Time to empty a ruptured pipeline 4

[JoshEng]

I'm struggling to determine the right equations for calculating the flow through a ruptured pipeline to determine the total amount of time before a 20-mile long section will completely empty following a shutdown of the pumps. The goal is to provide a safety plan that includes an estimate of the time available for repair of a pipe before the entire section empties out. The pipeline I'm evaluating runs downhill about 100 ft/ mile and I'm assuming the rupture happens near the end of the pipe, just before the receiving facility, to simulate a worst case scenario.

The point is to calculate the maximum time available to respond and repair the rupture before the pipe is emptied out. The pipe is a 6" ID with an average pipe pressure is around 800 psi and flow rate vary from 1-10 barrels per minute. Any guidance on how to calculate this (assuming the pressure changes over time as the head pressure of the remaining liquid column in the pipeline decreases). I have found some information in the forums on leaks from holes, but my goal is to determine the flow rate through a 6-12" long rupture, such as might happen following mechanical damage or if the pipe is otherwise embrittled and then ruptures.


I have a few specific questions, too:
1. Does the pressure in the pipe drop rapidly following shutdown of the pumps, and if so, how can I calculate or estimate the remaining pressure after the pump is off and the leak starts to dissipate the pressure?
2. Does the location of the leak on the pipe affect the flow rate? The pipe is buried under a few feet of typical rocky or porous soil conditions.
3. Are there any other considerations I need to consider?


Guidance with solving the problem or suggestions of sources to find the solution would be both be appreciated!

 

[PetroSynergy]

What's in the pipeline? Where are the pumps located?
As you know liquids are noncompressible. If you are talking abt pressure and not head pressure, upstream product will increase in pressure as velocity decreases due to the pumps shutting down. As you know from Newton, an object in motion stays in motion, etc etc. So downstream portion(ie from pump to rupture) will not be as rapid, more of a gradual pressure drop.

Does the location on the pipe of the rupture affect affect the flowrate? Yes, however, a rupture anywhere on the pipe at the pressures and flowrates you describe will be drastic no matter which way you look at it.

 

[jte]

Is this a homework problem? Sounds like a professor read the morning paper and decided a "real world" problem would be a good class exercise...

 

[JoshEng]

The pipeline is carrying crude. The system has a single pump at the input and flows down to the storage facility. This is a real-world problem but I rounded off the numbers to make it easier to analyze; this is going into an engineering and safety evaluation for the pipeline operator (and something similar might be a good idea across the industry to provide some guidelines to describe how response time affects leak rate?).

Thanks

 

[JoshEng]

There must also be some effect of the vaccum formed just after the shutdown valve on the total rate of leakage/flow. I assume the crude oil vaporizes and fills that vacuum at a rate that limits the leakage rate of the pipeline by reducing the pressure until it is filled by vaporization. Is that right?

 

[PetroSynergy]

Yes it will pull a slight vacuum, but only temporarily, and the amount of vaporization will be minimal, if not negligible. The pressure will drop and very quickly, the level of crude oil in the pipleine will not cover "top to bottom". Therefore- no vacuum. However, your assumption that the vaporization will slop the leaking is wrong. When it vaporizes, it expands, therefore increasing pressure in the piping and increasing the leak flowrate drastically. Probably 100 fold, if not more. That's best case scenario. Worst case- blows out another rupture in the pipleline(absulute worst case).

 

[zdas04]

The leak you are considering is best modeled as a guillotine fracture that leaves the entire 6-inch diameter open to flow.

Pressure supplying the leak is:
100 ft/mile * 20 miles=2000ft*0.43 psi/ft*0.7 SG = 600+800=1400 psig.

I would approach this as a simple pipe flow problem. L=20 miles (to start), P1=1400 psig, P2=0 psig, D=6.0 inches Using the Hazen Williams formula, this gives you about 16 bbl/min. So I would calculate how much would flow in ten minutes (160 bbl), and determine how much pipe was drained in that ten minutes (4400 ft), and calculate a new hydrostatic head (you dropped 83 ft, or 25 psi, so the new pressure is 1375 psig), then use Hazen Williams again to get a new flow rate of 15.8 bbl/min with the new "length" of 19.8 miles. Repeat until the oil is gone. Probably take 20 iterations or around 4 hours.

I wouldn't be concerned about pulling a vacuum and flashing crude at the top--this is self limiting since the flashed gas will quickly raise the pressure back up.

