Multidimensional transient heat conduction

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Two-dimensional transient heat conduction
Two-dimensional transient heat conduction

The transient heat conduction problems discussed in the preceding subsection are for the case that the transient temperature varies in one dimension only. Analytical solution of multidimensional transient heat conduction using product solution will be discussed in this subsection. More complicated multi-dimensional transient heat conduction can be solved using the numerical method to be discussed in the next section.

Let us consider transient heat conduction in a rectangular bar with dimensions of 2L1×2L2 and an initial temperature of Ti (see figure). At time t= 0, the rectangular bar is immersed in a fluid with temperature T. The convective heat transfer coefficient on the left and right sides of the rectangular bar is h1, while on the top and bottom it is h2. It is assumed that the thermal conductivity is independent from the temperature and there is no internal heat source. Since this is an axisymmetric problem, we can study a quarter of it. The energy equation for this problem is

\frac{{{\partial ^2}T}}{{\partial {x^2}}} + \frac{{{\partial ^2}T}}{{\partial {y^2}}} = \frac{1}{\alpha }\frac{{\partial T}}{{\partial t}},{\rm{  }}0 < x < {L_1},{\rm{ }}0 < y < {L_2},{\rm{ }}t > 0    \qquad \qquad(1)

with the following boundary and initial conditions:

\frac{{\partial T}}{{\partial x}} = 0,{\rm{  }}x = 0,{\rm{ }}0 < y < {L_2},{\rm{ }}t > 0    \qquad \qquad(2)

 - k\frac{{\partial T}}{{\partial x}} = h(T - {T_\infty }),{\rm{  }}x = {L_1},{\rm{ }}0 < y < {L_2},{\rm{ }}t > 0    \qquad \qquad(3)
\frac{{\partial T}}{{\partial y}} = 0,{\rm{  y}} = 0,{\rm{ }}0 < x < {L_1},{\rm{ }}t > 0    \qquad \qquad(4)
 - k\frac{{\partial T}}{{\partial y}} = h(T - {T_\infty }),{\rm{  y}} = {L_2},{\rm{ }}0 < x < {L_1},{\rm{ }}t > 0    \qquad \qquad(5)
T = {T_i}{\rm{,  }}0 < x < {L_1},{\rm{ }}0 < y < {L_2},{\rm{ }}t > 0    \qquad \qquad(6)

Defining the following dimensionless variables

\theta  = \frac{{T - {T_\infty }}}{{{T_i} - {T_\infty }}}{\rm{,  }}X = \frac{x}{{{L_1}}},{\rm{ }}Y = \frac{y}{{{L_2}}},{\rm{  F}}{{\rm{o}}_{\rm{1}}} = \frac{{\alpha t}}{{L_1^2}},{\rm{ F}}{{\rm{o}}_{\rm{2}}} = \frac{{\alpha t}}{{L_2^2}}    \qquad \qquad(7)

where both Fo1 and Fo2 are Fourier numbers based on different characteristic lengths. The dimensional time will be related to both Fourier numbers, i.e.,

t = t({\rm{F}}{{\rm{o}}_1},{\rm{F}}{{\rm{o}}_2})    \qquad \qquad(8)

therefore, the right-hand side of eq. (1) becomes

 \frac{{\partial T}}{{\partial t}} = ({T_i} - {T_\infty })\frac{{\partial \theta }}{{\partial t}} = ({T_i} - {T_\infty })\left[ {\frac{{\partial \theta }}{{\partial {\rm{F}}{{\rm{o}}_1}}}\frac{{\partial {\rm{F}}{{\rm{o}}_1}}}{{\partial t}} + \frac{{\partial \theta }}{{\partial {\rm{F}}{{\rm{o}}_2}}}\frac{{\partial {\rm{F}}{{\rm{o}}_2}}}{{\partial t}}} \right] \\ 
 {\rm{     }} = ({T_i} - {T_\infty })\left[ {\frac{{\partial \theta }}{{\partial {\rm{F}}{{\rm{o}}_1}}}\frac{\alpha }{{L_1^2}} + \frac{{\partial \theta }}{{\partial {\rm{F}}{{\rm{o}}_2}}}\frac{\alpha }{{L_2^2}}} \right] \\ 
 \end{array}    \qquad \qquad(9)

