Nodal Aproach SAP

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ABSTRACTUNSOLKXTEE~MAR 51979

A NODAL APPROACH FOR APPLYING SYSTEMS ANALYSIS TO THEFLOWING AND ARTIFICIAL LIFT OIL OR GAS WELL . .by

Joe Eduardo Proano~~rmit E, Ezown .

A nodal and new approach is presented for applying systemsanalysis to the complete well system from the outer boundary ofthe reservoiz to the sand face, across the perforations and com-pletion section to the tubing intake, up the ~~bing string in-cluding any restrictions and down hole safety valves, the surfacechoke, the flow line and separator. .

Fig. 1 shows a schematic of a simple producing system. Thissystem consists of three phases: .t]

(1) F1OW through porous medium.(2) Flow through vertical or directional conduit.(3) FIow through horizontal pipe.

.


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2

The v::.OUSwell configurations may vary from the very simplesystem of Fig. 1 to the more complex system of Fig. 2, or any Com-bination thereof, and present day completions more realisticallyinclude the various configurations of Fig. 2.

This paper will discuss the manner in which to interrelatethe various pressure losses. In particular, the ability of thewell to produce fluids willbe interfaced with the ability of thepiping system to take these fluids. The manner in which to treatthe effect of the various components will be shown by a new nodalconcept.

In order to solve the total producing system problem, nodesare placed to se.,znenthe portion defined by different equationsor correlations.

Figure 3 has been prepared showing locations of the variousnodes. This figure is the same as Figure 2 except only the nodepositions are shown. The node is classified as a functional nodewhen a pressure differential exists across it and the pressure orflow rate response can be represented by some mathematical or phys- ,ical function. -,

Node 1 represents the separator pressure which is usually reg-ulated at a constant value. There are two pressures that are nota function of flow rate. They are F= at Node 8 and PSEP at Node 1.For this reason, any trial and error solution to the total system


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.., .. ,. 3

to effectively evaluate a complete producingsystem. All of thecomponents in the well, starting fxom the static pressure (~=)and ending at the separator, are considered. This includes flowthrough the porous medium, flow across the perforations and comple-tion, flow up the tubing string with passage through a possib2edown-hole restriction and safety valve, flow in the horizontalflow line with passage through a surface choke and to the sep-arator. .

Various positions and/or components are selected as nodes andthe pressure losses are converged on that point from both direc-tions. Nodes can be effectively selected to better show the effectof inflow ability, perforations, restrictions, safety valves,surface chokes, tubing strings, flow lines and separator pressures.

The appropriatemultiphase flow correlations and equationsfor restrictions, chokes, etc. must be incorporated in the solution.An effective means of analyzing an existing well, making rec-ommended changes or planning properly for a new well can be accom-plished by the nodal systems analysis. This procedure offers ameans to more economically optimize producing wells.


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,.q,if~t,i.f4-.A NODAL APPROACH FOR APPLYING SYSTEMS ANALYSIS TO THE , i c

FLOWING AND ARTIFICIAL LIFT OIL OR GAS WELLby Joe Mach, Eduardo pro~~o,Kermit E. Brown

1.1 INTRODUCTIONA nodal and naw approach is presented for applying systemsanalysis to the complete wel I

system from the outer boundary of the reservoir to the sand face, across the perforations and completionsection to the tubing intake, up the tubing string including any restrictions and down h~le safetyva Ives, the surfsce choke, the flow Ii ne and separator.

Fig. 1 showsa schematic of a simple producing system. This system consistsof three phases:(1) Flow thraugh porous medium.(2) Flow through vertical or directional conduit.(3) Flow through horizontal pipe.

Fig. 2 showsthe various pressure lossesthat can occur in the system from the reservoir to the separator.Beginning Fromthe reservoir these arenoted as:

API = F - Pwfs = Pressure LQss n PorousMediumrAP2 = Pwfs - Pwf = Pressure LossAcross Completion~p3 = puR. pDR = Pressure LossAcross Regulator, Choke or Tubing Nipple


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. . . .,,.. . -20 This paper will d[scuss themanner in which tointerrelate the various pressure losses. In

particular the ability of the well toproduce fluids will beinterfaced with the ability of the pipingsysternt otakethesefluids. Themanner inwhich totreat theeffect oftkevarious components willbeshown byanew nocial conceptas explained infhe next section.1.2 NODAL CONCEPT

1.21 IntroductionIn order to solve the tot~ producing system

the portion defined by different equations or correlations.problem, nodes are placed to segment

Figure 3 has been prepared showing Iocationsof the various nodes, This figure isthe same as Fig. 2 except ordy the node positions are shown. The node is classified as a functionalnode when a pressure differential exists across it and the pressure or flow rate response can berepresented by some mathemati coI or physics I function.

