Example 1.46 Design a pulse test using the followingapproximate properties:

/x = 3 cp, 0 = 0.18, k = 200 md

A = 25 ft, r = 600 ft, ct = 10 x 10~6 psr1

£ = lbbl/STB, Q = 100bbl/day, F^ =0.6

Solution

Step 1. Calculate (fL)D from Equation 1.6.24 or 1.6.25. SinceF^ is 0.6, the odd pulses should be used and thereforefrom Equation 1.6.24:

(*L)D= 0.09+ 0.3(0.6) =0.27

Step 2. Selecting the first odd pulse, determine the dimen-sionless cycle period from Figure 1.114 to get:

[«L)D A£]Fig = 0.106

Step 3. Determine the dimensionless response amplitudefrom Figure 1.110 to get:

[ApD(tL/Atc)2]Flg = 0.00275

Step 4. Solve for t^, Ate, and Ap by applying Equations1.6.26 through 1.6.28, to give:Time lag:

F(0.18) (3) (10 x IQ-6) (66O)2I

L (0.0002637) (200) J / '

= 4.7 hours

Cycle time:

Ak = ~y~ = ^ 7 =17.5 hours(fL) D 0.Z7

Pulse length (shut-in):

AfP = AfcFN = (17.5) (0.27) « 5 hours

Flow period:

Aft = Ate - Afp = 17.5 - 4.7 « 13 hours

Step 5. Estimate the pressure response from Equation1.6.28:

This is the expected response amplitude for odd-pulse anal-ysis. We shut in the well for 5 hours and produced for 13hours and repeated each cycle with a period of 18 hours.

The above calculations can be repeated if we desire toanalyze the first even-pulse response.

1.7 Injection Well Testing

Injectivity testing is a pressure transient test during injec-tion into a well. Injection well testing and the associatedanalysis are essentially simple, as long as the mobility ratiobetween the injected fluid and the reservoir fluid is unity.Earlougher (1977) pointed out that the unit-mobility ratio isa reasonable approximation for many reservoirs underwaterfloods. The objectives of injection tests are similar to those

Figure 1.118 Pulse pressure response for Example 1.45.

Time, hr

Inje

ctio

n R

ate,

bw

/d

Off

set

Pre

ssur

e, p

sia

Previous Page

of production tests, namely the determination of:

• permeability;• skin;• average pressure;• reservoir heterogeneity;• front tracking.

Injection well testing involves the application of one or moreof the following approaches:

• injectivity test;• pressure falloff test;• step-rate injectivity test.

The above three analyses of injection well testing are brieflypresented below.

1.7.1 Injectivity test analysisIn an injectivity test, the well is shut in until the pressureis stabilized at initial reservoir pressure p\. At this time,the injection begins at a constant rate q-mj, as schematicallyillustrated in Figure 1.119, while recording the bottom-holepressure wf. For a unit-mobility ratio system, the injectivitytest would be identical to a pressure drawdown test exceptthat the constant rate is negative with a value of qmy How-ever, in all the preceding relationships, the injection rate willbe treated as a positive value, i.e., q-^ > 0.

For a constant injection rate, the bottom-hole pressure isgiven by the linear form of Equation 1.3.1 as:

P^ =/>ihr +WlOg(O [1.7.1]

The above relationship indicates that a plot of bottom-hole injection pressure versus the logarithm of injectiontime would produce a straight-line section as shown inFigure 1.119, with an intercept of pi^r and a slope m asdefined by:

_ 162.6qm]B /xM~ kh

where:

#inj = absolute value of injection rate, STB/daym = slope, psi/cyclek = permeability, mdh = thickness, ft

Sabet (1991) pointed out that, depending on whether thedensity of the injected fluid is higher or lower than the reser-voir fluid, the injected fluid will tend to override or underridethe reservoir fluid and, therefore the net pay h which shouldbe used in interpreting injectivity tests would not be thesame as the net pay which is used in interpreting drawdowntests.

Earlougher (1977) pointed out that, as in drawdown test-ing, the wellbore storage has great effects on the recordedinjectivity test data due to the expected large value of thewellbore storage coefficient. Earlougher recommended thatall injectivity test analyses must include the log-log plot of(Pvrf —Pi) versus injection time with the objective of determin-ing the duration of the wellbore storage effects. As definedpreviously, the beginning of the semilog straight line, i.e., theend of the wellbore storage effects, can be estimated fromthe following expression:

(200 OOP + 12 QQOs) C

kh/n [ J

where:

t = time that marks the end of wellbore storage effects,hours

k = permeability, mds = skin factor

C = wellbore storage coefficient, bbl/psi\x = viscosity, cp

Once the semilog straight line is identified, the permeabil-ity and skin can be determined as outlined previously by:

ft=162-6^ [1.7.3]YHn

s = 1.1513 P l h r ~ A - log (—^) + 3.2275] [1.7.4]

The above relationships are valid as long as the mobilityratio is approximately equal to 1. If the reservoir is underwater flood and a water injection well is used for the injec-tivity test, the following steps summarize the procedure ofanalyzing the test data assuming a unit-mobility ratio:

Step 1. Plot (/>wf —pi) versus injection time on a log-log scale.Step 2. Determine the time at which the unit-slope line, i.e.,

45° line, ends.Step 3. Move 1 \ log cycles ahead of the observed time in

step 2 and read the corresponding time which marksthe start of the semilog straight line.

