Fluid Conversions in Production Log Interpretation

 

Fluid Conversions in Production Log Interpretation

1974 EDITION

DOCUMENT

 

Schlumberger Fluid Conversions in Production Log Interpretation

0 Copyright 1974 SCHUJ~ERGER LIMITED All Rights Reserved Printed in U.S.A.

 

CONTENTS

SumJEcI

SEcTlOt

iii

A

Symbols, Subscripts , an d Abbreviations

B

INTRODUCTION

”

1

SOURCES OF DATA

1

2

A. Client-Supplied 1. surface Flow Information Rates 2. Downhole Fluid Parameters 3. Mechanical Feanues 4. Intrinsic Fluid Parameters 5. Gened

1 1 1 1 1 2

B. Values Calculated Before Logging

2

C. Log-Derived Data 1. Fluid Phases 2. Flow Regimes 3. The Conversion Problem

2 2 2 2

FLUID CONVERSIONS A. Gas Formarion Volume Factor, B, B. Gas Deviation

Factor, z

3 3 4

C. Gas Deosiy Graviy, yg (air = 1.0) 1. Spec&c 2. Density, & ( gm/cc or lb/a ft )

6 6 6

D. Gas Viiosiy,

6

R

E. Producing Gas-Oil Ratio, R

7

F. Bubble-Point

Pressure, ph

7

G. Solution Gas-Oil Ratio, R.

8

I. Oil Formation V olume Factor, B,

9 10

J. Oil Density, P.

12

K

13

H

Free Producing G as-Oil Ratio, R,

Oil Viscosiy,

p,

L. Solution Gas-Water Ratio, R,,

14

M. Warer Densicy, pw

15

N. Water Formation Volume Factor, B,

15

0. Water Viscosity, lh

16 16

P. Viscosity of Miies

 

SUEJECT

SECTtON

3

4

CONVERSION

(metric unis)

18

B. Example 2, Gas Production

(English)

18

C. Example 3, Gas (metric)

20

D. Enample 4, Gas and Water (English)

22

E. Example 5, Oil and Gas, no free gas @q$ish) F. Example 6, Oi Oill and Gas, free gas present (English)

24 26

G. Example 7, Oil, Gas, and Water (metric)

28

H. Example 8, Oil, Gas, and Wafer (English)

30

NOMOGRAPHS

33

A. Fg-1

l/B,

35

B. Fg-2

T,

C. Fg-3

Gas compressibil compressibiliy iy

D. Fg-4

l/B,

E. Fg-5

pFmf from yg and l/B,

39

F. Fg-6

h from p,,= and Tti

40

G. Fgo-I

pb 01 p+ from TM, Rsb a R,, R.. 01 ya. and I+ or Rgb from pa, T-r, P..< or yo, and yg

41

H. Fgo-2 I. Fgo-3

k from pwr/pn Bob from Rsh yp, yo, and TM

42 43

B, ffrom rom yg, R, and T,,

44

B, from pob or Q ( pwi - pi,), and Bob

45

L. Fgod M Fgo-7

Pownrrom yp, R,, Yose r Pa and Bo pob and yosc (‘API) from Tti and Rsb

46

N. Fgw-1

R’,, and FBo from be

48

0. Fw-1

pwr 01: .. from Csac,, pwi or p.., and TM or T,

49

P. Fw-2

pwM from Csaol and T,t

50

CHARTS AND

K

R

17

EXAMPLES

A. Example 1, Water Injection

J. Fgo-4

5

Pm3

Fgo-5

AUXILIARY

from yg, p-r, and TW (typica (typicall gasa)

36

and ppe from yg factor, z, from T,. and pur

from pwf, TM, and z

Twr, and total solids in brine

37 38

41

51

Pt CHARTS

A. A-1

Water analysis by Re

52

B. A-2

Resistivicy of NaCl Solutions

53

C. A-3

v, “S qr, no tool

54

D. A4

V, YS gr, 1 1*/16-in.

E. A-5

vr ys qr, 2 l/S-in.

References

tool

tool

55 56 57

 

SYMBOLS, SUBSCRIPTS, AND ABBREVIATIONS FOR PRODUCTION LOGGING

SYMBOLS

A

area

AC,

amplit%

B

formation volume factot

c

compressibility

C

concentration

G

specific heat of well fluid

G

fluid-flow-v&&y

Gradiomanome tez reading

correction factor

d F

Gradiomanometer friction factor

FBC

brine correction factor (for solubiliy

h

height

h

fluid head

h

vertical dice

K

Gradimnanometer kinetic factor

k

gas soluhility correction factor; Chart Fgo-2

m

of gas)

t

time

V

vohlme

v

velociy

7

average (su@cial)

W

weight

I

holdup

z

gas-compressibiliy

Y

gravity: y. is gravity of oil in ‘API; yp is spechic gravity of gas, referred to air (= 1.0)

P

density

I

viscosiy

vekiy

deviation factor

SUI~~~RIPTS

between RTS detectors

air

ai+

b

bubble point

f

fluid

mass

Fg

free gas

NBe

Reynolds Number

g

gas

P

p*eSlxe

Gr

Gradiomanometer

9

flow tare

h

hole

R

producing gas-oil ratio (GOR) : R = q&q-

Iiq

liquid

R’

producing ratio of free gas plus gas dissolved in oil,tooil: R’= (49.e-qaacL)/qoae

m

mixture

N&l

salt (sodium chloride)

free gas-water ratio

0

oil

solution gas-water rati ratioo in fresh water

K

pseudo-critical (temperature or pressure of gas)

temperaNte

P*

pseudo-reduced. p,,r = pwi/pw, T,r = T&T,

iii

 

slippage surface solution gas solution gas at bubble po int standard conditions solurion gas/water total water

well flowing well static threshold speed, Flowmeter spinner

ABBREVIATIONS

cf/D or aft/D

cubic feet per day

FBS

FulIboreSpinner

Flowmeter

Fw

fomution

GOR

gas-oil ratio (R )

GWR Mcf/D

gas-water ratio thousands cubic feet per day

MMscf/D

millions standard cubic feet per day

PFM

Packer Flowmeter

PI

productivity

PSia

lb per sq in. (absolute)

Psk

lb pa s q in. (gage)

PVT

pI~SW-VOl~~-teElpW-LU~

rPs

revolutions per second

RTS

Radioactive Tracer Survey

SC

standard conditions 14.7 psia, 60°F or 520-R I 1.033 k&q. cm., 15.56”C ox 288.56”K

volume factor

index

B/D

bards pa day

BHYP

bottom-hole pressure

BHT

bottom-hole temperature

BOPD

barrels oil per day

BWPD

barrels water per day

scf

cubic feet of gas, a SC

cf/B

cubic feet pa barrel

TC

time consrant

 

INTRODUCTION Ii

Production-Log interpret Production-Log interpretation ation consists of determining fluid flow rates by means of logs. The fluids are oil, g as, and water, all of which vary considerably in their properties. Each typ of Auid is variousl variouslyy atfeaed by changes in temperature and pressure. Moreover, the multi multiphase phase case (two or three fluids in contan) is complicate d by variations in solubility, particularly of gas in liquid. These solubility changes are also greatly influenced by temperature and pressure. The Production Logging Document’ shows how fflow low rates at existi existing ng pressures and temperatures are derived from Production Logs. For these data m be put to use, rhe downhole quantities quantities must be converred m surface condi conditions, tions, or vice vessa (as for injecti injection on wells). There is also the common production problem which makes use of both downhole and surface flow measurements as inputs; all the fluids must be converted ro rbe same conditions prior to the analysi analysis. s. This manual is devoted to providing solutions m these 0uidanversion problems.

 

SOURCES OF DATA I1 A.

CLIENT-SUPPLIED

INFORMATION

it will be seen rhar reliable fluid flow analyses, parriculaxly in multiphasic situations, require m ore input data rhan can be deri derived ved from logs alone. Much data must be obtained in advance of rhe job. The clienr’s representatives should rherefore consult with the Schlumberger engineer well ahead of the logging logging operation, in or order der m make plans an andd ro specify the rypes of da ta considere d n ecessary or desirable for Production Log interpretation. The information information ro be obtained from the well operator falls into three priority classifications:

l

T,

. well flowing rempaarwe Thermometer)

l

pp.

gas density at SC or use local knowledge)

l

pm.

oil density at SC or use local knowledg e)

l

pwec

water density at SC 01 derive from Rw)

1. Required.

3. Mechanical

2. Needed, but can be approximated from lcal knowledge, indirecr masuremems, and charrs, if direct measuremenrs cannor be provided.

l

Permanent

l

Casing

sizes

and weights

l

Tubing

sizes

and weighrs.

3. Infrequend Infrequendyy

l

Minimum

l

Depth,

available available,, but desirable for the mosr reliable interpretations.

Even rhough charts, nomogmphs, and average-case values are provided to all allow ow estimarions or approximati approximations ons of non-avail non-available able par parameter ameter m-emenrs, the use of rhae shortcut methods is discouraged except as a last ~esor ~esorr. r. Production-loggin g analyses are capable of high r&ability, provided the measu~emenrs on which they are based are accurat accuratee and complete. Pre-job planning planning should rhus provide for ob obtaining taining the following data (lisrings shown in boldface are essential) : 1. Surface qme

l

Flow

Rates at Standard

. . . oil flow

rate

. q,

. . . gas flow

.

. . . water

q,,

(stock-rank

rate,

flow

Conditions* oil)

M GOR, or GWR

rate,

or wafer cur

2. ments Downhole Fluidbut Parameters desired, approximations(Dirscf canmeasurebe obtained es shown.)

(or, m-e

with

Features depth

datum

restriction

of the

in tubing

size, and orientation

well.

string.‘ string.‘**

of perforations .

Production method being used. . Depth and narure narure of packers, plugs, or any other downhole mechanical devices. l

4.

Intrinsic Fluid Parameters (Sometimes laboratory data are avrdlable; this is preferable.)

. pob

density of oil ac bubble point pressure, pb

. pb

bubble-point bubble-point p~e.wre

l

Rsb

solution GOR at pb

. Bob

oil Formation Volume Factor* * * at pb

l

oil compressib ility

co

. R.

