## Generalized 1-D heat conduction

(a)Plane Wall (b)Hollow cylinder or sphere
Fig. 1 Heat conduction in different geometric configurations

One-dimensional heat conduction can occur in different geometric configurations. The figure on the right shows heat conduction in a plane wall and in a hollow cylinder or sphere. The energy equationin different geometric configurations can be expressed as:
Plane wall:

$\frac{d}{{dx}}\left( {k\frac{{dT}}{{dx}}} \right) = 0,\qquad {x_1} < x < {x_2} \qquad \qquad(1)$

Hollow cylinder:

$\frac{1}{r}\frac{d}{{dr}}\left( {kr\frac{{dT}}{{dr}}} \right) = 0 \qquad {r_1} < r < {r_2}{\rm{)}} \qquad \qquad(2)$

Hollow sphere:

$\frac{1}{{{r^2}}}\frac{d}{{dr}}\left( {k{r^2}\frac{{dT}}{{dr}}} \right) = 0 \qquad {r_1} < r < {r_2}{\rm{)}} \qquad \qquad(3)$

Equations (1) – (3) can be represented using the following general form (Arpaci, 1966):

Fig. 2 Generalized 1-D heat conduction problem.
$\frac{d}{{ds}}\left( {kA(s)\frac{{dT}}{{ds}}} \right) = 0,{\rm{ }}{s_1} < s < {s_2} \qquad \qquad(4)$

where s is space variable and A(s) can be viewed as the heat transfer area. The heat transfer area in a plane wall, hollow cylinder and sphere are A(s) = LW, 2πsL, and 2πs2, respectively (where W is the width of the plane wall, and L is length of the plane wall or cylinder). Figure 2 represents the general one-dimensional heat conduction problem.

Equation (4) can also be rewritten as

$\frac{{dq}}{{ds}} = 0,{\rm{ }}{s_1} < s < {s_2} \qquad \qquad(5)$

or

$q = - kA(s)\frac{{dT}}{{ds}} = {\rm{const}},{\rm{ }}{s_1} < s < {s_2} \qquad \qquad(6)$

which means the heat transfer rate at different cross-sections is a constant. It should be pointed out that the heat flux q''(s) = − kdT / ds is not a constant unless the cross-sectional area A(s) is constant. If the thermal conductivity is independent from temperature, the temperature distribution can be obtained by integrating eq. (4) twice:

$T = a\int_{{s_1}}^s {\frac{1}{{A(s)}}ds} + b \qquad \qquad(7)$

where a and b are unspecified integral constants. If the temperatures of both inner and outer surfaces are specified, i.e.,

$T = {T_1},{\rm{ }}s = {s_1} \qquad \qquad(8)$
$T = {T_2},{\rm{ }}s = {s_2} \qquad \qquad(9)$

The integral constants a and b in eq. (7) can be determined from eqs. (8) and (9), and the temperature becomes

$T = \frac{{{T_2} - {T_1}}}{{\int_{{s_1}}^{{s_2}} {\frac{1}{{A(s)}}ds} }}\int_{{s_1}}^s {\frac{1}{{A(s)}}ds} + {T_1} \qquad \qquad(10)$

## Thermal Resistance

The rate of heat transfer can be obtained using Fourier’s law:

$q = - kA(s)\frac{{dT}}{{ds}} = \frac{{{T_1} - {T_2}}}{{\frac{1}{k}\int_{{s_1}}^{{s_2}} {\frac{{ds}}{{A(s)}}} }} \qquad \qquad(11)$

which can be rewritten as

$q = \frac{{{T_1} - {T_2}}}{{{R_c}}} \qquad \qquad(12)$

where

${R_c} = \frac{1}{k}\int_{{s_1}}^{{s_2}} {\frac{{ds}}{{A(s)}}} \qquad \qquad(13)$
Figure 3: Analogy between conductions of electricity and heat

is the thermal resistance for heat conduction. Equation (12) is similar to Ohm’s law

$I = \frac{{{V_1} - {V_2}}}{R} \qquad \qquad(14)$

where V1 and V2 are, respectively, the electrical potential at two ends of the electrical resistance, R, and I is the current through the electrical resistance (see Fig. 3).

