In this book, we solve the heat equation partial differential equation by first transforming it into an integral equation. We use exponential temperature profiles which satisfy the boundary conditions and also the initial condition. We also look at cases where ther is natural convection and go ahead and solve for both the transient and steady state solution. We also go ahead an solve the heat equation in cylindrical coordinates. We explain alot of phenomena observed experimentally for example the melting of wax on the sides of a metal rod when heat is applied on one end. For updated information about heat flow, follow the link below: https://www.slideshare.net/Wasswaderrick3/analytic-solutions-to-the-heat-equation-using-integral-methods-with-experimental-resultspdf

1. THE ANALYTICAL SOLUTION TO THE HEAT EQUATION USING AN INTEGRAL METHOD
By Wasswa Derrick

2. Contact
wasswaderricktimothy7@gmail.com

3. TABLE OF CONTENTS
SEMI INFINITE WALL ANALYTICAL SOLUTION TO THE HEAT EQUATION................3
ALTERNATIVE SOLUTION TO THE SEMI-INFINITE WALL PROBLEM ...........................5
HOW DO WE EXPLAIN THE EXISTENCE OF THE FOURIER LAW IN STEADY
STATE?......................................................................................................................................................8
HOW DO WE DEAL WITH CONVECTION AT THE SURFACE AREA OF THE METAL
ROD...........................................................................................................................................................11
EQUAL FIXED TEMPERATURES AT THE END OF AN INSULATED METAL ROD....14
UNEQUAL FIXED TEMPERATURES AT THE END OF AN INSULATED METAL ROD.
....................................................................................................................................................................17
HOW DO WE DEAL WITH OTHER TYPES OF BOUNDARY CONDITIONS?..................20
HOW DO WE DEAL WITH CONVECTION AT THE SURFACE AREA OF THE METAL
ROD FOR FIXED END TEMPERATURE......................................................................................22
WHAT HAPPENS WHEN THE INITIAL TEMPERATURE IS A FUNCTION OF X? .......31
HOW DO WE DEAL WITH CYLINDRICAL CO-ORDINATES?..............................................34

4. SEMI INFINITE WALL ANALYTICAL SOLUTION TO THE
HEAT EQUATION.
The differential equation to be solved is
𝜕𝑇
𝜕𝑡
= 𝛼
𝜕2
𝑇
𝜕𝑥2
Where the initial and boundary conditions are
𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒕
𝑻 = 𝑻∞ 𝒂𝒕 𝒙 = ∞ 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒕
𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒙
We postulate:
𝑌 =
𝑇 − 𝑇∞
𝑇𝑠 − 𝑇∞
And
𝜂 =
𝑥
2√𝛼𝑡
We get
𝑑2
𝑌
𝑑𝜂2
+ 2𝜂
𝑑𝑌
𝑑𝜂
= 0 (1)
With the transformed boundary and initial conditions
𝑌 → 0 𝑎𝑠 𝜂 → ∞
And
𝑌 = 1 𝑎𝑡 𝜂 = 0
The first condition is the same as the initial condition 𝑇 = 𝑇∞ 𝑎𝑡 𝑡 = 0 and the
boundary condition
𝑇 → 𝑇∞ 𝑎𝑠 𝑥 → ∞
Equation 1 may be integrated once to get
𝑙𝑛
𝑑𝑌
𝑑𝜂
= 𝑐1 − 𝜂2
𝑑𝑌
𝑑𝜂
= 𝑐2𝑒−𝜂2

5. And integrated once more to get
𝑌 = 𝑐3 + 𝑐2 ∫ 𝑒−𝜂2
𝑑 𝜂
Applying the boundary conditions to the equation, we get
𝑌 = 1 − erf (
𝑥
2√𝛼𝑡
)
𝑻 − 𝑻∞
𝑻𝒔 − 𝑻∞
= 𝟏 − 𝐞𝐫𝐟 (
𝒙
𝟐√𝜶𝒕
)
Or
𝑻𝒔 − 𝑻
𝑻𝒔 − 𝑻∞
= 𝐞𝐫𝐟 (
𝒙
𝟐√𝜶𝒕
)

6. ALTERNATIVE SOLUTION TO THE SEMI-INFINITE WALL PROBLEM
The problem of the semi-infinite wall could also be solved as below:
Given the boundary and initial conditions
𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒕
𝑻 = 𝑻∞ 𝒂𝒕 𝒙 = ∞
𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎
And the governing equation
𝜕𝑇
𝜕𝑡
= 𝛼
𝜕2
𝑇
𝜕𝑥2
We assume an exponential temperature profile that satisfies the boundary
conditions:
𝑇 − 𝑇∞
𝑇𝑠 − 𝑇∞
= 𝑒
−𝑥
𝛿
We can satisfy the initial condition if we assume that 𝛿 will have a solution as
𝛿 = 𝑐𝑡𝑛
Where c and n are constants so that at 𝑡 = 0, 𝛿 = 0 and the initial condition is
satisfied as shown below.
𝑇 − 𝑇∞
𝑇𝑠 − 𝑇∞
= 𝑒
−𝑥
0 = 𝑒−∞
= 0
𝑇 = 𝑇∞ 𝑎𝑡 𝑡 = 0
We then transform the heat governing equation into an integral equation as:
𝛼 ∫ (
𝜕2
𝑇
𝜕𝑥2
) 𝑑𝑥
𝑙
0
= ∫
𝜕𝑇
𝜕𝑡
𝑑𝑥
𝑙
0
𝛼 ∫ (
𝜕2
𝑇
𝜕𝑥2
) 𝑑𝑥
𝑙
0
=
𝜕
𝜕𝑡
∫ 𝑇𝑑𝑥
𝑙
0
You notice that the integral
𝜕
𝜕𝑡
∫ 𝑇𝑑𝑥
𝑙
0
=
𝜕
𝜕𝑡
∫ (𝑇 − 𝑇∞)𝑑𝑥 =
𝜕
𝜕𝑡
∫ 𝑇𝑑𝑥
𝑙
0
−
𝜕(𝑙𝑇∞)
𝜕𝑡
𝑙
0
Since 𝑙 and 𝑇∞ are constants independent of time
𝜕(𝑙𝑇∞)
𝜕𝑡
= 0

7. So
𝜕
𝜕𝑡
∫ 𝑇𝑑𝑥
𝑙
0
=
𝜕
𝜕𝑡
∫ (𝑇 − 𝑇∞)𝑑𝑥
𝑙
0
We go ahead and find
𝜕2
𝑇
𝜕𝑥2
=
(𝑇𝑠 − 𝑇∞)
𝛿2
∫ (
𝜕2
𝑇
𝜕𝑥2
) 𝑑𝑥
𝛿
0
= −
(𝑇𝑠 − 𝑇∞)
𝛿
(𝑒
−𝑙
𝛿 − 1) =
(𝑇𝑠 − 𝑇∞)
𝛿
(1 − 𝑒
−𝑙
𝛿 )
𝜕
𝜕𝑡
∫ (𝑇 − 𝑇∞)𝑑𝑥
𝛿
0
=
𝑑𝛿
𝑑𝑡
[(𝑇𝑠 − 𝑇∞) (1 − 𝑒
−𝑙
𝛿 )]
Substituting into the integral equation, we get
𝛼
(𝑇𝑠 − 𝑇∞)
𝛿
(1 − 𝑒
−𝑙
𝛿 ) =
𝑑𝛿
𝑑𝑡
(1 − 𝑒
−𝑙
𝛿 )
The boundary conditions are
𝛿 = 0 𝑎𝑡 𝑡 = 0
We find
𝛿 = √2𝛼𝑡
We substitute in the temperature profile and get
𝑻 − 𝑻∞
𝑻𝒔 − 𝑻∞
= 𝒆
−𝒙
√𝟐𝜶𝒕
You notice that the initial condition is satisfied by the temperature profile
above i.e.,
At 𝑡 = 0
𝑻 − 𝑻∞
𝑻𝒔 − 𝑻∞
= 𝒆
−𝒙
√𝟐𝜶𝒕
Becomes
𝑻 − 𝑻∞
𝑻𝒔 − 𝑻∞
= 𝒆
−𝒙
𝟎 = 𝒆−∞
= 𝟎
Hence 𝑻 = 𝑻∞ throughout the rod at 𝑡 = 0

8. Observation.
The two equations
𝑻 − 𝑻∞
𝑻𝒔 − 𝑻∞
= 𝒆
−𝒙
√𝟐𝜶𝒕
And
𝑻 − 𝑻∞
𝑻𝒔 − 𝑻∞
= 𝟏 − 𝐞𝐫𝐟 (
𝒙
𝟐√𝜶𝒕
)
Should give the same answer. Indeed, they give answers that are the same with
a small error since the error function is got from tables after rounding off yet in
the exponential temperature profile there is no rounding off.

