# One-dimensional transient heat conduction in finite slab

For the case that the Biot number is greater than 0.1, the temperature distribution can no longer be treated as uniform and the knowledge about the temperature distribution is also of interest. We will now consider the situation where temperature only varies in one spatial dimension. Both homogeneous and nonhomogeneous problems will be considered.

## Homogeneous Problems

Figure 1 Transient conduction in a finite slab

Figure 1 shows a finite slab with thickness of L and a uniform initial temperature of Ti. At time t= 0, the left side of the slab is insulated while the right side of the slab is exposed to a fluid with temperature of ${T_\infty }$ (${T_\infty } < {T_i}$). The convective heat transfer coefficient between the fluid and finite slab is h. In contrast to the lumped capacitance method that assumes uniform temperature, we will present a more generalized model that takes non-uniform temperature distribution in the slab into account. It is assumed that there is no internal heat generation in the slab. The energy equation for this one-dimensional transient conduction problem is

$\frac{{{\partial ^2}T}}{{\partial {x^2}}} = \frac{1}{\alpha }\frac{{\partial T}}{{\partial t}},{\rm{ }}0 < x < L,{\rm{ }}t > 0 \qquad \qquad(1)$

subject to the following boundary and initial conditions

$\frac{{\partial T}}{{\partial x}} = 0,{\rm{ }}x = 0 \qquad \qquad(2)$
$- k\frac{{\partial T}}{{\partial x}} = h(T - {T_\infty }),{\rm{ }}x = L \qquad \qquad(3)$
$T = {T_i},{\rm{ }}0 < x < L,{\rm{ }}t = 0 \qquad \qquad(4)$

It should be pointed out that eqs. (1) – (4) are also valid for the case that both sides of a finite slab with thickness of 2L are cooled by convection. This is a nonhomogeneous problem because eq. (3) is not homogeneous. By introducing the excess temperature, $\vartheta = T - {T_\infty }$, the problem can be homogenized, i.e.,

$\frac{{{\partial ^2}\vartheta }}{{\partial {x^2}}} = \frac{1}{\alpha }\frac{{\partial \vartheta }}{{\partial t}},{\rm{ }}0 < x < L,{\rm{ }}t > 0 \qquad \qquad(5)$
$\frac{{\partial \vartheta }}{{\partial x}} = 0,{\rm{ }}x = 0 \qquad \qquad(6)$
$- k\frac{{\partial \vartheta }}{{\partial x}} = h\vartheta ,{\rm{ }}x = L \qquad \qquad(7)$
$\vartheta = {\vartheta _i} = {T_i} - {T_\infty },{\rm{ }}0 < x < L,{\rm{ }}t = 0 \qquad \qquad(8)$

To express our solution in a compact form so that it can be used for all similar problems, one can define the following dimensionless variables

$\theta = \frac{\vartheta }{{{\vartheta _i}}},{\rm{ }}X = \frac{x}{L},{\rm{ Fo}} = \frac{{\alpha t}}{{{L^2}}} \qquad \qquad(9)$

and eqs. (5) – (8) will be nondimensionalized as

$\frac{{{\partial ^2}\theta }}{{\partial {X^2}}} = \frac{{\partial \theta }}{{\partial {\rm{Fo}}}},{\rm{ }}0 < X < 1,{\rm{ Fo}} > 0 \qquad \qquad(10)$
$\frac{{\partial \theta }}{{\partial X}} = 0,{\rm{ }}X = 0 \qquad \qquad(11)$
$- \frac{{\partial \theta }}{{\partial X}} = {\rm{Bi}}\theta ,{\rm{ X}} = 1 \qquad \qquad(12)$
$\theta = 1,{\rm{ }}0 < X < 1,{\rm{ Fo}} = 0 \qquad \qquad(13)$

This problem can be solved using the method of separation of variables. Assuming that the temperature can be expressed as

$\theta (X,{\rm{Fo}}) = \Theta (X)\Gamma ({\rm{Fo}}) \qquad \qquad(14)$

where ΘandΓ are functions of X and Fo, respectively, eq. (10) becomes

$\frac{{\Theta ''(X)}}{{\Theta (X)}} = \frac{{\Gamma '({\rm{Fo}})}}{{\Gamma ({\rm{Fo}})}}$

