Pipeline Design.pdf

Surface Facilities

Pipelines Design, Operation and Maintenance Leonardo Montero R., M.Sc.

Copyright 2006, NExT, All rights reserved

S c h l u m b e r g e r P r i v a t e

Engineering Required Before Designing a Pipeline Exploration and Production Well System Definition

Reservoir Simulation

Reservoir Geology Geophysics

Facility Engineering

Drilling

Reservoir Description

Pipelines

Manifolds

Reservoir Management

Controls Exploration Geology

Geoscience

Production Management

Petroleum Engineering

Process Definition

Host Engineering

Pipeline Design

2 Copyright 2006, NExT, All rights reserved


Pipel Pi pelin ines es - Te Term rmin inol olog ogyy Fl Flow owli line nes s

& Gath Gather erin ing g Lin Lines es – The lines travel

short distances within an area. They gather products and move them to processing facilities. 





Flowlines Flowline s are usually usually small, small, e.g. e.g. 2- 4 in diameter diameter,, and gather gathering ing lines lines bigger bigger (say 4-12” 4-12” ) They carry together.

many

products,

often


mixed

Feeder Lines - Th Thes ese e pipe pipeli line nes s mov move prod produc ucts ts

from processing facilities, storage, etc., to the main transmission lines 

Typically 6-20 in diameter



Carry variety of products, sometimes ‘batched’.


Pipel Pi pelin ines es - Te Term rmin inol olog ogyy Transmission

Lines - These These are the main main cond conduits uits of of oil

and gas transportation. 

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



Natural gas transmission lines deliver to industry or ‘distrib ‘distributio ution’ n’ system. system.



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



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

Distribution


Lines - These These lines lines allow allow local local distr distribut ibution ion

from the transmission system. 

These lines can be large diameter, but most are under 6 in diameter


Pipe Pi peli line ness - Sy Syst stem em Storage Commercial Wells

Residential

Gathering

Production Wells Platforms Surface Facilities

Industrial & Utilities

Pipeline Storage

Transportation Metering Equipments Compression Station Plants

IT System Drawdown Analysis Linepack Linepacking ing Analysi Analysiss

Distributors

Distribution Gate Station Metering Equipment Compression Compression Stations



Oil and Gas Transportation by Pipelines Offshore Receiving Facilities J- Tu Tub bes Risers Process Equipment


Wyes Tees Hot Taps

Flowlines Cables

Land Pipeline Transmission Shore Approaches Crossings

Manifolds

Distribution Lines Trunk Lines


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 gas transportation



Economical: Is the most efficient method to transport high volume



Reliability



Pipelines Around the World 750 625 500

) s e s l i d M n a 375 ( s h u t o g h n T 250 e L

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 > 157.000 157.000 miles miles

125 0 UK

Western Europe USA

Rest of The World




Gathering Lines Lines Gathering


Flowli Flo wlines nes and Gath Gatheri ering ng Li Lines nes These lines travel short distances within an area. S c h l u m b e r g e r P r i v a t e

Gathers products and moves them to processing facilities. Flowlines Flowli nes are usuall usually y small, small, e.g. e.g. 2- 4in diameter, Gathering Gathering lines bigg bigger er (say (say 4-12” ) They carry many products, often mixed together. 10 Copyright 2006, NExT, All rights reserved

Flowli Flo wlines nes and Gath Gatheri ering ng Li Lines nes Grapa a nivel sublacustre

LL-83 LL-33 M-LH-9 20"

2 4 " - 0 . 3 7 5 ; 2 0 0 0 0 ' ( 1 9 8 6 )

LL-37 ) 9 6 9 1 ' ( 0 0 5 0 " ; 1 6

) 4 1 9 7 3 ' ( 3 6 3

12"

8"

LL-34

16"

16"-0.375

663' (1 63' (198 980) 0)

4047' 4047' (1977) (1977)

6"

1331' 1331' (1989) (1989)

) 8 8 ( 1 9 0 ' 5 0

12"-0.44;

13,5%

) 8 9 8 1 ( ' 7 8 7

M-LH-8

Flow Station

4 6 5 1 ' ( 1 9 9 0 )

0 ) 91 ' (1 9 9 5; 5 5 91 . 3 7 5; 0 3 2 0 " - 0.

Oil Manifold

) 4 7 9 1 ' ( 7 8 7 2

1 2 " 0 . 3 7 5 ;

v 2 i s 0 i , b l 3 e %

) 2 9 9 1 ' ( 0 0 6 9 ; 8 ) 3 . 9 7 9 0 1 " - 7 ' ( 4 5 1 2 6

M-LH-7

) 3 7 9 1 ' ( 2 7 2 3

LL-87

12"-0.44"; 10"-0.365;

16" ) 6 9 9 1 ( ' 0 8 0 5 ; 5 7 3 . 0 " 6 1

Linea Linea de 8" que debe debe ser desactivada

8 ) 1 9 8 0 ' ( 1 3 3

1 1 2 - 0 2 " " - 0 . 3 . 3 7 7 5 5

1 1 4 8 ' ( 1 9 8 8 )

0 ) 1 9 8 ' ( 0 ' ( 1 0 7

5 8 4 2 6 6 7 9 ' ' ( ( 1 1 9 9 7 9 9 0 ) )

Multiphase Manifold

) 0 9 9 1 ( ' 9 9 5 1

) 0 8 9 1 ' ( 8 4 6 1

) 0 9 9 1 ' ( 0 4 8 6

LL-29

Oil Pipeline Copyright 2006, NExT, All rights reserved

LL-16

) ) 6 6 9 9 9 9 1 1 ( ( ' ' 3 3 3 3 0 0 4 4 1 1 ; ; 5 5 7 7 3 3 . 0 0 - " " 6 6 1 1

) 9 7 9 1 ' ( 0 6 3

5 6 3 0 . " ) 0 7 8 1 9 1 ' ( 9 8 4 3

4 6 2 1 ' ( 1 9 8 1 )

4000' 000'

LL-41 (nueva)

Vertical deteriorado (corroido)

Multiphase Pipeline

1 6 " - 0 .3 8 ; 6 2 0 0 ' ( 1 9 9 5 )

6"

) 8 8 9 1 ' ( 6 2 4

LL-35

Grapa colocada para corregir corrosión en vertical.

24"

Macolla 3

31,8%

LL-39

Tank Farm

) 9 8 9 1 ( ' 8 2 0 1

4 4 . 4 " - 0 1 2

2 3 2 ' ( 1 9 8 8 )


LL-20

0 ) 9 9 ' ( 1 1 7 6 7

) 9 3 1 9 6 ' ( 4 3 1 5 ; 1 3 3 7 - . " - 0 2 4

12" 24"

2 4" -0 .3 7 7 5 5; ; 2 4 2 3 0 0' ( ' ( 19 9 93 ) 3

LL-41 LL-47

11

Gas Gathering System: Example The gas gathering system consists of several interconnected interconnected pipelines with diameter between 4 and 12 inches and low pressure line (< 500 psi). Equations:Bernoulli *Beggs & Brill * Moody o Darcy *Weymouth * Panhandle A/B * AGA

Gas Plant

FS-5-9


FS-1-8

Gas Plant FS 2-6

PE 8-3

Evaluation with Simulators: *Pipephase, Stationary State * Pipesim, Stationary State * TGNET, Dynamic State

FS 5-6

MG-CL-1 PA EM-2

EM-1

FS 16-5 FS 9-5

Gas Plant FS 22-5

FS 21-5 PC-VII

FS-23-5 Low Pressure System

FS 1-5

High Pressure System 12


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 well as piping and valves.


Gas Gathering System: Types

Wellhead S c h l u m b e r g e r P r i v a t e

Flowlines Header

Axial Gathering System In the axial gathering system, several wells produce into a common flowline.


Gas Gathering System: Types Wellhead


Compression Compression Station

Radial Gathering System Flowlines Flowline s emanati emanating ng from from several several differen differentt wellhe wellheads ads converge to a central point where facilities are located. Flowline Flo wlines s are usually usually terminat terminated ed at a header, header, which which is essentially a pipe large enough to handle the flow of all the flowlines


Gas Gathering System: Types Loop Gathering System


Compression Station

Wellhead

Separator


Gathering System: Well Center Well Center Gathering System The well center gathering system uses radial philosophy at the local level for individual wells, brings all the flowline flowlines s to a central central head header er Well Center


Central Gathering Section 17 Copyright 2006, NExT, All rights reserved

Gathering System: Trunk Line Trunk Line

Well Head Header

Uses an axial gathering scheme for the groups of wells. Uses several remote headers to collect fluid. Is more applicable to relatively large leases, and no cases where it is undesirable or impractical to build the field processing facilities at a central point. 18



Gathering System: Decision

The choice between the gathering systems is usually economic. S c h l u m b e r g e r P r i v a t e

The cost of the several small sections of pipe in well-center system is compared to the cost of single large pipe for the trunkline system. Technical criterion.

feasibility

may

be

another

The production characteristics of the field


Gathering System: Pipeline System

Series Pipelines LA

LB

Parallel Pipelines LC A


B A

B

C

C Loopless Loop less Pipeline Pipeline Systems Systems

Looped Pipelines LA

LC

A

q2 q1

C

NCE 1

Node 1 number pressure p1

q3 3

2

2 p2

qn-1

qn n-1

qn + 1

n

3

n-1

n

p3

pn-1

pn

n+1 pn+1

B 20 Copyright 2006, NExT, All rights reserved

Gathering System: Pipeline System Series Pipelines LA

A

LB

B

LC

C

Series pipelines: The inlet and outlet pressures for the system are different, For this system, the flow rates through each of the pipe legs are equal


qA = qB = qC = qt

∆pt = ∆pA + ∆pB + ∆pC

∆pA ≠ ∆pB ≠ ∆pC

Le = LA + LeBA + LeCA 21 Copyright 2006, NExT, All rights reserved

Gathering System: Pipeline System Parallel Pipelines

Parallel

A

Because the pipelines are in parallel with a common inlet and outlet, the pressure drop through each of them is the same, but the flow rates are not.

B C

qA ≠ qB ≠ qC

∆pA = ∆pB = ∆pC


From Weymouth equation 5

∆pA = ∆pB = ∆pC

pipelines:

Le =

de fe Le

0.5

=

5

dA fA L A

0.5

+

5

0.5

5

dB dC + fB LB fC LC

0.5

qt = qA + qB + qC 22 Copyright 2006, NExT, All rights reserved

Gathering System: Pipeline System Looped pipelines LA

LC

A C B

Le = LC + (Le )AB

Looped

pipeline:

A looped pipeline is one in which only a part of the line has a parallel segment. The original pipeline is looped to some distance with another line to increase the flow capacity.