Step 4. Estimate the wellbore storage coefficient C byselecting any point on the unit-slope line and read-ing its coordinates, i.e., Ap and t, and applying thefollowing expression:

Step 5. Plot />wf vs. t on a semilog scale and determinethe slope m of the straight line that represents thetransient flow condition.

Step 6. Calculate the permeability k and skin factor fromEquations 1.7.3 and 1.7.4 respectively.

Step 7. Calculate the radius of investigation rmv at the end ofinjection time. That is:

rinv = 0.0359 l-p- [1.7.6]Y $№

Step 8. Estimate the radius to the leading edge of thewater bank rwb before the initiation of the injectivity

Figure 1.119 Idealized rate schedule and pressureresponse for injectivity testing.

Time, t

Time, t

INJECTING

SHUTIN

Rat

e, q

Bot

tom

-Hol

eP

ress

ure,

Pw

test from:

_ / 5.615PFini /5.615W ini

wb ~ V nH(Sv - S ^ ) - V 7^0(ASw) l J

where:

rwb = radius to the water bank, ftWinj = cumulative water injected at the start of the

test, bblSw = average water saturation at the start of the

testSwi = initial water saturation

Step 9. Compare rwb with r-mv: i£r-mv < rwb, the unit-mobilityratio assumption is justified.

Example 1.47" Figures 1.120 and 1.121 show pressureresponse data for a 7 hour injectivity test in a water-floodedreservoir in terms of log(/>wf —p\) vs. log(0 and log(/>wf) vs.log(0 respectively. Before the test, the reservoir had beenunder water flood for 2 years with a constant injection rate of100 STB/day. The injectivity test was initiated after shuttingin all wells for several weeks to stabilize the pressure atp\.The following data is available:

ct = 6.67 x 10~6 psi-1

£=1.0bbl/STB, M = 1.0cp

Sw = 62.41b/ft3, 0 = 0.15, tfinj = 100 STB/day

h = 16 ft, rw = 0.25 ft, p\ = 194 psig

ASW = 0.4, depth = 1002 ft, total test time = 7 hours

The well is completed with 2 inch tubing set on a packer.Estimate the reservoir permeability and skin factor.

a After Robert Earlougher, Advances in Well Test Analysis, 1977.

Solution

Step 1. The log-log data plot of Figure 1.120 indicates thatthe data begins to deviate from the unit-slope line atabout 0.55 hours. Using the rule of thumb of moving1 to 1 \ cycles in time after the data starts deviatingfrom the unit-slope line, suggests that the start ofthe semilog straight line begins after 5 to 10 hoursof testing. However, Figures 1.120 and 1.121 clearlyshow that the wellbore storage effects have endedafter 2 to 3 hours.

Step 2. From the unit-slope portion of Figure 1.120, selectthe coordinates of a point (i.e. ,Ap and t) end calcu-late the wellbore storage coefficient C by applyingEquation 1.7.5:

Ap = 408 psig

£ = 1 hour

c = QmBt24Ap

Step 3. From the semilog plot in Figure 1.121, determine theslope of the straight line m to give:

m — 710 psig/cycle

Step 4. Calculate the permeability and skin factor by usingEquations 1.7.3 and 1.7.4:

k = 162.6qmiBnmh

(162.6) (100) (1.0) (1.0)

= Wm 12-7md

Injection Time, t, hr

Figure 1.120 Log-log data plot for the injectivity test of Example 1.47. Water injection into a reservoir at staticconditions (After Earlougher, R. Advances in Well Test Analysis) (Permission to publish by the SPE, copyrightSPE, 1977).

UNITSLOPE

APPROXIMATE ENDOF WELLBORESTORAGE EFFECTS

Pre

ssur

e D

iffer

ence

, Pw

f-P/,

psig

_j / 12/7 \° g V (0.15) (1.0) (6.67 x 10-6) (0.25)2)

+ 3.2275~| =2 .4

Step 5. Calculate the radius of investigation after 7 hours byapplying Equation 1.7.6:

rinv =0.0359 -^L

- 0 0359 / (12-7) (7)" ^~ °' 0359V (0.15)(1.0)(6.67xl0-6) " 3 3 8 ft

Step 6. Estimate the distance of the leading edge of the waterbank before the start of the test from Equation 1.7.7:

Wn = (2) (365) (100) (1.0) = 73 000 bbl

/5.615Wlnj

_ I (5.615X73000)"V 7r (16) (0.15) (0.4)

Since rmv < r^, the use of the unit-mobility ratioanalysis is justified.

1.7.2 Pressure falloff testA pressure falloff test is usually preceded by an injectiv-ity test of a long duration. As illustrated schematically inFigure 1.122, falloff testing is analogous to pressure builduptesting in a production well. After the injectivity test thatlasted for a total injection time of tp at a constant injection

Figure 1.122 Idealized rate schedule and pressureresponse for falloff testing.

rate of q-m^ the well is then shut in. The pressure data takenimmediately before and during the shut in period is analyzedby the Horner plot method.

