Chapter 3 - Flow Through Tubing

Production Engineering

CHAPTER 3: FLOW THROUGH TUBING

Petroleum Production Engineering SKPP 3513

FLOW THROUGH TUBING & FLOWLINES Mohd Fauzi Hamid Department of Petroleum Engineering Faculty of Petroleum & Renewable Engineering Universiti Technologi Malaysia

Chapter 3: Flow Through Tubing & Flowlines


Objective

Students Students will able to:  Calculate the static & flowing bottomhole pressure

velocity, density & viscosity for multiphase  Calculate the velocity, flow  Identify & calculate three components of pressure losses in tubing & flowlines  Use pressure traverses curves  Construct the VLP curve using Method I & II  Construct the CP line



Objective

Students Students will able to:  Calculate the static & flowing bottomhole pressure

velocity, density & viscosity for multiphase  Calculate the velocity, flow  Identify & calculate three components of pressure losses in tubing & flowlines  Use pressure traverses curves  Construct the VLP curve using Method I & II  Construct the CP line



CONTENTS 

Introduction



Vertical Lift Performance (VLP) 

Basic Theory of Fluid Flow in Pipe



Gilbert Method   

 

Determination of Pwf Determination of THP Selection of Optimum Tubing Size

Factors Affecting Affecting VLP

Choke Performance (CP)



Introduction  In order to analyze the performance of a conventionally completed well, it is necessary to recognize that there are three distinct phases, which have to be studied separately and then finally linked together before an overall picture of a flowing well’s behavior can be obtained.  These phases are:  Inflow performance: performance: the flow of fluid from the formation into the bottom of the well – IPR IPR..  Vertical lift performance (VLP): (VLP): involves a study of pressure losses in vertical pipes carrying two-phase mixture (gas and liquid). Also known as tubing performance (TP). (TP). (CP): a study of pressure losses across the  Choke performance (CP): choke in surface flow-line.



Introduction



Vertical Lift Performance (VLP)  Flow characteristics @ tubing (pressure losses) or relates to pressure-rate relationship @ wellbore as fluid flow from bottomhole to surface.  Directly affected by 

Tubing size & depth



GLR



Water production



Separator pressure



Surface flow line size & length



Fluid properties (density, surface tension, viscosity)



Production problems (scaling, sand & paraffin)

 Also known as: tubing performance, wellbore flow performance.



 The question: Is Pwf – ∆Pt > Pwh (or THP)? If ‘yes’, the well will flow. where:

∆Pt -

pressure losses or differential pressure in tubing Pwh- well head pressure or tubing head pressure (THP)

 Need a knowledge about the fluid flow through vertical pipe (tubing) which involve the “energy or pressure equilibrium”.  The result is the flowing pressure distribution along the tubing which can be used for the production planning of the well.  Basic requirement:  Dimensional analysis  Fluid properties: density, viscosity, compressibility, surface



 Gas properties: density, viscosity, compressibility, gas law  Other factors: Bo, Bg, Rs, etc

 Basic information:       

∆P pure water: 0.433 psi/ft ∆P brine @ SG = 1.07: 0.464 psi/ft ∆P 42 oAPI oil (SG = 0.815): 0.352 psi/ft Density = mass/volume SG oil: 141.5/(131.5+ oAPI) SGL = ρ L/ ρ W (density of water, ρ W = 62.4 lb/cuft) Hydrostatic pressure, Ph = ρ gh. If ρ in ppg and h in ft, Ph = 0.052 ρ h



Density  Mixture Density: Three types of density of liquid and gas mixture ( ρ m):  Slip density, s  No-slip density,  Kinetic density,

n

=

k

ρ s

=

ρ L H L + ρ g (1 − H L )

ρ n

=

ρ L λL + ρ g λ g

ρ k

=

2 L

ρ L λ H L

+

m

ρ g λ g2 1 − H L

where : H L

λ L H g

= =

liquid hold − up

no − slip liquid hold − up

=

gas hold − up



 When there is great density difference phenomenon

 slip & hold-up

 Slip:  Less dense (lighter) phase ability to flow at greater velocity than denser (heavier) phase

