2. Content of
the Course
Unit No. 01
• Introduction to Heat Transfer
• Steady State Heat Conduction
Unit No. 02
• Overall Heat Transfer
• Heat Source System
• Extended Surfaces
• Unsteady state heat Conduction
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
3. Prof. G. P. Badarkhe
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
• Explain the thermal radiation black body, emissivity and reflectivity and evaluation of view factor
and radiation shields
5. 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”
8. 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
9. Difference Between Heat and Temperature
HEAT
Heat is energy in transit.
Heat flows from hot body to cold body
Temperature diff. is the driving force
Usually denoted by Q
Unit is joules or calorie
TEMPERATURE
Temperature is a measure of the amount of
energy possessed by the molecules of a
substance.
Degree of Hotness
can be used to predict the direction of heat
transfer
Usually denoted by T
Unit is degree Celsius or Kelvin
10. 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.
11. Difference Between Thermodynamics and Heat
Transfer
Thermodynamics tells us:
•How much heat is transferred
How much work is done
Final state of the system
Heat transfer tells us:
How (with what modes) Q is transferred
At what rate Q is transferred
Temperature distribution inside the body
Example: Hot steel bar placed in a cold water bath
12. 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
14. 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.
15. 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
16. Modes of Heat Transfer
Conduction Convection Radition
17. 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
18. 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
19. Types of Convection
Natural Convection
Fluid motion occurs due to density variation
caused by temperature difference
Forced Convection
Fluid motion caused by
external agency
21. • Liquids gets converted
into vapours
Boiling
• Vapours gets converted
into liquids
Condensation
22. 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
25. 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
𝑭 =
𝒅
𝒅𝒕
(𝒎𝑽)
• 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
27. 𝑹𝒂𝒕𝒆 𝒐𝒇 𝑯𝒆𝒂𝒕 𝑪𝒐𝒏𝒅𝒖𝒄𝒕𝒊𝒐𝒏 ∝
𝑨𝒓𝒆𝒂)(𝑻𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆 𝑫𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆
𝑻𝒉𝒊𝒄𝒌𝒏𝒆𝒔𝒔
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
𝑸 ∝
𝑨 ∗ 𝒅𝑻
𝐝𝒙
𝑸
𝑨
= 𝒒 ∝
𝐝𝑻
𝐝𝒙
𝒒 = − 𝒌
𝐝𝑻
𝐝𝒙
Where q is heat flux
- Sign indicates heat flow in the direction of decreasing temp.
28. ThermalConductivity
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
𝒌 =
𝑸 ∗ 𝒅𝒙
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
30. 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
K(Liquids) α 1/T
Water as exception
Determination of Thermal
Conductivity
31. Heat Capacity and Thermal Diffusivity
Specific Heat Cp: J/kg.K
Cp(water)=4.18 kJ/kg K
Cp(iron)=0.45 kJ/kg K
Cp is the capacity to store thermal
energy
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
Thermal Diffusivity
∝=
𝐻𝑒𝑎𝑡 𝐶𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛
𝐻𝑒𝑎𝑡 𝑆𝑡𝑜𝑟𝑎𝑔𝑒
=
𝑘
𝜌𝐶𝑝
Unit of ∝ : Sq. m /Sec
32. 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. = ℎ 𝐴𝑠 𝑇𝑠 − 𝑇∞
𝑄𝐶𝑜𝑛𝑣. ∝ 𝐴𝑠 𝑇𝑠 − 𝑇∞
𝑯𝒆𝒂𝒕 𝑻𝒓𝒂𝒏𝒔𝒇𝒆𝒓 𝑪𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕:
ℎ =
𝑄𝐶𝑜𝑛𝑣.
𝐴𝑠 𝑇𝑠− 𝑇∞
= 𝑊
𝑚2𝑘 𝑜𝑟 ℃
Whose value depends upon:
Surface Geometry
Nature of Fluid Motion
Properties of Fluid
Bulk Fluid Velocity
33. Stefan-Boltzmann's Law
It sates that emissive power of a black body is directly
proportional to fourth power of its absolute temperature.
𝑄
𝐴
= 𝐸𝑏𝑙𝑎𝑐𝑘 ∝ 𝑇4
𝐸𝑏𝑙𝑎𝑐𝑘 = 𝜎 𝑇
4
or𝑄𝐵𝑙𝑎𝑐𝑘 𝑜𝑟 𝑚𝑎𝑥= 𝜎𝐴𝑠 𝑇𝑠
4
Where 𝜎 is Stefan-Boltzmann constant and
has the value of 5.67 ∗ 10−8 𝑊
𝑚2𝑘4
For Real Surfaces:
𝑄𝑅𝑒𝑎𝑙 = 𝜀𝐴𝑠 𝜎 𝑇𝑠
4
36. 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
37. 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
𝑄𝑥 = − 𝑘𝑥. 𝐴.
