Pipeline Design.pdf

8/16/2019 Pipeline Design.pdf

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Copyright 2006, NExT, All rights reserved

Surface Facilities

Pipelines Design, Operation and Maintenance

Leonardo Montero R., M.Sc.


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Engineering Required Before Designing a Pipeline

Exploration and Production

Reservoir Geology

Geoscience Petroleum Engineering

FacilityEngineering

PipelineDesign

Geophysics

ExplorationGeology

Drilling

ReservoirSimulation

ReservoirDescription

Reservoir Management

ProductionManagement

Well SystemDefinition

ProcessDefinition

Pipelines

Manifolds

Controls

HostEngineering


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Pipelines - Terminology

Flowlines & Gathering Lines – The lines travelshort distances within an area. They gather productsand move them to processing facilities.

Flowlines are usually small, e.g. 2- 4 in diameter,and gathering lines bigger (say 4-12” )

They carry many products, often mixedtogether.

Feeder Lines - These pipelines move productsfrom processing facilities, storage, etc., to the maintransmission lines

Typically 6-20 in diameter Carry variety of products, sometimes ‘batched’.


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Pipelines - TerminologyTransmission Lines - These are the main conduits of oil

and gas transportation.

These lines can be very large diameter (up to 56 in)

Natural gas transmission lines deliver to industry or‘distribution’ system.

Crude oil transmission lines carry different types ofproducts, sometimes batched, to refineries or storage

Petroleum product lines carry liquids such as refinedpetroleum products or natural gas liquids.

Distribution Lines - These lines allow local distributionfrom the transmission system.

These lines can be large diameter, but most are under6 in diameter


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Pipelines - System

ProductionProductionWells

PlatformsSurface Facilities

Transportation DistributionGate Station

Metering EquipmentCompression Stations

Metering Equipments

Compression Station

IT System

Drawdown AnalysisLinepacking Analysis

Wells

Gathering Pipeline

Storage

Commercial

Residential

Industrial & Utilities

Storage Distributors

Plants


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Oil and Gas Transportation by Pipelines

Offshore

Receiving Facilities

Land PipelineTransmission

J- TubesRisersProcess Equipment

Shore Approaches

Distribution Lines

Trunk Lines

Crossings

WyesTeesHot Taps

Manifolds

FlowlinesCables


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Pipelines are Preferred

Pipeline is the main mode of transportation for liquid

and gas, for several reasons: Less damaging to the environment

Safety: It is the safest the for oil and gastransportation

Economical: Is the most efficient method totransport high volume

Reliability


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Pipelines Around the World

0

125

250

375

500

625

750

UK Western Europe USA Rest of The World

L e n g

t h ( M i l e s

)

T h o u s a n

d s

Onshore Gas Trans > 300.000 miles

Offshore Gas Trans > 6.000 miles

Onshore Gas Gathering > 21.000 miles

Offshore Gas Gathering > 6.000 miles

Onshore Distribution > 1.000.000 miles

Liquid Trans. Lines > 157.000 miles


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Gathering LinesGathering Lines


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These lines travel short distances within an

area.Gathers products and moves them toprocessing facilities.

Flowlines are usually small, e.g. 2- 4indiameter,

Gathering lines bigger (say 4-12” )

They carry many products, often mixedtogether.

Flowlines and Gathering Lines


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Flowlines and Gathering Lines

Flow Station

Multiphase Manifold

Tank Farm

Multiphase Pipeline

Oil Pipeline

LL-33

LL-3416"-0.375

1 2 " - 0

. 4 4

M-LH-7

10"-0.365; 31,8%

12"-0.44; 13,5%

M-LH-8

LL-39

LL-35

LL-29

LL-41LL-47

LL-20

2 4 " -

0 . 3 8

; 9 6 0

0 ' ( 1 9

9 2 )

1 2 " - 0 . 3

7 5

1 2 " - 0

. 3 7 5

1 2 " - 0 . 3 7 5 ; 2 0 , 3 %

v i s i b l e

24"

16"

24"

1 0 " - 0 .

3 6 5

12"-0.44"; LL-16

2 0 " - 0. 3 7 5; 5 5 91

' (1 9 9 0 )

Macolla 3

LL-87

M-LH-9

2 4" -0 .3 7 5 ; 2 43 0 ' ( 19 9 3 )

2 4 " - 0

. 3 7 5 ; 1

1 4 3 6 ' ( 1

9 9 3 )

12"

16"

6"

20" 12"

Linea de 8" que debe ser

desactivada

8"

6 8 0 4

' ( 1 9 9 0

)

3 8 4 9 ' (

1 9 8 7

)

6 3 0 '

( 1 9 7

9 )

6 1 8 4

' ( 1 9 8 0

)

1 5 9 9

' ( 1 9 9

0 )

5 4 6 7

' ( 1 9 7 9 )

1 1 4 8 '

( 1 9 8 8 )

4 6 2 1 ' ( 1

9 8 1 )

8 2 6

9 ' ( 1

9 9 0 )

Vertical deteriorado (corroido)

1 3 3 0 ' ( 1

9 8 8 )

1 0 7 0 ' ( 1

9 8 0 )

4 6 5 1 ' ( 1 9 9 0 )

2 7 8 7 ' (

1 9 7 4 )

7 7 8 ' (

1 9 8 8 )

3 2 7

2 ' ( 1 9 7

3 )

2 3 2 ' ( 1 9 8 8 )

5 0 0 '

( 1 9 8 8

)

6 7 1 7 ' ( 1

9 9 0 )

3 6 3 3 ' (

1 9 7 4 )

1331' (1989)4047' (1977)

663' (1980)

6"

4 2 6 ' (

1 9 8

8 )

6 1 5 7 '

( 1 9 7

9 )

1 0 2 8 '

( 1 9 8

9 )

LL-41(nueva)

1 6

" - 0 .

3 7 5 ;

1 4 0

3 3 '

( 1 9

9 6 )

1 6 " - 0 . 3 8 ;

6 2 0 0 ' (

1 9 9 5

)

4000'

LL-83

LL-37

2 4 " - 0

. 3 7 5 ; 2

0 0 0 0 ' ( 1

9 8 6 )

1 6 " ; 5

0 0 0 '

( 1 9 9

6 )

1 6 " -

0 . 3

7 5

; 5 0 8 0 ' ( 1

9 9 6

)

Grapa a nivel sublacustre

Grapa colocada para corregir

corrosión en vertical.

