• Author / Uploaded
• Mohammed Yousif

Citation preview

Chapter Three Petroleum Department

LectureFourth Stage Multiphase Flow in Pipe

- -2015 Production Engineering II

Multiphase Flow in Pipes  Multiphase flow in pipes is the concurrent movement of free gases and liquids in the pipes.  Multiphase flow problem can be divided into four groups: 1. Vertical Multiphase Flow. 2. Horizontal Multiphase Flow. 3. Inclined Multiphase Flow. 4. Directional Multiphase Flow.

Fig. (3-1): Over-all production system

 Petroleum Production System  Stage-1: The fluid must first enter the well-bore (inflow performance).  Stage-2: The fluids traveling through the tubing and/or annulus to the surface may be either in vertical or directional flow.  Stage-3: At the surface, the fluids may or may not pass through the "choke".  Stage-4: The fluids traveling from the well-head to the separator may be either horizontal or inclined flow. 1

Chapter Three Petroleum Department

LectureFourth Stage Multiphase Flow in Pipe

- -2015 Production Engineering II

 Pressure Losses in Well System  Fluid flows from the reservoir to the stock tank occur because of the pressure gradients within the system.  The total pressure drop(loss) from the reservoir to the separator is the sum of the individual pressure drops through three different segments: ∆Pt = ∆P1+∆P2+∆P3 ∆P1 = Pr - Pwf (Inflow Performance) ∆P2 = Pwf – Pth (Vertical/directional flow Performance) ∆P3 = Pth – Psep (Horizontal/inclined flow Performance) Where: Pr= average reservoir pressure, psia Psep= separator pressure, psia Pwf= bottomhole pressure, psia Pwh(Pth)= wellhead (or tube head) pressure, psia.  The Multiphase Flow problems involve a study of the pressure losses in the pipes containing two-phase fluids (gas and liquid).  Multiphase flow correlations to predict pressure loss in the pipes is based on general energy equation.  The General Energy Equation  The theoretical basis for many fluid flow equations is the general energy equation. The general energy equation expresses an energy balance between two points in a fluid flow system.  General Energy Equation states that; {The energy of a fluid entering point 1 of a pipe …… plus…. any additional work done on the fluid between point 1 and 2…..minus…. any energy losses by the system between point 1 and 2, is ….equal…. to the energy of the fluid leaving point 2}.

2

Chapter Three Petroleum Department

LectureFourth Stage Multiphase Flow in Pipe

- -2015 Production Engineering II

Fig. (3-2): Flow diagram

General energy equation after a series of derivatives simplified as; 𝐝𝐩 𝐝𝐳

=

𝐠 𝐠𝐜

𝛒 𝐬𝐢𝐧𝛉 +

𝐟𝛒 𝐯 𝟐 𝟐𝐠 𝐜 𝐝

+𝛒

𝐯 𝐝𝐯 𝐠 𝐜 𝐝𝐳

--------------------------------------------------------------------------- (3-1)

Where: ρ= Density (lbm/ft³) d = Pipe Diameter (ft), v = Velocity (ft/sec) 2 g = Acceleration of Gravity= 32.2 ft/sec . ,sinθ = deviation of tubing. dp/dz = Total Pressure Gradient(

Ibf /ft2 ft

).

f = Friction Factor, function of Reynolds number and roughness. gc = Gravitational Constant= 32.2ft/sec2. { lbforce= 32.2 Poundal = 32.2lbmass. ft/sec2= Slug .ft/sec2}

3

Chapter Three Petroleum Department

LectureFourth Stage Multiphase Flow in Pipe

- -2015 Production Engineering II

 Pressure Gradient Equation The total pressure gradient can be considered to be composed of three distinct components, that is; Total gradient = elevation gradient + friction gradient + acceleration gradient 𝐝𝐩

𝐝𝐩

𝐝𝐩

𝐝𝐩

( 𝐝𝐳 ) = ( 𝐝𝐳 ) + ( 𝐝𝐳 ) + ( 𝐝𝐳 ) 𝐭

𝐞𝐥

𝐟

𝐚𝐜𝐜

------------------------------------------------------------------------- (3-2)

Where: 𝐝𝐩

( 𝐝𝐳 ) : Total pressure drop. 𝐭

𝐝𝐩

𝐠

( 𝐝𝐳 ) = 𝐠 𝛒 𝐬𝐢𝐧𝛉 = is the component due to elevation change. 𝐞𝐥

𝐜

𝐟𝛒𝐯 𝟐

𝐝𝐩

( 𝐝𝐳 ) = 𝟐𝐠 𝐟

𝐝𝐩

( 𝐝𝐳 )

𝐚𝐜𝐜.