𝜕𝑇
𝜕𝑥
𝑄𝑥 = − 𝑘𝑥. 𝑑𝑦. 𝑑𝑧 .
𝜕𝑇
𝜕𝑥
38. Quantity of heat leaving the element in x direction per unit time i.e.
Heat efflux
𝑄𝑥+𝑑𝑥 = 𝑄𝑥 +
𝜕
𝜕𝑥
𝑄𝑥 𝑑𝑥
Heat accumulated in the element due to heat flow in x direction
per unit time
𝑑𝑄𝑥 = 𝑄𝑥 − 𝑄𝑥 +
𝜕
𝜕𝑥
𝑄𝑥 𝑑𝑥
𝑑𝑄𝑥 = −
𝜕
𝜕𝑥
𝑄𝑥 𝑑𝑥
𝑑𝑄𝑥 = −
𝜕
𝜕𝑥
− 𝑘𝑥 𝑑𝑦. 𝑑𝑧
𝜕𝑇
𝜕𝑥
𝑑𝑥
𝑑𝑄𝑥 =
𝜕
𝜕𝑥
𝑘𝑥
𝜕𝑇
𝜕𝑥
𝑑𝑥. 𝑑𝑦. 𝑑𝑧
39. Similarly heat accumulated due to heat flow by conduction along y
and z direction will be:
𝑑𝑄𝑦 =
𝜕
𝜕𝑦
𝑘𝑦
𝜕𝑇
𝜕𝑦
𝑑𝑥. 𝑑𝑦. 𝑑𝑧
𝑑𝑄𝑧 =
𝜕
𝜕𝑧
𝑘𝑧
𝜕𝑇
𝜕𝑧
𝑑𝑥. 𝑑𝑦. 𝑑𝑧
Net heat accumulated in the element due to conduction of heat from all
the direction considered per unit time =
𝜕
𝜕𝑥
𝑘𝑥
𝜕𝑇
𝜕𝑥
+
𝜕
𝜕𝑦
𝑘𝑦
𝜕𝑇
𝜕𝑦
+
𝜕
𝜕𝑧
𝑘𝑧
𝜕𝑇
𝜕𝑧
𝑑𝑥. 𝑑𝑦. 𝑑𝑧
--------------------(i)
B. Total Heat generated within the element per unit time
𝑄𝑔 = 𝑞𝑔
′′′ ∗ 𝑑𝑥. 𝑑𝑦. 𝑑𝑧
--------------- (ii)
40. C. Rate of change of energy content of the element
𝑚. 𝐶𝑝.
𝜕𝑇
𝜕𝜏
𝜌. 𝑉. 𝐶𝑝 .
𝜕𝑇
𝜕𝜏
𝜌. 𝐶𝑝.
𝜕𝑇
𝜕𝜏
. 𝑑𝑥. 𝑑𝑦. 𝑑𝑧
------------------(iii)
Putting value of (i), (ii),(iii) in equation no. (1) the heat balance equation
will be
𝝏
𝝏𝒙
𝒌𝒙
𝝏𝑻
𝝏𝒙
+
𝝏
𝝏𝒚
𝒌𝒚
𝝏𝑻
𝝏𝒚
+
𝝏
𝝏𝒛
𝒌𝒛
𝝏𝑻
𝝏𝒛
𝒅𝒙. 𝒅𝒚. 𝒅𝒛 + 𝒒𝒈
′′′
∗ 𝒅𝒙. 𝒅𝒚. 𝒅𝒛 = 𝛒. 𝑪𝒑.
𝝏𝑻
𝝏𝝉
. 𝐝𝐱. 𝐝𝐲. 𝐝𝐳
41. 𝝏
𝝏𝒙
𝒌𝒙
𝝏𝑻
𝝏𝒙
+
𝝏
𝝏𝒚
𝒌𝒚
𝝏𝑻
𝝏𝒚
+
𝝏
𝝏𝒛
𝒌𝒛
𝝏𝑻
𝝏𝒛
+ 𝒒𝒈
′′′
= 𝛒. 𝑪𝒑.
𝝏𝑻
𝝏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
𝑘
𝜕
𝜕𝑥
𝜕𝑇
𝜕𝑥
+
𝜕
𝜕𝑦
𝜕𝑇
𝜕𝑦
+
𝜕
𝜕𝑧
𝜕𝑇
𝜕𝑧
+ 𝑞𝑔
′′′ = 𝜌. 𝐶𝑝.
𝜕𝑇
𝜕𝜏
𝝏𝟐
𝑻
𝝏𝒙𝟐 +
𝝏𝟐
𝑻
𝝏𝒚𝟐 +
𝝏𝟐
𝑻
𝝏𝒛𝟐 +
𝒒𝒈
′′′
𝒌
=
𝝆. 𝑪𝒑
𝒌
.
