The following temperature measurements were taken down a borehole: At the locati

Question

The following temperature measurements were taken down a borehole:

At the location of the borehole, 3m of clay soil are present at surface and are underlain by 197m of interbedded shale and limestone. Beneath this lies a sandstone layer more than 300 m thick. The shale beds have a thermal conductivity of 2.5W m^-1 C^-1, the limestone beds have a thermal conductivity of 1.4W m^-1 C^-1, and the sandstone has a thermal conductivity of 3.0W m^-1 C^-1.

a)Calculate the geothermal gradient in the area where measurements were taken.

b)Calculate the heat flow in sandstone.

c)Suppose the sandstone were instead a basalt. Would the heat flow in this layer be higher or lower (assuming the same geothermal gradient)? What if it were a quartzite?

Explanation / Answer

Answer:

a) Thermal gradient of the area= temperature difference / depth difference

= (19-18.3) c /(410 – 370) m = 0.7/40 c/m = 0.0175 c.m-1 (Ans)

b) sandstone layer presents from 200 m to 500 m. thus the point 370 m & 410 m are in sandstone with thermal gradient = 0.0175 c.m-1

According to the equation,

Q/t = kA T/d where Q,t,k,A,d are heat amount, time, thermal conductivity, Area, depth

Q/(t.A) = k. (T/d) …….(i)

= 3 Wm-1 c-1 x 0.0175 c.m-1

= 0.0525 W.m-2

Here, (T/d) = thermal gradient and Q/(t.A)= heat flow

Hence, heat-flow of sandstone is 0.0525 W.m-2 (Ans)

c) Typical Quartzite, Sandstone, and Basalt have thermal conductivity of 3.5, 2.9, and 2 Wm-1 c-1 . this shows that basalt has less conductivity and quartzite has higher conductivity than sandstone.

Now, we can see from the equation (i), heat-flow directly proportional to conductivity. Thus,

If sandstone is instead of basalt, heat-flow will be less and high for the case of quartzite. (Ans)

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