Since the left hand side is a function of X only and the right-hand side of the above equation is a function of Fo only, both sides must be equal to a separation constant, μ, i.e.,

$\frac{{\Theta ''(X)}}{{\Theta (X)}} = \frac{{\Gamma '({\rm{Fo}})}}{{\Gamma ({\rm{Fo}})}} = \mu \qquad \qquad(15)$

The separation constant μ can be either a real or a complex number. The solution of Γ from eq. (15) will be Γ = eμFo. If μ is a positive real number, we will have $\Gamma \to \infty$ when ${\rm{Fo}} \to \infty$, which does not make sense, therefore, μ cannot be a positive real number. If μ is zero, we will have Γ = const, and Θ is a linear function of X only. The final solution for θ will also be a linear function of X only, which does not make sense either. It can also be shown that the separation constant cannot be a complex number (see Problem 3.24), therefore, the separation variable has to be a negative real number. If we represent this negative number by μ = − λ2, eq. (15) can be rewritten as the following two equations

$\Theta '' + {\lambda ^2}\Theta = 0 \qquad \qquad(16)$
$\Gamma ' + {\lambda ^2}\Gamma = 0 \qquad \qquad(17)$

The general solutions of eqs. (16) and (17) are

$\Theta = {C_1}\cos \lambda X + {C_2}\sin \lambda X \qquad \qquad(18)$
$\Gamma = {C_3}{e^{ - {\lambda ^2}{\rm{Fo}}}} \qquad \qquad(19)$

where C1,C2, and C3 are integral constants. Substituting eq. (14) into eqs. (11) and (12), the following boundary conditions of eq. (16) are obtained

$\Theta '(0) = 0 \qquad \qquad(20)$
$- \Theta '(1) = {\rm{Bi}}\Theta (1) \qquad \qquad(21)$

Substituting eq. (18) into eq. (20) yields

Θ'(0) = − C1λsin(0) + C2λcos(0) = C2λ = 0

Since λ cannot be zero, C2 must be zero and eq. (18) becomes

$\Theta = {C_1}\cos \lambda X \qquad \qquad(22)$

Applying the convection boundary condition, eq. (21), one obtains

$\frac{{{\lambda _n}}}{{{\rm{Bi}}}} = \cot {\lambda _n} \qquad \qquad(23)$

where n is an integer. Equation (23) indicates that many possible values for λ – termed eigenvalues – can satisfy the convection condition. The eigenvalue λn can be obtained by solving eq. (23) using an iterative procedure. The dimensionless temperature with eigenvalue λn can be obtained by substituting eqs. (22) and (19) into eq. (14), i.e.,

${\theta _n} = {C_n}\cos \left( {{\lambda _n}X} \right){e^{ - \lambda _n^2{\rm{Fo}}}} \qquad \qquad(24)$

where Cn = C1C3. Equation (24) is a solution that satisfies eqs. (10) – (12). At this point, the constant Cn is still unspecified and eq. (13) is unused, however, if we substitute eq. (24) into (13), the constant Cn that satisfies the initial condition cannot be found. Since the one-dimensional transient heat conduction problem under consideration is a linear problem, the sum of different θn for each value of n also satisfies eqs. (10) – (12).

$\theta = \sum\limits_{n = 1}^\infty {{C_n}\cos \left( {{\lambda _n}X} \right){e^{ - \lambda _n^2{\rm{Fo}}}}} \qquad \qquad(25)$

Substituting eq. (25) into eq. (13) yields

$1 = \sum\limits_{n = 1}^\infty {{C_n}\cos \left( {{\lambda _n}X} \right)}$

Multiplying the above equation by $\cos \left( {{\lambda _m}X} \right)$ and integrating the resulting equation in the interval of (0, 1), one obtains

$\int_0^1 {\cos ({\lambda _m}X)dX} = \sum\limits_{n = 1}^\infty {{C_n}\int_0^1 {\cos ({\lambda _m}X)\cos ({\lambda _n}X)dX} } \qquad \qquad(26)$

