As we drill down into the rocks of Earth\'s crust, the temperature typically inc

Question

As we drill down into the rocks of Earth's crust, the temperature typically increases by 3.00 C for every 100 m of depth. Oil wells are commonly drilled to depths of 1830 m. If water is pumped into the shaft of the well, it will be heated by the hot rock at the bottom and the resulting heated steam can be used as a heat engine. Assume that the surface temperature is 20.0 C

a) Using such a 1830-m well as a heat engine, what is the maximum efficiency possible? (Express your answer to three significant figures.)

b) If a combination of such wells is to produce a 2.60-MW power plant, how much energy will it absorb from the interior of Earth each day? (Express your answer to three significant figures.)

Explanation / Answer

Given

Surface temperature Ti = 20.0oC = 293.15 K

Temperature gradient dT/dy = 3oC / 100 m = 276.15 K / 100 m

Depth of the well Y = 1830 m

Output power P = 2.60 MW = 2.60 x 106 W

Output duration t = 1 day = 1 x 24 x 60 x 60 s = 86400 s

Solution

A)

Temperature at the bottom of the well

T o = Ti (Y x dT/dy)

To = 293.15 + (1830 x 276.15/100)

To = 5346.695 K

Efficiency = 1- Ti / To

= 1 – (293.15/5346.695)

= 0.945 or 94.5%

B)

Output power E = P x t

E = 2.60 x 106 x 86400

E = 224640 x106 J

This is 94.5% of the produced energy the absorbed energy by earth is 100 – 94.5 = 5.5%

so energy absorbed Eabsorb = 5.5 x 224640 x 106 / 94.5

Eabsorb = 13074.2 x 106 J

Eabsorb = 1.3 x 1010 J

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