A Course on

HEAT TRANSFER
 Unit No. 01
• Introduction to Heat Transfer
• Steady State Heat Conduction
Unit No. 02
• Overall Heat Transfer
• Heat Source System

Content of
• Extended Surfaces
• Unsteady state heat Conduction

the Course Unit No. 03

• Principles of Convection
• Free and Forced Convection
Unit No. 04
• Heat Exchanger
• Parallel and Counter flow HE
• TEMA Standards

Unit No. 05
• Radiation Heat Transfer
Course Outcomes
At the end of the unit you will be able to :
CO1
• Explain the laws of heat transfer and deduce the general heat conduction equation and to explain it
for 1-D steady state heat transfer in regular shape bodies

CO2
• Describe the critical radius of insulation, overall heat transfer coefficient, thermal conductivity and
lumped heat transfer

CO3
• Interpret the extended surfaces

CO4
• Illustrate the boundary layer concept, dimensional analysis, forced and free convection under
different conditions
CO5
• Describe the Boiling heat transfer, Evaluate the heat exchanger and examine the LMTD and NTU
methods applied to engineering problems
CO6
Prof. G. P. Badarkhe

• Explain the thermal radiation black body, emissivity and reflectivity and evaluation of view factor
and radiation shields
 Continuous Assessment (20 Marks)
• CAI- 10 Marks
Assessment • CAII- 10 Marks
Strategy Mid Sem Exam (20 Marks)

End Sem Exam (60 Marks)

 Books to Refer for Heat Transfer

1. S. P. Sukhatme, “A Textbook on Heat Transfer”, Tata McGraw Hill Publications,

3rd edition.
2. Y. A. Cengel, “Heat Transfer A Practical Approach”, Tata McGraw Hill Publication
s,3rd edition, 2006.
3. F. P. Incoropera, D. P. Dewitt, “Fundamentals of Heat and Mass Transfer”, John-
Wiley, 5th edition, 1990.
4. J. P. Holman, “Heat Transfer”, Tata McGraw Hill Publications, 9th edition, 2004.
5. R. K. Rajput, “Heat and Mass Transfer”
Pre-Requisites for Heat transfer

Calculus

Fluid Mechanics

Thermodynamics
Approach towards the Course

Theoretical

Experimental

Computational
Macroscopic and Microscopic
Point of View
 Macro means Big or at Bulk
• Considers certain quantity of matter(Overall behaviour)
• Requires simple mathematical Formulae
• Properties of the system are average values
• Few properties are needed to describe the system
 Micro means Small or at Molecular level
• Concerned directly with the structure of matter
• Analysis requires statistical method as the number of molecules are
very large
• Properties can not be easily measured by instruments
• Large number of variables are needed to describe the system
 Difference Between Heat and Temperature
 HEAT  TEMPERATURE

 Heat is energy in transit.  Temperature is a measure of the amount of

energy possessed by the molecules of a
 Heat flows from hot body to cold body
substance.
 Temperature diff. is the driving force
 Degree of Hotness
 Usually denoted by Q
 can be used to predict the direction of heat
 Unit is joules or calorie transfer

 Usually denoted by T

 Unit is degree Celsius or Kelvin

 Heat Transfer
 Heat, which is the form of energy that can be
transferred from one system to another as a
result of temperature difference. The science
that deals with the determination of the rates
of such energy transfers is heat transfer.
 Temperature difference is the driving force.
 Heat always flows in the direction of
decreasing temperature.
 Heat transfer rate is expressed in terms of
Watt or Kilowatt.
Difference Between Thermodynamics and Heat
Transfer

Thermodynamics tells us: Heat transfer tells us:

•How much heat is transferred How (with what modes) Q is transferred
How much work is done At what rate Q is transferred
Final state of the system Temperature distribution inside the body

Example: Hot steel bar placed in a cold water bath

Applications of Heat Transfer
 Energy production and conversion
Steam Power plant, Solar energy conversion etc.
 Refrigeration and air conditioning
 Domestic applications
Ovens, Stoves, Toaster
 Cooling of electronic equipment
 Manufacturing / materials processing
Welding, Casting, Soldering, Laser machining
 Automobiles / aircraft design
 Nature (weather, climate etc)
 Chemical Industries
 Human Body
Applications of Heat Transfer Contd..
 Rating and Sizing type of Problem

 Rating type of problem deal with the determination of heat transfer

rate for an existing system at a specified temperature difference.

 Sizing type of problem deal with the determination of the size of a

system in order to transfer heat at a specified rate for a specified
temperature difference.
 Outcome of Unit No. 01

At the end of the unit you will be able to :

• Explain the laws of heat transfer and deduce the general heat
conduction equation and explain it for 1-D steady state heat transfer
in regular shape bodies
• Identify different modes of heat transfer in engineering applications
 Modes of Heat Transfer

Conduction Convection Radition

 Conduction
It is the transfer of energy from the more
energetic particles of substance to the adjacent
less energetic ones as a result of interaction
between the particles
 Convection
Mode of energy transfer between a
solid surface and the adjacent fluid
that is in motion and it involves the
combined effect of conduction and
fluid motion

In the absence of any bulk fluid

motion, heat transfer between the solid
surface and adjacent fluid is by pure
conduction
 Types of Convection

Natural Convection Forced Convection

Fluid motion occurs due to density variation Fluid motion caused by
caused by temperature difference external agency
Practical Examples of Natural and Forced Convection
 Boiling

• Liquids gets converted

into vapours

Condensation

• Vapours gets converted

into liquids
 Radiation
Energy transfer in the form of electromagnetic waves because of
vibrational rotational motion of molecules and atoms which makes up
the matter or substance
• All physical matter emits thermal radiation at a particular level of temperature
• It does not require medium for the energy transfer to occur
Examples of radiation mode of heat transfer
Combined example of all modes of
heat transfer
Laws of Heat Transfer
Fundamental Laws
• Law of conservation of mass
Mass of a closed system is constant in the absence of nuclear reaction
For a control volume : Mass entering –Mass leaving = increase of mass
• Newton’s law of motion
Rate of change of momentum is directly proportional to applied force
𝒅
𝑭= (𝒎𝑽)
𝒅𝒕

• First law of thermodynamics

Principle of conservation of energy
dE = dQ - dW
Subsidiary Laws
• Fourier’s law of Heat Conduction
• Newton’s law of Cooling
• Laws of thermal radiation
Fourier’s law of Heat Conduction

𝟏 𝑸∝ 𝑨
𝑸 ∝ ∆𝑻 𝑸∝
∆𝒙
Fourier’s law of heat conduction states that rate of heat flow
through a simple homogeneous solids is directly
proportional to the area which is normal to the direction of
heat flow and the change in temperature with respect to the
length of the path of the heat flow
𝑨𝒓𝒆𝒂)(𝑻𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆 𝑫𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆
𝑹𝒂𝒕𝒆 𝒐𝒇 𝑯𝒆𝒂𝒕 𝑪𝒐𝒏𝒅𝒖𝒄𝒕𝒊𝒐𝒏 ∝
𝑻𝒉𝒊𝒄𝒌𝒏𝒆𝒔𝒔

𝑨 ∗ 𝒅𝑻
𝑸 ∝
𝐝𝒙
𝑸 𝐝𝑻
=𝒒 ∝
𝑨 𝐝𝒙
𝐝𝑻
𝒒=−𝒌
𝐝𝒙
Where q is heat flux
- Sign indicates heat flow in the direction of decreasing temp.
Thermal Conductivity
Ability of a material to conduct heat

Rate of heat transfer through a unit thickness of the material

per unit area per unit temperature difference

𝑸 ∗ 𝒅𝒙
𝒌=
𝑨 ∗ 𝒅𝑻
Unit of thermal conductivity : 𝑾 𝒎𝒌 𝒐𝒓 𝑾 𝒎℃

For Example Copper has Thermal Conductivity of 401 𝐖 𝐦𝐤

Thermal Conductivity depend on

 Material Structure
 Density of Material
 Moisture Content
 Temperature
Insulator Conductor
Effect of Alloying:

k(Copper)= 401 W/m.K

K(Nickel)=91 W/m.K

K(constantan: 55% Cu & 45% Ni)= 23 w/m.K

Effect of Temperature:

K(Metals) α 1/T
Aluminium and Uranium as exception
(Mercury)
K(Non Metals) α T

K(Gases) α T Determination of Thermal

Conductivity
K(Liquids) α 1/T
Water as exception
 Heat Capacity and Thermal Diffusivity
Specific Heat Cp: J/kg.K Thermal Diffusivity

Cp(water)=4.18 kJ/kg K 𝐻𝑒𝑎𝑡 𝐶𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑘

∝= =
Cp(iron)=0.45 kJ/kg K 𝐻𝑒𝑎𝑡 𝑆𝑡𝑜𝑟𝑎𝑔𝑒 𝜌𝐶𝑝
Cp is the capacity to store thermal
energy Unit of ∝ : Sq. m /Sec

k(water)=0.608 W/m K
k(iron)=80.2 W/m K
k is the capacity to conduct thermal
energy