Gathering System: Pipeline System Loopl Lo opless ess Pi Pipe pelin lines es q2 q1

NCE 1

Node 1 number pressure p1

q3 2

2 p2

qn

qn-1 3

Loop Lo ople les ss Pi Pipe pellin ine es : A

n-1

qn + 1

n

3

n-1

n

p3

pn-1

pn

n+1 pn+1

loopless pipeline sys system tem, defined as one where the NCE's (node connecting elements) joined by nodes form no closed loop



Gathering System: Equations for Complex Gas Flow




Flow Flow of of Fluid Fluid


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: flow: vertical, inclined and horizontal Overall production system



Flow of Fluid Possible Pressure Losses ∆p8= Pwh-Psep Gas Flowlines

Pwh-Pds = ∆p5 P wh

∆p6= Pds-Psep P ds Flowlines

P sep

Surface Choke

Tanks

Pdv h w

P f

w

∆p4= Puv-Pdv

P = 7 p

Puv

Safety Valves

∆p1=Pm- Pwfs ∆p2=Pwfs- Pwf ∆p3=Pur- Pdr ∆p4=Puv- Pdv ∆p5=Pwh- Pds ∆p6=Pds- Psep ∆p7=Pwf- Pwh ∆p8=Pwh- Psep

Well Bottom Hole P Restricción dr

∆ ∆

∆p3= Pur-Pdr Reservoir



Separator

Pur

P wfs

P wf Pwfs-Pwf =

Loss in porous medium Loss across completion Loss across restrictions Loss across safety valves Loss across surface choke Loss in flowlines Total loss in tubing Total loss in flowlines

∆p2

P m

P e

∆p1=Pm-Pwfs Source: Handbook of Petroleum and Gas Engineering, William Lyons

28

Flow of Fluid Production Pressure Profile Drainage Boundary

Wellbore Wellhead & (Perforations) Choke

Separator

Stock Tank

Po


Pwf

e r u s s e r P

Pwf Psp ro

PST

W

Reservoir

Tubing

Flowline

Transfer Line


Source: Handbook of Petroleum and Gas Engineering, William Lyons

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

q

A

u

u=q/A 30 Copyright 2006, NExT, All rights reserved


Pipeline Fluid Flow pipeline


velocity pipeline

velocity

A) Laminar Flow Laminar Flow

B) Turbulent Flow

⇒ Re < 2000

Turbulent Flow ⇒ Re > 2100 Copyright 2006, NExT, All rights reserved

R = Duρ/µ 31

Flow of Fluid Single-Phase Flow: Liquid Pressure Drop Calculation dp + udu + gc ρ

g dz + 2 f u2 dL = 0 D gc gc

g ρ∆z + ∆p = p1 – p2 = g c

(Energy Equation)

ρ ∆u2 + 2f ρu2 L (ρ = constant) 2gc

D gc Where:

∆p = ∆pPE + ∆pKE + ∆pF

∆pPE : pressure drop due to potential energy change ∆pKE : pressure drop due to kinetic energy change ∆pF : frictional pressure drop u D L f

ρ Copyright 2006, NExT, All rights reserved

: velocity of the fluid :pipeline internal diameter :Length of the pipe : friction factor : liquid density

32


Flow of Fluid Single-Phase Flow: Liquid ∆pPE , the pressure drop due to potential energy change 2

1 L

θ

∆z

L

∆z

1


2

θ q

q

(a) Upward flow

(b) Downward flow

∆z = z2 – z1 = L sin θ ∆pPE = (g/gc)ρL sin θ Horizontal Flow Copyright 2006, NExT, All rights reserved

θ=0

∆pPE = 0 33

Flow of Fluid Single-Phase Flow: Liquid ∆pKE = the pressure drop due to kinetic energy change Is the pressure drop resulting from the change in the velocity of the fluid between positions 1 and 2. S c h l u m b e r g e r P r i v a t e

∆pKE = (ρ/2gc) ∆u2 = (ρ/2gc) (u22- u12) ρ = constant , A = constant

∆pKE = 0

q = constant u = q/A , A = πD2/4 ∴ u = 4q/πD2

Where:

∆pKE = 8ρq2/π2gc(1/D24 – 1/D14)

u = Velocity of the fluid, ft/sec. q = Volu Volumet metric ric flow flow rate, rate, ft ft 3/sec. D = Pipelin Pipeline e interna internall diamete diameter, r, ft ft 3 ρ = Liquid density, lbm/ft A = Pipel Pipeline ine cross-se cross-sectio ctional nal area, area, ft2 34


Flow of Fluid Single-Phase Flow: Liquid ∆pPE , the pressure drop due to potential energy change Example No. 1 Suppose that 1000 bbl/d of brine ( ρw= 1.05) is being injected through 2 7/8-in, 8.6-lbm/ft (I.D. = 2.259 in.) tubing in a well that is deviated 50º from vertical. Calculate the pressure pressure drop over over 1000 ft of tubing due to the potential energy change.


Solution:

∆pPE = (g/gc)ρ L sinθ For downward downward flow flow in a well deviated 50º 50º from vertical, vertical, the flow direction is -40º from horizonta horizontal, l, so so θ is -40º:

∆pPE = (32.17/32.17) (32.17/32.17) (1.05) (62.4) (62.4) (1000) sin (-40º) = - 292 psi (lbf/lbm) (lbm/ft3) (ft) (ft2/144 in2) = lbf/in2 = psi 35 Copyright 2006, NExT, All rights reserved

Flow of Fluid Single-Phase Flow: Liquid ∆pKE = the pressure drop due to kinetic energy change For oilfield units bbl/d for flow rate, lbm/ft3 for density, and in. for diameter, the constants and unit conversions can be combined to yield:

∆pKE = 1.3x10-8ρq2

1 4 D2

–


1 4 D1

Where q = Volumetric flow rate, bbl/d D = Internal pipeline diameter, in.

ρ = Liquid density, lbm/ft3 36 Copyright 2006, NExT, All rights reserved

Flow of Fluid Single-Phase Flow: Liquid ∆pKE = the pressure drop due to kinetic energy change Example Suppose that 2000 bbl/d of oil with a density of 58 lbm/ft3 is flowing through a horizontal pipeline having a diameter reduction from 4 in. to 2 in., as illustrated in the figure. Calculate the kinetic energy pressure drop caused by the diameter change.

q

q

u1

u2 D1

D2



Flow of Fluid Single-Phase Flow: Liquid ∆pKE = the pressure drop due to kinetic energy change Solution: Since ρ = constant, then ∆pKE = 8ρq2/π2(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 =

8(58 lbm/ft3)(0.130 ft3/sec.)2 (π2 x 32.17 ft-lbm/lbf-sec2)]

[

1 1 ] 4 4 (0.3333) (0.16667)

∆pKE = 0.28 psi


Flow of Fluid Single-Phase Flow: Liquid ∆pf = the pressure drop due to friction The frictional pressure drop is obtained from the equation:

∆pf =

fρu2L 2gcD


Wher Where: e: f = is the the Moo Moody dy’s ’s fric fricti tion on fact factor or.. In laminar flow

< 2100 NRe <

> 2100 In turb turbul ulen entt flo flow w NRe > where

f = 64/NRe f = f(NRe,ε)

NRe : is the Reynolds number

ε

: is the relative pipe roughness

which are given by: NRe = ρud/µ

ε Copyright 2006, NExT, All rights reserved

= k/D

(k = Absolute roughness, in)

39

Flow of Fluid Single-Phase Flow: Liquid where

ρ = Liquid density, lbm/ft3 u = Velocity, ft/s D = Internal pipeline diameter, ft µ = Liquid viscosity, lbm/ft-s S c h l u m b e r g e r P r i v a t e

Other expresions:

NRe = 1488 ρuD/µ where:

ρ : Liquid density, lbm/ft 3 u : Velocity, ft/s D : Internal pipeline diameter, ft

µ : Liquid viscosity, cP


Flow of Fluid Single-Phase Flow: Liquid ∆pf = the pressure drop due to friction In oilfield units

NRe = 1.48 qρ/Dµ = 92.35 γ γL q/Dµ


Where: ρ : Liquid density, lbm/ft 3 γ γ : Liquid specific gravity L q : Volumetric flow rate, bbl/d D : Internal pipe diameter, in. µ: Liquid viscosity, cP

NRe = 1.722 x 10-2 w D/A µ Where: w : Mass flow rate, lbm/d A : Pipeline cross-sectional area, ft 2 µ : Liquid viscosity, cP Copyright 2006, NExT, All rights reserved

41

Flow of Fluid Single-Phase Flow: Liquid ∆pf = the pressure drop due to friction Equations to calculate the friction factor f = f(NRe,ε) 

High precision



Intermediate precision



Low Lo w prec precis isio ion n


(Zigrang (Zigrang and Silvestre, Silvestre, Transacti Transactions ons of the ASME, 280/vol. 107, June 1985)

- The Colebroo Colebrook-Wh k-White ite equation equation (implicit (implicit in f): 1/√f = -2 log[(ε/3.7065) + 2.5226/(NRe√f)]

(Needs iteration to solve for f)

- The Chen Chen equatio equation n (explici (explicitt in f): 1/√f = -2 log{(ε/3.706 /3.7065) 5) – (5.045 (5.0452/ 2/ NRe) log [(ε1.1098/2.8257) + (7.149/NRe)0.8981]} - The Moody Moody’s ’s fricti friction on fact factor or diag diagra ram m f = f(NRe,ε) 42 Copyright 2006, NExT, All rights reserved

Flow of Fluid

Click to edit Master title style

Single-Phase Flow: Liquid ∆pf = the pressure drop due to friction Moody friction factor diagram S c h l u m b e r g e r P r i v a t e


Flow of Fluid


Relative Roughness of Common Piping Material.




