MIDDLE EAST TECHNICAL UNIVERSITY
PETROLEUM AND NATURAL GAS ENGINEERING
332 PRODUCTION ENGINEERING-2
Homework– 4
Name: Batuhan
Surname: KIZILKOCA
Student Number: 2373983
1
Table of Content
Table of Content.........................................................................................................................2
Table Of Figures.........................................................................................................................3
QUESTIONS:.............................................................................................................................4
1. PCP Selection:....................................................................................................................4
2. Gas Well Unloading:...........................................................................................................5
3. Separator Capacity:.............................................................................................................5
Question 4...............................................................................................................................7
ANSWERS.................................................................................................................................9
Question 1...............................................................................................................................9
Question 2.............................................................................................................................16
Question 3.............................................................................................................................17
Question 4.............................................................................................................................23
References.................................................................................................................................27
2
Table Of Figures
Figure 1 Available Pump Catalogue...........................................................................................4
Figure 2 Seperator Table.............................................................................................................6
Figure 3 Thermal measurements of the region...........................................................................7
Figure 4 Flow measurements of the well....................................................................................7
Figure 5 Pump diameters for the available pumps......................................................................9
Figure 6 Major diameters for the available pumps.....................................................................9
Figure 7 Efficiency values for each pump................................................................................10
Figure 8 Displacement rate values............................................................................................11
Figure 9 Phyton Equation Solver..............................................................................................13
Figure 10 Pressure Rating Values.............................................................................................14
Figure 11 Phyton code for diameter value................................................................................17
Figure 12 Seperator Alternatives..............................................................................................18
Figure 13 K value table.............................................................................................................18
Figure 14 Retention time for the separator...............................................................................19
Figure 15 Liquid Capacity Table..............................................................................................20
Figure 16 Table for K Value.....................................................................................................20
Figure 17 Retention time for second separator.........................................................................21
Figure 18 Liquid Capacity Value for Second Seperator...........................................................22
Figure 19 Tubular Data Inputs..................................................................................................23
Figure 20 Heat Transfer Inputs.................................................................................................24
Figure 21Reservoir Fluid Properties.........................................................................................24
Figure 22 Rate Properties..........................................................................................................25
Figure 23 First Scenario Noddle Analysis Graph.....................................................................25
Figure 24 Second Scenario Noddle Analysis Graph.................................................................26
3
QUESTIONS:
1. PCP Selection:
The PCP pumps given in the below table are available in your Company’s stocks. Select a
pump for the given conditions and calculate torque and power requirements for the selected
pump. For that question , student number is 2373983 which corresponds to;
a=2 , b=3 , c=7 , d=3 , e=9 , f=8 , g=3
Desired liquid pumping rate: 624 stb/d
Produced Fluid: 35 API oil with 21% water cut (produced water s.g. = 1.05)
Casing: 7 inch OD, 6.276 inch ID Tubing: 2.875 inch OD, 2.441 inch ID, 2.347 inch drift ID
Perforation depth: 6400 ft, Pump submergence: 300 ft
Pump speed should not exceed: 350 rpm Required well head pressure: 200 psia pressure
Assume Frictional pressure drop in tubings: 5 psi/100 ft
The well is in a saturated oil reservoir with following properties:
Porosity: 0.20 Effective horizontal permeability: 105 md
Pay zone thickness: 50 ft Drainage area: 640 acres (re= 2980 ft)
Reservoir Pressure: 2320 psia (@ the perforation depth)
Bubble point pressure: 2600 psia
Fluid formation volume factor: 1.25 Fluid viscosity: 3.0 cp
Total compressibility: 0.00002 psi-1
Wellbore diameter: 8.5 inches Skin factor: +3
Note: For calculating bottom hole flowing pressure, assume pseudo steady state flow and use
Vogel’s empirical equation. You can see the available pumps on your catalogue (Figure 1) as
follows:
Figure 1 Available Pump Catalogue
4
Assume that: The sum of friction and viscous torque will be 20% of the hydraulic torque at
the pump’s rated displacement and pressure. The power transmission system efficiency is
80%.
Note: If more than one pump can be selected, then choose the one with the highest
displacement rate.
2. Gas Well Unloading:
A gas well is producing 1900 Mscf/day gas with some water. The flowing well head pressure
is 180 psia and surface flowing temperature is 90 F. Tubing length is equal to perforation
depth. Assume the gas deviation factor (z): 0.85. In your stockyard, you have tubings with
different diameters. If a tubing size from the available alternatives can be used, what will be
the maximum tubing size which can be used to prevent water accumulation?
