fetkovich method
DESCRIPTION
Fluid Flow, Reservoir, OilTRANSCRIPT
'.i,:i
iI,
1'1
~,.iIii
'IP,. t'"I i i.. '!.\i
.
~ J
j!1~i! ~i
!:i"
f+I:"
PG E - q6rO( - A, G(\ w i ~ ~I.' I
:.',II.
242 GAS RESERVOIR ENGINEERING
IL/zg t
TABLE 10.10-AQUIFER PI's
Type of OuterAquifer Boundary
0.00708kh(/J/360)J""
p. In(r.Ir,)
0.00708kh(/J/360)j""
p. InJo.0142kt/q,p.c tr~
j for Radial Flow
(STBID-psi)
0.00708kh(/J/360)J""
p.[ln(r .Ir,) - 0.75]
'.:Finite, no flow
,I".!
Finite, constantpressure
;iInfinite j""
0.001127kwhj""
p.L
kwh
1,OOOp.JO.0633kt/q,p.c t
For example, at n=l,
[B.6.PI-WeOPO(tDl) ]W"l =W.o+(tDl-tDO) PD(tDl)-tDOPO(tDl)
[454.3(5)-0
]=0+(15.1)
1.83-0
=18,743 RB.
For n=2,
[B.6.P2- W"lPo(tm) ]W,,2= W"I +(tm -tDl) PD(tm)...,.tDlPo(tm)
[454.3(19)-18,743(0.0155) ]= 18,743(15.1)2.15 -15.1(0.0155)
=84,482 RB.
6. Table' 10.9 gives the final results.
..:::::::t> FetkovichI6 Method. To simplify water influx calculations fur-ther, Fetkovich proposed a model that uses a pseudosteady-stateaquifer PI and an aquifer material balance to represent the systemcompressibility. Like the Carter-Tracy method, Fetkovich's modeleliminates the use of superposition and therefore is much simplerthan van the Everdingen-Hurst method. However, because Fet-kovich neglects the early transient time period in these calculations,the calculated water influx will always be less than the values pre-dicted by the previous two models.
Similar to fluid flow from a reservoir to a well, Fetkovich usedan inflow equation to model water influx from the aquifer to thereservoir. Assumingconstantpressure at the original reservoir/aqui-fer boundary, the rate of water influx is
dW"qw=-=J(Paq-p,)n, (10.62)dt
where n=exponent for inflow equation (for flow obeying Darcy'slaw, n=l; for fully turbulent flow, n=0.5).
Assuming that the aquifer flow behavior obeys Darcy's law andis at pseudosteady-state conditions, n = 1. Based on an aquifer ma-terial balance, the cumulative water influx resulting from aquiferexpansiolkjs
We=CtWi(Paq,i-Paq)' """""""""""'" .(10.63)
Eq. 10.63 can be rearranged to yield an expression for the aver-age aquifer pressure,
- (W,, ) (
We )Paq=Paq.i 1- =Paq.i 1-- , (10.64)CtPaq.;W; W,,;
where We;=CtPaq.iWi """""""""""""" (10.65)
j for Linear Flow
(STB/D-psi)
3(O.001127)kwhJ""
p.L
is defined as the initial amount of encroachable water and repre-~sents the maximum possible aquifer expansion. After differentiat- J
ing Eq. 10.64 with respect to time and rearranging, we have'
dW" =- We; dPaq . . . . . . . . . . . . . . . . . . . . . . . . . . . (10.66)dt Pi dt
Combining Eqs. 10.62 and 10.66 and integrating yields
J faq -dPaq =- J t Jpi dt' (10.67)Paq.i Paq-P, 0 W,,;
(-Jp.t
)or Paq-P,=(Paq.;-p,)exp ~ . (10.68).W",
Table 10.10 summarizes the equations for calculating the aquiferPI for various reservoir/aquifer boundary conditions and aquifergeometries. Note that we must use the aquifer properties to calcu- "
~~ JFrom Eq. 10.67, we can derive an expression for (Paq -p,), and ,
following substitution into Eq. 10.68 and rearranging, we have
dW
(-JP.t
)!...=J(Paq.; -p,)exp ~ , (10.69)dt W,,;
which is integrated to obtain the cumulative water influx, We:
~ W,,; [.
( JPaq.;,
)]W,,=~(Paq.i-P') l-exp -- (10.70)Paq,l W"i
Recall that we derived Eq. 10.70 for constant pressure at the reser-voir/aquifer boundary. In reality, this boundary pressure changes '
as gas is produced from the reservoir. Rather than using superpo- ,sition, Fetkovich assumed that, if the reservoir/aquifer boundarypressure history is divided into a finite number of time intervals,the incremental water influx during the nth interval is
Wei - - [ ( JPaq,i.6.tn
)].6.W"n=~(Paq,n-l-Pm) l-exp .'P~~ ~,
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (10.71)
TABLE 10.11-PRESSURE HISTORY AT THE RESERVOIR!AQUIFER BOUNDARY, EXAMPLE 10.8
t(days)
091.5
183.0274.5366.0457.5549.0
p,(psia)3,7933,7883,7743,7483,7093,6803,643
GAS VOLUMES AND MATERIAL-BALANCE CALCULATIONS
. - ( We.n-l )wherepaq.n-l=Paq,i 1-- (10.72). Wei
, and P- - Pm-l +pm- . m
2.(10.73)
Althoughit was developedfor finite aquifers, Fetkovich's methodcanbeextended to infinite-actingaquifers. For infinite-actingaqui-fers,the method requires the ratio of water influx rate to pressure
, drop to be approximately constant throughout the productive lifeofthe reservoir. Under these conditions, we must use the aquifer
-- PI for an infinite-acting 2/'J)if~r.Th:: fc:"h'.-ing calculation pwc<.'uure illustrates this method.1. Calculate the maximum water volume, Wei, from the aquifer
~ thatcould enter the gas reservoir if the reservoir pressure were re-duced to zero.
Wei=CIPaq,iWi, .(10.65)
where Wi depends on the reservoir geometry and the PV availableto store water.
2. CalculateJ, Note that the equations summarized in Table 10.10depend on the boundary conditions and aquifer geometry.
3. Calculate the incremental water influx, lIWen, from the aqui-fer during the nth time interval.
Wei - -[ ( JPaq,illtn
)]~Wen=~(Paq,n-l-Pm) I-exp - . .Paq,t Wet
"""""""""""""""" (10.71)
4. Calculate Wen:n
Wen= L lIWei'i=1
, Example10.8-Estimating Water Influx With the FetkovichMethod.Calculate the water influx for the reservoir/aquifer sys-temdescribed below. Assume a finite radial aquifer with an areaof250,000acres and having a no-flow outer boundary. The esti-matedaquifer properties are given below; Table 10.11 summar-
.izesthe pressure history at the reservoirlaquifer boundary. Note~that the aquifer is a sector of a cylinder where 8= 180°.~'.. <P = 0.209.t rr = 5,807 ft.w'~. JL= 0.25 cpo
k = 275 md.e = 1800.
CI = 6xlO-6 psia-I.II = 19.2 ft.
,}t,'5<,~,.
..~i Solution.
~ I. Calculate the maximum volume of water from the aquifer,-§Wti,that could enter the reservoir if the reservoir pressure were~r reducedto zero. Note that the aquifer shape is a sector of a cylin-~~der, so the initial volume of water in the aquifer is
if! 1I'(rJ-r~)II<p(e/360)b:! lVi= ,
~~ 5.615f:':: .:_. \<.. r
"( ,.:. ,
(43,:60A )C:O)=
. =83,260ft.i, Therefore,~-
~: .(83,2602 -5,8072)(19.2)(0.209)(180/360)~11V.=5\i.: r 5.615
I: =7.744x 109 RB.""'.
~:t:J.
(43,560)(250,000) (360
)11' 180
243
From Eq. 10.65,
We; =CIPaq.i Wi =(6 x 10 -6)(3,793)(7.744 X 109)= 176.3x 106 RB.
2. CalculateJ. For radial flow in an aquiferwith a finite no-flowouter boundary, from Table 10.10,
0.00708kh(8/360)J=
JL[ln(ra/r r) -0.75]
(0.00708)(275)( 19.2)( 180/360)
" . c°L" ,,»),(,,;'..,00 i j -0. ijj
=39.1 STB/D-psi.
