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MINISTRY OF SCIENCE AND TECHNOLOGY
DEPARTMENT OF
TECHNICAL AND VOCATIONAL EDUCATION
Sample Questions & Worked Out Examples
For
PE-03023
PETROLEUM PRODUCTION ENGINEERING
B.Tech(First Year)
Petroleum Engineering
53
Ministry of Science and Technology
Department of Technical and Vocational Education
Petroleum Engineering
Sample Questions for
PE 03023 PETROLEUM PRODUCTION ENGINEERING Part ( I )
Chapter 1 Inflow Performance Relationship
1.*** Discuss the importance of knowing the IPR of a well with necessary illustrations. (10 marks)
2.** Using the Vogel's equation, construct the generalized IPR curve.
(10 marks)
3.*
The following test was conducted on a solution-gas drive well:
Static reservoir pressure = 2000 psi Flowing bottom-hole pressure = 500 psi
Producing rate = 146 bpd
Find (a) potential of the well, (b) flow rate for a flowing pressure of 1500 psi.
(10 marks)
4.** An oil well produced 500 bpd with a flowing BHP of 2500 psi. If the average reservoir
pressure is 3000 psi find the maximum possible rate from the well. Also find the production
rate when the producing pressure dropped to 1200 psi. Also find the maximum rate if a linear
relationship is assumed. The bubble point pressure is 3000 psi.
(15 marks)
5.** Construct the IPR curve for the well of problem (4).
(10 marks)
6.** Given data:
Average reservoir pressure, psi
Flowing bottom-hole pressure, psi
Production rate, bpd
Oil viscosity, cp
Oil formation volume factor
Oil saturation
Residual oil saturation
Residual water saturation
7.**
Present
2500
2300
400
3.0
1.18
0.76
Future
2000
1500
?
3.6
1.16
0.67
20%
22%
Find the production when the reservoir pressure drops to 2000 psi and for a flowing pressure
of 1500 psi, and construct the future IPR for the well.
(20 marks)
From a solution-gas drive well the following information was obtained:
Static reservoir pressure, psi
Flowing bottom-hole pressure, psi
Production rate, bpd
Oil viscosity, cp
Oil formation volume factor
Average oil saturation
Residual oil saturation
Residual water saturation
Present
1800
1600
250
4.0
1.16
0.75
Future
1500
4.5
1.145
0.72
18%
22%
Construct an IPR curve for the future time. (20 marks)
8.*** A well is flowing at 1120 bpd through 2⅞-in. tubing. There is zero water cut, and the GLR is
820 cu ft/bbl. A pressure survey on the well shows that the flowing pressure at 6470 ft (the
foot of the tubing) is 675 psig, while the pressure buildup survey gives a static pressure of
2080 psig at a datum level of 6500 ft.
Using Vogel's method draw the IPR curve and estimate the well's potential.
54
Reservoir analysis indicates that the ratio of the value of kro/µoBo now to its value at the static
pressure of 1500 psig is 1.57. Estimate what the well's potential rate will be when the static
pressure has dropped to 1500 psig. And construct the future IPR curve for the well on the
same graph paper.
(20 marks)
9.**
A series of tests is made on a certain well with the following results.
Oil rate, bpd
40
56
61
70
WOR
0
0.785
1.440
1.855
Flowing BHP, psi
2360
1950
1524
1000
Draw the curve of water cut versus gross production rate and the oil, water, and gross IPRs.
Might any harm result from shutting this well in for a few days?
(15 marks)
Chapter 2 Multiphase Flow in Pipes
10.*** Discuss the Gilbert's empirical approach to the vertical two-phase flow in oil wells.
(10 marks)
11.* Explain the slippage and flow resistance and discuss the optimum GLR.
(15 marks)
Chapter 3 Flowing Well Performance
(Use square graph paper and Gilbert's families of curves only)
12.* A well utilizing 2⅜-in. tubing is producing against 200 psi surface pressure at a rate of 200
bbl/day with a constant GOR of 600 from a depth of 8000 ft. Determine the flowing BHP
required to produce 200 bbl/day.
(5 marks)
13.* A 5000-ft well is equipped with 2⅜-in. tubing and produces into a constant surface tubing
pressure of 100 psi. The static BHP is 2150 psi. The following production test was made on
this well.
Production rate = 550 bwpd and 50 bpd, GOR = 1200 scf/bbl
Find the PI of this well.
(8 marks)
14.* A 6000-ft well is equipped with 1.9-in. tubing. A recent survey shows the following
information.
TEST 1 : THP = 250 psi
GOR = 400 Production = 100 bbl/day
TEST 2 : THP = 100 psi
GOR = 750 Production = 200 bbl/day
Determine the PI from these two flowing tests.
(10 marks)
15.** A well 6000 ft deep with a static BHP of 3000 psi is equipped with 1.9-in. tubing. This well
makes clean oil and has a PI of 0.2. The wellhead tubing pressure remains constant for all
rates of 200psi. What GLR will be required to produce 400 bbl/day from this well?(10 marks)
16.** A flowing well with 5540 ft of tubing in the hole is completed without a casing-tubing
packer. The CHP is 480 psig when the production rate is 750 bbl/day and 760 psig when the
production rate is 525 bbl/day. What are the PI, static pressure, and potential of the well?
(10 marks)
17.** A well completed over the interval 2994 to 3032 ft has 2⅜-in. tubing hung at 3000 ft. The
well is flowing 320 bbl/day, zero water cut, at a GOR of 400 cu ft/bbl with a THP of 500
psig. The static pressure is 1850 psig at 3000 ft. What would be the effect of changing the
choke size to ½-in.?
(15 marks)
18.** A certain well is completed with 7500 ft of 3½-in. tubing in the hole, the tubing shoe being
located just above the top perforations. The well is flowing 130 bbl/day of oil with a water
cut of 25 per cent and a GOR of 1200 cu ft/bbl. If the well's static pressure is 2800 psig and
its gross PI is 0.32, estimate the size of choke in the flow line. At what rate would the well
flow if a ½-in. bean were substituted for the current one?
(15 marks)
19.** A well was completed with 7-in. casing perforated from 7216 to 7253 ft with 7000 ft of 2⅜in. tubing in the hole. The well was flowing steadily at 320 bbl/day of clean oil, GOR 800 cu
ft/bbl against an 11/64-in. choke when a smaller choke was accidentally inserted in the flow
line. When the well stabilized against the new choke, it was flowing with a THP of 300 psig
and a CHP of 993 psig. Determine the new bean size, the well's static pressure and the PI.
(15 marks)
55
20.** A new flowing well completed with 2⅜-in. tubing hung at the top of the perforations at 5500
ft was initially produced on ¼-in. choke, the THP stabilizing at 400 psig. After a few days'
production the choke size was increased to ½-in., and the THP stabilized at 270 psig. One
week later the choke size was again increased, and the well then gauged at 600 bbl/day of
clean oil, GOR 800 cu ft/bbl, THP 140 psig. Estimate the well's static pressure and its
pumped-off potential.
(15 marks)
21.** The completion data of a well are as follows:
Total depth
4052 ft
7-in. casing surface to 4020 ft
3½-in. tubing hung at 4000 ft
Casing-tubing packer installed just above tubing shoe.
The well was flowing at 300 bbl/day of clean oil, a GOR of 600 cu ft/bbl, and a THP of 300
psi when it was decided to try the effects of an acidizing treatment. During this treatment,
10000 gals of acid was squeezed into the formation. A surface pressure of 3200 psi was
needed to overcome the static reservoir pressure of 1800 psi and to achieve the desired
injection rate of 2 bbl/min. After the treatment the well's production rate stabilized at 340
bbl/day of clean oil through 3½-in. tubing with a GOR of 1000 cu ft/bbl and a THP of 300
psi. (a) Determine whether or not the treatment was successful, and give an explanation of
the results obtained. (b) What would have been the production rate of the well at a THP of
300 psi if, instead of the acidizing treatment, the 3½-in. tubing string had been replaced with
a 2⅜-in. tubing (assuming that the change could have been made without damaging the
producing formation) ?
(15 marks)
22.*** A flowing well with 5000 ft of 2⅜-in. tubing in the hole is completed with a 7-in. casing. The
static pressure is 1450 psig at 5000 ft. The well is currently flowing with a CHP of 738 psig
and THP of 360 psig. The production is clean oil and GLR is 0.45 mcf/bbl. Construct the IPR
and determine the PI.
Which size of bean should be inserted if it is desired to produce at the maximum possible rate
with a drawdown of 980 psi?
(15 marks)
23.* A flowing well is completed with 7332 ft of 2⅜-in. tubing, the tubing shoe being located
opposite the top of the perforations. At what rate will the well flow against a ¼-in. choke
under each of the following conditions?
( ) Static pressure, 3000 psig; PI, 0.42 bbl/day/psi; GOR, 200 cu ft/bbl.
( ) Static pressure, 2500 psig; PI, 0.33 bbl/day/psi; GOR, 330 cu ft/bbl.
( ) Static pressure, 2000 psig; PI, 0.29 bbl/day/psi; GOR, 500 cu ft/bbl.
( ) Static pressure, 1500 psig; PI, 0.27 bbl/day/psi; GOR, 1000 cu ft/bbl.
On the same sheet of graph paper plot the pumped-off potential and the actual flow rate
against the static BHP?
Chapter 4 The Principles of Gas Lift
24.*
25.*
Describe the two major methods of gas-lift operation.
(10 marks)
Discuss the classification of wells to decide whether continuous or intermittent gas lift
operation is applicable to a well.
(10 marks)
26.* What are the three main types of gas-lift installations?
(10 marks)
27.** What type of installation would you recommend for the wells specified in the following
table?
(20 marks)
Well
A
B
C
D
E
F
G
Depth
(ft)
7000
2200
5000
6000
12000
1500
500
Static BHP
(psig)
3000
250
200
2500
1500
65
10
Productivity
index
10
0.1
2
0.02
0.2
0.5
0.1
Separator
pressure, (psig)
50
10
30
100
100
10
0
Fluid gradient,
(psi/ft)
0.50
0.35
0.40
0.35
0.45
0.33
0.33
28.*
29.*
30.**
31.**
32.**
33.**
34.**
35.***
36.**
56
What is the force balance under operating conditions for the conventional fluid-operated
valve of Fig.4.11? The notation is as follows: Ab = effective bellows area; As = area of stem;
