reservoir engineering aspects of horizontal...

Reservoir Engineering

Aspects of Horizontal Wells

PET 472 lecture notes

Spring 2012

Candidate selection criteria

• Natural fractures

• Reservoir thickness

• Anisotropy ratio, kv/kh

• Coning

• Well spacing

Effect of Reservoir Thickness

vk

hk

LehrLa

L

Laa

wrhLhoBo

phhk

hq

5.04)/2(25.05.0)2/(

2/

2)2/(2

)]2/(ln[/2]ln[2.141

Steady-state solution

0

1

2

3

4

5

6

7

8

0 500 1000 1500 2000 2500

Horizontal well length,ft

Pro

du

cti

vit

y r

ati

o, J

h/J

v

h=25

h=50

h=100

h=200

h=400

Joshi

Horizontal Wells

Advantage: Horizontal well orientation to natural fracture

direction

GRM-Engler-09

kmin =

0.03 mD

kmax =

3.34 mD

kMatrix =

0.01 mDN

10 m

Horizontal well

paths

A

B

kmin =

0.03 mD

kmax =

3.34 mD

kMatrix =

0.01 mDN

10 m

kmin =

0.03 mD

kmax =

3.34 mD

kMatrix =

0.01 mDN

10 m

Horizontal well

paths

A

B

Impact of Anisotropy Ratio

0

1

2

3

4

5

6

7

8

0 500 1000 1500 2000 2500

Horizontal well length, ft

Pro

du

cti

vit

y R

ati

o, J

h/J

v

kv/kh=.1

kv/kh=.5

kv/kh=152

hk

vk

h

LDL

Drainage Area

Joshi

Drainage Area

Joshi

Drilling along the high permeability direction

Coning

Function of drawdown

Critical rate –

maximum rate only oil is produced

Joshi

Joshi

Horizontal Well Applications

1. In low permeability reservoirs, horizontal wells

enhance the drainage area for a given time period

2. In high permeability reservoirs; horizontal wells reduce

near wellbore turbulence and thus improves the well’s

deliverability

3. Single dominant pay zones, good vertical permeability

GRM-Engler-09

Horizontal Wells

• Given 400 acre lease

• 10 Vertical wells (40 ac/well)

• 6-1000-ft long horizontal

wells (74 ac/well)

• 4-2000-ft long horizontal

wells (108 ac/well)

GRM-Engler-09

Joshi, 1991

Advantage: Well spacing and location

Horizontal well

• Single phase (liquid) flow

• Pseudosteady state…bounded reservoir

• Horizontal well is located arbitrarily within the bounded drainage

area

• Horizontal well is assumed to have infinite conductivity

Constraints

Wellbore pressure drop

Uniform wellbore pressure

(Infinite-conductivity model)

Uniform flux entry

Observed triangular

profiles

Pseudosteady state equations

skin mechanicalmS

fracturety conductivi-infinite

gpenetratin-fully toduefactor skin ),4/ln(fS

factorskin related shape

386.175./ln2.141

/

wrL

CAS

where

mSfSCASwrer

oBohk

hJ

1. Based on infinite-conductivity horizontal well

Mutalik,etal


e2xL when 0rS plane, areal

in then penetratio partial toduefactor skin rS

plane verticalin the area drainage h,e2y A

factor shape

)ln(75./ln2.141

/)2(

hC

where

rShCwrA

oBovkykex

hJ

2. Based on uniform flux horizontal well

Babu & Odeh


direction zin factor pseudoskinzS

geometry offunction

/)/2(2.141

/2

F

where

zSvkhkLhF

oBohhk

hJ

3. Based on pressure averaging along the horizontal length

Kuchuk et al

Pseudosteady state flow equation including non-Darcy

Flow

DqcCASmShSAwrerlnoBo2.141

wfprpkhq

Sh = equivalent negative skin factor due to either well stimulation or due

to the horizontal well. Also included are partial penetration and

perforation skin terms.

Sm = mechanical skin damage

SCA = shape related skin factor

c’ = shape factor conversion constant = 1.386

A´ = 0.75 for circular drainage area and 0.738 for square or rectangular

drainage areas.

r´e =

Horizontal well Solution

ft,/A

wr4

LlnhS

SCA - horizontal well shape-related skin factor

hk

vk

h2

LDL

Joshi,1991


For square drainage area

L/2xe

A horizontal well drilled in an oil reservoir has the following parameters.

Area = 160 acres rw = 0.365 ft

h = 50 ft kv/kh = 0.1

o = 0.5 cp kh = 1 md

Bo = 1.2 rb/stb Sm = 0

D = 0

Calculate the pss productivity index of a 2000-ft long horizontal well.

Step 1:

Horizontal well Example

ft1489/)43560)(160(/Aer

Step 2:

22.7)365.0(4

2000ln

wr4

LlnhS

SCA - horizontal well shape-related skin factor

32.6

1.0)50(2

2000

hk

vk

h2

LDL

Joshi,1991


For square drainage area

L/2xe=2000/2640

= 0.757

Step 3:

SCA = 2

Pseudosteady state flow equation,

psi/stb61.0

0386.10.2022.7738.0365.01489ln)2.1)(5.0(2.141

)50(1

p

q

Horizontal well Example

Jh, stbd/psi, Horizontal well productivity

kv/kh = 0.1 kv/kh = 0.5 kv/kh = 1.0

0.61 0.75 0.80

Horizontal Well Performance

• PIhorizontal > xPIvertical

• If less than expected, possible cause is

Lproductive < Ldrilled.

