Gs Lift Design Oourse

8/13/2019 Gs Lift Design Oourse

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2007 Workshop Clegg & Smith 1

API Gas Lift Design

• API RP 11V6: Recommended Practice for Design

of Continuous Flow Gas Lift Installations ---- Using Injection Pressure Operated Valves

• By Sid Smith




INTRODUCTIONS

PLEASE TELL US THE FOLLOWING

INFORMATION ABOUT YOURSELF:

Name

Work LocationJob (role)

Number of Years Experience

Gas Lift Background- hands on- previous training




Design Outline• Introduction (20)• General Design (20)

• Inflow & Outflow &…….Tubing (45)• Facilities (15)

• Gas Inj. Pressure (15)• Mandrels & Valves (30)• Temperature (15)• Gas Passage (15)

• Design Methods:.Constant Rate (30).Variable Rate (30).Intermittent (15).Equilibrium Curve (45)

• API Example # 1 (45)



• Summary (15)




0. Introduction :API RP 11L6

• RP to provide guidelines, proceduresand recommendations. See other APIRP’s. (Also ISO documents)

• 1 Scope: Guidelines for continuousflow using injection pressure valves

• 2 Intent: Maximize production and. Minimize costs

• 3 Definitions• 4 General Design Considerations




S.V.

Continuous Intermittent




Typical continuous flow gas lift installation

Injection gas into wing valve and then downthe casing-tubing annulus.

Well equipped with tubing, side pocket mandrels,

wireline retrievable gas lift valves and asingle production packer located justabove the producing zone.

Note: Upper gas lift valves closed and

gas enters gas lift valve near bottom.




4.1 General

• Complete system!

• Combination of concepts and experience

• Continuous flow gas lift has advantages

and limitations.




Continuous GL Strengths

• Flexible lift capacity

• Handles sand OK• Deviated holes OK

• Permits wire line op

• Tubing fully open

• High GLR beneficial

• Low well R&M

• Low surface profile

• Compatible /SSSV’s• Permits sounding

• Easy BHP surveys

• Permits PL surveys

• Dual lift feasible

• Tolerates bad design




Continuous GL Weakness• High back pressure imposed

• Needs uninterrupted high pressure gas

• Compressor expenses often high• Heading problem with low rates

• Potential gas freezing & hydrate problem

• Increased friction w/low gravity crude

• Valve interference & high inj. point

• Corrosion, Scale, & paraffin

• Efficient dual lift difficult

• Requires excellent data for good design




• Inject gas as deep as feasible

• Conserve injection pressure• Ensure upper valves stay closed

• Be able to work down to bottom• Check for ample gas passage

• Plan for changes in rate

• Avoid heading conditions

• Minimize costs & Maximize rates

GL Design Guidelines




Types of Installations• Conventional (Tubing): Inject gas down

annulus & produce up tubing.• Annulus: Inject gas down tubing & produce

up annulus• Special: Slim-hole, Dual, Concentric, etc

• Open installation: No packer or SV• Semi-closed: Packer but no SV




IPV UNLOADING

How do we inject gas into a well in the first place?How do we inject gas into a well in the first place?

This process of replacing the completion brine with injection gas is calledunloading and it is done only once after the initial completion and after any wellservicing where the casing to tubing annulus is filled with liquid.

Pressure

D e p t h

SBHP




IPV UNLOADINGGas is injected into the casing - tubing annulus and the pressure pushes the brinethrough each of the gas lift valves which are wide open. This is a particularly dangeroustime for the valves. If the differential is too high the liquid velocity can be enough to cutthe valve seat. Then, the valve will not be able to close and the design will not work.

Pressure

D e p t h

SBHP






IPV UNLOADINGOnce the brine level is below the top valve, gas will enter thetubing and begin lifting the well. If the tubing pressure is less thanthe SBHP the reservoir will begin to contribute. The firstproduction from the reservoir is normally recovered completionbrine.

Pressure

D e p t h

SBHP




IPV UNLOADINGWhen the second valve is uncovered, gas will begin to enter

the tubing at the second

valve.

Pressure

D e p t h

SBHP




IPV UNLOADINGIn the case of IP valves, the injection gas rate into the well at the surface must beregulated to control the gas entry to approximately the design rate of one valve. Sincetwo valves are passing injection gas, the pressure in the casing annulus will fall .

Gas from

casing 500

mcfd

Gas to

casing 500

mcfd

Gas from

casing 500

mcfd

+500

-500

-500

With more gas leaving casing than entering, the injection pressure mustfall.




IPV UNLOADING

When the casing pressure falls enough, the top valve will close basedon valve mechanics in a good design.

Pressure

D e p t h

SBHP




IPV UNLOADING

Since there is still more casing pressure than tubing pressure at thebottom valve and the bottom valve is still open, the injection gas willcontinue to displace the brine in the annulus until the third valve isuncovered.

Pressure

D e p t h

SBHP




IPV UNLOADINGOnce again with more gas leaving the casing through two valves, thecasing pressure will fall until the second valve closes. Obviously if therewere more valves deeper the unloading process would continue.

Pressure

D e p t h

SBHP

What would happen if the third

valve injected too much gas?






4.2 Well Performance

(Inflow and Outflow)

Well Productivity: A well’s ability to

produce fluids related to a reduction in

bottom hole pressure




Inflow:Pseudo-Steady-State-Radial

• Qo = C*k*h*(Pr-Pwf) .

. Bo*µ*[Ln(Re/Rw)-0.75+S+Dq]

• Qo = J *(Pr-Pwf) or J = Qo/(Pr-Pwf)

• Where J = PI = Productivity Index

• Specific PI = J/h

• C= 0.00708 bpd Oilfield Units• 1/C = 141

(After Darcy)




Inflow:

Flow into Well from the Reservoir • PI = Productivity Index=J (in bfpd/psi)

• Note: PI for single phase flow• PI=Change in Rate/Change in Pressure

• PI=Rate/(Pr-Pwf) in bfpd/psi• Rate in bfpd =Ql = PI * (Pr-Pwf)

• Drawdown = ∆P=(Pr-Pwf) = Rate/PI




PI Problem # 1B• Given: Pr = 2500 psig (172.4 bar)=Pb

• Pwf = 1750 psig (120.7 bar)

• Ql = 1500 BPD (238.5 m^3)

• Find: PI, Qmax, &Ql @ 500 psig (34.5 bar) &

Pwf if Ql = 3000 BPD (476.9 m^3)




IPR_VOG: VOGEL OIL WELL IPR

0

500

1000

1500

2000

2500

3000

0 1000 2000 3000 4000 5000

PRODUCTION RATE (BPD) OR (M^3/D)

P W F ; F L O W I N G

P R E S S U R E ( P S

I A ) O R ( k P a )

NO SKIN WITH SKIN

Pr

Pwf

Qmax

PI = 1500/(2500-1750) = 2 bpd/psi

o

PI Problem # 1B when Pb= 0




It is more accurate to describeWell Productivity in terms of:

Multi-phase, radial-flow

This means it handles flow ofboth liquids AND gas,

which changes the curve

This method is called:

Inflow Performance Relationshipor IPR

INFLOW PERFORMANCE RELATIONSHIP



0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

00.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Producing Rate (Q/Qmax) ratio

I n t a k e P r e s s u r e ( P w f / P r ) r a t i o

Vogel IPRQ=1-0.2*(Pwf/Pr)-0.8*(Pwf/Pr)̂ 2

Initial slope = -1.8

x




Inflow: Vogel IPR

• Ql/Qmax=1-0.2(Pwf/Pr)-.8(Pwf/Pr)^2

• Ql/Qmax=1-v*(Pwf/Pr)-(1-v)*(Pwf/Pr)^2

• For multiphase flow

• Need well test rate (Ql), Pr & Pwf

• Find pressure ratio: Pwf/Pr

• Find production ratio: Ql/Qmax =(x1)

from graph• Calculate Qmax: Qmax = Ql/(x1)

• Once Qmax known, find other rates

IPR F tk i h Q /Q = [1 P f2 /P 2]n n=1





0

100

200

300

400

500

600

700

800

900

1000

0 100 200 300 400 500 600 700 800 900 1000


P

W

F ; F L O W

I N G P

R E S S U R E ( P S I A

) O R

( k P a )

NO SKIN WITH SKIN

x

IPR Fetkovich: Qo/Qm = [1- Pwf2 /Pr2]n n=1

Initial Slope = -2.0

Ratio 1/1000

n>.5n< 1

TypicallyN=.8




Combination PI & IPR Problem• Given:

