assignment

1. HEAT TRANSFER (ME 721)
RectangularCoordinates

2. We will assume that the temperature in the material is a function onlyof the x-coordinate and
time;that is, T = T(x,t),and the conductivityk, density𝝆and specificheatc of the solidare all
constant
We will use Fourier’s law to express the two conduction terms and define the Symbol qG as
the rate of energy generation per unit volume inside the control volume. Then the word
equation (Eq. 2.1) can be expressed in mathematical form:
−𝒌𝑨(
𝝏𝑻
𝝏𝒙
)
𝒙
+ 𝒒𝑮𝑨∆𝒙 = −𝒌𝑨((
𝝏𝑻
𝝏𝒙
)
𝒙+∆𝒙
+ 𝝆𝑨∆𝒙𝒄
𝝏𝑻(𝒙+
∆𝒙
𝟐
,𝒕)
𝝏𝒕
(2.2)

3. Divide equation (2.2) by 𝑨∆𝒙
(−𝑘𝐴 (
𝜕𝑇
𝜕𝑥
)
𝑥
+ 𝑞𝐺𝐴∆𝑥 = −𝑘𝐴((
𝜕𝑇
𝜕𝑥
)
𝑥+∆𝑥
+ 𝜌𝐴∆𝑥𝑐
𝜕𝑇(𝑥+
∆𝑥
2
,𝑡)
𝜕𝑡
)
1
𝐴∆𝑥
(−𝑘𝐴 (
𝜕𝑇
𝜕𝑥
)
𝑥
+ 𝑞𝐺𝐴∆𝑥 = −𝑘𝐴((
𝜕𝑇
𝜕𝑥
)
𝑥+∆𝑥
+ 𝜌𝐴∆𝑥𝑐
𝜕𝑇(𝑥+
∆𝑥
2
,𝑡)
𝜕𝑡
)
1
𝐴∆𝑥
−𝑘𝐴(
𝜕𝑇
𝜕𝑥
)
𝑥
∆𝑥
+
𝑘𝐴((
𝜕𝑇
𝜕𝑥
)
𝑥+∆𝑥
∆𝑥
+ 𝑞𝐺 = 𝜌𝑐
𝜕𝑇(𝑥+
∆𝑥
2
,𝑡)
𝜕𝑡
𝒌(−(
𝝏𝑻
𝝏𝒙
)
𝒙
+(
𝝏𝑻
𝝏𝒙
)
𝒙+∆𝒙
)
∆𝒙
+ 𝒒𝑮 = 𝝆𝒄
𝝏𝑻( 𝒙+
∆𝒙
𝟐
,𝒕)
𝝏𝒕
(2.3)
In the Limit as ∆𝒙 → 𝟎, from the left side
(
𝝏𝑻
𝝏𝒙
)
𝒙+∆𝒙
= (
𝝏𝑻
𝝏𝒙
)
𝒙
+
𝝏
𝝏𝒙
(
𝝏𝑻
𝝏𝒙
)
𝒙
𝒅𝒙 = (
𝝏𝑻
𝝏𝒙
)
𝒙
+ (
𝝏 𝟐
𝑻
𝝏𝒙 𝟐
) 𝒙 𝒅𝒙 (𝟐. 𝟒)
In the right side of the equation2.3 using the taylor series
𝜕𝑇(𝑥 +
∆𝑥
2
, 𝑡)
𝜕𝑡
=
𝜕𝑇
𝜕𝑡
((𝑥 +
∆𝑥
2
), 𝑡) = (
𝜕𝑇
𝜕𝑡
)
𝑥
+
𝜕
𝜕𝑥
(
𝜕𝑇
𝜕𝑡
) 𝑥
∆𝑥
2
. . . . . . .. ..
= (
𝝏𝑻
𝝏𝒕
)
𝒙
+ (
𝝏 𝟐
𝑻
𝝏𝒕𝝏𝒙
) 𝒙
∆𝒙
𝟐
. .. . . . . . ..
Thus to the order of ∆𝒙
𝑘(−(
𝜕𝑇
𝜕𝑥
)
𝑥
+ (
𝝏𝑻
𝝏𝒙
)
𝒙
+ (
𝝏 𝟐 𝑻
𝝏𝒙 𝟐) 𝒙 𝒅𝒙)
∆𝑥
+ 𝑞𝐺 = 𝜌𝑐 (
𝝏𝑻
𝝏𝒕
)
𝒙
+ (
𝝏 𝟐 𝑻
𝝏𝒕𝝏𝒙
) 𝒙
∆𝒙
𝟐
………
𝒌(
𝝏 𝟐 𝑻
𝝏𝒙 𝟐) 𝒙 + 𝒒𝑮 = 𝝆𝒄 (
𝝏𝑻
𝝏𝒕
)
𝒙
eq. (2.5)

