Spe Nomenclature And Symbols.pdf

6.0 0.268

SOLUTIONS TO EXAMPLES

319 100

\

80

\,\

,,

60

Slope'

\\

40

20

LL

._,, , Intercept at '" =1 ,0. a = 0.774 ,, , '0\ ,,

= -1.53

o

30

t

length Faxis -17.95 m = length axis =

10

0,\

8 6

'0

'\,

,, ,

4

\,

3 2

,, ,, ,

\

1 0.8

\

\,

,,

--a

0.6 0.5 '-------'--'---'--L.L--'-l0=-'.OO:-:1-----'----'---'---'----'---L--L..11,J.0

Fig.A5.1 Fvs. From plot m = - 1.53 a = 0.774 Substitute back into laboratory data to calculate check values of F. Check Calculate



F

0.092 29.8

0.120 19.8

0.165 12.2

If the true resistivity is 1.29 Qm and water resistivity is 0.056 Qm then Ro 1.29 F = Rw = 0.056 = 23.04

= 0.109 If I =

1

where exp n = 2

w

R/ = 11.84 Qm Ro 11.84 then I = 1.29 = 9.18 Sw =

= 1.29 Qm

[I1]0.5 = 0.330

If exp

= 1.8

Sw

= 0.292

Ifexp

= 2.2

Sw = 0.365

0.205 8.7

0.268 5.80

PETROLEUM ENGINEERING: PRINCIPLES AND PRACTICE

320

Solution 5.2 (a) Data values

Calculated values

Log values

GR

FDC

SNP

C1LD

R 1Ld

V SHGB

VSHDas:

CPDIN

Shale Zone A B C

102 52 72 20

2.52 2.22 2.37 2.20

29.0 22.5 20.5 21.0

1100 150 350 4650

0.91 6.67 2.86 0.215

1.00 0.39 0.63 0.00

1.00 0.00 0.31 0.00

0.26 0.14 0.25

Bul k density grams/cc

Porosity %

Correction

r-------T------

-0.5 2.0

2.5

0

+0.5 3.0

Fig.A5.2.1

Sidewall

SOLUTIONS TO EXAMPLES

321

20

Gamma ray API units

Resistivity Ohms mlm

120 Depth

10 divisions

Conductivity Millimhos 1m Induction conductivity 40" spacing

lS"normol

o

o

Radiation intensity increases

o

Oil bose mud Temp =226

Induction resistivity

0____

__19

4000

o

8000

4000

0_____________ 1<2.0 I I

I

,

I

I

I

I

: I

I I

I

I I I

,

\

" ... _--- .....

A

B

,,/ I

I

I I

I

I

I

I

I I I I

C

"

..

,, ,

Fig. AS.2.2 '--_ _---"'"--'

I

2.0 Pr=1.0g/cc

2.2

Shale Matrix point

2.8

Fig. AS.2.3 Density/SNP crossplot.

()

'"

Sidewall neutron apparent limestone porosity (%)

322

PETROLEUM ENGINEERING: PRINCIPLES AND PRACTICE

(b) For zone C, point plots close to clean sandstone line with cJ> Ro = FRw = 1/cJ>2 . {Rw} Rw = cJ>2Ro = 0.262 X 0.215 = 0.0145 (taking R 1Ld as Ro)

= 0.25. Assuming C to be water bearing

(c) Shale values are listed above. (d)

GRclean

= 20, GRshale = 102

GR - GRclean GR-20 VshGR = - - - - - - - -

82

GRshale - GRclean

VshGR values calculated are tabulated above.

(e) See Fig. A5.2.3 for shale point. Only level B shows a significant displacement from clean line. Graphically Vsh for zone B = XB/XS = 1.25/4 = 0.31. (f) Taking the minimum shale indication (from DIN) gives only B as shaly. Presumably there are radioactive minerals in the sands (such as feldspar) so the GR overestimates shale content. As above graphically for level B, Vsh = 0.31. The porosity is given by point Yon the clean sandstone line where BY is parallel to the matrix shale line, i.e. cJ> = 0.14. The graphical construction is complicated by the curve on the sandstone line. More rigorously convert density and neutron values to sandstone matrix cJ>D = 16.5, cJ>N = 24.2. cJ>NSH = 32, cJ>DSH = 7.5, cJ> = cJ>N - V SH cJ>NSH, cJ> = cJ>D - VSH cJ>DSH Solving the equations for unknown VSH cJ>N - cJ>D 24.2 - 16.5 V SH = = = 0.31 cJ>NSH - cJ>DSH 32 - 7.5

cJ> = cJ>N - VSHcJ>NSH = 24.2 - 0.31 x 7.5 = 0.14 (g) Saturation calculations LevIe!

eq)uatiO:S reduce

/ Rw

1

'\ /0.0145

V6.67 =0.18 Level B with n =2, Rw =0.0145, R =2.86, RSH =0.91, V SH =0.31, =0.14.

:·Rr

FRw ·Sw :.Sw=

=

VR;

()VSH

0.26



t

Archie

.Sw' :. 0.35 :. Sw

1.352

S.'

= 0.51

S.' +

:. 035 :. Sw

R t

Si)mandzoux(VSH )

FR

w

. Sw + R

SH

.

=

I.352S.' + 0341 0.082 _

z

. Sw .. 0.35 - 1.352 Sw + 0. 341Sw . . Solvmg quadratIc :. Sw = 0.376

+ ve root only

323

SOLUTIONS TO EXAMPLES

Poupon and Leveaux (Indonesia) 1 1 V SH (1-VsH/2) 'I!Rr = YFRw Sw + VRsH . Sw :. 0.592 = 1.163 Sw

+ 390 Sw

:. Sw =0.38 where 1 1 -=Rt 2.86 V SH = 0.341 ; VSH(l

1

VRt

V

- VSH/2)

(1-VsHI2)

= 0.372;

RSH

1 1 FRw = Rw = 1.352 ; YFRw= 1.163

= 0.390 SH

Thus the modified Simandoux and Indonesia equations give similar Sw's which are less than the Archie Sw. The shale conductance in the basic Simandoux is already near to the measured conductance so the solution gives an unlikely optimistic value for a shaly sand.

Solution 5.3 Waxman and Thomas equation with a = 1, m = 2, n = 2

Rt

FR w Sw

= -1 { -1

F Rw

2+BQv Sw F

Sw 2 + BQvSw )

BQv = 0.046 x 0.3 mho.cm2 .meq-!.meq/cc = 0.0138 mho cm-! or ohm-! cm-! = 100 x 0.0138 = 1.38 ohm-) m- l 1 1 :. Rt = F (10 Sw 2 + 1.38 Sw) F=

= 1/0.262 = 14.79 R t = 5

:. 0.2 = 0.0676 (10 S} + 1.38 Sw) = 0.676 S} + 0.0933 Sw

:.0.676 Sw 2 + 0.0933 Sw - 0.2 = 0 Solving the quadratic _ (-0.0933 ± \1[0.0933 2 Sw 2(0.676) (see Archie solution, Sw =

-

4.676 . ( -0.2)

'\ 1FRw

VIi; = 0.544)

Modified Simandoux model

Rt

2

VSH

FRw Sw + RSH . Sw

Comparing with the Waxman Thomas equation

1) = 0.4

324

PETROLEUM ENGINEERING: PRINCIPLES AND PRACTICE VSH

BQv

BQv

p= RSH VSH

=

:. VSH=RsHp

1.5 X 1.38 14.79 = 0.140

i.e. it would take 14% shale with resistivity 1.5 ohm-m to get the same result as the Waxman Thomas equation .. ! 2 VSH.. _ 0.0676 2 0.14 BaslcSlmandoux R t - FRw Sw + RSH , .. 0.2 - 0.1 Sw + 1.5 0.2 = 0.676 Sw2

Sw =

+ 0.0933

1 1[0.2 - 0.0933]

V

0.676

= 0.397 Solution 5.4

(a) Prove From Darcy's law: -kA JP q=--!.t Jx

Assuming Boyle's law:

QscPo = qP and Po

3000ff

= 1 atm.

Hence: -kA

1000 ft:-I______----'

JP

Qsc = P!.t Jx kA

orQsc (b)

k

=-;

p/- P12 2L

Qsc!.t2 L p/)

= A(P12 -

6.2 x 2 x 0.018 x 2.54 XX

-

1)

127'

=0.2D

Solution 5.5 The problem requires correction of pressure so that the linear Darcy law can be used. In field units: kAAP q = 1.127 X lO-3 - ; BBLId

L

Assuming average water gradient of 0.45 psi/ft (0.433 x 1.038) and referring to a HWC datum of 5250 ft SS, static pressure at the outcrop is: PS2so = 0.45 x 5250

But pressure

= 2362.5 psi = 1450 psi at 5250 3

Hence, q = 1.127 x lO- x

q

= 2848.5 BBLId

750 x 3000 x 65 (2362.5 - 1450) 1 x 52 800 _-------

--\

j Poutcrop

---------- ---------P HWC . : : : : / / / - /

f Poutcrop at HWC datum

....."";------10 miles - - - - "

SOLUTIONS TO EXAMPLES

325

Solution 5.6 Using the equation: Qsc 2 ilL k = A(P I2 - pl)

(6.4/60) x 2 x 0.018 x 2.54 3t

(861)2 - 12) 1.272 (760

Sc for rate 1 = 0.0068 D = 6.8 mD Scfor rate 2

= 6.02mD

Scforrate 3

=5.0mD

This is because of the Klinkenberg effect. Plotting k against 11Pmean gives k L as 11Pmean

0 as 3 mD.

Solution 5.7 Assume cross-sectional area A. dh q = -A dt where q is flow rate and h is current height measured from bottom of core plug. Flow across core is: -kA I1P q=--

L

Il

But I1P = datum correction pressure difference, so: -kA pgh dh q=--=-AIl L dt L dh kPg'Jt so- = - dt h ilL

J

ho

0

ho kpg' or log., It = L t

so k = IlL pg'

Il

-J

ho

(holh)] =

ho

(lOge -h2 - loge hI ilL pg'

I1 t

Note:

pg' has to be in units such that pg' h = atm. 1 x 2 X 106 loge84 -loge 15.5 Hence k = 1.02 x 981 x 4500

=0.8D Note:

a plot of log.,h against t would be best.

