Pipeline emptying + liquid flashing at subatmospheric pressure

[Peter C]

It seems easy, yet it’s quite complicated.

Considering a full pipeline (e.g., water) close on both ends and without any venting valve. A drain is opened at a location of the pipe allowing fluid to flow out just by gravity. It is clear that all product in the vicinity and in a higher level than the hole will drain out, but what about the rest? Is there a simple way to evaluate it as well as the final liquid configuration in the pipe?
Furthermore, I’ve thinking about two other points so far: siphon effect and, most importantly, at what point the moved column will generate enough sub-atmospheric pressure to vaporize some of the fluid, like in the example of figure 1 (

https://files.engineering.com/getfi...ae70-9b8d4f95eddc&file=Flashing_induced.png).

Regards

 

[1503-44]

Flashing occurs when the highest fluid level reaches vapor pressure. Around 0.3 psi at normal temperatures.

Calculate pressure at any elevation above the drain by assuming atmospheric pressure (14.7 psia) at the drain outlet. If you reach 0.3 psia or so, the column of water can break away and drain independently from any water above. To avoid that and drain uniformly as possible, the drain rate needs to be slowed down so that water above the potential breakpoint has enough time to flow into the breakpoint area and replace any water withdrawn from the drain..

The subject was recently discussed here

https://www.eng-tips.com/viewthread.cfm?qid=510627

Have a read and come back here if you need more help.


--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[Peter C]

Hey thank you for your reply! I still have a lot of questions, given that in the example, our fellow colleague had some inlet air valves, while in my case the pipe is fully closed.

Please, let's consider the case bellow:

Suppose I open the drain located at x = 6.

1) At first, pipe would drain completly from points 5 to 10. They are all "above" the drain and in its vicinity. I am ignoring - for now - tensile forces. Therefore, points 5 to 10 will be air filled and at p = 14,7 psi

2) from x = 1 to x = 4 there is some column that will push more liquid to overcome the x = 5 "peak" and then be drained since that at x = 3 the pressure is 14,7 psi + the 15m column head

3) at this moment, flashing will happen in the vicinity of x = 1. I see like a piston effect, when the draining fluid starts creating some "vaccum" until preassure reaches the vapor pressure.

I still find it hard to evaluate the amount of liquid drained in "2" and specially the flashing quantity in "3"

Thank you in advance

 

[1503-44]

6 to 10 evacuated through the drain at 6. As there is only 3 m (is it meters?) of elevation difference, the pressure at x=10 is 1bar-0.3b = 7b, not close to vapor pressure, so no column separation occurs anywhere between x=6 to 10. Air must enter through the drain to displace that water. That will slow the water's movement outward through the drain glug/glue.

Left side. Reducing pressure as you climb in elevation, about a 10m rise will potentially drop absolute pressure to 0b, break the column, leaving vapor pressure at the break point, say at x= 2.3

We calculate the flow rate of that break away column and assume that is also the drain rate of that water from the left side at drain, x=6. When calculating that flow rate use 0.3 bar at x=2.3 and 1b at the drain. If water above x=2.3 can flow at that same rate, then there would not be a break away at 2.3, (any vacant space would be immediately replaced with water from above), in which case we could assume that head - some friction loss, since it is moving, is available to push all water down to the drain. Recalculate the pressure at x=2.3 = 7m = 0.7b and 0b at the high point x=1

As the water between x=1 to x=4 drain down, with vapor pressure at that surface, there will be sufficient pressure to almost reach x=3; same elevation as x=5. If that happens at a velocity sufficient to set up siphon action, the V between 3 and 5 might be cleared, but if not, the pressure at 5 will be roughly equal to 0b and a new break away column will initiate there. That will drain the x=5 to x=6 section leaving vapor pressure behind it. The "V may be cleared if siphoning is sufficient, but any water that is in there at the time break away occurs at x=5 will remain there, as well as all water between x=0 to 1.

The amount of water in vapor form can be found within the pipe assuming all the "vacant" space in the pipe is filled with saturated water vapor at a pressure = water vapor pressure at whatever the temperature of the water in the pipe happens to be.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[Snickster]

) At first, pipe would drain completly from points 5 to 10. They are all "above" the drain and in its vicinity. I am ignoring - for now - tensile forces. Therefore, points 5 to 10 will be air filled and at p = 14,7 psi

I dont think that any fluid will drain from 6 to 10 initially but will only drain from this section as the water is drained and level drops down on the left side of 6.