The integral in the right-hand side of eq. (26) can be evaluated as

$\int_0^1 {\cos ({\lambda _m}X)\cos ({\lambda _n}X)dX} = \left\{ {\begin{array}{*{20}{c}} {\frac{{{\lambda _m}\sin {\lambda _m}\cos {\lambda _n} - {\lambda _n}\cos {\lambda _m}\sin {\lambda _n}}}{{\lambda _m^2 - \lambda _n^2}},{\rm{ }}m \ne n} \\ {\frac{1}{{2{\lambda _m}}}\left( {{\lambda _m} + \frac{{\sin 2{\lambda _m}}}{2}} \right),{\rm{ }}m = n} \\ \end{array}} \right. \qquad \qquad(27)$

Equation (23) can be rewritten as

Bi = λntanλn

Similarly, for eigenvalue λm, we have

Bi = λmtanλm

Combining the above two equations, we have

λmtanλm − λntanλn = 0

or

λmsinλmcosλn − λncosλmsinλn = 0

therefore, the integral in eq. (27) is zero for the case that $m \ne n$, and the right hand side of eq. (26) becomes

$\sum\limits_{n = 1}^\infty {{C_n}\int_0^1 {\cos ({\lambda _m}X)\cos ({\lambda _n}X)dX} } = \frac{{{C_m}}}{{2{\lambda _m}}}\left( {{\lambda _m} + \frac{{\sin 2{\lambda _m}}}{2}} \right) \qquad \qquad(28)$

Substituting eq. (28) into eq. (26) and evaluating the integral at the left-hand side of eq. (26), we have

$\frac{1}{{{\lambda _m}}}\sin {\lambda _m} = \frac{{{C_m}}}{{2{\lambda _m}}}\left( {{\lambda _m} + \frac{{\sin 2{\lambda _m}}}{2}} \right)$

i.e.,

${C_m} = \frac{{4\sin {\lambda _m}}}{{2{\lambda _m} + \sin 2{\lambda _m}}}$

Changing notation from m to n, we get

${C_n} = \frac{{4\sin {\lambda _n}}}{{2{\lambda _n} + \sin 2{\lambda _n}}} \qquad \qquad(29)$

The dimensionless temperature, therefore, becomes

$\theta = \sum\limits_{n = 1}^\infty {\frac{{4\sin {\lambda _n}}}{{2{\lambda _n} + \sin 2{\lambda _n}}}\cos \left( {{\lambda _n}X} \right){e^{ - \lambda _n^2{\rm{Fo}}}}} \qquad \qquad(30)$

If the Biot number becomes infinite, the convection boundary condition becomes

$\theta = 0,{\rm{ X}} = 1 \qquad \qquad(31)$

which is an isothermal condition at the right-hand side of the wall. Equation (23) becomes

$\cos {\lambda _n} = 0 \qquad \qquad(32)$

and the eigenvalue is therefore

${\lambda _n} = \left( {n\pi - \pi /2} \right),{\rm{ }}n = 1,2,3,... \qquad \qquad(33)$

The temperature distribution for this case is then

$\theta = \sum\limits_{n = 1}^\infty {\frac{{4\sin (n - \pi /2)}}{{2(n - \pi /2) + \sin 2(n - \pi /2)}}\cos \left[ {(n - \pi /2)X} \right]{e^{ - {{(n - \pi /2)}^2}{\rm{Fo}}}}} \qquad \qquad(34)$

When Fourier’s number is greater than 0.2, only the first term in eq. (25) is necessary and the solution becomes

$\theta = {\theta _1} = {C_1}\cos \left( {{\lambda _1}X} \right){e^{ - \lambda _1^2{\rm{Fo}}}} \qquad \qquad(35)$

which is source of Hiesler’s charts (Incropera et al., 2007).

## Nonhomogeneous Problems

The transient one-dimensional conduction problems that we discussed so far are limited to the case that the problem is homogeneous and the method of separation of variables works. When the problem is not homogeneous due to a nonhomogeneous energy equation or boundary condition, the solution of a nonhomogeneous problem can be obtained by superposition of a particular solution of the nonhomogeneous problem and the general solution of the corresponding homogeneous problem.