David

 

[BigInch]

You mention average pressure, which really means nothing to pipe flow. You must be more specific about the conditons you are trying to describe. You will need to know the pressure at the inlet of this 20 mi segment (at the high point(????)) and at the outlet too. The inlet elevation seems to be 2000 ft higher than the outlet, so I really only have some idea of the overall static pressures involved. The pressures that initiate your leak condition remain too nebulous.

Knowing you have a crude pipeline, it might be reasonable to assume you are pumping up to the highest point at around 2000 ft elevation and the pressure at that high point is some nominal pressure around 50 to 100 psig. Or maybe you have an oil well up there and flow downhill with little pressure needed at the oil well site.

If you are indeed at some nominal small pressure at the 2000 ft elevation, the 2000 foot drop after the high point would have a static pressure of 2000 * 0.88 SG * 62.4/144 = 763 psi at the outlet, but that's not an average pressure. You also don't mention the sg of your oil, which could be anywhere from .85 to .9 or even higher. With sg=.88 that would give an outlet pressure with the outlet valve is closed of 763 psig. Which I would say the average pipeline pressure would be somewhere around 763/2 = 381.5 psi, not 800. That would mean my idea of what you're doing might not be quite right, or there is some higher pressure at the high point that you didn't tell us about. Are you confusing average pipeline pressure with average pressure at the outlet with no flow?

You mention 1 to 10 bbls/m, which would indicate up to 14,400 bbls/d flowrate in this pipeline, but I'm not sure if that is your design flowrate, or not, or if you're talking about some estimated leak rate. I do know that, if you want a design flowrate of something more than whatever gravity can do by itself alone, you will need more pressure up there than just some minimal amount to overcome the addtional head losses, but that's all I really know so far.

I can estimate the flow losses for that 20 mi segment for oil, with some kind of oil, 60 cP viscosity ?, and see that you could be the roughly 800 psi gain you'd get from elevation difference, if your flowrate maybe around 13,000 BOPD (377 gpm), equal to near gravity flow (flowing head loss equals the elevation head gain of 100 feet per mile) making the outlet pressure also near zero when at maximum flow, so maybe the 14,400 BOPD is your design flowrate. Its almost making sense... by accident, or some kind of coincidence perhaps, but in any case I think I have established some justifications for reaching a reasonable initiation condition.

I'll assume 13 MBOPD is the flowrate, and that you have some kind of low pressure at the high point and, since it does look like near gravity flow, I will also assume that there is another relatively low pressure at the outlet when the pipeline breaks and things start to happen.

I don't think the scenario can be calculated like zdas' example above, basically because he did not assume gravity flow is the "prime mover", thus he has a lot of pressure at the high point to drive his big leak rate. 1400 psig vs my nominal near-zero pressure. Talking average pressures, maybe his 700 vs 380 or so, but talking actual pressures, near minimum along the basically gravity flow line, there is no driver other than gravity and actual pressures gained = lost, so its near zero all the way down. Also keeping in mind that the gravity flow leak rate can only be sustained, if the head loss from the outlet leak rate is less than the sum of the elevation gain plus the head lost by flowing that leak rate to the outlet through that 6" pipe. If it's less, then the leak rate will drop as the sustaining inflow could not be as high as the outflow. I would tend to think more along the lines of a sloped tank 20 miles long with an outlet flow through a nozzle at its bottom, find the "nozzle flowrate" from the nozzle discharge coefficient, but limit it to the maximum possible sustaining flowrate through the pipe by gravity flow alone. Since any initial pressures at the high point will tend to drop fast with outflow from the leak position, any pressure at the high point will be quickly lost and you will have near zero very soon anyway and wind up with gravity flow in any case. When, if the pressure goes below the vapor point, the portion above the crude will fill with crude vapor and stay at that pressure and leak rates will continue with the same gravity driver.

In any case, it seems that the design flowrate would appear to be just about the same as gravity flow would provide, so I'll go with that.

That would mean you'll have a leak flow rate of around 377 gpm (or whatever your gravity flowrate actually is) until the line drains. With the volume in that 20 mile segment, I think you'll have a velocity of around 3.7 fps at the leak and it will take about 8 hours to drain.

