Ch 00 - Introduction

8/3/2019 Ch 00 - Introduction

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PETE 486

Petroleum Production Engineering

Boyun Guo, Ph.D.

Professor in Petroleum Engineering

[email protected]


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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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Ch 00 - Introduction