2) from x = 1 to x = 4 there is some column that will push more liquid to overcome the x = 5 "peak" and then be drained since that at x = 3 the pressure is 14,7 psi + the 15m column head

The fluid to the left of 6 will drain first until the pressure is equalized with the fluid to the right of 6 then they both will drain. Keep in mind that without any vents a vaccuum will form on both sides at the highest points with a maximum value equal to the vapor pressure of the liquid.

3) at this moment, flashing will happen in the vicinity of x = 1. I see like a piston effect, when the draining fluid starts creating some "vaccum" until preassure reaches the vapor pressure.

Flashing will commennce immediately at x=1 as liquid is drawn down left of 6 while no liquid is drawn down to the right of 6. As more fluid to the left of 6 is drawn down there will reach a point where the static height of liquid on left of 6 equals the static head of liquid on right of 6 (with vapor pressure at same time acting at top of both of these water columns)then flow will stop. Fluid will not draw down at all to right of 6 since if you take vapor pressure of water of about 0.5 psia acting at top and 14.7 acting at drain it will be in equilibrium with about the 10 meter liquid column static head on right of 6 so a little liquid say a few feet on right of 6 will drain, but remaining liquid will have a height of 144/62.4 x 14.2 (feet), assuming 0.5 psia vapor pressure, and this will be the resulting height of the liquid to the left of 6 too. This is because in the end static equilibrium of forces will exists so vapor pressure at top of water column x crosssection area of water column plus weight of water column acting down = 14.7 psia x crosssectional area of water column acting up at the bottom.

I still find it hard to evaluate the amount of liquid drained in "2" and specially the flashing quantity in "3"

 

[1503-44]

"The fluid to the left of 6 will drain first until the pressure is equalized with the fluid to the right of 6 then they both will drain. ..."

Before the drain is opened, pressure on both left and right sides at any given elevation are already equal and, yes, both sides may not drain much at all, ever, unless you somehow get air into them, as 1barA atmospheric pressure at the drain is indeed sufficient to hold the water up to a potentially 10m height. X= 2.3 You have to get air in to get the water out after reaching that 10m elevation.
No vacuum forms on the right as there is not a 10m elevation difference.


--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[LittleInch]

Sorry but without units of height data no one can figure this out. Plus the OP said NO AIR VENTS. So nothing will drain....

Unless air is allowed in at 5 and 10 nothing will drain off the height from drain to bought point is less than 10m.

Our it might start glugging as air bypasses the drain line.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[1503-44]

We're off on the PeterC question.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[LittleInch]

I know.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Peter C]

Hello guys, thank you for all your comments. I am taking the time now to read all of them.

Just to clarify, the line profile showed is just for example, the height values (y axe) should be ignored. I am not trying to solve a specific example, but more general one. In fact, I want to understand what are the concepts governing what may happen: at what level water (or any liquid, giving its vpressure) would flash and when would siphon happen (draining even more liquid).

In resume, what would be the final line fluid profile (liquid, vapor, air and where) for any closed line and with a drain.

As I said, I am taking a moment to read your comments

 

[LittleInch]

A totally closed line with less than 10m between drain point and the high point won't drain more than a spoonful.

More than 10m to the high point it will pull a vacuum, start to boil slowly but drain flow will stop once the liquid level gets to 10m higher than the drain.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Peter C]

That's a very interesting point, but I don't always agree that line won't drain: if the drain diameter is big enough so air can "overcome" surface tension (that would be a second part of that problem...) then a slug flow may happen. Of course it's not optimum since flow will not be constant...
I am therefore assuming that drain is big enough to allow this "slug" flow.

 

[Peter C]

Fellow 1503-44, I agree with all your points. They will be very helpful.

However I have a question and a point that i disagree :

1) Calculating velocity would be the only way to check if siphon will occur or not? There is no way i could trae a line saying: above that height there will be vapor, under that line pipe will be water filled, there pipe will be air filled...?

