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SKPP 3513 – PRODUCTION ENGINEERING
Chapter 2: Reservoir System Mohd Fauzi Hamid N01a-18 07-5535616
[email protected]
SKPP 3513: PRODUCTION ENGINEERING
Learning Outcome
Students will able to:
Understand the concept of IPR Differentiate the method used for calculating single phase flow and two phase flow IPR
Construct the IPR curve Identify the fluid phase for a particular given condition
CHAPTER 2: RESERVOIR SYSTEM
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SKPP 3513: PRODUCTION ENGINEERING
Contents Productivity Index (PI) Hydrocarbon Phase Behavior Inflow Performance Relationship (IPR)
CHAPTER 2: RESERVOIR SYSTEM
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MOHD FAUZI HAMID
SKPP 3513: PRODUCTION ENGINEERING
Fluid Inflow Fluid inflow is the flow of liquid from formation to the bottom hole. The flow of liquids are affected by 4 main factors: Reservoir fluid properties (physical & chemical properties). P, T, composition, etc. Physical & Geometrical properties of reservoir rocks. P, T, composition, structure, over burden, cementing, compaction, etc. Pressure differential (DP = Pf – Pw) Well geometry, spacing, production area @ Pf
CHAPTER 1: INTRODUCTION
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Productivity Index (PI) The PI express the ability of production of the well. Provide a measure of the capability of a reservoir to deliver fluids to the bottom of a wellbore for production.
It defines the relationship between the surface production rate and the pressure drop across the reservoir, known as the drawdown.
PI J where: PI q Ps Pwf
= = = =
q Ps Pwf
……………………… (2-1)
productivity index, STB/(day) (psi) producing flow rate, STB/day reservoir static pressure, psi flowing bottom hole pressure, psi
CHAPTER 2: RESERVOIR SYSTEM
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SKPP 3513: PRODUCTION ENGINEERING
Productivity Index (ctd) For radial flow and strong water drive or gas cap reservoir where flow tend to reach steady-state flow (from Eqn 1-3):
PI J
q 0.007082kh .………………. (2-2) re Ps Pwf Bo m ln( ) rw
PI will be a constant if m, Bo & k remain constant. A plot of P vs q should be a straight line of slope -1/PI with an intercept on the ordinate axis of Pe.
CHAPTER 2: RESERVOIR SYSTEM
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Productivity Index (ctd) Pe Slope = - 1/PI P
q Productivity Index CHAPTER 2: RESERVOIR SYSTEM
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SKPP 3513: PRODUCTION ENGINEERING
Productivity Index (ctd)
For pseudosteady-state flow (from Eqn 1-4) : PI J
q 0.007082kh Ps Pwf B m (ln re 1 ) o rw 2
………… (2-3)
In terms of average reservoir pressure, Pavg or Pr: PI J
CHAPTER 2: RESERVOIR SYSTEM
q 0.007082kh Pr Pwf B m (ln re 3 ) o rw 4
(8)
………… (2-4)
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Productivity Index (ctd)
Again, provided the fluid and rock properties (m, B and k) are constant, the PI should be constant, irrespective of the degree of depletion. Thus, as for the steady state case, a straight line relationship exists between P and q.
For gas well: PI J
qg ( Ps 2 Pwf 2 )
CHAPTER 2: RESERVOIR SYSTEM
0.703k g h r T ( m g z ) avg ln( e ) rw
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…… (2-5)
MOHD FAUZI HAMID
SKPP 3513: PRODUCTION ENGINEERING
Productivity Index (ctd)
Where the gas flow rate, qg (MSCF/day): qg
k g h( Pe 2 Pw2 ) r 1422T ( m g z ) avg ln( e ) rw
PI will remain constant if no change in the fluid and reservoir properties. If a graph of P2 vs q is drawn, a straight line should result of slope -1/PI.
CHAPTER 2: RESERVOIR SYSTEM
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Productivity Index (ctd) Pe2 Slope = - 1/PI
P2 Absolute Open Flow Potential (AOFP)
qo
qg
Gas well productivity relationship CHAPTER 2: RESERVOIR SYSTEM
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SKPP 3513: PRODUCTION ENGINEERING
Productivity Index (ctd)
PI specific, PIs (STB/(day)(psi)(ft):
PI s J s
PI q ………… (2-6) h h( Ps Pwf )
Example 2.1 A field is drilled up on a 80-acre spacing. The reservoir pressure (Ps) is 1000 psi, k = 50 md, h = 20 ft, m = 3 cp, Bo = 1.25 bbl/STB. The wells are completed with 7-in casing. If the producing pressure at the bottom of the well is 500 psi, (a) what is the production rate of the well, (b) the PI, and (c) the specific PI. CHAPTER 2: RESERVOIR SYSTEM
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Productivity Index (ctd)
If there is gas in fluid flow, the fluid is no longer incompressible, PI not constant and this situation happen when Ps < Pb. PI becomes:
PI
dq dPwf
………… (2-7)
If there is water in oil flow (oil and water mix): (q qw ) ………… (2-8) PI o ( Ps Pwf )
CHAPTER 2: RESERVOIR SYSTEM
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SKPP 3513: PRODUCTION ENGINEERING
Productivity Index (ctd)
For oil-water mix & radial flow (steady-state):
PI
(qo qw ) 0.007082h ko k ( w ) …… (2-9) r Ps Pwf mo Bo m w Bw ln( e ) rw
For two production condition: ( PI )1 ( ko / mo Bo )1 ( PI ) 2 ( ko / mo Bo ) 2
……………….... (2-10)
To estimate the PI, Lewis & Horner equation can be used:
PI 5.9*10 4 ( CHAPTER 2: RESERVOIR SYSTEM
(14)
kh ) mo Bo
……………….... (2-11) MOHD FAUZI HAMID
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Productivity Index (ctd)
Factors influencing PI.