𝝏𝑻
𝝏𝝉
𝝏𝟐𝑻
𝝏𝒙𝟐
+
𝝏𝟐𝑻
𝝏𝒚𝟐
+
𝝏𝟐𝑻
𝝏𝒛𝟐
+
𝒒𝒈
′′′
𝒌
=
𝟏
𝜶
.
𝝏𝑻
𝝏𝝉
𝜶=
𝒌
𝝆.𝑪𝒑
Thermal diffusivity
42. Simplified forms of heat conduction equation in cartesian
coordinate system
Unsteady state heat flow with no internal heat generation
(Fourier’s Equation)
𝝏𝟐𝑻
𝝏𝒙𝟐
+
𝝏𝟐𝑻
𝝏𝒚𝟐
+
𝝏𝟐𝑻
𝝏𝒛𝟐
=
𝟏
𝜶
.
𝝏𝑻
𝝏𝝉
Steady state heat flow with heat generation(Poisson’s equation)
𝝏𝟐𝑻
𝝏𝒙𝟐 +
𝝏𝟐𝑻
𝝏𝒚𝟐 +
𝝏𝟐𝑻
𝝏𝒛𝟐 +
𝒒𝒈
′′′
𝒌
= 𝟎
Steady state heat flow without internal heat generation(Laplace
Equation)
𝝏𝟐𝑻
𝝏𝒙𝟐
+
𝝏𝟐𝑻
𝝏𝒚𝟐
+
𝝏𝟐𝑻
𝝏𝒛𝟐
= 𝟎
43. Steady state, one dimensional heat flow with heat generation
𝝏𝟐
𝑻
𝝏𝒙𝟐
+
𝒒𝒈
′′′
𝒌
= 𝟎
Steady state, one dimensional heat flow without heat generation
𝝏𝟐
𝑻
𝝏𝒙𝟐
= 𝟎
Unsteady state, one dimensional heat flow without heat
Generation
𝝏𝟐𝑻
𝝏𝒙𝟐 =
𝟏
𝜶
.
𝝏𝑻
𝝏𝝉
44. 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
45. 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
)
Net heat accumulated in the
element due to conduction of
heat from all the direction
considered
Heat Generated within
the element
Rate of change of
energy content of the
element
+ =
46. 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
𝑸𝒓 = −𝒌. 𝑨.
𝝏𝑻
𝝏𝒓
𝑸𝒓 = −𝒌. 𝒓𝒅𝝓. 𝒅𝒛 .
𝝏𝑻
𝝏𝒓
Quantity of heat leaving the element in r direction per unit time i.e.
Heat efflux
𝐐𝐫+𝐝𝒓 = 𝐐𝒓 +
𝛛
𝛛𝒓
𝐐𝒓 𝐝𝐫
47. Net Heat accumulated in the element due to heat flow in radial
direction per unit time
𝐝𝐐𝒓 = 𝐐𝒓 − 𝐐𝒓 +
𝛛
𝛛𝒓
𝐐𝒓 𝐝𝒓
𝐝𝐐𝒓 = −
𝛛
𝛛𝒓
𝐐𝒓 𝐝𝐫
𝒅𝑸𝒓 = −
𝝏
𝝏𝒓
− 𝒌 𝒓𝒅𝝓. 𝒅𝒛
𝝏𝑻
𝝏𝒓
𝒅𝒓
𝒅𝑸𝒓 = 𝒌 𝒅𝒓. 𝒅𝝓. 𝒅𝒛
𝝏
𝝏𝒓
𝒓
𝝏𝑻
𝝏𝒓
𝒅𝑸𝒓 = 𝒌(𝒅𝒓. 𝒅𝝓. 𝒅𝒛) 𝒓
𝝏𝟐𝑻
𝝏𝒓𝟐
+
𝝏𝑻
𝝏𝒓
𝒅𝑸𝒓 = 𝒌(𝒅𝒓. 𝒓𝒅𝝓. 𝒅𝒛)
𝝏𝟐𝑻
𝝏𝒓𝟐
+
𝟏
𝒓
𝝏𝑻
𝝏𝒓
48. 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
𝑸𝝓 = −𝒌. 𝑨.
𝝏𝑻
𝒓. 𝝏𝝓
𝑸𝝓 = −𝒌. 𝒅𝒓. 𝒅𝒛 .
𝝏𝑻
𝒓. 𝝏𝝓
Quantity of heat leaving the element in 𝝓 direction per unit time i.e.
