On the integral-balance approach to the transient heat conduction with linearly temperature-dependent thermal diffusivity

Abstract

Closed form approximate solutions to nonlinear transient heat conduction with linearly temperature-dependent thermal diffusivity have been developed by the integral-balance integral method under transient conditions. The solutions uses improved direct approaches of the integral method and avoid the commonly used linearization by the Kirchhoff transformation. The main steps in the new solutions are improvements in the integration technique of the double-integration technique and the optimization of the exponent of the approximate parabolic profile with unspecified exponent. Solutions to Dirichlet and Neumann boundary condition problems have been developed as examples by the classical Heat-balance integral method (HBIM) and the Double-integration method (DIM). Additional examples with HBIM and DIM solutions to cases when the Kirchhoff transform is initially applied have been developed.

Appendices

Appendix 1: Fixed temperature problem: the numerical solutions used

• Finite difference solution

In order to validate the approximate integral-balance solutions, two numerical solutions were developed for the benchmarking procedures. The first numerical solution is found thanks to the method of finite difference, where an explicit scheme, due to its numerical stability and good convergence, was used. In additions, with a time step $\Delta t = 1/100$ and a space step $\Delta x = 1/50$ we got a good accuracy. In this contexts, in the approximation of Eq. (15) (with initial and boundary conditions $u\left( {x,0} \right) = 0$ and $u\left( {0,t} \right) = 1$ q $u\left( {\delta ,t} \right) = 0$) the error due to the numerical approximation is relatively high when for x is close to 0 but decreased rapidly when x increases. This high error close to 0 is due to the impossibility to calculate the derivative of u at 0.

• Runge–Kutta solution

With the Boltzmann transform $\eta = {x \mathord{\left/ {\vphantom {x {\sqrt {a_{0} t} }}} \right. \kern-0pt} {\sqrt {a_{0} t} }}$ and $X = x/\delta = \eta /f(n)$ we may express Eq. (15) in the forms

$$X\frac{{f(n)^{2} }}{2}\frac{\partial u(X)}{\partial X} + \frac{\beta }{1 + \beta u(X)}\left( {\frac{\partial u(X)}{\partial X}} \right)^{2} + \frac{{\partial^{2} u(X)}}{{\partial X^{2} }} = 0,\quad f\left( n \right) \ne 1$$

(84)

$$\begin{aligned} &\frac{1}{2}\frac{\partial u(\eta )}{\partial \eta } + \frac{\beta }{1 + \beta u(\eta )}\left( {\frac{\partial u(\eta )}{\partial \eta }} \right)^{2} + \frac{{\partial^{2} u(\eta )}}{{\partial \eta^{2} }} = 0,\\&\quad f\left( n \right) = 1\;{\text{and}}\;\eta = X \end{aligned}$$

(85)

The normalizing function $f(n)$ is introduced for consistency with the concept of the finite penetration depth $\delta$ which is missing in the classical solution of the linear equation expressed by the Gaussian error function. In fact, with $f\left( n \right) \ne 1$ the initial problem is transformed to a boundary value problem with $u(X = 0) = 1$ and $u(X = 1) = 0$ allowing to compare the integral-balance solutions with the numerical ones in the domain $0 \le X \le 1$. The solutions were developed by Maple 13 where Runge–Kutta solutions of 4th order are possible with absolute error less than $10^{ - 6}$. The normalizing function $f(n)$ for each β is expressed through the optimal n developed by minimization of the residual function (see Tables 2, 3) and it is equal either to $F_{HBIM}^{T} \left( {n,\beta } \right)$ or $F_{DIM}^{T} \left( {n,\beta } \right)$ (see Eqs. 27, 28), depending on the integration method applied.

Appendix 2: Fixed flux problem: the derivations of the approximations (43) and (44) HBIM solution

The equation about the penetration depth is developed by HBIM solution is (42), namely

$$\delta^{3} - \delta \left( {a_{0} t} \right)n\left( {n + 1} \right) - \beta \frac{{q_{0} }}{{k_{0} }}\left( {a_{0} t} \right)^{2} n\left( {n + 1} \right)^{2} = 0$$

(42)

For β = 0, it reduces to $\delta_{{0\left( {HBIM} \right)}}^{q} = \delta_{0}^{{}} = \sqrt {\left( {a_{0} t} \right)n\left( {n + 1} \right)}$ which is the classical HBIM solution [3, 26]. Denoting $\delta_{0}^{2} = \left( {a_{0} t} \right)n\left( {n + 1} \right) = \alpha_{1} > 0$ and $\beta \frac{{q_{0} }}{{k_{0} }}\left( {a_{0} t} \right)^{2} n\left( {n + 1} \right)^{2} = \alpha_{2} > 0$ we get a depressed cubic equation about $\delta \left( t \right)$, namely

$$\delta^{3} - \alpha_{1} \delta - \alpha_{2} = 0$$

(86)

The coefficients of (84) are related as $\alpha_{2} = \alpha_{1}^{2} \beta \left( {{{q_{0} } \mathord{\left/ {\vphantom {{q_{0} } {k_{0} n}}} \right. \kern-0pt} {k_{0} n}}} \right)$.