4. If T = T(x,y,z,t) and 𝜶 =
𝒌
𝝆𝒄
(𝒕𝒉𝒆𝒓𝒎𝒂𝒍 𝒅𝒊𝒇𝒇𝒖𝒔𝒊𝒗𝒊𝒕𝒚)
𝒌
𝝏 𝟐
𝑻
𝝏𝒙 𝟐
+ 𝒒𝑮 =
𝒌
𝜶
𝝏𝑻
𝝏𝒕
(𝟐. 𝟓)
𝝏 𝟐 𝑻
𝝏𝒙 𝟐
+
𝝏 𝟐 𝑻
𝝏𝒚 𝟐
+
𝝏 𝟐 𝑻
𝝏𝒛 𝟐
+
𝒒𝑮
𝒌
=
𝟏
𝜶
𝝏𝑻
𝝏𝒕
(2.6)

5. If the temperature of a material is not a function of time, the system is in the steady state and
does not store any energy. The steady-state form of a three-dimensional conduction equation
in rectangular coordinates is
𝝏 𝟐
𝑻
𝝏𝒙 𝟐
+
𝝏 𝟐
𝑻
𝝏𝒚 𝟐
+
𝝏 𝟐
𝑻
𝝏𝒛 𝟐
+
𝒒𝑮
𝒌
=
𝟏
𝜶
𝝏𝑻
𝝏𝒕
𝝏 𝟐
𝑻
𝝏𝒙 𝟐
+
𝝏 𝟐
𝑻
𝝏𝒚 𝟐
+
𝝏 𝟐
𝑻
𝝏𝒛 𝟐
+
𝒒𝑮
𝒌
= 𝟎 (𝟐. 𝟖)
If the system is in the steady state and no heat is generated internally, the conduction equation
further simplifies to
𝝏 𝟐
𝑻
𝝏𝒙 𝟐
+
𝝏 𝟐
𝑻
𝝏𝒚 𝟐
+
𝝏 𝟐
𝑻
𝝏𝒛 𝟐
+
𝒒𝑮
𝒌
= 𝟎
𝝏 𝟐
𝑻
𝝏𝒙 𝟐
+
𝝏 𝟐
𝑻
𝝏𝒚 𝟐
+
𝝏 𝟐
𝑻
𝝏𝒛 𝟐
= 𝟎 (𝟐. 𝟗)
Equation (2.9) is known as the Laplace equation, in honor of the French mathematician Pierre
Laplace. It occurs in a number of areas in addition to heat transfer, for instance, in diffusion of
mass or in electromagnetic fields. The operation of taking the second derivatives of the
potential in a field has therefore been given a shorthand symbol, 𝛁 𝟐
, called the Laplacian
operator. For the rectangular coordinate system Eq. (2.9) becomes
𝝏 𝟐
𝑻
𝝏𝒙 𝟐
+
𝝏 𝟐
𝑻
𝝏𝒚 𝟐
+
𝝏 𝟐
𝑻
𝝏𝒛 𝟐
= 𝛁 𝟐
= 𝟎 (𝟐. 𝟏𝟎)

