# Thermo: how should enthalpy change during cooling/heating?

1. Jan 28, 2009

### bumblebee77

Hi...I would very much appreciate thermo help. I am using software that requires enthalpy data to be in units of J/m^3 (i.e., enthalpy per unit volume). The software also requires that enthalpy always increases with increasing temperature.

My problem is that my enthalpy data in J/m3 do not always increase with increasing temperature. The data are output from calculations of multiphase, multicomponent cooling/crystallization. Specifically, the system (liquid + solids produced) AND the liquid AND the solids enthalpies vary as T decreases.

I am embarrassed to ask, but should 1) system and 2) liquid or solids enthalpies in units of J/unit volume stay constant, or only decrease, or decrease and increase during multiphase, multicomponent cooling/crystallization? What about in a one-component system? Would this be different?

My system, solids, and liquid enthalpy data do increase AND decrease as T decreases (i.e., during crystallization). I need to know whether they should. Then I can figure out how to deal with getting my enthalpy data into the software that requires enthalpy only decrease with decreasing T.

Thank you very, very much for any help!

2. Jan 28, 2009

### Mapes

Hi bumblebee77, welcome to PF. It sounds like the software cannot handle multiphase systems. It's not at all unusual for the total enthalpy to decrease or increase during a phase change; in fact, it has to decrease for one of the phase change directions (unless the system is at equilibrium at constant entropy and pressure, in which case the molar enthalpy of each phase is equal).

For a single-phase system, $\left(\frac{\partial H}{\partial T}\right)_P$ must be positive. It wouldn't break any laws of thermodynamics for $\left(\frac{\partial H}{\partial T}\right)_V<0$, but it would be extremely notable; I don't know of any actual examples.

Last edited: Jan 28, 2009
3. Jan 28, 2009

### bumblebee77

Hi Mapes:

Thank you very much for your quick and patient reply! I was afraid that the multiphase might be a problem....

If you have time, would you mind clarifying something? This might be a really basic question/concept, so feel free to direct me to a good text if it's too lame (I'm not a thermo person!)...

The thermo results are based on isobaric calculations. The volume is changing because P is constant, right? And it's this change in volume of all the different phases as they stabilize at different T that is the reason for increase as well as decrease in enthalpy per unit volume, correct? I am just trying to understand the reason that enthalpy can increase and decrease when the enthalpy units are heat per unit volume rather than specific H.

For example, I don't understand why the enthalpy per unit volume of the LIQUID phase, which is a SINGLE phase, can increase and decrease with decreasing T. I understand this (what you said): it has to decrease for one of the phase change directions (unless the system is at equilibrium at constant entropy and pressure, in which case the molar enthalpy of each phase is equal), but the liquid, for example, is a single phase, why can its enthalpy per unit volume increase AND decrease under isobaric conditions with decreasing T?

4. Jan 29, 2009

### Mapes

OK, for the volumetric enthalpy we're asking whether

$$\left(\frac{\partial \rho H}{\partial T}\right)_P=\rho \left(\frac{\partial H}{\partial T}\right)_P+H\left(\frac{\partial \rho }{\partial T}\right)_P$$

(where H is the specific enthalpy) can be less than zero for a liquid (or any single phase). The first term must be positive, but the second term can be positive or negative. So at first glance, there doesn't seem to be anything restricting the volumetric enthalpy from having a negative temperature coefficient, but it's something I'd like to think about more. Are you observing negative values experimentally?

5. Jan 29, 2009

### bumblebee77

Hi Mapes:

Thanks again! Please don't hesitate to let me know when you're out of time for dealing with this!

Yes I am observing negative values experimentally (i.e., in my calculation results). To clarify, when I run a multiphase crystallization calculation, I get H and other physical parameters for each solid phase, plus the average H of solids, the liquid phase, and any supercritical fluid exsolved. For each phase, H [J/m^3] increases and decreases with decreasing T. Is this what you want to know by am I seeing negative experimental values? If I can figure out how to attach stuff to a post here, I can send you an example of the data/output.

My goal is to use the H vs T (and density, viscosity, etc, output of the thermo modeling) to run flow-type calculations.

Many thanks.

6. Jan 29, 2009

### Mapes

The most likely conclusion is that something is wrong with the measurements, calculations, and/or interpretation. Here's another way of looking at things: if volumetric enthalpy is increasing during a temperature decrease for a single phase, it implies that one could transfer thermal energy into a certain volume of a certain material yet see the temperature drop. I've never heard of such a material.

EDIT: On the other hand, the volumetric condition may be screwing with my intuition. Would be interesting to look at liquid water, for example, which is very well characterized and has a density maximum.

Last edited: Jan 29, 2009
7. Jan 29, 2009

### bumblebee77

Hi Mapes:

If volumetric enthalpy is increasing during a T decrease for a single phase, then could it be that the volume is changing (in an isobaric calculation)?

I have attached a sheet that shows D enthalpy output from the thermo multiphase calculations. Column A shows T, column B shows system enthalpy in J. Column G shows system enthalpy in J/m^3. Column S shows liquid H in J/m^3.

At the bottom, there are a couple of plots. On the left is system volumetric enthalpy and on the right is liquid volumetric enthalpy (y axes versus T on x axis). Both increase and decrease as T decreases.

#### Attached Files:

• ###### H_sheet.xls
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Last edited: Jan 29, 2009
8. Jan 29, 2009

### Mapes

Perhaps this is the case I outlined above, where the second term is negative and larger than the first term.

9. Jan 29, 2009

### bumblebee77

Thanks Mapes. It's such a nice thing you're doing, to spend so much time helping people with this stuff. You've really helped a lot.