HMT Week 8.pdf convection convection radiation

HEAT AND MASS
TRANSFER
STEADY STATE CONDUCTION-
MULTIPLE DIMENSIONS
Arranged By
PROF. DR. ASAD NAEEM SHAH
anaeems@uet.edu.pk

TWO-DIMENSIONAL, STEADY-
STATE CONDUCTION
o When the boundaries of a system are irregular or the
temperature along a boundary is non-uniform, a 1-D treatment
may not be satisfactory. Temperature here may be a function of
two or even three coordinates.
o Some of the examples of multidimensional heat conduction
(HC) are as follows:
• Air conditioning ducts
• Cooling of IC engine block
• Heat treatment of metallic parts of different shapes
• Composite bodies
• Chimneys, etc.
Arranged by Prof. Dr. Asad Naeem Shah

TWO-DIMENSIONAL, STEADY-STATE
CONDUCTION Cont.
o Consider a solid in which there is two-dimensional heat
conduction (Fig.1). Two surfaces (upper & lower) of the solid
are insulated, while the other two are maintained at
temperatures 𝑇1 & 𝑇2 such that 𝑇1 > 𝑇2. So, heat transfer by
conduction will take place from surface 1 to surface 2.
Fig. 1

o According to the Fourier’s law (in general):
𝑞 = −𝑘𝛻𝑇 = −𝑘 𝒊
𝜕𝑇
𝜕𝑥
+ 𝒋
𝜕𝑇
𝜕𝑦
+ 𝒌
𝜕𝑇
𝜕𝑧
i.e., the local heat flux (consisting of 𝑞𝑥, 𝑞𝑦, 𝑞𝑧) in the solid is a
vector that is everywhere perpendicular to lines of constant
temperature (isotherms) as shown in Fig. 1.
o Since the heat flow lines (HFL) are, by definition, the lines along
which the heat flow occurs, so no heat can be conducted across
a heat flow line, and HFL are therefore termed as adiabats.
Thus, adiabatic surfaces (or adiabats or symmetry lines) are
known as HFL.
o The heat flux vector itself is the resultant of heat flux
components in the x and y direction (i.e., in 2- D case).
TWO-DIMENSIONAL, STEADY-STATE
CONDUCTION Cont.

MATHEMATICAL ANALYSIS OF
TWO-DIMENSIONAL HEAT CONDUCTION
o There are two major objectives in any conduction analysis. The first
objective is to determine the temperature distribution (TD) in the
medium, which represents how temperature varies with position in
the medium and thus necessitates determining T(x, y).
o Once TD is known, the heat flux at any point in the medium or on its
surface may be computed by using the Fourier law i.e., 2nd major
objective of the analysis is achieved. Thus, considering the 2 − 𝐷,
steady-state (S.S), no heat generation, and constant thermal
conductivity conditions, we have the following equation:
𝜕2
𝑇
𝜕𝑥2 +
𝜕2
𝑇
𝜕𝑦2 = 0 → (1)
o To solve the equation (1), the following important methods may be
used: Arranged by Prof. Dr. Asad Naeem Shah

TWO-DIMENSIONAL HEAT CONDUCTION Cont.
➢ ANALYTICAL METHOD (yielding an exact solution): Involves separation
of variables and Orthogonality
➢ APPROXIMATE METHODS (provide only approximate results at
discrete points)
• The Integral Method
• The method of scale analysis or order of magnitude analysis
• The Graphic Method
➢ NUMERICAL METHODS
• Finite Difference Method/The Energy Balance Method
• Finite element method,
• Finite volume method
The developed equations then may be solved either by “The Matrix
Inversion Method” or “The Gauss-Seidel Method”.

o The use of the analytical method is restricted to only relatively simple
geometric shapes. Similarly, the graphical method gives only an
approximate solution as it is based on plotting by trials, constant
temperature contours, constant heat flow paths, and using the BCs as
the initial guide.
o So, most of the practical (research-oriented) 2-D heat transfer problems
are solved by numerical techniques. The main advantage of these
techniques is their application to any shape, irrespective of its
complexity or boundary conditions.
o The major difference between the analytical solution and the numerical
solution is that the former gives an equation from which the
temperature may be obtained anywhere in the solid, whereas the latter
gives the temperature values at certain specific (discrete) points only.
These techniques are well suited for use with high-speed digital
computers.
TWO-DIMENSIONAL HEAT CONDUCTION Cont.

