PET 504 E Adv. Well Test Analysis Final Project 23.05.2015 505132501 EMRE CENGZ

Sayfa 1 / 18

Eq. 1 in the question booklet inverted using Stehfest Algorithm and compared with the time domain

solution and it shows pretty good match. All these calculations can be found in both matlab codes

and excel calculations. Excel sheet and matlab codes provided as attachment.

Delta pressure and pressure derivative calculations are done by using the pressure solution

hereafter, compared with Ecrin test design solution.

Figure 1:Calculated pressure and pressure derivative compared with test design data from ecrin. It shows pretty good match for early and middle time. As the model does not have WBS and skin there is no 1 slope and hump appared in the log log graph.

0

1

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9

0,E+00 2,E+06 4,E+06 6,E+06 8,E+06 1,E+07 1,E+07

pD

tD

PDFWS (RD=1)

PDFWS (RD=2)

PDFWS (RD=5)

PDFWS (RD=10)

PDFWS (RD=20)

Inversion (RD=1)

Inversion (RD=2)

Inversion (RD=5)

Inversion (RD=10)

Inversion (RD=20)

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10

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1000

1,00E-03 1,00E-02 1,00E-01 1,00E+00 1,00E+01

dp

&d

p',

psi

time,hr

Ecrin DP

Ecrin DP'

Calc DP

Calc DP'

Sayfa 2 / 18

There are 6 different flow period in during the test and during 2 of them the well is shut in.

Therefore, the pressure change in the reservoir is calculated and compared with simulated synthetic

data with ecrin.

Figure 2: Calculated pressure history during the test compared with simulated data with ecrin. Plots validate each other for the all flow periods. There is tiny contention in the final flow period. It has been tried several times to fix this error. However, MatLAB wins.

In the second step, WBS and Skin added to pressure model and compared with ecrin data then log

log derivative response are compared. First shut in period is calculated and evaluated in this

example.

Figure 3: Good match is observed with ecrin dp and calculated dp. Also, Caclulated ecrin delta pressure derivative and calculated pressure derivative shows good match. As the model defines wellbore storage:0.015 bbl/psi and 2.5 skin factor, there is hump and 1 slope is observer in the early time behaviour of the pressure derivative data.

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Pre

ssu

re, p

si

time, hr

Ecrin

Calc

0,1

1

10

100

1000

1,00E-03 1,00E-02 1,00E-01 1,00E+00 1,00E+01

Ecrin Dp

Ecrin Dp'

Calc DP

Calc DP'

Sayfa 3 / 18

Regarding the the pressure behaviour due to production, and shut in periods, simulated data from

ecrin and calculated data shows a good match during all flow periods except the latest one.

Calculated pressure derivative analysis of the reservoir system with 0.015 bbl/psi WBS and 2.5 skin

and a fault located at 200 ft shows good match with the ecrin simulated data. The unconformity at

the late time region of the pressure derivative as a cause of miscalculated pressure data after 5

hours.

For the last part of the problem a single sealing fault is located at the 200 ft away from the well.

Fairly good pressure match is observed with ecrin pressure history.

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4550

0 20 40 60 80 100 120

Ecrin

Calc

0,1

1

10

100

1000

1,00E-03 1,00E-02 1,00E-01 1,00E+00 1,00E+01

DP

an

d D

p'

time, hr

Ecrin DP

Ecrin DP'

Calc. DP

Calc. DP'

Sayfa 4 / 18

As we can see pressure history has a good work, however, like previous pressure histories last build

up data is problematic. Ecrin solution and calculated solution match at the end however during build

up period there is a small gap between them.

Variable Rate Analysis 2

Production rate is given as a function q(t)=500-180ln(t). It was not possible to input this function into

the Ecrin. Instead, we calculated rates for each one hour period. After 17 hours rate function

indicates, well start injection. Thus, interesting and complicated results are revealed.

