Transient heat-conduction-Part-I

Lectures on Heat Transfer --
TRANSIENT HEAT
CONDUCTION: Part-I
by
Dr. M. Thirumaleshwar
formerly:
Professor, Dept. of Mechanical Engineering,
St. Joseph Engg. College, Vamanjoor,
Mangalore
India

Preface
• This file contains slides on Transient Heat
conduction: Part-I
• The slides were prepared while teaching
Heat Transfer course to the M.Tech.
students in Mechanical Engineering Dept.
of St. Joseph Engineering College,
Vamanjoor, Mangalore, India, during Sept.
– Dec. 2010.
Aug. 2016 2MT/SJEC/M.Tech.

• It is hoped that these Slides will be useful
to teachers, students, researchers and
professionals working in this field.
• For students, it should be particularly
useful to study, quickly review the subject,
and to prepare for the examinations.
• M. Thirumaleshwar
August 2016
Aug. 2016 3MT/SJEC/M.Tech.

Aug. 2016 MT/SJEC/M.Tech. 4
References
• 1. M. Thirumaleshwar: Fundamentals of Heat &
Mass Transfer, Pearson Edu., 2006
• https://books.google.co.in/books?id=b2238B-
AsqcC&printsec=frontcover&source=gbs_atb#v=onepage&q&f=false
• 2. Cengel Y. A. Heat Transfer: A Practical
Approach, 2nd Ed. McGraw Hill Co., 2003
• 3. Cengel, Y. A. and Ghajar, A. J., Heat and
Mass Transfer - Fundamentals and Applications,
5th Ed., McGraw-Hill, New York, NY, 2014.

References… contd.
• 4. Incropera , Dewitt, Bergman, Lavine:
Fundamentals of Heat and Mass Transfer, 6th
Ed., Wiley Intl.
• 5. M. Thirumaleshwar: Software Solutions to
Problems on Heat Transfer – CONDUCTION-
Part-II, Bookboon, 2013
• http://bookboon.com/en/software-solutions-problems-on-heat-
transfer-cii-ebook

TRANSIENT HEAT CONDUCTION: Part-I ..
Outline:
• Lumped system analysis – criteria for
lumped system analysis – Biot and Fourier
Numbers – Response time of a
thermocouple - One-dimensional transient
conduction in large plane walls, long
cylinders and spheres when Bi > 0.1 –
one-term approximation - Heisler and
Grober charts- Problems

TRANSIENT HEAT CONDUCTION:
• In transient conduction, temperature depends
not only on position in the solid, but also on
time.
• So, mathematically, this can be written as T =
T(x,y,z,), where  represents the time
coordinate.
• Typical examples of transient
conduction:
• heat exchangers
• boiler tubes
• cooling of I.C.Engine cylinder heads

Examples (contd.):
• heat treatment of engineering
components and quenching of ingots
• heating of electric irons
• heating and cooling of buildings
• freezing of foods, etc.

Lumped system analysis
(Newtonian heating or cooling):
• In lumped system analysis, the internal
conduction resistance of the body to heat
flow (i.e. L/(k.A)) is negligible compared to
the convective resistance (i.e. 1/(h.A)) at
the surface.
• So, the temperature of the body, no doubt,
varies with time, but at any given instant,
the temperature within the body is uniform
and is independent of position. i.e. T = T()
only.

Lumped system analysis
(Newtonian heating or cooling):
• Practical examples of such cases are:
heat treatment of small metal pieces,
measurement of temperature with a
thermocouple or thermometer etc, where
the internal resistance of the object for
heat conduction may be considered as
negligible.

Analysis:
• Consider a solid body of arbitrary shape, volume V,
mass m, density , surface area A, and specific heat
Cp. See Fig. 7.1.
• To start with, at  = 0, let the temperature throughout
the body be uniform at T = Ti. At the instant  = 0, let
the body be suddenly placed in a medium at a
temperature of Ta, as shown.