Heat Capacity 𝝆𝑪𝒑 : J/cubic m. k

Newtons Law of Cooling
Rate of heat transfer is directly proportional
to the temperature difference between
surface & surrounding and area which is in
contact.
𝑄𝐶𝑜𝑛𝑣. ∝ 𝐴𝑠 𝑇𝑠 − 𝑇∞
𝑄𝐶𝑜𝑛𝑣. = ℎ 𝐴𝑠 𝑇𝑠 − 𝑇∞
 𝑯𝒆𝒂𝒕 𝑻𝒓𝒂𝒏𝒔𝒇𝒆𝒓 𝑪𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕:
𝑄𝐶𝑜𝑛𝑣.
ℎ= = 𝑊 𝑚2𝑘 𝑜𝑟 ℃
𝐴𝑠 𝑇𝑠 − 𝑇∞

 Whose value depends upon:

 Surface Geometry
 Nature of Fluid Motion
 Properties of Fluid
 Bulk Fluid Velocity
 Stefan-Boltzmann's Law
It sates that emissive power of a black body is directly
proportional to fourth power of its absolute temperature.
𝑄
= 𝐸𝑏𝑙𝑎𝑐𝑘 ∝ 𝑇 4
𝐴
= 𝜎 𝑇 or 𝑄𝐵𝑙𝑎𝑐𝑘 𝑜𝑟 𝑚𝑎𝑥 = 𝜎𝐴𝑠 𝑇𝑠 4
4
𝐸𝑏𝑙𝑎𝑐𝑘

Where 𝜎 is Stefan-Boltzmann constant and

has the value of 5.67 ∗ 10−8 𝑊 𝑚2 𝑘 4
 For Real Surfaces:
𝑄𝑅𝑒𝑎𝑙 = 𝜀𝐴𝑠 𝜎 𝑇𝑠 4
Co-ordinate systems
Generalised Heat Conduction equation in
Cartesian Co-ordinate systems

Consider a infinitesimal small elemental volume of side dx, dy & dz as

shown in figure in a medium in which temperature is varying with
location and time
Let kx ky kz be the thermal conductivities in x, y, and z direction
respectively
Energy balance equation for volumetric element (control volume) is given by:
𝑁𝑒𝑡 ℎ𝑒𝑎𝑡 𝑎𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑑𝑢𝑒 𝑡𝑜 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 ℎ𝑒𝑎𝑡 𝑓𝑟𝑜𝑚 𝑎𝑙𝑙 𝑡ℎ𝑒
𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑒𝑑(𝐴) + ℎ𝑒𝑎𝑡 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑤𝑖𝑡ℎ𝑖𝑛 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡(𝐵) =
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓𝑒𝑛𝑒𝑟𝑦 𝑐𝑜𝑛𝑡𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡(C)………………………………..(1)
A. Net heat accumulated in the element due to conduction of heat from all the
direction considered
Quantity of heat entering the element in x direction per unit time i.e. Heat
influx
𝜕𝑇
𝑄𝑥 = − 𝑘𝑥 . 𝐴.
𝜕𝑥
𝜕𝑇
𝑄𝑥 = − 𝑘𝑥 . 𝑑𝑦. 𝑑𝑧 .
𝜕𝑥
Quantity of heat leaving the element in x direction per unit time i.e.
Heat efflux
𝜕
𝑄𝑥+𝑑𝑥 = 𝑄𝑥 + 𝑄𝑥 𝑑𝑥
𝜕𝑥
Heat accumulated in the element due to heat flow in x direction
per unit time
𝜕
𝑑𝑄𝑥 = 𝑄𝑥 − 𝑄𝑥 + 𝑄𝑥 𝑑𝑥
𝜕𝑥

𝜕
𝑑𝑄𝑥 = − 𝑄𝑥 𝑑𝑥
𝜕𝑥

𝜕 𝜕𝑇
𝑑𝑄𝑥 = − − 𝑘𝑥 𝑑𝑦. 𝑑𝑧 𝑑𝑥
𝜕𝑥 𝜕𝑥

𝜕 𝜕𝑇
𝑑𝑄𝑥 = 𝑘𝑥 𝑑𝑥. 𝑑𝑦. 𝑑𝑧
𝜕𝑥 𝜕𝑥
Similarly heat accumulated due to heat flow by conduction along y
and z direction will be:
𝜕 𝜕𝑇
𝑑𝑄𝑦 = 𝑘𝑦 𝑑𝑥. 𝑑𝑦. 𝑑𝑧
𝜕𝑦 𝜕𝑦

𝜕 𝜕𝑇
𝑑𝑄𝑧 = 𝑘 𝑑𝑥. 𝑑𝑦. 𝑑𝑧
𝜕𝑧 𝑧 𝜕𝑧
Net heat accumulated in the element due to conduction of heat from all
the direction considered per unit time =
𝜕 𝜕𝑇 𝜕 𝜕𝑇 𝜕 𝜕𝑇
𝑘 + 𝑘 + 𝑘 𝑑𝑥. 𝑑𝑦. 𝑑𝑧
𝜕𝑥 𝑥 𝜕𝑥 𝜕𝑦 𝑦 𝜕𝑦 𝜕𝑧 𝑧 𝜕𝑧
--------------------(i)
B. Total Heat generated within the element per unit time

𝑄𝑔 = 𝑞𝑔′′′ ∗ 𝑑𝑥. 𝑑𝑦. 𝑑𝑧

--------------- (ii)
C. Rate of change of energy content of the element
𝜕𝑇
𝑚. 𝐶𝑝 .
𝜕𝜏

𝜕𝑇
𝜌. 𝑉. 𝐶𝑝 .
𝜕𝜏

𝜕𝑇
𝜌. 𝐶𝑝 . . 𝑑𝑥. 𝑑𝑦. 𝑑𝑧
𝜕𝜏
------------------(iii)

Putting value of (i), (ii),(iii) in equation no. (1) the heat balance equation
will be

𝝏 𝝏𝑻 𝝏 𝝏𝑻 𝝏 𝝏𝑻 𝝏𝑻
𝒌 + 𝒌 + 𝒌 𝒅𝒙. 𝒅𝒚. 𝒅𝒛 + 𝒒′′′ ∗ 𝒅𝒙. 𝒅𝒚. 𝒅𝒛 = 𝛒. 𝑪𝒑 . . 𝐝𝐱. 𝐝𝐲. 𝐝𝐳
𝝏𝒙 𝒙 𝝏𝒙 𝝏𝒚 𝒚 𝝏𝒚 𝝏𝒛 𝒛 𝝏𝒛 𝒈
𝝏𝝉
 𝝏 𝝏𝑻 𝝏 𝝏𝑻 𝝏 𝝏𝑻 𝝏𝑻
𝒌 + 𝒌 + 𝒌 + 𝒒′′′ = 𝛒. 𝑪𝒑 .
𝝏𝒙 𝒙 𝝏𝒙 𝝏𝒚 𝒚 𝝏𝒚 𝝏𝒛 𝒛 𝝏𝒛 𝒈
𝝏𝝉
This is known as generalised heat conduction equation in cartesian co-ordinate
system
If material or medium is Isotropic then kx=ky=kz=k then
above equation becomes

𝜕 𝜕𝑇 𝜕 𝜕𝑇 𝜕 𝜕𝑇 𝜕𝑇
𝑘 + + + 𝑞𝑔′′′ = 𝜌. 𝐶𝑝 .
𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑦 𝜕𝑧 𝜕𝑧 𝜕𝜏

𝝏𝟐 𝑻 𝝏𝟐 𝑻 𝝏𝟐 𝑻 𝒒′′′
𝒈 𝝆. 𝑪𝒑 𝝏𝑻
𝟐 + 𝟐+ 𝟐+ = .
𝝏𝒙 𝝏𝒚 𝝏𝒛 𝒌 𝒌 𝝏𝝉

𝝏𝟐 𝑻 𝝏𝟐 𝑻 𝝏𝟐 𝑻 𝒒′′′
𝒈 𝟏 𝝏𝑻
𝟐
+ 𝟐+ 𝟐+ = .
𝝏𝒙 𝝏𝒚 𝝏𝒛 𝒌 𝜶 𝝏𝝉
𝒌
𝜶= Thermal diffusivity
𝝆.𝑪𝒑
Simplified forms of heat conduction equation in cartesian
coordinate system

 Unsteady state heat flow with no internal heat generation

(Fourier’s Equation)
𝝏𝟐 𝑻 𝝏𝟐 𝑻 𝝏𝟐 𝑻 𝟏 𝝏𝑻
+ + = .
𝝏𝒙𝟐 𝝏𝒚𝟐 𝝏𝒛𝟐 𝜶 𝝏𝝉
 Steady state heat flow with heat generation(Poisson’s equation)