Oil Manifold


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FS-5-9FS-1-8

FS 21-5

FS 2-6

FS 16-5

FS 9-5

FS 1-5FS 22-5

FS 5-6

PE 8-3

PA

EM-2

EM-1

GasPlant

PC-VII

GasPlant

Gas

Plant

FS-23-5

MG-CL-1

High Pressure System

Low Pressure System

Evaluation with Simulators:

*Pipephase, Stationary State* Pipesim, Stationary State* TGNET, Dynamic State

Equations:Bernoulli*Beggs & Brill * Moody o Darcy*Weymouth* Panhandle A/B

* AGA

Gas Gathering System: Example

The gas gathering system consists of several interconnected pipelines withdiameter between 4 and 12 inches and low pressure line (< 500 psi).


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Gas Gathering System: Types

The smallest gathering system consists simply of two

or more gas wells interconnected by piping and tied

directly into a distribution system.

For large fields and for several interconnected fields

involving hundreds of miles of piping, gathering

systems may include such equipment as drips,

separators, meters, heaters, dehydrators, gasoline

plant, sulfur plant, cleaners and compressors, as wellas piping and valves.


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Axial Gathering SystemIn the axial gathering system, several

wells produce into a common flowline.

Flowlines

Wellhead

Header


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Radial Gathering SystemFlowlines emanating from several different wellheads

converge to a central point where facilities are located.

Flowlines are usually terminated at a header, which is

essentially a pipe large enough to handle the flow of all

the flowlines

Wellhead

Compression Station


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Wellhead

Loop Gathering System

CompressionStation

Separator


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Gathering System: Well Center

Well Center Gathering System

Central Gathering Section

Well Center

The well center gathering system uses radial philosophyat the local level for individual wells, brings all theflowlines to a central header


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Gathering System: Trunk Line

Uses an axialgathering scheme for

the groups of wells.Uses several remoteheaders to collectfluid.Is more applicable to

relatively largeleases, and no caseswhere it isundesirable orimpractical to build

the field processingfacilities at a centralpoint.

Trunk Line

Well Head

Header


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Gathering System: Decision

The choice between the gathering systems

is usually economic.The cost of the several small sections ofpipe in well-center system is compared tothe cost of single large pipe for the trunk-

line system.

Technical feasibility may be anothercriterion.

The production characteristics of the field


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Gathering System: Pipeline System

Series Pipelines

LA LB LC

A B C

Looped Pipelines Loopless Pipeline Systems

q1

q2 q3 qn-1 qnqn + 1

p1 p2 p3 pn-1 pn pn+1

NCE

1 2 3 n-1 n n+1Nodenumber

pressure

1 2 3 n-1 nA

B

C

LA LC

Parallel Pipelines

A

B

C


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Looped pipeline: Alooped pipeline is one in

which only a part of the linehas a parallel segment. Theoriginal pipeline is looped tosome distance with anotherline to increase the flowcapacity.Le = LC + (Le )AB

Looped pipelines

A

B

C

LA LC


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Loopless Pipelines : Aloopless pipeline system,

defined as one where theNCE's (node connectingelements) joined by nodesform no closed loop

Loopless Pipelines

q1

q2 q3 qn-1 qnqn + 1

p1 p2 p3 pn-1 pn pn+1

NCE

1 2 3 n-1 n n+1Nodenumber

pressure

1 23 n-1 n


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Gathering System: Equations for Complex Gas Flow


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Flow of Fluid

Fluid is defined as a single phase of gas or liquid or both.Each sort of flow results in a pressure drop.

Three categories of fluid flow: vertical, inclined andhorizontal

Overall production system


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Flow of Fluid

Possible Pressure Losses Possible Pressure Losses ∆∆∆∆∆∆∆∆pp88== PPwhwh--PPsepsep

PPwfswfs--PPwfwf == ∆∆∆∆∆∆∆∆pp22

P P m m P P e e P P wfs wfs P P wf wf

Pur

Puv

Pdv

P wh P ds P P sep sep

Gas Flowlines

Separator

Tanks

Reservoir

Flowlines

Well

∆∆∆∆∆∆∆∆pp11=P=Pmm--PPwfswfs

∆∆∆∆∆∆∆∆pp33== PPurur--PPdrdr

∆∆∆∆∆∆∆∆pp44== PPuvuv--PPdvdv

Pdr

∆∆∆∆∆∆∆∆pp66== PPdsds--PPsepsepPPwhwh--PPdsds == ∆∆∆∆∆∆∆∆pp55

∆ ∆∆ ∆ ∆ ∆∆ ∆ p p 7 7 = = P P w f w f - - P P w h w h

Surface Choke

Safety Valves

Bottom Hole

Restricción

∆∆∆∆p1=Pm- Pwfs Loss in porous medium∆∆∆∆p2=Pwfs- Pwf Loss across completion∆∆∆∆p3=Pur- Pdr Loss across restrictions∆∆∆∆p4=Puv- Pdv Loss across safety valves∆∆∆∆p5=Pwh- Pds Loss across surface choke

∆∆∆∆p6=Pds- Psep Loss in flowlines∆∆∆∆p7=Pwf- Pwh Total loss in tubing∆∆∆∆p8=Pwh- Psep Total loss in flowlines

Source: Handbook of Petroleum and Gas Engineering, William Lyons


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Flow of Fluid

Production Pressure Profile Production Pressure Profile

Source: Handbook of Petroleum and Gas Engineering, William Lyons

Reservoir Tubing Flowline Transfer Line

DrainageBoundary

Wellbore(Perforations)

Wellhead &Choke Separator

StockTank

P r e

s s u r e

ro W

Pwf

Po

Pwf

PspPST


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Flow of Fluid

SINGLE-PHASE FLOW: Liquid and gas velocity in a pipeline

q A u

u = q / A

It is the flow rate (q), at pressure and temperature in the pipe,divided by cross-sectional area of the pipe (A). It is calculated by

the following equation:


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A) Laminar Flow B) Turbulent Flow

Laminar Flow ⇒⇒⇒⇒ Re < 2000

Turbulent Flow⇒⇒⇒⇒

Re > 2100

R = Duρρρρ/µµµµ

Pipeline Fluid Flow

pipeline

velocity

pipeline

velocity


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Flow of Fluid

∆∆∆∆p = ∆∆∆∆pPE + ∆∆∆∆pKE + ∆∆∆∆pF∆∆∆∆pPE : pressure drop due to potential energy change