For that to be true, you must have a uniform slope with no local elevation points higher than 40 ft or so than any other local point, or siphoning will be broken at some time or another and some oil may be left within the 20 mile segment. Siphon balance and vapor lock in a localized subsegment may occur whenever upstream pressure is not sufficient to flow over any localized "hill" or series of smaller elevation rises downstream, so you'd have to have a very good elevation profile to tell when that might happen, how much of the line might be affected and if flow would be stopped in some segments of the line, but in others might still continue to flow towards the break.

If you have anything but a straight pipeline with accumulated "localized" deviations in elevations of more than your vapor pressure head, an accurate leak study isn't a trivial task by any means.

"I am sure it can be done. I've seen it on the internet." BigInch's favorite client.

"Being GREEN isn't easy." Kermit

http://www.youtube.com/watch?v=hpiIWMWWVco

http://virtualpipeline.spaces.live.com

 

[BigInch]

P.S.

"I assume the crude oil vaporizes and fills that vacuum at a rate that limits the leakage rate of the pipeline by reducing the pressure until it is filled by vaporization. Is that right?"

Vaporization is faster than your runout rate, so don't expect any holdback factors from "the vacuum" being left behind the runout fluid column to hold that back. If not, you would just create multiple vapor pockets and new columns would open up. The lower density of the vapor pockets located in downhill segments will lower the upstream pressure of any liquid columns just downstream of them trying to climb the next hill, so that would result in more potential for vapor locks to occur, which might be the "holdback" factor to which you are referring. Good for reducing the leak volume, if they hold.

"I am sure it can be done. I've seen it on the internet." BigInch's favorite client.

"Being GREEN isn't easy." Kermit

http://www.youtube.com/watch?v=hpiIWMWWVco

http://virtualpipeline.spaces.live.com

 

[kkoenig]

BigInch, you said:

"That would mean you'll have a leak flow rate of around 377 gpm (or whatever your gravity flowrate actually is) until the line drains. With the volume in that 20 mile segment, I think you'll have a velocity of around 3.7 fps at the leak and it will take about 8 hours to drain."

Do you assume constant flow velocity throughout the entire draining process or is this figure (3.7 fps) an average? I have seen elsewhere that the solution to this problem requires an iterative process. It seems to me that during each iteration, the height of the oil changes, thereby changing the leak pressure, friction head loss, flow velocity, and flow rate. What equations did you use if you avoided this assumption?

Any clarification would be greatly appreciated.

Thanks

 

[kkoenig]

My problem differs in that my longitudinal crack is in the middle of the length of the pipeline (15 miles along a 25 mile pipeline). My two main concerns are:

1) How to account for the varying volume flow rate as the column of crude descends. (I assume only gravitational flow and associated head losses of pipeline length)

2) How to calculate the leak flow rate from a pipeline. I have used the formula for leakage out of a pressurized vessel: Q_i=C*A*((2*d^2*g*h)+2d*(P-P_a))^(1/2). My situation is different because the fluid continues to flow past the spot, correct?

Thanks for any and all help.

 

[BigInch]

kk,

If you did that, you'd be considering this as flow from a tank nozzle with pressure from hydraulic head only. That would work only if you had the classic "discharge from a tank" example problem, which is a tank with an infinite diameter, meaning you would have no flow losses to account for as the fluid moves down to the nozzle and the flow could be Q = C_d*A*(2gH)^1/2; where H is the height of the fluid surface over the nozzle. For a draining tank, H would vary and, as you've mentioned, you'd have to iterate as outflow changes the value of H continuously.

This problem requires that you consider the tank as having a diameter of 6", implying that there would be significant flow losses associated as the fluid column moved towards the nozzle and H would not be the height to the fluid surface. To view the problem from a similar perspective, you would have to subtract the head lost to fluid friction from H.

Steady state gravity flow results when the pressure at the fluid's surface becomes very small (vapor pressure) and you have essentially no or very little driving pressure remaining on the column (in fact a vacuum). Note that this assumaption wouldn't be valid for propane and other fluids where VP is relatively very high. So now this 6" pipeline would go into gravity flow. Steady state gravity flow for a pipeline flowing full basically means that the gravity head available to drive flow is exactly equal to the friction head being consumed by whatever that flowrate happens to be. In this case it is 100 feet of head per mile; same as the assumed slope of the pipe, 0.0189 feet/foot. So now, if it was water, you could just look in an appropriate water pipe flow table for flow in a 100 ft pipe length with a head loss of 1.89 feet, or a pressure loss of 1.89 * 62.4/144 = 0.82 psi and get your gravity flow rate. Its oil, so we have to find the flowrate using that head and the specific crude's density and viscosity values.