Solution 5.8 Poil

50 = 50 lb/fe = .144 psi/ft = 0.3472 psi/ft

(a) Correct well pressures to 5750 ft = 1750 + 0.3472 x 750 = 2010.4 psi (b) Flowing gradient kA I1P q = -;- L 1.127 X 10-3 I1P =

1000 x 1.135 x 7 x 3000 10-3 x k x A = 1.127 X 10-3 x 150 x 150 x 1000

qllL

1.127

= 94 psi

X

326

PETROLEUM ENGINEERING: PRINCIPLES AND PRACTICE So, Powc = 2010.4 + 94 PV res = 3000

= 2104.4 psi

1000

X

X

150

X

cp

Equating production Vp = VTc· llP, then 3000 X 1000 X 150 x PVaquifer = (2104.4 _ 500) 3 X 10-6 = 93.5

X

109 x ft3

Solution 5.9 Q

Darcy's equation A

k dP

= - -; dx

for non-compressible flow

(a) Linear beds - parallel flow

P,

Q = qi + q2 + q3 Assume infinitely thin barriers between layers

llP llP .=kIAIf..tL +k2A 2 f..tL + ...

Q=qI+q2+"

q,

Q

---fIto-\

- -... Q

k,

llP =k'Af..tL

where k' is the apparent permeability and A the total area. Hencek'A = kiAI

+ k2A2 + ...

n

Therefore k' =

Lk;A; LA;

I

or if beds all same width =

fLkh·

(b) Series flow

Assume equal areas Al =A2 q, = qi

P,

=. - .

Now PI - P4

= (PI

- P2 )

Using Darcy's law L f..t LI f..t qtAk'

o

= q2 = q3 - .. + (P2 - P3 ) + (P3 - P4 )

D

P2

L2 f..t ki

B

P3

..• L,

= qi Aki + q2 A

P2

L2

+ ...

Since flow rates, cross-sections and viscosities are equal in all beds

(c) Radial flow parallel From the figure, it is noted that the same terms appear in the radial flow network as in the linear system. 2Jtkh (Pe - Pw) Q= f..tln(re/rwJ e - external boundary

w - internal boundary

P3

B

p.

327

SOLUTIONS TO EXAMPLES

hi h2

h3

qi

k,

q2

k2

q3

k3

1 ht

j

The only difference in the two systems is the manner of expressing the length over which the pressure drop occurs. All these terms are the same in each case. "[k·h· Therefore k' = - - '-' hI

(d) Radial flow series By same reasoning as in the linear case

k'

-

=

't In (r/rj_l) j=1

kj

Bed

Depth/ Length of bed

Horizontal permeability; mD

1

250 250 500 1000

25 50 100 200

2 3 4

For radial systems, wellbore = 6", and radius of effective drainage 2000' and bed 1 is adjacent to wellbore. Linear flow - parallel, and radial flow - parallel, take data lengths as bed depths and bed lengths and radii to be equal.

Linear flow in parallel k' = 250 x 25 + 250 x 50 + 500 x 100 + 1000 x 200 = 134.4 mD 2000

Radial flow in parallel "[kh

k'=T

250 x 25

+ 250 x 50 + 500 x

k' = k'

=

2000 268750 2000 = 134.4 mD

100 + 1000 x 200

328

PETROLEUM ENGINEERING: PRINCIPLES AND PRACTICE

Linear flow in series 2000 k' = 250 250 500 1000 2s +50 +100+ 200

2000 -=80mD 25

Radial flow in series In (200010.5) In 25010.5 + In 500/250 + In 10001500 + In 200011000 25 50 100 200 = 30.4mD i.e. permeability near wellbore most important.

Chapter 6 Solution 6.1 0 100 0

Pc

Sw

h

(h

=

Pc

4.4 100 33.3

5.3 90.1 40.2

5.6 82.4 42.4

7.6 60.0 57.5

10.5 43.7 79.6

15.7 32.2 119.0

)

(Pw - Po)/144

200 -

.e

a; >

...

c

150 rl-

50-

c

'CD

\0

I-

100 I-

:I:

> 0

.Q

=0.31

o

I-

1: c>

CI)

\\-+-\-----samPle location Sw

I-

-

CI)

Crest

"-----0-0

OWC at 33 ft relative

0.1

I

0.2

J

0.3

I

0.4

I

0.5

I

0.6

I

0.7

Sw (fraction)---

Fig. A6.1 Saturation distribution.

I

0.8

I

0.9

J

1.0

35.0 29.8 265.3

329

SOLUTIONS TO EXAMPLES

Note that the oil-water contact is at Sw = 1.0, not at Pc =0. At 100 ft above owe, Sw = 0.31 (135 ft relative) ISw dh S =--

h

w

From area under Sw against h curve: Sw = 0.37

Solution 6.2 100 0 0

100 4.4 65.1

90.1 5.3 78.4

82.4 5.6 82.9

60.0 7.6 112.5

43.7 10.5 155.4

32.2 15.7 232.4

29.8 35.0 518.0

f(J)=PcYf

0

1534.4

1847.9

1954.0

2651.7

3662.8

5477.7

12209.0

(PC)Hg

0

110.7

133.3

140.9

191.2

264.1

395.0

880.4

Sw (Pc)O-w (PdHg

for25mD and 0 = 0.13

Solution 6.3 For the laboratory data YkTcj>c = (150/0.22) 0.5 ](sw) vs Sw relationship is calculated.

= 26.11 and using ](Sw) =

CJ cos

Vi cj>

with CJ cos e = 72 dyne/cm the

] (Sw) = 0.363 Pc (Sw)Jab 1.0

o

1.0 0.363

0.9 1.451

0.8 2.176

0.7 2.901

0.6 3.445

0.5 4.862

0.4 4.968

0.3 5.984

0.2 8.341

0.2 36.27

0.3 3.451

0.2 4.846

0.2 21.07

cos e v'kicj>

](sw> CJ

At reservoir conditions PC(Sw)", =

for CJ cos e = 26 and v'kTcj> = 44.72 Pc (Sw)", = 0.581] (Sw) and the reservoir condition Pc curve is therefore calculated as 1.0

o

1.0 0.211

0.9 0.843

0.8 1.264

0.7 1.685

0.6 2.00

0.5 2.823

0.4 2.886

For the reservoir specific gravity of oil and water given

Ap = (1.026-0.785) = 0.241 The relationship between capillary pressure and height H above FWL is, in the units required, pc(sw) = 0.433 HAp :. H= Pc(sw) 0.104 Using the threshold value of pc(sw) (= Pct) as the observed oil water contact, then 0.211 = 2 ft above the FWL 0.104 4.85 = 0-- = 46.5 ft above the FWL .104

Howe = - -

H TIZ

330

PETROLEUM ENGINEERING: PRINCIPLES AND PRACTICE

Chapter 7

Solution 7.1 From Darcy's law modified for effective permeability in horizontal linear flow qo!-loL qw!-lw L Ko (s) = A I1P and Kw (s) = A I1P w

o

Assuming zero capillary pressure (Pc atmospheres for I1P, then:

Ke (md)

= 0 = Po - Pw) so I1Po = I1P w = I1P, and using Darcy units of eels for rate and

q!-l [(4) (9) (1000)] n: (3.2)2 3600

= I1P

For oil Ko =

(9.14)

qw For water Kw = I1P (5.0) Ko Kw For Kro= Krw =

o (cw)

O(cw)

90 Ko(cw) = 49.25 (9.14) = 16.7 md

15.0 19.8 25.1 32.1 41.0 54.9 68.1

o

1.0 0.452 0.30 0.20 0.12 0.05

0.017 0.025 0.049 0.075 0.156 0.249

o

These data are plotted in Fig. A 7.1

1.0

0

0.9 0.8

t

0.7 0.6 0.5

II

OA 0.3 0

0.2 0.1

I

0.8

swFig. A7.1 Steady-state relative permeability.

I

10

331

SOLUTIONS TO EXAMPLES

Solution 7.2 For pressure maintenance, the oil rate in RB/D is

10 000 x 1.2765 = 12765 RB/D The end points of the relative permeability curve are K ro ' Krw '

= 0.9 at Swi = 0.28 = 0.7 at Sor = 0.35

The ratio Ilw is then calculated from the given end point mobility ratio of 2.778. flo k rw ' flo flw krw' 0.7 Since M' = flw • k ro ' , then flo = M' k ro ' = 2.778 (0.9) = 0.28 The fractional flow curve can now be calculated for the horizontal reservoir: 1

fw

=

1 + 0.28

fk:: k } l 0.28

0.30

0.35

0.45

0.55

0.60

0.65

o

0.082

0.295

0.708

0.931

0.984

1.00

A line tangential to the fractional flow curve from Sw = 0.28 gives the tangent at Swf = 0.4 (fw withfw = 1 at Sw = 0.505. The gradient of this tangent[dfwfdSwlswis 4.44.

= 0.535) and the intercept

From Buckley-Leverett theory the constant rate frontal advance of the 40% saturation front is:

Xflday

=

Xflday =

q(t) (5.615) (A)(
[df

]

dS w

swf

(12765) (5.615) (4.44) (5280) (50) (0.25) = 4.82 ftlday

For a system of 5280 ft, breakthrough therefore occurs in 1095 days (= 3 years) At year 4 the pore volume injected is _ _4-,-(3_65...:.)-,-(1_2_7_65.:....)..:.-(5_.6_15..;...)_ =0.3PV (5280) (50) (5280) (0.25) and

dflds] Swe

= _1_ = 3.33 0.3

The tangent of gradient 3.33 to the fractional flow curve at saturations greater than frontal occurs at Swe dfw (from a plot of dS vs Sw)· w

At this saturation (Swe),fwe

= 0.71, the reservoir condition water cut.

