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THE ANALYTICAL SOLUTIONS TO THE HEAT EQUATION USING AN INTEGRAL
METHOD
Book · January 2023
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Heat Conduction
TABLE OF CONTENTS
PREFACE ............................................................................................................................................ 3
WHAT DO WE OBSERVE EXPERIMENTALLY WHEN HEATING A CYLINDRICAL
METAL ROD AT ONE END WITH WAX PARTICLES ALONG ITS SURFACE AREA? .. 4
HOW DO WE DEAL WITH NATURAL CONVECTION AT THE SURFACE AREA OF A
SEMI-INFINITE METAL ROD FOR FIXED WALL TEMPERATURE ................................... 6
SO HOW DO WE PRODUCE THE SEMI-INFINITE OBSERVED ROD SOLUTION? . 16
HOW DOES HEAT FLOW MANIFEST ITSELF FOR FINITE METAL RODS? ............... 26
CASE 1: CONVECTION AT THE END OF A FINITE METAL ROD ............................... 26
HOW DO WE INVESTIGATE THE NATURE OF 𝒉𝑳 EASILY? ..................................... 31
DERIVATION OF THE GENERAL EXPRESSION FOR HEAT TRANSFER
COEFFICIENT 𝒉𝑳........................................................................................................................ 39
CASE 2: ZERO FLUX AT THE END OF THE METAL ROD?.......................................... 44
HOW DO WE DEAL WITH CYLINDRICAL COORDINATES? ............................................. 50
HOW DO WE DEAL WITH CYLINDRICAL CO-ORDINATES FOR A SEMI-INFINITE
RADIUS CYLINDER? ..................................................................................................................... 51
HOW DO WE DEAL WITH NATURAL CONVECTION AT THE SURFACE AREA OF A
SEMI-INFINITE CYLINDER FOR FIXED END TEMPERATURE ....................................... 54
REFERENCES .................................................................................................................................. 58
PREFACE
In this book we go ahead and investigate the nature of heat conduction in a
metal rod heated at one end while the other end is free. We do this by sticking
wax particles along the surface area of the metal rod at known distances x from
the hot end and then record the time taken for each wax to melt since the
introduction of the flame at the hot end. We first of all look at the case of heat
conduction in a semi-infinite metal rod and solve the heat equation analytically
using the integral transform approach and compare the solution got in the
transient state to experimental observations. We make deductions and
conclusions from both the solution and the experimental values.
We then look at the case of a finite length metal rod heat conduction with
convection at the free end. We use the hyperbolic function solutions known in
literature to interpret experimental data. One fact that we get to learn from the
experimental values is that the heat transfer coefficient β„ŽπΏ at the end of the
metal rod is not a constant but varies with length L as shall be shown later. We
note that in deriving the solution for the convection boundary condition, the
solution derived should reduce to the semi-infinite rod solution as the length
tends to infinity.
We then look at the case of zero flux at the end of a finite metal rod and also
derive the governing equation.
Finally, we use the integral approach to solve the heat equation in cylindrical
co-ordinates for radial heat conduction and use the same techniques we used
before to solve for observed phenomena.
WHAT DO WE OBSERVE EXPERIMENTALLY WHEN
HEATING A CYLINDRICAL METAL ROD AT ONE END
WITH WAX PARTICLES ALONG ITS SURFACE AREA?
The situation we are talking about looks as below:
First of all, let us call the distance π‘₯ to be the distance of the wax particle from the hot
end and 𝑑 to be the time taken for the wax to melt since the introduction of the flame
at the hot end.
For a semi-infinite rod(𝑙 = ∞), it is observed that a graph of π‘₯ against time 𝑑 is a curve
as shown below for an aluminium rod of radius 2mm:
A semi-infinite cylindrical rod means that the length of the metal rod extends to
infinity but the radius is finite.
The graph below is for an aluminium rod of length 75cm and radius 2mm and it can
be treated as a semi-infinite metal rod.
A Graph of x against time t
0.3
x(metres)
0.25
0.2
0.15
0.1
0.05
0
0
100
200
300
t(seconds)
400
500
HOW DO WE DEAL WITH NATURAL CONVECTION AT
THE SURFACE AREA OF A SEMI-INFINITE METAL ROD
FOR FIXED WALL TEMPERATURE
A semi-infinite cylindrical rod means that the length of the metal rod extends to
infinity but the radius of the metal rod is finite. The governing heat equation is:
πœ•2𝑇
β„Žπ‘ƒ
πœ•π‘‡
(𝑇 − 𝑇∞ ) =
𝛼 2−
πœ•π‘₯
𝐴𝜌𝐢
πœ•π‘‘
We shall use the integral transform approach to solve the heat equation above.
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∞ = π’“π’π’π’Ž π’•π’†π’Žπ’‘π’†π’“π’‚π’•π’–π’“π’†
First, we assume a temperature profile that satisfies the boundary conditions as:
−π‘₯
𝑇 − 𝑇∞
=𝑒𝛿
𝑇𝑠 − 𝑇∞
where 𝛿 is to be determined and is a function of time t and not x.
for the initial condition, we assume 𝛿 = 0 at 𝑑 = 0 seconds so that the initial
condition is satisfied i.e.,
Since at 𝑑 = 0, 𝛿 = 0 we get
−π‘₯
𝑇 − 𝑇∞
= 𝑒 0 = 𝑒 −∞ = 0
𝑇𝑠 − 𝑇∞
Hence
𝑇 = 𝑇∞
Which is the initial condition.
The governing partial differential equation is:
𝛼
πœ•2𝑇
β„Žπ‘ƒ
πœ•π‘‡
(𝑇 − 𝑇∞ ) =
−
2
πœ•π‘₯
𝐴𝜌𝐢
πœ•π‘‘
Let us change transform the heat equation into an integral equation as below:
𝑙
𝛼∫ (
0
πœ•2𝑇
β„Žπ‘ƒ 𝑙
πœ• 𝑙
(
)
∫
∫ (𝑇)𝑑π‘₯ … … . . 𝑏)
)
𝑑π‘₯
−
𝑇
−
𝑇
𝑑π‘₯
=
∞
πœ•π‘₯ 2
𝐴𝜌𝐢 0
πœ•π‘‘ 0
πœ• 2 𝑇 (𝑇𝑠 − 𝑇∞ ) −π‘₯
=
𝑒𝛿
πœ•π‘₯ 2
𝛿2
𝑙
πœ•2𝑇
−(𝑇𝑠 − 𝑇∞ ) −𝑙
∫ ( 2 ) 𝑑π‘₯ =
(𝑒 𝛿 − 1)
πœ•π‘₯
𝛿
0
But for a semi-infinite cylindrical rod, 𝑙 = ∞, upon substitution, we get
𝑙
(𝑇𝑠 − 𝑇∞ )
πœ•2𝑇
∫ ( 2 ) 𝑑π‘₯ =
𝛿
0 πœ•π‘₯
𝑙
−𝑙
∫ (𝑇 − 𝑇∞ )𝑑π‘₯ = −𝛿 (𝑇𝑠 − 𝑇∞ )(𝑒 𝛿 − 1)
0
But 𝑙 = ∞, upon substitution, we get
𝑙
∫ (𝑇 − 𝑇∞ )𝑑π‘₯ = 𝛿 (𝑇𝑠 − 𝑇∞ )
0
−π‘₯
𝑇 = (𝑇𝑠 − 𝑇∞ )𝑒 𝛿 + 𝑇∞
𝑙
−𝑙
∫ (𝑇)𝑑π‘₯ = −𝛿(𝑇𝑠 − 𝑇∞ )(𝑒 𝛿 − 1) + 𝑇∞ 𝑙
0
Substitute 𝑙 = ∞ and get
πœ• 𝑙
𝑑𝛿
πœ•
(𝑇𝑠 − 𝑇∞ ) + (𝑇∞ 𝑙)
∫ (𝑇)𝑑π‘₯ =
πœ•π‘‘ 0
𝑑𝑑
πœ•π‘‘
Since 𝑇∞ π‘Žπ‘›π‘‘ 𝑙 are constants
πœ•
(𝑇 𝑙 ) = 0
πœ•π‘‘ ∞
πœ• 𝑙
𝑑𝛿
(𝑇 − 𝑇∞ )
∫ (𝑇)𝑑π‘₯ =
πœ•π‘‘ 0
𝑑𝑑 𝑠
Substituting the above expressions in equation b) above, we get
𝛼−
β„Žπ‘ƒ 2
𝑑𝛿
𝛿 =𝛿
𝐴𝜌𝐢
𝑑𝑑
We solve the equation above with initial condition
𝛿 = 0 π‘Žπ‘‘ 𝑑 = 0
And get
𝛿=√
−2β„Žπ‘ƒ
π›Όπ΄πœŒπΆ
𝑑
(1 − 𝑒 𝐴𝜌𝐢 )
β„Žπ‘ƒ
−2β„Žπ‘ƒ
𝐾𝐴
𝑑
𝛿 = √ (1 − 𝑒 𝐴𝜌𝐢 )
β„Žπ‘ƒ
Substituting for 𝛿 in the temperature profile, we get
−𝒙
−πŸπ’‰π‘·
𝑲𝑨
𝒕
√ (𝟏−𝒆 𝑨𝝆π‘ͺ )
𝑻 − 𝑻∞
𝒉𝑷
=𝒆
𝑻𝒔 − 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 (𝑑 = ∞) it
reduces to
𝒉𝑷
𝑻 − 𝑻∞
−√( )𝒙
𝑲𝑨
=𝒆
𝑻𝒔 − 𝑻∞
What are the predictions of the transient state?
For transient state the governing solution is:
−π‘₯
−2β„Žπ‘ƒ
𝑇 − 𝑇∞
=𝑒
𝑇𝑠 − 𝑇∞
𝑑
𝐾𝐴
√ (1−𝑒 𝐴𝜌𝐢 )
β„Žπ‘ƒ
Let us make π‘₯ the subject of the equation of transient state and get:
π‘₯ = [ln (
−2β„Žπ‘ƒ
𝑇𝑠 − 𝑇∞
𝐾𝐴
𝑑
)] × √ (1 − 𝑒 𝐴𝜌𝐢 )
𝑇 − 𝑇∞
β„Žπ‘ƒ
To measure the value of h, we use trial and error method in Microsoft excel and
2β„Žπ‘ƒ
choosing (𝐴𝜌𝐢 = 0.005) , plotting a graph of π‘₯ against √(1 − 𝑒 −0.005𝑑 ) for a semiinfinite aluminium metal rod of radius 2mm gave a straight-line graph with a
negative intercept as shown below for all times contrary to the equation
above i.e.,
π‘₯ = −𝑐 + [ln (
−2β„Žπ‘ƒ
𝑇𝑠 − 𝑇∞
𝐾𝐴
𝑑
)] × √
(1 − 𝑒 𝐴𝜌𝐢 )
𝑇 − 𝑇∞
β„Žπ‘ƒ
Let:
−2β„Žπ‘ƒ
𝑑
−0.005𝑑
)
π‘Œ = √(1 − 𝑒
= √(1 − 𝑒 𝐴𝜌𝐢 )
𝒙 = −𝒄 + [π₯𝐧 (
𝑻𝒔 − 𝑻∞
𝑲𝑨
) (√ )] × π’€
𝑻 − 𝑻∞
𝒉𝑷
A Graph of X against Y for radius 2mm semiinfinite aluminium rod
0.3
X
0.25
y = 1.515x - 0.0528
R² = 0.9967
0.2
0.15
Series1
0.1
Linear (Series1)
0.05
0
0
0.05
0.1
0.15
0.2
Y
Varying the radius of the aluminium metal rod to π‘Ÿ = 1π‘šπ‘š the graph looked as
below:
X
A Graph of X against Y for radius 1mm semiinfinite aluminium rod
0.18
0.16
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
y = 1.3556x - 0.0139
R² = 0.987
Series1
Linear (Series1)
0
0.05
0.1
0.15
Y
From the graph above, it is observed that the intercept c is directly proportional
to radius squared.
i.e.
𝒄 = πŸπŸ‘πŸ—πŸŽπŸŽπ’“πŸ
The heat transfer coefficient is calculated from
−2β„Žπ‘ƒ
𝑑
π‘Œ = √(1 − 𝑒 −0.005𝑑 ) = √(1 − 𝑒 𝐴𝜌𝐢 )
2β„Žπ‘ƒ
= 0.005
𝐴𝜌𝐢
h for aluminium was found to be
β„Ž=
6.075π‘Š
π‘š2 𝐾
Using the graph and the equation below:
π‘₯ = [ln (
−2β„Žπ‘ƒ
𝑇𝑓 − 𝑇∞
𝐾𝐴
𝑑
)] × √
(1 − 𝑒 𝐴𝜌𝐢 ) − π›½π‘Ÿ 2
𝑇 − 𝑇∞
β„Žπ‘ƒ
−2β„Žπ‘ƒ
𝑑
𝑇𝑓 −𝑇∞
From the gradient of the graph of x against √(1 − 𝑒 𝐴𝜌𝐢 )above the ratio (
𝑇−𝑇∞
)
was measured and was found to be:
(
𝑻 − 𝑻∞
) ≈ 𝟎. πŸπŸπŸ—πŸ–πŸ
𝑻𝒇 − 𝑻∞
From the graph above it is deduced that 𝑇𝑓 is a constant temperature
independent of time.