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Pre

ssu

re, p

si

time, hr

Ecrin

Calc

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-100

0

100

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500

600

0 20 40 60 80 100 120Flo

w r

ate

, STB

/d

time, hr

q(t)=500-180ln(t)

Sayfa 5 / 18

Constant Rate Analysis

NOWBS NO SKIN

Constant Rate Analysis

This part of the study is intended to analyse the pressure drop and model validation of a system

which produce 500 STB/d constant rate during 100 hours. Three model has been validated in this

part. The calculations fort he graph can be found in the MatLab code and results can be found in the

excel file. Model verification includes

1. A well with No WBS and No Skin

2. A well with C:0.015 bbl/psi WBS and S=2,5

3. A well with C:0.015 bbl/psi WBS, S=2,5 and A fault located 200ft away from the well.

Figure 4: Production rate is 500STB/D during 100 hours test for all 3 cases.

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0 50 100

Pre

ssu

re, p

si

time, hr

q=500-180ln(t) for NO WBS and No Skin

EcrinPressureHistory

Calc.PressureHistory

350

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450

500

550

0 20 40 60 80 100

flo

w r

ate

, STB

/d

time, hr

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4400

4500

4600

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4800

0 50 100

Pre

ssu

re, p

si

time, hr

q(t)=500-180ln(t) for 0.015 WBS +2,5 Skin

EcrinPressureHistory

Calc.PressureHistory

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Pre

ssu

re, P

si

time,hr

WBS+2,5Skin, A fault 200ft away

Ecrin Pressure History

Calc.Pressure History

Sayfa 6 / 18

1

10

100

1.000

1,00E-03 1,00E-02 1,00E-01 1,00E+00 1,00E+01 1,00E+02

Pre

ssu

re, p

si

time, hr

Ecrin pD

Ecrin pD'

Calc. pD

Calc. pD'

1. A well with NO WBS and NO Skin

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4550

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Pre

ssu

re,p

si

time, hr

Figure 5: Ecrin pD and pD' shows a good match with the calculated pD and pD'.

Figure 6: Pressure history is some how problematic for all cases especially in the late time region.

Sayfa 7 / 18

2. A well with C:0.015bbl/psi, and S=2,5

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100

1000

1,00E-03 1,00E-02 1,00E-01 1,00E+00 1,00E+01 1,00E+02

Pre

ssu

re, p

si

time, hr

Ecrin dP

Ecrin dP'

Calc. dP

Calc. dP'

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Pre

ssu

re, p

si

time, hr

Sayfa 8 / 18

3.WBS=0.015, S=2,5 a fault located at 200 ft away.

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10

100

1000

1,00E-03 1,00E-02 1,00E-01 1,00E+00 1,00E+01 1,00E+02

Pre

ssu

re, p

si

time,hr

Ecrin dP

Ecrin dP'

Calc. dP

Calc dP''

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ssu

re, p

si

time, hr

Sayfa 9 / 18

APPENDIX

Excel files and MatLAB codes will be provided by mail. Also, MatLab code can be found as follows.

Code used in Excel File to get inverted PD from laplace domain by using stehfest.

Function fact(n As Double) As Double

fact = 1

For t = 1 To n

fact = fact * t

Next t

End Function

Function v(i As Double, n As Double) As Double

Dim g As Double

Dim k As Double

Dim h As Double

Dim start1 As Integer

Dim finish As Integer

g = 0

k = 0

h = 0

If i > (n / 2) Then

finish = (n / 2)

Else

finish = i

End If

If ((i + 1) Mod 2) = 0 Then

start1 = (i + 1) / 2

Else

start1 = (i / 2)

End If

h = (-1) ^ ((n / 2) + i)

For k = start1 To finish

up = 0

down = 0

up = (k ^ (n / 2)) * fact(2 * k)

Sayfa 10 / 18

down = fact((n / 2) - k) * fact(k) * fact(k - 1) * fact(i - k) * fact((2 * k) - i)

g = g + (up / down)

Next k

v = h * g

End Function

Matlab Script

%% Inputs Pi=4500;