Node 1 represents the separator pressure which is usuaIIy regulated at a constantvalue. The pressure ~ node 1A is usually constant at either gassuction pressure. The pressure at node 1B is usually constant atpressurewi 11be held constant at the higher of the two pressures

soIes Ii nes pressure or gas compressorO psig. Therefore, the separatorneeded to flow singIe phase gas


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. . -3: . . .. .1.22 Example Problem l . .

Using Node f~ to Find the Flow Rate Possible ( ~ade 8 = ~)

Given Data: Flowing oil well. .Separator pressure: 100 psi

Flow line: 2, 300.0 ft longWOR: ODepth: 5000 ft mid perf.GOR: 400 scf/BF: 2200 psirIPR: PI = 1.0. B/D/psi (assumeconstant)Tubing size: 2-3/8

Find the oil flow rate using node f$asthe solution point.Procedure:1. Select flow rates foratrial and error procedure: Assume flow rates of200, 400,

600, 800, 1000, and 1500 B/D.2. For each rate start at PSEP= 100 and dci al I the pr~ssure lossesuntil reaching ~r


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, .. .

.3.

4.,5.

1.23

from node 8 to

.

4-Plot thecreated pressure vs. flowrate (Fig. 5). This represents th~systemperformance from the separator to ~r.Plot ~r at the given 2200 psi (Fig. 5).The intersection of the reservoir pressure Iineand the system performance l;negives the predicted flow rate (900 BOPD). ;

,.Example Problem 2Using scdution node 6 to find the flow rate (fl.w;g b : hol~?r-wtGiven data: Same as Example Problem lFor thfs solution pressure drops must be added from node 1 to node 6 and subtractednode 6.Procedure:(1)

(2)

Since ~be prix ieied flow rate is already known from Example 1, the same flowrates will be assumed: 200, 400, 600, 800, 1000, 1500 B/D.Determine the pressure Iossfrom node 1 (sepamtor) tonode6(Pw,,). For eachassumed flow rate stortat node 1 (PSEP) and add 4P 3-1 + P6-3

The following Table 1.23 shows these results.


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AssumedRate

200400600800

10001500

4.

5...

1

Fr220022002200220022002200

TABLE 1.23(B)

P8-6 200400600800

10001500

6= wf20001800160014001200700i .

Piot P6 vs. q from both step 2 and step 3 (Fig. 6). Node 6 is called the intakenode since this pwnt is the intake from the reservoir into the production tubing.The intersection of the PI ine and the so-called intake curve is the predictedflow rate for this system (900 BOPD) (Fig. 6). The presentation based on theselection of node 6 as the solution node is good if it is desired to evaluatechanging Prs or different IPR curves. Notice the answer is the same as Example1 and this is true regardless of the node selection.

1~24 Example Problem f3Using solution node 3 to find the flow rate [l~ew,tij wet/l?~4d pass+.Given Data:


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. -69-,,, TABLE 1.24(A)PRESSURELOSSES IN FLOWLINE FOR EXAMPLE PROBLEM %

20040060080010001500

SEP1001001001001001(H3

tP3-1 or

Hdz. Multiphase F ow.. ----1540 9

I 175I 329Ii --

3 = wh.. . 111514fl180230

.-.1

3. Determine t~e pressure lossfrom node 8 (~r) to node 3 (pwh). For each assumedrate start at ~r and add AP8-6+4P6-3. These values are tabulated in Table1.24(B).

TABLE 1.24(B)PRESSURELOSSES FROM NODE 8 (~/.TO NODE 3 (Pwh)

EXAMPLE PROBLEM 3

I F-200400

22002200

h

6 P*6 32000 2001800 400

610440

P6-313901250


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7=91.25 Example Problem 4 , ,,.