### Partial Solution

Figure 2: Heat conduction under boundary condition of the first kind

Let us consider a finite slab with thickness of L and a uniform initial temperature of Ti as shown in the figure on the right. At time t= 0, the temperature on the left side of the slab is suddenly increased to T0 while the temperature on the right side of the slab is maintained at Ti. Assuming that there is no internal heat generation in the slab and the thermophysical properties of the slab are constants, the energy equation is

$\frac{{{\partial ^2}T}}{{\partial {x^2}}} = \frac{1}{\alpha }\frac{{\partial T}}{{\partial t}},{\rm{ }}0 < x < L,{\rm{ }}t > 0 \qquad \qquad(36)$

subject to the following boundary and initial conditions

$T = {T_0},{\rm{ }}x = 0 \qquad \qquad(37)$
$T = {T_i},{\rm{ }}x = L \qquad \qquad(38)$
$T = {T_i},{\rm{ }}0 < x < L,{\rm{ }}t = 0 \qquad \qquad(39)$

By defining the following dimensionless variables

$\theta = \frac{{T - {T_i}}}{{{T_0} - {T_i}}},{\rm{ }}X = \frac{x}{L},{\rm{ Fo}} = \frac{{\alpha t}}{{{L^2}}} \qquad \qquad(40)$

eqs. (36) – (39) will be nondimensionalized as

$\frac{{{\partial ^2}\theta }}{{\partial {X^2}}} = \frac{{\partial \theta }}{{\partial {\rm{Fo}}}},{\rm{ }}0 < X < 1,{\rm{ Fo}} > 0 \qquad \qquad(41)$
$\theta = 1,{\rm{ }}X = 0 \qquad \qquad(42)$
$\theta = 0,{\rm{ X}} = 1 \qquad \qquad(43)$
$\theta = 0,{\rm{ }}0 < X < 1,{\rm{ Fo}} = 0 \qquad \qquad(44)$

It can be seen that the problem is still nonhomogeneous after nondimensionalization because eq. (42) is not homogeneous. When $t \to \infty$(i.e., $Fo \to \infty$), the temperature distribution can reach to steady state. If the steady state temperature is represented by θs, it must satisfy the following equations:

$\frac{{{\partial ^2}{\theta _s}}}{{\partial {X^2}}} = 0,{\rm{ }}0 < X < 1 \qquad \qquad(45)$
${\theta _s} = 1,{\rm{ }}X = 0 \qquad \qquad(46)$
${\theta _s} = 0,{\rm{ X}} = 1 \qquad \qquad(47)$

which have the following solution:

${\theta _s} = 1 - X \qquad \qquad(48)$

To obtain the generation of the problem described by eqs. (41) – (44), a method of partial solution (Myers, 1987) will be employed. In this methodology, it is assumed that the solution of a nonhomogeneous problem can be expressed as

$\theta (X,{\rm{Fo}}) = {\theta _s}(X) + {\theta _h}(X,{\rm{Fo}}) \qquad \qquad(49)$

where θh represent the solution of a homogeneous problem. Substituting eqs. (41) – (44) and considering eqs. (45) – (47), we have

$\frac{{{\partial ^2}{\theta _h}}}{{\partial {X^2}}} = \frac{{\partial {\theta _h}}}{{\partial {\rm{Fo}}}},{\rm{ }}0 < X < 1,{\rm{ Fo}} > 0 \qquad \qquad(50)$
${\theta _h} = 0,{\rm{ }}X = 0 \qquad \qquad(51)$
${\theta _h} = 0,{\rm{ X}} = 1 \qquad \qquad(52)$
${\theta _h} = X - 1,{\rm{ }}0 < X < 1,{\rm{ Fo}} = 0 \qquad \qquad(53)$

which represent a new homogeneous problem. This problem can be solved using the method of separation of variables (see Problem 3.31) and the result is

${\theta _h} = - \frac{2}{\pi }\sum\limits_{n = 1}^\infty {\frac{{\sin (n\pi X)}}{n}{e^{ - {{(n\pi )}^2}{\rm{Fo}}}}} \qquad \qquad(54)$

The solution of the nonhomogeneous problem thus becomes

$\theta = 1 - X - \frac{2}{\pi }\sum\limits_{n = 1}^\infty {\frac{{\sin (n\pi X)}}{n}{e^{ - {{(n\pi )}^2}{\rm{Fo}}}}} \qquad \qquad(55)$