Bo=1.2; h=50;

rw=0.3;

ct=0.00001; viscosity=2.3; porosity=0.18; k=150; qs=500;

skin=2.5; wbs=0.015; cd=5.615*wbs/(2*pi*porosity*ct*h*rw^2);

t=[logspace(-

3,1,100),mylog(10.001,15,100),mylog(15.001,25,1000),mylog(25.001,40,100),my

log(40.001,50,100),mylog(50.001,100,100)]; t=t'; tt=numel(t); a=[1 100 200 1200 1300 1400 1500]; td=(0.000264*k*t)/(porosity*ct*viscosity*rw^2); %dimensionless time rd=[1 2 5 10 20 50 100]; rr=numel(rd); %dimensionless radius pdfws=zeros(tt,rr); pdsth=zeros(tt,rr); pdsths=zeros(tt,rr); dp1=zeros(tt,rr); dp2=zeros(tt,rr); dp3=zeros(tt,rr); dp12=zeros(tt,rr);

dp22=zeros(tt,rr);

q=zeros(tt,1); q(1:a(2))=500; q((a(2)+1):a(3))=300; q((a(3)+1):a(4))=0; q((a(4)+1):a(5))=200; q((a(5)+1):a(6))=600; q((a(6)+1):a(7))=0;

q2=2583.3-180*log(td);

% q1, q2 and qs are flowrate functions % dp1: pressure difference with no wellbore storage & skin in q1 function % dp2: pressure difference with wellbore storage & skin in q1 function % dp3: pressure difference with wellbore storage & skin & 200ft fault in q1

function % dp501: pressure difference with no wellbore storage & skin in 500 stb/day % dp502: pressure difference with wellbore storage & skin in 500 stb/day % dp503: pressure difference with wellbore storage & skin & 200ft fault in

500 stb/day % pdsth: dimensionless pressure function with no wellbore storage & skin % pdsths: dimensionless pressure function with wellbore storage & skin % pdfws: dimensionless pressure function of finite wellbore solution in

time domain with no wellbore storage and skin

Sayfa 11 / 18

% dp12: pressure difference with no wellbore storage & skin in q2 function % dp22: pressure difference with wellbore storage & skin in q2 function

%% Calculation for i=1:rr for j=1:tt if j>=1 && j=a(2)+1 && j=a(3)+1 && j

Sayfa 12 / 18

(Stehfest((td(j)-td(a(2))),(4000/3),8)*(q(a(3))-

q(a(2))))+(pdsths(j,i)*(q(j)-q(a(3))))+(Stehfest((td(j)-

td(a(3))),(4000/3),8)*(q(j)-q(a(3)))));

dp12(j,i)=((141.2*Bo*viscosity)/(k*h))*((Stehfest(td(j),rd(i),8)*q2(a(2)))+

(Stehfest((td(j)-td(a(2))),rd(i),8)*(q2(a(3))-

q2(a(2))))+(pdsth(j,i)*(q2(j)-q2(a(3)))));

dp22(j,i)=((141.2*Bo*viscosity)/(k*h))*((Stehfest2(td(j),rd(i),8)*q2(a(2)))

+(Stehfest2((td(j)-td(a(2))),rd(i),8)*(q2(a(3))-

q2(a(2))))+(pdsths(j,i)*(q2(j)-q2(a(3)))));

dp3f(j,i)=((141.2*Bo*viscosity)/(k*h))*((Stehfest2(td(j),rd(i),8)*q2(a(2)))