Heat efflux
𝐐𝝓+𝐝𝝓 = 𝐐𝝓 +
𝛛
𝒓. 𝛛𝝓
𝐐𝝓 𝒓𝐝𝝓
Net Heat accumulated in the element due to heat flow in tangential
direction per unit time
𝐝𝐐𝝓 = 𝐐𝝓 − 𝐐𝝓+𝒅𝝓
49. Net Heat accumulated in the element due to heat flow in
tangential direction per unit time
𝐝𝐐𝝓 = 𝐐𝝓 − 𝐐𝝓 +
𝛛
𝒓. 𝛛𝝓
𝐐𝝓 𝒓𝐝𝝓
𝐝𝐐𝝓 = −
𝛛
𝒓. 𝛛𝝓
𝐐𝝓 𝒓𝐝𝝓
𝒅𝑸𝝓 = −
𝝏
𝒓. 𝛛𝝓
− 𝒌 𝒅𝒓. 𝒅𝒛
𝝏𝑻
𝒓. 𝛛𝝓
𝒓𝐝𝝓
𝒅𝑸𝝓 = 𝒌 𝒅𝒓. 𝒅𝝓. 𝒅𝒛
𝝏
𝝏𝝓
𝟏
𝒓
𝝏𝑻
𝝏𝝓
𝒅𝑸𝝓 = 𝒌 𝒅𝒓. 𝒓𝒅𝝓. 𝒅𝒛
𝟏
𝒓𝟐
𝝏
𝝏𝝓
𝝏𝑻
𝝏𝝓
𝒅𝑸𝝓 = 𝒌 𝒅𝒓. 𝒓𝒅𝝓. 𝒅𝒛
𝟏
𝒓𝟐
𝝏𝟐
𝑻
𝝏𝝓𝟐
50. 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
𝑸𝒛 = −𝒌. 𝑨.
𝝏𝑻
𝝏𝒛
𝑸𝒛 = −𝒌. 𝒓𝒅𝝓. 𝒅𝒓 .
𝝏𝑻
𝝏𝒛
Heat accumulated in the element due to heat flow in axial
direction per unit time
𝐝𝐐𝒛 = 𝐐𝒛 − 𝐐𝒛 +
𝛛
𝛛𝒛
𝐐𝒛 𝐝𝒛
𝐝𝐐𝒛 = −
𝛛
𝛛𝒛
𝐐𝒛 𝐝𝐳
Quantity of heat leaving the element in r direction per unit time i.e.
Heat efflux
𝐐𝐳 = 𝐐𝒛 +
𝛛
𝛛𝒛
𝐐𝒛 𝐝𝐳
51. 𝐝𝐐𝐳 = −
𝛛
𝛛𝐳
− 𝐤 𝐫𝐝𝛟. 𝐝𝐫
𝛛𝐓
𝛛𝐳
𝐝𝐳
𝐝𝐐𝐳 = 𝐤(𝐝𝐫. 𝐫𝐝𝛟. 𝐝𝐳)
𝛛𝟐
𝐓
𝛛𝐳𝟐
Net heat accumulated in the element due to conduction of
heat from all the direction considered per unit time =
𝝏𝟐𝑻
𝝏𝒓𝟐
+
𝟏
𝒓
𝝏𝑻
𝝏𝒓
+
𝟏
𝒓𝟐
𝝏𝟐𝑻
𝝏𝝓𝟐
+
𝛛𝟐𝐓
𝛛𝐳𝟐
𝐤(𝐝𝐫. 𝐫𝐝𝛟. 𝐝𝐳)
--------------------(i)
B. Total Heat generated within the element per unit time
𝑸𝒈 = 𝒒𝒈
′′′ ∗ 𝐝𝐫. 𝐫𝐝𝛟. 𝐝𝐳
--------------- (ii)
52. C. Rate of change of energy content of the element
𝐦. 𝐂𝐩.
𝛛𝐓
𝛛𝛕
𝛒. 𝐕. 𝐂𝐩 .
𝛛𝐓
𝛛𝛕
𝛒. 𝐂𝐩.
𝛛𝐓
𝛛𝛕
. 𝐝𝐫. 𝐫𝐝𝛟. 𝐝𝐳
------------------(iii)
Now
A + B = C
Net heat accumulated in the
element due to conduction of
heat from all the direction
considered
Heat Generated within
the element
Rate of change of
energy content of the
element
55. 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
56. 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 𝜃d𝜙
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
element due to conduction of
heat from all the direction
considered
Heat Generated within
the element
Rate of change of
energy content of the
element
+ =
57. 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
𝑄𝑟 = −𝑘. 𝐴.
𝜕𝑇
𝜕𝑟
𝑄𝑟 = −𝑘. rd𝜃. 𝑟𝑠𝑖𝑛 𝜃𝑑𝜙 .