Alternatives: Tubings with ID of 3.826, 3.068, 2.75, 2.441, 1.61. 1.049 inches
Note: Assume all other operating parameters (gas rate, surface flowing pressure, wellhead
flowing temp,gas deviation factor) remain constant when the tubings are changed.
For that question , student number is 2373983 which corresponds to;
a=2 , b=3 , c=7 , d=3 , e=9 , f=8 , g=3
3. Separator Capacity:
The production data and separator conditions are given below for an oil field:
For that question , student number is 2373983 which corresponds to;
a=2 , b=3 , c=7 , d=3 , e=9 , f=8 , g=3
Oil Flow Rate : 1044 stb/day,
Producing Gas Oil Ratio = 10600 scf/bbl
Water Cut = %45
Oil Gravity: 52 deg API
Z : 0.87 (at separator conditions)
Gas Density : 1 lbm/ft3 (at separator conditions)
Produced Water Density: 64.5 lbm/ft3 (at separator conditions)
Separator Operating Pressure: 400 psi
Separator Operating Temperature: 70 oF for g={2,3}
There are two available separators in the stock, and their sizes are determined with the last
digit ofyour ID number 3:
5
Figure 2 Seperator Table
Please analyze the given data and determine whether you can use the given separators in your
field. Evaluate the separators individually, and report if they can be used or not.
If you cannot use a certain separator (if you can use both of them, suppose that you cannot use
one of the separators), please briefly comment on how the operating temperature, desired oil
rate,and the separator size (diameter and height) may be changed to make a viable selection
for this field. Also, briefly comment on what can be done for this field if none of the
parameters can be changed.
Notes:
• For the retention time, take the average of corresponding oil/water/gas separation values.
• For K values, take the average of the values given in the table.
• For the horizontal separator, assume 1/3 full for g={3,4,5}
• For K values, retention times and Separator Properties, use Tables 10.1, 10.2, 10.3, 10.4,
10.5 and 10.6 from Guo’s Petroleum Production Engineering Book (2nd Edition), Chapter 10,
pages 253-257.
6
Question 4
Your task is to simulate whether a PCP application would be viable at a well. The field is
known to produce heavy oil with a producing GOR is 350 SCF/STB with a 10% water cut
(Hint: You may use the default black oil model in Pipesim as your starting point). The
laboratory measurements indicate that gas and water specific gravities are 0.7875 and 1.05,
respectively. The oil API is known to be 15oAPI. It is known that the field contains 10% and
15% mole fractions of CO2 and H2S, respectively.
The well is vertical, and cased entirely with API 28 lbm/ft 6 5/8in C95 casing for the entire
11030 ft of the well length, As the production tubing, API 4in 11 lbm/ft N80 tubing for
8590 ft into the well. The well features a production packer located at 7580 ft, and a PCP
pump at the bottom of the production tubing. The field exhibits variable temperature behavior
at different depths; hence, multiple temperature measurements have been taken at various
depths. The thermal transmittance (U-value) of the measurement
zones have also been measured, and the recordings have been provided in Figure 3.
Figure 3 Thermal measurements of the region
While the well may be represented with the productivity index, the exact value of the PI has
not been measured. To overcome this, a multipoint flow test has been conducted. The results
of the test are provided in Figure 4.
Figure 4 Flow measurements of the well
The reservoir pressure has been identified as 4000 psi, and the reservoir temperature has been
measured to be 250 F. The reservoir fluid is known to follow Vogel equation below the
bubble point, and the PI is to be estimated using the multipoint test data.
.
7
You should create a PIPESIM simulation to perform the nodal analysis of this well at various
conditions. Don’t forget to include your simulation files into the homework files.
Use the pump Weatherford BRZ 151-4100. Please perform the sensitivity nodal analyses for
the reservoir pressure of 4000, 3500, 3000 psi with pump speed of 300 rpm and for the pump
speed of 100, 200, 300, 400 rpm at the reservoir pressure of 4000 psi.
8
ANSWERS
Question 1
There are many points to consider when choosing PCP. These steps are explained in detail
under subheadings below. At the beginning of the choosing procedure pump diameters should
be considered whether pump diameter exceeds the casing inner diameter or not. For that
question casing inner diameter is 6.276 inches. In the following table (Figure 2), pump
diameters can be seen.
Figure 5 Pump diameters for the available pumps
As indicated in the graph any of the pump diameter does not exceed the casing inner diameter
value. As a result of this, for that criterion all of them is proper.
Moreover, for the given available pumps, major diameter values should be checked. As
indicated in the following graph (Figure 3), major diameter for pump E exceeds the 2.347
inch which is tubing drift ID. As a result of this pump E can not be used fort hat operation.
