COVENANT UNIVERSITYcovenantuniversity.edu.ng/content/download/51751... · pet522 question & marking...

COVENANT UNIVERSITY

OMEGA SEMESTER TUTORIAL KIT (VOL. 2)

P R O G R A M M E :

P E T R O L E U M E N G I N E E R I N G

500 LEVEL

DISCLAIMER

The contents of this document are intended for practice and learning purposes at the

undergraduate level. The materials are from different sources including the internet and the

contributors do not in any way claim authorship or ownership of them. The materials are also not

to be used for any commercial purpose.

LIST OF COURSES

PET522: Computer Applications in Petroleum Engineering II

PET523: Rock Physics and Formation Engineering

PET524: Reservoir Modeling and Simulation

PET525: Petroleum Refining Engineering

PET526: Petroleum Economics

PET528: Industrial Hazards and Environmental Pollution

PET527: Petroleum Engineering Laboratory II

*EDS521: Entrepreneurial Development Studies X

*TMC521: Total Man Concept X

*Not included

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PET522 QUESTION & MARKING GUIDE

COVENANT UNIVERSITY

CANAANLAND, KM 10, IDIROKO ROAD

P.M.B 1023, OTA, OGUN STATE, NIGERIA.

TITLE OF EXAMINATION: B.Eng EXAMINATION

COLLEGE: ENGINEERING

DEPARTMENT: PETROLEUM ENGINEERING

SESSION: 2015/2016 SEMESTER: OMEGA

COURSE CODE: PET 522 CREDIT UNIT: 3

COURSE TITLE: COMPUTER APPLICATIONS IN PETROLEUM ENGINEERING II

INSTRUCTION: Answer Question #1and any other 3 Questions TIME: 3 HOURS

Question 1

a. A gas-phase hydrocarbon system exists at 4000psia and 1600

F. Write/draw a flow chart to solve the cubic equation of state (Eq.1) form of the

Patel & Teja Equation of State (Eq.2) to compute gas compressibility. Use Newton-Raphson iteration scheme to solve the cubic equation of

state. Input data are natural gas composition (mole fraction), molecular weight, gas gravity-γ, critical pressure-Pc and temperature-Tc, and

acentric factor-ω. [9 marks]

b. Write a program based on the Newton-Raphson scheme to solve Eq. 1 given A, B and C as 0.827, 0.272 and 0.09 respectively. [9 marks]

𝑍 3 + (𝐶 − 1)𝑍2 + (𝐴 − 3𝐵𝐶 − 𝐵 − 𝐶 − 𝐵2)𝑍 = 0 (Eq. 1)

𝑃 =𝑅𝑇

𝑣−𝑏−

𝑎

𝑉2+(𝑏+𝑐)𝑉−𝑏𝑐 (Eq. 2)

Question 2

McKeon, Zagarola & Smits developed a friction factor for fully developed pipe flow, Eq. 3.Write a program in FORTRAN 90/95 to compute friction

factor with a tolerance of 1.0-10

as the stopping criteria of the program. The output of the program should include number of iterations, successive

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friction factor computed, difference between successive friction factor based on the following known variables: e is 0.00197in; q, 90592.77bbl/day

(600m3/hr); γo, 0.895; D, 15.1181in; µ, 38.7759cp. [13 marks]

1

√𝑓= 1.920 𝑙𝑜𝑔(𝑁𝑅𝑒√𝑓) − 0.475 −

7.04

(𝑁𝑅𝑒√𝑓)0.55 (Eq. 3)

where: e is relative roughness; D, diameter of pipe; and NRe, Reynolds number.

𝑁𝑅𝑒 =92.24𝑞(𝑏𝑏𝑙/𝑑𝑎𝑦)𝛾𝑜

𝐷(𝑖𝑛)𝜇(𝑐𝑝) (Eq. 4)

Question 3

Answer the following questions based on program List 1:

a. Identify the line on the main program where the function sub-program is called. [2 marks]

b. What type of do-loop is in the program? [2 marks]

c. Is it necessary to have the same variable name as arguments in the call statement of a subroutine (sub-program). [2 marks]

d. Re-write, edit and run the program without sub-programs (sub-routine & function) to calculate gas compressibility. [7 marks]

Question 4

a. Write a short note on the following:

i. Low level languages [1 marks]

ii. High level languages [2 marks]

iii. Flow charting [2 marks]

b. (i) Explain what Algorithm means in FORTRAN (ii) Highlight the steps involved in solving a problem with a computer. [4 marks]

c. (i) Give five (5) different features of FORTRAN 90/95 (ii) Highlight the detailed structure of a complete FORTRAN program. [4 marks]

Question 5

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Write the output(s) of the FORTRAN 90 program listing given below: [13 marks]

PROGRAM TRIDIAGONAL

IMPLICIT NONE

! THIS IS A SOLUTION OF TRIDIAGONAL SYSTEM OF EQUATIONS

! W, B, C, G: COEFFICIENTS OF THE EQUATION

! W(I)*X(I-1)+B(I)*X(I)+C(I)*X(I+1)=G(I)

! P=SOLUTION VECTOR

! N=NUMBER OF UNKNOWNS

INTEGER N,I

REAL, DIMENSION (4):: B=(/0.,2.,3.,2./)

REAL, DIMENSION (4):: C=(/3.,2.,1.,0./)

REAL, DIMENSION (4):: W=(/4.,1.,1.,1./)

REAL, DIMENSION (4):: G=(/25.,15.,12.,5./)

REAL, DIMENSION (4):: P

N=4

! DECOMPOSITION

DO I = 2, N

B(I) = B(I)/W(I-1)

W(I) = W(I) – B(I)*C(I-1)

END DO

! FORWARD SUBSTITUTION

DO I = 2, N

G(I) = G(I) – B(I)*G(I-1)

END DO

! BACK SUBSTITUTION

P(N)=(G(N)/W(N))

DO I = N-1, 1, -1

P(I)=((G(I)-C(I)*P(I+1))/W(I))

END DO

WRITE(*,*) (P(I),I=1,N)

STOP

END PROGRAM TRIDIAGONAL

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Question 6

For slightly compressible fluids such as crude oil, the volume at any pressure is given as; 𝑉 = 𝑉𝑏[𝑒𝑐(𝑝𝑏−𝑝)]

where c is compressibility coefficient; p, pressure, Vb, volume at low base pressure and pb is low base pressure.

𝑒𝑥 = ∑𝑥𝑛

𝑛!

∞

𝑛=0

= 1 + 𝑥 +𝑥2

2!+

𝑥3

3!+ ⋯

The expression, 𝑒𝑐(𝑝𝑏−𝑝)is normally approximated by the first two terms of the above expression as 𝑒𝑐(𝑝𝑏−𝑝) ≈ [1 + 𝑐(𝑝𝑏 − 𝑝)]. Write/Draw a flow chart to

compute volume ‘V’ at a pressure of 5000psia given that volume at a low base pressure of 1500psia is 2000ft3

. The fluid compressibility is 0.23×10-5

. ex should be

used based on the approximation to the first hundred terms of the above mathematical series. [13 marks]

Patel & Teja Equation of State – Variables of Eq. 1

A =𝑎mP

(RT)2 B =

bmP

RT C =

bmP

RT

𝑎m = [∑ 𝑦𝑖𝑎𝑖0.5

𝑖

]

2

𝑏𝑚 = ∑(𝑦𝑖𝑏𝑖)

𝑖

𝑐𝑚 = ∑(𝑦𝑖𝑐𝑖)

𝑖

𝑎 = Ω𝑎

𝑅2𝑇𝑐2

𝑃𝑐

[1 + 𝑚(1 − 𝑇𝑟0.5)]2 𝑏 = Ω𝑏

𝑅𝑇𝑐

𝑃𝑐 𝑐 = Ω𝑐

𝑅𝑇𝑐

𝑃𝑐

Ω𝑐 = 1 − 3𝜉𝑐

Ω𝑎 = 3𝜉𝑐2 + 3(1 − 2𝜉𝑐)Ω𝑏 + Ω𝑏

2 + (1 − 3𝜉𝑐) 𝜉𝑐 = 0.329032 − 0.0767992𝜔

+ 0.0211947𝜔2

𝑚 = 0.452413 + 1.30982𝜔 − 0.295938𝜔2

Ω𝑏3 + (2 − 3𝜉𝑐)Ω𝑏

2 + 3𝜉𝑐2Ω𝑏 − 𝜉𝑐

3=0

Ω𝑏 = 0.32429𝑍𝑐 − 0.002005 ,initial guess for Ωb above 𝑍𝑐 =

𝑃𝑐𝑉𝑐𝑀𝑤

𝑅𝑇𝑐

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Watansiri-Owens-Sterling Correlation for Vc

ln(𝑉𝑐) = 76.313887 − 129.8038𝛾 + 63.1750𝛾2 − 13.175𝛾3 + 1.101018 ln(𝑀𝑤)

+ 42.1958𝑖𝑛(𝛾)

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COVENANT UNIVERSITY








COURSE TITLE: COMPUTER APPLICATIONS IN PETROLEUM ENGINEERING II

INSTRUCTION: Answer Question #1and any other 3 Questions TIME: 3 HOURS

Question 1a [9 marks]

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yi, Mi, γi, Pci, Tci, ω, P, T, R

ξci=0.329032-0.0767092ωi+0.0211947ωi2

mi=0.452413+1.30982ωi-0.29593ωi2

Ωci=1-3ξci

lnVci=76.313887-

129.80386γi+63.17506γi2+1.101018ln(Mi)+42.1958ln(γi)

Zci=PciVciMi/RTci

apply 𝑦𝑘+1 = 𝑦𝑘 −𝑓(𝑦𝑘)

𝑓 (𝑦𝑘)

to solve for Ωbi of each natural gas component based on

Ωbi3+(2-3ξci) Ωbi

2+3ξci

2Ωbi-ξci

3=0

Tri=T/Tci; 𝑎𝑖 = Ω𝑎𝑖𝑅2𝑇𝑐𝑖

2

𝑃𝑐𝑖[1 + 𝑚𝑖(1 − 𝑇𝑟𝑖

0.5)]2;

𝑏𝑖 = Ω𝑏𝑖𝑅𝑇𝑐𝑖

𝑃𝑐𝑖; 𝑎m = [ 𝑦𝑖𝑎𝑖

0.5𝑖 ]

2; 𝑐𝑖 = Ω𝑐𝑖

𝑅𝑇𝑐𝑖

𝑃𝑐𝑖;

𝑏𝑚 = (𝑦𝑖𝑏𝑖)𝑖 ; 𝑐𝑚 = (𝑦𝑖𝑐𝑖)𝑖 ;

A =𝑎mP

(RT)2; B =bmP

RT; C =

cmP

RT;

apply 𝑦𝑘+1 = 𝑦𝑘 −𝑓(𝑦𝑘)

𝑓 (𝑦𝑘) to solve for Z based on;

𝑍 3 + (𝐶 − 1)𝑍2 + (𝐴 − 3𝐵𝐶 − 𝐵 − 𝐶 − 𝐵2)𝑍 = 0

Z

Stop

Start

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Question 1b

program Zfactor

real:: A=0.827, B=0.272, C=0.09

real: :Zassume=1, Znew, Zold

real:: tolerance=1e-5

integer:: interationCount=0

Zold=Zassume

do

Znew=Zold-f(A,B,B,Zold)/fD(A,B,C,Zold)

if(abs(Znew-Zold)<tolerance)exit

if(interationCount>=1000)exit

interationCount=interationCount+1

write(*,*)interationCount, Znew, abs(Znew-Zold)

Zold=Znew

end do

end program Zfactor

real function f(A,B,C,Zold)

f=(Zold**3)+((C-1)*Zold**2)+(A-3*B*C-B-C-B*B)*Zold

return

end

real function fD(A,B,C,Zold)

fD=3*(Zold**2)+2*(C-1)*Zold+(A-3*B*C-B-C-B*B)

return

end

Answer

1 0.827011 0.172989

2 0.761035 6.597590E-02

3 0.747172 1.386291E-02

4 0.745532 1.640737E-03

5 0.745372 1.593828E-04

6 0.745357 1.507998E-05

3marks

3marks

3marks

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Question 2

program friction_factor

real::flowrate=90592.77,specgravity=0.895,diameter=15.1181,viscosity=38.7759

real::finitial=0.1 ! initial guess of friction factor

real::f1,f2 !previous and current computed friction factor

real::reynoldsnumber

real::df !difference between previous and current computed friction factor

real::tolerance=1E-10

integer::k !counter

integer::ii !counter for exit notification by given tolerance

integer::maxiter=30 !maximum number of iterations

reynoldsnumber=(92.24*flowrate*specgravity)/(diameter*viscosity)

f1=finitial

ii=1 !initialization of counter for exit notification by tolerance

do k=1,maxiter

f2=1/(1.920*log10(reynoldsnumber*sqrt(f1))-0.475-(7.04/(reynoldsnumber*sqrt(f1))**0.55))**2

df=f2-f1

write(11,’(i4,4x,f13.10,2x,f17.15)’)k,abs(df),f2

if(abs(df)<tolerance)exit

ii=ii+1

f1=f2

end do

if(ii<maxiter)write(*,*)Tolerance met (1E-10)’ 2Marks

write(11,*)’Reynolds Number=’,reynoldsnumber 1Mark

if(reynoldsnumber>2500) then

write(11,*) ‘Flow in pipe is under Turbulent Flow’

else

write(11,*) ‘Flow in pipe is under Laminar Flow’

end if

close(11)

end program fricition_factor

Answer

1 0.0753933936 0.024606609717011

2 0.0055772681 0.030183877795935

3 0.0009327773 0.029251100495458

4 0.0001402758 0.029391376301646

5 0.0000214484 0.029369927942753

6 0.0000032727 0.029373200610280

7 0.0000004992 0.029372701421380

8 0.0000000782 0.029372779652476

9 0.0000000149 0.029372764751315

10 0.0000000037 0.029372768476605

11 0.0000000000 0.029372768476605

Tolerance met (1E-10)

Reynolds Number= 12757.8

Flow in pipe is under Turbulent Flow

5Marks

2Marks

1Marks

3Marks

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Question 3

Question 3a:

Line:6 1Marks

Question 3b:

Do-While-Loop 1Marks3Question 6c:

Question 3b:

NO 1Marks

Question 3d

1 PROGRAM hall_yarborough

2 REAL::sg=0.72, temp=640,pre=5000,counter=1,fy,f1y,t,pr,tr,pc,tc,x1,x2,x3.x4,y,y1,y2,Z

1Marks

3 OEPN(10,FILE=’ITERATIONS’) 1Marks

4 !

5 !

6 !

7 tc=169.0+(314.0*sg)

8 pc=708.75-(57.5*sg)

9 tr=temp/tc

10 pr=pres/pc

11 t=1/tr

12 !

13 !

14 !

15 !

16 X1=-0.06125*pr*t*exp(-1.2*(1-t)**2)

17 X2=(14.7*t)-(9.76*(t**2))+(4.58*(t**3))

18 X3=(90.7*t)-(242.2*(t**2))+(42.4*(t**3))

19 X4=2.18+(2.82*t)

20 !

21 !

22 !

23 y1=0.239916681

24 y=y1

25

26

27 DO counter=1,2000

28 fy=x1+((y+(y**2)+(y**3)-(y**4))/((1-y)**3))-(x2*(y**3))+(x3*(y**x4))

29 f1y=((1+(4*y)+(4*(y**2))-(4*(y**3))+(y**4))/((1-y)**4__-(2*x2*y)+(x3*x4*(y**(x4-1)))

30 y2=y-(fy/f1y)

31 WRITE(10,*)y2

32 y1=y

33 END DO

34

1Mark

1Mark

3Mark

of 15

35 Close(10) 1Marks

36

37

38 Z=(0.06125*t*pr/y)*exp(-1.2*(1-t)**2) 1Marks

39

40

41 WRITE(*,*) The Hall-Yarborough Z-factor is@, Z 1Marks

42 END PROGRAM hall_yarborough

Question 4

Question 4(a)(i) [1 mark]

The machine language is a low level interface native or natural to the hardware of the computer. Its

communication symbols consist of series of 0’s and 1’s otherwise known as the binary digit (Bit). Every

computer model has its own machine language which is defined by its hardware structure. Thus, this

language depends on the particular computer model, hence it is said to be machine dependent.

Question 4(a)(ii) [2 marks]

High level computer languages evolved to describe problems independently of the machine code of a

particular computer. HLL (also known as problem oriented languages) use English like phrases to represent

computer instructions. In this way a programmer can write computer instructions (codes) in a way that

resemble his thinking process using English-like expressions. The language is procedure oriented rather

than machine oriented.

They focus on the data processing procedure to be accomplished and not on the coding requirements of a

particular computer machine. Thus, it is machine independent and very easy to use. This means that

programs written in high level language are not restricted to be used with particular computer models.

Therefore, high level languages are said to be generally portable, but require appropriate translators to

translate them into the machine language. Examples of HLL include BASIC, Ada, C, COBOL, Java, Pascal

and FORTRAN

Question 4(a)(iii) [2 marks]

A flowchart is a schematic diagram of the successive steps in an algorithm for solving a problem which the

programmer wishes to have undertaken by the computer or simply put, a flowchart is a symbolic or

diagrammatic representation of the algorithm. As with all schematic approaches, it is necessary to describe

the set of elements from which flowcharts will be built.

An examination of the problems undertaken to date indicates that programs are made up of assignments,

interrogations, stops, directions to move to another step in the program and print/write.

Question 4(b)(i) [2 marks]

An algorithm is a step by step procedure to solve a particular problem or an unambiguous definition of the

actions to be carried out in solving a problem with a programming language.

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Question 4(b)(ii) [2 marks]

A clear knowledge of what inputs there are to be processed and outputs required from the process

as well as the format of the output.

An idea of the best strategy to be used in solving the problem.

Developing an algorithm for the problem and checking to see that the algorithm is executable, and

that it will lead to the appropriate solution or clearly indicate that the problem has no solution.

Turning the algorithm into appropriate high level computer language statement.