3. For each time period, calculate the incremental water influxusing Eq. 10.71.
Wei - -
[ ( JPaq.il1ln
)]lIWen=--:-(Paq,n-l-Pm) I-exp - f. 'Paq,t - I-Jel
where (from Eq. 10.72)
( We.n-l )Paq,n-I =Paq,i I-~ 'el
and from Eq. 10.73,
Pm-l +PmPm=
2
For example, at n= I,
- (WeO
) (0
) .Paqo=Paq,;lI-- =3,793 1- 6 =3,793 pSI.\ Wei 176.254xlO
The average pressure at the aquiferlreservoir boundary is
PrO+PrlPri = 3,793+3,788 =3,790.5 psia.
2 2
Therefore, the incremental water influx during Timestep I is
Wei - -
[ ( JPaq,il1l1
)]l1Wel =--:-(PaqO-PrI) I-exp - .Paq.l Wet
176.3x 106
[= (3,793-3,790.5) I-exp
3,793
X[
- (39.1)(3,793)(91.5)]1
176.3x106 5=8,587 RB.
Similarly, at n=2,
(Wel
) (8,587
)Paql =Paq.i 1-- =3,793 I 6 =3,792.8 psi.Wei 176.3 x 10
The average pressure at the aquifer/reservoir boundary is
- Prl +Pr2 3,788+3,774 .Pr2 =3,781 pSla.
2 2
The incremental water influx during Timestep 2 isOJ'i1,A!'.L: !.
:11
Wei
[ ( JPaq i1112
)]11We2 = --:-(Paql -Pr2) I-exp - . .Paq,t Wet
- 176.3x106 (3,792.8-3.781)[1-- 3,793
I
;1 r
'Fi':d:i".
'
..
'
j,I,
i,!;HiIii:I
II{tir
244.
GAS RESERVOIR ENGINEERING:.'~
[ (39.1)(3,793)(91.5)
]Jexp X -
176.3x106
=40,630 RB.4. Calculatethecumulativewaterinfluxduringeach timestep.
For example,at the end of the first timestep,n=1
l.'i,
'
1
..
!:It[I,t{I'Kit
,1,'iI"I!.!'l l';i,,bI'~.I'
I
~I
~jjltl".1
'IIIf!il
11
iliii'\
i,1,.j, :I.
liiiI
;iii
I!I j.
We! = E ~We;=~Wel =8,587 RB.;=1
Similarly, the cumulative water influx at the end of the secondtimestep is
n=2
We2= E ~We;=~Wel +~We2=8,587+40,630=49,217 RB.;=1
Table 10.12 summarizes the final results.Estimating Original Gas in Place With Material Balance for
a Dry-Gas Reservoir With Water Influx. Once the water influxhas been calculated, we can estimate the original gas in place withmaterial-balanceconcepts.Recall from Sec. 10.3.2 the general formof the material-balance equation including water influx:
GBgi=(G-Gp)Bg+We-BwWp'., (10.74)
which can be rearranged to yield
GpBg+ WpBw We=G+ . (10.75)(Bg-Bgi) (Bg-Bgi)
If we define a water influx constant, C, in terms of the cumula-tive water influx as
We=Cf(p,t), , (10.76)
then Eq. 10.75 becomes
GpBg+WpBw Cf(p,t)=G+ . (10.77)
(Bg-Bg;) (Bg-Bgi)
The form of Eq. 10.77 suggests that, if water influx is thepredominant reservoir drive mechanism, then a plot of
[(GpBg)+(WpBw)]/(Bg-Bgi) vs. f(p,t)/(Bg-Bgi) will be astraight line with a slope equal to C and an intercept equal to G.The functional form of f(p,t) varies according to the water influxmodel used. Any water influx model, such as steady state, unsteadystate, or pseudosteadystate, can be used with Eq. 10.77. Note that,if the incorrect water influx model is assumed, the data may notexhibit a straight line. Example 10.9 illustrates the application ofEq. 10.77 for an unsteady-state water influx model.
Example 10.9-Estimating Original Gas in Place With Materi-al Balance for a Dry-Gas Reservoir With Water Influx. Estimatethe originalgas in place and water influxconstantusing the material-balance equation developed for water influx in a dry-gas reservoir.Assume an unsteady-state, infinite-acting aquifer. According toMcEwen,22 the volumetric estimate of original gas in place is200 X106 Mscf. Table 10.13 gives the pressure and productionhistories; the estimated aquifer properties are summarized below.
rJ>= 0.24.p. = 1.0 cporr = 3,383 ft.
k = 50 md.c, = 6x 10-6 psia-I.e = 360°.h = 20.0 ft.
Bw= 1.0RB/STB.
Solution. Using a procedure similar to that illustrated for thevanEverdingenand Hurst method, estimatewater influx for eachtimestep as described below.
1. Calculate ~Pn' where
~Pn= 'h.(Pn-2 -PII)'
For example, at n=1 (i.e., t=182.5 days),
~PI = 'h.(PO-PI)= 'h.(5,392-5,368)= 12.0 psi.
At n=5 (i.e., t=912.5 days),
~Ps = 'h.(P3-Ps)= 'h(5,245 -5, 147)=49.0 psi.
2.' Calculate dimensionless times for each of the real times given.Use the dimensionless time defined by Eq. 10.37 for a radialge-ometry .
0.00633kl (0.00633)(50)1lD= = =0.0191.
rJ>p.c,r~ (0.24)(1.0)(6 x 10-6)(3,383)2
For example, at n=l (i.e., 1=182,5 days), lD=(0.019)(l82.5)=3.5.
3. For each dimensionless time computed in Step 2, calculateadimensionless cumulative water influx. We are assuming an infInite-'acting aquifer, so we can use Eq. 10.49.
QpD(ID)=
1.28381/1 + 1.193281D +0.2698721 iF +0.0085529415
I +0.6 I65991b' +0.0413oo81D
For example, at 11=I,
QpD(ID\ )=[1.2838(3.5) 'h + 1.19328(3.5)+0.269872(3.5)3/2
+0.00855294(3.5)2]/[1+0.616599(3.5)'h+0.0413008(3.5)]jQpD=3.68. ;4. Now, estimate the original gas in place and water influxcon-i
stant using the material-balance plotting method. Calculate the pial-:
,J
TABLE 10.12-SUMMARY OF FINAL RESULTS, EXAMPLE 10.8
Pm Pm P aq,n Paq.n-1 -Pm We Wen (psi a) (psia) (psia) (psia) (RB)-0 3,793 3,793.0 3,793 0 0 01 3,788 3,790.5 3,792.8 2.5 8,587 8,5872 3,774 3,781.0 3,791.9 11.8 40,630 49,2173 3,748 3,761.0 3,789.7 30.9 106,399 155,6164 3,709 3,728.5 3,785.1 61.2 210,285 365,9015 3,680 3,694.5 3,778.4 90.6 31'I ,642 677,5436 3,643 3,661.5 3,769.8 116.9 402,060 1,079,603
TABLE 10.13-PRESSURE AND PRODUCTION HISTORIES,EXAMPLE 10.9
t P Gp Wp Bg(days) (psia) (MMscf) (STB) z (RB/MscQ-
0 5,392 0 0 1.0530 0.6775182.5 5,368 677.7 3 1.0516 0.6796365 5,292 2,952.4 762 1.0470 0.6864547.5 5,245 5,199.6 2,054 1.0442 0.6907730 5,182 7,132.8 3,300 1.0404 0.6965912.5 5,147 9,196.9 4,644 1.0383 0.6999
1,095 5,110 11,171.5 5,945 1.0360 0.70331,277.5 5,066 12,999.5 7,148 1.0328 0.70721,460 5,006 14,769.5 8,238 1.0285 0.71271,642.5 4,994 16,317.0 9,289 1.0276 0.71381,825 4,997 17,868.0 10,356 1.0278 0.71362,007.5 -4,990 19,416.0 11,424 1.0273 0.71422,190 4,985 21,524.8 12,911 1.0270 0.7147
:GASVOLUMES AND MATERIAL-BALANCE CALCULATIONS 245
! lingfunctionsdefined in the previous section, [(GpBg)+(WpBw)]l(Bs.-Bgi)vs. f(p,t)/(Bg-Bgi)' where f(p,t)=ELlpQpD' '
ror example, at n= I the plotting function for the vertical axis,y, is
~GpBg+WpBw
'Y=J (Bg-Bg;)
(677.7x 103)(0.6796)+(3)(1.0)
(0.6796 -0.6775)
, =219.3x106 Mscf=219.3 Bscf.
