑟4
−𝑡 )=
𝑔
3 4
2𝜋𝑘3𝑙 𝑒
𝑟3

1 𝑟2 𝑟3 𝑟4
 From summation of these equalities;

11 12 3
𝑡1 − 𝑡4 = 𝑄 + +
[ 1𝑙
𝑙𝑜𝑔𝑒 2𝜋𝑘 𝑙𝑜𝑔𝑒 2𝜋𝑘 𝑙𝑜𝑔𝑒 ]
2𝜋𝑘 𝑙 𝑙 𝑟
𝑟 2
𝑟 3

𝑡1 − 𝑡4
 Thus the heat flow rate through a composite cylindrical wall is

𝑄 = 1 𝑙𝑜𝑔 𝑟2 + 1 + 𝑟4 − − − − − − − (2.73)
𝑟3 1

𝑙𝑜𝑔 𝑙𝑜𝑔
2𝜋𝑘1𝑙 𝑒 2𝜋𝑘2 𝑒 2𝜋𝑘3 𝑒 𝑟
𝑟1 𝑙 𝑟2 𝑙
3

 The quantity in the denominator is the sum of the thermal resistance of the different

𝑡1 − 𝑡4
layers comprising the composite cylinder.
𝑄= − − − − − − − (2.74)
𝑅
𝑡

 Where, Rt is the total resistance

Fig. 2.12 Heat conduction through cylindrical wall with convection coefficient
 If the internal and external heat transfer coefficients for the composite cylinder as
shown in figure 2.12 are hi and ho respectively, then the total thermal resistance to

𝑅𝑡 = 1 1 𝑟2 1 𝑟3 1
heat flow would be:

2𝜋𝑟 𝑙ℎ + 2𝜋 𝑘 𝑙𝑟
𝑙𝑜𝑔𝑒 + 2𝜋 𝑙𝑜𝑔𝑒 + 2𝜋𝑟 𝑙ℎ
𝑘 𝑙𝑟
1 𝑖 1 1 2 3 𝑜
2
 and heat transfer is given as
𝑄 = (𝑟𝑡𝑖 − 𝑡1𝑜)
1
+
1
𝑙𝑜𝑔 2
+ 𝑟3 1 − − − − − − − (2.75)
𝑙𝑜𝑔 +
2𝜋𝑟1𝑙ℎ𝑖 2𝜋𝑘1 𝑒 2𝜋𝑘2 𝑒 2𝜋𝑟3𝑙ℎ𝑜
𝑙 𝑟1 𝑙 𝑟2

 Overall Heat Transfer Coefficient U

𝑄 = 𝑈𝐴(𝑡𝑖 − 𝑡𝑜) − − − − − − − (2.76)

 The heat flow rate can be written as:

 Since the flow area varies for a cylindrical tube, it becomes necessary to specify the
area on which U is based.
 Thus depending upon whether the inner or outer area is specified, two different

𝑄 = 𝑈𝑖𝐴𝑖(𝑡𝑖 − 𝑡𝑜) = 𝑈𝑜𝐴𝑜(𝑡𝑖 − 𝑡𝑜)

values are defined for U.

 Equating equations of heat transfer

𝑈 2𝜋𝑟 𝑙(𝑡 (𝑡𝑖 − 𝑡𝑜)
− )=
𝑡
𝑖 1 𝑖 𝑙𝑜 𝑙𝑜
𝑜 1
+
1 𝑔 𝑟2
+ 𝑔 𝑟3
+
1
1
2𝜋𝑟1𝑙ℎ𝑖 2𝜋𝑘1 𝑒 2𝜋𝑘2 𝑒 2𝜋𝑟3𝑙ℎ𝑜
𝑟1 𝑟2
(𝑡𝑖 −
𝑙 𝑙

∴ 𝑈𝑖 = 𝑡𝑜)
+ − − − − − − − (2.77)
1 𝑟1
+
𝑟2 𝑟1
𝑟3 𝑟1
𝑙𝑜𝑔 𝑙𝑜𝑔 +

ℎ𝑖 𝑘 𝑒 𝑘 𝑒 𝑟3ℎ𝑜
 Similarly 𝑟1 𝑟2

(𝑡𝑖 − 𝑡𝑜)
1 2
𝑈𝑜 =
+ 𝑙𝑜𝑔 + − − − − − − − (2.78)
𝑟3 𝑟3 𝑟2
𝑟3 1
𝑟3
𝑙𝑜𝑔 +
𝑟1ℎ𝑖 𝑘 𝑒 𝑘 𝑒 ℎ𝑜
𝑟1 𝑟2
1 2

1
 Overall heat transfer coefficient may be calculated by simplified equation as follow

𝑈𝑖𝐴𝑖 = 𝑈𝑜𝐴𝑜 − − − − − − − (2.79)

= �
� 𝑡

2.11 Conduction Through a Sphere

spherical element at radius r and thickness dr.

 In the alternative approach to determine heat flow, consider an infinitesimal thin

 The change of temperature across it be dt. According to Fourier law of heat

𝑑𝑡 𝑑𝑡
conduction
𝑄 = −𝑘𝐴 = −𝑘(4𝜋𝑟2)
𝑑𝑟 𝑑𝑟

𝑄 𝑟2 𝑑𝑟
 Separating the variables and integrating within the boundary conditions
𝑡2
∫ = − ∫ 𝑑𝑡
4𝜋𝑘 𝑟1 � 2 𝑡1
�
 𝑄 1
− ) = (𝑡1 − 𝑡2)
1
(
4𝜋𝑘 𝑟1 𝑟2
4𝜋𝑘(𝑡1 − 𝑡2)𝑟1𝑟2 ( 𝑡1 −
∴𝑄 = )
𝑡2=
(𝑟 − 𝑟 ) (𝑟 − 𝑟 )
1 ⁄
2 1 2
4𝜋𝑘𝑟
1 𝑟
2
 Heat conduction through composite sphere can be obtained similar to heat
conduction through composite cylinder. Heat conduction through composite sphere
will be:

𝑄= (𝑡1 − 𝑡2)
𝑅𝑡1 + 𝑅𝑡2 +
𝑅𝑡3

𝑄= ( 𝑡1 −
) 𝑡2) − 𝑟3 ⁄) − − − − − (2.83)
(𝑟 2 − 𝑟1⁄ 4𝜋 4𝜋 𝑟
𝑘 𝑟𝑟 +
(𝑟3 − 𝑟2 ⁄) 𝑟
𝑟 +
(𝑟4 𝑘 𝑟
4𝜋
𝑘
1 1 2 2 2 3 3 3 4

(𝑡1 − 𝑡2)
 Further, if the convective heat transfer is considered, then

𝑄=𝑅 +𝑅 +𝑅 +𝑅
𝑡𝑖 𝑡1 𝑡2 𝑡3

+ 𝑅𝑡𝑜 (𝑡1 − 𝑡2)

𝑄 1 − 𝑟 ⁄) − 𝑟 ⁄) + 1⁄
= ⁄4𝜋𝑟2 − 𝑟1)
+ ⁄4𝜋 𝑟 +
2 4𝜋 𝑟
+
3 4𝜋 𝑟 4𝜋𝑟2ℎ
(𝑟2 𝑘 𝑟 (𝑟3 𝑘 𝑟 (𝑟4 𝑘 𝑟
ℎ
1 𝑖 1 1 2 2 2 3 3 3 4 4 𝑜

− − − − − − −(2.84)
2.12 Critical Thickness of Insulation
 There is some misunderstanding about that addition of insulating material on a
surface always brings about a decrease in the heat transfer rate.
 But addition of insulating material to the outside surfaces of cylindrical or spherical
walls (geometries which have non-constant cross-sectional areas) may increase the

To establish this fact, consider a thin walled metallic cylinder of length l, radius 𝑟𝑖
heat transfer rate rather than decrease under the certain circumstances.

and transporting a fluid at temperature 𝑡𝑖 which is higher than the ambient



𝑡𝑜.
temperature

 Insulation of thickness (𝑟−𝑟𝑖) and conductivity k is provided on the surface of the

cylinder.

