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General heat conduction equation in Cartesian coordinates

Short answer

General equation: ∇²T + q̇/k = (1/α) ∂T/∂t. Reduced forms: Laplace ∇²T = 0, Poisson ∇²T + q̇/k = 0, Fourier ∇²T = (1/α) ∂T/∂t. Cylindrical form: (1/r)∂/∂r(r ∂T/∂r) + (1/r²)∂²T/∂φ² + ∂²T/∂z² + q̇/k = (1/α) ∂T/∂t.

Derive the general three-dimensional heat conduction equation in Cartesian coordinates for an isotropic solid with internal heat generation. Write its reduced forms for (a) steady state with no generation (Laplace), (b) steady state with generation (Poisson), and (c) unsteady state with no generation (Fourier). State the cylindrical-coordinate form.

Derivation

Take a small element dx dy dzdx\,dy\,dz in a solid of density ρ\rho, specific heat cc and constant conductivity kk. Let q˙g\dot q_g be the heat generated per unit volume (W/m3^3). Apply the energy balance:

(heat in) + (heat generated) = (heat out) + (increase of internal energy).

Heat conducted in the xx direction at face xx (Fourier's law):

Qx=−k dy dz ∂T∂xQ_x = -k\,dy\,dz\,\frac{\partial T}{\partial x}

Heat leaving at x+dxx+dx (Taylor series):

Qx+dx=Qx+∂Qx∂xdx=Qx−k dy dz ∂2T∂x2dxQ_{x+dx} = Q_x + \frac{\partial Q_x}{\partial x}dx = Q_x - k\,dy\,dz\,\frac{\partial^2 T}{\partial x^2}dx

Net heat gained in xx direction is k ∂2T∂x2 dx dy dzk\,\dfrac{\partial^2 T}{\partial x^2}\,dx\,dy\,dz. Similarly for yy and zz. Adding generation q˙g dx dy dz\dot q_g\,dx\,dy\,dz and equating to the rate of increase of stored energy ρc ∂T∂t dx dy dz\rho c\,\dfrac{\partial T}{\partial t}\,dx\,dy\,dz:

k(∂2T∂x2+∂2T∂y2+∂2T∂z2)+q˙g=ρc∂T∂tk\left(\frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2}+\frac{\partial^2 T}{\partial z^2}\right)+\dot q_g = \rho c\frac{\partial T}{\partial t}

Dividing by kk and defining thermal diffusivity α=k/ρc\alpha = k/\rho c (m2^2/s):

∇2T+q˙gk=1α∂T∂t\nabla^2 T + \frac{\dot q_g}{k} = \frac{1}{\alpha}\frac{\partial T}{\partial t}

Reduced forms

ConditionEquation
Steady, no generation (Laplace)∇2T=0\nabla^2 T = 0
Steady, with generation (Poisson)∇2T+q˙g/k=0\nabla^2 T + \dot q_g/k = 0
Unsteady, no generation (Fourier)∇2T=1α∂T∂t\nabla^2 T = \dfrac{1}{\alpha}\dfrac{\partial T}{\partial t}

Cylindrical coordinates (r,ϕ,z)(r,\phi,z)

1r∂∂r(r∂T∂r)+1r2∂2T∂ϕ2+∂2T∂z2+q˙gk=1α∂T∂t\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial T}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2 T}{\partial \phi^2}+\frac{\partial^2 T}{\partial z^2}+\frac{\dot q_g}{k}=\frac{1}{\alpha}\frac{\partial T}{\partial t}

For one-dimensional radial steady flow without generation this becomes ddr(rdTdr)=0\dfrac{d}{dr}\left(r\dfrac{dT}{dr}\right)=0.

✓ Answer checked by Claude Sonnet 5.5

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