### Variation of Parameter

Figure 3: Heat conduction under boundary condition of the second kind

The partial solution only works if the steady-state solution exists. If the steady-state solution does not exist, we can use the method of variation of parameters to solve the problem. Let us consider a finite slab with thickness of L and a uniform initial temperature of Ti. At time t= 0, the left side is subject to a constant heat flux while the right side of the slab is adiabatic (see Fig. 3). Assuming that there is no internal heat generation in the slab and the thermophysical properties of the slab are constants, the energy equation is

$\frac{{{\partial ^2}T}}{{\partial {x^2}}} = \frac{1}{\alpha }\frac{{\partial T}}{{\partial t}},{\rm{ }}0 < x < L,{\rm{ }}t > 0 \qquad \qquad(56)$

subject to the following boundary and initial conditions

$- k\frac{{\partial T}}{{\partial x}} = {q''_0},{\rm{ }}x = 0 \qquad \qquad(57)$
$\frac{{\partial T}}{{\partial x}} = 0,{\rm{ }}x = L \qquad \qquad(58)$
$T = {T_i},{\rm{ }}0 < x < L,{\rm{ }}t = 0 \qquad \qquad(59)$

By defining the following dimensionless variables

$\theta = \frac{{T - {T_i}}}{{{{q''}_0}L/k}},{\rm{ }}X = \frac{x}{L},{\rm{ Fo}} = \frac{{\alpha t}}{{{L^2}}} \qquad \qquad(60)$

eqs. (56) – (59) will be nondimensionalized as

$\frac{{{\partial ^2}\theta }}{{\partial {X^2}}} = \frac{{\partial \theta }}{{\partial {\rm{Fo}}}},{\rm{ }}0 < X < 1,{\rm{ Fo}} > 0 \qquad \qquad(61)$
$\frac{{\partial \theta }}{{\partial X}} = - 1,{\rm{ }}X = 0 \qquad \qquad(62)$
$\frac{{\partial \theta }}{{\partial X}} = 0,{\rm{ }}X = 1 \qquad \qquad(63)$
$\theta = 0,{\rm{ }}0 < X < 1,{\rm{ Fo}} = 0 \qquad \qquad(64)$

This nonhomogeneous problem does not have a steady-state solution, and therefore the partial solution cannot be applied. We will use the method of variation of parameters (Myers, 1987) to solve this problem. This method requires the following steps:

1. Set up a homogeneous problem by dropping the nonhomogeneous terms.

2. Solve the homogeneous problem to get eigenvalue λn and eigenfunctions Θn(X)

3. Assuming the solution of the original nonhomogeneous problem has the form of $\theta (X,{\rm{Fo}}) = \sum\limits_{n = 1}^n {{A_n}({\rm{Fo}}){\Theta _n}(X)}$

4. Solve for An(Fo) using orthogonal property of Θn

5. Obtain an ordinary differential equation (ODE) for An(Fo) and solve for An(Fo) from the ODE

6. Put together the final solution.

We will solve this nonhomogeneous problem by following the above procedure. The corresponding homogeneous problem is:

$\frac{{{\partial ^2}{\theta _h}}}{{\partial {X^2}}} = \frac{{\partial {\theta _h}}}{{\partial {\rm{Fo}}}},{\rm{ }}0 < X < 1,{\rm{ Fo}} > 0 \qquad \qquad(65)$
$\frac{{\partial {\theta _h}}}{{\partial X}} = 0,{\rm{ }}X = 0 \qquad \qquad(66)$
$\frac{{\partial {\theta _h}}}{{\partial X}} = 0,{\rm{ }}X = 1 \qquad \qquad(67)$
${\theta _h} = 0,{\rm{ }}0 < X < 1,{\rm{ Fo}} = 0 \qquad \qquad(68)$

Assuming the solution of the above homogeneous problem is

${\theta _h} = \Theta (X)\Gamma ({\rm{Fo}}) \qquad \qquad(69)$

eq. (65) becomes

$\frac{{\Theta ''(X)}}{{\Theta (X)}} = \frac{{\Gamma '({\rm{Fo}})}}{{\Gamma ({\rm{Fo}})}} = - {\lambda ^2} \qquad \qquad(70)$

Since the objective here is to get the eigenvalue and eigen functions, we do not need to solve for Γ and only need to solve for Θ. The eigenvalue problem is

$\Theta '' + {\lambda ^2}\Theta = 0 \qquad \qquad(71)$
$\Theta '(0) = 0 \qquad \qquad(72)$
$\Theta '(1) = 0 \qquad \qquad(73)$