Fig. 2.14 Critical thickness of pipe insulation

 With assumption
a. Steady state heat conduction
b. One-dimensional heat flow only in radial direction
c. Negligible thermal resistance due to cylinder wall
d. Negligible radiation exchange between outer surface of insulation and
surrounding
 The heat transfer can be expressed as
(𝑡𝑖 − 𝑡𝑜)
𝑄=
𝑅𝑡1 + 𝑅𝑡2 +
𝑅𝑡3
 (𝑡𝑖 − 𝑡𝑜)
𝑄 = 1 𝑟 − − − − − − − (2.85)
1
2𝜋𝑟 𝑙ℎ+ 2𝜋𝑘𝑙
𝑙𝑜𝑔𝑒 𝑟 +2𝜋𝑟𝑙ℎ
1
𝑖 𝑖 𝑖 𝑜

 Where ℎ𝑖 and ℎ𝑜 are the convection coefficients at inner and outer surface
respectively.

The value of 𝑘, 𝑟𝑖, ℎ𝑖 and ℎ𝑜 are constant; therefore the total thermal resistance
 The denominator represents the sum of thermal resistance to heat flow.

will depend upon thickness of insulation which depends upon the outer radius of the
arrangement.
 It is clear from the equation 2.85 that with increase of radius r (i.e. thickness of
insulation), the conduction resistance of insulation increases but the convection
resistance of the outer surface decreases.
 Therefore, addition of insulation can either increase or decrease the rate of heat
flow depending upon a change in total resistance with outer radius r.

resistance 𝑅𝑡 with respect to r and equating to zero.

 To determine the effect of insulation on total heat flow, differentiate the total

𝑑𝑅𝑡
= 𝑑 [ 1 + 1 𝑙𝑜𝑔 � + 1 ]
𝑑𝑟 𝑑 2𝜋𝑟𝑖𝑙 2𝜋 � 2𝜋𝑟𝑙ℎ𝑜
𝑟 ℎ 𝑘𝑙 𝑟
𝑒 1
𝑖

1 1
= − 𝑖

2𝜋𝑘𝑙 𝑟 2𝜋𝑟2𝑙ℎ�
1 1 1
∴ − = 0
�

2𝜋𝑘𝑙 𝑟 2𝜋𝑟2𝑙ℎ
�
1 1 1
=
�

2𝜋𝑘𝑙 𝑟 2𝜋𝑟2𝑙ℎ
�

𝑘
∴𝑟=
�

− − − − − − − (2.86)
ℎ
𝑜

 To determine whether the foregoing result maximizes or minimizes the total

resistance, the second derivative need to be calculated
𝑑2𝑅𝑡 =𝑑 [ 1 1 − 1 ]
𝑑𝑟2 𝑑𝑟 2𝜋𝑘𝑙 𝑟 2𝜋𝑟2 𝑙ℎ𝑜
1 1 1
=− +
2𝜋𝑘𝑙 𝑟2
𝜋𝑟3𝑙ℎ
�

𝑘
𝑎𝑡 𝑟 =
�

ℎ
𝑜

𝑑2𝑅𝑡 1 2𝑜 2 1
=− ( ) 𝜋
𝑑𝑟 2𝜋 ℎ
+ 𝑘3 𝜋𝑙ℎ𝑜
2

𝑘𝑙 𝑘2 𝑙
= ℎ𝑜
2
 3
(
ℎ
)
𝑜

𝑘3
 which is positive, so 𝑟 = 𝑘⁄ represent the condition for minimum resistance and
ℎ𝑜

consequently maximum heat flow rate.
 The insulation radius at which resistance to heat flow is minimum is called critical

The critical radius, designated by 𝑟𝑐 is dependent only on thermal quantities 𝑘 and

radius.

ℎ𝑜.


𝑘
∴ 𝑟 = 𝑟𝑐 =
ℎ𝑜

transfer rate increases and reaches the maximum at 𝑟 = 𝑟𝑐 and then it will
 From the above equation it is clear that with increase of radius of insulation heat

decrease.
 Two cases of practical interest are:
 When 𝑟𝑖 < 𝑟𝑐
 It is clear from the equation 2.14a that with addition of insulation to bare pipe
increases the heat transfer rate until the outer radius of insulation becomes equal to
the critical radius.
 Because with addition of insulation decrease the convection resistance of surface of
insulation which is greater than increase in conduction resistance of insulation.

Fig. 2.14 Dependence of heat loss on thickness of insulation

 Any further increase in insulation thickness decreases the heat transfer from the

of insulation (𝑟*).
peak value but it is still greater than that of for the bare pipe until a certain amount

 So insulation greater than (𝑟* − 𝑟𝑖) must be added to reduce the heat loss below
the bare pipe.
 This may happen when insulating material of poor quality is applied to pipes and
wires of small radius.
 This condition is used for electric wire to increase the heat dissipation from the wire
which helps to increase the current carrying capacity of the cable.
  When 𝑟𝑖 <
Fig. 2.15 Critical radius of insulation for electric wire

𝑟𝑐
 It is clear from the figure 2.14b that increase in insulation thickness always decrease
the heat loss from the pipe.
 This condition is used to decrease the heat loss from steam and refrigeration pipes.
 Critical radius of insulation for the sphere can be obtain in the similar way:

𝑅𝑡 = 1 1 1 1
[ − ]
4𝜋 𝑟 + 4𝜋𝑟 ℎ𝑜
𝑟
𝑘 1 2

𝑑𝑅𝑡
= 𝑑 1 1 1 1
𝑑𝑟 [ [ − ] ]
𝑑 4𝜋 𝑟 + 4𝜋𝑟 ℎ𝑜
𝑑𝑅𝑡 𝑟 𝑘 1 𝑟 2

1 2
= − = 0
𝑑𝑟 4𝜋𝑘𝑟2 4𝜋𝑟3ℎ�
�
∴ 𝑟3ℎ𝑜 = 2𝑘𝑟2
2𝑘
∴ 𝑟 = 𝑟𝑐 = − − − − − − − (2.87)
ℎ𝑜
2.13 Solved Numerical

A 30 cm thick wall of 5 𝑚 𝑋 3 𝑚 size is made of red brick (𝑘 = 0.3 𝑊⁄𝑚

Ex 2.1.