+(Stehfest(td(j),(4000/3),8)*q2(a(2)))+(Stehfest2((td(j)-

td(a(2))),rd(i),8)*(q2(a(3))-q2(a(2))))+... (Stehfest((td(j)-td(a(2))),(4000/3),8)*(q2(a(3))-

q2(a(2))))+(pdsths(j,i)*(q2(j)-q2(a(3))))+(Stehfest((td(j)-

td(a(3))),(4000/3),8)*(q2(j)-q2(a(3))))); else if j>=a(4)+1 && j

Sayfa 13 / 18

else if j>=a(5)+1 && j=a(6)+1 && j

Sayfa 14 / 18

dp12(j,i)=((141.2*Bo*viscosity)/(k*h))*((Stehfest(td(j),rd(i),8)*q2(a(2)))+

(Stehfest((td(j)-td(a(2))),rd(i),8)*(q2(a(3))-q2(a(2))))+(Stehfest((td(j)-

td(a(3))),rd(i),8)*(q2(a(4))-q2(a(3))))+... (Stehfest((td(j)-

td(a(4))),rd(i),8)*(q2(a(5))-q2(a(4))))+(Stehfest((td(j)-

td(a(5))),rd(i),8)*(q2(a(6))-q2(a(5))))+(pdsth(j,i)*(q2(j)-q2(a(5)))));

dp22(j,i)=((141.2*Bo*viscosity)/(k*h))*((Stehfest2(td(j),rd(i),8)*q2(a(2)))

+(Stehfest2((td(j)-td(a(2))),rd(i),8)*(q2(a(3))-

q2(a(2))))+(Stehfest2((td(j)-td(a(3))),rd(i),8)*(q2(a(4))-q2(a(3))))+... (Stehfest2((td(j)-

td(a(4))),rd(i),8)*(q2(a(5))-q2(a(4))))+(Stehfest2((td(j)-

td(a(5))),rd(i),8)*(q2(a(6))-q2(a(5))))+(pdsths(j,i)*(q2(j)-q2(a(5)))));

dp3f(j,i)=((141.2*Bo*viscosity)/(k*h))*((Stehfest2(td(j),rd(i),8)*q2(a(2)))

+(Stehfest(td(j),(4000/3),8)*q2(a(2)))+(Stehfest2((td(j)-

td(a(2))),rd(i),8)*(q2(a(3))-q2(a(2))))+... (Stehfest((td(j)-

td(a(2))),(4000/3),8)*(q2(a(3))-q2(a(2))))+(Stehfest2((td(j)-

td(a(3))),rd(i),8)*(q2(a(4))-q2(a(3))))+(Stehfest((td(j)-

td(a(3))),(4000/3),8)*(q2(a(4))-q2(a(3))))+... (Stehfest2((td(j)-

td(a(4))),rd(i),8)*(q2(a(5))-q2(a(4))))+(Stehfest((td(j)-

td(a(4))),(4000/3),8)*(q2(a(5))-q2(a(4))))+(Stehfest2((td(j)-

td(a(5))),rd(i),8)*(q2(a(6))-q2(a(5))))+... (Stehfest((td(j)-

td(a(5))),(4000/3),8)*(q2(a(6))-q2(a(5))))+(pdsths(j,i)*(q2(j)-

q2(a(6))))+(Stehfest((td(j)-td(a(6))),(4000/3),8)*(q2(j)-q2(a(6)))));

dp1(j,i)=((141.2*Bo*viscosity)/(k*h))*((Stehfest(td(j),rd(i),8)*q(a(5)))+(S

tehfest((td(j)-td(a(6))),rd(i),8)*-q(a(5))));

dp2(j,i)=((141.2*Bo*viscosity)/(k*h))*((Stehfest2(td(j),rd(i),8)*q(a(5)))+(

Stehfest2((td(j)-td(a(6))),rd(i),8)*-q(a(5))));

dp3(j,i)=((141.2*Bo*viscosity)/(k*h))*((Stehfest2(td(j),rd(i),8)*q(a(5)))+(

Stehfest(td(j),(4000/3),8)*q(a(5)))+(Stehfest2((td(j)-td(a(6))),rd(i),8)*-

q(a(5)))+... (Stehfest((td(j)-

td(a(6))),(4000/3),8)*-q(a(5)))); end end end end end end end end

for i=1:rr for j=1:tt

dp501(j,i)=((141.2*Bo*viscosity)/(k*h))*(Stehfest(td(j),rd(i),8)*qs);