(obtained from p,,b)

. ..solurionGOR ..solurionGOR

English 14.7 pia and 60°F M 520’R Metric.. .I.033 kg/sqaand 15.56”Cor288.56” 15.56”Cor288.56”K K

.weU flowing flowing Manomera

l pwi

pressure (or, m easure with 01 Amerada Gage)

 

l

Bo

l

pm

l

Tw

.7.

. ..oilFvF

gas pseudo-crir pseudo-crirical ical pressure . gas pseudo&&xl

temperature

. . gas deviation factor

. Bs

gas formation volume factor

l R, . B,

. solution gas-water ratio water formation volume factor

and

. a chemical analysi analysiss of gas

since measuremerlu al-e not always available, some of values can be estimated from “average-pro “average-property” perty” nomographs and charts in Section 4. 5. Geneml . All open-hole

logs.

l

Driller’s, geologist’s strip logs.

l

Core descriptio descriptions. ns.

l

Drillstem Drillst em and pmducrion test results.

. Knowledge of flow regi regime me ro be expected: singlephase, bubble, slug, froth, emulsion, or mist flow. l

Any available mechanical-con mechanical-condition dition logs: Cement Band, Variable Density, Casing Collar, Pipe Inspetion, etc.

B. VALUES CALCULATED BEFORE LOGGING - NEEDED FOR PROPER SELECTION OF PRODUCTION LOGGING TOOLS These data are determined beforehand from flow mres and ocher surface surface measwemencs, using techniques and charrs given in this book They include flow rare, density, and viscosiq of oil, gzs, and water, ali ac well-flowing conditions. The symbols are shown in Table 1-L

TABLE 1-I

C. LOG-DERIVED DATA Production Logs’ provide measurement.s or derivat derivations ions of average fluid density, temperature, pressure, and flow rate

at any or zl zlll points in a well. LJltimately, the imerpreta6on of these logs consists of identifying and drawing logical inferences from my de-es of measured flow rates from those those assumed to exist. Such assumprions are ideal ideally ly based on suface flow rats converted to downhole conditions. (There are exceptions, such as dump-flood wells in which there is no surface fluid flow.) The selection selection of the bar logging mire to meet a given set of condition s should be made on the basis of ail available fore alowledge. 1. Fluid Phases It iiss important to know before hand w hether the fluid flow in the producing or injecting zone5 is monoph asic (liquid), m onophas ic (gas), biphasic (liquid-liquid), biphasic (liquid-gas), or aiphasic. It must be rremembered emembered char, except in wells having very high Eow rates, water is nearly always a factor, even tbougb no uuter is being produced. Also, wells wh ich flow g as, oil, and water at the surface will be flowing oil and water down hole, but may or may not be flowing gas. If the well flowing pressure ( pwi) is above bubble-point pressure ar existing conditions, the gas will still be in solution in the producing zone. The flow w ill be biphasic, facilitating the imerprerario n. 2. Flow Regimes Foreknowledge of the fflow low regimes to be encomxered in multiphase flows is also a factor in the selection of Prcducrion Log types. Chapter 1, Ref. 1, conrains a discussion of the interacting variables that govan Bow regimes. 3. The Conversion

Problem

Tne conversion of surface to downhole flow rates (or vice vasa) must be propaIy related to pressures, wmperatwes, and solubilities. The following section explains the sreps involved in each type of conversion Then, in Section 3, examples and flow &arcs are given for each of the general type of fluid conv ersion problems . Section 4 is made up of the chars and nomographs used in in these ope operations, rations, and Section 5 is a collection of supplemen tary charts which have ken found useful in Production Log interpretation work

2

 

F,“id con*enionr,*

FLUID CONVERSIONS )2 A. GAS FORMATION

Using Fig . 2-l (lines drawn in) this is solved as follows:

VOLUME FACTOR, 6.

Throughout the indust industry, ry, gas volumes are referred to a standard temperature of GOoF and a pressure of one atmosphere (14.7 psia). These conditio conditions ns are refened to as “standard con ditions”, abbreviated S C. Standard cub ic feei’, the English English unit of volume measured under these conditio conditions, ns, is sometimes sometimes abbreviated scf. Bh is the ratio of the volume occupied by a certain weight of gas at a given rempesatwe and prffsur prffsuree to the volume occupied by the same weight of gas at sstandard tandard conditions: B = gas volume at a given temp. and press. (V,) 8 gas volume at standard con conditions ditions (V,)

1. Select the group of temperatute lines for yp = 0.70. Entering on the abscis sa at 2,000 psia, go vertically to the 200’F line, then move left m the ordinate value of 125 for l/B,.

EXAMPLE-

(2-l)

In field usage, the gas formation volume factor is viewed as a function of flow rafes, no r volumes. Thus t x definition used is: B, is the ratio o f the flow rate at we&flowing temperatwe and pr pressure essure to the flflow ow rate at standard cond&ns: B, = gas flow rate ar weUlowing temp. and press. (qgaf) gas flow rate at standard conditi conditions ons (qgsC)

(2-Z)

As this number m ay be quite small, its reciprocal is commody used There are are &se general ways to compute Bz, or II&. &. In the order presented, they are increasingly accurare and (inevitably) increasingly lengthy. The fkst fkst (approximate) method assumes a typical gas. Referring to Fig. 2-1 (reproduced from Chart Fg-1 in SKtion 4), l/B, is found by entering the well-flowing pr-e, pat, on the abscissa, going vertically co both the proper gas speciiic gravity, y=, and the well-flowing temperatu re, TM , and then left to rhe value of l/B, on the ordiite. This usually requires a double level of iinterpolation: nterpolation: first between temperature lines, and second, between gas speci6c gravit gravities. ies.

Interpol Interpolation ation needed.

Find the volume occupied by 600 scf of gas where: yp = 0.74, Tm = 175’F, and pai = 2,000 psi psia. a. Using Fig. 2-1 (lies follows:

not drawn in) this is solved as

1 Select the group of tem perarure lines for l(g = 0.70. Entering on the abscissa ar 2,000 psia, go vertically to the 175°F position, then move left to the ordinate value of 133 for l&. 2 Selm the group of temperacure lines for yp = 0.80. Entering on the abscissa at 2,000 psia, go vertically to rhc 175°F position, then move lleft eft to the ordinate value of 138 for l/B.. 3 The interpolation as follows: YP 0.70 0.74 0.80

between the l/B,

1 BZ 133 X 138

valuff is solved

EXAMPLE - No interpol interpolation ation needed. Find the volume occupied by 400 scf of gas when yg = 0.70, T,, = 200°F, and pti = 2,000 psia.

El 1 -=133+0.4X %

(138-133)

= 135

 

ENGLISH Presum, psi*

Gas gravity,

yg

(air

= 1.0)

critical pressure, pnC, and the pseudo-critical pseudo-critical emperatwe, T,,., are available. This method should be used when the produced gas contains sigrG.fi~t amounts of some non-typical non-typical component: helium, carbon dioxide, h ydrogen sulfide, nitrogen, etc. The method may be used with typical gases by usi using ng “average” values of pPCand pPCand T,, as derived from Fig. 2-2 (Chart Fg-2), but this is just a laborious w ay m obtain the information already shown in Fig. 2-I. 6. TH E GAS DEVIATION

FACTOR, I

The gas deviation facmr (z) represents the degree m which a natural gas deviates from an iideal deal gas under the effeces of pressure and remperarure. The ideal gas law stares: 11

12

where pressure is in absol absolute ute units (psia, kg/q

cm, atmo-

4.

1 V,, g=135=---V,,,

_

600scf V,

600 135

V ,,t=-=4.44cuft The second (exact) method is used when the pseudo-

spheres), volume is in consi consistent stent units (cu fr, cu m), and temper%ure is in absolute units (“R, OK). Incorporati Incorporating ng the gas deviation factor, the relationship for namral gases becomes: ~ .cvezv., _ &kfVM T8C 2Tn.i (Z-3)

 

or, related ro the reciprocal of the gas formation factor: &=$t

=(+)(2(E)

volume

(2-4)

Hence, l/B, can be deermined only as accurately as z is known. Fig. 2-3 (Chart Fg-3) is the chart for determining the gas deviation factor, z., as a function of normalized (pseudoreduced) pressure and remperarure. This nnormalizat ormalization ion is accomplished by comparing the well-fl well-flowing owing pre~ure and rernperarure to rhe pseudo-crit pseudo-critical ical pressure (p, (p,)) and remperamre (T,) ; these are laboratory-derive d character istics of a given natural gas. The pseudo-reduced pressure ( pPr) is the rario of the well-flowing pressure (p-r) ro rhe pseudocritical pressu re (p,) : pur = pri/pPc Likewise, T,, = T.&T,, Absolute pressures and rernperarures rnusr be used. The pseudo-reduced pressure and temperature are, of course, dimension less. The gas deviation factor, z., is obtained from Fig. 2-3, by entering ppr on rhe abscissa, going vezrically to T,,, then horizaxally to z on the ordinate.

Fig. 2-4 -

(Chart

Fg4):

the equation:

Nomograph

for solsing

EXAMPLE Find the gas deviation factor, z, and rhe reciprocal of rhe gas formation volume factor, l/B,, for the folIowing condipKf = 2,000 psia ppe =

65Opsia

T,, =

ZOOOF r 660”R

T,

410”R

=

1. per = 2,000 psia psia /650 psia = 3.07 2. T,, = 660°R/4100R

= 1.61

3. From Fig. 2-3, using

Pr = 3.07 and T,. = 1.61, z =

0.828

4. Using the nomograph in Fig. 2-4 (Chart obrain the solution as follows:

Fg4),

1. Determine the swing point on Line A by drawing a line from 2,000 psia ro 200OF. 2. Determine (l/B,) by drawing a line from the swing point ro a 5 of 0.82 8. l/B,= 135

The third merhod (also exact) is used when a complete ga analysi analysiss has been run. If the mole fractions of the various gas componenrs are known, pw and T,, can be computed as shown in Table 2-l. With these two values computed, rhe derivations of z and l/B, are identical to those in the second “mhcd.