The average saturation remaining in the reservoir is given by the Welge equation as: foe Sw = Swe + [dfwfdSwls we

_ Sw =

The

0.45 +

(1-0.71)_ 3.33 :. Sw = 0.537

factor is thus: S -S . RF = W WI = 0.36 1- Swi

= 0.45

332

PETROLEUM ENGINEERING: PRINCIPLES AND PRACTICE

Solution 7.3 The critical injection rate for gas is given in field units of SCF/D as: 4.9 x 10-4 k k r / A (Yg - Yo) sin a: q SCFID = Ilg Bg (M - 1) where Bg is in units of RBISCF and a: is negative for updip injection. The density difference in terms of specific gravity is: 17 -48 Ay = 62A = -0.4968 0.5 (1.8)

M'

= 0.028 (0.9) = 35.71

sin (-10°) = -0.1736 4.9 x 10-4 (800) (0.5) (8000) (100) (-0.4968) (-0.1736) :. qcrit = (0.028) (35.71 - 1) (7.5 x 10-4)

= 18.589 MMSCF/D The rate of injection proposed (15 MMSCFID) is less than the critical rate and might almost lead to a stable displacement. The oil rate expected prior to breakthrough is therefore: 15 x 106 x 7.5 X 10-4 Qo = 1.125 = 10 MSTB/D

Solution 7.4 1.0 I

I

O8 \

t .

. :--1:."

0.6

Distribution after 0.5 yrs

\

I

i ./ Calculated frontal

iY

i i i i j

0.4

0.2

From the given data the saturation is plotted as shown in Fig. A 7.2 h

= 9434 rbld = 100'

Dip = 6° k = 276 mD = 0.215 A= 800 000 ft2

w = 8000' = 0.04

Ay

The fractional

curv(e is calculated as

=

1 + 1.127

fw

Ilw

kro

k rw

110

1+-·-

I

---.l....-_!.-

Fig. A7.2 Saturation distributions

ql

position

110 = 1.51 cp Ilw = 0.83 cp

X 10-3

[ qt 110

-

]1 0.4335 Ay sina:

Initio I distribution

SOLUTIONS TO EXAMPLES

333

The results are shown in Fig. A 7.3

/ .....

1.0 0.9 0.8 0.7

t

-

I

•

0.6 0.5 0.4

/

0.3 0.2

/

•

/.

0.1

• ...,.. I 0.2

0

I 0.4

Sw

I 0.8

I 0.6

I

1.0

Fig. A7.3 Fractional flow curve.

Therefore:

Sw

0.16

0.25

0.35

0.45

0.55

0.65

0.75

0.79

fw

0

0.036

0.127

0.344

0.64

0.88

0.98

1.0

Since there is no uniform saturation distribution initially a material balance solution is used:

(LlX)s"'j

= 5.615 Aq, At
[Afw 1 LlS w

S"'j

= 0.308 At

[Llfw LlS

w

1 for At in days S"'j

2.5

t

2.0

1.5 ;-', 1.0

-(/)

-0

0.5

Sw

1.0

Fig. A7.4 Slopes of fractional flow curve. The slope of the fractional flow curve as a function of saturation is plotted in Fig. A 7.4. Selecting saturations ForSw = 0.79 t(yrs)

X

0.5 1.0 2.0

10ft 10 + 23.5 = 33.5 10 + 47 = 57 10 + 94 = 104

o

ForSw

= 0.75

t (yrs)

X

0.5 1.0 2.0

12 ft 12 + 36.5 = 48.5 12 + 73.0 = 85 12 + 146 = 158

o

ForS w

= 0.7

t(yrs)

X

0.5 1.0 2.0

15 ft 15 + 56.2 = 71.2 15 + 112.4 = 127.4 15 + 225 = 240

o

334

PETROLEUM ENGINEERING: PRINCIPLES AND PRACTICE

At 0.5 years the saturation distribution is shown on Fig. A7.2 and is represented in 10' increments. 5.615 qt Llt ) <j>A = 56.22

( Note:

Llx

LLlx

Swi

Sw (0.5 yr)

Llx (Sw - SWi)

LLlX (Sw - SWi)

10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10

10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170

0.79 0.70 0.56 0.45 0.375 0.33 0.30 0.278 0.254 0.24 0.23 0.215 0.205 0.20 0.195 0.190 0.183

0.79 0.79 0.79 0.78 0.755 0.730 0.710 0.690 0.675 0.650 0.640 0.630 0.620 0.613 0.605 0.600 0.595

0 0.9 2.3 3.3 3.8 4.0 4.10 4.12 4.21 4.10 4.10 4.15 4.15 4.13 4.10 4.10 4.12

0 0.9 3.2 6.5 10.3 14.3 18.4 22.52 26.72 30.83 34.93 39.08 43.23 47.36 51.46 55.56 59.68

Interpolation

:. X f

From Fig. A 7.2, at X f

{ 56.22 - 55.56)

= 160 + 10 59.68 _ 55.56 = 161.6 ft from owe

= 161.6 ft, Swf = 0.60 Solution 7.5

For the particular example the problem reduces to the following tabulation, numbering layers n, from n 5, bottom to top. n 5 - _ 0.7n + 0.15 (5 - n) . _ 0.5 k Swn 5 ,Krwn - _ _ _ , K ron - 0.9_

= 0 to n = N =

f. j . - _ 'hf'j

5

5

kj 5

where: L kj = 50 + 500 + 1500 + 2000 + 500 1

kj

= 4550 mD. n

n 0 1 2 3 4 5

n

N

1

n+l 1

L kj

0 50 550 2050 4050 4550

Lk·

4550 4550 4000 2500 500 0

f. k j

N

5

5

1

1

L kj

0 0.0110 0.1209 0.4505 0.8901 1.00

Lk·

n+l 1

Lkj

krwn

kron

SWn

1.000 0.989 0.879 0.5494 0.1099 0

0 0.0055 0.0605 0.2253 0.4451 0.50

0.900 0.8901 0.7911 0.4940 0.0989 0

0.15 0.26 0.37 0.48 0.59 0.70

The resultant pseudo-relative permeability is plotted as Sw n vs j(rwn and j(ron

335

SOLUTIONS TO EXAMPLES

ChapterS

Solution 8.1 Using the relationship h + 139 = 164/sinh x the saturation vs height relation is calculated as follows: X (frac) sinh x h (ft)

0.33 0.3360 349

0.40 0.4108 260.2

0.50 0.5211 175

0.60 0.6367 118

0.70 0.7586 77

0.80 0.8881 45

0.90 1.0265 20.8

1.0 1.1752 0.55

Fig. A 8.1 shows the plot of water saturation and porosity as a function of depth. Fig. A. 8.2 shows the plot of isopach value vs area contained within the contour. In the absence of a phinimeter to measure area use metric graph paper in a simplified approach. Take 50 ft intervals from base to crest. Count squares to determine volume for each interval. Assign appropriate value of

0.20 320 280

-.-t... -... ti c

c: 0 u

240 200

-

160

>

Q)

120

.s::

80

Q)

c

-

0 .J:l C

CI

·CP

::I:

40 0 Water saturation (Sw)

Fig. AS.1

Sw and vs. depth.

PETROLEUM ENGINEERING: PRINCIPLES AND PRACTICE

336

Area within contour (acres)

Interval

No. ofsquares

Gross rock volume (lffacreft)

Saturation (Sw)

Porosity

Hydrocarbon (volume x UP HHLs)

0- 50 50-100 100-150 150-200 200-250 250-300 300-350

46 35 26.7 21.5 17.0 11.3 3.0

115.00 87.50 66.75 53.75 42.50 28.25 7.50 L =401.250

0.875 0.70 0.57 0.49 0.43 0.39 0.36

0.160 0.178 0.197 0.215 0.234 0.252 0.271

16.73 36.25 43.87 45.72 43.98 33.69 10.09 L =230.326

Hydrocarbon in place = 230326250 BBLs reservoir oil = 170 x 106 BBLs stock tank/oil

Solution 8.2 The oil in place at stock tank conditions is evaluated using the relationship 7758A h cjl So Hoi

where N is in STB A is in acres h is in feet cjlSo is a fraction Hoi is in RBISTB The recoverable reserve is N.(RF) where RFis the recovery factor (fraction). Deterministically, the minimum, 'most likely', and maximum values are calculated as: minimum 'most likely' maximum

43 x 106 STB 116 x 106 STB 274 x 106 STB

SOLUTIONS TO EXAMPLES

337

The distribution functions of the reservoir parameters are shown in Fig. A 8.3. These data are interrogated randomly using a Monte Carlo approach in the recoverable reserve calculation. The resulting cumulative frequency greater than a given value plot is shown in Fig. A 8.4. The values associated with the 90%, 50% and 10% levels are as follows: at 90% the recoverable reserve is at least 72 x 106 STB at 50% the recoverable reserve is at least 120 x 106 STB at 10% the recoverable reserve is at least 185 x 106 STB

.-i

100

100 hne!

,

Area

t

•

a.

I-'l

100

t a.

I-'l

50

0

.,

•

0

\

\ 0\

•

100

cf>So

t 0: 50 I-'l

100

t a.

50

0

•

•\

100

I-'l

0\

RF

50

•

\0

0

Fig. A8.3 Distribution functions.

cf>

\

•

338

PETROLEUM ENGINEERING: PRINCIPLES AND PRACTICE

100 -0

.l!! c 0 'ii .E c:

c

90 80

.,.....

C

c>

70

.!!!

., 60 =>

C

E: £i

50

..c

40

c>

30

c

.,a. 0

c

., .,ea. .,

C

C

:;

E => u

20 10 260

20 10 6 STS----

Fig. A8.4 Recoverable reserves distribution.