𝜕𝑇
𝜕𝑟
Quantity of heat leaving the element in r direction per
unit time i.e. Heat efflux
𝑄r+𝑑𝑟 = 𝑄𝑟 +
𝜕
𝜕𝑟
𝑄𝑟 𝑑r
58. Net Heat accumulated in the element due to heat flow in radial direction
per unit time
𝑑𝑄𝑟 = 𝑄𝑟 − 𝑄𝑟 +
𝜕
𝜕𝑟
𝑄𝑟 𝑑𝑟
𝑑𝑄𝑟 = −
𝜕
𝜕𝑟
𝑄𝑟 𝑑r
𝑑𝑄𝑟 = −
𝜕
𝜕𝑟
−𝑘 𝑟𝑑𝜃. 𝑟 sin 𝜃𝑑𝜙
𝜕𝑇
𝜕𝑟
𝑑𝑟
𝑑𝑄𝑟 = 𝑘 𝑑𝑟. 𝑑𝜃. sin 𝜃𝑑𝜙
𝜕
𝜕𝑟
𝑟2
𝜕𝑇
𝜕𝑟
𝑑𝑄𝑟 = 𝑘 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙
1
𝑟2
𝜕
𝜕𝑟
𝑟2
𝜕𝑇
𝜕𝑟
59. 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
𝑄𝜙 = −𝑘. 𝐴.
𝜕𝑇
𝑟. sin 𝜃 𝜕𝜙
𝑄𝜙 = −𝑘. 𝑑𝑟. 𝑟𝑑𝜃 .
𝜕𝑇
𝑟. sin 𝜃 𝜕𝜙
Quantity of heat leaving the element in 𝜙 direction per
unit time i.e. Heat efflux
𝑄𝜙+𝑑𝜙 = 𝑄𝜙 +
𝜕
𝑟. sin 𝜃 𝜕𝜙
𝑄𝜙 𝑟. sin 𝜃 𝑑𝜙
Net Heat accumulated in the element due to heat flow in
𝜙 per unit time
𝑑𝑄𝜙 = 𝑄𝜙 − 𝑄𝜙+𝑑𝜙
60. Net Heat accumulated in the element due to heat flow in 𝜙 per unit time
𝑑𝑄𝜙 = 𝑄𝜙 − 𝑄𝜙 +
𝜕
𝑟 sin 𝜃 𝜕𝜙
𝑄𝜙 𝑟 sin 𝜃. 𝑑𝜙
𝑑𝑄𝜙 = −
𝜕
𝑟 sin 𝜃 𝜕𝜙
𝑄𝜙 𝑟 sin 𝜃. 𝑑𝜙
𝑑𝑄𝜙 = −
𝜕
𝑟 sin 𝜃 𝜕𝜙
− 𝑘 𝑑𝑟. 𝑟𝑑𝜃
𝜕𝑇
𝑟 sin 𝜃 𝜕𝜙
𝑟 sin 𝜃. 𝑑𝜙
𝑑𝑄𝜙 = 𝑘 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙
1
𝑟2 sin2 𝜃
𝜕2
𝑇
𝜕𝜙2
61. 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
𝑄𝜃 = −𝑘. 𝐴.
𝜕𝑇
𝑟𝜕𝜃
𝑄𝜃 = −𝑘. 𝑑𝑟. 𝑟sin 𝜃𝑑𝜙 .
𝜕𝑇
𝑟𝜕𝜃
Quantity of heat leaving the element in 𝜃 direction per
unit time i.e. Heat efflux
𝑄𝜃+𝑑𝜃 = 𝑄𝜃 +
𝜕
𝑟. 𝜕𝜃
𝑄𝜃 𝑟𝑑𝜃
Net Heat accumulated in the element due to heat
flow in 𝜃 direction per unit time
𝑑𝑄𝜃 = 𝑄𝜃 − 𝑄𝜃+𝑑𝜃
62. Net Heat accumulated in the element due to heat flow in 𝜃direction per unit
time
𝑑𝑄𝜃 = 𝑄𝜃 − 𝑄𝜃 +
𝜕
𝑟𝜕𝜃
𝑄𝜃 𝑟. 𝑑𝜃
𝑑𝑄𝜃 = −
𝜕
𝑟𝜕𝜃
𝑄𝜃 𝑟𝑑𝜃
𝑑𝑄𝜃 = −
𝜕
𝑟𝜕𝜃
− 𝑘 𝑑𝑟. 𝑟sin 𝜃𝑑𝜙
𝜕𝑇
𝑟𝜕𝜃
𝑟𝑑𝜃
𝑑𝑄𝜃 = 𝑘 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙
𝜕
𝑟2 sin 𝜃 𝜕𝜃
sin 𝜃
𝜕𝑇
𝜕𝜃
63. Net heat accumulated in the element due to conduction of heat from all the direction
considered per unit time =
𝑘 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙
1
𝑟2
𝜕
𝜕𝑟
𝑟2
𝜕𝑇
𝜕𝑟
+ 𝑘 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙
1
𝑟2 sin2 𝜃
𝜕2𝑇
𝜕𝜙2
+ 𝑘 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙
𝜕
𝑟2 sin 𝜃 𝜕𝜃
sin 𝜃
𝜕𝑇
𝜕𝜃
1
𝑟2
𝜕
𝜕𝑟
𝑟2
𝜕𝑇
𝜕𝑟
+
1
𝑟2 sin2 𝜃
𝜕2
𝑇
𝜕𝜙2 +
𝜕
𝑟2 sin 𝜃 𝜕𝜃
sin 𝜃
𝜕𝑇
𝜕𝜃
𝑘 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙
--------------------(i)
B. Total Heat generated within the element per unit time
𝑄𝑔 = 𝑞𝑔
′′′ ∗ 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙
--------------- (ii)
64. C. Rate of change of energy content of the element
𝑚. 𝐶𝑝.