Production GOR Pressure depletion Cumulative Production Oil Saturation (So will affect the effective and relative permeability of the fluid)
Generally, PI will decrease with time & P because: Turbulence due to increasing of flow rate.
Oil permeability decrease due to present of gas when P decrease near the well bore.
mo increase with pressure depletion below Pb. Permeability decrease due to formation compression. CHAPTER 2: RESERVOIR SYSTEM
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SKPP 3513: PRODUCTION ENGINEERING
Productivity Index (ctd)
Future PI:
kro ) mo Bo f PI f PI p k ( ro ) p mo Bo (
………… (2-11)
where:
PI p
CHAPTER 2: RESERVOIR SYSTEM
1.8PI P 1 0.8( wf ) Pavg
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Hydrocarbon Phase Behavior HC reservoir fluid : complex mixture of HC molecules. Composition depend on: Source rock Maturation degree P&T Phase change from HPHT cool & low pressure effect to performance Single phase in reservoir (liquid phase) Remains single phase (liquid) at wellbore (significant reduction in P & small change in T @ flow in reservoir (single phase region)) Starts to evolve gas as T & P reduced @ flow up in tubing (bubble point) Evolves increasing amount of gas until separator (two phase region) CHAPTER 2: RESERVOIR SYSTEM (17) MOHD FAUZI HAMID
SKPP 3513: PRODUCTION ENGINEERING
Hydrocarbon Phase Behavior (ctd)
1. Single phase in reservoir (a) 2. Remains single phase liquid at wellbore (significant reduction in P & small change in T @ flow in reservoir (a – b) 3. Starts to evolve gas as T & P reduced @ flow up in tubing (c) 4. Evolves increasing amount of gas (d-e) until separator (f) CHAPTER 2: RESERVOIR SYSTEM
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Hydrocarbon Phase Behavior (ctd) Possible Phase Changes in Production Flow System
CHAPTER 2: RESERVOIR SYSTEM
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SKPP 3513: PRODUCTION ENGINEERING
Inflow Performance Relationship (IPR) The IPR is defines as the functional relationship between the production rate and the flowing bottom hole pressure.
Always termed as “IPR Curve”; a plot of Pwf vs q (Pwf on vertical axis and q on horizontal axis).
From the definition of PI q Pwf Ps PI Ps is constant, and if PI also assumed to be constant, the relationship between Pwf & q is linear with a negative slope of 1/PI.
CHAPTER 2: RESERVOIR SYSTEM
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IPR (ctd)
Slope = - 1/PI
Typical IPR Curve for Incompressible Fluid CHAPTER 2: RESERVOIR SYSTEM
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SKPP 3513: PRODUCTION ENGINEERING
IPR (ctd) When q = 0, Pwf = Ps (no flow). When Pwf = 0, q = Ps * PI = qmax (absolute open flow potential – AOFP)
AOFP is the maximum rate that the formation can deliver fluid to the bore hole.
Ps Slope = - 1/PI
Pwf
The figure is a typical IPR curve for
Absolute Open flow potential
radial flow equation for single phase, incompressible fluid.
From the figure: q tan max PI Ps CHAPTER 2: RESERVOIR SYSTEM
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q
qmax
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Example 2-2 Given:
Permeability, k Pay thickness, h Average reservoir pressure, Pr Reservoir temperature, T Well spacing, A Drilled hole size, D Formation volume factor, B Oil viscosity, m Skin factor, St
= = = = = = = = =
30 mD 40 ft 3000 psi 200oF 160 acres 12 ¼ in (open hole) 1.2 bbl/STB 0.8 cp 0
Calculate: Absolute Open Flow Potential and PI Plot IPR curve. CHAPTER 2: RESERVOIR SYSTEM
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SKPP 3513: PRODUCTION ENGINEERING
IPR (ctd) For most well, IPR curve is not a linear relationship, because PI decrease when q increase and pressure depleted.
PI decrease with pressure depletion. This method (PI method) of predicting IPR is only can be used for single phase flow (Ps > Pb). For solution gas reservoir at Ps < Pb, Vogel method is applicable.
CHAPTER 2: RESERVOIR SYSTEM
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IPR – Gas Flow Compressible fluid IPR not straight line
Can be derived from Darcy Law by using average properties between reservoir & wellbore (low q): q = C (Pr2 – Pwf2) C = constant For higher flow rate due non-Darcy (turbulent flow) effect field observation Bureau of Mines eq: q = C (Pr2 – Pwf2)n n = 0.5 – 1.0 Log-log plot of q vs (Pr2 – Pwf2) straight line with : Slope = n Intercept = C Standard test for gas well: measure Pwf at 4 production rate. CHAPTER 2: RESERVOIR SYSTEM
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SKPP 3513: PRODUCTION ENGINEERING
IPR – 2 Phase (Gas-Liquid) Flow Saturated reservoir Vogel equation for solution gas-drive wells can be used: q/qmax = 1 – 0.2(Pwf/Pr) -0.8(Pwf/Pr)2 qmax = AOF (q @ Pwf = 0) If multirate test data available: q/qmax = {1 – (Pwf/Pr)2}n Plot log-log q vs (Pwf/Pr)2 straight line with slope = n .