– Reservoir heterogeneity

– Wellbore pressure drop

– Formation damage

Estimate the fluid invasion and damage in heterogeneous

reservoirs

A heterogeneous reservoir is generated (zone A to E)

Permeability, porosity, and relative permeability curves are different in each zone

Invasion at 11 nodes from heel to toe are simulated

0 1 2 3 4 5 6 7 8 9 1

0

Reservoir type

A B C D E

k = 300 md 80 md 10 md 200 md 70 md

= 0.23 0.12 0.09 0.22 0.15

Kro, max = 0.75 0.7 0.6 0.75 0.7

no = 1.5 2 3 1.5 2

Supalak’s dissertation

Skin factors => 4.6 – 19.6 along the well

Zones B and C have severe damage

Zones D and E have the least damage

0

20

40

60

80

012345678910

Node no.

Da

ma

ge

ra

diu

s, in

.

0

10

20

30

40

50S

kin

fa

cto

r

rs, in.

s node

Skin damage and damage radius in a

horizontal well with a heterogeneous reservoir

A B

C

D E

Reservoir heterogeneity


Formation damage in horizontal wells

• Larger contact area

• Longer contact time

Is the damage more vulnerable

in horizontal wells?

Fig. 2: Fluid filtration in vertical and horizontal wells

Reservoir

rock !!!


L

Le av, max

aH, max

Heel

Formation damage in horizontal wells


Horizontal Wells

Pseudosteady state flow equation including non-Darcy

Flow

GRM-Engler-09

DqcCASmSS75.wrerlnizT1422

2wf

p2rpkh

q

S = equivalent negative skin factor due to either well stimulation or due to the

horizontal well. Also included are partial penetration and perforation skin

terms.

Sm = mechanical skin damage

SCA = shape related skin factor

c’ = shape factor conversion constant

Horizontal Well Example

GRM-Engler-09

An oil company recently signed an offshore reservoir concession. The lease

concession lasts only for a period of 5 years. The gas is to be delivered to a

pipeline operated at 300 psia. To meet this high-pressure requirement, it is

important to maintain a wellhead pressure of 500 psia. Before testing, the test

well, which is vertical, was cemented, perforated and cleaned using acid. The

perforated interval in the vertical well was 60 ft. The reservoir has a bottom

water zone separated by a 10-ft thick layer of shale (kv/kh = ?). It appears the

reservoir is not in communication with the bottom water.

An engineer suggests drilling a 2000-ft horizontal well not only to reduce near-

wellbore turbulence but also to ensure against water coning. Compare the IPR

curves for a vertical and horizontal well in this reservoir.


GRM-Engler-09

60’ =14%, k =6 md, kv/kh=.1

Tr=185°F, Pr = 3400 psi, g = 0.605

7870’

5.5” ID

Ppipeline = 300 psi Pwh = 500 psi

Twh=75°F

Sw=30%


Depth = 7870 ft

Well spacing = 640 acres

Wellbore radius = 0.25 ft.

Wellhead pressure = 500 psia

Wellhead temperature = 75°F

Tubing ID = 5.5 in.

GRM-Engler-09

Well Properties

60’ =14%, k =6 md, kv/kh=.1

7870’

5.5” ID

Ppipeline = 300 psi Pwh = 500 psi

Twh=75°F

L=2000 ft


Vertical Well Performance

GRM-Engler-09

S = 0, fully penetrating vertical well

Sm = 0, no mechanical skin damage, well was cleaned with acid

SCA = 0, well centrally located in the drainage area

c’ = 0, shape factor conversion constant for vertical well.

pwf2phwr

kh1510x222.2D

g

201.1k

1010x33.2

Laminar Steady state components

Non-Darcy components

where

aC

62.31ln

caS


Horizontal Well Performance

GRM-Engler-09

Laminar Steady state components

Non-Darcy components

S = -7.6, negative skin due to horizontal well

Rwa = L/4 = 2000/4 = 500 ft

6.725.0

500ln

rw

warlnS

Sm = 0, no mechanical skin damage, well was cleaned with acid

c’ = 1.386, shape factor conversion constant for horizontal well.

Replace hp with L

Horizontal Wells

GRM-Engler-09

SCA = 1.8, well centrally located in the drainage area

2xe = 5280 ft

L/(2xe) = 2000/5280 = 0.38

kv/kh = 0.6/6 = 0.1 3.51.0

)60(2

2000

hk

vk

h2

LDL

Joshi,1991


GRM-Engler-09

0

500

1000

1500

2000

2500

3000

3500

4000

0 50 100 150 200 250

gas rate, mmscfd

pre

ssu

re,

psi

B&B

C&S

IPR-horizontal

IPR-vertical

w/turbulence w/turbulence

reservoir engineering aspects of horizontal...

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