• Pr = 2800 psi ; Pb = 1800 psi• Pwf = 2300 psi ; Ql1 = 500 bpd

• Find: Qb, Qa, Qmax &Ql2 @ Pwf = 900 psig

• Solution: Just divide into a• PI + IPR problem

F P > Pb)




IPR_VOG: VOGEL OIL WELL IP

0

500

1000

1500

2000

2500

3000

0 500 1000 1500 2000 2500 3000

PRODUCTIONRATE (BPD) OR(M̂3/D)

P W F ; F L O W

I N G P R E S

S U R E ( P S I A )

O R

k P a

NOSKIN WITHSKIN

Pr= 2800

Pb=1800

Qmax

Pwf=2300PI = 1.0

Slope = - 1.8

For Pr > Pb)

Qb

Qa

ox

PI + IPR Vogel



Now youknow how tofind the

well’s inflow

Use PI for single-phase flow

Use IPR for multi-phase flow




Outflow Introduction

• Flow from perforations to storage tanks

• Requires a good vertical flow correlation• Also a horizontal flowline correlation

• Learn to use pressure-depth (gradient)curves

• Draw tbg outflow performance curves




Gas Sales

Oil

Pwf

PI or IPR Inflow

Outflow

Pr

Pwh

Psep

Inflow & Outflow Analysis

Flowline:Gradient Curves

Tubing Performance:Gradient Curves

X X

Tank

STB




Outflow: Multiphase Vertical Flow

• Empirical Models

• Gilbert (CA oil wells)-developed 1940 to1950 but published in 1954

• Poettmann & Carpenter (no slip) -1952

• Baxendell & Thomas (high rate extension of P&C)-1961• Duns & Ros (lab data)-1961

• Ros & Gray (improved D&R)-1964

• Hagedorn & Brown (most used--slip?)-1964

• Orkiszewski (Exxon composite)-1967• Beggs & Brill (incline flow)--1973

• MMSM ( Moreland-Mobil-Shell-Method)-1976

• Mechanistic Models $

• Aziz, Grover & Fogarasi-1972• OLGA –Norwegian- 1986

• Ansari. Et al. – 1990

• Choksi, Schmidt & Doty-1996• Brill, et al-ongoing

Shell : Zabaras-1990



Flow




FlowRegimes

Single Phase Flow

Bubble(y)

Plug or Piston

Slug or Churn

Annular

Mist

Above Bubble Point

Slightly below BP

Bubbles grow

Bubbles connect

and expand

Oil up tubing wallwith gas at higher

velocity up center

Gas with oil droplets

(Surface)




Water Cut Effect on Gradient

0 35 65 95 100 Water Cut (%)

γo

0 42 psi/ft

γw

As per ROS with Shell

Gradient

psi/ft

(.46+)

(.38+)

(Lab tests)




Outflow: Find Pwf: Case 1

• Given: Tubing ID = 1.995 inchesRate = 800+ bpd

• Cut = 50%+,GOR = 1200, GLR = 600• Pwh=440 psig, Flow Surf. Temp =100 ‘F

• Well Depth = 5100’,• BH Temp = 180 ‘F, Water SG = 1.074

• Oil Gravity = 35 ‘API, Gas SG = 0.65• Static BHP = 2060 psig

• Find Pwf & PI (Find the correct chart)

440 1560




440

5100’Gradient =0.42 psi/ft

2800’

7900’




Outflow Example• Case 1: Find Pwf = 1560 psig

from 800 BPD graph• Calculate PI = Prod/Drawdown

• PI = 800/(2060-1560) = 1.6 BPD/psi

• Problem B: Find Pwh for a Pwf of 1200psig?




0 100 200 300 400 500 600 700 800 900 1000 1100 12

0

1

2

3

T h o u

s a n d s

PRODUCTIOPN RATE (BPD) OR (M̂ 3/D) P W F ; F L O W I N G P R E

S S U R E ( P S I A ) O R ( k P a )

IPR TBG-1 TBG-2 TBG-3

OIL WELL INFLOW & OUTFLOW PERFORMANCEA 6000 ft flowing well with GLR = 500 and 10% WOR

Production Rate in BPD

1.995” 2.441” 2.992”

Pwf




Outflow: Summary• Essential to have a good multiphase flow

program or gradient charts

• Use well test data & correlation to find Pwf

• Calculate PI or IPR for each well

• Construct tbg perf curves• Select most profitable tbg size

• Size flowline (Typically same as tubing)

• Minimize Back-pressure and Maximize Rate

T bi Si G id li




Tubing Size Guideline

• In both flowing & gas lift wells, the size oftubing is critical.

•Too large a size results in heading, loadingup, and unstable flow.

• Too small a size results in excessive

friction and loss of production.• For best results, use the following:

1.995” ID-- 200 to 1000 bfpd

2.441” ID-- 500 to 1500 bfpd2.992” ID-- 1000 to 3000 bfpd3.958” ID-- > 3000 bfpd

TUBINGPERFORMANCE(OUTFLOW)CURVES

Typical Tubing Curves




.

0 1 2 3 4 5

1

2

3

4

5

Thousands

T h o

u s a n d s

RATE(BFPD)

P w f : B H

F l o w i n g T u b i n g P r e s s u r e ( P S

I G )

1.995"

2.441"

2.992"

3.467"

3.958"

TUBING PERFORMANCE (OUTFLOW) CURVESFOR10,000 FT WELL W/ 1000 GLR& 50%CUT

Typical Tubing Curves

Rate in 1000 BFPD

Pwf

in1000psi

PI’s

Pwh = 100 psig




4.4 Facilities

• John Martinez

GL S f E i t



GL Surface Equipment

API Manual Chapter 4

API RP 11V7

Consultant

Testing

Treating CompressionDehydration

DistributionMetering Miscellaneous

XX




A TypicalGas Lift

System

A TypicalGas Lift

System

DehydratorO




GL Surface Gas Facilities (49)

• GL is a system type AL; thus, all componentsmust operate efficiently

• Freezing often a problem requiringdehydration, heaters, or methanol injection

• Most important & expensive are the

compressors.• Proper piping & good meters are essential for

accurate gas measurement

• Manifolds to distribute the gas & adequateseparation/treating are also important

• Plus controls--all are some of our favoritethings




Simple SystemSimple System

Inflow

Outflow

Stock

tank

sales

GL Compression




GL Compression

• Reciprocating & Centrifugal• One --economical but all eggs in one basket

Two - most practical; Three -- allows better

maintenance; >3 -- too expensive

• Need high reliability (>96 %) [< 1

day/month]• Low suction Pressure (< 100 psia)

• Adequate discharge pressure !• Adequate cooling System

• Good maintenance important




Piping, Distribution, Metering

• Provide good operating and maintenance

plus minimize investment• Keep back-pressure low!

• Adequate separation & scrubbing• Follow good piping practices

• Provide for pigging and traps• Install gas meters properly (GPSA)

Choke-Regulation Control




Choke Regulation Controlfor Gas Lift Well

Meter Run

Pg Pio(0)

4 5 Gas Injection Pressure




4.5 Gas Injection Pressure• Has a large effect on efficiency and

operation of continuous flow GL wells

• Too high a pressure results in

needless investment of compressors &lines

• Need enough pressure to inject nearbottom ( 100 ft above perforations) atthe planned rate.

• Request suction pressure < 100 psig

• See paper by J.R. Blann, JPT Aug. 84




Depth,(ft)

Pressure (psig)

14001000600

x400 bpdx500 bpd

x600 bpd

x700 bpd

Gas Injection Pressures, psig

Equilibrium Curve

Benefit from Higher Gas Injection Pressure

0

Dw

x 200 bpd

Gas Injection Pressure




Gas Injection Pressure

• For the system, select an injection gas pressurethat will permit well gas injection just above the

producing zone.• Install pressure recorder at well and record

pressure for a minimum of 24 hours. The

pressure variation should be less than 100 psi.

• Use average gas injection pressure recorded at

the well for gas lift design. No safety factorshould be necessary.




Kick-Off Injection Gas Pressure

• If available, allows deeper lift.

• Normally not practicable in multi-wellinstallations.




