Dear Reader,

I have been slaving away with my pencil and a paper writing equations and now I have some answers for you.

If we have a slab with a cross sectional area of A and a thickness of x which has a thermal conductivity of λ and the temperture difference between the two sides is ΔT, then the thermal power passing through the slab will be

P = A ΔT λ / x

So as the temperature gradient in the slab is uniform we can now write it as

P = A λ ΔT / Δ x

This is the equation which is used for Searle’s bar, now the fun can start.

If we have an infinite slab of uniform thickness x in which heat is being produced evenly throughout all of the volume, and we lay the slab on a perfect insulator so heat can leave at only the top surface then the heat flux going up slab will be given by the equation.

P = ρ A *x*

This is where ρ is the power density in watts per cubic meter, A is the area of the slab and *x* is the distance from the bottom of the slab. We can combine the two equations to give us

dT/dx = ρ A *x* / λA = ρ *x* / λ

Now we integrate (reverse the differentiation) to give us

T = c + *x*^{2} ρ / 2λ

This gives us a curve which suggests that the top of the slab is hotter than the bottom so we have to add a – sign to give us

T = c – *x*^{2} ρ / 2λ

If c is zero we now have an equation for telling us what the temperture difference is at the different depths in the slab.

We can now add another term for the overall thickness (x) of the slab. The temperture at the top surface of the slab is x, and this then finishs off our equation.

T = c + (*x*^{2} ρ / -2λ) + (x^{2} ρ / 2λ) = c + {ρ(x^{2} – *x*^{2})/2λ}

Now the slab laying on the perfect insulator is equivilent to a two sided slab which has a thickness of 2x, now *x* is the distance from the centre line of the slab. This is the perfect equation to deal with a plate type fuel in a research reactor.

If we were to assume that the floor of the containment is a perfect thermal insulator then we also have the basis of the perfect equation.

As the volume of a cylinder is given by

V = 2 *l *π r^{2}

and the area is given by A = 2 *l* π r

We can make the equation

ρ r^{2} / 2 λ r = dT / dr = ρ r / 2 λ

Which gives us

T = c + ρ r^{2} / 4 λ

Which leads us to the same equation as the nuclear chemistry textbook on my desk has for this system.

T = c + {ρ(r^{2} – *r*^{2})/4λ}

Using the equation for the volume of a sphere (V = 4 π r^{3} / 3 ) and the area of a sphere ( A = 4 π r^{2}) I was then able to get equations which lead me to

T = c + {ρ(r^{2} – *r*^{2})/6λ}

This equation will explain to us how a sphere which has uniform heat production throughout it will be hotter in the middle than the surface.

Now we have some fun equations, I suggest that we go an understand the world a bit better using our equations.

Filed under: heat, nuclear, nuclear chemistry | Tagged: temperature gradient, thermal conductivity, thermal insulator |

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