Calculation of temperature drop after a pressure release

1. Jan 7, 2009

lari

I have a problem when calculating the temperature drop after a pressure release in a water sample. I use the following expression for the adiabatic temperature change:
(delta T/delta P)= alpha*T*V/Cp
where alpha is the thermal expansion coefficient, T is the temperature in Kelvin, V is the specific volume and Cp is the specific heat. I have appropriate equations to calculate all these parameters (all dependent on both pressure and temperature).
I calculate the temperature drop numerically integrating this equation with 0.1 MPa pressure increments. Nevertheless, for expansions from 100 MPa and -10ºC, I obtain surprising results: a little temperature increase (sample temperature after expansion is about -9.8ºC). That is due to the negative sign of the thermal expansion coefficient at low pressures and temperatures. However, in practice, I record a temperature drop in the sample (about -10.3ºC). Is the equation employed valid for negative thermal expansion coefficients??? How can I explain this behaviour??? Thanks in advance.

2. Jan 7, 2009

Astronuc

Staff Emeritus
Are you sure the equation is valid for what one is considering and the conditions.

"for expansions from 100 MPa and -10ºC," to what pressure is one reducing? This is basically ice at high pressure ~1000 atm (1 atm = 0.101325 MPa). That's very high pressure.

By reducing the pressure, one is unloading the ice. I'm not sure the temperature should change just by reducing the pressure (basically decompressing ice), although one would release the stored mechanical energy, and the amplitude of atomic vibrations would increase by very little.

Also consider when it is appropriate to use CV and CP.

3. Jan 8, 2009

lari

Sorry! Pressure release is to atmospheric conditions, that is, 0.1 MPa. I forgot to say it in my previous message. Sample initially is at liquid state. I proved several conditions: 100 MPa/-10ºC, 200 MPa/-10ºC, 200 MPa/-20ºC and 450 MPa/-10ºC. All these conditions correspond to liquid state according to the phase diagram of pure water, except 100 MPa/-10ºC. But, these conditions remain close to the melting curve of water and, taking into account that a considerable supercooling is needed under pressure to initiate nucleation, I have experimentally observed that water remained in liquid state at these conditions. Also I have experimentally observed a temperature drop in the samples after the pressure release. My problem is that the above mentioned equation yields a positive temperature increment after the pressure release. That is due to the fact that the thermal expansion coefficient has negative values at relatively "low pressures", that is 50 MPa and lower, and low temperatures. Nevertheless, experimentally, I always recorded a temperature drop in the sample after expansion. Thanks again.

4. Jan 8, 2009

scienture

'All these conditions correspond to liquid state according to the phase diagram of pure water, except 100 MPa/-10ºC.'
I agree with lari on this point. Let's try a simpler thought to prove this. Since the density of ice is lower then that of water, the volume of ice is always larger than that of water in same mass.So ice will melt under great presure.
There, I have some suggestions for lari.I hope those may help you.
First,what I think the most possible one, is that maybe your conditions of experiment are not so ideal. I mean perhaps your didn't consider some effective factors such as the outside environment may had heat exchange with your water sample, etc.
Then,maybe your equations are not correct or have bigger errors in some conditions such as high presure.