Chapter 9 Solution 9.1

Kt

to

=
with (a) to = 1481 (b) to = 14815 (c) to = 7.4 X 10-3

[-+

p;-p=

Solution 9.2

1


For (a) x = 4Kt = 4.2 x 10-3 as x is small

- E;(-x) = - 0.5772 -Jogex

= 4.895 Hence AP = 22.72 atmospheres For (b) x = 0.4375

From graph - E; (-x) = 0.62

Hence AP = 2.875 atmospheres For (c) x = 0.49

From graph - E; (-x) = 0.55

Hence AP = 64 atmospheres

339

SOLUTIONS TO EXAMPLES

Solution 9.3 From the plot shown in Fig. A 9.1, of P wi vs 10glOt

m

= 18 psi/cycle

ThenKh

=

162.6 (500) (0.5) (1.7535) 18

= 3960 mD ft 3960 KO=60=66mD

4940

m = 18 psi /cycle

4930

•

t

."

4920

•

Fig. A9.1

PwtVS

1091Ot.

Solution 9.4

f HAt) with the points in the table calculated, the slope is determined as

From a graph of P vs llog ---;;;:r 21. 7 psi/cycle ( = m).

For a reservoir rate q of 500 (1.454) rb/d (= 727 rb/d) 162.6q(..t Then, kh = = 3800 mD.ft m For h

= 120 ft then Ko = 32 mD.

The value of S = 1.151

= +7

P{h' corresponding to a Homer time function of 3.16 is 4981 psi 4981 - 4728 21. 7

32 10glO (0.135)(0.7)(17 X

10-6)(0.5)2 + 3.23

)

340

PETROLEUM ENGINEERING: PRINCIPLES AND PRACTICE

!!.Ps = 0.87 m S = 132 psi Efficiency =

4981 - 4728 - 132 4981 - 4728 = 0.5 (approx.)

t+!!.t

Data points for Horner plot on semi-log paper, P vs t

[t+!!.t]

At or a plot of p vs log At

on linear scales are as follows:

= 60 x 24 = 1440 hours Time (h)

t+!!.t !!.t

logJO {(t+ !!.t)/!!.t}

P (psi)

0.25 0.5 1.0 1.5 2.0 3.0 6.0 9.0 18.0 36.0 48.0

5761 2881 1441 961 721 481 241 161 81 41 31

3.76 3.46 3.16 2.98 2.88 2.68 2.38 2.21 1.91 1.61 1.49

4967 4974 4981 4984 4987 4991 4998 5002 5008 5014 5017

Solution 9.5 Examination of the data shows that: !!.P/day = 3 psi Assuming 1 - Sw We have NBoi and (co)e and Np N

=0.7

= NpBo/(co).!!.P

= 15 x 10-6/0.7 = 21.4 x 10-6

= 500 bId. For Bo = Boi then 500 6 10-6 x 3 = 7.8 x 10 BBL

= 21.4 X

Solution 9.6 Rate 1 2 3 4

Q (MSCFID)

(!!.p2) total

7290 16737 25724 35522

42181 126120 237 162 391616

shut in

6

HenceKh

=

Bg =

162.6 (q Bg) m

0.7404 0.6201 0.5119 0.4472 0.3979 0.3274 0.2788 0.2430

IA.

0.00504zT P BBLlscf = 0.00103·

2 2_14241A.zTQ { ) NowPe -Pw Kh InO.606re/rw +Sl

= 0.855 Q {8.93 + Sl}

P

Time since

5

= 7 psi/cycle from Homer plot

Kh = 14 500 K= 72mD

Assume tflow prior to build up is 4.5 hours:

1 1.5 2 2.5 3 4

Slope

2509.7 2510.7 2511.3 2511.7 2512.1 2512.5 2513.0 2513.2

Or Sl

=

Pe2 - Pw2 0.855 Q - 8.93

Rate 1 2 3 4

Q -2.16 -0.12 +1.85 +3.96

7290 35522

SOLUTIONS TO EXAMPLES D

= AS/AQ = 2.16 x

S

= -3.7

f3

=

341

10.4

DhWw = 2.865 x 109 2.22510 15 KYg 48211

f3theoretical

= <1>5.5 v'K

= 1.80 X

109

This is order of magnitude agreement. The inertial pressure term Ap2inertial is calculated from B as follows: 3.16 x1O-1ZygTzf3

B=

2

h rw

= 0.000185

Hence (Ap2)inertial is as follows: Q(MSCFID)

Rate 1 2 3 4

7290 16737 25724 35522

9851 51928 122665 233906

42181 126120 237162 391616

Comparison between the numbers shows that at high rates the inertial drop is over half the total drop, and that in this case only the inertial drop is close to the total drop of the previous rate. The AOF plot is shown in Fig. A 9.2 and when Ap2 is equal to Pe 2 (6.32X106psi2) then QAOF = 220 X 106 SCFld

AOF= 220mm SCF/D

C B -----------------e--------------------------------------------.

i

I

I I

I I I

I I

I I

o'"

I

n

=0.65 [= distance AB] ! distance Be

I I I I I I

I I I

I I

I

tA I

Fig. A9.2 AOF determination.

342

PETROLEUM ENGINEERING: PRINCIPLES AND PRACTICE

Chapter 10 Solution 10.1 Volume ofreservoir

= V = 100 x

(5280? x 500 cu.ft

Volume ofreservoir available for fluid

= (1 - Sw) cj>V = Vr = 0.65 x 0.12 x V

_ 2000 l520 _ 12 Vsc - 14.7 0.825595 - 15.6 x 10 SCF (1) Assume no water influx,

Vr Initial moles in place ni = Pi.RT Z,

r

= (PVi) RT

SId

Vi - gas in place measured at standard conditions. Abandonment moles left in place na

Pa Vr (PVa) =--n= RT Za r

SId

Gas recovered !1n = -Vr (Pi - - -Pj = -Ps !1 V RTr Zi Za RTs Recoverable gas measured at stan dar conditions =Vr -Ts -

(Pi - - Pa) TrPs Zi Za

500 At 500 psi, reduced pressure Ppr = 67 6

1.

Therefore recoverable gas

Z

= 0.94

Vr TsPi Z, Pa) =P T 1 - - -P r

s Zl

_

- 15.6

Recovery factor

= 0.75 Za

X 12( 10

I

0.825 500) 1 - 0.94 ·2000

=

15.6

X

10 12 (1 - 0.219)

=

12.2

X

1012 SCF

12.2

= 15.6 = 78%

Solution 10.2 Radial flow of oil q0

2nkoh

= --Bflo

0

Radial flow of gas qg =

2nk h

=..:::£.::B flg g

!1Pg re

log e

rw

and if the capillary pressure gradient is negligible, and the pressure drop over the same radii are considered, _ kgfloBo qo - ko flgBg

343

SOLUTIONS TO EXAMPLES

To this must be added the gas evolved from solution in the oil. The total measured gas-oil ratio will then be: kg flo Bo + Rs ko flg Bg For the figures given: (96)(0.8)(1.363) (1000)(0.018)(0.001162) + 500

= 5005 + 500 = 5505 SCF/STB

Solution 10.3

= Np B, + Bg(Rp - R si ) - N(B, - Bo;)

We

(i) At cumulative 1.715 x 106 BBL (P

Wei = (1.715

X

106) [1.437

=

1600)

+ 0.0015(878 - 690)] - 14.5 X 106[1.437 - l.363]

= 1.875 X 106

(ii) At cumulative 3.43 We2

= (3.43 = 4.112 x 106

X 106)

106 BBL (P = 1300)

X

[1.594 + 0.0019(996 -690)] - 14.5 [1.594 - l.363]

At P = 1000 estimated water influx = 6.375 x 106 (from trend) N(B, - B oi ) + We Np= B, + Bg(Rp - R si ) 14.5(1. 748 - l.363) x 106 + 6 375 000 1.748

+ 0.0025(1100- 690)

= 4.312 X 106 BBL Solution 10.4 Total hydrocarbon in place = i:n: ,-2h<j> (1 - Sw) 9 750 x 0.17 x 0.76 = :n: (528W 5.615 = 4.54 x 109 BBL

3

Since bubble-point is 1850 psi, this must be pressure at any gas-oil contact. Elevation of gas-oil contact above oil-water contact is: (1919 - 1850) 144 = 229 ft 43.4 This is less than hydrocarbon column so gas-oil contact exists at 4031 ft SS Height of gas zone = 750 - 229 = 521 ft 2

2

Therefore, oil in place

=

Ratio gas/total

= r h = h = (520)3 750 = 0.34 3

0.66 x 4.54 x 109 1.363

= 2.198 x

109 STB

344

PETROLEUM ENGINEERING: PRINCIPLES AND PRACTICE

m = 0.5 (= 113 +- 2/3) Material balance Np[B/ + BiRp - Rs;)] - We

N=

+ Wp Bw

mBoi (B/ - Bo;) + B. (Bg - Bgi) gl

At 1600 psi:

B/ + BiRp - Rs;)

= 1.437 + 0.00150(1100 = 2.0520

690)

m Boi 0.5 (1.363) B/ - Boi + Bgi (Bg - Bg;) = 1.437 - 1.363 + 0.00124 (0.0015 - 0.00124)

= 0.0740 + 0.1429 = 0.2169 We

= (2.052 X 3.1 X 108 + 31 = 1.904 X 108 BBL

X

106)

-

2.198

X

109

X

0.2169

At 1300 psi:

B/ + Bg(Rp - R si) = 1.594 + 0.0019(1350 - 690) = 2.8480 m Boi 0.5 (1.363) B/ - Boi + Bgi (Bg - Bgi) = 1.594 - 1.363 + 0.00124 (0.0019 - 0.00124)

= 0.5937

We

= 5.5 X 108 X 2.8480 + 55 X 106 = 3.164 X 109

2.198

X

109

X

0.5937

This is not simply linear with pressure but extrapolation is reasonably straightforward and water influx at 1000 psi is estimated at 3.75 X 109 BBL

B/ + BiRp - Rs;)

= 1.748 + 0.0025(1800 - 690) = 4.523

m Boi (0.5)1.363 B/ - Boi + Bgi (Bg - Bgi) = 1.748 - 1.363 + 0.00124 (0.00250 - 0.00124)

= 1.0775 (denominator term)

+ We - Wp Bw N=------'-----'-P B/ + Bg (Rp - R si ) 2.198 X 109 X 1.0775 + 3.75 X 108 - 63 X 106 N

X

denom.)