− 𝑑𝑒𝑔). It is covered on both sides by layers of plaster, 2 cm thick (𝑘 = 0.6

𝑊⁄𝑚 − 𝑑𝑒𝑔). The wall has a window size of 1 𝑚 𝑋 2 𝑚. The window door is
made of 12 mm thick glass (𝑘 = 1.2 𝑊⁄𝑚 − 𝑑𝑒𝑔). If the inner and outer
surface temperatures are 15 and 40 , make calculation for the rate of heat flow
through the wall.
Solution:

Plaster: 𝑘1 = 𝑘3 = 0.6 𝑊⁄𝑚 − 𝑑𝑒𝑔, 𝑋1 = 𝑋3 = 2 𝑐𝑚 = 2 ×

Given data:

10−2 𝑚 Red brick: 𝑘2 = 0.3 𝑊⁄𝑚 − 𝑑𝑒𝑔, 𝑋2 = 30 𝑐𝑚 = 30

× 10−2 𝑚 Glass: 𝑘4 = 1.2 𝑊⁄𝑚 − 𝑑𝑒𝑔, 𝑋4 = 12 𝑚𝑚 = 12
× 10−3 𝑚
𝑡𝑖 = 15℃, 𝑡𝑜 = 40℃, Total Area A = 5 𝑚 𝑋 3 𝑚 =
15 𝑚2, Area of glass Window 𝐴g𝑙𝑎𝑠𝑠 = 1 𝑚 𝑋 2 𝑚 =
2 𝑚2
 Total heat transfer from the given configuration is sum of the heat transfer
from composite wall and glass window. So,
𝑄𝑡𝑜𝑡𝑎𝑙 = 𝑄w𝑎𝑙𝑙 + 𝑄g𝑙𝑎𝑠𝑠
 Heat transfer from the composite wall 𝑄w𝑎𝑙𝑙
Area of the wall, 𝐴w𝑎𝑙𝑙 = 𝐴 − 𝐴g𝑙𝑎𝑠𝑠 = 15 − 2 = 13 𝑚2


Resistance of inner and outer plaster layers, 𝑅1 = 𝑅3

𝑋1 2 × 10−2
= = 2.564 ×
−3
𝑅1 = 𝑅3 = ℃⁄𝑊
𝑘1𝐴w𝑎 10
Resistance of brick work, 𝑙𝑙 0.6 × 13
𝑋2 30 × 10−2
= = 76.92 × 10 ℃⁄𝑊
−3
𝑅2 =
2𝐴w𝑎𝑙𝑙 0.3 × 13
𝑘 40 − 15
𝑡𝑜 −
𝑡𝑖
∴ 𝑄w𝑎𝑙𝑙 = + + = 2.564 × 10−3 + 76.92 × 10−3 + 2.564 × 10−3
1
𝑅 𝑅2 𝑅3 𝑄w𝑎𝑙𝑙 = 304.70 𝑊
 Heat transfer from glass window 𝑄g𝑙𝑎𝑠𝑠

𝑋4 12 × 10−3
Resistance of glass,

= = 5×
−3
𝑅4 = ℃⁄𝑊
𝑘 4𝐴g𝑙𝑎𝑠𝑠
1.2 × 2
10
 𝑡𝑜 − 40 − 15
∴ 𝑄g𝑙𝑎𝑠𝑠 =
𝑡𝑖 = = 5000 𝑊
5 × 10−3
𝑅4

𝑄𝑡𝑜𝑡𝑎𝑙 = 304.7 + 5000 = 5304.7 𝑊 = 5.304 𝑘𝑊

So total heat transfer is given by

Ex 2.2.
A cold storage room has walls made of 200 mm of brick on the outside, 80 mm of

temperatures are 25℃ and −3℃ respectively. If the outside and inside
plastic foam, and finally 20 mm of wood on the inside. The outside and inside air

convective heat transfer coefficients are respectively 10 and 30 𝑊⁄𝑚2℃, and

the thermal conductivities of brick, foam and wood are 1.0, 0.02 and 0.17 𝑊⁄𝑚
℃ respectively. Determine:
(i) Overall heat transfer coefficient
(ii) The rate of heat removed by refrigeration if the total wall area is 100𝑚2
(iii) Outside and inside surface temperatures and mid-plane temperatures of
composite wall.
Solution:

Brick: 𝑘1 = 1.0 𝑊⁄𝑚℃, 𝑋1 = 200 𝑚𝑚 = 0.2 𝑚

Given data:

Plastic foam: 𝑘2 = 0.02 𝑊⁄𝑚℃, 𝑋2 = 80 𝑚𝑚 = 80 × 10−3 𝑚

Wood: 𝑘3 = 0.17 𝑊⁄𝑚℃, 𝑋3 = 20 𝑚𝑚 = 20 × 10−3 𝑚
𝑡𝑖 = −3℃, 𝑡𝑜 = 25℃, ℎ𝑜 = 10 𝑊⁄𝑚2℃, ℎ𝑖 = 30 𝑊⁄𝑚2℃, 𝐴 = 100 𝑚2
i. Over all heat transfer co-efficient U

1 1
 Convection resistance of outer surface

𝑅 = = = 1 × 10−3 ℃⁄𝑊
ℎ𝑜 10 ×
𝑜
𝐴 100
𝑋 0.2
Resistance of brick,
= 2 × 10−3 ℃⁄𝑊

𝑅 = 1 =
1
𝑘1𝐴 1.0 × 100

𝑋2 80 × 10−3
 Resistance of plastic foam,

= = 40 ×
−3
𝑅2 = ℃⁄𝑊
𝑘 10
2𝐴 0.02 × 100

𝑋3 20 × 10−3
 Resistance of wood,

= = 1.176 ×
−3
𝑅3 = ℃⁄𝑊
𝑘 10
3𝐴 0.17 × 100
 Convection resistance of inner surface

𝑅𝑖 1 1
= = = 0.333 × 10−3 ℃⁄𝑊
ℎ𝑖 30 ×
𝐴 100

= 𝑅𝑜 + 𝑅1 + 𝑅2 + 𝑅3 + 𝑅𝑖
1

𝑈𝐴 −3
= 1 × 10−3 + 2 × 10 + 40 × 10−3 + 1.176 × 10−3 + 0.333 × 10−3
= 44.509 × 10−3
1
∴𝑈= = 0.224 𝑊⁄𝑚2℃
44.509 × 10−3 ×
100