Av = area of port; pb = bellows charge pressure; pc = injection pressure; pt = tubing pressure.
(10 marks)
Write the force balance under operating conditions for the partially balanced, fluid-operated
valve of Fig.4.11.
(10 marks)
Calculate the casing pressure opposite a valve at 6000 ft using Eq.(4.15), and check with
Fig.4.10 for the following well conditions:
(10 marks)
Geothermal gradient = 1.6°/100 ft
Wellhead pressure = 500 psia
Mean surface temperature = 70°F
Gas gravity = 0.60
A pressure-operated valve without a spring has an opening pressure in a valve tester of 600
psig at 60°F. The valve is to be placed in a well at 6000 ft where the temperature opposite the
valve is 160°F.
(15 marks)
( ) If the ratio Av/Ab is 0.112 what is the bellows pressure at 60°F?
( ) What is the bellows pressure at 160°F?
( ) What is the surface closing pressure of the valve?
An operator needs the surface closing pressure of a pressure-operated flow valve without
spring to be 600 psig when the valve is installed in a well at 6000 ft at a temperature of
160°F. The valve ratio Av/Ab is 0.112.
(15 marks)
( ) Determine the gas pressure in the casing opposite the valve at 6000 ft for a surface casing
pressure of 600 psig.
( ) What is the bellows pressure at 60°F?
( ) Calculate the valve opening pressure in a tester at 60°F.
A pressure-operated flow valve has a 75 psi spring pressure effect. The valve opening
pressure at 60°F in a tester is 600 psig. The valve is to be placed in a well at 6000 ft where
the temperature in the well opposite the valve is 160°F.
(20 marks)
( ) Calculate the pressure inside the bellows at 60°F if Av/Ab is 0.112.
( ) What is the bellows pressure at 160°F?
( ) Calculate the closing pressure of the valve at 6000 ft.
( ) What is the surface closing pressure of the valve?
An operator needs the surface closing pressure of a pressure-operated flow valve with a 75
psi spring effect to be 600 psig when the valve is installed in a well at 6000 ft where the
temperature is 160°F.
(20 marks)
( ) Using again the value 0.112 for Av/Ab calculate the pressure opposite the valve at 6000 ft
for a surface casing pressure of 600 psig. This is the closing pressure of the valve at the
operating temperature of 160°F.
( ) What is the pressure inside the bellows at 160°F?
( ) What is the pressure inside the bellows at 60°F?
( ) Calculate the opening pressure of the valve at 60°F which would be specified in ordering
from the manufacturer.
(a) Define the spread of a gas-lift valve.
(2 marks)
(b) What is the spread of a bellows valve having 0.75 sq in. of effective bellows area, o.10 sq
in. of seat area, 800 psig pressure in the dome, and zero back pressure? (4 marks)
(c) What is the spread with a back pressure of 700 psig?
(4 marks)
What are the tubing effect and tubing effect factor?
Calculate the T.E. and T.E.F. if there is a pressure in the tubing opposite the flow valve of
300 psig and the valve ratio is 0.11.
(15 marks)
Chapter 5 Gas Lift Design
37.* Explain the sequences of continuous unloading of a gas-lift well.
(10 marks)
38.** The following data are given on a well to be placed on intermittent flow.
Kick-off pressure = 750 psig
Working pressure = 700 psig
Production desired = 50 STB/day
Separator pressure = 50 psig
Spring tension = 100 psig
Valve ratio = 0.10
Gradient of well loaded with water = 0.5 psi/ft
Injection gravity and temperatures as Fig. 4.10.
Desired closing pressures of valves = 660, 650, 640, and 600 psig
57
( ) Calculate the valve spacing on the basic of closing pressure of the valve above.(10 marks)
( ) Using Eq.4.17 to calculate the temperatures opposite the valves, make an opening
pressure analysis in tabular form. (10 marks)
39.** It is desired to design a continuous gas-lift installation from the following well data:
Depth to top of perforations = 5800 ft
Operating pressure = 650 psig
Wellhead tubing pressure = 50 psig
Unloading gradient = 0.04 psi/ft
Static gradient = 0.5 psi/ft
Casing size = 5½-in., 20 lb/ft
Static fluid level = 2500 ft from the surface
Tubing size = 2-in.
Valve opening pressures at 60°F = 650, 625, 600, 575, and 550 psig
( ) What are the setting depths of each valve based on valve opening pressures at 60°F?
( ) A closing pressure analysis of the valve string calculated above should be made to be sure
that the valves as spaced will have closing pressures decreasing with depth. Valve ratio =
0.112 and spring effect = 75 psig. (20 marks)
40.** It is desired to design an intermittent gas-lift installation from the following well data:
Depth to top of perforations = 4750 ft
Depth to packer = 4700 ft
Tubing size = 2 in.
Casing size = 5½-in., 20 lb/ft
Surface kick-off pressure = 650 psig
Working gas pressure = 600 psig
Wellhead back pressure = 50 psig
Well full of water
Unloading gradient = 0.5 psi/ft
Valve opening pressures at 60°F = 650, 625, 600, 575 and 550 psig
( ) What are the setting depths of each valve based on valve opening pressures at 60°F?
(5 marks)
( ) A closing pressure analysis of the valve string calculated above should be made to be sure
that the valves as spaced will have closing pressures decreasing with depth. Complete the
closing pressure analysis given in the table.
(15 marks)
Closing Pressure Analysis
(1)
Valve
opening
pressure at
60°F,
psig
650
625
600
575
550
(2)
(3)
Depth of
valve,
ft
Temperature
at valve,
°F
89
106
121
133
145
(4)
(5)
Bellows
pressure at
60°F,
psig
511
Bellows
pressure at
Tv,
psig
541
(6)
Valve
closing
pressure at
depth,
psig
608
(7)
Surface
closing
pressure,
psig
592
41.** The following data are given for a well:
Depth to center of perforation = 5000 ft
Static BHP = 1600 psia
Specific gravity of oil = 0.802 (45°API)
Specific gravity of water = 1.07
Specific gravity of gas = 0.60
Desired production = 100 bbl/day
Tubing size = 2 in. nominal
Producing GOR = 500 SCF/bbl
Initial surface tubing pressure = 350 psia
Productivity index = 4
Surface flowing temperature = 114°F
Average flowing temperature = 124°F
Bottom-hole temperature = 134°F
(a) What is the flowing BHP when the well is producing 50 per cent water? (5 marks)
(b) Now the water-oil ratio has increased to 2:1 and the well is off production. The well is to
be placed on continuous-flow gas-lift with a surface operating injection pressure of 700 psig
and a kick-off pressure of 800 psig. Determine the optimum point of gas injection by the
graphical method, with a differential across the valve of 100 psi and a 200 psia surface tubing
pressure.
(15 marks)
42.*
A well is to be placed on continuous-flow gas lift. The well data are as follows:
Well depth = 4900 ft
Tubing size = 2 in.
Productivity index = 2.0
Static BHP = 2000 psig
58
Production = 500 bbl/day
Flowing gradient = 0.5 psi/ft
Tubing pressure = 50 psig
Unloading gradient = 0.12 psi/ft
Surface kick-off and operating pressure = 650 psig
Temperature at wellhead = 100°F
Temperature at 4900 ft = 149°F
Av/Ab = 0.11
Spring effect = 75 psig
( ) Using valves with surface closing pressure of 600, 580, and 550 psig, calculate the setting
depth for each valve, using the valve closing pressure at valve depth.
( ) What is the temperature at each valve depth?
( ) What are the pressure inside the bellows at valve depths?
( ) What are the pressure inside the bellows at 60°F?
( ) What are the valve opening pressures at 60°F?
(15 marks)
43.** A low productivity index, relatively high BHP well is to be placed on intermittent-flow gaslift. The well data are as follows:
Depth to middle of perforations = 9000ft
Surface operating pressure = 900 psig
Tubing size = 2 in. (capacity 0.00387 bbl/ft)
Casing size = 5½-in., 20 lb/ft
Gravity of oil = 34°F (0.855)
Static BHP = 2500 psig
Well productivity index = 0.20
Tubing pressure = 30 psig
Oil production desired = 100 bbl/day
Wellhead temperature = 100°F
Average tubing temperature = 142°F
Minimum cycle time = 35 min
( ) Assuming that 40 per cent of the starting slug is produced, what will be the depth to the
operating valve?
( ) What is the desired casing pressure opposite the valve?
( ) What is the minimum injection gas-oil ratio?
(15 marks)
44.** A well is to be placed on intermittent-flow gas-lift. The well data are as follows:
Well depth = 8000 ft
Tubing size = 2-in., Capacity: 0.00387 bbl/ft
Specific gravity of oil = 0.87
Casing size = 5½-in., 20 lb/ft
No water production.
Well PI = 0.08 bbl/day/psi
Static BHP = 2200 psig
Wellhead pressure = 50 psig
Oil production desired = 120 bbl/day Average tubing temperature = 127 °F
Cycle time = 48 min
Gas deviation factor = 0.87
Calculate the depth to the operating valve and theoretical minimum injection gas-oil ratio,
assuming 70 per cent of starting slug is produced.
(15 marks)
45.*** What are the two common types of chambers?
(10 marks)
46.* Calculate the chamber length for the following well data:
(10 marks)
Injection pressure = 400 psig
Wellhead tubing pressure = 50 psig
Fluid gradient = 0.4 psi/ft
Insert chamber = 1-in. tubing inside 4-in. OD tubing
Capacity of 1-in. tubing = 1.07 bbl/1000 ft
Annular capacity between 1-in. dip tubing and 4-in. tubing = 13.54 bbl/1000 ft
* = Must know, ** = Should know, *** = Could know
59
Ministry of Science and Technology
Department of Technical and Vocational Education
Petroleum Engineering
Worked Out Examples for
PE 03023 PETROLEUM PRODUCTION ENGINEERING Part ( I )
1. An oil well produced 500 bpd with a flowing BHP of 2500 psi. If the average reservoir pressure
is 3000 psi find the maximum possible rate from the well. And find the production rate when the
producing pressure dropped to 1200 psi. Also find the maximum rate if a linear relationship is
assumed. The bubble point pressure is 3000 psi.
Solution
The bubble point pressure is 3000 psi and therefore, flowing BHP of 2500 psi is well below the
bubble point pressure ,and Vogel's equation or generalized IPR curve can be used.
p wf
2500
=
= 0.833
pR
3000
 p wf
qo
= 1 − 0.2
(q o ) max
 pR
p