ANALYTICAL APPROACH
o This approach is applied only to simple problems as the technique is
cumbersome and it is difficult to get the solution always. This
method involves the ‘separation of variables’ & ‘orthogonality’.
o Let us consider a system consisting of a rectangular plate with three
sides maintained at a constant temperature 𝑇1 (or 𝑇∞) , while the
fourth side kept at a constant temperature 𝑇2 ≠ 𝑇1 (Fig. 1).
Fig. 1: The 2 − 𝐷 conduction in
a thin- rectangular plate.
As the 2 − 𝐷, S.S with no heat generation leads to:
𝜕2𝑇
𝜕𝑥2
+
𝜕2𝑇
𝜕𝑦2
= 0 … 𝑬𝒒 𝟏
Following are the boundary conditions:
1. 𝜃 0, 𝑦 = 0 3. 𝜃 𝑥, 0 = 0
2. 𝜃 𝐿, 𝑦 = 0 4. 𝜃 𝑥, 𝑊 = 1

ANALYTICAL APPROACH Cont.
o To get the temperature distribution 𝑇(𝑥, 𝑦) , the temperature
transformation 𝜃 is introduced to simplify the Boundary Conditions
(BCs) & solution as:
𝜃 =
𝑇 − 𝑇1
𝑇2 − 𝑇1
… 𝑬𝒒(𝟐)
o By Eqn.(2) & Eqn.(1), the transformed differential equation is:
𝜕2
𝜃
𝜕𝑥2 +
𝜕2
𝜃
𝜕𝑦2 = 0 … 𝑬𝒒(𝟑)
o It is important to note that through the transformation of Eqn. (2),
three of the four boundary conditions are now homogeneous, and
thus the value of 𝜃 is restricted to the range from 0 to 1.

o Applying the separation of variables technique by assuming that
the desired solution can be expressed as the product of two
functions i.e.,
𝜃 𝑥, 𝑦 = 𝑋 𝑥 𝑌(𝑦)
⇒
𝜕2
𝜃
𝜕𝑥2
= 𝑌
𝜕2
𝑋
𝜕𝑥2
and
⇒
𝜕2
𝜃
𝜕𝑦2 = 𝑋
𝜕2
𝑌
𝜕𝑦2
o Thus Eqn.(3) becomes:
⇒ 𝑌
𝜕2𝑋
𝜕𝑥2
+ 𝑋
𝜕2𝑌
𝜕𝑦2
= 0

o Dividing by XY
⇒
1
𝑋
𝜕2𝑋
𝜕𝑥2 +
1
𝑌
𝜕2𝑌
𝜕𝑦2 = 0
−
1
𝑋
𝜕2
𝑋
𝜕𝑥2
=
1
𝑌
𝜕2
𝑌
𝜕𝑦2
= + 𝜆2 𝑠𝑎𝑦 … 𝑬𝒒(𝟒)
i.e., 𝜆2
> 0, so
−
1
𝑋
𝜕2𝑋
𝜕𝑥2 = + 𝜆2
⇒
𝜕2𝑋
𝜕𝑥2
+ X 𝜆2
= 0
i.e., a homogeneous second order differential equation is obtained
leading to: (𝐷2
+ 𝜆2
)𝑋 = 0

