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Heat transfer

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The main article is for now Modeling with discrete exterior calculus but the geometry of the model is incomplete...


The amount of heat exchanged between two rooms is proportional to:

  1. the temperature difference,
  2. the conductance of the wall,
  3. the length of the wall that separates them,
  4. and, inversely, to the distance between the centers of mass of the rooms.

But there is more!

The amount of heat depends on the angle at which it passes through the wall.

Suppose $a$ is the wall (edge) and $b=a^*$ is the pipe between the centers (its Hodge dual). Suppose as vectors they are: $a=(x,y),b=(u,v)$. Then $c=(y,-x)$ is perpendicular to $a$ and has the same length.

Duality.PNG

Then the heat flow is proportional to the cosine of the angle between $b$ and $c$. To take items 3 and 4 above into account we have the "adjustment coefficient": $$A=\frac{|<c,b>|}{\|b\|^2},$$ where $|<c,b>|$ is the dot product. Note that when the geometry is Euclidean, $a$ and $b$ are perpendicular and equal in length. In that case, $A=1$.


This derivation follows Diffusion...

The preservation of the material in cell $\sigma$ is given by $$d_t U(\sigma,t)=−\int_{∂\sigma} *F(·,t),$$ where $d_t$ is the exterior derivative with respect to time (just the difference since the dimension is $1$) and $*$ is the Hodge duality. The flow $F$ through face $a$ of cell $\sigma$ is proportional to the difference of amounts of material in $\sigma$ and the other adjacent to $a$ cell. So, $$F(a)=-kd_x(*U)(a^*,t).$$ Substitute: $$d_t U(\sigma,t)=\int_{∂\sigma} *kd_x(*U)(a^*,t).$$ Integration is summation: $$d_t U(\sigma,t)=\sum \{*kd_x(*U)(a^*,t):a,∂\sigma =\sum a\}.$$