Nondimensionalizing the left-hand side of eq. (1) and considering eq. (9), the dimensionless energy equation of the problem becomes

\frac{{{\partial ^2}\theta }}{{\partial {X^2}}} + {\left( {\frac{{{L_1}}}{{{L_2}}}} \right)^2}\frac{{{\partial ^2}\theta }}{{\partial {Y^2}}} = \frac{{\partial \theta }}{{\partial {\rm{F}}{{\rm{o}}_1}}} + {\left( {\frac{{{L_1}}}{{{L_2}}}} \right)^2}\frac{{\partial \theta }}{{\partial {\rm{F}}{{\rm{o}}_2}}}    \qquad \qquad(10)

The boundary and initial conditions, eqs. (2) – (6) are nondimensionalized as

\frac{{\partial \theta }}{{\partial X}} = 0,{\rm{  X}} = 0,{\rm{ }}0 < Y < 1,{\rm{ F}}{{\rm{o}}_{\rm{1}}} > 0,{\rm{ F}}{{\rm{o}}_{\rm{2}}} > 0    \qquad \qquad(11)
 - \frac{{\partial \theta }}{{\partial X}} = {\rm{B}}{{\rm{i}}_1}\theta ,{\rm{  X}} = 1,{\rm{ }}0 < Y < 1,{\rm{ F}}{{\rm{o}}_{\rm{1}}} > 0,{\rm{ F}}{{\rm{o}}_{\rm{2}}} > 0    \qquad \qquad(12)
\frac{{\partial \theta }}{{\partial Y}} = 0,{\rm{  Y}} = 0,{\rm{ }}0 < X < 1,{\rm{ F}}{{\rm{o}}_{\rm{1}}} > 0,{\rm{ F}}{{\rm{o}}_{\rm{2}}} > 0    \qquad \qquad(13)
 - \frac{{\partial \theta }}{{\partial Y}} = {\rm{B}}{{\rm{i}}_2}\theta ,{\rm{  }}Y = 1,{\rm{ }}0 < X < 1,{\rm{ F}}{{\rm{o}}_{\rm{1}}} > 0,{\rm{ F}}{{\rm{o}}_{\rm{2}}} > 0    \qquad \qquad(14)
\theta  = 1{\rm{,  }}0 < X < 1,{\rm{ }}0 < Y < 1,{\rm{ F}}{{\rm{o}}_{\rm{1}}} = {\rm{F}}{{\rm{o}}_{\rm{2}}} = 0    \qquad \qquad(15)


{\rm{B}}{{\rm{i}}_1} = \frac{{{h_1}{L_1}}}{k},{\rm{  B}}{{\rm{i}}_2} = \frac{{{h_2}{L_2}}}{k}    \qquad \qquad(16)

are Biot numbers for different surfaces. The idea of product solution is that the solution of the two-dimensional problem can be expressed as the product of two one-dimensional problems, i.e.,

\theta  = \varphi (X,{\rm{F}}{{\rm{o}}_1})\psi (Y,{\rm{F}}{{\rm{o}}_2})    \qquad \qquad(17)

where \varphi is the solution of the following problem

\frac{{{\partial ^2}\varphi }}{{\partial {X^2}}} = \frac{{\partial \varphi }}{{\partial {\rm{F}}{{\rm{o}}_1}}}    \qquad \qquad( )
\frac{{\partial \varphi }}{{\partial X}} = 0,{\rm{  X}} = 0,{\rm{ F}}{{\rm{o}}_{\rm{1}}} > 0    \qquad \qquad(19)
 - \frac{{\partial \varphi }}{{\partial X}} = {\rm{B}}{{\rm{i}}_1}\varphi ,{\rm{  X}} = 1,{\rm{ F}}{{\rm{o}}_{\rm{1}}} > 0    \qquad \qquad(20)
\varphi  = 1{\rm{,  }}0 < X < 1,{\rm{ F}}{{\rm{o}}_{\rm{1}}} = 0    \qquad \qquad(21)

