Theory of heat diffusion

We have sample with form of cube. And source of  heat has form cube too.
Our goal to calculate thermal conductivity of the material.
Very important to understand is it possibility to use this material in Siberia.

Formula what defines thermal process:

 P = k* S*(dT/H)

P - value of heat flow;
k -  coefficient thermal conductivity - this is our;
S - area of loss of heat;
dT = (Tout - Tinside) - difference temperature inside and outside;
H - thickness of element;

Formula of thermal conductivity:

k =  (P*H)/(S*dT)

This formulas works for parallelepiped and perpendicular vector of heat flow.

It should be noted to things:


heat transfer occurs in three ways:

1. Thermal conductivity;
2. Сonvective heat transfer;
3. Heat transfer by radiation;

Convective heat transfer impossible because there are lot a very small volumes of air inside sample.
Radiation is excluded since sample is not transparent.
That why in out case we have only one way to loss of heat - thermal conductivity.
 

How we can define value of heat flow by electric energy?
We use electric bulb how source of heat.
According to this article most of power loss to heat.
Even radiation from bulb almost all lost inside.

We have sample in form cube and source of heat inside. Heat flowing isotropic in all dimensions in 3D. For calculations, we just need to consider one pyramid of cube.
And now we need to solve equation for define how heat flow through pyramid.

S0 -  area of base of the pyramid;
H - height of pyramid;

We assume that the temperature depends only on the height. The temperature is constant in each layer.
In reality, this is true only for "long" pyramids. The lower the pyramid, the more this condition is violated.

The dependence of the cross section of the pyramid on x:

Sx - area of x layer:

here x=0 -- base of pyramid.

All heat what comes in layer  with thickness dx will transfer to next layer

 

Let's introduce a constant:

T''(x) / T'(x) = S'(x) / S(x) 

here T'' -- second derivative, T' -- first derivative. 

Calculate the integral:

 

 And after that we have:

C1 -- unknown integration constant,

Needs to define constants C0 and C1 from conditions:
T(0) =T0
T(H) =T1

Now we compare how much heat passes through this pyramid with heat passing through box S1:

q2 - (current through parallelepiped) proportionally S1 * (T1-T0) / H
q1 - (through pyramid) proportionally: 

 And we have:

 

If named this ratio Kpir as result we have:

How we can use it coefficient?

Knowing the heating power and thermal resistance, you can find the temperature in the house at a given temperature on the street: 

Q=(Tin-Tout)/R

Tin = Q*R+Tout

R=h/k*S



S - total area of outside surface;
h - thickness wall;
k - it is that we calculated.
Tin = Q*h/k*S+Tout




After that if we know size of house, thickness of wall, power of heating and this coefficient of thermal conductivity we can find comfort balance of all this parameters.
For example for temperature +20 inside and -50 outside.

That's all! Good luck!