4 Engineering Example 1

4.1 Heat conduction through a furnace wall

Introduction

Conduction is a mode of heat transfer through molecular collision inside a material without any motion of the material as a whole. If one end of a solid material is at a higher temperature, then heat will be transferred towards the colder end because of the relative movement of the particles. They will collide with the each other with a net transfer of energy.

Energy flows through heat conductive materials by a thermal process generally known as ’gradient heat transport’. Gradient heat transport depends on three quantities: the heat conductivity of the material, the cross-sectional area of the material which is available for heat transfer and the spatial gradient of temperature (driving force for the process). The larger the conductivity, the gradient, and the cross section, the faster the heat flows.

The temperature profile within a body depends upon the rate of heat transfer to the atmosphere, its capacity to store some of this heat, and its rate of thermal conduction to its boundaries (where the heat is transferred to the surrounding environment). Mathematically this is stated by the heat equation

$\frac{\partial }{\partial x}\left(k\frac{\partial T}{\partial x}\right)=\rho c\frac{\partial T}{\partial x}$ (1)

The thermal diffusivity $\alpha$ is related to the thermal conductivity $k$ , the specific heat $c$ , and the density of solid material $\rho$ , by

$\alpha =\frac{k}{\rho c}$

Problem in Words

The wall (thickness $L$ ) of a furnace, with inside temperature $800^{\circ }$ C, is comprised of brick material [thermal conductivity $=0.02\text{W}{\text{m}}^{-1}{\text{K}}^{-1}$ )]. Given that the wall thickness is $12$ cm, the atmospheric temperature is $0^{\circ }$ C, the density and heat capacity of the brick material are $1.9{\text{gm cm}}^{-3}$ and $6.0\text{J}{\text{kg}}^{-1}{\text{K}}^{-1}$ respectively, estimate the temperature profile within the brick wall after $2$ hours.

Mathematical statement of problem

Solve the partial differential equation

$\frac{\partial }{\partial x}\left(k\frac{\partial T}{\partial x}\right)=\rho c\frac{\partial T}{\partial t}$ (2)

subject to the initial condition

$T\left(x,0\right)=800sin\frac{\pi x}{2L}$ (3)

and boundary conditions at the inner ( $x=L$ ) and outer ( $x=0$ ) walls of

$T=0\text{at}x=0$ (4)

and

$\frac{\partial T}{\partial x}=0\text{at}x=L$ (4b)

Find the temperature profile at $T=7200\text{seconds}=2\text{hours}.$

Mathematical analysis

Using separation of variables

$T\left(x,t\right)=X\left(x\right)\times Y\left(t\right)$ (5)

so Equation (2) becomes

$\frac{Y^{\prime }}{Y}=\alpha \frac{X^{″}}{X}=K$ (6)

Using values of $K$ which are zero or positive does not allow a solution which satisfies the initial and boundary conditions. Thus, $K$ is assumed to be negative i.e. $K=-{\lambda }^2$ . Equation (6) separates into the two ordinary differential equations

with solutions

and

$T=X\times Y=e^{-{\lambda }^{2}t}\left\{Acos\frac{\lambda }{\sqrt{\alpha }}x+Bsin\frac{\lambda }{\sqrt{\alpha }}x\right\}$ (8)

where $A=A^{\ast }\times C$ and $B=B^{\ast }\times C$ .

Setting $T=0$ where $x=0$ (Equation (4a)) gives $A=0$ i.e.

$T=B{e}^{-{\lambda }^{2}t}sin\frac{\lambda }{\sqrt{\alpha }}x$ (9)

and hence

$\frac{dT}{dx}=B\frac{\lambda }{\sqrt{\alpha }}{e}^{-{\lambda }^{2}t}cos\frac{\lambda }{\sqrt{\alpha }}x$ (10)

Setting $\frac{dT}{dx}=0$ where $x=L$ (and for all $t$ ), Equation (4b) gives one of the conclusions,

The first two possibilities ( $B=0$ and $\lambda =0$ ) can be discounted as they leave $T=0$ for all $x$ and $t$ and it is not possible to satisfy the initial condition (3). Hence $cos\frac{\lambda }{\sqrt{\alpha }}L=0$ so $\frac{\lambda }{\sqrt{\alpha }}L=\left(n+\frac{1}{2}\right)\pi$ and we deduce that

$\lambda =\frac{\sqrt{\alpha }}{L}\left(n+\frac{1}{2}\right)\pi$ (11)

and so the temperature $T$ satisfies

$T=B{e}^{-\frac{\alpha }{L^2}{\left(n+\frac{1}{2}\right)}^{2}t}sin\left\{\left(n+\frac{1}{2}\right)\frac{\pi x}{L}\right\}$ (12)

However, this must also satisfy Equation (2) i.e.

$800sin\frac{\pi x}{2L}=Bsin\left\{\left(n+\frac{1}{2}\right)\frac{\pi x}{L}\right\}$ (13)

Equating the arguments of the sine terms

$\frac{\pi x}{2L}=\left(n+\frac{1}{2}\right)\frac{\pi x}{L}\text{so}n=0$

Equating the coefficients of the sine terms

$800=B$

So the temperature profile is

$T=800e^{-\frac{\alpha t}{2L^2}}sin\frac{\pi x}{2L}$ (14)

where $\alpha =\frac{k}{\rho c}=\frac{0.02}{1900\times 6}=1.764\times 10^{-6}{m}^2{s}^{-1}$ .

After two hours, $t=7200$ so $-\frac{\alpha t}{2L^2}=-0.438$ so

$T=800\times e^{-0.438}sin\frac{\pi x}{2L}=516sin\frac{\pi x}{2L}$ (15)

so the inner wall of the furnace has cooled from $800^{\circ }$ C to $516^{\circ }$ C.

Interpretation

The boundary conditions (2) and (3) represent approximations to the true boundary conditions, approximations made to enable solution by separation of variables. More realistic conditions would be