ERT 216
HEAT & MASS TRANSFER
Sem 2/ 2014-2015
Dr Akmal Hadi Bin Ma’ Radzi
School of Bioprocess Engineering
University Malaysia Perlis
Conduction
1) Steady state conduction- One dimension
2) Steady state conduction- Multiple
dimension
3) Unsteady state conduction
Steady State
ConductionOne Dimension
(1D)
Point of view
Application of Fourier’s law of heat conduction to
calculation of heat flow in simple 1D system
(1) Plane Wall
(2) Cylinders
Radial Systems
(3) Spherical
1D The temp. in the body is a function only of
radial distance and independent of azimuth angle/
axial distance
(1) The Plane Wall
Integrated Fourier’s law 
# If k varies with temp. according linear relation
, the heat flow become;
# If 3 material (multilayer wall) involved, heat flow become;
Note: The heat flow must be SAME through all section
Electrical analog circuit
# The heat flow rate also can be represented as resistance network;
(Different conceptual view point for Fourier’s law)
 Thermal resistance (°C/W)
Consider Heat transfer rate as flow and combination of
as a
resistance to this flow. The temp. is the potential function of the
heat flow. So that, the Fourier equation may be written as:
3 wall side by side act as 3 thermal resistance in series
Electrical analog circuit: used to solve more complex problem
(series and parallel thermal resistance)
Electrical analog circuit
Parallel
Series
Insulation & R value
• The performance of insulation R value, define as
• Guide to choose insulating material in terms of their
application and allowable temperature range
(2) Cylinders
From Fourier’s Law 
Integrate this with the
boundary conditions 
The heat flow rate 
(2) Cylinders
Thermal resistant for cylinder is 
(2) Cylinders
Multiple cylindrical sections
Thermal-resistance concept for multiple-layer cylindrical walls
= Thermal-resistance concept for plane wall
So that,
the heat flow rate 
(3) Spheres
The heat flow rate 
Prove this equation!!
Convection Boundary Conditions
Newton rate equation
/Newton’s Law of cooling 
So that, an electric-resistance analogy for convection
process become:
Example 1:
(Multilayer plane wall conduction)
An exterior wall of a house may be approximated by a
4-in layer of common brick (k= 0.7 W/m.°C) followed
by a 1.5-in layer of gypsum plaster (k=0.48 W/m.°C).
What thickness of loosely packed rock wool insulation
(k=0.065 W/m.°C) should be added to reduce the heat
loss (or gain) through the wall by 80%?
Example 2:
(Multilayer Cylindrical System)
A thick-walled tube of stainless steel [18% Cr, 8% Ni, k=19
W/m.°C ] with 2 cm inner diameter (ID) and 4 cm outer
diameter (OD) is covered with a 3 cm layer of asbestos
insulation [k=0.2 W/m.°C ]. If the inside wall temperature
of the pipe is maintained at 600 °C, calculate the heat loss
per meter of length. Also calculate the tube-insulation
interface temperature.
The Overall Heat Transfer Coefficient, U
1/ h A represent the convection resistance;
∆x/ k A represent the conduction resistance
The overall heat transfer (conduction + convection) can be expressed in term
of an overall heat transfer coefficient, U defined by relation:
 A: Area for the heat flow
The Overall Heat Transfer Coefficient

U also related to the R-value:
Where,
The Overall Heat Transfer Coefficient, U
Consider:
Hollow cylinder exposed to a convection
environment on its inner and outer surfaces with TA
and TB the two fluid temp. The area for convection is
not same for both liquids (depend on the inside tube
diameter and wall thickness
The overall hate transfer rate become;
 Overall temp. difference
 The sum of the thermal resistances
The Overall Heat Transfer Coefficient, U
 Ai & Ao: Inside & outside
surface areas of the inner tube
The Overall Heat Transfer Coefficient (Hollow cylinder)
based on:
1) Inside area of the tube Ai 
2) Outside area of the tube, Ao 
The Overall Heat Transfer Coefficient, U
The general notion (plane wall or cylinder coordinate system)
is that;
 Rth Thermal resistance
Info
 Some typical value of U for heat exchanger are given in table.
 Some value of U for common types of building construction
system also given in table and employed for calculation involving
the heating and cooling buildings.
Example 3
A house wall may be approximated as two 1.2 cm
layers of fiber insulating board, an 8.0 cm layer of
loosely packed asbestos, and a 10 cm layer of common
brick. Assuming convection heat transfer coefficient of
12 W/m2. °C on both sides of the wall, calculate the
overall heat transfer coefficient for this arrangement.
Solution
Example 4
A wall is constructed of a section of stainless steel [k=16
W/m. °C] 4.0 mm thick with identical layers of plastic on
both sides of the steel. The overall heat transfer
coefficient, considering convection on both sides of the
plastic, is 120 W/m2.°C. If the overall temp. different
across the arrangement is 60 °C, calculated the
temperature difference across the stainless steel.
Solution
Critical Thickness of Insulation
Consider:
A layer of insulation which might be installed around
a circular pipe. The inner temp. of the insulation is
fixed at Ti and the outer surface is exposed to a
convection environment at T∞.
The heat transfer in the thermal
network term 
Critical Thickness of Insulation
Manipulated this equation to determine the outer radius of insulation, ro, which
will maximize the heat transfer. The maximization condition is
:
The result is 
(The critical-radius-of
insulation concept)
Critical Thickness of Insulation
 The critical-radius-of insulation
concept
Concept
1) If ro < critical radius value,
means:
The heat transfer will be increased by adding more insulation thickness
2) If ro > critical radius value,
The heat transfer will be decrease
means:
Example 5
A 1.0 mm diameter wire is maintained at a temp. of
400 °C and exposed to a convection environment at
40 °C with h= 120 W/m2. °C. Calculated the
thermal conductivity that will just cause an insulation
thickness of 0.2 mm to produce a “critical radius”.
How much of this insulation must be added to reduce
the heat transfer by 75 percent from that which would
be experienced by the bare wire?
Example 6
So the addition of
25 percent
Heat Source Systems
Situation:
The system generated heat internally. Confine
our discussion to 1D system (the temp. is a
function of only 1 space coordinate) which is:
1)Plane Wall
2)Cylinder
3)Hollow Cylinder
1) Plane wall with heat sources
Consider:
The plane wall with uniformly distributed heat sources.
The thickness of the wall in the x direction is 2 L
Assumed:
1)The heat flow as 1D
2)The heat generated per unit volume is
3)Thermal conductivity, k does not vary with temp.
*
From steady-state 1D heat flow with heat sources 
For the boundary conditions

