PCB3013-Well Test Analysis Tutorial# 1

Prof. Dr. Mustafa Onur, UTP, September 2013

PLOTTING DATA ON LOG-LOG, SEMI-LOG AND MILLIMETRIC GRAPH PAPERS Let’s consider pressure change functions as given below:

bwf attppp =−=Δ )(0 (1)

atbtppp wf +=−=Δ )ln()(0 (2)

and

abttppp wf +=−=Δ )(0 (3) In Eqs. 1-3, a and b are positive constants.

As we discussed in class, the Bourdet derivative (or pressure-derivative) of a function ∆p is

defined as the derivative of the function with respect to natural logarithm of time, given as:

( )dt

pdttd

pdtp Δ=

Δ=′Δ

ln (4)

Then, using Eq. 4 for the functions given by Eqs. 1, 2, and 3, we can show that the Bourdet

derivatives of the functions given by Eqs. 1-3 are given respectively by:

( ) babttp =′Δ (5)

( ) btp =′Δ (6)

and

bttp =′Δ )( (7)

(Mohamed Ali and/or Azeb, please show the students how Eqs. 5-7 are obtained from Eqs.

1-3, respectively.)

Suppose we make a log-log plot of ∆p versus t data based on the function given Eq. 1. It can be

shown that ∆p versus t based on Eq. 1 will yield a straight line with a slope equal to b: To show

this take a log of both sides of Eq. 1 to obtain:

( ) ( )

( ) )log(log)log()log(

loglog

atbta

atpb

b

+=+=

=Δ

(8)

As we plotting log(∆p) versus log(t) on a log-log plot, we will have a straight line with a slope

equal to b. Note that t = 1, log(∆p (t=1))=log(a), which is called the intercept at t = 1. Note that

on log-log plot, we do not need to take the log of (ab). The intercept at t = 1 will be directly

equal to the value of a. The beauty of log-log plot is that when we plot data we do not need to

take log of the numbers when we plot the numbers.

Suppose now, we plot the Bourdet derivative given by Eq. 5 versus on a log-log plot. What

would be slope of a line passing through data based Eq. 5? Again to determine the slope, we take

the logarithm of both sides of Eq. 5, which gives:

( ) ( )( )abtb

abtp b

log)log(loglog

+==′Δ

(9)

The slope will be equal to b as in the case of Eq. 8, but the intercept at t = 1 for Eq. 9 will be

equal to log(ab). Note that on log-log plot, we do not need to take the log of (ab). The intercept

at t = 1 will be directly equal to the value of ab. The beauty of log-log plot is that when we plot

data we do not need to take log of the numbers when we plot the numbers.

If we make log-log plot of ∆p data versus t based on Eq. 2, do you think we will get a straight

line? Mohamed and/or Azeb, Please explain to student that he/she will not get a straight

line. Show that you will get a straight line if you plot the same data on a semilog plot; i.e., plot of

dp versus log(t). It should be noted that log here is based 10, it is not the natural log (i.e., not ln)

In this case, the slope will be equal to 2.303*b. Why we have 2.303 in front of b? Please explain

the students Mohammed and/or Azeb?

Suppose we plot Bourdet derivative of ∆p function versus t given by Eq. 2, which is given by Eq.

6, on a log-log plot? Do you think we get a “straight” line? Yes, but the slope will be equal to

zero. A straight line with a zero-slope is called a constant with time. So, taking log of both sides

of Eq. 6 we see that the value of this constant line is equal to log(b):

( ) ( )bp loglog =′Δ (6)

Again, on a log-log when Eq. 6 applies, we just read the value of b not taking its log.

Then, please consider Eq. 3 and its Bourdet derivative given by Eq. 7?

Ask the student, what will be the slope of the line if ∆p versus t given by Eq. 3 is plotted on a

millimetric graph paper? The slope will be equal to b and intercept at t = 0, will be equal to a.

Ask the student what will be the slope of the line if ∆p versus t given by Eq. 3 is plotted on a log-

log graph paper? Actually, this is tricky because in this case, you will not have a straight line

on a log-plot unless a equal to zero. Explain student why?

Now consider Eq. 7; Ask the student what will be the slope of the line if Bourdet derivative of

∆p function versus t given by Eq. 7 is plotted on a log-log graph paper? The slope will be equal

to unity. What will be the in intercept at t = 1? It will be log(b). Show the students why?

Now let the students plot data on semi-log, log-log, and millimetric paper:

Problem 1: Plot the data given below on a semi-log plot and compute the slope and the intercept

at t = 1 hr. Pressure at t = 0 is equal to p0 = 4412 psia.

Problem 2: Consider the same data given above for problem 1, but this time construct dp data

using ∆p = p0-pwf. Then plot ∆p versus t on a semilog paper to compute slope and intercept at t =

1 hr.

Problem 3: Consider the data given in table below.

Pressure and derivative versus time data for the observation well.

Test duration

t, hours Gauge pressure, psi Delta pressure, Δp,

psi

Derivative, dΔp/dlnt, psi

0 3253 0 - 0.01 3251.3 1.72 1.6 0.1 3239.4 13.6 10.4 0.75 3198 55 26.4 1.416667 3180 73 29.9 3.2 3154 99 30.5 4.95 3141 112 31.3 6.95 3130 123 31.9 10.9 3116 137 27.7 16.38333 3106 147 26.2 22.451 3099 154 26.7 40.25 3083.6 169.4 26.2 48 3079 174 -

Plot delta p and delta p’ versus t data on log-log plot and identify straight lines if exist with their

appropriate slopes on delta p and delta p’ data.

Pressure Drawdown Test Data for Probem 1

t pwf t pwf 2 3510.3 18 3414.5 3 3492.7 24 3402.0 4 3480.1 30 3392.3 6 3462.4 36 3384.3 8 3449.9 48 3371.8 10 3440.2 60 3362.1 12 3432.2 72 3354.1 15 3422.5

0.001 0.01 0.1 1 10 100Time t (hr)

0.1

1

10

100

1000

Pres

sure

Cha

nge

and

its B

ourd

et D

eriv

ativ

e, p

si

Match Points(Δp)M = 10 psi,( t)M = 1 hr(pD)M = 0.2, (tD/CD) = 3.7(CDe2s)M = 3