 Hold up:  Consequence of slip  Volume fraction of pipe occupied by denser phase is greater than would be expected from (relative) in – and outflow of two phases, since its velocity slower than light phase





 Superficial phase velocities (VSL & VSG)  Liquid: VSL = qL / Ap  Gas : VSG = qg / Ap - q = phase volume flow rate - Ap = pipe cross sectional area

 In situ or actual velocity ( VL & VG)  Liquid : VL = qL / AL = qL / HL Ap  Gas : VG = qG / AG = = qG / HG Ap - AG = actual area of pipe occupied by gas - AL = actual area of pipe occupied by liquid



 For oil and water flow: Liquid density, ρ L

ρ L = ρo Fo + ρ w F w oil fraction

Fo =  For gas:

water fraction

qo qo + qw

ρ g =

Fw = 1 − F o

28.97γ g P ZRT

=

2.7γ g P ZT

oR (oF

+ 460)



Viscosity  Viscosity is a function of T, P, R s, ρ , composition.  Please refer to the reservoir fluid properties for determination of viscosity.  Mixture viscosity of multi-phase flow, m:

µm = µ L H L + µ g (1 − H L )  Viscosity of oil and water mixture, liquid viscosity,

L:

µ L = µo Fo + µ w F w where : H L Fo F

= =

liquid hold

−

oil fraction f

i

up



Hold-up Factor  Four types of hold-up factor involve when study on the twophase flow:    

Liquid Hold-up, HL No-slip liquid hold-up, λ L Gas hold-up, Hg No-slip gas hold-up, λ g

 Liquid hold-up, H L = volume of liquid in pipe/volume of the pipe. If HL = 0: 100% gas flow HL = 1: 100% liquid flow

 Gas hold-up, Hg



 No-slip liquid hold-up, λ L = comparison between the volume of liquid in pipe with the volume of the pipe when the gas and liquid move with the same velocity.

λ L =

q L q L + qg

 No-slip gas hold-up, λ g

λg = 1 − λ L =

qg

q L + qg


 Flow pattern @ tubing function of:  Gas & liquid flow rates  Tubing inclination angle  Tubing diameter  Phase densities




Basic Theory of Fluid Flow in Pipe (2) mgZ 2 mv22 U 2 , PV , 2 2, gc 2 gc

(- q) z2 2

mgZ1 mv1 , , U1 , PV 1 1 gc 2 gc

(1) z1

(+ W) Figure 3-2: Flow System in Vertical Pipe



Introduction  Based on Energy Equation which produce Energy Equilibrium: U1 +

2

mv1

2 gc

+

mgz1 gc

+ PV 1 1 − q +W = U2 +

2

mv2

2 gc

+

mgz2 gc

… (1) + PV 2 2

where: U

= internal energy carried with the fluid

 mv 2   = kinetic energy – energy due to velocity 2 g  c  mgz   = potential energy g  c  PV q

= pressure volume (also called energy of pressure) = transferred heat (heat energy)



 Solving Equation (1) by thermodynamic: dP dZ

=

g gc

ρ sin θ +

f ρ v 2

2 gc d

+

ρ vdv………………………... (2) gc dZ

where: f = friction factor = f(NRe, ε) NRe = Reynold number ε = absolute pipe roughness gc = 32.2lbm.ft/lbf .s2 NRe

 ρ m vm d  =    µ m 

NRe < 2100 : laminar flow NRe = 2100 – 4000 : transition flow NRe > 4000 : Turbulent flow In petroleum :



 Equation (2) can be rewrite in general form:

 dP   dP   dP  = +    +  ………………………... (3) dZ  dZ  ele  dZ  f  dZ  acc dP

component due to kinetic energy changes

total pressure gradient

=

g gc

= ρ sin θ

(component due to potential energy changes or elevation changes)