9. HOW DO WE EXPLAIN THE EXISTENCE OF THE
FOURIER LAW IN STEADY STATE?
The Fourier law states:
𝑄 = −𝑘𝐴
𝜕𝑇
𝜕𝑥
Under steady state.
It can be stated as:
𝜕𝑇
𝜕𝑥
= −
𝑄
𝑘𝐴
Under steady state.
To satisfy the Fourier law under steady state, we postulate the temperature
profile to be:
𝑻 − 𝑻∞ =
𝑸
𝒌𝑨
𝜹𝒆
−𝒙
𝜹
𝛿 is a function of time 𝑡 and not distance 𝑥
We believe that after solving for 𝛿, 𝛿 will be directly proportional to time t so
that 𝛿 = 𝑘𝑡𝑛
sothat at 𝑡 = ∞ , 𝛿 = ∞
And taking the first derivative of temperature with distance x at 𝑡 = ∞ , we get
𝜕𝑇
𝜕𝑥
|𝑡=∞ = −
𝑄
𝑘𝐴
𝑒
−𝑥
𝛿 = −
𝑄
𝑘𝐴
𝑒
−𝑥
∞ = −
𝑄
𝑘𝐴
𝑒0
𝝏𝑻
𝝏𝒙
= −
𝑸
𝒌𝑨
Hence the Fourier law is satisfied.
Now let us go ahead and solve for 𝛿.
Recall
PDE
𝜕𝑇
𝜕𝑡
= 𝛼
𝜕2
𝑇
𝜕𝑥2
The initial condition is
𝑇 = 𝑇∞ 𝑎𝑡 𝑡 = 0
The boundary conditions are

10. 𝑇 = 𝑇∞ 𝑎𝑡 𝑥 = ∞
𝜕𝑇
𝜕𝑥
|𝑥=0 = −
𝑄
𝑘𝐴
The PDE is
The temperature profile that satisfies the conditions above is
𝑇 − 𝑇∞ =
𝑄
𝑘𝐴
𝛿𝑒
−𝑥
𝛿
We transform the PDE into an integral equation
𝜕
𝜕𝑡
∫ (𝑇 − 𝑇∞)𝑑𝑥
𝛿
0
= 𝛼 ∫ (
𝜕2
𝑇
𝜕𝑥2
) 𝑑𝑥
𝛿
0
And using the temperature profile, we get
∫ (
𝜕2
𝑇
𝜕𝑥2
) 𝑑𝑥
𝛿
0
=
𝑄
𝑘𝐴
(1 − 𝑒
−𝑙
𝛿 )
𝜕
𝜕𝑡
∫ (𝑇 − 𝑇∞)𝑑𝑥
𝛿
0
=
𝑄
𝑘𝐴
𝛿2
(1 − 𝑒
−𝑙
𝛿 )
We then substitute into the integral equation
𝛼
𝑄
𝑘𝐴
(1 − 𝑒
−𝑙
𝛿 ) = 2𝛿
𝑑𝛿
𝑑𝑡
(
𝑄
𝑘𝐴
(1 − 𝑒
−𝑙
𝛿 ))
The boundary conditions are
𝛿 = 0 𝑎𝑡 𝑡 = 0
𝛿 = √𝛼𝑡
Substituting into the temperature profile, we get
𝑇 − 𝑇∞ =
𝑄
𝑘𝐴
𝛿𝑒
−𝑥
𝛿
𝑻 − 𝑻∞ =
𝑸
𝒌𝑨
× √𝜶𝒕 × 𝒆
−𝒙
√𝜶𝒕
You notice that the initial condition is satisfied

11. 𝝏𝑻
𝝏𝒙
|𝒕=∞ = −
𝑸
𝒌𝑨
Hence the Fourier law
So, our assumption of 𝛿 = 𝑘𝑡𝑛
is satisfied

12. HOW DO WE DEAL WITH CONVECTION AT THE
SURFACE AREA OF THE METAL ROD
Recall that the temperature profile that satisfies the Fourier law was
𝑇 − 𝑇∞ =
𝑄
𝑘𝐴
𝛿𝑒
−𝑥
𝛿
Recall
PDE
𝛼
𝜕2
𝑇
𝜕𝑥2
−
ℎ𝑃
𝐴𝜌𝐶
(𝑇 − 𝑇∞) =
𝜕𝑇
𝜕𝑡
The initial condition is
𝑇 = 𝑇∞ 𝑎𝑡 𝑡 = 0
The boundary conditions are
𝑇 = 𝑇∞ 𝑎𝑡 𝑥 = ∞
𝜕𝑇
𝜕𝑥
|𝑥=0 = −
𝑄
𝑘𝐴
We transform the PDE into an integral equation
𝛼 ∫ (
𝜕2
𝑇
𝜕𝑥2
) 𝑑𝑥
𝑙
0
−
ℎ𝑃
𝐴𝜌𝐶
∫ (𝑇 − 𝑇∞)𝑑𝑥
𝑙
0
=
𝜕
𝜕𝑡
∫ (𝑇 − 𝑇∞)𝑑𝑥
𝑙
0
Where:
We are dealing with a cylindrical metal rod.
𝑃 = 2𝜋𝑟 𝑎𝑛𝑑 𝜌 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑚𝑒𝑡𝑎𝑙 𝑟𝑜𝑑
And using the temperature profile, we get
∫ (
𝜕2
𝑇
𝜕𝑥2
) 𝑑𝑥
𝑙
0
=
𝑄
𝑘𝐴
(1 − 𝑒
−𝑙
𝛿 )
∫ (𝑇 − 𝑇∞)𝑑𝑥
𝑙
0
=
𝑄
𝑘𝐴
𝛿2
(1 − 𝑒
−𝑙
𝛿 )
Substituting in the integral equation above, we get
𝛼 −
ℎ𝑃
𝐴𝜌𝐶
𝛿2
= 2𝛿
𝑑𝛿
𝑑𝑡
The boundary condition is

13. 𝛿 = 0 𝑎𝑡 𝑡 = 0
We solve and get
𝛿 = √
𝐴𝜌𝐶𝛼
ℎ𝑃
(1 − 𝑒
−ℎ𝑃𝑡
𝐴𝜌𝐶 )
Substituting in the temperature profile, we get
𝑇 − 𝑇∞ =
𝑄
𝑘𝐴
𝛿𝑒
−𝑥
𝛿
𝑻 − 𝑻∞ =
𝑸
𝒌𝑨
× √
𝑨𝝆𝑪𝜶
𝒉𝑷
(𝟏 − 𝒆
−𝒉𝑷𝒕
𝑨𝝆𝑪 ) × 𝒆
−𝒙
√𝑨𝝆𝑪
𝒉𝑷
(𝟏−𝒆
−𝒉𝑷𝒕
𝑨𝝆𝑪 )
We notice that the initial condition and boundary conditions are satisfied.
For small time the term
ℎ𝑃𝑡
𝐴𝜌𝐶
≪ 1
And using binomial approximation of the exponential, we get
𝑒
−ℎ𝑃𝑡
𝐴𝜌𝐶 = 1 −
ℎ𝑃𝑡
𝐴𝜌𝐶
Then
(1 − 𝑒
−ℎ𝑃𝑡
𝐴𝜌𝐶 ) =
ℎ𝑃𝑡
𝐴𝜌𝐶
Upon substitution in the temperature profile, we get
𝑇 − 𝑇∞ =
𝑄
𝑘𝐴
× √𝛼𝑡 × 𝑒
−𝑥
√𝛼𝑡
Upon rearranging, we get
𝑥
√𝛼𝑡
= ln (
𝑄
𝑘𝐴
√𝛼𝑡) − ln (𝑇 − 𝑇∞)
𝑥
√𝑡
= √𝛼ln(√𝑡) + √𝛼 [ln (
𝑄
𝑘𝐴
√𝛼) − ln(𝑇 − 𝑇∞)]
What we observe is

