ch 00 - introduction
TRANSCRIPT
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PETE 486
Petroleum Production Engineering
Boyun Guo, Ph.D.
Professor in Petroleum Engineering
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What is petroleum ?
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Crude Oil and Natural Gas
What is oil?
What is natural gas?
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0 50 150 200 250 300 350100
1000
500
1500
2000
2500
3000
3500
4000
Reservoir Temperature (oF)
ReservoirPre
ssure(psia)
Liquid
Volum
e40%
20%
10%
80%
5% 0%
Bubble
Point
Gas ReservoirsRetrogradeCondensateReservoirs
CriticalPoint
pi, T
ptf, Ttf
pwf, Twf
Dew
Point
Cricondentherm
Point
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What is production ?
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A process of producing.
What?
From where?
Through what?
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What in engineering ?
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Designing, constructing
(manufacture & installation),monitoring, maintaining,analyzing, optimizing
What?
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Chapter 1
Petroleum Production System
- Reservoir
- Well
- Surface facilities
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Reservoir
- Fluid type
- Volume
- Pressure
- Flow mechanism
- OOIP & OGIP
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oi
wi
B
SAhOOIP
1
gi
wi
BSAhOGIP 1
How to estimate OOIP and OGIP?
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Reserves Estimate by Statistics
- Sample Mean
- Sample Standard Deviation- Distribution
- Probability
- Degree of Confidence
- Confidence Interval
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Sample Mean
n
i
ixn
x1
1
Sample No. Porosity
1 0.155
2 0.165
3 0.145
4 0.143
5 0.167
Sample Mean = 0.155
)(xE
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Sample Standard Deviation
Sample No. Porosity
1 0.155
2 0.165
3 0.145
4 0.1435 0.167
n
i
i xxn
S1
2
1
1 S2=Sample Variance
E[S2] =s2=population variance
Sample Variance = 0.000122
Sample Standard Deviation = 0.011045
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Distribution
- Normal distribution
- Log-normal distribution
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Number of Samples
From To Average
2 0.155 0.165 0.164 0.165 0.175 0.17
7 0.175 0.185 0.18
11 0.185 0.195 0.19
14 0.195 0.205 0.2
16 0.205 0.215 0.21
15 0.215 0.225 0.22
13 0.225 0.235 0.23
12 0.235 0.245 0.24
8 0.245 0.255 0.25
5 0.255 0.265 0.26
3 0.265 0.275 0.27
1 0.275 0.285 0.28
Porosity
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0
2
4
6
8
10
12
14
16
18
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35
Porosity
No.ofSamples
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0
2
4
6
8
10
12
14
16
18
0 100 200 300 400 500
Permeability (md)
No.of
Samples
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0
2
4
6
8
10
12
14
16
18
0.1 1 10 100 1000 10000
Permeability (md)
No.of
Samples
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Homework
Develop a
spreadsheet programand use it to calculatesample mean andstandard deviation of
the given data set.Does the data follow anormal distributionpattern?
Sample No. Porosity
1 0.155
2 0.164
3 0.1454 0.152
5 0.162
6 0.143
7 0.135
8 0.122
9 0.16110 0.181
11 0.195
12 0.157
13 0.154
14 0.161
15 0.158
16 0.147
17 0.162
18 0.154
19 0.149
20 0.172
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Probability
- Introduction example
- Definition
- Unit normal distribution function
- Exercises
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Introduction Example
Number of Samples
From To Average
2 0.155 0.165 0.16
4 0.165 0.175 0.17
7 0.175 0.185 0.18
11 0.185 0.195 0.19
14 0.195 0.205 0.2
16 0.205 0.215 0.21
15 0.215 0.225 0.22
13 0.225 0.235 0.23
12 0.235 0.245 0.24
8 0.245 0.255 0.25
5 0.255 0.265 0.26
3 0.265 0.275 0.27
1 0.275 0.285 0.28
Porosity
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0
2
4
6
8
10
12
14
16
18
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35
Porosity
N
o.ofSamples
Question: What is the probability that the nextsample has a porosity value of less than 0.2?
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(1) AveragePorosity
(2) Number ofvalues (n
i)
(3)Cumulative n
i
(4) CumulativeFrequency, (2)/(3)
0.16 2 2 0.0180180180.17 4 6 0.054054054
0.18 7 13 0.117117117
0.19 11 24 0.216216216
0.2 14 38 0.342342342
0.21 16 54 0.486486486
0.22 15 69 0.621621622
0.23 13 82 0.738738739
0.24 12 94 0.846846847
0.25 8 102 0.918918919
0.26 5 107 0.963963964
0.27 3 110 0.990990991
0.28 1 111 1
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0
0.2
0.4
0.6
0.8
1
1.2
0.15 0.17 0.19 0.21 0.23 0.25 0.27 0.29Porosity
Cumu
lativeFrequency
Note: This is like a probability function!
18
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0
2
4
6
8
10
12
14
16
18
0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3
x
Occur
rencef(x)
0
0.2
0.4
0.6
0.8
1
1.2
0.1 0.15 0.2 0.25 0.3
Porosity
Cum.
Fre
quency
Read Area / TotalArea
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Unit Normal Distribution Function
If random variable Xhas a normal distribution with meanand variance s, then variable
Z = (X-)/s
is a normal random variable with mean 0 and variance 1.
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X
Z
s
1
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Normal Distribution Function:
x
s 3 s 3
s
399.0
222/)(
2
1)(
s
s
xexf
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Unit Normal Distribution Function:
x3 3
399.0
2/2
2
1)(
xex
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Unit Normal Density Function:
x
dxxx )()(
x3 3
399.0
)(x
x
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x3 3
)(x
0.5
1.0
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Probability
=Degree of Confidence
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P{xx0} = 1-(x0)
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P{xb} = 1-(b)
P{a
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P{X
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P{X>a} = P{(X-)/s >(a-)/s}
=[-(a-)/s] = 1-[(a-)/s]
x
)(xf
a
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P{a
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n
i
i xxn
S1
2
1
1
S2=Sample Variance = 16
E[S2] =s2=population variance = 16
SD = 4
n
i
ixn
x1
13
Example-1:
Find the degrees of confidence for X < 11 and 2 < X < 7
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Solution:
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