The recorded pressure falloff data can be represented byEquation 1.3.11, as:

*„=,. + „ [log (*±^)]with:

162.6ginj£Mm~ kh

where p* is the false pressure that is only equal to the initial(original) reservoir pressure in a newly discovered field. As

Time, t

INJECTING

SHUTIN

Injection Time, t, hr

Figure 1.121 Semilog plot for the injectivity test of Example 1.47. Water injection into a reservoir at static conditions(After Earlougher, R. Advances in Well Test Analysis) (Permission to publish by the SPE, copyright SPE, 1977).

APPROXIMATE END •OFWELLBORESTORAGE EFFECTS

SLOPE = m = 80 PSIG/CYCLE

P1 h r = 770 PSIG

Bot

tom

-Hol

e P

ress

ure,

pw

f, ps

ig

Bot

tom

-Hol

e P

ress

ure,

pw

Rat

e, q

Figure 1.123 Homer plot of a typical falloff test.

shown in Figure 1.123, a plot of pws vs. log [(tp 4- Af) /At]would form a straight-line portion with an intercept ofp* at(tv + At) /At = 1 and a negative slope of m.

It should be pointed out that the log-log data plot shouldbe constructed to identify the end of the wellbore storageeffects and beginning of the proper semilog straight line.The permeability and skin factor can be estimated as outlinedpreviously by the expressions:

162.6qmiBLi

\m\h

5=1 .513 •—• log T + 3.2275

L M \<t>mrlJ JEarlougher (1977) indicated that if the injection rate varies

before the falloff test, the equivalent injection time may beapproximated by:

. _ 24Winjtp —

tfinj

where W1n] is the cumulative volume injected since the lastpressure equalization, i.e., last shut-in, and #inj is the injectionrate just before shut-in.

It is not uncommon for a falloff test to experience a changein wellbore storage after the test begins at the end of theinjectivity test. This will occur in any well which goes onvacuum during the test. An injection well will go on vacuumwhen the bottom-hole pressure decreases to a value which isinsufficient to support a column of water to the surface. Priorto going on vacuum, an injection well will experience storagedue to water expansion; after going on vacuum, the storagewill be due to a falling fluid level. This change in storage willgenerally exhibit itself as a decrease in the rate of pressuredecline.

The falloff data can also be expressed in graphical form byplottingpvfS vs. log (AO as proposed by MDH (Miller-Dyes-Hutchinson). The mathematical expression for estimatingthe false pressure p* from the MDH analysis is given byEquation 1.3.12 as:

P* =Pihr~ M log(fp + 1) [1.7.8]

Earlougher pointed out that the MDH plot is more prac-tical to use unless tp is less than about twice the shut-intime.

The following example, as adopted from the work ofMcLeod and Coulter (1969) and Earlougher (1977), is usedto illustrate the methodology of analyzing the falloff pressuredata.

Example 1.48a During a stimulation treatment, brinewas injected into a well and the falloff data, as reported byMcLeod and Coulter (1969), is shown graphically in Figures1.124 through 1.126. Other available data includes:

total injection time fp = 6.82 hours,total falloff time = 0.67 hoursqm] = 807 STB/day, B* = 1.0 bbl/STB,

cw = 3.0 x 10"6 psi"1

0 = 0.25, h = 28 ft, Atw = 1.0 cp

Ct = LOxIO-5PSi-1, rw = 0.4ft, Sw = 67.46 lb/ft3

depth = 4819 ft,hydrostatic fluid gradient = 0.4685 psi/ft

The recorded shut-in pressures are expressed in terms ofwellhead pressures pts with f at A*=O = 1310 psig. Calculate:

• the wellbore storage coefficient;• the permeability;• the skin factor;• the average pressure.

Solution

Step 1. From the log-log plot of Figure 1.124, the semilogstraight line begins around 0.1 to 0.2 hours aftershut-in. Using Ap = 238 psi at At = 0.01 hoursas the selected coordinates of a point on the unit-slope straight line, calculate the wellbore storagecoefficient from Equation 1.7.5, to give:

c = QinjBt24Ap

Step 2. Figures 1.125 and 1.126 show the Horner plot,i.e., "wellhead pressures vs. log \{tp + At) /At]"and the MDH plot, i.e., "wellhead pressuresvs. log(AO, respectively, with both plots giving:

m = 270 psig/cycle

Pihr = 85 psig

Using these two values, calculate k and s:

162.6qmiBfi

\m\h

(162.6) (807) (1.0) (1.0)= — w m r n — = 17-4md

s = L 5 1 3 [A* at A,=0 - Pl hr _ 1 Y_fe \ + 3 > 2 2 7 5 1

[1310-85 / V7A Y]

" L 270 g V (0.25) (1.0) (1.0 x 10-5) (0.4)2 / J

+ 3.2275 = 0.15

Step 3. Determine p* from the extrapolation of the Hornerplot of Figure 1.125 to (fp + At)/At = 1, to give:

p*ts = -151 psig

Equation 1.7.8 can be used to approximate p*:

p*=plhr-\m\\og(tp + l)

p*ts = 85 - (270) log (6.82 + 1) = -156 psig

aRobert Earlougher, Advances in Well Test Analysis, 1977.