=

f ρ v

ρ vdv g c dZ

2

2 g c d

component due to friction

Equation (3) above is a basic equation for the solution of the



 For vertical flow → θ = 90o, equation (2) become: dP dZ

=

g gc

ρ +

f ρ v

2

2 gc d

+

ρ vdv g c dZ

……………………... (4)

 For horizontal flow → θ = 0o: dP dZ

=

f ρ v

2

2 gc d

+

ρ vdv

……………………... (5)

g c dZ

 For multi-phase flow: dP dZ

=

g gc

ρ m sin θ +

f m ρm vm

2 gc d

2

+

ρ m vm dvm g c dZ

…………………… (6)



 All analytical methods using equation (3) as a basic calculation for the pressure distribution in pipes. The only differences are:  technique for determination of particular parameters.  assumption or approach used for solving equations (2) and (3).

 Generally, there are three groups of methods:  Group that does not consider the slip and the shape of flow. This includes:

• • •

Poettmann & Carpenter Baxendall Fancker & Brown



 Group that consider the slip but not the shape of flow. This include:

•

Hagedorn & Brown

 Group that consider the slip and the shape of flow. This includes:

• • • • • •

Ros Duns & Ros Okiszewski Aziz & Govier Beggs & Brill Chierici, Civcci & Scrocchi

 All the above methods are complex and difficult, especially for multi-phase flow.  For practical purpose, empirical method established by Gilbert



Gilbert Method  Gilbert accumulated a large amount of flowing well data, e.g: 

depth of tubing, ft



bottomhole flowing pressure (tubing intake pressure), psi



tubing head pressure, psi



production rate, BPD



gas-liquid ratio, Mcf/bbl



tubing size, in

 He correlate the above data and as a first attempt he chose wells with the same rate, GLR and tubing size, as shown in Figure 3-3.



0

0

Bottomhole flowing pressure, psig

A B t f , h t p e D

Figure 3-3: Flowing BHP as function of THP and tubing length: constant GLR,

C D

Figure 3-4: Pressure distribution curve: vertical two-phase flow



 Gilbert then assumed that all the curves of varying THP could be overlying as one curve with the THP converted to a depth equivalent, as shown in Figure 3-4.  He then continue his correlation to produce a pressure distribution chart (pressure traverse curve) for a specific tubing size and production rate. An example of this chart shown in Figure 3-5.  The pressure distribution charts can be used for:     

Selection of the optimum tubing size Prediction of a well life Prediction when the well need artificial lift Planning the artificial lift Planning the stimulation





Determination of Pwf of a Well

THP

Pwf

 The step are as follows: 

  

Locate/choose the Pressure distribution chart (PDC) that corresponds to the given nominal tubing size and oil rate. Find THP (given) on the x -axis of the chart. Draw a vertical line from THP to the given GLR (point A) Draw a horizontal line from point A to the y -axis. The intersection point is THP equivalent depth (zero

A

THP equivalent depth

Tubing depth

B Tubing equivalent depth (Pwf equivalent depth)



Determination of Pwf of a Well 

 

Determine tubing equivalent depth (Pwf equivalent depth) (= THP equivalent depth + tubing depth) Draw a horizontal line to the GLR (point B). Draw a vertical line from point B to the x-axis. The intersection point is a Bottomhole flowing pressure, Pwf.

THP

Pwf

A

THP equivalent depth

Tubing depth

B Tubing equivalent depth (Pwf equivalent depth)



Determination of THP of a Well

THP

Pwf

 The steps are as follow:  Locate/choose the Pressure

distribution chart (PDC) that corresponds to the given nominal tubing size and oil rate.  Find Pwf (given) on the x-axis of the chart.  Draw a vertical line from THP to given GLR (point C)  Draw a horizontal line from point C to the y-axis. The intersection point is tubing equivalent depth (Pwf

D THP equivalent depth

Tubing depth

C Tubing equivalent depth (Pwf equivalent depth)



Determination of THP of a Well

THP

Pwf

 Determine THP equivalent

depth (zero datum) (= tubing equivalent depth  tubing depth)  Draw a horizontal line to the GLR (point D).