To account for the intercept in the graph above for a semi-infinite rod, we have
to postulate that there’s convection at the hot end of the metal rod as below
i.e.,
−π’Œ
𝝏𝑻
|
= π’‰πŸŽ (𝑻𝒇 − 𝑻𝒔 )
𝝏𝒙 𝒙=𝟎
Recall:
−π‘₯
𝑇 − 𝑇∞
=𝑒𝛿
𝑇𝑠 − 𝑇∞
πœ•π‘‡ −(𝑇𝑠 − 𝑇∞ ) −π‘₯
=
𝑒𝛿
πœ•π‘₯
𝛿
πœ•π‘‡
−(𝑇𝑠 − 𝑇∞ )
|π‘₯=0 =
πœ•π‘₯
𝛿
Upon substitution, we get:
π‘˜
(𝑇𝑠 − 𝑇∞ )
= β„Ž0 (𝑇𝑓 − 𝑇𝑠 )
𝛿
(𝑇𝑠 − 𝑇∞ ) =
𝑇𝑠 (1 +
β„Ž0 𝛿
(𝑇𝑓 − 𝑇𝑠 )
π‘˜
β„Ž0 𝛿
β„Ž0 𝛿
)=(
) 𝑇𝑓 + 𝑇∞
π‘˜
π‘˜
β„Ž 𝛿
( 0 ) 𝑇𝑓 + 𝑇∞
π‘˜
𝑇𝑠 =
β„Ž 𝛿
(1 + π‘˜0 )
Subtracting 𝑇∞ from both sides, we get:
𝑇𝑠 − 𝑇∞ =
β„Ž 𝛿
( π‘˜0 ) 𝑇𝑓 + 𝑇∞
β„Ž 𝛿
(1 + π‘˜0 )
− 𝑇∞
We finally get:
β„Ž0 𝛿
𝑇𝑠 − 𝑇∞
π‘˜
=
𝑇𝑓 − 𝑇∞ 1 + β„Ž0 𝛿
π‘˜
Upon simplifying, we get:
𝑇𝑠 − 𝑇∞
𝛿
=
𝑇𝑓 − 𝑇∞ 𝛿 + π‘˜
β„Ž0
To explain the nature of β„Ž0 we postulate that the candle dissipates power
independent of area. i.e.,
π‘‘π‘š
𝐻 = π›Ύβ„Ž0 𝐴(𝑇𝑓 − 𝑇𝑠 )
𝑑𝑑 𝐢
Where:
𝛾 = π‘π‘Ÿπ‘œπ‘π‘œπ‘Ÿπ‘‘π‘–π‘œπ‘›π‘Žπ‘™π‘–π‘‘π‘¦ π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ π‘Žπ‘›π‘‘ π‘π‘Žπ‘› 𝑏𝑒 π‘‘π‘Žπ‘˜π‘’π‘› π‘‘π‘œ 𝑏𝑒 π‘’π‘žπ‘’π‘Žπ‘™ π‘‘π‘œ 1
And get
π’…π’Ž
𝑯 = π’‰πŸŽ 𝑨(𝑻𝒇 − 𝑻𝒔 )
𝒅𝒕 π‘ͺ
Where:
𝐻𝐢 = π‘’π‘›π‘‘β„Žπ‘Žπ‘™π‘π‘¦ π‘œπ‘“ π‘π‘œπ‘šπ‘π‘’π‘ π‘‘π‘–π‘œπ‘›
From experiment:
π‘‘π‘š
= π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘
𝑑𝑑
To explain the heat conduction phenomenon, we postulate that
𝐻𝐢 = πΆπ‘œ (𝑇𝑓 − 𝑇𝑠 )
Where:
πΆπ‘œ = 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 β„Žπ‘’π‘Žπ‘‘ π‘π‘Žπ‘π‘Žπ‘π‘–π‘‘π‘¦
Upon substitution, we have:
π‘‘π‘š
𝐢 (𝑇 − 𝑇𝑠 ) = β„Ž0 𝐴(𝑇𝑓 − 𝑇𝑠 )
𝑑𝑑 π‘œ 𝑓
Upon simplifying, we get:
π’‰πŸŽ =
π’…π’Ž π‘ͺ𝒐
×
𝒅𝒕
𝑨
Using the above result, we get
π‘˜
π‘˜π΄
=
β„Ž0 π‘‘π‘š 𝐢
𝑑𝑑 π‘œ
Upon substitution, we get:
𝑇𝑠 − 𝑇∞
𝛿
=
𝑇𝑓 − 𝑇∞ 𝛿 + π‘˜
β„Ž0
𝑇𝑠 − 𝑇∞
𝛿
=
𝑇𝑓 − 𝑇∞ 𝛿 + π‘˜π΄
π‘‘π‘š
𝑑𝑑 πΆπ‘œ
OR
𝑇𝑠 − 𝑇∞
𝛿
=
𝑇𝑓 − 𝑇∞ 𝛿 + π‘˜πœ‹ π‘Ÿ 2
π‘‘π‘š
𝐢
𝑑𝑑 π‘œ
Comparing with:
𝑻𝒔 − 𝑻∞
𝜹
=
𝑻𝒇 − 𝑻∞ 𝜹 + πœ·π’“πŸ
Where:
𝛽=
π‘˜πœ‹
π‘‘π‘š
𝐢
𝑑𝑑 π‘œ
We notice that 𝛽 is directly proportional to the thermal conductivity k.
The above expression shows that the temperature at π‘₯ = 0 varies with time also
and is not fixed until steady state is achieved
π’Œ
= πœ·π’“πŸ
π’‰πŸŽ
And we finally get:
𝑇𝑠 − 𝑇∞
𝛿
=
𝑇𝑓 − 𝑇∞ 𝛿 + π›½π‘Ÿ 2
For a semi-infinite rod,
𝒙
𝑻 − 𝑻∞
= 𝒆− 𝜹
𝑻𝒔 − 𝑻∞
Substituting the expression of (𝑇𝑠 − 𝑇∞ ) we get:
𝒙
𝑻 − 𝑻∞
𝜹
=(
) 𝒆− 𝜹
𝟐
𝑻𝒇 − 𝑻∞
𝜹 + πœ·π’“
As the general solution.
SO HOW DO WE PRODUCE THE SEMI-INFINITE OBSERVED ROD
SOLUTION?
As got before:
𝑇𝑠 − 𝑇∞
𝛿
=(
)
𝑇𝑓 − 𝑇∞
𝛿 + π›½π‘Ÿ 2
From
(
𝒙
𝑻 − 𝑻∞
) = 𝒆− 𝜹
𝑻𝒔 − 𝑻∞
Substituting the expression of (𝑇𝑠 − 𝑇∞ ) we get:
𝒙
𝑻 − 𝑻∞
𝜹
=(
) 𝒆− 𝜹
𝟐
𝑻𝒇 − 𝑻∞
𝜹 + πœ·π’“
Continuing with
𝒙
𝑻 − 𝑻∞
= 𝒆− 𝜹
𝑻𝒔 − 𝑻∞
Let us make π‘₯ the subject of the formula:
π‘₯ = [ln (
𝑇𝑠 − 𝑇∞
)] × π›Ώ
𝑇 − 𝑇∞
Where:
𝛿 = √2𝛼𝑑
For small time.
Or generally
−2β„Žπ‘ƒ
𝐾𝐴
𝑑
𝛿 = √ (1 − 𝑒 𝐴𝜌𝐢 )
β„Žπ‘ƒ
But we know from the above that:
(𝑇𝑠 − 𝑇∞ ) = (𝑇𝑓 − 𝑇∞ )(
𝛿
)
𝛿 + π›½π‘Ÿ 2
π‘₯ = [ln(𝑇𝑠 − 𝑇∞ ) − ln(𝑇 − 𝑇∞ )] × π›Ώ
𝑙𝑛(𝑇𝑠 − 𝑇∞ ) = 𝑙𝑛(𝑇𝑓 − 𝑇∞ ) + 𝑙𝑛(
𝛿
)
𝛿 + π›½π‘Ÿ 2
Upon substitution of 𝑙𝑛(𝑇𝑠 − 𝑇∞ ) in the equation of π‘₯ above, we get
π‘₯ = [𝑙𝑛(𝑇𝑓 − 𝑇∞ ) + 𝑙𝑛(
𝛿
) −ln(𝑇 − 𝑇∞ )] × π›Ώ
𝛿 + π›½π‘Ÿ 2
We get
𝒙 = 𝜹[𝒍𝒏 (
𝑻𝒇 − 𝑻∞
𝜹
) + 𝒍𝒏(
)]
𝑻 − 𝑻∞
𝜹 + πœ·π’“πŸ
Let us manipulate the equation above and get:
𝒙 = πœΉπ’π’ (
𝑻𝒇 − 𝑻∞
𝜹
) + πœΉπ’π’(
)
𝑻 − 𝑻∞
𝜹 + πœ·π’“πŸ
Factorizing out 𝛿 in the denominator we get:
π‘₯ = 𝛿𝑙𝑛 (
𝑇𝑓 − 𝑇∞
𝛿
1
)]
) + 𝛿𝑙𝑛[ (
π›½π‘Ÿ 2
𝑇 − 𝑇∞
𝛿
1+ 𝛿
𝑇𝑓 − 𝑇∞
π›½π‘Ÿ 2 −1
π‘₯ = 𝛿𝑙𝑛 (
) + 𝛿𝑙𝑛[(1 +
) ]
𝑇 − 𝑇∞
𝛿
Since
π›½π‘Ÿ 2
β‰ͺ 1 π‘“π‘œπ‘Ÿ 𝑑
𝛿
We can use the binomial first order approximation
(1 + π‘₯)𝑛 ≈ 1 + 𝑛π‘₯ π‘“π‘œπ‘Ÿ π‘₯ β‰ͺ 1
(1 +
π›½π‘Ÿ 2 −1
π›½π‘Ÿ 2
) =1−
𝛿
𝛿
π‘“π‘œπ‘Ÿ
π›½π‘Ÿ 2
β‰ͺ1
𝛿
And we get:
π‘₯ = 𝛿𝑙𝑛 (
𝑇𝑓 − 𝑇∞
π›½π‘Ÿ 2
) + 𝛿𝑙𝑛[(1 −
)]
𝑇 − 𝑇∞
𝛿
Again, we can expand the natural log as below:
ln(1 − π‘₯ ) ≈ −π‘₯ π‘“π‘œπ‘Ÿ π‘₯ β‰ͺ 1
𝑙𝑛 [(1 −
π›½π‘Ÿ 2
π›½π‘Ÿ 2
)] = −
𝛿
𝛿
π‘“π‘œπ‘Ÿ
π›½π‘Ÿ 2
β‰ͺ1
𝛿
Upon substitution we finally get
𝒙 = πœΉπ’π’ (
𝑻𝒇 − 𝑻∞
) − πœ·π’“πŸ
𝑻 − 𝑻∞
Which is what we got before.
𝒙 = πœΉπ’π’ (
𝑻𝒇 − 𝑻∞
) − πœ·π’“πŸ
𝑻 − 𝑻∞
Where:
𝛽=
π‘˜πœ‹
π‘‘π‘š
𝑑𝑑 πΆπ‘œ
We notice that the intercept above is proportional to the square of the radius as
demonstrated from experiment.
Looking at the general solution:
𝒙 = πœΉπ’π’ (
𝑻𝒇 − 𝑻∞
𝜹
) + πœΉπ’π’(
)
𝑻 − 𝑻∞
𝜹 + πœ·π’“πŸ
Plotting a graph of 𝒙 against 𝜹[𝒍𝒏 (
𝑻𝒇 −𝑻∞
𝑻−𝑻∞
) + 𝒍𝒏(
𝜹
𝜹+πœ·π’“πŸ
)] was found to give a
straight-line graph through the origin as stated by the equation above.