Now, let us suggest that the penetration depth for $\beta \ne 0$ is related to the $\delta_{0}^{2} = \left( {a_{0} t} \right)n\left( {n + 1} \right)$ by a correctional functional $f_{3} \left( {n,t} \right)$ that is $\delta^{2} = \delta_{0}^{2} f_{3}^{2}$. Now, we may rearrange (86) as

$$\delta_{0}^{3} f_{3}^{3} - \delta_{0}^{3} f_{3} - \delta_{0}^{4} \beta \frac{{q_{0} }}{{k_{0} }}\frac{1}{n} = 0 \Rightarrow f_{3}^{3} - f_{3} - \delta_{0}^{{}} \beta \frac{{q_{0} }}{{k_{0} }}\frac{1}{n} = 0$$

(87a, b)

For β = 0 it follows directly that $f_{3} = 1$ is a solution of Eq. (87a, b). The solution of the cubic equation $f_{3}^{3} = Af_{3} + B$ depends upon the sign of the determinant $D = \frac{{A^{3} }}{27} + \frac{{B^{2} }}{4}$. For D > 0 the equation has one real root and two imaginary roots. Since we need a unique real solution of the penetration depth, then D > 0 is the case. With A = 1 and $B = \delta_{0}^{{}} \beta \frac{{q_{0} }}{{k_{0} }}\frac{1}{n}$ we get

$$D = \frac{{A^{3} }}{27} + \frac{{B^{2} }}{4} = \delta_{0}^{2} \left( {\beta \frac{{q_{0} }}{{k_{0} }}\frac{1}{n}} \right)^{2} \left[ {1 - \frac{1}{27}\frac{1}{{\delta_{0}^{2} }}\left( {\frac{2}{\beta }\frac{{k_{0} }}{{q_{0} }}n} \right)^{2} } \right]$$

(88)

Therefore, the Cardano formula is $f_{3} = M + N$, where

$$M = \left( { - \frac{B}{2} + \sqrt D } \right)^{{{1 \mathord{\left/ {\vphantom {1 3}} \right. \kern-0pt} 3}}} = \left( {\delta_{0}^{{}} \beta \frac{{q_{0} }}{{k_{0} }}\frac{1}{n}} \right)^{{\frac{1}{3}}} \left[ {\frac{1}{2} + \sqrt {1 - \frac{1}{27}\frac{1}{{\delta_{0}^{2} }}\left( {\frac{1}{\beta }\frac{{k_{0} }}{{q_{0} }}n} \right)^{2} } } \right]^{{\frac{1}{3}}}$$

(89a)

$$N = \left( { - \frac{B}{2} - \sqrt D } \right)^{{{1 \mathord{\left/ {\vphantom {1 3}} \right. \kern-0pt} 3}}} = \left( {\delta_{0}^{{}} \beta \frac{{q_{0} }}{{k_{0} }}\frac{1}{n}} \right)^{{\frac{1}{3}}} \left[ {\frac{1}{2} - \sqrt {1 - \frac{1}{27}\frac{1}{{\delta_{0}^{2} }}\left( {\frac{1}{\beta }\frac{{k_{0} }}{{q_{0} }}n} \right)^{2} } } \right]^{{\frac{1}{3}}}$$

(89b)

$$\begin{aligned} f_{3} & = \left( {\delta_{0}^{{}} \beta \frac{{q_{0} }}{{k_{0} }}\frac{1}{n}} \right)^{{\frac{1}{3}}} \\ & \quad \times \left\{ {\left[ {\frac{1}{2} + \sqrt {1 - \frac{1}{27}\frac{1}{{\delta_{0}^{2} }}\left( {\frac{1}{\beta }\frac{{k_{0} }}{{q_{0} }}n} \right)^{2} } } \right]^{{\frac{1}{3}}} + \left[ {\frac{1}{2} - \sqrt {1 - \frac{1}{27}\frac{1}{{\delta_{0}^{2} }}\left( {\frac{1}{\beta }\frac{{k_{0} }}{{q_{0} }}n} \right)^{2} } } \right]^{{\frac{1}{3}}} } \right\} \\ \end{aligned}$$

(90a)

In (90a) the denominators of the second terms under the radicals grow in time rapidly because $\delta_{0}^{2} \equiv a_{0} t$. From this point of view, we may suggest that they could be neglected as smaller than 1 and this step allows to approximate f ₃ as

$$f_{3} \approx \left( {\delta_{0}^{{}} \beta \frac{{q_{0} }}{{k_{0} }}\frac{1}{n}} \right)^{{\frac{1}{3}}} \left[ {\left( {\frac{1}{2}} \right)^{{\frac{1}{3}}} + \left( {\frac{1}{2}} \right)^{{\frac{1}{3}}} } \right] \approx \left( {\delta_{0}^{{}} \frac{\beta }{4}\frac{{q_{0} }}{{k_{0} }}\frac{1}{n}} \right)^{{\frac{1}{3}}}$$

(90b)

Then the penetration depth $\delta \left( t \right)$ can be expressed as

$$\delta \left( t \right) = \delta_{0} f_{3} \approx \delta_{0} \left( {\delta_{0}^{{}} \frac{\beta }{4}\frac{{q_{0} }}{{k_{0} }}\frac{1}{n}} \right)^{{\frac{1}{3}}} \approx \delta_{0}^{{\frac{4}{3}}} \left( {\frac{\beta }{4}\frac{{q_{0} }}{{k_{0} }}\frac{1}{n}} \right)^{{\frac{1}{3}}}$$

(91a, b)

With the square-root expression of $\delta_{0} \left( t \right)$ inserted in (91b) we get (92) (that is the approximation (Eq. 43 in the main text)

$$\delta \left( t \right) \approx \left( {a_{0} t} \right)^{{\frac{2}{3}}} \left[ {n^{{\frac{1}{3}}} \left( {n + 1} \right)^{{\frac{2}{3}}} } \right]\left( {\frac{\beta }{4}\frac{{q_{0} }}{{k_{0} }}} \right)^{{\frac{1}{3}}}.$$

(92)