If the boundary conditions at the inner and outer surfaces are convective conditions, i.e.,

$- k\frac{{dT}}{{ds}} = {h_i}({T_i} - T),{\rm{ }}s = {s_1} \qquad \qquad(15)$
$- k\frac{{dT}}{{ds}} = {h_o}(T - {T_o}),{\rm{ }}s = {s_2} \qquad \qquad(16)$

where hi and ho are heat transfer coefficients at the inner and outer surfaces, respectively. The temperatures of the fluids that are in contact with the inner and outer surfaces are Ti and To, respectively. If the temperatures of the inner and outer walls are represented by T1 and T2, (both of them are unknown) eq. (12) is still valid. By multiplying eqs. (15) and (16) by the heat transfer areas at the inner and outer surfaces, A1 and A2, one obtains

$q = - kA({s_1})\frac{{dT}}{{ds}} = {h_i}{A_1}({T_i} - {T_1}),{\rm{ }}s = {s_1} \qquad \qquad(17)$
$q = - kA({s_2})\frac{{dT}}{{ds}} = {h_o}{A_2}({T_2} - {T_o}),{\rm{ }}s = {s_2} \qquad \qquad(18)$

Under steady state condition, the heat transfer rate obtained from eqs. (12), (17) and (18) are identical. By eliminating T1 and T2 from these equations, the following expression for heat transfer rate is obtained:

$q = \frac{{{T_i} - {T_o}}}{{{R_i} + {R_c} + {R_o}}} \qquad \qquad(19)$

where

${R_i} = \frac{1}{{{h_i}{A_1}}} \qquad \qquad(20)$
${R_o} = \frac{1}{{{h_o}{A_2}}} \qquad \qquad(21)$

are convective thermal resistances between inner fluid and inner wall and between outer fluid and outer wall, respectively. Equation (19) can be viewed as a system with three thermal resistances connected in series. Equation (19) can also be written as

$q = {U_o}{A_2}({T_i} - {T_o}) \qquad \qquad(22)$

where Uo is the overall- coefficient of heat transfer based on the outer surface area. Comparing eqs. (19) and (22) yields

$\frac{1}{{{U_o}{A_2}}} = {R_i} + {R_c} + {R_o} \qquad \qquad(23)$

or

$\frac{1}{{{U_o}}} = \frac{{{A_2}}}{{{h_i}{A_1}}} + \frac{{{A_2}}}{k}\int_{{s_1}}^{{s_2}} {\frac{{ds}}{{A(s)}}} + \frac{1}{{{h_o}}} \qquad \qquad(24)$

For the heat conduction in three different coordinates as shown in eqs. (1) – (3), the overall coefficients for heat transfer are

$\frac{1}{{{U_o}}} = \frac{1}{{{h_i}}} + \frac{L}{k} + \frac{1}{{{h_o}}}{\rm{ plane wall (}}{x_2} - {x_1} = L{\rm{)}} \qquad \qquad(25)$
$\frac{1}{{{U_o}}} = \frac{{{r_2}}}{{{h_i}{r_1}}} + \frac{{{r_2}}}{k}\ln \left( {\frac{{{r_2}}}{{{r_1}}}} \right) + \frac{1}{{{h_o}}},{\rm{ hollow cylinder}} \qquad \qquad(26)$
$\frac{1}{{{U_o}}} = \frac{{r_2^2}}{{{h_i}r_1^2}} + \frac{{{r_2}}}{k}\left( {\frac{{{r_2}}}{{{r_1}}} - 1} \right) + \frac{1}{{{h_o}}},{\rm{ hollow sphere}} \qquad \qquad(27)$

If the conducting wall shown in Fig. 1 has multiple layers and each layer has different thermal conductivity, there will be multiple conduction thermal resistances between two convection thermal resistances. If the number of layers is represented by N, the overall coefficient of heat transfer will be expressed as