• Writing an energy balance for this situation:
• Amount of heat transferred into the body in time
interval d =
Increase in the internal energy of the body in
time interval d
i.e. h A T a T ( ) d m C p
 dT  C p
 V dT ....(7.1) since m = . V
Now, since Ta is a constant, we can write: dT d T ( ) T a
Therefore,
d T ( ) T a
T ( ) T a
h A
 C p
 V
d .....(7.2)

Integrating between  = 0 (i.e. T = Ti) and
any , (i.e. T = T()),
ln
T ( ) T a
T i T a
h A 
 C p
 V
i.e.
T ( ) T a
T i T a
exp
h A 
 C p
 V
......(7.3)

• Now, let:
 C p
 V
h A
t
where, ‘t’ is known as ‘thermal time constant’ and has
units of time.
Therefore, eqn. (7.3) is written as:
T ( ) T a
T i T a
exp

t
.......(7.4)
Now denoting θ = (T() – Ta), we write eqn. (7.4)
compactly as:


 i
T ( ) T a
T i T a
exp

t
......(7.5)
Equation (7.5) gives the temperature distribution in a
solid as a function of time, when the internal resistance
of the solid for conduction is negligible compared to the
convective resistance at its surface. See Fig. 7.2 (a)

• Instantaneous heat Transfer:
• At any instant , heat transfer between the body and the
environment is easily calculated since we have the
temperature distribution from eqn. (7.4):
Q ( ) m C p
 dT ( )
d
 W......(7.6,a)
At that instant, heat transfer must also be equal to:
Q ( ) h A T ( ) T a
 W.....(7.6,b)
Total heat transfer:
Total heat transferred during  = 0 to  = , is equal to
the change in Internal energy of the body:
Q tot m Cp
 T ( ) T i
 J....(7.7,a)

• Qtot may also be calculated by integrating eqn.(7.6,a):
Q tot
0

Q ( )d J.....(7.7,b)
Max. heat transferred:
When the body reaches the temperature of the environment,
obviously maximum heat has been transferred:
Q max m C p
 T a T i
 J.....(7.8)
If Qmax is negative, it means that the body has lost heat,
and if Qmax is positive, then body has gained heat.

Criteria for lumped system analysis
(Biot number and Fourier number):
• Consider a plane slab as shown in Fig. 7.3.
• Let the surface on the left be maintained at
temperature T1 and the surface on the right is at
a temperature of T2 as a result of heat being lost
to a fluid at temperature Ta, flowing with a heat
transfer coeff. ha.
• Writing an energy balance at the right hand
surface,
k A
L
T1 T2( ) h A T2 T a


Criteria for lumped system analysis
(Biot number and Fourier number):
Rearranging,
T1 T2
T2 T a
L
k A
1
h A
R cond
R conv
h L
k
Bi ......(7.9)
The term, (h.L)/k, appearing on the RHS of
eqn. (7.9) is a dimensionless number, known
as ‘Biot number’.

Fig. 7.3(a) Biot number and temp. distribution in a plane wall
h, Ta
X
T1
T2
T2
T2
Ta
Bi << 1
Bi = 1
Bi >> 1
QconvQcond
L

• Note from Fig. (7.3, a) the temperature profile for
Bi << 1.
• It suggests that one can assume a uniform temperature
distribution within the solid if Bi << 1.
• Situation during transient conduction is shown in Fig.
(7.3,b). It may be observed hat temperature distribution
is a strong function of Biot number.
Fig. 7.3(b) Biot number and transient temp. distribution in a plane wall
h, Ta
Bi << 1 Bi = 1 Bi >> 1
T(x,0) = Ti T(x,0) = Ti

• For Bi << 1, temperature gradient in the solid is
small and temperature can be taken as a
function of time only.
• Note also that for Bi >> 1, temperature drop
across the solid is much larger than that across
the convective layer at the surface.
• Let us define Biot number, in general, as follows:
Bi
h L c

k
.....(7.10)

where, h is the heat transfer coeff. between
the solid surface and the surroundings, k is
the thermal conductivity of the solid, and Lc
is a characteristic length defined as the ratio
of the volume of the body to its surface area,
i.e.
Lc
V
A