(The temp. on either side of the wall)
Integrated
* equation with the boundary condition. So, the general solution become;
1) Plane wall with heat sources
 General solution
C1 = 0  Because the temp. must be the SAME on each side of the wall.
C2 = T0  The temp. T0 at the midplane (x=0)
The solution obtain the temp. distribution 
OR
 “Parabolic
distribution”
OR
1) Plane wall with heat sources
Midplane temp. T0  ??
(Obtained through an energy balance)
At steady state conditions:
The total heat generated = The heat lost at the faces
From parabolic
distribution :
So that 
 Differenced to get the temp. gradient at
the wall, dT/dx
1) Plane wall with heat sources
Simplify steady state condition become 

Midplane temp. T0 
The same result for midplane temp. could be obtain by substituting
into this equation (The temp. distribution):
2) Cylinder with heat sources
Consider:
Cylinder radius, R with uniformly distributed heat sources and constant thermal
conductivity.
Assumed:
The cylinder is sufficiently long so that the temp. as function of radius only.
The appropriate differential equation obtained by neglecting the axial, azimuth
and time dependent term
From steady-state 1D heat flow in
cylinder coordinates with heat sources

For the 1st boundary conditions 
At steady state conditions:
The total heat generated = The heat lost at the surface
2) Cylinder with heat sources
2nd boundary conditions
(The temp. function must
be continuous at the
centre of the cylinder), so
could specify that
From steady-state 1D heat flow
in cylinder coordinates with 
heat sources

Rewrite

Note that :
Substitute:
 Integrated
2) Cylinder with heat sources
**
*
Integration yield 
and
From 2nd boundary condition

*
Thus,
From 1st boundary condition

**
Thus,
The final solution for the temp. distribution is then

(substitute C1 & C2)
Or
Dimensionless form 
2) Cylinder with heat sources
The final solution for the temp. distribution is then
(substitute C1 & C2)

Where T0 is the temperature at r = 0. So, the temp. distribution become:
3) Hollow cylinder with heat sources
For hollow cylinder with uniformly distributed heat sources, the boundary condition
would be 
The general solution same with cylinder 
Used boundary condition to get C1 and C2 
Where the constant, C1 given by 
Basic steps involved in the solution of heat transfer problems
Example 7
Example 8
Heat is generated uniformly in a stainless steel plate having k = 20 W/m.oC. The thickness
of the plate is 1.0 cm and the hat generation rate is 500 MW/m3.If the two sides of the
plate are maintained at 100 and 200 oC, respectively, calculate the temperature at the
center of the plate.
Thermal Contact Resistance
Situation:
2 solid bars brought into contact, with the sides of the bars insulated so that
heat flow only in the axial direction. The material may have different thermal
conductivities, but if the side are insulated, the heat flux must be the same
through both materials under steady-state conditions.
 The temp. drop at plane 2 (the contact
plane between 2 material) because of
“Thermal Contact Resistance”
An energy balance on the 2 materials:
Thermal Contact Resistance
Where,
 Thermal Contact Resistance
This factor extremely important in a
no. of application because of the
many heat transfer situations involve
mechanical joining of 2 materials.
 Contact Coefficient
Thermal Contact Resistance
Physical mechanism
Examining a joint in more detail 
Concept:
No real surface is perfectly smooth and actual surface roughness is believed to play a
central role in determining the contact resistance.

 There are 2 principle contributions to the heat transfer at the joint:
1) The solid-to-solid conduction at the spots of contact
2) The conduction through entrapped gases in the void space created by the contact
2nd factor is believed to represent the major resistance to
heat flow because thermal conductivity of the gas is quite
small compare of solid
Thermal Contact Resistance
So, designating the contact area, AC and the void area, AV, the Heat
across the joint may write as:
: Thickness of the void space
: Thermal conductivity of the fluid which fills the void space
: Total cross-sectional area of the bars
Contact Coefficient 
Flow
Assignment 2
Book: J.P. Holman
1)2-1
2)2.2
3)2-5
4)2-6
5)2-7
6)2-8
7)2-9
8)2-17 and 2-18
9)2-26
10)2-62
Tutorial 1 
Book: J.P. Holman
1) 2-4
2) 2-13
3) 2-19
4) 2-20
5) 2-21
6) 2-22
7) 2-23
8) 2-27
9) 2-31
10) 2-60
Assignment 3
Book: J.P. Holman
1)2-30
2)2-55
3)2-117
Tutorial 2
Book: J.P. Holman
1)2-44
2)2-45
3)2-46