Let us call 𝒑 = 𝜹[𝒍𝒏 (
𝑻𝒇 −𝑻∞
𝑻−𝑻∞
) + 𝒍𝒏(
𝜹
𝜹+πœ·π’“πŸ
)]
Where:
π‘Š
π‘Š
πœ·π’“πŸ = 𝟎. πŸŽπŸ“πŸπŸ– for an aluminium rod of radius 2mm and 𝐾 = 238 π‘šπΎ , β„Ž = 6 π‘š2𝐾 𝜌 =
π‘˜π‘”
𝐽
𝑇−𝑇
2700 π‘š3 , 𝐢 = 900 π‘˜π‘”πΎ and 𝑇 −𝑇∞ = 0.21981
𝑓
∞
And
−2β„Žπ‘ƒ
𝐾𝐴
𝑑
𝛿 = √ (1 − 𝑒 𝐴𝜌𝐢 )
β„Žπ‘ƒ
Then a graph of x against p is a straight-line graph through the origin for a
semi-infinite rod as shown below for a semi-infinite aluminium rod of radius
2mm
A Graph of x against p
0.3
0.25
y = 1.0768x
x
0.2
0.15
0.1
0.05
0
0
0.05
0.1
0.15
0.2
0.25
p
Calling the solution below the approximated solution:
𝒙 = πœΉπ’π’ (
𝑻𝒇 − 𝑻∞
) − πœ·π’“πŸ
𝑻 − 𝑻∞
Or
π‘₯ = [ln (
−2β„Žπ‘ƒ
𝑇𝑓 − 𝑇∞
𝐾𝐴
𝑑
)] × √
(1 − 𝑒 𝐴𝜌𝐢 ) − π›½π‘Ÿ 2
𝑇 − 𝑇∞
β„Žπ‘ƒ
What are the lessons we have learnt?
ο‚·
ο‚·
We have learnt that in the approximated solution, we can measure off h
in the transient state.
We have learnt that knowing the thermo-conductivity and other physical
parameters of the metal rod, in the approximated solution, we can
measure off the ratio
ο‚·
𝑻−𝑻∞
𝑻𝒇 −𝑻∞
The intercept in the approximated solution can help us learn how its
nature varies with the radius of the rod. We can use the intercept to
measure the value of 𝜷.
The general solution is given by:
𝒙 = πœΉπ’π’ (
𝑻𝒇 − 𝑻∞
𝜹
) + πœΉπ’π’(
)
𝑻 − 𝑻∞
𝜹 + πœ·π’“πŸ
You notice that the initial condition is still satisfied.
From the general solution, we get:
𝒙
𝑻 − 𝑻∞
𝜹
−
𝜹
=(
)𝒆
𝑻𝒇 − 𝑻∞
𝜹 + πœ·π’“πŸ
For the initial condition; At 𝑑 = 0, you get
−𝒙
𝑻 − 𝑻∞
=𝟎×π’†πŸŽ =𝟎
𝑻𝒇 − 𝑻∞
Hence
𝑻 = 𝑻∞
Considering the approximated equation below:
𝒙 = [π₯𝐧 (
𝑻𝒇 − 𝑻∞
)] × √πŸπœΆπ’• − πœ·π’“πŸ
𝑻 − 𝑻∞
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 π‘₯ against √𝑑 is a straight-line
graph with an intercept as stated by the equation above when the times are
small. The equation is true because that is what is observed experimentally.
Looking at the approximate solution for a semi-infinite metal rod:
π‘₯ = −π›½π‘Ÿ 2 + [ln (
−2β„Žπ‘ƒ
𝑇𝑓 − 𝑇∞
𝐾𝐴
𝑑
)] × √ (1 − 𝑒 𝐴𝜌𝐢 )
𝑇 − 𝑇∞
β„Žπ‘ƒ
Let:
π‘Œ = √(1 − 𝑒 −0.005𝑑 )
A graph of x against Y looked as below:
A Graph of X against Y
0.3
0.25
y = 1.515x - 0.0528
R² = 0.9967
X
0.2
0.15
Series1
0.1
Linear (Series1)
0.05
0
0
0.05
0.1
0.15
0.2
Y
Since the graph above is a straight-line graph, it shows that 𝑇𝑓 IS NOT a
function of time as stated by the equation above of π‘₯ against Y.
Another point to note is that from experiment 𝑇𝑓 was found to be independent
of radius of the metal rod.
For aluminium
𝐾 = 238
π‘Š
π‘Š
,β„Ž = 6 2
π‘šπΎ
π‘š 𝐾
Another way to measure 𝑇𝑠1 is to consider the steady state equation and plot
the graph of
β„Žπ‘ƒ
𝑇 − 𝑇∞
𝛿
−√(𝐾𝐴)π‘₯
=(
)𝑒
𝑇𝑓 − 𝑇∞
𝛿 + π›½π‘Ÿ2
π₯𝐧(𝑻 − 𝑻∞ ) = 𝒍𝒏(𝑻𝒇 − 𝑻∞ ) + 𝒍𝒏(
And
𝛿=√
Upon substitution
𝐾𝐴
β„Žπ‘ƒ
𝜹
𝒉𝑷
√
)
−
(
)𝒙
𝜹 + πœ·π’“πŸ
𝑲𝑨
π₯𝐧(𝑻 − 𝑻∞ ) = 𝒍𝒏(𝑻𝒇 − 𝑻∞ ) + 𝒍𝒏(
√𝑲𝑨
𝒉𝑷
√𝑲𝑨 + πœ·π’“πŸ
𝒉𝑷
) − √(
𝒉𝑷
)𝒙
𝑲𝑨
𝑲𝑨
𝒉𝑷
√
A graph of ln(𝑇 − 𝑇∞ ) against x gives an intercept [𝒍𝒏(𝑻𝒇 − 𝑻∞ ) + 𝒍𝒏( 𝑲𝑨
√
𝒉𝑷
)] from
+πœ·π’“πŸ
which 𝑇𝑓 can be measured.
Also knowing the thermo-conductivity, from the gradient of the above graph the
heat transfer coefficient can be measured off.
From experiment, using an aluminium rod of radius 2mm and using a
thermoconductivity value of πŸπŸ‘πŸ– 𝑾⁄π’Žπ‘² , The heat transfer coefficient h of
aluminium was found to be πŸ” 𝑾⁄ 𝟐 .
π’Ž 𝑲
Therefore, for a semi-infinite rod, the equation obeyed for small times is:
𝒙 = [π₯𝐧 (
𝑻𝒇 − 𝑻∞
)] × √πŸπœΆπ’• − πœ·π’“πŸ
𝑻 − 𝑻∞
Looking at the steady state solution below:
β„Žπ‘ƒ
𝑇 − 𝑇∞
𝛿
−√(𝐾𝐴)π‘₯
=(
)
𝑒
𝑇𝑓 − 𝑇∞
𝛿 + π›½π‘Ÿ2
OR
√𝑲𝑨
𝒉𝑷
𝑻 − 𝑻∞
𝒉𝑷
−√( )𝒙
=(
)𝒆 𝑲𝑨
𝑻𝒇 − 𝑻∞
√𝑲𝑨 + πœ·π’“πŸ
𝒉𝑷
In most cases
𝑲𝑨
√
≫ πœ·π’“πŸ
𝒉𝑷
So, we observe:
𝒉𝑷
𝑻 − 𝑻∞
−√( )𝒙
= 𝒆 𝑲𝑨
𝑻𝒇 − 𝑻∞
Which is the usual solution we know.
From experiment, using a flame and candle wax on the aluminium rod, the
ratio below was found to be
(
𝑻 − 𝑻∞
) ≈ 𝟎. πŸπŸπŸ—πŸ–πŸ
𝑻𝒇 − 𝑻∞
So, when can we apply the semi-infinite rod solution?
Using the steady state equation of heat conduction
β„Žπ‘ƒ
𝑇 − 𝑇∞
−√( )𝐿
= 𝑒 𝐾𝐴 = 𝑒 −π‘šπΏ
𝑇𝑠 − 𝑇∞
Where:
π‘š = √(
β„Žπ‘ƒ
)
𝐾𝐴
And
From literature [1] the limiting length for use of semi-infinite model is got when
𝑇 − 𝑇∞
= 0.01
𝑇𝑠 − 𝑇∞
The corresponding value of π‘šπΏ = 4.6
πŸ’.πŸ”
𝑲𝑨
Hence as long as 𝑳 = π’Ž = πŸ’. πŸ”√( 𝒉𝑷 ) the equation can be applied accurately.
NB
A particular method we can use to predict which temperature profile to use in
solving the heat equation is by looking at the steady state equation below:
From
πœ•2𝑇
β„Žπ‘ƒ
πœ•π‘‡
(𝑇 − 𝑇∞ ) =
𝛼 2−
πœ•π‘₯
𝐴𝜌𝐢
πœ•π‘‘
In steady state
πœ•π‘‡
=0
πœ•π‘‘
So, the governing equation becomes:
𝛼
πœ•2𝑇
β„Žπ‘ƒ
(𝑇 − 𝑇∞ ) = 0
−
πœ•π‘₯ 2 𝐴𝜌𝐢
The general solution of the equation above is
(𝑇 − 𝑇∞ ) = 𝐢1 𝑒 −π‘šπ‘₯ + 𝐢2 𝑒 π‘šπ‘₯
Where:
π‘š=√
β„Žπ‘ƒ
𝐾𝐴
For the semi-infinite case: The boundary conditions are:
𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎
𝑻 = 𝑻∞ 𝒂𝒕 𝒙 = ∞
The second boundary condition makes 𝐢2 = 0
And the other boundary condition:
𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎
Leads to
𝑇 − 𝑇∞
= 𝑒 −π‘šπ‘₯
𝑇𝑠 − 𝑇∞
From what we learned earlier is that
π‘š=
1
𝛿
From now onwards we are going to use the fact that the temperature profile
below satisfies the heat equation
(𝑇 − 𝑇∞ ) = 𝐢1 𝑒 −π‘šπ‘₯ + 𝐢2 𝑒 π‘šπ‘₯
Or
−𝒙
𝒙
(𝑻 − 𝑻∞ ) = π‘ͺ𝟏 𝒆 𝜹 + π‘ͺ𝟐 π’†πœΉ
HOW DOES HEAT FLOW MANIFEST ITSELF FOR FINITE
METAL RODS?
CASE 1: CONVECTION AT THE END OF A FINITE METAL ROD
The boundary and initial conditions are:
𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎
−π’Œ
𝒅𝑻
= 𝒉𝑳 (𝑻 − 𝑻∞ ) 𝒂𝒕 𝒙 = 𝒍
𝒅𝒙
𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎
This same analysis can be extended to a metal rod with convection at the other
end of the metal rod where the temperature profile is given by: [2]
β„ŽπΏ
𝑇 − 𝑇∞ cosh[π‘š(𝐿 − π‘₯ )] + (π‘šπ‘˜) sinh[π‘š(𝐿 − π‘₯)]
=
β„ŽπΏ
𝑇𝑠 − 𝑇∞
cosh π‘šπΏ + (π‘šπ‘˜
) π‘ π‘–π‘›β„Žπ‘šπΏ
Or
𝑇 − 𝑇∞
=
𝑇𝑠 − 𝑇∞
cosh [
(𝐿 − π‘₯ )
β„ŽπΏ 𝛿
𝐿−π‘₯
𝛿 ] + ( π‘˜ ) sinh[( 𝛿 )]
𝐿
β„Ž 𝛿
𝐿
cosh + ( 𝐿 ) π‘ π‘–π‘›β„Ž
𝛿
π‘˜
𝛿
To show that the initial condition is satisfied we see from the above that π‘Žπ‘‘ 𝑑 =
0, 𝛿 = 0.
(𝐿 − π‘₯ )
β„ŽπΏ 𝛿
𝐿−π‘₯
𝑇 − 𝑇∞ cosh [ 𝛿 ] + ( π‘˜ ) sinh[( 𝛿 )]
=
𝐿
β„Žπ›Ώ
𝐿
𝑇𝑠 − 𝑇∞
cosh + ( ) π‘ π‘–π‘›β„Ž
𝛿
π‘˜
𝛿
Becomes:
(𝐿 − π‘₯ )
(𝐿−π‘₯)
−(𝐿−π‘₯)
𝑇 − 𝑇∞ cosh [ 𝛿 ] 𝑒 𝛿 + 𝑒 𝛿
=
=
𝐿
−𝐿
𝐿
𝑇𝑠 − 𝑇∞
𝛿 +𝑒 𝛿
cosh
𝑒
𝛿
𝑒
Similarly
−(𝐿−π‘₯)
(𝐿−π‘₯)
𝛿
= 𝑒 − 0 = 𝑒 −∞(𝐿−π‘₯) = 0
−𝐿
−𝐿
𝑒 𝛿 = 𝑒 0 = 𝑒 −∞𝐿 = 0
So, we are left with
𝑇 − 𝑇∞ 𝑒
=
𝑇𝑠 − 𝑇∞
(𝐿−π‘₯)
𝛿
𝐿
𝑒𝛿
−π‘₯
−π‘₯
= 𝑒 𝛿 = 𝑒 0 = 𝑒 −∞π‘₯ = 0
Hence at 𝑑 = 0, 𝑇 = 𝑇∞ and hence the initial condition.