Shut-In Time, Af, hr

Bo

tto

m-H

ole

Sh

ut-

In P

ress

ure

"p

wS"

This is the false pressure at the wellhead, i.e., thesurface. Using the hydrostatic gradient of 0.4685psi/ft and the depth of 4819 ft, the reservoir falsepressure is:

p* = (4819) (0.4685) - 151 = 2107 psig

and since injection time tp is short compared withthe shut-in time, we can assume that:

p=p* = 2107 psig

Pressure falloff analysis in non-unit-mobilityratio systemsFigure 1.127 shows a plan view of the saturation distributionin the vicinity of an injection well. This figure shows twodistinct zones.

Zone 1. represents the water bank with its leading edge ata distance of % from the injection well. The mobil-ity X of the injected fluid in this zone, i.e., zone 1,is defined as the ratio of effective permeability ofthe injected fluid at its average saturation to itsviscosity, or:

M = (k/fih

Zone 2. represents the oil bank with the leading edge at adistance of re from the injection well. The mobility A.of the oil bank in this zone, i.e., zone 2, is defined asthe ratio of oil effective permeability as evaluated atinitial or connate water saturation to its viscosity, or:

A2 = (k/iih

(tp + At)IAt

Figure 1.125 Horner plot of pressure falloff after brine injection, Example 1.48.

P1 hr = 85 PSIG

SLOPE= -m = 270 PSIG/CYCLE

Shut-in Time, Af, hr

Figure 1.124 Log-log data plot for a falloff test after brine injection, Example 1.48 (After Earlougher, R. Advances inWell Test Analysis) (Permission to publish by the SPE, copyright SPE, 1977).

UNITSLOPE

Wel

lhea

d P

ress

ure,

pt s,

psi

g

Figure 1.127 Schematic diagram of fluid distribution around an injection well (composite reservoir).

UNAFFECTED REGION

OIL BANK

INJECTED FLUID,BANK

Figure 1.126 Miller-Dyes-Hutchinson plot of pressure falloff after brine injection, Example 1.48.

Shut-In Time, Af, hr

SLOPE = m = -270 PSIG/CYCLE

Wel

lhea

d P

ress

ure,

pts,

psig

Figure 1.128 Pressure falloff behavior in a two-banksystem.

The assumption of a two-bank system is applicable if thereservoir is filled with liquid or if the maximum shut-in timeof the falloff test is such that the radius of investigation ofthe test does not exceed the outer radius of the oil bank.The ideal behavior of the falloff test in a two-bank systemas expressed in terms of the Horner plot is illustrated inFigure 1.128.

Figure 1.128 shows two distinct straight lines with slopesof m1 and W2, that intersect at At^. The slope m1 of the firstline is used to estimate the effective permeability to waterkw in the flooded zone and the skin factor s. It is commonlybelieved that the slope of the second line m2 will yield themobility of the oil bank X0. However, Merrill et al. (1974)pointed out that the slope m2 can be used only to determinethe oil zone mobility if r& > 10% and (0Ct)i = (0Ct) 2> anddeveloped a technique that can be used to determine the dis-tance Yi\ and mobility of each bank. The technique requiresknowing the values of (0ct) in the first and second zone,i.e., (0ct) i and (0ct)2. The authors proposed the followingexpression:

_ k _ 162.6QB

The authors also proposed two graphical correlations, asshown in Figures 1.129 and 1.130, that can be used with theHorner plot to analyze the pressure falloff data.

The proposed technique is summarized by the following:

Step 1. Plot Ap vs. A* on a log-log scale and determine theend of the wellbore storage effect.

Step 2. Construct the Horner plot or the MDH plot anddetermine Wi, m2, and At&.

Step 3. Estimate the effective permeability in the first zone,i.e., injected fluid invaded zone, "zone 1," and theskin factor from:

= itg.tyM

Im1Ih

s = 1.5tt\PvfiatAt=0~Plhr

L i»*ii- log T k\ 2) +3.2275]

where the subscript "1" denotes zone 1, the injectedfluid zone.

Mobility Ratio, XJX2

Figure 1.129 Relationship between mobility ratio, sloperatio, and storage ratio. (After Merrill, etal. 1974).

Step 4. Calculate the following dimensionless ratios:

^ and ^ imi (0ct)2

with the subscripts "1" and "2" denoting zone 1 andzone 2 respectively.

Step 5. Use Figure 1.129 with the two dimensionless ratiosof step 4 and read the mobility ratio XiA2.

Step 6. Estimate the effective permeability in the secondzone from the following expression:

H2 = (^)-^- [1.7.10]VMI/ X1 /X2

Step 7. Obtain the dimensionless time Afofx from Figure1.130.

Step 8. Calculate the distance to the leading edge of theinjected fluid bank rfl from:

yL (0^t) 1 JVAfofc/

To illustrate the technique, Merrill et al. (1974)presented the following example.

