# Heat conduction from buried object

When two-dimensional heat conduction occurs in a regular shape, the method of separation of variables can be easily employed. An infinitely long cylinder (in the direction perpendicular to the screen) buried in homogeneous medium extending to infinity (see Fig. 1), can find its application in buried large-current cable or steam pipe for residential or industrial heating. The cylinder surface temperature is uniformly T0 and the surface temperature is uniformly Ts. Our objective is to find the temperature distribution, T(x,y) and the total amount of heat transfer from the buried cylinder.

The energy equation for two-dimensional conduction $\frac{{{\partial ^2}T}}{{\partial {x^2}}} + \frac{{{\partial ^2}T}}{{\partial {y^2}}} = 0,{\rm{ }}0 < x < L,{\rm{ }}0 < y < H$

is still valid for this problem and the boundary conditions for this problem are $T = {T_s}, \qquad{\rm{ }}y = 0 \qquad \qquad(1)$ $T = {T_0},\qquad{\rm{ }}{x^2} + {(y - z)^2} = r_0^2 \qquad \qquad(2)$

Since the heat transfer domain is very irregular, it cannot be easily described using either Cartesian or cylindrical coordinate systems. A different method that uses heat source and sink (see Fig. 2) can be used to obtain the solution (Eckert and Drake, 1987). In this methodology, the cylinder is treated as a line heat source, q', located at a depth a(a < z). If the conduction medium extends infinitely in both the x and ydirection, the isotherms caused by this line heat source will be a series of circles with its center at y = a but eq. (1) cannot be satisfied. In order to fulfill the boundary condition at the surface y = 0, a fictitious heat sink, q', which is symmetric to the heat source, must be used. The combination of the heat source and sink will yield a constant surface temperature, Ts, at y = 0.

The heat generated from the heat source located at (0, a) will keep the surface temperature of the cylinder equal to T0,1. If the temperature at the point P caused by the heat source is T1(x,y), the amount of heat transfer from the surface of the cylinder is $q' = \frac{{2\pi k({T_{0,1}} - {T_1})}}{{\ln ({r_1}/{r_0})}}$

The temperature at point P caused by the heat source alone is then, ${T_1} = {T_{0,1}} - \frac{{q'}}{{2\pi k}}\ln \left( {\frac{{{r_1}}}{{{r_0}}}} \right) \qquad \qquad(3)$

Similarly, the heat sink located at (0, − a) will maintain the surface temperature of the fictitious cylinder at T0,2 and result in the temperature at the point P equal to T2(x,y), i.e., $- q' = \frac{{2\pi k({T_{0,2}} - {T_2})}}{{\ln ({r_2}/{r_0})}}$

Consequently, the temperature at point P caused by the heat sink alone is, ${T_2} = {T_{0,2}} + \frac{{q'}}{{2\pi k}}\ln \left( {\frac{{{r_2}}}{{{r_0}}}} \right) \qquad \qquad(4)$

If both heat source and sink are present, the temperature at point P is $T(x,y) = {T_1}(x,y) + {T_2}(x,y) \qquad \qquad(5)$

Substituting eqs. (3) and (4) into eq. (5) yields $T = ({T_{0,1}} + {T_{0,2}}) + \frac{{q'}}{{2\pi k}}\ln \left( {\frac{{{r_2}}}{{{r_1}}}} \right) \qquad \qquad(6)$

At the surface y = 0 where r1 = r2, eq. (1) requires T0,1 + T0,2 = Ts and eq. (6) becomes $T = {T_s} + \frac{{q'}}{{2\pi k}}\ln \left( {\frac{{{r_2}}}{{{r_1}}}} \right)$

Defining excess temperature θ = TTs, we have $\theta = \frac{{q'}}{{2\pi k}}\ln \left( {\frac{{{r_2}}}{{{r_1}}}} \right) \qquad \qquad(7)$