Theplotting function for the horizontal axis, x, is
ELlpQpD, x-.,1 -(Bg-Bgi)
(12)(3.68)
(0.6796-0.6775)
=21.0x 103 psi/RB-Mscf.
". 5. The material-balance plotting functions, summarized in Ta-[ble 10.14, are plotted in Fig. 10.11.e 6. From the slope of the line through the data points in Fig. 10.11,Cisestimated to be 1195 RB/psi, and the original gas in place esti-
.,matedfrom the intercept is G= 197 Bscf= 197x 106 Mscf, whichiagrees with the volumetric estimate of G=200x 106 Mscf."
4,~" Thegeneral problem that reservoir engineers face when analyz-~inggas reservoirs with water influx is simultaneous determination
. . .. . .. , .i ,"i ! .
"' I ~t-_. . ---f. .~
SI~pe. c= 1.1951rnfpsi I'. i
~t !-._---_._._..
' m L I L._____--.....i i !
'G =197'MMMscf i !! ! !
" "-"""1""-'-""---'---1""-'---'-"'---1"""---" '-"-'"
I I !
0 .0 50 100 150
u:"pQpD/(B,- B,i)' MMpsiIRB/ Mscf
Fig.10.11-Graphical solution to the material-balance equa-tionfor a dry-gas reservoir with water influx, Example 10.9.
of G, C, aquifer size or an aquifer/reservoir size ratio, ra/r" andthe relationship between real time, t, and dimensionless time, tD'Simultaneousdetermination of these variables that best fit a pres-sure/productionhistory is a complexproblem in regressionanalysis.
10.3.3 Volumetric Geopressured Gas Reservoirs. We developedthe material-balance equation for a volumetric dry-gas reservoirassuming that gas expansion was the dominant drive mechanismand that expansions of rock and water are negligible during theproductive life of the reservoir. These assumptions are valid fornormallypressured reservoirs (Le., reservoirs with initial pressuregradientsbetween 0.43 and 0.5 psilft) at low to moderate pressureswhen the magnitude of gas compressibilities greatly exceeds theeffects of rock and connate water compressibilities. However, forabnormallyor geopressured reservoirs, pressure gradients often ap-proachvaluesequal to the overburdenpressure gradient (i.e., "" 1.0psi/ft). In these and other higher-pressure reservoirs, the changesin rock and water compressibilities may be important and shouldbe considered in material-balance calculations.
Following Ramagost et al. 's23 derivation, we begin with Eq.10.26for a normallypressuredvolumetricgas reservoir and includethe effects of changing water volume, LlVw' and formation (rock)volume, LlVf' As Fig. 10.12 shows, the general form of thematerial-balanceequation for a volumetricgeopressured reservoir is
GBgi=(G-Gp)Bg+LlVw+LlVf' (10.78)
where G l!..8!..=reservoir volume occupied by gas at initial reservoirpressure, KH;(G-Gp)Bg = reservoir volume occupied by gas aftergas productionat a pressurebelowthe initialreservoir pressure, RB;LlVw=increase in reservoir PV occupiedby connate water and fol-lowing water expansion at a pressure below the initial reservoirpressure, RB; and LlVf=increase in reservoir PV occupied by for-mation(rock)at a pressure below the initial reservoir pressure, RB.
The expansion of the connate water following a finite pressuredecrease can be modeled with isothermal water compressibility:
G
200 Initial Conditions (p = pJ Later Conditions (p < Pi)
Fig. 10.12-Material-balance model showing reservoir PVoc-cupied by gas at initial and later conditions for a volumetricgeopressured gas reservoir.
TABLE 10.14-MATERIAL-BALANCE PLOTTING FUNCTIONS, EXAMPLE 10.9
t APnLApQ po/(B 13- B gi)
(GpBg + WpBw)/(Bg -Bgi)(days) ton Qpon (psia) (1,000 psi/R -Mscl) (Bscl)
0 - - 0 - -182.5 3.5 3.68 12.0 21.0 219.3365 6.9 5.82 50 28.5 227.7547.5 10.4 7.76 61.5 46.3 272.6730 13.9 9.56 55.0 55.8 260.9912.5 17.3 11.21 49.0 71.0 287.6
1,095 20.8 12.84 36.0 83.1 304.51,277.5 24.3 14.42 40.5 93.1 309.31,460 27.7 15.91 52.0 99.1 298.81,642.5 31.2 17.40 36.0 116.7 320.81,825 34.7 18.87 4.5 136.0 352.92,007.5 38.1 20.26 2.0 151.7 378.32,190 41.6 21.68 6.0 166.7 413.6
(G -Gp)Bg
GBgi
L1Vj
L1Vw
246
Cw=-~ (OVw) =-~ ilVw, (10.79)Vw op T Vwi ilp
where Vwi=initial reservoir volumeoccupiedby the connatewater . We can arrange Eq. 10.79 in terms of the change in watervolume,
ilVw=-cwVwiilp, .(10.80)
where ilp=P-Pi' We can also write the original water volume interms of the original gas in place:
4
SwiGBgiVwi= . (10.81)
(I-Swi)
Substituting Eq. 10.81 into Eq. 10.80 and noting that-ilp=Pi-P yields
SwiGBgiilVw=Cw(Pi-P) . . '" -.. -.. -. (10.82)
(I-Swi)
Similarly, the decrease in PV, ilVp' caused by a finite reduc-tion in pore pressure can be modeled with
Cr :p (o~ ) T (1O.83a)and thus,
I ilVp . . . . . . . . . . . . . . . . (1O.83b)cf=--' ...............Vpi ilp
7000i . I i -- i I- i Ii!
~---' T
---+ r-----! ! ------'ii I ,- .! ~___L L--
r
-I - . .------
--~ I ! iI , J J-- ! ! I 1 .--I i I i Ii - ! ' r---I i I I i-r--~- ~ I--i-, ; G=70.7Bcf -! ! ! I '
r---i ! ---T t--.-.-.-
..
~6000'}:2 5000"=::'
~4000.§
~3000+.
~ 2000
~ 1000.5
0 .0 20 40 60 80 100
Cumulative Gas Production, G, (Bet)
Fig. 10.13-Graphlcal solution to the material-balanceequation for a volumetric geopressured gas reservoir,Example 10.10.
GAS RESERVOIR ENGINEERING
TABLE 10.16-PLOTTING FUNCTIONS FOR EXAMPLE 10.10
Gp(MMscf)
0392.5
1,642.23,225.84,260.35,503.57,538.18,749.2
10,509.311,758.912,789.217,262.522,890.828,144.632,566.736,819.9
p/z(psia)
6,6026,5536,4676,3956,3316,2476,1366,0815,9545,8605,7995,5004,8784,6284,2093,802
p/z[1 -(CwSwi + Ct )(PI-p)/(1 -8wf))(psia)
6,6026,5156,3746,2396,1335,9985,8255,7405,5565,4245,3394,9514,2613,9853,5633,167
where cf=average formation compressibility over the finite pres-sure interval, ilp. In terms of the change in PV, Eq. 10.83 becomes
ilVp=cfVpiilp, (10.84)
where ilp=P-Pi' Again, we can write the original rock PVinterms of the original gas in place as
GBgiVpi= . - . . - - . - . - - . . . . . . . . ... .(10.85)
(l-Swi)~
Jhitw.
10.1
~~
=~Con,
1zlI
willicept~toPI:10.1\, J
Substituting Vpi from Eq. 10.85 into Eq. 10.84 and notingthat;-ilp=Pi-P and ilVp= -ilVf (the change in rock volume) yields.
- GBgi ,--ilVp=cf(Pi-P) -ilVf' .(10.86)~
(I-Swi)
Substituting Eqs. 10.82 and 10.86 into Eq. 10.78, we can nowwrite a general material-balance equation for a volumetric geopres-sured reservoir:
GB .= (G- G )B + Cw(Pi-P)SwiGBgi Cf (p.-p)GB.gl pg + 1 gl
(l-Sw) (l-S .)WI
After simplification, Eq. 10.87 becomes
GBgi(Pi-P) -GBgi=(G-Gp)Bg+ (cwSwi+cf)
(l-Sw)
.~0.