 − 0.8 wf
 pR

2

 = 1 − 0.2 × 0.833 − 0.8(0.833) 2 = 0.278

(qo)max = 500/0.278 = 1800 bbl/day
Maximum possible rate from the well = 1800 bbl/day.
When producing BHP drops to 1200 psi,
p wf
pR
=
1200
= 0.4
3000
 p wf
qo
= 1 − 0.2
(q o ) max
 pR

p
 − 0.8 wf

 pR
2

 = 1 − 0.2 × 0.4 − 0.8(0.4) 2 = 0.792

qo = 0.792×1800 = 1425 bbl/day
Production rate from the well at 1200 psi flowing BHP = 1425 bbl/day.
If linear relationship is assumed,
 p wf
qo
= 1 − 
(q o ) max
 pR

 = 1 − 0.833 = 0.167

(qo)max = 500/0.167 = 3000 bbl/day
Maximum possible rate from the well = 3000 bbl/day.
60
2. Given data:
Present
2500
2300
400
3.0
1.18
0.76
Average reservoir pressure, psi
Flowing bottom-hole pressure, psi
Production rate, bpd
Oil viscosity, cp
Oil formation volume factor
Oil saturation
Residual oil saturation
Residual water saturation
Future
2000
1500
?
3.6
1.16
0.67
20%
22%
Find the production when the reservoir pressure drops to 2000 psi and for a flowing pressure of
1500 psi, and construct the future IPR for the well.
Solution
Relative permeability to oil can be calculated using Corey type relationship.
For present day,
For future time,
4
4
4
4
4
4
 ( S o − S or ) 
 0 . 76 − 0 . 20 
 0 . 56 
k ro = 
=
 = 0 . 87
 = 

 1 − 0 . 20 − 0 . 22 
 0 . 58 
 (1 − S or − S wi ) 
 ( S o − S or ) 
 0 . 67 − 0 . 20 
 0 . 47 
k ro = 
=
 = 0 . 43
 = 

 1 − 0 . 20 − 0 . 22 
 0 . 58 
 (1 − S or − S wi ) 
The value of J at present day is
From Eq.(1.9),
J
*
p
=
J = 400/(2500–2300)=2.0 bbl/day/psi
1 .8 × 2 .0
= 2 . 2 bbl/day/ps
 2000 
1 + 0 .8 

