Archived How Does Fourier's Law Apply to Heat Flux in a Uniformly Heated Slab?

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The discussion focuses on applying Fourier's Law to derive the heat flux in a uniformly heated slab with constant thermal conductivity. The steady-state temperature profile is corrected to T = (gL^2/8λ)(1 - (2x/L)²) + (gL/2h) + Tf. To find the heat flux as a function of position x, the temperature profile is differentiated with respect to x, leading to dT/dx = -gx/λ. Consequently, the heat flux expression is derived as q'' = gx. This application of Fourier's Law illustrates the relationship between temperature gradient and heat flux in the slab.
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Homework Statement



a) A slab of thickness L and constant thermal conductivity \lambda generates heat at a constant rate throughout of g W m–3. The heat is dissipated from both sides of the slab by convection into the ambient air at a temperature Tf with a heat transfer coefficient h. The expression for the steady state temperature profile throughout the slab is given by

T(x) = \frac{g}{8 \lambda} L^{2}( 1- (\frac{2x^{2}}{L})) + \frac{gL}{2h} + T_{f}

where symbols have their usual meaning in this context.

(i) Derive an expression for the heat flux as a function of position x. Should I differentiate with respect to 'x'?
 
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There's a little error in that temperature profile, here's how it should look like
T = \frac{gL^2}{8 \lambda} \left[1 - \left(\frac{2x}{L}\right)^2 \right] + \frac{gL}{2h} + T_{f}
Fourier's Law of heat conduction states
q'' = - \lambda \frac{dT}{dx}
Where q'' is the heat flux in the x direction. So the way to go is to differentiate the temperature profile wrt x and multiply it by -λ.
\frac{dT}{dx} = - \frac{gx}{\lambda}
So the expression for the heat flux as a function of x is
q'' = gx
 
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