Figure 6 Major diameters for the available pumps
9
As a following procedure, required pump displacement rates should be calculated. The
equation is the following.
s=qa /(ω∗E)
Where;
q a=actual flow rate ( m3 /day )
s= pump displacement rate ¿
w=rotational speed ( rpm )
E=volumetric pumping efficiency ( fraction )
Then for each pump, s value should be calculated. Efficiency values obtained by the
following graph (Figure 4).
Figure 7 Efficiency values for each pump
stb stb
624 2.45
day day
sA = =
350rpm∗0.75 rpm
stb stb
624 2.1
day day
s B= =
350 rpm∗0.85 rpm
stb stb
624 2.38
day day
sC = =
350 rpm∗0.75 rpm
stb stb
624 2.23
day day
s D= =
350 rpm∗0.80 rpm
As indicated in the following graph (Figure 5), Pump A does not satisfy the condition. In
other words, calculated s value is bigger than the displacement values.
10
Figure 8 Displacement rate values
Therefore, pressure rating of the Pump B, C, and D should be calculated. For the pressure
calculation, specific gravity of produced fluid should be calculated. In this case, there is 35
API oil. To calculate its specific gravity, following procedure will be applied.
141.5 141.5
γ oil = = =0.85
131.5+ API ° 131.5+35
There is %21 water cut in the produced fluid, so it should be considered also. To calculate
specific gravity of produced fluid, following procedure will be applied.
γ fluid =( Water Cut )∗( γ water ) + (1−Water Cut )∗( γ oil ) =( 0.21 )∗( 1.05 ) + ( 1−0.21 )∗( 0.85 )=0.89
Then pressure gradient is;
Pressure Gradient=0.433∗0.89=0.38537
For that homework, since gas specific gravity was not supplied, casing is the vented to
atmosphere condition is assumed. As a result of this, gas column above the dynamic liquid
level does not consider in the pressure calculation.
Since the bottom hole pressure value will be used to calculate the dynamic liquid level value
in the following steps, the productivity index and related vogel equation (pseudo-steady state)
is applied in the following steps to calculate the bottom hole pressure.
For pseudo steady state flow, J can be calculated by the following formula
k∗h
J=
141.2∗B o∗μo∗¿ ¿
Where;
J= productivity index (bbl/day/psi)
k= permability (md)
μo = viscosity (cP)
r e = reservoir drainage are (ft)
11
r w = wellbore radius (ft)
S = Skin factor
Then,
105 md∗50 ft
J=
141.2∗1.25∗3 cP∗¿ ¿
For that formula, wellbore radius should be converted to ft. Also, it should be divided by two
since given value is diameter value.
As given in the question, reservoir pressure is less than the bubble point pressure. As a result
of this, Vogel’s Equation for two phase flow should be used.
Vogel’s Equation for two phase flow;
[ ( ( ) )]
2
Pbhf Pbhf
q=q max∗ 1−0.2∗ −0.8∗
Pr Pr
Where:
Pbhf =Bottom hole flowing pressure (psi)
q max= Maximum flow rate (stb/day)
Pr =Reservoir pressure (psi)
Then formula for the maximum flow rate is the given following:
( )
1.097 bbl
day
∗2320 psi
J∗Pr psi bbl
q max = = =1414
1.8 1.8 day
Then the previous formula takes the following form.
[ ( ( ) )]
2
P bhf Pbhf
q=(1414 bbl /day )∗ 1 −0.2∗ −0.8∗
2320 psi 2320 psi
q is the desired rate which is 624 stb/day. To calculate, bottom hole flowing pressure
following phyton code is scripted. The code makes an initial guess , then optimizes that value
fort he given equation.
12
import scipy.optimize as opt
# Define the function based on the given equation
def equation(pbhf):
return 1414 * (1 - 0.2 * (pbhf / 2320) - 0.8 * ((pbhf / 2320) ** 2))
- 624
# Initial guess for the root
initial_guess = 1000 #
# Solve the equation using fsolve
pbhf_solution, = opt.fsolve(equation, initial_guess)
# Print the solution
print(f"The solution for Pbhf is: {pbhf_solution}")
Figure 9 Phyton Equation Solver
The solution for Pbhf is: 1670.36. (Necessary code file is assigned to OdtuClass HW-4 Report
Submission). With that bottom hole flowing pressure value dynamic liquid level can be
calculated.