Coding, compiling and executing the program using actual data

Question 4(c)(i) [2 marks]

Implicit none

Logical if construct

Exclamation character

Double colon

Variable names up to 31 characters

Statement labels

Ampersand symbol

Free form source code

Question 4(c)(ii) [2 marks]

The order of FORTRAN statements in a program

1. PROGRAM statement

2. IMPLICIT NONE statement

3. Type declaration statements

REAL statement(s)

INTEGER statement(s)

LOGICAL statement(s)

CHARACTER statement(s)

4. Executable statements

Assignment statement(s)

READ statement(s)

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WRITE statement(s)

STOP statement(s)

Question 5

Decomposition [4 marks]

W B C G

4 0 3 25

1 2 2 15

1 3 1 12

1 2 0 5

W B C G

4 0 3 25

-1 2⁄ 12⁄ 2 15

1 3 1 12

1 2 0 5

W B C G

4 0 3 25

-1 2⁄ 12⁄ 2 15

13 -6 1 12

1 2 0 5

W B C G

of 15

4 0 3 25

-1 2⁄ 12⁄ 2 15

13 -6 1 12

1113⁄ 2

13⁄ 0 5

Forward Substitution [3 marks]

W B C G

4 0 3 25

-1 2⁄ 12⁄ 2 5

2⁄

13 -6 1 27

1113⁄ 2

13⁄ 0 1113⁄

Back Substitution [3 marks]

W B C G P

4 0 3 25

-1 2⁄ 12⁄ 2 5

2⁄

13 -6 1 27

1113⁄ 2

13⁄ 0 1113⁄ 1

W B C G P

4 0 3 25 4

-1 2⁄ 12⁄ 2 5

2⁄ 3

13 -6 1 27 2

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1113⁄ 2

13⁄ 0 1113⁄ 1

P=[4, 3, 2, 1] ; [3 marks]

Question 6 [13 marks]

c, pb, p, Vb

n=1; Nfactorial=1; eK=1

Nfactorial=Nfactorial×n

eKn+1=eK

n+c(Pb-P)n/Nfactorial

n>=100

no

yes

start

V=Vb×eK

stop

n=n+1

1


Covenant University Canaanland, Km 10, Idiroko Road

P.M.B 1023, Ota, Ogun State, Nigeria.

TITLE OF EXAMINATION: B.Eng. Omega Semester Examination

COLLEGE: Engineering

DEPARTMENT: Petroleum Engineering

SESSION: 2015/2016 SEMESTER: Omega

COURSE CODE: PET 523 ` CREDIT UNIT: 3

COURSE TITLE: Rock Physics and Formation Engineering

INSTRUCTION: Section A is compulsory, answer any question from sections B & C

TIME: 3 hrs.

SECTION A

QUESTION ONE

A rock with initial dimension, 50mm thick, 80mm wide and 200mm long is subjected to both axial

loading and confining stresses and the final dimension is 30mm thick, 72mm wide and 175mm long.

Given that the hydrostatic stress is 35,000Mpa; calculate.

i. The ratio of hydrostatic stress relative to volume. (2 Marks).

ii. The compressibility of the rock. (1 Mark).

iii. Explain briefly the problems associated with sand production. (4 Marks).

Given that:

and

iv. With the aid of diagram, show that: (8 Marks).

v. What is strain hardening? (1 Mark). vi. Explain the term pore collapse. (1 Mark). vii. List 4 effects of pore pressure on formation failure. (3 Marks).

oS

1 3

1sin 2

2

1 3 1 3

1 1cos 2

2 2

tan

1 3

1 sincos2

1 sin 1 sin

oS

2

QUESTION TWO

During a sonic logging application in a well drilled with a borehole fluid at Ozoro, Delta State, the

distance between the receiver and transmitter was 5m and the transmitter emitted a pulse which

generated a P- wave which travelled through the mud where it was received at the receiver as a reflected

wave, refracted wave and a converted wave (borehole mode). During this application, the following

parameters were obtained. Travel time for refracted wave (V) = 87.0857microsec/ft; Sin θcr = 0.7071,

distance between the receiver and the wall of the formation =100km; Ratio of density of the formation to

density of fluid in the borehole = 2.5; Shear velocity of the formation (VS) = 200,000cm/s.

a. Calculate the Length of the wave (Lw) travelled in the fluid. (2 Marks)

b. Calculate the length of the wave (L’) travelled in the formation. (2 Marks) c. Show that

𝑳𝒘

𝑽𝒘+

𝑳

𝑽

′=

𝑳

𝑽+ 𝟐𝑹(

𝟏

𝑽𝒘 √𝟏 − (𝑽𝒘

𝑽)𝟐

−𝑽𝒘

𝑽𝟐√𝟏 − (𝑽𝒘

𝑽)𝟐

) (7 Marks)

d. Using the answers obtained above, calculate the arrival time (t). (2 Marks) e. Calculate the TUBE WAVE velocity of the borehole mode. (1Mark) f. How does temperature, and chemical effects affect the elastic wave propagation in rocks? (2Marks) g. Write short notes on the following

viii. Thermally induced fracturing (TIF) (2Marks) ix. Fracturing fluids (2Marks)

SECTION B

QUESTION THREE

i. Elastic moduli, elastic limits and strengths of material all are quoted with the same units, Pascal.

Explain the differences between these three physical quantities. (6 Marks).

ii. Comment on the effect of pore fluid on rock failure. (2 Marks).

The state of plane stress at a point is represented by the stress element below.

3

iii. Draw the Mohr’s circle, (5 Marks).

iv. Determine the principal stresses and the maximum in plane shear stresses, (4 Marks).

v. And draw the corresponding stress elements. (3 Marks).

QUESTION FOUR

a) Given the following set of data for the Mohr-Coulomb failure criterion:

TEST 𝝈𝟑(𝑴𝒑𝒂) 𝝈𝟏 (𝑴𝒑𝒂) 𝝈𝟑′(𝑴𝒑𝒂) 𝝈𝟏

′ (𝑴𝒑𝒂)

1 34 329 10 305

2 44 442 20 418

3 54 555 30 531

4 64 668 40 644

5 74 50

𝛼 = 0.96 𝜌𝑓 = 25𝑀𝑝𝑎

i. calculate the angle . (2 Marks).

ii. Calculate the angle of internal friction, . (2 Marks).

iii. what is uniaxial compressive strength oC ? (1 Mark).

iv. Calculate the coefficient of internal friction, . (1 Mark).

v. Calculate the attraction, A . (2 Marks).

vi. Calculate the failure angle, . (2 Marks).

vii. what is the value of 𝜎1′ for Test 5? (3 Marks).

viii. Calculate 𝜎1 using the concept of effective stress. (3 Marks).

ix. Explain the term shale sloughing. (2 Mark).

x. Explain briefly the term differential sticking. (2 Mark).

SECTION C

QUESTION FIVE

For an oil field in the offshoe Dahomey Basin, where a vertical well is drilled to a maximum depth of

243,840cm, pore pressure gradient and average specific gravity (ϒ) are 72(pound/ft2

)/ft and 1.9

respectively, for the surrounding rock formation at the bottom of the well, calculate the following: Take

Biot’s constant = 1; Poisson’s ratio = 0.3.

a. Overburden stress(Psi) (2 Marks)

b. Least horizontal in-situ stress(Psi) (2 Marks)

4

c. Fracture initiation pressure (in psi) when the well is parallel to the intermediate principal

stress (if the well is a horizontal well). Assume Tensile strength = 3,000pound/inch2

(2Marks) d. Using the KGD – Model, calculate (in ft) the maximum width (Wm) of the fracture.

Young’s Modulus = 1,034,299mmHg; Fracture half-length = 20cm; Pressure in the

fracture = 8019 Psi (2 Marks) e. What causes hydraulic fracturing in rocks? (4 Marks) f. In weak rocks, what causes the discrepancy between static and dynamic moduli? (3 Marks) g. List the two classical models for fracture geometry in two dimensions, what are

their common assumptions and differences? (5 Marks)

QUESTION SIX

Consider the diagrams below (dimensions are in millimeters):

Calculate the following:

i. Volumetric strain, vol. (3 marks).

ii. Bulk modulus in GPa (assume hydrostatic stress = 40,000 MPa). (2 marks).

iii. Shear modulus in GPa(assume Young’s modulus = 28,000 MPa). (2 marks).

iv. Poisson’s ratio. (2 marks).

v. Lamé’s parameter in Gpa, . (2 marks).

vi. Rock compressibility (2 marks).

vii. While drilling an 8,000 ft formation of Fiyin-1 well in SPE, Student Chapter Covenant-Field, the

following data are given:

Total overburden stress = 7000 psi Pore fluid pressure = 4,500 psi

Total horizontal stress = 6,240 psi Tangential stress = 6,680 psi

Biot constant = 0.91 Failure angle = 60o

Uniaxial compressive strength = 850 psi

Estimate the minimum well pressure to drill the formation without experiencing collapse.

(2 marks).

viii. Explain the meaning of the terms tensile stress and tensile strain. (5 Mark).

5

Covenant University Canaanland, Km 10, Idiroko Road

P.M.B 1023, Ota, Ogun State, Nigeria.

TITLE OF EXAMINATION: B.Eng. Omega Semester Examination

COLLEGE: Engineering

DEPARTMENT: Petroleum Engineering

SESSION: 2015/2016 SEMESTER: Omega

COURSE CODE: PET 523 ` CREDIT UNIT: 3

COURSE TITLE: Rock Physics and Formation Engineering

INSTRUCTION: Section A is compulsory, answer any question from sections B & C

TIME: 3 hrs.

SECTION A

QUESTION ONE

a) A rock with initial dimension, 50mm thick, 80mm wide and 200mm long is subjected to both

axial loading and confining stresses and the final dimension is 30mm thick, 72mm wide and 175mm

long. Given that the hydrostatic stress is 35,000Mpa; calculate.

𝜀 =∆𝑙

𝑙

𝜀𝑥 =𝑥 − 𝑥′

𝑥

𝜀𝑦 =𝑦 − 𝑦′

𝑦

𝜀𝑧 =𝑧 − 𝑧′

𝑧

𝜺𝒙 =80 − 72

80= 𝟎. 𝟏

𝜺𝒚 =200 − 175

200= 𝟎. 𝟏𝟐𝟓

𝜺𝒛 =50 − 30

50= 𝟎. 𝟒

𝜺𝒗𝒐𝒍 = 0.1 + 0.125 + 0.4 = 𝟎. 𝟔𝟐𝟓

i. The ratio of hydrostatic stress relative to volume strain.

6

𝑩𝒖𝒍𝒌 𝒎𝒐𝒅𝒖𝒍𝒖𝒔, 𝑲 =𝜎𝑝

𝜀𝑣𝑜𝑙

𝑲 = 35,000

0.625= 𝟓𝟔, 𝟎𝟎𝟎𝑴𝒑𝒂 (2 Marks)

ii. The compressibility of the rock.

Compressibility 𝑪 =1

𝐾

𝑪 =1

𝐾

𝑪 = 156000⁄ = 𝟏. 𝟕𝟖𝟔 × 𝟏𝟎−𝟓 (1 Mark)

iii. Explain briefly the problems associated with sand production. (3 Marks)

1. Erosion of the production equipment. This is primarily a safety problem, as well as an

economical one. It is also a typical sand production problem, due to the abrasiveness of the

grains. The erosion caused by a given amount of sand depends strongly on the velocity of the

fluid in the pipeline.

2. Instability of the production cavities and the wellbore itself, which may in extreme cases result in

a complete filling of the borehole (“sand up”) so that the well has to be abandoned.

3. The necessity to handle large amounts of polluted solids at the rig.

b) Given that:

oS 1

1 3

1sin 2

2 2

Failu

M

tan

Slo

cottan

oo

SA S

7

1 3 1 3

1 1cos 2

2 2 3

tan 4

1 3 1 3 1 3

1 1 1sin 2 tan cos2

2 2 2oS

5

Recall 2β=90+φ and substituting into 5

1 3 1 3 1 3

1 1 1sin 90 tan cos 90

2 2 2oS

6

sin 90 sin90cos cos90sin cos

cos 90 cos90cos sin90sin sin

and

Therefore

Equation 6 becomes

1 3 1 3 1 3

1 1 1cos tan sin

2 2 2oS

7

Expanding and applying sin

tancos

into 7

2

1 3 1 3 1 3

1 1 sin 1 sincos

2 2 cos 2 cosoS

8

Multiplying through by 2cos , we have

2 2

1 3 1 3 1 3cos 2 cos sin sinoS 9

Re- arranging and applying 2 2sin cos 1

2 2

1 3 1 3cos sin 2 cos sin sinoS 10

1 3 1 32 cos sin sinoS 11

1 1 3 3sin 2 cos sinoS 12

8

1 31 sin 2 cos 1 sinoS 13

Therefore

(8 Marks)

ii. What is strain hardening? (1 Mark)

Strain hardening is a term used to describe a situation in triaxial compression test whereby with higher

confining pressures the specimen’s ability to support load is not lost, although its stiffness is clearly

reduced.

iii. Explain the term pore collapse. (1 Mark)

Pore collapse is a failure mode that is normally observed in highly porous materials, it is usually caused

by excessive hydrostatic stress. The term "pore collapse" is used to describe a drastic, irreversible

reduction in porosity due to the increase in effective stress on the rock matrix as a consequence of

withdrawal of oil and gas.

iv. List 4 effects of pore pressure on formation failure. (3 Marks)

Shear stress is unaffected by the pore pressure

Minimum & maximum principal stresses are decreased by the same amount.

Radius of the Mohr circle in unchanged.

Center of the circle has shifted to the left.

Circle moves towards the failure line when the fluid pressure is increased for a material obeying

the criterion.

QUESTION TWO

During a sonic logging application in a well drilled with a borehole fluid at Ozoro, Delta State, the

distance between the receiver and transmitter was 5m and the transmitter emitted a pulse which

generated a P- wave which travelled through the mud where it was received at the receiver as a reflected

wave, refracted wave and a converted wave (borehole mode). During this application, the following

parameters were obtained. Travel time for refracted wave (V) = 87.0857microsec/ft; Sin θcr = 0.7071,

distance between the receiver and the wall of the formation =100km; Ratio of density of the formation to

density of fluid in the borehole = 2.5; Shear velocity of the formation (VS) = 200,000cm/s

a. Calculate the Length of the wave (Lw) travelled in the fluid. (2 Mark)

Given, Sin θcr = 0.7071 R = 100km = 100/1000 = 0.1m

θcr = Sin-1

0.7071 = 45o

Lw = 2R/Cos θcr = 2 x 0.1/Cos 45o

= 0.2/0.7071 = 0.2828m = 0.283m = 0.2828 x 100 = 28.28cm = 28.3cm

1 3

1 sincos2

1 sin 1 sin

oS

9

b. Calculate the length of the wave (L’) travelled in the formation. (2 Marks)

L’ = L – 2R Tan θcr = 5 – (2 x 0.1 x 1) = 4.8m = 4.8 x 100 = 480cm

c. Show that (7 Marks) 𝑳𝒘

𝑽𝒘+

𝑳

𝑽

′=

𝑳

𝑽+ 𝟐𝑹(

𝟏

𝑽𝒘 √𝟏 − (𝑽𝒘

𝑽)𝟐

−𝑽𝒘

𝑽𝟐√𝟏 − (𝑽𝒘

𝑽)𝟐

)

Solution

Recall that L’ = L – 2R Tan θcr, Sin θcr = Vw/V, Lw = 2R/ Cosθcr,

𝐿𝑤

𝑉𝑤+

𝐿

𝑉

′ =

𝐿𝑤

𝑉𝑤+

𝐿 − 2𝑅 𝑇𝑎𝑛 θcr

𝑉=

𝐿𝑤

𝑉𝑤+

𝐿

𝑉−

2𝑅 𝑇𝑎𝑛 θcr

𝑉=

𝐿

𝑉+

𝐿𝑤

𝑉𝑤−

2𝑅 𝑇𝑎𝑛 θcr

𝑉

Recall that Tan θcr = Sinθcr / Cos θcr

= 𝐿

𝑉+

𝐿𝑤

𝑉𝑤−

2𝑅𝑆𝑖𝑛 θcr

𝐶𝑜𝑠 θcr

𝑉

= 𝐿

𝑉+

𝐿𝑤

𝑉𝑤−

𝐿𝑤 𝑆𝑖𝑛 θcr

𝑉

= 𝐿

𝑉+

𝐿𝑤

𝑉𝑤−

𝐿𝑤 𝑉𝑤

𝑉2

= 𝐿

𝑉+

2𝑅

𝑉𝑤 𝐶𝑜𝑠 θcr−

2𝑅𝑉𝑤

𝑉2𝐶𝑜𝑠 θcr

= 𝐿

𝑉+ 2𝑅(

1

𝑉𝑤 𝐶𝑜𝑠 θcr−

𝑉𝑤

𝑉2𝐶𝑜𝑠 θcr )

Recall that Sin2θ + Cos

2θ = 1

Cos2θcr = 1 - Sin

2θcr

Cosθ = √1 − 𝑆𝑖𝑛2θcr

Sinθ = 𝑉𝑤

𝑉

Sin2θ = Vw

2 / v

2 = (

𝑉𝑤

𝑉)2

Cosθ = √1 − (𝑉𝑤

𝑉)2

10

= 𝐿

𝑉+ 2𝑅(

1

𝑉𝑤 √1 − (𝑉𝑤

𝑉)2

−𝑉𝑤

𝑉2√1 − (𝑉𝑤

𝑉)2

)

Therefore,

𝑳𝒘

𝑽𝒘+

𝑳

𝑽

′=

𝑳

𝑽+ 𝟐𝑹(

𝟏

𝑽𝒘 √𝟏 − (𝑽𝒘

𝑽)𝟐

−𝑽𝒘

𝑽𝟐√𝟏 − (𝑽𝒘

𝑽)𝟐

)

d. Hence, using the answers obtained above, calculate the arrival time (t) (2 Marks)

𝑡 =𝐿𝑤

𝑉𝑤+

𝐿

𝑉

′

87.0857microsec/ft; = 11482.94 ft/sec = (11482.94 x 0.3048)m/s = 3,500m/s

Sin θcr = Vw/V

Vw = V x Sin θcr = 0.7071 x 3500 = 2,474.85m/s

t = (0.28/2,474.85) + (4.8/3,500) = 0.001486secs.

e. Calculate the TUBE WAVE velocity of the borehole mode. (1 Mark)

Density of the formation (Ƿ)/ Density of fluid in the borehole (Ƿw ) = 2.5

Density of fluid in the borehole (Ƿw )/ Density of the formation (Ƿ) = 1/2.5 = 0.4

Vst = 2,474/ √ [1 + (0.4 x 2474.852

/20002

)]

= 1,948.95m/s

f. How does Temperature, Chemical effects and Partial Saturation affect the elastic wave propagation in

rocks? (2 Marks)

Temperature

The effect on the velocity of propagation of elastic waves is minimal. For rock matrix with little or no

fluids in the pore spaces, the effect of an increase in temperature to 100o

C results in less than 5%

reduction in wave velocity. For rocks with significant amount of pore fluids and that undergo phase

changes within the range of temperature change, the effect of temperature change will be largely noticed.

Since temperature can affect the propagation of elastic waves, it will also affect the decrease in the energy

of waves as they spread out or are propagated radially. This is known as attenuation and it is reduced

with increasing temperature.

11

Chemical Effects

Mineral constituents of rocks can react with the pore fluids thereby causing alterations in rock properties

(hardness, compressive strength and porosity). This means that the moduli of the rock (elastic, bulk and

shear) will subsequently be affected by the chemical changes. Due to this change in the bulk moduli, the

velocity of propagation of the elastic wave will also be greatly affected.

Partial Saturation

The presence of pore fluids determines greatly the elastic response of the porous material. There will be

an increase in P-wave and S-wave velocities for loosely consolidated sediment saturated with water

compared to when it is dry. This is because pore fluid gives increased resistance against compression.