and ψ satisfies

\frac{{{\partial ^2}\psi }}{{\partial {Y^2}}} = \frac{{\partial \psi }}{{\partial {\rm{F}}{{\rm{o}}_2}}}    \qquad \qquad(22)
\frac{{\partial \psi }}{{\partial Y}} = 0,{\rm{  }}Y = 0,{\rm{ F}}{{\rm{o}}_{\rm{2}}} > 0    \qquad \qquad(23)
 - \frac{{\partial \psi }}{{\partial Y}} = {\rm{B}}{{\rm{i}}_2}\psi ,{\rm{  }}Y = 1,{\rm{ F}}{{\rm{o}}_{\rm{2}}} > 0    \qquad \qquad(24)
\psi  = 1{\rm{,  }}0 < Y < 1,{\rm{ F}}{{\rm{o}}_{\rm{2}}} = 0    \qquad \qquad(25)

It can be demonstrated that eqs. (10) – (15) can be satisfied by eq. (17) if \varphi and ψ are solutions of eqs. (18) – (2) and (22) – (25), respectively. The solution of eqs. (18) – (21) is:

\varphi  = \sum\limits_{n = 1}^\infty  {\frac{{4\sin {\lambda _n}}}{{2{\lambda _n} + \sin 2{\lambda _n}}}\cos \left( {{\lambda _n}X} \right){e^{ - \lambda _n^2{\rm{F}}{{\rm{o}}_{\rm{1}}}}}}     \qquad \qquad(26)


\frac{{{\lambda _n}}}{{{\rm{B}}{{\rm{i}}_{\rm{1}}}}} = \cot {\lambda _n}    \qquad \qquad( 27)

and the solution of eqs. (22) – (25) is

\psi  = \sum\limits_{m = 1}^\infty  {\frac{{4\sin {\nu _m}}}{{2{\nu _m} + \sin 2{\nu _m}}}\cos \left( {{\nu _m}Y} \right){e^{ - \nu _m^2{\rm{F}}{{\rm{o}}_{\rm{2}}}}}}     \qquad \qquad(28)


\frac{{{\nu _m}}}{{{\rm{B}}{{\rm{i}}_{\rm{2}}}}} = \cot {\nu _m}    \qquad \qquad(29)

Substituting eqs. (26) and (28) into eq. (17), the solution of the two-dimensional problem is obtained.

\theta  = \sum\limits_{n = 1}^\infty  {\sum\limits_{m = 1}^\infty  {\frac{{16\sin {\lambda _n}\sin {\nu _m}\cos \left( {{\lambda _n}X} \right)\cos \left( {{\nu _m}Y} \right)}}{{(2{\lambda _n} + \sin 2{\lambda _n})(2{\nu _m} + \sin 2{\nu _m})}}{e^{ - (\lambda _n^2{\rm{F}}{{\rm{o}}_{\rm{1}}} + \nu _m^2{\rm{F}}{{\rm{o}}_{\rm{2}}})}}} }     \qquad \qquad(30)

The product solution constructs the solution of a two-dimensional transient conduction as the product of the solutions of two one-dimensional problems. It is different from the method of separation of variables that changes the solution of a partial differential equation into the product of the solutions of ordinary differential equations. The product solution only works for transient problems and it will not work with steady-state problems. The only region that the product solution works is the intersection space of the two one-dimensional problems. The product solution will also work for a three-dimensional brick where the solution can be expressed as the solution of three one-dimensional problems. Another example that product solution can be applied to is transient conduction in a short cylinder where the solution can be expressed as the product of solutions of one-dimensional transient conduction in a plane wall and a cylinder.


Faghri, A., Zhang, Y., and Howell, J. R., 2010, Advanced Heat and Mass Transfer, Global Digital Press, Columbia, MO.

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