## Two-Dimensional Heat Conduction without Internal Heat Generation

Figure 1: Two-dimensional steady-state heat conduction

The problem under consideration is shown in Fig. 1. The left, right and bottom of the two-dimensional domain are kept at T1 while the top boundary has variable temperature f(x). It is assumed that the thermal conductivity is independent from the temperature. The energy equation for this problem is

$\frac{{{\partial ^2}T}}{{\partial {x^2}}} + \frac{{{\partial ^2}T}}{{\partial {y^2}}} = 0,{\rm{ }}0 < x < L,{\rm{ }}0 < y < H \qquad \qquad(1)$

with the following boundary conditions:

$T = {T_1},{\rm{ }}x = 0,L,{\rm{ }}0 < y < H \qquad \qquad(2)$
$T = {T_1},{\rm{ }}y = 0{\rm{, }}0 < x < L \qquad \qquad(3)$
$T = f(x),{\rm{ }}y = H,{\rm{ }}0 < x < L \qquad \qquad(4)$

Although eq. (1) is linear homogeneous, eqs. (2) – (4) are not homogeneous. Introducing excess temperature

$\vartheta = T - {T_1}$

eqs. (1) – (4) become

$\frac{{{\partial ^2}\vartheta }}{{\partial {x^2}}} + \frac{{{\partial ^2}\vartheta }}{{\partial {y^2}}} = 0,{\rm{ }}0 < x < L,{\rm{ }}0 < y < H \qquad \qquad(5)$
$\vartheta = 0,{\rm{ }}x = 0,L,{\rm{ }}0 < y < H \qquad \qquad(6)$
$\vartheta = 0,{\rm{ }}y = 0{\rm{, }}0 < x < L \qquad \qquad(7)$
$\vartheta = f(x) - {T_1},{\rm{ }}y = H,{\rm{ }}0 < x < L \qquad \qquad(8)$

It can be seen that there is only one non-homogeneous boundary condition, eq. (8). This problem can be solved using the method of separation of variables. In this methodology, it is assumed that the temperature can be expressed as

$\vartheta (x,y) = X(x)Y(y) \qquad \qquad(9)$

where X(x) and Y(y) are functions of x and y, respectively. Substituting eq. (9) into eq. (1), one obtains

$- \frac{1}{X}\frac{{{d^2}X}}{{d{x^2}}} = \frac{1}{Y}\frac{{{d^2}Y}}{{d{y^2}}} \qquad \qquad(10)$

Since the left hand side of (10) is a function of x only and the right-hand side is function of y only, the only way that eq. (10) is valid for any x and y is that both sides equal to a constant – termed separation constant. If this separation constant is represented by λ2, which is unknown at this point, eq. (10) will become two ordinary differential equations.

$\frac{{{d^2}X}}{{d{x^2}}} + {\lambda ^2}X = 0 \qquad \qquad(11)$
$\frac{{{d^2}Y}}{{d{y^2}}} - {\lambda ^2}Y = 0 \qquad \qquad(12)$

The general solutions of eqs. (11) and (12) are

$X = {C_1}\cos \lambda x + {C_2}\sin \lambda x \qquad \qquad(13)$
$Y = {C_3}{e^{ - \lambda y}} + {C_4}{e^{\lambda y}} \qquad \qquad(14)$

where C1,C2,C3 and C4 are integral constants to be determined by appropriate boundary conditions. Substituting eq. (9) into eqs. (6) and (7), the following boundary conditions of eqs. (11) and (12) are obtained

$X(0) = 0 \qquad \qquad(15)$
$X(L) = 0 \qquad \qquad(16)$
$Y(0) = 0 \qquad \qquad(17)$

Substituting eq. (13) into eq. (15) yields C1 = 0 and eq. (13) becomes

$X = {C_2}\sin \lambda x \qquad \qquad(18)$

Applying eq. (16) to (18), one obtains

${C_2}\sin \lambda L = 0 \qquad \qquad(19)$

While eq. (19) cannot be used to determine C2, the separation constant that satisfies eq. (16) can be obtained as

${\lambda _n} = \frac{{n\pi }}{L} \qquad \qquad(20)$

where n is an integer. Equation (20) indicates that there are many possible values for the separation constants. Substituting eq. (14) into eq. (17), we get C3 = − C4 and eq. (14) becomes

$Y = {C_4}({e^{\lambda y}} - {e^{ - \lambda y}}) = 2{C_4}\sinh \lambda y \qquad \qquad(21)$

Substituting eqs. (18) and (21) into eq. (9) and considering eq. (20), a solution for the excess temperature is obtained

${\vartheta _n} = {C_n}\sin \left( {\frac{{n\pi x}}{L}} \right)\sinh \left( {\frac{{n\pi y}}{L}} \right) \qquad \qquad(22)$

where Cn = 2C2C4. For any n, eq. (22) satisfies eqs. (5) – (7) but not eq. (8). Since the two-dimensional heat conduction problem under consideration is a linear problem, the sum of different ${\vartheta _n}$ for each value of n also satisfies eqs. (5) – (7).

