Topics: Kirchhoff's Transformation

Kirchhoff's Transformation

Description

Many non-linear conduction problems, where the properties are a function of temperature can be addressed with a variable transformation. In the foregoing demonstration, consider the conduction equation

$\nabla \cdot [k(T) \nabla T] + q''' = \rho c\frac{\partial T}{\partial t},$

where the conductivity is a function of temperature. The equation can be linearized with the following variable substitution.

$\theta = \frac{1}{k_0} \int_{T_0}^T k(T)\, dT,$

such that the reference conductivity $k 0 = k (T 0)$ . By taking the spatial and temporal derivative of both sides of this transformation with LeibnizRule, we obtain the following expressions.

$\frac{d\theta}{dt} = \frac{k}{k_0} \frac{dT}{dt},$

and

$\nabla \theta = \frac{k}{k_0} \nabla T.$

Using these in the original governing equations results in

$k_0 \nabla^2 \theta + q''' = k_0 \frac{\rho c}{k} \frac{\partial \theta}{\partial t}.$

Usually the temperature dependence of $α = k / ρ c$ is small compared to the dependence of either the product $ρ c$ or quantity $k$ . Therefore, the resulting equation can be solved for $θ$ using standard linear techniques.

$\nabla^2 \theta + \frac{q'''}{k_0} = \frac{1}{\alpha}\frac{\partial \theta}{\partial t}$

Once $θ$ is known, then the transformation can be used to convert back to temperature.

Example

In silicon, the thermal conductivity between room temperature and 800K is a strong function of temperature approximated as

$k(T) = k_r\exp\left (1-\frac{T}{T_r} \right),$

(Image(kvt.png) failed - File not found)

where $T_r=300\,{\rm K}$ , and $k_r=148\,{\rm W/mK}$ . In a SOI-MOSFET heat is generated in a localized spot underneath the drain. The transient conduction away from this region can be modeled as one-dimensional giving

$\frac{\partial}{\partial x} \left ( k \frac{\partial T}{\partial x} \right ) = \rho c \frac{\partial T}{\partial t},$

with boundary conditions $- k \partial T / \partial x | x = 0 = q''$ and $T (x \to\infty) = T r$ , and initial condition $T (t = 0) = T r$ .

Using the Kirchhoff transformation given above, the governing equation becomes

$\nabla^2 \theta = \frac{1}{\alpha} \frac{\partial \theta}{\partial t},$

and boundary conditions are $- k r \partialθ / \partial x | x = 0 = q''$ , $θ(x \to\infty) = 0$ , and $θ(t = 0) = 0$ .

The solution for the foregoing system, which can be obtained from any undergraduate heat transfer text, can be written as

$\theta(x,t) = \frac{q''}{2k} \left [ \sqrt{\frac{4\alpha t}{\pi}} \exp \left ( \frac{-x^2}{4\alpha t} \right ) + x {\rm erfc} \left ( \frac{x}{\sqrt{4 \alpha t}} \right ) \right ]$

Assuming our temperature dependent expression for conductivity, we can evaluate the integral in the transformation as

$\theta = \frac{1}{k_r} \int_{T_r}^{T} k_r\exp\left (1-\frac{T}{T_r} \right)\, dT = T_r \left [ 1 - \exp \left( 1 - \frac{T}{T_r} \right ) \right ]$

Now the temperature can be written as

$\frac{T(x,t)}{T_r} = 1 - \ln \left ( 1 - \frac{\theta(x,t)}{T_r} \right )$

The results below are shown for $q''=5\times 10^6\, {\rm W/m^2}$ , and $\alpha=1\times 10^{-5}\,{\rm m^2/s}$ after $t=1\,{\rm s}$ . The reference is the temperature distribution assuming constant conductivity at the reference value.

(Image(tvx.png) failed - File not found)

Acknowledgment

This development follows that of Arpaci (Conduction Heat Transfer, Addison Wesley, 1966), p. 129.

Last modified on 23 Jan, 2009

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Kirchhoff's Transformation

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