3 replies to this topic

[vinay]

Dear Friends,

I would like to know how to calculate the maximum vacuum possible in a tank when carrying out steam purging .

I understand that the volume of steam is quite large as compared to condensate & can calculate that based on difference in density of water at these two conditions, but how to translate it in to vacuum / pressure term .

Can we use gas law P1VI/T1 = P2V2/T2 even though water has condensed in the vessel ?

Regards,
vinay

[djack77494]

No, vinay. You cannot use the gas law P1VI/T1 = P2V2/T2 because the condensed water is not a gas and does not obey the gas law. As an approximation, its volume is zero (really a very small fraction of the steam's volume). If you know the final temperature of the tank, then the internal pressure is quickly & easily obtained by lookup in the Steam Tables (or the e-steam tables). Since that pressure is well below atmospheric, it is the pressure you seek. Nothing more to it.
Doug

[Art Montemayor]

vinay:

Your main question asks “how to calculate the maximum vacuum possible in a tank when carrying out steam purging”.

My response is based on the strict interpretation of your question. If you are purging with steam and you isolate the tank (for whatever reason) while it is filled 100% with saturated steam, the ultimate result will be that you will create a vacuum due to the fact that the captured steam will start to condense as it tries to meet atmospheric temperature conditions (assuming it is out and exposed to the atmosphere). If it happens to be raining on the un-insulated tank, this will happen quite rapidly – more rapidly and traumatically than you can imagine. If the atmospheric conditions are mild, with no wind effect, the condensation will be slow and so will the development of the condensation and subsequent partial vacuum. But the final result will be the same: tank mechanical failure – or collapse.

At this point, the degree or level of partial vacuum developed in the tank is rather academic and of little or no engineering importance since a conventional storage tank cannot withstand even a slight amount of vacuum – much less the partial vacuum levels developed by condensing steam. However, for the sake of pure academics and to prove that we know our physics and phase phenomena, we can discuss what would be the ultimate absolute pressure (partial vacuum) developed by the condensing steam.

As Doug has pointed out, the volume occupied by the condensed water is essentially “zero” (really a very small fraction of the steam's volume). Nevertheless, the volume occupied by the water condensate is irrelevant for the sake of calculating the ultimate partial vacuum possible. This ultimate vacuum, as Doug also notes, is in actuality the vapor pressure of the resultant steam condensate inside the tank – regardless of its volume or quantity inside the tank. The colder the steam condensate is, the lower will be its vapor pressure (the resulting partial vacuum).

If the resulting steam condensate temperature is 20 ^oC, the resulting tank pressure is 0.023393 bar;
If the resulting steam condensate temperature is 30 ^oC, the resulting tank pressure is 0.042470 bar.

For all practical purposes, you should design for Full Vacuum being developed.

[vinay]

Dear Art & Doug,

Thanks for your valuable explaination , We were designing the vessels (I mentioned it incorrectly as tanks) for full vacuum only, I however had a doubt & hence wanted to clarify .

It is efforts of veteran's like you who help us people like us learn so much. Thanks once again for your selfless sincere efforts for teaching young professionals like myself.

Regards,
Vinay