∆∆∆∆pKE : pressure drop due to kinetic energy change

∆∆∆∆pF : frictional pressure dropu : velocity of the fluid

D :pipeline internal diameter

L :Length of the pipe

f : friction factor

ρ : liquid density

Single-Phase Flow: Liquid

Pressure Drop Calculation

dp + udu + g dz + 2 f u2 dL = 0ρρρρ gc gc D gc (Energy Equation)

g ρ∆ρ∆ρ∆ρ∆z + ρρρρ ∆∆∆∆u2 + 2f ρρρρu2 L (ρρρρ = constant)

gc∆∆∆∆p = p1 – p2 =

2gc D gc

Where:


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Flow of Fluid

∆∆∆∆z = z2 – z1 = L sin θθθθ∆∆∆∆pPE = (g/gc)ρρρρL sin θθθθ

∆∆∆∆pPE , the pressure drop due to potential energy change

θθθθ = 0 ∆∆∆∆pPE = 0

q

∆∆∆∆z

L

1

2

θθθθ

q

∆∆∆∆z

L

2

1

θθθθ

(a) Upward flow (b) Downward flow


Horizontal Flow


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∆∆∆∆pKE = the pressure drop due to kinetic energy change

Is the pressure drop resulting from the change in the velocity of the fluidbetween positions 1 and 2.

∆∆∆∆pKE = (ρρρρ/2gc) ∆∆∆∆u2 = (ρρρρ/2gc) (u22- u12)

ρρρρ = constant , A = constant ∆∆∆∆pKE = 0

q = constant

u = q/A , A = ππππD2/4 ∴∴∴∴ u = 4q/ππππD2 ∆∆∆∆pKE = 8ρρρρq2/ππππ2gc(1/D24 – 1/D14)

Flow of Fluid


Where: u = Velocity of the fluid, ft/sec.q = Volumetric flow rate, ft3/sec.

D = Pipeline internal diameter, ft

ρρρρ = Liquid density, lbm/ft3

A = Pipeline cross-sectional area, ft2


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Flow of Fluid


Example

Suppose that 2000 bbl/d of oil with a density of 58 lbm/ft3 isflowing through a horizontal pipeline having a diameter reductionfrom 4 in. to 2 in., as illustrated in the figure. Calculate the kineticenergy pressure drop caused by the diameter change.

D1D2

q

u1

q

u2


Fl f Fl id


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Since ρρρρ = constant, then ∆∆∆∆pKE = 8ρρρρq2

/ππππ2

(1/D24

– 1/D14

)q = (2000 bbl/d)(5.615 ft3/bbl)(day/86400 sec.) = 0.130 ft3/sec.

D1 = (4/12) ft = 0.3333 ft

D2 = (2/12) ft = 0.16667 ft

∆∆∆∆pKE =

Flow of Fluid


Solution:

(ππππ2 x 32.17 ft-lbm/lbf-sec2)]

8(58 lbm/ft3)(0.130 ft3/sec.)2-

(0.3333)4 (0.16667)41 1[ ]

∆∆∆∆pKE = 0.28 psi


Fl f Fl id


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∆∆∆∆pf = the pressure drop due to friction

The frictional pressure drop is obtained from the equation:

Where: f = is the Moody’s friction factor.

In laminar flow NRe > 2100 f = f(NRe,εεεε)

where NRe : is the Reynolds number

εεεε : is the relative pipe roughness

which are given by:

NRe = ρρρρud/µµµµ

εεεε = k/D (k = Absolute roughness, in)

Flow of Fluid

∆∆∆∆pf =fρρρρu2L2gcD


Fl f Fl id


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Flow of Fluid

where ρρρρ = Liquid density, lbm/ft3

u = Velocity, ft/s

D = Internal pipeline diameter, ft

µµµµ = Liquid viscosity, lbm/ft-s

Other expresions:

NRe = 1488 ρρρρuD/µµµµ

where:ρρρρ : Liquid density, lbm/ft3

u : Velocity, ft/s

D : Internal pipeline diameter, ft

µµµµ : Liquid viscosity, cP


Fl f Fl id


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Flow of Fluid


In oilfield units

NRe = 1.48 qρρρρ/Dµµµµ = 92.35 γ γγ γ Lq/Dµµµµ

Where:ρρρρ : Liquid density, lbm/ft3γ γγ γ

L

: Liquid specific gravityq : Volumetric flow rate, bbl/dD : Internal pipe diameter, in.µ: Liquid viscosity, cP

NRe

= 1.722 x 10-2 w D/A µµµµ

Where:w : Mass flow rate, lbm/dA : Pipeline cross-sectional area, ft2

µµµµ : Liquid viscosity, cP



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Flow of Fluid


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Relative Roughness of Common Piping Material.

Flow of Fluid


Flow of Fluid


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Example

Calculate the frictional pressure drop for the 1000 bbl/d of brine injectiondescribed in Example No. 1. The brine has a viscosity of 1.2 cP, and thepipe relative roughness is 0.001.

Solution:

First, the Reynolds number must be calculated to determine if the flow islaminar or turbulent.

NRe = ρρρρuD/µµµµ = 1.48qρρρρ/Dµµµµ = (1.48)(1000bbl/d)(65.5 lbm/ft3)/(2.259 in.)(1.2 cP)

= 35,700 > 2100 ∴∴∴∴ the flow is turbulent

Using Chen equation:

1/√√√√f =

]}[log(0.001)1.1098

2.8257+ 7.194

3.57 x104( )0.89810.001

3.7065-

3.57 x 1045.0452-2log{

Flow of Fluid

Single-Phase Flow: LiquidHasta aquí vamos


Flow of Fluid


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f = 0.0252

u = q/A = 4q/ππππD2 = 4(1000 bbl/d)(5.615 ft3/bbl)(1day/86,400 s)

ππππ[(2.259/12) ft]2= 2.33 ft/s

∆∆∆∆pF =(0.0252)(65.5 lbm/ft3)(2.33 ft/s)2 (1000 ft)

2(32.17 ft-lbm/lbf-s2)[(2.259/12) ft]

= (740 lbf/ft2)(ft2/144 in2) = 5.14 psi

Notice that the frictional pressure drop is considerable less than the potentialenergy or hydrostatic pressure drop, which it was calculated to be -292 psi inExample No. 1

Flow of Fluid



Flow of Fluid


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Example

The 1000 bbl/d of injection water described in Examples 1 and 3 is suppliedto the wellhead through a 3000 ft long, 1 ½ in. I.D. flow line from a centralpumping station. The relative roughness of the galvanized iron pipe is0.004. If the pressure at the wellhead is 100 psia, what is the pressure at thepumping station, neglecting any pressure drops through valves or other

fittings?

