Skin Effect and Formation Damage

By

Dr. Mohammed A. Khamis

02-02-2016

Well Stimulation and Sand Production

Management (PGE 489 )

Skin Effect and Formation Damage

Skin Factor

• A formation damage model is a dynamic relationship expressing the

fluid transport capability of porous medium undergoing various

alteration processes.

• Modeling formation damage in petroleum reservoirs has been of

continuing interest. Although many models have been proposed,

these models do not have the general applicability.

Hawkins’ Formula

idealreals ppp

Hawkins’ Formula

w

s

s

realwfsrealr

r

hk

qppp ln

2,

w

sidealwfsideal

r

r

kh

qppp ln

2,

skh

qps

2

idealreals ppp

Steady-state pressure drop due to skin is given by:

w

s

w

s

s r

r

kh

q

r

r

hk

qs

kh

qln

2ln

22

w

s

w

s

s r

r

kr

r

ks

kln

1ln

11

w

s

s r

r

k

ks ln1

Hawkins’

Formula

Ps

pwf

rsrw

Hawkins’ Formula

Ps

pwf

rsrw

s

w

r

r

realrrk

dr

h

qp

)(2

w

sideal

r

r

kh

qp ln

2

skh

qps

2

idealreals ppp

Steady-state pressure drop due to skin is given by:

w

s

r

rr

r

kh

q

rrk

dr

h

qs

kh

q s

w

ln2)(22

General

Hawkins’

Formula

w

s

r

rr

r

krrk

drs

k

s

w

ln1

)(

1

w

s

r

rr

r

rrk

drks

s

w

ln)(

Hawkins’ Formula - Example

Hawkins’ Formula - Example

Example: assume that a well has a radius rw equal to 0.25 ft, and the

damage beyond the well is 5 ft. What would be the skin effect if the

permeability impairment results in k/ks equal to 6 and 12, respectively.

Solution:

22.1525.0

25.5ln16ln1

w

s

s r

r

k

ks

49.3325.0

25.5ln112ln1

w

s

s r

r

k

ks

non-Darcy Skin (S)

2

1.05106

perfw

s

hr

hkD

Where, is the gas gravity, ks is the

near-wellbore permeability in md, h

and hperf are the net and perforated

thicknesses in ft and is the gas

viscosity in cp.

Dqs

Skin due to turbulence is additional

pressure drop caused by high gas

velocity near the wellbore and applies

only to gas wells.

This shows the apparent skin from well testing.

Dqss

non-Darcy Skin (S)

Example: Calculate the non-Darcy coefficient for well with wellbore radius

of 0.25 ft. Assume that ks is the same as the reservoir permeability (0.1 md)

and hperf is half of the reservoir thickness. Use a viscosity of 0.01 cp, gas

gravity of 0.6, and reservoir thickness of 80 ft. What will happen if the near-

wellbore permeability reduced by damage to one-fifth?

Solution:

19

2

1.05

2

1.05

/1006.9

)40)(25.0)(01.0(

)80()1.0()6.0(106106

dMSCF

hr

hkD

perfw

s

18

2

1.05

2

1.05

/1006.1

)40)(25.0)(01.0(

)80()02.0()6.0(106106

dMSCF

hr

hkD

perfw

s

Slanted/Partial Penetration

Partial Completion Skin

• Skin due to partial penetration

is always greater than 0 and

typically ranges from 0 to 30.

• Partial penetration with

vertical permeability (kv)

equal to zero is the limiting

case. Skin factor can be

estimated using Hawkins’

formula.


1

2cos71

1ln

5.0h

hh

rh

r

rs

ww

epp

Muskat’s Model

zw

hw h

h = Reservoir height

hw = Perforation height

zw = The elevation of the

perforation midpoint from the base

of the reservoir

Completion Ratio

This model did not consider the permeability anisotropy

h

hh w

95.1lnln1.049.0ln135.1

825.0

wc

V

H

V

H

w

pp rk

kh

k

kh

h

hs

Odeh Model

zw

hw h

0

0753.22126.0exp

yforr

yforh

zr

r

w

mw

wc

2

wm

hyz y



zw = The elevation of the perforation midpoint from

the base of the reservoir

y = The distance from top of reservoir to top of

perforation

Where,


Papatzacos Model

zw

hw h

y



zw = The elevation of the

perforation midpoint from the base

of the reservoir

y = The distance from top of

reservoir to top of perforation

Where,

5.0

1

1

1

1

2ln1

1

B

A

h

h

hrhs

pD

pD

pDDpD

pp

43

1

4

1

1

1

1

5.0

pDD

pDD

D

H

VwD

wpD

hhB

hhA

y

hh

k

k

h

rr

h

hh

All models assumed kx = ky



Ref. Yildiz (2006), SPE-82249


Example:

A well with a radius rw = 0.328 ft is completed in a 33-ft reservoir.

In order to avoid severe water coning problems, only 8 ft are completed

and the midpoint of the perforations is 29 ft above the base of the

reservoir.

1. Calculate the skin effect due to partial completion for a vertical well.

What would be the composite skin effect if Ɵ = 45°?

2. Repeat this problem for h = 330 ft, hw = 80 ft and zw = 290 ft.

Partial Completion Skin - Example

Solution:

The dimensionless reservoir thickness hD is h/rw = 33/0.328 = 100.

The elevation ratio is zw/h = 29/33 = 0.875

The completion ratio is hw/h = 8/33 = 0.25.

From Table 5-1 for a vertical well,

Ɵ = 0°and Sc + Ɵ = 8.6, Sc = 8.6 and SƟ = 0.

If Ɵ = 45° , then

Sc = 8.6 but SƟ = -2.7 resulting in Sc + Ɵ = 6.

If hD = 330/0.328 = 1000 and the other ratios are the same.

From Table 5-2

Sc + Ɵ = 15.7 for the vertical well and

Sc + Ɵ = 10.4 for the 45°slant well.

Partial Completion Skin - Example

Solution:

Ɵ hD zw/h hw/h S Ɵ + c Sc S Ɵ

• When the angle of inclination

through the formation is

significant (> 10°), a reduction in

pressure drop can occur due to

the angle of inclination.

• This pressure drop is defined as

skin due to inclination

• The skin is negative and the

larger the angle of slant, the

larger the negative contribution

to the total skin effect.

100log

5641

865.1'

06.2'

Dw hs

w

V

Hw

k

k tantan 1'

V

H

w

Dk

k

r

hh

Does s negative all the time?

Skin due to Inclination

Perforation Skin

Documents

Skin Effect and Formation Damage