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Enthalpy seems to be commonly used to characterize geothermal reservoirs, but how is it calculated? I have an example here where the enthalpy is cited as 800 $kJ kg^{-1}$ for a reservoir with a temperature of 190 °C. In that case the enthalpy was apparently just taken to be the specific heat of water (4200 $J kg^{-1} °C^{-1}$) multiplied with the temperature, something like $H = c\Delta T$ (meaning that the reference temperature must be 0 °C).

Is that the correct way to do it? And if yes, is this really the enthalpy in the thermodynamical sense and why is the reference temperature 0 °C?

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    $\begingroup$ They made the assumption of the reference temperature perhaps due to the location of where they were working or what they were trying to achieve. It may or may not suit you, in which case you may need to adjust the reference according to your assumptions. $\endgroup$ – Solar Mike Jun 11 '17 at 18:19
  • $\begingroup$ @SolarMike This was from a textbook under "Classifications of geothermal systems", so it sounded rather general. Could you elaborate "due to [...] what they were trying to achieve"? What could that be? $\endgroup$ – ye-ti-800 Jun 26 '17 at 13:34
  • $\begingroup$ Hi @ye-ti-800 is this still open? Did you find an answer (if yes it would be great if you would post it)? I agree that it is not obvious why one would use the specific heat of water to calculate the enthalpy, on the other hand: if all you can use to extract the heat would be water, then this is the heat capacity of your process fluid. $\endgroup$ – rul30 Jan 14 '18 at 10:41
  • $\begingroup$ @rul30 Hi, yes it is still open and I don't have a really good answer to it yet. To clarify: my main question is not, why they use the specific heat of water, but rather why the reference temperature is 0 °C. My best guess is that this is some kind of maximum possible value, a potential enthalpy so to say. But I never saw this being defined anywhere $\endgroup$ – ye-ti-800 Jan 15 '18 at 15:39
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Nitpicky Answer: you can't.

More Detailed Answer: While enthalpy is commonly treated as an absolute quantity it is a thermodynamic potential. This is why textbooks might show the definition as \1\:

${\displaystyle \mathrm{d}H=\mathrm{d}U+\mathrm{d}\left(pV\right)}$

This means (essentially) you need a reference-point and all values for "enthalpy" are the differences in thermodynamic potential with respect to this reference point.

Let's illustrate this with a thought experiment. Imagine a reservoir (artificial lake behind a dam) and let's pretend it contains 1000l. The question how much energy can be produced by a pelton-turbine can only be answered when you add information about the height (In this case its even called potential energy).

In order to quantify the thermodynamic potential of a geothermal resource it is necessary to define a meaningful reference point and this reference point needs to be disclosed otherwise the numbers are not meaningful.

\1\ https://itp.uni-frankfurt.de/~valenti/

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  • $\begingroup$ Thanks for the answer. About your last paragraph: So, in the example given in the question the reference point is 0°C, right? And the reason for not explicitly stating that might be... that this is assumably some kind of default value in the field? $\endgroup$ – ye-ti-800 Jan 16 '18 at 13:41
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    $\begingroup$ You are right, one possible explanation might be some kind of default. But it could also be due to the fact that a lot of tabulated data for water has 0degC as a reference point. The proper way would be to define a representative reference, e.g. average ambient temperature. $\endgroup$ – rul30 Jan 16 '18 at 14:58

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