Pbhf 1670.36
D= = =4334 ft
0.433∗γ fluid 0.433∗0.89
Also;
D=L−H
Where:
D: Dynamic liquid level (ft)
L: Perforation Depth (ft)
H: Liquid column length (ft)
Then:
D=6400 ft−4334 ft=2066 ft
Also;
Pump Depth=D+ Pump submergence=2066 ft +300 ft =2366 ft
After finding that value Pout ,∧P ¿ values can be calculated. For the out value following
formula should be applied.
13
Pout = ( Weight of Produced Fluid Column ) + Friction∈tubing+Tubing Head Pressure
Also weight of produced fluid column = (pressure gradient of fluid)*(fluid column height)
Then;
(
Pout =2366 ft∗ 0. 38537
psi
ft)+ 2366 ft∗ (
5 psi
100 ft)+ 200 psi=1230 psi
Additionally, 0.38537 value was obtained at the previous steps.
To calculate P¿ value , following formula should be applied.
P¿ =( Pump Submergence )∗( Pressure Gradient )=300 ft∗ 0.38537( psi
ft)=115.611 psi
Therefore, required pressure differential for the pump can be calculated by the following
formula.
Pout −P ¿=1230 psi−115.611 psi=1114.389 psi
As indicated at the following table (Figure 7), each pump is capable for that pressure
differential value.
Figure 10 Pressure Rating Values
For the torque requirement, pump choosing is required. Pump A, and E was eliminated, so
Pump D is chosen since it has the highest Displacement Rate (bbl/d/rpm). To calculate total
torque, the following formula should be applied.
T t=T h +T f +T v
14
Where:
T t = Total torque ft−lbf
T h = Hydraulic torque ft−lbf
T f = Friction torque ft−lbf
T v = Viscous Torque ft−lbf
T h can be calculated by the following formula.
T h=C∗s∗Plift
Where:
T h = Hydraulic torque
C = Constant
s= Pump displacement (bbl/day/rpm)
Plift = differential pump pressure
Then:
( )
bbl
3.7
day
T h=0.111∗ ∗( 1114.389 psi )=457.68 ft −lbf
rpm
As stated in the question T f +T v value will be %20 of T h value. So it is equivalent to 91.536
ft-lbf.
Then;
Total Torque=457.69 ft −lbf +91.536 ft−lbf =549.216 ft−lbf
Then prime mover power requirement can be calculated by the following formula.
C∗T pr∗ω
Pmo=
E pt
Where;
Pmo = Required prime mover output kW (hP)
C = Constant
T pr = Polished Rod Torque Nm (ft-lbf)
ω = Polished rod rotational speed (rpm)
E pt = Power Transmission System Efficiency (%)
Then:
−2
1.047∗10 ∗350 rpm∗(549.216 ft−lbf )
Pmo= =26.83 kW =35.96 hp
75
15
Question 2
For that question, critical velocity which is minimum gas velocity required for lifting oil and
water droplets should be calculated. Following equations can be used to calculate the critical
gas velocity for field applications.
1
4
5.62∗( 67−0.0031∗P )
u g water= 1
( 0.0031∗P ) 2
1
4
4.02∗( 45−0.0031∗P )
u g condensate= 1
( 0.0031∗P )2
Where;
P = well head flowing pressure, psi
u g = minimum gas velocity required for lifting oil and water droplets ft/sec
Then:
1
4
5.62∗( 67−0.0031∗180 psi )
u g water= 1
=21.48 ft / sec
2
( 0.0031∗180 )
1
4
4.02∗( 45−0.0031∗180 psi )
u g condensate= 1
=13.90 ft /sec
2
( 0.0031∗180 psi )
As can be seen in the calculations, ug (water) > ug (condensate) because water droplets are
bigger and heavier compared to condensate droplets. Therefore, they are more difficult to lift.
The gas velocity can be converted to gas rate by the following formula.
3.06∗p∗ug∗A
q g=
Tz
Where;
q g = Gas rate (mmscf/day)
p = Surface (well head ) flowing pressure (psia)
u g = Gas velocity (ft/sec)
A= Flow area (ft^2)
T = Surface (well head) flowing pressure (psia)
Z = Gas deviation factor at P and T
16
Then:
π∗d 2 in2
ft /sec∗ ∗144 2
4 ft
1.9=3.06∗180 psi∗21.48
( 90 F +460 )∗( 0.85 )
To find d value following phyton code is scripted.
import sympy as sp
# Define the variable
d = sp.Symbol('d')
# Create the equation
equation = sp.Eq(1.9, (3.06 * 180 * 21.48 * (sp.pi * d**2 / 4 / 144)) /
((90 + 460) * 0.85))
# Solve the equation for d
solution = sp.solve(equation, d)
# Print the solution
print(solution)
Figure 11 Phyton code for diameter value
Then d = 3.710
Among the given tubing alternatives Tubing ID 3.826 in alternative should be chosen.
