Well Performance Analysis
Adam Muhammad approved by Mr. Akhmad Sofyan, M.T.
Day – 4 (Part 2) eq Day - 5
Reference
1. Brown, Kermit E., Beggs, D. H. 1977.“The Technology of Artificial Lift
Methods”. Vol 1. Tulsa, Oklahoma. PennWell Publishing Company.
Outline – Day 4 (Part 2) Eq Day 5
❖ Limited Correlations
❖Development and Utilization of The Best Correlations
in Predicting Pressure Loss
1
Limited Correlations
Poettmann and Carpenter Method
• Assumptions :
1.
The correlations is applicable to the pipe sizes used in field study (2, 2 ½ and 3 in.) and
extension of the correlation to other size should be done with caution.
2.
The total output flow rate was used to calculate the density at any one pressure point
3.
Flow pattern were ignored
4.
The effect of viscosity were assumed to be negligible
5.
The acceleration term of the general energy equation was considered to be negligible.
This will be true for most cases but the term is significant at very high velocities and was
included in the friction factor
6.
Although the energy loss factor is assumed to be an average value over the length of
vertical tube it was actually found to vary from the bottom to the top of the tubing string
Poettmann and Carpenter Method Cont’d
• The Poettmann and Carpenter correlation is still used quite extensively in the field.
It can be used with confidence for the following conditions :
1. Tubing sizes 2 in., 2 ½ in., 3 in.
2. For viscosities less than 5 cp
3. For GLR less than 1500 SCF/Bbl
4. For rates greater than 400 Bbl/d
• Extension of Poettmann and Carpenter Method is Fancher and Brown
Method and Hagedorn and Brown Method to Account for Viscous Effects
Poettmann and Carpenter Method Cont’d
• Poettmann and Carpenter started with the general energy eq :
𝑝
‫𝑉 𝑝׬‬
1
∆𝑝 +
𝑔
∆ℎ
𝑔𝑐
+ ∆
𝑣2
2 𝑔𝑐
+ W + Iw = 0
Assume :
W = 0 and kinetic energy to be negligible and that 𝑔Τ𝑔𝑐 = 1, then:
𝑝2
‫𝑉 𝑝׬‬
∆𝑝 + ∆ℎ+ Iw = 0
Poettmann and Carpenter Method Cont’d
Assume :
Total losses du to irreversibilities can be correlated by the Fanning Eq :
Iw =
∆ℎ =
4 𝑓 ∆ℎ 𝑣 2
2 𝑔𝑐 𝑑
𝑝
1
4 𝑓 𝑣2
1+ 2 𝑔 𝑑
𝑐
− ‫ 𝑝׬‬2 𝑉 ∆𝑝
; where : f = correlating energy loss factor, then :
𝑝2
‫𝑉 𝑝׬‬
1
∆𝑝 + ∆ℎ+
𝑉𝑚
𝑚
𝑞 𝑉𝑚
𝐴
;V=
,v=
4 𝑓 ∆ℎ 𝑣 2
2 𝑔𝑐 𝑑
=0
; rewrite the abbove eq as follows :
Poettmann and Carpenter Method Cont’d
∆ℎ =
𝑝
2
4 𝑓 𝑞2 𝑉𝑚 2
1+
2 𝐴2 𝑔𝑐 𝑑
1
𝑚
∆ℎ =
‫ 𝑝׬‬1 𝑉𝑚 ∆𝑝
; A = ¼ 𝜋 𝑑 2 and converting to field units