Example 1.49 Figure 1.131 shows the MDH semilogplot of simulated falloff data for a two-zone water floodwith no apparent wellbore storage effects. Data used in thesimulation is given below:

rw = 0.25 ft, h = 20 ft, % = 30 ft

T12 = re = 3600 ft, (k/fih = m = 100 md/cp

(k/tih = m = 50 md/cp, (0ct)i = 8.95 x 10~7 psr1

(0ct)2 = 1.54 x 10-6 psr1, tfinj = 400 STB/day

Bw = 1.0bbl/STB

Calculate X1, X2, and % and compare with the simulationdata.

intersect at Affx

Bot

tom

-Hol

eS

hut-

In P

ress

ure

"pw

s"

Slo

pe R

atio

, /V

2Im^

Shut-In Time, At, hr

Figure 1.131 Falloff test data for Example 1.49. (After Merrill et al. 1974).

Bot

tom

-Hol

e P

ress

ure,

pw

s, p

siD

imen

sion

less

Int

erse

ctio

n T

ime,

Afo

f X

Slope Ratio, Hi2Zm1

Figure 1.130 Correlation of dimensionless intersection time, Afox, for falloff data from a two-zone reservoir. (AfterMerrill etal. 1974).

m2 = -60.1 PSI/CYCLE

m< = -32.5 PSI/CYCLE

THIS LINE ISUSEFUL FORAIR INJECTION

Solution

Step 1. From Figure 1.131, determine M1, m2, and At^ togive:

M1 — 32.5 psi/cycle

m2 — 60.1 psi/cycle

Ak = 0.095 hourStep 2. Estimate (JkZiI)1, i.e., mobility of water bank, from

Equation 1.7.9:(k\ _ 162.6qmiB _ 162.6(40O)(LO)W 1

= Im1Ih ~ (32.5) (20)

= 100 md/cpThe value matches the value used in the simulation.

Step 3. Calculate the following dimensionless ratios:

« • = = ^ = 1.85M1 -32.5(0^)1 _ 8.95 x IQ-7 _(0ct)2 "1 .54x10-6 - U ' D 5 i

Step 4. Using the two dimensionless ratios as calculated instep 4, determine the ratio A1A2 from Figure 1.129:

^ = 2.0

Step 5. Calculate the mobility in the second zone, i.e., oilbank mobility X2 = (k/'/J)2, from Equation 1.7.10:

/k\ JkZfI)1 100I - J = Z1 A x = TTT = 5 0 m d / C P

V fij 2 Ui A2) 2.0with the exact match of the input data.

Step 6. Determine AtDfx from Figure 1.130:Afax = 3.05

Step 7. Calculate rfl from Equation 1.7.11:/ (0.0002637) (100) (0.095) _ OA ^

ni ~ V (8.95 x 10-7) (3.05) " ™ ^

Yeh and Agarwal (1989) presented a different approach ofanalyzing the recorded data from the injectivity and fallofftests. Their methodology uses the pressure derivate Apand Agarwal equivalent time Ate (see Equation 1.4.16) inperforming the analysis. The authors defined the followingnomenclature:During the injectivity test period:

APvrf = Avf - Px

x d(AA*)Pvd d(lnf)

where:

P^i — bottom-hole pressure at time t duringinjection, psi

t — injection time, hoursIn t = natural logarithm of t

During the falloff test period:A/>w s = ^wf at At=O — Pws

d(A^ws)Pws~ d(In Ate)

with:

tp + Atwhere:

A£ = shut-in time, hoursJp = injection time, hours

Through the use of a numerical simulator, Yeh and Agar-wal simulated a large number of injectivity and falloff testsand made the following observations for both tests:

Pressure behavior during injectivity tests(1) A log-log plot of the injection pressure difference A wf

and its derivative Ap^ versus injection time will exhibita constant-slope period, as shown in Figure 1.132, anddesignated as (A^)const- The water mobility X1 in

Injection Time At, hrs

Figure 1.132 Injection pressure response and derivative (base case).

Constant Slope Period

Pre

ssur

e R

espo

nse

"Ap"

and

Der

ivat

ive

Apx

the floodout zone, i.e., water bank, can be estimatedfrom:

kl==(k\ 70.62qm]B

W l /?(A/> )COnst

Notice that the constant 70.62 is used instead of 162.6because the pressure derivative is calculated withrespect to the natural logarithm of time.

(2) The skin factor as calculated from the semilog analysismethod is usually in excess of its true value becauseof the contrast between injected and reservoir fluidproperties.

Pressure behavior during falloff tests(1) The log-log plot of the pressure falloff response in

terms of Ap and its derivative as a function of the falloffequivalent time A/e is shown in Figure 1.133. The result-ing derivative curve shows two constant-slope periods,(Aplfs) i and (A^8) 2, which reflect the radial flow in thefloodout zone, i.e., water bank, and, the radial flow in theunflooded zone, i.e., oil bank.These two derivative constants can be used to estimatethe mobility of the water bank Xi and the oil bank X2from:

70.62gmJ?

H(ApIs)1

70.62qm]B

(2) The skin factor can be estimated from the first semilogstraight line and closely represents the actual mechani-cal skin on the wellbore.