The radii r1 and r2 can be expressed as ${r_1} = \sqrt {{x^2} + {{(y - z)}^2}} \textstyle.\over= \sqrt {{x^2} + {{(y - a)}^2}} \qquad \qquad(8)$ ${r_2} = \sqrt {{x^2} + {{(y + z)}^2}} \textstyle.\over= \sqrt {{x^2} + {{(y + a)}^2}} \qquad \qquad(9)$

which are valid because a is very close to z as will become evident later. Thus, the excess temperature in the conducting medium can be expressed as $\theta = \frac{{q'}}{{2\pi k}}\ln \left[ {\frac{{\sqrt {{x^2} + {{(y + a)}^2}} }}{{\sqrt {{x^2} + {{(y - a)}^2}} }}} \right] = \frac{{q'}}{{4\pi k}}\ln \left[ {\frac{{{x^2} + {{(y + a)}^2}}}{{{x^2} + {{(y - a)}^2}}}} \right] \qquad \qquad(10)$

which can be rewritten as $\exp \left( {\frac{{4\pi k\theta }}{{q'}}} \right) = \frac{{{x^2} + {{(y + a)}^2}}}{{{x^2} + {{(y - a)}^2}}} \qquad \qquad(11)$

When the left hand side of eq. (11) is held at a constant C, the following equation for isotherms is obtained: $\frac{{{x^2} + {{(y + a)}^2}}}{{{x^2} + {{(y - a)}^2}}} = C$

which can be rewritten as ${x^2} + {\left[ {y + a\left( {\frac{{1 + C}}{{1 - C}}} \right)} \right]^2} = \frac{{4C{a^2}}}{{{{(1 - C)}^2}}} \qquad \qquad(12)$

It represents a series of circles with the center located at [0, − a(1 + C) / (1 − C)] and radius of $2\sqrt C a/(1 - C)$; both the center of radius of the circular isotherms depend on the constant C, or the temperature.

For the cylinder surface, the excess temperature is θ0 = T0Ts and the constant C is ${C_0} = \exp \left( {\frac{{4\pi k{\theta _0}}}{{q'}}} \right) \qquad \qquad(13)$

The depth of the cylinder measured from the horizontal surface to the center of the cylinder is $z = \frac{{1 + {C_0}}}{{1 - {C_0}}}a \qquad \qquad(14)$

which can be rearranged to $a = \frac{{1 - {C_0}}}{{1 + {C_0}}}z \qquad \qquad(15)$

It is evident from eq. (15) that the depth of the line heat source must be shallower than that of the center of the cylinder because (1 − C0) / (1 + C0) < 1.

The radius of the cylinder is related to the constant C0 by [see eq. (12)] ${r_0} = \frac{{2\sqrt {{C_0}} a}}{{(1 - {C_0})}} = \frac{{2\sqrt {{C_0}} z}}{{(1 + {C_0})}} \qquad \qquad(16)$

Solving C0 in term of r0 yields $\sqrt {{C_0}} = \frac{z}{{{r_0}}} + \sqrt {{{\left( {\frac{z}{{{r_0}}}} \right)}^2} - 1} \qquad \qquad(17)$

Substituting eq. (13) into eq. (17) and solving for the heat transfer rate, one obtains $q' = \frac{{2\pi k}}{{\ln \left[ {z/{r_0} + \sqrt {{{\left( {z/{r_0}} \right)}^2} - 1} } \right]}}({T_0} - {T_s}) \qquad \qquad(18)$

which can be rewritten as $q' = \frac{{2\pi k}}{{{{\cosh }^{ - 1}}\left( {z/{r_0}} \right)}}({T_0} - {T_s}) \qquad \qquad(19)$

For a cylinder with length of $L (L \gg 2{r_0})$, the total heat transfer rate can be expressed as $Q = q'L = kS\Delta T \qquad \qquad(20)$

where $S = \frac{{2\pi L}}{{\ln \left[ {z/{r_0} + \sqrt {{{\left( {z/{r_0}} \right)}^2} - 1} } \right]}} = \frac{{2\pi L}}{{{{\cosh }^{ - 1}}\left( {z/{r_0}} \right)}} \qquad \qquad(21)$