$\frac{1}{{{U_o}}} = \frac{{{A_{N + 1}}}}{{{h_i}{A_1}}} + {A_{N + 1}}\sum\limits_{i = 1}^N {\frac{1}{{{k_i}}}\int_{{s_i}}^{{s_{i + 1}}} {\frac{{ds}}{{A(s)}}} } + \frac{1}{{{h_o}}} \qquad \qquad(28)$

where A1 and AN + 1 are the areas of the inner and outer surfaces, respectively; and ki is the thermal conductivity of the ith layer (si < s < si + 1). The heat transfer rate becomes

$q = {U_o}{A_{N + 1}}({T_i} - {T_o}) \qquad \qquad(29)$

or

$q = \frac{{{T_i} - {T_o}}}{{\frac{1}{{{h_i}{A_1}}} + \sum\limits_{i = 1}^N {\frac{1}{{{k_i}}}\int_{{s_i}}^{{s_{i + 1}}} {\frac{{ds}}{{A(s)}}} } + \frac{1}{{{h_o}{A_{N + 1}}}}}} \qquad \qquad(30)$

## Contact thermal resistance

Figure 4: Contact thermal resistance.

In arriving at eq. (28), it is assumed that different layers are in perfect contact so that the temperatures across the interfaces between different layers are continuous. When two rough surfaces are in contact, there may be a temperature drop, ΔT, across the interface between different materials (see Fig. 4). The heat flux and this temperature drop is related by

$q'' = \frac{{\Delta {T_i}}}{{{{R''}_{ct,i}}}} \qquad \qquad(31)$

where R''ct is contact thermal resistance (m2 − K / W). Heat transfer across an imperfect contact interface is due to combined effects of conduction at the actual contact area, convection of the air entrapped in the gap, and radiation between the two surfaces that are not in direct contact. The contact thermal resistance can be reduced by applying pressure in the direction perpendicular to the interface, or applying conduction grease at the interface. More information about contact thermal resistance can be found in Fletcher (1988). When contact thermal resistances are present, eq. (30) can be modified as

$q = \frac{{{T_i} - {T_o}}}{{\frac{1}{{{h_i}{A_1}}} + \sum\limits_{i = 1}^N {\frac{1}{{{k_i}}}\int_{{s_i}}^{{s_{i + 1}}} {\frac{{ds}}{{A(s)}}} } + \sum\limits_{i = 2}^N {\frac{{{{R''}_{tc,i}}}}{{{A_i}}}} + \frac{1}{{{h_o}{A_{N + 1}}}}}} \qquad \qquad(32)$

## With Internal Heat Generation

The 1-D steady-state heat conductions that we discussed so far are limited to the case without internal heat generation. Heat conduction with internal heat generation can be encountered in many applications such as electrical heating, chemical reaction, or nuclear reaction in the conduction medium. Let us consider the generalized 1-D heat conduction problem shown in Fig. 2. With uniform internal heat source, q''', the energy equation is

$\frac{1}{{A(s)}}\frac{d}{{ds}}\left( {A(s)\frac{{dT}}{{ds}}} \right) + \frac{{q'''}}{k} = 0,{\rm{ }}{s_1} < s < {s_2} \qquad \qquad(33)$

where the thermal conductivity is assumed to be independent from temperature. Equation (33) is subject to boundary conditions specified by eqs. (8) and (9). Multiplying eq. (19) by A(s) and integrating the resultant equation in the interval of (s1, s) yields

$A(s)\frac{{dT}}{{ds}} = - \frac{{q'''}}{k}\int_{{s_1}}^s {A(s)ds} + {A_1}{\left. {\frac{{dT}}{{ds}}} \right|_{s = {s_1}}} \qquad \qquad(34)$

Dividing eq. (34) by A(s) and integrating the resultant equation in the interval of (s1, s), we have