Solution:

NRe = 1.48qρρρρ/Dµµµµ = 1.48(1000 bbl/d)(65.5 lbm/ft3)/(1.5 in.)(1.2 cP) = 53,900

1/√√√√f =

f = 0.0304

]}[log(0.004)1.1098

2.8257 + 7.1945.39 x104( )0.89810.0043.7065 - 5.39 x 1045.0452-2log{

Flow of Fluid



Flow of Fluid


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4(1000 bbl/d)(5.615 ft3

/bbl)(1day/86,400 s)ππππ[(1.5/12) ft]2

= 5.3 ft/s


u = q/A = 4q/ππππD2 =

∆∆∆∆pF = p1 – p2 =(0.0304)(65.5 lbm/ft3)(5.3 ft/s)2 (3000 ft)

2(32.17 ft-lbm/lbf-s2)[(1.5/12) ft]

= 20,864 lbf/ft2 = (20,864 lbf/ft2) (ft2/144 in.2) = 145 psi

p1= p2 + 145 = 100 + 145 = 245 psia

This is a significant pressure loss over 3000 ft. It can be reducedsubstantially by using larger pipe for this water supply, since thefrictional pressure drop depends approximately on the pipe diameter tothe fifth power

Flow of Fluid



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Flow of Fluid


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To determine de diameter of the pipe

The equation can not be solve directly

Assume a friction factor (start with 0.025)Determine the Reynolds number

Read the friction factor in figure and

compare.Iterate the solution until the friction factorconverge.

Flow of Fluid

∆∆∆∆pf =fρρρρu2L

2gcD=11.5x10-6

fQ2L γ γγ γ L

D5

Flow of Fluid


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Flow of Fluid

Hazen-Williams Formula: To avoid iteration

0.015

C

Q1.85

D

1.854.87

WhereHL : Head loss due to friction. ftQ : Liquid flow rate, bpd

C : friction factor constant: 140 for new steel pipe: 130 for Cast iron pipe: 100 for riveted pipe

L : Length of the pipe, ftD : Internal pipe diameter, in.

∆∆∆∆P =HLγ γγ γ L xρρρρw144144144144

HL =

Pressure Drop in Liquid Pipeline


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Exercise

p q p

A pipeline transport condensate (800 bpd) and water (230 bpd). Thecondensate and water specific gravity are 0.87 and 1.05, respectively.Viscosity = 3cP, Length of the pipeline 7,000 ft., Inlet pressure 900 psi

and temperature 80ºC.Determine the pressure drop for 2 inch, 4 inch and 6 inch I.D, usingthe general equation and Hazen Williams (Assume C=120. AssumeOld pipeline (εεεε=0.004)



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Solution

p q p

Mixture’s rule

γ γγ γ L= X1x γ γγ γ 1 + x2x γ γγ γ 2 =(230)

(230 +800)

1.05 +(800)

(230 +800)

0.87

γ γγ γ L= 0.91

NRe =92.35 γ γγ γ Lq

Dµµµµ

In oilfield units

=D3

92.35 x0.87x1030

D

28,853=

f = f(NRe,εεεε)

∆∆∆∆pf =fρρρρu2L

2gcD=11.5x10-6

fQ2L γ γγ γ L

D

5

Pressure Drop



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Solution

p q p

Mixture’s rule

f = f(NRe,εεεε)∆∆∆∆pf =fρρρρu2L

2gcD=11.5x10-6

fQ2L γ γγ γ L

D5

Pressure Drop

∆∆∆∆pf = 11.5x10-6 f (1030)

2x7000x 0.91

D5

∆∆∆∆pf =f 77,716

D5



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2 inch 4 inch 6 inch

Re 14427 7200 4809

εεεε/D 0.0020 0.0010 0.0007

f (from chart) 0.032 0.034 0.038∆∆∆∆P (psi) 77.7 2.6 0.4

Diameter


Flow of Fluid


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SolutionHazen-Williams

0.015 ∆∆∆∆P =HLγ γγ γ L xρρρρw144144144144HL = C

Q1.85

D 1.854.87L

2 inch 4 inch 6 inch

HL (ft) 192 6.6 1

∆∆∆∆P (psi) 75.6 2.6 0.4

Diameter


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Flow of Fluid

Single phase: Gas

Flow of Fluid

Single phase: Gas


Flow of Fluid


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Steady state flow in simple pipeline systems: Gas flow equations

The basic energy balance on a unit mass basis:

MW p

ZRT

28.97γ γγ γ g p

ZRT=

TTsc

pscp

ZRT28.97γ γγ γ gp

dp +g

gcsin θθθθ 8 f

ππππ2gcD5+ qsc Z dL = 0

(From the real gas law)

u = 4ππππ D2

qsc ZTTsc

pscp

(The velocity in terms of the volumetric flowrate at standard conditions)

dz = sin θθθθ dL and dWs = 0 (Neglecting for the time being anykinetic energy change)

Single-Phase Flow: Gas

dp + udu + g dz + 2 f u2 dL + dWs= 0

ρρρρ gc gc D gc(Energy Equation)

ρρρρ =

2


Flow of Fluid


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Where: qsc : gas flow rate measured at standard conditions, Mscfd

psc : pressure at standard conditions, psia

Tsc : temperature at standard conditions, ºR

p1 : upstream pressure, psia

p2 : downstream pressure, psia

D : diameter of pipe, in

γ γγ γ g : gas specific gravity

T : flowing temperature, ºR

Z:average gas compressibility

f : Moody friction factor

L : length of pipe, ft



Flow of Fluid


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61

S ch l um b er g er P r i v a

t e



To solve this equation notice that:

a) Z, T and p are functions of position, z

b) Rigorously solution need: T = T(z) and Z = Z(T,p) (Equation ofState)

c) This approach will likely require numerical integration

d) Alternatively,

e) Average values of Z and T can be assumed

f) Mean temperature (T1 + T2)/2 or Log-mean temperature

Tlm = (T2 – T1)/ln(T2/T1)

h) Solving for horizontal flow yields

p12 – p2

2 =(16)(28.97) γ γγ γ g f ZT

ππππ2gcD5R(

pscqscTsc

) L2



Flow of Fluid


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t e


For oilfield units:

p12 – p2

2 = 2.5175 x 10-5γ γγ γ g f ZT qsc2 L

D5

NRe = 20.09γ γγ γ g qscDµµµµ

Where: p : psia

q : Mscfd

D : in.

