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Two Phase Acceleration Pressure Drop


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#1 theskipper

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Posted 01 April 2012 - 10:15 PM

Hi All,

Want to ask, how can we calculate the Two Phase Acceleration Pressure Drop in the pipeline?

I have noticed that in two phase system we cannot neglected the acceleration pressure drop due to density change of the fluid.

Is there another solution beside using HYSYS Dynamic to calculate this subject?

And what data that i should have for this?

Some data:
1. 12" Subsea Pipeline. From Well Platform to Processing Platform.
2. Length : 25 km.
3. Flow rate : Gas 5 MMSCFD, Oil 5100 BOPD, Water 5800 BWPD.
4. P1 = 255 psi, T1 = 112 F, P2 = 195 psi, T2 = 82 F.


Thanks Before.

Best Regards,
Andi

#2 ankur2061

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Posted 04 April 2012 - 11:49 AM

Andi,

The term 2-phase acceleration pressure drop is entirely new to me. Can you provide some reference for this.

What I know is that in 2-phase flow, flow can be visualized to be happening in various flow regimes such as bubble flow, annular mist flow, stratified flow and slug flow.

For pressure drop calculations in multi-phase, several flow correlations are utilized such as Beggs and Brill, Lockhart & Martinelli, Dukler, Flanigan, Eaton etc. In certain cases a combination of correlations is used such Dukler correlation for horizontal pipe and Flanigan for vertical elevations. Superficail velocities for both gas and liquid phase need to be calculated based on the volume fractions of gas and liquid. Slip factor has to be considered for gas slippage over liquid.

As far as software is concerned HYSYS is not considered to be a real pipeline simulator. For multi-phase steady-state flow better to use PIPESIM or PIPEPHASE as a simulator. For transient analysis for multi-phase flow the most accurate software presently is OLGA. There are others available but they have not evolved as much as OLGA has.

However, you still need to explain, what is meant by "2-phase acceleration pressure drop" since I could not find this term in the extensive literature I have collected on "pipelines".

Regards,
Ankur.

#3 kkala

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Posted 04 April 2012 - 05:07 PM

In two phase flow gas usually develops much higher velocity than liquid, so gas accelerates the liquid, causing an additional pressure drop for the fluid. This is explained in Coulson - Richardson, Chemical Engineering, Vol 1 (1977), Chapter: Two phase flow / Pressure, Momentum and Energy relations. Correlation by Lockhart and Martinellyi estimates total pressure drop (friction+acceleration, not separately) . But this would not be accurate for a 12" pipeline (Perry, 7th edition, Fluid dynamics, Multiphase flow, p. 6-27).
Some estimate of ΔP due to acceleration can be found in Handbook of Heat transfer, by W M Rohsenow and J P Hartnett, McGrawHill - 1973, Section 14: Two-phase flow.
For the homogeneous model (of two phase flow) the momentum pressure drop can be expressed between points 1 and 2 as ΔP=0.5*ρ2*v22 - 0.5*ρ1*v12 = 0.5*G2*(1/ρ2-1/ρ1), where G=mass velocity of fluid in the pipe, ρ=fluid density, both referring to gas+liquid. Rohsenow seems to omit the factor of 0.5 in the book (probably I miss something).
Said equation is understood to express ΔP due to gas density decrease, since pressure is decreasing along the pipeline. Consequently it does not represent acceleration ΔP in the meaning written in Coulson-Richardson's book. Nevertheless Rohsenow suggests same equation for estimating "acceleration pressure drop" in Thom's model (non homogeneous). And for Martinelli's model, Rohsenow suggests use of said equation for "momentum pressure drop".
Rohsenow equation for acceleration ΔP is not clear (English units), at least to me. Somebody experienced in two phase flow calculations may obtain a result. But try to find more precise (and modern) correlations on the subject, before trying to locate Rohsenow's book. E.g. CHE reprint "Chemical Engineering aspects of two phase flow" (around 1980) probably has the data required (I have not used the reprint for at least 15 years).

Edited by kkala, 04 April 2012 - 05:22 PM.


#4 breizh

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Posted 04 April 2012 - 07:04 PM

http://www.energy.kt...essure_drop.pdf

http://www.enggcyclo...lculator-phase/


Will it help?

Breizh

Edited by breizh, 04 April 2012 - 07:05 PM.


#5 kkala

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Posted 05 April 2012 - 01:24 AM

Second link by Breizh presents a quick method to estimate two phase pressure drop according to Lockhard - Martinelli method. Acceleration ΔP is included in this method according to Coulson - Richardson (I believe it), but not according to Rohsenow. See post No 3 (by kkala) for the above. Mass flux G mentioned in first link by Breizh must be mass velocity of fluid in pipe per Rohsenow.

Relevant formula in Rohsenow book is exactly written as follows (p. 14-3 and on).
ΔPm = G2*(v2-v1)/go, where
ΔPm=momentum pressure drop, lbf/ft2. "Acceleration pressure drop" is evaluated using this formula in the method of Thom. "For unheated pipes it is normally small".
G=mass velocity in pipeline, lbm/hr2-ft2, total flow rate/pipe cross-sectional area.
go=gravitational constant, ft-lbm/hr2-lbf
v1=specific volume entering the pipe, ft3/lbm
v2=specific volume leaving the pipe, ft3/lbm
where in general v=vf+xvfg, vf=specific volume of the liquid, ft3/lbm, x=quality equal to weight fraction of gas flowing (dimensionless), vfg=vg-vf at saturation conditions, ft3/lbm, vg=specific volume of the gas phase, ft3/lbm.
No arithmetic example is presented.

Edited by kkala, 05 April 2012 - 01:28 AM.


#6 cuchuoi2005

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Posted 05 April 2012 - 02:17 AM

Theskipper


Based on my experience, for the such long pipeline, the accelerational pressure drop is very small and could be ignored. This pressure drop is normally high and should be considered for flare header system.



Regards


cuchuoi2005

Edited by cuchuoi2005, 05 April 2012 - 02:32 AM.





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