To explain the transient state provide we have to get the expression for
(𝑇𝑠 − 𝑇∞ ) from:
As we learned earlier in the semi-infinite case, we use
−π’Œ
𝝏𝑻
|
= π’‰πŸŽ (𝑻𝒇 − 𝑻𝒔 )
𝝏𝒙 𝒙=𝟎
Recall the compact temperature profile is:
β„ŽπΏ
𝑇 − 𝑇∞ cosh[π‘š(𝐿 − π‘₯ )] + (π‘šπ‘˜) sinh[π‘š(𝐿 − π‘₯)]
=
β„ŽπΏ
𝑇𝑠 − 𝑇∞
cosh π‘šπΏ + (π‘šπ‘˜
) π‘ π‘–π‘›β„Žπ‘šπΏ
Where:
π‘š=
1
𝛿
−2β„Žπ‘ƒ
𝐾𝐴
𝑑
𝛿 = √ (1 − 𝑒 𝐴𝜌𝐢 )
β„Žπ‘ƒ
As shall be shown later
β„Ž
π‘šπ‘‘π‘Žπ‘›β„Žπ‘šπΏ + π‘˜πΏ
πœ•π‘‡
|
= −(𝑇𝑠 − 𝑇∞ )(
)
β„ŽπΏ
πœ•π‘₯ π‘₯=0
1 + π‘šπ‘˜ π‘‘π‘Žπ‘›β„Žπ‘šπΏ
β„Ž
π‘šπ‘‘π‘Žπ‘›β„Žπ‘šπΏ + π‘˜πΏ
πœ•π‘‡
)
−π‘˜
|
= π‘˜(𝑇𝑠 − 𝑇∞ ) (
β„ŽπΏ
πœ•π‘₯ π‘₯=0
1 + π‘šπ‘˜ π‘‘π‘Žπ‘›β„Žπ‘šπΏ
β„Ž
π‘šπ‘‘π‘Žπ‘›β„Žπ‘šπΏ + π‘˜πΏ
)
β„Ž0 (𝑇𝑓 − 𝑇𝑠 ) = π‘˜(𝑇𝑠 − 𝑇∞ ) (
β„ŽπΏ
1 + π‘šπ‘˜
π‘‘π‘Žπ‘›β„Žπ‘šπΏ
β„Ž
π‘šπ‘‘π‘Žπ‘›β„Žπ‘šπΏ + π‘˜πΏ
β„Ž
π‘šπ‘‘π‘Žπ‘›β„Žπ‘šπΏ + π‘˜πΏ
) − π‘˜(𝑇∞ ) (
)
β„Ž0 (𝑇𝑓 ) − β„Ž0 (𝑇𝑠 ) = π‘˜ (𝑇𝑠 ) (
β„ŽπΏ
β„ŽπΏ
1 + π‘šπ‘˜
π‘‘π‘Žπ‘›β„Žπ‘šπΏ
1 + π‘šπ‘˜
π‘‘π‘Žπ‘›β„Žπ‘šπΏ
Collecting like terms we get:
β„Ž
π‘šπ‘‘π‘Žπ‘›β„Žπ‘šπΏ + π‘˜πΏ
β„Ž
π‘šπ‘‘π‘Žπ‘›β„Žπ‘šπΏ + π‘˜πΏ
) + β„Ž0 ) = β„Ž0 (𝑇𝑓 ) + π‘˜(𝑇∞ ) (
)
𝑇𝑠 (π‘˜ (
β„ŽπΏ
β„ŽπΏ
1 + π‘šπ‘˜ π‘‘π‘Žπ‘›β„Žπ‘šπΏ
1 + π‘šπ‘˜
π‘‘π‘Žπ‘›β„Žπ‘šπΏ
β„Ž
π‘šπ‘‘π‘Žπ‘›β„Žπ‘šπΏ + π‘˜πΏ
)
β„Ž0 (𝑇𝑓 ) + π‘˜(𝑇∞ ) (
β„ŽπΏ
1 + π‘šπ‘˜
π‘‘π‘Žπ‘›β„Žπ‘šπΏ
𝑇𝑠 =
β„Ž
π‘šπ‘‘π‘Žπ‘›β„Žπ‘šπΏ + π‘˜πΏ
(π‘˜ (
) + β„Ž0 )
β„Ž
1 + 𝐿 π‘‘π‘Žπ‘›β„Žπ‘šπΏ
π‘šπ‘˜
Subtracting 𝑇∞ from both sides we get:
β„Ž
π‘šπ‘‘π‘Žπ‘›β„Žπ‘šπΏ + π‘˜πΏ
)
β„Ž0 (𝑇𝑓 ) + π‘˜(𝑇∞ ) (
β„Ž
1 + 𝐿 π‘‘π‘Žπ‘›β„Žπ‘šπΏ
π‘šπ‘˜
𝑇𝑠 − 𝑇∞ =
− 𝑇∞
β„ŽπΏ
π‘šπ‘‘π‘Žπ‘›β„Žπ‘šπΏ + π‘˜
(π‘˜ (
) + β„Ž0 )
β„ŽπΏ
1 + π‘šπ‘˜
π‘‘π‘Žπ‘›β„Žπ‘šπΏ
Upon simplification, we get:
𝑇𝑠 − 𝑇∞
=
𝑇𝑓 − 𝑇∞
β„Ž0
β„Ž
π‘šπ‘‘π‘Žπ‘›β„Žπ‘šπΏ + π‘˜πΏ
(π‘˜ (
) + β„Ž0 )
β„ŽπΏ
1 + π‘šπ‘˜ π‘‘π‘Žπ‘›β„Žπ‘šπΏ
Dividing through by β„Ž0 we get:
𝑇𝑠 − 𝑇∞
=
𝑇𝑓 − 𝑇∞
1
β„ŽπΏ
π‘˜ π‘šπ‘‘π‘Žπ‘›β„Žπ‘šπΏ + π‘˜
( (
) + 1)
β„Ž0
β„ŽπΏ
1 + π‘šπ‘˜
π‘‘π‘Žπ‘›β„Žπ‘šπΏ
But
π‘š=
1
𝛿
Upon substitution and simplification, we get
𝑇𝑠 − 𝑇∞
=
𝑇𝑓 − 𝑇∞
π‘˜
β„Ž0
(
(
1
𝐿
π‘‘π‘Žπ‘›β„Ž 𝛿 β„Ž
𝐿
𝛿 + π‘˜
β„Ž 𝛿
𝐿
1 + π‘˜πΏ π‘‘π‘Žπ‘›β„Ž 𝛿
+1
)
)
Multiplying through by 𝛿 we get:
𝑇𝑠 − 𝑇∞
=
𝑇𝑓 − 𝑇∞
𝛿
𝐿 β„ŽπΏ 𝛿
π‘˜ π‘‘π‘Žπ‘›β„Ž 𝛿 + π‘˜
( (
)
β„Ž0
β„ŽπΏ 𝛿
𝐿 + 𝛿)
1+
π‘‘π‘Žπ‘›β„Ž
π‘˜
𝛿
But from the semi-infinite rod solution, we have
π‘˜
= π›½π‘Ÿ 2
β„Ž0
Upon substitution, we get:
𝑇𝑠 − 𝑇∞
=
𝑇𝑓 − 𝑇∞
𝛿
𝐿 β„Ž 𝛿
π‘‘π‘Žπ‘›β„Ž 𝛿 + π‘˜πΏ
2
(π›½π‘Ÿ (
)
β„Ž 𝛿
𝐿 + 𝛿)
1 + 𝐿 π‘‘π‘Žπ‘›β„Ž
π‘˜
𝛿
(𝑇𝑠 − 𝑇∞ ) = (𝑇𝑓 − 𝑇∞ )
Upon substitution of (𝑇𝑠 − 𝑇∞ ) in:
𝛿
𝐿 β„Ž 𝛿
π‘‘π‘Žπ‘›β„Ž + 𝐿
𝛿
π‘˜ ) + 𝛿)
(π›½π‘Ÿ 2 (
β„ŽπΏ 𝛿
𝐿
1+
π‘‘π‘Žπ‘›β„Ž
(
)
π‘˜
𝛿
(𝐿 − π‘₯ )
β„ŽπΏ 𝛿
𝐿−π‘₯
𝑇 − 𝑇∞ cosh [ 𝛿 ] + ( π‘˜ ) sinh[( 𝛿 )]
=
𝐿
β„Ž 𝛿
𝐿
𝑇𝑠 − 𝑇∞
cosh + ( 𝐿 ) π‘ π‘–π‘›β„Ž
𝛿
π‘˜
𝛿
We get:
𝑻 − 𝑻∞
=
𝑻𝒇 − 𝑻∞
( 𝑳 − 𝒙)
𝒉 𝜹
𝑳−𝒙
𝐜𝐨𝐬𝐑 [ 𝜹 ] + ( π’Œπ‘³ ) 𝐬𝐒𝐧𝐑 [( 𝜹 )]
𝜹
(
)
𝑳
𝒉 𝜹
𝑳
𝑳 𝒉 𝜹
𝐜𝐨𝐬𝐑 + ( 𝑳 ) π’”π’Šπ’π’‰
𝒕𝒂𝒏𝒉 𝜹 + π’Œπ‘³
𝜹
π’Œ
𝜹
(πœ·π’“πŸ (
)
𝒉𝑳 𝜹
𝑳 + 𝜹)
𝟏+
𝒕𝒂𝒏𝒉
(
)
π’Œ
𝜹
HOW DO WE INVESTIGATE THE NATURE OF 𝒉𝑳 EASILY?
For convection boundary condition, the temperature profile obeyed is:
𝑇 − 𝑇∞
=
𝑇𝑓 − 𝑇∞
(𝐿 − π‘₯ )
β„Ž 𝛿
𝐿−π‘₯
cosh [ 𝛿 ] + ( π‘˜πΏ ) sinh [( 𝛿 )]
𝛿
(
)
𝐿
β„ŽπΏ 𝛿
𝐿
𝐿 β„ŽπΏ 𝛿
cosh + (
) π‘ π‘–π‘›β„Ž
π‘‘π‘Žπ‘›β„Ž 𝛿 + π‘˜
𝛿
π‘˜
𝛿
(π›½π‘Ÿ 2 (
) + 𝛿)
β„ŽπΏ 𝛿
𝐿
1+
π‘‘π‘Žπ‘›β„Ž
(
)
π‘˜
𝛿
To investigate 𝒉𝑳 easily, we use this simple experiment:
Where:
−2β„Žπ‘ƒ
𝐾𝐴
𝑑
𝛿 = √ (1 − 𝑒 𝐴𝜌𝐢 )
β„Žπ‘ƒ
As shall be shown later when we solve the heat equation analytically using the
integral transform to get 𝛿 .
For a wax particle at
π‘₯=𝐿
As shown in the diagram above with convection allowed:
The temperature profile obeyed is:
𝑇 − 𝑇∞
=
𝑇𝑓 − 𝑇∞
𝛿
1
(
)
𝐿
β„ŽπΏ 𝛿
𝐿
𝐿 β„ŽπΏ 𝛿
cosh
+
(
)
π‘ π‘–π‘›β„Ž
π‘‘π‘Žπ‘›β„Ž 𝛿 + π‘˜
𝛿
π‘˜
𝛿
(π›½π‘Ÿ 2 (
)
β„ŽπΏ 𝛿
𝐿 + 𝛿)
1+
π‘‘π‘Žπ‘›β„Ž
(
)
π‘˜
𝛿
Mathematically, what form should the equation of β„ŽπΏ against length L take on?
First of all, we know that when the length L becomes zero(i.e., there is no metal
rod), the flux is due to only the flame and is given by
π‘ž = β„Žπ‘œ (𝑇𝑓 − 𝑇∞ )
The power then is
β„Žπ‘œ 𝐴(𝑇𝑓 − 𝑇∞ ) =
π‘‘π‘š
𝐻
𝑑𝑑 𝐢
The above is the condition to be satisfied.