D THP equivalent depth

 Draw a vertical line from point

D to the x-axis. The intersection point is a Tubing head pressure, THP.

Tubing depth

C Tubing equivalent depth (Pwf equivalent depth)





Example 3-1  Find the flowing pressure at the foot of 13,000 ft of 2 ⅜-in tubing if the well is flowing 100 BPD at a GLR of 1.0 Mcf/bbl with a THP of 200 psi.  Refer to suitable Pressure distribution chart (Figure 3-8);  THP equivalent depth = 2500 ft  Tubing equivalent depth = 2500 + 13000 = 15500 ft  Pwf = 1860 psi



THP 200 psi THP equivalent 2500 ft

Pwf equivalent 15500 ft

Pwf 1860 psi



Example 3-2  What is the THP of a well, completed with 8000 ft of 2 ⅜-in tubing, that is flowing at 600 BPD and a GLR of 0.4 Mcf/bbl if the pressure at the bottom of the tubing is 2250 psi?.  Refer to suitable Pressure distribution chart (Figure 3-9);  Pwf equivalent depth = 12,150 ft  Tubing equivalent depth = 12,150 - 8000 = 4,150 ft  THP = 610 psi



THP 610 psi

THP equivalent 4150 ft

Pwf equivalent 12150 ft

Pwf 2250 psi



Selection of Optimum Tubing Size  There are 2 methods available: Method 1 and Method 2 Method 1

 From Pwf , Ps and q, plot IPR curve.  Plot Pwf vs q for several size of tubing based on VLP on the same graph.     

select tubing diameter (available tubing size). assume q (assumption must be tally with chart). from the chart : given THP  THP equivalent depth  tubing equivalent depth (P wf equivalent depth)  Pwf . Repeat all steps above for new assumed q. Repeat all steps above for new tubing size.



 All the steps can be summarize in form of table (for each tubing size selected): (1) q (BPD)

(2) THP equivalent depth (ft)

(3) Pwf equivalent depth (ft)

(4) Pwf (psi)

q1

-

-

Pwf1

q2

-

-

Pwf2

q3

-

-

Pwf3

q4

-

-

Pwf4

assume according to chart

pressure distribution chart

(2) + tubing depth


 Intersection point between IPR and VLP corresponds to the optimum q for each particular tubing size.



) i s p (

f w

P

Pwf vs q A

IPR B C

0

qopt-A qopt-B qopt-C

q (BPD)



Method 2  Plot IPR curve.  Assume q and determine Pwf from IPR curve (or from PI formula).  By using a suitable pressure distribution chart, determine Pwf equivalent depth (tubing equivalent depth) (P wf from 2nd step).  Determine THP equivalent depth. (= Pwf equivalent depth – tubing depth), and then THP.

 Repeat the above steps for new tubing size.  From required THP, draw a horizontal line to the right, until intercept with the THP vs q curves. Intersection points will give the optimum q for that THP and tubing size.  The highest optimum q correspond to the optimum tubing



 All the steps can be summarize in form of table (for each tubing size selected): (4) THP equivalent depth (ft)

(5) THP (psi)

(1) q (BPD)

(2) Pwf (psi)


q1

-

-

-

THP1

q2

-

-

-

THP2

q3

-

-

-

THP3

q4

-

-

-

THP4


IPR or PI formula


(3) - tubing depth




) i s p ( P

IPR

Pwf A B C

THP vs q

THP q 0

qopt-C qopt-A qopt-B

q (BPD)



Example 3-3 A well producing from a pay zone between 5000 and 5052 ft is completed with 2⅞-in tubing hung at 5000 ft. The well has a static BHP of 2000 psi and a PI of 0.3 bb/day.psi and produces with a GOR of 300 cuft/bbl and a water cut of 10%. At what rate will the well flow with a THP of 100 psi? Assume a straight line IPR. qw q

= 0.1,

GLR =

and

gas q

=

gas qo

300qo q

=

= 300 300(q − qw ) q

= 300(1 −

qw q

)