L : ft

µµµµ : cP

T : ºR


Where: f = f(NRe,εεεε) Moody diagram

NRe =4(28.97) γ γγ γ g qsc psc

ππππ D µµµµ R Tscand εεεε = k/D



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Flow of Fluid


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65


t e



The equation is an implicit equation in p and must be solvediteratively. It can be solved first by neglecting the kinetic energy

term; then, if ln(p1/p2) is small compared with 6fL/D, the kinetic

energy pressure drop is negligible.

NRe = 20.09 D µµµµ

γ γγ γ g qsc



Flow of Fluid


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t e



Example

Gas production from a low-pressure gas well (wellhead pressure = 100

psia) to be transported through 1000 ft of a 3.in.-I.D., line (εεεε = 0.001) to acompressor station, where the inlet pressure must be at least 20 psia.The gas has a specific gravity of 0.7, a temperature of 100 ºF and anaverage viscosity of 0.012 cP. What is the maximum flow rate possiblethrough this gas line?

Solution:

Solving for q:

(p12 – p2

2) D4

(4.195 x 10-7) γ γγ γ g Z T [(6 f L/D) + ln(p1/p2)]qsc =

0.5

p12 – p2

2 = (4.195 x 10-7)D4

γ γγ γ g Z T qsc+ ln

D

6 f Lp2

p12



Flow of Fluid


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t e



Assuming (1) that the friction factor depends only on the piperoughness. Then from the Moody diagram, for high Reynolds numberand a relative roughness of 0.001

f = 0.0196

and (2) that Z = 1 at these low pressures. Then

4.73 x 109

39.2 + 1.61

Checking the Reynolds number,

NRe = (20.09)(0.7)(10,800)/[(3)(0.012)] = 4.2 x 106

qsc =(1002 – 202)(3)4

(4.195 x 10-7)(0.7)(1)(560) {[(6)(0.0196)(1000)/3] + ln(100/20)}

0.5

qsc = = 10,800 Mscfd

0.5



Flow of Fluid


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t e



So the friction factor based on fully rough wallturbulence is correct.

It is found that this line can transport over 10

MMscfd. Notice that even at this high flow rateand with a velocity five times higher at the pipeoutlet than at the entrance, the kinetic energycontribution to the overall pressure drop is still

small relative to the frictional pressure drop.



Flow of Fluid


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t e


Steady state flow in series pipeline systems: Gas flow equations

Waymouth Equation

f = 0.032/D1/2 and qsc = 1.11 D2.67

p12 – p22

L γ γγ γ g Z T1

0.5

Where:

qsc : gas flow rate, MMscfd

D : pipe internal diameter, in.

p1 : inlet pressure, psia

p2 : outlet pressure, psia

L : length of pipe, ftγ γγ γ g : gas gravity

T1 : temperature of gas at inlet, ºR

Z : compressibility factor of gas



Flow of Fluid


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t e



Waymouth Equation

Comments:

Moody friction factor is independent of the Reynoldsnumber and dependent upon the relative roughness.

For a given roughness, εεεε, the friction factor is merely afunction of diameter.

Industry experience indicates that Weymouth’sequation is suitable for most piping within the

production facility.

Good for short lengths of pipe with high pressure dropand turbulent flow



Flow of Fluid


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t e



Panhandle Equation

f =C

NReNRe = 5 x 10

6 to 11 x 106 n = 0.146

NRe >>>> 11 x 106 n = 0.039

Using this assumption and assuming a constant viscosity forthe gas,

A) qsc = 0.020 Ep1

2 – p22

γ γγ γ g0.853 Z T1 Lm

0.059

D2.62

qsc = 0.028 Ep1

2 – p22


0.51

D2.53B)

n



Flow of Fluid


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t e



Panhandle Equation

Where:

E : efficiency factor

= 1.0 for brand new pipe

= 0.95 for good operating conditions= 0.92 for average operating conditions

= 0.85 for unfavorable operating conditions

Lm : length of pipe, miles

In practice, Panhandle’s equations are commonly used for large

diameter, long pipelines where the Reynolds number is on the straightline portion of the Moody diagram.



Flow of Fluid


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t e



Spitzglass Equation

f = 1 + + 0.03 D

D

3.6 1

100

Assuming that:

T = 520ºR (60ºF)

p1 = 15 psi (near-atmospheric pressure lines)

Z = 1.0

∆∆∆∆p


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Spitzglass Equation

1 +

∆∆∆∆hw D5

+ 0.03 DD

3.6γ γγ γ g L

1/2

qsc = 0.09Where:

∆∆∆∆hw : pressure loss,inches of water

or expressing pressure drop in terms of inches of water, the Spitzglassequation can be written:

p12 – p2

2 = 2.5175 x 10-5γ γγ γ g f ZT qsc L

D5

2

∆∆∆∆p = 12.6γ γγ γ g qsc Z T1 f L

p1 D5

2



Flow of Fluid


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75



Example : Pressure Drop in Gas lineGiven: Gas flow rate = 23 MMscfd

Gas viscosity = 3 cPGas specific gravity = 0.85Length = 7,000 ft

Inlet pressure = 900 psiaTemperature = 80ºF

Z = 0.67εεεε = 0.004 (assume old steel)

Calculate: The pressure drop in a 4-in and 6-in I. D. line using the:

1. General equation2. Assumption of ∆∆∆∆P


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Solution:

1. General equation

p12 – p2

2 = 2.5175 x 10-5γ γγ γ g f ZT qsc2 L

D5

NRe = 20.09γ γγ γ g qsc

Dµµµµ

20.09(0.85)(23000)

D (0.013)= =

D

30,212,269

p12 – p22 = 2.5175 x 10-5 f(0.85)(0.67)(540)(23,000)2

(7,000)D5

p12 – p2

2 =2.87 x 1010 (f)

D5



Flow of Fluid


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37 psi395 psi∆∆∆∆p

863505p2

66 x 103555 x 103p12 – p2

2

0.01800.0198f (from Moody

diagram)

0.000660.001εεεε/D

5.0 x 1067.6 x 106NRe

6-in.4-in.DVariable

p1= 900 psia



Moody friction factor diagram

Flow of Fluid


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y g

0.01980.018


Flow of Fluid


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79




2. Approximate Equation

∆∆∆∆p = 12.6γ γγ γ g qsc2 Z T1 f L

p1 D5

(for ∆∆∆∆p


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80




3. Panhandle B equation

qsc = 0.028 Ep1

2 – p22


0.51

D2.53Lm = 7000/5280 = 1.33 miles

E = 0.95 (assumed)

23 = 0.028 (0.95)

(900)2 – p22

(0.85)0.961(0.67)(540)(1.33)