The reason for this is that this is the only value that the calculated d value can fit among the
alternatives.
Question 3
As given in the question statement, Oil Flow Rate: 1044 bbl/day. Also, there is %45 water
cut. So, Water Flow Rate can be calculated by the followi8ng formula.
0.45∗1044 stb
OilFlowRate∗(1−WaterCut ) day stb
Water flow rate= = =854.18
Water Cut 0.55 day
17
As a result of that the minimum required separator liquid capacity should be (854.18+1044)
stb/day=1898.12 stb/day. Then, for the separator choosing process will be applied. According
to my student id seperators which are indicated at the following graph was chosen.
Figure 12 Seperator Alternatives
Therefore, for that 2 seperator, capacity calculations should be applied. Calculation for the
first one which is 3-Phase Horizontal D=48in, H=10ft is the following. Gas capacity formula
is:
q st =
2.4∗D2∗K∗p
√
z∗(T + 460)
ρ −ρ
∗ l g
ρg
Where:
q st = gas capacity at standart conditions (MMscfd)
D= internal diameter of vessel (ft)
P= Operation pressure (psi)
T= Operation temperature, deg f
Z= gas compressibility factor at operating pressure and temperature
K=Empirical factor
ρl =Density of liquid at operating conditions (lbm/ft3)
ρ g=Density of liquid at operating conditions (lbm/ft3)
Then:
ρl =
62.4 lbm
ft
3 (
∗
141.5
131.5+52) lbm
=48.11 3
ft
18
Figure 13 K value table
For the K value , average value is taken from the table K=(0.4+0.5)/2=0.45. Also, ρliquid
should be computed.
3
ρliquid =0.55∗64.5+ 0.45∗48.11=57.12lbm /ft
Then;
q st =2.4∗¿ ¿ ¿
Therefore, separator gas capacity is higher than gas production rate. So, it is proper for gas
capacity. ( gas production rate = Oil Flow Rate : 1044 stb/day *Producing Gas Oil Ratio =
10600 scf/bbl = 11.06 MM scf/day.)
To calculate, liquid capacity of seperator following formula will be applied.
1440∗V L
q L=
t
Where:
q L = Liquid capacity bbl/day
V L = Liquid settling volüme bbl
T = retention time
Retention time value obtained by the following table (Figure 12). (Average value will be
used)
19
Figure 14 Retention time for the separator
20
Liquid Capacity value obtained by the following table (for 48”x10’ 1/3 full seperator).
Figure 15 Liquid Capacity Table
Then:
1440∗6.77
q L= =433.28 bbl/day
22.5 min
Since that liquid capacity value is less than total fluid production rate (1898.18 bbl/day) , it
can not be operated.
Fort he second seperator, following procedure will be applied. Firstly, gas capacity should be
calculated. To calculate it, K value is needed. K value can be obtained by the following table.
(Average value will be used)
Figure 16 Table for K Value
21
q st =2.4∗¿ ¿ ¿
Therefore, separator gas capacity is higher than gas production rate. So, it is proper for gas
capacity. ( gas production rate = Oil Flow Rate : 1044 stb/day *Producing Gas Oil Ratio =
10600 scf/bbl = 11.06 MM scf/day.)
To calculate, liquid capacity of seperator following formula will be applied.
1440∗V L
q L=
t
Retention time value obtained by the following table (Figure 12). (Average value will be
used)
Figure 17 Retention time for second separator
22
Liquid Capacity value obtained by the following table (Figure 16) (for 36”x10’ seperator).
Figure 18 Liquid Capacity Value for Second Seperator
1440∗5.24
q L= =335.36 bbl/day
22.5 min
So required liquid rate is not satisfied. Both separator cannot be operated.
23
Question 4
The steps we performed when starting the question are as follows: First of all, tubular (casing,
tubing) data was loaded into the program called PIPESIM. (Figure 19)
Figure 19 Tubular Data Inputs
24
Then, the relevant heat transfer data was entered into the program.
Figure 20 Heat Transfer Inputs
Then, the relevant fluid property data was entered into the program.
Figure 21Reservoir Fluid Properties
25
Then, the relevant rate property data was entered into the program.
Figure 22 Rate Properties
For the first scenario is the following (Figure 23)
Figure 23 First Scenario Noddle Analysis Graph
26
Then , second scenario is the following.(Figure 24)
Figure 24 Second Scenario Noddle Analysis Graph
27
References
Guo, B., Liu, X., & Tan, X. (2017). Petroleum Production Engineering (2nd ed.).
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