5

 

TABLE 2-l Computation

Carbon dioxide Methane Ethone Propane lsobutane

l---t

of pseudo-critical Crit temp, OR

MO1

Lb/

Y

wt

mole

Tc

=YTO 10.

0.0040 0.9432 0.0390 0.0117 0.0003

44 16 30 44 58 58

0.17 15.09 1.17 0.51 0.04

548 343 550 666 735

2.1 323.5 21.4 7.8 0.6 1.0 356.4

T

point Crit pressure, psia PC 1,073 673 708 617 528 551

=YPr PPC 4.3 634.7 27.6 7.2 0.4 0.7 674.9

C. GAS DENSITY 1. Gas specific gravity, yg, is widely used in the oil industry to characterize natural ga ses. Gas specilk gravity is d&ed as the ratio of the density of gas to

7, (Air=l.O) 0-l

PI

the den&y of air, both at standard conditions. Y==$=&

(2-5)

The we ight of any volume of a gas can be determined by multiplying rhe volume of gas times yz times p&i=. The density of air at standard conditions is 0.001223 p/cc or 0.0762 lb/cu ft 2. The density of gas at any temperature and pressure can be found from the gas formation volume factor, 4: 1 =__

BZ

Pti Pm.

~~=~,(.001223)) ~~=~,(.001223

X$

(g&cc)

(2-6)

EXAMPLE

Solution: 1. From Fig. 2-1, and as shown in a previous example, l/B,= 125.

Find the weight of 500 scf of gas with yg= 0.55

2.

This is solved as follows: Weight = V, X pair X Yp Weight = 500 cu ft X 0.0762 lb/cu ft X 0.55 = 20.95

Ibs

EXAMPLE Find the densiy of a gas at standard conditions when: yp = 0.70 Solution: Ps=YsX

Pair

-cc1

p

pwi (from paragraph C-2, above). B. P, 3. p,=y;xO.O01223 X (l/F&). pti=0.7 x O.O01223gm /ccX125 = O.l07gm/cc 0IY:

2. Using the nomograph

in Fig. 2-5 (Chati Fg-5), comrmc c a lime from yp = 0.7 to (l/B,) = 125 to obtaix

pti

pp = 0.7 x 0.001223 0.001223g&cc g&cc = 0.000856 gm/cc

= 0.11 gm/cc

EXAMPLE

D. GAS VISCOSITY

Find the density of a gas when:

At elevate d temperam es and low pressures , low-gravity gases closely resemble “perfect” gas in their behavior, while at low cem paatwes and high pressures, the heavier gases resemble liquids.

yg pm Td

=0.70 = 2,000 psia =ZOO’F

 

To clarify:

R = qdqosc R’ = (qmc - q.m R.,) /qme where R,, is the solution gas-wa ter rario. EXAMPLE Find the producing gas-oil ratio when: 90s = 300 B/D qpse= qpse = 150 Mcf/D (Note:

1 Mcf/D

is 1,000 cf/D.)

This is solved simply: R =

150,000 cf/D 300 B/D

=

500 cf/B

EXAMPLE Find the gas flow rare at standard conditions conditions when: qmc = 250 B/D R = 400 cf/B Solution: ‘31

ssc=RXq,., %=

= 400 X 250 = 100,000 cf/D

100 &F/D

F. B UBBLE-POINT PRESSURE, pb The bubble-point bubble-point pressure is tint pressure at which a given volume of gas, as measured ar srandard srandard conditions, will be completely soluble in a given volume of oil, as measured ar standard (wxk-tank) conditions conditions.. It iiss dependent on the oil gravity, the gas gravity, the rempe rarure, and the amour The charts charts in Fi Fig. g. 2-6 (Chart Fg-6) give gas viscosity as a function of temperature and pressure, for gases of five dtierem gcaviries. Nore &at aboveabout above about 1,500 psi, an increw in temperature decrerzses the gas viscosity, while below that poim incmsing the temperature bcreares the viscosity. viscosity. E. PRODUCING GAS-OIL RATIO, R R is the ratio of the production rate of gas at srandard conditions to the production fate of oil ar standard (smcktank) condirions. In equation form:

R is commooly referred ro as the gas gas-oil -oil ratio (GOR) In English units ir is expresse d in cf/B -cubic feet (of gas) pa barrel (of oil). In metri metricc u&s it is expressed in cu m/mm-cubic meters (of gas) per cubic meter (of oil). When gas is produced along with both oif and war warqq an ambiguity exists in the definition of R. The gas in the

Pb(P.f)

numeram of the fraction expressed by R (q&q& includes all all gas presenr in the produced fluids, free gas, solution gas in the oil, and solution gas in the water. As will be explained, the gas dissolved in water under well flowing conditions is dealt with separately, for practicality. Thus rhere is a need ffor or another symbol for GO R which rakes into account only the free gas and the solution gas in oil. This symbol is R’.

 

of gas. In English units it is expresse d in psia (pounds per square in), while cbe metric units are kg/sq cm (kilograms per square centimeter). The bubble-point pressure may have been measured by the cusfomet in the laborator laboratory. y. If not, * for “average” conditions may be determined by the nomogmph in Fig. 2-7 (Chart Fgo-1 ). As the nomograph shows, there is a practical limit on pb of approximate ly 7,000 psia Remember that in field asage the bubble-point pressure is a function of flow rates, not volumes. That is, the rer* is used as though it were wri written: tten: “The bubble-point pressure is chat pressure at which a given flow rare of gas will be completely soluble in a given flow rate of oil, both flow rates measured at standard conditions.” PracticaIly speaking , as the saturated oil flows up the hole, the bubble-point bubble-point pressure is char pressure awxiat awxiated ed with a certain depth, where gas bubbles are just starting to come out of solution. EXAMPLE-Fig.

2-7

Fiid the bubble-point conditions: q,,

pressure , pb, under the foIlowing

= 600 B/D

1. R=

The value of Rsb is also dependen dependentt upon yor ye, and T. It can be found with Chart Fgo-1 (Fig. 2-7, the same one as was wed co solve for p,,), providing pb is known or assumed. R.b may have been measured in the labor laboratory. atory. Such measurements should be used when available. Ocherwise, proceed as follows: EXAMPLE Find hb, the solution gas-oil ratio ratio at bubble-point pressure, under the following conditions: * T,f yg

=9OOpsia = 140°F =0.7 = 40" API

1. From right to left: draw a line throug throughh y0 = 40” and yg = 0.7 ro locate a point on Line B.

ir as follows:

*40,000~f~~ 600 B/D

Evidently, the definitions of pb and Rat, are mutually interdependent. If the quanriry interdependent. quanriry of dissolved gas is iincreased, ncreased, then Rsb has likewise increased, and a higher value of pa is required to hold the increased gas quantity in solurion.

Yo This is solved using Fig. 2-7 (lines not drawn in), as follows:

F f ==**OMcfD 180°F yg =0.75 Yo =4O”API Tbesohxion

pressure is bubble-poin bubble-pointt pressure, pb. for rhe specified condirions. R., the solution g as-oil ratio at bubble-poin t pressure , may be choughc of as an expression expression of the solubilitg solubilitg of gas in oil at that pressure.

2. From tit tit pains draw a line throug throughh pb = 900 psia to Line A.

=400cf/B

3. F rom there, draw a line m T,, = 14O’F.

2. Take I(, = R, that is, the solution gas-oil ratio at bubble-point pressure is defined as being equal ro rhe producing gas-oil ratio. The assumption is implicit that, for all practical pupxes, all the dissolved gas has come out of solurion when standard conditions are reached. 3. On the nomogmph (Fig. 2-7 ) , draw a line fmm the cempemture (T,.+ = 18O0F) through the ssolution olution gas-oil ratio at bubble-poin r prffsure (I& = 400 cf/B ) to Point a on Line A. 4. Draw a line from the gas gravity (vp = 0.75) through the oil gravity (7. = 40” API) to Point b on Line B. 5. Connect Points Points a and b aand nd read the answer: b

=

4. Read rhe answer: Rab = 220 cf/B Recalling chat bubble-paint

pressure was d&ed

in

terms of a given amount of available gas, the solution gas-oil ratio ar a pressure greater than bubbk-+rzt @SZW~ wiU be equal rroo tthe he solution solution gas-oil ratio ratio at bubble-poim bubble-poim pressure; or, R, = Rsb when plri > pb. It takes just so much pressure to dissolve the given volume of gas, and if pressures go higher WJ more gas iiss diss dissolved, olved, because all of the available gas is in solution. This is a case of ~rz&rsutur~~ted oil; it could dissolve more gas if mope gas were available. available. Fig. 2-8 is a graphical representati representation on of this case. In the shaded area where P,vr > pt,, % is consram and equal to RsbThe solution gas-oil ra ratio tio at a pressure lerr tbm bubblepoint pressure will be equaI co some fraction of rbe soluti solution on

1,560

psia.

G. SOLUTION GAS-OIL RATIO, R.

gas-oil ratio at bubble-point pressure; or R. < Rsb when put < pl,. There is not enough pressure to force the amount of gas that is present into solurion. This is a case of saturu#ed

R. is the ratio of the volume of dissolved gas at a specified pressure to tthe he volume of oil in which it is dissolved , both volumes convened to st standard andard condi conditions. tions. At any given pxssue, there iiss a maximum limit on the amount of gas which will dissolve in a certain quan tiy of oil; hence a maximum possible value of R,. This ratio is termed Rsb, and this

oil, and is accompanied by free gas. Fig. 2-8 represents this case, too. In the non-shaded area area where pti < b, the solusolution p-oil ratio is some fraction of the solution gas-o il ratio at bubble-point bubble-point pressure. As a rough approximation: approximation: R, = Rsb X (p,*/p,, (p,*/p,,)) A more accurate accurate solution for R. is through the relation, relation, R, = k hb, and Fig. 2-9 (Chart

8

 

Bubble

s r zl

Point

P-i T, qosc ti

Pressure

Yo

= 4,000 psia =170°F =3OOB/D = 120 Mcf/D = 40” API

yp = 0.70 The so lution is as follows: 1. R =

12;$O;zD

= 400 cf/B

2. Setri Setring ng Rab = R, and using the nomograph in Fig. 2.7, p,, = 1,600 ps psia. ia.

3

3. ~>phhence%=R.b=R=40Qd/B

8

The oil oil would flow up the casing/tubing casing/tubing wirh no free gas, until the pwssure decreased to less &an 1,600 psia; then gas bubbles would would begin to form Fig. 2-8 -

Solubil~ty

EXAMPLE -Saturated -Saturated

of gas in oil (gtwraz).