𝜕𝑇
𝜕𝜏
𝜌. 𝑉. 𝐶𝑝 .
𝜕𝑇
𝜕𝜏
𝜌. 𝐶𝑝.
𝜕𝑇
𝜕𝜏
. 𝑑𝑟. 𝑟𝑑𝜃. 𝑟sin 𝜃𝑑𝜙
------------------(iii)
Now
A + B = C
Net heat accumulated in the
element due to conduction of
heat from all the direction
considered
Heat Generated within
the element
Rate of change of
energy content of the
element
68. 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
69. 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
𝒒𝒔 = −𝒌
𝝏𝑻
𝝏𝒙 𝒙=𝟎,𝒕
70. 3. Adiabatic or insulated wall
Boundary Condition
i.e. Heat Flux(Heat rate per unit surface area) is
constant
𝒒𝒔 = 𝟎 = 𝒌
𝝏𝑻
𝝏𝒙 𝒙=𝟎,𝒕
𝝏𝑻
𝝏𝒙 𝒙=𝟎,𝒕
= 𝟎
4. Convection Boundary Condition
i.e. Heat Flux(Heat rate per unit surface area) is
constant
−𝒌
𝝏𝑻
𝝏𝒙 𝟎,𝒕
= 𝒉𝟏 𝑻∞𝟏
− 𝑻 𝟎,𝒕)
−𝒌
𝝏𝑻
𝝏𝒙 𝑳,𝒕
= 𝒉𝟐 𝑻 𝑳,𝒕) − 𝑻∞𝟐
71. Solution Procedure for solving heat conduction
problem
Heat
Transfer
Problem
Mathematical
Formulation
(Diff.
Equation &
Boundary
condition)
General
Solution of
Diff.
Equation
Application
of
Boundary
Condition
Solution of
the
Problem
72. 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
𝝏𝟐𝑻
𝝏𝒙𝟐
+
𝝏𝟐𝑻
𝝏𝒚𝟐
+
𝝏𝟐𝑻
𝝏𝒛𝟐
+
𝒒𝒈
′′′
𝒌
=
𝟏
𝜶
.
𝝏𝑻
𝝏𝝉
= 0
Final equation in case of steady, one dimensional and
without heat generation will be
𝝏𝟐𝑻
𝝏𝒙𝟐 = 𝟎 or
𝒅𝟐𝑻
𝒅𝒙𝟐 = 𝟎
73. On integrating twice
𝑑𝑇
𝑑𝑥
= 𝐶2 and 𝑇 = 𝐶1𝑥 + 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 = 𝐶1𝐿 + 𝐶2
Here 𝐶1 =
𝑻𝟐−𝑻𝟏
𝑳
Equation i reduced to
𝑻 =
𝑻𝟐 − 𝑻𝟏
𝑳
𝒙 + 𝑻𝟏
74. For Heat Flow through Slab/Wall
We have from Fourier’s law of heat conduction
𝑸 = −𝒌𝑨
𝒅𝑻
𝒅𝒙
𝒅𝑻
𝒅𝒙
=
𝒅
𝒅𝒙
𝑻𝟐 − 𝑻𝟏
𝑳
𝒙 + 𝑻𝟏
𝒅𝑻
𝒅𝒙
=
𝑻𝟐 − 𝑻𝟏
𝑳
∴ 𝑸 = −𝒌𝑨
𝑻𝟐 − 𝑻𝟏
𝑳
Or
𝑸 = 𝒌𝑨
𝑻𝟏 − 𝑻𝟐
𝑳
𝑸 =
𝑻𝟏 − 𝑻𝟐
𝑳
𝒌𝑨
75. Alternative Method
We have from Fourier’s law of heat conduction
𝑄 = −𝑘𝐴
𝑑𝑇
𝑑𝑥
𝑄. 𝑑𝑥 = −𝑘𝐴. 𝑑𝑇
Integrating on both side
𝑄 𝑑𝑥 = −𝑘𝐴 𝑑𝑇
𝑇2
𝑇1
𝐿
0
𝑄 𝑑𝑥 = 𝑘𝐴 𝑑𝑇
𝑇1
𝑇2
𝐿
0
𝑄. 𝑥 0
𝐿
= 𝑘𝐴 𝑇 𝑇2
𝑇1
𝑄. 𝐿 = 𝑘𝐴 𝑇1 − 𝑇2 ……….(i)
𝑸 = 𝒌𝑨
𝑻𝟏 − 𝑻𝟐
𝑳
𝑄 𝑑𝑥 = −𝑘𝐴 𝑑𝑇
𝑇
𝑇1
𝑥
0
𝑄. 𝑥 = 𝑘𝐴 𝑇1 − 𝑇 ……….(ii)
Dividing ii by I we get
𝑄. 𝑥
𝑄. 𝐿
=
𝑘𝐴 𝑇1 − 𝑇
𝑘𝐴 𝑇1 − 𝑇2
𝑻𝟏 − 𝑻
𝑻𝟏 − 𝑻𝟐
=
𝒙
𝑳
76. Variable Thermal Conductivity
The dependence of thermal conductivity on temperature
can be expressed as-
𝐤 = 𝐤𝟎 𝟏 + 𝛃𝐓
𝐤𝟎 - thermal conductivity at reference
𝛃 – is constant for given material
T – Temperature
According to Fourier’s Law
𝑸 = −𝒌𝑨
𝒅𝑻
𝒅𝒙
𝑸 = −𝒌𝟎 𝟏 + 𝜷𝑻 . 𝑨.