CHAPTER 2: RESERVOIR SYSTEM
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Vogel’s Equation Introduced by Vogel in 1968 and recommended for solution gas drive reservoir where Ps < Pb .
Vogel’s Equation; Pwf qo 1 0.2 (qo )max Ps
Pwf 0.8 Ps
Based on the equation, Vogel plot for solution gas drive reservoir.
2
Pwf qo vs (qo ) max Ps
known as dimensionless IPR Curve of Vogel Chart, Vogel Equation or Vogel Chart.
can be used to predict the IPR of solution gas drive reservoir CHAPTER 2: RESERVOIR SYSTEM
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SKPP 3513: PRODUCTION ENGINEERING
Vogel’s Equation (ctd) This IPR curve can be generated if either the qmax and reservoir pressure are known, or the reservoir pressure & a flow rate and the corresponding bottom hole flowing pressure are known.
For either case, a buildup test for the reservoir pressure, and a flow test with bottom hole gauge are required. Pwf qo 1 0, P ( q o ) max s Pwf 0, 1 Ps
From the equation; if and if
qo (qo ) max
Pwf Pwf qo 1 1 0.8 (qo ) max Ps Ps CHAPTER 2: RESERVOIR SYSTEM
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Example 2-3 Given qo Pwf Ps
= = =
300 BPD 1500 psi 1800 psi
Determine: (a) q maximum (b) q @ Pwf = 1000 psi by PI and Vogel methods.
CHAPTER 2: RESERVOIR SYSTEM
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SKPP 3513: PRODUCTION ENGINEERING
Example 2-4 Given qo Pwf Ps
= = =
300 BPD 1500 psi 1800 psi
Determine Pwf for qo = 600 BPD by PI and Vogel methods.
CHAPTER 2: RESERVOIR SYSTEM
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Construction of IPR
Based on Vogel Method, the data required are: Flow rate, q Bottom hole flowing pressure, Pwf Reservoir pressure, Ps. All these data can be obtained from: Production test Flow test PBU test The method of constructing the curve depend upon the condition of well, i.e. Single phase flow Two phase flow: Ps < Pb and Ps > Pb
CHAPTER 2: RESERVOIR SYSTEM
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SKPP 3513: PRODUCTION ENGINEERING
Single Phase Flow q ( Ps Pwf )
Determine PI: PI
Determine qmax : qmax = Ps * PI
Plot graph of Pwf vs q
Absolute Open flow potential
The plot will produce a straight line
CHAPTER 2: RESERVOIR SYSTEM
Slope = - 1/PI
Pwf
Pwf = Ps q = 0 Pwf = 0 q = qmax
Ps
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q
qmax
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Two Phase Flow
There are three methods depend upon the Ps, Pwf, and Pb. 1) Ps < Pb 2) Ps > Pwf > Pb 3) Ps > Pb > Pwf
CHAPTER 2: RESERVOIR SYSTEM
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SKPP 3513: PRODUCTION ENGINEERING
(1) Ps < Pb
Determine (qo)max from Vogel Equation Pwf Pwf qo 1 0.2 0.8 (qo ) max Ps Ps qo (qo ) max Pwf Pwf 1 0.2 0.8 Ps Ps
2
2
If production test data is not available, qo can be estimated from Darcy equation, and (qo)max from: (qo ) max
qo Ps PI * Ps 1.8( Ps Pwf ) 1.8
Derivation-1
q
CHAPTER 2: RESERVOIR SYSTEM
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0.007082kh( Pr Pwf ) r 3 Bo m (ln e ) rw 4 MOHD FAUZI HAMID
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Derivation - 1 Pwf Pwf qo 1 0.2 0.8 (qo ) max Ps Ps Pwf Pwf 1 1 0.8 Ps Ps
(qo ) max
qo Ps 1 Pwf ( Ps Pwf ) 1 0.8 Ps
2
Pwf Ps Pwf 1 0.8 Ps Ps
1 PI * P s Pwf 1 0.8 Ps
If Pwf to tend to Ps (with small drawdown – at initial stage) (qo ) max
qo Ps PI * Ps 1.8( Ps Pwf ) 1.8 Back
CHAPTER 2: RESERVOIR SYSTEM
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SKPP 3513: PRODUCTION ENGINEERING
(1) Ps < Pb (ctd)
Assume a numbers of Pwf (from 0 to Ps), determine q from Vogel Eqn or Vogel’s Reference IPR
Plot graph of Pwf vs qo
Ps
Pwf = Ps qo = 0
Pwf
Pwf = 0 qo = (qo)max
The plot will produce a curve q
qmax
2 Pwf Pwf qo (qo ) max 1 0.2 0.8 Ps Ps
CHAPTER 2: RESERVOIR SYSTEM
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(2) Ps > Pb
IPR curve will have two parts:
Ps > Pwf > Pb a straight line
4000
Pwf < Pb a curve
3500
Considered as generalized IPR – 2500 a combination of the PI method 2000 above the bubble point pressure and Vogel’s IPR below 1500 1000 the bubble point pressure.
Two situations:
Pwf(psi)
3000
500 0
Given Pwf > Pb
0
50
100
150
200
q(BPD)
Given Pwf < Pb CHAPTER 2: RESERVOIR SYSTEM
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SKPP 3513: PRODUCTION ENGINEERING
(2) Ps > Pb (ctd)
qb PI ( Ps Pb ) qPb qv 1.8 Ps Pwf
CHAPTER 2: RESERVOIR SYSTEM
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PI * P b 1.8
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(2-1) Pwf > Pb
Mark the point Ps @ q = 0, and Pwf @ given q. Connect the two points with a straight line.