14. 𝒙
√𝒕
= √𝜶𝐥𝐧(√𝒕) + √𝜶 [𝐥𝐧 (
𝑸
𝒌𝑨√𝜶
(𝑻 − 𝑻∞)
)]
That is what we observe for short times.
When the times become big, we observe
𝑻 − 𝑻∞ =
𝑸
𝒌𝑨
× √
𝑨𝝆𝑪𝜶
𝒉𝑷
(𝟏 − 𝒆
−𝒉𝑷𝒕
𝑨𝝆𝑪 ) × 𝒆
−𝒙
√𝑨𝝆𝑪
𝒉𝑷
(𝟏−𝒆
−𝒉𝑷𝒕
𝑨𝝆𝑪 )
And in steady state (𝑡 = ∞), we observe
𝑻 − 𝑻∞ =
𝑸
𝒌𝑨
× √
𝑨𝝆𝑪𝜶
𝒉𝑷
× 𝒆
−𝒙
√𝑨𝝆𝑪𝜶
𝒉𝑷
𝛼 =
𝑘
𝜌𝐶
We finally get
𝑻 − 𝑻∞ =
𝑸
𝒌𝑨
× √
𝒌𝑨
𝒉𝑷
× 𝒆
−𝒙
√𝒌𝑨
𝒉𝑷
the heat flow in steady state is given by:
𝜕𝑇
𝜕𝑥
= −
𝑄
𝑘𝐴
𝑒
−𝑥
√𝑘𝐴
ℎ𝑃
−𝒌𝑨
𝝏𝑻
𝝏𝒙
= 𝑸𝒆
−𝒙
√𝒌𝑨
𝒉𝑷

15. EQUAL FIXED TEMPERATURES AT THE END OF AN
INSULATED METAL ROD.
PDE
𝜕𝑇
𝜕𝑡
= 𝛼
𝜕2
𝑇
𝜕𝑥2
BCs
𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 𝟎 < 𝒕 < ∞
𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝒍 𝟎 < 𝒕 < ∞
IC
𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎 𝟎 ≤ 𝒙 ≤ 𝒍
we know a Fourier series solution exists given by
𝑻 − 𝑻𝒔
𝑻∞ − 𝑻𝒔
=
𝟒
𝝅
∑
𝟏
𝒏
∞
𝒏=𝟏
𝒔𝒊𝒏 (
𝒏𝝅𝒙
𝒍
) 𝒆
−(
𝒏𝝅
𝟐 )
𝜶𝒕
(
𝒍
𝟐
)𝟐
𝒏 = 𝟏, 𝟑, 𝟓, …
You notice that this solution is not entirely deterministic since it involves
summing terms up to infinity.
There is an alternative solution as shown below:
𝜕𝑇
𝜕𝑡
= 𝛼
𝜕2
𝑇
𝜕𝑥2
BCs
𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 𝟎 < 𝒕 < ∞
𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝒍 𝟎 < 𝒕 < ∞
IC
𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎 𝟎 ≤ 𝒙 ≤ 𝒍
We assume an exponential temperature profile that satisfies the boundary
conditions:
𝑇 − 𝑇∞
𝑇𝑠 − 𝑇∞
= 𝑒
−𝑥
𝛿
(1−
𝑥
𝑙
)

16. You notice that the temperature profile above satisfies the boundary
conditions. We can satisfy the initial condition if we assume that 𝛿 will assume
a solution as
𝛿 = 𝑐𝑡𝑛
Where c and n are constants so that at 𝑡 = 0, 𝛿 = 0 and the initial condition is
satisfied as shown below.
𝑇 − 𝑇∞
𝑇𝑠 − 𝑇∞
= 𝑒
−𝑥
0 = 𝑒−∞
= 0
𝑇 = 𝑇∞ 𝑎𝑡 𝑡 = 0
We transform the PDE into an integral equation
𝜕
𝜕𝑡
∫ (𝑇 − 𝑇∞)𝑑𝑥
𝑙
0
= 𝛼 ∫ (
𝜕2
𝑇
𝜕𝑥2
) 𝑑𝑥
𝑙
0
∫ (
𝜕2
𝑇
𝜕𝑥2
) 𝑑𝑥
𝑙
0
= [
(−𝑙 + 2𝑥)
𝛿𝑙
× 𝑒
−𝑥
𝛿
(1−
𝑥
𝑙
)
] 𝑙
0
=
2
𝛿
∫ (𝑇 − 𝑇∞)𝑑𝑥
𝑙
0
= [
𝛿𝑙
(−𝑙 + 2𝑥)
× 𝑒
−𝑥
𝛿
(1−
𝑥
𝑙
)
] 𝑙
0
= 2𝛿
Substituting in the integral equation above, we get:
𝛼 (
2
𝛿
) = 2
𝑑𝛿
𝑑𝑡
𝛿 = √2𝛼𝑡
Substituting back 𝛿 into the temperature profile, we get
𝑻 − 𝑻∞
𝑻𝒔 − 𝑻∞
= 𝒆
−𝒙
√𝟐𝜶𝒕
(𝟏−
𝒙
𝒍
)
Or
𝑻 − 𝑻𝒔
𝑻∞ − 𝑻𝒔
= 𝟏 − 𝒆
−𝒙
√𝟐𝜶𝒕
(𝟏−
𝒙
𝒍
)
You notice that the initial condition is satisfied.
You notice that when 𝑙 = ∞ , we reduce to the temperature profile we derived
before

17. 𝑻 − 𝑻∞
𝑻𝒔 − 𝑻∞
= 𝒆
−𝒙
√𝟐𝜶𝒕
you notice that in the temperature profile developed, we get an exact solution
to the problem not an approximate as the Fourier series.

18. UNEQUAL FIXED TEMPERATURES AT THE END OF AN
INSULATED METAL ROD.
𝜕𝑇
𝜕𝑡
= 𝛼
𝜕2
𝑇
𝜕𝑥2
The boundary conditions are:
𝑇 = 𝑇𝑠 𝑎𝑡 𝑥 = 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡
𝑇 = 𝑇1 𝑎𝑡 𝑥 = 𝑙 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡
The initial condition is
𝑇 = 𝑇∞ 𝑎𝑡 𝑡 = 0 0 ≤ 𝑥 ≤ 𝑙
The temperature profile that satisfies the boundary conditions is:
𝑻 − 𝑻∞
[
𝒙
𝒍
(𝑻𝟏 − 𝑻∞) + (𝑻𝒔 − 𝑻∞) (𝟏 −
𝒙
𝒍)]
= 𝒆−
𝒙
𝜹
(𝟏−
𝒙
𝒍
)
For now, we shall have a solution where 𝛿 is proportional to time t so that at
𝑡 = 0, 𝛿 = 0 and the initial condition will be satisfied.
We then transform the heat governing equation into an integral equation as:
𝛼 ∫ (
𝜕2
𝑇
𝜕𝑥2
) 𝑑𝑥
𝑙
0
= ∫
𝜕𝑇
𝜕𝑡
𝑑𝑥
𝑙
0
Where:
𝑙 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑒𝑡𝑎𝑙 𝑟𝑜𝑑
You notice that the integral
∫
𝜕𝑇
𝜕𝑡
𝑑𝑥
𝑙
0
=
𝜕
𝜕𝑡
∫ 𝑇𝑑𝑥
𝑙
0
=
𝜕
𝜕𝑡
∫ (𝑇 − 𝑇∞)𝑑𝑥
𝑙
0
Since 𝑙 and 𝑇∞ are constants independent of time
So, the integral equation becomes:
𝛼 ∫ (
𝜕2
𝑇
𝜕𝑥2
) 𝑑𝑥
𝑙
0
=
𝜕
𝜕𝑡
∫ (𝑇 − 𝑇∞)𝑑𝑥
𝑙
0
We go ahead and find