= 300(1 − 0.1) = 270 cuft / bbl PI

q

0 3* 2000

600 bbl / d



Method 1 Involves calculation of P wf at various values of q, and THP of 100 psi. (1) q (BPD)

(2) THP equivalent depth (ft)


(4) Pwf (psi)

50

500

5500

1275

100

700

5700

1150

200

800

5800

1050

400

800

5800

975

600

800

5800

910



(2) + 5000




2500

Method 1

Exampl e 3-3

P(psi)

q = 300 bbl/d qw = 30 bbl/d (10% WC) qo = 270 bbl/d Pwf = 1000 psi

2000

1500

1000

Method 2 500

0 0

100

200

300

400

500

q(BPD)

600

700

q = 280 bbl/d qw = 28 bbl/d (10% WC) qo = 252 bbl/d THP= 100 psi



Method 2 Involves calculation of THP at various values of q, using the value of Pwf from IPR. (4) THP equivalent depth (ft)

(5) THP (psi)

(1) q (BPD)

(2) Pwf (psi)


50

1833

7300

2300

450

100

1667

7500

2500

400

200

1333

6700

1700

250

400

667

4200

-

-

600

0

-

-

-

assume according to

IPR or PI

pressure distribution

(3) 5000

pressure distribution



Exercise  A well is to becompleted for which the following data has been derived:    

Gas liquid ratio = 1.0 Mcf/bbl Productivity index = 18 bbl/day/psi Reservoir pressure @ 8200 ft = 5500 psi 7” tubing (of 6.366” ID) is available

 A THP requirement of 750 psi is required for the well. At what rate will the well flow, assuming zero water cut.



PI =

q Ps − Pwf

q max =18 x 5500 = 99, 000 BPD (at Pwf = 0) q = 0 at Pwf = Ps = 5, 500 psia

 At THP = 750 psig (VLP Method-1) q (bbl/day) 5000 10000 20000 30000

THP Equi. Depth (ft) 5450 5800 4950 3800

Pwf Equi. Depth (ft) 13650 14000 13100 12000

Pwf (psig) 2770 3110 3000 3290

The well will flow at rate,



 VLP Method-2 q (bbl/day) 5000 10000 20000 30000

Pwf (psig)

Pwf Equi. Depth (ft)

THP Equi. Depth (ft)

THP (psig)



Determination of Producing Well  Plot IPR curve.  Plot Pwf vs q for several size of tubing based on VLP on the same graph (same procedure as Method 1)  select tubing diameter (available tubing size).  assume q (assumption must be tally with chart).  from the chart : given THP  THP equivalent depth tubing equivalent depth  Pwf .  Repeat all steps above for new assumed q.  Repeat all steps above for new tubing size. (use the same table as Method 1)



 If the Pwf vs q curve is located outside and not intercept to the IPR, the well is not producing (not flowing).



) i s p (

A – Producing B – Not Producing

f w

P

B

A

0

Pwf vs q

IPR

q (BPD)



Determination of Well Life  Plot IPR curve.  Plot Pwf vs q (VLP) – use same method as previously discussed.  Plot a few future IPR curve  A future IPR curve which not intercept with P wf vs q curve shows the dead well.  A future IPR curve which touching the P wf vs q curve shows the life of the well.



) i s p (

The well will dead at IPR no (3)

f w

P

(1) (2) (4) 0

Pwf vs q IPR

(3) q (BPD)



Determination of Pwf & q at Specific Tubing Size, GLR and THP  Plot IPR curve  Plot Pwf vs q (VLP) – use same method as previously discussed.  Intersection point between these plots will give the P wf and q for that specific tubing size, GLR and THP.