Determine qb from above plot or from equation qb = PI(Ps – Pb)
PI * Pb qb qb 1.8 Ps Pwf 1.8
Determine qmax from: qmax
qPb
Assume a numbers of Pwf (0 to Pb) and determine q from: 2 Pwf Pwf (q qb ) Other Form 1 0.2 0.8 (qmax qb ) Pb Pb
Plot Pwf vs q.
PI * Pb q qb 1.8
CHAPTER 2: RESERVOIR SYSTEM
2 Pwf Pwf 1 0.2 0.8 Pb Pb
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SKPP 3513: PRODUCTION ENGINEERING
P (q q b ) 1 0.2 wf (q max q b ) Pb q q b (q max
Pwf 0.8 Pb
2
2 Pwf Pwf q b ) 1 0.2 0.8 Pb Pb
PI * Pb qPb but (q max q b ) q v 1.8 P P 1.8 s wf Pwf qPb q q b 1 0.2 Pb 1.8 Ps Pwf PI * Pb q qb 1.8 CHAPTER 2: RESERVOIR SYSTEM
Pwf 1 0.2 Pb (40)
2 Pwf 0.8 Pb
2 Pwf 0.8 Pb
Back MOHD FAUZI HAMID
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(2-2) Pwf < Pb
Determine qb from equation
Pwf Pb qb q 1 1 0.2 1.8( Ps Pwf ) Pb
qmax
2 Pwf 0.8 Pb qPb q 1.8 Ps Pwf b
Determine qmax from:
Assume a numbers of Pwf (0 to Pb) and determine q from: 2 Pwf Pwf (q qb ) 1 0.2 0.8 (qmax qb ) Pb Pb
Plot Pwf vs q + a straight line from Pb to Ps. q qb
CHAPTER 2: RESERVOIR SYSTEM
PI * Pb 1.8
Pwf 1 0.2 Pb
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2 Pwf 0.8 Pb MOHD FAUZI HAMID
SKPP 3513: PRODUCTION ENGINEERING
4000
Ps 3500
Pwf(psi)
3000
2500
2000
Pb 1500
1000
500
qb
0 0
20
40
60
80
100
120
140
160
q(BPD)
180
qmax
200
Typical Composite IPR Curve CHAPTER 2: RESERVOIR SYSTEM
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Example 2-5
The Pulai No 1 was tested for eight hours at a rate of about 38 STB/day. Wellbore flowing pressure is 585 psi. After shutting the well in for 24 hours, the bottomhole pressure reached a static value of 1125 psi. Calculate: 1. Productivity index, PI 2. Absolute open hole flowing potential 3. Oil flow rate for a wellbore flowing pressure of 350 psi 4. Wellbore flowing pressure required to produce 60 STB/day
Draw the IPR curve, indicating the calculated quantities.
CHAPTER 2: RESERVOIR SYSTEM
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SKPP 3513: PRODUCTION ENGINEERING
Example 2-5: Solution
1200
1125 100 800 585 400 350 200 0
0 72.2 22.9 38.0 51.0 54.6 65.1 79.2
Example 2-5
1000
Pwf(psi)
Pwf (psi) q(BPD)
800
600
400
200
0 0
CHAPTER 2: RESERVOIR SYSTEM
10
20
30
(44)
40
50
q (STB/day)
60
70
80
90
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Example 2-6
A discovery well, Tanjung No. 3, was tested in the T-9 Sand at a rate of 200 STB/day with a bottomhole flowing pressure of 3220 psi. The estimated bubble point pressure of 4020 psi indicates that the well is draining saturated oil, since the static pressure was 4000 psi. Plot the IPR using the Vogel equation.
CHAPTER 2: RESERVOIR SYSTEM
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SKPP 3513: PRODUCTION ENGINEERING
Example 2-6: Solution
Vogel Equation: 2
Pwf Pwf qo 1 0.2 0.8 (qo )max Ps Ps qo (qo ) max 2 P P 1 0.2 wf 0.8 wf Ps Ps 200 (qo )max 2 3220 3220 1 0.2 0.8 4000 4000
624STB / day
Calculate several rates at specific drawdown to have enough points to plot IPR
CHAPTER 2: RESERVOIR SYSTEM
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Example 2-6: Solution (ctd)
(Pwf/Ps) 1 0.9375 0.8750 0.8125 0.7500 0.6875 0.6250 0.5625 0.5000 0.4375 0.3750 0.3125 0.2500 0.1875 0.125 0.0625 0
(Pwf/Ps)2 1 0.8789 0.7656 0.6602 0.5625 0.4727 0.3906 0.3164 0.2500 0.1914 0.1406 0.0977 0.0625 0.0352 0.0156 0.0039 0.0000
q(BPD) 0 68 133 193 250 302 351 396 437 474 507 536 562 583 601 614 624
4500
Example 2-6
4000 3500
3000
Pwf(psi)
Pwf (psi) 4000 3750 3500 3250 3000 2750 2500 2250 2000 1750 1500 1250 1000 750 500 250 0
2500 2000 1500 1000 500 0 0
100
200
300
400
500
600
700
q(BPD)
2 Pwf Pwf qo 624 1 0.2 0.8 4000 4000
CHAPTER 2: RESERVOIR SYSTEM
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SKPP 3513: PRODUCTION ENGINEERING
Example 2-7 Given: Static reservoir pressure, psi Bubble point pressure, psi Bottomhole flowing pressure, psi Production rate, BPD 1. 2. 3. 4.