19. ∫ (
𝜕2
𝑇
𝜕𝑥2
) 𝑑𝑥
𝑙
0
= [
𝜕𝑇
𝜕𝑥
]
𝑙
0
𝜕𝑇
𝜕𝑥
= (𝑇𝑠 − 𝑇∞)𝑒−
𝑥
𝛿
(1−
𝑥
𝑙
)
[
1
𝑙
(𝑇1 − 𝑇∞)
(𝑇𝑠 − 𝑇∞)
−
1
𝑙
] + (𝑇𝑠 − 𝑇∞)𝑒−
𝑥
𝛿
(1−
𝑥
𝑙
)
[
𝑥
𝑙
(𝑇1 − 𝑇∞)
(𝑇𝑠 − 𝑇∞)
− (1 −
𝑥
𝑙
)] (
−𝑙 + 2𝑥
𝛿𝑙
)
∫ (
𝜕2
𝑇
𝜕𝑥2
) 𝑑𝑥
𝑙
0
= [
𝜕𝑇
𝜕𝑥
]
𝑙
0
=
(𝑇𝑠 + 𝑇1 − 2𝑇∞)
𝛿
∫ (𝑇 − 𝑇∞)𝑑𝑥
𝑙
0
=
(𝑇1 − 𝑇∞)
𝑙
∫ 𝑥
𝑙
0
𝑒
−
𝑥
𝛿
(1−
𝑥
𝑙
)
𝑑𝑥 + (𝑇𝑠 − 𝑇∞) ∫ 𝑒
−
𝑥
𝛿
(1−
𝑥
𝑙
)
𝑑𝑥
𝑙
0
−
(𝑇𝑠 − 𝑇∞)
𝑙
∫ 𝑥
𝑙
0
𝑒
−
𝑥
𝛿
(1−
𝑥
𝑙
)
∫ 𝑥
𝑙
0
𝑒
−
𝑥
𝛿
(1−
𝑥
𝑙
)
𝑑𝑥 = 𝑙𝛿
∫ 𝑒
−
𝑥
𝛿
(1−
𝑥
𝑙
)
𝑑𝑥
𝑙
0
= 2𝛿
So
∫ (𝑇 − 𝑇∞)𝑑𝑥
𝑙
0
= 𝛿(𝑇𝑠 + 𝑇1 − 2𝑇∞)
Substituting ∫ (
𝜕2𝑇
𝜕𝑥2) 𝑑𝑥
𝑙
0
and ∫ (𝑇 − 𝑇∞)𝑑𝑥
𝑙
0
in the integral equation, we get
𝛼
𝛿
=
𝑑𝛿
𝑑𝑡
Where:
𝛿 = 0 𝑎𝑡 𝑡 = 0
𝛿 = √2𝛼𝑡
You notice that the initial condition is satisfied since after finding the solution
as done before to the PDE,
𝛿 = √2𝛼𝑡
Therefore substituting 𝛿 in the temperature profile, we get:
𝑻 − 𝑻∞
[
𝒙
𝒍
(𝑻𝟏 − 𝑻∞) + (𝑻𝒔 − 𝑻∞) (𝟏 −
𝒙
𝒍)]
= 𝒆
−
𝒙
√𝟐𝜶𝒕
(𝟏−
𝒙
𝒍
)
you notice that at steady state (𝑡 = ∞)
𝒆
−
𝒙
√𝟐𝜶𝒕
(𝟏−
𝒙
𝒍
)
= 𝒆−
𝒙
∞
(𝟏−
𝒙
𝒍
)
= 𝒆−𝟎
= 𝟏

20. The temperature profile becomes:
𝑻 − 𝑻∞ = [
𝒙
𝒍
(𝑻𝟏 − 𝑻∞) + (𝑻𝒔 − 𝑻∞) (𝟏 −
𝒙
𝒍
)]

21. HOW DO WE DEAL WITH OTHER TYPES OF BOUNDARY
CONDITIONS?
Consider the following types of boundary conditions and initial condition:
A)
𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎
𝒅𝑻
𝒅𝒙
= 𝟎 𝒂𝒕 𝒙 = 𝒍
𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎
B)
𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎
−𝒌
𝒅𝑻
𝒅𝒙
= 𝒉(𝑻 − 𝑻∞) 𝒂𝒕 𝒙 = 𝒍
𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎
Let us go about solving for the above boundary conditions but let us deal with set A
boundary conditions and then we can deal with set B later.
We start with a temperature profile below:
𝑇 − 𝑇∞ = (𝑇𝑠 − 𝑇∞)𝑒−
𝑥
𝛿
(1−
𝑥
𝑙
)
[
𝑥
𝑙
(𝑇1 − 𝑇∞)
(𝑇𝑠 − 𝑇∞)
+ (1 −
𝑥
𝑙
)]
we take the derivative
𝒅𝑻
𝒅𝒙
𝒂𝒕 𝒙 = 𝒍 and equate it to 0 and get:
𝑑𝑇
𝑑𝑥
|𝑥=𝑙 = (
(𝑇1 − 𝑇∞)
𝑙
−
(𝑇𝑠 − 𝑇∞)
𝑙
+
(𝑇1 − 𝑇∞)
𝛿
)
𝑑𝑇
𝑑𝑥
|𝑥=𝑙 = (
(𝑇1 − 𝑇∞)
𝑙
−
(𝑇𝑠 − 𝑇∞)
𝑙
+
(𝑇1 − 𝑇∞)
𝛿
) = 0
We finally get
(𝑇1 − 𝑇∞) = (𝑇𝑠 − 𝑇∞)(
𝛿
𝑙 + 𝛿
)
We substitute 𝑇1 − 𝑇∞ into the temperature profile and get
(𝑻 − 𝑻∞) = (𝑻𝒔 − 𝑻∞)𝒆−
𝒙
𝜹
(𝟏 −
𝒙
𝒍
)
[
𝒙
𝒍
(
𝜹
𝜹 + 𝒍
) + (𝟏 −
𝒙
𝒍
)]

22. So, the temperature profile above satisfies the set A) boundary and initial
conditions and we can go ahead and solve the governing equation using the
temperature profile above.
For set B) boundary conditions, we again start with the temperature profile
below:
𝑇 − 𝑇∞ = (𝑇𝑠 − 𝑇∞)𝑒−
𝑥
𝛿
(1 −
𝑥
𝑙
)
[
𝑥
𝑙
(𝑇1 − 𝑇∞)
(𝑇𝑠 − 𝑇∞)
+ (1 −
𝑥
𝑙
)]
we take the derivative
𝒅𝑻
𝒅𝒙
𝒂𝒕 𝒙 = 𝒍 and equate it to:
𝑑𝑇
𝑑𝑥
|𝑥=𝑙 = (
(𝑇1 − 𝑇∞)
𝑙
−
(𝑇𝑠 − 𝑇∞)
𝑙
+
(𝑇1 − 𝑇∞)
𝛿
)
𝒅𝑻
𝒅𝒙
|𝒙=𝒍 = −
𝒉
𝒌
(𝑻𝟏 − 𝑻∞)
We then find the required temperature profile which we can use to solve the
governing equation.

23. HOW DO WE DEAL WITH CONVECTION AT THE
SURFACE AREA OF THE METAL ROD FOR FIXED END
TEMPERATURE
𝛼
𝜕2
𝑇
𝜕𝑥2
−
ℎ𝑃
𝐴𝜌𝐶
(𝑇 − 𝑇∞) =
𝜕𝑇
𝜕𝑡
We shall use the integral approach.
The boundary and initial conditions are
𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒕
𝑻 = 𝑻∞ 𝒂𝒕 𝒙 = ∞
𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎
Where: 𝑻∞ = 𝒓𝒐𝒐𝒎 𝒕𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆
First, we assume a temperature profile that satisfies the boundary conditions
as:
𝑇 − 𝑇∞
𝑇𝑠 − 𝑇∞
= 𝑒
−𝑥
𝛿
where 𝛿 is to be determined and is a function of time t.
The governing equation is
𝛼
𝜕2
𝑇
𝜕𝑥2
−
ℎ𝑃
𝐴𝜌𝐶
(𝑇 − 𝑇∞) =
𝜕𝑇
𝜕𝑡
Let us change this equation into an integral as below:
𝛼 ∫ (
𝜕2
𝑇
𝜕𝑥2
) 𝑑𝑥
𝑙
0
−
ℎ𝑃
𝐴𝜌𝐶
∫ (𝑇 − 𝑇∞)𝑑𝑥
𝑙
0
=
𝜕
𝜕𝑡
∫ (𝑇 − 𝑇∞)𝑑𝑥
𝑙
0
… … . . 𝑏)
𝜕2
𝑇
𝜕𝑥2
=
(𝑇𝑠 − 𝑇∞)
𝛿2
𝑒
−𝑥
𝛿
∫ (
𝜕2
𝑇
𝜕𝑥2
) 𝑑𝑥
𝑙
0
=
−(𝑇𝑠 − 𝑇∞)
𝛿
(𝑒
−𝑥
𝑙 − 1)
∫ (𝑇 − 𝑇∞)𝑑𝑥
𝑙
0
= −𝛿(𝑇𝑠 − 𝑇∞)(𝑒
−𝑥
𝑙 − 1)