References/Bibliography• Clegg, JD; S.M. Bucaram; N.W. Hein Jr.:

Recommendations and Comparisons for Selecting

Artificial-Lift Methods,” JPT Dec 93• Neely, Clegg, Wilson, & Capps: “Selection of Artificial

Lift Methods: A Panel Discussion,” SPE 56th Annual

Fall Meeting Oct 81• Redden, Sherman, & Blann: “Optimizing Gas-Lift

Systems,” SPE 5150 , 1974

• Winkler & Smith: Camco Gas LIft Manual 1962

• Brown: “The Technology of Artificial Lift Methods”

• API Recommended Practices 11V8

API




API

MandrelSee API 11V1

And ISO 17078-1

API Mandrel Selection Guideline




G Lift M d l




Gas Lift Mandrels

• Conventional--Tubing Retrievable• Side-pocket--Wireline Retrievable (Oval or Round)

• Connections same as tubing (avoid crossovers)

• Material normally 4130

• Valve Receptacle 1” or 1.5”

• With Guard and Orienting Sleeve

• Drift to tbg size; Fluid Passage (S)

• Internal test pressure-to tbg rating

• External test pressure to max collapse case• Check clearance; Min spacing 90 ft

• Plastic coating (optional--Drift check after coating)

4 7 G Lift V l




4.7 Gas Lift Valves

• Conventional (Tubing Retrievable)

• Wireline Retrievable *

• Valve size : 0.625”, 1.0” & 1.5” *• Closing Force: Gas Charged*, Spring Loaded;

Combination Spring-Gas

• Valve Type: Injection Pressure Operated*;Dummy; ...Production Pressure Operated;Pilot; Orifice; Other

• Flow Configuration: Type 1*,2, 3, or 4• Service Class: Standard *; SCC

• Reference API Spec 11V1 & ISO 17078-2






Typical Gas Lift Valves

BK-1

BK

R20




Gas Lift Valves Guidelines




Gas Lift Valves Guidelines

• Use a 3-ply bellows-- single-element, unbalanced valvew/ a nitrogen charged dome and/or a spring

• Choice between: Injection Pressure* orProduction Pressure (Fluid) Operated.

• Use reverse flow (check) valve in each

• Age valve and shelf test

• Standard Monel Seats or Solid Carbide• Use Orifice Valve w/ check on bottom

• Dummy all unused mandrels

• Consider combination Gas-charged & Spring- loadedfor set pressures > 1500 psi (Field Experience)

• Use screened orifice/nozzle-Venturi on bottom






IPO G/L VALVE BEHAVIOR




In an IPO valve, highpressure nitrogen in thedome exerts pressure on

the inside of the bellows.This causes the bellows toextend down and pushesthe ball on the seat.

In an IPO valve, highpressure nitrogen in thedome exerts pressure onthe inside of the bellows.This causes the bellows toextend down and pushesthe ball on the seat.

Nitrogenpressure





Gas from the casing tries toget into the valve. Thepressure acts on the outsideof the bellows, trying tocompress the bellows.

Gas from the casing tries toget into the valve. Thepressure acts on the outside

of the bellows, trying tocompress the bellows.





Fluid from theproduction tubing triesto force the stem off theseat.

Fluid from theproduction tubing tries

to force the stem off theseat.





When the valve opensgas moves through thevalve and out the nose.

When the valve opensgas moves through thevalve and out the nose.






The valve is in the closed positionnow.

What it take to get this valve open?

Let’s look first at the forces trying toclose the valve.

The valve is in the closed positionnow.

What it take to get this valve open?

Let’s look first at the forces trying toclose the valve.

AbAb

PbPb

Closing force = Pb* (Ab)

where:

Pb = nitrogen pressure

Ab = area of the bellows





Now the opening forces:Now the opening forces:

Opening force = Ppd (Ap) + Piod(Ab - Ap)

where:

Ppd = tubing pressure

Piod = casing pressure

Ap = port area

Ab = bellows area

ApAp

PpdPpd

PiodPiod

The casing pressure onlyacts on this area when the

valve is closed.

Ap

Ab

IPO G/L VALVE OPENING BEHAVIOR




IPO G/L VALVE OPENING BEHAVIOR

The solutionThe solution

The valve begins to come open whenthe opening and closing forces are

equal.

Pb (Ab) = Ppd (Ap) + Piod(Ab - Ap)

For a given Pb we could solve for Piod

the pressure at which the valve shouldopen.

Or for a design case of Piod and Ppd,

we could solve for the correct Pb.

ApAp

PpdPpd

PiodPiod

AbAb

PbPb

Test RackSet Pressure




Piod

Ppd

Pb

Unbalanced pressure

charged valve

Ppd=0

Pb

Pvo

Set Pressure

Test Rack Set Pressures, Pvo




• Simple Injection Pressure Operated Valve wherewell forces ready to open valve:(1) Pb*Ab = Piod *(Ab-Ap) + Ppd * Ap

• For Test rack conditions where Ppd = 0:(2) Pb *Ab = Pvo * (Ab-Ap)

• Then by substitution:(3) Pvo * (Ab-Ap) = Piod *(Ab-Ap) + Ppd * Ap

• Or: (4) Pvo = Piod + Ppd *Ap/(Ab-Ap) and by

• definition Ap /(Ab-Ap)=Ap/Ab/(1-Ap/Ab) = PPEFCorrect for Shop temperature for bellows charged valve

Pvo = [Ppd*PPEF+Piod]*Ct

TABLE A OILFIELD UNITS)

TYPICAL INJECTION PRESSURE VALVES WITH CHARGED NITROGEN BELLOW




RETRIEVABLE GAS LIFT VALVES

VALVE Ab PORT (MONEL) Ap/Ab Ap/Ab

OD BELLOWS SIZE SIZE RATIO (1-Ap/Ab)

(IN) (IN^2) (IN) (1/64") Mfg PPEF

------- ------- ------- ------- (MONEL) (MONEL)

1.5 0.77 0.1875 12 0.0380 0.0395 0.77 0.2500 16 0.0670 0.0718

0.77 0.3125 20 0.1040 0.1161

0.77 0.3750 24 0.1480 0.1737

0.77 0.4375 28 0.2010 0.2516

0.77 0.5000 32 0.2620 0.3550

1.0 0.31 0.1250 8 0.0430 0.0449

0.31 0.1875 12 0.0940 0.1038 0.31 0.2500 16 0.1640 0.1962

0.31 0.2813 18 0.2070 0.2610

0.31 0.3125 20 0.2550 0.3423

0.31 0.3750 24 0.3650 0.5748

Test Rack Set Pressure: Example




Test Rack Set Pressure: Example

• Given: XXXXX 1-inch BK valve w/ 3/16” portw/monel seat. Find PPEF = 0.1038 (mfg. data)

• Pio=Csg pressure @ valve depth=1060 psig• Ppd =Tbg pressure @ depth = 420 psig

• Tv = Valve Temp @ depth = 121 ‘F• Tshop = 60 ‘F (Valve temp in shop)

• Find from Table 4-1 that Ct = 0.88

• Thus: Pvo = [PPEF*Ppd+Pio]*Ct

• Pvo = [0.1038*420+1060]*0.880= 971 psig

Well Name: Exampl joe PRESS URE (Pv) 1000 PSIA

TEMPERATURE COR. FACTORS TEMP.in Shop (Ts) 60 'F

(F) (Ct) (F) (Ct) (F) (Ct) (F) (Ct) (F) (Ct) (F) (Ct)

61 0.998 101 0.916 141 0.847 181 0.787 221 0.735 261 0.690

62 0.996 102 0.914 142 0.845 182 0.786 222 0.734 262 0.689

63 0 993 103 0 912 143 0 843 183 0 784 223 0 733 263 0 688

API Gas Lift ManualTable 4 1




63 0.993 103 0.912 143 0.843 183 0.784 223 0.733 263 0.688

64 0.991 104 0.910 144 0.842 184 0.783 224 0.732 264 0.687

65 0.989 105 0.909 145 0.840 185 0.781 225 0.730 265 0.686

66 0.987 106 0.907 146 0.839 186 0.780 226 0.729 266 0.685

67 0.985 107 0.905 147 0.837 187 0.779 227 0.728 267 0.68368 0.982 108 0.903 148 0.836 188 0.777 228 0.727 268 0.682

69 0.980 109 0.901 149 0.834 189 0.776 229 0.726 269 0.681

70 0.978 110 0.899 150 0.832 190 0.775 230 0.724 270 0.680

71 0.976 111 0.898 151 0.831 191 0.773 231 0.723 271 0.679

72 0.974 112 0.896 152 0.829 192 0.772 232 0.722 272 0.678

73 0.972 113 0.894 153 0.828 193 0.771 233 0.721 273 0.677

74 0.970 114 0.892 154 0.826 194 0.769 234 0.720 274 0.676

75 0.968 115 0.890 155 0.825 195 0.768 235 0.719 275 0.675

76 0.965 116 0.889 156 0.823 196 0.767 236 0.717 276 0.674

77 0.963 117 0.887 157 0.822 197 0.765 237 0.716 277 0.673

78 0.961 118 0.885 158 0.820 198 0.764 238 0.715 278 0.672

79 0.959 119 0.883 159 0.819 199 0.763 239 0.714 279 0.671

80 0.957 120 0.882 160 0.817 200 0.761 240 0.713 280 0.670

81 0.955 121 0.880 161 0.816 201 0.760 241 0.712 281 0.669

82 0.953 122 0.878 162 0.814 202 0.759 242 0.711 282 0.668

83 0.951 123 0.876 163 0.813 203 0.758 243 0.710 283 0.667

84 0.949 124 0.875 164 0.811 204 0.756 244 0.708 284 0.66685 0.947 125 0.873 165 0.810 205 0.755 245 0.707 285 0.665