𝝏𝟐 𝑻 𝝏𝟐 𝑻 𝝏𝟐 𝑻 𝒒′′′ 𝒈
𝟐 + 𝟐 + 𝟐 + =𝟎
𝝏𝒙 𝝏𝒚 𝝏𝒛 𝒌
 Steady state heat flow without internal heat generation(Laplace
Equation)
𝝏𝟐 𝑻 𝝏𝟐 𝑻 𝝏𝟐 𝑻
+ + =𝟎
𝝏𝒙𝟐 𝝏𝒚𝟐 𝝏𝒛𝟐
 Steady state, one dimensional heat flow with heat generation
𝝏𝟐 𝑻 𝒒′′′
𝒈
+ =𝟎
𝝏𝒙𝟐 𝒌

 Steady state, one dimensional heat flow without heat generation

𝝏𝟐 𝑻
=𝟎
𝝏𝒙𝟐

 Unsteady state, one dimensional heat flow without heat

Generation
𝝏𝟐 𝑻 𝟏 𝝏𝑻
𝟐 = .
𝝏𝒙 𝜶 𝝏𝝉
Generalised Heat Conduction equation in
Polar/Cylindrical Co-ordinate systems

Consider a infinitesimally small elemental volume having the

coordinates r,𝝓 & z as shown in figure in a medium in which
temperature is varying with location and time
 Let us assume that Thermal Conductivity k, Density 𝛒 and
specific heat Cp do not change with space
Volume of the element rd𝝓.dr.dz
Energy balance equation for volumetric element (control volume)
is given by:

𝐍𝐞𝐭 𝐡𝐞𝐚𝐭 𝐚𝐜𝐜𝐮𝐦𝐮𝐥𝐚𝐭𝐞𝐝 𝐢𝐧 𝐭𝐡𝐞 𝐞𝐥𝐞𝐦𝐞𝐧𝐭 𝐝𝐮𝐞 𝐭𝐨 𝐜𝐨𝐧𝐝𝐮𝐜𝐭𝐢𝐨𝐧 𝐨𝐟 𝐡𝐞𝐚𝐭 𝐟𝐫𝐨𝐦 𝐚𝐥𝐥 𝐭𝐡𝐞
𝐝𝐢𝐫𝐞𝐜𝐭𝐢𝐨𝐧 𝐜𝐨𝐧𝐬𝐢𝐝𝐞𝐫𝐞𝐝(𝐀) + 𝐡𝐞𝐚𝐭 𝐠𝐞𝐧𝐞𝐫𝐚𝐭𝐞𝐝 𝐰𝐢𝐭𝐡𝐢𝐧 𝐭𝐡𝐞 𝐞𝐥𝐞𝐦𝐞𝐧𝐭(𝐁) =
𝐑𝐚𝐭𝐞 𝐨𝐟 𝐜𝐡𝐚𝐧𝐠𝐞 𝐨𝐟𝐞𝐧𝐞𝐫𝐲 𝐜𝐨𝐧𝐭𝐞𝐧𝐭 𝐨𝐟 𝐭𝐡𝐞 𝐞𝐥𝐞𝐦𝐞𝐧𝐭 (C)………………………………..(1
) heat accumulated in the
Net
Rate of change of
element due to conduction of Heat Generated within
heat from all the direction + the element
= energy content of the
element
considered
A. Net heat accumulated in the element due to conduction of heat
from all the direction considered

Heat Flow in Radial Direction

Quantity of heat entering the element in r direction (z- 𝝓 plane) per

unit time i.e. Heat influx in r direction
𝝏𝑻
𝑸𝒓 = −𝒌. 𝑨.
𝝏𝒓
𝝏𝑻
𝑸𝒓 = −𝒌. 𝒓𝒅𝝓. 𝒅𝒛 .
𝝏𝒓

Quantity of heat leaving the element in r direction per unit time i.e.
Heat efflux
𝛛
𝐐𝐫+𝐝𝒓 = 𝐐𝒓 + 𝐐𝒓 𝐝𝐫
𝛛𝒓
Net Heat accumulated in the element due to heat flow in radial
direction per unit time
𝛛
𝐝𝐐𝒓 = 𝐐𝒓 − 𝐐𝒓 + 𝐐 𝐝𝒓
𝛛𝒓 𝒓
𝛛
𝐝𝐐𝒓 = − 𝐐𝒓 𝐝𝐫
𝛛𝒓
𝝏 𝝏𝑻
𝒅𝑸𝒓 = − − 𝒌 𝒓𝒅𝝓. 𝒅𝒛 𝒅𝒓
𝝏𝒓 𝝏𝒓
𝝏 𝝏𝑻
𝒅𝑸𝒓 = 𝒌 𝒅𝒓. 𝒅𝝓. 𝒅𝒛 𝒓
𝝏𝒓 𝝏𝒓
𝝏𝟐 𝑻 𝝏𝑻
𝒅𝑸𝒓 = 𝒌(𝒅𝒓. 𝒅𝝓. 𝒅𝒛) 𝒓 𝟐 +
𝝏𝒓 𝝏𝒓
𝝏𝟐 𝑻 𝟏 𝝏𝑻
𝒅𝑸𝒓 = 𝒌(𝒅𝒓. 𝒓𝒅𝝓. 𝒅𝒛) 𝟐
+
𝝏𝒓 𝒓 𝝏𝒓
 Heat Flow in Tangential i.e. 𝝓 (r-z plane) Direction

Quantity of heat entering the element in 𝝓 direction (r-z plane)

per unit time i.e. Heat influx in 𝝓 direction
𝝏𝑻
𝑸𝝓 = −𝒌. 𝑨.
𝒓. 𝝏𝝓
𝝏𝑻
𝑸𝝓 = −𝒌. 𝒅𝒓. 𝒅𝒛 .
𝒓. 𝝏𝝓

Quantity of heat leaving the element in 𝝓 direction per unit time i.e.
Heat efflux
𝛛
𝐐𝝓+𝐝𝝓 = 𝐐𝝓 + 𝐐 𝒓𝐝𝝓
𝒓. 𝛛𝝓 𝝓
Net Heat accumulated in the element due to heat flow in tangential
direction per unit time
𝐝𝐐𝝓 = 𝐐𝝓 − 𝐐𝝓+𝒅𝝓
Net Heat accumulated in the element due to heat flow in
tangential direction per unit time
𝛛
𝐝𝐐𝝓 = 𝐐𝝓 − 𝐐𝝓 + 𝐐 𝒓𝐝𝝓
𝒓. 𝛛𝝓 𝝓
𝛛
𝐝𝐐𝝓 = − 𝐐𝝓 𝒓𝐝𝝓
𝒓. 𝛛𝝓
𝝏 𝝏𝑻
𝒅𝑸𝝓 = − − 𝒌 𝒅𝒓. 𝒅𝒛 𝒓𝐝𝝓
𝒓. 𝛛𝝓 𝒓. 𝛛𝝓
𝝏 𝟏 𝝏𝑻
𝒅𝑸𝝓 = 𝒌 𝒅𝒓. 𝒅𝝓. 𝒅𝒛
𝝏𝝓 𝒓 𝝏𝝓
𝟏 𝝏 𝝏𝑻
𝒅𝑸𝝓 = 𝒌 𝒅𝒓. 𝒓𝒅𝝓. 𝒅𝒛 𝟐
𝒓 𝝏𝝓 𝝏𝝓
𝟏 𝝏𝟐 𝑻
𝒅𝑸𝝓 = 𝒌 𝒅𝒓. 𝒓𝒅𝝓. 𝒅𝒛 𝟐
𝒓 𝝏𝝓𝟐
 Heat Flow in Axial i.e. z (r- 𝝓 plane) Direction
Quantity of heat entering the element in z direction (r- 𝛟 plane)
per unit time i.e. Heat influx in 𝒛 direction
𝝏𝑻
𝑸𝒛 = −𝒌. 𝑨.
𝝏𝒛
𝝏𝑻
𝑸𝒛 = −𝒌. 𝒓𝒅𝝓. 𝒅𝒓 .
𝝏𝒛
Quantity of heat leaving the element in r direction per unit time i.e.
Heat efflux
𝛛
𝐐 𝐳 = 𝐐𝒛 + 𝐐𝒛 𝐝𝐳
𝛛𝒛
Heat accumulated in the element due to heat flow in axial
direction per unit time
𝛛
𝐝𝐐𝒛 = 𝐐𝒛 − 𝐐𝒛 + 𝐐𝒛 𝐝𝒛
𝛛𝒛
𝛛
𝐝𝐐𝒛 = − 𝐐𝒛 𝐝𝐳
𝛛𝒛
 𝛛 𝛛𝐓
𝐝𝐐𝐳 = − − 𝐤 𝐫𝐝𝛟. 𝐝𝐫 𝐝𝐳
𝛛𝐳 𝛛𝐳

𝛛𝟐 𝐓
𝐝𝐐𝐳 = 𝐤(𝐝𝐫. 𝐫𝐝𝛟. 𝐝𝐳)
𝛛𝐳 𝟐
Net heat accumulated in the element due to conduction of
heat from all the direction considered per unit time =