Flow of Fluid

Single-Phase Flow: Liquid ∆pf = the pressure drop due to friction

Hasta Has ta aquí aquí va vamos mos

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


Solution: First, the Reynolds number must be calculated to determine if the flow is laminar or turbulent. NRe = ρuD/µ = 1.48qρ/Dµ = (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 = -2log{ 0.001 3.7065 Copyright 2006, NExT, All rights reserved

(0.001)1.1098 5.0452 7.194 log [ +( 4 3.57 x 10 2.8257 3.57 x104

)0.8981 ]} 45


Flow of Fluid

Single-Phase Flow: Liquid ∆pf = the pressure drop due to friction f = 0.0252 u = q/A = 4q/πD2 =

4(1000 bbl/d)(5.615 ft 3/bbl)(1day/86,400 s)

= 2.33 ft/s

π[(2.259/12) ft]2

∆pF =

(0.0252)(65.5 lbm/ft 3)(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 potential energy energy or hydrostat hydrostatic ic pressure pressure drop, drop, which it was calculated calculated to be -292 -292 psi in Example No. 1 46 Copyright 2006, NExT, All rights reserved



Flow of Fluid

Single-Phase Flow: Liquid ∆pf = the pressure drop due to friction Example The 1000 bbl/d of injection water described in Examples 1 and 3 is supplied S to the wellhead through a 3000 3000 ft long, 1 ½ in. I.D. flow line from a central c uh l m pumping pum ping station. station. The relative relative roughne roughness ss of the galvaniz galvanized ed iron pipe is b e r e 0.004. If the pressure at the wellhead is 100 psia, psia, what is the pressure at the g r P pumping pum ping station, station, neglectin neglecting g any pressure pressure drops through through valves valves or otherr i v a t e fittings? Solution: NRe = 1.48qρ/Dµ = 1.48(1000 bbl/d)(65.5 lbm/ft3)/(1.5 in.)(1.2 cP) = 53,900 1/√f =

-2log{ 0.004 3.7065

(0.004)1.1098 5.0452 7.194 log [ +( 4 2.8257 5.39 x 10 5.39 x104

)0.8981]}

f = 0.0304 47 Copyright 2006, NExT, All rights reserved


Flow of Fluid

Single-Phase Flow: Liquid ∆pf = the pressure drop due to friction u = q/A = 4q/ πD2 =

∆pF = p1 – p2 =

4(1000 bbl/d)(5.615 ft3/bbl)(1day/86,400 /bbl)(1day/86,400 s)

π[(1.5/12) ft]2

= 5.3 ft/s S c

h l u m b e r g e r P r i v a t e

(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 over 3000 ft. It can can be reduced substantially by using larger pipe for this water supply, since the frictional pressure drop depends approximately on the pipe diameter to the fifth power Copyright 2006, NExT, All rights reserved

48

Flow of Fluid Single-Phase Flow: Liquid ∆pf = the pressure drop due to friction In oilfield units

fρu2L ∆pf = 2gcD

=11.5x10-6


fQ2L γ γL D5

Q : Liqu Liquid id flow flow rate rate,, bpd bpd

f

: Mod Moddy dy frictio friction n fact factor or L : Leng Length th of the the pipe pipe,, ft D : Internal Internal pipe diameter, in. γ γ : Liquid specific gravity L

The most common use of this equation is to determine the pipe diameter 49 Copyright 2006, NExT, All rights reserved

Flow of Fluid fρu2L ∆pf = 2gcD

=11.5x10

-6

fQ2L γ γL D5 S c h l u m b e r g e r P r i v a t e

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 factor converge. 50 Copyright 2006, NExT, All rights reserved

Flow of Fluid Hazen-Williams Formula: To avoid iteration 1.85

HL = 0.015

Q

4.87 1.85

D

C

Where HL : Head loss due to friction. ft Q : Liq Liqui uid d flo flow w rat rate, e, bp bpd d C : fric fricti tion on fac facto torr cons consta tant nt : 140 for new steel pipe : 130 for Cast iron pipe : 100 for riveted pipe L : Le Leng ngth th of the pipe pipe,, ft ft D : Int Inter erna nall p pip ipe e dia diame mete ter, r, in. in.


γ γL xρw ∆P =HL 144


Pressure Drop in Liquid Pipeline Exercise A pipeline transport condensate (800 bpd) and water (230 bpd). The condensate 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 drop for 2 inch, 4 inch and 6 inch I.D, using the general equation and Hazen Williams (Assume C=120. Assume Old pipeline (ε=0.004)



Pressure Drop in Liquid Pipeline Solution Mixture’s rule (230)

(800)

γ γL = X1x γ γ1 + x2x γ γ2 =(230 +800) 1.05 +(230 +800) 0.87 γ γL = 0.91 In oilfield units

92.35 x0.87x1030 28,853 92.35 γ q γ L NRe = = = Dµ D3 D f = f(NRe,ε)

Pressure Drop

∆pf = fρ

u2L

2gcD


=11.5x10

-6

fQ2L γ γL D5

53


Pressure Drop in Liquid Pipeline Solution Mixture’s rule Pressure Drop

fρu2L ∆pf = 2gcD

∆pf = 11.5x10-6 ∆pf =

=11.5x10

-6

fQ2L γ γL


f = f(NRe,ε)

D5 f (1030)2x7000x 0.91 5

D

f 77,716 D5 54


Pressure Drop in Liquid Pipeline

Re ε/D f (from chart) ∆P (psi)

2 inch

Diameter 4 inch

6 inch

14 42 7

7 20 0

4 809

0. 002 0

0.0010

0. 00 07

0.032

0.034

0.038

77.7

2. 6

0.4



Flow of Fluid


Single-Phase Flow: Liquid



Pressure Drop in Liquid Pipeline Solution Hazen-Williams 1.85

L Q HL = 0.015 4.87 1.85 D C

HL (ft) ∆P (psi)

γ γL xρw ∆P =HL 144

2 inch

Diameter 4 inch

6 inch

192

6.6

1

75.6

2.6

0.4




Flow Flow of of Fluid Fluid Single Single phase: phase: Gas Gas



Flow of Fluid

Single-Phase Flow: Gas Steady state flow in in simple pipeline systems: Gas Gas flow equations The basic energy balance on a unit mass basis:

dp + udu + gc ρ ρ= u=

MW p ZRT

=

g dz + 2 f u2 dL + dWs= 0 D gc gc

ZRT T Tsc

psc p

dz = sin θ dL and dWs = 0 ZRT dp + 28.97γ γg p


(From the real gas law)

28.97γ γg p

4 q Z π D2 sc

(Energy Equation)

g sin θ + gc

(The velocity in terms of the volumetric flow rate at standard conditions) (Neglecting for the time being any kinetic energy change)

8f 2 π gcD5

T Tsc

psc p

2 qsc Z

dL = 0



Flow of Fluid

Single-Phase Flow: Gas Steady state flow in simple pipeline systems: Gas flow equations 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 γg : gas specific gravity T : flowing temperature, ºR Z : average gas compressibility f : Moody friction factor L : length of pipe, ft 60 Copyright 2006, NExT, All rights reserved


Flow of Fluid

Single-Phase Flow: Gas Steady state flow in simple pipeline systems: Gas flow equations To solve this equation notice that: a)

Z, T an and d p are are fun funct ctio ions ns of po posi siti tion on,, z

b)

Rigoro Rigorousl usly y soluti solution on need: need: T = T(z) T(z) and and Z = Z(T Z(T,p) ,p) (Equa (Equatio tion n of State)

c)

This Th is appr approa oach ch will will like likely ly req requi uire re num numer eric ical al inte integr grat atio ion n

d)

Alternatively,

e)

Aver Averag age e val value ues s of of Z an and d T can can b be e ass assum umed ed

f)

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


Tlm = (T2 – T1)/ln(T2/T1) h)

Solv Solvin ing g for for ho hori rizo zont ntal al flo flow yie yield lds s

p12 – p22 = Copyright 2006, NExT, All rights reserved

(16)(28.97) γ γg f ZT π2gcD5R

(

pscqsc Tsc

2

)L

61


Flow of Fluid

Single-Phase Flow: Gas Steady state flow in simple pipeline systems: Gas flow equations Where: f = f(NRe,ε) NRe =

Moody diagram

4(28.97) γ γg qsc psc

π D µ R Tsc

and


ε = k/D

For oilfield units: p12 –

p22

= 2.5175 x

10-5

γ γg f ZT D5

qsc2

Where:

L

p : psia q : Mscfd D : in. L : ft

γ γg qsc NRe = 20.09 Dµ

µ : cP T : ºR 62



Flow of Fluid

Single-Phase Flow: Gas Steady state flow in in simple pipeline systems: Gas Gas flow equations In a high-rate, low-pressure line, the change in kinetic energy may be significant. In this case, for a horizontal horizontal line, the energy balance equation is: dp/ρ + udu/gc + 2 f u2 dL/gcD = 0 For a real gas:

ρ=

28.97 γ γg p ZRT

and

u=


4 q Z ( T )( psc ) π D2 sc Tsc p

The differential form of the kinetic energy term is udu =

(

4qscZ T

π D2

psc Tsc

2

)

dp p3

Substituting for ρ and udu, assuming average values of Z and T over the length of the pipeline, and integrating we obtain for oilfield units Copyright 2006, NExT, All rights reserved

63


Flow of Fluid

Single-Phase Flow: Gas Steady state flow in simple pipeline systems: Gas flow equations

p12 – p22 = (4.195 x 10-7)

2 γ γg Z T qsc

D4

6fL D

+ ln

p1 p2


Where: p1 and p2 are in psia T is in ºR qsc is in Mscfd D is in in. L

is in ft

The friction factor is obtained from the Reynolds number and pipe roughness, roughness, with the Reynolds number number given in oilfield units by

γ γg qsc NRe = 20.09 Dµ Copyright 2006, NExT, All rights reserved

64


Flow of Fluid


γ γg qsc NRe = 20.09 Dµ


The equation is an implicit equation in p and must be solved iteratively.

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.



Flow of Fluid

Single-Phase Flow: Gas Steady state flow in simple pipeline systems: Gas flow equations 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 (ε = 0.001) to a compress compressor or station, station, where where the inlet inlet pressure pressure must must be at least 20 psia. The gas has a specific gravity of 0.7, a temperature of 100 ºF and an average viscosity of 0.012 cP. What is the maximum flow rate possible through this gas line? Solution: p12 – p22 = (4.195 x 10-7)

γ γg Z T q2sc D4

6fL D

+ ln

p1 p2

Solving for q: qsc =

(p12 – p22) D4

0.5

(4.195 x 10-7) γ γg Z T [(6 f L/D) + ln(p1/p2)] 66




Flow of Fluid

Single-Phase Flow: Gas Steady state flow in in simple pipeline systems: Gas Gas flow equations Assuming (1) that the friction factor depends only on the pipe roughness. Then from the Moody diagram, for high Reynolds number and a relative roughness of 0.001


f = 0.0196 and (2) that that Z = 1 at these low pressures. pressures. Then qsc = qsc =

(1002 – 202)(3)4

0.5

(4.195 x 10-7)(0.7)(1)(560) {[(6)(0.0196)(1000)/3] + ln(100/20)} 4.73 x 109 39.2 + 1.61

0.5

= 10,800 Mscfd

Checking the Reynolds number, NRe = (20.09)(0.7)(10,800)/[(3)(0.012)] = 4.2 x 106 Copyright 2006, NExT, All rights reserved

67


Flow of Fluid


So the friction factor based on fully rough wall turbulence is correct.