$\vartheta = \sum\limits_{n = 1}^\infty {{C_n}\sin \left( {\frac{{n\pi x}}{L}} \right)\sinh \left( {\frac{{n\pi y}}{L}} \right)} \qquad \qquad(23)$

where the term for n = 0 is not included because sinh(0) = 0. Applying the boundary condition at the top of the domain, eq. (8), yields

$f(x) - {T_1} = \sum\limits_{n = 1}^\infty {{C_n}\sin \left( {\frac{{n\pi x}}{L}} \right)\sinh \left( {\frac{{n\pi H}}{L}} \right)}$

Multiplying the above equation by $\sin \left( {m\pi x/L} \right)$ and integrating the resulting equation in the interval of (0, L), one obtains

$\begin{array}{l} \int_0^L {[f(x) - {T_1}} ]\sin \left( {\frac{{m\pi x}}{L}} \right)dx \\ = \sum\limits_{n = 1}^\infty {{C_n}\sinh \left( {\frac{{n\pi H}}{L}} \right)\int_0^L {\sin \left( {\frac{{n\pi x}}{L}} \right)\sin \left( {\frac{{m\pi x}}{L}} \right)dx} } \\ \end{array} \qquad \qquad(24)$

The integral in the right-hand side of eq. (24) can be evaluated as

$\begin{array}{l} \int_0^L {\sin \left( {\frac{{n\pi x}}{L}} \right)\sin \left( {\frac{{m\pi x}}{L}} \right)dx} = \left[ {\frac{{\sin (n\pi x/L - m\pi x/L)}}{{2(n\pi x/L - m\pi x/L)}}} \right. \\ \left. { - \frac{{\sin (n\pi x/L + m\pi x/L)}}{{2(n\pi x/L + m\pi x/L)}}} \right]_0^L = 0{\rm{ }}({\rm{for }}m \ne n) \\ \end{array}$
$\int_0^L {{{\sin }^2}\left( {\frac{{m\pi x}}{L}} \right)dx} = \frac{L}{{2m\pi }}\left[ {\frac{{m\pi x}}{L} - \frac{1}{2}\sin \left( {\frac{{2m\pi x}}{L}} \right)} \right]_0^L = \frac{L}{2}{\rm{ }}({\rm{for }}m = n)$

This is referred to as the orthogonal property of the sine function. This property indicates that only one term (m = n) on the right hand of eq. (24) is not zero and eq. (24) becomes

$\int_0^L {[f(x) - {T_1}} ]\sin \left( {\frac{{m\pi x}}{L}} \right)dx = \frac{L}{2}{C_m}\sinh \left( {\frac{{m\pi H}}{L}} \right)$

i.e.,

${C_m} = \frac{1}{{\sinh \left( {m\pi H/L} \right)}}\frac{2}{L}\int_0^L {[f(x) - {T_1}} ]\sin \left( {\frac{{m\pi x}}{L}} \right)dx$

Changing notation from m to n, we get

${C_n} = \frac{1}{{\sinh \left( {n\pi H/L} \right)}}\frac{2}{L}\int_0^L {[f(x) - {T_1}} ]\sin \left( {\frac{{n\pi x}}{L}} \right)dx \qquad \qquad(25)$

Thus, the temperature profile, eq. (23), becomes

$\vartheta = \frac{2}{L}\sum\limits_{n = 1}^\infty {\frac{{\sinh \left( {n\pi y/L} \right)}}{{\sinh \left( {n\pi H/L} \right)}}\sin \left( {\frac{{n\pi x}}{L}} \right)\int_0^L {[f(x) - {T_1}} ]\sin \left( {\frac{{m\pi x}}{L}} \right)dx} \qquad \qquad(26)$

If the temperature at the top of the domain is constant, i.e,. f(x) = T2, the temperature distribution becomes

$\frac{{T - {T_1}}}{{{T_2} - {T_1}}} = \frac{2}{\pi }\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^{n + 1}} + 1}}{2}\frac{{\sinh \left( {n\pi y/L} \right)}}{{\sinh \left( {n\pi H/L} \right)}}\sin \left( {\frac{{n\pi x}}{L}} \right)} \qquad \qquad(27)$

## Two-Dimensional Heat Conduction with Internal Heat Generation

Figure 2: Two-dimensional steady-state heat conduction with internal heat generation

The condition under which the two-dimensional heat conduction can be solved by separation of variables is that the governing equation must be linear homogeneous and no more than one boundary condition is nonhomogeneous. For the case that an internal heat source is present, the energy equation is no longer homogeneous and the method of separation of variables will not work. Let us consider heat conduction in a rectangular domain with a dimension of $2L \times 2H$ with uniform heat generation (see Fig. 2). All four boundaries are kept at a constant temperature of T1. Since this is an axisymmetric problem, we can study one fourth of the problem. The governing equations and corresponding boundary conditions are