𝝏𝟐 𝑻 𝟏 𝝏𝑻 𝟏 𝝏𝟐 𝑻 𝛛𝟐 𝐓
𝟐
+ + 𝟐 𝟐
+ 𝟐
𝐤(𝐝𝐫. 𝐫𝐝𝛟. 𝐝𝐳)
𝝏𝒓 𝒓 𝝏𝒓 𝒓 𝝏𝝓 𝛛𝐳
--------------------(i)

B. Total Heat generated within the element per unit time

𝑸𝒈 = 𝒒′′′
𝒈 ∗ 𝐝𝐫. 𝐫𝐝𝛟. 𝐝𝐳
--------------- (ii)
 C. Rate of change of energy content of the element
𝛛𝐓
𝐦. 𝐂𝐩 .
𝛛𝛕
𝛛𝐓
𝛒. 𝐕. 𝐂𝐩 .
𝛛𝛕
𝛛𝐓
𝛒. 𝐂𝐩 . . 𝐝𝐫. 𝐫𝐝𝛟. 𝐝𝐳
𝛛𝛕
------------------(iii)
Now
A + B = C
Net heat accumulated in the
Rate of change of
element due to conduction of Heat Generated within
energy content of the
heat from all the direction the element
element
considered
 𝝏𝟐 𝑻 𝟏 𝝏𝑻 𝟏 𝝏𝟐 𝑻 𝝏𝟐 𝑻
+ + 𝟐 + 𝐤(𝐝𝐫. 𝐫𝐝𝛟. 𝐝𝐳)
𝝏𝒓𝟐 𝒓 𝝏𝒓 𝒓 𝝏𝝓𝟐 𝛛𝒛𝟐

+
𝒒′′′
𝒈 ∗ 𝐝𝐫. 𝐫𝐝𝛟. 𝐝𝐳

=
𝝏𝑻
𝛒. 𝑪𝒑 . . 𝐝𝐫. 𝐫𝐝𝛟. 𝐝𝐳
𝝏𝝉

𝝏𝟐 𝑻 𝟏 𝝏𝑻 𝟏 𝝏𝟐 𝑻 𝛛𝟐 𝐓
𝟐
+ + 𝟐 𝟐
+ 𝐤 𝐝𝐫. 𝐫𝐝𝛟. 𝐝𝐳 + 𝒒′′′
𝒈 ∗ 𝐝𝐫. 𝐫𝐝𝛟. 𝐝𝐳
𝝏𝒓 𝒓 𝝏𝒓 𝒓 𝝏𝝓 𝛛𝐳 𝟐
𝛛𝐓
= 𝛒. 𝐂𝐩 . . 𝐝𝐫. 𝐫𝐝𝛟. 𝐝𝐳
𝛛𝛕
Dividing both side by 𝑑𝑟. 𝑟𝑑𝜙. 𝑑𝑧
𝝏𝟐 𝑻 𝟏 𝝏𝑻 𝟏 𝝏𝟐 𝑻 𝝏𝟐 𝑻 𝝏𝑻
𝟐
+ + 𝟐 𝟐
+ 𝒌+ 𝒒′′′
𝒈 = 𝝆. 𝑪𝒑 .
𝝏𝒓 𝒓 𝝏𝒓 𝒓 𝝏𝝓 𝝏𝒛𝟐 𝝏𝝉
Or
𝝏𝟐 𝑻 𝟏 𝝏𝑻 𝟏 𝝏𝟐 𝑻 𝝏𝟐 𝑻 𝒒′′′
𝒈 𝝆. 𝑪𝒑 𝝏𝑻
𝟐
+ + 𝟐 𝟐
+ + = .
𝝏𝒓 𝒓 𝝏𝒓 𝒓 𝝏𝝓 𝝏𝒛𝟐 𝒌 𝒌 𝝏𝝉
𝝏𝟐 𝑻 𝟏 𝝏𝑻 𝟏 𝝏𝟐 𝑻 𝝏𝟐 𝑻 𝒒′′′
𝒈 𝟏 𝝏𝑻
𝟐
+ + 𝟐 𝟐
+ + = .
𝝏𝒓 𝒓 𝝏𝒓 𝒓 𝝏𝝓 𝝏𝒛𝟐 𝒌 𝜶 𝝏𝝉
For Steady state, without internal heat generation and one dimensional flow

𝝏𝟐 𝑻 𝟏 𝝏𝑻
𝟐
+ =𝟎
𝝏𝒓 𝒓 𝝏𝒓
Generalised Heat Conduction equation in
Spherical Co-ordinate systems

Consider a infinitesimally small elemental volume having the coordinates r, 𝜙 & 𝜃 as

shown in figure in a medium in which temperature is varying with location and time
 Let us assume that Thermal Conductivity k, Density 𝜌 and specific heat Cp
do not change with space
Volume of the element dr. rd𝜃. 𝑟𝑠𝑖𝑛 𝜃 d𝜙
Energy balance equation for volumetric element (control volume) is given by:

𝑁𝑒𝑡 ℎ𝑒𝑎𝑡 𝑎𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑑𝑢𝑒 𝑡𝑜 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 ℎ𝑒𝑎𝑡 𝑓𝑟𝑜𝑚 𝑎𝑙𝑙 𝑡ℎ𝑒
𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑒𝑑(𝐴) + ℎ𝑒𝑎𝑡 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑤𝑖𝑡ℎ𝑖𝑛 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡(𝐵) =
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓𝑒𝑛𝑒𝑟𝑦 𝑐𝑜𝑛𝑡𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡(C)………………………………..(1)

Net heat accumulated in the

Rate of change of
element due to conduction of Heat Generated within
heat from all the direction + the element
= energy content of the
element
considered
A. Net heat accumulated in the element due to conduction of heat from all the
direction considered

Heat Flow in Radial (r) Direction

Quantity of heat entering the element in r direction (𝜃 - 𝜙 plane) per unit time
i.e. Heat influx in r direction
𝜕𝑇
𝑄𝑟 = −𝑘. 𝐴.
𝜕𝑟
𝜕𝑇
𝑄𝑟 = −𝑘. rd𝜃. 𝑟𝑠𝑖𝑛 𝜃𝑑𝜙 .
𝜕𝑟

Quantity of heat leaving the element in r direction per

unit time i.e. Heat efflux
𝜕
𝑄r+𝑑𝑟 = 𝑄𝑟 + 𝑄𝑟 𝑑r
𝜕𝑟
Net Heat accumulated in the element due to heat flow in radial direction
per unit time
𝜕
𝑑𝑄𝑟 = 𝑄𝑟 − 𝑄𝑟 + 𝑄 𝑑𝑟
𝜕𝑟 𝑟

𝜕
𝑑𝑄𝑟 = − 𝑄𝑟 𝑑r
𝜕𝑟

𝜕 𝜕𝑇
𝑑𝑄𝑟 = − −𝑘 𝑟𝑑𝜃. 𝑟 sin 𝜃𝑑𝜙 𝑑𝑟
𝜕𝑟 𝜕𝑟

𝜕 2 𝜕𝑇
𝑑𝑄𝑟 = 𝑘 𝑑𝑟. 𝑑𝜃. sin 𝜃𝑑𝜙 𝑟
𝜕𝑟 𝜕𝑟

1 𝜕 2 𝜕𝑇
𝑑𝑄𝑟 = 𝑘 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙 2 𝑟
𝑟 𝜕𝑟 𝜕𝑟
Heat Flow in 𝜙 (r-𝜃 plane) Direction

Quantity of heat entering the element in 𝜙 direction (r-𝜃 plane) per unit
time i.e. Heat influx in 𝜙 direction
𝜕𝑇
𝑄𝜙 = −𝑘. 𝐴.
𝑟. sin 𝜃 𝜕𝜙
𝜕𝑇
𝑄𝜙 = −𝑘. 𝑑𝑟. 𝑟𝑑𝜃 .
𝑟. sin 𝜃 𝜕𝜙
Quantity of heat leaving the element in 𝜙 direction per
unit time i.e. Heat efflux
𝜕
𝑄𝜙+𝑑𝜙 = 𝑄𝜙 + 𝑄 𝑟. sin 𝜃 𝑑𝜙
𝑟. sin 𝜃 𝜕𝜙 𝜙
Net Heat accumulated in the element due to heat flow in
𝜙 per unit time
𝑑𝑄𝜙 = 𝑄𝜙 − 𝑄𝜙+𝑑𝜙
Net Heat accumulated in the element due to heat flow in 𝜙 per unit time