It is found that this line can transport over 10 MMscfd. Notice that even at this high flow rate and with a velocity five times higher at the pipe outlet than at the entrance, the kinetic energy contribution to the overall pressure drop is still small relative to the frictional pressure drop.



Flow of Fluid

Single-Phase Flow: Gas Steady state flow in series pipeline pipeline systems: systems: Gas flow equations equations

Waymouth Equation f=

0.032/D1/2

and

p12 – p22 qsc = 1.11

D2.67

0.5 S c h l u m b e r g e r P r i v a t e

L γ γg Z T1

Where: qsc : gas flow rate, MMscfd D : pipe internal diameter, in. p1 : inlet pressure, psia p2 : outlet pressure, psia L : length of pipe, ft

γ γg : gas gravity T1 : temperature of gas at inlet, ºR Z : compressibility factor of gas Copyright 2006, NExT, All rights reserved

69


Flow of Fluid

Single-Phase Flow: Gas Steady state flow in series pipeline pipeline systems: systems: Gas flow equations equations

Waymouth Equation S c h l u m b e r g e r P r i v a t e

Comments:

Moody friction factor is independent of the Reynolds number and dependent upon the relative roughness. For a given roughness, ε, the friction factor is merely a function of diameter. Industry experience indicates that Weymouth’s equation is suitable for most piping within the production facility. Good for short lengths of pipe with high pressure drop and turbulent flow 70 Copyright 2006, NExT, All rights reserved


Flow of Fluid

Single-Phase Flow: Gas Steady state flow in in series pipeline systems: systems: Gas flow equations equations

Panhandle Equation f=

C n NRe

NRe = 5 x 106 to 11 x 106

n = 0.146

> 11 x 106 NRe >

n = 0.039


Using this assumption and assuming a constant viscosity for the gas, 0.059 p12 – p22 D2.62 A) qsc = 0.020 E γ γg 0.853 Z T1 Lm B) qsc = 0.028 E

p12 – p22

γ γg 0.961 Z T1 Lm

0.51

D2.53 71



Flow of Fluid

Single-Phase Flow: Gas Steady state flow in in series pipeline systems: systems: Gas Gas flow equations

Panhandle Equation S c h l u m b e r g e r P r i v a t e

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 p practi ractice, ce, Panhandle Panhandle’s ’s equations equations are commonly commonly used for large large diameter, long pipelines where the Reynolds number is on the straight line portion of the Moody diagram.



Flow of Fluid

Single-Phase Flow: Gas Steady state flow in in series pipeline systems: systems: Gas Gas flow equations

Spitzglass Equation S c h l u m b e r g e r P r i v a t e

Assuming that:

f= 1+

3.6 D

+ 0.03 D

1 100

T = 520 520ºR ºR (60ºF (60ºF)) p1 = 15 psi (near-atmospheric (near-atmospheric pressure lines) Z = 1.0

∆p < 10%p1 73 Copyright 2006, NExT, All rights reserved


Flow of Fluid

Single-Phase Flow: Gas Steady state flow in in series pipeline systems: systems: Gas flow equations equations

Spitzglass Equation

p12 – p22 = 2.5175 x 10-5

γ γg f ZT q2sc L

∆p = 12.6

D5

2 Z T1 f L γ γg qsc

p1 D5

or expressing expressing pressure drop in terms terms of inches of water, the Spitzglass Spitzglass equation can be written: 1/2

∆hw D5

Where:

qsc = 0.09

γ γg L 1 + Copyright 2006, NExT, All rights reserved

3.6 D

∆hw : pressure loss, + 0.03 D

inches of water 74



Flow of Fluid

Single-Phase Flow: Gas Steady state flow in in simple pipeline systems: Gas Gas flow equations Example : Pressure Drop in Gas line Giv Given en:: Gas Gas flo flow rate rate = 23 MMsc MMscfd fd 1 Gas viscosity = 3 cP Gas Gas specific gravity = 0.85 Length = 7,000 ft L,D Inlet pressure = 900 psia Temperature = 80ºF Z = 0.67 ε = 0.004 (assume old steel) Calculate: The pressure drop in a 4-in and 6-in I. D. line using the: 1. Gen ene era rall eq equ uat atio ion n 2. Ass ssu umpti tio on of of ∆P < 10% p1 3. Pa Panh nhan andl dle e B eq equa uati tion on 4. Wey eym mouth eq equat atio ion n


2



Flow of Fluid

Single-Phase Flow: Gas Steady state flow in simple pipeline systems: Gas flow equations Solution:


1. General equation

γ γg qsc NRe = 20.09 Dµ =

20.09(0.85)(23000)

p12 – p22 = 2.5175 x 10-5 p12 – p22 = 2.5175 x 10-5 p12 –

p22

=


D (0.013)

=

30,212,269 D

γ γg f ZT qsc2 L D5

f(0.85)(0.67)(540)(23,000)2(7,000) D5

2.87 x 1010 (f) D5

76


Flow of Fluid

Single-Phase Flow: Gas Steady state flow in simple pipeline systems: Gas flow equations D

Variable

4-in. 7.6 x 10 6

NRe

ε/D f (from Moody diagram)

0.00066

0.0198

0.0180

555 x 103

p2

505 395 psi


5.0 x 106

0.001

p12 – p22

∆p

6-in.

66 x 10 3 863 37 psi 77

p1= 900 psia Copyright 2006, NExT, All rights reserved


Flow of Fluid

Moody friction factor diagram


0.0198 0.018



Flow of Fluid

Single-Phase Flow: Gas Steady state flow in in simple pipeline systems: Gas Gas flow equations 2. Approximate Equation

γ γg qsc2 Z T1 f L

∆p = 12.6 ∆p = 12.6

p1

(for ∆p < 10%p1)

D5

(0.85)(23)2(0.67)(540)(7000) f (900)

D5

p2 = p1 + ∆p

4-in.

∆p (psi) 308 P2 (psi)


592

=

1.59 x 107 (f) D5

6-in. 37 863 79



Flow of Fluid

Single-Phase Flow: Gas Steady state flow in in simple pipeline systems: Gas Gas flow equations 3. Panhandle B equation qsc = 0.028 E

γ γg 0.961 Z T1 Lm

23 = 0.028 (0.95)

p22

= 810 x

103

-

Lm = 7000/5280 = 1.33 miles

0.51

p12 – p22

D2.53

E = 0.95 (assumed) 0.51

(900)2 – p22

D2.53

(0.85)0.961(0.67)(540)(1.33) 235 x 106 D4.96

4-in.

6-in.

p2

753

882 psi

∆p

147

18 psi 80




Flow of Fluid

Single-Phase Flow: Gas Steady state flow in simple simple pipeline pipeline systems: systems: Gas flow equations equations 3.

Weymouth equation p12 – p22

qsc = 1.11 D2.67


0.5

L γ γg Z T1 (900)2 – p22

23 = 1.11 D2.667

1/2

(7000)(0.85)(0.67)(540) p22

= 810

x103

-

9.44 x 108 D5.33

4-in.

6-in.

P2

476

862 psi

∆p

424

38 psi 81



Flow of Fluid

Single-Phase Flow: Gas Steady state flow in simple simple pipeline pipeline systems: systems: Gas flow equations

Solution:

6 In

4 In

∆P

∆P

P2

(psia)

P2 (psia)

(psia)

(psia)

General Equation

395

505

37

863

∆P < 10% p1

308

592

37

863

Panhandle B Equation

147

753

18

882

Weymouth Equation

424

476

38

862




Flow of Fluid

Single-Phase Flow: Gas Steady state flow in simple simple pipeline pipeline systems: systems: Gas flow equations 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 equations are 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 low pressure vent lines less than 12 inches in diameter. When using gas flow equations for old pipe, attempt to derive the proper efficiency factor through field tests. Buildup of scale, corrosion, liquids, paraffin, etc. can have a large effect on gas flow efficiency. 83 Copyright 2006, NExT, All rights reserved


Flow of Fluid

Single Phase Flow Correlations Available. 

Moody



Dry Gas Equation (AGA)



Panhandle A



Panhandle B



Weymouth



Hazen Williams




Flow of Fluid

Applicability of Single Phase Correlations Vertical oil Flow

Horizontal Oil Flow

Vertical Gas Flow

Horizontal Gas Flow

Moody









AGA









Panhandle A









Panhandle B









Hazen Williams









Weymouth











Horizontal Pipeline

The pressure drop in horizontal pipe is basically caused by friction.


The friction factor is a function of Reynolds number and roughness.



Multiphase Multiphase Phase Phase Flow Flow


Multiphase Flow: Concepts and Definitions Multiphase Flow: Is the flow of several phases. The biphasic flow is the most simple of the multiphase flow There are different types of multiphase flow in the oil industry


Gas-Liquid, Liquid-Liquid, Liquid-Solid, Gas-Solid, Gas-Liquid-Solid, Gas-Liquid-Liquid. Immiscible Liquids: Liquids: Immiscible liquids are those that are not soluble. 88 Copyright 2006, NExT, All rights reserved

Multiphase Flow: Concepts and Definitions Flow Pattern or Flow Regime: is the geometric configuration of the phases in the pipeline. The Flow pattern is determined by the interface interaction or form.


Interface: is the surface that separates the two phases.

Phase Inversion of the two immiscible liquid dispersion: is the transition of a disperse to a continuous phase and vice versa.