𝑄 = 𝑈 × 𝐴 × (𝑡𝑜 − 𝑡𝑖) = 0.224 × 100 × (25 − (−3)) = 627.2 𝑊

ii. The rate of heat removed by refrigeration if the total wall area is A = 100𝑚2

iii. Outside and inside surface temperatures and mid-plane temperatures of

 Temperature of outer surface 𝑡1

composite wall

𝑄= 𝑡𝑜
− 𝑡1
𝑅𝑜
𝑡1 = 𝑡𝑜 − 𝑄 × 𝑅𝑜 = 25 − 627.2 × 1 × 10−3 = 24.37℃
 Temperature of middle plane 𝑡2
𝑄 = 𝑡1 −
𝑡2
𝑅1
𝑡2 = 𝑡1 − 𝑄 × 𝑅1 = 24.37 − 627.2 × 2 × 10−3 = 23.11 ℃
 Temperature of middle plane 𝑡3
𝑄 = 𝑡2 −
𝑡3
𝑅2
𝑡3 = 𝑡2 − 𝑄 × 𝑅2 = 23.11 − 627.2 × 40 × 10−3 = −1.97 ℃
 Temperature of inner surface 𝑡4
𝑄 = 𝑡3 −
𝑡4
𝑅3
𝑡4 = 𝑡3 − 𝑄 × 𝑅3 = −1.97 − 627.2 × 1.176 × 10−3 = −2.70 ℃
Ex 2.3.

with thermal conductivity of 1.65, k and 9.2 𝑊⁄𝑚℃ respectively. The inside is
A furnace wall is made up of three layer of thickness 250 mm, 100 mm and 150 mm

exposed to gases at 1250℃ with a convection coefficient of 25 𝑊⁄𝑚2℃ and the

convection coefficient of 12 𝑊⁄𝑚2℃. Determine:

inside surface is at 1100℃, the outside surface is exposed to air at 25℃ with

(i) The unknown thermal conductivity k

(ii) The overall heat transfer coefficient
(iii) All surface temperatures
Solution:

Layer 1: k1 = 1.65 W⁄m℃, X1 = 250 mm =

Given data:

0.25 m Layer 2: k2 = k W⁄m℃, X2 = 100 mm

= 0.1 m Layer 3: k3 = 9.2 W⁄m℃, X3 = 150
mm = 0.15 m
ti = 1250℃, to = 25℃, t1 = 1100℃ ho = 12 W⁄m2℃, hi = 25 W⁄m2℃, Take A = 1 m2

i. Unknown thermal conductivity k

1 1
 Convection resistance of inner surface

𝑅𝑖 = = = 0.04 ℃⁄𝑊
ℎ 𝐴� 25 × 1

𝑋 0.25
�
 Resistance of layer 1,
𝑅 = 1 = = 0.1515 ℃⁄𝑊
1
𝑘1𝐴 1.65 × 1
 𝑋2
 Resistance of layer 2,
= 0.1⁄ ℃⁄𝑊
0.1
𝑅 = =
2
𝑘2𝐴 𝑘2 × 𝑘2
1
𝑋 0.15
 Resistance of layer 3,
𝑅 = 3 = = 0.0163 ℃⁄𝑊
3
𝑘3𝐴 9.2 × 1

1 1
 Convection resistance of outer surface
𝑅𝑜 = = = 0.083 ℃⁄𝑊
ℎ𝑜 𝐴 12 × 1
Heat transfer by convection is given by

𝑄 = 𝑡𝑖 − 1250 −
𝑡1 = 1100 = 3750 𝑊
𝑅𝑖 0.04

𝑡𝑖 − 𝑡0
Heat transfer through composite wall is given by
𝑄=
𝑅𝑖 + 𝑅1 + 𝑅2 + 𝑅3 + 𝑅0
1250 − 25
3750 = 0.04 + 0.1515 + 𝑅2 + 0.0163 + 0.083

0.04 + 0.1515 + 1250 − 25

𝑅2 + 0.0163 + 0.083 =
𝑅2 = 0.0358
3750

∴ 0.1⁄ = 0.0358
𝑘2
∴ 𝑘2 = 0.1⁄0.0358 = 2.79 W⁄m℃
ii. Overall heat transfer co-efficient U

= 𝑅𝑖 + 𝑅1 + 𝑅2 + 𝑅3 + 𝑅𝑜
1

𝑈𝐴
= 0.04 + 0.1515 + 0.0358 + 0.0163 + 0.083 = 0.3103
1
𝑈= = 3.222 𝑊⁄𝑚2℃
0.3103 ×
1
 Temperature of inner surface 𝑡1 = 1100℃
iii. All surface temperature

 Temperature of middle plane 𝑡2

𝑄 = 𝑡1 −
𝑡2
𝑅1
𝑡2 = 𝑡1 − 𝑄 × 𝑅1 = 1100 − 3750 × 0.1515 = 531.87 ℃
 Temperature of middle plane t3
𝑄 = 𝑡2 −
𝑡3
𝑅2
𝑡3 = 𝑡2 − 𝑄 × 𝑅2 = 531.87 − 3750 × 0.0358 = 397.62 ℃
 Temperature of outer surface 𝑡4
 𝑄 = 𝑡3 −
𝑡4
𝑅3
𝑡4 = 𝑡3 − 𝑄 × 𝑅3 = 397.62 − 3750 × 0.0163 = 336.49 ℃
Ex 2.4.

and B. Slab A is 18 mm thick with 𝑘 = 55 𝑊⁄𝑚 𝐾. Slab B is 10 mm thick

A heater of 150 mm X 150 mm size and 800 W rating is placed between two slabs A

𝑘 = 0.2 𝑊⁄𝑚 𝐾. Convective heat transfer coefficients on outside surface of slab A

with

and B are 200 𝑊⁄𝑚2 𝐾 and 45 𝑊⁄𝑚2 𝐾 respectively. If ambient temperature

is 27℃, calculate maximum temperature of the system and outside surface
Solution:
temperature of both slabs.
Given data:
𝐴𝑟𝑒𝑎 𝑜𝑓 ℎ𝑎𝑡𝑒𝑟 𝐴 = 150 𝑚𝑚 × 150 𝑚𝑚 = 22500 𝑚𝑚2 = 22.5 × 10−3 𝑚2
𝑅𝑎𝑡𝑖𝑛𝑔 𝑜𝑓 ℎ𝑒𝑎𝑡𝑒𝑟 = 800 𝑊, toA = toB = to = 27℃
Slab A: kA = 55 W⁄mK, XA = 18 mm = 18 × 10−3 m, hoA = 200 W⁄m2K
Slab B: kB = 0.2 W⁄mK, XB = 10 mm = 10 × 10−3 m, hoB = 45 W⁄m2K

i. Maximum temperature of the system

So, maximum temperature 𝑡𝑚𝑎𝑥 = 𝑡1𝐴 = 𝑡1𝐵

 Maximum temperature exist at the inner surfaces of both slab A and slab B

 Under the steady state condition heat generated by the heater is equal to the

𝑄 = 𝑄𝐴 + 𝑄𝐵
heat transfer through the slab A and slab B.
  Heat transfer through the slab A, 𝑄𝐴∶