Fii dirions:

Oil and Free Gas

the solution gas-oil ratio under the following

con-

pti = 2,880 psia T,f = 145’F q0.c = 100 B/D ti = 140 Mcf/D Yo = 50” API .,= =0.60 The solution is: 1. R =

140,000 cm = 1,400 cf/B lOOB/D 2. Setting Setting &,, = R, and using the nomograph in Fig. 2-7, pb = 3,600 psia 3. Rough solution: R, = Rsb x (p&d R. = 1,400 x (2,880/3,600)

Pwf /Pb Fig. 2-9-

(Chart Fgo-2): ‘1”

us a fmctition

of

PldPa.

= 1,120 cf/B

4. Accurate solution: p,i/pb = 2,880/3,600 = 0.8 FromFig. 2-9: k = 0.83

~=kXR,~=0.83X1,400=1,160cf/B

Fgo-2) which relates k ro pwJpb These p&m marized in Table 2-K

TABLE 2-11

H. FREE PRODUCING GAS-OIL RATIO, RF~ are sum-

In the previous discussion of the solurion gas-oil ratio, it was shown that in the case of sartxared oil there is usually an occurrence of free g&s. The free producing gas-oil ratio, which occurs only with samrated oils, is equal m the producing gas-oil ratio (R) minus the solution gas -oil ratio (R,),

or, in equation form: RF~=R-R~,oIR=R~+RF. EXAMPLE Find the free producing the following conditions: EXAMPLE - Undersotumted Oil Find the solution gas-oil ratio under the following dirions:

gas-oil ratio, if it exists, under

pai =217kg/sqcm T,, = 66°C qm= 150cum/D

con-

 

qgse= 25,500 cum/D qgse= y. = 0.88 g&c yg =0.70 The solution is as follows: 1.

R=

25,500cumD

(2-S)

= 170cum/cum 150~11 m/D 2. Setting Setting Rsb = R, and using the nomogmph in Fig. Z-7, pa = 310 kg/q cm.

3. p,Jpb = 217/310 = 0.7, and from Fig. 22-9, 9, k = 0.77. Then R, = k x Rsb = 0.77 x 170 = 131 al m/cu m. 4. R > R,, so free gas exists. 5. RF= = R - R. = 170-131 = 39 cu m/w

m

I. OIL FORMATION VOLUME F ACTOR, B. B, is the ratio of the volume of a given mass of oil w ith its dissolved gas, at at a given pressure and remperanue, CO he volume of the same mass of oil at standard (stock-tank) con-

 

B,=-

VOWf V0.C

(2-9)

Compres sion Compression of liquid all gas is is in so lution

As in the case of gas formation volume factor, the fezm in oil-field usage is usually associated with flow rates, so that: B.=E

or rhe nomograph shown in Fig. Z-11 (Chart Fgo4), both based on “average” oils and gases. Char Fgo4 incorporates borh English and metric units. EXAMPLE - Chart Solution Find Bob under the fol following lowing

condirions:

Rsb = 260 cf cf/B /B T,i = 160°F yg =0.7 Yo = 36” API Using Fig. 2-10, the solurion is: 1. Start at Reb = 260 cf/B 2. Go right ro yg = 0.7 3. Go down m .,,, = 36” API 4. Go right m T,, = 160’F 5. Go down ro the answer: Bob = I .155 .155 EXAMPLE - Nomograph Solution Find Bob under the following conditions: %b Ta yg y.

increase in liquid volume due to

(Z-10)

As the well flowing pressure (p-r) increases up to the bubble-point pressure (pb), the oil formati formation on volume factor (B,) increaser from 1.0 m a maximum value as more and more gas is dissolved into the oil. Then, as pwi increases past pt,, B, decreases as the oil is ccompressed ompressed with pressure. Of particular particul ar interest in this conrimmm of B, values, corresponding m a continuum of pr pressure essure values, is the matimun oil formation volume factor (B at bubble-point pressure), BOh. Bob is a function of remperarue , oil gravity, gas gravity, and solution solution gas-oil rat ratio io at bubble-poin bubble-pointt pressure. Bob may have been measured in the labor laborarory. arory. If no< it can be determined from the chart shown in Fig. ZZ-10 10 (Chart Fgo-3)

= 400 cf/B = 180’F =0.65 =45”API

Using Fig. 2-11, the solution is:

after

i z > b-

Pressure Fig. Z-12 -

-

Oil FVF (B,) as n function (generrzl).

of j~resswe

the downhole downhole well-fl well-flowing owing pressure increases abcwe the bubble-poinr pressure, no mae gas is dissolved, because no mme is present; and the oil (with its consrant amounr of dissolved gas) is compressed ttoo a smaller volume. The shaded area of Fig. 2-12 shows this relationship. This is a case of an undersaturated oil. (In some reservoir work bot bothh oil and warer are rreared as incompressible fluids, but, strictly strictly speaking, rhis is nor uue.) The oil compres sibility factor (c,) is a function of tie densiry of the oil ar bubble-point bubble-point pressure (&). Fig. 2-13

1. Draw a line from the ga gass gravity (yg) of 0.65 rhmugh the solution gas-oil ratio at bubble-point Pressure (Rsb) of 4400 00 cf/ cf/B B to Line A. 2. Draw a line from Point a through Tri Bob, at 1.24.

(180OF) to

Note: This nomograph ignores rhe effect of tthe he oi oill gravity, y. A look at the chart for Buh Fig. 2-10, shows that the oil gravity lines are crowded together; hence, the nomograph is nearly correct for all oil gravities. B, is equal equal ro Bob ar bubbl bubble-point e-point pressure, but is less than Bob at eirher grearer or lesser pressures (Fig. 2-12) As

0 0.5

0.7

0.6

0.8

hb. v/cc

“co” as a function of &a, for for “‘merage” oils (after Calhoun, Ref. 5).

Fig. Z-13 -

 

2. Move this point m Line B, following (a.3 n small-de example).

3. Draw a line from &ii pint

the guide lines

through Bob of 1.22, to

the B. answer of I .20. EXAMPLE -Saturate -Saturatedd

Oil

Find B. under the following conditions: B, = 1.20 pwi = 1,800 psia pb = 2,400psia The solutions are as follows: Rough solution: 1. B, = 1.0 + (p&x,)

X (B,

2. B, = 1.0 + (1,800/2,400)

Fig. 2-14 -

(Chart Fgo-5): Nonto~r@h fo find&. Needed: pm,, pb, co or P.d, md Boa.

3. B,=

- 1 .0) X (1.20 - 1.0)

1.15

Accurate solution: 1. p,t/pn

= (1,800/2,400 )

= 0.75

2. k = 0.80 (from Fig. 2-9) graphs the due of co a.5a a.5a function of pob for “aveI “aveIa& a& Oils (given also on the right-most line of Fig. 2-14 ( Chart Fgo-5) ) At this pain< it is necessary to assume that Pa can be determined. For undersatua.red oil, the equation relating B, to Bob is: Bo=Bo~(l-co(pv-pb))

(2-11)

where c, is the oil compress ibility from Fig. 2-13. When m drops below pb the oil volume again is reduced, this time because dissolved gas comes out of solutio solutionn Again, B, is reduced to some fractio fractionn of Bob This is a case of a rdwated oil, accompanied by free gas. The unshaded area in Fig. 2-12 shows this relationship. As a rough appmximation: B, = Bob X ( p,t/pb) A more accurate soluti solution on for B, is through+ig. 2.9, which re lates k to pe’pb; then B. = 1

3. B,=l.O+k

(Bob-LO)

4. B, = 1.0 + 0.80 (1.20 - 1.0) 5. B,=

1.16

J. OIL DENSITY, p. At standard conditions the density of oil is equal m the weight divided by the volume; or, in equation form: Pm.. = W&V,,

At welKIowing conditions the density of oil is still equal to the weight divided by the volume. It is, howeve r, not quite straightforward because the weight of the oil has been increased by dissolved gas, and the volume of the oil has been increased by the oil formation volume factor:

w0.c + Wdiva8 mvf = Pmvf=

+k(B,,-l).* EXAMPLE - Undersaturated Oil Find B, under the following

01

conditions:

Bob = 1.22

pob = 0.66 gm/cc

pti = 3,000 psia pb = 2,000 psia

(2-12)

In English units:

Pd =

Vex X Bo

(2-13)

The solution is as follows: 1. FmmFig.2-13,c.=

2. B,=Bob(l-co(fif-pb)) B, = 1.22 (1 = 1.20

15 X 10-B (15 X 10-6) (3,000 -2, -2,000) 000)

141.5 + 0.0002178 ysR, 131.5 + y.

Pa’ =

)

B,

p/cc (Z-14)

And at bubble-point pressure, ph as used in determining An alternate solution is with Fig. 2-14 (Chart the nomograph for solving for B,:

Fgo-5),

1. Draw a line from 1,000 psia, the vvalue alue of pw - pb, to pob ar 0.66. Mark the intersection with Line A. “8. = qow./q.s. - (w,+ q.)/q.r= - R. + 1, and B,. - R.a + 1; thusk=R.,R~a=(B.-l)/(Ba~-l thusk=R.,R~a=(B.l)/(Ba~-l1).andB.=1+k(B~~1).andB.=1+k(B~~1). subject o the indicated approximatio ns.