For materials with higher compressive strength or those under high confining pressure, the frame

stiffness is higher and the contribution of the pore fluid is relatively less, hence, the effect of saturation is

much smaller. Since saturation refers to an increase in density and partial saturation, it can result in the

reduction of the P-wave velocity for a stiff rock.

g. Write short notes on the following (2 Marks)

i. Thermally induced fracturing (TIF)

ii. Fracturing fluids

Thermally induced fracturing (TIF)

This is normally observed during cold water injection, especially when there is a large temperature

difference between the water injected and the hot reservoir. The response expected will be a sudden

increase in injectivity after a significant period of stable injection, showing that the reservoir has been

gradually cooled during the injection process. The reservoir then shrinks due to this cooling and the

smallest in-situ stress is reduced to a level below the bottom hole injection pressure and fractures are

thus created. The length of these fractures are limited by the extent of the cooled zone and they provide

a much larger contact area with the formation and hence a dramatic increase in injectivity. This whole

process which leads to the formation of these fractures is termed TIF.

Fracturing fluids & types of proppants

Fracturing fluids contain two key parts – the base fluids and the chemical additives which impart a

predictable set of properties of the fluid including viscosity, friction etc. These give the fracturing fluids

the capacity to achieve the objectives of the fracturing operation. The base fluids can either be slick water

or linear gel, depending on the desired properties. The chemical agents are gelling agents, friction

reducers, emulsifiers, resin activator etc; and they work together to give the fracturing fluid its desired

characteristics.

Fracturing fluids are usually pumped into the well to create conductive fractures and bypass near-

wellbore damage in hydrocarbon bearing zones. The net result is usually an expansion in the productive

surface area of the reservoir compared to the unfractured formation.

12

Proppants are of different sizes and strength, depending on the purpose for which it is deployed and the

nature of the formation in which it is to be used. Types of proppants include, silica sand, resin coated

sand, ceramic proppants and resin-coated ceramic proppants.

SECTION B

QUESTION THREE

i. Elastic moduli, elastic limits and strengths of material all are quoted with the same units, Pascal. Explain

the differences between these three physical quantities. (6 Marks)

Elastic moduli are group of coefficients that describes or give an indication of rock’s resistance against

being compressed by a uniaxial stress, hydrostatic compression and sideway stresses. These parameters

are function of the rock composition and applied force. Elastic limit talks about the maximum stress or

force per unit area within a solid material that can arise before the onset of permanent deformation, it

gives an information about the threshold amount of deformation that a sample can accommodate

without significant change in structure. While strength of material is not a uniquely defined parameter, it

gives an information about the stress level at which a rock typically fails. Rock strength is therefore a

meaningful parameter only when the stress geometry is specified.

ii. Comment on the effect of pore fluid on rock failure. (2 Marks)

Pore fluid can affect the failure of the rock in 2 ways;

Mechanical effect of pore pressure.

Chemical interactions between the rock and the fluid

Some abundant components in rocks, are salts, some clay and carbonate minerals, these components

are dissolvable in water. Over geological time, the solid minerals in a rock will establish a chemical

equilibrium with the pore water, which implies that the minerals dissolve and precipitate at the same

rate. If the pore fluid is changed, for instance during drilling or production, the chemical equilibrium

may be disturbed, and a net dissolution or deposition of minerals may occur. This process may have a

strong effect on the rock properties, typically a reduction in strength.

i. Draw the Mohr’s circle. (5 Marks)

13

ii. Determine the principal stresses and the maximum in plane shear stresses. (4 Marks)

𝐶 = 𝜎𝑎𝑣𝑔 =𝜎𝑥 + 𝜎𝑦

2=

80 + (−50)

2= 𝟑𝟎 𝑴𝒑𝒂

𝑅 = √(𝜎1 − 𝜎𝑎𝑣𝑔)2

+ 𝜏𝑥𝑦2

𝑅 = √((−50) − 30)2 + 252

𝑅 = √6400 + 625

𝑹 = 𝟖𝟑. 𝟖

The Principal stresses are:

𝜎1,2 = 𝐶 ± 𝑅

𝜎1 = 𝐶 + 𝑅

𝜎1 = 30 + 83.8 = 𝟏𝟏𝟑. 𝟖 𝑴𝒑𝒂

𝜎2 = 𝐶 − 𝑅

𝜎2 = 30 − 83.8

𝝈𝟐 = −𝟓𝟑. 𝟖 𝑴𝒑𝒂

Maximum in plane shear stress

𝝉𝒙𝒚 = 𝑹

𝝉𝒙𝒚 = −𝟓𝟑. 𝟖𝑴𝒑𝒂

iii. And draw the corresponding stress elements. (3 Marks)

14

QUESTION FOUR

b) Given the following set of data for the Mohr-Coulomb failure criterion:

TEST 𝝈𝟑(𝑴𝒑𝒂) 𝝈𝟏 (𝑴𝒑𝒂) 𝝈𝟑′(𝑴𝒑𝒂) 𝝈𝟏

′ (𝑴𝒑𝒂)

1 34 329 10 305

2 44 442 20 418

3 54 555 30 531

4 64 668 40 644

5 74 50

𝑆𝑙𝑜𝑝𝑒 =∆𝑦

∆𝑥=

640 − 300

40 − 10= 11.333

a. 𝐭𝐚𝐧 𝜸 = 𝟏𝟏. 𝟑𝟑

0

100

200

300

400

500

600

700

0 10 20 30 40 50 60

Pri

nci

pal

Str

ess

1 M

pa

Principal 3 Mpa

Chart Title

15

𝛾 = 𝑡𝑎𝑛−111.33

𝜸 = 𝟖𝟒. 𝟗𝟓𝒐 (2 Marks)

b. Uniaxial compressive strength, 𝑪𝒐(1 Mark)

𝟏𝟗𝟕 𝑴𝒑𝒂

c. 𝑪𝒐𝒉𝒆𝒔𝒊𝒐𝒏 = 𝑺𝒐

Recall 𝐶𝑜 = 2𝑆𝑜𝑡𝑎𝑛𝛽

𝑡𝑎𝑛2𝛽 = 𝑠𝑙𝑜𝑝𝑒 = 11.33

tan 𝛽 = 3.365

𝛽 = 𝑡𝑎𝑛−13.365

𝛽 = 73.45𝑜

𝑆𝑜 =𝐶𝑜

2𝑡𝑎𝑛𝛽=

197

2(3.365)= 𝟐𝟖. 𝟗𝟕𝟓 𝑴𝒑𝒂

Recall 𝟐𝜷 = 𝟗𝟎 + 𝜽

𝜃 = 2𝛽 − 90

2(73.45) − 90 = 𝟓𝟔. 𝟗𝟎 (2 Marks)

d. Coefficient of Internal Friction. (1 Mark)

𝜇 = 𝑡𝑎𝑛𝜃

tan 56.9 = 𝟏. 𝟓𝟑𝟒

e. Attraction A (2 Marks)

𝑆𝑜 = 𝐴𝑡𝑎𝑛𝜃

𝐴 = 28.975

1.534= 𝟏𝟖. 𝟖𝟖𝟗 𝑴𝒑𝒂

f. Failure angle, (2 Marks)

2𝛽 = 90 + 𝜃

𝛽 = 45 + 𝜃2⁄

𝛽 = 45 + 56.92⁄ = 𝟕𝟑. 𝟒𝟓𝒐

g. The straight line equation is in the form (3 Marks)

𝐶𝑜 + 𝜎3′𝑡𝑎𝑛2𝛽 = 𝜎1

′

𝜎1′ = 197 + 𝜎3

′(11.33)

𝐹𝑜𝑟 𝑡𝑒𝑠𝑡 𝑝𝑜𝑖𝑛𝑡 5, 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑏𝑒𝑐𝑜𝑚𝑒𝑠; 𝜎1

′ = 197 + 50(11.33)

𝝈𝟏′ = 𝟕𝟔𝟏. 𝟎𝟔 𝑴𝒑𝒂

Using the effective stress concept, (3 Marks)

𝝈𝟏′ =?

16

Recall effective stress for saturated rocks;

𝝈𝟏′ = 𝜎1 − 𝛼𝜌𝑓

𝜌𝑓 = 𝑃𝑜𝑟𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒, 𝛼 = 𝐵𝑖𝑜𝑡′𝑠𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝜎1 = 𝜎′1 + 𝛼𝜌𝑓

= 761.06 + (0.96) × 25

= 𝟕𝟖𝟓. 𝟎𝟔 𝑴𝒑𝒂

ix. Explain the term shale sloughing. (2 Marks)

During drilling increase in borehole size occur due to brittle failure and caving of the wellbore wall. If

the cavings are not transported away, this poses a potential source of a stuck pipe situation. This

normally takes place in brittle rocks, but the borehole size may also increase by erosion (hydraulic or

mechanical) in a weak rock. “SLOUGHING SHALE” is often used (although not very well defined) to

describe fragments or “spallings” generated from the borehole wall.

x. Explain briefly the term differential sticking. (2 Marks)

Differential sticking occurs in a permeable reservoir rock. When differentially stuck, the drilling tool is

clamped against the borehole wall as a result of a differential pressure between the well and the

formation. This differential pressure is caused by an overpressure in the borehole, and maintained by a

mud-cake. Since shales have extremely low permeabilities, and mud-cakes do not form on shales, this

mechanism is not possible in shale zones.

SECTION C

QUESTION FIVE

1. For an oil field in the offshore Dahomey Basin, where a vertical well is drilled to a maximum

depth of 243,840cm, pore pressure gradient and average specific gravity (ϒ) are 72(pound/ft2

)/ft and 1.9

respectively, for the surrounding rock formation at the bottom of the well, calculate the following: Take

Biot’s constant = 1; Poisson’s ratio = 0.3.

σV = 0.434ϒd(ft) ------------------------------(x)

σh =[ (𝜈

1−𝜈)( σV – βPO) ] + βPO -----------(y)

PO = Pore pressure (Psi)

h. Overburden stress(σV) (2 Marks)

σV = 0.434ϒd(ft)

30.48cm = 1ft

243,840cm = 8,000ft

σV = 0.434 x 1.9 x 8,000ft

17

= 6,596.8 Psi.

i. Least horizontal in-situ stress (σh) (2 Marks)

σh =[ (𝜈

1−𝜈)( σV – βPO) ] + βPO

1pound/ft2

= 0.006944444Psi

72pound/ft2

= 0.5Psi

Po = 0.5 x 8000ft = 4,000

σh = [ (0.3/1-0.3)(6596.8 – 1 x (4000) ] + (1 x 4000)

σh = 5,112.914Psi

j. Fracture initiation pressure (in psi) when the well is parallel to the intermediate principal

stress (if the well is a horizontal well). Assume Tensile strength = 3,000pound/inch2

(2 Marks).

When the well is parallel to the intermediate principal stress

Fracture initiation pressure (Pfrac.

= 3σh – σv – Po + To)

Pfrac.

= 3 x 5112.914 – 6596.8 – 4000 + 3000

Pfrac.

= 7,741.943Psi.

k. Using the KGD – Model, calculate (in ft) the maximum width (Wm) of the fracture.

Young’s Modulus = 1,034,299mmHg; Fracture half-length = 20cm; Pressure in the

fracture = 8019 Psi (2 Marks).

(Wm) = [4(1- ν2

) LPe ]/ E

Pe = 8019 – 5112.914 = 2906.086Psi.

1mmHg = 0.01933677Psi

1,034,299mmH = 20,000Psi

L = Fracture length = 20cm = (20/30.48)ft = 0.6561ft

(Wm) = [4(1- 0.32

) x 0.6561 x 2906.086]/20,000Psi

(Wm) = 0.347ft = 0.35ft

l. What causes hydraulic fracturing in rocks? (4 Marks)

18

This occurs when fluid pressure within the rocks exceeds the smallest principal stress plus the tensile

strength of the rock. This usually results in tensile failure or splitting of the rock. Hydraulic fracture can

also be initiated by natural, geological processes in the earth whereby the fluid pressure increases and the

smallest principal stress decreases.

m. In weak rocks, what causes the discrepancy between static and dynamic moduli? (3 Marks)

In weak rocks (sandstones) discrepancy between static and dynamic moduli is due to series of local

failure processes on a microscopic scale occurring during the entire loading of the sequence.

n. List the two classical models for fracture geometry in two dimensions, what are their common

assumptions and differences? (5 Marks)

The two classical models are the PKN (Perkins-Kern-Nordgren) Model and KGD(Kristianovitch-

Geertsma-de Klerk) Model

Common Assumptions

The fracture height is constant and independent of the fracture length.

The net pressure at the fracture tip is zero.

Differences

The PKN model is more used in conventional hydraulic fracture model where the fracture is normally

short compared to the height unlike the KGD model.

The KGD model gives a better approximation than the PKN model.

The KGD model assumes strain to be confined to the horizontal plane while the PKN model assumes

vertical plane strain.

19

QUESTION SIX

ix. Volumetric strain, vol. (3 Marks)

𝜺𝒙 =∆𝒙

𝒙=

𝟒 − 𝟏

𝟒= 𝟎. 𝟕𝟓

𝜺𝒚 =∆𝒚

𝒚=

𝟔 − 𝟓

𝟔= 𝟎. 𝟏𝟔𝟔𝟕

𝜺𝒛 =∆𝒛

𝒛=

𝟓 − 𝟑

𝟓= 𝟎. 𝟒

𝜺𝒗𝒐𝒍 = 𝜺𝒙 + 𝜺𝒚 + 𝜺𝒛

𝜺𝒗𝒐𝒍 = 𝟏. 𝟑𝟏𝟔𝟕

x. Bulk modulus in Gpa (assume hydrostatic stress = 40,000 MPa). (2 Marks)

𝐵𝑢𝑙𝑘 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝐾 =𝜎𝑝

𝜀𝑣𝑜𝑙

𝑲 =𝟒𝟎, 𝟎𝟎𝟎 × 𝟏𝟎𝟎𝟎𝟎𝟎𝟎

𝟏. 𝟑𝟏𝟔𝟕 × 𝟏𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎= 𝟑𝟎. 𝟒 𝑮𝒑𝒂

xi. Shear modulus in GPa (assume Young’s modulus = 28,000 MPa). (2 Marks)

𝑬 =𝟗𝑲𝑮

𝟑𝑲 + 𝑮 , 𝑻𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆, 𝑮 =

𝟑𝑬 ∗ 𝑲

𝟗𝑲 − 𝑬

𝑮 =𝟑𝑬 ∗ 𝑲

𝟗𝑲 − 𝑬=

[𝟑 × (𝟐𝟖𝟎𝟎𝟎𝟏𝟎𝟎𝟎 ) × 𝟑𝟎. 𝟒]

[𝟗 × 𝟑𝟎. 𝟒 − (𝟐𝟖𝟎𝟎𝟎𝟏𝟎𝟎𝟎 )]

= 𝟏𝟎. 𝟒 𝑮𝒑𝒂

xii. Poisson’s ratio. (2 Marks)

𝒗 =𝟑𝑲 − 𝑬

𝟔𝑲=

[(𝟑 × 𝟑𝟎. 𝟒) − (𝟐𝟖𝟎𝟎𝟎/𝟏𝟎𝟎𝟎)]

𝟔 × 𝟑𝟎. 𝟒= 𝟎. 𝟑𝟒𝟔𝟒

xiii. Lamé’s parameter in Gpa . (2 marks).

𝝀 =𝑬𝒗

(𝟏 + 𝒗) × (𝟏 − 𝟐𝒗)= [

(𝟐𝟖𝟎𝟎𝟎 × 𝟎. 𝟑𝟒𝟔𝟒/𝟏𝟎𝟎𝟎)

(𝟏 + 𝟎. 𝟑𝟒𝟔𝟒) × (𝟏 − 𝟐 × 𝟎. 𝟑𝟒𝟔𝟒)] = 𝟐𝟑. 𝟒𝑮𝒑𝒂

xiv. Rock compressibility (2 marks).

𝑪 =𝟏

𝑲=

𝟏

𝟑𝟎. 𝟒= 𝟎. 𝟎𝟑𝟐𝟗

xv. While drilling an 8,000 ft formation of Fiyin-1 well in SPE, Student Chapter Covenant-Field, the

following data are given:

Total overburden stress = 7000 psi Pore fluid pressure = 4,500 psi

20

Total horizontal stress = 6,240 psi Tangential stress = 6,680 psi

Biot constant = 0.91 Failure angle = 60o

Uniaxial compressive strength = 850 psi

Estimate the minimum well pressure to drill the formation without experiencing collapse.

2

1

tanw v f o fp p C p

2

17000 0.91*4500 850 0.91*4500

tan 60

4780

w

w

p

p psi

(2 marks).

xvi. Explain the meaning of the terms tensile stress and tensile strain. (5 Marks).

TENSILE STRESS (or tension) is the stress state leading to expansion; that is, the length of a material

tends to increase in the tensile direction. The volume of the material stays constant. When equal and

opposite forces are applied on a body, then the stress due to this force is called tensile stress.

TENSILE STRAIN is the relative elongation of a material when subjected to stress. A strain is measure

of deformation representing the displacement between particles in the body relative to a reference

length.

1


COVENANT UNIVERSITY


P.M.B 1023, OTA, OGUN STATE, NIGERIA TITLE OF EXAMINATION: B.Eng. EXAMINATION



SESSION: 2015/2016 SEMESTER: ALPHA


COURSE TITLE: RESERVOIR MODELING AND SIMULATION

INSTRUCTION: ATTEMPT TO ANSWER ANY THREE (3) QUESTIONS IN SECTION A AND

ANY ONE (1) QUESTION IN SECTION B TIME: 3 HOURS

SECTION A: Attempt to answer any three (3) questions here.

The data in Table SEC-A is pertinent to Re`servoir TTOWG and may be useful in answering questions in this

section.

