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Let's say I have ground water at a depth of $3000m$ at $50^{o}C$.

If I now pump this water up to the surface, the water pressure decreases from the $\pm 300 atm$ at that depth to $1 atm$ at the surface.

Silly enough I thought of using the ideal gas law, but as it's name implies it's only valid for ideal gasses, of which a basically incompressible liquid is not one.

If I use the ideal gas law for a isovolumetric process where V is constant, I get:

$T_1=T_2\frac{P_1}{P_2}$

which yields a rediculously small value for $T_1$ which cannot be right.

How would I go about calculating the change in water temp as a function of depth?

I am sure it won't be massive, but I think in the calculation I need it for, where I have a heat engine that already has a Carnot efficiency of $9\%$ if I assume $\Delta T$ is negligible I think that because of the great depth involved the temperature might lower sufficiently as it is pumped up that this efficiency might be even lower.

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    $\begingroup$ You need to get yourself a table with the thermodynamic properties of water under pressure and use those values to calculate the total internal energy. The internal energy won't change if you take the pressure off without extracting work, so you can then apply that energy to water under normal pressure to find the final temperature. www.clarkson.edu/.../Thermodynamic_tables_SI.pdf may have what you need. $\endgroup$ – CuriousOne Jul 5 '15 at 20:49
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As you calculated, but seem to dismiss, the temperature change in water pumped up from deep, high-pressure zones to the surface depends very little on the change in pressure. Other factors would have far more effect:

  • Pumping energy loss into the water
  • Conduction from the pipe and its surroundings as the water rises
  • Friction as the water moves through the pipe.

If you're thinking to run a heat engine from the difference between the raised water and the local environment, you'll need to take these factors into account when calculating the efficiency.

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