𝒅𝑻
𝒅𝒙
𝑸. 𝒅𝒙 = −𝒌𝟎 𝟏 + 𝜷𝑻 . 𝑨. 𝒅𝑻
Integrating on both side and applying boundary condition
as
At x = 0, T=T1
At x = L, T=T2
We get
𝑸 𝒅𝒙 = − 𝒌𝟎 𝟏 + 𝜷𝑻 . 𝑨. 𝒅𝑻
𝑻𝟐
𝑻𝟏
𝑳
𝟎
𝑸 𝒅𝒙 = − 𝒌𝟎 𝟏 + 𝜷𝑻 . 𝑨. 𝒅𝑻
𝑻𝟐
𝑻𝟏
𝑳
𝟎
𝑸. 𝒙 𝟎
𝑳
= −𝒌𝟎𝑨 𝑻 + 𝜷
𝑻𝟐
𝟐 𝑻𝟏
𝑻𝟐
𝑸. 𝑳 = −𝒌𝟎𝑨 𝑻𝟐 − 𝑻𝟏 +
𝜷
𝟐
𝑻𝟐
𝟐
− 𝑻𝟏
𝟐
𝑸. 𝑳 = 𝒌𝟎𝑨 𝑻𝟏 − 𝑻𝟐
+
𝜷
𝟐
𝑻𝟏 − 𝑻𝟐 𝑻𝟏 + 𝑻𝟐
𝑸. 𝑳 = 𝒌𝟎𝑨 𝑻𝟏 − 𝑻𝟐 𝟏 +
𝜷
𝟐
𝑻𝟏 + 𝑻𝟐
𝒌𝒎 = 𝒌𝟎 𝟏 +
𝜷
𝟐
𝑻𝟏 + 𝑻𝟐
𝑸 = 𝒌𝒎𝑨
𝑻𝟏 − 𝑻𝟐
𝑳
77. Electrical Analogy
As per Fourier’s Law
𝑸 = −𝒌𝑨
𝒅𝑻
𝒅𝒙
𝑸 = 𝒌𝑨
𝑻𝟏 − 𝑻𝟐
𝑳
𝑸 =
𝑻𝟏 − 𝑻𝟐
𝑳
𝒌𝑨
𝑸 =
𝒅𝑻
𝑹𝒕𝒉
𝑹𝒕𝒉𝑪𝒐𝒏𝒅𝒖𝒄𝒕𝒊𝒐𝒏) = 𝑳
𝒌𝑨
𝑹𝒕𝒉𝑪𝒐𝒏𝒅𝒖𝒄𝒕𝒊𝒐𝒏) = 𝟏
𝒉𝑨
As per Ohm’s Law
𝑰 =
𝒅𝑽
𝑹
I – Current flow
dV – Voltage difference
R – Electrical Resistance
I is analogous to Q
dV is analogous to dT
R is analogous to 𝑹𝒕𝒉
83. 𝑸 =
𝑻∞𝟏 − 𝑻∞𝟐
𝟏
𝒉𝟏𝑨
+
𝑳𝟏
𝒌𝟏𝑨
+
𝑳𝟐
𝒌𝟐𝑨
+
𝟏
𝒉𝟐𝑨
Overall Heat Transfer Coefficient
𝑸 =
𝑨 𝑻∞𝟏 − 𝑻∞𝟐
𝟏
𝒉𝟏
+
𝑳𝟏
𝒌𝟏
+
𝑳𝟐
𝒌𝟐
+
𝟏
𝒉𝟐
When there is fluid to fluid heat transfer it usual practice
to adopt overall heat transfer coefficient
If U is the overall heat transfer coefficient
then
𝑸 = 𝑼. 𝑨. 𝑻∞𝟏 − 𝑻∞𝟐
𝑸 = 𝑼. 𝑨. 𝑻∞𝟏 − 𝑻∞𝟐 =
𝑨 𝑻∞𝟏 − 𝑻∞𝟐
𝟏
𝒉𝟏
+
𝑳𝟏
𝒌𝟏
+
𝑳𝟐
𝒌𝟐
+
𝟏
𝒉𝟐
84. 𝑼 =
𝟏
𝟏
𝒉𝟏
+
𝑳𝟏
𝒌𝟏
+
𝑳𝟐
𝒌𝟐
+
𝟏
𝒉𝟐
𝑼. 𝑨. 𝑻∞𝟏 − 𝑻∞𝟐 =
𝑻∞𝟏 − 𝑻∞𝟐
𝟏
𝒉𝟏𝑨
+
𝑳𝟏
𝒌𝟏𝑨
+
𝑳𝟐
𝒌𝟐𝑨
+
𝟏
𝒉𝟐𝑨
𝑼𝑨 =
𝟏
𝑹
0r 𝑼 =
𝟏
𝑨 𝑹
For unit area overall heat transfer coefficient
is reciprocal of total thermal resistance
85. Thermal Contact Resistance
Ideal or perfect contact Actual or imperfect contact
𝑄𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 = 𝑄𝐶𝑜𝑛𝑡𝑎𝑐𝑡 + 𝑄𝑔𝑎𝑝 Depends upon
Surface roughness
Material properties
Temperature
Pressure
In analogous to Newton’s law of cooling
It can be also expressed as
𝑄 = ℎ𝑐. 𝐴. ∆𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒
86. Thermal Contact Resistance