0.51

D2.53

p22 = 810 x 103 -

D4.96

235 x 106

4-in. 6-in.

p2 753 882 psi

∆∆∆∆p 147 18 psi

Single Phase Flow: Gas


Flow of FluidSingle-Phase Flow: Gas


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3. Weymouth equation

qsc = 1.11 D2.67

p12 – p2

2

L γ γγ γ g Z T1

0.5

23 = 1.11 D2.667 (900)2 – p22

(7000)(0.85)(0.67)(540)

1/2

p22 = 810 x103 -

D5.33

9.44 x 108

4-in. 6-in.P2 476 862 psi

∆∆∆∆p 424 38 psi

Single Phase Flow: Gas


Flow of FluidSingle-Phase Flow: Gas


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S ch l um b er g er P r i v a t

e



86238476424WeymouthEquation

88218753147Panhandle B

Equation

86337592308∆∆∆∆P


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83


e



Application of Gas Flow Equations: Recommended guidelines

The general gas flow equation is recommended for most general usage.If it is inconvenient to use the iterative procedure of the general equation

and it is not known whether the Weymouth or the Panhandle equationsare applicable,

Compute the results using both Weymouth and Panhandle equations and use

the higher calculated pressure drop.

Use the Weymouth equation only for small-diameter (3-6 in.)

Use the Panhandle equation only for large-diameter (10 ≤≤≤≤ D)

Use the Spitzglass equation for low pressure vent lines less than 12 inches in

diameter.

When using gas flow equations for old pipe, attempt to derive the properefficiency factor through field tests. Buildup of scale, corrosion, liquids,paraffin, etc. can have a large effect on gas flow efficiency.

g


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Flow of Fluid


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85


e


Applicability of Single Phase Correlations

Weymouth

Hazen Williams

Panhandle B

Panhandle A

AGA

Moody

HorizontalGas Flow

Vertical GasFlow

HorizontalOil Flow

Vertical oilFlow

Horizontal Pipeline


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The pressure drop in horizontal pipe isbasically caused by friction.

The friction factor is a function ofReynolds number and roughness.


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Multiphase Phase FlowMultiphase Phase Flow

Multiphase Flow: Concepts and DefinitionsMultiphase Flow:


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Is the flow of several phases.

The biphasic flow is the most simple of themultiphase flow

There are different types of multiphase flow in theoil industry

Gas-Liquid,

Liquid-Liquid,

Liquid-Solid,

Gas-Solid,

Gas-Liquid-Solid,

Gas-Liquid-Liquid.

Immiscible Liquids: Immiscible liquids are thosethat are not soluble.

Multiphase Flow: Concepts and Definitions


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Flow Pattern or Flow Regime: is the geometricconfiguration of the phases in the pipeline. TheFlow pattern is determined by the interface

interaction or form.

Interface: is the surface that separates the twophases.

Phase Inversion of the two immiscibleliquid dispersion: is the transition of a disperseto a continuous phase and vice versa.

Phase Inversion Point: is the volumetric fractionof the disperse phase that becomes a continuousphase.

Gas Gas - - liquid flow regimes: Horizontal Flow liquid flow regimes: Horizontal Flow

Multiphase Flow


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Stratified Smooth

Stratified Wavy

Plug

Slug

Annular

Bubble Flow

Spray

S t r a

t i f i e

d

I n t e r m

i t t e n

t

A n n u

l a r


S ifi d S h

Multiphase Flow


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Stratified Smooth: a distinct horizontal interfaceseparates the gas and liquid flows. This flow pattern isusually observed at relatively low rates of gas and liquidflow

Stratified Wavy : as the airflow rate is increased, surfacewaves appear on the stratified flow interface. The smoothinterface will become rippled and wavy

Plug : for increased airflow rates the air bubbles coalesceforming an intermittent flow pattern in which gas pocketswill develop. These pockets or plugs are entrapped in themain liquid flow and are transported alternately with theliquid flow along the top of the pipe

Slug : wave amplitudes are large enough to seal theconduit. The wave forms a frothy slug where it touches theroof of the conduit. The slug travels with a higher velocity

than the average liquid velocity.


A l

Multiphase Flow


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Annular : for high gas flow rates the liquid flows as afilm on the wall of the pipe (the annular zone), while the gasflows in a high-speed core down the central portion of the

pipe.

Bubble : the gas forms in bubbles at the upper surfaceof the pipe. The bubble and liquid velocities are about equal.If the bubbles are dispersed though the liquid, the flow istermed froth flow. Bubble flow pattern occurs at relatively large

liquid flow rates, with little gas flow

Spray: for very great gas flow rates the annular film isstripped from the pipe walls and is carried in the air as

entrained droplets.


Multiphase Flow


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TWO-PHASE FLOW: Gas-Liquids

Gas

Oil + Water

Oil/Water/Gas

Mixture

Most frequently encountered in:

• Well tubing• Flowlines

Mixing rules are used to

predict pressure drop inpipelines


a

Multiphase Flow


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TWO-PHASE FLOW: Gas-Liquid

Two-phase flow variables

Mass flow rate, w (lbm/s)

wL: Liquid mass flow rate

wg : Gas mass flow rate

w : Total mass flow rate

w : wL + wg

a-a

wg

wL

a

wL = ρρρρLALuL wg = ρρρρg Ag ug

a-a

a

a

w

w = ρρρρ A u ⇒⇒⇒⇒ u = W/ρρρρA

a

Volumetric flow rate, q (ft3/s)

qL : Liquid volumetric flow rate

qg : Gas volumetric flow rate

q : Total volumetric flow rate

q = qL + qg



Multiphase Flow


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TWO PHASE FLOW: Gas Liquid


Liquid Holdup, HL, Gas void Fraction, αααα, (-)

The liquid Holdup is the fraction of a volume element in the two-phase flowfield occupied by the liquid phase.