is referred to as the shape factor – a parameter that is related to the geometric configuration of the problem only. Since the isotherms for the above problem are circles at different centers, heat conduction in a cylinder with an eccentric circular bore (see Fig. 3) can be analyzed using eq. (18). In this methodology, the inner and outer cylinders can be viewed as two isotherms caused by a line heat source at (0,a) and a heat sink with the same intensity at (0,a). Under steady-state, the heat transfer rate across both the isotherms T1 and T2 must be identical, i.e., $q' = \frac{{({T_1} - {T_s})}}{{[1/(2\pi k)]\ln \left[ {{z_1}/{r_1} + \sqrt {{{\left( {{z_1}/{r_1}} \right)}^2} - 1} } \right]}} = \frac{{({T_1} - {T_s})}}{{[1/(2\pi k)]{{\cosh }^{ - 1}}\left( {{z_1}/{r_1}} \right)}} \qquad \qquad(22)$ $q' = \frac{{({T_2} - {T_s})}}{{[1/(2\pi k)]\ln \left[ {{z_2}/{r_2} + \sqrt {{{\left( {{z_2}/{r_2}} \right)}^2} - 1} } \right]}} = \frac{{({T_2} - {T_s})}}{{[1/(2\pi k)]{{\cosh }^{ - 1}}\left( {{z_2}/{r_2}} \right)}} \qquad \qquad(23)$

Equations (22) and (23) can be combined to eliminate Ts to yield $q' = \frac{{2\pi k({T_1} - {T_2})}}{{\ln \left[ {\frac{{{z_1}/{r_1} + \sqrt {{{\left( {{z_1}/{r_1}} \right)}^2} - 1} }}{{{z_2}/{r_2} + \sqrt {{{\left( {{z_2}/{r_2}} \right)}^2} - 1} }}} \right]}} = \frac{{2\pi k({T_1} - {T_2})}}{{{{\cosh }^{ - 1}}\left( {{z_1}/{r_1}} \right) - {{\cosh }^{ - 1}}\left( {{z_2}/{r_2}} \right)}} \qquad \qquad(24)$

It is desirable to eliminate z1 and z2 from eq. (24) so that the heat transfer rate will be related to the size and locations of cylinders only. It follows from Fig. 3 that $z_1^2 - r_1^2 = z_2^2 - r_2^2 = {a^2}$ ${z_2} - {z_1} = \varepsilon$

where ε is the distance between the centers of the two cylinders and is referred to as the eccentricity. Solving for z1 and z2 from the above two equations yields ${z_1} = \frac{{r_2^2 - r_1^2 - {\varepsilon ^2}}}{{2\varepsilon }} \qquad \qquad(25)$ ${z_2} = \frac{{r_2^2 - r_1^2 + {\varepsilon ^2}}}{{2\varepsilon }} \qquad \qquad(26)$

Substituting eqs. (25) and (26) into eq. (24), one obtains $q' = \frac{{2\pi k({T_1} - {T_2})}}{{{{\cosh }^{ - 1}}\left[ {(8r_1^2 + 8r_2^2 - {\varepsilon ^2})/(4{r_1}{r_2})} \right]}} \qquad \qquad(27)$

If the length of the cylinder and cylindrical bore is $L (L \gg 2{r_0})$, the total heat transfer rate can be obtained from eq. (20) with ΔT = T1T2 and the shape factor of $S = \frac{{2\pi L}}{{{{\cosh }^{ - 1}}\left[ {(8r_1^2 + 8r_2^2 - {\varepsilon ^2})/(4{r_1}{r_2})} \right]}} \qquad \qquad(28)$

Heat transfer rate for many multidimensional conduction problems can be calculated using eq. (21) and appropriate shape factors. The shape factors for selected geometric configurations can be found in Incropera et al. (2006). More complete results for shape factors can be found in heat transfer handbooks, e.g., Rohsenow et al. (1998).

## References

Eckert, E.R.G., and Drake, R.M., 1987, Analysis of Heat and Mass Transfer, Hemisphere, Washington, DC.

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

Rohsenow, W.W, Hartnett, J.P., and Cho, Y.I., 1998, Handbook of Heat Transfer, 3rd ed., McGraw-Hill, New York.