𝑄𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 = ℎ𝑐. 𝐴. ∆𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒
ℎ𝑐 =
𝑄𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒
𝐴. ∆𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒
Where ℎ𝑐 is thermal contact conductance 𝑊
𝑚2℃
Thermal contact resistance is the inverse of
thermal contact conductance
𝑅𝑐 =
1
ℎ𝑐
=
𝐴. ∆𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒
𝑄𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒
Unit of thermal contact resistance is 𝑚2℃
𝑊
Thermal contact resistance
can be minimized by
By applying thermal grease like
silicon oil
Removing air by better conducting
gas like Helium or Hydrogen
Inserting soft metallic foil like Tin,
Silver or copper
Increasing the interface pressure
87. 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
𝑟1and 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
𝝏𝟐𝑻
𝝏𝒓𝟐
+
𝟏
𝒓
𝝏𝑻
𝝏𝒓
+
𝟏
𝒓𝟐
𝝏𝟐𝑻
𝝏𝝓𝟐
+
𝝏𝟐𝑻
𝝏𝒛𝟐
+
𝒒𝒈
′′′
𝒌
=
𝟏
𝜶
.
𝝏𝑻
𝝏𝝉
= 0
𝝏𝟐𝑻
𝝏𝒓𝟐 +
𝟏
𝒓
𝝏𝑻
𝝏𝒓
= 𝟎 or
𝒅𝟐𝑻
𝒅𝒓𝟐 +
𝟏
𝒓
𝒅𝑻
𝒅𝒓
= 𝟎
88. 𝒅𝟐𝑻
𝒅𝒓𝟐
+
𝟏
𝒓
𝒅𝑻
𝒅𝒓
= 𝟎 ⟹
𝟏
𝒓
.
𝒅
𝒅𝒓
𝒓.
𝒅𝑻
𝒅𝒓
= 0
𝒅
𝒅𝒓
𝒓.
𝒅𝑻
𝒅𝒓
= 0
On integrating both side we get
𝒓.
𝒅𝑻
𝒅𝒓
= C ⟹
𝒅𝑻
𝒅𝒓
=
C
𝒓
On integrating both side we get
𝑻 = 𝑪. 𝒍𝒏 𝒓 + 𝑪𝟏 −−−− −(𝒊)
C & 𝐶𝟐 are the arbitrary constant and can be calculated by
applying boundary condition
At 𝒓 = 𝒓𝟏 ; 𝑻 = 𝑻𝟏
𝒓 = 𝒓𝟐 ; 𝑻 = 𝑻𝟐
𝑻𝟏 = 𝑪. 𝒍𝒏 𝒓𝟏 + 𝑪𝟏
𝑻𝟐 = 𝑪. 𝒍𝒏 𝒓𝟐 + 𝑪𝟏
-
𝑻𝟏 − 𝑻𝟐 = 𝑪 𝒍𝒏 𝒓𝟏 − 𝒍𝒏 𝒓𝟐
93. 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
94. Heat rate from the base surface is given by
𝑄𝐶𝑜𝑛𝑣. = ℎ 𝐴𝑠 𝑇𝑠 − 𝑇∞
95. 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
96. Types of Extended Surfaces or Fins
Rectangular Fins
IC Engines and Electronic Component
Trapezoidal Fins
Rocket Engines and Fishes