24. Substituting the above expressions in equation b) above, we get
𝛼 −
ℎ𝑃
𝐴𝜌𝐶
𝛿2
= 𝛿
𝑑𝛿
𝑑𝑡
We solve the equation above assuming that
𝛿 = 0 𝑎𝑡 𝑡 = 0
And get
𝛿 = √
𝛼𝐴𝜌𝐶
ℎ𝑃
(1 − 𝑒
−2ℎ𝑃
𝐴𝜌𝐶
𝑡
)
𝛿 = √
𝐾𝐴
ℎ𝑃
(1 − 𝑒
−2ℎ𝑃
𝐴𝜌𝐶
𝑡
)
Substituting for 𝛿 in the temperature profile, we get
𝑻 − 𝑻∞
𝑻𝒔 − 𝑻∞
= 𝒆
−𝒙
√𝑲𝑨
𝒉𝑷
(𝟏−𝒆
−𝟐𝒉𝑷
𝑨𝝆𝑪
𝒕
)
From the equation above, we notice that the initial condition is satisfied i.e.,
𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎
The equation above predicts the transient state and in steady state (𝑡 = ∞) it
reduces to
𝑻 − 𝑻∞
𝑻𝒔 − 𝑻∞
= 𝒆
−√(
𝒉𝑷
𝑲𝑨
)𝒙
What are the predictions of the transient state?
Let us make 𝑥 the subject of the equation of transient state and get:
𝑥2
= [ln (
𝑇𝑠 − 𝑇∞
𝑇 − 𝑇∞
)]2
×
𝐾𝐴
ℎ𝑃
(1 − 𝑒
−2ℎ𝑃
𝐴𝜌𝐶
𝑡
)
When the time duration is small and
2ℎ𝑃
𝐴𝜌𝐶
𝑡 ≪ 1

25. We use the binomial expansion approximation
𝑒
−2ℎ𝑃
𝐴𝜌𝐶
𝑡
= 1 −
2ℎ𝑃
𝐴𝜌𝐶
𝑡
Substituting in the equation of 𝑥2
as the subject, we get
𝑥2
= 2𝛼[ln (
𝑇𝑠 − 𝑇∞
𝑇 − 𝑇∞
)]2
× 𝑡
Where:
𝛼 =
𝐾
𝜌𝐶
What that equation says is that when you stick wax particles on a long metal
rod (𝑙 = ∞) at distances x from the hot end of the rod and note the time t it
takes the wax particles to melt, then a graph of 𝑥2
against 𝑡 is a straight-line
graph through the origin as stated by the equation above when the times are
small. The equation is true because that is what is observed experimentally.
How do we measure the heat transfer coefficient?
From,
𝑥2
= [ln (
𝑇𝑠 − 𝑇∞
𝑇 − 𝑇∞
)]2
×
𝐾𝐴
ℎ𝑃
(1 − 𝑒
−2ℎ𝑃
𝐴𝜌𝐶
𝑡
)
When the times concerned are big, we shall observe the above equation.
Let us call
𝐵 = [ln (
𝑇𝑠 − 𝑇∞
𝑇 − 𝑇∞
)]2
×
𝐾𝐴
ℎ𝑃
So, the equation above becomes
𝑥2
= 𝐵 (1 − 𝑒
−2ℎ𝑃
𝐴𝜌𝐶
𝑡
)
Let’s take the derivative of x against time t and get
𝑥
𝑑𝑥
𝑑𝑡
= 𝐵
2ℎ
𝑟𝜌𝐶
𝑒
−2ℎ𝑃
𝐴𝜌𝐶
𝑡

26. 𝑒
−2ℎ𝑃
𝐴𝜌𝐶
𝑡
=
(𝐵 − 𝑥2
)
𝐵
Substitute and get
𝑥
𝑑𝑥
𝑑𝑡
=
2ℎ
𝑟𝜌𝐶
(𝐵 − 𝑥2
)
𝒙
𝒅𝒙
𝒅𝒕
=
𝟐𝒉
𝒓𝝆𝑪
𝑩 −
𝟐𝒉
𝒓𝝆𝑪
𝒙𝟐
A graph of 𝑥
𝑑𝑥
𝑑𝑡
against 𝑥2
when the time is big has a negative gradient of
ℎ
𝑟𝜌𝐶
from which h can be measured experimentally.
From experiments using aluminium rod, h was found to be
ℎ = 5.7712𝑊𝑚−2
𝐾−1
h can also be got from Stefan’s law of cooling that reduces to the Newton’s law
of cooling.
Stefan’s law of cooling in natural convection states
𝒅𝑸
𝒅𝒕
= (𝟏 + 𝜷)𝑨𝝈𝜺[𝑻𝟒
− 𝑻∞
𝟒
]
From tables
𝜺 = 𝟎. 𝟑 𝒇𝒐𝒓 𝒂𝒍𝒖𝒎𝒊𝒏𝒊𝒖𝒎
Where:
𝜷 = 𝟒. 𝟓𝟕𝟗𝟕 = 𝒌𝑷𝒓𝒏
Considering
𝑇 = 𝑇∞ + ∆𝑇
𝑑𝑄
𝑑𝑡
= (1 + 𝛽)𝐴𝜎𝜀[(𝑇∞ + ∆𝑇)4
− 𝑇∞
4
]
Factorizing out 𝑇1, we get
𝑑𝑄
𝑑𝑡
= (1 + 𝛽)𝐴𝜎𝜀[𝑇∞
4
(1 +
(𝑇 − 𝑇∞)
𝑇∞
)4
− 𝑇∞
4
]
It is known from Binomial expansion that:
(1 + 𝑥)𝑛
≈ 1 + 𝑛𝑥 𝑓𝑜𝑟 𝑥 ≪ 1
So:

27. (1 +
(𝑇 − 𝑇∞)
𝑇∞
)4
≈ 1 + 4
(𝑇 − 𝑇∞)
𝑇∞
= 1 + 4
∆𝑇
𝑇∞
= 𝑓𝑜𝑟
∆𝑇
𝑇∞
≪ 1
Simplifying, we get Newton’s law of cooling i.e.
𝒅𝑸
𝒅𝒕
= 𝟒(𝟏 + 𝜷)𝑨𝝈𝜺𝑻∞
𝟑 (𝑻 − 𝑻∞)
𝒅𝑸
𝒅𝒕
= 𝒉𝑨(𝑻 − 𝑻∞)
Where:
𝒉 = 𝟒(𝟏 + 𝜷)𝝈𝜺𝑻∞
𝟑
Substitute for the above parameters of aluminium and get h theoretically and
compare as got experimentally.
How do we deal with metal rods of finite length 𝒍 ?
The boundary and initial conditions are
𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎

