Vapor Composition in a Closed Tank

[RJB32482]

Hello,

I'm trying to figure out the vapor composition in our expansion tank. The solution is a 30% propylene glycol solution (by weight) and it is at a temperature of 110 degrees Celsius. The total pressure in the tank is 26 PSIG.

There is a 5 PSIG nitrogen blanket on the tank too.

My calculation method was to use the vapor pressure of water at 110 degrees C (20.78 PSI) and solve:

20.78/(26+14.7)*100% = 51.05 mol % water
19.7/(26+14.7)*100% = 48.4% nitrogen
100-51.05-48.40 = 0.55 mol% PG

Or is there a better calculation method to figure out the composition of vapor in the headspace of this tank.

Thanks in advance.

 

[TD2K]

How are you determining that the N2 pad is 5 psig?

I think you have something else contributing to the vapor pressure

http://www.dow.com/PublishedLiterature/dh_0047/0901b803800479d9.pdf

The vapor pressure of 30% PG at 110C is about 1000 mmHg abs, that's about 5 psig at sea level. Add your nitrogen (if 5 psig is right) and that should be a total of 10 psig. If you have 26 psig, there's something else you haven't considered.

 

[25362]

Is it possible that the tank with the PG solution was blanketed with 5 psig N[sub]2[/sub] and then heated to 110[sup]o[/sup]C?

 

[25362]

Or, may be the tank contained N[sub]2[/sub] at 5 psig and at some lower temperature and then a hot PG solution was pumped in? In short, could it be the N[sub]2[/sub] pressure increased due to gas heating?

 

[RJB32482]

Thanks for the replies. This is an expansion tank so I forgot to mention that some of the pressure increase is due to the volume change of liquid from 20C to 110C. The tank starts with the 5 PSIG pad at 20C, then the solution is heated to 110C.

Thanks.

 

[25362]

As suspected, the total pressure is the sum of the vapor pressure of the solution at 110[sup]o[/sup]C plus that of nitrogen.

If the solution approaches ideality Raoult's law becomes applicable.

 

[petrochemical87]

I am not sure I am correct but I will give this a try. Assuming rault's law is valid in this case for both glycol and water, their partial pressures at 110 deg C can be calculated as
p(glycol)=vapor pressure of glycol*mole fraction in solution
p(water)=vapor pressure of water*mole fraction in solution
now
p(nitrogen)=P(total)-p(water)-p(glycol)
By knowing all three partial pressures the mole fractions may be calculated.
p(glycol)=P(total)*y(glycol) and so on

 

[RJB32482]

Thanks everyone.

The reasoning behind this is that we had a PSV discharge line freeze that discharges into this tank. The PSV discharges from the tube side of a shell and tube heat exchanger back into the tank. We have a flow switch on this discharge line that has not alarmed and it is working so the PSV has not discharged.

The discharge line is not a dip tube that discharges in the liquid level of the tank (tank is around 30% full before heating the glycol). The thought was the vapor from the 30% glycol is mostly water, it condensed in the discharge line and froze. The line is insulated but not heat traced.

That is the reasoning why I wanted to know if the vapor was mostly water since 30% glycol freezes around 7 degrees F and water is 32 F.

Thanks again for the responses.