1 𝑝1
‫𝑝∆ 𝑉 ׬‬
𝑚 𝑝2 𝑚
𝑓 𝑞2 𝑉𝑚 2
1+
7.413 1010 𝑑2
; 𝑉𝑚 =
𝑚
; Then
𝜌
, ∆ℎ =
1 𝑝1
‫𝑝∆ 𝑉 ׬‬
𝑚 𝑝2 𝑚
𝑚2
𝑓 𝑞2
𝜌
1+
7.413 1010 𝑑2
Where : w = qm = mass flow rate, or :
∆ℎ =
1 𝑝1
‫𝑝∆ 𝑉 ׬‬
𝑚 𝑝2 𝑚
𝑓 𝑤2
1+
7.413 1010 𝜌−2 𝑑2
; Ass :
𝑤2
𝑓
7.413 1010 𝜌−2 𝑑 2
= K ; Then, ∆ℎ =
1 𝑝1
‫𝑉 ׬‬
𝑚 𝑝2 𝑚
1+𝐾
∆𝑝
Poettmann and Carpenter Method Cont’d
Dividing the above eq into ∆𝑝 :
∆𝑝
∆𝑝
=
1 𝑝1
∆ℎ
𝑉 ∆𝑝
𝑚 ‫𝑝׬‬2 𝑚
1+𝐾
∆𝑝
∆𝑝 (1 + 𝐾)
=
1 𝑝1
∆ℎ
𝑉 ∆𝑝
𝑚 ‫𝑝׬‬2 𝑚
∆𝑝
∆ℎ
=
(𝑃1 − 𝑃2 )
1 𝑝1
‫𝑉 ׬‬
𝑚 𝑝2 𝑚
∆𝑝
+
𝑃1 − 𝑃2 𝐾
1 𝑝1
‫𝑉 ׬‬
𝑚 𝑝2 𝑚
∆𝑝
; It is noted that : 𝑣𝑚 =
1 𝑝1
‫𝑉 ׬‬
𝑚 𝑝2 𝑚
∆𝑝
(𝑃1 − 𝑃2 )
Poettmann and Carpenter Method Cont’d
Then,
∆𝑝
∆ℎ
=
𝑚
𝑣𝑚
+
∆𝑝
∆ℎ
𝑚𝐾
𝑣𝑚
; It is know that 𝜌 =
=𝜌+
𝑚
𝑉𝑚
𝑓 𝑤2
7.413 1010 𝜌 𝑑 2
Poettmann and Carpenter developed their energy loss factor correlation by
knowing
∆𝑝
∆ℎ
in psi/ft and solving the above eq for the energy loss factor as
follows :
∆𝑝
𝑓=
7.413 1010 𝜌 𝑑 2 144∆ℎ −𝜌
𝑤2
Example Problem Worked By the Poettman and
Carpenter Method
Given :
d
= 1.995 in ID
P1
= 500 psig
P2
= 1000 psig
T1
= 120 oF
T2
= 150 oF
SGg
= 0.65
SGw
= 1.07
Example Problem Worked By the Poettman and
Carpenter Method Cont’d
Given :
Density oil
= 22 OAPI
qo
= 400 STB/d
qw
= 600 STB/d
Viscosity Gas
= 0.018 cp
SFTo
= 30 dynes/cm
SFTw
= 70 dynes/cm
GLR
= 500 SCF/STB
Solution
1. SGo = 0.9218241042
2. qg = 500,000 SCF/d
3. GOR = 1250 cu ft/Bbl
4. m (massa per STB oil) = 350 (Sgo) + 0.0764 (GOR) (SGg) + 350 (Sgw)
(WOR) = 946,4634365 lbm/STB oil
5. Total mass flow rate/d = qo x m = 378585,3746 lbm/day
Solution Cont’d
7. P = 500 psig
1. Rs = 56,50240974
Where : Frick utilizes the following formula for calculating Rs for a particular set of conditions :
𝑅𝑠 = 𝑆𝐺𝑔
𝑝
18
𝑥
100.0125 𝐴𝑃𝐼
1Τ
0.83
100.00091 𝑡
2. Bo = 1,045192931
Where : Frick has used the following for calculated Bo :
𝐵𝑜 = 0.972 + 0.000147 𝑅𝑠
𝑆𝐺𝑔
𝑆𝐺𝑜
1.175
0.5
+ 1.25 𝑇
Solution Cont’d
3. Vm = 5.61 Bo + 5.61 (WOR) + vol free gas (
14.7
𝑇
𝑧
)( )( )
𝑝
520 1
Vm = 49.82700728 cu ft
4. 𝜌𝑚= 19,00575715 lbm cu ft
5. Flowing pressure gradient @ 500 psig = 0.176 psi/ft (From Fig 2.37)
8. P = 1000 psig
Bo = 1.083, Rs = 120 SCF/Bbl, z = 0.9, Vm = 31.8 cu ft, 𝜌𝑚 = 29.8 lbm/cu ft,
flowing pressure gradient = 0.236 psi/ft → Kermit E Brown Vol 1 Page 317
Solution Cont’d
9. Average pressure gradient = 0.206 psi/ft
10. The distance between 500 and 1000 psig = 2427,184466 ft
11. d v 𝜌 = 33.55917392 ; f = 0.008
12.