𝐜

𝐝

=𝛒

= is the component due to friction losses. 𝐯 𝐝𝐯

𝐠 𝐜 𝐝𝐳

= is the component due to velocity change.

 For multiphase flow, the terms of Eq. (3-1) is modified to take care of two phases (gas and liquid) flowing.  The pressure-gradient, Eq. (3-1), can be modified for multiphase flow by considering the fluids to be a homogeneous mixture.  In particular the following variables must take into account the mixture of gas and liquid as the following: 𝛒 = 𝛒𝐦 = 𝐝𝐞𝐧𝐬𝐢𝐭𝐲 𝐨𝐟 𝐦𝐢𝐱𝐭𝐮𝐫𝐞 𝐨𝐟 𝐠𝐚𝐬 + 𝐥𝐢𝐪𝐮𝐢𝐝 𝐯 = 𝐯𝐦 = 𝐯𝐞𝐥𝐨𝐬𝐢𝐭𝐲 𝐨𝐟 𝐦𝐢𝐱𝐭𝐮𝐫𝐞 𝐟 = 𝐟𝐦 = 𝐟𝐫𝐢𝐜𝐭𝐢𝐨𝐧 𝐟𝐚𝐜𝐭𝐨𝐫 𝐟𝐨𝐫 𝐟𝐥𝐨𝐰 𝐨𝐟 𝐠𝐚𝐬 + 𝐥𝐢𝐪𝐮𝐢𝐝 Where: (m) Denotes mixture, the Eq.(3-1)can then be modified to read: 𝐝𝐩 𝐝𝐳

=

𝐠 𝐠𝐜

𝛒𝐦 𝐬𝐢𝐧 𝛉 +

𝟐 𝐟𝐦 𝛒𝐦 𝐕𝐦

𝟐𝐠 𝐜 𝐝

+

𝛒𝐦 𝐕𝐦 𝐝𝐕𝐦 𝐠𝐜

𝐝𝐳

----------------------------------------------------------------- (3-3)

Where: m = symbol represents mixture fluids, ρm= density of mixture fluids(lbm/ft³) fm = friction factor of mixture fluids. Vm = velocity of mixture fluids (ft/sec) 4

Chapter Three Petroleum Department

LectureFourth Stage Multiphase Flow in Pipe

- -2015 Production Engineering II

 Pressure Gradient Calculations Calculation of pressure gradients requires values of flow conditions such as velocity and fluid properties such as density, viscosity, and others. 1. Liquid Hold-Up Liquid holdup is defined as the ratio of the volume of a pipe segment occupied by liquid to the volume of the pipe segment. 𝐇𝐋 =

𝐕𝐋 𝐕𝐩𝐢𝐩𝐞

------------------------------------------------------------------------------------------------------- (3-4)

 Liquid holdup is a fraction which varies from zero for all gas flow to one for all liquid flow.  The remainder of the pipe segment is occupied by gas, which is referred to as gas holdup or gas void fraction. That is; 𝐇𝐠 = 𝟏 − 𝐇𝐋 =

𝐕𝐠 𝐕𝐩𝐢𝐩𝐞

----------------------------------------------------------------------------------------- (3-5)

Liquid and gas holdup occurs due to;  Gas slippage. {In Gas-Liquid Flow, the gas has a tendency to flow faster than the liquids, i.e. gas slipping past the liquid….. OR…..  Density difference between phases. The liquid tends to accumulate and will occupy more of the pipe space, if no slippage was occurring. 2. No-Slip Liquid Holdup. No-slip holdup is defined as the ratio of the volume of liquid in a pipe segment divided by the volume of the pipe segment which would exist if the gas and liquid travelled at the same velocity (no-slippage). 𝛌𝐋 =

𝐪𝐋 𝐪𝐋 +𝐪𝐠

------------------------------------------------------------------------------------------------------ (3-6)