1.7.3 Step-rate testStep-rate injectivity tests are specifically designed to deter-mine the pressure at which fracturing could be induced inthe reservoir rock. In this test, water is injected at a con-stant rate for about 30 minutes before the rate is increasedand maintained for successive periods, each of which also

Injection Rate, STB/D

Figure 1.134 Step-rate injectivity data plot.

lasts for 30 minutes. The pressure observed at the endof each injection rate is plotted versus the rate. This plotusually shows two straight lines which intersect at the frac-ture pressure of the formation, as shown schematicallyin Figure 1.134. The suggested procedure is summarizedbelow:

Step 1. Shut in the well and allow the bottom-hole pressureto stabilize (if shutting in the well is not possible, ornot practical, stabilize the well at a low flow rate).Measure the stabilized pressure.

Step 2. Open the well at a low injection rate and maintainthis rate for a preset time. Record the pressure atthe end of the flow period.

Step 3. Increase the rate, and at the end of an interval oftime equal to that used in step 2, again record thepressure.

Step 4. Repeat step 3 for a number of increasing rates untilthe parting pressure is noted on the step-rate plotdepicted by Figure 1.134.

FRACTURE PRESSURE -1000 PSI AT SURFACE

Falloff Equivalent Time Ate, hr

Figure 1.133 Falloff pressure response and derivative (base case).

Late-Time ConstantSlope Period

Early-Time ConstantSlope Period

Pre

ssur

e re

spon

se "

Ap"

and

Der

ivat

ive

Apv

As pointed out by Horn (1995), data presented in graph-ical form is much easier to understand than a singletable of numbers. Horn proposed the following

Flow period

Infinite-acting radial flowdrawdown)

Infinite-acting radial flow (buildup)Wellbore storage

Finite conductivity fracture

Infinite conductivity fracture

Dual-porosity behavior

Closed boundary

Impermeable fault

Constant-pressure boundary

"toolbox" of graphing functions that is considered anessential part of computer-aided well test interpretationsystem:

Characteristic

Semilog straight line

Horner straight lineStraight line p vs. t, or unit-slope

log Ap vs. log AtStraight-line slope \, log Ap

vs. log A* plotStraight-line slope \, log Ap

vs. log At plotS-shaped transition between

parallel semilog straight linesPseudosteady state, pressure

linear with timeDoubling of slope on semilog

straight lineConstant pressure, flat line

on all p, t plots

Plot used

p vs. log At (semilog plot, sometimescalled MDH plot)

p vs. \og% + At)/At (Horner plot)log Ap vs. log At (log-log plot, type

curve)log Ap vs. log At, or Ap vs. At1/4

log Ap vs. log At, or Ap vs. At1/2

p vs. log At (semilog plot)

p vs. At (Cartesian plot)

p vs. log At (semilog plot)

Any

Chaudhry (2003) presented another useful "toolbox" thatsummarizes the pressure derivative trends for common flowregimes that have been presented in this chapter, as shownin Table 1-10.

Kamal et al. (1995) conveniently summarized; in tabulatedform, various plots and flow regimes most commonly used intransient tests and the information obtained from each testas shown in Tables 1-11 and 1-12.

Table 1.10 Pressure Derivative Trends for Common Flow Regimes.

Wellbore storage dual-porositymatrix to fissure flow

Dual porosity withpseudosteady-state interporosityflow

Dual porosity with transient inter-porosity flow

Pseudosteady state

Constant-pressure boundary(steady state)

Single sealing fault (pseudoradialflow)

Elongated reservoir linear flow

Wellbore storage infinite-actingradial flow

Wellbore storage, partialpenetration, infinite-acting radialflow

Linear flow in an infiniteconductivity vertical fracture

Bilinear flow to an infiniteconductivity vertical fracture

Semilog straight lines with slope 1.151Parallel straight-line responses are characteristics of naturally fractured reservoirsPressure change slope -> increasing, leveling off, increasingPressure derivative slope = 0, valley = 0Additional distinguishing characteristic is middle-time valley trend during more than

1 log cyclePressure change slope -> steepeningPressure derivative slope = 0, upward trend = 0Additional distinguishing characteristic -> middle-time slope doublesPressure change slope -> for drawdown and zero for buildupPressure derivative slope —> for drawdown and steeply descending for buildupAdditional distinguishing characteristic ->- late time drawdown pressure change and

derivative are overlain; slope of 1 occurs much earlier in the derivativePressure change slope -> 0Pressure derivative slope -» steeply descendingAdditional distinguishing characteristic -> cannot be distinguished from psuedosteady

state in pressure buildup testPressure change slope -+ steepingPressure derivative slope -> 0, upward trend —> 0Additional distinguishing characteristic ->• late-time slope doublesPressure change slope -> 0.5Pressure derivative slope -• 0.5Additional distinguishing characteristic -» late-time pressure change and derivative

are offset by factor of 2; slope of 0.5 occurs much earlier in the derivativePressure change slope = 1, pressure derivative slope = 1Additional distinguishing characteristics are: early time pressure change, and derivative

are overlainPressure change increases and pressure derivative slope = 0Additional distinguishing characteristic is: middile-time flat derivative

K(Xi)2 -> calculate from specialized plotPressure slope = 0.5 and pressure derivative slope = 0.5Additional distinguishing characteristics are: early-time pressure change and the

derivative are offset by a factor of 2Kf w ->• calculate from specialized plotPressure slope = 0.25 and pressure derivative slope = 0.25Additional distinguishing characteristic are: early-time pressure change and derivative

are offset by factor of 4

(continued)

Table 1.10 Pressure Derivative "

Wellbore storage infinite actingradial flow

Wellbore storageWellbore storage linear flow

Trends for Common Fiow Regimes (continued)

Sealing fault

No flow boundaryKb2 -» calculate from specialized plot

Table 1.11 Reservoir properties obtainable from various transient tests (After Kamal et al. 1995).