) i s p (

f w

P

IPR

Pwf vs q Pwf

0

q

q (BPD)



Other Correlation Year

Researcher

Type of work

Pipe size

Fluid type

Comments

1952

Poettmann & Carpenter

Semi-empirical & field data

2, 2.5, 3 inch

Oil, water, gas

Practical solution for 2, 2.5, 3 in. with GLR < 1500 scf/bbl and q > 420 BPD

1962

Winker & Smith

Practical

1 – 3.5 inch

Oil, water, gas

Curve for Poettmann & Carpenter

1960

US Industries

Practical

1 – 4.5 inch

Oil, water, gas

Curve for Poettmann & Carpenter

1954

Gilbert

Field data for practical use

2, 2.5, 3 inch

Oil, water, gas

Vertical multiphase flow traverses curve

1961

Ros

Lab. exp & field data

All

All

Good correlation for all ranges of flow

1961

Duns & Ros

Lab. exp & field data

All

All

Good correlation for all ranges of flow & easier to



Other Correlation Year

Researcher

196 5

Hagedorn & Brown

Field exp.

1 – 4 inch

Oil, water, gas

Generalized correlation to handle all ranges of multiphase flow

196 7

Okiszewski

Review all correlation

All

Oil, water, gas

General correlation to predict pressure losses for all ranges of flow by utilized Ros, Griffith & Wallis works

Laboratory & field data

All

All

Testing lab data with field data

Laboratory

1, 1.5 inch

Air, water

Generalized correlation to handle all ranges of multiple phase flow & for any pipe angle

197 2

Aziz & Govier

197 3

Beggs & Brill

Type of work

Pipe size

Fluid type

Comments



Choke Performance (CP)  Wellhead chokes or surface chokes are used to:     

limit production rates for regulations protect surface equipment from slugging avoid sand problems due to high drawdown control flow rate to avoid water or gas coning Produce reservoir at most efficient rates.

 Placing a choke at the wellhead means fixing the wellhead pressure and, thus, the P wf and production rate.  The choke therefore plays an important role in:  well control  reservoir depletion management



 Two types of surface chokes are used:  Positive (fixed) chokes – the orifice size is specified before installation.  Adjustable chokes – the orifice size can be adjusted after installation to suit the well and operational requirement.



Choke Flow Characterization  The main function of a choke is to dissipate large amounts of potential energy (i.e pressure losses) over a very short distance.  The design of a choke takes advantage of the flow regime resulting from a sudden disturbance in continuous flow through a circular conduit.

Figure 3-21: Flow Regime in a Fixed Choke

 Figure 3-21 gives a schematic of the normal flow character of fluid through a fixed choke.  It describes the combined effect of a sudden flow restriction,



 As the fluid approaches the orifice, it leaves the pipe wall and contracts to form a high-velocity jet.  The jet converges to a minimum called the throat or vena contracta, and then it expands toward the wall of the choke bore.  After leaving the choke, the stream of fluid expands and returns to a flow geometry similar to what it was before entering the choke.




 Total irreversible pressure losses are summarized in the following:  friction throughout the choke and near-choke areas  turbulence (associated with the eddy current) near the entrance and exit of the choke  slow eddy motions between the contracted jet and the pipe walls


 Abrupt expansion at the exit to the choke.



Choke Flow Correlation  The general correlation for the choke performance: THP =

B

A * GLR C

S

* q ………………………………..…... (1)

where: THP = tubing head pressure, psi (except Ros – psia) GLR = gas-liquid ratio, Mcf/bbl S = choke size, 1/64 inch q = flow rate, BPD A, B, & C = empirical constants related to fluid properties



 The value of A, B, and C are: Researcher

A

B

C

Gilbert

435

0.546

1.89

Baxendel

419

0.546

1.83

Achong

342

0.650

1.88

Ausens

427

0.680

1.97

ROS

550

0.500

2.00

 The value of the constant will depend upon:  the choke characteristics and dimensions  the gas and liquid properties



 Correlation by Gilbert: THP =

435(GLR) 0.546 1.89

S

……………………………..…... (2) *q

 Gilbert also presented the information as a nomograph (Choke Performance Chart – Figure 3.22).  The nomograph is split into two portions. The left hand side relates the production rate & GLR, and the right hand side utilized a 10/64 in choke for reference and this is related to choke size and THP.