= = = =
3620 1825 1980 108
Determine the Productivity Index that is valid if Pwf > Pb. Calculate the oil rate if Pwf = Pb. Calculate the maximum oil rate. Plot IPR
CHAPTER 2: RESERVOIR SYSTEM
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Example 2-7: Solution
Ps > Pwf > Pb:
1) PI:
PI (
q 108 ) 0.066 BPD / psi Ps Pwf 3620 1980
2) qb = PI(Ps – Pb) = 0.066(3620 – 1825) = 118.5 BPD 3) qmax:
4) IPR:
qmax
qPb (108)(1825) qb 118.5 185.3BPD 1.8( Ps Pwf ) 1.8(3620 1980)
2 Pwf Pwf qo (qmax qb ) 1 0.2 0.8 qb Pb Pb 2 Pwf Pwf qo (185.3 118.5) 1 0.2 0.8 118.5 (66.8) 1825 1825
CHAPTER 2: RESERVOIR SYSTEM
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SKPP 3513: PRODUCTION ENGINEERING
OR 1) PI:
PI (
q 108 ) 0.066 BPD / psi Ps Pwf 3620 1980
2) qb = PI(Ps – Pb) = 0.066(3620 – 1825) = 118.5 BPD 2 Pwf Pwf 1 0.2 0.8 Pb Pb 2 (0.066) * (1825) Pwf Pwf q o 118.5 0.8 1 0.2 1.8 1825 1825 2 P P q o 118.5 66.9 1 0.2 wf 0.8 wf 1825 1825
3) IPR: q q b
CHAPTER 2: RESERVOIR SYSTEM
PI * Pb 1.8
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Example 2-7: Solution (ctd) Pwf (psi) (Pwf/Pb) (Pwf/Pb)2 q(BPD)
1 0.9041 0.8219 0.6849 0.5479 0.411 0.274 0.137 0
1.0000 0.8174 0.6755 0.4691 0.3002 0.1689 0.0751 0.0188 0.0000
0 119 130 138 151 162 171 178 182 185
Example 2-7
3500 3000
Pwf(psi)
3620 1825 1650 1500 1250 1000 750 500 250 0
4000
2500 2000 1500 1000 500 0
2 20 40 Pwf Pwf 0 qo (qmax qb ) 1 0.2 0.8 qb Pb Pb 2 Pwf Pwf qo (185.3 118.5) 1 0.2 0.8 118.5 1825 1825 (66.8)
CHAPTER 2: RESERVOIR SYSTEM
60
80
100
120
140
160
180
200
q(BPD)
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SKPP 3513: PRODUCTION ENGINEERING
Example 2-8 Given: Static reservoir pressure, psi Bubble point pressure, psi Bottomhole flowing pressure, psi Production rate, BPD
= 1750 = 1200 = 900 = 600
1. Plot IPR 2. Determine well potential 3. Determine Pwf required if qo is 800 BPD
CHAPTER 2: RESERVOIR SYSTEM
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Example 2-8: Solution
Ps > Pb > Pwf: Determine qb: 2 Pwf Pwf Pb qb q 1 1 0.2 0.8 1.8( Ps Pwf ) Pb Pb
2 1200 900 900 qb 600 1 1 0.2 0.8 412BPD 1200 1200 1.8(1750 900)
Determine qmax: qmax
qPb 600*1200 qb 412 883BPD 1.8( Ps Pwf ) 1.8(1750 900)
Calculate several rates at specific drawdown to have enough 2 points to plot IPR P P wf wf qo (qmax qb ) 1 0.2 0.8 qb P P b b
CHAPTER 2: RESERVOIR SYSTEM
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MOHD FAUZI HAMID
SKPP 3513: PRODUCTION ENGINEERING
OR 1) PI:
PI (
q 600 ) 0.706 BPD / psi Ps Pwf 1750 900 Pwf 1 0.2 Pb
2 Pwf 2) qb : q b 0.8 Pb 2 0.706 *1200 900 900 q b 600 0.8 1 0.2 411.73BPD 1.8 1200 1200
PI * Pb q 1.8
3) IPR:
q o 411.73
2 (0.706) * (1200) Pwf Pwf 1 0.2 0.8 1200 1200 1.8
2 P P q o 412 471 1 0.2 wf 0.8 wf 1200 1200
CHAPTER 2: RESERVOIR SYSTEM
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Example 2-8: Solution (ctd) 2000
Example 2-8
1800 1600
Pwf (psi) (Pwf/Pb) (Pwf/Pb)2 q(BPD)
1400
1 0.8333 0.6667 0.5 0.3333 0.1667 0
1 0.6944 0.4444 0.2500 0.1111 0.0278 0.0000
412 543 653 742 810 857 883
1000 800
Pwf(psi)
1200
1200 1000 800 600 400 200 0
600 400 200 0 0
100
200
2 Pwf Pwf qo (qmax qb ) 1 0.2 0.8 qb Pb Pb
2 Pwf Pwf qo (883 412) 1 0.2 0.8 412 (471) 1200 1200
CHAPTER 2: RESERVOIR SYSTEM
300
400
500
600
700
800
900
1000
q(BPD)
(ii) Well Potential = qmax = 883 BPD (iii) Pwf @ q = 800 BPD: from IPR, Pwf = (55)
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Fetkovich Method
Multipoint back pressure test of gas wells is a common procedure to establish the performance curve of gas wells or deliverability.