Using solution mde l to find the flow rate. iepdabGiven Data:Same as Example Problem1.

a In this example the separator pressure is held constant at 100 psi and is designated asnode 1. Ther6fore all pressure lossesfrom node 8 (~r) to node 1 (separator) are determined and then,subtracted from node 8.

Procedure:1.2.

,,Assume flow rates of: 200, 400, 600, 800, 1000, 1500 B/D.For each rate, start at ~ = 2200psiandsubtractp8-6+ AP6-3+APr ~-1. Thisinformation is noted in Table 1.25.

TABLE 1.25- ----PRESSURELOSSES FROM NODE 8 @r) TO NODE 1 (PSEP)

+--, :0 ? Frs1 -@is

II1 r II IP8-6 II 3

From HorizontalMultiphase FlowP6-3 1 P3-I

00I

2200 2000 200409 6 102200 1800 400 Ssf) 1390125 )


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.,, -8-., .1.26 Discussion of Exomple Problems 1.22 Through 1.25

It is important to notice that when starting at the reservoir (node 8), the slope of theresuIting system curve on the pressure-flow rate diagram at the solution node is zero or negative, hiscan be observed clearly in Figures5 through 8. This is expected since any system curve developed bystarting at ~r (regardless of the solution node) incfudes reservoir performance in the form of PI ~r IPR.,.A pressure-flow rate curve generated by starting at F actually displays the required pressure at thesolutl on node for the reservoir to produce the stated flow rate. For example, the vertical and IPRcurve shown on Fig. 7 showsthat if a flowing we Ilhead pressure of 100 psi couId somehow be created,the reservoir and wel I would produce 1100 B/D. .

In contrast, notice that when starting at the separator pressure (node l), the slope ofthe resuIting systems curve on the pressure-flow rate diagram at the solution node is zero or negative.This is again sl,ewn clearly in Figures 5 through 8. Theat the sepurator pressure displays the created pressure

pressure-flow rate curve generated by startingat the solution node for each flow rate. For

example, the flowline curve shown on Figure 7 shows that for a production rate of 1100 BOPD thecreated wel Ihead pressure is 300 psi. .-..

The total producing system wi II produce only where the created pressure at any nodek equaltothe required pressureat that node for the stated producing rate. This occurs where the


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,,,.-9- .1.3 CHANGES IN FLOW CONDUIT SIZE

1.31 IntroductionThus far the discussion has pertained to the rather simple system shewn in Fig. 4.

Notice on this system there is only one flow line size and one tubing size. Of course it is possibleand sometimes advantageous to change one of these pipe sizes in the middle of the string ~ Toevaluate a system of this nature, the solutlon node could be placed at the point where the p pe sizechanges.

1.32 Example Problem 5 - Tapered Tubing Strings5uppcxe in the previous example that for some reason it was necess~ry to set o liner

from near 3S00 through the producing zone at 5000 and this liner was of such ID that 2-3/8 tubingwas the largest size tubing that couId be installed. Let us investigate the possibleincreases by insta I I ing larger than 2-3/8 tubing above the liner from 3500 to theto Figure 10.

Given Data: Same as Example 1.

production ratesurface. Refer I

The solution node (node 5) selected to solve this probienl is located at the tubingtaper (Fig. 10). In this example the pressure drops must be added from node 1 to ncxk 5 andsubtracted from node 8 to node 5. In keeping with the same nomenclature as Fig. 3, we have


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., . . . -1o-TABLE 1.26(A)

PRESSURELOSSES FROM NODE 1 TO NODE 5 (EXAMPLE PROBLEM 5)

qi 2Mi 400600I

80010001500I

SEP

SEP100lo f-)lqo100100109

(2-7/81 tubing) IHoriz. Multi has= Flow

? TlVertical Mu Itiohase

3 P3-I 5 P5-3;i-140 40 500 360180 89 690 429 230 13~ 718 488

275 175 820 545 .170 550 . .

1

4f) 40180 80230 130275 175420 320

3. Determine the pressure

(3 ID tubing) \Vertical Mu tjphase Flow5 P5-3

~~f) 3n5475 335::; ~~o43f)78~ . 505.900 4$0

q____ I

losses from node 8 to node 5. For each rate start at i (-;) ;{); j,?< , ..,. (* .?]]-,...