P0Wf=

141.5 + 0.0002178 yg Re, 131.5 + yo

BOb

c,:

p/cc (2-15)

Note char the 0.001223 is the density of air in gm/cc, and rhe 5.615 is rhe necessary conversion factor to co11vat

 

the English units of cf/B iato B/B. The 141.5/(131.5 + yJ, of course, converts y. in “API to pm. in g&cc. In metric units the above equations woul wouldd simplif simplifyy to: P WI= and

Pm. + 0.001223 yeR, BO

post + 0.001223 y,R,b Pa = BN,

= = = =

g&cc

30” API 0.75 350 cf/B 1.21

The so lution is as follows: 141.5 = 0.876 131.5 + 30.0

1. po..= 2. p

ow

=

Posc+o.~o~17~Y~,% BO

0.876 + 0.0002178 X 0.75 X 350.0 1.21 = 0.771 gm/cc

3. powf=

3. Draw a lie from Point b through through B, = 1.21 to the answer, porn = 017gm/cc. EXAMPLE-Metric

units

Find pob in the foIlowing

EXAMPLE -English units Find povi in the foIlowing simarion: y. ye R. B,

(2-14a)

2. Draw a line from Point a to y = 30” API and establish Point b.

(2.15a)

P... .,= Rsb B,,b

situations:

= O.&lgdcc =0.65 = 100 cum/cum = 1.26

The solution is: 1. pan = Pow + 0.001223 y,R.b BOh 0.84 + 0.001223 X 0.65 X 100 1.26 = 0.730 gill /cc

POb=

K. OIL VISCOSITY, p., The viscosity viscosity of a crud&l decreases with a temperature increase and wi with th an increase increase of dissolved gas. Heavier oils are genially more viscous than llighter ighter oils of the same hydrocarbon base. The question of which viscosit viscosityy units are co be used used may be rather confusing since there are several differenr differenr systems used in the oil field. The centipo ise is the unit used throughout this document. Conversions from ocher units ro cemipoises are given in Ref. 1. The chars given io Fig. 2.16 (Chart Fgo-7) correlate crude oil vi viscosity scosity wi with th stock-milk oil graviq, remperamre, and solurion gas-oil ratio as or below below bubble-point bubble-point pressure. If the pr pressure essure on the oil is above bubblebubble-point point pressure its viscosicy is increased by the amount given by the correccicm cufve. EXAMPLE Find the oil viscosity in cenripoixs, situation: YO = 30” API Tm= 200°F ph = 1,700 psia

in the following

pxvr = 2,700 psia Rab = 400 cf/B Referring m Fig. 2-16 (English units), the solution is made as foknv foknvs: s: F&.2-I5-(Chart

Fgo-6):

Nomogczph

80 find

1. Starr on the ordinate ar 30” API.

pm,. Needed: R , ya yo, and B, B,..

2. Go right right to 200°F. 200°F. 3. Drop m the Rsb value of 400 cf/B.

An alternate &ion is with Fi Fig. g. 2-15 (Chart the nomograph for determining powi.

4. Go left m read 1 cp, the viscosity at bubble-poi bubble-point nt pressure, and, al along ong the way, note the Point D.

Fgo-6),

5. Drop vertical vertically ly from Poim D to the absci abscissa. ssa. Read a viscosity gradient of 0.07 cp/l,OOO psi. This figure is used only when hi > pb, which is the case in this

1. Draw a line from yp = 0.75 through R, = 350 cf/B to Line A

 

Enalish

that will will dissolve in the same volume of oil under the same conditions. In practice, rhe water is always always assumed to be saturated with gas; that is, to contain the maximum amount of dissol dissolved ved gas, as though ir were were at bu bubble-point bble-point presswe. Thus, for OUT purposa, “R,C’ may be read *‘RSvb”. This ratio is a function of pressure, temperatu re, and water salinity. There is no esrablished relationship describing the S&Irion of gas in the presence of both oil an d water. It is thus necessary co a. a.ssume ssume hat any available available gas wiIl first dissolve into the wara until sacw arion (k- b) is reached, after which any additional gas wiU dissolve into the oil Although an ovasimplification, these assump tions facilitate the calculations and the error introduced is, at worst, quite sm.xK The calculation of the solution gas-wate r ratio is made by the use of Fi Fig. g. 2-17 (Chart Fgw-1 ). The main section of the chart is used m determine R’,, the solution gas-water ratio in fresh water. Then the bottom part of the chart is used to find FBQ the Brine Corre ction Factor. Finally, R, is Calculated: R ,, = R’., x FB C. t should be noted that the brine correction factor is plotted only for saliniriff up to 35,000

24,

Viscosity

increase, cp/l,OOO psi

I

ENGLISH

problem. The increa se in viscosity is from the value obtained on the ordinate (Step 4), so that: k = 1.0 cp +

1 0;;07psia (pwr-Pb)

p*=1.ocp+

,000 0.07psia ( 2,700 - 1,700 ) = 1.07 cp

1. SOLUTION GAS-WATER RATIO, R,, R, is rhe ratio, volume of the di dissolved ssolved gas M the volume of warer in which the gas is dissolved; both volumes converted to scandud condition conditions. s. In theory, the pressuresolubilicy relationship for gas in water resemb les chat for gas in oil (Fig. 2-12 ) , though the amoum of gas that will diilve in a given volum e of water is only about l/60& the amoutx

 

produced warer has sixty rimes de volume of rhe ooil, il, the volume of gas dissolved in the mater will be approximately equal 03 tit dissolved in de oil. M. WATER DENSITY, p1 The density of gas-free gas-free water is a fooaion of temperature, pxssore, and water salinity. A nomogmph of thi thiss relarionship is shown in Fig. 2-18 (Chart Fw-1 ) As the solubility of gas in water (R ,) is small, the e&a of the dissolved gas is ignored, and this than is used for the density of water at any 11,. EXAMPLE Find the density of wafer under the following following

condi-

tions: csm= 9woo ppm = 200’=F L, p& =2,ooopsia F&2-18-(Chart

Fw-1):

Nomograph

Using Fig. 2-18 , the solution is as follows:

to find

1. Starting from Csacl = 90,000 ppm, draw a line through T,, = 200’F to Poim a.

~zm,. Needed: C~Nocb C~Nocbm m and pm?

2. Starting from p-.1 = 2,000 psia, draw a line thro ugh Point a fo warer density (pIrKf). of 1.035 gm/cc. ppm Above 35,000 ppm the value of F, F,oo fades into oncertaimy, but seems IO remain less than onicy. ‘Thus, for any salinity, R., will probably be less rhan R’.w. One published extrapola extrapolation’ tion’ of FBC can be pm into the form: FBC = 1.0 -

0.079-o.,, 0.079-o.,,($J] ($J]

x [C$gqJ (2-16)

For instance, at. 250°F and wi with th Cyacl = 140,000 ppm: FBC = 1.0 -

0.079 - 0.019 (S)]

x [ 1;;;;;

FBC = 1.0 - ( [0.079 - 0.0475] X [ 141) = 0.56 EXAMPLE

] )

N. WATER FORMATION

VOLUME FACTOR, g.

B, is the rat ratio io of the volume of a given mass of water with its dissolved gas, a~ a gi given ven pressure and temperatore, to the volume of the sam e mass of wanx ar standard conditions In equation form: (2-17) As in the case of oil formation volume factor, in oil field usage the tterm erm is wdly associated with flow races, so chat: (2-18) In practice, however, the amoom of dissolved gas is

Find the solution g as-water ratio, R,, TVf Pm G&J

when:

= 1800F = 3,400 psia = 20,00 20,0000 pp

ignored because it is so small. This leaves tthe he m ass-flow-rate relationship:

or:

Referring to Fig. 2-17, the solution is as follows: 1. From the rap pxn of th thee than than R’,

= 16 cf/B

2. From the bot bottom tom pan of the chart chart,, FBo = 0.91 3. TheoR., TheoR., - FBo X R’., = 0.91 X 15 cf/B As the solution solution gas-water ratio is very much smaller than the solution gasoil ratio -roughly 1/6Oth - its e&t will only be noticed wh en the produced warer is gready io excess of the produced oil. For instance, if the produced stock-tank oil has ten times the volume of the produced water, the gas dissolved in water will be on the order of 1/600rh of the gas dissolved in oil. On the other hand, if the

 

Using Fig. 2-18, the solution is as foIlova: 1. Ar standard conditions conditions,, pISC = 1.088 g&cc 2. At well-flowing

conditions, pwwf = 1.060 gm/cc

3. Bw = ~w.dPwwc = 1.088/1.060 = 1.026 0.

WATER VISCOSITT, pw

The viscosity of water is primarily a function of remperanm and water salini salinity. ty. Fig. 2-19 ((Chcrri Chcrri Fw-2) shows rhis relationship. EXAMPLE tions:

Find the viscosity of water under tthe he following Cxac, = 150,000 ppm TM = 200°F Using Fig. 2-19, the solution is:

condi-

Hence, in practice, B, is equal lo the ratio of the density of the water at standard conditions to the density of the warer at welLflowing

conditions.

EXAMPLE Find the water formation volume factor under the following conditions: CN&., = 120,000 ppm TV, = 1800F PWf = 1,500 psia

1. ~-0.43cp P. VISCOSITY OF MIXTURES The viscosity of a water-in&l emulsion may be many times that of eithe r th e water o r the oil. As a practical limit, however, the viscosit viscosityy of a water-in4 emulsion does not exceed 5 cp.

 

CONVERSION EXAMPLES 13

These field field problems are chosen m familiarize you with rhe use of the fluidfluid-conversion conversion charts, nomographs, and rrelaelationships.. Bot tionships Bothh English-uni English-unitt and m&c-unit examples are given Each case gives, on the left-hand page, a work-sheer development of rhe problem, complete with small ilillusualusuations of the appropriate charts. On the right-hand page is shown a flow chart, with a block for each operation. Some cases involve making decisions as to which branch of rhe flow cha rt to follow; the chosen route is indicated by a heavy flow line (as in Examples 4,5,6,7, and 8). In each case, he solutions are given in bold-face type, or in heavy-lined boxes

EXAMPLE

FLUID PRODUCIION

“NITS

PAGE

1

Water (injection)

Metric

18

2

Gas

English

18

3

Gas

Metric

20

4

Gas and Water

English

22

5

Oil and Gas (No Free Gas )

English

24

6

Oil and Gas (Free Gas)

English

26

7

Oil, Gas, and Water

Metric

28

8

Oil, Gas, and Water

English

30

 

El I EXAMPLE

No.