4.5230

= 5.926 = 590 X 106 STB X 108

Solution 10.5 . . . _ GBgi _ 120.7 X 109 X 6.486 X 10-4 _ Gas cap. OIl zone ratio m - NBoi 300 X 106 X 1.3050 - 0.2

From PVT data the values of B o, Rs and Bg at 4300 psi can be estimated by linear interpolation as:

Bo = 1.228 RBISTB; Rs = 338 SCF/STB; Bg = 7.545

X

10-4 RB/SCF

345

SPE NOMENCLATURE AND UNITS

From production data the value of Rp is calculated as GpfNp to give the following table.

Time

pepsi)

1.1.80 1.1.81 1.1.82

5000 4300 4250

Using the relationship F = N(E T )

o

550 600

R.(SCFISTB) 500 338 325

Units

1.1.B1

1.1.B2

106 RB

30.28

62.41

RB/STB

0.0420

0.0435

RB/STB

0.2131

0.2302

RB/STB

0.0061

0.0065

RB/STB

0.0887

0.0960

106 BBL

3.67

8.06

o o

o

RpCSCFISTB)

21.9 43.8

25.55

+ We + WinjBwinj the following is calculated where:

E T = mEg + Eo + Efw

(a) F = Np Bo

+ (Rp - Rs)Bg

(b) Eo = (Bo - B oi ) + (Rsi - Rs) Bg

Bo;

(c) (d)

Efw = (1 + m) Boi t!..P

(e) ET = mEg

[c.$W + cf1 1 - Sw

+ Eo + Efw

(f) We = F - N (E T )

-

Winj

Bwinj

Solution 10.6 The dimensionless radius ratio is: r aquifer 81000 re = =--=9 D r oil zone 9000 The dimensionless time tD is related to real time by: 2.309 k t (years) 2.309 (707t) tD = ql, = (0.18)(7x 10-6) (0.4) (900W = 40t The instantaneous pressure drops which at the start of each year are equivalent to the continuous pressure declines are:

Pi -

PI

t!..Po = - - 2 - =

5870 - 5020 . 2 = 425 pSI

Pi - P2 2

5870 - 4310

PI - P3

5020 - 3850 2

t!..P I = - - - = t!..Pz = --2-=

2

= 780 psi 585 psi

The aquifer constant is:

U=

1.119 fh c rb

U = 1.119 x 1 x 0.18 x 200 x (7 x 10-6) x (900W U = 22841 BBLIpsi

From tables or charts for dimensionless influx at reo = 9 we have: 40 80 120

21 29 34

346

PETROLEUM ENGINEERING: PRINCIPLES AND PRACTICE j=n-I

From We = U

L APWD (TD -

(Dj)

j=O

Wei = 22841 [425 (21)] = 203.9 X 106 BBL

+ 780 (21)] = 655.7 X 106 BBL We3 = 22841 [425 (34) + 780 (29) + 585 (21)] = 1127.3 X 106 BBL We2 = 22841 [425 (29)

Chapter 11 Solution 11.1 1 [0.00708 k kro h PI=-

",,"0 In -re - 0.75 + S rw

1

For re = 1500 ft rw=0.5ft S= +4 K ro = 0.6 h = 100ft k= 1325mD 50

PI=!.to

..

0.5

5

50

500

5000

100

10

1

0.1

0.01

-----------------------------------------------PI

Solution 11.2 The injectivity index is given in field units by: 0.00708 k k rw h !'w In

- 0.75+

s]

Assuming all other factors equal then

Solution 11.3 Use is made of the plot in Fig. 11.4 which correlates areal sweep efficiency E A as a function of end point mobility ration (M') for different fractional injection volumes, VD. Kw' !.to 0.4 3.4 M'=-·- =--·-=4 !.tw Ko' 0.4 0.85 The volume of injected fluid, in reservoir barrels, after 10 years is: 10

X

365.25

X

53 000 x 1.005

= 1.945 x 108 RB

347

SOLUTIONS TO EXAMPLES The displaceable pore volume (= PV (I-SoT - Swi» is given in reservoir barrels as follows: (4

x 5280) (1 x 5280) (98) (0.25) 5.615

= 1.946

[1 - 0.3 - 0.3]

x lOS RB

1.945 X 108 V D = 1.946 X 108 - 1 From Fig. 11.4 the value of EA corresponding to M' = 4 and VD = 1 is 0.7

Solution 11.4 For stable cone formation = g' X (Pw - Po)

For

(in psi), and cone height X (in feet) and density difference as specific gravities then 62.4 = 144 (1.01 - 0.81) 50

= 4.33 psi Chapter 12 Solution 12.1 (a) In field units 1.25 (4000) U= 70(1500)

0.0476 BID - ft3

The viscous-gravity force ratio is calculated from 2050 UItaL R v_g = (

=

Po

_

Ps

)

kh

2050 (0.0476) (0.5) (1500) (0.8 - 0.4) (130) (70)

= 20 (b)

(a)

Solvent

Oil

Oil

Regions I and n

Region

m

Region I : Single gravity override tongue

(c)

Solvent

Region n : Single tongue but sweepout independent of RV- G for given M Regionill: Transition region with secondary fingers below main tongue RegionN: Multiple fingers with sweepout independent of RV- Gfor given M

Region N

Fig. A12.1 Displacement regimes.

348

PETROLEUM ENGINEERING: PRINCIPLES AND PRACTICE

For a mobility ratio, M represented by !1ot'1ls (= 25), Figure A 12.2 shows a breakthrough sweep efficiency of about 15% and a flow dominated by gravity tonguing. (Fig A 12.1) (b) In field units 1.25 (1000) u = 30 (2000)

= 0.0208 BID -

ft

2

The viscous gravity force ratio requires an approximation of permeability as:

k=VKv·Kh :. k = «1) (3»°·5 = 1. 73md Then 2050 (0.0208) (0.36) (2000)

R v_g = (0.75 - 0.64) (1.73) (30)

= 5378 For a mobility ratio of M (IlJIls = 0.36/0.055 = 6.55), Figures A 12.1 and A 12.2 show a breakthrough sweepout efficiency of around 50% and a flow dominated by viscous fingering. 100 >u c: Q)

·u

:::

Q) -; 60

t-Regionill-i------Region N

0 0Q) Q)

M=6.5

til

.£: CI> :::J

e

.£: 0

E co

M =27 Region N I-----Regionll---·I....• -----Region ill----------<-t-''-i

100

10 .

.

.

..

Viscous-gravity force ratiO (R V- G)' field units,

1000 2050UJ-L L _

At kh

0

,

10000

(B/O-FT 2 )(CP)(FT) '----,3:;--'-'----'--'----'

(G/cm )(md) (FT)

Fig. A12.2 Breakthrough sweep efficiency.

Solution 12.2 The tie lines for the system join the equilibrium compositions of systems A and B in the two phase region. The compositions are plotted in Figure A12.3 (a) The critical point (CP) is estimated where the limiting tie line becomes tangential to the phase envelope and has the composition, wt%, 21 % surfactant, 67% oil; 12% brine. (b) The point with the composition 4% surfactant and 77% oil is given on Figure A12.3 as point A. From the slope of tie lines in this region the equilibrium phase compositions are AI and A2 with weight percents estimated as: AI 10% oil; 10% surfactant; 80% brine A2 97% oil; 2% surfactant; 1% brine For an original 200 g mixture containing 8g surfactant, 154 g oil, 38 g brine

349

SOLUTIONS TO EXAMPLES 100% Surfactant

\

Wt% Brine

100% 100t Brine

0

30

40 50 Wt%

60

70

Wt% Surfactant

80

100% Oil

Fig. A12.3 Ternary diagram. The tie line ratios give: wt of AI phase 3/13 x 200 = 46 g wt of A2 phase 10/13 x 200 = 154 g :. Composition of AI

= 4.6 g oil

:. Composition of A2

= 149.5 g oil

4.6 g surfactant 36.8 g brine

3.0 g surfactant 1.5 g brine

(c) On Figure A 12.3 the composition 20% oil and 80% brine is shown at location B. A line from B to the 100% surfactant point leaves the two phase region at location B', having a composition oil 16.5%, surfactant 17.5%, brine 66%. The oil + brine weight is 100 g and would constitute 82.5% of the mixture, so surfactant needed is 0.175 (100/0.825) = 21.2 g. (d) On Figure A 12.3, location 1 is 10% oil, 40% surfactant and location 2 is 50% oil, 40% surfactant. They are in a single phase region and the resulting mixture contains 30% oil, 40% surfactant and 30% brine, as denoted by position 3. (e) On Figure A 12.3, location 4 is 12% surfactant, 5% oil and location 5 is 20% surfactant, 77% oil. The mixture weight is 200 g and contains 41 % oil, 16% surfactant, and 43% brine. It is shown as location 6. The mixture is in the two phase region and equilibriates to compositions C and D on the equilibrium tie line through location 6. The compositions are: C: 58% brine; 21.5% oil, 20.5% surfactant (146 g total) D: 94% oil; 5% surfactant; 1% brine (54 g total)

350

PETROLEUM ENGINEERING: PRINCIPLES AND PRACTICE

Solution 12.3 For conventional production

tJ.o In I

208.71AO. S

- 0.964

(0.003541 (1000)(60»

700

150 ...

H2O':': (3») - 0964]

= 161 rbld For 9 acre spacing and a 200 psi differential.