Table SEC-A

Length of the reservoir, in x-direction (ft) 3000

Length of the reservoir, in y-direction (ft) 3000

Thickness of the reservoir (ft) 50

Reservoir Porosity (fraction) (incompressible) 0.3

Reservoir Permeability in x-direction (mD) 200

Reservoir Permeability in y-direction (mD) 400

Initial Water Saturation (fraction) 0.2

Initial Reservoir Pressure (psi) 4000

Bubble point Pressure (psi) 3330

Fluid (Oil) Compressibility (/psi) (slightly-compressible) 0.0000113

Oil Formation Volume Factor at Initial Pressure (RB/STB) 1.2417

Oil Formation Volume Factor at Bubble-point Pressure (RB/STB) 1.2511

Fluid (Oil) Viscosity (cp) 0.5

Question 1: (15 marks)

a. Using relevant equation(s) highlight the source of the integral terms that featured in the course of

formulating reservoir fluid flow model. Using appropriate sketches and equations show two (2)

approximation methods by which these integrals are evaluated. (3 marks)

b. Highlight the occurrence of non-linearity on the LHS and the RHS of the flow equation for slightly

compressible fluids. How are these non-linearities treated on each sides of the equation? (2 marks)

c. At the South boundary, Reservoir TTOWG loses fluid at a rate 500STB/D while a strong aquifer

adjoining the East boundary maintains pressure thereat; other boundaries are sealed to flow. Give a well-

2

labelled sketch of a 4×5 block-centered 2-D (x-y) naturally-ordered discretized model of the reservoir. (3

marks)

d. Starting with the compact form flow model, write the flow model for Block 𝑛𝑥 in the discretized model

you proffered in 1c; be sure to calculate all possible LHS parameters of the model. (5 marks)

e. What range of index i would you specify in writing a for loop code to implement IDQ-1 below? Justify

your range. (2 marks)

∆𝑥𝑖−12⁄ = 1

2⁄ (∆𝑥𝑖 + ∆𝑥𝑖−1) − − − − − − − −𝐼𝐷𝑄 − 1


a. Using relevant equation(s) in the formulation of reservoir flow model highlight the source of the Area, A

term that featured in the flow transmissibilities. Give equation to compute such Area term given that the

fluid flows in the vertical direction. (3 marks)

b. Suggest features that may give rise to no-flow boundary condition and to specified boundary pressure

condition. (2 marks)

c. Given that Reservoir TTOWG is bounded on all sides by no-flow boundaries and is being produced by

wells J-1 described in Table TQ-2. Using engineering ordering, give a neat sketch of a 6×3 block-

centered 2-D (x-y) discretized model of the reservoir; indicate the locations of the well. (3 marks)

Table TQ-2

Well Name Well Location Well Operating Parameters

J-1 1250ft from East Boundary

1500ft from South Boundary

Constant Rate: 1000STB/D

Well Diameter: 7inch

d. Starting with the compact form flow model, write the flow model for Block 𝑛𝑥 + 4 in the discretized

model you proffered in 2c; be sure to calculate all possible LHS parameters of the model. (5 marks)

e. Code line CLQ-2 below is excerpted from the source codes of TTOWGSIM flow simulator and it is to

set the just-obtained (n+1)th pressures to nth pressures. What is the purpose of including this setting, and

at what point is this line to be executed? (2 marks)

Plast = Pnow; - - - - - - - - - - - - - -CLQ-2


a. Highlight two (2) common misapplications of reservoir simulation. (2 marks)

b. Using relevant equation(s) in the formulation of reservoir flow model highlight the source of the B (oil

FVF) parameter in the model. Justify the absence of the B parameter in the source/sink term of the

model. (3 marks)

c. If Reservoir TTOWG is known to contain a slightly-compressible fluid; produced by a single well and is

bounded on all sides by no-flow boundaries; give a well-labelled sketch of a 6×4 block-centered 2-D (x-y)

naturally-ordered discretized model of the reservoir. (3 marks)

d. Using time step of 10 days, write the implicit flow model for Block 𝑛𝑥 + 3 (this block host the well

producing at 400STB/D). Be sure to calculate ALL possible parameters of the model and to separate all

unknown terms to the LHS of the equation. (4 marks)

e. Assuming the solution of the model at a previous time step yielded a pressure value of 3980.75psi for

Block 𝑛𝑥 + 3, re-write the implicit flow model for this block taking into consideration this new

information. (1 mark)

f. What range of index i would you specify in writing a for loop code to implement IDQ-3 below? Justify


∆𝑥𝑖+12⁄ = 1

2⁄ (∆𝑥𝑖 + ∆𝑥𝑖+1) − − − − − − − −𝐼𝐷𝑄 − 3


a. How is the challenge of heterogeneity handled in reservoir simulation? What relief is granted upon

handling the challenge that manner? At what costs is the relief granted? (2marks)

3

b. Pressure is the unknown variable in the reservoir flow model. Using relevant equation(s), demonstrate

the source of the pressure variable and the outcome of introducing such equation(s) into the

formulation. (3 marks)

c. Given that reservoir TTOWG has no significant heterogeneity in y-direction and has therefore been

discretized into 1-D x-direction model; give a well-labelled sketch of a 4×1 block-centered naturally-

ordered discretized model of the reservoir. (1 mark)

d. Given that such model as you proffered in Question 4c has been simulated to yield pressure values for

all four blocks at April 14, 2016 as indicated in Table TQ-4. Compute the total cumulative oil produced

from the reservoir at April 14, 2016 Assume rock and water are of zero compressibility; also assume no

water is produced or injected. (7 marks)

Table TQ-4

Block Number Thursday April 14, 2016

1 3993.75

2 3980.75

3 3966.24

4 3949.10

e. What range of index i would you specify in writing a for loop code to implement IDQ-3 below? Justify


𝑥𝑖+1 = 𝑥𝑖 + ∆𝑥𝑖+12⁄ − − − − − − − − − − − 𝐼𝐷𝑄 − 4

SECTION B: Attempt to answer any one (1) question here.


a. Why is well placement important in reservoir engineering?

b. Outline and discuss five technical factors which affect well placement in reservoir engineering.

c. What are three advantages of using a horizontal well over a vertical well?

d. List three advantages of using a vertical well over a horizontal well in the development of a field.

e. Figure FQ-5 below shows the simulation result of a water flooded reservoir. Place 2 additional Producer

wells and 2 injector wells on the reservoir shown in figure FQ-5 below; do support your preferred well

locations with relevant engineering judgment.


The grid below (Figure FQ-6) represents an initially under saturated oil reservoir with an aquifer below. You

have been selected as a development petroleum engineer to optimally locate 5 production wells and 5 injection

wells on the structure.

a. The first development option proposes a balanced injection system such that the reservoir pressure is

maintained above the bubble point. This is ensured using water injection. Indicate your preferred

location for the producer wells using the symbol P and the injector wells using the symbol I. Support

your preferred locations with relevant technical discussion keeping in mind the anticipated recovery

mechanism

b. A second development option is to drill 6 producer wells and have the field produce by primary

depletion for five years down below the bubble point before gas injection using 3 gas injection wells

commences through a suitable location. Propose the locations and trajectories for the 6 producers and 3

gas injector wells with a detailed technical argument.

c. What keywords are used in placing wells in Eclipse? State three details contained under each keyword.

4

Best Wishes, always!!!

Figure FQ-5

Figure FQ-6

5

COVENANT UNIVERSITY



TITLE OF EXAMINATION: B.Eng. EXAMINATION




COURSE CODE: PET524 CREDIT UNIT: 3

COURSE TITLE: RESERVOIR MODELING AND SIMULATION

MARKING GUIDE

SECTION A


f.

Topic Covered:

Formulation (derivation of reservoir fluid flow model

Answer Keywords:

Conversion of mass terms to mass flowrate terms; implicit method and explicit method

Answer:

The source of the integral is in the step wherein mass terms are converted to mass rate terms as shown

below.

𝑚 = ∫ 𝑤𝑑𝑡𝑡𝑛+1

𝑡𝑛

The integral terms are evaluated by the following approximation methods.

i. Explicit method

∫ 𝐹𝑑𝑡𝑡𝑛+1

𝑡𝑛= 𝐹𝑛∆𝑡

6

ii. Implicit method

∫ 𝐹𝑑𝑡𝑡𝑛+1

𝑡𝑛= 𝐹𝑛+1∆𝑡

g.

Topic Covered:

Formulation of reservoir fluid models

Answer Keywords:

Viscosity, oil formulation volume factor, dependence on pressure

Answer:

Non-linearity on the LHS is due to the presence of pressure-dependent parameters such viscosity and oil

formation volume factor in the coefficients of pressure variable.

Non-linearity in on the RHS is due to the occurrence of pressure-dependent properties such as porosity

and oil formation volume factor on the RHS.

The non-linearity on the LHS is treated by assuming weak dependence of viscosity and oil formation

volume factor; hence both parameters are taken to be constant.

The non-linearity on the RHS is treated by substituting the following equations.

𝐵 =𝐵𝑜

[1 + 𝑐(𝑃 − 𝑃𝑜)]

𝜙 = 𝜙𝑜[1 + 𝑐𝑟(𝑃 − 𝑃𝑜)]

h.

Topic Covered:

Reservoir discretization; flow model in compact form; treatment of boundary conditions

Answer Keywords:

Not applicable

7

Answer:

17 18 19 20

13 14 15 16

9 10 11 12

5 6 7 8

1 2 3 4

i.

Topic Covered:

As in 1(c)

Answer Keywords:

Not applicable

Answer:

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑙𝑜𝑐𝑘𝑠 𝑖𝑛 𝑥 − 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛, 𝑛𝑥 = 4 ∴ 𝐵𝑙𝑜𝑐𝑘 𝑛𝑥 = 𝐵𝑙𝑜𝑐𝑘 4

Flow equation for Block 4:

∑ 𝑇𝑙,𝑛𝑚 (P𝑙

𝑚 − P𝑛𝑚)

𝑙∈𝜓𝑛

+ ∑ 𝑞𝑠𝑐𝑙,𝑛𝑚

𝑙∈𝜉𝑛

+ 𝑞𝑠𝑐𝑛𝑚 =

𝑉𝑏𝑛

𝛼𝑐Δ𝑡[(

𝜙

𝐵)

𝑛

𝑛+1

− (𝜙

𝐵)

𝑛

𝑛

]

𝑞𝑠𝑐𝑏𝑆,4𝑚 + 𝑇𝑥3,4

𝑚 (P3𝑚 − P4

𝑚)+𝑇𝑦8,4

𝑚 (P8𝑚 − P4

𝑚) + 𝑞𝑠𝑐𝑏𝐸,4𝑚 =

𝑉𝑏4

𝛼𝑐Δ𝑡[(

𝜙

𝐵)

4

𝑛+1

− (𝜙

𝐵)

4

𝑛

]

𝑞𝑠𝑐𝑏𝑆,4𝑚 =

−500

4= −125𝑆𝑇𝐵/𝐷

bE (PbE = 4000psi)

750ft

qscbs = 500STB/D

3000ft

8

𝑇𝑥3,4𝑚 = 𝛽𝑐

𝐾𝑥𝐴𝑥

𝜇𝐵∆𝑥

∆𝑥 =3000

4= 750𝑓𝑡

∆𝑦 =3000

5= 600𝑓𝑡

∆𝑧 =50

1= 50𝑓𝑡

𝐴𝑥 = ∆𝑦. ∆𝑧 = 600 × 50 = 30,000𝑓𝑡2

𝑇𝑥3,4𝑚 = 0.001127 ×

200 × 30,000

0.5 × 1.2417 × 750= 14.52𝑆𝑇𝐵/𝐷. 𝑝𝑠𝑖

𝑇𝑦8,4

𝑚 = 𝛽𝑐

𝐾𝑦𝐴𝑦

𝜇𝐵∆𝑦

𝐴𝑦 = ∆𝑥. ∆𝑧 = 750 × 50 = 37,500𝑓𝑡2

𝑇𝑦8,4

𝑚 = 0.001127 ×400 × 37,500

0.5 × 1.2417 × 600= 45.38𝑆𝑇𝐵/𝐷. 𝑝𝑠𝑖

𝑞𝑠𝑐𝑏𝐸,4𝑚 = 𝑇𝑏𝐸,4

𝑚(P𝑏𝐸𝑚 − P4

𝑚)

𝑇𝑏𝐸,4𝑚 = 𝛽𝑐

𝐾𝑥𝐴𝑥

𝜇𝐵 ∆𝑥2⁄

𝑇𝑏𝐸,4𝑚 = 0.001127 ×

200 × 30,000

0.5 × 1.2417 × 7502⁄

= 29.04𝑆𝑇𝐵/𝐷. 𝑝𝑠𝑖


−125 + 14.52(P3𝑚 − P4

𝑚) + 45.38(P8𝑚 − P4

𝑚) + 29.04(4000 − P4𝑚) =

𝑉𝑏4

𝛼𝑐Δ𝑡[(

𝜙

𝐵)

4

𝑛+1

− (𝜙

𝐵)

4

𝑛

]

14.52P3𝑚 − 88.94P4

𝑚 + 45.38P8𝑚 + 116,035 =

𝑉𝑏4

𝛼𝑐Δ𝑡[(

𝜙

𝐵)

4

𝑛+1

− (𝜙

𝐵)

4

𝑛

]

j.

Topic Covered:

Reservoir discretization

9

Answer Keywords:

Range i = 2 to nx; no left neighbor for Block 1.

Answer:

I would specify range i = 2 to nx.

Justification: ∆𝑥𝑖−12⁄ refers to the distance between a point i and the neighboring point to the left, i-1. The first

block, i.e. Block 1 has no neighbor to the left, hence IDQ-1 cannot be implemented for Block 1; therefore, the

implementation must start at Block 2.


a.

Topic Covered:

Formulation of reservoir fluid flow model

Answer Keywords:

Conversion of mass flowrate terms to mass flux terms

Answer:

The source of the Area, A term is in the step wherein mass flowrate terms are converted to mass flux terms.

𝑤 = �̇�𝐴

Area in vertical direction:

𝐴𝑧 = ∆𝑥. ∆𝑦

b.

Topic Covered:

Implementation of boundary conditions

Answer Keywords:

Sealing faults; meeting point of two drainage areas; impermeable layer; strong aquifer

Answer:

Features that may give rise to no-flow boundary conditions are:

1. Sealing faults

2. The meeting points of two drainage areas.

3. Impermeable layer.

Features that may give rise to no-flow boundary conditions are:

1. Presence of a strong aquifer

2. Wells producing at specified bottom-hole flowing pressure

c.

Topic Covered:

10

Reservoir discretization; flow model in compact form; well models

Answer Keywords:

Not applicable

Answer:

1,3 2,3 3,3 4,3 5,3 6,3

1,2 2,2 3,2 4,2

5,2 6,2

1,1 2,1 3,1 4,1 5,1 6,1

Location of Well J-1:

1250ft from East boundary = Column 4

1500ft from South boundary = Row 2

Hence, Well J-1 is located at Block 4,2.

d.

Topic Covered:

As in Question 2(c)

Answer Keywords:

Not applicable

Answer:

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑙𝑜𝑐𝑘𝑠 𝑖𝑛 𝑥 − 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛, 𝑛𝑥 = 6 ∴ 𝑛𝑥 + 4 = 10

𝐻𝑒𝑛𝑐𝑒𝐵𝑙𝑜𝑐𝑘 𝑛𝑥 + 4 = 𝐵𝑙𝑜𝑐𝑘 10 = 𝐵𝑙𝑜𝑐𝑘 4,2

Flow equation for Block 4,2:

3000ft

500ft

11

∑ 𝑇𝑙,𝑛𝑚 (P𝑙

𝑚 − P𝑛𝑚)

𝑙∈𝜓𝑛

+ ∑ 𝑞𝑠𝑐𝑙,𝑛𝑚

𝑙∈𝜉𝑛

+ 𝑞𝑠𝑐𝑛𝑚 =

𝑉𝑏𝑛

𝛼𝑐Δ𝑡[(

𝜙

𝐵)

𝑛

𝑛+1

− (𝜙

𝐵)

𝑛

𝑛

]

𝑇𝑦(4,1),(4,2)

𝑚 (P4,1𝑚 − P4,2

𝑚 ) + 𝑇𝑥(3,2),(4,2)𝑚 (P3,2

𝑚 − P4,2𝑚 ) + 𝑇𝑥(5,2),(4,2)

𝑚 (P5,2𝑚 − P4,2

𝑚 ) + 𝑇𝑦(4,3),(4,2)

𝑚 (P4,3𝑚 − P4,2

𝑚 ) − 1000

=𝑉𝑏4,2

𝛼𝑐Δ𝑡[(

𝜙

𝐵)

4,2

𝑛+1

− (𝜙

𝐵)

4,2

𝑛

]

𝑇𝑦4,2−12⁄

𝑚 (P4,1𝑚 − P4,2

𝑚 ) + 𝑇𝑥4−12⁄ ,2

𝑚 (P3,2𝑚 − P4,2

𝑚 ) + 𝑇𝑥4+12⁄ ,2

𝑚 (P5,2𝑚 − P4,2

𝑚 ) + 𝑇𝑦4,2+12⁄

𝑚 (P4,3𝑚 − P4,2

𝑚 ) − 1000

=𝑉𝑏4,2

𝛼𝑐Δ𝑡[(

𝜙

𝐵)

4,2

𝑛+1

− (𝜙

𝐵)

4,2

𝑛

]

𝑇𝑥4−12⁄ ,2

𝑚 = 𝑇𝑥4+12⁄ ,2

𝑚 = 𝛽𝑐

𝐾𝑥𝐴𝑥

𝜇𝐵∆𝑥

∆𝑥 =3000

6= 500𝑓𝑡

∆𝑦 =3000

3= 1000𝑓𝑡

∆𝑧 =50

1= 50𝑓𝑡

𝐴𝑥 = ∆𝑦. ∆𝑧 = 1000 × 50 = 50,000𝑓𝑡2

𝑇𝑥4−12⁄ ,2

𝑚 = 𝑇𝑥4+12⁄ ,2

𝑚 = 0.001127 ×200 × 50,000

0.5 × 1.2417 × 500= 36.3𝑆𝑇𝐵/𝐷. 𝑝𝑠𝑖

𝑇𝑦4,2−12⁄

𝑚 = 𝑇𝑦4,2+12⁄

𝑚 = 𝛽𝑐

𝐾𝑦𝐴𝑦

𝜇𝐵∆𝑦

𝐴𝑦 = ∆𝑥. ∆𝑧 = 500 × 50 = 25,000𝑓𝑡2

𝑇𝑦4,2−12⁄

𝑚 = 𝑇𝑦4,2+12⁄

𝑚 = 0.001127 ×400 × 25,000

0.5 × 1.2417 × 1000= 18.15𝑆𝑇𝐵/𝐷. 𝑝𝑠𝑖

Flow equation for Block 4,2:

18.15(P4,1𝑚 − P4,2

𝑚 ) + 36.3(P3,2𝑚 − P4,2

𝑚 ) + 36.3(P5,2𝑚 − P4,2

𝑚 ) + 18.15(P4,3𝑚 − P4,2

𝑚 ) − 1000

=𝑉𝑏4,2

𝛼𝑐Δ𝑡[(

𝜙

𝐵)

4,2

𝑛+1

− (𝜙

𝐵)

4,2

𝑛

]

18.15P4,1𝑚 + 36.3P3,2

𝑚 − 108.9P4,2𝑚 + 36.3P5,2

𝑚 + 18.5P4,3𝑚 − 1000 =

𝑉𝑏𝑛

𝛼𝑐Δ𝑡[(

𝜙

𝐵)

𝑛

𝑛+1

− (𝜙

𝐵)

𝑛

𝑛

]

12

e.

Topic Covered:

Floe equation for slightly compressible fluids

Answer Keywords:

Advancing simulation; end of each time step

Answer:

The purpose of setting the just-obtained (n+1)th pressures to nth pressures is to advance the simulation to the

next time node.

The line of code is executed at the end of each time step.


g.

Topic Covered:

Introduction: Application of reservoir simulation

Answer keywords:

As in answers below

Answer:

Common misapplications of reservoir simulation are as follows:

1. Use of inappropriate models.

2. Use of poor quality data

3. Over-definition of problem

4. Over-reliance on numerical results

5. Use of reservoir simulation in situations that can be handled by less-expensive tools

h.

Topic Covered:

Formulation of reservoir fluid flow models

Answer keywords:

Conversion of density terms to standard density terms

Answer:

The source of the B parameter is in the step wherein density terms are converted to standard density terms.