6. DimensionlessForm;
𝜽 =
𝑻
𝑻 𝒓
(𝟐. 𝟏𝟏) , 𝜺 =
𝒙
𝑳 𝒓
(𝟐. 𝟏𝟐), 𝝉 =
𝒕
𝒕 𝒓
(𝟐. 𝟏𝟑), 𝜶 =
𝒌
𝝆𝒄
(𝟐. 𝟕)
𝒌
𝝏 𝟐
𝑻
𝝏𝒙 𝟐
+ 𝒒𝑮 =
𝒌
𝜶
𝝏𝑻
𝝏𝒕
(𝟐. 𝟓)
Where the symbols Tr, Lr, and tr represent a reference temperature, a reference length, and a
reference time, respectively. Although the choice of reference quantities is somewhat arbitrary,
the values selected should be physically significant. The choice of dimensionless groups varies
from problem to problem, but the form of the dimensionless groups should be structured so
that they limit the dimensionless variables between convenient extremes, such as zero and one.
The value for Lr should therefore be selected as the maximum x dimension of the system for
which the temperature distribution is sought. Similarly, a dimensionless ratio of temperature
differences that varies between zero and unity is often preferable to a ratio of absolute
temperatures.
If the definitions of the dimensionless temperature, x coordinate, and time are substituted into
Eq. (2.5), we obtain the conduction equation in the nondimensional form
If the definitions of dimensionless will be substitute to eq. (2.5)
Thus;
𝒌
𝝏 𝟐
𝜃𝑇𝑟
𝝏𝜀 𝟐 𝐿 𝑟
𝟐 + 𝒒𝑮 =
𝒌
𝜶
𝝏𝜃𝑇𝑟
𝝏𝑡 𝑟 𝜏
(𝒌
𝝏 𝟐
𝜃𝑇𝑟
𝝏𝜀 𝟐 𝐿 𝑟
𝟐 + 𝒒𝑮 =
𝒌
𝜶
𝝏𝜃𝑇𝑟
𝝏𝑡 𝑟 𝜏
) (
𝐿 𝑟
𝟐
𝒌𝑇𝑟
)
𝝏 𝟐
𝜃
𝝏𝜀 𝟐
+
𝒒𝑮𝐿 𝑟
𝟐
𝒌𝑇𝑟
=
𝐿 𝑟
𝟐
𝜶𝑡 𝑟
𝝏𝜃
𝝏𝜏
(𝟐. 𝟏𝟒)
The reciprocal of
𝐿 𝑟
𝟐
𝜶𝑡 𝑟
is called the Fourier number,
𝑭𝒐 =
𝜶𝑡 𝑟
𝐿 𝑟
𝟐 =
(
𝒌
𝐿 𝑟
)
(
𝜌𝑐𝐿 𝑟
𝑡 𝑟
)
(2.15)
The other dimensionless group appearing in Eq. (2.14) is a ratio of internal heat generation
per unit time to heat conduction through the volume per unit time. We will use the symbol
𝑸 𝑮 to represent this dimensionless heat generation number:
𝑸 𝑮 =
𝒒𝑮𝐿 𝑟
𝟐
𝒌𝑇𝑟

7. The one-dimensionless form of the conduction equation expressed in dimensionless form
now becomes.
𝝏 𝟐
𝜃
𝝏𝜀 𝟐
+
𝒒𝑮𝐿 𝑟
𝟐
𝒌𝑇𝑟
=
𝐿 𝑟
𝟐
𝜶𝑡 𝑟
𝝏𝜃
𝝏𝜏
(𝟐. 𝟏𝟒)
𝝏 𝟐
𝜃
𝝏𝜀 𝟐
+ 𝑸 𝑮 =
𝟏
𝑭𝒐
𝝏𝜃
𝝏𝜏
(𝟐. 𝟏𝟕)