For thermal stimulation and steam injection a 5 fold improvement in flow resistance between producers and injectors would lead to rates around 800 bid. To determine the steady state production/injection time at which such rates will lead to 50% of the pattern volume being occupied by steam we can conduct the following analysis: The cumulative heat injected into the reservoir, Q;, can be calculated from heat injection rate, using the mass rate of injection Wi

tQi = Wi [ Cw

fsdhLYdh

[C

= qinj (5.615) (62.4)

w

]

(380 - 100)

+ 0.75 (845)]

The average specific heat, C w , over the temperature range 380 - 100°F is given by: Cw = hw(Ts) - hw(Tres) 355 - 69 1.02 Btullb m - degF Ts - T res 380 - 100 :. Qi = t·

qinj .

322118 Btu

The ratio of latent heat to total energy injected, fhv is calculated from: _ {

fhv -

Cw

1 + fsdb LVdb

)"1 _{ -

1+

(1.02 (380 - 100) 0.75 (845)

)"1

= 0.689 Figure A 12.4 can now be used to estimate the thermal efficiency of the steam zone, Ehs , at different values of The values of to are given from: to

= 4t

MR

h2

45]2 [0.75] = 4t [ 35 (60?

= 0.00138t days or 0.504 t years The following table may now be constructed usingfhv = 0.689 on Fig A 12.4. t (yr)

1.0 1.5 2.0 2.5

t(days) 365.25 547.9 730.5 913.1

to 0.5 0.75 1.0 1.25

The volume of a steam zone, V., is in general given by: QiEhs Vs = 43560MR AT

0.64 0.59 0.56 0.52

233.8 323.3 409.1 474.8

351

SOLUTIONS TO EXAMPLES

.

1.0

.c:.

W

oJ c:: 0

N

E c

2If)

15

0.6

fhv (ratio latent heat to total energy injected) =

A

>() c:: Q)

-

.<3

;;::

0.4

Q)

c

E

Q)

.c::

I-

1.O

0.50

0.33 0.23

0.167 0.091

0.2

0

0.01

100

0.1 Dimensionless time, tD

Fig. A12.4 Thermal efficiency For the case of 50% steam volume in the pattern of area A acres then _ (43560 MR Qi - 0.5Ah E

AT)

hs

Equating values of Qi we obtain the relationship 0.5 (9) (60) (43560) (35) (280) 322118 qinj . t = E hs

where t is in days 357817.5

That is

The injection rates needed to provide 50% pattern volume of steam at the following times are therefore as shown in the following table. t

(rblD) 1531 1107 875 753

(yr) 1.0 1.5 2.0 2.5

qinj

These data may be further evaluated in terms of steam injection equipment capacity and project economics.

Solution 12.4 The wet condensate gas volume is obtained from the volumetric calculation: Ah n (,5) V = g B .

sc

g.

In terms of standard cubic feet this is: 1 Vsc = [It(3 x 5280? 300 (0.18) (0.75)]

s. gl

352

PETROLEUM ENGINEERING: PRINCIPLES AND PRACTICE

Vsc

=

3.1937 X 1010 B SCF g;

In order to find Bg; we need the super compressibility factor z which can be obtained from Fig 4.7 using the reservoir condition molecular weight or gas gravity. The oil molecular weight is given by 44.3 PL Mo = (1.03 - PL) 141.5 NOWPL= API + 131.5 0 .75

:.Mo = 119 The weight associated with a stock tank barrel of liquid is given by: W

= (5.615

= 262.78

x 62.4

= 0.75)

+

5000 (0.58) (28.97) 379.4

+ 221.44

= 484.22 The number of moles associated with this weight is 5000 (62.4) (0.75) (5.615) n = 379.4 + 119

n

= 13.18 + 2.21

n = 15.39 W

484.22

:. MW(res)

= -;;= 15.39 = 31.46

and Yg(res)

= 28.97 = 28.97 = 1.086

i.e. Yg(res)

= 1.09

MW(res)

From Fig 4.7,

31.46

= 620 and Tpc = 465

P pc

From reservoir datum conditions 4500 670 P pr = 620 = 726 and Tpr = 465 = 1.44 So, from Fig 4.7 z

= 0.925

Then:

B g;=

(0.02829) (0.925) (670) 4500

= 3.8962 x

10-3 RCF/SCF 3.1937 x 1010 Vsc = 3.8962 X 10-3

= 8.197 X

1012 SCF

The dry gas volume G -- [ 8.197 x 10 12] [5000/379.4] 15.39 G

= (8.197 x

G = 7.019

X

J(p) (0.8563)

10 12 SCF

Similarly the oil volume Vsc 8.197 x 10 12 NX R = 5000 N

= 1.639 X 109 STB

SOLUTIONS TO EXAMPLES

353

Chapter 13 Solution 13.1 Using the relationship that the depth equivalent of the total head is equal to the sum of the depth equivalents of the well head pressure and the well depth, then:

DT = D whp + Dwell (a) From Fig. A 13.1 at a well head pressure of 400 psi then DwhQ = 3700 ft. Since Dwell = 6000ft then DT = 9700 ft. At the GOR of 200 scf/stb the pressure at a depth equivalent of Y700 ft is read as 2400 psi. (b) From Fig. A 13.2 at the bottom hole pressure of 1200 psi and GOR of 500 scflstb the depth equivalent Dr. is read as 8900 ft. Since Dwell is 5000 ft then Dwhp is 3900 ft. The well head pressure is read from the graph at 3900 ft as 360 psi.

Vertical flawing_pressure gradients (all oil)

3 Q; 0 0

$2 .!: Q)

..J

3

Tubing Size 4 in.I.D. Producing Rate 3000 Bbls/day Oil API Gravity 35° API Gas Specific Gravity 0.65 Average Flowing Temp. 140°F

Q;

4

0 0

$2

5

.!:

.c

C. c:

Vertical flowing_pressure gradients (all oill

Tubing Size 4 in.I.D. Producing Rate 2000 Bbls/day Oil API Gravity 35° API Gas Specific Gravity 0.65 Average Flowing Temp. 140° F

4 5

.c

C. c:

6

Q)

..J

7

6

7

8

8

9

9

10 10

Fig.A13.2

Fig. A13.1

Solution 13.2 The maximum production rate qrnax can be evaluated using the Vogel relationship, withp, the static pressure, i.e.

H.2

1 - 0.2

= 0.619 3315 therefore, qrnax = 0.619

08

] - 08

[= r

= 5355 bid

PETROLEUM ENGINEERING: PRINCIPLES AND PRACTICE

354

Pressure in 100 PSIG

Verlical flowing_p.ressure gradienls (all oil)

Verlical flowing_pressure gradienls (all oil)

3

Tubing Size 4 in.I.O. Producing Rale 4000 8bls/day Oil API Gravily 35° API Gas Specific Gravily 0.65 Average Flowing Temp. 140°F

4in.1.0. Tubing Size Producing Rale 1000 8bls/day Oil API Gravily 35° API Gas Specific Gravily 0.65 Average Flowing Temp. 140°F

3

Q; 0 0

$2

.S;

4

Q;

5

0 0

$2

.t::

"6> c Q)

...J

4 5

.S;

6

.t::

"6> 6 c Q)

...J

7

7 8 8 9 9

10

10

Fig. A13.3 Fig. A13.4

Verlical flowing_pressure gradienls (all oil) Size 4 in. 1.0. Producing Rale 5000 8bls/day Oil API Gravily 35° API Gas Specific Gravily 0.65 Average Flowing Temp. 140°F

3 Q;

0 0

$2 .S;

4 5

.t::

"6> c Q)

...J

6

7 8 9 10

Fig.A13.5

355

SOLUTIONS TO EXAMPLES

From Fig. A 13.1 to A 13.5 the different vertical flowing pressure gradient curves at different rates are found for 4 in. tubing and a GOR of 200 SCF/STB. The total head depth is obtained as the sum of the well depth and the depth equivalent to a tubing head pressure of 400 psig. The flowing bottom hole pressure equivalent to the total head depth is recorded as a function of flow rate. It can be seen that the bottom hole pressure is essentially independent of rate at this condition and is 2200 PSi[. Hence

q = qmax

=

(2200 ) 1 - 0.2 2600

(2200

)2]

- 0.8 2600

1400 bid

Solution 13.3 For a residence time of 3 min. the volume of oil in the separator will be: (1000) (3) 3 Vo = (24) (60) = 2.083 m At 40°C and 20 bar the volumetric rate of associated gas will be V (1000) (95) (313.15) (1) 3 --II. = (24) (60) (60) (273.15) (20) = 0.06303 m Is At separator conditions the gas density Pg is given by (273.15) Pg = 1.272 (0.75) (20) (313.15)

= 16.682 kg/m3 The maximum velocity equation is then used: Umax = 0.125 [ =

796 - 16.682]0.5 16.682 mls

0.8544m/s

Since cross-sectional area = volume ratelvelocity then for an interface half way up the separator we have: n D2 0.06303 (2) (4) :.D

0.8544

= 0.4334 m

Total volume of the separator is thus twice the oil volume for an interface half way up the separator :. Vsep = 2Vo = 4.166 m3

Design length for LID = 3 gives (4.166) (4) 3D = L = nD2 .

3_(4.166)(4) 3n

:. D

= 1.209m

.. D -

and L = 3.627 m Design length for LID = 4 gives D3 = (4.166)(4)

4n :. D = 1.099 and L = 4.396 m In practice the separator design would be based on a standard size selected to be nearest the size calculated.