• For solids such as a plane slab, long cylinder and
sphere, it is found that transient temperature distribution
within the solid at any instant is uniform, with the error
being less than about 5%, if the following criterion is
satisfied:
Bi
h Lc

k
0.1 ......(7.11)
Lc for common shapes:
(i) Plane wall (thickness 2L): L c
A 2 L
2 A
L = half thickness of wall
(ii) Long cylinder, radius, R: L c
 R
2 L
2  R L
R
2
(iii) Sphere, radius, R: L c
4
3
 R
3
4  R
2
R
3
(iv) Cube, side L: L c
L
3
6 L
2
L
6

• Therefore, we can write eqn. (7.3) as:

 i
T ( ) T a
T i T a
exp
h A 
 C p
 V
if Bi < 0.1.....(7.12)
Eqn. (7.12) is important.
Its application to a given problem is very simple and solution
of any transient conduction problem must begin with
examining if the criterion, Bi < 0.1 is satisfied to see if
eqn. (7.12) could be applied.
Now, the term (hA)/(Cp V) can be written as follows:
h A 
 C p
 V
h Lc

k
k 
 C p
 Lc
2

h Lc

k
 
Lc
2
 Bi Fo
where, Fo
 
Lc
2
= Fourier number, or relative time

• Fourier number, like Biot number, is an
important parameter in transient heat
transfer problems.
• It is also known as ‘dimensionless time’.
• Fourier number signifies the degree of
penetration of heating or cooling effect
through a solid.
• For small Fo, large  will be required to get
significant temperature changes.

• Now, we can rewrite eqn. (7.12) as:

 i
T ( ) T a
T i T a
exp Bi Fo( ) if Bi < 0.1.....(7.13)
Eqn. (7.13) is plotted in Fig. (7.4) below.
Remember that this graph is for the cases
where lumped system analysis is applicable,
i.e. Bi < 0.1.

Let X Bi Fo
X 0 0.1 5
Then

 i
exp X( )
exp X( )
X
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
1 10
3
0.01
0.1
1
Transient temp.distrib.in solids, Bi<0.1
Fig. (7.4) Dimensionless temperature distribution in solids during transient h
transfer, (Bi < 0.1)--for lumped system analysis

Response time of a thermocouple:
• Lumped system analysis is usefully
applied in the case of temperature
measurement with a thermometer or a
thermocouple. Obviously, it is desirable
that the thermocouple indicates the source
temperature as fast as possible.
• ‘Response time’ of a thermocouple is
defined as the time taken by it to reach the
source temperature.

Response time of a thermocouple:
• Consider eqn. (7.12):

 i
T ( ) T a
T i T a
exp
h A 
 C p
 V
if Bi < 0.1.....(7.12)
For rapid response, the term (h A )/( Cp V)
should be large so that the exponential term will
reach zero faster. This means that:
(i) increase (A/V), i.e. decrease the wire
diameter
(ii) decrease density and specific heat, and
(iii) increase the value of heat transfer coeff. h

• The quantity ( Cp V)/(h A) is known as ‘thermal time
constant’, t, of the measuring system and has units of
time.
• At  = t i.e. at a time interval of one time constant, we
have:
T ( ) T a
T i T a
e
1
0.368 .....(7.14)
From eqn. (7.14), it is clear that after an interval of time
equal to one time constant of the given temperature
measuring system, the temperature difference between the
body (thermocouple) and the source would be 36.8% of
the initial temperature difference. i.e. the temperature
difference would be reduced by 63.2%.

Time required by a thermocouple to attain
63.2% of the value of initial temperature
difference is called its ‘sensitivity’.
For good response, obviously the response
time of thermocouple should be low.
As a thumb rule, it is recommended that while
using a thermocouple to measure temperatures,
reading of the thermocouple should be taken
after a time equal to about four time periods has
elapsed.