8. Cylindrical and Spherical Coordinates
Equation (2.6) was derived for a rectangular coordinate system. Although the generation and
energy storage terms are independent of the coordinate system, the heat conduction terms
depend on geometry and therefore on the coordinate system. The dependence on the coordinate
system used to formulate the problem can be removed by replacing the heat conduction terms
with the Laplacian operator.
𝛁 𝟐
𝑻 +
𝒒 𝑮
𝒌
=
𝟏
𝜶
𝝏𝑻
𝝏𝒕
(𝟐. 𝟏𝟖)
In Laplacian form for Cylindrical
∆𝒇 = 𝛁 𝟐 𝒇 = 𝛁 ∗ 𝛁𝒇 =
𝟏
𝝆
𝝏
𝝏𝝆
(𝝆
𝝏𝒇
𝝏𝝆
)+
𝟏
𝝆 𝟐
𝝏 𝟐 𝒇
𝝏𝝋 𝟐
+
𝝏 𝟐 𝒇
𝝏𝒛 𝟐
If 𝑻 = 𝑻(𝒓, ∅, 𝒛, 𝒕)
𝛁 𝟐 𝑻 = 𝛁 ∗ 𝛁𝑻 =
𝟏
𝒓
𝝏
𝝏𝒓
( 𝒓
𝝏𝑻
𝝏𝒓
) +
𝟏
𝒓 𝟐
𝝏 𝟐 𝑻
𝝏∅ 𝟐 +
𝝏 𝟐 𝑻
𝝏𝒛 𝟐 +
𝒒 𝑮
𝒌
=
𝟏
𝜶
𝝏𝑻
𝝏𝒕
(𝟐. 𝟏𝟗)

9. IF the heat flow in a cylindrical shape is only in the radial direction, 𝑻 = 𝐓(𝐫, 𝐭)
𝛁 𝟐
𝑻 =
𝟏
𝒓
𝝏
𝝏𝒓
(𝒓
𝝏𝑻
𝝏𝒓
) +
𝟏
𝒓 𝟐
𝝏 𝟐
𝑻
𝝏∅ 𝟐
+
𝝏 𝟐
𝑻
𝝏𝒛 𝟐
+
𝒒 𝑮
𝒌
=
𝟏
𝜶
𝝏𝑻
𝝏𝒕
𝛁 𝟐
𝑻 =
𝟏
𝒓
𝝏
𝝏𝒓
(𝒓
𝝏𝑻
𝝏𝒓
) +
𝒒 𝑮
𝒌
=
𝟏
𝜶
𝝏𝑻
𝝏𝒕
(𝟐. 𝟐𝟎)
If the temperature distribution does not vary with time
𝛁 𝟐
𝑻 =
𝟏
𝒓
𝝏
𝝏𝒓
(𝒓
𝝏𝑻
𝝏𝒓
) +
𝒒 𝑮
𝒌
=
𝟏
𝜶
𝝏𝑻
𝝏𝒕
𝟏
𝒓
𝒅
𝒅𝒓
(𝒓
𝒅𝑻
𝒅𝒓
) +
𝒒 𝑮
𝒌
= 𝟎 (𝟐. 𝟐𝟏)
In this case the equation for the temperature contains only a single variable r and is therefore
an ordinary differential equation. When there is no internal energy generation and the
temperature is a function of the radius only, the steady-state conduction equation for cylindrical
coordinates is
𝟏
𝒓
𝒅
𝒅𝒓
(𝒓
𝒅𝑻
𝒅𝒓
) +
𝒒 𝑮
𝒌
= 𝟎
𝒅
𝒅𝒓
(𝒓
𝒅𝑻
𝒅𝒓
) = 𝟎 (𝟐.𝟐𝟐)