𝜕
𝑑𝑄𝜙 = 𝑄𝜙 − 𝑄𝜙 + 𝑄 𝑟 sin 𝜃. 𝑑𝜙
𝑟 sin 𝜃 𝜕𝜙 𝜙
𝜕
𝑑𝑄𝜙 = − 𝑄𝜙 𝑟 sin 𝜃. 𝑑𝜙
𝑟 sin 𝜃 𝜕𝜙
𝜕 𝜕𝑇
𝑑𝑄𝜙 = − − 𝑘 𝑑𝑟. 𝑟𝑑𝜃 𝑟 sin 𝜃. 𝑑𝜙
𝑟 sin 𝜃 𝜕𝜙 𝑟 sin 𝜃 𝜕𝜙

1 𝜕2𝑇
𝑑𝑄𝜙 = 𝑘 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙 2 2
𝑟 sin 𝜃 𝜕𝜙 2
Heat Flow in 𝜃(r-𝜙plane) Direction

Quantity of heat entering the element in 𝜃 direction (r-𝜙 plane) per unit
time i.e. Heat influx in 𝜃 direction
𝜕𝑇
𝑄𝜃 = −𝑘. 𝐴.
𝑟𝜕𝜃
𝜕𝑇
𝑄𝜃 = −𝑘. 𝑑𝑟. 𝑟sin 𝜃𝑑𝜙 .
𝑟𝜕𝜃
Quantity of heat leaving the element in 𝜃 direction per
unit time i.e. Heat efflux
𝜕
𝑄𝜃+𝑑𝜃 = 𝑄𝜃 + 𝑄 𝑟𝑑𝜃
𝑟. 𝜕𝜃 𝜃
Net Heat accumulated in the element due to heat
flow in 𝜃 direction per unit time
𝑑𝑄𝜃 = 𝑄𝜃 − 𝑄𝜃+𝑑𝜃
Net Heat accumulated in the element due to heat flow in 𝜃direction per unit
time

𝜕
𝑑𝑄𝜃 = 𝑄𝜃 − 𝑄𝜃 + 𝑄𝜃 𝑟. 𝑑𝜃
𝑟𝜕𝜃
𝜕
𝑑𝑄𝜃 = − 𝑄 𝑟𝑑𝜃
𝑟𝜕𝜃 𝜃
𝜕 𝜕𝑇
𝑑𝑄𝜃 = − − 𝑘 𝑑𝑟. 𝑟sin 𝜃𝑑𝜙 𝑟𝑑𝜃
𝑟𝜕𝜃 𝑟𝜕𝜃

𝜕 𝜕𝑇
𝑑𝑄𝜃 = 𝑘 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙 2 sin 𝜃
𝑟 sin 𝜃 𝜕𝜃 𝜕𝜃
Net heat accumulated in the element due to conduction of heat from all the direction
considered per unit time =
1 𝜕 2 𝜕𝑇 1 𝜕2𝑇
𝑘 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙 2 𝑟 + 𝑘 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙 2 2
𝑟 𝜕𝑟 𝜕𝑟 𝑟 sin 𝜃 𝜕𝜙 2
𝜕 𝜕𝑇
+ 𝑘 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙 2 sin 𝜃
𝑟 sin 𝜃 𝜕𝜃 𝜕𝜃
1 𝜕 2 𝜕𝑇 1 𝜕 2𝑇 𝜕 𝜕𝑇
𝑟 + 2 2 + 2 sin 𝜃 𝑘 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙
𝑟 2 𝜕𝑟 𝜕𝑟 𝑟 sin 𝜃 𝜕𝜙 2 𝑟 sin 𝜃 𝜕𝜃 𝜕𝜃
--------------------(i)
B. Total Heat generated within the element per unit time

𝑄𝑔 = 𝑞𝑔′′′ ∗ 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙

--------------- (ii)
 C. Rate of change of energy content of the element
𝜕𝑇
𝑚. 𝐶𝑝 .
𝜕𝜏
𝜕𝑇
𝜌. 𝑉. 𝐶𝑝 .
𝜕𝜏
𝜕𝑇
𝜌. 𝐶𝑝 . . 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙
𝜕𝜏
------------------(iii)
Now
A + B = C
Net heat accumulated in the
Rate of change of
element due to conduction of Heat Generated within
energy content of the
heat from all the direction the element
element
considered
 𝟏 𝝏 𝟐 𝝏𝑻 𝟏 𝝏𝟐 𝑻 𝝏 𝝏𝑻
𝒓 + + 𝐬𝐢𝐧 𝜽 𝒌 𝒅𝒓. 𝒓𝒅𝜽. 𝒓𝐬𝐢𝐧 𝜽𝒅𝝓
𝒓𝟐 𝝏𝒓 𝝏𝒓 𝒓𝟐 𝐬𝐢𝐧𝟐 𝜽 𝝏𝝓𝟐 𝒓𝟐 𝐬𝐢𝐧 𝜽 𝝏𝜽 𝝏𝜽

+
𝒒′′′
𝒈 ∗ 𝒅𝒓. 𝒓𝒅𝜽. 𝒓𝒔𝒊𝒏 𝜽𝒅𝝓

=
𝝏𝑻
𝝆. 𝑪𝒑 . . 𝒅𝒓. 𝒓𝒅𝜽. 𝒓𝒔𝒊𝒏 𝜽𝒅𝝓
𝝏𝝉

1 𝜕 2 𝜕𝑇 1 𝜕2𝑇 𝜕 𝜕𝑇 ′′′
𝑟 + + 𝑠𝑖𝑛 𝜃 𝑘 𝑑𝑟. 𝑟𝑑𝜃. 𝑟𝑠𝑖𝑛 𝜃𝑑𝜙 + 𝑞𝑔
𝑟 2 𝜕𝑟 𝜕𝑟 𝑟 2 𝑠𝑖𝑛2 𝜃 𝜕𝜙 2 𝑟 2 𝑠𝑖𝑛 𝜃 𝜕𝜃 𝜕𝜃
𝜕𝑇
∗ 𝑑𝑟. 𝑟𝑑𝜃. 𝑟𝑠𝑖𝑛 𝜃𝑑𝜙 = 𝜌. 𝐶𝑝 . . 𝑑𝑟. 𝑟𝑑𝜃. 𝑟𝑠𝑖𝑛 𝜃𝑑𝜙
𝜕𝜏
Dividing both side by 𝑑𝑟. 𝑟𝑑𝜃. 𝑟𝑠𝑖𝑛 𝜃𝑑𝜙

𝟏 𝝏 𝟐 𝝏𝑻 𝟏 𝝏𝟐 𝑻 𝝏 𝝏𝑻 𝝏𝑻
𝟐 𝒓 + 𝟐 𝟐 𝟐 + 𝟐 𝒔𝒊𝒏 𝜽 𝒌 + 𝒒′′′
𝒈 = 𝝆. 𝑪𝒑 .
𝒓 𝝏𝒓 𝝏𝒓 𝒓 𝒔𝒊𝒏 𝜽 𝝏𝝓 𝒓 𝒔𝒊𝒏 𝜽 𝝏𝜽 𝝏𝜽 𝝏𝝉
Or
𝟏 𝝏 𝟐 𝝏𝑻 𝟏 𝝏𝟐 𝑻 𝝏 𝝏𝑻 𝒒′′′
𝒈 𝝆. 𝑪𝒑 𝝏𝑻
𝟐 𝒓 + 𝟐 𝟐 𝟐 + 𝟐 𝒔𝒊𝒏 𝜽 + = .
𝒓 𝝏𝒓 𝝏𝒓 𝒓 𝒔𝒊𝒏 𝜽 𝝏𝝓 𝒓 𝒔𝒊𝒏 𝜽 𝝏𝜽 𝝏𝜽 𝒌 𝒌 𝝏𝝉

𝟏 𝝏 𝟐 𝝏𝑻 𝟏 𝝏𝟐 𝑻 𝝏 𝝏𝑻 𝒒′′′
𝒈 𝟏 𝝏𝑻
𝒓 + 𝟐 + 𝟐 𝒔𝒊𝒏 𝜽 + = .
𝒓𝟐 𝝏𝒓 𝝏𝒓 𝒓 𝒔𝒊𝒏𝟐 𝜽 𝝏𝝓𝟐 𝒓 𝒔𝒊𝒏 𝜽 𝝏𝜽 𝝏𝜽 𝒌 𝜶 𝝏𝝉
For Steady state, without internal heat generation and one dimensional flow