$\frac{{{\partial ^2}T}}{{\partial {x^2}}} + \frac{{{\partial ^2}T}}{{\partial {y^2}}} + \frac{{q'''}}{k} = 0,{\rm{ }}0 < x < L,{\rm{ }}0 < y < H \qquad \qquad(28)$

with the following boundary conditions:

$\frac{{\partial T}}{{\partial x}} = 0,{\rm{ }}x = 0,{\rm{ }}0 < y < H \qquad \qquad(29)$
$T = {T_1},{\rm{ }}x = L,{\rm{ }}0 < y < H \qquad \qquad(30)$
$\frac{{\partial T}}{{\partial y}} = 0,{\rm{ }}y = 0{\rm{, }}0 < x < L \qquad \qquad(31)$
$T = {T_1},{\rm{ }}y = H,{\rm{ }}0 < x < L \qquad \qquad(32)$

Although eq. (28) – (30) are all linear, eqs. (28), (30) and, (32) are nonhomogeneous. Introducing excess temperature

$\vartheta = T - {T_1}$

eqs. (28) – (32) become

$\frac{{{\partial ^2}\vartheta }}{{\partial {x^2}}} + \frac{{{\partial ^2}\vartheta }}{{\partial {y^2}}} + \frac{{q'''}}{k} = 0,{\rm{ }}0 < x < L,{\rm{ }}0 < y < H \qquad \qquad(33)$
$\frac{{\partial \vartheta }}{{\partial x}} = 0,{\rm{ }}x = 0,{\rm{ }}0 < y < H \qquad \qquad(34)$
$\vartheta = 0,{\rm{ }}x = L,{\rm{ }}0 < y < H \qquad \qquad(35)$
$\frac{{\partial \vartheta }}{{\partial y}} = 0,{\rm{ }}y = 0{\rm{, }}0 < x < L \qquad \qquad(36)$
$\vartheta = 0,{\rm{ }}y = H,{\rm{ }}0 < x < L \qquad \qquad(37)$

Equation (33) is still nonhomogeneous but all boundary conditions are homogeneous. Mathematically, the solution of a nonhomogeneous problem can be obtained by superposing a particular solution of the nonhomogeneous problem and the general solution of the corresponding homogeneous problem. Assuming the solution is

$\vartheta (x,y) = \varphi (x) + \psi (x,y) \qquad \qquad(38)$

where φ(x) is the solution of the following problem

$\frac{{{d^2}\varphi }}{{d{x^2}}} + \frac{{q'''}}{k} = 0,{\rm{ }}0 < x < L \qquad \qquad(39)$
$\frac{{d\varphi }}{{dx}} = 0,{\rm{ }}x = 0 \qquad \qquad(40)$
$\varphi = 0,{\rm{ }}x = L \qquad \qquad(41)$

and ψ(x,y) is the solution of the following problem

$\frac{{{\partial ^2}\psi }}{{\partial {x^2}}} + \frac{{{\partial ^2}\psi }}{{\partial {y^2}}} = 0,{\rm{ }}0 < x < L,{\rm{ }}0 < y < H \qquad \qquad(42)$
$\frac{{\partial \psi }}{{\partial x}} = 0,{\rm{ }}x = 0,{\rm{ }}0 < y < H \qquad \qquad(43)$
$\psi = 0,{\rm{ }}x = L,{\rm{ }}0 < y < H \qquad \qquad(44)$
$\frac{{\partial \psi }}{{\partial y}} = 0,{\rm{ }}y = 0{\rm{, }}0 < x < L \qquad \qquad(45)$
$\psi = - \varphi (x),{\rm{ }}y = H,{\rm{ }}0 < x < L \qquad \qquad(46)$

It can be shown that the summation of the solutions of the above two simpler problems gives the solution of the original problem. Integrating eq. (39) twice and determining the integral constants using eqs. (40) and (41), we obtain

$\varphi (x) = \frac{{q'''{L^2}}}{{2k}}\left( {1 - \frac{{{x^2}}}{{{L^2}}}} \right) \qquad \qquad(47)$

The solution of eqs. (42) – (46) can be obtained by using the method of separation of variables and the result is

$\psi (x,y) = - \frac{{2q'''{L^2}}}{k}\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{{{({\lambda _n}L)}^3}}}\frac{{\cosh ({\lambda _n}y)}}{{\cosh ({\lambda _n}H)}}\cos } ({\lambda _n}x) \qquad \qquad(48)$

where the eigenvalue is

${\lambda _n} = \frac{{(2n + 1)\pi }}{{2L}},{\rm{ }}n = 0,1,2,... \qquad \qquad(49)$

Substituting eqs. (47) and (48) into eq. (38), the solution of heat conduction in a rectangular domain with internal heat generation can be obtained. The solution of heat conduction with internal heat generation was obtained by solving a 1-D heat conduction problem in the x-direction with an internal heat source and a 2-D heat conduction problem without internal heat generation. Alternatively, the solution can also be assumed to have the following form

$\vartheta (x,y) = \varphi (y) + \psi (x,y) \qquad \qquad(50)$

and an equivalent but different solution can be obtained.

## References

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