Fetkovich applied these tests on oil well and found that oil well producing below the bubble point pressure behave just like gas well.
The general conclusion from the backpressure tests:
q PI '( Ps2 Pwf2 )n where: PI’ is the productivity index, the exponent “n” was found to be between 0.5 and 1.0. CHAPTER 2: RESERVOIR SYSTEM
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Fetkovich Method (ctd) A plot of (Ps2 – Pwf2) vs q on log-log paper results in a straight line with slope 1/n. [q vs (Ps2 – Pwf2) with slope n]
This equation often referred to as the backpressure equation.
The equation was derived from Evinger and Muskat for two phase radial flow:
q
(7.08kh) f ( P )dP re ln( ) rw
f ( P)
kro mo B0
CHAPTER 2: RESERVOIR SYSTEM
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Example 2-8-1 Given: Static reservoir pressure, psi
=
2000
Bottomhole flowing pressure, psi =
1400
Production rate, BPD
1020
=
1. Plot IPR when n = 1.0 2. Plot IPR when n = 0.5
CHAPTER 2: RESERVOIR SYSTEM
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Example 2-8-1: Solution (a) n = 1.0: [(b) n = 0.5]:
q PI '( Ps2 Pwf2 ) n q 1020 BPD 1.0 2 1.0 ( P Pwf ) (2000 psi ) 2 1400 psi 2 4 2 5 x 10 bbl /( day )( psi) 0.7141
PI '
2 s
2 q 5 x 104 bbl /( day )( psi) 2 2000 psi Pw2f
5 x 104 4 x 106 Pwf2 BPD 0.7141 4 x 106 Pw2f
CHAPTER 2: RESERVOIR SYSTEM
0.5
BPD
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Example 2-8-1: Solution (ctd) 2500
Example 2-8-1
Pwf(psi) 2000
Pwf (psi) q(BPD) 2000 0 1500 875 1200 1280 1000 1500 800 1680 600 1820 400 1920 200 1980 0 2000
Pwf (psi) 2000 1500 1200 1000 800 600 400 200 0
q(BPD) 0 944.67 1500 1142.56 1236.86 1308.97 1000 1362.42 1399.34 1421.04 500 1428.20
0 0
CHAPTER 2: RESERVOIR SYSTEM
500
(60)
1000
q(BPD)
1500
2000
2500
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Prediction of Future IPR
One of the method used to predict future IPR is Standing Method. Develop by Standing based on Vogel Equation.
Standing rewrote Vogel Equation as: Pwf Pwf q (1 )(1 0.8 ) qmax Ps Ps Pwf q q ( max )(1 0.8 ) Ps Pwf Ps Ps Pwf q PI ( max )(1 0.8 ) Ps Ps
……………. (1)
Let PI* be the initial value of PI (the PI for small drawdown. Allowing Pwf to tend to Ps:
CHAPTER 2: RESERVOIR SYSTEM
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Prediction of Future IPR (ctd) 1.8qmax ………….. (2) PI * ( ) Ps
Equation (1) & (2):
1.8PI P (1 0.8 wf ) Ps Vogel equation & (2): produce an equation for future IPR PI *
PI *f *( Ps ) f qf 1.8
CHAPTER 2: RESERVOIR SYSTEM
Pwf 1 0.2 ( Ps ) f
(62)
Pwf 0.8 ( Ps ) f
2
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Prediction of Future IPR (ctd) where:
( kro / mo Bo ) f PI *f PI *p ( kro / mo Bo ) p s s kro o or 1 sor swi
CHAPTER 2: RESERVOIR SYSTEM
n
(63)
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Prediction of Future IPR (ctd) procedure: i. PI: PI ii. PI*p :
iii. PI*f :
PI *p
Pwf (q ) q o max 1 0.8 Ps Pwf Ps Ps 1.8* PI Pwf 1 0.8 Ps
(kro / mo Bo ) f PI PI (kro / mo Bo ) p
iv. IPR :
CHAPTER 2: RESERVOIR SYSTEM
* f
* p
s sor (kro ) o 1 sor swi
P PI *f *( Ps ) f 1 0.2 wf qf (P ) 1.8 s f
(64)
Pwf 0.8 ( Ps ) f
2
n
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Example 2-9
Production test data for a well are as follows: Parameter Static reservoir pressure, psig Bottomhole flowing pressure, psig Buble point pressure, psig Production rate, BPD Oil saturation, % Residual oil saturation, % Initial water saturation, % Oil formation volume factor, bbl/STB Oil viscosity, cp Gas liquid ratio, MCF/STB Oil specific gravity, oAPI Gas specific gravity
CHAPTER 2: RESERVOIR SYSTEM
2017
2027
3000 2500 3500 5000 60 20 10
2000 45 15 30
1.3 0.8 0.8 30 0.6
1.01 0.8 0.8 30 0.6
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Example 2-9 (ctd) i.
Construct IPR Curve for the year of 2017 and 2027.
ii. Determine well potential for the year of 2017 and 2027. iii. Determine the bottomhole flowing pressure if the well should produced 14000 BPD in 2017. iv. Determine production rate in 2027 if the bottomhole flowing pressure is 750 psig.