𝟏 𝝏 𝟐 𝝏𝑻
𝟐
𝒓 =𝟎
𝒓 𝝏𝒓 𝝏𝒓
 For Cartesian Coordinate system

𝝏𝟐 𝑻 𝝏𝟐 𝑻 𝝏𝟐 𝑻 𝒒′′′
𝒈 𝟏 𝝏𝑻
+ + + = .
𝝏𝒙𝟐 𝝏𝒚𝟐 𝝏𝒛𝟐 𝒌 𝜶 𝝏𝝉

 For Polar/Cylindrical Coordinate system

𝝏𝟐 𝑻 𝟏 𝝏𝑻 𝟏 𝝏𝟐 𝑻 𝝏𝟐 𝑻 𝒒′′′
𝒈 𝟏 𝝏𝑻
𝟐
+ + 𝟐 𝟐
+ + = .
𝝏𝒓 𝒓 𝝏𝒓 𝒓 𝝏𝝓 𝝏𝒛𝟐 𝒌 𝜶 𝝏𝝉

 For Spherical Coordinate system

𝟏 𝝏 𝟐 𝝏𝑻 𝟏 𝝏𝟐 𝑻 𝝏 𝝏𝑻 𝒒′′′
𝒈 𝟏 𝝏𝑻
𝒓 + 𝟐 + 𝟐 𝒔𝒊𝒏 𝜽 + = .
𝒓𝟐 𝝏𝒓 𝝏𝒓 𝒓 𝒔𝒊𝒏𝟐 𝜽 𝝏𝝓𝟐 𝒓 𝒔𝒊𝒏 𝜽 𝝏𝜽 𝝏𝜽 𝒌 𝜶 𝝏𝝉
Initial and Boundary Condition
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 mathematical expressions of the thermal conditions at the boundaries are

called the boundary conditions

Thermal Condition which is usually specified at time t=0 is called initial Condition

As the generalized heat conduction equation is second order in space coordinate

thus the boundary condition may involve first derivative at the boundaries as well
as specified value of temperature

Heat conduction equation is first order in time thus the initial condition can not
involve any derivatives
1. Specified Temperature Boundary
Condition

i.e. Surface temperature is constant

𝑻 𝟎, 𝒕 = 𝑻𝟏
𝑻 𝑳, 𝒕 = 𝑻𝟐

2. Specified Heat Flux Boundary

Condition

i.e. Heat Flux(Heat rate per unit surface area) is

constant
𝝏𝑻
𝒒𝒔 = −𝒌
𝝏𝒙 𝒙=𝟎,𝒕
3. Adiabatic or insulated wall
Boundary Condition
i.e. Heat Flux(Heat rate per unit surface area) is
constant
𝝏𝑻
𝒒𝒔 = 𝟎 = 𝒌
𝝏𝒙 𝒙=𝟎,𝒕
𝝏𝑻
=𝟎
𝝏𝒙 𝒙=𝟎,𝒕
4. Convection Boundary Condition
i.e. Heat Flux(Heat rate per unit surface area) is
constant
𝝏𝑻
−𝒌 = 𝒉𝟏 𝑻∞𝟏 − 𝑻 𝟎,𝒕)
𝝏𝒙 𝟎,𝒕
𝝏𝑻
−𝒌 = 𝒉𝟐 𝑻 𝑳,𝒕) − 𝑻∞𝟐
𝝏𝒙 𝑳,𝒕
Solution Procedure for solving heat conduction
problem

Mathematical
Formulation General Application
Heat Solution of
(Diff. Solution of of
Transfer the
Equation & Diff. Boundary
Problem Problem
Boundary Equation Condition
condition)
One Dimensional Steady State Heat Conduction
Through Slab
Consider a plane wall of homogeneous material through
which heat is flowing in x direction
Let L – Thickness of slab/wall
k – Thermal Conductivity
A – Area perpendicular to heat flow
T1 & T2 are the temperatures at two faces
The general heat conduction equation in cartesian
coordinate system is given by
𝝏𝟐 𝑻 𝝏𝟐 𝑻 𝝏𝟐 𝑻 𝒒′′′
𝒈 𝟏 𝝏𝑻
𝟐
+ 𝟐+ 𝟐+ = .
𝝏𝒙 𝝏𝒚 𝝏𝒛 𝒌 𝜶 𝝏𝝉

=0
Final equation in case of steady, one dimensional and
without heat generation will be
𝝏𝟐 𝑻 𝒅𝟐 𝑻
= 𝟎 or =𝟎
𝝏𝒙𝟐 𝒅𝒙𝟐
On integrating twice

𝑑𝑇
𝑑𝑥
= 𝐶2 and 𝑇 = 𝐶1 𝑥 + 𝐶2 …………….(i)
𝐶𝟏 & 𝐶𝟐 are the arbitrary constant and can be calculated by applying boundary
condition
At x = 0, T=T1
At x = L, T=T2
Applying first boundary condition we get

𝑇1 = 𝐶2
Applying second boundary condition we get

𝑇2 = 𝐶1 𝐿 + 𝐶2
𝑻𝟐 −𝑻𝟏
Here 𝐶1 =
𝑳
Equation i reduced to
𝑻𝟐 − 𝑻𝟏
𝑻= 𝒙 + 𝑻𝟏
𝑳
For Heat Flow through Slab/Wall
We have from Fourier’s law of heat conduction
𝒅𝑻
𝑸 = −𝒌𝑨
𝒅𝒙

𝒅𝑻 𝒅 𝑻𝟐 − 𝑻𝟏
= 𝒙 + 𝑻𝟏
𝒅𝒙 𝒅𝒙 𝑳

𝒅𝑻 𝑻𝟐 − 𝑻𝟏
=
𝒅𝒙 𝑳

𝑻𝟐 − 𝑻𝟏
∴ 𝑸 = −𝒌𝑨
𝑳
Or
𝑻𝟏 − 𝑻𝟐
𝑸 = 𝒌𝑨
𝑳

𝑻𝟏 − 𝑻𝟐
𝑸=
𝑳
𝒌𝑨
Alternative Method
We have from Fourier’s law of heat conduction
𝑑𝑇
𝑄 = −𝑘𝐴
𝑑𝑥
𝑥 𝑇
𝑄. 𝑑𝑥 = −𝑘𝐴. 𝑑𝑇
𝑄 𝑑𝑥 = −𝑘𝐴 𝑑𝑇
0 𝑇1
Integrating on both side
𝐿 𝑇2
𝑄 𝑑𝑥 = −𝑘𝐴 𝑑𝑇 𝑄. 𝑥 = 𝑘𝐴 𝑇1 − 𝑇 ……….(ii)
0 𝑇1
Dividing ii by I we get
𝐿 𝑇1
𝑄 𝑑𝑥 = 𝑘𝐴 𝑑𝑇 𝑄. 𝑥 𝑘𝐴 𝑇1 − 𝑇
0 𝑇2 =
𝑄. 𝐿 𝑘𝐴 𝑇1 − 𝑇2
𝐿 𝑇1
𝑄. 𝑥 0 = 𝑘𝐴 𝑇 𝑇2
𝑻𝟏 − 𝑻 𝒙
=
𝑄. 𝐿 = 𝑘𝐴 𝑇1 − 𝑇2 ……….(i) 𝑻𝟏 − 𝑻𝟐 𝑳

𝑻𝟏 − 𝑻𝟐
𝑸 = 𝒌𝑨
𝑳
Variable Thermal Conductivity 𝑳 𝑻𝟐
The dependence of thermal conductivity on temperature 𝑸 𝒅𝒙 = − 𝒌𝟎 𝟏 + 𝜷𝑻 . 𝑨. 𝒅𝑻
can be expressed as- 𝟎 𝑻𝟏
𝐤 = 𝐤 𝟎 𝟏 + 𝛃𝐓 𝑻𝟐
𝐤 𝟎 - thermal conductivity at reference 𝑳 𝑻𝟐
𝑸. 𝒙 𝟎 = −𝒌𝟎 𝑨 𝑻 + 𝜷
𝛃 – is constant for given material 𝟐 𝑻𝟏
T – Temperature
According to Fourier’s Law 𝜷
𝒅𝑻 𝑸. 𝑳 = −𝒌𝟎 𝑨 𝑻𝟐 − 𝑻𝟏 + 𝑻𝟐 𝟐 − 𝑻𝟏 𝟐
𝟐
𝑸 = −𝒌𝑨
𝒅𝒙
𝒅𝑻 𝑸. 𝑳 = 𝒌𝟎 𝑨 𝑻𝟏 − 𝑻𝟐
𝑸 = −𝒌𝟎 𝟏 + 𝜷𝑻 . 𝑨.
𝒅𝒙 𝜷
+ 𝑻 − 𝑻𝟐 𝑻𝟏 + 𝑻𝟐
𝑸. 𝒅𝒙 = −𝒌𝟎 𝟏 + 𝜷𝑻 . 𝑨. 𝒅𝑻 𝟐 𝟏
Integrating on both side and applying boundary condition
as 𝜷
𝑸. 𝑳 = 𝒌𝟎 𝑨 𝑻𝟏 − 𝑻𝟐 𝟏 + 𝑻𝟏 + 𝑻𝟐
At x = 0, T=T1 𝟐
At x = L, T=T2 𝜷
We get 𝒌 𝒎 = 𝒌𝟎 𝟏 + 𝟐 𝑻 𝟏 + 𝑻 𝟐
𝑳 𝑻𝟐
𝑸 𝒅𝒙 = − 𝒌𝟎 𝟏 + 𝜷𝑻 . 𝑨. 𝒅𝑻 𝑻𝟏 − 𝑻𝟐
𝟎 𝑻𝟏 𝑸 = 𝒌𝒎 𝑨
𝑳
 Electrical Analogy
As per Fourier’s Law As per Ohm’s Law
𝒅𝑻 𝒅𝑽
𝑸 = −𝒌𝑨 𝑰=
𝒅𝒙 𝑹
I – Current flow
𝑻𝟏 − 𝑻𝟐
𝑸 = 𝒌𝑨 dV – Voltage difference
𝑳 R – Electrical Resistance
𝑻𝟏 − 𝑻𝟐
𝑸= I is analogous to Q
𝑳
𝒌𝑨 dV is analogous to dT
𝒅𝑻 R is analogous to 𝑹𝒕𝒉
𝑸=
𝑹𝒕𝒉

