Ocean water as a heat engine

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 4.4.

A different kind of proposed power plant uses the thermal gradient of the ocean drive a heat engine. If the temperature of the water at the surface of the ocean in some (presumably fairly tropical; it never reaches that temperature here in Scotland) location is {T_{h}=22^{\circ}\mbox{ C}=295\mbox{ K}} and at the bottom is {T_{c}=4^{\circ}\mbox{ C}=277\mbox{ K}}, then the maximum efficiency of an engine operating between these two reservoirs is

\displaystyle e_{max}=1-\frac{T_{c}}{T_{h}}=0.061 \ \ \ \ \ (1)

This is a pretty dire efficiency. If we want to produce 1 GW of electricity, so the plant is comparable to land-based power plants, how much surface water would we need to process to extract the required heat {Q_{h}}? We have

\displaystyle e_{max}=\frac{W}{Q_{h}} \ \ \ \ \ (2)

The amount of heat to be extracted per second is therefore

\displaystyle Q_{h}=\frac{10^{9}}{0.061}=1.64\times10^{10}\mbox{ J s}^{-1} \ \ \ \ \ (3)

The heat expelled to the cold reservoir is

\displaystyle Q_{c}=Q_{h}-W=1.54\times10^{10}\mbox{ J s}^{-1} \ \ \ \ \ (4)

To find the minimum volume of water that must be processed per second, we can try the following argument. [Disclaimer: I’m not 100% convinced this argument is correct, but it seems to make some sense. Comments welcome.]

The amount of water needed clearly depends on how much of a temperature drop the extraction of heat {Q_{h}} produces; the smaller the volume of water, the larger the temperature drop produced. The same is true (in reverse) when expelling heat {Q_{c}} into the cold layer; the smaller the volume of water, the larger the temperature increase.

Now, we can’t decrease the temperature of the upper layer to a final temperature {T_{f}} that is lower than the final temperature of the cold layer, so the maximum temperature drop of the upper layer is to a value of {T_{f}} that is equal to the final temperature of the lower layer after {Q_{c}} is dumped into it. If we take the masses (I’ll use mass instead of volume, since heat capacities are expressed in terms of mass) of water processed to be the same in both layers, then we have

\displaystyle T_{h}-T_{f} \displaystyle = \displaystyle \frac{Q_{h}}{mc}\ \ \ \ \ (5)
\displaystyle T_{f}-T_{c} \displaystyle = \displaystyle \frac{Q_{c}}{mc} \ \ \ \ \ (6)

where {m} is the mass of water processed in each layer and {c} is the specific heat capacity. We can add these two equations to get

\displaystyle T_{h}-T_{c} \displaystyle = \displaystyle \frac{Q_{h}+Q_{c}}{mc}\ \ \ \ \ (7)
\displaystyle m \displaystyle = \displaystyle \frac{Q_{h}+Q_{c}}{c\left(T_{h}-T_{c}\right)} \ \ \ \ \ (8)

Plugging in the numbers we get

\displaystyle m=\frac{\left(1.64+1.54\right)\times10^{10}}{4181\times18}=4.23\times10^{5}\mbox{ kg} \ \ \ \ \ (9)

This corresponds to a volume (since {1\mbox{ m}^{3}} of water has a mass of 1000 kg) of

\displaystyle V=423\mbox{ m}^{3} \ \ \ \ \ (10)

The total volume of water processed (hot + cold) is twice this, or

\displaystyle V_{total}=846\mbox{ m}^{3} \ \ \ \ \ (11)

One thought on “Ocean water as a heat engine

  1. dileep

    hello…i like this idea. the same effect can be viewed in the earth crusts. temperature in the earth’s crusts is far more times greater than the temperature at the surface of the earth. but the situation is the temperature gradient of the working fluid here in between the source(earth’s crust) and sink(earth surface) is very low. but in case of thermal power plants it’s value is very high. So i think these ways of producing energy are useful but not efficient.

    Reply

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