86 0.945 126 0.871 166 0.808 206 0.754 246 0.706 286 0.664

87 0.943 127 0.870 167 0.807 207 0.753 247 0.705 287 0.663

88 0.941 128 0.868 168 0.805 208 0.751 248 0.704 288 0.662

89 0.939 129 0.866 169 0.804 209 0.750 249 0.703 289 0.661

90 0.937 130 0.865 170 0.803 210 0.749 250 0.702 290 0.660

91 0.935 131 0.863 171 0.801 211 0.747 251 0.701 291 0.659

92 0.933 132 0.861 172 0.800 212 0.746 252 0.700 292 0.65893 0.931 133 0.860 173 0.798 213 0.745 253 0.698 293 0.657

94 0.929 134 0.858 174 0.797 214 0.744 254 0.697 294 0.656

95 0.927 135 0.856 175 0.795 215 0.743 255 0.696 295 0.655

96 0.925 136 0.855 176 0.794 216 0.741 256 0.695 296 0.654

97 0.924 137 0.853 177 0.793 217 0.740 257 0.694 297 0.654

98 0.922 138 0.851 178 0.791 218 0.739 258 0.693 298 0.653

99 0.920 139 0.850 179 0.790 219 0.738 259 0.692 299 0.652

100 0.918 140 0.848 180 0.788 220 0.736 260 0.691 300 0.651

(F) (Ct) (F) (Ct) (F) (Ct) (F) (Ct) (F) (Ct) (F) (Ct)

Table 4.1

Temperature Correctionfor

Nitrogen Charged Bellows

1000 psia & 60 ‘F

Table 4-1 Page 40 (Program CT TEMP)




Table 4 1 Page 40 (Program CT_TEMP)Temperature Correction Factors for Nitrogen

• Based of 60 F’ and Pbv = 1000 psig

• ‘F Ct• 121 .880

• Where: Ct =1/[1.0+ (Tv(n)-60) x M/Pbv]

• For Pbv<1238 psia :M=3.054xPbv^2/10000000+1.934xPbv/1000-2.26/1000

• For Pbv> 1238 psia :M=1.804xPbv^2/10000000+2.298xPbv/1000-2.67/10

AGL_SET:INJECTIONPRESSUREVALVEDESIG

E l

P 1 ?

Calculate Valve Set Pressures




Example

0

500

1000

1500

2000

0 1000 2000 3000 4000 5000 6000

DEPTH(ft) of (meters)

P R E S

S U R E

s i o r b a r

INJ GAS TBG Temp VALVESPACING

420

1060

121o150 o

: Pvo1= ?

121135

145

Gg=.03

Gf=.15




4 10 Temperature




4.10 Temperature

• Determine Surface Static and

…………...Reservoir Static Temperatures• Calculate Static Gradient

• Measure Flowing Temperature fordifferent rates

• Never use static for design basis

• Design on estimated production rate




AGL_TEMP: FLOWING TEMPERATURE PROFILE

0

50

100

150

200

250

300

0 1000 2000 3000 4000 5000 6000 7000 8000 9000DEPTH (ft) or (meters)

T

E

M

P

E

R

A

T

U

R

E

( ' F

)

o

f

' C

STATIC TEMP FLO W TEMP

Static

Flowing



IsothermalGradient Map




Gradient Map

1.2

Kirkpatrick Correlation




• Example:Ql =2250 bpd through 3.5” Tbg• Well Depth = 5500’ , BHT = 180 ‘F

• Geothermal Gradient =(180-75)/(5500/100) =1.9 ‘F/100’

• Solution: intersection of 2250/1.5 =1500 bpd & 1.9 GG find flow GG =

1.0 ‘F/100’• Flowing Surf Temperature =

180 - 1.0*5500/100 = 125 ‘F

Fig. 6-9 Kirkpatrick

Fig 6-9 KirkpatrickChart to be used directly for 2.5” tubing




Chart to be used directly for 2.5 tubing For 2” tubing, multiply rate by 2For 3” tubing, divide by 1.5

1.9

Temp Computer Program Solution




Temp Computer Program Solution

• “Predicting Temperature Profiles in aFlowing Well,” by Sagar, Doty &Schmidt

• Also for gas lift wells

• For multi-phase flow: Regressionanalysis--- many assumptions

• Check against real field data

VERSION 8.0db

Flowing Temperature

ExampleWell Name

2003**PREDICTING TEMPERATURE PROFILES**14-Aug-06




(OILFIELD = E; METRIC = M)E or MEUnit Selection15

mcfd0Qi:Gas Lift Inj. Rate14

'F125Twh =Flowingft*0Di: Depth of Gas Inj.*13

-0.0043SGM2:CORRECTIONft5,500Dw: Total Well Depth12

-0.0043SGM1:CORRECTIONin7.000CSG: Casing OD11

1.00E-04A2: COEf.in3.500OD: Tubing OD10

1.21E-04A1: COEF.air=10.700SGi: SG Gas (Air = 1)9

#N/AU2:HEAT t.c.sp.gr.1.060SGw: SG. Water 8

121.22U1:HEAT t.c.'API40.0API: Oil Weight7

lbm/sec8.23Wt2:MASS FLOWmcfd750Qg:Form. Gas Rate6

lbm/sec8.23Wt1:MASS FLOWbwpd250Qw:Water Rate5

sp.gr.0.825SGo: OIL wt.bopd2000Qo:Oil Rate4

't2.306f:DIM.TIME'F180Tf:Temp of Formation3

BTU/lbm0.542Cpl:SPEC. HEAT'F75Ts: Temp.Surf.Static2

'F/100'1.909Tg:TEMP GRADpsi100Pwh: Wellhead Pressure1

--CALCULATIONS----INPUT DATA---

p

Temperature data




Temperature data

• API Gas Lift Manual & API RP11V6• C.V. Kirkpatrick “The Power of Gas”

• K.E. Brown “The Technology of ArtificialLift Methods,” Volume 2a

• Sagar, Doty & Schmidt, “FlowingTemperature Profiles in a Flowing Well”

• Winkler & Eads, “ Algorithm for more accurately

predicting nitrogen-charged gas lift valve operationsat high pressures and temperatures”

4 12 Gas Passage




4.12 Gas Passage

• Use the minimum size port (choke) thatwill pass the desired rate of gas!

• Check Valve Port size for amount of gaspassage that is possible

• Use Thornhill-Craver Equation/Chart• Make Gas Gravity & Temp Correction

• Predict on high side

• Use next higher standard port/orifice

Upstream:i d




Piod

SGg=

Tv =

Pb

port

Ppd

square-edgeorifice

Thornhill-Craver Chart: Example




• Find corrected gas throughput (Qgi)• Given: Upstream Pressure (Piod)=1000 psig

• Downstream Pressure (Ppd)= 790 psig

• Orifice Size (Valve Port) = 12/64”

• Temperature of valve = 160 ‘F

• Gas SG = 0.75• Find from chart : Qgi = 660 MCFD

• From Correction Chart find: Cc = 1.17• Actual Qgi = 660/1.17 = 564 MCFD

Choke Chart

660






2007 Workshop Clegg & Smith 98Cc=

x

Orifice/Choke Problems




• Orifice/Choke Problem # 1

• Upstream Pressure = 1250 psig

• Downstream Pressure = 1150 psig• Valve Port Size = 8/64 inch

• GG = 0.7 & Temp @ Depth = 180 ‘F• How much gas can be Injected?

• What size orifice for Injection GAS Volumeof 850 MCFD?