The flux at x=L, is given by:
π‘ž = β„ŽπΏ (𝑇𝐿 − 𝑇∞ )
Upon substituting for (𝑇𝐿 − 𝑇∞ ), we get
𝒒 = 𝒉𝑳
𝜹
𝟏
(
) (𝑻𝒇 − 𝑻∞ )
𝑳
𝒉𝑳 𝜹
𝑳
𝑳 𝒉𝑳 𝜹
𝐜𝐨𝐬𝐑 + (
) π’”π’Šπ’π’‰
𝒕𝒂𝒏𝒉 𝜹 + π’Œ
𝜹
π’Œ
𝜹
(πœ·π’“πŸ (
) + 𝜹)
𝒉𝑳 𝜹
𝑳
𝟏+
𝒕𝒂𝒏𝒉
(
)
π’Œ
𝜹
We know that when L=0, the flux should reduce to
𝒒 = 𝒉𝒐 (𝑻𝒇 − 𝑻∞ )
Making the first guess that
β„ŽπΏ = β„Žπ‘œ 𝑒 −π‘šπΏ
Where:
π‘š=√
𝒒 = 𝒉𝑳
β„Žπ‘ƒ
𝐾𝐴
𝜹
𝟏
(
) (𝑻𝒇 − 𝑻∞ )
𝑳
𝒉 𝜹
𝑳
𝑳 𝒉 𝜹
𝐜𝐨𝐬𝐑 + ( 𝑳 ) π’”π’Šπ’π’‰
𝒕𝒂𝒏𝒉 𝜹 + π’Œπ‘³
𝜹
π’Œ
𝜹
(πœ·π’“πŸ (
) + 𝜹)
𝒉𝑳 𝜹
𝑳
𝟏+
𝒕𝒂𝒏𝒉
(
)
π’Œ
𝜹
β„ŽπΏ = β„Žπ‘œ 𝑒 −π‘šπΏ
Substituting for L=0 in the flux equation we get
β„Ž 𝐿 = β„Žπ‘œ
So
𝒒 = π’‰πŸŽ
𝜹
𝟏
(
) (𝑻𝒇 − 𝑻∞ )
𝟎 π’‰πŸŽ 𝜹
𝟎
π’‰πŸŽ 𝜹
𝟎
𝒕𝒂𝒏𝒉 𝜹 + π’Œ
𝐜𝐨𝐬𝐑 𝜹 + ( π’Œ ) π’”π’Šπ’π’‰ 𝜹
(πœ·π’“πŸ (
) + 𝜹)
𝒉 𝜹
𝟎
𝟏 + 𝟎 𝒕𝒂𝒏𝒉
(
)
π’Œ
𝜹
We get:
π‘ž = β„Ž0 (
𝛿
) (𝑇𝑓 − 𝑇∞ )
β„Ž 𝛿
(π›½π‘Ÿ 2 ( π‘˜0 ) + 𝛿)
π‘ž = β„Ž0 (
1
) (𝑇𝑓 − 𝑇∞ )
β„Ž0
2
(π›½π‘Ÿ ( π‘˜ ) + 1)
But
β„Ž0
1
= 2
π‘˜
π›½π‘Ÿ
π‘ž = β„Ž0 (
1
) (𝑇𝑓 − 𝑇∞ )
(1 + 1)
We finally get:
𝒒=
𝟏
𝒉 (𝑻 − 𝑻∞ )
𝟐 𝒐 𝒇
Which doesn’t satisfy the condition above.
Now choosing
β„ŽπΏ =
𝛾
𝐿𝑛
Which means that β„ŽπΏ is inversely proportional to length L to power n.
As length 𝐿 → 0, β„ŽπΏ → ∞
Upon substituting in the flux equation for L=0, we end up with:
π‘ž = β„ŽπΏ
𝛿
1
(
) (𝑇𝑓 − 𝑇∞ )
0
β„Ž 𝛿
0
0 β„Ž 𝛿
cosh + ( 𝐿 ) π‘ π‘–π‘›β„Ž
π‘‘π‘Žπ‘›β„Ž 𝛿 + π‘˜πΏ
𝛿
π‘˜
𝛿
(π›½π‘Ÿ 2 (
)
β„ŽπΏ 𝛿
0 + 𝛿)
1+
π‘‘π‘Žπ‘›β„Ž
(
)
π‘˜
𝛿
𝛿
) (𝑇𝑓 − 𝑇∞ )
π‘ž = β„ŽπΏ (
β„Ž 𝛿
(π›½π‘Ÿ 2 ( 𝐿 ) + 𝛿)
π‘˜
1
) (𝑇𝑓 − 𝑇∞ )
π‘ž = β„ŽπΏ (
β„Ž
(π›½π‘Ÿ 2 ( π‘˜πΏ ) + 1)
β„Ž
β„Ž
But π‘Žπ‘  𝐿 → ∞, β„ŽπΏ → ∞, π‘ π‘œ π›½π‘Ÿ 2 ( π‘˜πΏ) + 1 ≈ π›½π‘Ÿ 2 ( π‘˜πΏ)
We get:
1
π‘ž = β„ŽπΏ
(
β„Ž
(π›½π‘Ÿ 2 ( 𝐿 ))
π‘˜
(𝑇𝑓 − 𝑇∞ )
)
We get
π‘ž=
π‘˜
(𝑇 − 𝑇∞ )
π›½π‘Ÿ 2 𝑓
But
π‘˜
= β„Žπ‘œ
π›½π‘Ÿ 2
So, we end up with:
π‘ž = β„Žπ‘œ (𝑇𝑓 − 𝑇∞ )
Which is the required equation hence β„ŽπΏ takes on the form
β„ŽπΏ =
𝛾
𝐿𝑛
From experiment, it was found that 𝑛 = 1.
Going back to
𝑇 − 𝑇∞
=
𝑇𝑓 − 𝑇∞
𝛿
1
(
)
𝐿
β„ŽπΏ 𝛿
𝐿
𝐿 β„ŽπΏ 𝛿
cosh + (
) π‘ π‘–π‘›β„Ž
π‘‘π‘Žπ‘›β„Ž 𝛿 + π‘˜
𝛿
π‘˜
𝛿
(π›½π‘Ÿ 2 (
) + 𝛿)
β„ŽπΏ 𝛿
𝐿
1+
π‘‘π‘Žπ‘›β„Ž
(
)
π‘˜
𝛿
We go ahead and rearrange the equation above to get a quadratic equation in
β„ŽπΏ and investigate the nature of β„ŽπΏ by varying the length of the metal rod and
noting the time taken for the wax to melt.
Calling,
𝑇 − 𝑇∞
1
=
𝑇𝑓 − 𝑇∞ 𝐡
And
π›½π‘Ÿ 2 = 𝑐
Upon rearranging, we get:
𝛿 2𝑐
𝐿
𝛿3
𝐿
𝐿
𝛿𝑐
𝐿
𝛿𝑐
𝐿
𝐿
2𝛿 2
𝐿
𝐡𝛿 2
𝐿
𝐿
𝐿
π’‰πŸπ‘³ [ 2 π‘ π‘–π‘›β„Ž ( ) + 2 π‘ π‘–π‘›β„Ž ( ) π‘‘π‘Žπ‘›β„Ž ( )] + 𝒉𝑳 [ π‘π‘œπ‘ β„Ž ( ) + π‘ π‘–π‘›β„Ž ( ) π‘‘π‘Žπ‘›β„Ž ( ) +
π‘ π‘–π‘›β„Ž ( ) −
π‘‘π‘Žπ‘›β„Ž ( )] + [π‘π‘ π‘–π‘›β„Ž ( ) + π›Ώπ‘π‘œπ‘ β„Ž ( ) − 𝐡𝛿]
π‘˜
𝛿
π‘˜
𝛿
𝛿
π‘˜
𝛿
π‘˜
𝛿
𝛿
π‘˜
𝛿
π‘˜
𝛿
𝛿
𝛿
From experimental values, it was found that β„ŽπΏ varies inversely with length
taking on the form below:
𝒉𝑳 =
π’‰π‘³πŸŽ
π’Žπ’
Where:
π‘š=√
β„Žπ‘ƒ
𝐾𝐴
Taking natural logs, we get:
𝑙𝑛(β„ŽπΏ ) = ln (
β„ŽπΏ0
) − ln(𝐿)
π‘š
For aluminium rods of radius 2mm, the graph looked as below:
Ln(hL)
A graph of Ln(hL) against Ln(L) for AL rods radius
2mm
-4
10
9
8
7
6
5
4
3
2
1
0
y = -1.001x + 5.1071
R² = 0.9781
-3
-2
-1
Series1
Linear (Series1)
0
Ln(L)
π’‰π‘³πŸŽ = 𝜺 ×
𝑲
𝒓
Where:
πœ€ = π‘’π‘šπ‘–π‘ π‘ π‘–π‘£π‘–π‘‘π‘¦
πœ€ = 0.006671
β„ŽπΏ =
β„ŽπΏ0
π‘šπ‘™
𝑲
𝑲
×√
𝑳
πŸπ’‰π’“
𝒉𝑳 = 𝜺 ×
The emissivity πœ€ can be taken to be independent of nature of metal.
For aluminium rods of radius 1mm, the graph looked as below:
Ln(hL)
A graph of Ln(hL) against Ln(L) for AL rods radius
1mm
-4
10
9
8
7
6
5
4
3
2
1
0
y = -1.078x + 5.364
R² = 0.9919
-3
-2
-1
Series1
Linear (Series1)
0
Ln(L)
Looking at the solution at π‘₯ = 𝐿
𝑇 − 𝑇∞
=
𝑇𝑓 − 𝑇∞
𝛿
1
(
)
𝐿
β„ŽπΏ 𝛿
𝐿
𝐿 β„ŽπΏ 𝛿
cosh + (
) π‘ π‘–π‘›β„Ž
π‘‘π‘Žπ‘›β„Ž 𝛿 + π‘˜
𝛿
π‘˜
𝛿
(π›½π‘Ÿ 2 (
) + 𝛿)
β„ŽπΏ 𝛿
𝐿
1+
π‘‘π‘Žπ‘›β„Ž
(
)
π‘˜
𝛿
Rearranging the equation above, we get
(π›½π‘Ÿ 2 (
𝐿 β„Ž 𝛿
π‘‘π‘Žπ‘›β„Ž 𝛿 + π‘˜πΏ
1+
β„ŽπΏ 𝛿
𝐿
π‘‘π‘Žπ‘›β„Ž
π‘˜
𝛿
) + 𝛿) = (
𝑇𝑓 − 𝑇∞
𝛿
)
)×(
𝐿
β„ŽπΏ 𝛿
𝐿
𝑇 − 𝑇∞
cosh + (
) π‘ π‘–π‘›β„Ž
𝛿
π‘˜
𝛿
Calling
𝑦=
(π›½π‘Ÿ 2 (
𝐿 β„Ž 𝛿
π‘‘π‘Žπ‘›β„Ž 𝛿 + π‘˜πΏ
β„Ž 𝛿
𝐿
1 + 𝐿 π‘‘π‘Žπ‘›β„Ž
π‘˜
𝛿
) + 𝛿)
And
π‘₯=(
𝛿
)
𝐿
β„Ž 𝛿
𝐿
cosh + ( 𝐿 ) π‘ π‘–π‘›β„Ž
𝛿
π‘˜
𝛿
Plotting a graph of y against x for aluminium rods of radius 1mm looked as
below with:
β„ŽπΏ = πœ€ ×
𝐾
𝐾
×√
𝐿
2β„Žπ‘Ÿ
𝒉𝑳 = 𝟎. πŸŽπŸŽπŸ”πŸ”πŸ•πŸ ×
𝑲
𝑲
×√
𝑳
πŸπ’‰π’“
A graph of y against x
0.16
y = 4.3301x
0.14
0.12
y
0.1
0.08
0.06
0.04
0.02
0
0
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
x
The flux at π‘₯ = 𝐿 is given by:
𝒒 = 𝒉𝑳 (𝑻 − 𝑻∞ )
It can be shown that after substituting for temperature (𝑻 − 𝑻∞ ) and 𝒉𝑳 , the
maximum possible flux got is when length L tends to zero and is given by
π’’π’Žπ’‚π’™ = π’‰πŸŽ (𝑻𝒇 − 𝑻∞ )
Which is the flux of the hot flame.
DERIVATION OF THE GENERAL EXPRESSION FOR HEAT TRANSFER
COEFFICIENT 𝒉𝑳 .
Recall for cylindrical rods the expression was:
β„ŽπΏ = πœ€ ×
𝐾
𝐾
×√
𝐿
2β„Žπ‘Ÿ
OR
β„ŽπΏ = 0.006671 ×
𝐾
𝐾
×√
𝐿
2β„Žπ‘Ÿ
How could we arrive to that expression from a general expression?
The general expression is given by:
𝒉𝑳 𝑨𝑳 (𝑻𝑳 − 𝑻∞ ) = 𝜺 ×
π’Žπ‘²
×𝑸
πŸπ’‰
OR
𝒉𝑳 𝑨𝑳 (𝑻𝑳 − 𝑻∞ ) = 𝟎. πŸŽπŸŽπŸ”πŸ”πŸ•πŸ ×
π’Žπ‘²
×𝑸
πŸπ’‰
Where:
𝑄=
𝑇𝐿 − 𝑇∞
𝑅
𝑅 = π‘π‘œπ‘›π‘‘π‘’π‘π‘‘π‘–π‘£π‘’ π‘Ÿπ‘’π‘ π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’
𝐴𝐿 = π‘π‘œπ‘›π‘‘π‘’π‘π‘‘π‘–π‘œπ‘› π‘Žπ‘Ÿπ‘’π‘Ž π‘Žπ‘‘ π‘™π‘’π‘›π‘”π‘‘β„Ž 𝐿
β„Žπ‘ƒ
π‘š=√
𝐾𝐴
2β„Ž
π‘“π‘œπ‘Ÿ π‘π‘¦π‘™π‘–π‘›π‘‘π‘Ÿπ‘–π‘π‘Žπ‘™ π‘Ÿπ‘œπ‘‘ π‘š = √
πΎπ‘Ÿ
For cylindrical rod
𝑅=
𝐿
𝐾𝐴
For cylindrical metal rods,
𝐴𝐿 = 𝐴
So, we have
β„ŽπΏ 𝐴(𝑇𝐿 − 𝑇∞ ) = 0.006671 ×
𝐾 2β„Ž
𝑇 − 𝑇∞
√ × πΎπ΄( 𝐿
)
2β„Ž πΎπ‘Ÿ
𝐿
Upon simplification, we get the expected expression:
β„ŽπΏ = 0.006671 ×
𝐾
𝐾
×√
𝐿
2β„Žπ‘Ÿ
OR
𝒉𝑳 = 𝜺 ×
𝑲
𝑲
×√
𝑳
πŸπ’‰π’“
We can extend the above analysis to cylindrical co-ordinates heat conduction
knowing their conductive resistance.