𝜌 =𝜌𝑠𝑐

𝐵

13

Another source is in the conversion of well flowrate to standard well flowrate

𝑞 = 𝐵𝑞𝑠𝑐

Parameter B is absent in the source/sink term because there are two conversions done there involving the two

equations above. The B in the denominator helps to cancel the B in the numerator.

𝜌𝑞 =𝜌𝑠𝑐

𝐵. 𝐵𝑞𝑠𝑐 = 𝜌𝑠𝑐𝑞𝑠𝑐

i.

Topic Covered:

Reservoir discretization; flow model in compact form; flow models for slightly compressible fluids.

Answer Keywords:

Not applicable

Answer:

19 20 21 22 23 24

13 14 15 16 17 18

7 8 9

10 11 12

1 2 3 4 5 6

j.

Topic Covered:

As in Question 3c

Answer Keywords:

3000ft

500ft

14

Not applicable

Answer:

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑙𝑜𝑐𝑘𝑠 𝑖𝑛 𝑥 − 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛, 𝑛𝑥 = 6 ∴ 𝑛𝑥 + 3 = 9

𝐻𝑒𝑛𝑐𝑒 𝐵𝑙𝑜𝑐𝑘 𝑛𝑥 + 3 = 𝐵𝑙𝑜𝑐𝑘 9

Implicit flow model for Block 9:

∑ 𝑇𝑙,𝑛(P𝑙𝑛+1 − P𝑛

𝑛+1)

𝑙∈𝜓𝑛

+ ∑ 𝑞𝑠𝑐𝑙,𝑛𝑛+1

𝑙∈𝜉𝑛

+ 𝑞𝑠𝑐𝑛𝑛+1 =

𝑉𝑏𝑛𝜙𝑐

𝛼𝑐Δ𝑡𝐵𝑜𝑏

[𝑃𝑛𝑛+1 − 𝑃𝑛

𝑛]

𝑇𝑦3,9(P3

𝑛+1 − P9𝑛+1) + 𝑇𝑥8,9

(P8𝑛+1 − P9

𝑛+1) + 𝑇𝑥10,9(P10

𝑛+1 − P9𝑛+1) + 𝑇𝑦15,9

(P15𝑛+1 − P9

𝑛+1) − 400

=𝑉𝑏9

𝜙𝑐

5.615Δ𝑡𝐵𝑜𝑏

[𝑃9𝑛+1 − 𝑃9

𝑛]

𝑇𝑥8,9= 𝑇𝑥10,9

= 𝛽𝑐

𝐾𝑥𝐴𝑥

𝜇𝐵∆𝑥

∆𝑥 =3000

6= 500𝑓𝑡

∆𝑦 =3000

4= 750𝑓𝑡

∆𝑧 =50

1= 50𝑓𝑡

𝐴𝑥 = ∆𝑦. ∆𝑧 = 750 × 50 = 37,500𝑓𝑡2

𝑇𝑥8,9= 𝑇𝑥10,9

= 0.001127 ×200 × 37,500

0.5 × 1.2417 × 500= 27.23𝑆𝑇𝐵/𝐷. 𝑝𝑠𝑖

𝑇𝑦3,9= 𝑇𝑦15,9

= 𝛽𝑐

𝐾𝑦𝐴𝑦

𝜇𝐵∆𝑦

𝐴𝑦 = ∆𝑥. ∆𝑧 = 500 × 50 = 25,000𝑓𝑡2

𝑇𝑦3,9= 𝑇𝑦15,9

= 0.001127 ×400 × 25,000

0.5 × 1.2417 × 750= 24.2𝑆𝑇𝐵/𝐷. 𝑝𝑠𝑖

𝑉𝑏 = ∆𝑥. ∆𝑦. ∆𝑧 = 500 × 750 × 50 = 18,750,000𝑓𝑡3

𝜙 = 0.3

15

𝑐 = 0.0000113𝑝𝑠𝑖−1

∆𝑡 = 10 𝑑𝑎𝑦𝑠

𝐵𝑜𝑏 = 1.2511𝑅𝐵/𝑆𝑇𝐵


24.2(P3𝑛+1 − P9

𝑛+1) + 27.3(P8𝑛+1 − P9

𝑛+1) + 27.3(P10𝑛+1 − P9

𝑛+1) + 24.2(P15𝑛+1 − P9

𝑛+1) − 400

=18,750,000 × 0.3 × 0.0000113

5.615 × 10 × 1.2511[𝑃9

𝑛+1 − 𝑃9𝑛]

24.2P3𝑛+1 + 27.3P8

𝑛+1 − 102.86P9𝑛+1 + 27.3P10

𝑛+1 + 24.2P15𝑛+1 − 400 = 0.9[𝑃9

𝑛+1 − 𝑃9𝑛]

24.2P3𝑛+1 + 27.3P8

𝑛+1 − 103.76P9𝑛+1 + 27.3P10

𝑛+1 + 24.2P15𝑛+1 = 400 − 0.9𝑃9

𝑛

k.

Topic covered:

As is Question 3(c)

Answer Keywords:

Not applicable

Answer

Substituting 𝑃9𝑛 = 3980.75 into the RHS of the flow model obtained in Question 3(d):

24.2P3𝑛+1 + 27.3P8

𝑛+1 − 103.76P9𝑛+1 + 27.3P10

𝑛+1 + 24.2P15𝑛+1 = 400 − 0.9 × 3980.75

24.2P3𝑛+1 + 27.3P8

𝑛+1 − 103.76P9𝑛+1 + 27.3P10

𝑛+1 + 24.2P15𝑛+1 = −3,182.7

l.

Topic covered:

Reservoir discretization

Answer Keywords:

Range i = 1 to nx-1; no right neighbor for Block nx

16

Answer:

I would specify range i = 1 to nx-1.

Justification: ∆𝑥𝑖+12⁄ refers to the distance between point i and the neighboring point to the right, i=1. The last

block, i.e. Block nx has no neighbor to the left, hence IDQ-3 cannot be implemented for Block nx; therefore,

the implementation must end at Block nx-1.


a.

Topic covered:

Introduction: Handling Heterogeneity

Answer Keywords:

Discretization; rock properties no more function of space; flow equations to be solved for all blocks.

Answer:

Heterogeneity is handled by discretizing the reservoir into gridblocks such that for a given block, a single value

for a rock property is locally acceptable.

The relief granted upon discretization is that rock properties are no more functions of space.

The cost incurred upon discretization is that the flow equation would then need to be solved for each of the

several blocks.

b.

Topic covered:

Formulation of Reservoir Fluid Flow Model

Answer Keywords:

Introduction of Darcy equation through volumetric flux terms.

Answer:

The source of the pressure variable is in the step wherein Darcy equation is introduced into the formulation

through the volumetric flux terms.

17

𝑈𝑥 = 𝛽𝑐

𝐾𝑥(𝑃𝑖±1 − 𝑃𝑖)

𝜇∆𝑥

Outcome of introducing Darcy equation:

∫𝑈𝐴

𝐵|

𝑥𝑖−12⁄

𝑑𝑡𝑡𝑛+1

𝑡𝑛− ∫

𝑈𝐴

𝐵|

𝑥𝑖+12⁄

𝑑𝑡𝑡𝑛+1

𝑡𝑛− ∫ 𝑞𝑠𝑐𝑑𝑡

𝑡𝑛+1

𝑡𝑛=

𝑉𝑏𝑖

𝛼𝑐[(

𝜙

𝐵)

𝑖

𝑛+1

− (𝜙

𝐵)

𝑖

𝑛

]

𝑈𝑥𝑖−12⁄

= 𝛽𝑐

𝐾𝑥(𝑃𝑖−1 − 𝑃𝑖)

𝜇∆𝑥𝑖−12⁄

𝑈𝑥𝑖+12⁄

= 𝛽𝑐

𝐾𝑥(𝑃𝑖 − 𝑃𝑖+1)

𝜇∆𝑥𝑖+12⁄

∫ 𝛽𝑐

𝐾𝑥𝐴𝑥

𝜇𝐵∆𝑥|

𝑥𝑖−12⁄

(𝑃𝑖 − 𝑃𝑖+1) 𝑑𝑡𝑡𝑛+1

𝑡𝑛− ∫ 𝛽𝑐

𝐾𝑥𝐴𝑥

𝜇𝐵∆𝑥|

𝑥𝑖+12⁄

(𝑃𝑖 − 𝑃𝑖+1) 𝑑𝑡𝑡𝑛+1

𝑡𝑛− ∫ 𝑞𝑠𝑐𝑑𝑡

𝑡𝑛+1

𝑡𝑛

= 𝑉𝑏𝑖

𝛼𝑐[(

𝜙

𝐵)

𝑖

𝑛+1

− (𝜙

𝐵)

𝑖

𝑛

]

c.

Topic covered:

Reservoir Discretization

Answer Keywords:

Not applicable.

Answer:

18

d.

Topic covered:

Volumetric Computations in Reservoir Simulation

Answer Keywords:

Not applicable.

Answer:

𝑁𝑝𝑖=

𝑁𝑖𝐵𝑜𝑖𝑐𝑒(𝑃𝑖 − 𝑃𝑛𝑜𝑤)

𝐵𝑜

𝑁𝑖 = ∆𝑥∆𝑦∆𝑧𝜙(1 − 𝑆𝑤𝑖)

5.615𝐵𝑜𝑖

∆𝑥 = 3000

4= 750

∆𝑥 = 3000

1= 3000

∆𝑥 = 50

1= 50

1 2 3 4

750ft

3000ft

19

𝑁𝑖 = 750 × 3000 × 50 × 0.3(1 − 0.2)

5.615 × 1.2417

𝑁𝑖 = 3,872,553 𝑆𝑇𝐵

𝑐𝑒 =(1 − 𝑆𝑤𝑖)𝑐𝑜 + 𝑐𝑓 + 𝑆𝑤𝑖𝑐𝑤

1 − 𝑆𝑤𝑖

𝑐𝑓 = 𝑐𝑤 = 0

𝑐𝑒 =(1 − 0.2) × 0.0000113

1 − 0.2= 0.0000113

𝐵𝑜 = 𝐵𝑜𝑏[1 − 𝑐𝑜(𝑃 − 𝑃𝑏)]

𝐵𝑜 = 1.2511[1 − 0.0000113 × (𝑃 − 3330)]

𝐵𝑜 = 1.298 − 0.0000141𝑃 − − − − − − − −1

Using Equation 1 above, the 𝐵𝑜 corresponding to each of the pressure values is computed and tabulated below

Block Pressure (14/04/’16) Bo (14/04/’16)

1 3993.75 1.2417

2 3980.75 1.2419

3 3966.24 1.2421

4 3949.10 1.2423

𝑁𝑝𝑖=

𝑁𝑖𝐵𝑜𝑖𝑐𝑒(𝑃𝑖 − 𝑃𝑛𝑜𝑤)

𝐵𝑜

20

𝑁𝑝𝑖=

3,872,553 × 1.2417 × 0.0000113 × (4000 − 𝑃𝑛𝑜𝑤)

𝐵𝑜

𝑁𝑝𝑖=

217346.4 − 54.3𝑃𝑛𝑜𝑤

𝐵𝑜− − − − − − − − − 2

Equation 2 is used to compute cumulative oil produced in each block; the values are tabulated below

Block Pressure (14/04/’16) 𝑵𝒑𝒊 (14/04/’16)

1 3993.75 262.56

2 3980.75 831.34

3 3966.24 1466

4 3949.10 2215.49

Total Cumulative Oil Produced, NpT:

𝑁𝑝 = ∑ 𝑁𝑝𝑖= 262.56 + 831.34 + 1466 + 2215.49 = 4775.4 STB

Therefore, the Total Cumulative Oil Produced, NpT is 4775.4 STB

e.

Topic covered:

Reservoir Discretization

Answer Keywords:

Range i = 1 to nx-1; no right neighbor for Block nx

Answer:

I would specify range i = 1 to nx-1.

Justification:

𝑥𝑖+1 refers to the location of the neighbor of Block i to the right. The last block, Block nx has no neighbor

to the right; hence IDQ-4 cannot be implemented for Block nx; therefore, the implementation must end at

Block nx-1.

1


COVENANT UNIVERSITY CANAANLAND, KM 10, IDIROKO ROAD P.M.B 1023, OTA, OGUN STATE, NIGERIA.

TITLE OF EXAMINATION: B. Eng. Semester Examination. COLLEGE: Engineering. DEPARTMENT: Petroleum Engineering. SEMESTER: Omega. SESSION: 2015/2016. CREDIT UNIT: 3. COURSE CODE: PET525. TIME: 3 Hours. COURSE TITLE: Petroleum Refining Engineering and Petrochemical Technology. INSTRUCTION: Answer TWO Questions from Section A and ONE Each from Sections B and C

SECTION A:

QUESTION 1:

a) Define and Explain the following [5 Marks]

i. Sour crude

ii. Sweet crude

iii. Isomerization

iv. Polymerization

v. Hydrocracking

b) Describe and explain with suitable diagrams the first 2 processes in an oil refinery

[20 Marks]

QUESTION 2:

a) Define and explain the following [5 Marks]

i. Blending

ii. Lubrication

iii. Additives

iv. Greases

v. Base Oils

b) Describe and explain with suitable diagrams the catalytic cracking and alkylation

process of the oil refinery. [20 Marks]

QUESTION 3:

Write and elaborate on LDPE, HDPE, and PVC stating their uses. And with a suitable

diagram describe the PVC process.

[25 Marks]

2

SECTION B

QUESTION 4:

a) Calculate the RVP of a blend of LSR gasoline, HSR gasoline, Reformate and FCC

gasoline. Properties and available quantities of the components are given in the table

below: [15 Marks]

b) Determine the amount of n-butane required to produce a gasoline blend with RVP =

10 psi from the components listed in (a). The RVP of n-butane is 52 psi. [10 Marks]

NOTE:

𝐵𝐼𝑅𝑉𝑃𝑖 = 𝑅𝑉𝑃𝑖1.25

𝐵𝐼𝑅𝑉𝑃𝑖 = 𝑅𝑉𝑃𝑖1.25

SECTION C

QUESTION 5:

a) With suitable diagrams and equations, show the main products of coal gasification.

[10 Marks]

b) Describe how methanol is produced by the Oxo-Synthesis process. [7 Marks]

c) Starting from refinery petroleum residues and using equations only, how are the

following petrochemicals produced:

i. Butanol [2 Mark]

ii. Benzene hexachloride [2 Mark]

iii. Triethanolanine [2 Mark]

iv. Nylon 6.6 [2 Mark]

QUESTION 6:

a) How is polymerization of polyvinylchloride (PVC) monomers carried out in

industry?

[10 Marks]

b) The production technologies for High Density Polyethylene (HDPE) and

Polypropylene (PP) are similar but differ in their operating parameters. Discuss.

[5 Marks]

c) With a suitable diagram and equations, describe the process of manufacturing

polystyrene. [5 Marks]

d) Write shortly on diene polymerization. [5 Marks]

3

COVENANT UNIVERSITY CANAANLAND, KM 10, IDIROKO ROAD P.M.B 1023, OTA, OGUN STATE, NIGERIA.

MARKING SCHEME ON PET525

1. a)

i) Sour crude: Sour crude is a term used in describing crude oil with considerable quantity of sulphur compounds (>1% sulphur). Sour crude must be well-desulphurized before they are processed in the refinery. This is because the presence of sulphur in a crude oil type, when sent to the catalytic reforming unit, poisons the noble catalysts (platinum and rhenium) being used in the reformer to upgrade the quality of gasoline. Sour crude is not as expensive as sweet crude in the world market, because the presence of sulphur in them reduces their value, quality and yields, and also results in high refining cost.

Sour crude also damages the various refining equipment in the refinery. Example of sour crude is crude with ethanethiol (C2H5SH).

ii) Sweet crude is the crude oil type with minimal or no sulphur content. This definition of minimal depends on the required standard in terms of market and refinery to which the sulphur content percentage limit is set. Sweet crude cost more than sour crude in the market.

iii) Polymerization: In this reaction, two or more molecules combine into a single larger molecule, for instance, two molecules of ethylene C2H4 are polymerized into a higher-boiling hydrocarbon, butylene’s C4H8. Using polymerization, gaseous olefin hydrocarbons (ethylene, butylene’s) can be converted into liquid or ever solid hydrocarbons of a higher molecular mass.

iv) Isomerization: It is the chemical re arrangement of straight chain hydrocarbons (paraffin’s) so as to contain branches to the main chain. These branched chain compounds have better octane number and desirable in gasoline.

v) Hydrocracking: Hydrocracking is a two stage process combining catalytic cracking and hydrogenation, where in heavier feedstock is cracked in the presence of hydrogen to produce more desirable products. It is a catalytic chemical process used in petroleum refineries for converting the high-boiling constituent hydrocarbons in petroleum crude oils to more valuable lower-boiling products such as gasoline, kerosene, and jet fuel and diesel oil.

1b.

Crude oil separation processes

The first step in petroleum refining is the fractionation of crude oil in atmospheric and vacuum

distillation towers. Heated crude oil is physically separated into various fractions, or straight-run

cuts, differentiated by specific boiling-point ranges and classified, in order of decreasing

volatility, as gases, light distillates, middle distillates, gas oils and residuum. Fractionation works

because the gradation in temperature from the bottom to the top of the distillation tower causes the

4

higher-boiling-point components to condense first, while the lower-boiling-point fractions rise

higher in the tower before they condense. Within the tower, the rising vapours and the descending

liquids (reflux) mix at levels where they have compositions in equilibrium with each other.

Special trays are located at these levels (or stages) which remove a fraction of the liquid which

condenses at each level. In a typical two-stage crude unit, the atmospheric tower, producing light

fractions and distillate, is immediately followed by a vacuum tower which processes the

atmospheric residuals. After distillation, only a few hydrocarbons are suitable for use as finished

products without further processing.

a) Atmospheric distillation

In atmospheric distillation towers, the desalted crude feedstock is preheated using recovered

process heat. It then flows to a direct-fired crude charge heater, where it is fed into the vertical

distillation column just above the bottom at pressures slightly above atmosphere and at

temperatures from 343 °C to 371 °C, to avoid undesirable thermal cracking at higher

temperatures. The lighter (lower boiling point) fractions diffuse into the upper part of the tower,

and are continuously drawn off and directed to other units for further processing, treating,

blending and distribution.

Fractions with the lowest boiling points, such as fuel gas and light naphtha, are removed from the

top of the tower by an overhead line as vapours. Naphtha, or straight-run gasoline, is taken from

the upper section of the tower as an overhead stream. These products are used as petrochemical

and reformer feedstocks, gasoline blending stocks, solvents and LPGs.

Intermediate boiling range fractions, including gas oil, heavy naphtha and distillates, are removed

from the middle section of the tower as side streams. These are sent to finishing operations for use

as kerosene, diesel fuel, fuel oil, jet fuel, catalytic cracker feedstock and blending stocks. Some of

these liquid fractions are stripped of their lighter ends, which are returned to the tower as down

flowing reflux streams.

5

b) vacuum distillation

•To further distill the residuum or topped crude from the atmospheric tower without thermal

cracking, reduced pressure is required. The process takes place in one or more vacuum

distillation towers.