Typical VPC Gas-Lift Valve Performance PlotFig. 1

Camco BK with

12/64ths VPCPvoT= 964 Pcf=920 Temp=150

Camco BK with16/64ths VPCPvoT= 964 Pcf=920 Temp=150


F l o w r a t e - ( M s c f / d )

Downstream Pressure - (psig)

0

500

1000

1500

0 200 400 600 800 1000

(after Decker & Dunham)




Comparison of Gas-Lift Valve Performance Based on VPC ModelVs. Performance Based on Thornhill-Craver Model

Fig. 2


Camco BK with12/64thsThornhillPvoT= 964 Pcf=920 Temp=150

F l o w r a t e - ( M s c f / d )

Downstream Pressure - (psig)

0

200

400

600

800

0 200 400 600 800 1000 API

Pvo(n) = Test rack opening

pressure for nth valve

Pcf = Injection gas pressure

Nozzle-Venturiif l




Gas Lift Valve




GasInjectionRate

Production Pressure

Pressures Acting on anPressures Acting on an




Production Pressure

P ~ Tubing

Pressure

The pressure down-

stream of ball is nearly

equal to the tubing

(production) pressure.

Injection Pressure > P

> Tubing pressure

Injection Pressure. This

pressure is greater than

the pressure downstream

of the ball.

Large pressure

drop, large

suction force on

ball, small gap

between ball

and seat

The gas injection rate

through the valve is reducedas the valve throttles closed.

essu es ct g o ag

Unchoked ValveUnchoked Valve

Pressures Acting on aPressures Acting on a




P ~ Tubing

Pressure

P ~ Tubing

Pressure

The pressure down-stream of

the ball is held higher by the

choke.

Injection Pressure

Small pressure

drop, small

suction force on

ball, more gap

between ball

and seat

Large

pressure

drop

The gas injection rate

through the valve is higher

because the valve ball is

held off of the seat by the

higher pressure beneath

the ball.

Choke

gg

Choked ValveChoked Valve

Comparison Between Choked




Plot of Injection Rate vs. Pressure

Unloading Gas-Lift Valve with Choke vs. Valve with no ChokeUsing Gas-Lift Valve/Choke Model

Macco R-1D3/16" port10/64" chokePc = 1200 psiPt = 325 psi

Note that choked valve remainsopen over entire range andactually transmits much more gas.It "snaps" closed when closingpressure is reached.

Choked

Unchoked

and Unchoked Valves(After Dunham, Decker, & Waring)



Graphical Solution--Gas Lift Spacing




Gas Lift Spacing.........................

• Need to space mandrel/valves to permit working to thelowest possible depth

• You will learn to space using different techniques--depending on type valve

• Find the location of the first valve

• Injection Pressure Operated Valves

• (1) Constant Rate Design*

• (2) Variable Rate Design• (3) Intermittent Design: Other Designs

1st Mandrel/Valve...................................




• Depth of 1st Valve is the same for most designs• Strictly a U-tube case where the outlet and inlet

pressures are nearly balanced

• Outlet Po. = Pwh+Depth(1)*[Liquid Grad (Gs)]Inlet Pi. = Pg + Gg * Depth(1) -Psf

• Example: Pwh = 100 psig,Gs= .465 psi/ft, Pg = 1000 psig,

• Gg = .03 psi/ft, Psf = 20 psi or about 50’(min)

• 100+D(1)*.465=1000+.03*D(1)-20

• D(1) = [1000-100-20]/[.465-.03] = 2023 ft

.

Injection Gas GradientForTs=75'F&Tf=175'F

API GL ManualPage 44 Fig. 4.7



500 600 700 800 900 1000 1100 1200 1300 1400 150

00.01

0.02

0.030.04

0.05

0.060.07

0.08

0.090.1

Pg, Surface Gas Injection Pressure, psig

G a s G r a d i e n t , p s

i / f t SGg 0.6

SGg 0.7

SGg 0.8

SGg 0.9

For Ts= 75 F & Tf = 175 F

x

Pgd=Pg x e(0.1875xGxD/(TaxZ)

AGL_SPAC: GAS LIFT SPACING

26002800




0200400600800

10001200140016001800200022002400

2600

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000

DEPTH (ft) or (meters)

P R E S S U

R E

( p s i )

GAS INJ TBG SPACING

Gs = 0.465

Gg = 0.03Pio

Pwh

Gf = 0.1



Tgs Twh

Psep




SGiTgs Twh

InjectionPressureValve

Worksheet

Constant Rate Design:for Injection Pressure Operated Valve with good well data




j p g

• Draw Gas Injection Pressure Line (Pgd) & Grad.• Find location of 1st Valve

• Select desired rate

• Select two points (Dw1 & Dw2) from gradientcurve for desired rate. Plot gradient curve for

desired rate on graph paper.• Use unloading gradient to find intersection with

Pgd. Move back up-hole to achieve necessary PD.

• This is depth of 2nd Valve.

• Repeat until Dw is reached or min space

Constant Rate Design Problem• Pg = 1200 psig; SGg = 65; Gg = 0 033 psi/ft




• Pg = 1200 psig; SGg = .65; Gg = 0.033 psi/ft

• Gs = 0.465 psi/ft; Pwh = 100 psig; Psep = 50 psi;Dw=8000’• Max rate = 600 bpd; Cut 50 %; 35 API;

• Tgs = 75 ‘F; Twh = 100 ‘F; Tf = 180 ‘F (650 psi @ 4000’)

• Tbg = 1.995” ID; GLR = 1000; (1300 psi @ 8000’ )

• Min Space = 500’; PD = 20 psi

• Solution: Use program or gradient curves to find pressures @depth for desired rate. Draw Ppd(1)…Ppd(n)

• Find 1st Valve @ about 2600’ w/ Ppd(1) = 445 psig• Extend .465 psi/ft gradient to Pgd. Move back up-hole until

a 20 psi PD results. D(2) = about 4450’

• Continue until 500’ min spacing reached.• Find operating valve

AGL_SPAC: GASLIFT SPACIN

1500

1464Constant Rate Design




0

500

1000

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000

DEPTH(ft) or (meters)

P R E S S U R

E

s i

GASINJ TBG SPACING

1200

x

x



A. Mandrel Spacing: Constant Rate

for injection pressure valves




for injection pressure valves

• Given: Pwh=Psep =100 psig; Pg = 1400 psig

• Gs = 0.465 psi/ft; Twh = 120 F’; Tf=200 ‘F• SGi = 0.7; Dw= 10000’; Dmin = 500’

• PD = 25 psi; Tubing = 3.5” OD• Max rate = 2000 bpd @ max depth

• Total Rgl = 1000 CF/B• Space mandrels & find max injection depth



P d( )C t t R t P bl AWELLNAME




00.000200#N/A0.0184010000Dw

#N/A#N/A#N/A#N/A#N/A#N/A#N/A#N/A#N/A#N/A12




16261184#N/A13070.75020016121225182899508

16321210#N/A13200.75419616261250172095007

1634123557313310.76019216371275160390006

16371259118913420.76518816471300149085005

16371283152013520.77118416561325137479734

16101301180613510.78317616421350116469653

15501313189413350.8031631600137585353312

14551323183413020.8351441526140047229801

--120140014001000

----------------------------------------------------------------(INPUT)

psipsimscfdpsi-'Fpsipsipsift

@ depthSurf closeINJ GAST-RTEMP@ VALVEV openSURFTBG PRESDEPTHNO.V closePvc(n)QgiPvo(n)CTTvPiod(n)Pio(n)Ppd(n)D(n)

Pvcd(n)Constant Rate Problem ANAME:

S.O.DD

Variable Gradient Designfor limited data




for limited data• Draw Gas Injection Pressure Line (Pgd)

• Draw Upper Design Line (UDL) {Pgd-Pwh}

• Find location of 1st valve @ D(1). (Same approach)• Calculate Pseudo Tbg Pressure@ surface:

Ps = Pwh + 0.2 *(Pg-Pwh)

• Find Ppd(n) at total depth or total injection depth:Note: (Ppd(n) < Pgd-200 psi)

• Connect these two points: Lower Design Line (LDL)

• Find intersection of LDL @ D(1): Extend usingunloading gradient to intersection w/ Pig (UDL). Nopressure adjustment necessary. Locate D(2)

• Continue spacing until Dw reached or Min Space

Variable Rate Design Problem• Pg = 1200 psig; SGg = .65; Gg = 0.033 psi/ft




Pg 1200 psig; SGg .65; Gg 0.033 psi/ft

• Gs = 0.465 psi/ft; Pwh = 100 psig; Psep = 50 psi;Dw=8000’• Rate = 200-600 bpd; Cut 50 %; 35 ‘API

• Tgs = 75 ‘F; Twh = 100 ‘F; Tf = 180 ‘F

• Tbg = 1.995” ID; GLR = 1000;

• Min Space = 500’

• Calculate Pseudo Tbg Pressure@ surface:Ps = Pwh + 0.2 *(Pg-Pwh)= 100 + 0.2 (1200-100) = 320 psig

• Find Ppd(n) at total depth or total injection depth:• Ppd(n) = 1200 + .033x8000 -200 =1264 psig @ 8000’ [max]

• Connect these two points: Lower Design Line (LDL)

• Find D(1) {Same approach as before}• Find intersection of LDL...Ppd(1) @ D(1): Extend using unloading

gradient to intersection w/ Pig (UDL). Locate D(2), D(3)….D(n)

• Continue spacing until Dw reached or Min Space


1500 Bellows Injection Pressure Operated Valves



0

500

1000

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000


P R E S S U

R E

( p s i )

GAS INJ TBG SPACING

Variable RateInjection Pressure Valves

320

1264


1500 Spring Production Pressure Operated Valves



0

500

1000

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000


P R E S S U R E

( p s i )

GAS INJ TBG SPACING

Variable RateProducing Pressure (Fluid) Valves

Intermittent Design




• Draw Gas Injection Pressure Line, Pgd:UDL• Find location of 1st Valve.