Index

Abandonment pressure 159 absolute permeability 102 AFE (authorisation for expenditure) document 23, 24-S Amerada gauge 147, 148 API (American Petroleum Institute) gravity and oil density 14 aquifer characteristics correlation with model 167 determination of 165-6 aquifers and pressure change 165 areal sweep efficiency 176, 182-3 Back pressure equation 143-4, 221 barrel 14 bedforms, grain size and stream power 242 biocides and injection water 229 biopolymers 197 black oil reservoir modelling, uncertainties in 24&-7 black oil systems 42 blow-out 35 blow-out preventers 34-5 blowdown 210 Boltzmann transformation 134 bond number 191 BOPs see blow-out preventers bottlenecks 219 bottom-hole sampling 52 Boyle's law method and grain volume 73 Brent Sand reservoirs 10, 11 brine disposal 186 bubble-point 41, 51, 53, 54, 55,159,220,221 bubble-point pressure 52, 54-5, 56-7,160,163,221 in volatile oil reservoirs 211 Buckley-Leverett theory 105 Buckley-LeverettlWelge technique 107, 109

Footnote: Numbers in italic indicate figures; Numbers in bold indicate tables

Capillary number 191,193 capillary pressure 93 and residual fluids 111-12 defined 92 capillary pressure data (given rock type, correlation 99 capillary pressure hysteresis 97-8 capillary suction pressure see imbibition wetting phase threshold pressure carbon dioxide in miscible displacement 195, 196 casing a well, reasons for 2S casing eccentricity 35-6 casing selection 27 main design criteria 28 casings 23, 25, 26, 28 caustic solutions 196 cementation problems 35-6 chemical flood processes 196-200 choke assembly 146 Christmas tree 36 coalescer 227 Coates and Dumanoir equation 86 combination drive material balance equation 166 compaction drive 161 complete voidage replacement 173 completion 28,29 completion for production (permanent, normal) 36 composite cores 111 compressibility 42-3,55 Compton scattering 76 conceptual models 233,245 condensate analysis 208 condensate reservoirs and liquid drop-out 208 condensate systems 42 condensing gas drive 194-5 cone height, critical 182 coning 181-2 core analysis and permeability distribution 83-4 routine 69-71, 81 presentation of results 70,71

357

358

PETROLEUM ENGINEERING: PRINCIPLES AND PRACTICE

core data and palaeogeographical reconstruction 237-8 and recognition of sand body type 238 core-derived data 68 core floods and surfactant testing 200 core for special core analysis 67, 68 core length and imbibition processes 110-11 core log 64,68 core plug experiments, concern over laboratory-derived data 113-14 core plugs 68 analysis on 65 and effective permeability 109 and fluid saturation 93-4 and oil saturation 193 and permeability 81 and porosity 72 and residual saturation 174 core porosity, compaction corrected 131 core preservation 67-8 core recovery, fluids for 31 Core gamma surface logger 68 cores 62 composite 111 correlation with wireline logs 63,65,75 data obtainable from 63 diversity of information available 64 and geological studies 68-9 and heavy oil reservoirs 202 residual fluid saturation determination 69 coring the case for 65 conventional and oriented 66 of development wells 65--{i of exploration wells 65 coring decisions 64--{i coring mud systems 66-7 corresponding states, law of 44-5,47 Cricondenbar 41 , 42 Cricondentherm 41 critical displacement rate 177 critical displacement ratio 112 critical gas (equilibrium) saturation 159 critical production rate (coning) 182 crude oil flow of in wellbore 221, 223 metering of 229 processing 226-8 cushion 147 cuttings logs 31 cyclic steam stimulation 205 Darcy (def. )79 Darcy's equation 79 data acquisition during drilling 30-1 datum correction 79-80 deltaic environments, division of 238,240, 241,242 deltaic models, use of 238-43 deltaic system model 242, 244 demulsifiers and heavy oil processing 228 depositional processes and reservoir rocks 7 dew-point 41 dew-point locus 42 diamond coring 33 differential liberation at reservoir temperature 53 displacement calculations, validation of relative permeability data for 113-14

displacement principles 173-5 drawdown testing 138 drill bits 22, 32-3 drill collars 23 drill stem testing 145 testing tools and assemblies 145-7 drilling, turbine versus rotary 33 drilling costs 23, 24, 25 drilling fluid see drilling mud drilling logs 30 drilling mud pressure, excessive 29 drilling muds 22-3 control of 28-9 main constituents 67 drilling muds and cements, rheology of 29-30 drilling optimization 32-3 drilling, special problems in cementation problems 35--{i pressure control and well kicks 34-5 stuck pipe and fishing 33-4 drillstring 23 drive mechanisms 159 dry gas reservoirs 41-2 dual porosity systems 71,73 and gravity drainage 164-5 Early (transient) time solution 138 economic factors and oil production rates 180 effective permeability 102 and wettability 108 enhanced oii recovery schemes and uncertainty 247 equity, distribution of, petroleum reservoirs 130-1 exploration well drilling 7, 8 Faults, identification of 238 faults (in-reservoir), effect on injection/production well locations 180 field processing 224 filtration, injection water treatment 229 flash liberation at reservoir temperature 52-3 flash separation tests 53-4 flooding efficiency ratio 110 flow equations, linear and radial 80-1 flow string 145 fluid contacts 12-13 multiple 12 fluid flow in porous media 78-9 fluid pairs 93 fluid pressure and overburden load 11-12 fluid pressures, hydrocarbon zone 12-13 fluid saturation, laboratory measurements and relationship with reservoir systems 93--{i fluids, recovery of by depletion 211 Forcheimer equation 143 formation breakdown pressure 30 formation density logs and interpretation of porosity 202-3 formation density tool response 75--{i formation factor see formation resistivity factor formation interval tester (FIT) 148 formation resistivity factor 74 formation tester (FT) 148 formation volume factor 14, 55 two-phase 55--{i formation volume factors B 49-51 formation waters 14 fractional flow 104--{i analysis methods 105--{i effect of dip angle and wettability 175, 177 free water level (FWL) 12,95

359

INDEX Gas cap expansion drive 163-4 gas compressibilities 48-9 gas condensate, critical properties of 210 gas condensate and volatile oil reservoirs, uncertainties in 247 gas condensate reservoirs 207-11 production methods for 209-11 gas deviation factor Z 46, 47 gas expansion during production 157 gas flow and gradient 159 gas flow and permeability 81 gas flow rate, measurement of 150, 229 gas formation volume factor 157 gas formation volume factor Bg 49-50 gas properties 45 gas recycling, gas condensate reservoirs 210 gas reinjection 186 gas reservoirs, recovery from 157-9 gas viscosities 47-8 gas-kicks 12 gas-oil ratio 14,51-2,54,159 gas-oil systems and relative permeability 103-4 gas well testing 143-5 gases, behaviour of 43-4 gases, flow of in wellbore 221 geological model, development of 237-8 geothermal gradient and hydrocarbon generation 7, 9 geothermal gradient and reservoir temperature 13 GOR see gas-oil ratio grain density 71 grain volume and Boyle's law method 73 gravity drainage and dual porosity systems 164-5 gravity segregation and recovery efficiencies 164-5 gravity stabilization and reservoir dip 175 Head loss in wellbores 221 heavy crude oil characteristics of UKCS heavy crude oils 201 general classification 200 Yen classification 200, 201 heavy oil processing 228 heavy oil recovery 200-2 heavy oil reservoirs examples of 201 permeability increase and production improvement 204 production characteristics of 203-4 properties of 202-3 and thermal energy 204-7 and uncertainty 247 heavy oil systems and thermal energy addition 204 HKW (highest known water) 12,13 homogeneous reservoirs and coning 181-2 Horner analysis 13 hydrates 224 hydrocarbon accumulation and sedimentary basins 7 hydrocarbon accumulations and formation waters 14 hydrocarbon exploitation, types of interactions 16 hydrocarbon field 7 hydrocarbon generation and geothermal gradient 7, 9 hydrocarbon pore thickness (HPT) 126--7 hydrocarbon pore volume maps 126--7 hydrocarbon properties 47 hydrocarbon recovery, improved 191-211 hydrocarbon reservoir fluids 15 hydrocarbon systems volumetric and phase behaviour 40-1 applications to field systems 41-2

hydrocarbon volume in place calculations 127-8 hydrocarbons, migration of (modelled) 93-4 hydrocarbons (commercial reservoirs), geological characteristics 62 hydrostatic gradient, regional 10-11 Ideal gas law (and modification) 43 imbibition processes and core length 110-11 liquid 104 imbibition wetting phase threshold pressure 97 in-place volume 122 inflow performance relationship, 220 dimensionless, for oil wells 220-1 for gas wells 221 injection fluids, compatibility with reservoir fluids 183-4 injection fluids, quality of 183-6 injection water, viscosity of 184 injection water treatment 229 injectivity index 174, insert bits 33 isobaric thermal expansion coefficient 43 isocapacity maps 126 isochores 124 isochronal testing 144 isoliths 124 isopachs 124 isoporosity maps 125 isosaturation lines 99 isosaturation maps 126 isothermal compressibility 43 isothermal retrograde condensation 42 Kay's rule 45 kelly 23 kick 34-5 Kimmeridge Clay 7, 9 Klinkenberg correction 81, 82 Kolmogorov-Smirnoff test 84 Lasater correlation (bubble-point pressure) 55 leak off tests 30 Leverett J-function correlation 99 light oil processing 226 foaming problems 227-8 separator design considerations 227 wax problems 228 line source solution (fluid flowing in a porous medium) 134-5 development of 135-6 liquid drop out 208 liquids systems, generalized correlations 54-8 lithofacies representation 125 LKO (lowest known oil) 12,13 low interfacial tension (Iff) systems 193 Material balance, reservoirs with water encroachment or water injection 165-8 material balance calculations generation of data 52 sources of error 168-9 material balance equation 158 combination drive 166 gas cap expansion drive 163-4 solution gas drive 161-3 material balance residual oil saturation 174 mathematical models 233-4