28. −𝒌
𝒅𝑻
𝒅𝒙
= 𝒉(𝑻 − 𝑻∞) 𝒂𝒕 𝒙 = 𝒍
𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎
Let us go about solving for the above boundary conditions
We start with a temperature profile below:
𝑇 − 𝑇∞ = (𝑇𝑠 − 𝑇∞)𝑒−
𝑥
𝛿
(1−
𝑥
𝑙
)
[
𝑥
𝑙
(𝑇1 − 𝑇∞)
(𝑇𝑠 − 𝑇∞)
+ (1 −
𝑥
𝑙
)]
Which says
𝑇 = 𝑇𝑠 𝑎𝑡 𝑥 = 0
𝑇 = 𝑇1 𝑎𝑡 𝑥 = 𝑙
𝑇 = 𝑇∞ 𝑎𝑡 𝑡 = 0
Provided 𝛿 = 0 𝑎𝑡 𝑡 = 0 , then the initial condition above is satisfied
we take the derivative
𝒅𝑻
𝒅𝒙
𝒂𝒕 𝒙 = 𝒍 and equate it to −
ℎ
𝑘
(𝑇1 − 𝑇∞) and get:
𝑑𝑇
𝑑𝑥
|𝑥=𝑙 = (
(𝑇1 − 𝑇∞)
𝑙
−
(𝑇𝑠 − 𝑇∞)
𝑙
+
(𝑇1 − 𝑇∞)
𝛿
)
𝑑𝑇
𝑑𝑥
|𝑥=𝑙 = −
ℎ
𝑘
(𝑇1 − 𝑇∞)
We equate the two and get
(
(𝑇1 − 𝑇∞)
𝑙
−
(𝑇𝑠 − 𝑇∞)
𝑙
+
(𝑇1 − 𝑇∞)
𝛿
) = −
ℎ
𝑘
(𝑇1 − 𝑇∞)
We finally get
(𝑇1 − 𝑇∞) = (𝑇𝑠 − 𝑇∞)(
𝛿𝑘
𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿
)
We substitute 𝑇1 − 𝑇∞ into the temperature profile and get
(𝑻 − 𝑻∞) = (𝑻𝒔 − 𝑻∞)𝒆−
𝒙
𝜹
(𝟏 −
𝒙
𝒍
)
[
𝒙
𝒍
(
𝜹𝒌
𝜹𝒌 + 𝒍𝒌 + 𝒉𝒍𝜹
) + (𝟏 −
𝒙
𝒍
)]
This the temperature profile that satisfies the boundary and initial conditions
below

29. 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎
−𝒌
𝒅𝑻
𝒅𝒙
= 𝒉(𝑻 − 𝑻∞) 𝒂𝒕 𝒙 = 𝒍
𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎
Let us go ahead and solve for 𝛿
The governing equation is
𝛼
𝜕2
𝑇
𝜕𝑥2
−
ℎ𝑃
𝐴𝜌𝐶
(𝑇 − 𝑇∞) =
𝜕𝑇
𝜕𝑡
Let us change this equation into an integral as below:
𝛼 ∫ (
𝜕2
𝑇
𝜕𝑥2
) 𝑑𝑥
𝑙
0
−
ℎ𝑃
𝐴𝜌𝐶
∫ (𝑇 − 𝑇∞)𝑑𝑥
𝑙
0
=
𝜕
𝜕𝑡
∫ (𝑇 − 𝑇∞)𝑑𝑥
𝑙
0
… … . . 𝑏)
𝛼 ∫ (
𝜕2
𝑇
𝜕𝑥2
) 𝑑𝑥
𝑙
0
−
2ℎ
𝑟𝜌𝐶
∫ (𝑇 − 𝑇∞)𝑑𝑥
𝑙
0
=
𝜕
𝜕𝑡
∫ (𝑇 − 𝑇∞)𝑑𝑥
𝑙
0
∫ (
𝜕2
𝑇
𝜕𝑥2
) 𝑑𝑥
𝑙
0
= [
𝜕𝑇
𝜕𝑥
]
𝑙
0
𝜕𝑇
𝜕𝑥
= (𝑇𝑠 − 𝑇∞)𝑒−
𝑥
𝛿
(1−
𝑥
𝑙
)
[
1
𝑙
(𝑇1 − 𝑇∞)
(𝑇𝑠 − 𝑇∞)
−
1
𝑙
] + (𝑇𝑠 − 𝑇∞)𝑒−
𝑥
𝛿
(1−
𝑥
𝑙
)
[
𝑥
𝑙
(𝑇1 − 𝑇∞)
(𝑇𝑠 − 𝑇∞)
− (1 −
𝑥
𝑙
)] (
−𝑙 + 2𝑥
𝛿𝑙
)
∫ (
𝜕2
𝑇
𝜕𝑥2
) 𝑑𝑥
𝑙
0
= [
𝜕𝑇
𝜕𝑥
]
𝑙
0
=
(𝑇𝑠 − 𝑇∞)
𝛿
+
(𝑇1 − 𝑇∞)
𝛿
Substitute for
(𝑇1 − 𝑇∞) = (𝑇𝑠 − 𝑇∞)(
𝛿𝑘
𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿
)
∫ (
𝜕2
𝑇
𝜕𝑥2
) 𝑑𝑥
𝑙
0
= [
𝜕𝑇
𝜕𝑥
]
𝑙
0
= (𝑇𝑠 − 𝑇∞)(
2𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿
𝛿(𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿)
)
2ℎ
𝑟𝜌𝐶
∫ (𝑇 − 𝑇∞)𝑑𝑥
𝑙
0
= ℎ2𝜋𝑟 ∫ ((𝑇𝑠 − 𝑇∞)𝑒−
𝑥
𝛿
(1−
𝑥
𝑙
)
[
𝑥
𝑙
(𝑇1 − 𝑇∞)
(𝑇𝑠 − 𝑇∞)
+ (1 −
𝑥
𝑙
)])𝑑𝑥
𝑙
0
2ℎ
𝑟𝜌𝐶
∫ (𝑇 − 𝑇∞)𝑑𝑥
𝑙
0
= ℎ2𝜋𝑟[
(𝑇1 − 𝑇∞)
𝑙
∫ 𝑥𝑒−
𝑥
𝛿
(1−
𝑥
𝑙
)
𝑑𝑥
𝑙
0
+ (𝑇𝑠 − 𝑇∞) ∫ 𝑒−
𝑥
𝛿
(1−
𝑥
𝑙
)
𝑑𝑥
𝑙
0
−
(𝑇𝑠 − 𝑇∞)
𝑙
∫ 𝑥𝑒−
𝑥
𝛿
(1−
𝑥
𝑙
)
𝑑𝑥
𝑙
0
]

30. Integrating by parts shows that
∫ 𝑒
−
𝑥
𝛿
(1−
𝑥
𝑙
)
𝑑𝑥
𝑙
0
= [(
𝑙𝛿
−𝑙 + 2𝑥
)𝑒
−
𝑥
𝛿
(1−
𝑥
𝑙
)
]
𝑙
0
= 2𝛿
∫ 𝑥𝑒
−
𝑥
𝛿
(1−
𝑥
𝑙
)
𝑑𝑥
𝑙
0
= [(
𝑥𝑙𝛿
−𝑙 + 2𝑥
−
𝑙2
𝛿2
(−𝑙 + 2𝑥)2[1 −
2𝑙𝛿
(−𝑙 + 2𝑥)2]
) 𝑒
−
𝑥
𝛿
(1−
𝑥
𝑙
)
]
𝑙
0
= 𝑙𝛿
Substituting back into the heat loss equation we get
2ℎ
𝑟𝜌𝐶
∫ (𝑇 − 𝑇∞)𝑑𝑥
𝑙
0
=
2ℎ
𝑟𝜌𝐶
[(𝑇1 − 𝑇∞) + (𝑇𝑠 − 𝑇∞)]𝛿
substitute for (𝑇1 − 𝑇∞) and get
2ℎ
𝑟𝜌𝐶
∫ (𝑇 − 𝑇∞)𝑑𝑥
𝑙
0
=
2ℎ
𝑟𝜌𝐶
(𝑇𝑠 − 𝑇∞)(
2𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿
(𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿)
)𝛿
𝜕
𝜕𝑡
∫ (𝑇 − 𝑇∞)𝑑𝑥
𝑙
0
= (𝑇𝑠 − 𝑇∞)(
2𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿
(𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿)
)
𝑑𝛿
𝑑𝑡
Substituting into the integral equation we get
𝛼(𝑇𝑠 − 𝑇∞) (
2𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿
𝛿(𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿)
) −
2ℎ
𝑟𝜌𝐶
(𝑇𝑠 − 𝑇∞) (
2𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿
(𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿)
) 𝛿 = (𝑇𝑠 − 𝑇∞)(
2𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿
(𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿)
)
𝑑𝛿
𝑑𝑡
𝛼 −
ℎ𝑃
𝐴𝜌𝐶
𝛿2
= 𝛿
𝑑𝛿
𝑑𝑡
We solve the equation above assuming that
𝛿 = 0 𝑎𝑡 𝑡 = 0
And get
𝛿 = √
𝛼𝐴𝜌𝐶
ℎ𝑃
(1 − 𝑒
−2ℎ𝑃
𝐴𝜌𝐶
𝑡
)