𝑋𝐴 18 × 10−3
 Resistance of slab A,

𝑅𝐴 = = = 0.0145 𝐾⁄𝑊
�𝐴 𝐴 55 × 22.5 × 10
−3

Convection resistance of�

1 = 1 = 0.222 𝐾⁄𝑊
 outer surface of slab A
𝑅𝑜𝐴 = 𝐴 200 × ×
ℎ𝑜𝐴
−3

𝑡1𝐴 − 𝑡𝑜𝐴
22.5 10

∴ 𝑄𝐴 =
�𝐴 + 𝑅𝑜𝐴
�
𝑋𝐵 10 × 10−3
 Resistance of slab B,

𝑅𝐵 = = = 2.22 𝐾⁄𝑊
�𝐵 𝐴 0.2 × 22.5 × 10−3

Convection resistance of �
1 = 1 = 0.987 𝐾⁄𝑊
 outer surface of slab B
𝑅𝑜𝐵 = 𝐴 45 × 22.5 × 10−3
ℎ𝑜𝐵
𝑡1𝐵 − 𝑡𝑜𝐵
∴ 𝑄𝐵 =
� 𝐵 + 𝑅𝑜𝐵
𝑡�1𝐴 − 𝑡𝑜𝐴 𝑡1𝐵 − 𝑡𝑜𝐵
∴ 𝑄 = 𝑄𝐴 + 𝑄𝐵 +
= � + 𝑅 + 𝑅𝑜𝐵
�𝐴 𝑅𝑜𝐴 𝐵

𝑡𝑚𝑎𝑥 − 𝑡𝑜 + 𝑡𝑚𝑎𝑥 − 𝑡𝑜) = 1 1

∴
(𝑡𝑄 = ( −𝑡 ){ + }
0.0145 + 2.22 + 𝑚𝑎
𝑥
𝑜
3.207
0.222 0.987
0.2365
( )
𝑄 = 𝑡𝑚𝑎𝑥 − 𝑡𝑜 × 4.54
𝑄 800
𝑡𝑚𝑎𝑥 = + 𝑡𝑜 = + 27 = 203.21℃
4.54 4.54
ii. Outside surface temperature of both slabs

𝑡1𝐴 − 𝑡𝑜𝐴 203.21 − 27

 Heat transfer through slab A

𝑄𝐴 = = = 745.07 𝑊
�𝐴 + 0.0145 + 0.222
� 𝑅𝑜𝐴
 Outside surface temperature of slab A, 𝑡2𝐴
𝑡1𝐴 − 𝑡2𝐴
𝑄𝐴 =
𝑅𝐴
𝑡2𝐴 = 𝑡1𝐴 − 𝑄𝐴 × 𝑅𝐴 = 203.21 − 745.07 × 0.0145 = 192.4 ℃

𝑄 = 𝑄𝐴 + 𝑄𝐵
 Heat transfer through slab B

∴ 𝑄𝐵 = 800 − 745.07 = 54.93 𝑊

 Outside surface temperature of both slab B, 𝑡2𝐵
𝑡1𝐵 − 𝑡2𝐵
𝑄𝐴 =
𝑅𝐵
𝑡2𝐵 = 𝑡1𝐵 − 𝑄𝐵 × 𝑅𝐵 = 203.21 − 54.93 × 2.22 = 81.2 ℃

Ex 2.5.

insulation of thermal conductivity 0.092 𝑊⁄𝑚℃ and 50 mm low temperature

A 240 mm dia. steam pipe, 200 m long is covered with 50 mm of high temperature
 insulation of thermal conductivity 0.062 𝑊⁄𝑚℃. The inner and outer
surface temperatures are maintained at 340℃ and 35℃ respectively. Calculate:

(ii) The heat loss per 𝑚2 of pipe surface

(i) The total heat loss per hour

(iii) The heat loss per 𝑚2 of outer surface

(iv) The temperature between interfaces of two layers of insulation.
Neglect heat conduction through pipe material.
Solution:

240
Given data:
r1 = 120 mm = 0.12 m
= 2
r2 = 120 + 50 = 170 mm = 0.17 m, r3 = 170 + 50 = 210 mm = 0.21 m
k1 = 0.092 W⁄m℃ , k2 = 0.062 W⁄m℃
L = 200 m, t1 = 340 ℃, t3 = 35 ℃

i. Total heat loss per hour

𝑟
ln 2⁄1𝑟 ln(0.17
 Resistance of high temperature insulation
R = = ⁄ = 0.3012 × 10−3 ℃⁄W
0.1
)
2
1
2𝜋𝑘1𝐿 2𝜋 × 0.92 × 200

𝑟
ln 3⁄𝑟
 Resistance of low temperature insulation
0.21⁄ = 2.712 × 10−3 ℃⁄W
0.1
)
7
ln( 2
R = =
2
2𝜋𝑘2𝐿 2𝜋 × 0.062 × 200
𝑡1 − 𝑡3 340 −
35 = 101221.3 𝐽⁄𝑠
∴𝑄= =
𝑅1 + 𝑅2 0.3012 × 10−3 + 2.712 × 10−3
= 101221.3 × 3600⁄1000 = 364.39 × 103 𝐽⁄ℎ𝑟 = 364.39 𝑘𝐽⁄ℎ𝑟
ii. The heat loss per 𝑚2 of pipe surface
Heat Transfer (-) 2. Steady State Heat Conduction

𝑄 101221.3
= = = 671.24 𝑊⁄𝑚2
2𝜋𝑟1 2𝜋 × 0.12 ×
𝐿 200
iii. The heat loss per 𝑚 of outer surface
𝑄 101221.3
2

= = = 383.56 𝑊⁄𝑚2
2𝜋𝑟3 2𝜋 × 0.21 ×
𝐿 200

𝑡1 − 𝑡2
iv. The temperature between interfaces of two layers of insulation
∴𝑄 =
𝑅1
𝑡2 = 𝑡1 − 𝑄 × 𝑅1 = 340 − 101221.3 × 0.3012 × 10−3 = 309.51 ℃

Ex 2.6.
A hot fluid is being conveyed through a long pipe of 4 cm outer dia. And covered
with 2 cm thick insulation. It is proposed to reduce the conduction heat loss to the
surroundings to one-third of the present rate by further covering with some
insulation. Calculate the additional thickness of insulation.
Solution:

4
Given data:

r1 = = 2 cm = 0.02 m
2
r2 = 2 + 2 = 4 cm = 0.04 m, r3 =?