Phase Inversion Point: is the volumetric fraction of the disperse phase that becomes a continuous phase. 89 Copyright 2006, NExT, All rights reserved

Multiphase Flow Gas - -liquid liquid flow regimes: Horizontal Flow Stratified Smooth Stratified Wavy Plug Slug Annular

d e i f i t a r t S


t n e t t i m r e t n I r a l u n n A

Bubble Flow Spray 90 Copyright 2006, NExT, All rights reserved

Multiphase Flow Gas - -liquid liquid flow regimes: Horizontal Flow Stratified Smooth:

a distinct horizontal interface separates the gas and liquid flows. This flow pattern is usually observed at relatively low rates of gas and liquid flow S c h l u m b e r g e r P r i v a t e

Stratified Wavy :

as the airflow rate is increased, surface waves appear on the stratified flow interface. The smooth interface will become rippled and wavy

Plug :

for increased airflow rates the air bubbles coalesce forming an intermittent flow pattern in which gas pockets will develop. These pockets or plugs are entrapped in the main liquid flow and are transported alternately with the liquid flow along the top of the pipe

Slug :

wave amplitudes are large enough to seal the conduit. The wave forms a frothy slug where it touches the roof of the conduit. The slug travels with a higher velocity than the average liquid velocity. Copyright 2006, NExT, All rights reserved

91

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

for high gas flow rates the liquid flows as a film on the wall of the pipe (the annular zone), while the gas flows in a high-speed core down the central portion of the pipe. S c h l u m b e r g e r P r i v a t e

Bubble :

the gas forms in bubbles at the upper surface of the pipe. The bubble and liquid velocities are about equal. If the bubbles are dispersed though the liquid, the flow is termed 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 is stripped from the pipe walls and is carried in the air as entrained droplets.


Multiphase Flow


TWO-PHASE FLOW: Gas-Liquids


Gas Oil/Water/Gas Mixture Oil + Water

Most frequently encountered in: • Well Well tubi tubing ng • Flow Flowliline nes s

Mixing rules are used to predict pressure drop in pipelines


Multiphase Flow


a

TWO-PHASE FLOW: Gas-Liquid Two-phase flow variables

w a

Mass flow rate, w (lbm/s)

a-a w = ρ A u ⇒ u = W/ρA


wL: Liquid mass flow rate wg : Gas mass flow rate w : Total mass flow rate

a

w : wL + wg

wg

Volumetric flow rate, q (ft3/s) wL

qL : Liquid volumetric flow rate

a

qg : Gas volumetric flow rate q : Total volumetric flow rate

q = qL + qg

a-a

wL = ρLALuL Copyright 2006, NExT, All rights reserved

wg = ρg Ag ug

94

Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Two-phase flow variables Liquid Holdup, HL, Gas void Fraction, α, (-) The liquid Holdup is the fraction of a volume element in the two-phase flow field occupied by the liquid phase.

HL = HL = HL =


Liquid phase volume in pipe element Pipe element volume VL VL + Vg AL A

A = AL + Ag

Gas

HL + Hg = 1

Líquido 95


Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Two-phase flow variables Liquid Holdup, HL, Gas void Fraction, λ, (-) S c h l u m b e r g e r P r i v a t e

Similarly, the gas void fraction is the fraction of the volume element element that is occupied by the gas phase. For two-Phase flow 0 < HL or λ < 1, where for single-phase flow λ or HL are either 0 or 1. Ag = 1 - HL λ = Hg = A

λ=

qg qg + qL

Where:

qL = qo + qw

gg : is the actual gas flow rate a P y T 96 Copyright 2006, NExT, All rights reserved

Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Two-phase flow variables Superficial velocity (volumetric flux), (ft/s) S c h l u m b e r g e r P r i v a t e

The superficial velocity of a phase is the velocity which would occur if only that phase flows flows alone in the pipe. It is called also the volumetric volumetric flux, and represents the volumetric flow rate per unit area of each of the phases. Thus the superficial superficial velocities of the the liquid liquid and gas phases are: qL qg usL = and usg = A A The mixture velocity is the total volumetric flow rate of both phases per unit area, and is given by: uM =

qL + qg A

= usL + usg 97


Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Two-phase flow variables Mass Flux, G (lbm/ft2-s) S c h l u m b e r g e r P r i v a t e

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

GL =

Gg =

G=

wL A wg A

= Liquid mass flux

= Gas mass flux

w L + wg A

= Total mass flux


Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Two-phase flow variables Actual (in-situ) Velocity, u (ft/s) S c h l u m b e r g e r P r i v a t e

The superficial velocities defined above are not the actual velocities of the phases, as each phase occupies only a fraction of the pipe cross section. Thus the actual actual velocities velocities of of the liquid and gas phase are, respectively: qL uL = = AL

qL usL = A HL HL

qg ug = = Ag

qg usg = A Hg 1 - HL

Ag g

AL

L


Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Two-phase flow variables 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 two phases uslip = ug – uL Quality x, (-) The quality is the ratio of the gas mass flow rate to the total mass flow rate across a given area wg x= w +w = g L

wg w 100


Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Two-phase flow variables Example No. 7 Oil and and na natural tural gas flow flow in a 2” I.D. horizontal horizontal pipe. The in-situ flow flow rates of the oil and the natural gas are 0.147 ft3/s and 0.5885 ft3/s, respectively. respectively.


The corresponding liquid holdup is 0.35. Determine: 1. The gas and liquid liquid veloci velocities ties and the mixture mixture velocity velocity 2. The actual actual veloc velocitie ities s of of the the two two ph phase ases s 3. The slip velocity velocity between between de de gas phase phase and and the liquid liquid phase phase Solution:: Solution A = π(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 101 Copyright 2006, NExT, All rights reserved

Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Two-phase flow variables S c h l u m b e r g e r P r i v a t e

Solution (Cont.): 2. uL = usL/HL = 6.74/0.35 = 19.26 ft/s ug = usg/(1 – HL) = 27/(1 27/(1 – 0.35) 0.35) = 41.54 41.54 ft/s ft/s 3. Uslip = ug – uL = (41.54 (41.54 – 19.26) 19.26) = 22.28 22.28 ft/s ft/s


Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Fundamental phenomena in two-phase flow a qg

Ag

Gas

τi

ug

Gas

Liquid

uL Liquid

AL

qL

a-a

a



Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Fundamental phenomena in two-phase flow: Slippage and Holdup Holdup: When gas and liquid phases flow at the same velocity…. S c h l u m b e r g e r P r i v a t e

Ug UL

ug = uL

UL

Ug

∴ uslip = 0 (no-slip)

HL = λL = qL/(qg + qL) = usL/(usg + usL) 104 Copyright 2006, NExT, All rights reserved

Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Fundamental phenomena in two-phase flow: Slippage and Holdup Holdup: The velocity of the gas is greater than that of the liquid. thereby resulting in a liquid resulting liquid holdup holdup that not only affect affects s well friction friction loss losses es but also flowing density. Liquid holdup is defined as the in-situ flowing volume fraction of liquid, It depends of the flow pattern.

Ug UL

Ug UL

> uL ug > Copyright 2006, NExT, All rights reserved

∴ uslip ≠≠ 0 (slip)

HL > > λL = qL/(qg + qL)

105


Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid

Flow Pattern Prediction: Baker Flow Regime Map 10-1 105

1

10

102

103

104

Disperse 104 Wave

By =

Bubble

Annular

g λ λ

G

Gg


λ

GLλφ Bx = G g

Slug

λ= (

Stratified

103

Baker Parameters

ρg 0.075

)(

ρL

1/2

) 62.4 1/3

φ=

Plug 102 10-1

1

10 GLλφ Gg


102

103

104

73

σL

µL ( 62.4 )2 ρL

Gg = ρg usg GL = ρg usL 106

Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Flow Pattern Pattern Prediction: Prediction: Beggs and Brill flow regime regime map

uM2 NFr = gD

UM : Mixture velocity D : inside pipe diameter g : gravitational acceleration

λ: liquid input volume fraction



Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Flow Pattern Prediction: Prediction: Taitel-Dukler flow regime regime map map

75.0

Bubbly Bubbly


10.0 Intermittent U sL 1.00 (ft/s) 0.10

Annular

Stratified Smooth

0.01 0 .1

1 .0

Stratified Wavy 1 0 .0 U sG (ft/s)

100.0

9 0 0 .0


Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Flow Pattern Prediction: Prediction: Mandhan flow regime regime map Dispersed Flow

10.0 . s / t f

Bubble Flow

,


Slug Flow

L S

v 1.0 , y t i c o l e v l a i c 0.1 i f r e p u s d i u q i L 0.01

0.1

Annular Flow Wavy Flow Stratified Flow

1.0

10.0

Gas superficial velocity, v SG, ft/s Copyright 2006, NExT, All rights reserved

100 109

Multiphase Flow


TWO - -PHASE PHASE FLOW: Gas - -Liquid L iquid

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


(Plug)


Multiphase Flow


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

Flow Pattern Prediction: Example: Predicting horizontal gas-liquid flow regime


Using de Baker, Mandhane, Mandhane, and Beggs & Brill flow flow regime regime maps, determine the flow regime for the flow of 2000 bbl/d of oil oil and and 1 MMs MMscfd cfd of gas gas at at 800 800 psi psia a and 175ºF 175ºF in a 2 ½ in. I.D. pipe. The oil density and viscosity are 49.92 lbm/ft3 and 2 cP, respectively. The oil-gas surface tension is 30 dynes/cm and the gas density, viscosity and the compressibility factor are 2.6 lbm/ ft3, 0.01 0.0131 31 cP an and d 0.935 0.935 respectively. The pipe relative roughness is 0.0006.