HL =Liquid phase volume in pipe element

Pipe element volume

VL

VL + VgHL =

HL =AL

AA = AL + Ag

HL + Hg = 1

Gas

Líquido



Multiphase Flow


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Liquid Holdup, HL, Gas void Fraction, λλλλ, (-)

Similarly, the gas void fraction is the fraction of the volume element thatis occupied by the gas phase. For two-Phase flow 0


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Superficial velocity (volumetric flux), (ft/s)

The superficial velocity of a phase is the velocity which would occur ifonly that phase flows alone in the pipe. It is called also the volumetricflux, and represents the volumetric flow rate per unit area of each ofthe phases. Thus the superficial velocities of the liquid and gasphases are:

usL =qLA

and usg = A

qg

The mixture velocity is the total volumetric flow rate of both phasesper unit area, and is given by:

uM = A

qL + qg= usL + usg



Multiphase Flow


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TWO PHASE FLOW: Gas Liquid


Mass Flux, G (lbm/ft2-s)

The mass flux is the mass flow rate per unit area, and is given by

Gg = A

wg

GL = A

wL= Liquid mass flux

= Gas mass flux

G = A

wL + wg

= Total mass flux



Multiphase Flow


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Actual (in-situ) Velocity, u (ft/s)

The superficial velocities defined above are not the actual velocities ofthe phases, as each phase occupies only a fraction of the pipe crosssection. Thus the actual velocities of the liquid and gas phase are,respectively:

uL =qLAL

qLA HL

=usLHL

=

ug =

qg

Ag

qg

A Hg=

usg

1 - HL=

g

LA

L

Ag


Multiphase Flow


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Slip Velocity, uslip (ft/s)The actual velocities of the liquid and gas phases are usually different.The slip velocity represents the relative velocity between the twophases

uslip = ug – uLQuality x, (-)

The quality is the ratio of the gas mass flow rate to the total mass flowrate across a given area

x =wg

wg + wL

wgw

=



Multiphase Flow


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Example No. 7

Oil and natural gas flow in a 2” I.D. horizontal pipe. The in-situ flow ratesof the oil and the natural gas are 0.147 ft3/s and 0.5885 ft3/s, respectively.

The corresponding liquid holdup is 0.35. Determine:

1. The gas and liquid velocities and the mixture velocity

2. The actual velocities of the two phases

3. The slip velocity between de gas phase and the liquid phase

Solution:

A = ππππ(2/12)2/4 = 0.021821 ft2

1.usL

= qL

/A = (0.147 ft3/s) /(0.021821 ft2) = 6.74 ft/s

usg = qg/A = (0.5885 ft3/s)/(0.021821 ft2) = 27 ft/s

uM = usL + usg = 6.74 + 27 = 33.74 ft/s


Multiphase Flow


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Solution (Cont.):

2. uL = usL/HL = 6.74/0.35 = 19.26 ft/s

ug = usg/(1 – HL) = 27/(1 – 0.35) = 41.54 ft/s

3. Uslip = ug – uL = (41.54 – 19.26) = 22.28 ft/s



Multiphase Flow


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q

Fundamental phenomena in two-phase flow

ττττi

ug

uL

qg

qLAL

Ag

Gas

Liquid

LiquidGas

a

a

a - a



Multiphase Flow


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Fundamental phenomena in two-phase flow: Slippage and Holdup

ug = uL ∴∴∴∴ uslip = 0 (no-slip)HL = λλλλL = qL/(qg + qL) = usL/(usg + usL)

Holdup: When gas and liquid phases flow at the same velocity….

Ug

ULUL

Ug



Multiphase Flow


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Fundamental phenomena in two-phase flow: Slippage and Holdup

ug >>>> uL ∴∴∴∴ uslip ≠≠≠≠ 0 (slip)

HL >>>> λλλλ

L= q

L/(q

g+ q

L)

Holdup: The velocity of the gas is greater than that of the liquid. therebyresulting in a liquid holdup that not only affects well friction losses

but also flowing density. Liquid holdup is defined as the in-situflowing volume fraction of liquid, It depends of the flow pattern.

Ug

UL

UL

Ug



Flow Pattern Prediction: Baker Flow Regime Map

Multiphase Flow


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Gg

λλλλ

Gg

GLλφλφλφλφ

λλλλ =ρρρρg

0.075( )

ρρρρL62.4

( )1/2

φφφφ =( )2

ρρρρL62.4µµµµL

1/3

σσσσL

73

Gg = ρρρρg usg

GL = ρρρρg usL

G

g λ λλ λ

Gg

GLλφλφλφλφ

By =

Bx =

Baker Parameters

Slug

Plug

Stratified

AnnularWave

Disperse

Bubble

102

103

104

10510-1 1 10 102 103 104

10-1 1 10 102 103 104



Multiphase Flow


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Flow Pattern Prediction: Beggs and Brill flow regime map

NFr = gD

uM2

UM : Mixture velocity

D : inside pipediameter

g : gravitationalacceleration

λλλλ: liquid inputvolume fraction



Multiphase Flow


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Flow Pattern Prediction: Taitel-Dukler flow regime map

1.00

0.10

0.01

10.0

75.0

0.1 1.0 10.0 900.0100.0

Intermittent

Annular

StratifiedWavy

StratifiedSmooth

Bubbly

U sL

(ft/s)

U sG (ft/s)


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TWO TWO - - PHASE FLOW: Gas PHASE FLOW: Gas - - Liquid Liquid

Multiphase Flow


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Flow Pattern Prediction: Gregory -Mandhane-Aziz flow regime map

(Plug)



Multiphase Flow


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Flow Pattern Prediction:

Example: Predicting horizontal gas-liquid flowregime

Using de Baker, Mandhane, and Beggs & Brill flow regimemaps, determine the flow regime for the flow of 2000 bbl/d

of oil and 1 MMscfd of gas at 800 psia and 175ºF in a 2 ½in. I.D. pipe. The oil density and viscosity are 49.92 lbm/ft3

and 2 cP, respectively. The oil-gas surface tension is 30dynes/cm and the gas density, viscosity and thecompressibility factor are 2.6 lbm/ ft3, 0.0131 cP and 0.935respectively. The pipe relative roughness is 0.0006.




Multiphase Flow

λλλλ =ρρρρg

( ) ρρρρL( )

1/2


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Solution for Baker :

Baker’s parameters

Gg

λλλλ

GgGLλφλφλφλφ

By =

Bx =

λλλλ =0.075

( )62.4

( )

φφφφ =

( )2

ρρρρL

62.4µµµµL

1/3

σσσσL73

Gg = ρρρρg usg

GL = ρρρρg usLλλλλ = [(2.6/0.075)(49.92/62.4)]0.5 = 5.27

φ= (73/30)[(2)(62.4/49.92)2

]1/3

= 3.56

GL

= wL

/A = ρρρρL

qL

/A = ρρρρL

usL

,

= (49.92lbm/ft3)(0.130 ft3/s)/(0.0341ft2) (3600 s/hr) = 6.85 x 105 lbm/hr-ft2

A = ππππ (2.5/12)2 /4 = 0.0341 ft2

qL = (2,000bbl/day)(5.615 ft3/bbl)/(86,400 day/s) = 0.130 ft3/s



Flow Pattern Prediction:Solution for Baker :

