atmospheric pressure effect on a open end pipe 3

[USAeng]

so if you have a valve and oil flowing through it... then you shut off the valve... I assume that most of the oil will stop flowing and the pressure will basically be 0 at the valve and atmosphere at the end of the pipe...

so if from left to right you had a shutoff valve then a 5' hor. run then 10' vert. rise then 10' hor. run then a 15' vert. drop and then a 15' hor. run then there is essentially open end pipe into a furnace

oh and its 1/2" steel pipe

so do you think the atmospheric pressure will hold in most of the oil? I set this up with rubber hose and tried it... it wasnt the same size and it was rubber and water, but most of the water stayed in the line...

maybe the original velocity of the fluid will create a siphon or something I'm not thinking about? Maybe the oil will leak out over a long amount of time?

Trying to use my memory of 1 semester of fluid mechanics here... hopefully someone with more experience can help...

I asked this in another forum yesterday, so sorry if someone finds it here too... just looking for some opinions from some valve people.

Sorry for the long post

Thanks a lot

 

[katmar]

Given enough time (probably a few minutes, depending on the oil viscosity) the 10' hor run, the 15' vert drop and the 15' hor run will drain into the furnace. Air will have to flow from the open end into the pipe to replace the volume of the drained oil. If the valve is shut tight no siphon can form.

The pressure upstream of the closed valve will be the supply pressure, plus or minus any changes in static head from the source of pressure. The pressure downstream of the valve will be 10' of your oil.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[USAeng]

I hate to sound dumb... but I dont want to say I understand this when I still dont...

According to the senario I wrote above there is a difference of 5' vertical... every 2.31' = 1psi so thats a little over 2 psi which should be easily overcome by the friction in the pipe, pipe fittings, and atmospheric pressure which I think is like 14.7psi or around there...

the valve before the piping is closed completely tight... so I guess I just dont understand how fluid can leak out if that valve is closed - because of the difference in pressure between after the valve and the atmosphere... but I dont have any real world experience here... Im still pretty fresh out of college, and I'm the only engineer at my company

everytime I tried doing this with my rubber hose test I had 5' horizontal then 5' vertical and then 5' horizontal and then another 5' vertical..... I filled the hose with water and then closed the top of the line and only about 1-2 foot of line at the bottom horizontal part would drain out....

very confused here... help me out if you can... thanks a lot

 

[katmar]

Do not worry about the friction in the pipe. Friction is only a factor when there is flow. At the very slow rate at which the oil will leak out of the pipe the friction will be nothing.

The two important forces are gravity and surface tension. If you take a drinking straw, with a diameter of about 1/8", fill it with water and hold it vertically with your finger closing the upper end the water will not run out. Gravity is trying to pull the water out of the bottom of the drinking straw and surface tension is holding it in the pipe. With the small diameter surface tension wins.

If you try the drinking straw experiment with a 6" pipe (imagine the top is closed with a blank flange - fingers aren't that big!) the water will not be held in the pipe. The weight of the water is too much for the surface tension. Somewhere between 1/8" and 6" there is a point where the switch occurs. I don't know for sure which side of the switch your 1/2" pipe with oil will be, but I suspect it will be more like the 6" pipe.

If the pipe/straw is horizontal there is less head for gravity to operate on and the point where gravity wins over surface tension will probably occur at a larger diameter than for a vertical pipe.

The oil is not leaking through the closed valve. It drips out of the open pipe (assuming gravity wins over surface tension) and air flows into the pipe to replace the lost oil.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[tr1ntx]

A slight correction to your thinking:

2.31' head = 1 psi is for water

Oil has a lower specific gravity (it floats on water, right?), so it will produce less pressure for a given head height. We have an oil that is 7.65 (vs. 8.32 lb/gal for water) so its sp. gr. is 0.92.

It's like katmar said, the viscosity/surface tension of the oil is what will (might) hold it in the pipe. Your 1/2" steel pipe, assuming it is Sch. 40, will have an ID of 0.622". I think that will mostly drain, i.e., gravity will win over surface tension. I don't remember fluid mechanics well enough to whip out a calculation.

For a simple trial, make up a 15' hor. run with a 5' vert. rise on one end. This will give you the same net head as your full system (delta H). Fill it with oil (using appropriate stoppers, plugs or valves), cap/plug the top, open the end and catch what comes out in a bucket. Measure it and you have your answer. You probably won't get the full 20' worth of volume out. If it has high viscosity, you won't get but a portion of the horizontal run.

CAUTION: There is a potential pitfall on your system. Let's suppose you do the trial and determine that the post-closure drainage is only so many feet of the hor. run and that's acceptable and you are going to use this piping/valve arrangement. If your oil is such that it gives off vapor, especially if it's hot, you will get additional drainage as the vapor collects in the top hor. run.