360 mercury injection and porosimetry 73,96,97 meters 229 microemulsion 198 middle (late transient) time solution 139 miscible displacement mechanisms 194-5 miscible displacement processes 193 miscible floods 194 applications 195-6 examples 196 miscible fluids, properties of 195 mobility ratio 104-5, 107, 175,176 and polymers 197 modelling of reservoirs 130-1 models 233--4 mole (def.) 44 Monte Carlo approach, probabilistic estimation 127 technique and recoverable reserves estimate 130 movable hydrocarbon formula (MHV) 130 mud cake 36 mud circulation system 22, 23 mud composition, general limitations on 67 mud logging 30-1 mud systems, bland (unreactive) and core recovery 31-2,67 multicomponent systems, phase behaviour 41 multimodal porosity 78 multirate data, analysis of 144-5 multiphase flow, equations of 234-5 Natural gas calorific value 226 dehydration 224-5 onshore processing 225-6 sales specification 224 sweetening 225 natural gas processing 224-6 nitrogen in miscible displacement 195, 196 non-wetting phase fluid 94 non-wetting phase saturation 102 North Sea, heavy oil reservoirs 202 North Sea, hydrocarbon fields Beryl field 196 Brent field 196 Buchan field 37 Dunlin field 131,178 Forties field 249 Fulmar field 249-51 Magnus field 184 Maureen field 187 Montrose reservoir (RFf data) 151 Murchison field 125 Rough gas field 123, 124, 126, 127 Statfjord field 196, 245, 246 Thistle oil reservoir 122, 123, 125 North Sea, oil correlations, recent 56-8 North Sea, reservoirs, fluid choice for miscible displacement 196 North Sea, reservoirs and surfactants 198, 199 ODT (oil down to) 13 offshore production/injection system, principle components of 184,185,186 offshore system 21 oil bank formation 195 oil density 14 oil flow rate, measurement of 150 oil formation factor Bn 51

PETROLEUM ENGINEERING: PRINCIPLES AND PRACTICE oil saturation, local, influences on 191 oil viscosity 56 oil-water contact (OWC) 96, 98-9 oil-water systems and relative permeability 102-3 open-hole tests 145 optimal salinity 198 orifice meters 229 overpressure 11, 12 Packer 146 Peng and Robinson equation 44 permeabilities, averaging of 83 permeability 7, 78-86 and critical displacement ratio 112 anistropy 82-3 distributions 83--4 improvement 193--4 laboratory determination of 81-2 ratios 104-5 variation, effects of 106-8 permeameter 81 petroleum migration of 9-10 origin and formation of 7 recovery 5 petroleum engineering function of 1 problem solving in 3 phase (def.) 14 phase inversion temperature (PIT) 198 physical models 233 piston displacement, stratified reservoirs 107-8 planimeter 124, 127 polyacrylamides 197 polymer fluids 193 polymer systems and adsorption 197 pool see reservoir pore fluid pressures 11 pore pressure, significance in drilling and well completion 26, 28 pore size distribution 96-7 pore space characteristics and equilibrium saturation distribution 92-3 pore volume compressibility 160 of reservoir rocks 203 poro-perm data, validity of 242 porosity 7, 71-8 and permeability, relationship between 84-6 cut-off 124 distributions 77-8 logs 75-7 main logging tools for 75 measurement of 72-3 potential gradient 174 pressure (abnormal) and d-exponent 25-6 pressure build-up analysis 139-40 pressure build-up (testing) 149 pressure control and well kicks 34-5 pressure decline, rates of 137 pressure depletion 210 pressure drawdown and reservoir limit testing 142-3 pressure equilibrium, static system 12 pressure gauges 137, 147 (downhole), characteristics of 136 pressure gradients and heterogeneity of reservoir pore space 129 pressure maintenance 173 pressure regimes, abnormal 11-12 primary recovery, oil reservoirs 159-64

361

INDEX probabilistic estimation 127-8, 129, 130 produced fluids and offshore processing 184-{5 produced water treatment 228 producing rates (well inflow equations/pressure loss calculations) 174-5 production costs, significance of 1, 3 production engineering, and well performance 220-1 production engineering described 218 production operations, influencing factors 218-29 production rate effects 180-2 production rates, technical and economic factors 219 production system 218-19 production testing 150-1 productivity index (PI) 245 and inflow performance 220 pseudo-critical temperatures and pressures 45-7 pseudo-relative permeability in dynamic systems 115 pseudo-relative permeability functions 177,178, 243,245 static 115-16 pseudo-relative permeability relationships and thicker sands 107 PVT analysis 52-4 PVTrelationships, single and multicomponent systems 40-1 Radial equations in practical units 136 radial flow in a simple system 134-5, 137 recombination sampling 52 recovery efficiency, water reservoirs 168 recovery factors and reserves 128-30 recovery string 34 recovery targets 191 Redlich-Kwong equation 44 relative permeability 102-4,106-7 effect of temperature 204 relative permeability data, laboratory determination of 109-11 from correlations 112-13 improvement, heavy oil reservoirs 204 relative spreading concept 93 repeat formation tester (RFf) 148-50 reservoir behaviour in production engineering 220-1 reservoir condition material balance techniques 160 volumetric balance techniques 160-1 reservoir data, sources 14-15, 17 reservoir (def.) 7 reservoir description in modelling 237-45 uncertainty in 245-7 reservoir development, costs of3, 4 reservoir dip angle 175,177 reservoir flow rate, effect of 181 reservoir fluid properties, measurement and prediction of 43-9 reservoir fluids and compressibility 42-3 nature of 14 properties of 40-58 reservoir geometry and continuity 180, 238-45 reservoir heterogeneity 177-80 reservoir mapping and cross-section interpretation 245-6, 247 reservoir modelling analysis and data requirements 237 application in field development 248-51 concepts in 233-48 reservoir performance analysis 157-68 reservoir pore volume and change in fluid pressure 42-3 reservoir pressures 10-12

reservoir rocks, characteristics of 62-86 pore volume compressibility 203 reservoir simulation modelling 233-7 reservoir simulation and vertical communication 243, 245 reservoir temperatures 13 reservoirs 7-18 areal extent of 122-4 residual oil 53, 191 influence of recovery mechanism 191, 193 residual oil saturation 192 average 174 and material balance 174 measurement of 191, 192 residual saturations 111-112 resistivity factor see formation resistivity factor resistivity index 74 retrograde condensation 208 reverse circulating sub 146 rotary table 23 Safety joints and jars 147 salinity and water viscosity 56 samplers 147 sand body continuity 180 importance of 238,239-40 sand body type effect on injected water and oil displacement 178-80 recognition of 238 saturation distributions in reservoir intervals 98-9 saturation gradients 164 saturation pressure see bubble-point pressure scribe shoe 66 sea water as injection water 184 seawater floods (continuous) and low surfactant concentration 199-200 secondary recovery and pressure maintenance 173-86 secondary recovery techniques 173 sedimentary basins and hydrocarbon accumulation 7 origin of7 worldwide 2 segregated displacement 177 sensitivity studies 246-7 shaliness, effect of 13 Shinoda diagrams 198 simulators applications 235 classification of 235,236 single component systems, phase behaviour 40-1 skin effect 140-2 negative factors 142 skin zone 194 slabbing 68 solution gas drive, analysis by material balance 159-63 solution gas-oil ratio 53, 54, 55 Standing-Katz correlations 46, 47 Standing'S data (bubble-point correlation) 55 STB (stock tank barrel) 14 steady state permeability tests 110 steam flooding 205 steam properties 206, 207 steamdrive analysis, example data requirements 207 Stiles technique 107-8 stock tank oil 54 and retrograde condensation 208 stock tank oil in place and equity

362 determination 130 stock tank units 14 stock tank volume 53 Stratapax bits 33 stratified reservoir analysis 106 stripping 191 structure contour maps 122 stuck pipe and fishing 33-4 summation of fluids and porosity 72-3, 74 superposition technique 140 surfactant concentration (low) and continuous seawater floods 199-200 surfactant flooding 198-200 surfactant phase systems 197-8 surfactant processes 197-200 surfactants 193 synthetic 199 sweetening, natural gas 225 Tester valve 146 thermal energy 204-7 thermal injection processes 204-6 thickness maps 124 threshold capillary pressure (reservoir rocks) 95 threshold pressure 94 traps (structural and stratigraphic) 10 tricone bits 32, 33 trip gas 34 turbine meters 229 Ultimate recovery formula see movable hydrocarbon formula uncertainty in reservoir model description 245-8 unitization 130-1 universal gas constant, values of 43 unsteady state relative permeability tests 109-10 USA, heavy oil resource distribution 202 Van der Laan method (volume in place) 128 vaporizing gas drive 194, 195 vapour phase 42 vertical bed resolution 76 vertical permeability variation and fractional flow curve 177 vertical pressure logging 148-50 Viking Graben area (N North Sea) 10 Vogel dimensionless IPR 220-1 volatile oil reservoirs 211 volatile oil systems 42 volumetric balance techniques 160 vugular carbonates and whole core analysis 69

PETROLEUM ENGINEERING: PRINCIPLES AND PRACTICE Walther's law offacies 238 water drive and gas condensate reservoirs 209,210 water drive reservoirs 167 recovery efficiency of 168 water formation factor Bw 50-1 water influx 165, 166 water influx, gas reservoir 158-9 water injection 166, 178 water saturation distribution, homogeneous reservoir 96 water viscosity 56 waterflooding 178, 179,180 Welge analysis 106 Welge's equations 174 well arrangements, dipping reservoirs 181 well classification 20 well description log 31, 32 well drilling operations 20-3 well locations and patterns 182-3 well performance, radial flow analysis of 134-51 well productivity improvement 193-4 well test methods, applications of analytical solutions 136-9 well test procedures 145-50 data analysis 147-8 well testing and pressure analysis 150-1 well/reservoir responses, different reservoir systems 139 wellbore, altered zone 141 wellbore flow 221-3 wellbore inflow equations 174 wellsite controls and core recovery 68 wettability 175 change in 67, 196 degree of 93 wettability control, in situ 112 wettabilityeffects 108 wettability preference 93 wetting phase fluid 93 wetting phase saturation 94 wetting preference 175 wireline logs, correlation with cores 63, 65, 75 wireline testing 148-50 WUT (water up to) 13 Xanthan gums 197 Zonation 99, 131,242,243,245 Forties reservoir 249 and geological core study 68-9 and histogram analysis 84 and permeability distributions 84