{ 3000
120
,,Exairial 1.'the fi(the!usinigas ~norn:mate:ducti[
0 ,0 20 40 60 80 100
Cumulative Gas Production, G, (Bet)
Fig. 10.1~-lncorrect graphical analysis ofgeopressured gas reservoir, Example 10.10.
TABLE10.15-pRESSURE ANDPRODUCTIONHISTORIES,EXAMPLE 10.10
P Gp(psia) z (MMscf). -9,507 1.440 09,292 1.418 392.58,970 1.387 1,642.28,595 1.344 3,225.88,332 1.316 4,260.38,009 1.282 5,503.57,603 1.239 7,538.17,406 1.218 8,749.27,002 1.176 10,509.36,721 1.147 11,758.96,535 1.127 12,789.25,764 1.048 17,262.54,766 0.977 22,890.84,295 0.928 28,144.63,750 0.891 32,566.73,247 0.854 36,819.9
GASVOLUMES AND MATERIAL-BALANCE CALCULATIONS
TABLE 10.17-PRESSURE AND PRODUCTION HISTORIES,EXAMPLE 10.11
p(psia)
9-,5079,2928,9708,5958,3328,0097,6037,4067,0026,7216,5355,7644,7664,2953,7503,247
Gp(MMscf)
0392.5
1,642.23,225.84,260.35,503.57,538.18,749.2
10,509.311,758.912,789.217,262.522,890.828,144.632,566.736,819.9
z
1.4401.4181.3871.3441.3161.2821.2391.2181.1761.1471.1271.0480.9770.9280.8910.854
~:
.
'
q(CWSwi+Cf)(Pj-P)
]Bgj =G-G p . (10.89)~or 1-
~ (l-Swi) Bg
~ SubstitutingBg/Bg=pz/pjZ into Eq. 10.89 and rearranging, we~ have
~ ~[
(cwSwi+cf)(Pi-P)
]- Pi - pj GpI -. (10.90)
z (l-Swj) Zj Zj G
Notethat, when the effects of rock and water compressibility are,negligible, Eq. 10.90 reduces to the material-balance equation inr.whichgas expansion is the primary source of reservoir energy (Eq.10.29).Failure to include the effects of rock and water compress i-bilitiesin the analysis of high-pressure reservoirs can result in errors
,inboth original gas in place and subsequent gas reserve estimates.;Wepresent two analysis techniques based on Eq. 10.90 for volu-imetric geopressured reservoirs.(i Estimating Original Gas in Place When Average Fonnation
~ CcmpressibiliJyIs Known. If cf is assumed to be constant with
[time,the form of Eq. 10.90 suggests that a plot of
.eo""P
[
f (CwSWj+Cf)(Pj-P)
]~ - I vs. Gpr. z (l-Swi)
.willbe a straight line with slope equal to -p/ZjG and an inter-
,ceptequal to p/Zj. At p/z=O, Gp =G, so extrapolation of the line',~plz=O providesan estimateof originalgas in place. Example;10.10illustrates this analysis technique.
[Example10.10-Estimating Original Gas in Place With Mate-:rIaIBalance for a Volumetric Geopressured Gas Reservoir. For'!befollowing data taken from an abnormally pressured reservoiri(theAnderson "L" sand24), estimate the original gas in placehlsingthe material-balance equation developed for a high-pressurefgasreservoir.In addition,use thematerial-balanceequationfor ainonnallypressured gas reservoir, and compare the initial gas esti-imatesfrom both equations. Table 10.15 gives the pressure and pro-fductionhistories.~i
Pi = 9,507 psia.Cw = 3.2xlO-6 psi-I.Swi = 0.24.
r: Original pressure gradient=0.843 psi/ft.
~ Cj=19.5 x 10 -6 psi -1 (assumed constant).II
~Solution.fff1. Calculate the geopressured and normally pressured pressure"lottingfunctions for each data point (Table 10.16).
2. From the two plots, estimate original gas in place from therceptwith the horizontal axis: high-pressure reservoir analysis
247
150
"'"~
! I I
I.
i -6-1
i SIOIf =13.3!x 10 <1>fMSCO I
100-I'''''':''''''''''''!
,:"'-""-"i t-.~! l-.........I .
~5O~--~~1-----.J. . !
i ' I i Ii I I I I
0 -I""'''''''';-'''''''''''''-1''''''''''''''''''''r''''''''''''''''''''[''''''''''-'' 1--..-....-6 -1 ' ,
I~tercept ~ ..18.5 xi10 psi i
I
II I I! I
)(
'U!8-::;
.:,-<>.
~
g~~
-50 ,0 2 4 6 8 10 12
[(G,I(P; p)](P,zl ZiP),MMscflpsi
14
Fig. 10.15-Graphlcal solution to the material-balanceequation for simultaneous estimation of original gas In placeand formation compressibilityin volumetric geopressured gasreservoirs, Example 10.11.
(Fig. 10.13)-G=70.7 Bcf; normally pressured reservoir analysis(Fig. 10.14)-G=89.3 Bcf.
The results showthat, if we analyze this geopressured reservoirusing techniques for normallypressured reservoirs, we will over-estimate the original gas in place by more than 25%. In addition,note that the data in Fig. 10.14 are beginning to trend downward,which indicates that a normally p'ressuredanalysis of the availabledata is not valid.
We developedthe material-balanceequation for a high"pressuregas reservoir usinga singlevalue for formationcompressibilityoverthe life of the reservoir. In reality, formation compressibility mayvary during pressure depletion, especially at the highestpressures.Further, the previous derivationassumed that values for formationcompressibility are readily available. However, these values arevery difficult to measure accurately in the laboratory, especiallyas a function of changes in pore pressure. Therefore, in the nextsection, we present a graphical technique for simultaneously es-timating original gas in place and an average value of formationcompressibility.
Simultaneous Determination of Average Formation Compress-ibility and Original Gas in Place. Roach25 developeda material-balance technique for simultaneously estimating formation com..pressibility and original gas in place in geopressured reservoirs.Beginning with Eq. 10.90, Roach presented the material-balanceequation in the following form:
1
(pjZ
)1
[Gp PiZ
]
Swicw+cf
(Pj-p) PZi -I = G (Pj-p) PZj - l-Swj .
.. . . . . . . . . . . . ... . . . . . . . . . ... ... . (10.91)
Again, if cf is constant, the form of Eq. 10.91 suggests that aplot of
1
(PiZ ) [Gp PiZ
]- --I vs --(Pi-P) PZi (Pi-P) PZj
will be a straight line with a slope that equals lIG and an interceptthat equals -(S"icw +cI'I-Swi)' We can then calculate the origi-nal gas in place, G, and the average formation compressibility, cf'using the slope and intercept, respectively. Poston and Chen26 ap-plied this method to the geopressured gas reservoir data presentedin Example 10.10. Their analysis is reproduced in Example 10.11.
Example 10.H-Simultaneous Determination of Average For-mation Compressibility and Original Gas in Place With Mate-rial Balance in a Volumetric Geopressured Gas Reservoir. Forthe followingdatatakenfromtheAnderson"L" sand,24 estimate
248 GAS RESERVOIR ENGINEERING
TABLE 10.18-PLOTTING FUNCTIONS, EXAMPLE 10.11
original gas in place using Eq. 10.91 and an average value for for-mation compressibility. Table 10.17 gives the pressure and pro-duction histories.
Pi = 9,507 psia.Cw = 3.2xI0-6 psi-I.Original pressure gradient = 0.843 psi/ft.Swi = 0.24.
Solution.I. First, we must generate the plotting functions developed by
Roach.25 ~ample calculations for Gp=392.5 MMscf=3.925 x105 Mscf follow. For the variable on the vertical axis, we plot
1
(PiZ )(Pi-P) PZi-l
I
[(9,507)( 1.418)
]- (9,507-9,292) (9,292)(1.440)-1
=34.9xlO-6 psi-I.
For the variable on the horizontal axis, we plot
[Gp PiZ
](Pi-P) PZi
[
(3.92x 105) (9,507)(1.418)
]= (9,507-9,292) (9,292)(1.440)
= 1.84X 103 Mscf/psi.