𝑹𝒕𝒉 𝑪𝒐𝒏𝒅𝒖𝒄𝒕𝒊𝒐𝒏 ) = 𝑳 𝒌𝑨
𝑹𝒕𝒉 𝑪𝒐𝒏𝒅𝒖𝒄𝒕𝒊𝒐𝒏 ) = 𝟏 𝒉𝑨
Resistances in series and parallel

𝑹 = 𝑹𝒄𝒐𝒏𝒗𝟏 + 𝑹𝟏 + 𝑹𝟐 + 𝑹𝑪𝒐𝒏𝒄𝟐 𝟏 𝟏 𝟏 𝟏 𝟏 𝟏
= + = +
𝟏 𝑳𝟏 𝑳𝟐 𝟏 𝑹𝒆𝒒. 𝑹𝟏 𝑹𝟐 𝑹𝒆𝒒. 𝑳𝟏 𝑳𝟐
𝑹= + + + 𝒌𝟏 𝑨𝟏 𝒌𝟐 𝑨𝟐
𝒉𝟏 𝑨 𝒌𝟏 𝑨 𝒌𝟐 𝑨 𝒉𝟐 𝑨 𝟏
𝑹𝒆𝒒. =
𝟏 𝟏
𝟏 𝟏 𝑳 𝟏 𝑳𝟐 𝟏 +𝑳
𝑹= + + + 𝑳𝟏 𝟐
𝑨 𝒉𝟏 𝒌𝟏 𝒌𝟐 𝒉𝟐 𝒌𝟏 𝑨𝟏 𝒌𝟐 𝑨𝟐
Resistances in combination
𝑹 = 𝑹𝒆𝒒. + 𝑹𝟑 + 𝑹𝒄𝒐𝒏𝒗

𝟏 𝟏 𝟏
= +
𝑹𝒆𝒒. 𝑳𝟏 𝑳𝟐
𝒌𝟏 𝑨𝟏 𝒌𝟐 𝑨𝟐
𝟏
𝑹𝒆𝒒. =
𝟏 𝟏
𝑳𝟏 +𝑳
𝟐
𝒌𝟏 𝑨𝟏 𝒌𝟐 𝑨𝟐
𝟏 𝑳𝟑 𝟏
𝑹= + +
𝟏 𝟏 𝒌𝟑 𝑨𝟑 𝒉𝑨𝟑
𝑳𝟏 +𝑳
𝟐
𝒌𝟏 𝑨𝟏 𝒌𝟐 𝑨𝟐
Heat flow through the composite wall/slab
 Heat flow through the composite wall/slab
In steady state conduction heat transfer rate is constant
through out this composite slab so we can write
𝒌𝟏 𝑨 𝑻 𝟏 − 𝑻 𝟐 𝒌𝟐 𝑨 𝑻 𝟐 − 𝑻 𝟑
𝑸 = 𝒉𝟏 𝑨 𝑻∞𝟏 − 𝑻𝟏 = = = 𝒉𝟐 𝑨 𝑻𝟑 − 𝑻∞𝟐
𝑳𝟏 𝑳𝟐

𝑻∞𝟏 − 𝑻𝟏 𝑻𝟏 − 𝑻𝟐 𝑻𝟐 − 𝑻 𝟑 𝑻𝟑 − 𝑻∞𝟐
𝑸= = = =
𝟏 𝑳𝟏 𝑳𝟐 𝟏
𝒉𝟏 𝑨 𝒌𝟏 𝑨 𝒌𝟐 𝑨 𝒉𝟐 𝑨

𝑻∞𝟏 − 𝑻𝟏 = 𝑸 ∗ 𝟏 𝒉𝟏 𝑨------------------(1)

𝑻𝟏 − 𝑻𝟐 = 𝑸 ∗ 𝑳𝟏 𝒌𝟏 𝑨-------------------(2)

𝑻𝟐 − 𝑻𝟑 = 𝑸 ∗ 𝑳𝟐 𝒌𝟐 𝑨-------------------(3)

𝑻𝟑 − 𝑻∞𝟐 = 𝑸 ∗ 𝟏 𝒉𝟐 𝑨------------------(4)
 Adding 1, 2, 3 & 4 we get
𝑳 𝑳
𝑻∞𝟏 − 𝑻∞𝟐 = 𝑸 𝟏 𝒉 𝑨 + 𝟏 𝒌 𝑨 + 𝟐 𝒌 𝑨 + 𝟏 𝒉 𝑨
𝟏 𝟏 𝟐 𝟐

𝑻∞𝟏 − 𝑻∞𝟐
𝑸=
𝟏 𝑳 𝑳 𝟏
+ 𝟏 + 𝟐 +
𝒉𝟏 𝑨 𝒌𝟏 𝑨 𝒌𝟐 𝑨 𝒉𝟐 𝑨
𝑻∞𝟏 − 𝑻∞𝟐
𝑸=
𝟏 𝑳 𝟏
+ 𝒏𝟏 𝒏 +
𝒉𝟏 𝑨 𝒌𝒏 𝑨 𝒉𝟐 𝑨
𝑻∞𝟏 − 𝑻∞𝟐
𝑸=
𝟏 𝟏 𝑳 𝟏
+ 𝒏𝟏 𝒏 +
𝑨 𝒉𝟏 𝒌𝒏 𝒉𝟐

𝑨 𝑻∞𝟏 − 𝑻∞𝟐 𝑻𝟏 − 𝑻𝒏+𝟏

𝑸= 𝑸=
𝟏 𝑳 𝟏 𝟏 𝒏 𝑳𝒏
+ 𝒏𝟏 𝒏 + 𝑨 𝟏𝒌
𝒉𝟏 𝒌𝒏 𝒉𝟐 𝒏
Overall Heat Transfer Coefficient
When there is fluid to fluid heat transfer it usual practice
to adopt overall heat transfer coefficient
𝑻∞𝟏 − 𝑻∞𝟐
𝑸=
𝟏 𝑳 𝑳 𝟏
+ 𝟏 + 𝟐 +
𝒉𝟏 𝑨 𝒌𝟏 𝑨 𝒌𝟐 𝑨 𝒉𝟐 𝑨
𝑨 𝑻∞𝟏 − 𝑻∞𝟐
𝑸=
𝟏 𝑳𝟏 𝑳𝟐 𝟏
+ + +
𝒉𝟏 𝒌𝟏 𝒌𝟐 𝒉𝟐
If U is the overall heat transfer coefficient
then
𝑸 = 𝑼. 𝑨. 𝑻∞𝟏 − 𝑻∞𝟐
𝑨 𝑻∞𝟏 − 𝑻∞𝟐
𝑸 = 𝑼. 𝑨. 𝑻∞𝟏 − 𝑻∞𝟐 =
𝟏 𝑳𝟏 𝑳𝟐 𝟏
+ + +
𝒉𝟏 𝒌𝟏 𝒌𝟐 𝒉𝟐
 𝟏
𝑼=
𝟏 𝑳𝟏 𝑳𝟐 𝟏
+ + +
𝒉𝟏 𝒌𝟏 𝒌𝟐 𝒉𝟐

