petroleum production engineering

UNIVERSITY OF KIRKUKCollege of Engineering - Pet. Eng. Department

PETROLEUM PRODUCTION ENGINEERING

BOYUN GUO – WILLIAM C. LYONS – ALI GHALAMBOR

CHPTER -3-

RESERVOIR DELIVERABILITY

Solution of Problems page 3/43

By: Hawar AbdulKhaliq Hamma Gul

3.1 Construct IPR of a vertical well in an oil reservoir .

Consider (1) transient flow at 1 month , (2) steady – state flow , and (3) pseudo – steady state flow. the following data are given:

Porosity , ϕ = 0.25

Effective horizontal permeability , k = 10 md

Pay zone thickness , h = 50 ft

Reservoir pressure , pe = 5000 psi

Bubble point pressure , pb = 100 psi

Fluid formation volume factor , Bo = 1.2

Fluid viscosity , µo = 1.5 cp.

Total compressibility , Ct = 0.0000125 psi-1

Drainage area , A = 640 acre (re =2.980 ft)

Wellbore radius , rw = 0.328 ft

Skin factor , S = 5

Solution:

time = 30 day = 24 x 30 = 720 hr.

1. For transient flow,

J*=kh

162.6µB[ logt+ log( k

ϕ µCt r w2 )−3.23+0.87 S] =

(10 )(50)

162.6 (1.2 )(1.5)[ log (720)+log( 10(0.25 ) (1.5 ) (0.0000125 )(0.328)2 )−3.23+0.87 (5)]

J* = 0.1515215 bbl/day/psi

Calculated data points are:qo = J* (Pe – pwf)

Pwf (psi) qo (stb/day)0 757.60751000 606.0862000 454.56453000 303.0434000 151.52155000 0

01002003004005006007008000

1000

2000

3000

4000

5000

6000

qo (stb/day)

pwf (p

si)

"Transient IPR curve"

2. For steady state flow,

J*=kh

141.2µB [ ln ( ℜrw )+S] =

(10 )(50)

141.2(1.2)(1.5)[ ln( 29800.328 )+5]

J* = 0.13938 bbl/day/psi

Calculated points are:

Pwf (psi) qo (stb/day)0 696.9

1000 557.522000 418.143000 278.764000 139.385000 0

01002003004005006007008000

1000

2000

3000

4000

5000

6000

qo (stb/day)

pwf (

psi)

"Steady state IPR curve"

3. For pseudo steady state,

J*=kh

141.2µB ¿¿ =(10 )(50)

141.2(1.2)(1.5)¿¿

J*= 0.147202 bbl/day/psi

Calculated points are:


1000 588.8082000 441.83000 294.4044000 147.2025000 0

01002003004005006007008000

1000

2000

3000

4000

5000

6000

qo (stb/day)

pwf (

psi)

Pseudo-steady state IPR curve""

3.2 Construct IPR of a vertical well in an saturated oil reservoir

Using Vogel's equation .the following data are given:











Skin factor , S = 2

Solution:

Assume pseudo-steady state flow,

J*=kh

141.2µB ¿¿= (80 )(55)

141.2(1.8)(1.1)¿¿

1.518476 STB/day/psi=

q max =J∗Pe

1.8

=(1.518476 )(4500)

1.8

=3796.2 STB/day.

Calculated points by Vogel's equation:

q = q max [1−0.2( pwfpe )−0.8( pwfpe )2]


1000 3477.5072000 2858.8673000 1940.284000 721.74674500 0

050010001500200025003000350040000

500

1000

1500

2000

2500

3000

3500

4000

4500

5000

qo (stb)

pwf (p

si)

IPR curve""

3.3 Construct IPR of a vertical well in an unsaturated oil reservoir

Using generalized Vogel's equation .the following data are given:











Skin factor , S = 5.5

Solution:

Assume pseudo-steady state flow,

J*=kh

141.2µB ¿¿= (100 )(55)

141.2(1.8)(1.2)¿¿

J* = 1.300687 STB/day/psi

qb = J* (Pe – Pb)

) = 1.300687) (5000 – 3000 = (2601.4 STB/day.

qv = J ¿Pb1.8

=)1.300687) (3000 / (1.8 = 2167.8 STB/day.