Where: qL and qg are the in-situ liquid and gas flow rates, respectively. The no-slip gas holdup or gas void fraction is defined as; 𝛌𝐠 = 𝟏 − 𝛌𝐋 =

𝐪𝐠 𝐪𝐋 +𝐪𝐠

----------------------------------------------------------------------------------------- (3-7)

3. Density  For Liquid Mixture (Oil-Water) The total liquid density (𝛒𝐋 ) may be calculated from the oil and water densities and flow rates if no slippage between the oil and water phases is assumed. 5

Chapter Three Petroleum Department

LectureFourth Stage Multiphase Flow in Pipe

- -2015 Production Engineering II

𝛒𝐋 = 𝛒𝐨 𝐟𝐨 + 𝛒𝐰 𝐟𝐰 -------------------------------------------------------------------------------------------- (3-8) fo= Oil fraction(or oil cut) fw= water fraction (or water cut) 𝐟𝐨 =

𝐐𝐨 𝐐𝐨 +𝐐𝐰

=

𝟏 𝐁 𝟏+𝐖𝐎𝐑( 𝐰 )

------------------------------------------------------------------------------------ (3-9)

𝐁𝐨

𝐟𝐰 = 𝟏 − 𝐟𝐨 ---------------------------------------------------------------------------------------------------- (3-10)  For Two Phase Mixture (Liquid-Gas) Calculation of the two-phase density requires knowledge of the liquid holdup. Two equations for two-phase density are used in two-phase flow. 𝛒𝐬 = 𝛒𝐋 (𝐇𝐋 ) + 𝛒𝐠 (𝟏 − 𝐇𝐋 ) ---------------------------------------------------------------------------- (3-11) 𝛒𝐧 = 𝛒𝐋 (𝛌𝐋 ) + 𝛒𝐠 (𝟏 − 𝛌𝐋 ) ------------------------------------------------------------------------------ (3-12) Where: ρs = slip density of mixture (Ibm / ft3) ρn = non slip density of mixture (Ibm / ft3) Eq. (3-11) is used to determine the pressure gradient due to elevation change. Some correlations are based on the assumption of no-slippage and therefore use Eq. (3-12) for twophase density. 4. Velocity:  Many two-phase flow correlations are based on a variable called superficial velocity (Vs).  The superficial velocity of a fluid phase is the velocity which those phases would exhibit if it flowed through the total cross section of the pipe alone.  For Gas and Liquid 𝐕𝐬𝐋 = 𝐕𝐬𝐠 =

𝐪𝐋 𝐀 𝐪𝐠 𝐀

------------------------------------------------------------------------------------------------------- (3-13) ------------------------------------------------------------------------------------------------------- (3-14)

VsL, Vsg = superficial liquid and gas velocity, respectively.  For Two Phase Mixture (Liquid-Gas) 𝐕𝐦 = 𝐕𝐬𝐋 + 𝐕𝐬𝐠 ---------------------------------------------------------------------------------------------- (3-15) 𝐪𝐦 = 𝐪𝐋 + 𝐪𝐠 ------------------------------------------------------------------------------------------------ (3-16)

6

Chapter Three Petroleum Department

LectureFourth Stage Multiphase Flow in Pipe

- -2015 Production Engineering II

The actual velocity is calculated from;  (For Gas and Liquid); 𝐕𝐋 = 𝐕𝐠 =

𝐪𝐋 𝐀 𝐇𝐋 𝐪𝐠 𝐀 𝐇𝐠

------------------------------------------------------------------------------------------------------ (3-17) ------------------------------------------------------------------------------------------------------- (3-18)

Where: A is the pipe area. HL , Hg = Liquid and gas holdup, respectively. 5. Slip Velocity The slip velocity is defined as the difference in the actual gas and liquid velocities. 𝐕𝐬 = 𝐕𝐠 − 𝐕𝐋 =

𝐕𝐬𝐠 𝐇𝐠

−

𝐕𝐬𝐋 𝐇𝐋

----------------------------------------------------------------------------------- (3-19)

Using the above definitions for velocity, an alternate equation for no-slip holdup is; 𝛌𝐋 =