1 (Metric)

In a warer inject injection ion well, iind the downhole waxer flow rare (qwwi) under the following conditions: qwsc = 100 N m/D

pti = 135 kg/sqcm

Crac, = 120,000 ppm

Tti = 55°C

El

The solution is as follows: 1)

Chart Fw-1 : pws. = 1.090 gm/cc

2)

Chart Fw-1 : ~,,-f = 1.077 &cc

3)

B, = &,Jp,,,

4)

q-f

EXAMPLE

= Bwqm = 1.01 X 100 = 101 cum/D

No.

At well flowing

2 (English) conditions,

find the gas flow rate (q*f)

density (PpRf) at the following conditions: q,,, = 2,500 Mcf/D pRf = 6,000 psia T,r

“IMI

= 1.090/1.077 = 1.01

= 360°F

y*

= 0.70

and the gas

El

The solution is: 1)

Chart Fg-1 : l/B,

= 235

2)

~r=q,/(2,500, ~r=q,/(2,500,000/235) 000/235)

= 10,640&D

3)

&,t=

=0.001223

0.001223 (yp) (l/B,)

X 0.70 X 235

(or: ChartFg-5) 4)

PM’

0.201

gm/cc

 

I. Surface

I

Production: Production: WATER WATE R

Must know or estimate 1

Metric 100 cum/o

qwsc

629

120,00O/pm

CNoCl

120,OOOppm

135 kg/sqcm

Pwf

1,925

55oc

Twf

131 OF

(OK= oC+Z73D)

English BID

1 This example

uses metric

units

PSia

(“R=oF+4600)

I

GAS

I

/Must know or estimate 1

example

English

units

I

implified

route

‘yFp?l 7 10,64Ocf/D

 

EXAMPLE

No.

3 (Metric)

At well flowing conditions, what is rhe gas flow rare , q,r, density, pgwf, at the following conditions: qegc= qegc = 39,500 cum/D

yg

T,

pwr = 200 kg/q

= 217°K

P ~ =46kg/sqcm

=0.65 cm

T,, = 70”Cor 343°K

The solution is as foflows: 1)

pp. = pM/pPe = zoo/46 = 4. 4.35 35

2)

T,,=T,*/Tw

3)

Chart Fg-3: z = 0.82

4)

Chart Fg-4, or from:

5)

qgvf = qggc/ ( l/B,)

6)

P,,=

= 343/217=

1.58

= 39,500/199 = 198 cu m/D 0 001223 yg l/B, = 0.001223 X 0.65 X 199.0 =

and the gas

=0201

gm /cc

0.158 gm/cc or: Chart Fg-5: &wf

= 0.155 gin/cc

 

I

Surface

Production : GAS w or estimate

Simplified

f:

Chort

Fg-I

1. ;Chort Bg

Fg-4

=I99

I

-I-

This example uses metric units

.-.

I

I

%I= qFod=/BP = 198 cu m/D

I

t

I PFgvf :

Chart

Fg-5

= 0.155 gmkc

 

r

EXAMPLE No. 4 (English) At well flowing conditions, find the gas and water flow rates, qFti

and

g-f, and the gas and water densit densities, ies, pEgRt and paaa at the following conditions: qgsr = 600 Mcf/D q,wc = 500 B/D ye

=0.70

Cm, PWf Tw1

= 30,000 ppm = 2,800 psia = IGOOF

The solutio n is as follows: 1)

Chart Fgw-1 : R’., = 14.5 cf/B, and FBO FBO= = 0.85, hence R,, = R’., X FBC = 14.5 X 0.8 0.855 = 12.3 cf/B

2)

Chart Fw-1 : p-c = 1.027 &cc

3)

Chart Fw-1 : pypwf = 1.007 &cc

4)

B, = P.&J-‘,,

5)

qarr=qa3c XB,=510B/D

6)

Rm= R+

= 1.027/1.007 = 1.02

(qgse-qaseX%-n)/qwsc = (600,000 - 500 X 12.3) /500 = 593,850/500

=

l,lBBcf/B where RFga is the ratio of free gas to water 7) 8)

qQse = qwsc X Rpm = 593,850 cf/‘D Chart Fg-1 : l/B, = 200

9)

qrg~t = q&

10)

ChartFg-5:

( l/B,)

= 593,850,‘ 593,850,‘200 200

= 2,970

cf/D

pFgWr=0.17gm/cc

 

I

Surface

Producti Production on

: GAS

and WATER

I

Must know or estimate F Metric English 16992 79.5

cum ID cum/D

0.70 30000 lS’kg/sq 71-c

PPm cm

qgac

600

Mcf/D

qwsc

500

8/D

Y9

0.70

CN.CI

30.000

pm

I

Prf

2,GM)

pda

1

Twf

1

16OOF I I

R,,:Chart

Fgw-I

=12.3

cf/S

I

= 1.007

This example uses English units

qmkc

=1.02

qwf = %rc& f-l=

I

RFg =

I

Free

_I

99?ic-‘&rcRrr

510 S/D

=I,188

Gas Pesent

1

9v.c

No

Insignificant gas:

treat

Cf/S

Free Gas

amount as water

of Yell.

I

Cha : Fg.

jgi&iq =2,970

 

EXAMPLE

No. 5 (English)

AC well flowing mnditions, is the oil saturated or undersa mraced? If saturated, find qp,r and 40-i. Find /&vr and &wf. If undersanxared, fmd qoti and powt. The following is known: q,,

=450B/D

yp =0.70

%c = 200 Mcf

pui = 2,600 psia

YO = 35” API

T,, = 142OF

The solution is: 1)

R = q&q-c

= 200,000/450 = 444 cf/B

2)

Chart Fgo-1 : pb = 1,900 psia

3)

pm. = 141.5/ (131. (131.55 + 35.0) = 141.5/166.5 = 0.85 g&cc

4)

pb - p.pt = 1,900 - 2,600 = - 700 psia Conclusion : undersatura ted

oil, no free gas

5)

R.=R(sinceall R.=R(sinceallrhegasisdiss rhegasisdissolvedinoil olvedinoil))

6)

ThenlerRgb=Rs=444cf/B

=444cf/B

cf/D

=0.17 gm/cc

7)

Chart Fgo-3 or Fgo-4: B,,t, = 1.23

8)

Chari Fgo-6: &b = 0.75 gm,‘cc

9)

Chart Fgo-5: co = 10 x 1W

10)

Bo=Bots[l-co(pwf-pb)l B, = I.23 [ 1.0 - IO X 1W” X (2,600 - 1,900) ] B.= 1.23 (1.0 - 0.007) = 1.221 Or from Chart Fgo-5: B, = 1.22

11)

Chart Fgo-6: powr = 0.755 gm/cc

12)

qowr =q,,.c B, = 450 X 1.221 = 550B/D

13)

q,,,=O.Ocf/D

 

I

Surface

II

Production: Production:

OIL

1i

Must know or estimate I

Metric

1

English

71.7 cu m/D

qosc

450

5,680

agsc

200 MCf/D

0.84

cu m/D

PO

gmkc

0.70 163 kg/y

cm

61 OC

and Saturated

B/O

35O API

Yn

yg

0.70

Tuf

142-F

2,600

Pwf

psio I

1 R= qgsc/qosc

/=444cf/6

I

I

I 1 pb: Chart

This example uses English units. Undersaturated I

,

I

oil and no free gas

Fgo-I

I

if oil in *API

&

-

F‘b-p”,,

/

+

)=l,SOOPsia I

I

=0.85 gm/cc

GAS oil and free (10s

I

=444 cf/B Saturated oil and no free gas

1 t’ob: Chart Fgo-6

/=0.75gm/cc

 

EXAMPLE

No. 6 (English)

At well flowing conditions, is the oil sanuated or mdersa mrated? If satnrated, what are qpm and q&? pFw and &wr? If undersaturat undersaturated, ed, what are qm and PM? The following is known: q-

= 200 B/D

bI

&

= 360 Mcf/D

Tm = 195°F = 655%

=45”ApI YO yz =0.65

= 4,300 psia

b

=68Opsia

Tw = 345%

The s olution is as follows: 1)

R = 360,000/ 360,000/200 200

2)

Chart Fgo-1 : pb = 5,500 psia

3)

pm=

4)

*

- *

= 1,800 cf/B

141.5/(131.5

+45.0)

=0.802gdcc

= 5,500 - 4,300 = 1,200 psia

Conclusion : saturated

oil with

5)

LetR.b = R = 1,800 cf/B

6)

pal/pb = 4,300/5,50 4,300/5,5000

= 0.78

Chart Fgo-2: k = 0. 0.825 825

free gas

R. = k X Rsb = 0.825 X 1,800 = 1,485 d/B 7) R~z=R-RR.=1,800-l,4?35=315cf/B 8) 9)

Chart Fgo-2 or Fgo4: k=0.825 (fromb)

B.b = 1.93

B,=l.O+k(B,-1.0) B, = 1 O + 0.825 X 0.93 = 1.77 10)

Chart Fgo-6: &t

= 0.57 &cc

11)

q,t

12)

qrgse = q,, Rpg = 200 X 315 = 63,000 cf cf/D /D

13)

hr = p~t/p~. = 4,300/680 = 6.32

14)

T,r = T,,/T,

15)

Chart

Fg-3: z = 0.978

16)

Chart

Fg4:

17)

qrmf = qFme/ ( l/B,)

18)

Chari Fg-5: pFd

= qm B, = 200 X 1.7 1.777 = 354 g/D

= 655/345 = 1.90 l/B,

= 237 = 63,000/237 = 266 cf/D

= 0.188 gm/cc

 

I

Surface

Production: Production:

OIL

Must know or estimate Metric 31.9 C” m/o 10.200~~

0.802 0.65 303 so-c

qosc

m/O

20 0 - B/D-

(

360

qgsc

pm ICC kg/q

Enalish

GAS

1

I 1

1

Saturated

Mcf/O

oil

45O API

PO---Y. 7,

cm

and

0.65 4.300 psi0 195 OF

Pwf Twf

k: Chart

Fgo-2

,Chor;

Fgo-3

I

This example English units

uses

I

1 Pb:Chart if oil in O/LPI Fgo-,

oil and no free gas _

Saturated

“ps’a

I

PO,, :Ct,mt

Undersaturated

= 1,485 cf/B

oil and

=oB02

gm’cc

1

[ D

( ~. ^_

no free gas

A

I-

= 354

B/O

R*=R=Rsb

= 237

P,

:Chort

Fgo-6

= 266 cf/O

=o.im

gm/cc

27

 