𝐷2
+ 𝜆2
= 0 ⟹ 𝐷2
= − 𝜆2
⟹ 𝐷 = ± 𝑖𝜆
o Therefore,
𝑋 = 𝐶1 cos 𝜆𝑥 + 𝐶2 sin 𝜆𝑥 … 𝑬𝒒(𝟓)
o Also, considering
1
𝑌
𝑑2
𝑌
𝑑𝑦2
= +𝜆2
⇒
𝑑2𝑌
𝑑𝑦2
− 𝜆2𝑌 = 0
⇒ 𝐷2 − 𝜆2 𝑌 = 0
⇒ 𝐷2
− 𝜆2
= 0 ⟹ 𝐷 − 𝜆 𝐷 + 𝜆 = 0 ⟹ 𝐷 = −𝜆, +𝜆
o Therefore,
𝑌 = 𝐶3𝑒−𝜆𝑦
+ 𝐶4 𝑒𝜆𝑦
… 𝑬𝒒(𝟔)

⇒ 𝜃 = 𝜃 𝑥, 𝑦 = 𝐶1 cos 𝜆𝑥 + 𝐶2 sin 𝜆𝑥 𝐶3𝑒−𝜆𝑦
… 𝑬𝒒(𝟕)
o Applying the boundary condition 𝜃 0, 𝑦 = 0, Eqn.(7) gives:
0 = (𝐶1 + 0)( 𝐶3𝑒−𝜆𝑦
)
⇒ 𝑪𝟏 = 𝟎
o Also, applying the boundary condition 𝜃 𝑥, 0 = 0, Eqn.(7) leads to:
0 = 𝐶1 cos 𝜆𝑥 + 𝐶2 sin 𝜆𝑥 𝐶3𝑒−𝜆(0) + 𝐶4 𝑒𝜆(0)
o Since 𝐶1 = 0,
⇒ 0 = 𝐶2 sin 𝜆𝑥 (𝐶3 + 𝐶4)
o If 𝐶2 = 0, the equation is satisfied but equality eliminates the 𝑥
dependence and hence no acceptable solution is available.
o However, if 𝐶3 = −𝐶4, equation is satisfied without eliminating the 𝑥
dependence. Thus,
⇒ 𝑪𝟑 = −𝑪𝟒

o Now applying the boundary condition 𝜃 𝐿, 𝑦 = 0 to Eqn.(7):
0 = (𝐶1 cos 𝜆𝐿 + 𝐶2 sin 𝜆𝐿) −𝐶4𝑒−𝜆𝑦
∵ 𝑪𝟑 = −𝑪𝟒
o Also, 𝑪𝟏 = 𝟎,
⇒ 0 = 𝐶2𝐶4 sin 𝜆𝐿 (𝑒𝜆𝑦
− 𝑒−𝜆𝑦
)
o Since (𝑒𝜆𝑦
− 𝑒−𝜆𝑦
) ≠ 0,
⇒ sin 𝜆𝐿 = 0
⇒ 𝜆𝐿 = 𝑛𝜋
⇒ 𝜆 =
𝑛𝜋
𝐿
; 𝑛 = 1,2,3, …
o Integer 𝑛 = 0 is prevented to be used because it provides an
unacceptable solution.
o The solution of Eqn. (7) now converges to:
⇒ 𝜃 = 𝐶2𝐶4 sin
𝑛𝜋
𝐿
𝑥 𝑒
𝑛𝜋
𝐿 𝑦
− 𝑒−
𝑛𝜋
𝐿 𝑦