dp502(j,i)=((141.2*Bo*viscosity)/(k*h))*(Stehfest2(td(j),rd(i),8)*qs);

dp503(j,i)=((141.2*Bo*viscosity)/(k*h))*(Stehfest2(td(j),rd(i),8)*qs+Stehfe

st(td(j),(4000/3),8)*qs); end end

Sayfa 15 / 18

dp1=Pi-dp1; dp2=Pi-dp2; dp3=Pi-dp3; dp12=Pi-dp12; dp22=Pi-dp22; dp3f=Pi-

dp3f; dp501=Pi-dp501; dp502=Pi-dp502; dp503=Pi-dp503;

%% Plots f1=plot(t,dp1(:,1)); f2=plot(t,dp2(:,1)); f3=plot(t,dp3(:,1)); f4=plot(t,dp501(:,1)); f5=plot(t,dp502(:,1)); f6=plot(t,dp503(:,1)); f7=plot(t,dp12(:,1)); f8=plot(t,dp22(:,1)); f9=plot(t,dp3f(:,1));

% DP501, DP502 & DP503 are delta pressures for 2nd buildup section for

functions dp501, dp502 & dp503 % DDP501, DDP502 & DDP503 are differential delta pressures for 2nd buildup

section for functions dp501, dp502 & dp503 % DP1, DP2 & DP3 are delta pressures for 2nd buildup section for functions

dp1, dp2 & dp3 % DDP1, DDP2 & DDP3 are differential delta pressures for 2nd buildup

section for functions dp1, dp2 & dp3 % DP31 & DP32 are delta pressures for 1st buildup section for functions dp1

& dp2 % DDP31 & DDP32 are differential delta pressures for 1st buildup section

for functions dp1 & dp2

%% Delta Pressure and Pressure Derivative for Drawdown of 500 stb/day

constant production for u=1:tt DP501(u)=Pi-dp501(u); DP502(u)=Pi-dp502(u); DP503(u)=Pi-dp503(u); dt(u)=t(u); end

for u=2:tt-1 DDP501(u-1)=(DP501(u+1)-DP501(u-1))/(log(t(u+1)/t(u-1))); DDP502(u-1)=(DP502(u+1)-DP502(u-1))/(log(t(u+1)/t(u-1))); DDP503(u-1)=(DP503(u+1)-DP503(u-1))/(log(t(u+1)/t(u-1))); end

%% Delta Pressure and Pressure Derivative for Buildup#1 and Buildup#2 m=0; for i=a(6):a(7) m=m+1; DP1(m)=dp1(i)-dp1(a(6)); DP2(m)=dp2(i)-dp2(a(6)); DP3(m)=dp3(i)-dp3(a(6)); dte(m)=t(i)-t(a(6)); end

for m=2:(a(7)-a(6)) DDP1(m-1)=(DP1(m+1)-DP1(m-1))/(log(dte(m+1)/dte(m-1))); DDP2(m-1)=(DP2(m+1)-DP2(m-1))/(log(dte(m+1)/dte(m-1))); DDP3(m-1)=(DP3(m+1)-DP3(m-1))/(log(dte(m+1)/dte(m-1))); end

ss=0;

Sayfa 16 / 18

for i=a(3):a(4) ss=ss+1; DP31(ss)=dp1(i)-dp1(a(3)); DP32(ss)=dp2(i)-dp2(a(3)); DP33(ss)=dp3(i)-dp3(a(3)); dte3(ss)=t(i)-t(a(3)); end

for ss=2:(a(4)-a(3)) DDP31(ss-1)=(DP31(ss+1)-DP31(ss-1))/(log(dte3(ss+1)/dte3(ss-1))); DDP32(ss-1)=(DP32(ss+1)-DP32(ss-1))/(log(dte3(ss+1)/dte3(ss-1))); DDP33(ss-1)=(DP33(ss+1)-DP33(ss-1))/(log(dte3(ss+1)/dte3(ss-1))); end

dte(1)=[]; dte(end)=[]; dte3(1)=[]; dte3(end)=[]; DP1(1)=[]; DP1(end)=[];