EXAMPLE No. 7 (Metric) At well flowing conditions, is the oil saturated or undersatw ared? If sateraced, what are qrmi, qowt q~, pm-r, po.cn and P,W? If mdersanuated, v&at are qowf and qwwr? povi and pwar? The followi following ng is known: qoS =4

cu m/D_

%c =34Ocum/D q,vae= q,vae = 160 N m /D

pa.. = 0.85 gdcc The solution, following 1)

= 0.75 Ys C mc, = 30,000 *pm P-i T,

= 120 kg/sqcm =55”C

the flow diagram, is as follows:

Chart Fgw-1 : R’,, = 2.0cum/cnm, R,, = R’,,

FBC= 0.84

X FBc = 2.0 X 0.84 = 1.7 N m /cum

2)

Chart Fw-1 : p,,... = 1.028 gm/cc

3)

ChartFw-l:p,,,=1.013gm/cc

4)

B, = p&p-

5 ) q-i

= q,,

= 1.015 B, = 160 X 1.015 = 162 cum/D

6) R’=R,+R,=(q,,,R,,)/q,,=(340-160X1.7)/4 7)

=17cum/cum FromChart Fgo-l,pb

8)

pb-pwi=33-120=-87kg/sqcm Gmchion:

9)

= 33 kg&cm

undersaturated

oil and no free gas

LetRab=R’=17cum/cum

IO)

Chart Fgo-4: B.b = 1.08

11)

Chart Fgo-6: pob = 0.80 p/cc

12)

Chart Fgo-5:

13)

Chart Fgo-6: pow = 0.81 gm/cc

14)

qowr = qoac qoacB, B, = 4 X 1.07 = 4.28 cu m/D

15)

qFgswi = 0.0 cf/D

B, = 1.07

Fgo-6

u rE  

I

Surface

Production:

OIL,

GAS. and

Itumtad

I R,.,:Chart Fgw-I

= 1.7cu 1.7cu m,cu m

I = I.026 pm/cc Pw.. : PVW‘: Chart Fw-’ = 1.013 gm/cc I

This example metric units

uses

1

WATE WATER R oil and fm

gor

R, R’=R,b

17 cum/cum

I &:

Chart Fgo-6

I&: Chart

Fgo-5

=0.80 pm/cc

I= 1.07

 

EXAMPLE No. 8 (English) At well flowing conditions, is the oil saturated or undersatu rated? If sacw ==d, find qrgnt qowi, q-t; and PP,~, pm + pmt. poti, qwwf, Powr, and pwwf.The following following is known:

If undersaturated, find

qcm = 50 B/D

C NiaCI 80,000 ppm

qeae =80 Mcf/D

p-i TWf

= 3,200 psia

=45”Ap1

PC+

= 660 psia

=0.65

L

= 390”R

qwac= qwac = 300 B/D 70 .,g

= 170°F = 630”R

The solution is: 1)

Chart Fgw-1 : R’s, = 15.5 cf/B, and FBC = 0.65 (estimated)

2)

%, = 15.5 X 0.65 = 10.0 cf/B

4)

B, = p,../&,r

5)

qM=q,,B,

6)

R’=R.+&=

= 1.060/1.037 = 1.022 = 300 X 1.022 =3076/D (Gc--qw,.L)/qo. (Gc--qw,.L)/qo..= .=

(80,000-300

X llO.O,/SO O.O,/SO = 77,000/50 = 1,540 cf/B 7)

Chart Fgo-1 : pb = 4,500 psia

8)

pb -p-t

= 4,550 - 3,200 = 1,350 psia

Conclusio n: saturated 9) 10)

oil and free gas

LetR.,=R= 1,54Ocf/B p&pb = 3,200/4,550 = 0.703: Chart Fgo.2: k = 0.775 R. = kR.b = 0.773 x 1,540 = 1,193 cf/B

11)

RF. = R’ - R, = 1,540 - 1,193 = 347 cf/B

12)

Chart Fgo-3 or Fgo4:

13)

FromlO,k=0.775 B,=

Bob = 1.78

1.0 + k (Bob - 1.0) = 1.0 + 0.775 (0.78) = 1.60

14)

ChartFgo-6:

pow,=0.610gm/cc

15 )

goti = qw< B,=50~1.60=80B/D

16)

qF=e = qow Rpg = 50 x 347 = 17,350 cf/D

17)

pp. = 3,200/&O

18)

TDr = 630/390 = 1.62

19)

Chart Fg-3: z = 0.848

20)

Chart Fg4:

21)

qpwr=qr~e,‘(l/ qpwr=qr~e,‘(l/Bg) Bg)

22)

&se=

= 4.85

l/B,

= 212 =17,350/212=81.8cf/ =17,350/212=81.8cf/D D

0.001223 yz (l/B,)

= 0.001223 X 0.65 X 212

= 0.169 gm/cc

 

I

Surface

I

Must

I

I

Production: Production:

know or estimate

7.95 2,266 47.7

cu m/D cu m/D cu m/D

0.8 gm/cc

I 4

Saturated

oil and free

English

Metric qosc qgsc

50 B/D 80

Mcf/D

qwsc

300

B/D

POJC * 7”

To

(

45’API

[

0.65

1

OIL, GAS, and WATER WATE R gas = 1,540

cf/B

=I,193

cf/B

-I=347

cf/B

ml=,.78 mm

R,,:Chert

I I

This example uses English units.

Fgw-I

= 10.0 cf/B

WI=

0.610

gm/cc

Undersaturated

gas 1

II

P

(

oil, no free

[*I=

307

B/D

1

R,=R’=Ptb

1

I

I I PPr=FwPw PPr=FwPw 11T.r=Twt T.r=Twt1T.c 1T.c

Chart

)=I.6 = 0.848

i

hwf:

I= 4.85

Fgo-6

pq-jq =8l.B

 

cf/D

=O.l69gm/cc

32

 

Ch.rtr

and

Nomog..phs,4

CHARTS AND NOMOGRAPHS 14

C”mn CODE

Fg-1

35

Fg-2

36

b-3

37

b-4

38

Fg-5

39

b-6

40

Fgo-1

41

Fgo-2

42

Fgo-3

43

Fgo-4

44

Fgo-5

45

Fgo-6

46

Fgo-7

47

Fgw-1

48

Fw-1

49

Fw-2

50

 

GAS ORMATION OLUME A CTOR ACTOR After S tanding and Katz, Ref. 3.

ENGLISH prerarre. psia

Find V@. Given:

V,

= 400 cu ft = 0.70 YP T we = 200’F pwt =2,000 psia

1.

Select “ys = 0.70” section. Enter absassa at 2,000 psia, go vert vertically ically to 200’F.

2.

Goleftm

L= %

125. V &=3.2cuft

 

PSEUDO CRITICALATURAL A TURALAS A S ARAMETERS After Brown et at. Ref. 4.

yg

Gas grav ‘ity, Find T, and pw

1.

Given: yg = 0.75, average gases

(air

= 1.0)

Enter abscissa at 0.75. Go up to: T, = 406OR and pp. = 664 psia

 

NATURALAS EVIATION ACTOR After Standing

and Katz, Ref. 3.

Pseudo-Reduced

Pressure,

ppr

Pseudo-Reduced Find z. Given:

1.

ppr

par/pw

Pressure,

ppr

= 2,000/ 650 = 3.07.

Pti = 2,000 psia

2. Tpr = TM/T,

pm = 650 psia T,, = 200°F (660”R ) Tw = 410’R

3.

Enter abscissa (top) at 3.07 (p,,). 1.6 and 1.7 lines.

4.

r=0.828

= 660/410 = 1.61 Go down ro Tpr of 1.61, between

 

GAS ORMATI ORM ATION ON OLUME A CTOR ACTOR Nomograph) @974CHL”MBErGER Pwf psia

kg/sq cm

\ l~ooo-

2

/

10,000----500 5,000-

-B,

Twf OF

~2,000

‘C

0.3’ loo-.

~I.000 0.4-

‘-50

-500 0.5-

--zoo

150-

-200

2,000---loo

0.7-

_I00

l,ooo~-

Solving

-50

-40 500-

for

-0,

250-’

300--150

= IO

200-

--loo

l.O-

-20 --20

200-

1.5-

-5

350--

--10 IOO-

2.0-

-2

400-

--200

2.2450-I-250 500-

Find l/B,. Given:

pwf = 140 kg/q T,, = 93°C z = 0.828

cm

1.

Enter pwr scale at 140 kg/q

2.

l/B,

cm. Follow lines as n small di diagram. agram.

= 135

 

GAS ENSITY @974CHLUMBERGER

yg (Air =l.O) 0.5

P

gwf

0. s-

I Bg

o-2-

0.9

0. I-

im

0.05 0.7

O.M20 O.Ol0.6

Find PF+ Given:

= 0.75 l/B, = 140

yp

1. Connect yg = 0.75 and l/B, = 140, as n small diagram 2.

pFmf=0.13

gm/cc

 

GAS ISCOSITY

Courtesy of Oil and Gas Journal, May 12,1949

:

10

Pressure,

1,000

Pressure,

22 z ‘;: E 0 a. $

1,000 psi a

0.07 0.06 0.05 0.04 ,“z 0.0, 0

I

2

3

Pressure, Find F@. Given:

psia

yg =0.70 pwi = 2,000 psia T,i = 200°F

4

1,000 psi0 1.

Enter yg = 0.70 chart at put = 2,000 psia.

2.

GouproT~=200°F

3.

pm = 0.018

centipoise centipoises. s.

 

BUBBLE-POINT RESSURE Rsb

:g/sqcm

(R,) cu m/cu m 700 500

Pbhvf)

Cf/B

1000

psi0

15,000

f

4000 3000 2000

Oil Gas Pose Yo Yg

300 300

4000

250

150 too

%

70

-1.2 ii?-

-20

-1.0

0.9:

1000

--30 0 60

\

40

500 30

400

20

300

15 40 i

-1.1

0.8-

-0.9 -0.8 4r -0.7

--50 -0.6 --60

200

-0.5

150

0.7--70 -0.4 --60 0.65-

Find pa Given: T, = 180°F qm = 600 B/D ~c = 240 Mif/D yg =0.75 y. =40” API 1, R = 240,00Ocf/D = 400 cf/B. 600 B/D

2. Rab Rab= = R, since he field-wage definition of pb stipulatesgiven stipulatesgiven flow ratesof rates of oil and gas, aken here to be qosc nd use (above). 3.