2) I don't think there is anychance that the "V" may be completly drained... if that happens, air at x = 5 (1 atm) would recondensate all the vapor left behind. That's one of the main challenges for me in understand this problem.

 

[LittleInch]

That's a massive assumption/ leap of faith and really shouldn't be part of this theoretical discussion.

Just stick a vent in the right place.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[1503-44]

So we kind of set the elevation in meters.


Agree. Probably not going to drain the V. With a vent at the top, maybe a small chance.

1) no magic line, but calculating velocity isn't hard. Pipe velocity It only has to be >= drain rate. Estimate that as the lowest of the sqrt(2gh) h=head above any elevation, or a lower velocity of your controlled drain rate.
2) air at atm will recondence the vapor, which will pull in a lot of air to replace the vapor volume as that happens.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[Snickster]

No vacuum forms on the right as there is not a 10m elevation difference.

Oh sorry I was mistakenlhy looking at the bottom numbers of the graph and saw 10. That was dumb lol. Anyway I think the fluid to the left of 6 will stop flowing when the vapor pressure plus static height of that fluid column equals 14.7 psia acting at the bottom of the column so that static equillibrium occurs unless somehow air is able to bubble up into the pipe from the drain - like holding a jug upside down. The right side of 6 will stay packed with liquid. In this case if you start with 14.7 psia and subtract static liquid level on that side you will get the gauge pressure on top of that column.

 

[1503-44]

It does seem that way. Right, it may stop at x=2.3. The siphon might not form at all when the level is at x=3, 7m over the drain, because atm might push the level back to 2.3. I think you'd have to put some suction at the drain.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[Peter C]

So considering that the values given in the image are in meters, what would be the final configuration in pipe?
I am having troubles to model the siphon phenomena and, therefore, the flashing one.

Besides, is Bernoulli applicable in that situation given that it's a transitory state and not something continuous?

Regards,

 

[1503-44]

Theoretically.. without any vents at the high points ... Water will exit the drain on its own until its surface level is 10m above the drain. Once it reaches 10m above the drain, any water drained out must be displaced by air entering the pipe... through the drain. Then bubble flow up and water flow down will occur. Clearing the V of water may be impossible without introducing additional measures, tapping a vent, or drain, an air compressor or pump and hoses.

"Modified" Bernoulli always works. At any given time P1/λ +(v1^2)/2g -f*L + z = P2/λ +(v2^2)/2g
Although with air and water inside, the height and location of each phase must be known.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[Snickster]

Yes, the Bernoulli equation is good but remember that the Bernoulli equation is just an equation that describes SUM Forces = Mass x Acceleration, in other words an equation of state of equilibrium of forces – when forces are equal there is no more acceleration. So, pressure head accelerates the fluid to a velocity such that at that velocity the force of friction resistance equal the pressure force (minus the kinetic energy that went into the velocity) so the fluid no longer accelerates. Most of mechanics, structural, electrical design etc., is based on the equilibrium of driving forces versus resisting forces under conditions of zero acceleration.

I believe the way to look at it is that at any given time and any given cross sectional cut of the piping there must be an equilibrium of forces. So at time = 0 you take a cut at the drain point 6 before the drain is open you would have the pressure equal to the static height of the liquid to the left of point 6 plus add 14.7 psia considering that the pipe was initially open to atmosphere (at high point 1) on the left so that 14.7 psia also acts at top of fluid column on left.

P = 20(3.28)(62.4/144 = 28.43 psig = 43.13 psia. [note that the pressure at point 0 = 14.7 – 5(3.28)(62.4/144)]

So this has to be the pressure at the bottom of the water column to the right of point 6 also. So if you work back upwards to the right from point 6 you would subtract the static height of fluid then add the pressure acting on top of that column to get the total pressure. In other words, the column to the right is being compressed by the column on the left and there is a force of the water on the pipe at the top of the right side of 6:

P = 43.13 psia – 6(3.28)62.4/144 = 34.6 psia (acting at top of fluid column to right of point 6 @ T = 0)

The drain is then opened to atmosphere at 14.7 psia. Water begins to flow out of the drain from the left side of point 6. Assuming the velocity in the main pipeline is very low (to simplify this analysis), then pressure drop in main pipe will be negligible and all pressure drop will be across the drain piping. So when drain is first opened the pressure upstream of the drain is 43.13 psia per above and 14.7 downstream of the drain with pressure drop of 28.43 psi across the drain piping. The flowrate out of the drain will be such that at the resulting velocity developed, the 28.13 total head will be converted to friction losses plus velocity head (per Bernoulli equation).