𝑄1
i. Heat loss with existing insulation

𝑟2
ln ⁄𝑟
=
 Resistance of existing insulation
1
R
1
2𝜋𝑘1
𝐿
𝑄1 = 𝑡1 −
𝑡2
ii. Heat loss with additional insulation 𝑄2
 Resistance of existing insulation 𝑅1

𝑟3
ln ⁄ 𝑟
2
R =
2
2𝜋𝑘1𝐿
 𝑡1 −
𝑄2 = 𝑡2
𝑅2
But, 𝑄1 = 1⁄3
𝑄2
𝑡1 − 1 𝑡1 − 𝑡2
𝑡2 = ×
3 𝑅2
𝑅1

1
𝑟 3 𝑅2 =
𝑅1 3 𝑟2
𝑟
ln ⁄𝑟2 1 ln ⁄ 1
= ×
𝑟3 2𝜋𝑘1 3
𝐿 1 2𝜋𝑘1𝐿
0.04
1 𝑟2
ln ⁄𝑟2 = × ⁄𝑟1 = × ln ⁄0.02 = 0.231
ln 3 3
𝑟3
⁄𝑟2 = 𝑒 = 1.259
0.231

∴ 𝑟3 = 1.259 × 𝑟2 = 1.259 × 0.04 = 0.0503 𝑚 = 5 𝑐𝑚

∴ 𝑎𝑑𝑖𝑡𝑖𝑜𝑛𝑎𝑙 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑡 = 𝑟3 − 𝑟2 = 5 − 4 =
1 𝑐𝑚
A hot gas at 330℃ with convection coefficient 222 𝑊⁄𝑚2 𝐾 is flowing through a
Ex 2.7.

insulating material of thickness 2 cm, having conductivity of 0.2 𝑊⁄𝑚 𝐾. The outer
steel tube of outside diameter 8 cm and thickness 1.3 cm. It is covered with an

55 𝑊⁄𝑚2 𝐾.
surface of insulation is exposed to ambient air at 25℃ with convection coefficient of

Calculate: (1) Heat loss to air from 5 m long tube. (2) The temperature drop due to

air. Take conductivity of steel = 50 𝑊⁄𝑚 𝐾.

thermal resistance of the hot gases, steel tube, the insulation layer and the outside

Solution:

8
Given data:

r2 = = 4 cm = 0.04 m
2
r1 = 4 − 1.3 = 2.7 cm = 0.027 m, r3 = 4 + 2 = 6 cm = 0.06 m
k1 = 50 W⁄mK , k2 = 0.2 W⁄mK, ℎ𝑖 = 222 𝑊⁄𝑚2 𝐾, ℎ𝑜 = 55 𝑊⁄𝑚2 𝐾
L = 5 m, ti = 330 ℃, to = 25 ℃
 i. Total heat loss to air from 5 m long tube, Q

1 1 1
 Convection resistance of hot gases
Ri = = = = 5.31 × 10−3 ℃⁄W
ℎ𝑖𝐴 ℎ𝑖2𝜋𝑟1 222 × 2𝜋 × 0.027
1 𝐿 ×5
𝑟2 0.04⁄ )
 Resistance of steel
𝑟 0.027
ln ⁄ 1 = 0.25 × 10−3 ℃⁄W
ln( =
1R =
2𝜋𝑘1𝐿 2𝜋 × 50 × 5
𝑟3 0.06⁄ )
 Resistance of insulation
ln ⁄𝑟2 0.04
ln( = 64.53 × 10−3 ℃⁄W
R = =
2
2𝜋𝑘2𝐿 2𝜋 × 0.2 × 5

1 1 1
 Convection resistance of outside air
Ro = = = = 9.64 × 10−3 ℃⁄W
ℎ𝑜𝐴 ℎ𝑜2𝜋𝑟3 55 × 2𝜋 × 0.06
𝑜 𝐿 ×5

∴𝑄= 𝑡𝑖 − 𝑡𝑜
𝑅𝑖 + 𝑅1 + 𝑅2
+ 𝑅𝑜
330 − 25
= = 3825.4 𝑊
5.31 × 10−3 + 0.25 × 10−3 + 64.53 × 10−3 + 9.64 × 10−3
ii. Temperature drop

𝑡𝑖 − 𝑡1
 Temperature drop due to thermal resistance of hot gases
𝑄=
𝑅𝑖
∴ 𝑡𝑖 − 𝑡1 = 𝑄 × 𝑅𝑖 = 3825.4 × 5.31 × 10−3 = 20.31 ℃

𝑡1 − 𝑡2
 Temperature drop due to thermal resistance of steel tube
𝑄=
𝑅1
∴ 𝑡1 − 𝑡2 = 𝑄 × 𝑅1 = 3825.4 × 0.25 × 10−3 = 0.95 ℃

𝑡2 − 𝑡3
 Temperature drop due to thermal resistance of insulation
𝑄=
𝑅2
∴ 𝑡2 − 𝑡3 = 𝑄 × 𝑅2 = 3825.4 × 64.53 × 10−3 = 246.85 ℃

𝑡3 − 𝑡𝑜
 Temperature drop due to thermal resistance of outside air
𝑄=
𝑅𝑜
∴ 𝑡3 − 𝑡𝑜 = 𝑄 × 𝑅𝑜 = 3825.4 × 9.64 × 10−3 = 36.87 ℃
Ex 2.8.
A pipe carrying the liquid at −20℃ is 10 mm in outer diameter and is exposed to
ambient at 25℃ with convective heat transfer coefficient of 50 W⁄m2 K. It is

0.5 W⁄m K. Determine the thickness of insulation beyond which the heat gain will
proposed to apply the insulation of material having thermal conductivity of

be reduced. Also calculate the heat loss for 2.5 mm, 7.5 mm and 15 mm thickness of
insulation over 1m length. Which one is more effective thickness of insulation?
Solution:

10
Given data:

r1 = = 5 mm = 0.005 m, r2 = 5 + 2.5 = 7.5 mm = 0.0075 m

r3 = 52+ 7.5 = 12.5 mm = 0.0125 m, r4 = 5 + 15 = 20 mm = 0.02 m
k = 0.5 W⁄mK, ℎ𝑜 = 50 𝑊⁄𝑚2 𝐾
L = 1 m, ti = −20 ℃, to = 25 ℃
i. Thickness of insulation beyond which heat gain will be reduced

rc = k⁄ℎ = 0.5⁄50 = 0.01 m =

 Critical radius of insulation

�
10 mm t = rc − r1 = 10 − 5
�

= 5 mm
ii. Heat loss for 2.5 mm thickness of insulation, Q1

ln 𝑟⁄2𝑟 ln(0.0075 ⁄ )
0.005
 Resistance of insulation

R = = = 0.129 K⁄W
1
1
2𝜋𝑘𝐿 2𝜋 × 0.5 × 1

1 1 1
 Convection resistance of outside air
Ro = = = = 0.424 K⁄W
ℎ𝑜 𝐴 ℎ 2𝜋𝑟2𝐿 50 × 2𝜋 × 0.0075 × 1

𝑡𝑜 − 𝑡𝑖 25 − (−20)
𝑜 𝑜

∴ Q1 = = = 81.37 𝑊
𝑅1 + 0.129 + 0.424
𝑅𝑜
iii. Heat loss for 7.5 mm thickness of insulation, Q2

ln𝑟⁄3𝑟 ln(0.0125 ⁄ )
0.005
 Resistance of insulation

R = = = 0.291 K⁄W
1
2
2𝜋𝑘𝐿 2𝜋 × 0.5 × 1

1 1 1
 Convection resistance of outside air
Ro = = = = 0.254 K⁄W
ℎ𝑜 𝐴 ℎ 2𝜋𝑟3𝐿 50 × 2𝜋 × 0.0125 × 1
𝑜 𝑜
 𝑡𝑜 − 𝑡𝑖 25 − (−20)
∴ Q2 = = = 82.56 𝑊
𝑅2 + 0.291 + 0.254
𝑅𝑜
iv. Heat loss for 15 mm thickness of insulation, Q3

ln𝑟⁄4𝑟 ln(0.02⁄ )
0.005
 Resistance of insulation

R = = = 0.441 K⁄W
1
3
2𝜋𝑘𝐿 2𝜋 × 0.5 × 1

1 1 1
 Convection resistance of outside air
Ro = = = = 0.159 K⁄W
ℎ𝑜 𝐴 ℎ 2𝜋𝑟4𝐿 50 × 2𝜋 × 0.02 × 1