 2500 
i
From Eq.(1.12)
 k ro
J *f = J *p 
 µ o Bo



f
 k ro
/ 
 µ o Bo

0 . 43
  0 . 87
=
 = 2 . 2 
 /

 3 . 6 × 1 . 16   3 . 0 × 1 . 18 
p
0.921 bbl/day/psi
From Eq.(1.11),
qo =
 p
J * pR 
1 − 0 . 2  wf
1 .8 
 pR

qo =
0 . 921 × 2000
1 .8
 p

 − 0 . 8  wf
 pR




2



2

 1500 
 1500  
 − 0 .8
  = 409 bbl/day
1 − 0 . 2 
 2000  
 2000 

Production at flowing BHP 1500 psi at future is 255 bbl/day.
To construct the future IPR, prepare the following table and plot the data on the square graph
paper.
Assumed
flowing BHP, psi
0
500
1000
1500
2000
Production,
bbl/day
1023
920
716
409
0
61
Bottom-hole pressure, psi
2000
1500
1000
500
0
0
500
1000
1500
Production rate, bbl/day
Problem (2)
3.
A well 6000 ft deep with a static BHP of 3000 psi is equipped with 1.9-in. tubing. This well
makes clean oil and has a PI of 0.2. The wellhead tubing pressure remains constant for all rates of
100psi. What GLR will be required to produce 400 bbl/day from this well?
Solution
Static BHP = 3000 psi and PI = 0.2 bbl/day/psi, to produce 400 bbl/day, flowing BHP will be
pwf = 3000 – (400/0.2) = 1000 psi
Using the 1.9 in., 400 bpd pressure-curve, with FBHP of 1000 psi, observe various GLR at
various THP.
GLR,
mcf/bbl
0.3
0.4
0.6
Equiv. FBHP, ft
Equiv. THP, ft
6700
7250
8000
700
1250
2000
THP,
psi
80
140
200
Plot GLR vs. THP, from the graph GLR = 0.33 mcf/bbl at 100 psi THP.
THP, psi
200
150
100
50
0.3
0.4
0.5
GLR, mcf/bbl
Problem (3)
0.6
62
4. A flowing well with 5540 ft of tubing in the hole is completed without a casing-tubing packer.
The CHP is 480 psig when the production rate is 750 bbl/day and 760 psig when the production
rate is 525 bbl/day. What are the PI, static pressure, and potential of the well?
Solution
Using Eq.(3.3):
p wf = p c (1 +
D 1 .5
)
100
D1.5 = (5.54)1.5 = 13,
p wf = p c (1 +
13
) = 1 .13 p c
100
When CHP = 480 psig = 495 psia, p wf = 1.13 ×495 = 560 psia = 545 psig
When CHP = 760 psig = 775 psia, p wf = 1.13 ×775 = 875 psia = 860 psig
Assuming straight line IPR, and using p wf = p s −
For case I
545 J = Jp s − 750
For case II
860 J = Jp s − 525
Solving these two unknown equations,
potential is 1130 bbl/day. #
q
J
(1)
(2)
PI = 0.71 psi/bbl/day, ps = 1600 psig, and the well's
5. A well was completed with 7-in. casing perforated from 7216 to 7253 ft with 7000 ft of 2⅜-in.
tubing in the hole. The well was flowing steadily at 320 bbl/day of clean oil, GOR 800 cu ft/bbl
against an 11/64-in. choke when a smaller choke was accidentally inserted in the flow line. When
the well stabilized against the new choke, it was flowing with a THP of 300 psig and a CHP of
993 psig. Determine the new bean size, the well's static pressure and the PI.
Solution
First case: choke size was 11/16-in. and, so S = 44 for the Eq.(3.5),
0 . 546
0 .546
Tubing head pressure, p tf = 435 R1.89 q = 435 × 0 .851.89 × 320 = 100 psig
S
44
Second case: At THP = 300 psig, CHP = 993 psig = 1008 psia
p wf = 1008 (1 +
7 1 .5
) = 1195 psia = 1180 psig
100
Prepare the following table using the pressure gradient curve for 2⅜-in. tubing,
Assumed flow
rates, bbl/day
50
100
200
400
600
Flowing BHPs
at THP of 100 psig, psig
1000
820
760
800
840
Flowing BHPs
at THP of 100 psig, psig
1460
1200
1100
1160
1180
Assuming a straight line IPR, draw the IPR connecting the points 1180 psi on the first curve and
320 bbl/day on the second curve. From the IPR: static BHP = 1400 psig and the well's potential is
725 bbl/day. So PI is 725/1400 = 0.52 bbl/day/psi.
With new choke flowing BHP is 1180 psig and production is 110 bbl/day with THP 300 psig.
S 1.89 =
435 × 0.85 0.546 × 110
= 146 , S = 14 and 14/64 = 7/32-in. choke.
300
63
1600
Bottom-hole pressure, psi
1400
1200
1000
800
600
400
200
0
0
200
400
600
800
Production rate, bbl/day
Problem (5)
6. What is the force balance under operating conditions for the conventional fluid-operated valve of
Fig.4.11? The notation is as follows: Ab = effective bellows area; As = area of stem; Av = area of
port; pb = bellows charge pressure; pc = injection pressure; pt = tubing pressure.
Solution
Referred to Fig. 4.11, conventional fluid-operated valve
Closing force, Fc = pb Ab
Opening force, Fo = pt ( Ab − Av ) + p c Av
Force balance, Fc = Fo
pb Ab = pt Ab − pt Av + p c Av
7. Calculate the casing pressure opposite a valve at 6000 ft using Eq.(4.15), and check with Fig.4.10
for the following well conditions:
(10 marks)
Geothermal gradient = 1.6°/100 ft
Wellhead pressure = 500 psia
Mean surface temperature = 70°F
Gas gravity = 0.60
Solution
Temperature at 5000 ft, Tv = 70°F + 5000 ×(1.6/100)= 150°F
Therefore,
T avg =
70 + 150
= 110 ° F (or) 570 ° R
2
Increase in casing pressure,
∆ p ≅ 0 .25 ( p wh / 100 )( D / 100 ) = 0 .25 × (500 / 100 )( 5000 / 100 ) = 62 .5 psi
The pressure at the valve depth = 500 + 62.5 = 562.5 psia
p avg =
500 + 562.5
= 531 psia (or) 516 psig
2
From Fig. 4.9, z = 0.994
p v = p wh 10
γD / 122.7 zTavg
= 500 × 10 0.6×5000 / 122.7×0.944×570 = 555 psia= 540 psig
64
500 + 540
= 520psig
2
From Fig. 4.9 , z = 0.942 which gives pv = 555 psia (or) 540 psig
Again
p avg =
From Fig.4.10, pv = 550 psig
8. An operator needs the surface closing pressure of a pressure-operated flow valve with a 75 psi
spring effect to be 600 psig when the valve is installed in a well at 6000 ft where the temperature
is 160°F.
(20 marks)
(a)Using again the value 0.112 for Av/Ab calculate the pressure opposite the valve at 6000 ft for
a surface casing pressure of 600 psig. This is the closing pressure of the valve at the operating
temperature of 160°F.
(b)What is the pressure inside the bellows at 160°F?
(c)What is the pressure inside the bellows at 60°F?
(d)Calculate the opening pressure of the valve at 60°F which would be specified in ordering
from the manufacturer.
Solution
( ) R = 0.112, from Fig. 4.10, pvc @ Dv = 685 psig
(b)
p
p vc = (1 − R)[ bT + S t ]
1− R
p
685 = 0.888[ bT + 75]
0.888
pbT = 618 psig
(c)From Fig.4.10, pb = 515 psig
(d)
p
515
p vo = b + S t =
+ 75 = 655psig
1− R
0.888
9. It is desired to design a continuous gas-lift installation from the following well data:
Depth to top of perforations = 5800 ft
Operating pressure = 650 psig
Wellhead tubing pressure = 50 psig
Unloading gradient = 0.04 psi/ft
Static gradient = 0.5 psi/ft
Casing size = 5½-in., 20 lb/ft
Static fluid level = 2500 ft from the surface
Tubing size = 2-in.
Valve opening pressures at 60°F = 650, 625, 600, 575, and 550 psig
(a)What are the setting depths of each valve based on valve opening pressures at 60°F?
(b)A closing pressure analysis of the valve string calculated above should be made to be sure
that the valves as spaced will have closing pressures decreasing with depth. Valve ratio =
0.112 and spring effect = 75 psig.
Solution
From Table 5.1, Aa/At = 4.31 and, since the static fluid level is known as 2500 ft from the surface.
The depth to the top valve is
D1 = D s +
p v1 − p wh
650 − 50
= 2500 +
= 2726 ft (Depth to top valve)
(1 + R ) G s
(1 + 4 .31) 0 .5
65
D1− 2 =
p v 2 − G f D1 − p wh
Gs
=
625 − 0 .04 × 2726 − 50
= 932 ft
0 .5
D 2 = 2726 + 932 = 3658 ft (Depth to 2nd valve)
600 − 0 .04 × 3658 − 50
= 808 ft, D 3 = 3658 + 808 = 4466 ft (Depth to 3rd valve)
0 .5
575 − 0 .04 × 4466 − 50
=
= 692 ft, D 4 = 4466 + 692 = 5158 ft (Depth to 4th valve)
0 .5
550 − 0 .04 × 5158 − 50
=
= 588 ft, D 5 = 5158 + 588 = 5746 ft (Depth to 5th valve)
0 .5
D 2 −3 =
D 3− 4
D 4 −5
Surface Closing Pressure Analysis
(1)
Valve
opening
pressure at
60°F,
psig
650
625
600
575
550
(2)
(3)
(4)
(5)
Depth of
valve,
ft
2726
3658
4466
5158
5746
Temperature
at valve,
°F
114
129
141
153
162
Bellows
pressure at
60°F,
psig
511
488
466
444
422
Bellows
pressure at
Tv,
psig
570
560
540
525
515
(6)
Valve
closing
pressure at
depth,
psig
637
627
607
592
582
(7)
Surface
closing
pressure,
psig
