VLP AND WORKING CHART LECTURE #2.pdf

1. Dr. MOHAMMED ABDUL AMEER ALHUMAIRI
MISAN UNIVERSITY
COLLEGE OF ENGINEERING – PETROLEUM DEPARTMENT
VFP
dr.alhumairi@uomisan.edu.iq

2. VERTICAL LIFT RELATIONSHIP
VFP

3. WORKING CHARTS FOR TUBING SIZE 1 INCH AND DIFFERENT FLOW RATES

5. EXAMPLE : GIVEN DATA ARE ,FLOW RATE =500 BBL /DAY,
TUBING SIZE =1.5 INCH, THE PRODUCTION ALL OIL, GLR=200
SCF/BBL, WELL DEPTH 5000 FT, PTH =200 PSI.
CALCULATE PWF FROM THE ABOVE GIVEN DATA.
SOLUTION :
1- CHOOSE THE WORKING CHART FOR QO=500 BBL/DAY AND
TUBING SIZE =1.5 INCH.
2- INTER THE CHART WITH THE VALUE OF PTH=200 PSI FROM X-
AXIS.
3- USE THE GLR =200 SCF/BBL CURVE TO FIND THE EQUIVELENT
DEPTH FOR PTH ON Y-AXIS.
4- THE EQUIVELENT DEPTH FOR PTH= 1300 FT.
5- FIND THE EQUIVELENT DEPTH FOR
PWF=EQUVELENT DEPTH FOR PTH+WELL DEPTH
= 1300+5000=6300FT.
6-USE THE SAME GLR =200 SCF/BBL TO FIND THE VALUE OF
PWF= 1600 PSI
PTH PWF

6. Vertical Working Charts : are defined as graphical presentation of multi phase correlation
( vertical and horizontal flow), that were constructed as relationship between pressure
and depth for different flow rates, pipes size, GLR gas liquid ratio
Pressure
depth
Flow rate
Pipe size
GLR

7. We have vertical and horizontal working chart , these chart build depend on oil API gravity
and gas specific gravity and average flowing temperature, through these charts , we can
constructing vertical flow performance with IPR.
The cross point between IPR and vertical flow performance (VFP) represent the allowable
flow rate.
Cross point

8. Example depth =6000 ft, tubing size=2 inch, Pr=3000 psi, GLR=400 scf/bbl, Pth=200 Psi,
(well head pressure), Qo=400 bbl/day ( all oil).
Determine Pwf from working chart.
Solution:
Pth= 120 psi
Equivalent depth to Pth= 2200 psi .
Equivalent depth to pwf= 2200+6000=8200 ft
Pwf =1230 Psi
Now find Pth from Pwf=1230 psi
Solution :
Equivalent depth to Pwf= 8200 ft
Equivalent depth to Pth= 8200-6000 = 2200 ft
So Pth from working charts = 120 Psi

11. Pth=200 psi
Equivalent depth to
Pth =2200 ft
Equivalent depth to Pwf
=2200+ 6000=8200 ft
Pwf=1230
psi

12. Tubing in fig(1) greater than tubing in fig(2)
Heriot Watt
Fig 1 Fig 2 Fig 3

13. Example
Pr=2000 Psi, PI= 1.5 bbl/day/psi, qo=400 bbl/day, all oil, depth =6000 ft, GLR=400 scf/bbl, Pth=200 Psi tubing
size=2 inch.
Solution :
Find Pwf from Pr=2000 psi, qo=400 bbl/day, PI=1.5 bbl/day/psi
𝑃𝐼 =
𝑞𝑜
𝑝𝑟 − 𝑝𝑤𝑓
= 1.5 =
400
2000 − 𝑝𝑤𝑓
=== 𝑝𝑤𝑓 = 1733 𝑝𝑠𝑖
Now IPR can be constructed from Pr and (pwf, qo),( two test points)
To construct the vertical flow performance need to assume qo ( available from working charts ) to find pwf
depending on Pth, as follows :
From working charts for tubing size 2 inches we have different flow rate ( 200, 400, 600, 800, 1000, 1500), from
these charts find Pwf for each flow rate:
Flow rate (
available )
Pwf (from working charts for Pth =120 psi)
200 1100
400 1300
600 1380
800 1450
1000 1570
1500 1800
Pwf
qo
2000
0
1733
400
1000
1500

14. IPR
VFP
Qo= 820
bbl/day
Pwf =1450 Psi

15. Example
Depth 8000 ft, Pr=2000 psi, PI=2 bbl/day/psi, assume linear , Pth=120 psi, tubing diameter 2.5 inch, GLR= 600
scf/bbl Determine the flow rate, assume all oil.
Solution :
Construct IPR
Assume any value for Pwf =1000 psi less than Pr.
𝑃𝐼 =
𝑞𝑜
𝑝𝑟 − 𝑝𝑤𝑓
= 2 =
𝑞𝑜
2000 − 1000
𝑞𝑜 = 2000 𝑏𝑏𝑙/𝑑𝑎𝑦
Now construct vertical flow curve for Pth =120 psi, by assuming qo available for tubing size =2.5 inch:
From the vertical flow curve and IPR:
Qo=1150 bbl/day
Pwf=1410 psi
Flow rate ( available ) Pwf (from working charts for Pth =120 psi)
400 950
800 1255
1000 1360
1200 1440
1500 1570
Pwf
qo
2000
0
1000
2000