2. Prepare a plot (Fig. 10.15) of the plotting functionssummar-ized in Table 10.18.
3. Estimate the original gas in place and average formation com-pressibility from Fig. 10.15.
A. The original gas in place, G, is estimated from the slope ofthe line, m:
I- = slope = 13.3x 10-6 (MMscf) -I,G
or G= =75,190 MMscf.13.3x 10-6 MMscf-!
Note that, in Example 10.10, we estimated the original gas inplace assumingan averagevalue of formationcompressibility,whilein this example we calculated the original gas in plaC'eand formaction compressibility simultaneously. As a result, the two estimatesof gas in place are slightly different.
B. The average formation compressibility is determined fromtheintercept of the line, b:
b=- (SWiCW+Cf)l-Swi
= -18.5 x 10-6 psi -1 = - [ (0.24)(3.2 x 10-6) +Cf](1-0.24) .
Cf=13.3xlO-6 psi-!..
The data plotted in Fig. 10.15 do not lie completely on a straightline. Poston and Chen26 concluded that, initially, most ofthefor-'malion resistance to the overburden pressure is provided by!)1e'fluidsin the pore spaces. However, as fluids are withdrawn-from these.pore spaces, the formation compacts, resulting in more resistance:to the overburden being transferred to the rock matrix. Under these'
conditions, the formation compressibility is not a constantbutchanges with time, as indicated by the initial nonlinear portionofthe data in Fig. 10.15.
10.3.4 Volumetric Gas-Condensate Reservoirs. In this section,wedevelop material-balance equations for a volumetric gas reservoirwith gas condensation during pressure depletion. We also includethe effectsof connatewater vaporization. Both phenomenaaremostprevalent in deep, high-temperature, high-pressure gas reservoirs]and must be included for accurate material-balance calculations.
Depending on whether the pressure is above or below the dew-point, two or three fluid phases may be present in a gas-condensate
Gn.N
Vhevi Vhcv
Vhel
Initial Conditions YJ > pdJ
Fig. 10.16-Material-balance model showing reservoir PV'occupied by gas and liquid hydrocarbon phases at Initialandlater conditions for a gas-condensate reservoir.
p/Z Gp 1/(p,-p)[( Piz/PZ,) -1] Gp/(Pi -P)(Piz/PZi)(psia) (MMscf) 10-6psi-1 (Mscf/psi)
6,602.1 0 - -6,552.9 392.5 34.9 1,8406,467.2 1,642.2 38.9 3,1206,395.1 3,225.8 35.5 3,6506,331.3 4,260.3 36.4 3,7806,247.3 5,503.5 37.9 3,8806,136.4 7,538.1 39.9 4,2606,080.5 8,749.2 40.8 4,5205,954.1 10,509.3 43.4 4,6505,859.6 11,758.9 45.4 4,7605,798.6 12,789.2 46.6 4,9005,500.0 17,262.5 53.5 5,5404,878.2 22,890.8 74.5 6,5304,628.2 28,144.6 81.8 7,7004,208.8 32,566.7 98.8 8,8703,802.1 36,819.9 117.6 10,210
.~ .0__1I.ESL.',c t\~.c~,-,---=:~.<,.,,::=: :"-:.-'-:"-':>'-=
m'oir. Aoo\.e the dewpoim. the \.apor phase Ll'nsisrs l'f nl'[ ,'nlyJrocarbon and inert gases but also water vapor. As the reservoirssure declines, the water in the liquid phase continues to vapo-~ to remain in equilibrium with the existing water vapor, thusreasing the saturation of the liquid water in the reservoir and'easing the PV occupied by the vapor phases. As the reservoir;sure declines further, the amount of water vapor present in thephase may increase significantly. However, as the reservoir.sure decreases below the dewpoint. the fraction of PV avail-. for the vapor phases decreases as liquids condense from the:ocarbon vapor phase.J develop a material-balance equation that considers the effects1Scondensation and water vaporization requires that we include;hanges in reservoir PV resulting from these phenomena. Wen witn a material-balance equation for gas-condensate reser-s. We then extend this equation to include the effects of con-water vaporization. In addition, because changes in fonnationpressibilities often are significant in these deep, high-pressurereservoirs, we include geopressured effects.u-Condensate Reservoirs. We derived the material-balance
.tions in previous sectioBS for dry gases with the inherent as-)tion that no changes in hydrocarbon phases occurred during;ure depletion. Unlike dry-gas reservoirs, gas-condensate reser-; are characteristically rich with intennediate and heavierocarbon molecules. At pressures above the dewpoint, gas con-atesexist as a single-phase gas; however, as the reservoir pres-decreases below the dewpoint, the gas condenses and formslid hydrocarbon phase. Often, a significant volume of this COll-ate is immobile and remains in the reservoir. Therefore, cor-application of material-balance concepts requires that weider the liquid volume remaining in the reservoir and any liq-produced at the surface.,suming that the initial reservoir pressure is above the dew-, the reservoir PV is O£cupied initially by hydrocarbons in theJUSphase (Fig. 10.16), or
i=Vhi' (10.92)
e reservoir PV occupied by hydrocarbons in the gaseous phasecan be written as
vi=GTBgi, (10.93)
e GT includes gas and the gaseous equivalent of produced con-ttes and Bgi is defined by Eq. 10.7.later conditions following a pressure reduction below the dew-.the reservoir PV is now occupied by both gas and liquid!Carbon phases, or
=Vhv+VhL .(10.94)
~ Vp=reservoir PV at later conditions, RB; Vhv=reservoir'lCoccupiedby gaseous hydrocarbons at later conditions,RB;'hL= reservoir PV occupied by liquid hydrocarbons at latertions,RB.10.94assumes that rock expansion and water vaporization
~gligibl-e.In terms of the condensate saturation, So' we can
=(I-SoWp (10.95)
'hL=SoVp' """""""""""""""'" (10.96)
.ddition, the hydrocarbon vapor phase at later conditions is
=(GT-GpT)Bg, ","""""""""""'" (10.97)
Bg is evaluated at later conditions.ating Eqs. 10.95 and 10.97, the reservoir PV is
(GT-GpT)Bg, . (10.98)
(1-So)
itituting Eq. 10.98 into Eq. 10.95 and combining with Eq.yields an expression for the reservoir PV at later conditions:
So(GT-GpT)Bg . . . . . . . . . . . . . (10.99)-G )B + . ....pT g I-SQ
24'-~
:\"c'W. Ll'mbmm:: Eqs. Il).9': anJ ll). IN yidJs [he folllmingmaterial-balance equation:
S,,(GT-GpT)BgGTBgi=(GT-GpT)Bg+ .(10.100)
I-So
or, if we substitute Bg/Bg =(pz;)/( PiZ) into Eq. 10.100 and re-arrange,
.., ~i ~, ;
fIP Pi
( GpT )(I-So)-=- 1-- , (10.101)Z Zi GT
which suggests that a plot of (I-So)(Plz) vs. GpT will be a straightline from which GT can be estimated. Correct application of Eq.10.101, however, requires estimates of the liquid hydrocarbonvolumes formed as a function of pressure below the dewpoint. Themost accurate source of these estimates is a laboratory analysis ofthe reservoir fluid samples. Unfortunately, laboratory analyses offluid samples often are not available.
An alternative material-balance technique is
GTBzgi=(GT-GpT)BZg' """"""""""'" (10.102)
where GTBzgi=reservoir PV occupied by the total gas, which in-cludes gas and the gaseous equivalent of the produced condensates,at the initial reservoir pressure above the dewpoint, RB;(GT-GpT)BZg=reservoir PV occupied by hydrocarbon vaporphase and the vapor equivalent of liquid phase after some produc-tion at a pressure below the initial reservoir pressure and dewpoint
pressure, RB; and Bz i and Bzg =gas FVF's based on two-phaseZ factors at initial anJ later conditions, respectively, RB/Mscf.
If we substitute Bzg/BZg =(pzzi)/(PiZZ) into Eq. 10.102 and re-arrange, we have
ii,i!1
!
j
j ;
:A
~=~(i- GpT). (10.103)Zz ZZi GT
where ZZiand Zz=two-phasegas deviation factors evaluated at in-itial reservoir pressure and at a later pressure, respectively.