𝐕𝐬𝐋 𝐕𝐦

=

𝐕𝐬𝐋 𝐕𝐬𝐋 +𝐕𝐬𝐠

------------------------------------------------------------------------------------------- (3-20)

6. Viscosity:  For Liquid Mixture (Oil-Water) 𝛍𝐋 = 𝛍𝐨 𝐟𝐨 + 𝛍𝐰 𝐟𝐰 ----------------------------------------------------------------------------------------- (3-21)  For Two Phase Mixture (Liquid-Gas) 𝛍𝐬 = 𝛍𝐋 𝐇𝐋 + 𝛍𝐠 𝐇𝐠 ------------------------------------------------------------------------------------------ (3-22) 𝛍𝐧 = 𝛍𝐋 𝛌𝐋 + 𝛍𝐠 𝛌𝐠 ----------------------------------------------------------------------------------------- (3-23) 7. Surface Tension 𝛔𝐋 = 𝛔𝐨 𝐟𝐨 + 𝛔𝐰 𝐟𝐰 ----------------------------------------------------------------------------------------- (3-24) σo , σw= oil and water surface tension respectively (dyne/cm) Problem (3-1): An oil well is flowing 10000 STB/D with a producing gas/oil ratio of 1000 scf/STB or a gas-production rate of 10 MM scf/D. At a location in the tubing where the pressure and temperature are 1700 psi and 180°F, calculate the in-situ volumetric flow rates and superficial velocities of the liquid and gas phases. Also calculate the mixture velocity and the no-slip liquid holdup. Additional given data are; Bo= 1.197 bbl/STB Bg= 0.0091 ft3/scf Rs= 281 scf/STB d= 6 in. 7

Chapter Three Petroleum Department

LectureFourth Stage Multiphase Flow in Pipe

- -2015 Production Engineering II

Solution: π

6 2

π

Ap = d2 = ( ) ( ) = 0.19625 ft 2 4 4 12 1) in-situ volumetric flow rates calculated from; 𝐪𝐨 = 𝐪𝐨𝐬𝐜 𝐁𝐨 -------------------------------------------------------------------------------------------------- (3-25) 𝐪𝐰 = 𝐪𝐰𝐬𝐜 𝐁𝐰 ------------------------------------------------------------------------------------------------- (3-26) 𝐪𝐠 = (𝐪𝐠𝐬𝐜 −𝐪𝐨𝐬𝐜 𝐑 𝐬 ) 𝐁𝐠 ----------------------------------------------------------------------------------- (3-27) Rs = Gas solubility: The gas solubility Rs is defined as the number of standard cubic feet of gas that will dissolve in one stock-tank barrel of crude oil at certain pressure and temperature. (𝐯𝐨𝐥𝐮𝐦𝐞) 𝐫𝐞𝐬𝐞𝐫𝐯𝐨𝐢𝐫 𝐏,𝐓

B: Formation volume Factor = (𝐯𝐨𝐥𝐮𝐦𝐞)

𝐬𝐮𝐫𝐟𝐚𝐜𝐞 𝐏,𝐓 ,(𝐚𝐭 𝐬𝐭𝐚𝐧𝐭𝐚𝐫𝐝 𝐜𝐨𝐧𝐝𝐢𝐭𝐢𝐨𝐧𝐬)

qo = qosc Bo = 10000 ∗ 1.197 = 11970 bbl/day bbl ft3 )∗5.615( ) day bbl sec 86400 ( ) day

11970(

qo =

= 0.778

ft3 sec

2) superficial velocities of the liquid and gas phases VsL =

qL

0.778

=

A

0.19625

= 3.97 ft/sec

qg = (qgsc −qosc R s ) Bg = [10x106 − (10000)(281)](0.0091) = 65429 qg =

ft3 day sec 86400 ( ) day

65429

Vsg =

qg A

=

= 0.757

0.757 0.19625

ft3 sec

= 3.86 ft/sec

3) mixture velocity Vm = VsL + Vsg = 3.97 + 3.86 = 7.83 ft/sec 4) no-slip liquid holdup λL =

qL qL +qg

=

0.778 0.778+0.757

= 0.507

8

ft 3 day

Chapter Three Petroleum Department

LectureFourth Stage Multiphase Flow in Pipe

- -2015 Production Engineering II

Vertical Multiphase Flow  The vertical lift performance involves a study of the pressure losses in vertical pipes carrying two-phase mixtures (gas and liquid) (∆P = Pwf – Pth).  For Vertical Flow; 1) θ= 90° , and sin 90° = 1. 2) g/gc= 1. 3) dz = dL. 4) The pressure-drop component caused by friction losses requires evaluation of a twophase friction factor. 𝐝𝐩