𝑻∞𝟏 − 𝑻∞𝟐
𝑼. 𝑨. 𝑻∞𝟏 − 𝑻∞𝟐 =
𝟏 𝑳 𝑳 𝟏
+ 𝟏 + 𝟐 +
𝒉𝟏 𝑨 𝒌𝟏 𝑨 𝒌𝟐 𝑨 𝒉𝟐 𝑨

𝟏 𝟏
𝑼𝑨 =
𝑹
0r 𝑼 = 𝑨 𝑹

For unit area overall heat transfer coefficient

is reciprocal of total thermal resistance
Thermal Contact Resistance

Ideal or perfect contact Actual or imperfect contact

𝑄𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 = 𝑄𝐶𝑜𝑛𝑡𝑎𝑐𝑡 + 𝑄𝑔𝑎𝑝 Depends upon
In analogous to Newton’s law of cooling Surface roughness
It can be also expressed as Material properties
𝑄 = ℎ𝑐 . 𝐴. ∆𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 Temperature
Pressure
 Thermal Contact Resistance
𝑄𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 = ℎ𝑐 . 𝐴. ∆𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 Thermal contact resistance
can be minimized by
𝑄𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒
ℎ𝑐 =  By applying thermal grease like
𝐴. ∆𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 silicon oil
 Removing air by better conducting
Where ℎ𝑐 is thermal contact conductance 𝑊 𝑚2 ℃ gas like Helium or Hydrogen
 Inserting soft metallic foil like Tin,
Silver or copper
Thermal contact resistance is the inverse of  Increasing the interface pressure
thermal contact conductance
1 𝐴. ∆𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒
𝑅𝑐 = =
ℎ𝑐 𝑄𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒
2
Unit of thermal contact resistance is 𝑚 ℃ 𝑊
 One Dimensional Steady State Heat Conduction
Through Cylinder
Consider a hollow cylinder of homogeneous material through
which heat is flowing in r direction
Let L – length of Cylinder
k – Thermal Conductivity
𝑟1 and 𝑟2 be the inner and outer radius of cyl. respectively.
T1 & T2 are the temperatures at inner and outer surface
of cylinder respectively
The general heat conduction equation in Cylindrical /Polar
coordinate system is given by
𝝏𝟐 𝑻 𝟏 𝝏𝑻 𝟏 𝝏𝟐 𝑻 𝝏𝟐 𝑻 𝒒′′′
𝒈 𝟏 𝝏𝑻
𝟐
+ + 𝟐 𝟐
+ + = .
𝝏𝒓 𝒓 𝝏𝒓 𝒓 𝝏𝝓 𝝏𝒛𝟐 𝒌 𝜶 𝝏𝝉

=0

𝝏𝟐 𝑻 𝟏 𝝏𝑻 𝒅𝟐 𝑻 𝟏 𝒅𝑻
+ =𝟎 or + =𝟎
𝝏𝒓𝟐 𝒓 𝝏𝒓 𝒅𝒓𝟐 𝒓 𝒅𝒓
𝒅𝟐 𝑻 𝟏 𝒅𝑻 𝟏 𝒅 𝒅𝑻
+
𝒅𝒓𝟐 𝒓 𝒅𝒓
=𝟎 ⟹ .
𝒓 𝒅𝒓
𝒓.
𝒅𝒓
=0

𝒅 𝒅𝑻
𝒓. =0
𝒅𝒓 𝒅𝒓
On integrating both side we get

𝒅𝑻 𝒅𝑻 C
𝒓.
𝒅𝒓
=C ⟹ =
𝒅𝒓 𝒓
On integrating both side we get
𝑻𝟏 = 𝑪. 𝒍𝒏 𝒓𝟏 + 𝑪𝟏
𝑻 = 𝑪. 𝒍𝒏 𝒓 + 𝑪𝟏 −−−− −(𝒊) - 𝑻𝟐 = 𝑪. 𝒍𝒏 𝒓𝟐 + 𝑪𝟏
C & 𝐶𝟐 are the arbitrary constant and can be calculated by
applying boundary condition
𝑻𝟏 − 𝑻𝟐 = 𝑪 𝒍𝒏 𝒓𝟏 − 𝒍𝒏 𝒓𝟐
At 𝒓 = 𝒓𝟏 ; 𝑻 = 𝑻𝟏
𝒓 = 𝒓𝟐 ; 𝑻 = 𝑻𝟐
 𝑻𝟏 − 𝑻𝟐 𝑻𝟏 − 𝑻𝟐
𝑪= 𝒓𝟏 =− 𝒓𝟐
𝒍𝒏 𝒓𝟐 𝒍𝒏 𝒓𝟏

Again we can write

𝑻𝟏 − 𝑻𝟐
𝑻𝟏 = − 𝒓𝟐 . 𝒍𝒏 𝒓𝟏 + 𝑪𝟏
𝒍𝒏 𝒓𝟏

𝑻𝟏 − 𝑻𝟐 𝒓𝟐
𝑪𝟏 = 𝑻𝟏 + . 𝒍𝒏 𝒓𝟏 𝑻 − 𝑻𝟏 × 𝒍𝒏 𝒓𝟏 = 𝑻𝟏 − 𝑻𝟐 . 𝒍𝒏 𝒓𝟏 − 𝑻𝟏 − 𝑻𝟐 . 𝒍𝒏 𝒓
𝒓𝟐
𝒍𝒏 𝒓𝟏
= 𝑻𝟐 − 𝑻𝟏 . 𝒍𝒏 𝒓 − 𝑻𝟐 − 𝑻𝟏 . 𝒍𝒏 𝒓𝟏
By putting value of 𝑪 and 𝑪𝟏 in equation no. (i)
= 𝑻𝟐 − 𝑻𝟏 𝒍𝒏 𝒓 − 𝒍𝒏 𝒓𝟏
𝑻𝟏 − 𝑻𝟐 𝑻𝟏 − 𝑻𝟐
𝑻 = 𝑻𝟏 + . 𝒍𝒏 𝒓 − . 𝒍𝒏 𝒓
𝒍𝒏
𝒓𝟐
𝒓𝟏
𝟏
𝒍𝒏
𝒓𝟐
𝒓𝟏 𝑻 − 𝑻𝟏 𝒍𝒏 𝒓 𝒓𝟏
=
𝑻𝟐 − 𝑻𝟏 𝒍𝒏 𝒓𝟐 𝒓
𝟏
For Heat Flow through Slab/Wall
We have from Fourier’s law of heat conduction
𝒅𝑻
𝑸 = −𝒌𝑨
𝒅𝒓

𝒅𝑻 𝒅 𝑻𝟏 − 𝑻𝟐 𝑻𝟏 − 𝑻𝟐
= 𝑻 + . 𝒍𝒏 𝒓𝟏 − . 𝒍𝒏 𝒓
𝒅𝒓 𝒅𝒓 𝟏 𝒍𝒏 𝒓𝟐 𝒓 𝒍𝒏
𝒓𝟐
𝒓𝟏
𝟏

𝒅𝑻 𝑻𝟏 − 𝑻𝟐
=− 𝒓
𝒅𝒓 𝒓. 𝒍𝒏 𝟐 𝒓 𝟏

𝑻𝟏 − 𝑻𝟐 𝑻𝟏 − 𝑻𝟐
∴ 𝑸 = −𝒌𝑨 − 𝒓
𝑸= 𝒓
𝒓. 𝒍𝒏 𝟐 𝒓𝟏 𝒍𝒏 𝟐 𝒓𝟏
Or 𝟐𝝅𝒌𝑳
𝑻𝟏 − 𝑻𝟐
𝑸 = 𝒌. 𝟐𝝅𝒓𝑳 𝒓𝟐
𝒓. 𝒍𝒏 𝒓𝟏
Alternative Method
We have from Fourier’s law of heat conduction

𝑑𝑇 𝑟2 𝑇1
𝑄 = −𝑘𝐴 𝑄. 𝑙𝑛 𝑟 𝑟1 = 2𝜋𝑘𝐿 𝑇 𝑇2
𝑑𝑟
𝑑𝑇 𝑟2
𝑄 = −𝑘. 2𝜋𝑟𝐿. 𝑄. 𝑙𝑛 = 2𝜋𝑘𝐿 𝑇1 − 𝑇2
𝑑𝑟 𝑟1
𝑄
. 𝑑𝑟 = −𝑘. 2𝜋𝐿. 𝑑𝑇 2𝜋𝑘𝐿 𝑇1 − 𝑇2
𝑟 𝑄=
𝑟
𝑙𝑛 2
Integrating on both side 𝑟1

𝑟2
1 𝑇2 𝑇1 − 𝑇2
𝑄 𝑑𝑟 = −𝑘. 2𝜋𝐿 𝑑𝑇 𝑄=
𝑟1 𝑟 𝑇1 𝑟2
𝑙𝑛
𝑟1
𝑄
𝑟2
1
𝑑𝑟 = 𝑘. 2𝜋𝐿
𝑇1
𝑑𝑇
2𝜋𝑘𝐿
𝑟1 𝑟 𝑇2
Heat flow through the composite cylinder
 EXTENDED SURFACES - FINS
 Mechanical Attachment to the base surface of a system for enhancing
the heat transfer rate by increasing the effective area of the body is
called as extended surfaces of fins
Combine Conduction through the fin and Convection to/from the fin
Heat rate from the base surface is given by

𝑄𝐶𝑜𝑛𝑣. = ℎ 𝐴𝑠 𝑇𝑠 − 𝑇∞
Applications of Extended Surfaces or Fins
Heat Exchangers used in Process Industries
Economiser used to heat the feed water in steam
power plant
Radiators of automobiles
Air Cooled Engine Cylinder and Head
Cooling coils and condenser coils in refrigerator
and air conditioners
Small Capacity Compressor
Electric motor bodies
Transformer and electronic equipment's
Handle of ladle used to pour the molten metal
Types of Extended Surfaces or Fins
Rectangular Fins Trapezoidal Fins
IC Engines and Electronic Component Rocket Engines and Fishes
Splines Annular Fins
Electrical Generator Shaft Fin Tube heat exchanger
Longitudinal Fins Pin Fins
Electrical motor bodies Electrical motor bodies