11 replies to this topic

[GTE]

Good Morning all,

I have a question regarding the application of henry's law and henry constant.

in the graph below you will see that theoretically at 100°C, the oxgen content of water is 0.

However, when I try to apply the formulea, I do not find zero because exp (-x ) is only zero for x = infinity.

So how can you find a zero oxygen content. so my question is. where is my mistake in reasoning.

 

Many thanks in anticipation,

Regards,

GTE

 

 

 

                   

                   

Attached Files

•  oxygen content.gif   74.31KB   3 downloads

Edited by GTE, 12 August 2013 - 09:47 AM.

[Steve Hall]

Empiracal equations are not physical laws, they have limits in accuracy and range of application. Pretty clearly, if water is brought to 100C at atmospheric pressure (assumed unless stated otherwise), it will be boiling and its very unlikely that any general correlation will correctly predict the oxygen concentration. Your mistake is taking the correlation too literally.

[GTE]

Hi Steve,

Thanks. I am looking at this for a presurised deaerator which will be set at 0.2 bar (g). I think T vap sat is 105°C. Here I am just confused by the fact that the graph clearly shows the oxygen content at 0 ppm when the equation shows the contrary.

do you know of a better equation?

Regards,

thanks again,

GTE

[MrShorty]

I'm not sure I am quite following your question.

 

One note of discrepancy: your plot is showing solubility (xO2) as a function of temperature (at an assumed total pressure of 1 atm, since the pressure is not specified in the document). Your equation is computing the henrys constant (kH). Can I assume you are using the simple expression kH=PO2/xO2 (where PO2 is the partial pressure of O2) for kH? Note that, while kH and xO2 are related, they are not the same quantity.

 

Perhaps if I better understood the steps you are taking in getting this non-zero result.

 

Reading between the lines -- are you expecting kH to be 0? a kH of 0 implies infinite solubility -- an impossible feat for something like O2 in H2O.

In many ways, kH is meaningless at the boiling point of the solvent (at the specified pressure). At the boiling point, the Ptot=PH2O meaning PO2=0. If PO2=0, xO2=0 independent of kH.

[thorium90]

Will this website help in your query?

 

http://www.spiraxsar...-deaerators.asp

[GTE]

Mr Shorty,

"At the boiling point, the Ptot=PH2O meaning PO2=0. If PO2=0, xO2=0 independent of kH."

So that equation of van't hoff is misleading.

how come kH is independant of xO2 at boiling temperature? kH=PO2/xO2

 

PO2 equals 0. I guess is because it is saturated with water vapour?!!

Please help me a bit more. I would like to plot that curve above on excel with equations.

Regards,

GTE

[MrShorty]

So that equation of van't hoff is misleading.

No, I don't think so. van't Hoff's equations tell us the relationship between kH and temperature. They do not (directly) tell us the relationship between xO2 and temperature at a specified pressure. I expect the trouble you are having is making the right connection between kH and xO2. Can you describe the steps you are following to get xO2 from kH? If we understand exactly how you are trying to get xO2 from kH, we should be able to help you understand where your calculation is going wrong.

[GTE]

thanks,

how I calculate xO2 (ppm)

 

 

XO2= (Partial pressure of O2, ((which is appx 0.21 at 1 atm)))/ kH

where kH is taken from the vant' hoff formulae.

in the van't hoff formulea, I use the relevant constant C(K)=1700, Temperature =100+273 (K) at the relevant standard kH(at standard temperature).

 

But I can see that you are telling me that Partial pressure figure is wrong?

thanks for time btw

GTE

[MrShorty]

If you are assuming the partial pressure of O2 is 0.21 atm at all temperatures (which I assume means we are talking about air=21%O2+79%N2, not just O2), then, yes, that is going to be incorrect. At low temperatures assuming PO2+PN2=Ptot=1atm is a good assumption, because the vapor pressure of water is so small near room temperature. But, as you approach the boiling point of water, this assumption becomes less valid, and it becomes necessary to factor in PH2O as well. In a generic sense, here's how I see this kind of calculation being done:

 

1) Dalton's law: Ptot=PH2O+PO2+PN2=1 atm (or 1.2 atm or whatever your specified total pressure is)

2) Using Raoult's law, calculate PH2O: PH2O=P0H2O*xH2O where P0H2O is the vapor pressure of water at T. As a first approximation, assume xH2O is 1 (which should be very nearly true) and compute PH2O.

3) From Dalton's law PO2+PN2=Ptot-PH2O; PO2=(Ptot-PH2O)*0.21; PN2=(Ptot-PH2O)*0.79. Now you will have a good value for PO2!

4) Calculate kHO2 (and maybe kHN2, if you want to be rigorous and include N2 in the computation) from van't Hoff equations.

5) Calculate xO2 (and xN2) from xO2=PO2/kHO2.

6) At this point, if desired, you can take xO2 and xN2, go back to step 2, calculate a new xH2O, and loop through 2-6 until the calculated x's stop changing.

[GTE]

I think I can see it now. I should spend more time on it...

 

Thanks (very much) for your time,

GTE.

thanks again

[GTE]

Mr Shorty,

I have changed my calcs, it does look much better, I did not do the iteration. how is xH20 changing? once I have xO2.

[MrShorty]

For this, I'm assuming x is mole fraction and that O2, N2, and H2O are the only species present.

 

xi=moles of i/total moles of solution -- basic definition of composition expressed as mole fraction.

With no other species present. xO2+xN2+xH2O=1

For the first guess, we assume xO2 and xN2 are 0, so xH2O=1

Once we have our first guess at xO2 and xN2, we can now compute a new value for xH2O=1-xO2-xN2. Because xO2 and xN2 are very small (graph indicates 1E-6 to 1e-5, though your graph doesn't specify if this is molarity, molality, mole fraction, or other), this will make xH2O slightly smaller than 1.