∆𝑝
@
∆ℎ
∆𝑝
@
∆ℎ
500 =
1
144
1000 =
1
144
𝜌+
𝑓 𝑤2
7.413 1010 𝜌 𝑑 2
𝜌+
𝑓 𝑤2
7.413 1010 𝜌 𝑑 2
= 0.177 psi/ft
= 0.235 psi/ft
13. The average gradient is 0.206 psi/ft → giving a distance of 2427, 18 ft between
500 and 1000 psig
2
Development and Utilization
of The Best Correlations in
Predicting Pressure Loss
Generalized correlation of Hagedorn and Brown
• An effort was made by Hagedorn and Brown to determine a generalized
which would include all practical ranges of flow rates, a wide range of
GLR, all ordinarilly used tubing sizes and the effects of fluid
properties.
• Starting with the basic flow equation in symbolic differential form based on
one pound mass of flowing fluid :
144
𝑔𝑐
𝑔
𝑉 𝑑𝑝 + 𝑑ℎ +
𝑣 𝑑𝑣
𝑔
+ 𝑑𝑊𝑓 + 𝑑𝑊𝑒 = 0
Generalized correlation of Hagedorn and Brown
Cont’d
• Assumes steady state flow and its units are in ft.
• Hagedorn and Brown chose to treat the mixture of gas and liquid as a
homogeneous mixture of combined properties.
• Assuming that no work is done by the flowing fluid, the equation becomes :
144
𝑔𝑐
𝑔
𝑉 𝑑𝑝 + 𝑑ℎ +
𝑣 𝑑𝑣𝑚
𝑔
+ 𝑑𝑤𝑓 = 0
• Assuming that the frictional loss can be defined by the Darcy Weisbach Eq :
𝑓 𝑑ℎ 𝒗𝒎 𝟐
𝑑𝑊𝑓 =
2𝑑𝑔
Generalized correlation of Hagedorn and Brown
Cont’d
• The following eq for the mixture results :
𝑔𝑐
𝑣 𝑑𝑣𝑚 𝑓 𝑑ℎ 𝒗𝒎 𝟐
144
𝑉 𝑑𝑝 + 𝑑ℎ +
+
=0
𝑔
𝑔
2𝑑𝑔
Where ∶ vm is an average of the mixture
• Writing this eq in integral form to solve:
𝑃2
ℎ2
𝑣𝑚2
ℎ2
𝑔𝑐
𝑣𝑑𝑣𝑚
𝑓 𝑑ℎ 𝒗𝒎 𝟐
144
න 𝑉 𝑑𝑝 + න 𝑑ℎ + න
+ න
=0
𝑔
𝑔
2𝑑𝑔
𝑃1
ℎ1
𝑣𝑚1
ℎ1
Generalized correlation of Hagedorn and Brown
Cont’d
• Evaluating three of terms :
𝑃2
𝑔𝑐
𝑣𝑚2 2 − 𝑣𝑚 2
𝑓 (ℎ2 − ℎ1) 𝒗𝒎 2
144
න 𝑉 𝑑𝑝 + ℎ2 − ℎ1 +
+
=0
𝑔
2𝑔
2 𝑑𝑔
𝑃1
• Solving h2 – h1, we have :
𝑔𝑐 𝑃1
𝑣𝑚2 2 − 𝑣𝑚 2
144 ‫𝑃׬‬2 𝑉 𝑑𝑝 −
𝑔
2𝑔
ℎ2 − ℎ1 =
𝑓 𝒗𝒎 𝟐
1+
𝑑2𝑔
Generalized correlation of Hagedorn and Brown
Cont’d
• Similar to the method of Poettmann and Carpenter,
𝑃1
‫𝑃׬‬2 𝑉 𝑑𝑝
𝑉𝑚 =
𝑃1 − 𝑃2
• Substiuting Vm (P1 –
𝑃1
P2) for ‫𝑃׬‬2 𝑉
ℎ2 − ℎ1 =
𝑔𝑐
144 𝑔
𝑑𝑝, then :
Vm ∆𝑝 −∆
𝑓 𝒗𝒎 𝟐
1+ 𝑑2𝑔
𝑣𝑚 2
2𝑔
; vm = vsl + vsg
Generalized correlation of Hagedorn and Brown
Cont’d
∆ℎ =
𝑔𝑐
144
𝑔
Vm ∆𝑝 −∆
1+
𝑓(
(
vsl + vsg)2
2𝑔
vsl + vsg)2
2 𝑔𝑐 𝑑
• The average integrated density of the mixture is:
1
𝜌𝑚 =
𝑉𝑚
• Solving for the average pressure gradient :
∆𝑝
𝑔
𝑓𝝆𝒎 (𝒗𝒔𝒍 + 𝒗𝒔𝒈)𝟐 𝝆𝒎 ∆ (𝑣𝑠𝑙 + 𝑣𝑠𝑔)2
144
=
𝝆𝒎 +
+
∆ℎ
𝑔𝑐
2 𝑔𝑐 𝑑
∆ℎ
Generalized correlation of Hagedorn and Brown
Cont’d
• If g = gc and the conversion to units commonly used in the field is made:
• Based on 1 bbl of liquid, the mass is :
• Where, w = qm
Generalized correlation of Hagedorn and Brown
Cont’d
• In this procedure we need to detemine the average integrated density.
