# Heat conduction from buried object

### From Thermal-FluidsPedia

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 *T*_{0} and the surface temperature is uniformly *T*_{s}. 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

is still valid for this problem and the boundary conditions for this problem are

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 *y* − direction, 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, *T*_{s}, at *y* = 0.

The heat generated from the heat source located at (0, a) will keep the surface temperature of the cylinder equal to *T*_{0,1}. If the temperature at the point P caused by the heat source is *T*_{1}(*x*,*y*), the amount of heat transfer from the surface of the cylinder is

The temperature at point P caused by the heat source *alone* is then,

Similarly, the heat sink located at (0, − *a*) will maintain the surface temperature of the fictitious cylinder at *T*_{0,2} and result in the temperature at the point P equal to *T*_{2}(*x*,*y*), i.e.,

Consequently, the temperature at point P caused by the heat sink *alone* is,

If both heat source and sink are present, the temperature at point P is

Substituting eqs. (3) and (4) into eq. (5) yields

At the surface y = 0 where *r*_{1} = *r*_{2}, eq. (1) requires *T*_{0,1} + *T*_{0,2} = *T*_{s} and eq. (6) becomes

Defining excess temperature θ = *T* − *T*_{s}, we have

The radii *r*_{1} and *r*_{2} can be expressed as

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

which can be rewritten as

When the left hand side of eq. (11) is held at a constant C, the following equation for isotherms is obtained:

which can be rewritten as

It represents a series of circles with the center located at [0, − *a*(1 + *C*) / (1 − *C*)] and radius of ; 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 θ_{0} = *T*_{0} − *T*_{s} and the constant *C* is

The depth of the cylinder measured from the horizontal surface to the center of the cylinder is

which can be rearranged to

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 − *C*_{0}) / (1 + *C*_{0}) < 1.

The radius of the cylinder is related to the constant *C*_{0} by [see eq. (12)]

Solving *C*_{0} in term of *r*_{0} yields

Substituting eq. (13) into eq. (17) and solving for the heat transfer rate, one obtains

which can be rewritten as

For a cylinder with length of , the total heat transfer rate can be expressed as

where

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 *T*_{1} and *T*_{2} must be identical, i.e.,

Equations (22) and (23) can be combined to eliminate *T*_{s} to yield

It is desirable to eliminate *z*_{1} and *z*_{2} 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

where ε is the distance between the centers of the two cylinders and is referred to as the eccentricity. Solving for *z*_{1} and *z*_{2} from the above two equations yields

Substituting eqs. (25) and (26) into eq. (24), one obtains

If the length of the cylinder and cylindrical bore is , the total heat transfer rate can be obtained from eq. (20) with Δ*T* = *T*_{1} − *T*_{2} and the shape factor of

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*, 3^{rd} ed., McGraw-Hill, New York.