Determination of Flow Rate

 Required data: THP, S, and GLR



Determination of THP

 Required data: q, GLR and S



Determination of Choke Size (S)

 Required data: q, GLR and THP



Example 3-4 Calculate the choke size required to flow a well at 200 bbl/day with a GLR of 1.0 Mcf/bbl and a tubing head pressure of 400 psi.

THP = S

435(GLR)0.546 1.89

S

 435(GLR ) = THP  = 17.25 = 17 / 64 in.

*q

0.546

 * q 

(1/1.89)

 435(1) =  400

0.546

 * 200  

(1/1.89)



Example 3-5 Given data: o API = 40 THP = 400 psi Choke size = 14/64 inch GLR= GLR= 1000 cuft/bbl Down stream pressure = 100 psi Find the flow rate in BPD THP = q

435(GLR )0.546 1.89

S

*q

 THP * S 1 .89   400(14)1.89  = = = 134.82 BPD 0.546  0.546   435 * GLR   435(1) 



Example 3-6 A well is producing through through a ¼-in choke at 100 bbl/day with a THP of 150 psi. What is the GLR as calculated from nomograph and from Eq n. (2)? What would be the calculated GLR if, all other things being equal, the choke size were 17/64 in.? THP =

435(GLR )0.546 1.89

S

 THP * S  =  4 3 5 * q   1.89

GLR16

150(17) =

1.89

GLR

*q (1 / 0.546 )

 

150(16)  =  4 3 5 ( 1 0 0 )   1.89

(1/ 0.546 )

(1/0.546)

= 0 562Mcf / bbl

= 0.455Mcf / bbl



Determination of Optimum Choke Size (Without VLP) 1. Plot IPR curve. 2. Assume several values of q, find relationship between THP vs q (Eqn 2 or nomograph) for each of choke size. 3. Plot choke performance lines on the same graph as IPR. 4. Intersection point between the two plots (step 1 and 3) give the optimum flow rate for choke size. 5. Choose the highest optimum q as optimum choke size.



) i s p (

f w

Optimum choke size is S3 (highest q)

IPR

P

S1

CP

S2 S3

0

q

q

q

q (BPD)



Determination of Optimum Choke Size (With VLP) 1. Plot IPR curve. 2. Plot VLP (THP vs q) curve (Method 2) on the same graph 3. Assume several values of q, find relationship between THP vs q (Eqn 2 or nomograph) for each of choke size. 4. Plot choke performance lines on the same graph as IPR & VLP. 5. Intersection point between the VLP and CP (step 2 and 4) give the optimum flow rate for choke size. 6. Choose the highest optimum q as optimum choke size.



Optimum choke size is S3 (highest q)

) i s p (

f w

IPR

P

S1

CP

S2

VLP

S3

0

q

q2

q3

q (BPD)



Determination of Optimum Tubing Size ) i s p (

f w

IPR

P

A B C

0

Optimum tubing size is B (highest q)

CP VLP

S

q (BPD)



Example 3-7 A well producing from a pay zone between 5000 and 5052 ft is completed with 2⅞-in tubing hung at 5000 ft. The well has a static BHP of 2000 psi and a PI of 0.3 bb/day.psi and produces with a GOR of 300 cuft/bbl and a water cut of 10%. What size of choke is required in the flow line to hold aTHP of 100 psi? What would be the production rate on a ¼-in. choke? Assume a straight line IPR. qw q

= 0.1,

GLR =

gas q

gas

and

=

qo

300qo q

=

= 300 300(q − qw ) q

= 300(1 −

qw q

)

= 300(1 − 0.1) = 270 cuft / bbl = 0.27 Mcf / bbl



 Based on Method 1: q = 300 bbl/day

THP = S

435(GLR )0.546 1.89

S

 435(GLR) = THP  = 30 = 30 / 64 in.

*q

0.546

 * q 

(1/1.89)

 435(0.27) = 100 

0.546

 *300  

(1/1.89)

 To determine the q on a ¼-in choke, note that THP and q are

unknowns. Substituting 0.27 for GLR and 16 for S in Eq n. 2 give the result:

Chapter 3 - Flow Through Tubing

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