It cannot be calculated directly because of the slippage or fallback of
the liquid phase.
• If Ve = volume of a section of pipe, then Hl Ve = (1-Hl) Ve, is that
portion occupied by gas
Generalized correlation of Hagedorn and Brown
Cont’d
• By substituting than :
• Hl → 1, vsl → 0, vsl → 0 (This eq will reduce to single phase liquid flow)
Generalized correlation of Hagedorn and Brown
Cont’d
• Hl → 0, vsl → 0, vsl → 0 (This eq will reduce to single phase gas flow)
Example Problem By Generalized Method of
Hagedorn and Brown
Given :
d
= 1.995 in ID
P1
= 500 psig
P2
= 1000 psig
T1
= 120 oF
T2
= 150 oF
SGg
= 0.65
SGw
= 1.07
Example Problem By Generalized Method of
Hagedorn and Brown Cont’d
Given :
Density oil
= 22 OAPI
qo
= 400 STB/d
qw
= 600 STB/d
Viscosity Gas
= 0.018 cp
SFTo
= 30 dynes/cm
SFTw
= 70 dynes/cm
GLR
= 500 SCF/STB
Solution
Solution Cont’d
Solution Cont’d
Solution Cont’d
Example Problem By Duns and Ros Method
Given :
d
= 1.38 in ID
P1
= 500 psig
P2
= 1000 psig
T1
= 120 oF
T2
= 150 oF
SGg
= 0.65
SGw
= 1.07
Example Problem By Duns and Ros Method Cont’d
Given :
Density oil
= 22 OAPI
qo
= 400 STB/d
qw
= 600 STB/d
Viscosity Gas
= 0.02 cp
SFTo
= 30 dynes/cm
SFTw
= 70 dynes/cm
GLR
= 500 SCF/STB
Solution
Solution Cont’d
Solution Cont’d
Solution Cont’d
Homework # 1 Calculate with Poettmann and
Carpenter Method and Hagedorn and Brown
Method
Given :
d
= 1.380 in ID @ 1 ¼ in Tubing
P1
= 500 psig
P2
= 1000 psig
T1
= 120 oF
T2
= 150 oF
SGg
= 0.65
SGw
= 1.07
Homework # 1 Calculate with Poettmann and
Carpenter Method and Hagedorn and Brown
Method Cont’d
Given :
Density oil
= 22 OAPI
qo
= 400 STB/d
qw
= 600 STB/d
Viscosity Gas
= 0.018 cp
SFTo
= 30 dynes/cm
SFTw
= 70 dynes/cm
GLR
= 500 SCF/STB
Homework # 2 Calculate with Duns and Ros Method
Given :
d
= 1.995 in ID @ 2 in Tubing
P1
= 500 psig
P2
= 1000 psig
T1
= 120 oF
T2
= 150 oF
SGg
= 0.65
SGw
= 1.07
Homework # 2 Calculate with Duns and Ros Method
Given :
Density oil
= 22 OAPI
qo
= 400 STB/d
qw
= 600 STB/d
Viscosity Gas
= 0.02 cp
SFTo
= 30 dynes/cm
SFTw
= 70 dynes/cm
GLR
= 500 SCF/STB
Thank You
END OF SLIDES – FOR DAY 4 (PART 2) eq DAY 5