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e4. Plot P5 vs. q from both step 2 and step 3 (Fig. 11). ,

5. The intersection the two performance curves ~t the taper connection predict aflow rate of about 1020 BOPD for 2.5 ID tubing and 1045 BOPD for 3 ID tubing. Remember for a2.0 ID tubing string the predicted rate was 900 BOPD. 2.fY IDtubing string the predicted rate was 900 BOPD. Notice the increase in rate from 2.0 ID to 2.5 IDis much more significant than the increase in rate from 2.5 ID to 3 ID. As pointed out previouslythis problem could have been solved by placing the solution node at any point in the system. However,this approach can simplify the procedure depending on the manner in which the curves or computerprograms avai fable are formated. This same procedure couId be used if a change in flow line con-figuration occurs at some point along1.4 THE FUNCTIONAL NODE

1.41 Introduction

the path of the horizonta I system.

In the previous discussion it has been assumed that no pressurediscontinuity existsacross the so ution node. However, in a total producing system there is usually at least one point ornode where this assumption is not true. When ais termed a functional node since the pressure

pressure differential exists across a node, that nodeflow rate response can be represented by some physical

or mathematical function. Figure 3 shows examples of some common system parameters which are ,. .,


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-12-That is, the pressure drop, AP, is proportiona I to the flow rate. In fact, there

are many equations avai labIe in the litemture to describe these pressure drops for common systemrestrictions. It is not the purpose of the paper to discuss the merit of the different equations butrather to show how to use them once the selection has been made, conside;i ng the entire producingsystem.

1.42 Surface Wellhead ChokeRefer to Figure 12 for a physical description of the wel

, The same nodes as set out in Figure 3 are maintained.with a surface choke installed.

Since the wel head choke is usuaI Iy placed at node 2, this wi 1I be the solution nodeselected to solve the problem. It is necessary to solve this problem in two parts. The first part ofthe solution is exactly the same as previously shown in Example 3. For the given data used in theprevious examples thethat the vertica I and(Pwh, Fig. 7) and the

resuIts of this analysis are shovn in Fig. 7. Inspection of Figures 12 and 7 showIPR performance curve actuul Iy represents the upstream pressure from node 2horizontal system performance curve actua IIy represents the downstream

pressure from node 2 ((PD5C, Fig. 7). Thus far, we have considered no pressure drop across the nodeand therefore the predicted rate is where upstream pressure equals the downstream pressure (Pwh =p~s~ .However, we know the wellhead choke wi 1I create a pressure drop across functional node


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is the sam~ as Figure 7 with 4Ps displayed. )These results are noted in Table 1.27(A).

TABLE 1.27(A)RESULTSOF EXAMPLE PROBLEM f6

I 1Ap = ~wh - ~sc q, B/DIflo 800200 69 f)300 560

1400 410

J,,

3. From step 2 plot the required AP vs. q as shown on Figure 14.

-13-.

4. Calculate the created pressure drop vs. flow. rate forchoke beans of interest.The equation used for these calculation sis:

P =LK q (from Gi lbert)z ------.,wh ,2 -A. - zPwh =R =

-

Flowing we lheadGLR, MCF,/STB

pressure, psi

Gross liquid rate, STB/D

., . . . . . .b -14-:


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.TABLE 1.27(B)

AP VS RATE FOR DIFFERENT CHOKE SIZES (PROBLEM 6)

= 2= JDs BOPD FromFig. 13 FromEa.2

242354457561

128140160180

370494617741

.35.28

.26.24

AP =~ -PD5CwhFig. i.3 Eclo 2

3005007WI900..

128160 237395553711

.54.41.36.35

199235353461200250

b,

Wh ~-From Ap = DscrwhscFromCi Fig. 13 ??0. 2500700 160200 274384 .58 114

, -1.5-, .


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.

The 6Ps calculated are unique to the example sys~emsince the downstreampressureswere calculated fortheexomple system. Notice that in each casea

check was made to ensure P s 0;7so that Gilberts equation wouldDSCpwhapply. If this is not the case a subcritical flowcalculate ~? across the choke.

5. From the tables generated, plot the choke bean

Fig. 15.

equation must be used to

performance as shown on

6. Overlay t~e results shown on Figure 14 and Figure 15 (Fig. 16).Figure 16 c@Ays the total system performance for different wel lhead choke sizes.