CHAPTER 2: RESERVOIR SYSTEM
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Example 2-9: Solution
For 2017: Ps
(qo ) max
qo P 1 0.2 wf Ps
Pwf 0.8 Ps
2
5000 2500 2500 1 0.2 0.8 3000 3000
2
18, 000 BPD
Calculate several present rates at specific drawdown to have enough points to plot 2017 IPR. 2 Pwf Pwf qo (qo ) max 1 0.2 0.8 Ps Ps 2 Pwf Pwf qo 18000 1 0.2 0.8 3000 3000
CHAPTER 2: RESERVOIR SYSTEM
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Example 2-9: Solution (ctd) Present IPR Curve (Pwf/Ps) 1 0.9167 0.8333 0.75 0.6667 0.5833 0.5 0.4167 0.3333 0.25 0.1667 0.0833 0.0333 0
(Pwf/Ps)2 1 0.8403 0.6944 0.5625 0.4444 0.3403 0.2500 0.1736 0.1111 0.0625 0.0278 0.0069 0.0011 0
q(BPD) 0 2600 5000 7200 9200 11000 12600 14000 15200 16200 17000 17600 17864 18000
2017 IPR 3500 3000 2500 2000 1500
2017 IPR
Pwf(psi)
Pwf (psi) 3000 2750 2500 2250 2000 1750 1500 1250 1000 750 500 250 100 0
1000 500 0 0
2000
4000
CHAPTER 2: RESERVOIR SYSTEM
6000
8000
10000
12000
14000
16000
18000
20000
q(BPD)
2 Pwf Pwf qo (qo ) max 1 0.2 0.8 Ps Ps
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Example 2-9: Solution (ctd)
To predict future IPR, use Standing Approach: (q ) Pwf PI o max 1 0.8 Ps Ps PI *p
where
18000 2500 1 0.8 3000 10 3000
1.8* PI 1.8*10 10.8 Pwf (2500) 1 0.8 1 0.8 (3000) Ps
(kro / mo Bo ) f PI *f PI *p (kro / mo Bo ) p n 4 s sor 0.45 0.15 (kro ) f o 0.0885 1 sor swi f 1 0.15 0.3 n
4
s sor 0.60 0.20 (kro ) p o 0.1066 1 sor swi p 1 0.20 0.1
CHAPTER 2: RESERVOIR SYSTEM
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Example 2-9: Solution (ctd) (kro / mo Bo ) f (0.0885 /(0.8 x1.01)) PI *f PI *p 10.8 11.53 (kro / mo Bo ) p (0.1066 /(0.8 x1.3))
P PI *f *( Ps ) f 1 0.2 wf qf (P ) 1.8 s f qf
Pwf 0.8 ( Ps ) f
2
Pwf Pwf 11.53 * 2000 1 0.2 0.8 1.8 2000 2000
2
2 Pwf Pwf q f 12,811 1 0.2 0.8 2000 2000
q f 12,8111 1*104 (Pwf ) 2 *107 (Pwf )2
CHAPTER 2: RESERVOIR SYSTEM
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Example 2-9: Solution (ctd)
Calculate several future rates at specific drawdown to have enough points to plot 2027 IPR. Future IPR Curve 3500
Pwf (psi) (Pwf/Ps) (Pwf/Ps 2000 1750 1500 1250 1000 750 500 250 0 qf
1 0.875 0.75 0.625 0.5 0.375 0.25 0.125 0
1 0.7656 0.5625 0.3906 0.2500 0.1406 0.0625 0.0156 0.0000
P PI *f *( Ps ) f 1 0.2 wf (P ) 1.8 s f
qf(BPD) 0 2722 5124 7206 8968 10409 11530 12331 12811
Pwf(psi)
Example 2-9
)2
3000
2500
2017 IPR
2000
2027 IPR 1500
1000
Pwf 0.8 ( Ps ) f
500
2
0
0
2000
4000
6000
q f 12,8111 1*104 (Pwf ) 2 *107 (Pwf )2 CHAPTER 2: RESERVOIR SYSTEM
8000
10000
12000
14000
16000
18000
20000
q(BPD) (71)
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Predicting Future IPR by Fetkovich Method
From Muskat: PI at any specific time is related to the PI at other time:
PI1 (kro / mo Bo )1 PI 2 (kro / mo Bo ) 2
Fetkovich: kro is linear towards the pressure:
kro @ Ps Ps kroi Psi
From backpressure equation:
q PI '( Ps2 Pwf2 )n …………. (1)
CHAPTER 2: RESERVOIR SYSTEM
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Predicting Future IPR by Fetkovich Method (ctd)
Based on these three equations:
Ps 2 ( Ps Pwf2 ) n Psi OR: q PI ' Ps 2 ( P 2 P 2 ) n 1 s2 wf Ps1 q PI '
………… (2)
Determine PI’ by backpressure equation. Find the relationship q vs Pwf by Eqn. (2) Assume a numbers of Pwf, and use Eqn (2) to find q vs Pwf at specific Ps.
Plot Pwf vs q. CHAPTER 2: RESERVOIR SYSTEM
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Example 2-10 Given: Static reservoir pressure Bubble point pressure Flowing bottom hole pressure Production rate
= = = =
3000 psi 3000 psi 2000 psi 500 BPD
If n=1.0, construct IPR when static reservoir pressure decline to 2000 psi.
CHAPTER 2: RESERVOIR SYSTEM
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Predicting Future IPR by Simple Method
Collect several PI data from wells which known Ps.
Plot PI/(PI)i vs Ps. (PI)i = PI at early stage of production PI @ Pi).
Find the best straight line.
For any specific value of Ps, find PI.
Based on the obtained PI, future IPR curve can be plotted using previous methods.