Let us solve the heat equation to get the expression for 𝛿.
We use this temperature profile which satisfies the initial condition to solve the
heat equation and get 𝛿 as shown below:
Boundary and initial conditions are:
𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎
−π’Œ
𝒅𝑻
= 𝒉𝑳 (𝑻 − 𝑻∞ ) 𝒂𝒕 𝒙 = 𝒍
𝒅𝒙
𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎
The governing temperature profile is:
(𝐿 − π‘₯ )
β„ŽπΏ 𝛿
𝐿−π‘₯
𝑇 − 𝑇∞ cosh [ 𝛿 ] + ( π‘˜ ) sinh[( 𝛿 )]
=
𝐿
β„Ž 𝛿
𝐿
𝑇𝑠 − 𝑇∞
cosh + ( 𝐿 ) π‘ π‘–π‘›β„Ž
𝛿
π‘˜
𝛿
The governing equation is
𝛼
πœ•2𝑇
β„Žπ‘ƒ
πœ•π‘‡
(
)
−
𝑇
−
𝑇
=
∞
πœ•π‘₯ 2 𝐴𝜌𝐢
πœ•π‘‘
Let us change this equation into an integral equation as below:
𝑙
πœ•2𝑇
β„Žπ‘ƒ 𝑙
πœ• 𝑙
∫ (𝑇 − 𝑇∞ )𝑑π‘₯ = ∫ (𝑇)𝑑π‘₯ … … . . 𝑏)
𝛼 ∫ ( 2 ) 𝑑π‘₯ −
𝐴𝜌𝐢 0
πœ•π‘‘ 0
0 πœ•π‘₯
𝑙
𝛼∫ (
0
πœ•2𝑇
2β„Ž 𝑙
πœ• 𝑙
(
)
∫
∫ (𝑇)𝑑π‘₯
)
𝑑π‘₯
−
𝑇
−
𝑇
𝑑π‘₯
=
∞
πœ•π‘₯ 2
π‘ŸπœŒπΆ 0
πœ•π‘‘ 0
β„ŽπΏ 𝛿
𝐿 β„ŽπΏ 𝛿
𝐿
πœ•2𝑇
πœ•π‘‡ 𝑙 (𝑇𝑠 − 𝑇∞ ) − π‘˜ + (π‘ π‘–π‘›β„Ž 𝛿 + π‘˜ π‘π‘œπ‘ β„Ž 𝛿 )
∫ ( 2 ) 𝑑π‘₯ = [ ] =
(
)
𝐿 β„ŽπΏ 𝛿
𝐿
πœ•π‘₯ 0
𝛿
0 πœ•π‘₯
π‘π‘œπ‘ β„Ž +
π‘ π‘–π‘›β„Ž
𝛿
π‘˜
𝛿
𝑙
𝑙
∫ (𝑇 − 𝑇∞ )𝑑π‘₯ = |−𝛿(𝑇𝑠 − 𝑇∞ ) (
0
sinh [
(𝐿 − π‘₯ )
(𝐿 − π‘₯ )
β„ŽπΏ 𝛿
𝛿 ] + π‘˜ cosh [ 𝛿 ] 𝑙
)|
𝐿 β„Ž 𝛿
𝐿
0
π‘π‘œπ‘ β„Ž + 𝐿 π‘ π‘–π‘›β„Ž
𝛿
π‘˜
𝛿
β„Ž 𝛿
𝐿 β„Ž 𝛿
𝐿
− π‘˜πΏ + (π‘ π‘–π‘›β„Ž 𝛿 + π‘˜πΏ π‘π‘œπ‘ β„Ž 𝛿 )
∫ (𝑇 − 𝑇∞ )𝑑π‘₯ = 𝛿 (𝑇𝑠 − 𝑇∞ ) (
)
𝐿 β„Ž 𝛿
𝐿
0
π‘π‘œπ‘ β„Ž + 𝐿 π‘ π‘–π‘›β„Ž
𝛿
π‘˜
𝛿
𝑙
(𝐿 − π‘₯ )
β„ŽπΏ 𝛿
𝐿−π‘₯
𝑇 − 𝑇∞ cosh [ 𝛿 ] + ( π‘˜ ) sinh[( 𝛿 )]
=
𝐿
β„Ž 𝛿
𝐿
𝑇𝑠 − 𝑇∞
cosh + ( 𝐿 ) π‘ π‘–π‘›β„Ž
𝛿
π‘˜
𝛿
cosh [
𝑇=
(𝐿 − π‘₯ )
β„ŽπΏ 𝛿
𝐿−π‘₯
𝛿 ] + ( π‘˜ ) sinh[( 𝛿 )]
(𝑇𝑠 − 𝑇∞ ) + 𝑇∞
𝐿
β„ŽπΏ 𝛿
𝐿
cosh + (
) π‘ π‘–π‘›β„Ž
𝛿
π‘˜
𝛿
β„Ž 𝛿
𝐿 β„Ž 𝛿
𝐿
− π‘˜πΏ + (π‘ π‘–π‘›β„Ž 𝛿 + π‘˜πΏ π‘π‘œπ‘ β„Ž 𝛿 )
πœ• 𝑙
πœ•
πœ•(𝑙𝑇∞ )
∫ (𝑇)𝑑π‘₯ = [𝛿(𝑇𝑠 − 𝑇∞ ) (
)] +
𝐿 β„Ž 𝛿
𝐿
πœ•π‘‘ 0
πœ•π‘‘
πœ•π‘‘
π‘π‘œπ‘ β„Ž + 𝐿 π‘ π‘–π‘›β„Ž
𝛿
π‘˜
𝛿
πœ•(𝑙𝑇∞ )
=0
πœ•π‘‘
Upon substitution of all the above in the heat equation, we get:
𝛼
β„ŽπΏ 𝛿
𝐿 β„ŽπΏ 𝛿
𝐿
β„Ž 𝛿
𝐿 β„Ž 𝛿
𝐿
β„Ž 𝛿
𝐿 β„Ž 𝛿
𝐿
− 𝐿 + (π‘ π‘–π‘›β„Ž + 𝐿 π‘π‘œπ‘ β„Ž )
− 𝐿 + (π‘ π‘–π‘›β„Ž + 𝐿 π‘π‘œπ‘ β„Ž )
(𝑇𝑠 − 𝑇∞) − π‘˜ + (π‘ π‘–π‘›β„Ž 𝛿 + π‘˜ π‘π‘œπ‘ β„Ž 𝛿 )
2β„Ž
𝛿
π‘˜
𝛿 ) = πœ• [𝛿(𝑇 − 𝑇 ) ( π‘˜
𝛿
π‘˜
𝛿 )]
(
)−
𝛿(𝑇𝑠 − 𝑇∞ ) ( π‘˜
𝑠
∞
𝐿 β„ŽπΏ 𝛿
𝐿
𝐿 β„ŽπΏ 𝛿
𝐿
𝐿 β„ŽπΏ 𝛿
𝐿
𝛿
π‘ŸπœŒπΆ
πœ•π‘‘
π‘π‘œπ‘ β„Ž +
π‘ π‘–π‘›β„Ž
π‘π‘œπ‘ β„Ž +
π‘ π‘–π‘›β„Ž
π‘π‘œπ‘ β„Ž +
π‘ π‘–π‘›β„Ž
𝛿
π‘˜
𝛿
𝛿
π‘˜
𝛿
𝛿
π‘˜
𝛿
We notice that the term(𝑇𝑠 − 𝑇∞ ) (
β„Ž 𝛿
π‘˜
𝐿 β„Ž 𝛿
𝛿
π‘˜
𝐿 β„Ž 𝛿
𝐿
π‘π‘œπ‘ β„Ž + 𝐿 π‘ π‘–π‘›β„Ž
𝛿
π‘˜
𝛿
𝐿
𝛿
− 𝐿 +(π‘ π‘–π‘›β„Ž + 𝐿 π‘π‘œπ‘ β„Ž )
) is common and can be
eliminated and what this signifies is that the nature of (𝑇𝑠 − 𝑇∞ ) doesn’t matter
and so we get:
𝛼
2β„Ž
𝑑𝛿
−
𝛿=
𝛿 π‘ŸπœŒπΆ
𝑑𝑑
We go ahead and solve for 𝛿 provided 𝛿 = 0π‘Žπ‘‘ 𝑑 = 0 and get the expression
−2β„Žπ‘ƒ
𝐾𝐴
𝑑
𝛿 = √ (1 − 𝑒 𝐴𝜌𝐢 )
β„Žπ‘ƒ
So, the final solution for the finite metal rod with convective flux at the end of
the metal rod is:
𝑻 − 𝑻∞
=
𝑻𝒇 − 𝑻∞
( 𝑳 − 𝒙)
𝒉 𝜹
𝑳−𝒙
𝐜𝐨𝐬𝐑 [ 𝜹 ] + ( π’Œπ‘³ ) 𝐬𝐒𝐧𝐑 [( 𝜹 )]
𝜹
(
)
𝑳
𝒉𝑳 𝜹
𝑳
𝑳 𝒉𝑳 𝜹
𝐜𝐨𝐬𝐑
+
(
)
π’”π’Šπ’π’‰
𝒕𝒂𝒏𝒉 𝜹 + π’Œ
𝜹
π’Œ
𝜹
(πœ·π’“πŸ (
)
𝒉𝑳 𝜹
𝑳 + 𝜹)
𝟏+
𝒕𝒂𝒏𝒉
(
)
π’Œ
𝜹
The solution reduces to the semi-infinite rod solution when the length L of the
metal rod tends to infinity.
Using the above solution, it was shown experimentally that the ratio
𝑻 − 𝑻∞
= 𝟎. πŸπŸ‘πŸŽπŸ—πŸ’
𝑻𝒇 − 𝑻∞
As for the semi-infinite metal rod.
The above completes our analysis.
CASE 2: ZERO FLUX AT THE END OF THE METAL ROD?
In reality, it is hard to achieve zero flux.
The boundary and initial conditions are:
𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎
𝒅𝑻
= 𝟎 𝒂𝒕 𝒙 = 𝒍
𝒅𝒙
𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎
The governing equation is
𝛼
πœ•2𝑇
β„Žπ‘ƒ
πœ•π‘‡
(𝑇 − 𝑇∞ ) =
−
2
πœ•π‘₯
𝐴𝜌𝐢
πœ•π‘‘
Recall that the temperature profile we are going to use is:
(𝑇 − 𝑇∞ ) = 𝐢1 𝑒 −π‘šπ‘₯ + 𝐢2 𝑒 π‘šπ‘₯
Or
−𝒙
𝒙
(𝑻 − 𝑻∞ ) = π‘ͺ𝟏 𝒆 𝜹 + π‘ͺ𝟐 π’†πœΉ
First of all, to satisfy the boundary conditions above, the temperature profile
becomes [2]:
𝑇 − 𝑇∞
𝑒 π‘šπ‘₯
𝑒 −π‘šπ‘₯
=
+
𝑇𝑠 − 𝑇∞ 1 + 𝑒 2π‘šπΏ 1 + 𝑒 −2π‘šπΏ
Or
The equation above can be rearranged to get [3]
𝑇 − 𝑇∞ cosh[π‘š(𝐿 − π‘₯)]
=
𝑇𝑠 − 𝑇∞
cosh π‘šπΏ
In terms of 𝜹 we get
(𝐿 − π‘₯)
𝑇 − 𝑇∞ cosh[ 𝛿 ]
=
𝐿
𝑇𝑠 − 𝑇∞
cosh 𝛿
Or using the first equation, we get:
𝑇 − 𝑇∞
=
𝑇𝑠 − 𝑇∞
π‘₯
𝑒𝛿
−π‘₯
𝑒𝛿
2𝐿 +
−2𝐿
1+𝑒𝛿 1+𝑒 𝛿
Let us examine the initial condition,
It can be shown that after solving the heat equation 𝛿 will take on the form:
−2β„Žπ‘ƒ
𝐾𝐴
𝑑
𝛿 = √ (1 − 𝑒 𝐴𝜌𝐢 )
β„Žπ‘ƒ
At
𝑑 = 0, 𝛿 = 0 π‘Žπ‘›π‘‘ π‘š =
1
=∞
𝛿
Upon substitution in
𝑇 − 𝑇∞
𝑒 π‘šπ‘₯
𝑒 −π‘šπ‘₯
=
+
𝑇𝑠 − 𝑇∞ 1 + 𝑒 2π‘šπΏ 1 + 𝑒 −2π‘šπΏ
We get
𝑇 − 𝑇∞
𝑒 ∞π‘₯
𝑒 −∞π‘₯
=
+
𝑇𝑠 − 𝑇∞ 1 + 𝑒 2∞𝐿 1 + 𝑒 −2∞𝐿
For a given π‘₯
We get:
𝑇 − 𝑇∞
𝑒 π‘šπ‘₯
𝑒 π‘šπ‘₯
=
≈
= 𝑒 −π‘š(𝐿−π‘₯) = 𝑒 −∞(𝐿−π‘₯) = 0
𝑇𝑠 − 𝑇∞ 1 + 𝑒 2∞𝐿 𝑒 2π‘šπΏ
Since
(𝐿 − π‘₯ ) > 0
Hence the initial condition is satisfied.
