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udaveu (Mechanical) (OP)
20 Jul 07 8:42
I have a problem to calculate a free flooding of offshore subsea pipeline. I would like to calculate it with as minimum aproximations as possible.The calculation is not so simple since the pressure in pipeline is changing as the pipeline is filled with water. The concept is to fill the pipeline from one side (deeper side)and to have a check valve at the other side.
Any equation for a start or a spreadsheet?
Thanks in advance,
udaveu
BigInch (Petroleum)
20 Jul 07 8:51
There could be several parameters that affect how the calculations could be done.

Contents:  
What will the pipeline contain at the time it is flooded?  Such as natural gas, crude oil, gas and oil?

Internal Pressures:  
What will be the pipeline's internal pressure profile at the time the valve opens?

External Pressures:
What is the pipeline's external pressure profile (depths of the pipeline)?

Temperatures:
Internal product temperature profile.
External ocean temperatures profile.

http://virtualpipeline.spaces.msn.com

udaveu (Mechanical) (OP)
20 Jul 07 9:33
Thanks For a quick reply :)
So, at the begining, before opening, pipeline is full of air and closed so we can consider the pressure is 1 barA. The pipeline will be free flooded by opening a 4" ball valve with sea water. Temperature of water will not make a big impact but we can assume its about 20 deg C. The problem is general, but we can also take the depth 100m and lets say 70 m from another side....
BigInch (Petroleum)
20 Jul 07 9:58
Before it couldn't be answered generally, since if the air pressure inside at time of opening was greater than the sea water's pressure outside, the calculations would have to include a step for pressure equalization from escaping air from the lowest valve, before you could assume the remaining air would be displaced to the highest valve.

You didn't mention the profile.  I think its possible that an undulating pipeline profile may trap air at local high points.  That air might lock at the high points and not free flow to the exit point.  Can you assume there is a continuous constant slope from low to high?

http://virtualpipeline.spaces.msn.com

udaveu (Mechanical) (OP)
20 Jul 07 10:10
Ok. First, the pressure inside pipeline is 1 barA and static pressure outside at the entering point is ~10 barG (at 100m water depth). It is definitely lower pressure inside pipeline before start of free flooding.
Secondly, I agree your point for a profile but we can assume the continious constant slope of pipeline...
BigInch (Petroleum)
20 Jul 07 10:47
Assuming a continuous slope, that's enough water pressure to compress the air to roughly 11 barA, or 1/11th of the pipeline's air filled inside volume at 1 BarA, if you opened the lower valve first without opening the air escape valve.  Infow rate would be whatever the full open Cv is of the valve times the differential pressure across the valve at any given time, which would tend to reduce as the pipe filled, with an inside pressure (P2) = appx = P1 x V1 / V2, where V1 is initial volume and V2 is the volume of air at any given time.  Since pressures arn't really too significant (compressibility factor of air = 1) I think you could just assume some reasonable time steps and incrementally calculate the inside volume, the inside pressure, outside pressure = 10 barG, and get the flowrate across the valve for the next timestep and arrive at a reasonable approximation to the water filling rate through that 4" valve.  Once the pressures had equalized, you could do the same thing (kinda' in reverse) with opening the air escape valve and calculating the airflow rate out as the water displaced it.

http://virtualpipeline.spaces.msn.com

udaveu (Mechanical) (OP)
20 Jul 07 10:58
Hm, I don't think I would get a realistic timing for free flooding this way.Basically the water should fill the line and traped air pressure will be continually increased until it reaches ~7 bar and check valve on another side opens and air will start to discharge out of the pipeline. The flow of water on entering point will not be constant but will be slowed as the pressure is rising in the pipeline.  
udaveu (Mechanical) (OP)
20 Jul 07 11:00
sorry, biginch, I am leaving the office untill Monday...than we continue...have a nice weekend and thanks
BigInch (Petroleum)
20 Jul 07 11:45
That's OK.  Just assume small time steps.  If it doesn't look right, make the timesteps smaller.  So, fill to 7 bars, then change your calculations to include air releasing on the other side.  Non-constant air pressure is exactly why you must use time steps.  Calcualte the air pressure at each step and use that pressure differential for the valve flow calculations during the next step.  If it doesn't look reasonable, make the timestep smaller.  

http://virtualpipeline.spaces.msn.com

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