The principles of vacuum distillation resemble those of fractional distillation except that larger

diameter columns are used to maintain comparable vapor velocities at the reduced pressures.

The internal designs of some vacuum towers are different from atmospheric towers in that

random packing and demister pads are used instead of trays.

A typical first‐phase vacuum tower may produce gas oils, lubricating‐oil base stocks, and heavy

residual for propane de asphalting. A second‐phase tower operating at lower vacuum may distill

surplus residuum from the atmospheric tower, which is not used for lube‐stock processing, and

surplus residuum from the first vacuum tower not used for de asphalting.

Vacuum towers are typically used to separate catalytic cracking feedstock from surplus

residuum.

2.

6

i) Blending: Blending is the physical mixture of a number of different liquid hydrocarbons to

produce a finished product with certain desired characteristics. Products can be blended in‐line

through a manifold system, or batch blended in tanks and vessels. In‐line blending of gasoline,

distillates, jet fuel, and kerosene is accomplished by injecting proportionate amounts of each

component into the main stream where turbulence promotes thorough.

ii) Lubrication: Lubrication is the science of reduction of friction by using oils, greases or other

special materials between the moving parts and the stationary parts of machines.

iii) Additives: They are chemicals added to lubricating blends and feed stocks to improve their

quality and meet required standard specifications and also performance. Their amounts vary

from 1-25% and the major markets are transportation (trucks, buses, trains, ships etc.). Additives

protect metal surfaces, rings, bearings, and gears, extend the range of lubricant applicability and

also extend lubricant life.

iv) Grease: is a two phase system thicker than liquid lubricant. It is a lubricating fluid which has

been gelled with thickening agents so that the lubricants can be retained more readily in the

required area. The functions are to prolong life and efficiency of the machine by reducing friction

and wear, heat dissipation, corrosion protection, contaminant removal, power transmission.

v) Hydrocracking: Hydrocracking is a two stage process combining catalytic cracking and

hydrogenation, where in heavier feedstock is cracked in the presence of hydrogen to produce

more desirable products. It is a catalytic chemical process used in petroleum refineries for

converting the high-boiling constituent hydrocarbons in petroleum crude oils to more valuable

lower-boiling products such as gasoline, kerosene, and jet fuel and diesel oil.

2. b.

i) Catalytic process.

The most common process is FCC, in which the oil is cracked in the presence of a finely divided

catalyst which is maintained in an aerated or fluidized state by the oil vapors. Fluid catalytic

cracking or "cat cracking," is the basic gasoline-making process. Using intense heat (about 1,000

degrees Fahrenheit), low pressure and a powdered catalyst (a substance that accelerates

chemical reactions), the cat cracker can convert most relatively heavy fractions into smaller

gasoline molecules. Catalyst used includes zeolite, bauxite and silica.

A typical FCC process involves mixing a preheated hydrocarbon charge with hot, regenerated

catalyst as it enters the riser leading to the reactor. The charge is combined with a recycle

stream within the riser, vaporized, and raised to reactor temperature (900°-1,000° F) by the hot

catalyst. As the mixture travels up the riser, the charge is cracked at 10-30 psi. In the more

modern FCC units, all cracking takes place in the riser. The "reactor" no longer functions as a

7

reactor; it merely serves as a holding vessel for the cyclones. This cracking continues until the oil

vapors are separated from the catalyst in the reactor cyclones. The resultant product stream

(cracked product) is then charged to a fractionating column where it is separated into fractions,

and some of the heavy oil is recycled to the riser.

During the trip up the riser, the cracking catalyst is "spent" by reactions which deposit coke on

the catalyst and greatly reduce activity and selectivity. The "spent" catalyst is disengaged from

the cracked hydrocarbon vapors and sent to a stripper where it is contacted with steam to

remove hydrocarbons remaining in the catalyst pores. The "spent" catalyst then flows into a

fluidized-bed regenerator where air (or in some cases air plus oxygen) is used to burn off the

coke to restore catalyst activity and also provide the necessary heat for the next reaction cycle,

cracking being an endothermic reaction. The "regenerated" catalyst then flows to the base of the

riser, repeating the cycle.

The gasoline produced in the FCC unit has an elevated octane rating but is less chemically stable

compared to other gasoline components due to its olefinic profile. Olefins in gasoline are

responsible for the formation of polymeric deposits in storage tanks, fuel ducts and injectors.

The FCC LPG is an important source of C3-C4 olefins and isobutene that are essential feeds for

the alkylation process and the production of polymers such as polypropylene.

Fresh catalyst is added and worn-out catalyst removed to optimize the cracking process.

8

b) Ii. Alkylation process: Is one of the conversion processes used in the petroleum refineries.

It is used to convert iso-butane and low-molecular-weight alkenes (primarily a mixture of

propene and butene) into an alkylate, a high octane gasoline component. The process

occurs in presence of a strong acting acid as catalyst. The acid can be either sulfuric acid

or hydrofluoric acid (HF). Depending on the acid used as catalyst the unit takes the name

of SAAU (Sulphuric Acid Alkylation Unit) or HFAU (Hydrofluoric Acid Alkylation Unit). The

product of the unit, the alkylate, is composed of a mixture of high-octane, branched-

chain paraffinic hydrocarbons (mostly isoheptane and isooctane). Alkylate is a premium

gasoline blending stock because it has exceptional antiknock properties and is clean

burning. The octane number of the alkylate depends mainly upon the kind of alkenes

used and upon operating conditions.

A HSAU can be divided into five major sections: reaction, refrigeration, effluent treating,

fractionation and blowdown.

In the reaction section the reacting hydrocarbons (olefin feed with both fresh and recycled

isobutane) are brought into contact with sulfuric acid catalyst under controlled conditions and at a

temperature of 15.6 °C (60 °F). The feeds are treated to remove impurities, especially water in

order to reduce corrosion.

The heat of reaction is removed in the refrigeration section and the light hydrocarbons are purged

from the unit. In the effluent treating Section the free acid, alkyl sulfates and di-alkyl sulfates are

removed from the net effluent stream to avoid downstream corrosion and fouling using a settler.

The sulfuric acid present in the reaction zone serves as a catalyst to the alkylation reaction.

Theoretically, a catalyst promotes a chemical reaction without being changed as a result of that

reaction. In reality, however, the acid is diluted as a result of the side reactions and feed

contaminants. To maintain the desired spent acid strength, a small amount of fresh acid is

continuously charged to the acid recycle line from the acid settler to the reactor and an equivalent

amount of spent acid is withdrawn from the acid settler. In the fractionation section the unreacted

isobutane is recovered for recycle to the reaction section and remaining hydrocarbons are

separated into the desired products.

The spent acid is degassed in an acid blow down drum, waste water pH is adjusted and acid vent

streams are neutralized with caustic in a scrubber before being flared. Spent acid goes to storage

and periodically removed.

9

3. i) Low Density Polyethylene (LDPE)

LDPE have its own gravity in the range of 0.913 to 0.935 and is thus called low density

polyethylene to distinguish it from the other main type of polyethylene, which has a specific

gravity of 0.96 and is called HDPE. LDPE is made from cracker grade ethylene which is

compressed to 1 000 to 3 000 atmospheres (15 000 psi to 45 000 psi) and polymerized at a

temperature of 100 – 3000c to give solid but flexible waxy like polymer.

LDPE is a tough, flexible, translucent polymer with low tensile strength but outstandingly good

dielectric properties. Its first use was as an insulant for high frequency cable (RADAR). Its main

use now is for films, for packaging and agriculture. A further important use is for coating paper to

make it liquid proof and heat sealable, the familiar tetrapak.

ii) High density polyethylene

HDPE has a higher density than LDPE because it does not have the highly branched structure

referred to earlier. This results in HDPE having a higher melting point and stiffer properties but in

many respects it compares with LDPE. Unlike LDPE however it is not transparent in thin films and

so has not displaced LDPE in these packaging applications but rather replaced waxed paper and

tissue paper. HDPE has good melt strength when extruded, so is widely used for pipe and blow

mouldings where it has largely replaced LDPE on account of its better stiffness. The toughness of

HDPE, combined with its stiffness, has made it ideal for boxes and crates made by injection

moulding. It has inferior long term creep resistance to polypropylene, but better low

temperature impact strength. Thus milk bottle crates and fish boxes that have to be stored for

short periods at low temperatures tend to use HDPE, whereas beer bottle crates that have to be

steam sterilized and stacked to great heights for long periods tend to use PP.

iii) Poly vinyl chloride

10

PVC is made by the polymerization of vinyl chloride, which is a gas under normal conditions of

temperature and pressure and which is prepared by reacting hydrogen chloride gas and

acetylene in the presence of a catalyst or by the chlorination of ethylene.

Vinyl chloride monomer is a member of a class of chemical compounds, known as the vinyl’s or

ethenoids, which possesses the grouping CH2 = CH – Ethylene, CH2 = CH2, is the parent substance

of the class, and although the many “vinyl” plastics are not necessarily prepared from ethylene

itself, the chemical relationship to the parent can be traced. By substitution in the CH2 = CH2

molecule a whole range of polymer sable monomers is obtainable. For vinyl chloride, one

hydrogen atom is replaced by one of chlorine, giving CH2 = CHC1.

The main process for PVC, accounting for over 80% of capacity is t he suspension process (SPVC).

In the suspension polymerization process a monomer soluble catalyst is used and a suspension

agent prevents the newly formed particles of PVC coagulating. The reactants are held in a stirred

vessel until most of the VCM monomer has polymerized.

The particles of PVC are large enough to be separated from the water by filtration and they can

be washed to remove most of the granulating agent. After washing, the polymer is dried in ovens

or tunnel driers.

Vinyl chloride polymers are fine, white free-flowing thermoplastic powers which soften at about

160oC and decompose fairly rapidly at temperatures above that. PVC is invariably compounded

with heat stabilizers (because some decomposition takes place at lower temperatures), and

frequently with other ingredients such as plastics, fillers and pigments to yield compound which

range from soft elastomers to rigid materials.

The main application for PVC is for pipe and guttering, other applications in the construction

industry such as cable and flooring mean that over 50 percent of PVC usage is linked to the

construction industry. It is also used for window frames (Germany) and water bottles (France).

11

4. the RVP of the blend is calculated through the following steps:

I. The volume fraction of each component 𝑥𝑣𝑖 is calculated

II. The RVP index of each component is calculated

III. The volume fraction is multiplied by the index for each component

The summary of the RVP calculations is listed below:

12

From the summation, BIRVP, Blend = 14.305 then

RVPBlend = 14.3051 1.25⁄

RVPBlend = 8.4psi

b) Assume VButane is the volume flow rate (BPD) of n-butane needed to be added to the

blend. RVP indices are calculated for all components including n-butane and listed in

the table below:

The volume is first multiplied by the index and then divided by the total volume. The

summation yields the RVP index for the blend.

BIRVP, Blend = 101.25 = 314,912 + 139.64VButane

22,000 + VButane

The above equation is solved for VButane to be 625.7 BPD, which is the amount of n-

butane needed to adjust the RVP of the blend to 10 psi.

of 5

COVENANT UNIVERSITY








COURSE TITLE: PETROLEUM ECONOMICS

INSTRUCTION: Answer 4 Questions and at least 1 Question from each Section TIME: 3 HOURS

SECTION A

Question 1

a. A 320-acre lease has to be evaluated. Data on the lease are as follows:

Average porosity 15%

Net pay 15ft

Water Saturation 30%

Reservoir oil Formation Volume Factor 1.2

Initial Producing rate 1714 Stb/month

Producing rate after a month 1658 Stb/month

Economic Limit 200 Stb/month

Recovery Factor 20%

Determine by calculations the following:

i. STBs initially in place (2 marks)

ii. Recoverable reserves (2 marks)

iii. Exponential decline rate required to recover the reserves (2 marks)

iv. Life of field in years (2 marks)

b. Given that a well has declined from 180 stb/day to 172 stb/day during a 1-month period, use the

exponential decline model to perform the following tasks:

i. Predict the production rate after 12 months (1 mark)

ii. Calculate the amount of oil produced during the first year (1 mark)

iii. Project the yearly production for the well for the next 6 years (7 marks)

of 5

Question 2

i. For the data given in table below, identify a suitable decline model (7 marks)

ii. Determine the model parameters, (6 marks)

iii. Project production rate until a marginal rate of 40 stb/day is reached. (4 marks)

T (mo) Q (stb/day) T (mo) Q (stb/day)

1.0 904.84 13.0 272.53

2.0 818.73 14.0 246.60

3.0 740.82 15.0 223.13

4.0 670.32 16.0 201.90

5.0 606.53 17.0 182.68

6.0 548.81 18.0 165.30

7.0 496.59 19.0 149.57

8.0 449.33 20.0 135.34

9.0 406.57 21.0 122.46

10.0 367.88 22.0 110.80

11.0 332.87 23.0 100.26

12.0 301.19 24.0 90.72

SECTION B

Question 3

An OPL at a water depth of 1000-1263m, which is located offshore in the central Niger Delta and about 55-

67 Km from the Bonga field is scheduled to be converted to OML and oil production would commence by

2017. There are 7 participants with different stakes. Oil 0

API is 32-35 and recoverable reserves estimated at

200mmbbl (proven + probable).

Production Profile

Year 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009

000 bbl/d 34.5 77 77 65.5 55.6 47.3 40.2 34.2 29 24.7

Cost

Exploration and Appraisal Cost (includes 4 exploratory and appraisal wells): USD220MM

Operating Cost: fixed cost – USD54MM/year; variable cost – USD2/bbl.

Capital Cost (USD MM)

Year 2015 2016 2017 2018 2019 2031

Prod. Facility - 400 - - - -

Subsea 200 750 750 - - -

Dev. Drilling - 500 600 600 300 -

Abandonment cost - - - - - 240

Royalty

Royalty rate is at 20%

PPT (Petroleum Profit Tax)

Taxable income is at 38.5%

of 5

Assumptions

Discount rate, 10% and end of year discounting is applicable

Neglect Exploration and appraisal cost in the cash flow

Oil price, USD79.59/bbl

Use 5years SLD and all capital cost for capital allowance computation

a) Compute Net present value (NPV)

b) Explain the consequences of the present oil price challenge faced in the world on our NPV

c) Explain the procedure to carry out sensitivity analysis on NPV using the variable parameters

d) List three factors that determines the choice of a depreciation method to use

Question 4

a. define the following economic terms

i. Royalty and bonus

ii. Petroleum fiscal system

iii. Depreciation and depletion

b. For each of the following investment projects, calculate:

i. The net present values using a 15% discount rate

ii. The payback periods

Based on the two profitability measures calculated, which project(s) should be accepted?

0 1 2 3 4 5 6 7

A (4564) 1000 1000 1000 1000 1000 1000 1000

B (2000) 524.7 524.7 524.7 524.7 524.7 524.7 524.7

C (21000) 3000 3000 3000 3000 3000 3000 3000

c. Show a mathematical relationship between the effective and nominal interest rate

SECTION C

Question 5

a. The results that can be expected from a proposed water flood, in terms of additional oil recovery per

acre-foot, are represented by a triangular distribution. The lower limit has been estimated to be 50,

the upper limit 200, and the most likely result to be 100bbl of oil per acre-foot. Calculate the mean

and standard deviation of this distribution. What is the probability that the recovery will be less than

75 bbl/acre-ft? what is the probability that the recovery will be greater than 100 bbl/acre-ft? What is

the probability that the recovery will be greater than 150 bbl/acre-ft?

of 5

b. The table shows representative sizes of possible discoveries by exploratory wells.

Size of Discovery (bbl) Probability

0 0.60

2×104 0.20

20×104 0.15

50×104 0.05 A company drills 3 wells. What is the probability that this group (sample) contains one dry well,

one discovery of 2×104

bbl, and one discovery of 20×104

bbl?

Question 6

Question 6

A company (Company-A) owns lease on 120 acres. An adjoining lease by another company (Company-B) is

on 520 acres. Company-B has proposed a test well. Well spacing is 640 acres and combined leases make up

one drilling unit, in which one well can be drilled. Company-A has its choice of either participating in the

working interest on an acreage basis, or it can farm-out its interest, for 1/16 of 7/8 overriding royalty interest

under its 120 acres. Ownership interests as a fraction on the total gas produced are based on 5% state tax.

What decision should Company-A take?

Probable reserves of a successful well; estimates

Maximum 5,30```````````````0 mmscf(30% of productive wells)

Average 3,750 mmscf(50% of productive wells)

Minimum 2,500 mmscf(20% of productive wells)

Regional wildcat success ratio 35%

Regional development ratio 85%

Spacing 640wells/acre

Dry hole costs $40,000

Completion costs $32,000

Operating costs $150 per well per month

Gas price $150 per mmscf

Producing life 18 years

• Management decision, based on PV at

15% yearly discount rate

• Estimate discount factor

• Production assumed uniform for 18 years

• Consider ⅛ Land owner’s royalty

Note: use end-of-year discounting and consider delay of 2 years from the date of well completion until gas

sales commences.

of 5

Equations

Decline type Exponential

decline

Harmonic decline Hyperbolic decline

Decline

exponent

(n)

0n 1n 0 1n

Nominal

Decline rate

(D)

)1ln( eDD

11

ee DDD iQ

QiQD

Effective

Decline rate i

QQ

QD i

iQ

QiQD

DeeD 1

Oil production

rate(Q)

Dt

ieQQ

11

tDQQ ii

1

1 ni iQ Q nD t

𝑃 = 𝐴(1 + 𝑖)0.5 [(1 + 𝑖)𝑛 − 1

𝑖(1 + 𝑖)𝑛]

𝑚

= 𝑝1𝑋1 + (1

− 𝑝1)𝑎 + 𝑏 + 𝑐

3

𝑚 =𝑎 + 𝑏 + 𝑐

3

𝑝2 = (1 − 𝑝1)𝑏 − 𝑎

𝑐 − 𝑎 𝑝3 =

𝑐 − 𝑏

𝑐 − 𝑎 𝑝2 =

𝑏 − 𝑎

𝑐 − 𝑎

𝐹 = 1 − 𝑃3 (𝑐 − 𝑥

𝑐 − 𝑏)

2

𝑝3 = (1 − 𝑝1)𝑐 − 𝑏

𝑐 − 𝑎 𝜎2 =

(𝑐 − 𝑎)2

18(1 − 𝑝2𝑝3)

𝐹 = 𝑃2 (𝑥 − 𝑎

𝑏 − 𝑎)

2

𝜎2 =𝑝1

1 − 𝑝1(𝑚 − 𝑋1)2 +

(𝑐 − 𝑎)2

18(1 − 𝑝1 −

𝑝2𝑝3

1 − 𝑝1)

𝑝(𝑠) =𝑁!