⇒ 𝜃 = 𝐶2𝐶4 sin
𝑛𝜋
𝐿
𝑥 𝑒
𝑛𝜋
𝐿 𝑦
− 𝑒−
𝑛𝜋
𝐿 𝑦
×
2
2
o Combining constants so that 2𝐶2𝐶4 = 𝐶𝑛, so a general solution is:
⇒ 𝜃 𝑥, 𝑦 = ෍
𝑛=1
∞
𝐶𝑛 sin(
𝑛𝜋𝑥
𝐿
) sinh(
𝑛𝜋𝑦
𝐿
) … 𝑬𝒒(𝟖)
o Since 𝐶𝑛 is unknown, applying the 4th boundary condition i.e., 𝜃 𝑥, 𝑊 = 1
to determine 𝐶𝑛.
𝜃 𝑥, 𝑊 = 1 = ෍
𝑛=1
∞
𝑪𝒏 𝒔𝒊𝒏𝒉(
𝒏𝝅𝑾
𝑳
) sin(
𝑛𝜋𝑥
𝐿
) … 𝑬𝒒(𝟗)
o Any function 𝑓(𝑥) may be expressed in terms of an infinite series of orthogonal
functions, if:
𝑓 𝑥 = ෍
𝑛=1
∞
𝐴𝑛𝑔𝑛(𝑥) … 𝑬𝒒(𝟏𝟎)
Thus, from Eqns. (9) & (10): 𝑓 𝑥 = 1; 𝑨𝒏 = 𝑪𝒏 𝒔𝒊𝒏𝒉
𝒏𝝅𝑾
𝑳
; and 𝑔𝑛 𝑥 = sin(
𝑛𝜋𝑥
𝐿
)

o An infinite set of functions 𝑔1 𝑥 , 𝑔2 𝑥 , … , 𝑔𝑛(𝑥) is said to be orthogonal in the
domain 𝑎 ≤ 𝑥 ≤ 𝑏 if:
න
𝑎
𝑏
𝑔𝑚 𝑥 𝑔𝑛 𝑥 𝑑𝑥 = 0 ; 𝑚 ≠ 𝑛 … 𝑬𝒒(𝟏𝟏)
o So, multiplying each side of Eqn.(10) by 𝑔𝑛(𝑥) and integrating
between a and b
න
𝑎
𝑏
𝑓 𝑥 𝑔𝑛 𝑥 𝑑𝑥 = න
𝑎
𝑏
𝑔𝑛 𝑥 ෍
𝑛=1
∞
𝐴𝑛𝑔𝑛 𝑥 𝑑𝑥
⇒ න
𝑎
𝑏
𝑓 𝑥 𝑔𝑛 𝑥 𝑑𝑥 = 𝐴𝑛 න
𝑎
𝑏
𝑔𝑛
2
𝑥 𝑑𝑥
⇒ 𝐴𝑛 =
‫׬‬
𝑎
𝑏
𝑓 𝑥 𝑔𝑛 𝑥 𝑑𝑥
‫׬‬
𝑎
𝑏
𝑔𝑛
2
𝑥 𝑑𝑥

o From Fig. 1, since 𝑎 = 0 and 𝑏 = 𝐿:
𝐴𝑛 =
‫׬‬0
𝐿
(1)(sin
𝑛𝜋
𝐿
𝑥) 𝑑𝑥
‫׬‬0
𝐿
sin2 𝑛𝜋
𝐿 𝑥 𝑑𝑥
⇒ 𝐴𝑛 =
𝐿
𝑛𝜋
− cos
𝑛𝜋
𝐿
𝑥
𝐿
0
‫׬‬0
𝐿 1 − cos 2
𝑛𝜋
𝐿
𝑥
2
𝑑𝑥
⇒ 𝐴𝑛 =
−
𝐿
𝑛𝜋
cos 𝑛𝜋 − 1
1
2
𝐿 −
𝐿
4𝑛𝜋
sin 2
𝑛𝜋
𝐿
𝑥
0
𝐿
⇒ 𝐴𝑛 =
−
𝐿
𝑛𝜋
cos 𝑛𝜋 − 1
𝐿
2
=
−1
1
𝑛𝜋
−1 𝑛 − 1
1
2