590
570
550
540
530
Column (3): Tv = 0.016 D + 70 ° F
Column (4): p vo =  p d  + S t − p t  R 
1− R 
1− R 
Column (5)and (7): Fig.4.10
Column (6): p vc = p d @ T v + S t (1 − R )
10. A well is to be placed on intermittent-flow gas-lift. The well data are as follows:
Well depth = 8000 ft
Tubing size = 2-in., Capacity: 0.00387 bbl/ft
Specific gravity of oil = 0.87
Casing size = 5½-in., 20 lb/ft
No water production.
Well PI = 0.08 bbl/day/psi
Static BHP = 2200 psig
Wellhead pressure = 50 psig
Oil production desired = 120 bbl/day Average tubing temperature = 127 °F
Cycle time = 48 min
Gas deviation factor = 0.87
Calculate the depth to the operating valve and theoretical minimum injection gas-oil ratio,
assuming 70 per cent of starting slug is produced.
Solution
Static gradient = 0.433 × 0.87 = 0.377 psi/ft
Drawdown, ∆p =q/PI =120/0.08 =1500 psi
Depth to static fluid level = 8000 – [(2200 – 50)/0.377] = 2297 ft
Equivalent hydrostatic head of ∆p = 1500/0.377 = 3979 ft
Depth to working fluid level = 2297 + 3979 = 6276 ft which is mid point of the slug buildup
66
bbl/cycle =
120 bbl/day × 48 min/cycle
= 4 bbl/cycle
1440 min/ day
Since 70% of the starting slug is produced,
Volume of starting slug = 4/0.70 = 5.7 bbl
Length of starting slug = 5.7/0.00387 = 1473 ft
Depth to operating valve = (1473/2)+6276 = 7012 ft #
Tubing pressure opposite the operating valve , pt = 50 + (1473 × 0.377) = 605 psig
Minimum casing pressure at operating valve = 605 + (605/2) = 908 psig
Space for gas injected = 7012 × 0.00387 – 5.7 = 21.44 bbl = 120.3 cu ft
Tubing pressure under slug the surface = 50 + (4/0.00387) × 0.377 = 440 psig
pavg = (908 + 440)/2 = 674 psig = 688.7 psia
Tavg = 127 + 460 = 587 °R
V sc = 120 .3 ×
688 .7 520
1
×
×
= 5739 SCF/cycle
14 .7
587 0 .87
Minimum injection GOR = 5739 SCF/cycle / 4 bbl/cycle = 1435 SCF/STB #
****************
67
Ministry of Science and Technology
Department of Technical and Vocational Education
Petroleum Engineering
Sample Questions for
PE 03023 PETROLEUM PRODUCTION ENGINEERING Part ( II )
Chapter 1 Sucker Rod Pumping
1.*
2.**
3.*
4.*
5.**
6.*
7.*
8.**
9.**
10.**
11.*
12.*
13.**
14.**
15.**
Explain the counterbalance effect of the counterbalancing system of a beam pumping unit.
(10 marks)
Define the factors which contribute to the net polished rod load and give their expressions.
(15 marks)
Describe the two possible approaches to the design of a tapered sucker rod string and explain
the four assumptions made to simplify the design.
(10 marks)
Explain why volumetric efficiency is extremely important in sucker rod pumping design.
(10 marks)
Discuss the factors considered in sucker rod and tubing failure.
(10 marks)
Draw a dynamometer card for an idealized pumping system and explain the six conditions.
(10 marks)
Draw a dynamometer card for the actual pumping system and explain the eight load factors
contribute to the shape of the card.
(15 marks)
Describe "fluid pound" and "gas pound" of the pumping operation. (10 marks)
For a well polished rod stroke length is 68 in., pumping speed is 16 spm, and length of the
sucker rod string is 7352 ft. What would be the increase in effective plunger stroke if
pumping speed were increased to 20 spm?
(10 marks)
A well has a sucker rod string consisting of 4600 ft of ¾-in. rods and 1950 ft of ⅞-in. rods.
The plunger diameter is 1½-in., polished rod stroke length is 64 in., and average specific
gravity of the well fluid is 0.825. At what pumping speed will polished rod stroke length and
effective stroke length be equal if the tubing is anchored?
(10 marks)
For a well, the plunger diameter is 2 in., pumping speed is 23 spm, the calculated effective
plunger stroke is 30.8 in., and production is 20 bbl of oil and 180 bbl of water per day.
Calculate the pump constant and the volumetric efficiency of the pump in this well. (5 marks)
For a well, the 1¾-in. plunger is set in 2-in. tubing on 4275 ft of ¾-in. sucker rods. There is
no tubing anchor, and the working fluid level is known to be low. When the well is pumped
at 18 spm with a 64-in. stroke, the production is 283 bbl per day of a fluid of specific gravity
0.825. From these data calculate the volumetric efficiency of the pump, peak polished rod
load, ideal counterbalance effect, and polished rod horsepower.
(20 marks)
For a well, a 1¾-in. plunger is to be set with 3200 ft of ⅝-in. sucker rods. It is estimated that
a pumping speed of 17 spm and a polished rod stroke length of 42 in. will give the desired
production of 157 bbl per day. It is further estimated that the working fluid level will be
about 2000 ft, and that the specific gravity of the produced fluid will be 0.825. Assuming a
30 per cent reduction due to cyclic loading, calculate the minimum nameplate rating for the
electric motor to be used as the prime mover. (10 marks)
For a certain well, a 2-in. plunger is set on a 3300-ft string of ⅞-in. rods. The tubing size is 3
in., and there is no tubing anchor or gas anchor. Production is 120 bbl/day of oil and 267
bbl/day of water, the average specific gravity of the tubing fluid being 0.83. An echometer
shows the pumping fluid level to be at 2066 ft. The pumping speed is 18 spm at a polished
rod stroke length of 64 in. Calculate the effective plunger stroke and peak polished rod load.
(15 marks)
For a certain pumping well the following data are available: pump setting depth = 3500 ft;
rod size = ¾ in.; tubing size = 2½ in. (tubing not anchored); plunger size = 1¾ in.; polished
rod stroke length = 54 in.; specific gravity of produced fluid = 0.85; pumping speed = 18
68
spm; working fluid level = 3150 ft. Calculate the volumetric efficiency of the pump if the
well is producing 180 bbl/day.
(15 marks)
16.** A well is producing 240 bbl/day of a fluid having specific gravity of 0.80. The pump is set at
3500 ft on ¾-in. rods, and the plunger size is 1¾ in. Calculate the ideal counterbalance effect
for the unit. (5 marks)
17.** A certain crank counterbalance has its center of gravity 33 in. from the center of the
crankshaft when stroke length is 44 in. In order to reduce pumping speed, and thereby
decrease peak polished rod load, it was decided to change the stroke length to 64 in. What
will be the new distance from the center of gravity of the counterbalance to the center of the
crankshaft if it is desired to keep the counterbalance effect the same? (10 marks)
18.** A dynamometer test was run on the well of Prob.(12). An analysis of the resulting
dynamometer card shows a peak load of 11,500 lb, a minimum polished rod load of 3800 lb,
a polished rod horsepower of 11.02. From results of the dynamometer test, calculate the ideal
counterbalance effect, maximum sucker rod stress, and peak torque if the unit is
counterbalanced to within 5 per cent of the ideal value.
(10 marks)
19.** For a certain pumping installation production is 200 bbl per day of a fluid having a specific
gravity of 0.845 from a depth of 2500 ft. What is the below-ground efficiency for this
installation if a dynamometer test shows a polished rod horse power of 7.4 hp?
(5 marks)
20.** The area within the loop of a certain dynamometer card is 2.97 sq in.; and the area between
the loop and the zero load line is 3.82 sq in. The length of the loop is 4.56 in.; and the
instrument constant for the dynamometer is 5550 lb per in. Calculate the correct
counterbalance effect for the unit.
(5 marks)
21.** For the well of Prob.(20), the pumping speed is 20 spm and stroke length is 48 in. Calculate
the above-ground efficiency if the wattmeter (on the line to the electric motor being used as
the prime mover) shows an average power input of 14.5 kw.
(7 marks)
22.** For a well the sucker rod string consists of 3575 ft of ¾-in. rods and 2625 ft of ⅞-in. rods. A
