This document discusses artificial lift and well performance. It begins by explaining that as reservoir pressure declines, artificial lift is needed to produce fluids from the well. It then provides background on Darcy's law, which describes the relationship between flow rate and pressure differences for fluid flow through porous media. The document discusses how Darcy's law is applied to model radial flow into a wellbore, and how the productivity index (PI) is calculated. It also covers factors that affect well performance, such as water cut, gas interference, skin damage, pressure depletion over time, and tubing size.

1. 1
Why do we need Artificial Lift?
Gordon Kappelhoff

2. 2
PRODUCED FLOWRATE
WELL OUTFLOW
RELATIONSHIP
WELL INFLOW (IPR)
SURFACE PRESSURE
At Wellhead
Pwf Pwf
WELL FACE
PRESSURE
Reservoir Pressure- Pr
Required Po to produce desired rate
P
o
Typical Oil Well – Two Parts
Part 1 – The Well
Part 2 – The Reservior

3. 3
PrPwf
Pwh
Psep
P
PI
P
Q
P
P
0
500
1000
1500
2000
2500
3000
3500
0 500 1000 1500 2000 2500 3000 3500 4000 4500
Production rate, STB/D
Flowingbottomholepressure,psi
Tubing Curve
Po
Part 1 – The Well
Part 2 – The Reservior

4. 4
Depth
Pressure
Pwh = ?
Po = ?
Pr = ?
Part 1 – The Well
Part 2 – The Reservior

5. 5
Well Bore Fluid Calculations
As we can see from the formula’s the most relevant
parameter to well bore calculation is pressure. Therefore
we will spend some time looking at the basics of what
pressure is.
What is pressure?
What is Force?

6. 6
Well Bore Fluid Calculations
In english units:
Mass = lbm
Acceleration = gravity
In English Units lbm = lbf
This is not the case in metric units

7. 8
Well Bore Fluid Calculations
What exerts more force?
a. 1000 ft of water in 2 3/8” tubing
b. 1000 ft of water in 2 7/8” tubing
What exerts more pressure?
a. 1000 ft of water in 2 3/8” tubing
b. 1000 ft of water in 2 7/8” tubing

8. 10
When dealing with fluid in a tube what is the standard
Pressure calculation?
P = Force/Area
Area = π x (ID of Tubing/2)2 = ID Area
Force = Mass x Acceleration
Mass = volume of fluid x density
Volume of Fluid = ID Area x H
Acceleration = gravity
Force = ID Area x H x density x gravity
P = (ID Area x H x density x gravity)/ ID Area = ρ x g x h

9. 11
P = ρ x g x h
For pure water the english units are as follows:
ρ = 62.3 lbm/ft3
As mentioned 1 lbm = 1lbf at standard gravity
So ρ x g for water = 62.3 lbf/ft3
Gradient pressure is pressure divided by height
Rearranging the formula P/h = ρ x g
So therefore the pressure gradient for water is 62.3 lbf/ft3
Does that look right?

10. 12
We know that P = psi = lbf / in2
We know that 1 ft = 12 in
Water Grad = 62.3 lbf/ft3
= 62.3 lbf x 1 ft x 1 ft
(ft3) 12 in 12 in
= 0.433 __lbf__
(in2 x ft)
= 0.433 psi/ft
Does that sound right?

11. 13
So for Pure Water
P(psi) = 0.433 x h(ft)
For all other fluids we use specific gravity = sg
Sg = density of a fluid / density of pure water
Therefore the standard in English is:
P(psi) = 0.433 x sg x h(ft)

12. 14
Specific Gravity
Often specific gravity comes in the form of API, to covert
the following is used:
sg = 141.5
131.5+API
When two liquids of different density make one fluid, the
Specific gravity is calculated as follows:
Sp. Gr. = w f o+ ×( )sg× )( of w sg

13. 15
Formulas So far
sg = 141.5
131.5+API
Sp. Gr. =
w
f o+ ×
( )
×
)(
of
w
P(psi) = 0.433 x sg x h(ft)
Pressure due to fluid:
API to sg:
Composite sg:
sg sg

14. 1
Exercise 1a
Oil Density: 30 API
Water cut: 0%
Water Density: 1.026 sg
Pres: 3765 psi
P whead: 100 psi
PI: 10 stb/d/psi
Bo: 1.33 rb/stb
TVD: 9183 feet
Find Poutflow for the above conditions
(assume no friction)

15. 2
Exercise 1b
Oil Density: 30 API
Water cut: 30%
Water SG: 1.026 sg
Pres: 3765 psi
P whead: 100 psi
PI: 10 stb/d/psi
Bo: 1.33 rb/stb
TVD: 9183 feet
Find Poutflow for the new water cut
(assume no friction)

16. 18
Well Performance Pressure gradient plots
Depth
Pressure Po
(0%)
Po
(30%)
Pwh
Po (30%) Required for 100 psi
wellhead pressure = psi
Po (0%)Required for 100 psi
wellhead pressure = psi