Multiphase Flow


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Solution for Baker :

Gg

λλλλBy =

qsc Z (TTsc

) )(p

pscqg =

qg = (106 ft3/day)(0.935)(635ºR/520ºR)(15psia/800psia) 1day/86400s=

0.2478 ft3/s

Gg = wg/A = ρρρρgqg/A = ρρρρgusg= (2.6 lbm/ ft3 x 0.2478 ft3/s)/(0.0341 ft2)x(3600s/hr)=

Gg =6.8x 104 lbm/hr-ft2

By = = 6.8x 104 lbm/hr-ft2/ 5.27= 1.29x104

Bx

GLλφλφλφλφ/G

g= (6.85 x 105)(5.27)(3.56)/(6.8 x 104) = 188

Flow Pattern: Bubblethough the conditions arevery near the boundaries

with slug flow and

annular mist flow




Multiphase Flow


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G

g λ λλ λ

Gg

GLλφλφλφλφ

Slug

Plug

Stratified

AnnularWave

Disperse

Bubble

102

103

104

10510-1 1 10 102 103 104

10-1 1 10 102 103 104




Multiphase Flow


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Multiphase Flow


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Solution for Mandhane map :

The Mandhane map is simply a plot of superficial liquid velocity versussuperficial gas velocity. For our values usL = 3.81 ft/s and usg = 7.27 ft/s, theflow regime is predicted to be slug flow.

UsL = qL/A = 0.130 ft3/s/(0.0341 ft2) = 3.81 ft/s

Usg = qg/A = 0.2478 ft3/s/(0.0341 ft2) = 7.27 ft/s



Flow Pattern Prediction: Mandhan flow regime map

Multiphase Flow


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Gas superficial velocity, USG, ft/s

L i q u i

d s u p e r f

i c i a l v e l o c

i t y ,

U S L ,

f t / s

.

0.1 1.0 10.0 100

0.01

0.1

1.0

10.0

StratifiedFlow

Dispersed Flow

BubbleFlow

Slug Flow

AnnularFlow

WavyFlow


TWO TWO

- - PHASE FLOW: Gas PHASE FLOW: Gas

- - Liquid Liquid

Flow Pattern Prediction: Gregory -Mandhane-Aziz flow regime map

Multiphase Flow


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(Plug)


TWO TWO - - PHASE FLOW: Gas PHASE FLOW: Gas - - Liquid Liquid Flow Pattern Prediction:

Multiphase Flow


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Solution:

The Beggs & Brill map. The parameters are

uM = usL + usg = 3.81 + 7.27 = 11.08ft/s

NFr = (11.08ft/s)/[(32.17ft2/s)(2.5in/12in/ft)] = 17.8

λλλλL = usL/uM = 3.81/11.08 = 0.35

From the Beggs & Brill flow regime map, the flow regime is predicted to be

intermittent.

Slug flow is the likely flow regime.

NFr = gD

uM2



Flow Pattern Prediction: Beggs and Brill flow regime map

Multiphase Flow


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NFr = gD

uM2

UM : Mixture velocityD : inside pipediameter

g : gravitationalacceleration

λλλλ: liquid inputvolume fraction



Fundamental phenomena in two-phase flow: Pressure drop correlations

Multiphase Flow


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Fundamental phenomena in two phase flow: Pressure drop correlations

Three main components for predicting pressure los are:

1. Elevation or static component

2. Friction component

3. Acceleration component

Total Loss Loss LossPressure = Caused by + Caused by + Caused byloss Elevation Friction acceleration

General energy flow equation

2 gc Df ρρρρ u2dp

dz =ggc

ρρρρ sin θθθθ + ρρρρ ugcdudz+


TWO-PHASE FLOW: Gas-LiquidFundamental phenomena in two-phase flow: Pressure drop correlations

Multiphase Flow


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Energy equation for horizontal flow

dp

dz

dp

dzf

dp

dz acc= +

or neglecting the kinetic energy effects

dp

dz

dp

dzf

=

dpdz = 2 gc D

f ρρρρ u2 ρρρρ ugc

dudz+



Fundamental phenomena in two-phase flow: Pressure drop correlations

Multiphase Flow


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p p p

Pressure Loss Components

Where:

ρρρρ : Density, lbm/ft3

u : velocity, ft/s

D : pipe diameter, ft

g : acceleration caused by gravity, ft/s2

gc : conversion factor, lbm-ft/lbf-s2

f : friction factor

dp/dz : pressure gradient, psi/ft



Horizontal Pressure Loss Prediction Methods

Multiphase Flow


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Over the years, numerous correlations have been developed tocalculate the pressure gradient in horizontal gas-liquid flow. The most

commonly used in the oil and gas industry today are those of Beggsand Brill (1973), Eaton et al. (1967), and Dukler (1969). Thesecorrelations all include a kinetic energy contribution to the pressuregradient; however, this can be considered negligible unless the gasrate is high and the pressure is low.

Correlations most widely used

1. Beggs and Brill (JPT, 607-617, May 1973)

2. Dukler (AGA, API, Vol. 1, Research Results , May 1969)

3. Eaton et al. (Trans. AIME , 240: 815-828, 1967)




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Beggs and Brill correlation

Correlating parameters:

NFr = um2 / gD

λλλλL = usL/um

L1

= 316 λλλλL

0.302

L2 = 0.0009252 λλλλ-2.4684

L3 = 0.10 λλλλL- 1.4516

L4 = 0.5 λλλλL-6.738




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The flow regime transitions are given by the following:

Segregated flow exist if

λλλλL


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The flow regime transitions are given by the following:

Transition flow

If the flow regime is transition flow, the liquid holdup is calculated

using both the segregated and intermittent equations and interpolatedusing the following

HL = A λλλλL(segregated) + B λλλλL(intermittent)

Where: A =L3 - NFRL3 – L2

and B = 1 - A




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Liquid holdup, and hence, the average densityHL(φφφφ) = HL(0) x ψ ψψ ψ

HL(0) = a λλλλLb / NFRc

With the constraint that HL(0)

≥≥≥≥ λλλλL

and

ψ ψψ ψ = 1 + C[sin (1.8θθθθ) – 0.333 sin3(1.8θθθθ)]

Where

C = (1 - λλλλL)ln(d λλλλLe NLVf NFRg)

Where: a, b, c, d, e, f, and g depend on the flow regime and are given in thefollowing tables.

C must be ≥≥≥≥ 0 and NLV = usL(ρρρρL/g σσσσ)1/4




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