• Example 7.1 (M.U.): A steel ball 5 cm
dia, initially at an uniform temperature
of 450 C is suddenly placed in an
environment at 100 C. Heat transfer
coeff. h, between the steel ball and the
fluid is 10 W/(m2.K). For steel, cp = 0.46
kJ/(kg.K), ρ = 7800 kg/m3, k = 35
W/(m.K). Calculate the time required for
the ball to reach a temperature of 150
C. Also find the rate of cooling after 1
hr. Show graphically how the temp. of
the sphere falls with time.

• First, calculate the Biot number:
Since Bi < 0.1, lumped system analysis is applicable, and the
temperature variation within the solid will be within an error of 5%.
Applying eqn. (7.12), we get:

 i
T ( ) T a
T i T a
exp
h A 
 C p
 V
if Bi < 0.1.....(7.12)
i.e.
T T a
T i T a
exp

t
where t is the time constant.

• And, time constant is given by:
t
 c p
 V
A h
 c p

h
R
3
 ..since for sphere, V/A = R/3
i.e. t
 c p

h
R
3
 ...define time constant, t
i.e. t 2990 s....time constant
Therefore, we write:
150 100
450 100
exp

2990
where is the time required to reach 150 C

i.e. ln
50
350

2990
or,  2990ln
50
350
 s....define, the time reqd. to reach 150 C
i.e.  5.81810
3
 s... time reqd. to reach 150 C...Ans.
i.e.  1.616 hrs.....Ans
Rate of cooling after 1 hr.:
i.e.  3600 s
From eqn. (7.12), we have:
T ( ) T i T a exp
h A 
 c p
 V
 T a ....define T()

i.e.
dT
d
T i T a
h A
 V c p

 exp
h A 
 V c p

 C/s.....rate of cooling

T ( )
d
d
0.035 C/s....rate of cooling after 1 hr..Ans.
i.e.
-ve sign indicates that as time increases, temperature falls.
To sketch the fall in temp. of sphere with time:
Temp. as a function of time is given by eqn. (7.12):

 i
T ( ) T a
T i T a
exp
h A 
 c p
 V
if Bi < 0.1.....(7.12)
i.e. T ( ) T a T i T a exp
h A 
 c p
 V
 ....eqn. (A)

• We will plot eqn. (A) against different times, :
T  3600( )

0 0.5 1 1.5 2 2.5 3 3.5 4
100
150
200
250
300
350
400
450
500
Trans. cooling of sphere-lumped system
 in hrs. and T()
in deg.C
Fig. Ex. 7.1 Transient cooling of a sphere considred as a lumped system
Note from the fig. how the cooling progresses with time.
After about 4 hrs. duration, the sphere approaches the temp.
of the ambient.
You can also verify from the graph that the time required for the
sphere to reach 150 C is 1.616 hrs, as calculated earlier.

• Example 7.4 (M.U.): A Thermocouple (TC)
junction is in the form of 8 mm sphere.
Properties of the material are: cp = 420 J/(kg.
K), ρ = 8000 kg/m3, k = 40 W/(m.K), and heat
transfer coeff., h = 45 W/(m2.K). Find, if the
junction is initially at a temp. of 28 C and
inserted in a stream of hot air at 300 C:
(i) the time const. of the TC
(ii) The TC is taken out from the hot air after
10 s and kept in still air at 30 C. Assuming ‘h’
in air as 10 W/(m2.K), find the temp. attained
by the junction 15 s after removing from hot
air stream.

• First, calculate the Biot number:
Bi
h L c

k
h
k
V
A
 h
k
4
3
( ) R
3
4  R
2

i.e. Bi
h
k
R
3
 ...define Biot number
i.e. Bi 1.5 10
3
 ...Biot number
Since Bi < 0.1, lumped system analysis is applicable, and the
temperature variation within the solid will be within an error of 5%.
See Fig. Ex. 7.4 (a).
Time constantis given by:
t
 c p
 V
A h
 c p

h
R
3
 ..since for sphere, V/A = R/3
i.e. t
 c p

h
R
3
i.e. t 99.556 s....time constant.....Ans.