• Select design rate -Use small ported unloading valves

• Find Int. Spacing Factor (Fs):API GLM - Page 106, Fig 8-4

• Pdp(n) =Fs*Dw+Pwh; Connect Pwh & Pdp(n): LDL

• From intersection of LDL & D(1) extend unloadinggradient (Sg) to UDL. Move back up-hole until the PD isreached. This is location of D(2). Find D(3)...

• Continue same procedure until Dw reached.• Select large ported or pilot operating valve



Intermittent Gas LiftSpacing Factors

Fig. 8.4

Intermittent Gas Lift



0 100 200 300 400 500

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

Rate in BPD

S p a

c i n g F a c t o r ( S F ) i n

p s i / f t

1.61"ID

1.995"ID2.441"ID

2.992"ID

p g

o

oo


Intermittent Lift Spacing




0100200300400

500600700800900

100011001200130014001500

0 1000 2000 3000 4000 5000 6000 7000 8000 9000


P

R E S S U R E ( p s i )

GAS INJ TBG SPACING

Spacing Factor = .1

1192

100 +8000x.1

Mandrel Spacing Summary




Mandrel Spacing Summary• You have learned methods for

spacing valves:• (2)Constant Rate for Injection Pressure Valves;

(3) Variable Gradient for Inj. & Prod. Pressure Valves;(4) Intermittent design for Inj. Pressure Valves.

• Good spacing is essential for goodoperation & being able to work down

to the lowest possible valve

• Plan ahead for changing conditions.

B. Mandrel Spacing: Variable

Gradient for injection pressure operated Valves




Gradient for injection pressure operated Valves

Given: Psep = 100 psig; Pg = 1400 psig

Gs = 0.465 psi/ft; Twh = 100 ‘F; Tf=160 ‘FSGi = 0.65; Dw= 7000’; Dmin = 500’

PD = 25 psi; Tubing = 3.5” ODRate ? = 1500+ bpd from 7000’+

Total Rgl = 750 CF/BCalculate Pseudo Tubing Pressure

Space mandrels & find max injection depth

C. Mandrel Spacing: Intermittent Gas Liftfor injection pressure valves




. for injection pressure valves

Given: Psep= 50 psig; Pwh = 50 psig; Pg = 800 psig

Gs = 0.465 psi/ft; Tgs=75’F;Twh =100‘F; Tf=150‘F

SGi = 0.70; Dw= 7,000’ ; Gg = 0.024 psi/ft

PD = 25 psi; Tubing = 2.875” OD

Rate = 200 bpd from 7,000’

Use intermittent lift spacing factors for LDL (.055)

Space mandrels

Equilibrium Curve• Definition: A curve that connects the intersection of the




Definition: A curve that connects the intersection of the

natural flowing gradients with the gas lift producinggradients.

• Draw graph of Pressure Vs Depths. Plot upper flowing

gradient curves for selected rates for GL conditions.

• Find from PI or IPR various Pwf’s for different rates

at total depth.• Draw the lower gradient curves and find the

intersection with the appropriate upper gradient curve

• Maximum production rate will be about 200 psi less

than the gas injection pressure.

GasLiftManual




Gf 0.42 psi/ft

120 ‘F

Manual

Upper Gradient Curves (Trace)




PrPwf

o

E.C.

pp ( )

Lower Gradient Curves

(0.42 psi/ft above BP)200

200 bpd200 bpd




E.C.

After Jack Blann

Equilibrium Curve Problem




• Find the rate and the lift depth for the followingplanned gas lift well:

• Well Depth = 10,000’; Tubing size = 3.5” OD

• Gas Injection Pressure = 1400 psig; Pwh = 100 psig

• Temp @ Surface = 75 F’; BH Temp = 200 F’

• Pr= 3000 psig; Pb = 500 psig• PI = 2.0

• Lower flowing gradient = 0.42 psi/ft (if Pwf > Pb)

• Planned GLR = 1000 during gas lift

GAS LIFT GRADIENT CURVES

FOR 2.992" ID TUBING & 1000 GLR2500




0

500

1000

1500

2000

2500

0 2000 4000 6000 8000 10000

DEPTH (FEET)

F L O W I N G

T B G

P R

E S S U R E

( P S I G )

AGL_DSN: EQUILIBRIUM PROGRAM

3000




0

500

1000

1500

2000

2500

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000


P R E S S U R E ( p s i ) o r ( b a r )

NEEDED INJ PRESSURE GASINJ PRESSURE INJ GAS-SF

5. Continuous Flow Problems




API RP 11V6• 5.1 Example Problem No. 1

• 5.2 Example Problem No. 2• 5.3 Example Problem No. 3

5. API RP 11V6: Injection PressureOperated Valves:Example Problem No. 1:




Typical Well with Good Data• Pwh = 100 psig

• Pg = 1250-1200 psig

• Ts = 78 ‘F

• Twh = 108 ‘F

• Tf = 178 ‘F• Gs = 0.465 psi/ft

• Dw = 8000 ft.

• Pgd at Dw = 1440 psig• Water Cut 50%

• Pr = PB* = 2125 psig

• Rgo = 700 & Rgl = 350

• Psp = 75 psig

• Sgi = 0.6 (.65)

• Pl1 = 400 psig @ 2500’

• Pl2 = 880 psig @ 6000’• PD = 25 psig

• Dmin = 250’

• Tbg. = 2.441” ID• Valve = 1” w/3/16” port

• Rate = 200 BPD @ 1941 psig


2500

API RP 11V6: Example # 1




0

500

1000

1500

2000

0 200 400 600 800 1000 1200 1400


P W F ; F L O W I N G P R E S S U R E ( P S I A ) O R ( k P a )

NO SKIN WITH SKIN

p

PI = 1000/1000 =1.0

1941 psig

1125 psi

Pr=2125 psig

Qmax = 1325

or NOpsipsiftpsi/ftftpsibfpd

OKPgdPRES.DEPTHGFALEVELPwf RATE

RemarksCSGTBGINJ.FLUID




14428000

#N/A#N/A#N/A#N/A0.1455730953100010

#N/A#N/A#N/A#N/A0.139534311169009

OK1431110876340.132499712618008

OK141298070150.126468113947007

OK139586864550.119438815176006

OK138076959450.113411516325005

OK136668154770.106385717404004

OK135360250450.100361218433003

OK134053246450.093337919412002

OK132947042730.087315520351001

120010000.080000

---------------------------------------------------------


0

500

1000

1500

2000

2500

3000

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000


P R E S S U R E ( p s

i ) o r

( b a r )

NEEDED INJ PRESSURE GAS INJ PRESSURE INJ GAS-SF

Optimum Injection Gas




• Use the available injection gas to make themost oil production and greatest profit

• Split the gas between wells to achieve suchresults

• However; installation of additionalcompressors to achieve max rate seldom

justified! {Parkinson’s Gas Law}• Experience: Maximum profit @ about 50% of

max injection to achieve max rate

2500P S I G

Gas Lift PerformanceFor 2.992"Tbg & 2000 BPD




.