DP2(1)=[]; DP2(end)=[]; DP3(1)=[]; DP3(end)=[]; DP31(1)=[]; DP31(end)=[]; DP32(1)=[]; DP32(end)=[]; DP33(1)=[]; DP33(end)=[]; DP501(1)=[];

DP501(end)=[]; DP502(1)=[]; DP502(end)=[]; DP503(1)=[]; DP503(end)=[];

dt(1)=[]; dt(end)=[];

figure=loglog(dte,DP1,dte,DDP1); figure=loglog(dte,DP2,dte,DDP2); figure=loglog(dte,DP3,dte,DDP3); figure=loglog(dte3,DP31,dte3,DDP31); figure=loglog(dte3,DP32,dte3,DDP32); figure=loglog(dte3,DP33,dte3,DDP33); figure=loglog(dt,DP501,dt,DDP501); figure=loglog(dt,DP502,dt,DDP502); figure=loglog(dt,DP503,dt,DDP503);

% Comparison of PDSTH (Stehfest Inversion Algorithm) and PDFWS (Finite

wellbore solution in time domain) % for i=1:rr % for j=1:tt % pdfws(j,i)=(2/pi)*integral(@(u)

(((bessely(0,rd(i)*u).*besselj(1,u)-

besselj(0,rd(i)*u).*bessely(1,u))/(besselj(1,u).^2+bessely(1,u).^2))*((1-

exp(-td(j)*u.^2))/u.^2)),0,10^3,'ArrayValued',true); % pdsth(j,i)=Stehfest(td(j),rd(i),8); % end % end

Stehfest2 Function.

function [InverseLT]=Stehfest2(t,rd,L) a=L/2; b=a+1; fh=@pdlss;

for n=1:L z=0.0; for k=floor((n+1)/2):min(n,a) z=z+((k^a)*factorial(2*k))/ ... (factorial(a-k)*factorial(k)*factorial(k-1)* ... factorial(n-k)*factorial(2*k-n)); end v(n)=(-1)^(n+a)*z; end

Sayfa 17 / 18

sum=0.0; ln2ot=log(2.0)/t;

for n=1:L p=n*ln2ot; sum=sum+v(n)*fh(p,rd); end

InverseLT=sum*ln2ot;

Stehfest Function function [InverseLT]=Stehfest(t,rd,L) a=L/2; b=a+1; fh=@pdls;

for n=1:L z=0.0; for k=floor((n+1)/2):min(n,a) z=z+((k^a)*factorial(2*k))/ ... (factorial(a-k)*factorial(k)*factorial(k-1)* ... factorial(n-k)*factorial(2*k-n)); end v(n)=(-1)^(n+a)*z; end

sum=0.0; ln2ot=log(2.0)/t;

for n=1:L p=n*ln2ot; sum=sum+v(n)*fh(p,rd); end

InverseLT=sum*ln2ot;

pdls function

function f=pdls(s,rd) f=(besselk(0,rd*sqrt(s))/((s^1.5)*besselk(1,sqrt(s))));

end

pdlss function

function f=pdlss(s,rd) skin=2.5; cd=1654.9; %

f=1/(s*(s*cd+(1/(skin+(besselk(0,rd*sqrt(s))/(sqrt(s)*besselk(1,sqrt(s)))))

))); f=(1/s)*((skin+s*(besselk(0,rd*sqrt(s))/((s^1.5)*besselk(1,sqrt(s)))))/(1+s

*cd*(skin+s*(besselk(0,rd*sqrt(s))/((s^1.5)*besselk(1,sqrt(s)))))));

end

Sayfa 18 / 18

mylog function

function y = mylog(d1, d2, n)

if nargin == 2 n = 50; end

if d2 == pi || d2 == single(pi) d2 = log10(d2); end

y = 10.^ linspace(log10(d1), log10(d2), n);

end