The system performance curve shows the required AP for various flow rates considering the entiresystem from reservoir to separator. The choke performance curves show the created AP forvarious flow rates considering choke performance for different choke sizes. The intersection pointsof the created and required APs repr~sent the possible solutions. For example the rate will dropfrom 900 BOPD to715 BOPD with the installation of a 24/64 welihead choke.

Figure 17 showsanother presentation that is often used to evaluate wellhead chokes,The.presentation shows the entire systemperformance which sometimes is advantageous. The same


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1.5 Summary and ConclusionsA new Cnodal) system has

evaluate a complete producingbeen presented in order to effectivelysystem. All of the components in the

well, starting from the static pressure (~r) and ending at the sepa-rator, are considered. This includes flow through ~he porous medium,flow across the perforations and completion, flow up the tubi~gstring with passage through a possible down-hole restrict-on andsafety valve, flow in the horizontal flow line with passage througha surface choke and on to the separator.

Various positions and/or components are selected as nodes andthe pressure losses are converged on that point from both directions.Nodes can be effectively selected to better show the effect of in-flow ability, perforations, restrictions, Safety valves, surfacechokes, tubing strings, flowlines and separator pressures.

The appropriate multiphase flow correlations and equations forrestrictions, chokes, etc. must be incorporated in the solution.

In conclusion, an effective means of analyzing an existing well,

. ,. .


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s .

I

,.4 .

zUJ1-U)>(nc )z

I

Uto

mC@0

@lA


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-

ii? = (~sv-pDSC)

t?w;PihBOTTOM HOLERESTRICTION, /DF1

API = Pr - P~fs = LOSS IhlPOROUSMEDIUMAP* = P~f~-P~f = LOSS ACROSSCOMPLETIONAP~ = pu~- po~ =

IIRESTRICTIONAl?$ 01 II= p~v po~v = SAFETY VALVE

APiJ = p~h- ~o~~ = II SURFACE CHOKE&p6 = pose-p~~p = IN FLOWLINEAPT = P~f -P~h = TOTAL LOSS IN TUBINGAP~ = Pwh- p~ep = II FLOWLINE

...

.-. FIG. 2 POSSIBLE PRESSURE LOS-SES IN COMPLETE...

1


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v)wn0zu)30aa>L0z0


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nHORIZONTAL FLOWLINE

bp6.3=p~h)

*

NODE LOCATION@ SEPARATOR@ p~h@) Pwf@ F,

. .

.

.

.-. .. . . .FIG. 4 NOD E-S FOR SIMPLE- PRODUCING S-YSTE-iih

.,, .


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HORIZONTAL FLOWLINE ~~

LINER

I+++++t

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12-7/8OR 3TUBING

2-318TUBINGdP

NODE LOCATION%ep

@ TAPER CONNECTION

5

@

1 I. -. ..... .... .. . . ..- .. . ... ....FIG. 10 TAPERED STRINGS


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. 2500.P

2000

500

0

.

-

TAPERED STRING5000-3500 2 TUBING3500- o 2-7/8 TUBING3500- o 3 TUBING

TUBING~ 2- 7/8m

1020 BOPD -Ji 1045 BOP )I I0 500 1000 1500

q., BOPDFIG. ;l- TAPERED STRING SOLUTION (EXAMP-iE NO. 5)

:


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,., ..- r

LLl

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cLLlIC2

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+

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0=

Ptf Psl

mo0

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,-00 -P00 cm00

410BOPD +- %-q = 560 04BOPDq = 690BOPD @q = 800BOPD

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500

400

300

200

00

. ,.

.

9

t

.

.

.v. ,

q.= 900 BOPD AT AP = O

1000qo BOPD

1500

., . ..-. -. .. . .. . ---- -. .. .... . .+- . ... ---- . . - .FIG. 14 TOTAL SYSTEMS PERFORMANCE CURVE FOR


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400

300

200

~100

0

28/64

*

500 1000 1500oqo, BOPD.. ..... .. . .-..-----.FIG. .15 CHOKE BEAN PERFORMANCE

,


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500

400

300

200

loo

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.- .,.,D28/64

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o 500 1000. q *, BOPD.-.FIG. 16 SYSTEMS PERFORMANCE

i500. .

.-=- - .-.FOR VARIOUS


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2000

-CnQwmmU&QD

1500

1000

50C

o I I Io 500 1000 1500q., BOPD.-~lGe 17- &jR~AcE - CHOKE ~vA~j~TloN

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