Calculated points by:

q = J* (Pe – Pb) + J¿Pb1.8

× [1−0.2( pwfpb )−0.8( pwfpb )2]


500 4648.7671000 4431.9871500 4118.862000 3709.3872500 3203.567

3000 2601.45000 0

01000200030004000500060000

1000

2000

3000

4000

5000

6000

qo (stb)pw

f (ps

i)

3.4 Construct IPR of a vertical well in an unsaturated oil reservoir

Using generalized Vogel's equation .the following data are given:



Tested following bottom-hole pressure in well A , pwf 1 = 4000 psi

Tested production rate from well A , q1 = 400 stb / day

Tested following bottom-hole pressure in well B, pwf 1 = 2000 psi

Tested production rate from well B , q1 = 1000 stb/day

Solution:

Well A : Pwf1 > pb

J* = q1

( pe−pwf 1)

=400) / 5500 - 4000 = (0.2667 stb/day/psi.

qb = J* (Pe – Pb)

= 0.2667) 5500 – 3500 = (533.4 stb/day

qv = J ¿Pb1.8

) = 0.2667) (3500 / (1.8 = 518.6 stb/day


q = J* (Pe – Pb) + J ¿Pb1.8

× [1−0.2( pwfpb )−0.8( pwfpb )2]

Pwf (psi) qo (stb/day)0 1052

500 1028.7161000 988.4981500 931.34612000 857.26042500 766.24083000 658.28733500 533.45000 0

0200400600800100012000

1000

2000

3000

4000

5000

6000

qo (stb/day)

pwf (

psi)

Well B : Pwf1 < pb

J* = q 1

(( pe−pb )+ pb1.8 [1−0.2( pwfpb )−0.8( pwfpb )

2])

= 1000

((5500−3500 )+ 35001.8 [1−0.2( 2000

3500 )−0.8( 20003500 )

2])

= 0.3111 stb/day/psi

qb = J* (Pe – Pb) = 0.3111 (5500 – 3500) = 622.2 stb/day

qv = J ¿Pb1.8

) = 0.3111) (3500 / (1.8 = 604.92 stb/day


qo= J* (Pe – Pb) + J¿Pb1.8

× [1−0.2( pwfpb )−0.8( pwfpb )2]

qo = 622.2 + 604.92× [1−0.2( pwf3500 )−0.8( pwf3500 )2]


500 1199.961000 1153.0481500 1086.3842000 999.96642500 893.79673000 767.87463500 622.25000 0

02004006008001000120014000

1000

2000

3000

4000

5000

6000

qo (stb/day)

pwf (

psi)

3.5 Construct IPR of a well in saturated oil reservoir using both Vogel's equation

and Fetkovich's equation.

The following data are given:



Tested following bottom-hole pressure , pwf 1 = 2500 psi

Tested production rate at pwf 1 , q1 = 600 stb / day

Tested following bottom-hole pressure , pwf 2 = 1500 psi

Tested production rate at pwf 2, q2 = 900 stb/day

Solution:

Vogel's equation: qomax. =q1

1−0.2( pwf 1pe )−0.8( pwf 1

pe )2

qomax.=600

1−0.2( 25003500 )−0.8( 1500

3500 )2 ¿602.46

stbday

.

calculated data points are:


500 575.41081000 528.68941500 462.29582000 376.23012500 270.49223000 145.08223500 0

Fetkovich's equation:

n= log( q1

q2 )log( pe2−pwf 12

pe2−pwf 22 )=

log( 600900 )

log(35002−25002

35002−15002 )=0.8

c= q1

(pe2−pwf 12 )n= 600

(35002−25002 )0.8=0.0023 stb /day . psi2n

calculated data points are:

q= 0.0023 (35002−pwf 2 )0.8


500 1059.4311000 1006.121500 915.64652000 785.03862500 608.48353000 372.59293500 0