𝑘1! 𝑘2! … . . 𝑘𝑚!𝑃1

𝑘1𝑝2𝑘2 … … 𝑝𝑚

𝑘𝑚

𝑝(𝑠) =(

𝑀1

𝑘1) (

𝑀2

𝑘2) … … . (

𝑀𝑚

𝑘𝑚)

(𝑀𝑁

)

COVENANT UNIVERSITY








COURSE TITLE: PETROLEUM ECONOMICS

INSTRUCTION: Answer 4 Questions and at least 1 Question from each Section TIME: 3 HOURS

Marking Guide

QUESTION ONE

i. STBs initially in place

𝑉 = 7758𝐴ℎ𝜑(1 − 𝑆𝑤𝑖)

𝐵𝑜=

7758 ∗ 320 ∗ 15 ∗ 0.15 ∗ (1 − 0.3)

1.2= 𝟑, 𝟐𝟓𝟖, 𝟑𝟔𝟎 𝑺𝑻𝑩

(2 marks for the formulae, substitution and presentation of final answer)

ii. Recoverable reserves

𝑅𝑒𝑐𝑜𝑣𝑒𝑟𝑎𝑏𝑙𝑒 𝑟𝑒𝑠𝑒𝑟𝑣𝑒𝑠 = 𝑂𝑂𝐼𝑃 ∗ 𝑅𝐹 = 3258360 ∗ 0.2 = 𝟔𝟓𝟏, 𝟔𝟕𝟐 𝑺𝑻𝑩


iii. Exponential Decline rate

𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑑𝑒𝑐𝑙𝑖𝑛𝑒, 𝐷 =𝑄𝑖 − 𝑄

𝑄𝑖=

1714 − 1658

1714= 𝟎. 𝟎𝟑𝟐𝟕 /𝒎𝒐𝒏𝒕𝒉

𝑛𝑜𝑟𝑚𝑖𝑛𝑎𝑙 𝑑𝑒𝑐𝑙𝑖𝑛𝑒, 𝑎 = − ln(1 − 𝐷𝑒) = − ln(1 − 0.0327) = 𝟎. 𝟎𝟑𝟑𝟐𝒑𝒆𝒓 𝒎𝒐

= 𝟎. 𝟑𝟗𝟖𝟔𝒑𝒆𝒓 𝒚𝒓


iv. Life of field

𝑡 =1

𝐷𝑙𝑛

𝑄𝑖

𝑄=

1

0.3986𝑙𝑛

1714

200= 𝟓. 𝟑𝟖𝟗 𝒚𝒆𝒂𝒓𝒔


b.

i. Production rate after 12 months

𝐷 =1

𝑡𝑙𝑛

𝑄𝑖

𝑄=

1

(1)𝑙𝑛

180

172= 𝟎. 𝟎𝟒𝟓𝟒𝟔 𝒑𝒆𝒓 𝒎𝒐

Rate at the end of one year

𝑄 = 𝑄0𝑒−𝐷𝑡 = 180𝑒−0.04546(12) = 𝟏𝟎𝟒. 𝟑𝟏𝟒 𝒔𝒕𝒃/𝒅𝒂𝒚

(1 mark for the formulae, substitution and presentation of final answer)

ii. The amount of oil produced during the first year

𝐷𝑦 = 0.04546 ∗ 12 = 𝟎. 𝟓𝟒𝟓𝟓𝟓 𝒑𝒆𝒓 𝒚𝒓

𝑁𝑝, 1 =𝑄𝑜 − 𝑄1

𝐷𝑦= (

180 − 104.314

0.54555) 365 = 𝟓𝟎, 𝟔𝟑𝟕. 𝟔 𝒔𝒕𝒃

(1 mark for the formulae, substitution and presentation of final answer)

iii. Yearly production in the next 6 years

𝑄2 = 𝑄1𝑒−𝐷𝑡 = 104.314𝑒−0.04546(12) = 𝟔𝟎. 𝟒𝟓 𝒔𝒕𝒃/𝒅𝒂𝒚

(1/2 mark for the formulae, substitution and presentation of final answer)

𝑁𝑝, 2 =𝑄1 − 𝑄2

𝐷𝑦= (

104.314 − 60.45

0.54555) 365 = 𝟐𝟗, 𝟑𝟒𝟓. 𝟕 𝒔𝒕𝒃


𝑄3 = 𝑄2𝑒−𝐷𝑡 = 60.45𝑒−0.04546(12) = 𝟑𝟓. 𝟎𝟑 𝒔𝒕𝒃/𝒅𝒂𝒚


𝑁𝑝, 3 =𝑄2 − 𝑄3

𝐷𝑦= (

60.45 − 35.03

0.54555) 365 = 𝟏𝟕, 𝟎𝟎𝟔. 𝟓 𝒔𝒕𝒃


𝑄4 = 𝑄3𝑒−𝐷𝑡 = 35.03𝑒−0.04546(12) = 𝟐𝟎. 𝟑𝟎𝟑 𝒔𝒕𝒃/𝒅𝒂𝒚


𝑁𝑝, 4 =𝑄3 − 𝑄4

𝐷𝑦= (

35.03 − 20.303

0.54555) 365 = 𝟗, 𝟖𝟓𝟓. 𝟕 𝒔𝒕𝒃


𝑄5 = 𝑄4𝑒−𝐷𝑡 = 20.303𝑒−0.04546(12) = 𝟏𝟏. 𝟕𝟔𝟔 𝒔𝒕𝒃/𝒅𝒂𝒚


𝑁𝑝, 5 =𝑄4 − 𝑄5

𝐷𝑦= (

20.303 − 11.766

0.54555) 365 = 𝟓, 𝟕𝟏𝟏. 𝟔 𝒔𝒕𝒃


𝑄6 = 𝑄5𝑒−𝐷𝑡 = 11.766𝑒−0.04546(12) = 𝟔. 𝟖𝟏𝟖 𝒔𝒕𝒃/𝒅𝒂𝒚


𝑁𝑝, 6 =𝑄5 − 𝑄6

𝐷𝑦= (

11.766 − 6.818

0.54555) 365 = 𝟑, 𝟑𝟏𝟎 𝒔𝒕𝒃


In summary,

Year Rate at the End of year (stb/day)

Yearly Production (stb)

0 180 -

1 104.3 50637.6

2 60.45 29345.7

3 35.03 17006.5

4 20.3 9855.7

5 11.7 5711.6

6 6.8 3310.0

115867.2

(1 mark for summary of answers)

TOTAL = 17 MARKS

QUESTION TWO

i. Identification of a suitable model

A graph of log (q) versus t is plotted. It gives a straight line. Based on this fingerprint, the system is

undergoing an exponential decline.

(1 mark for identifying the model)

(6 marks for plotting on a SEMI-LOG graph!)

ii. Determination of model parameters, a and b

Select any two points on the trend line.

𝑡1 = 5 𝑚𝑜𝑛𝑡ℎ𝑠, 𝑞1 = 607𝑠𝑡𝑏

𝑑

𝑡2 = 20 𝑚𝑜𝑛𝑡ℎ𝑠, 𝑞2 = 135𝑠𝑡𝑏

𝑑

Decline rate can be calculated with the equation;

𝑏 =1

𝑡1−𝑡2𝑙𝑛 (

𝑞1

𝑞2) =

1

5−20𝑙𝑛 (

135

607) = 0.11/𝒎𝒐𝒏𝒕𝒉


a, can be determined as the intercept of the graph, i.e. when 𝑡 = 0

Making use of the equation;

𝑄 = 𝑄0𝑒−𝑏𝑡

𝑚𝑎𝑘𝑖𝑛𝑔 𝑄𝑜 𝑡ℎ𝑒 𝑠𝑢𝑏𝑗𝑒𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎,

𝑄0 =𝑄

𝑒−𝑏𝑡

Or Qo can be determined by extending the graph to the intercept at 𝑡 = 0

𝑎 = 𝑄𝑜 = 𝟏𝟎𝟎𝟎 𝒔𝒕𝒃/𝒅𝒂𝒚

y = 1000e-0.1x R² = 1

10

100

1000

0 5 10 15 20 25

Q, s

tb/d

ay

t, months

A semilog graph pg Q versus t


iii. Projection of production rate until a marginal rate of 40 stb/day is reached.

(4 marks for plotting the graph on a Cartesian graph properly indicating the extension)

TOTAL = 17 MARKS

0

100

200

300

400

500

600

700

800

900

1000

0 5 10 15 20 25 30 35 40 45 50

Q-t

40 stb/d

33

Question Three

(a) The net present value is as shown below [10 marks]

The net present value is $MM 3683

(b) A fall in the oil price will cause a decrease the revenue and also on the Net present

value expected from an oil venture, this has caused a decrease on the employment and

retrenchment of staff in the oil industry. [2½ marks]

(c) The procedures are:

Varying one parameter such as the oil price, CAPEX, OPEX, tax rate etc. at a time

on the calculated NPV [1 mark]

Varying two or more parameters (combined effect) e.g. CAPEX & OPEX;

Production & OPEX etc. on the NPV [1 mark]

Variation of parameters & their effect on NPV is shown on a spider diagram: NPV

(variation ) plotted against percentage change in each parameter [1 mark]

The steeper the straight line, the higher the sensitivity [1 mark]

(d) The factors that determine the choice of depreciation method are:

Type and function of property, lifetime, salvage value [½ mark]

Time value of money [½ mark]

Simplicity [½ mark]

N Year

Annual

Prod rate

bbl/yr

Price

$/bbl

Revenue

$MM

capex

$MM

Total

Capital

Allowanc

e/Depre

fixed

cost(

$MM

)

$2/bbl,

Variable

cost($M

M)

Royal

ty

$MM

Incom

e Tax

$MM

Net

Cash

Flow

$MM

Disco

untin

g

Facto

r

NPV/

Disc

NCF

$MM

0 2015 0 79.59 0 200 54 0 0 0 -254 1 0

1 2016 0 79.59 0 1650 54 0 0 0 -1704 0.909 0

2 2017 12592500 79.59 1002.237 1350 54 25.185 200.4 278.2 -906 0.826 0

3 2018 28105000 79.59 2236.877 600 $760.00 54 56.21 447.4 353.9 725.4 0.751 545

4 2019 28105000 79.59 2236.877 300 $820.00 54 56.21 447.4 330.8 1048 0.683 716.1

5 2020 23944000 79.59 1905.703 $820.00 54 47.888 381.1 232 1191 0.621 739.3

6 2021 20294000 79.59 1615.199 $820.00 54 40.588 323 145.4 1052 0.564 593.9

7 2022 17264500 79.59 1374.082 $820.00 54 34.529 274.8 73.43 937.3 0.513 481

8 2023 14673000 79.59 1167.824 $60.00 54 29.346 233.6 304.5 546.4 0.467 254.9

9 2024 12483000 79.59 993.522 54 24.966 198.7 275.6 440.2 0.424 186.7

10 2025 10585000 79.59 842.4602 54 21.17 168.5 230.5 368.3 0.386 142

11 2026 9015500 79.59 717.5436 240 54 18.031 143.5 193.3 68.73 0.35 24.09

1.77E+08 14092.32 4340 648 3683

OPEX($MM)

Question Four

(a) Royalty is the percentage of gross value and amount of production, it is paid in cash or kind and

its tax deductible. [1 marks]

Bonus is the payment made upon conclusion of negotiation of a contract or exploration within s

specified time after the signing of the contract. [1 marks]

Petroleum fiscal system defines the relationship between the legislative, tax, contractual and fiscal

elements under which petroleum operations are conducted in petroleum region or nation. [1

marks]

Depreciation is defined as loss in value of asset over time, it is deducible non-cash expense for

income tax purpose. Depreciation is on tangible assets. [1 marks]

Depletion is the actual physical deduction of natural resources, depletion is on non-renewable

resources. [1 marks]

(b) Table [10 marks]

t NCF (A) NCF (B) NCF © CNCF (A) CNCF (B) CNCF © 1/(1+i)^t NPV (A) NPV (B) NPV ©

0 -4564 -2000 -21000 -4564 -2000 -21000 1 -4564 -2000 -21000

1 1000 524.7 3000 -3564 -1475.3 -18000 0.869565217 869.5652174 456.2609 2608.7

2 1000 524.7 3000 -2564 -950.6 -15000 0.756143667 756.1436673 396.7486 2268.43

3 1000 524.7 3000 -1564 -425.9 -12000 0.657516232 657.5162324 344.9988 1972.55

4 1000 524.7 3000 -564 98.8 -9000 0.571753246 571.7532456 299.9989 1715.26

5 1000 524.7 3000 436 623.5 -6000 0.497176735 497.1767353 260.8686 1491.53

6 1000 524.7 3000 1436 1148.2 -3000 0.432327596 432.3275959 226.8423 1296.98

7 1000 524.7 3000 2436 1672.9 0 0.37593704 375.9370399 197.2542 1127.81

TOTAL -403.58027 182.9722 -8518.7

Project B should be accepted because it has higher NPV than others, it also has a better

payback period of 31

2 years [2 marks]

(c) Relationship between effective and nominal interest rate [1 mark]

(1 + 𝑖)𝑛 = (1 + 𝑗/𝑚)𝑚𝑛

1 + 𝑖 = (1 + 𝑗/𝑚)𝑚

𝑖 = (1 + 𝑗/𝑚)𝑚 − 1

Question Five

Question 5a

[2 marks]

-25000

-20000

-15000

-10000

-5000

0

5000

0 1 2 3 4 5 6 7 8

CN

CF

Time

payback period

payback period (A) payback period (B) payback period (C)

50 100 200bbl/acre

𝑚 =𝑎 + 𝑏 + 𝑐

3=

50 + 100 + 200

3= 116.67𝑏𝑏𝑙/𝑎𝑐𝑟𝑒

[1 mark]

𝑝2 =𝑏 − 𝑎

𝑐 − 𝑎=

100 − 50

200 − 50= 0.33

𝑝3 = 1 − 𝑝2 = 1 − 0.33 = 0.67

[2 marks]

𝜎 = √(𝑐 − 𝑎)2

18(1 − 𝑝2𝑝3) = √

(200 − 50)2

18(1 − 0.33 × 0.67) = 31.203

[1 mark]

F(Recovery < 75) = 𝑃2 (𝑥 − 𝑎

𝑏 − 𝑎)

2

= 0.33 (75 − 50

200 − 50)

2

= 0.0092

[2 marks]

F(Recovery > 100) = 1 − F(Recovery < 100) = 1 − [1 − 𝑝3 (𝑐 − 𝑥

𝑐 − 𝑏)

2

]

= 0.67 (200 − 100

200 − 100)

2

= 0.67

[2 marks]

F(Recovery > 150) = 1 − F(Recovery < 150) = 1 − [1 − 𝑝3 (𝑐 − 𝑥

𝑐 − 𝑏)

2

]

= 0.67 (200 − 150

200 − 100)

2

= 0.1675

[2 marks]

Question 5b

N=3

k1=1; k2=1; k3=1; k4=0 [1 marks]

P1=0.6; P2=0.2; P3=0.15; P4=0.05 [2 marks]

𝑝(𝑠) =𝑁!

𝑘1! 𝑘2! … . . 𝑘𝑚!𝑃1

𝑘1𝑝2𝑘2 … … 𝑝𝑚

𝑘𝑚

𝑝(𝑠) =3!

1! 1!1!0!0.61 0.21 0.151 0.050 = 0.108 [2 marks]

Question Six

Ownership interests as a fraction on the total gas produced are based on 5% state tax

120acres/640 acres = 3/16 of acreage

1/8 land owner’s royalty

For partial working interest

(3/16)(7/8)(0.95)=0.1559 of total [1 mark]

For overriding royalty interest

(3/16)(1/16 ORI × 7/8)(0.95)=0.009741 of total [1 mark]

Operating costs

Total operating costs:

($150 per month)(12 months/yrs)(18yrs)=$32,400 [1 mark]

Net share of operating costs:

(3/16)($32,400)=$6,080 [1 mark]

Investment

Net additional investment for the 3/16 working interest

Completed well:

(3/16)($40,000+$32,000)=$13,500 [1 mark]

Dry hole:

(3/16)($40,000)=$7,500 [1 mark]

Discount Factor

𝑃 = 𝐴(1 + 𝑖)0.5 [(1 + 𝑖)𝑛 − 1

𝑖(1 + 𝑖)𝑛 ] (1 + 𝑖)−2

A = equal yearly amount

E = sum of equal yearly amount

for this case E= 18 * A; A = E/ 18

𝑃 =𝐸

18(1 + 𝑖)0.5 [

(1 + 𝑖)𝑛 − 1

𝑖(1 + 𝑖)𝑛 ] (1 + 𝑖)−2

for

i=0.15; n=18; P=0.276056 E [2 mark]

Calculation of net PV of possible terminal events for partial working interest (Operating Income =

Reserves × 0.1559 × $150; PV = 0.276 × undiscounted value) [3 marks]

Total

Reserves

MMscf

Operating

Income

$

Operating

Cost

$

Undiscounted

Value

$

PV

15%

$

Investment

$

NPV

$

5300 123940 6080 117860 32600 13500 19100

3750 87690 6080 81610 22610 13500 9110

2500 58460 6080 52380 14510 13500 1010

0 0 0 0 0 7500 -7500

Calculation of net PV of possible terminal events for overriding royalty interest (Overriding Royalty

Income = Reserves × 0.009741 × $150; NPV = 0.276 × undiscounted value)) [3 marks]

Total

Reserves

mmscf

Overriding

royalty

Income

$

NPV

@

15%

$

5300 7710 2140

3750 5450 1520

2500 3640 1010

0 0 0

Computation of EMV [2 marks]

Outcome

Event

Conditional

Pro.

PV, $ Expected PV,

$

For partial working interest, producing well

5300mmscf 0.3 19100 5730

3750mmscf 0.5 9110 4560

2500mmscf 0.2 1010 200

10490

For partial working interest, prior to drilling

Producing

well

0.35 10490 3670

Dry hole 0.65 -7500 -4880

-1210

For overriding royalty interest, producing well

5300mmscf 0.3 2140 640

3750mmscf 0.5 1520 760

2500mmscf 0.2 1010 200

1600

For overriding interest, prior to drilling

Producing

well

0.35 1600 560

Dry well 0.65 0 0

560

Hence Overriding Royalty Interest ($560) is a preferable choice over Partial Working Interest (-

$1210

Figure showing results of computed Terminal NPV for each branch [1 mark]

0.2

0.5

0.3

0.2

0.5

0.3

0.35

0.35

0.65 0.65

0.65

$19,100

$9,110

$1,000

-$7,500

$2,140

$1,520

$1,010

$0

Terminal NPV

1


COVENANT UNIVERSITY CANAANLAND, KM 10, IDIROKO ROAD

P.M.B 1023, OTA, OGUN STATE, NIGERIA. TITLE OF EXAMINATION: B.Eng EXAMINATION

COLLEGE: Engineering.

SCHOOL: School of Petroleum and Chemical Engineering

DEPARTMENT: Petroleum Engineering.

SESSION: 2015/2016. SEMESTER: OMEGA

COURSE CODE: PET 528. CREDIT UNIT: 2

COURSE TITLE: Industrial Hazards and Environmental Pollution.

