Wednesday, 11 May 2016

Temperature Distributions in the Presence of Heat Sources

Temperature Distributions in the Presence of Heat Sources

There are a number of situations in which there are sources of heat in the domain of interest. Examples are:
  1. Electrical heaters where electrical energy is converted resistively into heat.
  2. Nuclear power supplies.
  3. Propellants where chemical energy is the source.
These situations can be analyzed by looking at a model problem of a slab with heat sources $ \alpha$ (W/m3) distributed throughout. We take the outside walls to be at temperature $ T_w$ and we will determine the maximum internal temperature.
Figure : Slab with heat sources
[Overall configuration] Image fig10SlabHeatSourcesOverall_web [Infinitesimal slice] Image fig10SlabHeatSources_web
With reference to Figure (b), a steady-state energy balance yields an equation for the heat flux, $ \dot{q}$ :
$\displaystyle \dot{q} + \alpha dx - \left(\dot{q}+\frac{d\dot{q}}{dx}dx\right)=0,$
or
$\displaystyle \frac{d\dot{q}}{dx}=\alpha.$
There is a change in heat flux with $ x$ due to the presence of the heat sources. The equation for the temperature is
$\displaystyle \frac{d^2T}{dx^2} +\frac{\alpha}{k} = 0.$

Equation  can be integrated once,
$\displaystyle \frac{dT}{dx} = -\frac{\alpha}{k}x + a,$
and again to give
$\displaystyle T = -\frac{\alpha}{2k}x^2 +a x + b,$

where $ a$ and $ b$ are constants of integration. The boundary conditions imposed are $ T(0)= T(L) = T_w$ . Substituting these into Equation  gives $ b =T_w$ and $ a = \alpha L/2k$ . The temperature distribution is thus
$\displaystyle T = -\frac{\alpha}{2k}x^2 +\frac{\alpha}{2k}L x + T_w.$

Writing  in a normalized, non-dimensional fashion gives a form that exhibits in a more useful manner the way in which the different parameters enter the problem:
$\displaystyle \frac{T-T_w}{\alpha L^2/k} = \frac{1}{2}\left(\frac{x}{L}-\frac{x^2}{L^2}\right).$

Figure 18.2: Temperature distribution for slab with distributed heat sources
Image fig10HeatSourceTempDist_web
This distribution is sketched in Figure . It is symmetric about the mid-plane at $ x = L/2$ , with half the energy due to the sources exiting the slab on each side.
The heat flux at the side of the slab, $ x = 0$ , can be found by differentiating the temperature distribution and evaluating at $ x = 0$ :
$\displaystyle -k\frac{dT}{dx}\biggr\vert _{x=0} = - k\frac{\alpha L^2}{2 k}\frac{1}{L} = -\frac{\alpha L}{2}.$
This is half of the total heat generated within the slab. The magnitude of the heat flux is the same at $ x = L$ , although the direction is opposite.
A related problem would be one in which there were heat flux (rather than temperature) boundary conditions at $ x = 0$ and $ x = L$ , so that $ T_w$ is not known. We again determine the maximum temperature. At $ x = 0$ and $ L$ , the heat flux and temperature are continuous so
$\displaystyle -k\frac{dT}{dx} = h(T-T_\infty)\quad\textrm{ at }\quad x=0,L.$

Referring to the temperature distribution of Equation  gives for the two terms in Equation ,
$\displaystyle k\frac{dT}{dx} = k\left(-\frac{\alpha x}{k}+a\right)=(-\alpha x + k a).$
$\displaystyle h(T-T_\infty) = h\left(-\frac{\alpha x^2}{2k}+a x + b - T_\infty\right).$

Evaluating  at $ x = 0$ and $ L$ allows determination of the two constants $ a$ and $ b$ . This is left as an exercise for the reader.

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