31. 𝛿 = √
𝐾𝐴
ℎ𝑃
(1 − 𝑒
−2ℎ𝑃
𝐴𝜌𝐶
𝑡
)
We go ahead and substitute for 𝛿 in the temperature profile below
(𝑻 − 𝑻∞) = (𝑻𝒔 − 𝑻∞)𝒆−
𝒙
𝜹
(𝟏 −
𝒙
𝒍
)
[
𝒙
𝒍
(
𝜹𝒌
𝜹𝒌 + 𝒍𝒌 + 𝒉𝒍𝜹
) + (𝟏 −
𝒙
𝒍
)]
When the time is small, 𝛿 using binomial approximation becomes
𝛿 = √
𝐾𝐴
ℎ𝑃
(1 − 𝑒
−2ℎ𝑃
𝐴𝜌𝐶
𝑡
)
2ℎ𝑃
𝐴𝜌𝐶
𝑡 ≪ 1
𝑒
−ℎ𝑃
𝐴𝜌𝐶
𝑡
= 1 −
2ℎ𝑃
𝐴𝜌𝐶
𝑡
1 − 𝑒
−2ℎ𝑃
𝐴𝜌𝐶
𝑡
=
2ℎ𝑃
𝐴𝜌𝐶
𝑡
𝛿 = √2𝛼𝑡
We substitute for 𝛿 in the temperature profile.
What happens when the length is big or tends to infinity?
(𝑻 − 𝑻∞) = (𝑻𝒔 − 𝑻∞)𝒆−
𝒙
𝜹
(𝟏 −
𝒙
𝒍
)
[
𝒙
𝒍
(
𝜹𝒌
𝜹𝒌 + 𝒍𝒌 + 𝒉𝑳𝒍𝜹
) + (𝟏 −
𝒙
𝒍
)]
Becomes
(𝑻 − 𝑻∞) = (𝑻𝒔 − 𝑻∞)𝒆−
𝒙
𝜹
(𝑻 − 𝑻∞) = (𝑻𝒔 − 𝑻∞)𝒆
−
𝒙
√𝑲𝑨
𝒉𝑷
(𝟏−𝒆
−𝟐𝒉𝑷
𝑨𝝆𝑪
𝒕
)
Which is what we got before.

32. WHAT HAPPENS WHEN THE INITIAL TEMPERATURE IS
A FUNCTION OF X?
The governing equation is
𝛼
𝜕2
𝑇
𝜕𝑥2
=
𝜕𝑇
𝜕𝑡
BCs
𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 𝟎 < 𝒕 < ∞
𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝒍 𝟎 < 𝒕 < ∞
IC
𝑻 = ∅(𝒙) 𝒂𝒕 𝒕 = 𝟎 𝟎 ≤ 𝒙 ≤ 𝒍
We assume an exponential temperature profile that satisfies the boundary
conditions:
𝑇 − ∅
𝑇𝑠 − ∅
= 𝑒
−𝑥
𝛿
𝑇 = ∅ + 𝑇𝑠𝑒
−𝑥
𝛿 − ∅𝑒
−𝑥
𝛿
The PDE becomes an integral equation given by:
𝛼 ∫ (
𝜕2
𝑇
𝜕𝑥2
) 𝑑𝑥
𝛿
0
=
𝜕
𝜕𝑡
∫ 𝑇𝑑𝑥
𝛿
0
You notice the limits of the integral become from 0 𝑡𝑜 𝛿 as this eliminate
exponentials which are functions of 𝛿 and this will make the integration
simpler.
Let us give an example say
∅ = 𝑥
We make T the subject of the formula and get
𝑇 = ∅ + 𝑇𝑠𝑒
−𝑥
𝛿 − ∅𝑒
−𝑥
𝛿
∅ = 𝑥
𝑇 = 𝑥 + 𝑇𝑠𝑒
−𝑥
𝛿 − 𝑥𝑒
−𝑥
𝛿

33. ∫ (
𝜕2
𝑇
𝜕𝑥2
) 𝑑𝑥
𝛿
0
= [
𝜕𝑇
𝜕𝑥
]
𝛿
0
𝜕𝑇
𝜕𝑥
= 1 −
𝑇𝑠
𝛿
𝑒
−𝑥
𝛿 −
𝑥
𝛿
𝑒
−𝑥
𝛿 + 𝑒
−𝑥
𝛿
∫ (
𝜕2
𝑇
𝜕𝑥2
) 𝑑𝑥
𝛿
0
= [1 −
𝑇𝑠
𝛿
𝑒
−𝑥
𝛿 −
𝑥
𝛿
𝑒
−𝑥
𝛿 + 𝑒
−𝑥
𝛿 ]
𝛿
0
=
𝑇𝑠
𝛿
(1 − 𝑒
−1
)
∫ 𝑇𝑑𝑥
𝛿
0
= ∫ [𝑥 + 𝑇𝑠𝑒
−𝑥
𝛿 − 𝑥𝑒
−𝑥
𝛿 ]𝑑𝑥
𝛿
0
= [
𝑥2
2
− 𝛿𝑇𝑠𝑒
−𝑥
𝛿 + 𝑥𝛿𝑒
−𝑥
𝛿 + 𝛿2
𝑒
−𝑥
𝛿 ]
𝛿
0
= [−
𝛿2
2
+ 𝛿𝑇𝑠(1 − 𝑒−1) + 2𝛿2
𝑒−1
]
𝜕
𝜕𝑡
∫ 𝑇𝑑𝑥
𝛿
0
= −2𝛿
𝑑𝛿
𝑑𝑡
+ 𝑇𝑠(1 − 𝑒−1)
𝑑𝛿
𝑑𝑡
+ 4𝛿𝑒−1
𝑑𝛿
𝑑𝑡
Substituting ∫ (
𝜕2𝑇
𝜕𝑥2
) 𝑑𝑥
𝛿
0
and
𝜕
𝜕𝑡
∫ 𝑇𝑑𝑥
𝛿
0
into the integral equation, we get
𝑇𝑠
𝛿
(1 − 𝑒
−1
) = [−2𝛿 + 𝑇𝑠(1 − 𝑒−1) + 4𝛿𝑒−1
]
𝑑𝛿
𝑑𝑡
The boundary conditions are 𝛿 = 0 𝑎𝑡 𝑡 = 0
𝑇𝑠(1 − 𝑒−1
) ∫ 𝑑𝑡
𝑡
0
= [−2 ∫ 𝛿2
𝑑𝛿
𝛿
0
+ 𝑇𝑠(1 − 𝑒−1
) ∫ 𝛿𝑑𝛿
𝛿
0
+ 4𝑒−1
∫ 𝛿2
𝛿
0
𝑑𝛿]
We end up with
6𝑇𝑠(1 − 𝑒
−1
)𝑡 = (8𝑒−1
− 4)𝛿
3
+ 3𝑇𝑠(1 − 𝑒−1
)𝛿2
From which we can get 𝛿
Where 𝛿 is in the form
𝑎𝛿3
+ 𝑏𝛿2
+ 𝑑 = 0
The above is a cubic equation whose solutions exist.
Where:
𝑎 = (8𝑒−1
− 4)
𝑏 = 3𝑇𝑠(1 − 𝑒−1
)
𝑑 = −6𝑇𝑠(1 − 𝑒
−1
)𝑡

34. There are three roots of 𝛿 but we choose those which reduce to zero when time
is zero.
The two roots that satisfy the above condition are
𝛿1 = −
𝑏
3𝑎
+
1 + 𝑖√3
6𝑎
√(
1
2
) [2𝑏3 + 27𝑎2𝑑 + √(2𝑏3 + 27𝑎2𝑑)2 − 4𝑏6)]
3
+
1 − 𝑖√3
6𝑎
√(
1
2
) [2𝑏3 + 27𝑎2𝑑 − √(2𝑏3 + 27𝑎2𝑑)2 − 4𝑏6)]
3
Or
𝛿2 = −
𝑏
3𝑎
+
1 − 𝑖√3
6𝑎
√(
1
2
) [2𝑏3 + 27𝑎2𝑑 + √(2𝑏3 + 27𝑎2𝑑)2 − 4𝑏6)]
3
+
1 + 𝑖√3
6𝑎
√(
1
2
) [2𝑏3 + 27𝑎2𝑑 − √(2𝑏3 + 27𝑎2𝑑)2 − 4𝑏6)]
3
After getting the solution of 𝛿 ,we go ahead and substitute it into the
exponential temperature profile.
We shall be faced with more scenarios where 𝛿 is a cubic equation with time
but the root or solution of 𝛿 to choose is the one for which 𝛿 = 0 𝑎𝑡 𝑡 = 0.
Using this analytical method, we can also go ahead and solve PDES like
𝜕𝑇
𝜕𝑡
= 𝛼
𝜕2
𝑇
𝜕𝑥2
+ 𝑓(𝑥)
Again, we take limits from 0 𝑡𝑜 𝛿 when solving the integral equation to eliminate
exponentials with a function of 𝛿 in order to make solving for the solution easy.