𝑡𝑜 − 𝑡𝑖 25 − (−20)
𝑜 𝑜

∴ Q3 = = = 75 𝑊
𝑅3 + 0.441 + 0.159
𝑅𝑜
Hence the insulation thickness of 15 mm is more effective

2.14 References:
[1] Heat and Mass Transfer by D. S. Kumar, S K Kataria and Sons Publications.
[2] Heat Transfer – A Practical Approach by Yunus Cengel & Boles, McGraw-Hill
Publication.
[3] Principles of Heat Transfer by Frank Kreith, Cengage Learining.
HEAT TRANSFER FROM EXTENDED
SURFACES

Course Contents
3.1 Introduction
3.2 Steady flow of heat along a
rod (governing differential
equation)
3.3 Heat dissipation from an
infinitely long fin
3.4 Heat dissipation from a fin
insulated at the tip
3.5 Heat dissipation from a fin
losing heat at the tip
3.6 Fin performance
3.7 Thermometric well
3.8 Solved Numerical
3.9 References
3.1 Introduction

Newton’s cooling law: 𝑄𝑐𝑜𝑛𝑣 = ℎ𝐴𝑠(𝑇0 − 𝑇𝑎), where 𝑇0 is the surface

 Heat transfer between a solid surface and a moving fluid is governed by the

temperature and 𝑇𝑎 is the fluid temperature.

i Increase the temperature difference (𝑇0 − 𝑇𝑎) between the surface and the fluid.
 Therefore, to increase the convective heat transfer, one can

ii Increase the convection coefficient ℎ. This can be accomplished by increasing the

fluid flow over the surface since h is a function of the flow velocity and the higher

iii Increase the contact surface area 𝐴𝑠

the velocity, the higher the h.

 Many times, when the first option is not in our control and the second option (i.e.
increasing h) is already stretched to its limit, we are left with the only alternative of
increasing the effective surface area by using fins or extended surfaces.
 Fins are protrusions from the base surface into the cooling fluid, so that the extra
surface of the protrusions is also in contact with the fluid.
 Most of you have encountered cooling fins on air-cooled engines (motorcycles,
portable generators, etc.), electronic equipment (CPUs), automobile radiators, air
conditioning equipment (condensers) and elsewhere
3.2 Steady Flow of Heat Along A Rod (Governing Differential
Equation)
 Consider a straight rectangular or pin fin protruding from a wall surface (figure 3.1a
and figure 3.1b).

Ac and the circumferential parameter P.

 The characteristic dimensions of the fin are its length L, constant cross-sectional area

Fig. 3.1a Schematic diagram of a rectangular fin protruding from a wall

 Fig. 3.1b Schematic diagram of a pin fin protruding from a wall

𝐴𝑐 = 𝑏𝛿 ; 𝑃 = 2(𝑏 + 𝛿)
 Thus for a rectangular fin

π 2
and for a pin fin
Ac = d ; P = πd
4
The temperature at the base of the fin is T0 and the temperature of the ambient
fluid into which the rod extends is considered to be constant at temperature Ta.


 The base temperature T0 is highest and the temperature along the fin length goes
on diminishing.
 Analysis of heat flow from the finned surface is made with the following
assumptions:
i Thickness of the fin is small compared with the length and width; temperature
gradients over the cross-section are neglected and heat conduction treated one
dimensional
ii Homogeneous and isotropic fin material; the thermal conductivity k of the fin
material is constant
iii Uniform heat transfer coefficient h over the entire fin surface
iv No heat generation within the fin itself

temperature at root or base of the fin is uniform and equal to temperature 𝑇0 of

v Joint between the fin and the heated wall offers no bond resistance;

the wall
vi Negligible radiation exchange with the surroundings; radiation effects, if any, are
considered as included in the convection coefficient h
 vii Steady state heat dissipation
 Heat from the heated wall is conducted through the fin and convected from the
sides of the fin to the surroundings.
 Consider infinitesimal element of the fin of thickness dx at a distance x from base
wall as shown in figure 3.2.

Fig. 3.2 Heat transfer through a fin

𝑑𝑡
 Heat conducted into the element at plane x

𝑄𝑥 = −𝑘 𝐴𝑐 ( ) − − − − − − − −(3.1)
𝑑𝑥 ��
 Heat conducted out of the element at plane (𝑥 + 𝑑𝑥)
𝑑𝑡
𝑄𝑥+𝑑𝑥 = −𝑘 𝐴𝑐 )
(
𝑑
𝑥+𝑑𝑥

𝑥
𝑑 𝑑𝑡
= −𝑘 𝐴𝑐 (𝑡 + 𝑑𝑥) − − − − − − − −(3.2)
𝑑𝑥 𝑑𝑥
 Heat convected out of the element between the planes x and (𝑥 + 𝑑𝑥)
𝑄𝑐𝑜𝑛𝑣 = ℎ (𝑃 𝑑𝑥)(𝑡 − 𝑡𝑎) − − − − − − − −(3.3)
 Here temperature t of the fin has been assumed to be uniform and non-variant for
the infinitesimal element.
 According to first law of thermodynamic, for the steady state condition, heat

𝑄𝑥 = 𝑄𝑥+𝑑𝑥 + 𝑄𝑐𝑜𝑛𝑣
transfer into element is equal to heat transfer from the element
 𝑑𝑡 𝑑 𝑑𝑡
−𝑘 𝐴𝑐 = −𝑘 𝐴𝑐 (𝑡 + 𝑑𝑥) + ℎ (𝑃 𝑑𝑥)(𝑡 − 𝑡𝑎)
𝑑𝑥 𝑑𝑥 𝑑𝑥
𝑑𝑡
𝑑𝑡 𝑑2𝑡
−𝑘 𝐴𝑐 = −𝑘 𝐴𝑐 − 𝑘 𝐴𝑐 𝑑𝑥 + ℎ (𝑃 𝑑𝑥)(𝑡 − 𝑡𝑎)
𝑑𝑥 𝑑𝑥 𝑑𝑥2
 Upon arrangement and simplification
𝑑2𝑡 ℎ𝑃
− (𝑡 − 𝑡𝑎) = 0 − − − − − − − −(3.4)
𝑑𝑥2 𝑘 𝐴�
�
Let, 𝜃(𝑥) = 𝑡(𝑥) − 𝑡𝑎
 As the ambient temperature is constant, so differentiation of the equation is