Fig. Ex.7.4 (a) Temperature measurement, with thermocouple
placed in the air stream
Ta = 300 C
h = 45 W/(m2
.K)
Thermocouple, D = 8 mm
Ti = 28 C
Air

• Temp. of TC after 10s:
 10 s....time duration for which TC is kept in the stream at 300 C
We use eqn. (7.12). i.e.

 i
T ( ) T a
T i T a
exp
h A 
 C p
 V
if Bi < 0.1.....(7.12)
i.e.
T T a
T i T a
exp

t
where t is the time constant.
Therefore, T T i T a exp

t
 T a C....define temp. of TC after 10 s in the
stream
i.e. T 53.994 C....temp. of TC 10 s after it is placed in the stream at 300
(b) Now, this TC is removed from the stream at 300 C and placed
in still air at 30 C. So, the temp. of 53.994C becomes initial temp. Ti
for this case:

i.e. new Ti: T i 53.994 C....initial temp. when the TC is placed in still
And, new:  15 s....duration for which T C is kept in still air
And, new Ta: T a 30 C....new temp. of ambient
And, new h: h 10 W/(m2.K)....heat tr. coeff. in still air
See Fig. Ex. 7.4 (b).
Fig. Ex.7.4 (b) Temperature measurement, with thermocouple
placed in still air
Ta = 30 C
h = 10 W/(m2
.K)
Thermocouple, D = 8 mm
Ti = 53.994 C
Still air

And, new time constant:
i.e. t
 c p

h
R
3
i.e. t 448 s....time constant
Therefore, T T i T a exp

t
 T a C....define temp. of TC after 15 s in still a
i.e. T 53.204 C....temp. of TC 15 s after it is placed in still air
at 30 C...Ans.

One-dimensional transient conduction in large
plane walls, long cylinders and spheres when
Biot number > 0.1:
• When the temperature gradient in the solid is not
negligible (i.e. Bi > 0.1), lumped system analysis
is not applicable.
• One term approximation solutions:
• Fig. 7.6 shows schematic diagram and
coordinate systems for a large, plane slab, long
cylinder and a sphere.
• Consider a plane slab of thickness 2L, shown in
Fig (a) above. Initially, i.e. at  = 0, the slab is at
an uniform temperature, Ti.

One-dimensional transient conduction in large
plane walls, long cylinders and spheres when
Biot number > 0.1:
• Suddenly, at  = 0, both the surfaces of the slab
are subjected to convection heat transfer with an
ambient at temperature Ta , with a heat transfer
coeff. h, as shown.
• Because of symmetry, we need to consider only
half the slab, and that is the reason why we
chose the origin of the coordinate system on the
mid-plane.
• Then, we can write the mathematical formulation
of the problem for plane slab as follows:

d
2
T
dx
2
1

dT
d
 in 0<x <L, for > 0.......(7.23, a)
dT
dx
0 at x = 0, for > 0.......(7.23, b)
k
dT
dx
 h T T a
 at x = L, for > 0 ......(7.23, c)
T T i for = 0, in 0 < x < L .....(7.23, d)
The solution of the above problem, however, is rather
involved and consists of infinite series. So, it is more
convenient to present the solution either in tabular form or
charts.
For this purpose, we define the following dimensionless parameters:

• While using the tabular or chart solutions, note one
important difference in defining Biot number:
• Characteristic length in Biot number is taken as half
thickness L for a plane wall, Radius R for a long
cylinder and sphere instead of being calculated as
V/A, as done in lumped system analysis.
(i) Dimensionless temperature:  x ( )
T x ( ) T a
T i T a
(ii) Dimensionless distance from the centre: X
x
L
(iii) Dimensionless heat transfer coefficient: Bi
h L
k
...Biot number
(iv) Dimensionless time:
 

Fo
 
L
2
....Fourier number

• It is observed that for Fo > 0.2, considering only the first
term of the series and neglecting other terms, involves
an error of less than 2%.
• Generally, we are interested in times, Fo > 0.2.
• So, it becomes very useful and convenient to use one
term approximation solution, for all these three cases,
as follows:
Plane wall:  x ( )
T x ( ) T a
T i T a
A 1 e
 1
2
Fo
 cos
 1 x
L
 ...Fo > 0.2....(7.24, a)
Long cylinder:  x ( )
T r ( ) T a
T i T a
A 1 e
 1
2
Fo
 J 0
 1 r
R
 ...Fo > 0.2....(7.24, b)
sphere:  x ( )
T r ( ) T a
T i T a
A 1 e
 1
2
Fo

sin
 1 r
R
 1 r
R
 ...Fo > 0.2....(7.24, c)