Drill item tests

Repeat/multiple-formationtests

Drawdown tests

Buildup tests

Reservoir behavior Step-rate testsPermeabilitySkinFracture length Falloff testsReservoir pressureReservoir limitBoundariesPressure profile

Reservoir behavior Interference and pulsePermeability testsSkinFracture lengthReservoir limitBoundariesReservoir behavior Layered reservoir testsPermeabilitySkinFracture lengthReservoir pressureBoundaries

Formation parting pressurePermeabilitySkinMobility in various banksSkinReservoir pressureFracture lengthLocation of frontBoundariesCommunication between wells

Reservoir type behaviorPorosityInterwell permeabilityVertical permeabilityProperties of individual layersHorizontal permeabilityVertical permeabilitySkinAverage layer pressureOuter Boundaries

Table 1.12 Plots and flow regimes of transient tests (After Kamal et al. 1995)

Flow regime Cartesian

Wellbore storage Straight lineSlope -* CIntercept -» Atc

Linear flow

Bilinear flow

First IARF a (high-& Decreasinglayer, fractures) slope

Transition More decreasingslope

Second IARF Similar slope to(total system) first IARF

Single no-flow boundary

Outer no-flow Straight lineboundaries Slope = m* -> 4>Ah(drawdown test only) p-m^ -» C^

aIARF = Infinite-Acting Radial Flow.

Straight lineSlope = mf -> IfIntercept = fracturedamage

Plot

Wt

Straight lineSlope = 0ibf -* Cfd

Log-log

Unit slope on Ap and /AAp and p \ coincide

Slope = ^ on p \ and onAp if s = 0Slope < \ on A j i f s ^ Op \ at half the level of ApSlope = Jp \ at \ level of App \ horizontal at p ^ = 0.5

Ap = Xe~2s orB\P^ = 0.25 (transition)

=<0.25 (pseudo-steady state)p \ horizontal at p ^ = 0.5

/A horizontal at p ^ = 1.0

Unit slope for Ap and /AAp and p \ coincide

Semilog

Positive sNegative s

Straight lineSlope = m^ khAp1 h r -> sStraight lineSlope = tn/2 (transition)

= 0 (pseudo-steady state)Straight lineSlope = m -> kh, p*Ap1 hr - • s

Straight lineSlope = 2mIntersection withIARF-> distance toboundaryIncreasing slope

Problems

1. An incompressible fluid flows in a linear porous mediawith the following properties.

L = 2500 ft, A = 30 ft, width = 500 ft, & = 50md,

0 = 17%, /x = 2 cp, inlet pressure = 2100 psi,

Q = 4bbl/day, p = 451b/ft3

Calculate and plot the pressure profile throughout thelinear system.

2. Assume the reservoir linear system as described in prob-lem 1 is tilted with a dip angle of 7°. Calculate the fluidpotential through the linear system.

3. A gas of 0.7 specific gravity is flowing in a linear reser-voir system at 1500F. The upstream and downstreampressures are 2000 and 1800 psi, respectively. Thesystem has the following properties:

L = 2000 ft, W = 300 ft, A = 15 ftk = 40 md, 0 = 15%

Calculate the gas flow rate.4. An oil well is producing a crude oil system at 1000

STB/day and 2000 psi of bottom-hole flowing pressure.The pay zone and the producing well have the followingcharacteristics.

^ = 35 ft, rw = 0.25 ft, drainage area = 40 acresAPI = 45°, yg = 0.72, Rs = 700 scf/STBk = 80 md

Assuming steady-state flowing conditions, calculate andplot the pressure profile around the wellbore.

5. Assuming steady-state flow and an incompressiblefluid, calculate the oil flow rate under the followingconditions:

pe = 2500 psi, />wf = 2000 psi, re = 745 ftrw = 0.3 ft, /X0 = 2 cp, B0 = IA bbl/STB/2 = 30 ft, k = 60md

6. A gas well is flowing under a bottom-hole flowingpressure of 900 psi. The current reservoir pressure is1300 psi. The following additional data is available:

T = 1400F, yg = 0.65, rw = 0.3ftk = 60 md, H = 40 ft, re = 1000 ft

Calculate the gas flow rate by using

(a) the real-gas pseudopressure approach;(b) the pressure-squared method.

7. After a period of shut-in of an oil well, the reservoir pres-sure has stabilized at 3200 psi. The well is allowed to flowat a constant flow rate of 500 STB/day under a transientflow condition. Given:

B0 = 1.1 bbl/STB, /xo = 2cp, Ct = I S x I O - 6 P s H£ = 50md, /z = 20 ft, 0 = 20%r w =0.3f t , pi = 3200 psi

calculate and plot the pressure profile after 1, 5,10, 15,and 20 hours.