$T = - \frac{{q'''}}{k}\int_{{s_1}}^s {\left[ {\frac{1}{{A(s)}}\int_{{s_1}}^s {A(s)ds} } \right]ds} + {A_1}{\left. {\frac{{dT}}{{ds}}} \right|_{s = {s_1}}}\int_{{s_1}}^s {\frac{1}{{A(s)}}} ds + {T_1} \qquad \qquad(35)$

where the temperature gradient at s = s1 is still unknown. Equation (35) already satisfies eq. (8) but not eq. (9). Substituting eq. (35) into eq. (9), one obtains

${T_2} = - \frac{{q'''}}{k}\int_{{s_1}}^{{s_2}} {\left[ {\frac{1}{{A(s)}}\int_{{s_1}}^s {A(s)ds} } \right]ds} + {A_1}{\left. {\frac{{dT}}{{ds}}} \right|_{s = {s_1}}}\int_{{s_1}}^{{s_2}} {\frac{1}{{A(s)}}} ds + {T_1}$

which can be rearranged to get

${A_1}{\left. {\frac{{dT}}{{ds}}} \right|_{s = {s_1}}} = \frac{{({T_2} - {T_1}) + \frac{{q'''}}{k}\int_{{s_1}}^{{s_2}} {\left[ {\frac{1}{{A(s)}}\int_{{s_1}}^s {A(s)ds} } \right]ds} }}{{\int_{{s_1}}^{{s_2}} {\frac{1}{{A(s)}}} ds}}$

Therefore, the temperature profile becomes

$\begin{array}{l} T = - \frac{{q'''}}{k}\int_{{s_1}}^s {\left[ {\frac{1}{{A(s)}}\int_{{s_1}}^s {A(s)ds} } \right]ds} \\ {\rm{ }} + \frac{{({T_2} - {T_1}) + \frac{{q'''}}{k}\int_{{s_1}}^{{s_2}} {\left[ {\frac{1}{{A(s)}}\int_{{s_1}}^s {A(s)ds} } \right]ds} }}{{\int_{{s_1}}^{{s_2}} {\frac{1}{{A(s)}}} ds}}\int_{{s_1}}^s {\frac{1}{{A(s)}}} ds + {T_1} \\ \end{array} \qquad \qquad(36)$

The heat transfer rate can be obtained by Fourier’s law

$q = - kA(s)\frac{{dT}}{{ds}} \qquad \qquad(37)$

Substituting eq. (34) into eq. (37), the heat transfer rate becomes

$q = q'''\int_{{s_1}}^s {A(s)ds} + q({s_1}) \qquad \qquad(38)$

where

$q({s_1}) = - k{A_1}{\left. {\frac{{dT}}{{ds}}} \right|_{s = {s_1}}} = \frac{{k({T_1} - {T_2}) - q'''\int_{{s_1}}^{{s_2}} {\left[ {\frac{1}{{A(s)}}\int_{{s_1}}^s {A(s)ds} } \right]ds} }}{{\int_{{s_1}}^{{s_2}} {\frac{1}{{A(s)}}} ds}} \qquad \qquad(39)$

is the heat transfer rate at s = s1. It is evident from eq. (38) that the heat transfer rate is no longer independent from s when there is internal heat generation. It should be pointed out that eqs. (36) and (38) are valid for any coordinate system. For a Cartesian coordinate system with the origin of the coordinate on the left surface (i.e., x1 = 0 in Fig. 1), A(s) = A is a constant and eqs. (36) and (38) become

$T(x) = \frac{{q'''}}{{2k}}\left[ {(L - x)x} \right] + \frac{{({T_2} - {T_1})}}{L}x + {T_1} \qquad \qquad(40)$
$q = q'''AL\left( {x - \frac{L}{2}} \right) + \frac{{kA({T_1} - {T_2})}}{L} \qquad \qquad(41)$

For the heat conduction in a cylindrical and spherical coordinate system, the general solution, eqs. (36) and (38), can be simplified by considering the variation of conduction area (see Problem 3.7 and 3.8).