0 1 2 3 4 5

500

1000

1500

2000

2500

Thousands

Total Gas (Formation+Injection) in MCFD

P t

: T u b i n g P r e s s u r e @

5 0 0 0 f t i n

Excessive GasOpti-mum

Gas@ 5000’

Max Rate

Excessive Gas

InsufficientGas

MMCFD

Typical

AGL_GRAD: GAS LIFT GRADIENT CURVESEXAMPLE FOR 1000 BPD UP 2.875" OD TUBING




0 1 2 3 4 5 6 7 8

0

500

1000

1500

2000

2500

Thousands


P R E S S U R E ( p s

i ) o r ( b a r )

GLR=250(44.5) 500(89) 750(134)

1000(178) 1250(223) 1500(267)

0

1000

BOPDTYPICAL CASE




Gas Injection Rate in MCFD

Max Oil Rate

Max OCI

Max Profit



5.1.6 Injection Gas Required @ Depth




• For a production rate of 800 BFPD

For a Rgl of : (Correction Factor @ 178 ‘F & Gg of .65 = 1.11)

1500: (1500x800 – 350x800)/1000 = 920 x1.11= 1021 MCFD

1200: (1200x800 - 350x800)/1000 = 680 x1.11= 755 MCFD

1000: (1000x800 – 350x800)/1000 = 520 x1.11=577 MCFD

800: (800x800 – 350x800)/1000 = 360 x1.11=400 MCFD

AGL_TEMP: FLOWING TEMPERATURE PROFILE

3005.1.7 Temperature




0

50

100

150

200

250

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 1000

0


T E M P E R

A T U R E ' F

o f ' C

STATIC TEMP FLOW TEMP

Tf=178 ‘F

Ts = 78 ‘F

Twh= 108 ‘F

.

Injection Gas GradientFor Ts= 75 'F & Tf = 175 'F

API GL ManualPage 44 Fig. 4.7

5.1.8 Gas Gradient



500 600 700 800 900 1000 1100 1200 1300 1400 150

0

0.01

0.02

0.030.04

0.05

0.060.07

0.08

0.090.1

Pg, Surface Gas Injection Pressure, psig

G a s G

r a d i e n t , p s i / f t SGg 0.6

SGg 0.7

SGg 0.8

SGg 0.9

Pgd=Pg x e(0.1875xGxD/(TaxZ)

oo

5.1.10 Valve Setting Depths




• First Valve Setting Depth

• Tubing pressure = casing pressure – Psf

• Max unloading flowing pressure = .gas injection pressure- Psf

• Pwh + gs x D(1) = Pg + gg x D(1) – Psf

• 100+0.465 x D(1)= 1200+ .03xD(1) -20

• (.465-.03)xD(1) = 1200-100-20=1080

• D(1) = 1080/.435 = 2483 ft• Adjusted D(1) = 2475 ft (close as you can read chart)

0.465




0.42

2325x

o

o

.

5.1.12 Subsequent Valve Setting Depths




• Ppd(n)+gsxDbv=(pg-nxPD)+ ggx(D(n)+Dbv-Psf

• For valve(2)

• 400+.465xDbv=(Pg-25)+.03x(2483+Dvb)-20

• Dbv =1907 ft

• D(2) = 2483+1907 = 4390 ft about 4375 ft

• D(3) = 5797 about 5800 ft (as close as can read chart)

• D(4) = 6783 about 6775 ft

• D(5) = 7434 about 7425 ft• D(6) = 7820 Adjustment to 7690 ft (half-way between D(5) & D(7)

• D(7) = 8000+ Adjustment to 7940 ft (30 ft above packer)






5.1.14 Valve Selection




• In this case, one-inch, unbalanced,

nitrogen-charged bellows valves without a

spring was selected. (A common practice

in some areas.)

Piod

Ppd

Pb

Port Size = ?

PPEF = ?

755 Thornhill-Craver Equation




I” Inj. Pressure w/ 12/64” port




2475’ 130 .104 400 42 1275 1317 .869 1144

WELL NAME: API RP 11V6: Example # 1

Summary: Table 2 – Test Rack Pressure Calculation--Modified




D(n) Ppd(n) Pio(n) Piod(n) Tv CT Pvo(n) QgiNO. DEPTH TBG PRES SURF V open @ VALVE TEMP T-R INJ GAS

ft psi psi psi 'F - psi mscfd

(INPUT) ------- ------- ------- ------- ------- ------- ------- 0 100 1200 1200 108 - -

1 2475 397 1200 1273 130 0.863 1134 931

2 4375 648 1175 1302 146 0.835 1143 9403 5800 851 1150 1315 159 0.815 1145 919

4 6775 996 1125 1315 167 0.803 1138 832

5 7425 1095 1100 1304 173 0.795 1126 709

6 7690 1137 1075 1282 175 0.792 1108 604

7 7940 1176 1050 1258 177 0.789 1089 46514/64” SO

5.1.16 Summary for Example Problem No. 1




• Predicted rate of about 800 BFPD

• Lift from Screened Orifice near bottom

• Gas injection rate of about 680(755) MCFD

• Flowing surface temperature of about 108 “F

• Anticipated operating surface gas injection of 1075 psig

• After installation, production tests should be run to optimize

production and injection gas rates.

.




• .

*•*max=7100x.465=

3300 psig

160

**Pi from 0.1 to 1.0 bpd/psi

Min PI = 0.1

Av. PI = 0.4Max PI = 1.0

Note: 2 3/8 inch tbg




API Example 2: Assume PI = 0.1 bpd/psi




#N/A#N/A#N/A#N/A0.10052477782504

OK102858854050.094400513002003

OK99041538010.088281418001502

OK95326522540.082162423001001

9008000.070000

---------------------------------------------------------




7000 700 1060


API Example 2: Assume PI = 200/500=0.4 bpd/psi & Pr = 3300 psi




NO1064123269370.166400513008007

NO103595556840.154341015507006

OK100872545400.142281418006005

OK98353434900.130221920505004

OK96037825230.118162423004003

OK93925316310.106102925503002

OK9191568040.09443328002001

9008000.070000

---------------------------------------------------------





RemarksCSGTBGINJ.FLUIDAPI Example 2: PI = 1 bpd/psi




10687100

NO1011107946700.21421002100120010

OK99487939540.2021862220011009

OK97970933130.1901624230010008

OK96556727360.178138624009007

OK95244722130.166114825008006

OK94134717370.15491026007005

OK93126513020.14267127006004

OK9211979030.13043328005003

OK9131435360.11819529004002

OK9051011970.106030003001

9008000.070000

---------------------------------------------------------


1500

API Example 2: Variable Gradient Design




0

500

1000

0 1000 2000 3000 4000 5000 6000 7000DEPTH (ft) or (meters)

P R E

S S U R E

( p s

i ) o r ( b a r )

GAS INJ TBG SPACING

Pg= 900 psi

1060

860

Pwh’= 160+.2x(900-160)=308 psi




Pwh= .2x(900-160)+160

Min. Spacing = 200 ft

1060860

WELL




853687#N/A7110.8021738537208667000

8547113587220.81316585774078060478727344297390.8151638767607625847

8897564897550.8181618947807425610

9037785437710.8221599108007135281

9147995897860.8271559238206764846

9228206287990.8341519348406304290

9268416548110.8421459408605743599

9268616658210.8531389438805072757

9218806688300.8671309409004321753

--1159009003080

----------------------------------------------------------------(INPUT)

psipsimscfdpsi-'Fpsipsipsift

@ depthSurf closeINJ GAST-RTEMP@ VALVEV openSURFTBG PRESDEPTH

V closePvc(n)QgiPvo(n)CTTvPiod(n)Pio(n)Ppd(n)D(n)

Pvcd(n)Example #2NAME:

DummyDummy

¼” s. orifice



A.Summary: Problem # 2




• The design results are for Injection PressureOperated Valves.

• This spacing is too wide for ProductionPressure Operated (PPO) Valves.

• For PPO Valves, the top of the design line

should be based on 30 or 40% of thedifference between Pio1 and Pwh.

• The resulting closer spacing permits the

uppermost PPO Valves to close as unloadingprogresses deeper in the well.

B.Summary: Problem # 2




• Original mandrel spacing for new wellsmust be carefully thought out.

• When little or no well productivity infois available, mandrel spacing should be

closer, mainly in the upper part of thestring.

• Mandrel spacing must be sufficient tolast for the lifetime of the completion.

