9 replies to this topic

[dinesh]

hii i m a chemical engineering student
for my project i would like to ask that
How to calculate a barometric leg when the system is under vacuum?

[djack77494]

Hi Dinesh,

When you ask "how to calculate barometric leg", I assume you are asking how do you calculate the height of the barometric leg. Consider a vacuum system where you are attempting to generate and maintain nearly a complete vacuum (in Imperial Units, 1 atmosphere = 14.7 psia; complete vacuum = -14.7 psig.) Typically, you would use steam jets or eductors or ejectors (I've heard them called all three) to generate your vacuum. You'd be condensing the steam and condensible vapors in heat exchange equipment using either direct contact (Barometric Condensers) or indirect contact (Surface Condensers). With either, you produce liquid water, perhaps with other condensibles, at the pressure at your jet's discharge nozzle. For simplicity's sake, let's just say it could be as low as the system pressure or -14.7 psig.

After the Condenser, you normally want to drop down into a Receiver (horizontal vessel) which is often vented to atmosphere or to a low pressure vent system. So you have condensed water in the Condenser at -14.7 psig and you have the destination (Receiver) at zero psig. The pressure difference goes the WRONG way. How are you going to get water at lower pressure to flow into the Receiver at higher pressure?

The answer is very simple. Just locate the Condenser a suitable distance above the Receiver, and the water will flow downhill against the pressure difference. The calculation of what constitutes a "suitable distance" is what you call "calculate barometric leg", since the piping between the Condenser and the Receiver is called the barometric leg.

The difference in elevation between the Condenser and the Receiver must be such that the static head of the water in the barometric leg exceeds the pressure difference. Since water at standard conditions (60F or 20C) exerts a force of about 2.31 psi per foot, you would need an elevation difference of 14.7*2.31 = nearly 34 ft (10 m) to overcome the pressure differential. Divide this number by the specific gravity of the condensate at your expected temperature to arrive at a more accurate estimate. In designing your system, I would add a couple of feet to allow for some hydraulic loss in the barometric leg and the vent pipe. Be sure that the barometric leg extends into the Receiver and is submerged enough that atmospheric air/vent gases cannot be "pulled" in the piping. This forms your very necessary seal. If you vent into a system under any pressure, the situation is more complicated as your pressure differential will increase.

Hope that helps,
Doug

[psia = pounds per square inch absolute]
[psig = pounds per square inch gauge]
Imperial to SI Unit conversions if needed --> That's up to you!

[dinesh]

dear doug,
thnks for replying me and giving me such guidance,
for me, can u suggest any refrence or any side for vacuum related problems

DINESH

[djack77494]

Dinesh,

You might try visiting the websites of some of the major vendors of such equipment. I would try Graham and Croll Reynolds. For more (and I think there are quite a few) do a google search. Also, don't forget to check your standard handbooks. As I recall, Perry's had a section on steam jet vacuum production.

Good Luck,
Doug

[yergan]

I was under the impression that 11 metres is sufficient to obtain a vacuum for most flows.

[djack77494]

Yes, 11 meters vertical height of condensate should be adequate. Just remember that the specific gravity of the condensate is less than one (since temperature is elevated). Also, watch out for any backpressure in the venting system.

[Art Montemayor]

Yergan:

Doug's math is absolutely correct when you read through his basis and assumptions. First and foremost, Doug is indicating that the definition of a vacuum is the elimination of 14.7 lb/in² of atmospheric pressure. He explicitly states: "(in Imperial Units, 1 atmosphere = 14.7 psia; complete vacuum = -14.7 psig.)" Therefore, when you proceed on to convert the equivalent height of a water column at 60 ^oF, you should do the following math:

(14.7 lb/in²) (144 in²) (ft²/62.31 lb) = 33.972 ft

Your impression of 11 meters (36.0892 ft) of height being the equivalent height is related to taking the density of the water as 58.65 lb/ft³ - which is for much hotter water. In fact this height, when related to water at 14.7 psia, does not make any sense. The density of saturated water at 14.7 psia is 59.829 lb/ft³ - which is heavier than your "required" density. That means your 11 meters cannot be correct for water at 14.7 psia. Perhaps you are assuming another value for the atmospheric pressure.

[dinesh]

Dinesh,

You might try visiting the websites of some of the major vendors of such equipment. I would try Graham and Croll Reynolds. For more (and I think there are quite a few) do a google search. Also, don't forget to check your standard handbooks. As I recall, Perry's had a section on steam jet vacuum production.

Good Luck,
Doug

hi Dough
thnks for giving me relply but im quite satisfy abt calculatig baro metric leg.
here u mention for steam ejector. but i wanna know abt other. i
give u the discription as

-system under vaccum(1.3 mbarA)
- having Eg (ethylene glycol ) ejector.
now say , how to deside the lenth of baromertic leg.

regards,
DINESH
---------------------------------

[djack77494]

Dinesh,

You have moved well beyond my area of expertise. If I understand your question correctly, you are refering to a liquid motivated ejector. I would think that the same principles apply and that you would calculate the barometric leg requirement as before, BUT taking into account the fluid's density. The number Art offered, 62.31, is the density of water in pounds mass per cubic foot. Substitute the density of your fluid for an estimate of the height required. Beyond that suggestion, I have no experience with your system.
Doug

[tomerbu]

Thanks , i singed up to this site just to say this: the explanation was clear thank you!