Multiphase Flow



Flow Pattern Prediction: Solution for Baker : Baker’s parameters

By =

Gg

λ= (

λ = [(2.6/0.075)(49.92/62.4)]0.5 = 5.27

0.075

)(

ρL

1/2

) 62.4 1/3

λ

GLλφ Bx = Gg

ρg

φ=

73

σL

µL ( 62.4 )2 ρL


Gg = ρg usg GL = ρg usL

φ= (73/30)[(2)(62.4/49.92)2]1/3 = 3.56 A = π (2.5/12)2 /4 = 0.0341 ft2 qL = (2,000bbl/day)(5.615 ft3/bbl)/(86,400 day/s) = 0.130 ft3/s GL = wL/A = ρLqL/A = ρLusL , = (49.92lbm/ft3)(0.130 ft3/s)/(0.0341ft2) (3600 s/hr) = 6.85 x 105 lbm/hr-ft2 112 Copyright 2006, NExT, All rights reserved

Multiphase Flow



Flow Pattern Prediction: Solution for Baker : T )( psc ) q Z ( qg = sc p Tsc


qg = (106 ft3/day)(0.935)(635ºR/520 /day)(0.935)(635ºR/520ºR)(15psia/800psia) ºR)(15psia/800psia) 1day/86400s= 0.2478 ft3/s

Gg = wg/A = ρgqg/A = ρgusg= (2.6 lbm/ ft3 x 0.2478 ft3/s)/(0.0341 ft2)x(3600s/hr)= Gg =6.8x 104 lbm/hr-ft2 By =

Gg

λ

Flow Pattern: Bubble = 6.8x 104 lbm/hr-ft2/ 5.27= 1.29x104

Bx GLλφ/Gg = (6.85 x 105)(5.27)(3.56)/(6.8 x 104) = 188

though the conditions are very near the boundaries with slug flow and annular mist flow


Multiphase Flow



Flow Pattern Prediction: Baker Flow Regime Map 10-1

105

1

10

102

103

104 S c h l u m b e r g e r P r i v a t e

Disperse 104

Wave

Bubble

Annular

g λ λ

G

Slug Stratified

103

Plug 102 10-1

1

10 GLλφ Gg


102

103

104 114

Multiphase Flow



Flow Pattern Prediction: Baker Flow Regime Map



Multiphase Flow



Flow Pattern Prediction: Solution Solu tion for Man Mandhan dhane e map : The Mandhan Mandhane e map is simply simply a plot of superfic superficial ial liquid liquid velocity velocity versus versus superficial superfici al gas velocity. For our values usL = 3.81 ft/s and usg = 7.27 ft/s, the flow 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



Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Flow Pattern Pattern Prediction: Prediction: Mandhan flow regime regime map map Dispersed Flow

10.0 . s / t f

Bubble Flow

,


Slug Flow

L S

U 1.0 , y t i c o l e v l a i c 0.1 i f r e p u s d i u q i L 0.01

0.1

Annular Flow Wavy Flow Stratified Flow

1.0

10.0

Gas superficial velocity, U SG, ft/s Copyright 2006, NExT, All rights reserved

100 117

Multiphase Flow



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


(Plug)


Multiphase Flow



Flow Pattern Prediction: 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

uM2 NFr = gD

λL = usL/uM = 3.81/11.08 = 0.35 From the Beggs Beggs & Brill flow regime regime map, the flow regime is predicted to be be intermittent. Slug flow is the likely flow regime. 119 Copyright 2006, NExT, All rights reserved

Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Flow Pattern Pattern Prediction: Prediction: Beggs and Brill flow regime regime map

uM2 NFr = gD

UM : Mixture velocity D : inside pipe diameter g : gravitational acceleration

λ: liquid input volume fraction



Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Fundamental phenomena in two-phase flow: Pressure drop correlations General energy flow equation dp dz =

g ρ sin θ gc

+

f ρ u2 2 gc D

+

ρu gc


du dz

Three main components for predicting pressure los are: 1. Elev Elevat atio ion n or stat static ic com compo pone nent nt 2. Fric Fricti tion on co comp mpon onen entt 3. Accel cceler erat atio ion n compo compone nent nt Total Loss Loss Loss Pressure = Caused by + Caused by + Caused by loss Elevation Friction acceleration Copyright 2006, NExT, All rights reserved

121

Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Fundamental phenomena in two-phase flow: Pressure drop correlations Energy equation for horizontal flow dp dz =

dp dz

f ρ u2 2 gc D

=

ρu

+

dp dz

gc


du dz

+ f

dp dz acc

or neglecting the kinetic energy effects dp dz

=

dp dz

f 122


Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Fundamental phenomena in two-phase flow: Pressure drop correlations Pressure Loss Components S c h l u m b e r g e r P r i v a t e

Where: ρ : Density, lbm/ft3 u : velocity, ft/s D : pipe diameter, ft g : acceleration caused by gravity, ft/s 2 gc : conversion factor, lbm-ft/lbf-s2 f : friction factor dp/dz : pressure gradient, psi/ft


Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Horizontal Pressure Loss Prediction Methods Over the years, numerous correlations have been developed to calculate the pressure gradient in horizontal horizontal gas-liquid gas-liquid flow. The most most commonly used in the oil and gas industry today are those of Beggs and Brill Brill (1973 (1973), ), Eaton Eaton et al. (1967) (1967),, and Du Dukle klerr (1969) (1969).. These Th ese correlations all include a kinetic energy contribution to the pressure gradient; however, this can be considered negligible unless the gas rate is high and the pressure is low.


Correlations most widely used 1. Beggs Beggs and and Bril Brilll (JPT (JPT,, 607-6 607-617, 17, May 1973) 1973) 2. Du Dukl kler er (AGA (AGA,, API API,, Vol Vol.. 1, 1, Research Results , May 1969) 3. Eaton et al. (Trans. AIME , 240: 815-828, 1967) 124 Copyright 2006, NExT, All rights reserved

Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Horizontal Pressure Loss Prediction Methods Beggs and Brill correlation S c h l u m b e r g e r P r i v a t e

Correlating parameters:

NFr = um2 / gD

λL = usL/um L1 = 316 λL0.302 L2 = 0.0009252 λ-2.4684 1.4516 16 L3 = 0.10 λL- 1.45

L4 = 0.5 λL-6.738


Multiphase Flow



The flow regime transitions are given by the following: Segregated flow exist if

λL << 0.01 and NFR < L1 or λL ≥≥ 0.01 and NFR << L2 Transition flow occurs when

λL ≥≥ 0.01 and L2 << NFR ≤≤ L3 Intermittent flow exist when

≥ 0.4 and L3 < NFR ≤≤ L4 0.01 ≤ λL < 0.4 and L3 < NFR ≤ L1 or λL ≥ Distributed flow occurs if

λL < 0.4 and NFR ≥≥ L1 or λL ≥≥ 0.4 and NFR > L4 126 Copyright 2006, NExT, All rights reserved

Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Horizontal Pressure Loss Prediction Methods Beggs Beggs and Brill correlati correlation on


The flow regime transitions are given by the following: Transition flow

If the flow f low regime is transition flow, the liquid holdup is calculated using both the segregated and intermittent equations and interpolated using the following HL = A λL(segregated) + B λL(intermittent) Where: A =

L3 - NFR L3 – L2

and

B=1-A 127


Multiphase Flow



Liquid holdup, and hence, the average density HL(φ) = HL(0) x ψ ψ HL(0) = a λLb / NFRc

≥ λL and With the constraint that HL(0) ≥ 0.333 3 sin sin3(1.8θ)] ψ ψ = 1 + C[sin (1.8θ) – 0.33 Where

λL)ln(d λLe NLVf NFRg) C = (1 - λ Where: Where: a, b, c, d, d, e, f, and and g depend depend on the flow flow regime regime and are are given given in the following tables. C must be ≥ 0 and NLV = usL(ρL/g σ)1/4 Copyright 2006, NExT, All rights reserved

128

Multiphase Flow



Liquid holdup, and hence, the average density Beggs and Brill holdup constant Flow pattern Segregated

a

b

c

0.98

0.4846

0.0868

Intermittent

0.845

Distributed

1.065

0.5351 0.5824

0.0173 0.0609

≥ λL With the constraint that HL(0) ≥ 129 Copyright 2006, NExT, All rights reserved



Liquid holdup, and hence, the average density Beggs and Brill holdup constant Horizontal flow pattern

d

e

f

g

Segregated uphill

0.011

-3.768

3.539

-1.614

Intermittent uphill

2.960

0.305

-0.4473

0.0978

Distributed uphill All flow pattern downhill

C = 0, ψ ψ = 1, HL ≠≠ f(θ)

No correction

4.70

-0.3692

0.1244

-0.5056


Multiphase Flow



Friction factor The frictional pressure gradient is calculated from

dp dz

f

=

ρn um2 ftp ρ 2 gc D

λL + ρg λλg Where: ρn = ρL λ ftp = fn (ftp / fn) 131 Copyright 2006, NExT, All rights reserved

Multiphase Flow



Friction factor The non-slip friction factor is determined from from the smooth pipe pipe curve on a Mood Moody y diag diagra ram m or fro from m fn = 1/ [2 log(NRen / (4.5223 log NRen – 3.8215))]2 Where: NRen = ρn um D / µn and

µn = µL λλL + µg λλg

The ratio of the two-phase to no-slip friction factor is calculated from ftp / fn = eS 132 Copyright 2006, NExT, All rights reserved

Multiphase Flow



Friction factor Where: S = [ln(y)] / {-0.0523 {-0.0523 + 3.182 3.182 ln(y) – 0.8725[ln(y 0.8725[ln(y)] )]2 + 0.01853[ln(y)]4} and y = λL / [HL(θ)]2

< y <1.2, for The value of S becomes unbounded at a point in the interval 1 < this interval, S = ln( ln(2. 2.2y 2y – 1.2) 1.2)


Multiphase Flow



Acceleration term The kinetic energy contribution to the pressure gradient is accounted for with a parameter Ek as follows: dp dz

ρs um usg acc

=

gc p

dp dz

= Ek

dp dz

Where:

Ek =

ρs um usg gc p

ρs = ρL HL + ρg (1 – HL) The total pressure gradient can be calculate from

dp dz

=

dp dz

+

el (1 – Ek)

dp dz

f

=

dp dz f (1 – Ek)

(For horizontal flow) 134


Multiphase Flow



Exer Ex erci cise se No No.. 9 Given the following information for a pipe transporting a wet gas, qg’ = 400 MMscfd qo’ = 4000 stb/D γ D = 16 in. = 1.333 ft = 0.70 γg API = 40º p = 2500 psia ε T = 60 ºF = 0.0006 ft Calculate the pressure gradient using the Beggs and Brill correlation Solution: 1. Prel Prelim imin inar ary y calc calcul ulat atio ions ns RP = 400 x 106/4000 = 100,000 scf/stb γ γo = 141.5/(131.5 + 40) = 0.825 A = π(1.333)2/4 = 1.396 ft2 From empirical correlations 135 Copyright 2006, NExT, All rights reserved

Multiphase Flow



Rs = 919 scf/STBO Bo = 1.419 bbl/STBO Z = 0.666

µo = 1.359 cP µg = 0.0233 cP = 1.566 x 10-5 lbm/ft-s σo = 4.608 dyne/cm Using fluid physical property values:

ρo = 42.45 lbm/ft3 136 Copyright 2006, NExT, All rights reserved

Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Horizontal Pressure Loss Prediction Methods Beggs and Brill correlation

ρg = 13.96 lbm/ft3 Bg = 0.003916 ft3/scf qo = qo’ Bo = (4,000)(1.419)(5.614)/86 (4,000)(1.419)(5.614)/86,400 ,400 = 0.369 ft3/s qg = qo’ (Rp – Rs)Bg = (4,000)(100, (4,000)(100,000 000 – 919)(0.003 919)(0.003916) 916)/86,40 /86,400 0 = 17.963 17.963 ft3/s usL = qo/A = 0.369/1.396 = 0.264 ft/s usg = qg/A = 17.963/1.396 = 12.867 ft/s um = usL + usg = 13.131 ft/s

