Notes on Steady State Conduction in Multiple Dimensions for GATE Mechanical Engineering

During our preparation for GATE Mechanical Engineering, we have all studied One Dimensional Steady State Heat Conduction. In this article, we have created short notes useful for GATE Mechanical Engineering aspirants, on the topic of Steady State Heat Conduction in Multiple Dimensions.

It is known that for a steady state conduction with no heat generation, the Laplace Equation holds true as given below,

$\frac{\partial&space;^{2}T}{\partial&space;x^{2}}+\frac{\partial&space;^{2}T}{\partial&space;y^{2}}=0&space;........(1)$

In the above equation we assume that the thermal conductivity of the material is constant. You can solve the above equation by analytical, numerical or graphical methods.

The main objective of any heat transfer analysis is to estimate the heat flow or the temperature conditions that results from that certain type of heat flow. If we solve the above equation (1), we will obtain a temperature value that is a function of two space variables, x and y. The calculation of heat flow in the x and y direction can be found out by the below given Fourier Equations.

$q_{x}=-kA_{x}\frac{\partial&space;T}{\partial&space;x}.......(2)$

$q_{y}=-kA_{y}\frac{\partial&space;T}{\partial&space;y}........(3)$

The total quantity of heat flow at any point in a given material is the resultant of the values of $q_{x}$ and $q_{y}$ at that point. This means that the total heat flow vector is perpendicular to the isotherms in the material. Like this, if the temperature distribution in a material is known we can easily establish the heat flow in the material. Refer the figure given below to understand this paragraph easily.

While deriving the solution for the two-dimensional heat flow problem, it must be kept in mind that analytical solutions are not always possible to obtain. Instead, when the analytical methods fail we make use of the numerical methods.

Let us consider a rectangular flat plate, as shown the figure below.

Three sides of this plate is maintained at a constant temperature of $T_{1}$, and the upper side of the plate has a temperature distribution as a function of x. To solve equation (1), let us assume the differential equation in a product form as,

$T=XY$ where $X=X(x)$ and $Y=Y(y)$.....(4)

The temperature distribution in the figure shown above can be a constant temperature or a more complex like a sine wave distribution. The assumption in equation (4) is ONLY justified if the solution of this form satisfies the boundary conditions. Let us consider the boundary conditions with a sine wave distribution on the upper edge of the plate.

$T=T_{1}$ at $y=0$

$T=T_{1}$ at $x=0$

$T=T_{1}$ at $x=W$........(5)

Let us assume the sine wave distribution on the upper edge as,

$T=T_{m}sin(\frac{\pi&space;x}{W})+T_{1}$ at $y=H$

where $T_{m}$ is the amplitude of the sine function.

Let us now substitute (4) in (1), and we will get,

$-\frac{1}{X}&space;\frac{d^{2}X}{dx^{2}}=\frac{1}{Y}&space;\frac{d^{2}Y}{dy^{2}}......(6)$

Upon observing closely, we see that both sides of equation (6) are independent of each other as both variables $x$ and $y$ are independent of each other. So, due to this, we need to equate the equation (6) to some constant, let us say $\lambda^{2}$, which is called the separation constant. Thus, the equation (6) becomes,

$\frac{d^{2}X}{dx^{2}}+\lambda^{2}X=0.......(7)$

$\frac{d^{2}Y}{dy^{2}}-\lambda^{2}Y=0.......(8)$

The value of the separation constant can be found out from the boundary conditions. For different values of $\lambda^{2}$, the equations (7) and (8) will have different possible solutions.

For $\lambda^{2}=0$:

$X=C_{1}+C_{2}x$

$Y=C_{3}+C_{4}y$

$T=(C_{1}+C_{2}x)(C_{3}+C_{4}y)$ ......(9)

The above solution does not fit the sine wave boundary condition, so we will discard the solution of $\lambda^{2}=0$.

For $\lambda^{2}<0$:

$X=C_{5}e^{-\lambda&space;x}+C_{6}e^{\lambda&space;x}$

$Y=C_{7}cos&space;\lambda&space;y+C_{8}&space;sin&space;\lambda&space;y$

$T=(C_{5}e^{-\lambda&space;x}+C_{6}e^{\lambda&space;x})(C_{7}cos&space;\lambda&space;y+C_{8}&space;sin&space;\lambda&space;y)$ .......(10)

Again, in the above solution the sine wave function condition is not satisfied, so this solution is also discarded.

For $\lambda^{2}>0$:

$X=C_{9}cos&space;\lambda&space;x+C_{10}&space;sin&space;\lambda&space;x$

$Y=C_{11}e^{-&space;\lambda&space;y}+C_{12}e^{\lambda&space;y}$

$T=(C_{9}cos&space;\lambda&space;x+C_{10}sin&space;\lambda&space;x)(C_{11}e^{-&space;\lambda&space;y}+C_{12}e^{\lambda&space;y})$.....(11)

Now, the above condition satisfies the sine wave temperature function of the upper edge, so we accept this solution and try to satisfy the other boundary conditions with it. To make calculations easier, let us make a substitution, $\theta&space;=&space;T-T_{1}$. With this substitution, the differential equations retain the original form, but the boundary conditions are transformed as follows:

$\theta&space;=&space;0$ at $y=0$

$\theta&space;=&space;0$ at $x=0$

$\theta&space;=0$ at $x=W$

$\theta&space;=&space;T_{m}&space;sin&space;\frac&space;{\pi&space;x}{W}$ at $y=H$.......(12)

After applying the above boundary conditions in the solution accepted, we achieve $C_{11}=-C_{12}$ and $C_{9}=0$.

Upon carrying out algeebraic manipulations using Fourier Series and Bessel Functions, we get the final solution as,

$T=T_{m}\frac{sinh(\pi&space;y/W)}{sinh(\pi&space;H/W)}sin(\frac{\pi&space;x}{W})+T_{1}$........(13) and the expression for the temperature distribution is given as,

$\frac{T-T_{1}}{T_{2}-T_{1}}=\frac&space;{2}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}+1}{n}sin(\frac{n\pi&space;x}{W})\frac{sinh(n\pi&space;y/W)}{sinh(n&space;\pi&space;H/W)}.....(14)$.