(𝐝𝐋) = 𝐟

𝐟𝐦 𝛒𝐬 𝐯𝐦 𝟐 𝟐𝐠 𝐜 𝐝

5) The pressure drop caused by elevation change depends on the density of the two-phase mixture. 𝐝𝐩

𝐠

(𝐝𝐋) = 𝐠 𝛒𝐬 𝐬𝐢𝐧𝛉 = 𝛒𝐬 𝐞𝐥

𝐜

6) The pressure-drop component caused by acceleration is normally negligible and is considered only for cases of high flow velocities. 7) Applying above conditions, the total pressure gradient in pipe for multiphase vertical flow becomes; 𝐝𝐩 𝐝𝐋

= 𝛒𝐬 +

𝟐 𝐟𝐦 𝛒𝐬 𝐕𝐦

𝟐𝐠 𝐜 𝐝

---------------------------------------------------------------------------------------- (3-28)

 Flow patterns in vertical two-phase flow As the pressure on a crude oil containing gas in solution is steadily reduced, free gas is evolved; as a consequence, the liquid volume decreases. This phenomenon affects the relative volumes of free gas and oil present at each point in the tubing of a flowing well. For instance, if the flowing BHP in a particular well is above the bubble point of the crude being produced, liquid only is present in the lower part of the tubing. As the liquid moves up the tubing, the pressure drops and gas bubbles begin to form. Therefore, the flow regimes changed and that effect on calculations of pressure drops. Figure (3-3) shows the flow patterns.

9

Chapter Three Petroleum Department

LectureFourth Stage Multiphase Flow in Pipe

- -2015 Production Engineering II

Fig. (3-3): shown flow patterns

Single-phase liquid: the flowing BHP in a well is above bubble point pressure, then singlephase liquid produce. Bubble flow: the liquid moves up the tubing, the pressure drops under bubble point pressure and gas bubbles begin to form. Bubbles of gas dispersed in continuous liquid medium. Slug or plug flow: as the fluid moves further up the tubing, the gas bubbles grow and become more numerous. Thus, the larger bubbles grow by entrainment of the smaller bubbles they overtake. A stage is reached in which these large bubbles extend across almost the entire diameter of the tubing, so the flow regime has become one in which slugs of oil containing small gas bubbles. Annular flow: still higher in tubing, the small gas bubbles are separated from each other by gas pockets, this is, at lower pressure, the gas pockets may have grown and expanded to such an extent that they are able to break through the more viscose oil slugs, with the result that the gas forms a continuous phase near the center of h tubing, carrying droplets of oil up with it. Along the walls of the tubing there is an upward-moving oil film.

10

Chapter Three Petroleum Department

LectureFourth Stage Multiphase Flow in Pipe

- -2015 Production Engineering II

Mist flow: continued decrease in pressure with resultant increase in gas volume results in a thinner and thinner oil film, until finally the film all disappears and the flow regime has become a continuous gas phase in which oil droplets are carried along with the gas. In addition to the flow regimes themselves, the viscosities of the oil and gas, the variations of these viscosities with temperature and pressure, the PVT characteristics of the reservoir fluids, the flowing BHP, and the tubing-head pressure (THP) all directly affect the pressure gradient at a particular point of the tubing. Figure (3-4) shown effect of pressure and temperature in tubing flow patterns.

Fig. (3-4): shown effect of pressure and temperature in tubing flow patterns

 Vertical two-phase flow Methods  Many methods have been developed to predict two-phase, flowing- pressure gradients. They differ in the method used to calculate the three components of the total pressure gradient.  The empirical correlations used to predict pressure gradient in vertical multiphase flow can be classified into three groups; 11

Chapter Three Petroleum Department

LectureFourth Stage Multiphase Flow in Pipe

- -2015 Production Engineering II

A.    