• In the above equations, A1 and λ1 are functions of Biot
number only.
• A1 and λ1 are calculated from the following relations:
(Values are available in Tables, Ref: Cengel)
Functions J0 and J1are the zeroth and first order Bessel functions of
the first kind, available from Handbooks.

• Now, at the centre of a plane wall, cylinder and sphere,
we have the condition x = 0 or r = 0.
• Then, noting that cos(0) = 1, J0 (0) = 1, and limit of
sin(x)/x is also 1, eqns. (7.24) become:
• At the centre of plane wall, cylinder and sphere:
Centre of plane wall:
(x = 0)
 0
T 0 T a
T i T a
A 1 e
 1
2
Fo
 ....(7.25, a)
Centre of long cylinder:
(r = 0)
 0
T 0 T a
T i T a
A 1 e
 1
2
Fo
 ...(7.25, b)
Centre of sphere:
(r = 0)
 0
T 0 T a
T i T a
A 1 e
 1
2
Fo
 ...(7.25, c)

• The one-term solutions are presented in chart
form in the next section.
• But, generally, it is difficult to read these charts
accurately.
• So, relations given in eqns. (7.24) and (7.25)
along with Tables for A1 and λ1 should be
preferred to the charts.
Therefore, first step in the solution is to
calculate the Biot number;
• Once the Biot number is known, constants A1
and λ1 are found out from Tables and then use
relations given in eqns. (7.24) and (7.25) to
calculate the temperature at any desired
location.

• Calculation of amount of heat
transferred, Q:
• Let Q be the amount of heat lost (or gained) by
the body, during the time interval  = 0 to  = ,
i.e. from the beginning upto a given time.
• Let Qmax be the maximum possible heat transfer.
• Obviously, maximum amount of heat has
been transferred when the body has
reached equilibrium with the ambient.

• i.e.
Q max  V Cp
 T i T a
 m Cp
 T i T a
 J.....(7.26)
where  is the density, V is the volume, (V) is the mass,
Cp is the specific heat of the body.
Based on the one term approximation discussed above,
(Q/ Qmax) is calculated for the three cases, from the
following:
Plane wall:
Q
Q max
1  0
sin  1
 1
 ......(7.27, a)
Cylinder:
Q
Q max
1 2  0

J 1  1
 1
 .....(7.27, b)
Sphere:
Q
Q max
1 3  0

sin  1  1 cos  1

 1
3
 .....(7.27, c)

• Note:
• (i) Remember well the definition of Biot
number- i.e. Bi = (hL/k), where L is half
thickness of the slab, and Bi = (hR/k), where
R is the outer radius of the cylinder or the
sphere.
• (ii) Foregoing results are equally applicable to
a plane wall of thickness L, insulated on one
side and suddenly subjected to convection at
the other side. This is so because, the
boundary condition dT/dx = 0 at x = 0 for the
mid-plane of a slab of thickness 2L (see eqn.
7.23, b), is equally applicable to a slab of
thickness L, insulated at x = 0.

• Note (contd.):
• (iii) These results are also applicable to
determine the temperature response of a
body when temperature of its surface is
suddenly changed to Ts . This case is
equivalent to having convection at the
surface with a heat transfer coeff., h = ∞;
now, Ta is replaced by the prescribed
surface temperature, Ts.
• Remember that these results are valid
for the situation where Fourier number,
Fo > 0.2.

Heisler and Grober charts:
• The one term approximation solutions
(eqn. (7.25)) were represented in graphical
form by Heisler in 1947. They were
supplemented by Grober in 1961, with
graphs for heat transfer (eqn. (7.27)).
• These graphs are shown in following
slides for plane wall, long cylinder and a
sphere, respectively.