10. The general form of the conduction equation in spherical coordinates is then;
∆𝒇 = 𝛁 𝟐
𝒇 = 𝛁 ∗ 𝛁𝒇 =
𝟏
𝒓 𝟐
𝝏
𝝏𝒓
(𝒓 𝟐
𝝏𝒇
𝝏𝒓
) +
𝟏
𝒓 𝟐 𝒔𝒊𝒏𝜽
𝝏
𝝏𝜽
(𝒔𝒊𝒏𝜽
𝝏𝒇
𝝏𝜽
) +
𝟏
𝒓 𝟐 𝒔𝒊𝒏𝜽
𝝏 𝟐
𝒇
𝝏𝝋 𝟐
=
𝟏
𝒓
𝝏 𝟐
𝝏 𝟐
( 𝒓𝒇) +
𝟏
𝒓 𝟐 𝒔𝒊𝒏𝜽
𝝏
𝝏𝜽
(𝒔𝒊𝒏𝜽
𝝏𝒇
𝝏𝜽
) +
𝟏
𝒓 𝟐 𝒔𝒊𝒏𝜽
𝝏 𝟐
𝒇
𝝏𝝋 𝟐
If the temperature is a function of the three space coordinates 𝒓, 𝜽, ∅ and time t, that is,
𝑻 = 𝐓(𝒓, 𝜽, ∅ 𝐭), then
𝛁 𝟐 𝑻 = 𝛁 ∗ 𝛁𝑻 =
𝟏
𝒓 𝟐
𝝏
𝝏𝒓
( 𝒓 𝟐
𝝏𝑻
𝝏𝒓
) +
𝟏
𝒓 𝟐 𝒔𝒊𝒏𝜽
𝝏
𝝏𝜽
(𝒔𝒊𝒏𝜽
𝝏𝑻
𝝏𝜽
) +
𝟏
𝒓 𝟐 𝒔𝒊𝒏𝜽
𝝏 𝟐 𝑻
𝝏∅ 𝟐 +
𝒒 𝑮
𝒌
=
𝟏
𝜶
𝝏𝑻
𝝏𝒕
=
𝟏
𝒓
𝝏 𝟐
𝝏 𝟐
( 𝒓𝑻) +
𝟏
𝒓 𝟐 𝒔𝒊𝒏𝜽
𝝏
𝝏𝜽
(𝒔𝒊𝒏𝜽
𝝏𝒇
𝝏𝜽
) +
𝟏
𝒓 𝟐 𝒔𝒊𝒏𝜽
𝝏 𝟐
𝒇
𝝏𝝋 𝟐
+
𝒒 𝑮
𝒌
=
𝟏
𝜶
𝝏𝑻
𝝏𝒕
(𝟐. 𝟐𝟑)

11. Plane wall with and without Heat Generation
From eq. 2.6
𝝏 𝟐 𝑻
𝝏𝒙 𝟐
+
𝝏 𝟐 𝑻
𝝏𝒚 𝟐
+
𝝏 𝟐 𝑻
𝝏𝒛 𝟐
+
𝒒𝑮
𝒌
=
𝟏
𝜶
𝝏𝑻
𝝏𝒕
(2.6)
If for steady state
𝝏𝑻
𝝏𝒕
= 𝟎 and also T is onlythe functionof x,
𝝏𝑻
𝝏𝒚
= 𝟎 and
𝝏𝑻
𝝏𝒛
= 𝟎 and there
is no internal generation, qG = 0
𝝏 𝟐
𝑻
𝝏𝒙 𝟐
+
𝝏 𝟐
𝑻
𝝏𝒚 𝟐
+
𝝏 𝟐
𝑻
𝝏𝒛 𝟐
+
𝒒𝑮
𝒌
=
𝟏
𝜶
𝝏𝑻
𝝏𝒕
Eq. (2.6) reduces to
𝒅 𝟐
𝑻
𝒅𝒙 𝟐
= 𝟎 (𝟐. 𝟐𝟒)
Integrating this ordinary differential equation twice yields the temperature
distribution
∬
𝒅
𝟐
𝑻
𝒅𝒙 𝟐
= 𝟎
∫
𝒅𝑻
𝒅𝒙
= 𝑪
𝑻( 𝒙) = 𝑪 𝟏 𝒙 + 𝑪 𝟐 (𝟐. 𝟐𝟓)

12. For a wall with 𝑻( 𝒙 = 𝟎) = 𝑻 𝟏 𝒂𝒏𝒅 𝑻( 𝒙 = 𝑳) = 𝑻 𝟐
𝑻( 𝒙) = 𝑪 𝟏 𝒙 + 𝑪 𝟐 𝒊𝒇 ( 𝒙 = 𝟎) = 𝑻 𝟏
𝑻( 𝒙) = 𝑪 𝟏 𝟎 + 𝑪 𝟐
𝑻( 𝒙) = 𝑪 𝟐 = 𝑻 𝟏 𝒆𝒒. 𝟏
𝑻( 𝒙) = 𝑪 𝟏 𝒙 + 𝑪 𝟐 𝒊𝒇 𝑻( 𝒙 = 𝑳) = 𝑻 𝟐
𝑻( 𝒙) = 𝑪 𝟏 𝑳 + 𝑪 𝟐 = 𝑻 𝟐
𝑻 𝟐 − 𝑻 𝟏
𝑳
= 𝑪 𝟏
𝑻( 𝒙) =
𝑻 𝟐 − 𝑻 𝟏
𝑳
𝒙 + 𝑻 𝟏 (𝟐. 𝟐𝟔)