35. HOW DO WE DEAL WITH CYLINDRICAL CO-
ORDINATES?
We know that for an insulated cylinder where there is no heat loss by
convection from the sides, the governing PDE equation is
𝛼 [
𝜕2
𝑇
𝜕𝑟2
+
1
𝑟
𝜕𝑇
𝜕𝑟
] =
𝜕𝑇
𝜕𝑡
The boundary conditions are
𝑇 = 𝑇𝑠 𝑎𝑡 𝑟 = 0
𝑇 = 𝑇∞ 𝑎𝑡 𝑟 = ∞
The initial condition is:
𝑇 = 𝑇∞ 𝑎𝑡 𝑡 = 0
The temperature profile that satisfies the conditions above is
𝑇 − 𝑇∞
𝑇𝑠 − 𝑇∞
= 𝑒
−(𝑟−𝑟1)
𝛿
We transform the PDE into an integral equation
𝛼 [
𝜕2
𝑇
𝜕𝑟2
+
1
𝑟
𝜕𝑇
𝜕𝑟
] =
𝜕𝑇
𝜕𝑡
𝛼 ∫ (
𝜕2
𝑇
𝜕𝑟2
) 𝑑𝑟
𝛿+𝑟1
𝑟1
+ 𝛼 ∫ [
1
𝑟
(
𝜕𝑇
𝜕𝑟
)]𝑑𝑟
𝛿+𝑟1
𝑟1
=
𝜕
𝜕𝑡
∫ 𝑇𝑑𝑟
𝛿+𝑟1
𝑟1
We then go ahead to solve and find 𝛿 as before.
∫ (
𝜕2
𝑇
𝜕𝑟2
) 𝑑𝑟
𝛿+𝑟1
𝑟1
= [
𝜕𝑇
𝜕𝑟
]
𝛿 + 𝑟1
𝑟1
= −
𝑇𝑠 − 𝑇∞
𝛿
[𝑒
−(𝑟−𝑟1)
𝛿 ]
𝛿 + 𝑟1
𝑟1
=
𝑇𝑠 − 𝑇∞
𝛿
(1 − 𝑒−1
)
𝜕𝑇
𝜕𝑟
= −
𝑇𝑠 − 𝑇∞
𝛿
𝑒
−(𝑟−𝑟1)
𝛿
∫ [
1
𝑟
(
𝜕𝑇
𝜕𝑟
)] 𝑑𝑟
𝛿+𝑟1
𝑟1
= −
(𝑇𝑠 − 𝑇∞)
𝛿
∫
1
𝑟
𝑒
−(𝑟−𝑟1)
𝛿 𝑑𝑟
𝛿+𝑟1
𝑟1
∫
1
𝑟
𝑒
−(𝑟−𝑟1)
𝛿 𝑑𝑟
𝛿+𝑟1
𝑟1
= 𝑢𝑣 − ∫ 𝑣
𝑑𝑢
𝑑𝑟
𝑑𝑟

36. 𝑢 =
1
𝑟
𝑑𝑣
𝑑𝑟
= 𝑒
−(𝑟−𝑟1)
𝛿
𝑣 = −𝛿𝑒
−(𝑟−𝑟1)
𝛿
∫
1
𝑟
𝑒
−(𝑟−𝑟1)
𝛿 𝑑𝑟
𝛿+𝑟1
𝑟1
=
−𝛿𝑒
−(𝑟−𝑟1)
𝛿
𝑟
−
𝛿
𝑟
∫
1
𝑟
𝑒
−(𝑟−𝑟1)
𝛿 𝑑𝑟
𝛿+
𝑟1
[1 +
𝛿
𝑟
] ∫
1
𝑟
𝑒
−(𝑟−𝑟1)
𝛿 𝑑𝑟
𝛿+𝑟1
𝑟1
=
−𝛿𝑒
−(𝑟−𝑟1)
𝛿
𝑟
∫
1
𝑟
𝑒
−(𝑟−𝑟1)
𝛿 𝑑𝑟
𝛿+𝑟1
𝑟1
= [
−𝛿
𝑟 + 𝛿
𝑒
−(𝑟−𝑟1)
𝛿 ]
𝛿 + 𝑟1
𝑟1
=
𝛿
𝛿 + 𝑟1
−
𝛿𝑒−1
2𝛿 + 𝑟1
Substituting in
∫ [
1
𝑟
(
𝜕𝑇
𝜕𝑟
)]𝑑𝑟
𝛿+𝑟1
𝑟1
= −
(𝑇𝑠 − 𝑇∞)
𝛿
∫
1
𝑟
𝑒
−(𝑟−𝑟1)
𝛿 𝑑𝑟
𝛿+𝑟1
𝑟1
= −
(𝑇𝑠 − 𝑇∞)
𝛿
(
𝛿
𝛿 + 𝑟1
−
𝛿𝑒−1
2𝛿 + 𝑟1
)
We get
∫ [
1
𝑟
(
𝜕𝑇
𝜕𝑟
)]𝑑𝑟
𝛿+𝑟1
𝑟1
=
(𝑇𝑠 − 𝑇∞)
𝛿
(
𝛿𝑒−1
2𝛿 + 𝑟1
−
𝛿
𝛿 + 𝑟1
)
𝜕
𝜕𝑡
∫ 𝑇𝑑𝑟
𝛿+𝑟1
𝑟1
𝑇 = (𝑇𝑠 − 𝑇∞)𝑒
−(𝑟−𝑟1)
𝛿 + 𝑇∞
∫ 𝑇𝑑𝑟
𝛿+𝑟1
𝑟1
= ∫ ((𝑇𝑠 − 𝑇∞)𝑒
−(𝑟−𝑟1)
𝛿 + 𝑇∞)𝑑𝑟
𝛿+𝑟1
𝑟1
= 𝛿((𝑇𝑠 − 𝑇∞)(1 − 𝑒−1) + 𝑇∞)
𝜕
𝜕𝑡
∫ 𝑇𝑑𝑟
𝛿+𝑟1
𝑟1
=
𝑑𝛿
𝑑𝑡
((𝑇𝑠 − 𝑇∞)(1 − 𝑒−1) + 𝑇∞)
substituting all the above in the integral equation, we get
𝛼 ∫ (
𝜕2
𝑇
𝜕𝑟2
) 𝑑𝑟
𝛿+𝑟1
𝑟1
+ 𝛼 ∫ [
1
𝑟
(
𝜕𝑇
𝜕𝑟
)]𝑑𝑟
𝛿+𝑟1
𝑟1
=
𝜕
𝜕𝑡
∫ 𝑇𝑑𝑟
𝛿+𝑟1
𝑟1

37. 𝛼
(𝑇𝑠 − 𝑇∞)
𝛿
(1 − 𝑒−1
) + 𝛼(𝑇𝑠 − 𝑇∞)(
𝑒−1
2𝛿 + 𝑟1
−
1
𝛿 + 𝑟1
) =
𝑑𝛿
𝑑𝑡
((𝑇𝑠 − 𝑇∞)(1 − 𝑒−1) + 𝑇∞)
We go ahead and solve for 𝛿
We can also go ahead and look at situations where there is natural convection
and other situations where the radius r is finite and not infinite.