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Heat Transfer

Lecturer : Dr. Rafel Hekmat Hameed University of Babylon

Subject : Heat Transfer College of Engineering

Year : Third B.Sc. Mechanical Engineering Dep.

Steady-State Conduction-Multiple Dimensions

For a 2 - D, steady state situation with no heat generation and constant thermal conductivity,

the heat equation is simplified to

……….. (1)

it needs two boundary conditions in each direction.

There are three approaches to solve this equation:

• Analytical Method: The mathematical equation can be solved using techniques like the method of

separation of variables.

• Graphical Method: Limited use. However, the conduction shape factor concept derived under this

concept can be useful for specific configurations.

• Numerical Method: Finite volume, Finite difference or finite element schemes, usually will be

solved using computers.

Analytical Method

It is worthwhile to mention here that analytical solutions are not always possible to obtain;

indeed, in many instances they are very cumbersome and difficult to use. Consider the rectangular

plate shown in figure below. Three sides of the plate are maintained at the constant temperature T1,

and the upper side has some temperature distribution impressed upon it. This distribution could be

simply a constant temperature or something more complex, such as a sine-wave distribution

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T is a function of X and Y

T=X.Y ; where X=X(x) , and Y=Y(y)

Solve by Separation Variables Method

𝜕2(𝑋.𝑌)

𝜕𝑥2+

𝜕2(𝑋.𝑌)

𝜕𝑦2= 0 ;

𝜕2𝑋

𝜕𝑥2𝑌 +

𝜕2𝑌

𝜕𝑦2𝑋 = 0 divided by XY

𝜕2𝑋

𝜕𝑥21

𝑋+

𝜕2𝑌

𝜕𝑦21

𝑌= 0 ;

𝜕2𝑋

𝜕𝑥21

𝑋= −

𝜕2𝑌

𝜕𝑦21

𝑌

Since the variables are separated , each side is constant taking this constant to be 2 so

𝜕2𝑋

𝜕𝑥2= 𝑋𝜆2 �̿� − 𝜆2𝑋 = 0 ; −

𝜕2𝑌

𝜕𝑦2= 𝜆2𝑌 �̿� + 𝜆2𝑌 = 0

The general solution will be

𝑋 = 𝐶1𝑒−𝜆𝑥 + 𝐶2𝑒

𝜆𝑥 𝑌 = 𝐶3 cos 𝜆𝑦 + 𝐶4 sin 𝜆𝑦

T=X.Y T=(𝑪𝟏𝒆−𝝀𝒙 + 𝑪𝟐𝒆

𝝀𝒙) (𝑪𝟑 𝐜𝐨𝐬 𝝀𝒚 + 𝑪𝟒 𝐬𝐢𝐧 𝝀𝒚)

If 2 =0

�̿� = 0 ; 𝑋 = 𝐶1 + 𝐶2𝑥 ; 𝑌 = 𝐶3 + 𝐶4𝑦

T=X.Y T=(𝑪𝟏 + 𝑪𝟐𝒙) (𝑪𝟑 + 𝑪𝟒𝒚) solution may be excluded

For 2 < 0

𝜕2𝑋

𝜕𝑥2= −𝜆2𝑋 �̿� + 𝜆2𝑋 = 0 ; −

𝜕2𝑌

𝜕𝑦2= −𝜆2𝑌 �̿� − 𝜆2𝑌 = 0

𝑋 = 𝐶1 sin 𝜆𝑥 + 𝐶2 cos 𝜆𝑥 ; 𝑌 = 𝐶3𝑒𝜆𝑦 + 𝐶4𝑒

−𝜆𝑦

T=X.Y T=(𝑪𝟏 𝐬𝐢𝐧 𝝀𝒙 + 𝑪𝟐 𝐜𝐨𝐬 𝝀𝒙)(𝑪𝟑𝒆𝝀𝒚 + 𝑪𝟒𝒆

−𝝀𝒚)

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The values of C1 , C2 , C3 , C4 and 𝜆 are determined from the boundary conditions

a) T=0 at x=0

b) T=0 at y=0

c) T=0 at x=L

d) T=Tm 𝑠𝑖𝑛𝜋𝑥

𝐿 at y=W

Applying B.C. (a)

0 = 𝐶2(𝑪𝟑𝒆𝝀𝒚 + 𝑪𝟒𝒆

−𝝀𝒚) ……(*)

C2 = 0 ; 𝐂𝟑𝐞𝛌𝐲 = −𝐂𝟒𝐞

−𝛌𝐲

Applying B.C. (b)

0 = (𝐶1 sin 𝜆𝑥 + 𝐶2 cos 𝜆𝑥)(𝐶3 + 𝐶4) ……(**)

Applying B.C. (c)

0 =(𝐶1 sin 𝜆𝐿 + 𝐶2 cos 𝜆𝐿)(𝐶3𝑒𝜆𝑦 + 𝐶4𝑒

−𝜆𝑦) ……(***)

Applying B.C. (d)

𝑇𝑚𝑠𝑖𝑛𝜋𝑥

𝐿= (𝐶1 sin 𝜆𝑥 + 𝐶2 cos 𝜆𝑥)(𝐶3𝑒

𝜆𝑊 + 𝐶4𝑒−𝜆𝑊) ……(****)

From eq. (**) (𝐶3 + 𝐶4) = 0 𝐶3 = −𝐶4 ; sub. in eq. (***)