The form of Eq. 10.103suggests that a plot of p/zz vs. GpTwillbe a straight line for a volumetric gas-condensate reservoir whentwo-phase gas deviation factors are used.
Two-phase gas deviation factors account for both gas and liquidphases in the reservoir. Fig. 10.17 is an example of the relation-ship between the equilibrium gas (Le., single-phase gas) and two-phase deviation factorsfor a gas-condensatereservoir. At pressuresabove the dewpoint, the single- and two-phase Zfactors are equal;at pressures below the dewpoint, however, the two-phase Zfactorsare lower than those for the single-phase gas.
Ideally, two-phase gas deviation factors are determined from alaboratory analysis of reservoir fluid samples. Specifically, thesetwo-phase Z factors are measured from a constant-volume deple-tion study.27-Z9However, in the absence of a laboratory study,correiationsZ9are availablefor estimatingtwo-phaseZfactors fromproperties of the well-stream fluids.
Gas-Condensate Reservoirs With Water VaporiZiltion. In thissection, we developa material-balanceequation for gas-condensatereservoirs in which both phase changes and water vaporizationoccur. Similar to Humphreys'30 work, we include the effects ofrock and watercompressibilities,whichare often significantin deep,high-pressure reservoirs. The reservoir PV is occupied initially byhydrocarbon and water vapor phases as well as the connate liquidphase, or
Vpi= V\'i+ V...i (10.104)
where V'.i= initial reservoir PV occupied by hydrocarbon andwater vapors, RB, and V...i=initial reservoir PV occupied by theliquid water, RB.
If the reservoir pressure is above the dewpoint. connate wateris the only liquid phase present. From the definition of water satu-ration, we can write the initial reservoir PV occupiedby the liquidphase as
Vw;=Sw;Vp;' (10.105)
i
!
j!ii;:.ji"
]\.
Gas-Plus<:
~~
0.7Two-Ph1se
0.6
Q.$0 1000 2000 3000 ~ooo
Pressure, psia
Fig. 10.17-Example of equilibrium and two-phase gas devI-ation factors for a gas-condensate reservoir. 29
Similarly, we can express the initial reservoir volume of the vaporphases as
Vvi=(1-SwiWpi- """""""""""""'" (10.106)
Now, we define the fraction of the initial vapor phase volumethat is water vapor as
Ywi=Vwv/Vvi (10.107)
and the fraction occupied by the hydrocarbon gases as
(1-Yw;)=Vhv/Vvi (10.108)
where Vwvi=initialreservoir PV occupiedby water vapor, RB, andVhvi=initial reservoir PV occupied by hydrocarbon vapor, RB.
SubstitutingEq. 10.106 into Eq. 10.108 gives an expression forthe hydrocarbon vapor-phase volume in terms of the initial reser-voir PV:
Vhvi=Vpi(1-Swi)(I-Ywi)' .(10.109)Finally, becausethe initialhydrocarbonvaporphase is the original
gas in place,
Vhvi= GBgi, """""""""""""""'" .(10.110)
GBgithen Vpi= . ..(10.111)
(1-Swi)(1-ywi)
The form of the material-balanceequationat some pressure lowerthan the initial reservoir pressure depends on the value of the dew-point. Therefore, we will develop material-balance equations fordepletion at pressures above and below the dewpoint.
Depletion at Pressures Above the Dewpoint. Because the reser-voir pressure is still above the dewpoint, no hydrocarbon gas hascondensed. However, as the pressure declines, mOreof the liquidwater vaporizes, thus reducing the liquid water saturation. There-fore, the volume of liquid phase becomes (Fig. 10.18)
Vw=SwVp' """"""""""""""""" (10.112)
where Sw=current value of connate water saturation. Similarly,the volume of the vapor phase is
Vv=(I-SwWp' (10.113)In addition, we define the fractionof the vaporphase that is water
vapor as
Yw=Vw)Vv """""""""""""""'" .(10.114)
and the fraction of vapor phase that is hydrocarbon_~s
(1-Yw)=Vhv/Vv' (10.115)
Gn.N
Initial Conditions ~ >pJ
Fig. -10.18-Materiai-balance model showing reservoir PV0c-cupied by hydrocarbons and water at Initial and later condI-tions for a gas-condensate reservoir with water vaporization.
If we substitute Eq. 10.I 13 into Eq. 10.115, we can writeanexpression for the hydrocarbon vapor phase in terms of thecur-rent reservoir PV:
Vhv=Vp(1-Sw)(l-yw)' , (10.116);
The current hydrocarbon vapor phase is
Vhv=(G-Gp)Bg' (l0.1I7)~Combining Eqs. 10. I 16 and 10.117 gives the current reservoir,1
PV:
(G-Gp)BgVp= .
(1-Sw)(1 -Yw)
Like geopressuredgas reservoirs, deep, high-pressuredgas-~condensate reservoirs often experience significant changes inPV~during pressure depletion. Therefore, using a method similar to thai~presented in the section on geopressured gas reservoirs, wecanexpress the change in reservoir formation (rock) volume in tennsof the formation compressibility as
Cj(p--p) GB - jI P 1
dVf . '"'''''''''''''''''''' (10.119);(1-Swi)(I-ywi) - ~
In terms of Eq. 10.119, the material-balance equation forpres-~sures above the dewpoint becomes ~
GBgi (G-Gp)Bg ci<Pi-p)GBgi- + .(I-Swi)(I-Ywi) (I-Sw)(l-yw) (1-Swi)(1 -Yw;)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (10.120)
Rearranging terms gives
(l-Sw) (l-yw) BgiG -[I-Cj(Pi-P)]=G-Gp'
(1-Swi) (1-Ywi) Bg
Substituting Bg/Bg=pz/p;'<. into Eq. 10.121 and rearrangill&yields .
(1-Sw) (l-yw) - P Pi Pi Gp[1-Cj(pi-p)]-= ,
(I -Swi) (l-yw;) Z Zi Zi G
The form of Eq. 10.122 suggests that a plot of
(I-Sw) (1-yw) P
[I-ci<pi-p)]- vs. Gp 1(1-Swi) (1-Ywi) Z:
jwill be a straight line with a slope equal to -P/ZiG and an in!er~cept equal to P/Zi' At p/z=O, Gp=G, so extrapolation oCme:
straight line to plz =0 provides an estimate of original gas in place.1N~te that, if t~e ":,,ater saturation remains constant during the~~of the reservOIr (I.e., Sw=Swi and Yw=Ywi) and when formatic8jcompressibility is negligible, Eg. 10.122 reduces to Eg. 10.29a volumetric dry-gas reservoir.
250
1.0
0.9
50It Q.Jc:
'0.;;1cd.>8a0
VhcvVhcvi
Vwv
VwVwvi
Vwi
VOLUMESAND MATERIAL-BALANCE CALCULATIONS
Gp,Np
V hcvi
Vhcv
Vwv
Vw
VwviVhcl
L1VVwi
Later Conditions ~< Pd)
FIg.10.19-Material-balance model showing reservoir PVoc-cupiedby hydr9carbons and water at initial and later condi-lionsfora gas-condensatereservoirwithwatervaporization.
,Depletionat PressuresBelow the Dewpoint. Whenreservoirpres-decrease below the dewpoint, the gas phase condenses. Ingas-condensate reservoirs, the liquid hydrocarbons formed
ithereservoirremain immobile. Therefore, we must modify Eq.U20to include this additional liquid phase (Fig. 10.19).Addingthe liquid phase gives
(G-Gp)Bg Cf(Pi-P)GBgi+ ,
(I-Swi)(I-yw) (l-Sw-So)(l-yw) (l-Swi)(l-Ywi)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (10.123)
GBgi
lereSo=liquid hydrocarbon phase (i.e., condensate) saturation.lerrearranging Eq. 10.123, we write a material-balance equa-
'usimilar in form to Eq. 10.122:
(I-Sw-So) (l-yw) - P Pi Pi Gp[1-CjCPi -p)]- =- - - -.
