Assume that geothermal heat transfer (at least near the surface of the earth) is by conduction through the crust rocks. For an average geothermal heat flow of $0.087$ $\mathrm { W } / \mathrm { m } ^ { 2 }$ and typical thermal gradient of $100 ^ { \circ } \mathrm { C } / \mathrm { km }$, calculate the thermal conductivity of the rock. Compare with the known thermal conductivities of similar materials given in Chapter 8.

Question

Quizplus · Accepted Answer

The Answer is Equation (8.4) gives

P = \frac { k A \Delta T } { l }

where

k

is the thermal conductivity,

A

is the area,

\Delta T

is the temperature difference and

l

is the distance. Solving for

k

gives

k = \frac { P l } { A \Delta T }

So in this case

P / A = 0.087 \mathrm {~W} / \mathrm { m } ^ { 2 }

and

\Delta T / l = 100 { } ^ { \circ } \mathrm { C } / \mathrm { km }

or

0.001 { } ^ { \circ } \mathrm { C } / \mathrm { cm }

. The thermal conductivity is then

k = \frac { 0.087 \mathrm {~W} / \mathrm { m } ^ { 2 } } { \left( 0.001 ^ { \circ } \mathrm { C } / \mathrm { cm } \right) } = 87 \frac { \mathrm { W } \cdot \mathrm { cm } } { \mathrm { m } ^ { 2 } \cdot { } ^ { \circ } \mathrm { C } }

From Table 8.1, the values for concrete and brick are

170 \frac { \mathrm { W } \cdot \mathrm { cm } } { \mathrm { m } ^ { 2 } \cdot { } ^ { \circ } \mathrm { C } }

and

71 \frac { \mathrm { W } \cdot \mathrm { cm } } { \mathrm { m } ^ { 2 } \cdot { } ^ { \circ } \mathrm { C } }

, respectively, so the calculated value for rock is in the proper range.

Assume That Geothermal Heat Transfer (At Least near the Surface 0.0870.0870.087

Assume That Geothermal Heat Transfer (At Least near the Surface $0.087$