8. An oil well is producing at a constant flow rate of 800STB/day under a transient flow condition. The followingdata is available:

B0 = 1.2bbl/STB, /x0 = 3 cp, ct = 15 x 10~6 ps r 1

k = 100 md, A = 25 ft, 0 = 15%rw = 0.5, pi = 4000 psi,

Using the Ei function approach and the pD method, cal-culate the bottom-hole flowing pressure after 1, 2, 3, 5,and 10 hours. Plot the results on a semilog scale andCartesian scale.

9. A well is flowing under a drawdown pressure of 350 psiand produces at a constant flow rate of 300 STB/day. Thenet thickness is 25 ft. Given:

re = 660 ft, rw = 0.25 ft/X0 = I- 2 cp, B0 = 1.25 bbl/STB

calculate:

(a) the average permeability;(b) the capacity of the formation.

10. An oil well is producing from the center of a 40 acresquare drilling pattern. Given:

0 = 20%, h = 15ft, & = 60md

/*0 = 1.5cp, B0 = 1.4 bbl/STB, rw = 0.25ft

pi = 2000 psi, />wf = 1500 psi

calculate the oil flow rate.11. A shut-in well is located at a distance of 700 ft from one

well and 1100 ft from a second well. The first well flowsfor 5 days at 180 STB/day, at which time the second wellbegins to flow at 280 STB/day. Calculate the pressuredrop in the shut-in well when the second well has beenflowing for 7 days. The following additional data is given:

Pi = 3000 psi, B0 = LSbWZSTB, jxo = 1.2cp,

h = 60 ft, ct = 15 x 10~6 psi"1, 0 = 15%, k = 45 md

12. A well is opened to flow at 150 STB/day for 24 hours. Theflow rate is then increased to 360 STB/day and lasts foranother 24 hours. The well flow rate is then reduced to310 STB/day for 16 hours. Calculate the pressure dropin a shut-in well 700 ft away from the well, given:

0 = 15%, h = 20 ft, k = 100 md

At0 = 2 cp, B0 = 1.2 bbl/STB, rw = 0.25 ft

pi = 3000 psi, ct = 12 x 10~6 psi-1

13. A well is flowing under unsteady-state flowing conditionsfor 5 days at 300 STB/day. The well is located at 350 ftand 420 ft distance from two sealing faults. Given:

0 = 17%, ct = 16 x 10-6 psi-1, k = 80 md

pi = 3000 psi, B0 = 1.3 bbl/STB, /x0 = 1.1 cp

rw = 0.25 ft, A = 25 ft

calculate the pressure in the well after 5 days.14. A drawdown test was conducted on a new well with

results as given below:

t (hr) Avf (psi)

1.50 29783.75 29497.50 2927

15.00 290437.50 287656.25 286375.00 2848

112.50 2810150.00 2790225.00 2763

Given:

pi = 3400 psi, ft = 25 ft, Q = 300 STB/day

ct = 18 x 10"6 psi-1, /Z0 = 1.8 cp,B0 = 1.1 bbl/STB, rw = 0.25 ft, <f> = 12%,

and assuming no wellbore storage, calculate:

(a) the average permeability;(b) the skin factor.

15. A drawdown test was conducted on a discovery well.The well was allowed to flow at a constant flow rate of175 STB/day. The fluid and reservoir data is given below:

5wi = 25%, 0 = 15%, A = 30 ft, ct = 18 x 10"6 psi-1

rw = 0.25 ft, pi = 4680 psi, ^ 0 = 1.5 cp,B0 = 1.25 bbl/STB

The drawdown test data is given below:

t (hr) /\rf (psi)

0.6 43881.2 43671.8 43552.4 43443.6 43346.0 43188.4 4309

12.0 430024.0 427836.0 426148.0 425860.0 425372.0 424984.0 424496.0 4240

108.0 4235120.0 4230144.0 4222180.0 4206

Calculate:

(a) the drainage area;(b) the skin factor;(C) the oil flow rate at a bottom-hole flowing pressure

of 4300 psi, assuming a semisteady-state flowingconditions.

16. A pressure buildup test was conducted on a well thathad been producing at 146 STB/day for 53 hours.

The reservoir and fluid data is given below.

B0 = 1.29 bbl/STB, /x0 = 0.85 cp,

ct = 12 x 10-6 psi"1, <p = 10%, Avf = 1426.9 psig,A = 20 acres

The buildup data is as follows:

Time pws (psig)

0.167 1451.50.333 1476.00.500 1498.60.667 1520.10.833 1541.51.000 1561.31.167 1581.91.333 1599.71.500 1617.91.667 1635.32.000 1665.72.333 1691.82.667 1715.33.000 1736.33.333 1754.73.667 1770.14.000 1783.54.500 1800.75.000 1812.85.500 1822.46.000 1830.76.500 1837.27.000 1841.17.500 1844.58.000 1846.78.500 1849.69.000 1850.4

10.000 1852.711.000 1853.512.000 1854.012.667 1854.014.620 1855.0

Calculate:

(a) the average reservoir pressure;(b) the skin factor;(c) the formation capacity;(d) an estimate of the drainage area and compare with

the given value.