standing valve test on the well gave a static rod load of 10,190 lb. What is the effective
specific gravity of the fluid in the tubing?
(5 marks)
23.** For the well of Prob.(22), plunger size is 1¾-in. If a traveling valve test shows a load of
14,725 lb, what pumping BHP is indicated? (7 marks)
24.** What is the net lift for the well of Probs.(22) and (23)?
(3 marks)
25.*** In the design of a certain well, it is anticipated that production will be 375 bbl per day with a
pump setting depth of 6500 ft. Assuming a volumetric efficiency of 75 per cent for the pump,
calculate the rod size and the length of each section of the tapered string if sucker rod are
available in lengths of 25 ft. (15 marks)
26.*** A dynamometer card for a pumping well is shown in Fig.(1.29). Stroke length for the unit is
44 in., and pumping speed is 19.5 spm. The dynamometer used for the measurements has an
instrument constant of 8000 lb per in. Calculate the torque on the gear reducer when the
crank has turned through an angle of 330° from the beginning of the upstroke.
(10
marks)
27.*** For the well of Prob.(26), the depth to the plunger is 5000 ft, and production is 225 bbl/day at
the pumping speed of 19.5 spm. Because of an increase in water cut, it is desired to obtain
more production by increasing pumping to a value greater than 19.5 spm. At what
approximate production rate would a synchronous pumping speed be encountered?(10 marks)
28.*** Construct the torque curve over a complete pumping cycle for the well of Prob. (26).
(15 marks)
29.*** The overall efficiency of a certain pumping unit (utilizing an electric motor as the prime
mover) is 39 per cent. If electric power costs $ 0.03 per kilowatt-hour, calculate the cost (for
power only) of lifting a barrel of fluid of specific gravity 0.855 from a depth of 5400 ft.
(10 marks)
30.*** A well is producing 584 bbl per day of a fluid having a specific gravity of 0.90. Pump setting
depth is 5125 ft, pumping fluid level is 3200 ft, and tubing pressure is 50 psi. What is the net
lift of the fluid?
(5 marks)
31.*** A pumping unit has a single reduction gear with a gear ratio of 9:6. The unit sheave has a
pitch diameter of 34 in., and the sheave on the multicylinder gas engine has a pitch diameter
69
of 8 in. What is the engine speed when the pumping speed is 23 spm?
(5
marks)
32.*** It is desired to put on pump a well having a static fluid level of 1000 ft. The well fluid has a
40 per cent water cut, with oil of gravity 40°API and water of specific gravity 1.015. The
installation is to be designed for an oil production of 125 bbl/day. Reservoir studies and well
fluids analyses indicate a probable pump volumetric efficiency of at least 80 per cent.
(b) Determine the API rating of the pumping unit assembly to be purchased for this
installation.
(b) Determine, for the propose installation, polished rod stroke length, optimum plunger size,
and resulting tubing size and pumping speed.
(b) Design the sucker rod string, assuming sucker rods to be available in lengths of 25 ft.
(b) Calculate the actual oil production for the assumed volumetric efficiency and the
operating conditions of part (b).
(b) An example of a pumping unit assembly which meets the requirements of parts (a) and
(b) is the Bethlehem 160D-20S-64. This unit has a beam load rating of 20,700 lb, an
available counterbalance effect (without auxiliary counterweights) of 11,770 lb, and a
peak torque rating of 160,000 in-lb. Compare these values with calculated values under
the selected operating conditions.
(b) What is the maximum sucker rod stress under the selected operating conditions?
(b) Determine the minimum brake horsepower rating for the electric motor to be used as the
prime mover.
(b) The standard unit sheave for the Bethlehem 160D-20S-64 pumping assembly has a pitch
diameter of 24 in., and the double reduction gear has a reduction ratio of 26.4. If motor
speed is 1200 rpm, what should be the pitch diameter of the motor sheave to obtain the
desired pumping speed?
(20 marks)
33.** For a certain pumping installation, the pump setting depth is 515 ft and the working fluid
level is 3200 ft. The well is making 580 bbl per day of a fluid of specific gravity 0.90.
Assuming negligible tubing pressure, calculate the overall efficiency of the installation if the
wattmeter on the line to the electric motor being used as the prime mover shows an average
power input of 23 kw.
(10 marks)
34.*** What is the average plunger speed for a well being pumped at 20 spm if the effective plunger
stroke is 58 in.?
(5 marks)
35.*** It is desired to design a pumping installation to obtain a production of 360 bbl per day at a
pump setting depth of 3600 ft and an assumed volumetric efficiency of 80 per cent. The
appropriate design data shows two possible plunger sizes for the conditions. Considering the
possibility of excessive sucker rod failure resulting from synchronous pumping speed, which
plunger size should be selected?
(10 marks)
Chapter 2 Electrical Submersible Centrifugal Pumps
36.* Explain briefly about the components of a electrical submersible pumping system. (15 marks)
37.** What are the accessory items for the submersible pumping system? (10 marks)
38.** Draw a sketch of standard performance curve for a submersible pump and explain its usage.
(15 marks)
39.* What is total dynamic head (TDH) and how can it be determined? (10 marks)
40.** Calculate the total dynamic head for the given well: (10 marks)
Required wellhead pressure = 200 psig
Pump setting depth = 12000 ft
Tubing size = 3½-in. EUE
Pumping rate = 2000 bpd
Fluid pumped = 70% 40°API oil with 30% water of specific gravity 1.05
Average density = 54.79 lbm/ft3
Fluid over pump intake = 650 ft
41.* Describe the summary of pump sizing procedure in sizing a submersible pumping unit.
(10 marks)
Chapter 3 Hydraulic Pumping
42.* Describe the advantages of the piston-type hydraulic pumping system.
(10 marks)
43.** Describe the basic types of power fluid system and explain them.
(15 marks)
70
44.* Explain the considerations and decisions that must be made when designing a hydraulic
pumping installation. (10 marks)
45.** How do you determine the power fluid rate? (10 marks)
* = Must know, ** = Should know, *** = Could know
*********************
71
Ministry of Science and Technology
Department of Technical and Vocational Education
Petroleum Engineering
Worked Out Examples for
PE 03023 PRODUCTION ENGINEERING Part ( II )
1. For a well polished rod stroke length is 68 in., pumping speed is 16 spm, and length of the sucker
rod string is 7352 ft. What would be the increase in effective plunger stroke if pumping speed
were increased to 20 spm?
Solution
Acceleration factors for respective pumping speeds are first calculated,
α =
SN 2
70500
α1 =
68 × 16 × 16
= 0 .247
70500
α2 =
68 × 20 × 20
= 0 .386
70500
40.8 L2α
only of Eq.1.31, therefore
E
40 .8 L2α 1
40 .8 L2α 2
–
Increase in effective plunger stroke, ∆ S p =
Increase in N will affect the term
E
E
=
40 . 8 L 2
40 . 8 × 7325
(α 2 − α 1 ) =
E
30 × 10 6
2
( 0 . 386 − 0 . 246 ) = 10 . 1 in. #
2. A well has a sucker rod string consisting of 4600 ft of ¾-in. rods and 1950 ft of ⅞-in. rods. The
plunger diameter is 1½-in., polished rod stroke length is 64 in., and average specific gravity of
the well fluid is 0.825. At what pumping speed will polished rod stroke length and effective
stroke length be equal if the tubing is anchored?
Solution
For ¾-in. rods, Ar1 = 0.442 sq in., M1 = 1.63 lb/ft
For ⅞-in. rods, Ar2 = 0.601 sq in., M2 = 2.16 lb/ft
For 1½-in. plunger, Ap = 1.767 sq in.
When S = Sp , Eq.1.30 can be rewritten as,
 L
L 
40 .8 L2α = 5 .2GDA p  1 + 2  , tubing is anchored.
 A r 1 Ar 2 
5 .2 × 0 .825 × 6550 × 1 .767  4600
1950 
α =
+