Group "A" No slip consideration, i.e. the gas and liquid are assumed to travel at the same velocity. No flow pattern consideration. Required friction factor determination. The pressure gradient equation Eq. (3-3) represents by;  Hydrostatic pressure gradient.  Friction pressure gradient. Example: Poettmann and Carpenter Correlation. B. Group "B"  Slip considered, i.e. the liquid and gas can travel at different velocities  No flow pattern considered.  Required both liquid holdup and friction factor calculations.  The pressure gradient equation Eq. (3-3) represents by;  Hydrostatic pressure gradient.  Friction pressure gradient. Example: Hagedorn and Brown Correlation. C. Group "C"  Slip considered.  Flow pattern considered.  Required liquid holdup, friction factor and flow pattern calculations.  The pressure gradient equation Eq. (3-3) represents by;  Hydrostatic pressure gradient.  Friction pressure gradient.  Acceleration pressure gradient Example: Beggs and Brill Correlation & Duns and Ros Correlation.  Poettmann and Carpenter Correlation Poettmann and Carpenter developed an empirical method which is based on the following assumptions;  No slip consideration, i.e. the gas and liquid are assumed to travel at the same velocity.  No flow pattern consideration. 12

Chapter Three Petroleum Department     𝐝𝐩 𝐝𝐋

LectureFourth Stage Multiphase Flow in Pipe

- -2015 Production Engineering II

The correlation is applicable to the pipe sizes of (2, 21/2, and 3 in.). The effects of viscosity were assumed to be negligible. The acceleration term of the general energy equation was considered to be negligible. Based on above conditions, the total pressure gradient in pipe for multiphase vertical flow; = 𝛒𝐧 +

𝟐 𝐟𝐦 𝛒𝐧 𝐕𝐦

------------------------------------------------------------------------------------------- (3-29)

𝟐𝐠 𝐜 𝐝

Where: ρn= no slip density of mixture, Ibm/ft3 fm = two phase friction factor, calculated using Fig.1 . Problem (3-2): Given: Vsg = 4.09 ft/sec VsL = 2.65 ft/sec ρg = 2.48 Ibm/ ft3 d = 0.249 ft Wm = 7.87 Ibm/sec P = 720 psia Calculate the flowing pressure gradient at these conditions. Solution:

ρL = 56.6 Ibm/ ft3

1. Determine no-slip density λL =

VsL VsL +Vsg

=

2.65 2.65+4.09

= 0.393

λg = 1 − λL = 1 − 0.393 = 0.607 ρn = ρL (λL ) + ρg (1 − λL ) = (56.6)(0.393) + (2.84)(0.607) = 23.97 lbm /ft 3 2. Determine friction factor Vm = VsL + Vsg = 2.65 + 4.09 = 6.74 ft/sec ρn vm d = (23.97)(6.74)(0.249) = 40.23 From Fig. 1, f =0.021 3. Determine total pressure gradient dp dz

= ρn +

fm ρn V2m 2gc d

= 23.97 +

(0.021)(23.97)(6.742 ) (2)(32.2)(0.249)

13

= 25.386

Ibf /ft2 ft

Chapter Three Petroleum Department dp dz

=

25.386 144

= 0.176

LectureFourth Stage Multiphase Flow in Pipe

psi ft

(where psi =

Ib in2

- -2015 Production Engineering II

)

 Hagedorn and Brown Correlation Hagedorn and Brown developed this pressure gradient equation for vertical multiphase flow; 𝐝𝐩 𝐝𝐋

= 𝛒𝐬 +

𝟐 𝐟𝐦 𝛒𝟐𝐧 𝐕𝐦

𝟐 𝐠 𝐜 𝛒𝐬 𝐝

+

𝟐) 𝛒𝐬 ∆(𝐕𝐦

𝟐 𝐠 𝐜 𝐝𝐳

----------------------------------------------------------------------------- (3-30)

 In this method the slip is considered, i.e. the liquid and gas can travel at different velocities; a liquid-holdup value must be determined to calculate the pressure-gradient component that results from a change in elevation.  To calculate the liquid-holdup values, Hagedorn and Brown used four dimensionless groups: Liquid velocity number: 𝛒