λL = usL/um = 0-264/13.131 = 0.02 ρn = ρo λλL + ρg (1 - λλL) = (42.45)(0.02 (42.45)(0.02)) + (13.66)(1 – 0.02) 0.02) = 14.236 lbm/ft lbm/ft3 µn = µL λλL + µg (1 - λλL) = (1.359)(0.0 (1.359)(0.02) 2) + (0.0233)(1 (0.0233)(1 – 0.02) = 0.05 0.05 cP 137 Copyright 2006, NExT, All rights reserved


Multiphase Flow



2. Dete Determ rmin ine e flo flow w reg regim ime e NFR = um2/gD = (13.131)2/(32.2 x1.333) = 4.02 L1

= 316 λL0.302 = 96.96

L2

= 0.0009252 λL-2.4684 = 14.45

≥ =.01 and NFR < L2 Since λL ≥ Flow regime is segregated 3. Dete Determ rmin ine e liq liquid uid ho hold ldup up HL(0) = (0.98) (0.02)0.4846/(4.02)0.0868 = 0.13 4. Dete Determ rmin ine e fric fricti tion on fact factor or NRen = 1488ρnumD/µn = 7.416 x 106 138 Copyright 2006, NExT, All rights reserved

Multiphase Flow



fn = 1 / [2 log(NRen/ (4.5223 log NRen – 3.8215))]2 = 0.00846 y = λL /[HL(0)]2 = 0.02 /(0.13)2 = 1.18343 Since 1 < y < 1.2, then S = ln(2. ln(2.2 2 y – 1.2) 1.2) = ln(2.2 ln(2.2 x 1.183 1.18343 43 – 1.2) 1.2) = 0.339 0.339 ftp/fn = eS = e0.339 = 1.4035 ftp = fn (ftp/fn) = (0.00846)(1.4035) = 0.01187


Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Horizontal Pressure Loss Prediction Methods Beggs and Brill correlation 5) Determ Determine ine (dp/ (dp/dz) dz)f ρn um2 (dp/dz)f = ftp ρ = 2 gc D


(0.01187)(14.236)(13.131)2 (2)(32.17)(1.333)

= 0.340 psf/ft 6) Determine Ek Ek = ρs um usg/ gc p

ρs = ρL HL + ρg (1 – HL) = (42.45)(0.1 (42.45)(0.13) 3) + 13.66)(1 13.66)(1 – 0.13) 0.13) = 17.4 lbm/ft lbm/ft3 Ek = (17.4)(13.131)(12.867)/[(32.17)(2500)(144)] = 0.00025 Copyright 2006, NExT, All rights reserved

140

Multiphase Flow



7) Determ Determine ine the the total total pres pressur sure e gradie gradient nt dp/dz = (dp/dz)f / (1 – Ek) = 0.340 0.340/(1 /(1 – 0.0002 0.00025) 5) = 0.340 psf/ft = 0.00236 psi/ft


Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Horizontal Pressure Loss Prediction Methods Dukler et al Dukler al Corre Correlatio lation n (case (case II – Cons Constant tant slip) The Dukler et al correlation was based on similarity analysis analysis and the friction factor and the liquid holdup correlations were developed from field data. This correlation is recommended in a design manual published jointly by the AGA and APIº Dukler Dukler Fricti Friction on Factor Factor (dp/dz)f = f ρk um2 / 2 gc D

λL2 / HL + ρg λλg2 / Hg Where ρk = ρL λ f / fn = 1 + y / [1.281 [1.281 – 0.478y 0.478y + 0.444y 0.444y2 – 0.094y3 + 0.00843y4] Where y = –ln(λL) 142 Copyright 2006, NExT, All rights reserved


Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Horizontal Pressure Loss Prediction Methods Duklerr et al Correlation Dukle Correlation (case (case II – Cons Constant tant slip) slip) S c h l u m b e r g e r P r i v a t e

Dukler Friction Factor fn = 0.0056 + 0.5 NRek-0.32 Where NRek = ρk um D / µn um = usL + usg

µn = µL λλL + µg λλg Dukler Liquid Holdup An iterative or trial and error error procedure procedure is required required to obtain a value value of liquid holdup using Dukler’s method. That is 143 Copyright 2006, NExT, All rights reserved

Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Horizontal Pressure Loss Prediction Methods Duklerr et al Correlation Dukle Correlation (case (case II – Cons Constant tant slip) slip) Dukler Liquid Holdup HL = f(λL, NRek)

and

NRek = f(HL)

The correlation is given in a graphical form with liquid holdup plotted versus no-slip holdup with Reynolds number as a parameter. The procedure for obtaining a holdup value is


1) Calculate λL 2) Estimate HL 3) Calculate NRek 4) Obtain HL from graph 5) Compare values of HL from step 2 and 4. If they are not sufficiently sufficiently close, set the value obtained in step 4 as the new value and return to step 3. Agreement within 5% is considered close enough 144 Copyright 2006, NExT, All rights reserved

Multiphase Flow



Dukler Acceleration Term dp dz

or

= acc Ek =

1

gc dz 1

gc dp

∆

∆

ρg usg2 Hg

ρg usg2 Hg

The total pressure gradient is

+

+

ρL usL2 HL

ρL usL2 HL

dp = dz

dp dz 1 - Ek

f 145


Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Horizontal Pressure Loss Prediction Methods Duklerr et al Dukle al Corre Correlatio lation n (case (case II – Cons Constant tant slip) Example No. 10 Calculate the the pressure pressure gradient gradient for the the problem of Example Example No. 9 using the Dukler et al. correlation correlation and neglecting kinetic energy energy effects. Solution:


1)Determine Liquid Holdup Assume HL = 0.02

ρk = ρL λλL2 / HL + ρg λλg2 / Hg = (42.45)(0.02)2/(0.02) + (13.66)(0.98)2/(0.98) = 14.236 lbm/ft3 NRek = 1488 ρk um D/µn = 7.416 x106 146 Copyright 2006, NExT, All rights reserved

Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Horizontal Pressure Loss Prediction Methods Duklerr et al Correlation Dukle Correlation (case (case II – Cons Constant tant slip) slip) From the figure for NRek = ∞ first iteration.

HL = 0.02 ∴ convergence is obtain in the

2) Determine friction factor fn = 0.0056 + 0.5 NRek-0.32 = 0.00877 y = -ln(λL) = -ln(0.02) = f/fn = 1 + y / [1.281 [1.281 – 0.478y + 0.444y 0.444y2 – 0.094y3 + 0.00843y4] = 2.57 f = fn (f/fn) = (0.00877)(2.57) = 0.0225 147 Copyright 2006, NExT, All rights reserved


Multiphase Flow



3) Determine pressure gradient

dp/dz = (dp/dz)f = f ρk um2 / 2 gc D = (0.0225)(14.236)(13.131) (0.0225)(14.236)(13.131)2 / [(2)(32.174)(1.333)] = 0.6439 psf/ft = 0.00447 psi/ft


Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Horizontal Pressure Loss Prediction Methods Eaton et al Correlation Eaton Friction Factor dp f ρn um2 = 2g D d c f z


f wm2 = 2 gc D A2 ρn

The two-phase friction factor is correlated with the group f

wL wm

0.1

ψ = ψ

0.057 (wg wm)0.5

µg D2.25

This group is dimensionless for the units of lbm, ft, s. The correlation is shown in graphical graphical form. form. The viscosity viscosity has the units of lbm/ft-s Copyright 2006, NExT, All rights reserved

149

Multiphase Flow


Horizontal Pressure Loss Prediction Methods Friction Factor Eaton et al. Correlation


1 . 0 L

m

w w f

0.057 (wg wm)0.5 150

µg Copyright 2006, NExT, All rights reserved

D2.25

Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Horizontal Pressure Loss Prediction Methods Eaton et al. Correlation Eaton Liquid Holdup Eaton’s analysis for acceleration loss requires a value of liquid holdup since the acceleration term is based on change in actual gas and liquid velocities. Liquid holdup was correlated with the following dimensionless group. 1.84 NLv0.575 Ngv Nd0.0277

p 0.05 0.1 NL pb

ρL /gσ)1/4 Where: NLv = usL ( ρ ρLg /σ)1/2 Nd = D ( ρ

ρL /gσ)1/4 Ngv = usg ( ρ NL = µL ( g/ρLσ3)1/4

pb = 14.65 psi HL is needed only for calculating the acceleration term 151 Copyright 2006, NExT, All rights reserved


Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Horizontal Pressure Loss Prediction Methods Eaton et al. Correlation Eaton et al. liquid Holdup Correlation Correlation



Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Horizontal Pressure Loss Prediction Methods Eaton et al. Correlation Eaton Acceleration Term


The pressure gradient due to acceleration is given by dp dz

acc

=

wL ∆ ∆uL2 + wg ∆ ∆ug2 2 gc qm dz

Where ∆uL2 = uL2(p1,T1) – uL2(p2,T2) ∆ug2 is similarly defined and

If we define Ek as Ek = dz dp

dp dz

ac c

=

∆uL2 + wg ∆ ∆ug2 wL ∆ 2 gc qm dp

The total pressure gradient can be calculated from:

dp = dz

dp dz 1 - Ek

f 153


Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Horizontal Pressure Loss Prediction Methods Example No. 11 Calculate the pressure pressure gradient for the problem of Example No. No. 9 using the Eaton et al. correlation and neglecting neglecting kinetic energy effects. Solution: 1)Determine friction factor wL = ρL qL = (42.45)(0.369) = 15.664 lbm/ft3 wg = ρg qg = (13.66)(17.963) = 245.375 lbm/ft3 wm = wL + wg = 261.039 lbm/ft3 The correlation parameter 154 Copyright 2006, NExT, All rights reserved


Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Horizontal Pressure Loss Prediction Methods Example No. 11 Calculate the pressure gradient for the problem of Exercise No. 9 using the Eaton et al. correlation and neglecting kinetic energy effects. 0.057 (wg wm)0.5

µg

D2.25

=

(0.057)[(245.375)(261.039)]0.5 (1.566 x

10-5)(1.333)2.25


= 4.822 x 105

From figure, Ordinate = 0.011 f = Ordinate/(w L/wm)0.1 = 0.011/(15.664/ 261.039)0.1 = 0.01457 2) Determine liquid holdup holdup (Note – this is not required required since since kinetic energy effects are neglected.) NLv = 1.938 usL (ρL / σL)1/4 = (1.938)(0.264)(42.45/4.6 (1.938)(0.264)(42.45/4.608) 08)1/4 = 0.891 Ngv = 1.938 usg (ρL / σL)1/4 = (1.938)(12,867)(42.45/4.608) (1.938)(12,867)(42.45/4.608)1/4 = 43.439 155 Copyright 2006, NExT, All rights reserved