 = 0 .386
40 .8 × 6550 × 6550
 0 .442 0 .601 
SN 2
= α = 0 .386 and N =
70500
0 .386 × 70500
= 20 .6 spm #
64
3. A certain crank counterbalance has its center of gravity 33 in. from the center of the crankshaft
when stroke length is 44 in. In order to reduce pumping speed, and thereby decrease peak
polished rod load, it was decided to change the stroke length to 64 in. What will be the new
distance from the center of gravity of the counterbalance to the center of the crankshaft if it is
desired to keep the counterbalance effect the same?
72
Solution
Assume l1 = l2 , r1 = S/2 = 22 in., and r2 = 32 in.
From Eq.6.57,
d1 =
r1 ( C − C s )
Wc
where
(C - C s )
= constant
Wc
Then
d1 d 2
=
r1
r2
d2 = (33/22)×32 = 48 in. # The new distance from the center of gravity of the counterbalance to
the center of the crankshaft.
4. For a well the sucker rod string consists of 3575 ft of ¾-in. rods and 2625 ft of ⅞-in. rods. A
standing valve test on the well gave a static rod load of 10,190 lb. What is the effective specific
gravity of the fluid in the tubing?
For this well, plunger size is 1¾-in. If a traveling valve test shows a load of 14,725 lb, what
pumping BHP is indicated?
What is the net lift for the well ?
Solution
For ¾-in. rods, M1 = 1.63 lb/ft
For ⅞-in. rods, M2 = 2.16 lb/ft
From standing valve test, Wrm = 10190 lb
Calculated rod load, Wrc = 3575 × 1.63 + 2625 × 2.16 = 11497 lb
From Eq.1.84,
Wrm = Wrc (1 – 0.127 G)
1 – 0.127 G = 10190/11497 = 0.886
G = 0.90 #
For 1¾-in. plunger, Ap = 2.405 sq in.
From traveling valve test, Wrfm = 14725 lb
Measured fluid load , Wfm = Wrfm – Wrm = 14725 – 10190 = 4535 lb
Calculated fluid load, Wfc = 0.433 GLAp = 0.433×0.90×6200×2.405 = 5810 lb
Pumping BHP,
pw = (Wfc – Wfm)/Ap = (5810 – 4535)/2.405 = 530 psi #
Net lift,
LN = L × Wfm/Wfc = 6200 × 4535 /5810 = 4840 ft #
5. In the design of a certain well, it is anticipated that production will be 375 bbl per day with a
pump setting depth of 6500 ft. Assuming a volumetric efficiency of 75 per cent for the pump,
calculate the rod size and the length of each section of the tapered string if sucker rod are
available in lengths of 25 ft.
Solution
Pump displacement, V = 375/0.75 = 500 bbl/day
From Fig.(1.6), pump displacement of 500 bbl/day and pump setting depth 6500 ft give an API
size 640 with stroke length 144 in.
73
From Table 1-12, for pump setting depth 6500 ft, pump size is 1¾-in. (2.405 sq in.) and rod sizes
are ¾-⅞-1.
Using the equations from Table 1-4,
R1 = 0.664 – 0.0894 × 2.405 = 0.449
R2 = 0.181 + 0.0478 × 2.405 = 0.296
R3 = 0.155 + 0.0416 × 2.405 = 0.255
L1 = 0.449 × 6500 = 2918 ft
L2 = 0.296 × 6500 = 1924 ft
L3 = 0.255 × 6500 = 1657 ft
Sucker rods are available in lengths of 25 ft, therefore
¾-in. rods 2925 ft #
⅞-in. rods 1925 ft #
1-in. rods 1650 ft #
6. For a well, a 1¾-in. plunger is to be set with 3200 ft of ⅝-in. sucker rods. It is estimated that a
pumping speed of 17 spm and a polished rod stroke length of 42 in. will give the desired
production of 157 bbl per day. It is further estimated that the working fluid level will be about
2000 ft, and that the specific gravity of the produced fluid will be 0.825. Assuming a 30 per cent
reduction due to cyclic loading, calculate the minimum nameplate rating for the electric motor to
be used as the prime mover.
Solution
Hydraulic horsepower, Hh = 7.36×10–6qGD
= 7.36×10–6×157×0.825×2000 = 1.95 hp
Frictional horsepower, Hf = 6.31×10–7WrSN
= 6.31×10–7×3720×42×17 = 1.68 hp
Polished rod horsepower = 1.95 + 1.68 = 3.63 hp
Brake horsepower = 3.63 × 1.5 = 5.44 hp
Minimum nameplate rating with 30 per cent power reduction due to cyclic loading is
= 5.44/0.70 = 7.7 hp #
7. The area within the loop of a certain dynamometer card is 2.97 sq in.; and the area between the
loop and the zero load line is 3.82 sq in. The length of the loop is 4.56 in.; and the instrument
constant for the dynamometer is 5550 lb per in. Calculate the correct counterbalance effect for
the unit.
For the above well, the pumping speed is 20 spm and stroke length is 48 in. Calculate the aboveground efficiency if the wattmeter (on the line to the electric motor being used as the prime
mover) shows an average power input of 14.5 kw.
Solution
The correct counterbalance effect,
A
Y
5550
2 .97
[ Al + u ] =
[3 .82 +
] = 6450 lb#
Ll
2
4 .56
2
14 .5 kw
Average power input =
= 19 .43 hp
0.746 kw/hp
Ci =
Polished rod horsepower from dynamometer card,
74
Hp =
SYA u N
48 × 5550 × 2097 × 20
=
= 8 .76 hp
396000 L l
396000 × 4 .56
Above-ground efficiency =8.76/19.43 = 45 per cent #
8. For the well of Prob.(26), the depth to the plunger is 5000 ft, and production is 225 bbl/day at the
pumping speed of 19.5 spm. Because of an increase in water cut, it is desired to obtain more
production by increasing pumping to a value greater than 19.5 spm. At what approximate
production rate would a synchronous pumping speed be encountered?
Solution
Order of synchronization, n = 237000 / NL = 237000 / 19 . 5 × 5000 = 2 . 43
To obtain a greater pumping speed, n must be assumed to be 2
N = 237000/2×5000 = 23.7 spm
q = KSpNEv
q/N = KSpEv = Constant
q1/N1 = q2/N2
q2 = 225×23.7/19.5 = 273 bbl per day #
9. A well is producing 240 bbl/day of a fluid having specific gravity of 0.80. The pump is set at
3500 ft on ¾-in. rods, and the plunger size is 1¾ in. Calculate the ideal counterbalance effect for
the unit.
Solution
For ¾-in. rods, M1 = 1.63 lb/ft
For 1¾-in. plunger, Ap = 2.405 sq in.
Weight of rod string,
Wr = 3500 × 1.63 lb/ft = 5705 lb
Fluid load,
Wf = 0.433G (LAp – 0.294 Wr)
= 0.433×0.8(3500×2.405 – 0.294×5705)= 2334 lb
Ideal counterbalance effect,
Ci =0.5 Wf +[Wr (1 – 0.127G)]
= 0.5×2334 + 5705(1–0.127×0.8)
= 6290 lb #
10. It is desired to put on pump a well having a static fluid level of 1000 ft. The well fluid has a 40
per cent water cut, with oil of gravity 40°API and water of specific gravity 1.015. The installation
is to be designed for an oil production of 125 bbl/day. Reservoir studies and well fluids analyses
indicate a probable pump volumetric efficiency of at least 80 per cent.
(a) Determine the API rating of the pumping unit assembly to be purchased for this
installation.
(b) Determine, for the propose installation, polished rod stroke length, optimum plunger size,
and resulting tubing size and pumping speed.
(c) Design the sucker rod string, assuming sucker rods to be available in lengths of 25 ft.
(d) Calculate the actual oil production for the assumed volumetric efficiency and the
operating conditions of part (b).
(e) An example of a pumping unit assembly which meets the requirements of parts (a) and
(b) is the Bethlehem 160D-20S-64. This unit has a beam load rating of 20,700 lb, an
available counterbalance effect (without auxiliary counterweights) of 11,770 lb, and a
75
peak torque rating of 160,000 in-lb. Compare these values with calculated values under
the selected operating conditions.
( ) What is the maximum sucker rod stress under the selected operating conditions?
( ) Determine the minimum brake horsepower rating for the electric motor to be used as the
prime mover.
(h) The standard unit sheave for the Bethlehem 160D-20S-64 pumping assembly has a pitch
diameter of 24 in., and the double reduction gear has a reduction ratio of 26.4. If motor
speed is 1200 rpm, what should be the pitch diameter of the motor sheave to obtain the
desired pumping speed?
Solution
(a) Production is 125 bbl/day of oil with 40% water cut and if gross production rate is q
Water cut = (q – qo)/q = 0.40
Gross production, q = 125/0.60= 208 bbl/day
Pump displacement, V = 208/0.80 = 260 bbl/day
From Fig. 1.6, with pump setting depth 4500 ft, API unit size 160 is obtained (curve E).
API rating of 160,000 in-lb.#
( ) From Table 1-9 for API 160, polished rod stroke length, S = 64 in. and rod string: ⅝-¾-⅞
Optimum plunger size = 1½ in.
Tubing size = 2 in.
Pumping speed, N = 20 spm (by interpolation)
(c) Sucker rod string design, for 1½-in. plunger, Ap = 1.767 sq in.
R1 = 0.627 – 0.1393×1.767 = 0.380
R2 = 0.199 + 0.0737×1.767 = 0.329
R3 = 0.175 + 0.0655×1.767 = 0.291
L1 = 0.380×4500 = 1710 ft ≈ 1725 ft #
L2 = 0.329×4500 = 1480 ft ≈ 1475 ft #
L3 = 0.291×4500 = 1309 ft ≈ 1300 ft #
(d) Oil gravity = 141.5/(131.5+40) = 0.825
Water gravity = 1.015
Gravity of well fluid, G = 0.825× 3 + 1.015× 2 = 0.901
5
5
For ⅝-in. rods, M1 = 1.16 lb/ft, A1 =0.307 sq in.
For ¾-in. rods, M1 = 1.63 lb/ft, A2 =0.442 sq in.
For ⅞-in. rods, M2 = 2.16 lb/ft, A3 =0.601 sq in.
α = 64 × 20 × 20 / 70500 = 0 . 363
ep =
40 .8 L 2α
40 .8 × 4500 2 × 0 .363
=
= 10 in.
E
30 × 10 6
et + e r =
5 .2 GDA p  L
L 
L
L