𝟒

𝐍𝐋𝐯 = 𝟏. 𝟗𝟑𝟖 𝐕𝐬𝐋 √ 𝐋 ------------------------------------------------------------------------------------- (3-31) 𝛔 𝐋

Gas velocity number: 𝟒

𝐍𝐠𝐯 = 𝟏. 𝟗𝟑𝟖 𝐕𝐬𝐠 √

𝛒𝐋

-------------------------------------------------------------------------------------- (3-32)

𝛔𝐋

Pipe diameter number: 𝐍𝐝 = 𝟏𝟐𝟎. 𝟖𝟕𝟐 𝐝 √

𝛒𝐋

𝛔𝐋

-------------------------------------------------------------------------------------- (3-33)

Liquid viscosity number: 𝟒

𝐍𝐋 = 𝟎. 𝟏𝟓𝟕𝟐𝟔 𝛍𝐋 √ Where: VsL and Vsg= ft/sec

𝟏 𝛒𝐋𝛔𝟑𝐋

---------------------------------------------------------------------------------- (3-34)

ρL= Ibm/ ft3 14

Chapter Three Petroleum Department

LectureFourth Stage Multiphase Flow in Pipe

σL = liquid surface tension =dyne/cm

- -2015 Production Engineering II

µL= cpd =ft

   

Fig.2 shows the correlation for liquid holdup divided by a secondary correction factor, ψ. The correlating function requires a value of CNL, which is correlated with NL in Fig.3. Fig.4 shows the secondary correction-factor ψ correlation. When a value for liquid holdup has been determined from Figs.2 through 4, the slip density (ρs) can be calculated from Eq. (3-11).  When a value for slip density (ρs) has been determined, the pressure gradient due to elevation change can be calculated from Eq. (3-30). 𝐝𝐩 ( ) = 𝛒𝐬 𝐝𝐋 𝐞𝐥𝐯  Friction-Factor Prediction. The two-phase friction factor (fm) is correlated with a two-phase Reynolds number and relative roughness (ε/d) using a standard Moody diagram, Fig.5.  The Reynolds number is calculated from; 𝐍𝐑𝐞 = 𝟏𝟒𝟖𝟖

𝛒𝐧 𝐯𝐦 𝐝 𝛍𝐬

---------------------------------------------------------------------------------------- (3-35)

A) Laminar Flow NRe < 2000 → Laminar Flow 𝐟𝐦 =

𝟔𝟒 𝐍𝐑𝐞

----------------------------------------------------------------------------------------------------- (3-36)

B) TurbulentFlow 3000 < NRe < 3x1010 For smooth wall pipes;

→ Turbulent Flow

𝐟𝐦 = 𝟎. 𝟎𝟎𝟓𝟔 + 𝟎. 𝟓 𝐍𝐑𝐞 −𝟎.𝟑𝟐 --------------------------------------------------------------------------- (3-37)  The inside wall of a pipe is not normally smooth, and, in turbulent flow, the roughness can have a definite effect on the friction factor, and thus the pressure gradient. Therefore; The two-phase friction factor (fm) is correlated with a two-phase Reynolds number and relative roughness (ε/d) using a standard Moody diagram, Fig.5. 15

Chapter Three Petroleum Department

LectureFourth Stage Multiphase Flow in Pipe

- -2015 Production Engineering II

ε = absolute roughness of a pipe, (ft) = It is mean protruding height of uniformly distributed and sized packed sand grains that would give the same pressure gradient behavior as the actual pipe]. d = Pipe diameter, (ft). ε/d = relative roughness, dimensionless.  When a value for friction factor has been determined, the pressure gradient due to friction loss can be calculated from Eq. (3-30). 𝟐 𝐝𝐩 𝐟𝐦 𝛒𝟐𝐧 𝐕𝐦 ( ) = 𝐝𝐋 𝐟 𝟐 𝐠 𝐜 𝛒𝐬 𝐝  Acceleration Term. The pressure gradient resulting from acceleration is given by; 𝟐) 𝐝𝐩 𝛒𝐬 ∆(𝐕𝐦 ( ) = 𝐝𝐋 𝐚𝐜𝐜 𝟐 𝐠 𝐜 𝐝𝐳 Where; 𝟐 𝟐 𝟐) ∆(𝐕𝐦 = 𝐕𝐦𝟏 − 𝐕𝐦𝟐 and 1, 2 designate downstream (P1, T1) and upstream (P2, T2) ends of a calculation increment, respectively.  If we define parameter as; 𝐄𝐤 =