17. 19
For this course we are going to make the
assumption that fluid always flows from high
pressure toward low pressure.
Some of you may recognize that this is not exactly
true.
The exactly true expression is fluid always flows
from high potential toward low potential.
Well Productivity

18. 21
Inflow – Darcy’s Experiments
The relationship between pressure and Flow rate was first
studied extensively by the scientist Henry Darcy (1803-
1858).
He created pressure differentials across a porous media and
measured the resulting flow rates that resulted from those
pressures.
His experiments resulted in what is now known as ‘Darcy’s
Law’ (1856) and are the benchmark for permeability. In fact,
the unit of permeability is called the ‘Darcy’ (D).
P0P1
Direction of Flow
Permeable Medium:
Area, Length, Permeability
Fluid Properties:
Viscosity, Volume Factor

19. 22
Darcy’s Law
For general flow through porous Media:
0 1* *( )
*
k A P P
Q
Lµ
−
=
But we’re working with oil reservoirs, not general
porous media…

20. 23
Darcy's Law for radial flow into a wellbore:
Pwf
Pr
Pr
Pr
Q=?
Reservoir
Outer "drainage"
boundary
Fluid FlowFluid Flow

21. 24
Darcy's Law for radial flow into a wellbore:
For the system just described, Darcy's Law looks
like:
qo = flow rate ko = effective permeability
h = effective feet of pay µo = average viscosity
Pr = reservoir pressure Pwf = wellbore pressure
re = drainage radius rw = wellbore radius
Bo = formation volume factor
Note: (Pr - Pwf) is the drawdown pressure
q
k h PP
B
r
o
o r wf
oo
e
w
=
7.08 x 10
S
-3
ln
( )
µ
r

22. 25
Darcy's Law for radial flow into a wellbore:
If we make the assumption that ko, h, re, rw, Bo and
µο are constant for a particular well the equation
becomes:
q
k k PP
k
k
o
1 r wf
54
6
7
=
lnk
k
2 3k
k 8
( )
Simplifying...
q K PPo r wf
= −( )

23. 26
Darcy's Law for radial flow into a wellbore:
Q - Flow Rate (BPD)
Pressure - PSI
Intercept = Pr
Slope = -1/K
0
0
Pwf

24. 27
Darcy's Law for radial flow into a wellbore:
The Productivity Index (PI) is equal to the flow
rate divided by the "drawdown":
PI
qo
=
PPr wf−( )
PI xqo
= PPr wf−( )

25. 28
Example
Darcy's Law for radial flow into a wellbore:
Consider the following example:
Pr = 2,300 psi, and
Pwf = 1,200 psi @ qo = 1,150 bpd
What is the Productivity Index (PI) of the well?
PI =
2300 - 1200( )
1150
= 1.046 bbl/day/psi

26. 29
Darcy's Law for radial flow into a wellbore:
What is the maximum flow rate the well will produce?
The maximum flow rate occurs at the maximum
drawdown (Pwf = 0).
PI =
qmax
0Pr −( )
or qmax Pr
PI= x
2300 x 1.046 = 2406 BPDqmax =

27. 30
Darcy's Law for radial flow into a wellbore:
The straight-line PI works great for single phase fluid
(i.e. water, oil, or water/oil*) flowing into a wellbore, but
what happens if gas comes "out of solution" in the
reservoir?
* Even though water and oil are two separate phases,
they are considered single phase since they are both
liquid.

28. 31
Darcy's Law for radial flow into a wellbore:
What happens when the gas comes out of solution?
Darcy's law works just as well for a single phase gas
as it does for a single phase oil.
Let's look qualitatively at what will happen to the flow
rate of gas.
q
k h PP
B
r
g
g r wf
gg
e
w
=
7.08 x 10
0.75
-3
lnµ
r

29. 31
Gas will begin
to form here
Pr
Pr
Pressure drops as we
move toward the
wellbore
Pb

30. 33
Darcy's Law for radial flow into a wellbore:
Q - Flow Rate
(BPD)
Pressure - PSI
0
0
Pwf
Graphically it would look like this:
Pr < Pb
Darcy's law
predicted
Qmax
Actual
Qmax

31. 34
We use instead Vogel's IPR curve. The equation
is:
where qo(max) is the maximum flow rate the well
can produce.
Inflow Performance Relationship - IPR:
Q(max)
= 1 - 0.2 - 0.8
2
P
wf
r
PQ
P
wf
r
P

32. 35
Consider our previous example…
Pr = 2,300 psi
Pwf = 1,200 psi @ qo = 1,150 bpd
Inflow Performance Relationship - IPR:

33. 36
First we need to calculate Q/Qmax:
Inflow Performance Relationship - IPR:
Q(max) =
1 - 0.2 - 0.8
2
1200
2300
1200
2300
1150-bpd
= 1696 bpd
Q(max)
= 1 - 0.2 - 0.8
2
P
wf
r
PQ
P
wf
r
P
Q(max)
=
Q =
0.678
0.678
Then…