0002004006008000100210

005

0001

0051

0002

0052

0003

0053

0004

)ledom s'hcivoktef(fwp)ledom s'legov(fwp

(yad/bts) oq

3.6 Determine the IPR for a well at the time when the average reservoir pressure

will be 1500 psig. The following data are obtained from laboratory tests of well fluid samples:

futurepresentReservoir pressure15002200Average pressure (psi)-------1.25Productivity index J* (stb/day-psi)3.853.55Oil viscosity (cp)1.151.20Oil formation volume factor (rb/stb)0.650.82Relative permeability to oil

Solution :

J*f =J¿ p( K ro

Bo∗μo )F( K ro

Bo∗μo )P=1.25

( 0.653.85∗1.15 )

F

( 0.823.55∗1.20 )

P

=0.9534 stb/day-psi

Vogel's equation for future IPR:

q= (J¿¿¿F)( pef )

1.8 [1−0.2( pwfpef )−0.8( pwfpef )2]¿

¿(0.9534 )(1500)

1.8×[1−0.2( pwf1500 )−0.8 ( pwf1500 )

2]… ..1

Vogel's equation for present IPR:

q= (J¿¿¿P)( pep)

1.8 [1−0.2( pwfpep )−0.8( pwfpep )2]¿

¿(1.25 )(2200)

1.8×[1−0.2( pwf2200 )−0.8( pwf2200 )

2]….2

Calculated data points are:

Reservoir press. = 1500 psiReservoir press. = 2200 psiq (stb/day)Pwf (psi)q (stb/day)Pwf (psi)

0150002200136.6541350262.77821980260.5961200501.11181760371.8261050715.0011540470.344900904.44581320556.157501069.4461100629.2446001210.002880689.6264501326.113660737.2963001417.78440772.2541501485.002220794.501527.780

00010020030040050060070080090

002

004

006

008

0001

0021

0041

0061

)tneserp( fwp)erutuf( fwp

(yad/bts) oq

*in the above table, col. 2 calculated from eq.2 & col.4 from eq.1

3.7 Using Fetcovich's method , plot the IPR curve for a well in which pi is 3000 psia and J'

o = 4×10-4 stb/day-psi . Predict the IPRs of the well at well shut-in static pressures of 2500 psia, 2000 psia, 1500 psia, and 1000 psia.

Solution :

The value of J'O at 2500 psia is:

J'O = J'

i pepi………eq. (3.61 ) page3/ 41

J'O = 4*10-4 )

30002500

¿=4.8∗10−4 stb /day ( psi)2

The value of J'O at 2000 psia is:

J'O = 4*10-4 )

30002000

¿=6∗10−4 stb /day ( psi)2

and the value of J'O at 1500 psia is :

J'O = 4*10-4 )

30001500

¿=8∗10−4 stb /day ( psi)2

and the value of J'O at 1000 psia is :

J'O = 4*10-4 )

30001000

¿=1.2∗10−4 stb /day ( psi)2

Calculated data points are:

qo = J'O ¿ ( pe2−pwf 2 )… ...eq . (3.60 ) page3 /41

Pe = 2000 psiPe = 2500 psiPe = 3000 psiq(stb/day)pwf (psi)q(stb/day)pwf (psi)q(stb/day)pwf (psi)

02000025000300045618005702250684270086416001080200012962400

12241400153017501836210015361200192015002304180018001000225012502700150020168002520100030241200218460027307503276900230440028805003456600237620029702503564300240003000036000

Pe = 1000 psiPe = 1500 psiq(stb/day)pwf (psi)q(stb/day)pwf (psi)

010000150022890034213504328006481200612700918105076860011529009005001350750

100840015126001092300163845011522001728300118810017821501200018000

000500010051000200520003005300040

005

0001

0051

0002

0052

0003

0053

)isp 0003=ep()isp 0052=ep()isp 0001-ep()isp 0002=ep()isp 0051=ep(

(yad/bts) oq

(isp)

fwp

"IPR curve , problem 3.7"

petroleum production engineering

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