Getting back to business, we noticed that in the semi-infinite rod solution there
was convection at the hot end of the rod. So, to solve for what is observed in
the finite metal rod solution with zero flux at the end of the rod, we have to use
that fact as stated below:
−π’Œ
𝝏𝑻
|
= π’‰πŸŽ (𝑻𝒇 − 𝑻𝒔 )
𝝏𝒙 𝒙=𝟎
Recall the compact temperature profile for zero flux at the end of a finite metal
rod is:
𝑇 − 𝑇∞ cosh[π‘š(𝐿 − π‘₯)]
=
𝑇𝑠 − 𝑇∞
cosh π‘šπΏ
Where:
π‘š=
1
𝛿
πœ•π‘‡
|
= −(𝑇𝑠 − 𝑇∞ )π‘šπ‘‘π‘Žπ‘›β„Žπ‘šπΏ
πœ•π‘₯ π‘₯=0
−π‘˜
πœ•π‘‡
|
= π‘˜(𝑇𝑠 − 𝑇∞ )π‘šπ‘‘π‘Žπ‘›β„Žπ‘šπΏ
πœ•π‘₯ π‘₯=0
β„Ž0 (𝑇𝑓 − 𝑇𝑠 ) = π‘˜ (𝑇𝑠 − 𝑇∞ )π‘šπ‘‘π‘Žπ‘›β„Žπ‘šπΏ
β„Ž0 (𝑇𝑓 ) − β„Ž0 (𝑇𝑠 ) = π‘˜(𝑇𝑠 )π‘šπ‘‘π‘Žπ‘›β„Žπ‘šπΏ − π‘˜(𝑇∞ )π‘šπ‘‘π‘Žπ‘›β„Žπ‘šπΏ
Collecting like terms we get:
𝑇𝑠 (π‘˜π‘šπ‘‘π‘Žπ‘›β„Žπ‘šπΏ + β„Ž0 ) = β„Ž0 (𝑇𝑓 ) + π‘˜(𝑇∞ )π‘šπ‘‘π‘Žπ‘›β„Žπ‘šπΏ
𝑇𝑠 =
β„Ž0 (𝑇𝑓 ) + π‘˜(𝑇∞ )π‘šπ‘‘π‘Žπ‘›β„Žπ‘šπΏ
(π‘˜π‘šπ‘‘π‘Žπ‘›β„Žπ‘šπΏ + β„Ž0 )
Subtracting 𝑇∞ from both sides we get:
𝑇𝑠 − 𝑇∞ =
β„Ž0 (𝑇𝑓 ) + π‘˜(𝑇∞ )π‘šπ‘‘π‘Žπ‘›β„Žπ‘šπΏ
− 𝑇∞
(π‘˜π‘šπ‘‘π‘Žπ‘›β„Žπ‘šπΏ + β„Ž0 )
Upon simplification, we get:
𝑇𝑠 − 𝑇∞
β„Ž0
=
𝑇𝑓 − 𝑇∞ π‘˜π‘šπ‘‘π‘Žπ‘›β„Žπ‘šπΏ + β„Ž0
But
π‘š=
1
𝛿
Upon substitution and simplification, we get
𝑇𝑠 − 𝑇∞
𝛿
=
π‘˜
𝐿
𝑇𝑓 − 𝑇∞
π‘‘π‘Žπ‘›β„Ž
+𝛿
β„Ž0
𝛿
But from the semi-infinite rod solution, we have
π‘˜
= π›½π‘Ÿ 2
β„Ž0
Upon substitution, we get:
𝑇𝑠 − 𝑇∞
𝛿
=
𝑇𝑓 − 𝑇∞ π›½π‘Ÿ 2 π‘‘π‘Žπ‘›β„Ž 𝐿 + 𝛿
𝛿
(𝑇𝑠 − 𝑇∞ ) = (𝑇𝑓 − 𝑇∞ )(
𝛿
𝐿
π›½π‘Ÿ 2 π‘‘π‘Žπ‘›β„Ž 𝛿 + 𝛿
)
Upon substitution in the temperature profile
(𝐿 − π‘₯)
𝑇 − 𝑇∞ cosh[ 𝛿 ]
=
𝐿
𝑇𝑠 − 𝑇∞
cosh 𝛿
We get:
(𝑳 − 𝒙)
𝐜𝐨𝐬𝐑[ 𝜹 ]
(𝑻 − 𝑻∞ ) = (𝑻𝒇 − 𝑻∞ )(
)(
)
𝑳
𝑳
πœ·π’“πŸ 𝒕𝒂𝒏𝒉 𝜹 + 𝜹
𝐜𝐨𝐬𝐑 𝜹
𝜹
The above temperature profile satisfies the initial condition and the boundary
conditions provided the temperature at the hot end varies with time.
Let us now solve the heat equation using the above temperature profile:
Recall the compact temperature profile is:
(𝐿 − π‘₯)
𝑇 − 𝑇∞ cosh[ 𝛿 ]
=
𝐿
𝑇𝑠 − 𝑇∞
cosh 𝛿
The boundary and initial conditions are:
𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎
𝒅𝑻
= 𝟎 𝒂𝒕 𝒙 = 𝒍
𝒅𝒙
𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎
The governing equation is
πœ•2𝑇
β„Žπ‘ƒ
πœ•π‘‡
(𝑇 − 𝑇∞ ) =
𝛼 2−
πœ•π‘₯
𝐴𝜌𝐢
πœ•π‘‘
Let us change this equation into an integral equation as below:
𝑙
𝛼∫ (
0
πœ•2𝑇
β„Žπ‘ƒ 𝑙
πœ• 𝑙
(
)
∫
∫ (𝑇)𝑑π‘₯ … … . . 𝑏)
)
𝑑π‘₯
−
𝑇
−
𝑇
𝑑π‘₯
=
∞
πœ•π‘₯ 2
𝐴𝜌𝐢 0
πœ•π‘‘ 0
𝑙
𝛼∫ (
0
πœ•2𝑇
2β„Ž 𝑙
πœ• 𝑙
(
)
∫
∫ (𝑇)𝑑π‘₯
)
𝑑π‘₯
−
𝑇
−
𝑇
𝑑π‘₯
=
∞
πœ•π‘₯ 2
π‘ŸπœŒπΆ 0
πœ•π‘‘ 0
𝐿
tanh(𝛿 )
πœ• 2𝑇
πœ•π‘‡ 𝑙
∫ ( 2 ) 𝑑π‘₯ = [ ] = (𝑇𝑠 − 𝑇∞ )
πœ•π‘₯ 0
𝛿
0 πœ•π‘₯
𝑙
𝑙
𝐿
∫ (𝑇 − 𝑇∞ )𝑑π‘₯ = (𝑇𝑠 − 𝑇∞ )𝛿tanh( )
𝛿
0
(𝐿 − π‘₯ )
𝛿 ]
(𝑇𝑠 − 𝑇∞ ) + 𝑇∞
𝐿
cosh 𝛿
cosh [
𝑇=
πœ• 𝑙
πœ•
𝐿
πœ•(𝑙𝑇∞ )
∫ (𝑇)𝑑π‘₯ = [𝛿(𝑇𝑠 − 𝑇∞ ) tanh ( )] +
πœ•π‘‘ 0
πœ•π‘‘
𝛿
πœ•π‘‘
πœ•(𝑙𝑇∞ )
=0
πœ•π‘‘
Upon substitution of all the above in the heat equation, we get:
𝛼(𝑇𝑠 − 𝑇∞ )
𝐿
tanh (𝛿 )
𝛿
−
2β„Ž
𝐿
πœ•
𝐿
(𝑇𝑠 − 𝑇∞ )𝛿 tanh ( ) = [𝛿(𝑇𝑠 − 𝑇∞ ) tanh ( )]
π‘ŸπœŒπΆ
𝛿
πœ•π‘‘
𝛿
𝐿
We notice that the term (𝑇𝑠 − 𝑇∞ )tanh(𝛿 ) is common and can be eliminated and
what this signifies is that the nature of (𝑇𝑠 − 𝑇∞ ) doesn’t matter and so we get:
𝛼
2β„Ž
𝑑𝛿
−
𝛿=
𝛿 π‘ŸπœŒπΆ
𝑑𝑑
We go ahead and solve for 𝛿 provided 𝛿 = 0π‘Žπ‘‘ 𝑑 = 0 and get the expression
−2β„Žπ‘ƒ
𝐾𝐴
𝑑
𝛿 = √ (1 − 𝑒 𝐴𝜌𝐢 )
β„Žπ‘ƒ
So, the final solution for the finite metal rod with zero flux at the end of the
metal rod is:
(𝑳 − 𝒙)
𝐜𝐨𝐬𝐑[ 𝜹 ]
(𝑻 − 𝑻∞ ) = (𝑻𝒇 − 𝑻∞ )(
)(
)
𝑳
𝑳
πœ·π’“πŸ 𝒕𝒂𝒏𝒉 𝜹 + 𝜹
𝐜𝐨𝐬𝐑 𝜹
𝜹
The solution reduces to the semi-infinite rod solution when the length L of the
metal rod tends to infinity.
Using the above solution, it was shown experimentally that the ratio
𝑻 − 𝑻∞
= 𝟎. πŸπŸπŸ—πŸ–πŸ
𝑻𝒇 − 𝑻∞
As for the semi-infinite rod.
HOW DO WE DEAL WITH CYLINDRICAL
COORDINATES?
We know that for an insulated cylinder where there is no heat loss by
convection from the sides, the governing PDE equation is
𝜢 𝝏
𝝏𝑻
𝝏𝑻
(𝒓 ) =
𝒓 𝝏𝒓 𝝏𝒓
𝝏𝒕
In steady state
πœ•π‘‡
=0
πœ•π‘‘
We end up with
πœ•
πœ•π‘‡
(π‘Ÿ ) = 0
πœ•π‘Ÿ πœ•π‘Ÿ
We can then integrate once to get
∫(
πœ•
πœ•π‘‡
(π‘Ÿ )) π‘‘π‘Ÿ = ∫(0)π‘‘π‘Ÿ
πœ•π‘Ÿ πœ•π‘Ÿ
And get
π‘Ÿ
πœ•π‘‡
= 𝐢1
πœ•π‘Ÿ
Therefore
πœ•π‘‡ 𝐢1
=
πœ•π‘Ÿ
π‘Ÿ
We can go ahead and find the temperature profile as a function of radius r.
HOW DO WE DEAL WITH CYLINDRICAL CO-ORDINATES
FOR A SEMI-INFINITE RADIUS CYLINDER?
The governing PDE is:
𝜢 𝝏
𝝏𝑻
𝝏𝑻
(𝒓 ) =
𝒓 𝝏𝒓 𝝏𝒓
𝝏𝒕
The boundary conditions are
𝑇 = 𝑇𝑠 π‘Žπ‘‘ π‘Ÿ = π‘Ÿ1
𝑇 = 𝑇∞ π‘Žπ‘‘ π‘Ÿ = ∞
The initial condition is:
𝑇 = 𝑇∞ π‘Žπ‘‘ 𝑑 = 0
The temperature profile that satisfies the conditions above is
−(π‘Ÿ−π‘Ÿ1 )
𝑇 − 𝑇∞
=𝑒 𝛿
𝑇𝑠 − 𝑇∞
We transform the equation above into an integral equation and take integrals
with limits from π‘Ÿ = π‘Ÿ1 to π‘Ÿ = 𝑅 = ∞.
𝛼 πœ•
πœ•π‘‡
πœ•π‘‡
(π‘Ÿ ) =
π‘Ÿ πœ•π‘Ÿ πœ•π‘Ÿ
πœ•π‘‘
We take integrals and get
𝑅
𝛼 πœ•
πœ•π‘‡
πœ• 𝑅
(π‘Ÿ )]π‘‘π‘Ÿ = ∫ (𝑇)π‘‘π‘Ÿ
πœ•π‘Ÿ
πœ•π‘‘ π‘Ÿ1
π‘Ÿ1 π‘Ÿ πœ•π‘Ÿ
∫ [
𝑅
𝛼 πœ•
πœ•π‘‡
πœ• 𝑅
∫ [
(π‘Ÿ )]∂r
/ = ∫ (𝑇)π‘‘π‘Ÿ
πœ•π‘Ÿ
πœ•π‘‘ π‘Ÿ1
/
π‘Ÿ1 π‘Ÿ ∂r
𝝏
You notice that 𝝏𝒓 cancels out with 𝒅𝒓 and we get:
𝛼
πœ•π‘‡
πœ• 𝑅
= ∫ (𝑇)π‘‘π‘Ÿ
πœ•π‘Ÿ πœ•π‘‘ π‘Ÿ1
But
𝑅
πœ•π‘‡
πœ•2𝑇
= ∫ ( 2 )π‘‘π‘Ÿ
πœ•π‘Ÿ
π‘Ÿ1 πœ•π‘Ÿ
So, the PDE becomes:
𝑹
𝝏𝟐 𝑻
𝝏 𝑹
𝜢 ∫ ( 𝟐 )𝒅𝒓 = ∫ (𝑻)𝒅𝒓
𝝏𝒕 π’“πŸ
π’“πŸ 𝝏𝒓
We then go ahead to solve and find 𝛿 as before.