On the nomograph locate Point ca y a ine through through T,t = 180°F and Rsb= Rsb= 400.

4.

LocatePoint Locate Point b by a ine through yp = 0.75 and y. = 40” API

5.

Connect a and b: pb = 1,560 psia

 

SOLUTION O R ORRECTI OR O RRECTION ACTOR ON 0 1574 C HWMBERGER CHWMBERGER R, = k x R,,

.-0.35

B, = I + k( B,,- I )

k

0.9

pwf

hb

Find k Given: pti= 250 kg/q cm pb = 185 kg/sq cm 1. pwf/Pb = 0.74. 2. Enter abscissa r 0.74 3. k=O.SlO.

 

FORMATION OLUME A CTORTp,,,011 ACTOR 0 11 After S tanding, Ref. 6.

I

Find Bob Given:

Rsb = 260 cf/B T, = 16O’F yg = 0.7 70

= 36” API

1.

Enter upper left scale ar R.a = 260.

2.

Go right m yp = 0.7

3.

Go down to yo = 36.

4.

Go right ro T,r = 160.

5.

Go down to the answer: Bob= 1.155

 

FORMATION OLUME A CTORTp,,, ACTOR p,,,011 011 (Nomograph)

YU

B b - z5 - 2.4

(air = 1.0) I.7 1.6 1.5

#

I.4 1.3

Temp. ‘C

1.2

OF

I.1

1.6

1.0 0

I.4

I

1.2 I.1 I.0

Find Bob Given: Rsb = 400 cf cf/B /B Tw = 180°F yg = 0.65 y. =45”APl 1. LocatePoint LocatePoint a by lie &rough 7% 0.65 andRat, and Rat,= = 400 2. Draw a ine from a thmugh T,, = NOoF, to the answer: B, = 1.24

 

FORMATION OLUME A CTOR, ACTOR, 11

&f-b)

psia

kg/sq cm

B O\

\

Bob

Pob

co

-1

.-1 20 0. ..2

ii; 1.9. 1.6. 1.7,

SO1m

,*5

1.6

..

1.5,

.-IO

t

1-U.

200 -.

SOLVING FOR 80

:20

1.3,

(P

1.2. 1.1. l.O-

Find B, Given:

Bob = 1.22

pob= pob = 0.66 @n/cc pti = = 2,000 3,000 psia pa psia

.w

1.

Since pwi > pb, oil is undersaturate undersaturated. d. Connect pRI - pb = 1,000 psia to f&b = 0.66. Mark the point 50 located, asshown in small diagram.

2.

Move this point as shown, ffollowing ollowing mend lines.

3.

Connect this point ro Boa = 1.22 and read: B,=1.20

 

OILDENSITYTWELL ONDITIONS

P 2500

a 60-70 I 0.6

, 0.59

I 0.56

I 0.57

0.7

I 0.56

I 0.55

Yi (air = 1.0) Find pd. Given: y. yp Rs B,

1. LocatePoint LocatePoint a by a ie ie from yz = 0.75 through R. = 350. = 30” API =0.75 = 350 d/B = 1.21

2. LocatePoint LocatePoint b by a ine from Point a coy,, co y,, = 30” API. 3. Draw a ine from Point b through B, = 1.21. Poai = 0.77 gm/cc

 

011 ISCOSITY @ 574CHNMSERGER C HNMSERGER English

- 1.0

Metric

s-

S7?I

o=

IO

0,

Viscosity

Viscosity increase, cp/lOO kg/sq cm

increase, cp/l,OOO cp/l,OOO psi

Find poti. Given:

y. = 30” API Td = ZOOOF *ppt == 2,700 1JOOpsia psia % = 400 cf/B

1.

Enter odinate at yosc yosc= = 30” API.

2.

Go right to Tm = 200.

3.

GadovmtoRst,=400

4.

Go left toanswer, 1catingPoi 1catingPointD ntD

5.

Since bf

6.

pLonr= pa,, + Ap(p,, pLonr=

> ph hti

ontheway:~~=l.Ocent ontheway:~~=l.Ocentipoire. ipoire.

> bb From Point D, go down ttoo read: viscos viscosity ity increase = 0.07 cp/l,OOO psi - p,,)/l, p,,)/l,OOO OOO

= I.0 + 0.07(2,7@0 -

1,700) /l,OOO = 1.07 centipoises centipoises..

 

SOLUTION A S-WATER AS-WATER A TIO ATIO After Dodson

24,

and Standing, Ref.

ENGLISH

2.

METRIC I

4.0 -1

d8

j 60

Cl000 I

j-.y&%y~ Ioa

I40

180

Temperature Cunxtion

220

i

260

OF

for brine salinit salinityy

20

40

1.0 P IL QS

Total solidsin

OBO

brine,ppm IO-J

Tezperatu:

“C

Cccection for brim salinity p 50-z’ IO 20 30 T&l skids in brine, ppm IO-J

Fiid &, Given:

:

TwI

= 180’F

E=.,,

== 3,400 20,000psia ppm

1.

(Top) Enter rhe abscissa abscissaat at Twf = 180.

2.

Goupto

pm = 3,400.

3.

R&R’.,

= 16 cf/B.

4.

(Bottom)

Enter the abscissa abscissaat at 20 (20,000 ppm)

5.

R., =R’,,FBc

= 16 x 0.91=

100

Go up ttoo T = 180. Read FBC = 0.91

15 d/B.

 

DENSITIES FNaCl OLUTIONS 0,974CHL”MBERCER 0,974 CHL”MBERCER

120

&f

Q$

01 Psc ,

CNaCl 300

F cc 0.95

2501.00

f 1.05

I.10

I .I5

Sto;ldard Conditions

-I

Given:

CNsCl= 90,000 ppm CNsCl= Tm =ZOO’=F w = 2,ooopsiz

1.

Locate Point a by a line from CNa a = 90,000 through Twr = 200

2.

Draw a line from Pwf = 2,000 through a. Read: pwwr = 1.032 gm/cc.

 

WATERISCOSITY WATERISCOSITY @974CHUlMBERGER

.20

Temperature,

T,f : “C

Temperature,

Twr: OF

Find pwur. Given: CNacl = 150,000 ppm T Vf = 200°F

 

I.

Enter abscissa abscissaat at Twr = 200.

2.

Go up to CN,C, = 150,000.

3.

hi

= 0.43 cp

AUXILIARY AUXILI ARY

PL CHARTS 15

CHART

DESCRlPTlON

MBE

A-1

TO determine proportion of mud fifiluare luare to formarion wara from resistiviry of rec recovered overed water water..

52

A-2

Resistivity of NaCl Solutions

53

A-2

Flow rate vs average flfluid uid v&&y: no Production Logging tool in stream.

54

A-4

Flow rare vs average fluid velocity: 1 11/16-h tool in stream.

55

A-5

Flow rate vs average fluid velocity: 2 l/S-in. tool in stream.

56

 

DETERMINATION F ILTRATEONTAMINATION

Rrf

= Resistivity

Rw

I Rcsirtivity Rcsirtivity

Rmf=

Recovered Formation

Resistivity Resistivity

Rrf - Rw-

Fluid Water

Mud Filtrate

Rmf

must

so

so

be at

I Rmf Rr(

so / STATIC fp, SC 810

1’; -100 L

/

/ ,/

40

/ I

//

/

’ 8a/ 80

’ 10

I/

/ so

/

/

/ /

se

70

110

.o

/

/Kw) 110 IS0 I80 1.0

ICO .

IS0 Is0

Ire

 

RESISTIVITY FNaCl OLUTIONS

180

/-1.0

Isa

I40

9

 

AVERAGE LUIDELOCITY LUID ELOCITY SFLOW FLOWATE ATE

FOR ARIOUS IPE IZES IPE WITH OPL PLTOOL TOOLNFLOW FLOWTREAM TREAM

100000 5 2

10000 5

2

5 2

100 5 2

10 AVERAGE FLUID VELOCITY

.v 0

IN FEET/MINUTE

 

AVERAGE LUIDELOCITY LUID ELOCITY SFLOW ATE

FOR ARIOUS I PE IZES IPE WITH-11/16” -11/16”PL PLTOOL TOOLNFLOW FLOWTREAM TREAM

 

AVERAGE LUIDELOCITY LUID ELOCITY SFLOW FLOWATE ATE FOR ARIOUS IPE IZES IPE WITH -14” PL PLTOOL TOOLNFLOW FLOWTREAM TREAM

100000 5

2

10000 5

2

5

. AVERAGE

 

FLUID

I I I , I, 5 J2 57 00 % VELOCITY IN FEET/MI FEET/MINUTE NUTE

REFERENCES

1. “Schlumberger Production Log Interpretat Interpretation”, ion”, Schlumberger Limited, New York, 1973 revision. 2. Dodson, C. R. and and Standi Standing, ng, M. B.: “Pressure-VolumeTempe rarure and Solubiliry Relations fo r NaturalGas-Warer Mixrures”, Dr&ng and Production Practice, API, 1944, p. 173. 3. Standing, M. B. and Katz, D. L.: “Density of Crude Oils Saturated with Narural Gas”, Tram. AIME, Vol. 146 (1942). 4. Brown, G. G., Kar Karz, z, D. L., Ober Oberfell, fell, G. C., and Alden Alden,, R. C.: “Natural Gasoline and the Volatile Hydrocarbons”, Natural Gas Association of America, Tulsa, Okla., 1948. 5. Calhoun: “Fundamen~alr of Rawvow Engineering” Engineering”,, versiry of Oklahoma Press, 1953.

Uni-

6. Standing, M. B.: “A Pressure-Volume-Temperature Correlation for Mixtures of California Oils and Gases ”, API Drilling md Productimz P,ani P,anice, ce, 1947, p. 275.

 

.-..

-