INSTRUCTION: Answer question One (1) and any other two (2). TIME: 2 hours

QUESTION ONE

a) As an air pollution control engineer you have been requested to evaluate the

gross heating value of a natural gas of a given composition: (3 marks)

Component Mole Fraction xi Gross Heating Value (Btu/scf)

N2 0.0515 0

CH4 0.8111 1013

C2H6 0.0967 1792

C3H8 0.0351 2590

C4H10 0.0056 3370

∑ 1.0000

Enthalpy of combustion gases (Btu/Ibmol)

T (oF) N2 Air (MW=28.97) CO2 H2O

32 0 0 0 0

60 194.9 194.6 243.1 224.2

77 312.2 312.7 392.2 360.5

100 473.3 472.7 597.9 545.3

200 1,170 1,170 1,527 1,353

300 1,868 1,870 2,509 2,171

400 2,570 2,576 3,537 3,001

500 3,277 3,289 4,607 3,842

600 3,991 4,010 5,714 4,700

700 4,713 4,740 6,855 5,572

800 5,443 5,479 8,026 6,460

900 6,182 6,227 9,224 7,364

1000 6,929 6,984 10,447 8,284

1200 8,452 8,524 12,960 10,176

1500 10,799 10,895 16,860 13,140

2000 14,840 14,970 23,630 18,380

2

2500 19,020 19,170 30,620 23,950

3000 23,280 23,460 37,750 29,780

b) You are also to determine the heat required for the combustion of the contaminated

air to a given temperature. Temperature of dry natural gas = 600

F. Volumetric flowrate of

contaminated air = 5000 scf (600

F, 1 atm). It is required to heat the contaminated air from

200o

Fto 16000

F. (15 marks)

ii) Calculate the enthalpy change of heating air from 200o

Fto 16000

F

1 Ibmol = 379 scf (4 marks)

c) Estimate the required landfill area (in acres) for a community with a population of 260,000.

Assume that the following conditions apply:

i. Solid waste generation per person in a day = 7.6 lb/day

ii. Compacted specific weight of solid wastes in landfill = 830 lb/yd3

iii. Average depth of compacted solid wastes = 18.3 m

iv. Landfill process took 18 months (8 marks)

QUESTION TWO

a) Write briefly on wet scrubbers and explain why they are used for handling

industrial particulate emission problem. (5 marks)

b) With the aid of a diagram, describe one of them and give four disadvantages

of wet scrubbers. (9 marks)

c) Briefly describe the absorption and adsorption processes of controlling air

pollution. (6 marks)

QUESTION THREE

a) Describe Land-filling and Land-farming and state the major differences

between them. (8 marks)

b) What are the disadvantages of Land-filling and Land-farming as solid

waste management tools? (6 marks)

c) Rank the performance of the two methods in the environmental

management of drilling cuttings; explain your choice. (6 marks)

QUESTION FOUR

a) Describe briefly the following terms:

i. DO ii. BOD III. COD (6 marks)

b) Write fully on primary, secondary and tertiary wastewater treatment methods

explaining the differences (6 marks)

c) Describe fully two basic methods of controlling sulfur oxides (SOX) emissions

in industrial facilities using appropriate illustrations. (8marks)

QUESTION FIVE

a) Write briefly on air pollutants and classify them. (6 marks)

b) Name four examples of each class of air pollutants normally produced in

the Upstream Sector of Oil and Gas Industry. (4 marks)

c) Name and describe two global environmental impacts of air pollution

explaining how they can be minimized. (10 marks)

3

MARKING SCHEME FOR PET 528 2015/2016 EXAMINATIONS

QUESTION ONE

Component Mole

Fraction xi

Gross Heating

Value (Btu/scf)

N2 0.0515 0

CH4 0.8111 1013

C2H6 0.0967 1792

C3H8 0.0351 2590

C4H10 0.0056 3370

∑ 1.0000

a) The gross heating value of the gas mixture, HVG in Btu/scf is

𝐻𝑉𝐺 = ∑ 𝑥𝑖𝐻𝑉𝐺𝑖

=(0.0515)(0) + (0.8111)(1013) + (0.0967)(1792) +(0.0351)(2590)+(0.0056)(3370)

= 𝟏𝟏𝟎𝟓 𝑩𝒕𝒖/𝒔𝒄𝒇 of natural gas (3 marks)

b)

The balanced chemical combustion equations for each of the four components of the natural gas

using 1 scf of natural gas as a basis is

0.8111CH4 + 1.6222O2 → 0.8111CO2 + 1.6222H2O

0.0967C2H6 + 0.3385O2 → 0.1934CO2 + 0.2901H2O

0.0351C3H8 H- 0.1755O2 → 0.1053CO2 + 0.1404H2O

0.0056C4H10 + 0.0364O2 → 0.0224CO2 + 0.0280H2O (1 mark for each equation)

The number of standard cubic feet for each of the following components of combustion may now

be determined.

For O2: 1.6222 + 0.3385 + 0.1755 + 0.0364 = 2.172 scf/scf natural gas

For CO2: 0.8111 + 0.1934 + 0.1053 + 0.0224 = 1.132 scf/scf natural gas

For H2O: 1.6222 + 0.2901 + 0.1404 + 0.0280 = 2.081 scf/scf natural gas

For N2: 0.0515 + (79/21)(2.172) = 8.222 scf/scf natural gas

(1 mark for each equation)

The total cubic feet of combustion products per scf of natural gas burned are the sum of scf

CO2/scf of natural gas, scf H2O/scf of natural gas, and scf N2/scf of natural gas:

Total cubic feet of combustion products = 1.132 + 2.081 + 8.222

= 11.435 scf of products/scf of natural gas (1 mark)

From the table in the problem statement, the following values of enthalpies at 600

F

and 16000

F are obtained.

For CO2:

4

ΔHCO2 = H at 16000

F - H at 600

F

H at 16000

F is estimated by interpolation between H at 15000

F and H at 20000

F

= 18214-243.1 = 17,971 Btu/lbmol

Since there are 379 scf per lbmol of any ideal gas and 1.132 scf CO2/scf natural gas,

ΔHCO2 =(17,971)(1.132)/(379)

= 53.68 Btu/scf of natural gas (1 mark (1/2 mark for interpolation))

For N2:

ΔHN2= (11607.2 - 194.9)(8.222)/(379)


For H2O :

ΔHH2o= (14188 - 224.2)(2.081)/(379)


Determine the amount of heat required to take the products of combustion from 60 to 16000

F

(∑ΔH).

Therefore,

∑ΔH = ΔHCO2 + ΔHN2 + ΔHH2o

= 53.68 + 247.6 + 76.67

= 377.95 Btu/scf of natural gas (2 marks)

The available heat (HA) of natural gas at 16000

F in Btu/scf natural gas is

HA = HVG - ∑ΔH

= 1105-377.95

= 727.05 Btu/scf of natural gas (1 mark)

The enthalpy change of air going from 200 to 16000

F is

ΔH air = ΔH at 1600°F - ΔH at 2000

F

ΔH at 1600°F by interpolation = 11633 Btu/lbmol

= 11633 - 1170

= 10463.2 Btu/lbmol (4 marks)

5

c)

Determine the daily solid wastes generation rate in tons per day:

𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 =260000 ∗ 7.6

2000

= 988 𝑡𝑜𝑛/𝑑𝑎𝑦 (2 marks)

The required area is determined as follows:

𝑉𝑜𝑙𝑢𝑚𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑝𝑒𝑟 𝑑𝑎𝑦 = (988ton/day)(20001b/ton)

830 𝑙𝑏/𝑦𝑑3

= 2381yd3

/day (2 marks)

Area required per yr =(2381yd3/day)(

365day

yr)(27

ft3

yd3)

(20ft)(43,650ft2

acre)

= 26.88 acre/yr (2 marks)

Time to refill =18 months = 1.5 yr

Area in acres = 40.32 acres (2 marks)

TOTAL = 30 marks

QUESTION TWO

a) The basic function of wet scrubbers is to provide contact between the scrubbing liquid

(usually water) and the particulates to be collected. The natural mechanism by which

atmospheric particles are removed is “wet precipitation”.

This contact can be achieved in variety of ways, thus the major types of Wet Scrubbers are:

Spray Towers

Centrifugal Scrubbers

Packed Beds and Plate Columns

Venturi Scrubbers. (5 marks)

b) SPRAY TOWERS

The Spray Tower is the smallest type of wet scrubber into which water is introduced by means of

spray nozzles. The polluted gas flows upward and particles collection results because of inertial

impaction and interception on the droplets.

6

CENTRIFUGAL SCRUBBERS

Centrifugal Scrubbers are normally used to recover particles smaller than those recovered in Spray

Towers.

The simplest type of centrifugal scrubber can be assembled by inserting banks of nozzles inside a

conventional dry cyclone. The spray acts on the particles in the outer vortex, and the droplets

loaded with particles are thrown outward against the wet inner wall of the cyclone

PACKED BEDS AND PLATE COLUMNS

Packed beds and plate columns widely known as “absorbers” or “fractionators” can also be used to

scrub particulate matter from polluted gases. In a typical Countercurrent-flow Packed bed

scrubber, the polluted gas stream flows upward and comes in contact with the scrubbing liquid

stream which is flowing downward over the packing in a film.

They are normally used for scrubbing particles that are soluble in the scrubbing liquid

7

VENTURI SCRUBBERS

This is designed to effectively use the energy from the inlet gas stream to atomize the liquid being

used to scrub the gas stream

Venturi scrubber consists of three sections: a converging section, a throat section, and a diffusing

section. The inlet gas stream enters the converging section and, as the area decreases, gas velocity

increases (in accordance with the Bernoulli equation).

Liquid is introduced either at the throat or at the entrance to the converging section

(5 marks, for description and diagram for ANY ONE)

Demerits

1. Relatively high energy costs.

2. Problem of wet sludge disposal.

3. Corrosion problems.

4.Visible wet plume, reduction in buoyancy

8

5.Very small particles(< 1µm) may not be captured

(4 marks, 1 for each disadvantage)

C) ABSORPTION involves the transfer of the pollutant from the gas phase to the liquid

phase across an interface in response to a concentration gradient, with the concentration

decreasing in the direction of mass transfer. (3 marks)

ADSORPTION is a surface phenomenon by which gas or liquid molecules adhere to the surface

of a solid adsorbent. The molecules are held onto the adsorbent surface by attractive forces; when

the forces are physical then phenomenon is physical adsorption, while it is chemical adsorption

(chemisorption) when the forces are chemical in nature. (3 marks)

TOTAL = 20 marks

QUESTION THREE

a) LAND-FILLING

This is the simplest and widely used waste disposal management method for solid, non-

hazardous and non-radioactive wastes.

The operation is basically a biological method of waste management.

It is basically an engineered structure complete with underground drains to remove

water percolating through the waste so as to prevent leachates from flowing into

sub-surface water.

Land-filling is the only method that can handle all wastes alone but may give rise to

significant environmental problems. (4 marks)

While LAND-FARMING

is a bioremediation treatment process that is performed in the upper soil zone or in bio-

treatment cells.

Contaminated soils, sediments, or sludge are incorporated into the soil surface and

periodically turned over (tilled) to aerate the mixture. Contaminated soils are mixed with

soil amendments such as soil bulking agents and nutrients, and then they are tilled into the

earth.

Contaminants are degraded, transformed, and immobilized by microbiological processes

and by oxidation. (4 marks)

b) Disadvantages of landfills

9

Landfills that are poorly designed or operated share more problems that are faced at the

uncontrolled dumping areas.

The areas surrounding the landfills become heavily polluted.

Landfill can pollute air, water and also the soil.

In a poorly developed landfill it is difficult to keep the dangerous chemicals from leaching

out into the surrounding area.

Dangerous chemicals can seep into the ground water system.

Many insects and rodents are attracted to landfills and can result in dangerous diseases.

It can cause diseases and illness in the communities living around the landfill.(3 marks)

Disadvantages of land-farming

large space requirements

the conditions advantageous for biological degradation of contaminants are largely

uncontrolled, which increases the length of time to complete remediation, particularly for

recalcitrant compounds.

inorganic contaminants are not biodegraded

the potential of large amounts of particulate matter released by operations

the presence of metal ions may be toxic to microbes and may leach from the contaminated

soil into the ground. (3 marks)

c) Land farming is best for the environmental management of drilling cuttings. Land farming

or land application is a process whereby drilling fluids are spread on the land surface. The

rate of application depends on the characteristics of the soil and the chemical composition

of the drilling fluid. Land application is often preferred to reduce the costs of disposal, and

may be conducted either on-site (where the drilling occurred) or at an off-site location. This

process also is used to treat and/or dilute potentially harmful constituents in the drilling

fluid, when present. These constituents may include petroleum hydrocarbons, salts and/or

heavy metals. (6

marks)

TOTAL = 20 marks

QUESTION FOUR

i. DO - Dissolved oxygen (DO) is very essential for the sustenance of plant and

animal life in aquatic systems.

Processes which affect the level of DO in water include re-aeration, photosynthesis,

respiration and oxidation of wastes. Saturation concentration of oxygen in water decreases

with increasing temperature. The primary cause of de-oxygenation of aquatic systems is the

presence of organic substances collectively called oxygen demanding wastes.

10

A fall in DO level is one of the first indications that a water body is polluted by organic

matter. (2 marks)

ii. BOD - is a measure of the oxygen utilized by micro-organisms during oxidation of

organic materials.

It is the most widely known parameter used for assessing water pollution potential of an

organic waste. A standard method for measuring BOD is by determining the amount of

oxygen required by the bacteria during the first five days of decomposition at 20o

C.

(2 marks)

iii. Chemical oxygen demand (COD) is the amount of oxygen required to cause chemical

oxidation of organic matter. (2 marks)

b. i. Primary Treatment Processes

Primary treatment removes identifiable suspended solids and floating matter.

It is a combination of simple physical processes designed to remove large floating or

suspended solids or solids that can readily settle under gravity.

Primary wastewater treatment usually involves screening, grit removal and sedimentation,

pre-treatment for coagulation, flocculation, and settling and sludge removal.

(2 marks)

ii. Secondary Treatment Processes

Secondary treatment removes organic matter that is soluble or is in the colloidal form.

Secondary treatment involves the bio-degradation of organic matter in solution or

suspension, in the presence of added oxygen, until the BOD of the water has been adjusted

to acceptable levels.

The principal accomplishment of secondary wastewater treatment lies in the removal of

suspended matter by settling, flotation, biochemical degradation and adsorption on the

sludge by the primary treatment processes followed by biodegradation processes.

(2 marks)

iii. Tertiary Treatment Processes

Tertiary treatment removes residual soluble non-biodegradable organic

compounds, including surfactants, inorganic nutrients and salts, trace contaminants of

various types, and dissolved inorganic salts.

The Tertiary treatment processes are expensive, and are used only when higher quality

water than that produced by conventional secondary treatment is required, which can be

put to direct re-use.

11

The processes normally applied in Tertiary treatment are:

Removal of suspended solids – micro straining, coagulation and filtration

Removal of dissolved solids – adsorption on activated carbon, ion exchange, reverse osmosis,

electro dialysis (2 marks)

c. OZONE DEPLETION

i. Desulfurization of fossil fuels

Desulfurization is a process whereby sulfur (organic and inorganic) is removed through

reaction with hydrogen to form H2S, which is collected and used for the production of sulfur.

This reaction is normally conducted on commercial catalysts packed in bed at temperatures of

300-500o

C and pressures of about 40-150 atm.

Hydro-desulfurization is by far the most effective method for sulfur removal from fuels

ii. Sulfur Reduction During Combustion

In this method finely ground limestone (CaCO3) is injected directly into the combustion

chamber, where it is calcined to CaO. The latter reacts with SO2 contained in the flue gas to

form sulfites and sulfates:

3.....2/1

2......

1......

422

32

23

CaSOOSOCaO

CaSOSOCaO

COCaOCaCO

The solid reaction products, unreacted materials and fly ash are removed by dry or wet

collectors.

iii. Treatment of Flue Gases

Flue gases can be treated through desulfurization processes, which are classified as dry or wet.

Dry process

Dry processes take place at elevated temperatures.

Adsorption of SO2 by metal oxides to form stable sulfites or sulfates with

subsequent regeneration of the oxide and recovery of sulfur, and

Adsorption on activated carbon followed by regeneration and conversion of

concentrated SO2 to H2SO4 or elemental S.

The metal oxides include aluminium sodium oxide, manganese oxide, cobalt and copper

oxides.

4......2

1. 324222322 OAlSONaOSOOAlONa

Wet process

SO2 can also be removed from a flue gas using wet processes.

12

One of such method is the scrubbing of flue gas with slurry of magnesium oxide in a

venture scrubber, which absorbs SO2 and yields magnesium sulfite and sulfate according to

reaction

8.....2

1

7.....2

6.....

5.....

423

2323

23223

32

MgSOOMgSO

OHMgSOMgOHSOMg

HSOMgOHSOMgSO

MgSOSOMgO

(4 marks for any ONE, total of 8 marks)

TOTAL = 20 marks

QUESTION FIVE

a) Air pollution is the presence of any solid, liquid or gaseous substance including

noise and radioactive radiation in the atmosphere in such concentration that may

be directly and indirectly injurious to humans or other living organisms, plants,

property or interferes with the normal environmental processes. Air pollutants are

of two types

(1) suspended particulate matter, and

(2) gaseous pollutants like carbon dioxide (CO2 ), NOx (6 marks)

b) The five primary criteria pollutants include the gases- Carbon Monoxide (CO),

nitrogen oxides (NOx) , sulfur dioxide (SO2), and solid or liquid particulates

(smaller than 10 µm), and particulate lead. (4 marks)

c) OZONE DEPLETION Substances (ODS) are compounds which possess high

ozone depletion potentials. As O3 absorbs UV rays it is broken down into O2

molecules and radicals:

CFCs are inert, non-toxic halogenated hydrocarbons, which have excellent

industrial applications such as Refrigeration and Air-conditioning, foam production,

fire-fighting, agro-chemicals (pesticides and fungicides).CFCs also cause depletion

of the ozone layer leading to gaping holes in the stratosphere, thus allowing ultra-

violet (UV) rays from the sun to reach the earth thereby causing the following:

Eye cataracts

Human immune deficiency

Stunted plant growth and by implication low agricultural productivity

ODS have been and are still being replaced by non-ozone depleting substitutes such as hydro-

chloro-fluoro-carbons (HCFCs) and hydrofluorocarbons (HFCs) as well as new technologies.

For e.g. refrigerators now use natural substances such as n-butane and i-butane as refrigerants.

13

2

23

2

23

OClOClO

OClOOCl

ClCFCluvCFCl

ClCFCluvCFCl

Ways of minimization

Limit private vehicle driving

Use eco-friendly household cleaning products

Avoid using pesticides

Developing stringent regulations for rocket launches

Banning the use of dangerous nitrous oxide (5 marks)

CLIMATE CHANGE

Climate Change is the result of Global Warming resulting from increasing concentrations of

“Greenhouse Gases” in the atmosphere, they include CO2, CH4, N2O, O3, and CFCs.

These gases create a condition in the upper atmosphere causing a trapping of infra-red (IR)

rays – heat, which in turn leads to increased surface and lower tropospheric temperatures.

Greenhouse gases give rise to Global Warming, which is responsible for world-wide Climate

Change

Ways of minimization

Be energy efficient

Choose renewable power

Eat wisely

Trim your waste

Get informed

Cleaner alternative energy sources

Reduce Reuse Recycle practices

Re-forestation (5 marks)

TOTAL = 20 marks

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