0 =(𝐶1 sin 𝜆𝐿 + 0 cos 𝜆𝐿)(𝐶3𝑒𝜆𝑦 − 𝐶3𝑒

−𝜆𝑦)

0 =(𝐶1 sin 𝜆𝐿)(𝐶3𝑒𝜆𝑦 − 𝐶3𝑒

−𝜆𝑦)

0 =(𝐶1 𝐶3𝑠𝑖𝑛 𝜆𝐿)(𝑒𝜆𝑦 − 𝑒−𝜆𝑦)

T =(𝐶1 𝐶3 𝑠𝑖𝑛 𝜆𝐿)(𝑒𝜆𝑦 − 𝑒−𝜆𝑦) ; We know that sinh(𝜆𝑦) =

𝑒𝜆𝑦−𝑒−𝜆𝑦

2 so

T =(2𝐶1 𝐶3 𝑠𝑖𝑛 𝜆𝑥) sinh 𝜆𝑦 T =(𝑪𝒏 𝒔𝒊𝒏 𝝀𝒙) 𝐬𝐢𝐧𝐡 𝝀𝒚

From B.C. ( c )

0 =(𝐶𝑛 𝑠𝑖𝑛 𝜆𝐿) sinh 𝜆𝑦 ; sin (L) = 0 for all y ; =𝑛𝜋

𝐿 for n=(1,2,3,..)

𝑻 = 𝑪𝒏 𝐬𝐢𝐧 (𝒏𝝅𝒙

𝑳) 𝐬𝐢𝐧𝐡 (

𝒏𝝅𝒚

𝑳)

For each integer n there exists a different solution, so

𝑇 = ∑∞𝑛=1 𝐶𝑛 sin (𝑛𝜋𝑥

𝐿) sinh (

𝑛𝜋𝑦

𝐿)

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from B.C. (d)

𝑇𝑚sin (𝜋𝑥

𝐿) = ∑∞𝑛=1 𝐶𝑛 sin (

𝑛𝜋𝑥

𝐿) sinh (

𝑛𝜋𝑊

𝐿)

Which holds only if C2= C3= C4=0

𝐶𝑛 = 𝐶1 =𝑇𝑚sin (

𝜋𝑥

𝐿)

sin(𝜋𝑥

𝐿) sinh(

𝜋𝑊

𝐿)

=𝑇𝑚

sinh(𝜋𝑊

𝐿)

The solution therefore, becomes

𝑇 =𝑇𝑚 sin(

𝜋𝑥

𝐿) sinh(

𝜋𝑦

𝐿)

sinh(𝜋𝑊

𝐿)

THE CONDUCTION SHAPE FACTOR

This approach applied to 2-D conduction involving two isothermal surfaces, with all other surfaces

being adiabatic. We may define a conduction shape factor S such that

q=kS Toverall

The values of S have been worked out for several geometries and are summarized in Table 3-1 (page.78

in Holman) For a three-dimensional wall, as in a furnace, separate shape factors are used to calculate

the heat flow through the edge and corner sections, with the dimensions shown in Figure below. When

all the interior dimensions are greater than one-fifth of the wall thickness,

where

A= area of wall

L= wall thickness

D= length of edge

NUMERICAL METHOD OF ANALYSIS

Due to the increasing complexities encountered in the development of modern technology, analytical

solutions usually are not available. For these problems, numerical solutions obtained using high-speed

computer are very useful, especially when the geometry of the object of interest is irregular, or the

boundary conditions are nonlinear. In heat transfer problems, the finite difference method is used more

often and will be discussed here.The finite difference method involves:

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* Establish nodal networks

* Derive finite difference approximations for the governing equation at both interior and exterior nodal

points

* Develop a system of simultaneous algebraic nodal equations

* Solve the system of equations using numerical schemes

The Nodal Networks

• The basic idea is to subdivide the area of interest into sub- volumes with the distance between

adjacent nodes by Δx and Δy as shown.

• If the distance between points is small enough, the differential equation can be approximated locally

by a set of finite difference equations.

• Each node now represents a small region where the nodal temperature is a measure of the average

temperature of the region.

`

Finite Difference Approximation

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approximate the second order differentiation at m,n

Similarly, the approximation can be applied to the other dimension y

To model the steady state, no generation heat equation:∇2𝑇 = 0. This approximation can be simplified

by specify Δx=Δy and the nodal equation can be obtained as

This equation approximates the nodal temperature distribution based on the heat equation. We can also

devise a finite-difference scheme to take heat generation into account.We merely add the term ˙q/k

into the general equation and obtain

Then for a square grid in which x =y,

A very simple example is shown in figure below, and the four equations for nodes 1, 2, 3, and 4 would

be

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These equations have the solution

Of course, we could recognize from symmetry that T1 =T2 and T3 =T4 and would then only need two

nodal equations,

Numerical Solutions By using MATLAB

Matrix form: [A][T]=[C].

From linear algebra: [A]-1[A][T]=[A]-1[C], [T]=[A]-1[C]

where [A]-1 is the inverse of matrix [A]. [T] is the solution vector.

With internal generation where g is the power generated per unit volume (W/m3). G= g.V Based on

the energy balance concept:

When the solid is exposed to some convection boundary condition as shown in figure below, the

temperatures at the surface must be computed differently from the method given above

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If x=y

See table (3-2) in Holman book

Radiation heat exchange with respect to the surrounding (assume no convection, no generation to

simplify the derivation). Given surface emissivity ε, surrounding temperature Tsurr.

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Non-linear term, can solve using the iteration method

Biot number Bi =