Solving eqs. (71) – (73) yields the following eigenvalues and eigen functions

${\lambda _n} = n\pi \qquad \qquad(74)$
${\Theta _n}(X) = \cos (n\pi X),{\rm{ }}n = 0,1,2,... \qquad \qquad(75)$

Now, let us assume that the solution of the original nonhomogeneous problem is

$\theta (X,{\rm{Fo}}) = \sum\limits_{n = 0}^\infty {{A_n}({\rm{Fo}})\cos \left( {n\pi X} \right)} \qquad \qquad(76)$

Multiplying eq. (76) by $\cos \left( {m\pi X} \right)$ and integrating the resulting equation in the interval of (0, 1), one obtains

$\int_0^1 {\theta (X,{\rm{Fo}})\cos (m\pi X)dX} = \sum\limits_{n = 1}^\infty {{A_n}\int_0^1 {\cos (m\pi X)\cos (n\pi X)dX} } \qquad \qquad(77)$

The integral on the right-hand side of eq. (77) can be evaluated as

$\int_0^1 {\cos (m\pi X)\cos (n\pi X)dX} = \left\{ {\begin{array}{*{20}{c}} {0, m \ne n} \\ \begin{array}{l} 1/2, m = n \ne 0 \\ 1, m = n = 0 \\ \end{array} \\ \end{array}} \right. \qquad \qquad(78)$

thus, eq. (77) becomes

${A_{\rm{0}}}({\rm{Fo}}) = \int_0^1 {\theta (X,{\rm{Fo}})dX} ,{\rm{ }}m = 0 \qquad \qquad(79)$
${A_{\rm{m}}}({\rm{Fo}}) = 2\int_0^1 {\theta (X,{\rm{Fo}})\cos (m\pi X)dX} ,{\rm{ }}m \ne 0 \qquad \qquad(80)$

Differentiating eq. (79) with respect to Fo, one obtains:

$\frac{{d{A_{\rm{0}}}}}{{{\rm{dFo}}}} = \int_0^1 {\frac{{\partial \theta }}{{\partial {\rm{Fo}}}}dX} \qquad \qquad(81)$

Substituting eq. (61) into eq. (81) and integrating with respect to X yields

$\frac{{d{A_{\rm{0}}}}}{{{\rm{dFo}}}} = \int_0^1 {\frac{{{\partial ^2}\theta }}{{\partial {X^2}}}dX = } {\left. {\frac{{\partial \theta }}{{\partial X}}} \right|_{X = 1}} - {\left. {\frac{{\partial \theta }}{{\partial X}}} \right|_{X = 0}} = 1 \qquad \qquad(82)$

Integrating eq. (82) with respect to Fo, we have

${A_0}({\rm{Fo}}) = {\rm{Fo}} + {C_1} = \int_0^1 {\theta (X,{\rm{Fo}})dX} \qquad \qquad(83)$

When Fo = 0, eq. (83) becomes

${A_0}(0) = {C_1} = \int_0^1 {\theta (X,0)dX} = 0 \qquad \qquad(84)$

thus, we have

${A_0}({\rm{Fo}}) = {\rm{Fo}} \qquad \qquad(85)$

Differentiating eq. (80) and considering eq. (61) yield

$\frac{{d{A_{\rm{m}}}}}{{d{\rm{Fo}}}} = 2\int_0^1 {\frac{{\partial \theta }}{{\partial {\rm{Fo}}}}\cos (m\pi X)dX} = 2\int_0^1 {\frac{{{\partial ^2}\theta }}{{\partial {X^2}}}\cos (m\pi X)dX} \qquad \qquad(86)$

Using integration by parts twice, the following ODE is obtained:

$\frac{{d{A_{\rm{m}}}}}{{d{\rm{Fo}}}} = 2 - {(m\pi )^2}{A_m} \qquad \qquad(87)$

Multiplying eq. (87) by an integrating factor ${e^{{{(m\pi )}^2}{\rm{Fo}}}}$, we have