CHAPTER 2: RESERVOIR SYSTEM
(75)
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Determination of PI or IPR at Field
Field determination is depend on the conditions of the well.
a) Simple Case. Data required are Pwf, q and Ps. Shut-in well for 24 – 72 hours PBU With pressure gauge at bottom, flow the well at lowest q and record Pwf (stable – may be 24 hours) Flow the well at highest rate and record Pwf (as step 2) Repeat for another flow rate minimum 3 reading Ps from pressure buildup test. PI can be determine by method explained earlier.
CHAPTER 2: RESERVOIR SYSTEM
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Determination of PI or IPR at Field (ctd) b) Special Case For well without packer. During production, some of the gas will come out and accumulated at the top of the annulus. At equilibrium, Pwf = Pc + pressure due to gas column Gilbert: Pressure due to gas column = PcD1.5/100 psia Pwf = Pc(1+D1.5/100 ) psia where Pc = D = CHAPTER 2: RESERVOIR SYSTEM
casing head pressure, psia tubing depth, per 1000 ft (77)
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Example 2-11 Given: Tubing depth q
= 3000 ft = 42 BPD @ Pc = 550 psig = 66 BPD @ Pc = 320 psig By assuming constant PI, determine: 1. PI 2. Ps 3. Well potential
CHAPTER 2: RESERVOIR SYSTEM
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Example 2-11: Solution D1.5 31.5 Pwf Pc 1 565 1 595 psia 580 psig 100 100
Pwf = 580 psig @ q = 42 BPD D1.5 31.5 Pwf Pc 1 335 1 353 psia 338 psig 100 100
Pwf = 338 psig @ q = 66 BPD
CHAPTER 2: RESERVOIR SYSTEM
(79)
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Pwf (psig)
Example 2-11: Solution (ctd)
1000
PI = 100/1000 = 0.1 BPD/psi Ps = 1000 psig qmax = 100 BPD
580 338
0
CHAPTER 2: RESERVOIR SYSTEM
From the plot:
42
66
100 q (BPD)
(80)
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IPR – Method Selection
Dissolved Gas Drive Reservoir Above Pb: • Constant GOR - No gas breaking out of solution and no interfering with oil flow. • PI method is applicable
Below Pb: • Gas break out of solution & impedes the oil flow • Vogel method
CHAPTER 2: RESERVOIR SYSTEM
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IPR – Method Selection (ctd)
Water Drive Reservoir Has: • Constant producing GOR • Pr is maintained by water influx • No gas breaking out of solution • PI method can be used. However, if water influx rate is exceeded, then the reservoir will perform like a dissolved gas drive reservoir, therefore Vogel method will be applicable .
CHAPTER 2: RESERVOIR SYSTEM
(82)
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IPR – Method Selection (ctd)
Gas Cap Drive Reservoir Wells producing from the oil column below the GOC • Produce with a constant GOR • PI method can be used. But, after gas breaks through to a well, neither the PI nor a vogel method is applicable. In most instances, the well should be shut in after the gas cap break through in order to conserve gas cap energy to produce down-structure well.
CHAPTER 2: RESERVOIR SYSTEM
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IPR – Method Selection Summary
PI Method Use to evaluate dissolved gas-drive wells above the bubble point Use to evaluate gas cap expansion drive wells prior to gas cap break through Use to evaluate water drive wells before exceeding water influx rate Vogel Method
Use to evaluate dissolved gas drive wells below the bubble point
Use to evaluate water drive wells after water influx rate is exceeded
CHAPTER 2: RESERVOIR SYSTEM
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Gas Well Performance
Gas well productivity determined with deliverability testing Two basic relationship in used: • Rawlins & Schellhardt (empirical backpressure method) • Houpeurt (theoretical )
Deliverability test method: • Flow-after-flow test (four-point test) • Isochronal test • Modified isochronal test
CHAPTER 2: RESERVOIR SYSTEM
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Gas Well Performance (ctd)
Rawlins & Schellhardt’s deliverability eq: qg = C (Ps2 – Pwf2)n where:
C = flow coefficient n = deliverability exponent non-Darcy: n = 0.5 – 1.0 Darcy flow: n = 1.0
A plot of qg vs (Ps2 – Pwf2) on log-log paper results in a straight line with slope 1/n.
This equation often referred to as the backpressure equation IPR can be plotted by: qg/qg max = [ 1 – (Pwf/Ps)2]n CHAPTER 2: RESERVOIR SYSTEM
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Gas Well Performance (ctd)
Houpeurt relationship: Deliverability eq: Ps2 – Pwf2 = aqg + bqg2 where; a = laminar flow coefficient b = turbulence coefficient From plot of (Ps2 – Pwf2 ) / qg vs qg (straight line) a = intercept with y axis b = slope
CHAPTER 2: RESERVOIR SYSTEM
(87)
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Gas Well Performance (ctd)
OR: a = 1422mzT[ln (re/rw) – ¾ + s] / kh b = (1422mzT/kh)D D = non-Darcy flow coefficient, D/Mscf D = [2.715x10-15 b kMPsc] / hmgrwTsc
b = turbulence factor, ft-1: k
m Psc Tsc M
f
s h CHAPTER 2: RESERVOIR SYSTEM
= = = = = = = =
b = 1.88x1010k-1.47f-0.53
md cp standard pressure, psia standard temperature, oR molecular weight, lbm/lbm-mole porosity, fraction skin factor formation thickness, ft (88)
MOHD FAUZI HAMID
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