As flow continues out of the drain the level of liquid to left of point 1 wants to drop but only way this is possible is to separate. This initial separation will happen at the highest point to the left of 6 which is at point 1 because this is where the pressure is the lowest. This is because as you work back from the drain up to point 1 by subtracting the static height you get atmospheric pressure at point 1 at T=0.

The separation process is itself a matter of balance/equilibrium of forces. To the right of point 1 the weight of the liquid column of 20 meters (65.8 feet) is acting down, at the open end of the drain atmospheric 14.7 psia (times area) is acting up, at top of column at 1 vapor pressure of about 0.5 psia (times area) is acting down. So you have .5 psia acting at top of column, plus weight of column (worth a force of 28.43 psi times area of pipe acting down) balanced by a pressure force of 14.7 psia acting up. Therefore, you still end up with a positive pressure across the drain and the water will still move down and out drain pipe due to imbalance of force. Vapor pressure will result at the top of the water column since if a true vacuum of 0 psia would try to form it would be filled with the evaporating liquid at its vapor pressure (assuming the rate of evaporation of liquid keeps up with the drop in level of liquid).

So the liquid level on the left of drain keeps dropping increasing the vapor space above. Since separation initially occurs at point 1, I believe that the liquid between 0 and 1 will remain in the pipe. This is because if vacuum/vapor first starts forming at point 1 it would tend to try to pull the liquid over the hump at 1 but if this happens then the fluid at point 0 would need to separate and vacuum vapor will need to form at 0 for the fluid to be pulled over 1, but if that happens then on both sides of the liquid section between 0 and 1 there will be same vacuum pressure for a net force of zero. So again by balance of forces you would have about equal pressure forces acting on top and bottom of liquid 0 to 1 but the weight of liquid acting down towards 0 so fluid cannot move up as net force is down.

So the fluid to left of drain/to the right of point 1 keeps moving down and draining out, with vapor space increasing above until there is another point where forces balance. This is the point where the forces acting down are balanced with the forces acting up. Assuming 0.5 psia in vapor pressure space above liquid column on left then forces down are balances with forces up when:

(Vapor pressure x Pipe Area) + Weight of Liquid = (14.7 psia x Pipe Area)

Or Weight of Liquid = 14.7 psia – 0.5 psia = 14.2 psi (in equivalent pressure head)

14.2 psi in terms of liquid static height of water = 14.2(144)/62.4 = 32.8 FT. (this is the resulting liquid level on the left of drain based on static equilibrium of forces)

Note that at this equilibrium/balance of forces the velocity = 0 and therefore pressure at drain exit = pressure upstream of drain at point 6 which is 14.7 psia and this acts on the entire cross sectional area of pipe or the entire cross section of the fluid column.

Now if you have 14.7 psia on upstream side of drain then fluid to right of drain will want to move down. But if it does then a vacuum will form with resulting vapor pressure of 0.5 psia. So on the right the Sum of Forces acting down versus acting up would be same as given above except water column is shorter and therefore:

Down = (0.5 x Pipe Area) + Weight of Liquid = 0.5 + 6(3.28)(62.4)/144 = 0.5 + 8.528 = 9.028 psia

Acting up = 14.7 psia

So net force is up and liquid cannot move down. However since forces must balance on the right of drain with 14.7 psia acting up, then what will result is a 14.7 psia pressure pushing up, a 8.528 liquid static pressure pushing down and a pressure of compression at top of liquid column of 14.7-8.528=6.172 psia pushing down so no separation of liquid will occur at top of column on right.

This is all theoretical based on hydraulics and balance of forces as I see it. It is what is called a thought experiment. haha. Possible air may bubble up into drain over time and liquid will further exit the pipe, or some other factor may cause unequilibrium conditions so what happens may not exactly follow this balance of force/static equilibrium analysis. What do you think?