Multiphase Flow


TWO-PHASE FLOW: Gas-Liquid Horizontal Pressure Loss Prediction Methods Nd = 120.872 D (ρL / σL)1/2 = (120.872)(1.333)(42.45 ( 120.872)(1.333)(42.45/4.608) /4.608)1/2 = 489.05 NL = 0.15726 µL [1 / (ρL σL3)] = (0.15726)(1.359)[1 / (42.45 x 4.6083)] = 0.0266 1.84 NLv0.575 Ngv Nd0.0277

p 0.05 0.1 1.84 (0.891)0.575 NL = pb (43.439) (489.05)0.0277

2500 14.65

0.05

(0.0266)0.1 = 0.03

From the the figure HL = 0.09 3) Determine pressure gradient dp = dz

dp dz

f wm2 = 2 ρ = 2 g D A c n f

(0.01457)(261.039)2 (2)(32.174)(1.333)(1.396)2(14.236)

= 0.417 psf/ft = 0.0029 psi/ft 156 Copyright 2006, NExT, All rights reserved


Flow of Fluid Flow in Horizontal Pipes


Example of the simulation of the horizontal stratified two-phase flow using the lattice Boltzmann method: It is observed that the wave, which is generated at the flow inlet, moves downstream and is deformed by the flow velocity velocity difference between between the two phases. phases. The color distribution represents the distribution of the flow velocity in the flow direction


Flow of Fluid Multiphase Flow Correlations Industry Standard Correlation. 

Dan ans s & Ro Ros



Orkiszewsk



Hagedo Hagedorn rn & Brown Brown



Beggs Beggs & Brill (Original (Original & Revis Revised) ed)



Govier, Govier, Aziz Aziz & Forgar Forgarasi asi



Noslip



AGA & Flanigan



Oliemans



Gray



Flow of Fluid Applicability of Two Phase Correlations Vertical oil Well

Vertical Gas Condensate

Oil Pipeline

Gas Condensate

Dans Dans & Ro Ros s









Orkiszewsk











Hage Hagedo dorn rn & Brown Brown











Begg Beggs s & Bril Brilll (Original & Revised)









Govie Govier, r, Aziz Aziz & Forgarasi









Noslip









AGA & Flanigan



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Oliemans

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Gray

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Effect of Flow Pattern on Pipeline corrosion The presence of water in pipeline can cause severe corrosion Corrosion Rate depends on the flow pattern S c h l u m b e r g e r P r i v a t e

Pipeline internal inspection shows that the highest corrosion rate is obtained when low flow velocity generates phase segregation (stratified phases) The deposition of water with solids generates pitting corrosion and colonies of SRB (sulfate reducing bacteria) at the bottom of the pipeline 160 Copyright 2006, NExT, All rights reserved

Effect of Flow Pattern on Pipeline corrosion Flow Pattern is is one the main consideration consideration for corrosion corrosion risk on pipelines Flow Diagram for Corrosion Risk evaluation on pipeline IS LIQUID WATER POSSIBLE?

START

STOP!! NO CORROSION POSSIBLE

STEADY STATE CORROSION

YES

YES

DETERMINE FLOW REGIME


NO STOP!! NO CORROSION POSSIBLE

IS FLOW DISRUPTION PRESENT? NO

NO

CAN LIQUID WATER CONTACT WALL? YES

TURBULENT

DEGREE OF WATER TURBULENT

EQUILIBRIUM CORROSION

STAGNANT OR LAMINAR


Effect of Flow Pattern on Pipeline corrosion Type of Corrosion Under Multiphase Flow Flow Pattern 2 phases gas/water flow Stratified

Free water localization

Water Turbulence

Bottom

Stagnant to Laminar

Generally in the bottom, circumferential mixed

Very Turbulent

Slug

Turbulent

Type Of Corrosion Corrosion under deposit Corrosion induced by flow Corrosion induced by flow


Annular 2 phases oil/water flow Segregated (Stratified) Mixed

Bottom of the pipeline

Stagnant to Laminar

Generally in the bottom, mixed.

Laminar to Turbulent Turbulent

Dispersed

Mixed

3 phases gas/oil/water flow Stratified

Bottom of the pipeline, separated

Slug Annular Copyright 2006, NExT, All rights reserved

Stagnant to Laminar Very Turbulent

Generally in the bottom, mixed Circumferential

Possible Turbulent

Corrosion under deposit Corrosion under deposit Corrosion induced by flow Corrosion under deposit Corrosion induced by flow Corrosion induced by flow

162

Standards and Practices Minimum pipeline transition velocity for different mixing elements, ASTM-4177



Use of of Software Software Use Liquid Line Line Sizing Sizing --Liquid Gas Line Line Sizing Sizing --Gas Multiphase Line Line Sizing Sizing --Multiphase



Pipesim Session Session Pipesim Gathering Network Network Modeling Modeling Gathering Gas Transmission Transmission Modeling Modeling Gas




PIPESIM Network Analysis

1 day session


Network Simulation Rigorous and comprehensive steady-state multiphase network simulator Combines the detailed well modeling capability of the single branch model with the ability to solve large complex networks Networks of any size and topology (loops, multiple sources & sinks, parallel flowlines, crossovers) In-line flashing, black oil/compositional Rigorous thermodynamic calculations All single branch components can be included in a network



Model Example




• –


Model Example Network Model has no Maximum Node Limit S c h l u m b e r g e r P r i v a t e

791 Wells 977 Branches 949 Nodes 169 Copyright 2006, NExT, All rights reserved

Types of Networks Gather Gatherin ing g flowli flowline ne system systems s Distribution (including water injection and gas lift distribution)


Looped networks (calculations in  • flow direction around the system) –

Transmission Transmission lines systems


Steps in building a model n Set units & job title n Define components in the model:


n branches (flowline or trunklines) n Enter physical data for each component n Define global/local fluid models and flow correlations



Basic Design Design Basic Considerations Considerations


General Consideratio Considerations ns The pipe design takes into consideration pipe diameter and wall thickness selection The idea is to find the internal diameter that satisfies the flow regime for a given pressure.


The procedure is by try and error (iterative solution). Select the diameter and the pressure drop is calculated for a required flow.


Basic Design Considerations Pressure drop through pipeline components Pipe (Procedure). S c h l u m b e r g e r P r i v a t e

For a diameter and flow, determine the Reynolds number Estimate the friction factor Calculate the friction pressure drop Obtain the total pressure drop


Basic Design Considerations Pressure drop through pipeline components Elbows (Procedure). Determine Reynolds number for a given diameter and flow


Estimate the friction factor

∆P =2.16x10-4

KρU2 2gc

∆P =1.08x10-4 KρU2 ∆P =2.16x10

-4

KρQ2 d4

Where ∆P : Pre Press ssur ure ed dro rop, p, ps psia ia K : Re Resistance co coefficient ρ : Density, lbm/ft3 U : Flow velocity, ft/s d : Internal diameter, in Q : Flow rate, gpm


Basic Design Considerations Resistance Coefficient for elbows









Basic Design Considerations Pressure drop through pipeline components Valve Valves s (Proce (Procedur dures) es).. S c h l u m b e r g e r P r i v a t e

Obtain the resistance coefficient using L/D from table and figures, same procedure than elbows.


Basic Design Considerations Valve Equivalent Lengths L y L/D y and Resistance coefficient k



Basic Design Considerations













Pipeline Design Design Pipeline


Pipelines Around the World



Pipeline Pipel ine Desig Design n - Fun Fundame damental ntal Elem Elements ents 



Pipeline design includes: 

Selection of the route traversed by the pipe



Determination fluid to be transported and the operational conditions



Calculation of pressure gradient



Selection of pumps and other equipment



Determination material



Engineering economic analysis and a market analysis to determine the optimum system based on alternate designs

Each design considerations:

of

must

pipe

have

thickness

the



Safety



Leak and damage prevention



Government regulations



Environmental concerns


and

following


Pipeline Legislation Pipeline

Legal/Statutory Position

 The

operation of transmission pipelines is usually controlled by national regulations or laws. The selection of a design code, or design calculations are often limited by these regulations/laws.


USA  Regulations

based heavily on the ASME B31 standards.


Pipeline Design Codes  Oil

and gas pipeline codes:  ANSI/ASME

B31.4 Petroleum Transportation Transportation

Liquid

 ANSI/ASME

B31.8 Gas and Distribution

Transmission Pipeline System


 Others.  IGE/TD/1  In

is for methane gas only

1993 BS 8010 was introduced

 International

(ISO)

ISO

13623 - This is an international pipeline standard, covering oil and gas lines.


Pipeline Design Codes Pipeline Pipel ine Design Design Codes - Oil and and Gas Gas  Pipeline  For

Codes treat oil and gas pipelines different

oil pipelines:



No account is taken of population density in the location of the pipelines



There is no specified distance to occupied buildings



It can generally build an oil pipeline with a high design factor (0.72) in most locations.

 For


gas pipelines:



Account is taken of population density



Minimum distance from occupied buildings is specified



Design factor is lowered in populated areas (0.3 in UK, 0.4 in USA) 190


Pipeline Design Codes American Lifelines Alliance Matrix of Standards and Guidelines for Natural Hazards: Oil Product System Component System Reliability Buried Pipelines Aboveground Piping

Pumping Station Piping Well Facilities Refineries Storage Tanks

Natural Hazard Provisions

Guide/Standard

Loading

Design

ASME/ANSI B31.4 ASME/ANSI B31.4 ASME/ANSI B31.3 API 2510 API 2508 ASME/ANSI B31.4 ASME/ANSI B31.3 API 2510 ASME/ANSI B31.4 ASME/ANSI B31.3 API RP 14E ASME/ANSI B31.3 ASME BPV API 2508 API 620 API 650 API 2508

None

None

None None

None Earthquake, Wind, Ice

None None

None Earthquake, Wind, Ice

None None

None Earthquake, Wind

None None

Earthquake, Wind, Ice Earthquake, Wind, Ice

Earthquake, Wind Earthquake, Wind

Earthquake, Wind Earthquake, Wind



Pipeline Design Codes American Lifelines Alliance Matrix of Standards and Guidelines for Natural Hazards:



Pipeline Design.pdf

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