Heisler and Grober charts:
• About the charts:
• First chart in each of these figures gives the non-
dimensionalised centre temperature T0. i.e. at x
= 0 for the slab of thickness 2L, and at r = 0 for
the cylinder and sphere, at a given time .
• Temperature at any other position at the same
time , is calculated using the next graph, called
‘position correction chart’.
• Third chart gives Q/Qmax.

Temperature of Plate at mid-plane:

Temperature of Plate at any plane:

Heat transfer for Plate:

Temperature of Cylinder at mid-point:

Temperature of Cylinder at any radius:

Heat transfer for Cylinder:

Temperature of Sphere at mid-point:

Temperature of Sphere at any radius:

Heat transfer for Sphere:

How to use these charts?
• Procedure of using these charts to solve
a numerical problem is as follows:
• First of all, calculate Bi from the data, with the
usual definition of Bi i.e. Bi = (h.Lc)/k, where Lc
is the characteristic dimension, given as: Lc =
(V/A).
• i.e. Lc = L, half-thickness for a plane wall, Lc =
R/2 for a cylinder, and Lc=R/3 for a sphere.
• If Bi < 0.1, use lumped system analysis;
otherwise, go for one term approx. or chart
solution.

How to use these charts?
• If Bi > 0.1, calculate the Biot number
again with the appropriate definition of Bi
i.e. Bi =(hL/k) for a plane wall where L is
half-thickness, and Bi = (hR/k) for a
cylinder or sphere, where R is the outer
radius. Also, calculate Fourier number,
Fo = ./L2 for the plane wall, and Fo =
./R2 for a cylinder or sphere.
• To calculate the centre temperature, use
the first chart in the set given, depending
upon the geometry being considered.

• Fo is on the x-axis; draw a vertical line to intersect the
(1/Bi) line; from the point of intersection, move
horizontally to the left to the y-axis to read the value of
o = (To –Ta)/(Ti – Ta). Here, To is the centre
temperature, which can now be calculated since Ti
and Ta are known.
• To calculate the temperature at any other position, use
second fig. of the set, as appropriate:
• Enter the chart with 1/Bi on the x-axis, move vertically
up to intersect the (x/L) (or (r/R)) curve, and from the
point of intersection, move to the left to read on the y-
axis, the value of  = (T –Ta)/(To – Ta). Here, T is the
desired temperature at the indicated position.

• To find out the amount of heat transferred Q,
during a particular time interval  from the
beginning (i.e.  = 0), use the third fig. of the
set, depending upon the geometry.
• Enter the x-axis with the value of (Bi2 .Fo) and
move vertically up to intersect the curve
representing the appropriate Bi, and move to
the left to read on the y-axis, the value of
Q/Qmax.
• Q is then easily found out since Qmax =
m.Cp.(Ti – Ta).
• And, Q = (Q/Qmax ). Qmax .

• Note the following in connection with these
charts:
• These charts are valid for Fourier number Fo >
0.2
• Note from the ‘position correction charts’ that at
Bi < 0.1 (i.e. 1/Bi > 10), temperature within the
body can be taken as uniform, without
introducing an error of more than 5%. This was
precisely the condition for application of ‘lumped
system analysis’.
• It is difficult to read these charts accurately,
since logarithmic scales are involved; also, the
graphs are rather crowded with lines. So, use of
one term approximation with tabulated values of
A1 and 1 should be preferred.

• Example 7.7: A steel plate ( = 1.2 x 10-5
m2/s, k = 43 W/(m.C)), of thickness 2L = 10
cm, initially at an uniform temperature of
250 C is suddenly immersed in an oil bath
at Ta = 45 C. Convection heat transfer
coeff. between the fluid and the surfaces is
700 W/(m2.C).
• How long will it take for the centre plane to
cool to 100 C?
• What fraction of the energy is removed
during this time?
• Draw the temp. profile in the slab at
different times.