+ 1 + 2 + 3 
E
A1 A2
A3 
 At
5.2 × 0.901 × 4500 × 1.767  4500 1725 1475 1300 
+
+
+

 = 18 in.
30 × 10 6
 1.304 0.307 0.442 0.601 
Sp = 64+10–18 = 56 in.
=
76
V = 0.1484×1.767×56×20 = 294 bbl/day
q = 294×0.80 = 235 bbl/day (oil and water)
Oil production = 235×0.60 = 141 bbl/day #
(e) Bethlehem 160D-20S-64 unit
Beam loading = 20700 lb, CBE = 11770 lb, Peak torque = 160,000 in.-lb
Under operating conditions:
Wr = (1725×1.16)+(1475×1.63)+(1300×2.16) = 7210 lb
Wf = 0.433×0.901(4500×1.767 – 0.294×7210) = 2275 lb
Maximum polished rod load, Wmax = 2275 ×1.363 = 12100 lb #
Minimum polished rod load, Wmin = 7210(1 – 0.363 – 0.127×0.901) = 3770 lb
Ideal counterbalance effect, Ci = 12100+3770/2 = 7935 lb #
Peak torque = (12100 – 0.95×7935) 64/2 = 145,976 in.-lb #
(f) Maximum sucker rod stress = Wmax / A3 = 12100/0.601 = 20100 psi #
(g) Hh = 7.36×10–6×235×0.901×4500 = 7 hp
Hf = 6.31×10–7×7210×64×20 = 5.8 hp
Brake horsepower = 1.5 (7 + 5.8) = 20 hp #
(h) du = 24 in., Z = 26.4, Ne = 1200 rpm, N = 20 spm
de = Zdu(N/Ne) = 26.4×24(20/1200) = 10.6 in. # Pitch diameter of the motor sheave.
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