13. Next consider a similar problem, but with heat generation throughout the system, as shown in
Fig. 2.6. If the thermal conductivity is constant and the heat generation is uniform, Eq. (2.5)
reduces to
𝒌
𝝏 𝟐 𝑻
𝝏𝒙 𝟐 + 𝒒𝑮 = 𝜶
𝝏𝑻
𝝏𝒕
(𝟐. 𝟓)
Thus;
𝒌
𝒅 𝟐 𝑻
𝒅𝒙 𝟐 = −𝒒𝑮 (𝟐. 𝟐𝟕)
Integrating this equation once gives
∫ 𝒌
𝝏 𝟐
𝑻
𝝏𝒙 𝟐
= ∫−𝒒𝑮
𝒅𝑻( 𝒙)
𝒅𝒙
= −
𝒒𝑮𝒙
𝒌
+ 𝑪 𝟏 (𝟐. 𝟐𝟖)
And a second integration yields
∫
𝒅𝑻( 𝒙)
𝒅𝒙
= ∫−
𝒒𝑮𝒙
𝒌
+ 𝑪 𝟏
𝑻( 𝒙) =
−𝒒𝑮𝒙 𝟐
𝟐𝒌
+ 𝑪 𝟏𝒙 + 𝑪 𝟐 (𝟐. 𝟐𝟗)
Where C1 and C2 are constants of integration whose values are determined by the boundary
conditions. The specified conditions require that the temperature at x = 0 be T1 and at x = L be
T2. Substituting these conditions successively into the conduction equation gives
IF 𝑻𝟏 = 𝑪 𝟐( 𝒙 = 𝟎) (𝟐. 𝟑𝟎) 𝒂𝒏𝒅 𝑻 𝟐 = −
𝒒𝑮𝑳 𝟐
𝟐𝒌
+ 𝑪 𝟏 𝑳 + 𝑻𝟏 ( 𝒙 = 𝑳)(𝟐. 𝟑𝟏)
From
𝑻( 𝒙) =
−𝒒𝑮𝒙 𝟐
𝟐𝒌
+ 𝑪 𝟏𝒙 + 𝑪 𝟐 (𝟐. 𝟐𝟗)
If 𝑻𝟏 = 𝑪 𝟐( 𝒙 = 𝟎)
𝑻𝟏 = 𝑻( 𝒙) =
−𝒒𝑮𝟎 𝟐
𝟐𝒌
+ 𝑪 𝟏 𝟎 + 𝑪 𝟐 (𝟐. 𝟐𝟗)

14. 𝑻𝟏 = 𝑻( 𝒙) = 𝑪 𝟐
If 𝑻 𝟐 = −
𝒒𝑮𝑳 𝟐
𝟐𝒌
+ 𝑪 𝟏 𝑳 + 𝑻𝟏 (𝒙 = 𝑳)
𝑻 𝟐 = 𝑻( 𝒙) =
−𝒒𝑮𝑳 𝟐
𝟐𝒌
+ 𝑪 𝟏 𝑳 + 𝑪 𝟐 (𝟐. 𝟐𝟗)
𝑪 𝟏 =
𝒒𝑮𝑳 𝟐
𝟐𝒌
+ 𝑻 𝟐 − 𝑪 𝟐
𝑳
Combine the two equation
𝑻( 𝒙) =
−𝒒𝑮𝒙 𝟐
𝟐𝒌
+ (
𝒒𝑮𝑳 𝟐
𝟐𝒌
+ 𝑻 𝟐 − 𝑻𝟏
𝑳
)𝑿 + 𝑻𝟏(𝟐. 𝟐𝟗)
𝑻( 𝒙) =
−𝒒𝑮𝒙 𝟐
𝟐𝒌
+
𝒒𝑮𝑳
𝟐𝒌
+ (
𝑻 𝟐 − 𝑻𝟏
𝑳
)𝑿 + 𝑻𝟏