9 replies to this topic

[chemks2012]

Hi all,
There is an existing 1m3 storage tank containing DMS [dimethyl sulphate]. This tank has open overflow/vent line [2” line open in atmosphere]. As I understand this overflow/vent line serves two purposes i.e. normal vent for the vapour release during tank filling etc plus acts as a liquid overflow line. Am I correct? Please comment.

I am sure 2” line is sufficient for the liquid overflow i.e. fill rate is only 10m3/hr and therefore, 2” line is OK. However I am not sure how to verify this line for the vapour/gas? Can I use the Spitzglass Formula [please see attached file] for the Low Pressure? Could I use vapour pressure as differential pressure as the overflow line is open in atmosphere. Please let me know if my calculation is correct.

Thanks

Attached Files

•  overflow-vent line verification.xls   40KB   74 downloads

[latexman]

In addition, the vent line lets air into the tank when pumping liquid out of the tank, so excessive vacuum is not created. It also lets the tank breathe due to temperature fluctuations from day to night.

Your liquid calculations do not verify the vent line size, only the fluid velocity is calculated. And, your vapor calculations contain a fatal flaw - vapor pressure is NOT the driving force for fluid flow in your case. So no, you cannot use vapour pressure as differential pressure. The differential pressure that creates (or is created by) fluid flow is due to pipe wall friction, elevation changes, and velocity changes.

I suggest you open your fluid flow textbook and review. For a vent line, you want to calculate the "backpressure" in the top head of the tank due to the flow from your worst case scenarios (liquid and vapor) through the vent line. Then see if this backpressure exceeds the design pressure (or vacuum) of the tank and it's overpressure allowed by Code.

Is there an experienced relief designer that can mentor you on this?

[chemks2012]

Hi latexman,

Thank you very much for your reply/input.
If the maximum flowrate in to the tank and out of the tank is 10m3/hr, could you please let me know how to verify the existing 2” line for both liquid and vapour? And how to check the back pressure? Any amendment/comments will be highly appreciated.

Thanks

[latexman]

Calculate the pressure needed in the top head of the tank to push 10 m3/hr of liquid through the vent line to atmosphere. Call this P_L. Then, calculate the pressure needed in the top head of the tank to push 10 m3/hr of vapor through the vent line to atmosphere. Call this P_G. Is P_L and P_G less than the allowable overpressure of the tank? If so, it is okay. If not, there is more work to do.

[chemks2012]

Hi latexman,

I am sorry but I don't know what exactly you meant.

What do you mean by top head of vessel? Do yo mean, I need to assume that my vessel is closed I.e. no overflow/vent line to verify if the existing 2" line is sufficient? Do I need to apply Darcy's equation for pressure drop?
Thanks

[S.AHMAD]

Hi Chem
1. The top pressure is the pressure at the inlet to the vent line.
2. You can use Darcy equation for liquid and assume constant density for vapor (or use average density for vapor between vent inlet and outlet).
3. For liquid, elevation is significant parameter but for vapor you can assume negligible.

[breizh]

http://www.gl-group....sch/anhanga.pdf

Consider this resource to support your query.
Breizh

[chemks2012]

Thanks Ahmad,Breizh

Assuming liquid as the worst case, I have now calculated total pressure drop and compared with the design pressure of the tank. Could anybody look at the calculation in the attached file and let me know if it’s OK. My special thanks to ‘Latexman’ for his input!

Thanks

Attached Files

•  Overflow Line Check - Liquid Rev.A.xlsx   21.02KB   71 downloads

[latexman]

The K you used for the elbow is about 4X what I use. 1.1 versus 0.3 What kind of elbow is it? If it is a mitered elbow, your value is correct. My value is for a r/d = 1.5 pipe bend elbow.

The zero in cell C59 should be a 1 so the K for an entrance and exit loss in cell F59 is applied to the calculation. Right now no entrance and exit loss are included. That's not correct.

Previously we did not know the static head above the vent line was 0.2 m. So, my previous comments did not cover this. What Code was this tank built to? It appears the static head occurs on the straight side portion. Since you need the backpressure in the top (or top head or top cone) of the tank so it is comparable to the "design pressure" (and it's allowable overpressure), I believe you should subtract the static head from the dynamic pressure drop.

Edited by latexman, 13 May 2012 - 08:02 PM.

[S.AHMAD]

Chem
1. If you include k=1.5 for the head loss coefficient for entrance & exit loss, your tank design is inadequate. The design pressure should be higher. The design pressure = tank height + vent height + line pressure drop
2. Your hydraulic balance is correct:
Pressure of tank = static head + line pressure drop.

Edited by S.AHMAD, 13 May 2012 - 07:30 PM.