(l-Sw) (l-yw) Z Zi Zi G
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (10.124)
Again, the form of Eq. 10.124 suggests that a plot of
(I-S -S) (l-y ) Pwow [1-cf(Pi-P)]-vs.Gp
(I-Sw) (I-Ywi) ZI;iillbe a straight line with a slope equal to -P/ZiG and an inter-'reptequalto P/Zi' Aat plz=O, Gp=G, so extrapolationof themight line to plz=O provides an estimate of original gas in place..gain,the gas deviation factors in Eqs. 10.122 and 10.124 should:two-phase Z factors representing both gas and liquid hydrocar-)0phases in the reservoir. In addition, gas production should in-
idudenotonly productionfromall separatorsand the stocktankI!1talso the gaseous equivalent of the produced condensates.Thewater vapor content of a gas has been shown31 to be de-
i~dent on pressure, temperature, and gas composition. Gas com-iPJSition also has more effect on water vapor content at higher'f'tSsures.Unfortunately, laboratory analyses of gas usually do not'~tify the amount of water vapor; however, as discussed in Chap.t,empirical methods32.33 are available for estimating the waterIIpOrcontent of a gas.Correct application of Eq. 10.124 also requires estimates of the
I
~Uidhydrocarbon volumes formed at pressures below the dew-fOint.The most accurate source of these estimates is a laboratory
lmalysisof the reservoir fluid samples. These liquid saturations are«Jtainedfrom a constant-volume depletion study.27.28 Note thatlibistypeof laboratory fluid study assumesthat the liquid hydrocar-'Ionsformedin the reservoir are immobile.This assumptionis validlormost gas-condensate reservoirs; however, some very rich gas-rondensatefluids may be characterized by mobile liquid saturations.Forthese conditions, compositional simulators are required to model!bemultiphase flow and predict future performanceaccurately.
i
251
6000
5000
4000
.~~ 3000~""
2000
1000
00 10 20 30
Gp. MMscf
40 50
Fig. 10.20-plz plot for Exercise 10.3.
10.4 Summary
Reading this chapter should prepare you to do the following..Calculate original gas in place in a volumetric dry-gas reser-voir using volumetric methods..Calculate gas reserves and recovery factor for a gas reservoirwith water influx using volumetric methods..Calculate original gas in place and condensate in place for avolumetric wet-gas reservoir using volumetric methods..State the principle of conservation of mass and derive a gener-al material-balance equation from that principle..Calculate original gas in place using material balance for a volu-metric dry-gas reservoir..Estimate water influx using the van Everdingen-Hurst, Carter-Tracy, and Petkovich methods and state the assumptions and limi-tations of each method..Estimate original gas in place using material balance for a dry-gas reservoir with water influx..Estimate original gas in place using material balance for a volu-metric geopressured gas reservoir..Determine average formation compressibility and original gasin place simultaneously using material balance in a volumetric ge-opressured gas reservoir..Derive material-balance relationships for volumetric gas-condensate reservoirs, gas-condensate reservoirs with water vapori-zation, gas-condensatereservoirswith depletion at pressures abovethe dewpoint, and gas-condensate reservoirs with depletion at pres-sures below the dewpoint.
Questions for Discussion
I. Your company is planning to drill a wildcat well in an areabelieved to have great potentialfor natural gas. In planning for thewell(andassuminggas is discovered),what data shouldbe gathered?Why?
2. The well does discover gas. Using information from the dis-covery well alone, you are asked to estimate reserves and forecastfutureproduction.Whatmethodswillyou use? What additionaldatawill you need?
3. Several additional wells are planned. What sort of informa-tion should be obtained from these wells? Why?
4. Two years after the discovery well was drilled, the field ishighly developed. You are again asked to estimate reserves andforecast future production. What methods will you use? What datawill you need?
5. How would your strategy be affected if the reservoir is ge-opressured? Is a retrograde gas reservoir? Has a waterdrive? Is thegas cap of a oil reservoir, a fact discovered only much later? Howcan you determine whether any of these possibilities is a fact?
6. How would you determine optimal well spacing? Optimaldepletionplan? In answer to both these questions, what data wouldyou need, and what study methods would you use?
7. What are the potential errors in the volumetric method? Thematerial-balance method?
252
TABLE 10.19-PRODUCTION AND PRESSURE HISTORIES,EXERCISE 10.4
Cumulative GasProduction
(MMscf)0
1020304050
Pressure(psia)
2,7512,3812,2232,0851,9401,801
zfactor
0.7420.7390.7370.7420.7470.755
TABLE 10.20-PRESSURE HISTORY, EXERCISE 10.5
Time(days)
091.5
183.0274.5366.0457.5549.0
Reservoir/Aquifer Pressure(psi a)3,7933,7883,7743,7483,7093,6803,643
TABLE 10.21-PRESSURE HISTORY, EXERCISE 10.6
Time(days)
091.5
183.0274.5366.0457.5549.0
Reservoir/Aquifer Pressure(psi a)
3,7933,7883,7743,7483,7093,6803,643
TABLE 10.22-PRESSURE HISTORY, EXERCISE 10.7
Time
(days)
091.5
183.0274.5366.0457.5549.0
Reservoir/Aquifer Pressure(psia)
3,7933,7883,7743,7483,7093,6803,643
8. What are typical recovery factors for volumetric gas reser-voirs? How may we estimate abandonment pressure to calculaterecovery factor? At what point does abandonment pressure enterthe recovery factor calculation?
9. Compare typical recoveries in waterdrive gas reservoirs withthose from volumetric gas reservoirs and explain the reasons forthe difference.
10. Explain two methods for improving the recovery fromwaterdrive gas reservoirs.
II. Describe the different types of aquifer that may be in associ-ation with a reservoir.
12. How does the aquifer offsetpressure decline in the reservoir?13. If the time elapsed sinee production began doubled, do you
think the amount of water influx into a gas reservoir would dou-ble? Or would it be more or less than double? Why? If additionalinformationis requiredto answerthisquestion,what is thataddi-tional information?
Exercises
10.1 For the reservoir data given below, calculate the total cu-
.-8=60°
GAS RESERVOIR ENGINEERING
SealingFault
SealingFault
Fig. 10.21-Reservoir configuration for Exercise 10.10.
mulative gas production and the recovery factor at Po=500psia, where Bgo=6A5 RB/Mscf.
Pi = 3,150 psia.A = 640 acres.h = 10 ft.G = 8.882 X106 Mscf.<p = 0.22.
S"'i = 0.23.Bgi = 0.947 RB/Mscf.
10.2 For the reservoir data givenbelow, estimatethe gas reserves,;and the recovery factor using the volumetric method.
Pi = 3,150psia.A = 640 acres.h = 10 ft.
Bgi = 0.947 RB/Mscf.<p = 0.22.
S"'i = 0.23.Sgr = 0040.Bgo = 1.48 RB/Mscf.
10.3 Using the data givenbelow and the plot of p/z vs. GpinFig.10.20, estimate the cumulative gas production at an aban-donmentpressureof 500 psia. In addition,estimatetherecov-ery factor at the abandonment pressure.
~. i:~i
~: !Pi = 4,000psia.Po = 500 psiaZi = 0.80 psia.Zo = 0.94 psia.
lOA Using the reservoir pressure and cumulative gas productiongiven below, estimate the original gas in place, gas reservesat Po= 500 psia, and the gas recovery factor.
Pi = 2.751psia.za = 0.92.Po = 663 psia.
Table 10.19 gives the reservoir production and pressurehistories.
10.5 Calculate the water influx at timestep n=3 (i.e., (=274.5days) for the reservoir/aquifer system describedbelow. Usethe van Everdingen-Hurst method and assume an infinite-acting aquifer.
'",
&~ i
~~..
,.,.
<p = 0.209.JL = 0.25 cpoh = 19.2 ft.e = 180°.
GA~VOLUMES AND MATERIAL-BALANCE CALCULATIONS 253
IJI
. \!
k = 275 md.CI = 6xlO-6 psia-l.rr = 5,807 ft.
Table 10.20 gives the pressure history at the initialreservoir-aquifer interface.
r 10.6 Calculate the water influx at timestep n=3 (i.e., t=274.5days) for the reservoir/aquifer system given below. Use theCarter-Tracy method and assume an infinite-actingaquifer.
<P = 0.209.Jl.= 0.25 cpoh = 19.2 ft.0 = 180°.k = 275 md.
CI = 6xlO-6 psia-l.rr = 5,807 ft.
Table 10.21 gives the pressure history at the initialreservoir-aquifer interface.
.JO.7 Calculate the water influx at timestep n=3 (i.e., t=274.5,- days)for the reservoir/aquifer system givenbelow. Use the
Fetkovich method and assume a finite aquifer.