𝐝𝐋 𝐝𝐩

( ) 𝐝𝐏 𝐝𝐋

𝐚𝐜𝐜

=

𝟐) 𝛒𝐬 ∆(𝐕𝐦

𝟐 𝐠 𝐜 𝐝𝐳

-------------------------------------------------------------------------------- (3-38)

The total pressure gradient can be calculated from; 𝐝𝐩 𝐝𝐋

=

𝐝𝐩 𝐝𝐋 𝐞𝐥

𝐝𝐩 𝐝𝐋 𝐟

( ) +( ) 𝟏−𝐄𝐤

---------------------------------------------------------------------------------------------- (3-39)

Problem (3-3): Given Vsg = 4.09 ft/sec VsL = 2.65 ft/sec d = 0.249 ft P = 720 psi

ρg = 2.48 Ibm/ ft3 T = 128o F

ρL = 56.6 Ibm/ ft3 μo = 18 cp μg = 0.018 cp

NLV = 6.02

NL = 0.08

Nd = 41.34

NgV = 9.29

ε/d = 0.0006

Neglecting acceleration, calculate the flowing pressure gradient at these conditions. 16

Chapter Three Petroleum Department

LectureFourth Stage Multiphase Flow in Pipe

- -2015 Production Engineering II

Solution: 1. Determine liquid holdup and two-phase density. A. Determine CNL from Fig.3 for NL = 0.08 to be 0.0055 B. Determine (HL/ψ) from Fig.2. P 0.1 CNL (N0.575) (P ) ( N ) a d gv NLv

6.02

=(

9.290.575

720 0.1 0.0055

( 41.34 ) = 0.00032

) (14.7)

From Fig.2, (HL/ψ) = 0.52 C. Determine secondary correction factor, ψ from Fig.4. Ngv N0.38 L N2.14 d

=

(9.29)(0.08)0.38 (41.34)2.14

= 0.00124

From Fig.4, ψ = 1 D. Determine liquid holdup from: H

HL = ( L) . ψ = (0.52)(1) = 0.52 ψ ρs = ρL (HL ) + ρg (1 − HL ) = (56.5)(0.52) + (2.84)(1 − 0.52) = 30.79 2. Determine elevation term. dp

(dL)

elv

= ρs = 30.79

Ibf ft3

3. Determine frictional term. ρn =

ρL VsL +ρg Vsg Vm

=

(56.6)(2.65)+(2.84)(4.09) 6.74 0.52

μs = μL HL + μg Hg = (18) NRe = 1488

ρ n vm d μs

=

= 23.98

Ibm ft3

+ (0.018)0.48 = 0.65 cp

(1488)(23.98)(4.09+2.65)(0.249) 0.65

= 92135

From Fig.5, fm = 0.021 , for ε/d = 0.0006 dp

(dL) = f

fm ρ2n V2m 2gc ρs d s

=

(0.021)(23.98)2 (4.09+2.65)2 (2)(32.2)(30.79)(0.249)

= 1.11

4. Calculate total pressure gradient dp dL

= ρs +

fm ρ2n V2m 2 gc ρs d

= 30.79 + 1.11 = 31.9

Ibf ft3

17

Ibf /ft2 ft

Ibm ft3

Chapter Three Petroleum Department

dp dL

Ib

=

31.9 3f ft 144 in3 /ft3

= 0.221

LectureFourth Stage Multiphase Flow in Pipe psi ft

or 0.221

- -2015 Production Engineering II

Ib/in2 ft

H.W-1: Given: µo = 0.97 cp, σo= 8.41 dynes/cm, µg = 0.016 cp, ε= 0.00006 ft. Using the Modified Hagedorn and Brown Method, Calculate the Vertical, Multiphase-Flow Pressure Gradient for problem (3-1). (Answer: dp/dL = 0.193 psi/ft) Procedure: 1. Determine Duns and Ros dimensionless groups: NLv = 11.87 Ngv = 11.54 Nd = 143.8 NL = 0.0118 2. Determine liquid holdup: HL = 0.3 3. Check validity of HL: Because HL