$\frac{d}{{d{\rm{Fo}}}}\left[ {{A_{\rm{m}}}{e^{{{(m\pi )}^2}{\rm{Fo}}}}} \right] = 2{e^{{{(m\pi )}^2}{\rm{Fo}}}} \qquad \qquad(88)$

which can be integrated to get

${A_m} = \frac{2}{{{{(m\pi )}^2}}} + {C_2}{e^{ - {{(m\pi )}^2}{\rm{Fo}}}} \qquad \qquad(89)$

where C2 is an integral constant that needs to be determined by an initial condition. For Fo = 0, eq. (80) becomes

${A_{\rm{m}}}({\rm{0}}) = 2\int_0^1 {\theta (X,{\rm{0}})\cos (m\pi X)dX} = 2\int_0^1 {\cos (m\pi X)dX} = 0 \qquad \qquad(90)$

Substituting eq. (89) into eq. (90), one obtains

${C_2} = - \frac{2}{{{{(m\pi )}^2}}}$

therefore, we have

${A_m} = \frac{2}{{{{(m\pi )}^2}}} - \frac{2}{{{{(m\pi )}^2}}}{e^{ - {{(m\pi )}^2}{\rm{Fo}}}}$

Changing m back to n for notation,

${A_n} = \frac{2}{{{{(n\pi )}^2}}} - \frac{2}{{{{(n\pi )}^2}}}{e^{ - {{(n\pi )}^2}{\rm{Fo}}}} \qquad \qquad(91)$

Substituting eqs. (85) and (91) into eq. (76), the solution becomes

$\theta (X,{\rm{Fo}}) = {\rm{Fo + }}\frac{{\rm{2}}}{{{\pi ^{\rm{2}}}}}\sum\limits_{n = 1}^\infty {\frac{{\cos \left( {n\pi X} \right)}}{{{n^2}}}} - \frac{{\rm{2}}}{{{\pi ^{\rm{2}}}}}\sum\limits_{n = 1}^\infty {\frac{{\cos \left( {n\pi X} \right)}}{{{n^2}}}{e^{ - {{(n\pi )}^2}{\rm{Fo}}}}} \qquad \qquad(92)$

When the time (Fourier number) becomes large, the last term on the right-hand side will become zero and the solution is represented by the first two terms only. To simplify eq. (92), let us assume the solution at large Fo can be expressed as

$\theta (X,{\rm{Fo}}) = {\rm{Fo + }}f(X) \qquad \qquad(93)$

which is referred to as an asymptotic solution and it must satisfy eqs. (61) – (63). Substituting eq. (93) into eqs. (61) – (63), we have

$f''(X) = 1\qquad \qquad(94)$
$f'(0) = - 1,{\rm{ }}f'(1) = 0\qquad \qquad(95)$

Integrating eq. (94) and considering eq. (95), we obtain

$f(X) = \frac{{{X^2}}}{2} - X + C\qquad \qquad(96)$

where C cannot be determined from eq. (95) because both boundary conditions are for the first order derivative. To determine C, we can expand f(X) defined in eq. (96) into cosine Fourier series, i.e.

$f(X) = \frac{{{X^2}}}{2} - X + C = {a_0} + \sum\limits_{n = 1}^\infty {{a_n}\cos (n\pi X)} \qquad \qquad(97)$

After determining a0 and an, and considering that f(X) is identical to the second term on the right-hand side of eq. (92), we have

$\frac{{{X^2}}}{2} - X + \frac{1}{3} = \frac{{\rm{2}}}{{{\pi ^{\rm{2}}}}}\sum\limits_{n = 1}^\infty {\frac{{\cos \left( {n\pi X} \right)}}{{{n^2}}}} \qquad \qquad(98)$

Substituting eq. (98) into eq. (92), the final solution becomes

$\theta (X,{\rm{Fo}}) = {\rm{Fo + }}\frac{{{X^2}}}{2} - X + \frac{1}{3} - \frac{{\rm{2}}}{{{\pi ^{\rm{2}}}}}\sum\limits_{n = 1}^\infty {\frac{{\cos \left( {n\pi X} \right)}}{{{n^2}}}{e^{ - {{(n\pi )}^2}{\rm{Fo}}}}} \qquad \qquad(99)$

## References

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

Incropera, F.P., Dewitt, D.P., Bergman, T.L., and Lavine, A.S., 2007, Fundamentals of Heat and Mass Transfer, John Wiley & Sons, Hoboken, NJ.