34. 37
Compare this to the Qmax we got from Darcy's equation of 2406
bpd. The well has lost 710 bpd (~-30%) in capability due to gas
interference.
Inflow Performance Relationship - IPR:
Vogel vs. PI for given test point
0
500
1000
1500
2000
2500
0 500 1000 1500 2000 2500 3000
Q (bpd)
Pwf(psi)

35. 38
We saw that we could use Darcy's law when gas was not a
problem (Pwf > Pb).
We also saw how to use Vogel's IPR for cases where Pwf <
Pb.
What about a case where Pr is above Pb and Pwf is less
than Pb?
Combined IPR

36. 40
Combined IPR:
0
500
1000
1500
2000
2500
0 500 1000 1500 2000
Flow Rate - BPD
Pressure - psi
Pr=2300
Pb=1800
We use a straight line PI above Pb
We use VOGEL below Pb
Qtot-max = Qb + Qv
QvQb
Qb = PI x (Pr-Pb)
Qv = PI x Pb / 1.8
Pwf =0 .125x Pb {-1+[81-80(q-qb)/(qtmx-qb)]^.5}

37. 41
Vogel's relationship works reasonably well for water cuts
below 50%.
For higher water cuts, a method has been developed
which takes an arithmetic average of the PI and IPR
equations to yield a "composite IPR“.
For a given PWF, therefore, Composite predicts more
flow than Vogel but less flow than straight-line PI.
Composite Vogel IPR:

38. 42
qo(max)Flow Rate - BPD
Pressure
Water PI
Oil
IPR
Composite
IPR
qw(max)qt(max)
 Finally, we can consider both combined (straight-line plus curve) and
composite on the same IPR.
 Graphically it would look like this, where qt is the composite flow:
Composite and Combined IPR:

39. The “Skin” effect
(van Everdingen & Hurst)
Skin is a wellbore phenomenon, that causes an additional pressure drop in
the near-wellbore region:
S
hk
Bq
S
hk
q
p
o
oo
o
o
skin
µ
π
µ 2.141
2
)( ==∆

40. 44
Darcy's Law for radial flow into a wellbore:
Why is removing skin so important?
skin

41. Effect of Skin on IPR
Outflow
Flowrate
PressureatNode
5 0 -1 -3
SKIN
Inflow
(IPR)
qo α 1/ ln re +S
rw
Note : Log effect
10

42. 46
Darcy's Law for radial flow into a wellbore:
In some cases, the PI can also be improved slightly by
acidizing or fracturing. Acidizing cleans up "skin" on
the perforations and can improve porosity in limestone
reservoirs by making larger holes for oil flow.
Skin Damage Acid
Before After

43. 47
Darcy's Law for radial flow into a wellbore:
Fracturing can also improve permeability by making
large cracks near the wellbore.
Before After

44. Effect of Pressure Depletion on IPR
Outflow
Flowrate
PressureatNode
Reservoir with no pressure support
Inflow
Decreasing reservoir pressure

45. Effect of Tubing Size on Outflow
Inflow
(IPR)
Outflow
Flowrate (stb/d)
PressureatNode
2 3/8”
2 7/8”
4 1/2”
3 1/2”

46. 50
Well Performance Pressure gradient plots
Depth
Pressure Po
(0%)
Po
(30%)
Pwh
Po (30%) Required for 100 psi
wellhead pressure = 3761 psi
Po (0%)Required for 100 psi
wellhead pressure = 3582 psi
This is outflow
Now let’s include inflow
Pres

47. 51
If the desired flow rate is 1000 BPD do
we need artificial lift?
Calculate Pwf at 1000 BPD

48. 52
Remember our Data - Exercise 1a
Oil Density: 30 API
Water cut: 0%
Water Density: 1.026 sg
Pres: 3765 psig
P whead: 100 psia
PI: 10 stb/d/psi
Bo: 1.33 rb/stb
TVD: 9183 feet
Find Pwf at a flow rate of 1000 BPD
(assume no friction)

49. 3
Well Performance Pressure gradient plots
Depth
Pressure PresPo
(0%)
Po
(30%)
Pwh
Po (30%) Required for 100 psi
wellhead pressure = psi
Po (0%)Required for 100 psi
wellhead pressure = psi
Pwf available at 1000 BPD
= psi
Pwf

50. 54
HOW?
Artificial Lift
Pressure
or
gas lift the well
Introduce a pump to reduce Pwf

51. 55
Artificial Lift Options
ESP -Creates head (P) to lower Pwf
GAS LIFT -Reduces fluid column gradient to lower Pwf
PCP -Creates head (P) to lower Pwf
JET PUMP -provides pressure drop in venturi to lower Pwf
ROD PUMP -Intermittently sucks fluid from well bore lowering Pwf
ALL INCREASE DRAWDOWN TO PRODUCE FLOW

52. 56
CashFlow
Time Artificial
Lift
• Make good wells better
• Generate more revenue earlier in the life
of a project
Field Development

53. END of MODULE One