⇒ 𝑨𝒏 =
2
𝑛𝜋
−1 𝑛+1
+ 1
o Therefore, Eqn.(10) becomes:
1 = ෍
𝑛=1
∞
𝟐
𝒏𝝅
−𝟏 𝒏+𝟏 + 𝟏 sin
𝑛𝜋
𝐿
𝑥 … 𝑬𝒒(𝟏𝟐)
o Comparing Eqn.(9) and Eqn.(12):
⇒ 𝐶𝑛 =
2
𝑛𝜋
−1 𝑛+1 + 1
sinh
𝑛𝜋𝑊
𝐿
𝑛 = 1,2,3, … ,
o Finally, Eqn.(8) gives:
𝜽 𝒙, 𝒚 =
𝟐
𝝅
෍
𝒏=𝟏
∞
[ −𝟏 𝒏+𝟏
+ 𝟏]
𝒏
𝒔𝒊𝒏(
𝒏𝝅
𝑳
𝒙)
𝒔𝒊𝒏𝒉
𝒏𝝅𝒚
𝑳
𝒔𝒊𝒏𝒉
𝒏𝝅𝑾
𝑳
… 𝑬𝒒(𝟏𝟑)

o Above Eqn. (13) is a convergent series, from which the value of 𝜃 may be
determined for any value of x and y. The results are shown in terms of
isotherms in the Fig. 2.
o Also, the temperature 𝑇 corresponding to a value ′𝜃′ may be obtained
from Eqns.(2) & (13).
Fig. 2: Isotherms and heat flow lines for 2-D conduction in a rectangular plate.

PROBLEM
Consider a two-dimensional problem shown in Fig. The linear dimensions
are 𝐿 = 𝑊 = 2𝑚, and the temperature on three sides is 𝑇1 = 280𝐾. The
temperature along the upper surface is 𝑇2 = 320𝐾 . Determine the
temperature at the center of the plate.
SOLUTION:
𝜃 =
𝑇 − 𝑇1
𝑇2 − 𝑇1
=
𝜃∗
𝜃𝑐
𝜃∗(𝑥, 𝑦) = 𝜃𝑐
2
𝜋
෍
𝑛=1
∞
−1 𝑛+1 + 1
sinh
𝑛𝜋𝑦
𝐿
sinh
𝑛𝜋𝑊
𝐿
sin
𝑛𝜋𝑥
𝐿
∵
𝜋𝑦
𝐿
=
𝜋𝑥
𝐿
=
𝜋 1
2
=
𝜋
2
(∵ 𝑎𝑡 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑒𝑟
𝑥
2
,
𝑦
2
)
𝑎𝑛𝑑
𝜋𝑊
𝐿
=
𝜋 2
2
= 𝜋
𝑇2
𝑇𝑐

∵ 𝜃𝑐 = 𝑇2 − 𝑇1 = 320 − 280 = 40𝐾
We construct the following table to find the required temperature for first three
values of “n” :
n −𝟏 𝒏+𝟏 + 𝟏
𝒏
𝒔𝒊𝒏𝒉 (
𝒏𝝅
𝟐
)
𝒔𝒊𝒏𝒉 (𝒏𝝅)
𝒔𝒊𝒏
𝒏𝝅
𝟐
1 2 2.30130 11.54870 1.0000
2 2/3 55.65440 6195.82000 -1.0000
3 2/5 1287.98000 3.31871 × 106 1.0000

So,
𝜃∗
(𝑥, 𝑦) = 40
2
𝜋
2 sinh
𝜋
2
sinh 𝜋
sin
𝜋
2
+
2
3
sinh
3𝜋
2
sinh 3𝜋
sin
3𝜋
2
+
2
5
sinh
5𝜋
2
sinh 5𝜋
sin
5𝜋
2
+ ⋯
⇒ 𝜃∗
(1, 1) =
80
𝜋
0.398538 − 0.005988 + 0.001555 − ⋯ = 10.0002𝐾
∵ 𝑇 1,1 = 𝜃∗
(1, 1) + 𝑇1
∴ 𝑇 1,1 = 10.0002 + 280 ≅ 290𝐾 Answer

HMT Week 8.pdf convection convection radiation

HMT Week 8.pdf convection convection radiation

Recommended

Recommended

More Related Content

Recently uploaded

Recently uploaded (20)

Featured

Featured (20)

HMT Week 8.pdf convection convection radiation