• Data:

• It is noted that Biot number is > 0.1; so, lumped system analysis is
not applicable.
• We will adopt Heisler chart solution and then check the results from
one term approximation solution.
• To find the time reqd. for the centre to reach 100 C:
• For using the charts, Bi = hL/k, which is already calculated.
Fourier number: Fo
 
L
2

• Now, with this value of , enter the y-axis of Fig. (7.7,a). Move
horizontally to intersect the 1/Bi = 1.229 line; from the point of
intersection, move vertically down to x-axis to read Fo = 2.4.

• Surface temperature:
• At the surface, x/L =1. Enter
Fig. (7.7, b) on the x-axis
with a value of 1/Bi = 1.229,
move up to intersect the
curve of x/L = 1, then move
to left to read on y-axis the
value of Θ = 0.7

• Fraction of max. heat transferred, Q/Qmax:
• We will use Grober's chart, Fig. (7.7, c):
• We need Bi2Fo to enter the x-axis:
We get: Bi
2
Fo 1.59
With the value of 1.59, enter the x-axis of Fig. (7.7, c), move vertically
up to intersect the curve of Bi = 0.814, then move horizontally to
read Q/Qmax = 0.8

i.e. from Fig. (7.7, c), we get:
Q
Q max
0.8
i.e. 80% of the energy is remov ed by the time the centre temp. has reached
100 C.....Ans.

• Verify by one term approximation solution:
and
Therefore, eqn. (7.25, a) becomes:

Compare this value with the one got from Heisler's chart,
viz. 500 s. The error is in reading the chart.

Compare this with the value of 83.5 C obtained earlier.

Fraction of max. heat transferred, Q/Qmax:
i.e. 75.9% of the energy is removed by the time the
centre temp. has reached 100 C.....Ans.
Compare this with the value of 80% obtained earlier;
again, the error is in reading the charts.

• To draw temp. profile in the plate at different
times:
• We have, for temp. distribution at any location:
Plane wall:  x ( )
T x ( ) T a
T i T a
A 1 e
 1
2
Fo
 cos
 1 x
L
 ...Fo > 0.2....(7.24, a)
And, Centre of plane wall:
(x = 0)
 0
T 0 T a
T i T a
A 1 e
 1
2
Fo
 ....(7.25, a)
Fourier number as a function of: Fo ( )
 
L
2
...for slab

• By writing Fourier no. as a function of , and incuding it
in eqn. (A) below as shown, it is ensured that for each
new , the corresponding new Fo is calculated.
Then, T x ( ) T a T i T a A 1 e
 1
2
Fo ( )
 x 0if
T a T i T a A 1 e
 1
2
Fo ( )
 cos
 1 x
L
 otherwise
.....eqn. (A)

• Note that this
graph shows
temp.
distribution for
one half of the
plate; for the
other half, the
temp.
distribution will
be identical.
• (ii) See from the
fig. how cooling
progresses with
time. After a
time period of 25
min. the
temperatures in
the plate are
almost uniform
at 45 C.

• Example 7.8: A long, 15 cm diameter cylindrical shaft
made of stainless steel 304 (k = 14.9 W/(m.C), r = 7900
kg/m3, Cp = 477 J/(kg.C), and a = 3.95 x 10-6 m2/s),
comes out of an oven at an uniform temperature of 450
C. The shaft is then allowed to cool slowly in a chamber
at 150 C with an average heat transfer coeff. of 85
W/(m2.C).
• (i) Determine the temperature at the centre of the shaft
25 min. after the start of the cooling process.
• (ii) Determine the surface temp.at that time, and
• (iii) Determine the heat transfer per unit length of the
shaft during this time period.
• (iv) draw the temp. profile along the radius for different
times

Working with Heisler Charts is left a an exercise to students.
However, we shall solve the problem with one-term
approximation formulas for a cylinder:

Note:
(i) see from the fig. how
cooling progresses
with time. After a time
period of 2 hrs. the
temperatures along
the radius are almost
uniform, but is yet to
reach ambient temp.
of 150 C.
(ii) observe that after 25
min. temp. at the
centre (r = 0) is 296.7
C and at the surface (
r = 0.075 m), the
temp. is 269.9 C as
already calculated.

Transient heat-conduction-Part-I

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