πœ•π‘‡
𝑇𝑠 − 𝑇∞ −(π‘Ÿ−π‘Ÿ1 )
=−
𝑒 𝛿
πœ•π‘Ÿ
𝛿
𝑅
∫ (
π‘Ÿ1
(𝑇𝑠 − 𝑇∞ ) −(π‘Ÿ−π‘Ÿ1 ) 𝑅 = ∞ (𝑇𝑠 − 𝑇∞ )
πœ•2𝑇
πœ•π‘‡ 𝑅
) π‘‘π‘Ÿ = [ ] = −
[𝑒 𝛿 ]
=
2
π‘Ÿ1
πœ•π‘Ÿ
πœ•π‘Ÿ π‘Ÿ1
𝛿
𝛿
𝑇 = (𝑇𝑠 − 𝑇∞ )𝑒
𝑅=∞
𝑅=∞
∫
π‘‡π‘‘π‘Ÿ = ∫
π‘Ÿ1
π‘Ÿ1
((𝑇𝑠 − 𝑇∞ )𝑒
−(π‘Ÿ−π‘Ÿ1)
𝛿
+ 𝑇∞
𝑅=∞
−(π‘Ÿ−π‘Ÿ1 )
𝛿
)π‘‘π‘Ÿ + ∫
𝑇∞ π‘‘π‘Ÿ = 𝛿 (𝑇𝑠 − 𝑇∞ ) + 𝑇∞ (𝑅 − π‘Ÿ1 )
π‘Ÿ1
πœ• 𝑅
𝑑𝛿
𝑑(𝑇∞ (𝑅 − π‘Ÿ1 ))
(𝑇𝑠 − 𝑇∞ ) +
∫ π‘‡π‘‘π‘Ÿ =
πœ•π‘‘ 𝛿
𝑑𝑑
𝑑𝑑
But
𝑑(𝑇∞ (𝑅 − π‘Ÿ1 ))
=0
𝑑𝑑
Since 𝑇∞ , 𝑅, π‘Ÿ1 are constants independent of time.
So
πœ• 𝑅
𝑑𝛿
(𝑇 − 𝑇∞ )
∫ π‘‡π‘‘π‘Ÿ =
πœ•π‘‘ 𝛿
𝑑𝑑 𝑠
substituting all the above in the integral equation, we get
𝑅
πœ•2𝑇
πœ• 𝑅
𝛼 ∫ ( 2 )π‘‘π‘Ÿ = ∫ (𝑇)π‘‘π‘Ÿ
πœ•π‘‘ π‘Ÿ1
π‘Ÿ1 πœ•π‘Ÿ
𝛼
𝑑𝛿
(𝑇𝑠 − 𝑇∞ ) =
(𝑇 − 𝑇∞ )
𝛿
𝑑𝑑 𝑠
Divide through by (𝑇𝑠 − 𝑇∞ ) and get
𝛼 𝑑𝛿
=
𝛿 𝑑𝑑
𝛼 𝑑𝛿
=
𝛿 𝑑𝑑
The boundary conditions are:
𝛿 = 0 π‘Žπ‘‘ 𝑑 = 0
𝛿 = √2𝛼𝑑
We substitute 𝛿 in the temperature profile and get:
−(π‘Ÿ−π‘Ÿ1 )
𝑇 − 𝑇∞
=𝑒 𝛿
𝑇𝑠 − 𝑇∞
−(𝒓−π’“πŸ )
𝑻 − 𝑻∞
= 𝒆 √πŸπœΆπ’•
𝑻𝒔 − 𝑻∞
You notice that the initial condition is satisfied for the above temperature
profile.
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.
HOW DO WE DEAL WITH NATURAL CONVECTION AT
THE SURFACE AREA OF A SEMI-INFINITE CYLINDER
FOR FIXED END TEMPERATURE
The governing equation is:
𝜢 𝝏
𝝏𝑻
𝒉𝑷
𝝏𝑻
(𝑻 − 𝑻∞ ) =
(𝒓 ) −
𝒓 𝝏𝒓 𝝏𝒓
𝑨𝝆π‘ͺ
𝝏𝒕
𝑃 = 2πœ‹π‘Ÿ
𝐴 = 2πœ‹π‘Ÿπ‘‘
Where:
𝑑 = β„Žπ‘’π‘–π‘”β„Žπ‘‘ π‘œπ‘“ π‘π‘¦π‘™π‘–π‘›π‘‘π‘’π‘Ÿ
The boundary conditions are:
𝑇 = 𝑇𝑠 π‘Žπ‘‘ π‘Ÿ = π‘Ÿ1
𝑇 = 𝑇∞ π‘Žπ‘‘ π‘Ÿ = 𝑅 = ∞
The initial condition is:
𝑇 = 𝑇∞ π‘Žπ‘‘ 𝑑 = 0
The temperature profile that satisfies the boundary conditions above is:
−(π‘Ÿ−π‘Ÿ1 )
𝑇 − 𝑇∞
=𝑒 𝛿
𝑇𝑠 − 𝑇∞
𝛼 πœ•
πœ•π‘‡
β„Žπ‘ƒ
πœ•π‘‡
(𝑇 − 𝑇∞ ) =
(π‘Ÿ ) −
π‘Ÿ πœ•π‘Ÿ πœ•π‘Ÿ
𝐴𝜌𝐢
πœ•π‘‘
We transform the PDE into an integral equation and take the limits to be from
π‘Ÿ = π‘Ÿ1 to π‘Ÿ = 𝑅 = ∞
𝑅=∞
𝛼∫
[
π‘Ÿ1
𝑅=∞
∫
π‘Ÿ1
𝑅=∞
𝛼 πœ•
πœ•π‘‡
β„Ž
πœ• 𝑅=∞
(𝑇 − 𝑇∞ )π‘‘π‘Ÿ = ∫
∫
(π‘Ÿ )]π‘‘π‘Ÿ −
(𝑇)π‘‘π‘Ÿ
π‘Ÿ πœ•π‘Ÿ πœ•π‘Ÿ
π‘‘πœŒπΆ π‘Ÿ1
πœ•π‘‘ π‘Ÿ1
𝑅=∞
𝛼 πœ•
πœ•π‘‡
β„Ž
πœ• 𝑅=∞
(𝑇 − 𝑇∞ )π‘‘π‘Ÿ = ∫
∫
[
(π‘Ÿ )]∂r
/ −
(𝑇)π‘‘π‘Ÿ
π‘Ÿ ∂r
πœ•π‘Ÿ
π‘‘πœŒπΆ π‘Ÿ1
πœ•π‘‘ π‘Ÿ1
/
𝝏
You notice that 𝝏𝒓 cancels out with 𝒅𝒓 and we get:
𝑅=∞
πœ•π‘‡
β„Ž
πœ• 𝑅=∞
(𝑇 − 𝑇∞ ) π‘‘π‘Ÿ = ∫
∫
𝛼
−
(𝑇)π‘‘π‘Ÿ
πœ•π‘Ÿ π‘‘πœŒπΆ π‘Ÿ1
πœ•π‘‘ π‘Ÿ1
But
𝑅
πœ•π‘‡
πœ•2𝑇
= ∫ ( 2 )π‘‘π‘Ÿ
πœ•π‘Ÿ
π‘Ÿ1 πœ•π‘Ÿ
So, the PDE becomes:
𝑹
𝑹=∞
𝝏𝟐 𝑻
𝒉
𝝏 𝑹
(
)
∫
∫ (𝑻)𝒅𝒓
)𝒅𝒓
−
𝑻
−
𝑻
𝒅𝒓
=
∞
𝟐
𝒅𝝆π‘ͺ π’“πŸ
𝝏𝒕 π’“πŸ
π’“πŸ 𝝏𝒓
𝜢∫ (
𝑅=∞
β„Ž
π›Ώβ„Ž
(𝑇 − 𝑇∞ )π‘‘π‘Ÿ =
∫
(𝑇 − 𝑇∞ )
π‘‘πœŒπΆ π‘Ÿ1
π‘‘πœŒπΆ 𝑠
From the derivations above, we get:
𝑅
∫ (
π‘Ÿ1
πœ•2𝑇
(𝑇𝑠 − 𝑇∞ )
) π‘‘π‘Ÿ =
2
πœ•π‘Ÿ
𝛿
πœ• 𝑅
𝑑𝛿
(𝑇 − 𝑇∞ )
∫ π‘‡π‘‘π‘Ÿ =
πœ•π‘‘ π‘Ÿ1
𝑑𝑑 𝑠
Upon substitution of the above expressions in the integral equation, we get:
𝛼
β„Ž
𝑑𝛿
−
𝛿=
𝛿 π‘‘πœŒπΆ
𝑑𝑑
𝛼
β„Ž
𝑑𝛿
−
𝛿=
𝛿 π‘‘πœŒπΆ
𝑑𝑑
𝛼−
β„Ž 2
𝑑𝛿
𝛿 =𝛿
π‘‘πœŒπΆ
𝑑𝑑
The boundary conditions are:
𝛿 = 0 π‘Žπ‘‘ 𝑑 = 0
The solution of the equation above is
2β„Žπ‘‘
π›Όπ‘‘πœŒπΆ
−
𝛿=√
(1 − 𝑒 π‘‘πœŒπΆ )
β„Ž
𝛼=
𝐾
𝜌𝐢
And get
2β„Žπ‘‘
𝐾𝑑
−
𝛿 = √ (1 − 𝑒 π‘‘πœŒπΆ )
β„Ž
The temperature profile becomes:
π‘Ÿ−π‘Ÿ1
𝑇 − 𝑇∞
= 𝑒 −( 𝛿 )
𝑇𝑠 − 𝑇∞
Substitute for 𝛿 and get
𝒓−π’“πŸ
−(
𝑻 − 𝑻∞
=𝒆
𝑻𝒔 − 𝑻∞
√𝑲𝒅(𝟏−𝒆
𝒉
−
)
πŸπ’‰π’•
𝒅𝝆π‘ͺ )
You notice that the initial condition is satisfied by the above temperature
profile.
What do we observe for short time in transient state?
For short time, the exponential is small and it becomes:
−
2β„Žπ‘‘
𝑒 π‘‘πœŒπΆ = 1 −
2β„Žπ‘‘
π‘‘πœŒπΆ
After using binomial expansion of the exponential in the above
And
−
2β„Žπ‘‘
(1 − 𝑒 π‘‘πœŒπΆ ) =
2β„Žπ‘‘
π‘‘πœŒπΆ
2β„Žπ‘‘
𝐾𝑑
−
𝛿 = √ (1 − 𝑒 π‘‘πœŒπΆ )
β„Ž
Becomes
𝛿 = √2𝛼𝑑
Where:
𝛼=
𝐾
𝜌𝐢
Let us make r the subject of the temperature profile above:
π‘Ÿ−π‘Ÿ1
𝑇 − 𝑇∞
= 𝑒 −( 𝛿 )
𝑇𝑠 − 𝑇∞
π‘Ÿ − π‘Ÿ1
𝑇𝑠 − 𝑇∞
= [ln(
)]
𝛿
𝑇 − 𝑇∞
(π‘Ÿ − π‘Ÿ1 ) = 𝛿 [ln (
𝑇𝑠 − 𝑇∞
)]
𝑇 − 𝑇∞
Where:
𝛿 = √2𝛼𝑑
For small time.
We can get an expression for 𝑇𝑠 as we did before.
We can extend the above analysis to finite radius metal rods and even to
rectangular metal rods.
REFERENCES
[1] C.P.Kothandaraman, "HEAT TRANSFER WITH EXTENDED SURFACES(FINS)," in Fundamentals of Heat
and Mass Transfer, New Delhi, NEW AGE INTERNATIONAL PUBLISHERS, 2006, p. 131.
[2] C. E. W. R. E. W. G. L. R. James R. Welty, "HEAT TRANSFER FROM EXTENDED SURFACES," in
Fundamentals of Momentum, Heat, and Mass Transfer 5th Edition, Corvallis, Oregon, John Wiley &
Sons, Inc., 2000, pp. 236-237.
[3] C. E. W. R. E. W. G. L. R. James R. Welty, "HEAT TRANSFER FROM EXTENDED SURFACES," in
Fundamentals of Momentum, Heat and Mass Transfer 5th Edition, Oregon, John Wiley & Sons, Inc.,
2008, p. 237.
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