𝑑𝜃 𝑑𝑡 𝑑2𝜃 𝑑2𝑡
= ; =
𝑑𝑥 𝑑𝑥 𝑑𝑥2 𝑑𝑥2
Thus
𝑑2𝜃

− 𝑚2𝜃 = 0 − − − − − − − −(3.5)
Where 𝑑𝑥2

ℎ𝑃
𝑚=√
𝑘 𝐴𝑐
 Equations 3.4 and 3.5 provide a general form of the energy equation for one
dimensional heat dissipation from an extended surface.
 The general solution of this linear homogeneous second order differential equation

θ = C1emx + C2e−mx − − − − − − − −(3.6)

is of the form

 The constant C1 and C2 are to be determined with the aid of relevant

boundary conditions. We will treat the following four cases:
i Heat dissipation from an infinitely long fin
ii Heat dissipation from a fin insulated at the tip
iii Heat dissipation from a fin losing heat at the tip

3.6 Fin Performance

 It is necessary to evaluate the performance of fins to achieve minimum weight or
maximum heat flow etc.
 Fin effectiveness and fin efficiency are some methods used for performance
evaluation of fins
 Efficiency of fin:
 It relates the performance of an actual fin to that of an ideal or fully effective fin.
 In reality, temperature of fin drop along the length of fin, and thus the heat transfer
from the fin will be less because of the decreasing temperature difference towards
the tip of fin.
 A fin will be most effective, i.e., it would dissipate heat at maximum rate if the entire
fin surface area is maintained at the base temperature as shown in figure 3.7

𝑎𝑐𝑡𝑢𝑎𝑙 ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑟𝑎𝑡𝑒 𝑓𝑟𝑜𝑚 𝑓𝑖𝑛

Fig. 3.7 Ideal and actual temperature distribution in a fin

𝜂𝑓 =
𝑚𝑎𝑥𝑚𝑖𝑚𝑢𝑚 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑟𝑎𝑡𝑒 𝑓𝑟𝑜𝑚 𝑓𝑖𝑛
 Thus for a fin insulated at tip

𝜂𝑓 = √𝑃ℎ𝑘 𝐴𝑐 (𝑡0 − 𝑡𝑎) tanh(𝑚𝐿)

ℎ(𝑃𝐿)(𝑡0 − 𝑡𝑎)
 The parameter PL represents the total surface area exposed for convective

tanh(𝑚𝐿) tanh(𝑚𝐿)
heat flow. Upon simplification,
𝜂𝑓 = − − − − − − − −(3.29)
𝑚𝐿
=
√𝑃ℎ⁄𝑘 𝐴𝑐
𝐿
 Following poins are noted down from the above equation

tanh(𝑚 1
i For a very long fin
→
𝐿) 𝑙𝑎𝑟𝑔𝑒 𝑛𝑢𝑚𝑏𝑒𝑟
𝑚𝐿

For small values of ml, the fin efficiency increases. When the length is reduced to
 Obviously the fin efficiency drops with an increase in its length.


tanh(𝑚𝐿) 𝑚𝐿
zero, then,
→ =1
𝑚𝐿 𝑚𝐿

𝐿 = 0, i.e., no fin at all.

 Thus the fin efficiency reaches its maximum vlaue of 100% for a tgrivial value of

 Actually efficiency of fin is used for the design of the fin but it is used for comparision
 of the relative merits of fin of different geometries or material.
 Note that fins with triangular and parabolic profiles contain less material and are
more efficient than the ones with rectangular profiles, and thus are more suitable for
applications requiring minimum weight such as space applications.
 An important consideration in the design of finned surfaces is the selection of the
proper fin length L.
 Normally the longer the fin, the larger the heat transfer area and thus the higher the
rate of heat transfer from the fin.
 But also the larger the fin, the bigger the mass, the higher the price, and the larger
the fluid friction.
 Therefore, increasing the length of the fin beyond a certain value cannot be justified
unless the added benefits outweigh the added cost.
 Also, the fin efficiency decreases with increasing fin length because of the decrease
in fin temperature with length.
 Fin lengths that cause the fin efficiency to drop below 60 percent usually cannot be
justified economically and should be avoided.
 The efficiency of most fins used in practice is above 90 percent.
 Effectiveness of fin (𝛜f):
 Fins are used to increase the heat transfer. And use of fin can not be recommended
unless the increase in heat transfer justifies the added cost of fin.
 In fact, use of fin may not ensure the increase in heat transfer. Effectiveness of fin
gives the increase in heat transfer with fin relative to no fin case.
 It represents the ratio of the fin heat transfer rate to the heat transfer rate that

heat transfer with

would exist without a fin.

ϵf =
fin heat transfer
without fin
 Figure 3.8 shows the base heat transfer surface before and after the fin has been

Heat transfer through the root area Ac before the fin attached is:
attached.

Q = hAc(t0 − ta)


Fig. 3.8 Heat dissipation with and without fin

 After the attachment of an infinitely long fin, the heat transfer rate through the root
area becomes:
𝑄𝑓𝑖𝑛 = √𝑃ℎ𝑘 𝐴𝑐 (t0 − ta)
So, effectiveness of fin is given as

∴ 𝜖𝑓 =
√𝑃ℎ𝑘 𝐴𝑐 (t0
− ta) ℎ𝐴𝑐(t0
− ta)
 𝑃𝑘
𝜖𝑓 = √ − − − − − − − −(3.30)
ℎ𝐴��
 Following conclusions are given from the effectiveness of the fin
i If the fin is used to improve heat dissipation from the surface, then the fin
effectivenss must be greater than unity. That is,
𝑃𝑘
√ >
1 ℎ𝐴𝑐

𝑃𝑘⁄ℎ𝐴𝑐 is greater than 5.

But literature suggests that use of fins on surrface is justified only if the ratio

ii To improve effectiveness of fin, fin should be made from high conductive

manterial such as copper and aluminium alloys. Although copper is superior to
aluminium regarding to the thermal conductivity, yet fins are generally made of
aluminium because of their additional advantage related to lower cost and
weight.
iii Effectiveness of fin can also be increased by increasing the ratio of perimeter to
the cross sectional area. So it is better to use more thin fins of closer pitch than
fewer thicker fins at longer pitch.
iv A high value of film coefficient has an adverse effect on effectiveness. So fins are
used with the media with low film coefficient. Therefore, in liquid – gas heat
exchanger,such as car radiator, fins are placed on gas side.

√𝑃ℎ𝑘 𝐴𝑐 (t0 − ta) tanh(𝑚𝐿)

 Relation between effeciency of fin and effectiveness of fin

𝜂𝑓 =
ℎ(𝑃𝑙) − ta)
(t0

𝜖𝑓 = √𝑃ℎ𝑘 𝐴𝑐 (t0 − ta)

tanh(𝑚𝐿)
ℎ𝐴𝑐(t0 − ta)

∴ 𝜖𝑓 = 𝜂𝑓 ℎ(𝑃𝐿)(t0 − ta)
ℎ𝐴𝑐(t0 − ta)

= 𝜂𝑓 (𝑃𝐿)
𝐴𝑐
𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑓𝑖𝑛
∴ 𝜖𝑓 = 𝜂𝑓 − − − − − − − −(3.31)
𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑓𝑖𝑛
 An increase in the fin effectiveness can be obtained by extending the length of fin
but that rapidly becomes a losing proposition in term of efficiency.