ME 651Advanced Heat Conduction

M.M. Yovanovich

CONDEQS1.MWS

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Derivation of Conduction Equations in Cartesian Coordinates.

The solid has volumetric heat sources which are the result

of a variety of processes such as chemical, electrical, nuclear,

biological, mechanical, etc.

All thermophysical properties are constant.

Obtain the following special cases from the general

three-dimensional equation:

1) Two-dimensional transient without heat sources: T(x, y, t) and P = 0;

2) One-dimensional transient with heat sources: T(x, t) and P > 0;

3) Two-dimensional, steady-state with sources: T(x, y) and P > 0;

4) Two-dimensional, steady-state without sources: T(x, y) and P = 0;

5) One-dimensional, steady-state with heat sources: T(x) and P > 0;

6) One-dimensional, steady-state without heat souces: T(x) and P = 0.

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> restart:

Conduction rates into the control volume through the faces at .

> Q[x]:= - k*dA[x]*diff(T(x, y, z, t), x);

> Q[y]:= - k*dA[y]*diff(T(x, y, z, t), y);

> Q[z]:= - k*dA[z]*diff(T(x, y, z, t), z);

Conduction rates out of the control volume through surfaces at . Take the first two terms of the Taylor series expansion.

> Q[x + dx]:= Q[x] + diff(Q[x], x)*dx;

> Q[y + dy]:= Q[y] + diff(Q[y], y)*dy;

> Q[z + dz]:= Q[z] + diff(Q[z], z)*dz;

Define the differential conduction areas.

Differential conduction area normal to x-axis

> dA[x]:= dy*dz;

Differential conduction area normal to y-axis

> dA[y]:= dx*dz;

Differential conduction area normal to z-axis.

> dA[z]:= dx*dy;

Net heat conduction rate into the differential control volume through the six

faces.

Q[cond_net]:=
Q[x] + Q[y] + Q[z] - Q[x + dx]
- Q[y + dy] - Q[z + dz];

Heat generation rate within the differential control volume due to distributed

volumetric heat sources. Volumetric heat source is with units .

Q[gen]:= P*dV;

Rate of energy storage within the differential control volume .

Q[storage]:=
rho*c[p]*diff(T(x, y, z, t), t)*dV;

> dV:= dx*dy*dz;

Energy balance over the differential control volume to obtain the general three-dimensional conduction equation. Divide by the product to simplify.

> eq:=
expand((Q[cond_net] + Q[gen] =
Q[storage])/(k*dV));

Introduce the thermal diffusivity into the equation.

eq:= subs(rho = k/alpha/c[p], eq);

The above equation is the three-dimensional diffusion equation

within a substance that has uniformly distributed volumetric heat sources.

The thermophysical properties are constant.

This partial differential equation is second-order in space and

first order in time.

It therefore requires one initial condition and six boundary

conditions to obtain its solution.

The above general equation can be used to obtain special cases.

A few special cases will be obtained below.

1) Two-dimensional transient without heat sources: and .

> DiffusionEq_2D:=
expand(
subs(T(x, y, z, t) = T(x, y, t), P = 0, eq));

2) One-dimensional transient with heat sources: and .

> DiffusionPoissonEq_1D:=
expand(
subs(T(x, y, z, t) = T(x, t), eq));

3) Two-dimensional, steady-state with sources: and .

> DiffusionPoissonEq_2D:=
expand(
subs(T(x, y, z, t) = T(x, y, t), eq));

4) Two-dimensional, steady-state without sources: and .

> LaplaceEq_2D:=
expand(
subs(T(x, y, z, t) = T(x, y), P = 0, eq));

5) One-dimensional, steady-state with heat sources: and .

> PoissonEq_1D:=
expand(
subs(T(x, y, z, t) = T(x), eq));

6) One-dimensional, steady-state without heat souces: and .

> LaplaceEq_1D:=
expand(
subs(T(x, y, z, t) = T(x), P = 0, eq));

>