C.Summary: Problem # 2




• Flexible design from 250 to 1100 BFPD

• Minimum of 10 mandrels• “Valve” to lift from near total depth

• Drop injection pressure 20 psi on eachlower valve to deter multi-point injection

• Use ¼-inch screened orifice on bottom

API RP 11V6: 5.3 Example No. 3 - Fixed MandrelUsing Injection Pressure Operated Valves

P P 3350 i G 0 465 i/f




• Pr = Pws = 3350 psig• Pb = 1500; (Standing)

• Test Rate = 200 BFPD

• Pwh = 120 psig• API Oil = 35o; GOR = 400

• Cut = 50 %:Water

SG=1.074• Tbg = 1.995” ID

• Dw=8000’TVD/9936’MD

• Pwf = 2550 psig (Fig 21)

• Ts = 74 ‘F

• Tf = 180 ‘F

• Gs = 0.465 psi/ft• Pg = 1150 to 1250 psig

• Sgi = 0.7

• Mandrel = Oval Side Pocket• Min Spacing = 500’

• PD= 20 psig

• GL Valve = 1” w/ 3/16” port• Casing = &’ OD

• Directional Well

• Gg = 0.032• Twh = Measured: 100 ‘F

• Use all known information

5.3.2 Well Data

P P (D th SFL) P h




• Pr = Pws = gsx(Depth-SFL) + Pwh• Pws = 0.465 x (8000 – 900*) + 50 = 3350 psig

…..from sonic fluid level measurement

• Pwf = 2550 psig from Fig. 21

• Pb = Bubble Point Using Standing’s = 1500 psia

• Gg

= 0.032 psi/ft Fig. 5

• Temp. gradient = 100x(180-100)/8000= 1 ‘F/100’

API RP 11V6Example # 3

2550 psig




GOR 400

Bubble Point




GOR = 400SGg = 0.85Oil Gr. = 35Ff = 180

BP = 1500 psia


3500

4000

R ( k P a )

5.3 API Example Problem No. 3

Pws = 3350 psig




0

500

1000

1500

2000

2500

3000

0 100 200 300 400 500 600 700 800 900 1000


P W F ; F L O

W I N G P R E S S U

R E ( P S I A ) O R

NO SKIN WITH SKIN

Pwf = 2550 psig

PI = 200/(3350-2550) = .25 bpd/psi

Qmax = 670 bpd

BP




API 11V6 Example # 3: Equilibrium Curve Data




15018000

#N/A#N/A#N/A#N/A#N/A#N/A#N/A#N/A10




#N/A#N/A#N/A#N/A0.14260768086006

OK1477107873700.130480313435005

OK141579657280.118383317504004

OK135956942360.106288121503003

OK130738828530.094192925502002

OK125924915680.08297629501001

120012000.070000

---------------------------------------------------------

Maximum Production Rate 0f 500+ bfpd from 7370 ft


3500




0

500

1000

1500

2000

2500

3000

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000


P R E S S U R E

( p s

i ) o r ( b a r )

NEEDED INJ PRESSURE GAS INJ PRESSURE INJ GAS-SF

200 BPD500 BPD

Well Data: Mandrel Spacing:Well straight to 1500’ then Directional

N Dt / D




• No. Dtv / Dm• 0 0’ 0’

• 1 2350’ 2450’

• 2 3460’ 3921’• 3 4345’ 5094’

• 4 5000’ 5962’

• 5 5500’ 6624’• 6 6000’ 7287’

• 7 6500’ 7949’

• 8 7000’ 8612’• 9 7500’ 9274’

• 10 7900’ 9804’

41+ degree angle from

2450’ to total depth

Dtv TD Dm

o 1500’

*




* Fluid rate---50% cut

Gas Passage Chart







PPEF = 0.104

API RP 11V6 Example # 3

ORIGINAL

SPACING




SPACINGDESIGN

AGL_SPAC: GAS LIFT SPACING2000

API RP 11V6: Example # 3- Ideal spacing

Requires pulling tubing to re-space




0

500

1000

1500

0 1000 2000 3000 4000 5000 6000 7000 8000 9000DEPTH (ft) or (meters)

P R E S S U R

E

( p s i )

GAS INJ TBG SPACING

500 bfpd

Requires pulling tubing to re-space

AGL_SET:INJECTION PRESSURE VALVE DESIGN

Example2000

Valve installed in each mandrel




0

500

1000

1500

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000

DEPTH (ft) of (meters)

P R E S S U R E ( p

s i ) o r ( b a r )

INJ GAS TBG Temp VALVE SPACING

API RP 11V6 Example Problem No. 3




Qgi

MCFD




MCFD



WELL NAME: API RP 11V6: Fixed Mandrels-Example # 3

D(n) Ppd(n) Pio(n) Piod(n) Tv CT Pvo(n) Qgi

NO DEPTH TBG PRES SURF V open @ VALVE TEMP T R INJ GAS

API RP 11V6 Example # 3:Table 7 - Mandrel/Valve Summary




4345 D5500 D

D

NO. DEPTH TBG PRES SURF V open @ VALVE TEMP T-R INJ GAS

ft psi psi psi 'F - psi mscfd

(INPUT) ------- ------- ------- ------- ------- ------- -------

0 120 1200 1200 100 - -

1 2350 406 1200 1275 124 0.873 1150 917

2 3460 562 1180 1288 135 0.854 1151 918

3 5000 800 1160 1315 150 0.829 1159 918

4 6000 968 1140 1323 160 0.813 1158 848

5 6500 1056 1120 1315 165 0.806 1148 761

6 7000 1147 1100 1307 170 0.799 1139 626

7 7500 1240 1080 1298 175 0.792 1129 393

8 7900 1317 1060 1286 179 0.786 1118 #N/A

1/4” SODummy

4345 D5500 D

D

Summary Example Problem No. 3




• Calculated a PI = 0.25 bfpd

• Used Equilibrium Curve to predict a rate of 500+ bfpd

from about 7500 ft.• Spaced valves using TVD

• Recommended valves @ 2350, 3460, 5000, 6000,6500 & 7000 ft with dummies @ 4345, 5500 & 7900 ftplus a 14/64 inch screened orifice @ 7500 ft to pass500 MCFD

Summary for Design

• The better the data the more specific the design




• The better the data, the more specific the design.• The poorer the data, the more flexible the design.

• If feasible, design to lift from near bottom.• Carefully select the tubing size.

• For better valve performance, use 1 ½-inch valves.

Select the smallest port that will pass the requiredinjection gas.

• For most wells, use a screened orifice as the bottom

injection “valve.”

Impact of Significant Variables

• INJECTION PRESSURE: The higher the injection gas pressure, theid th d l i d th d th i lift d th




INJECTION PRESSURE: The higher the injection gas pressure, thewider the mandrel spacing and the deeper the maximum lift depthcan be.

• FLOWING WELLHEAD TUBINE PRESSURE: The higher the

pressure, the lower the maximum producing rate and the higherthe injection gas requirement will be.

• TUBING SIZE: Larger tubing sizes permit higher producing ratesand wider mandrel spacing.

• UNLOADING GRADIENTS: Higher gradients mean closer mandrelspacing and shallower maximum lift depths.

• INJECTION GAS GRAVITY: Higher gas gravity means wider

mandrel spacing and higher test rack opening pressures for gaslift valves.

• CLOSER MANDREL SPACING: Permits near optimum gas liftperformance and unloading the well with less injection gas

volume.

DESIGN PRINCIPLES




• CLOSER MANDREL SPACING IS

PREFERRED.

• HIGHER PRODUCTIVITY WELLS

REQUIRE CLOSER MANDREL SPACING

NEAR TOP OF WELL.• MANDREL SPACING SHOULD BE

BASED ON WELL LIFE CYCLEESPECIALLY OFFSHORE.

GOOD GAS-LIFT PRACTICES

• Streamlined Wellhead




• Streamlined Wellhead

• Flowline Size• Separator Pressure

• Well Conditioningf/Unloading

• Unloading Precautions

UNLOADING PRECAUTIONS

• Don’t Cut Out the Valves!




• Don t Cut Out the Valves!• Buildup Injection Pressure Slowly

• 5 psi/min. to 400 psi• 10 psi/min. Until Gas Production

• Then increase Gas Injection Rateto the Desired Value

Preferences

SINGLE COMPLETIONS PREFERRED OVERDUALS



2007 Workshop Clegg & Smith 198……………….GasLift Workshop - Smith & Clegg 2/3/2006 Page 82

DUALS.

LARGER OD (1.5”) GAS LIFT VALVES PERMITSMALLER INJECTION PRESSURE DROPS FORCONTINUOUS FLOW IN SMALLER PORT SIZES.

MANDRELS W/ORIENTING SLEEVES BETTERFOR HIGHLY DEVIATED WELLS.

RuleOfThumb: FLOWLINE SIZE SHOULD BE ONESTANDARD SIZE LARGER THAN TUBING SIZE.

STREAMLINING WELLHEADS REDUCESFLOWING WELLHEAD BACKPRESSURE &INJECTION GAS REQUIREMENT.

Summary




• You have learned to do a complete gas liftdesign using an equilibrium curve to determine

the max rate, various graphical methods forspacing, and how to calculate the test rackpressures for injection pressure operatedvalves.

• You how know more than most people about

gas lift---hopefully!

This is to certify that



________________

Completed the

API Gas-Lift

Design Course

id mith

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