6 replies to this topic

[niallmacdowell]

Hi all,

I have a question regarding the prediction of a boiling point of an oil at varying pressures.

I have a natural/essential oil whose thermodynamic properties I know at 1atm. The properties I have are density, viscosity, specific heat capacity, latent heat of vapourisation and its vapour pressure.

What I want to do is to predict the boiling point at 1.5atm. I would imagine that there should be some expression to predict this, but Im having difficulty in finding it.

Note: I do not know the critical temperature or critical pressure of this substance (it is D-Limonene, is that helps anyone)

If anyone knows a good method of predicting the boiling point at a different temperature, that is suitable for use by hand, please let me know.

All help/input appreciated.

Niall

[gvdlans]

Naill,

Following link may be useful: http://age-web.age.u...84(7-8)Yuan.pdf

Basically, you need an equation that relates vapour pressure with temperature (= boiling curve or vapour pressure curve). Example equations are given as eq. 1 - 3 in the linked document. For example the Clapeyron equation:

Log Pv = A - B/T

Note that at T = Tboil, Pv = system Pressure or in words, a liquid will boil if the vapour pressure is equal to the system pressure. For example, for water the vapour pressure at 100 °C is 1 atm, since the normal boiling point of water is 100 °C.

For the Clapeyron equation, you need at least two vapour pressure/temperature combination in order to estimate the parameters A and B. You can then estimate the vapour pressure at a different temperature or alternatively estimate the boiling point at a different pressure. You will find better results when you have more vapour pressure/temperature combinations and/or when you use a more sophisticated equation (e.g. Antoine's Equation).

According to your posting you have only one vapour pressure/temperature combination (vapour pressure at 20°C?). Or do you also know the boiling point at 1 atm?

[mbeychok]

Niall:

If your oil is a single-component oil (as differentiated from multi-component petroleum oils), then you may be able to find the Antoine equation coefficients for your oil. To learn more about the Antoine equation coefficients and where to find lists of them, read http://antoine.frost...chem/senese/101 and search for "Antoine equation".

You might also find the Antoine coefficients for your oil in the NIST web site at
http://webbook.nist.gov/chemistry/

Hope this helps.

[niallmacdowell]

gvdlans,

Thank you for the link, I have previously looked at that paper, but ran into the problem of not having the requisite constants for the expressions .

I should have mentioned that I do have the boiling point of this material for P=1atm, it is 449.15K.

I also know that this is an natural/organic oil with a molecular weight of 136.23. Do you think that I could go to a book like Coulson & Richardson Vol 6 and take an average of the constant values for organic molecules with a similar molecular weight, and still get a reasonably accurate answer?

Do you have any opther suggestions as to how I should proceed?

Thanks for your help,

Niall

[niallmacdowell]

Guys

Thanks for the info. From gvdlans paper, I take an expression log Pv = A - B/T. From mbeychoks links I have found some different physical properties, namely vapour pressure: 400 Pa at 287.55K. So now I have two sets of data points, Pv1:266pa, T1: 293.15 and Pv2: 400Pa, T2: 287.55.

Substituting these data into the above equation and rearranging I get two expressions of the form:

(1) B = 287.55A - 748.22235
(2) B = 293.15A - 710.85405

(2) - (1) : 0 = 5.6A + 37.368299 => A = -6.67291 and thus B = -2667.01776.

As the pressure at which I want to find my boiling point is 2BarA = 200,000Pa I now have an expression of the form:

log (200,000) = -6.67291 + 2667.01776/T where T should be my boiling point.

Solving gives me: 6.67291 + log(200,000) = 2667.01776/Tb = 11.97394

Thus Tb = 2667.01776/11.97394 = 222.73K which is much less than the boiling point at 1 atm of 449.15K.

I wonder should I be using the clausius clapeyron expressions, i.e. ln(P2/P1) = -DH/R (1/T2 - 1/T1) as has been suggested elsewhere. I would have thought that this expression was for phase change? Or does it apply anyway as boiling suggests a phase change?

All input appreciated.

Niall

[gvdlans]

I have a couple of remarks on your last post:

1) Boiling is clearly a phase change, since phase changes from liquid to vapour...

2) Vapour pressure should increase with temperature. Similarly, boiling temperature should increase with pressure...

3) The two sets are at almost the same temperature (293 vs. 288 K). This will not give good results when you are trying to fit a curve (estimate parameters in an expression).

4) I expect better results if you use the 3rd set you have: vapour pressure is 1 atm at 449.5 K. You would still have to extrapolate if you want to find the boiling point at 2 bara... So the result will only be an approximate value.

[niallmacdowell]

gvdlands

thanks for the input;

In response to your points,

1) Yes, that makes sense. Im not sure why I asked that question...wood for trees, maybe.....

2)Thats what I thought also, the data point I got from www2.siri.org seemed to imply that at 288 the vapour pressure was greater than that at 293...I have my doubts about this to say the least.

3)I thought that it was a good idea to use values that are close to eachother, you are saying not so if Im trying to estimate the constants?

4) I have used this third data point with both the Clapeyron eqn and the Clausius-Clapeyron equation and got approximately the same results, (478.93K and 477.25K respecively) so Im reasonably happy that I have a decent answer. Incidentally these results are also very close to what I got via approximation of the Antoine constants as I outlined in one of my earlier posts, the answer I got here was ~479.2K

Thanks for all the help,

Regards

Niall