Second question.

Leak rate can be calculated by flow across an equivalent circular, sharp-edged orifice, probably something like the equation you have there (I'm too lazy to check it). You can do that as long as long as you have some significant driving head across the orifice. The difficult part is often to determine what value to use for the upstream head, or pressure equivalent. In most situations it does tend to change with fluid level and volume remaining inside the container. For liquid pipelines it only takes a very small drop in volume to get a relatively large drop in pressure, because the compressibility of liquids are very small in relation to gases; almost nonexistant.

http://hyperphysics.phy-astr.gsu.edu/hbase/tables/compress.html

Gas pipelines, on the other hand and depending on initial pressure, might need 100 to 1000 times or more of the same volume to be released to see an equivalent pressure drop of a oil or water pipeline. That is the reason that most liquid pipeline leaks quickly reach depressurizd conditions at the high point and lose driving pressures.

Assuming gradual slopes, after that, its gravity flow in a full pipe. Any tendency to outflow faster than what gravity can supply results in pressures near the outlet dropping to vapor pressure in the faster running segment and a new vapor bubble forming within, so you'd get fluid, vapor, fluid, vapor as the still higher pressure column above breaks to vapor each time there was a faster outflow than gravity could supply and returning to liquid flow as the liquid phase outflow slows and allows the column to catch up.

The other possibility is that you may have supercritical or subcritical flows. Super critical flow will result, essentially pressured pipe flowing flow, when the slope of the pipeline is greater than critical slope, critical flow, flowrate when gravity and headloss are equal with depth of flow roughly equalling pipe diameter, and subcritical flow will result when the pipe slope is less than critical slope (pipe not flowing full, you must use hydraulic radius instead of diameter to calculate friction). I estimated the slope of 100 ft in 1 mile would be subcritical, or near critical and result in flows not exceeding steady state constant gravity flow, where pressure head lost = gravity head gain. If you have a steep slopes, faster outflows can be reached. If you have shallower slopes, outflows faster than critical flowrate could not be sustained. Of course if the slope varies in various subsegments of your pipeline, you may have any one, or all, of these types of flow trying to occur in any given segment of pipeline, even changing as draining changes the hydraulic profile of sub segments. It can become a transient closed-open-closed-open channel flow problem. Seeing an elevation profile of your line would make it easier to get some idea of what type of hydraulics might develop... but not much.

"I am sure it can be done. I've seen it on the internet." BigInch's favorite client.

"Being GREEN isn't easy." Kermit

http://www.youtube.com/watch?v=hpiIWMWWVco

http://virtualpipeline.spaces.live.com

 

[BigInch]

Gives you some idea of what the gravity flow sewer designers face. I think they tend to design for minimum-maximum flow capacities, rather than worry about the time to drain the system from a full condition. Now the reason for that is pretty obvious.

"I am sure it can be done. I've seen it on the internet." BigInch's favorite client.

"Being GREEN isn't easy." Kermit

http://www.youtube.com/watch?v=hpiIWMWWVco

http://virtualpipeline.spaces.live.com

 

[kkoenig]

BigInch,

How would you proceed with a gravity flow analysis given a detailed elevation profile?

Sewer designing sounds like a messy bunch of friction factors...

 

[BigInch]

Pretty much the same way as you would for a pressure driven flow analysis, except use absolute units. Your pressure at the supposed inlet, assume its the high point, is provided by atmosphere, as is the backpressure at any outlet. As air enters the line, or if there is no inlet, when vapor pressure is reached at any point, the liquid level starts dropping. From then on you could assume that same pressure follows the dropping fluid elevation and the pipe lengths effectively change to equal the length of liquid still remaining. As long as pipe slopes are greater than critical slope for gravity flow in the pipe, friction loss = elevation gain, the pipe will flow full, if it began full, otherwise it will flow at constant depth. Complications arise for gently sloped pipes that flow partially full, but they can be analyzed using open channel flow methods. Keep track of the total energy gradient at all times to find out if the, a, or a segment of pipe is full or partially full.

A few other open channel flow analysis tips in my comment here just a couple of days ago,

http://www.eng-tips.com/viewthread.cfm?qid=284611&page=1

"I am sure it can be done. I've seen it on the internet." BigInch's favorite client.

"Being GREEN isn't easy." Kermit

http://www.youtube.com/watch?v=hpiIWMWWVco

http://virtualpipeline.spaces.live.com