<P = 0.209.Jl. = 0.25 cpoh = 19.2 ft.0 = 180°.k = 275 md.CI= 6xlO-6 psia-l.rr = 5,807 ft.A = 250,000acres.
Table 10.22 gives the pressure history at the initial reser-voir/aquifer interface. The cumulative water influx after 183
TABLE 10.25-PRESSURE HISTORY, EXERCISE 10.10
Time(days)
010
200400500
Reservoir/Aquifer Pressure
(psia)4,0003,9903,9003,8203,760
days was 40,630+8,587=49,217 RB.
10.8 For the following gas reservoir offshore Louisiana,23 esti-mate the original gas in place using the material-balancemethod for geopressured reservoirs. Also use the material-balance method for a normally pressured gas reservoir, andcompare the initial gas estimates from both equations. Theplotting functions for both material-balance methods are tabu-late in Table 10.23.
Pr = 11,444psia.cf = 19.5xlO-6 psia-l.
Swi = 0.22.Cw= 3.2xlO-6 psia-I.
10.9 For the following gas reservoir offshore Louisiana,23 esti-mate the original gas in place and formation compressibilityusing the material-balance method for geopressured reser-voirs. In addition, use the material-balancemethod present-ed for a normally pressured gas reservoir, and compare theinitial gas estimates from both equations. The plotting func-tions are tabulated in Table 10.24.
TABLE 10.23-PLOTTING FUNCTIONS, EXERCISE 10.8
Reservoir Cumulative GasPressure Production p/z p/z[1-(CWSwi +C,)(Pi -p)/(1-Swi)]
(psia) (MMscf) (psia) (psia)
11,444 0 7,650 7,65010,674 9,920 7,423 7,27610,131 28,620 7,252 7,0089,253 53,600 6,957 6,5618,574 77,670 6,698 6,2027,906 101 ,420 6,428 5,8397,380 120,360 6,191 5,5396,847 145,010 5,933 5,2296,388 160,630 5,693 4,9505,827 182,340 5,375 4,5945,409 197,730 5,132 4,3305,000 215,660 4,840 4,0354,500 235,740 4,478 3,6744,170 245,900 4,221 3,425
TABLE 10.24-PLOTTING FUNCTIONS, EXERCISE 10.9
Reservoir Cumulative GasPressure Production 1/(Pi -p)[(PiP/zz,) -1] Gp/(Pi -P)(PiP/ZZi)
(psia) (MMscf) 10 -6psi -1 (Mscf/psi)11 ,444 0 - -10,674 9,920 39.7 13,30010,131 28,620 41.8 12,9009,253 53,600 45.4 26,9008,574 77,670 49.5 30,9007,906 101,420 53.7 34,1007,380 120,360 58.0 36,6006,847 145,010 63.0 40,7006,388 160,630 68.0 42,7005,827 182,340 75.4 46,2005,409 197,730 81.3 48,8005,000 215,660 90.1 52,9004,500 235,740 102.0 58,0004,170 245,900 111.7 61,300
254
Pr = 11,444psia.Cw = 3.2xlO-6 psia-I.
Swi = 0.22.
10.10 The reservoir shown in Fig. 10.21 is undergoing edgewater, influx. The pressure has been measuredat the reservoir/aqui-
fer boundary after several time periods, and this pressurehistory is given in Table 10.25. The reservoirradius is 1,000ft and the aquifer radius is 10,000 ft. The aquifer has a porosi- '
ty of 10%, a permeability of 10 md, and a thickness of 600ft. The water has a compressibility of 1x 10 -5 psia -I anda viscosity of 0.6323 cpoA. Calculate B.
B . Calculate the relationship between dimensionless time andtime in days.
C. Calculate f:J.Pl for 10 days.D. Calculate f:J.P4for 500 days.E. Calculate water influx in reservoir barrels after 500 days.
Nomenclature
A = well drainage area, LZ, acresb = interceptB = van Everdingen-Hurst constant, RB/psi
Bg = gas FVF, RB/MscfBgo = gas FVF at reservoir abandonment pressure and
temperature conditions, RB/MscfBgi = gas FVF at initial reservoir pressure and
temperature, RB/MscfBw = water FVF, RB/STB
BZg = gas FVF based on two-phase gas deviation factor,RB/Mscf
BZgi = gas FVF based on two-phase gas deviation factor atinitial reservoir pressure and temperature,RB/Mscf
C = water influx constant from material-balancecalculations, RB/psi
cf = in-situ formation compressibility, Ltz/m, psia-lC( = total aquifer compressibility, cf+cw, Ltz/m, psia-l
Cw = water compressibility, LtZ/m, psia-lE V = volumetric sweep efficiency, fractionfg = fraction of reservoir gas produced as liquid at
surface, fractionF = gas recovery factorG = original gas in place, L3, Mscf
Go = gas in place at reservoir abandonment, L3, MscfGp = cumulative gas production, L3, Mscf
Gpo = additional gas production from secondary separatorand stock tank, L3/L3, scf/STB
GpT = cumulative gas production from primary andsecondary separators, stock tank, and the gaseousequivalent of produced condensates, L3, Mscf
Gpl = cumulative gas production from primary separator,L3, Mscf
Gpz = cumulative gas production from secondary separator,L3, Mscf
Gp3 = cumulative gas production from stock tank, L3, MscfG( = reservoir volume occupied by gas trapped by
encroaching water, L3, MscfGT = total initial gas in place, including gas and gaseous
equivalent of produced condensates, L3, Mscfh = net formation thickness, L, ftJ = aquifer PI, L3t/m, STB/D-psik = reservoir permeability, L2, mdL = length of linear-shaped reservoir, L, ftm = slope
Mo = molecular weight of stock tank liquids, m,lbm/lbm-mol
n = number of moles of gasN = original oil (condensate) in place, L3, STB
GAS RESERVOIR ENGINEERING
Np = cumulative oil (condensate) production, L3, STBP = reservoir pressure, m/Lt2, psia
Po = abandonment pressure, m/Lt2, psiaPaq = aquifer pressure, m/Lt2, psia
Paq,i = initial aquifer pressure, m/Lt2, psiaPd = dewpoint pressure of gas-condensate reservoir,
m/Ltz, psiaPD = dimensionless pressurePD = dimensionlessrressure derivativePi = initial reservoir pressure, m/Lt2, psiaPn = pressure at time interval n, m/Lt2, psiaPr = pressure at aquifer/reservoir interface, m/LtZ, psia
Psc = pressure at standard conditions, m/LtZ, psiaf:J.p= difference between initial aquifer pressure and
pressure at original reservoir/aquifer boundary,m/Ltz, psia
qD = dimensionless water influx rateqw = water influxrate, L3/t, STB/D
QpD = dimensionless cumulative water influxro = radiusof aquifer, L, ftre = outer radius of reservoir, L, ftrr = radius to aquifer/reservoir interface, L, ftR = universal gas constant, 10.73 psia-ft3/Ibm-mol-oR
R( = total gas/stock-tank liquid ratio, scf/STBR 1 = ratio of high-pressure separator gas volume to stock-
tank liquid volume, L3/L3, scf/STBRz = ratio of low-pressure separator gas volume to stock-
tank liquid volume, L3/L3, scf/STBR3 = ratio of stock-tank gas volume to stock-tank liquid
volume, L3/L3, scf/STB
Sgi = initial gas saturation, fractionSgr = residual gas saturation, fractionSo = condensate saturation, fractionSw = connate water saturation, fraction
Swo = water saturation at abandonment conditions, fractionSwi = initial connate water saturation, fraction
t = time, t, daystD = dimensionless timeT = reservoir temperature, T, oR
Tsc = temperature at standard conditions, T, oRVeq = vapor equivalent of primary separator liquid, L3/LJ,
scf/STB
Vf = reservoir formation (rock) volume, L3, res bblVgi = gas volume at initial conditions, L3, res bblVhL = volume of hydrocarbon liquids, L3, res bblVhv = volume of hydrocarbon vapors, L3, res bblVhvi = initial volume of hydrocarbon vapors, L3, res bbl
Vp = reservoir PV, L3, res bblVpi = initial reservoir PV, L3, res bblVsc = gas volume at standard conditions, L3, res bblVv = volume of vapor phase, L3, res bbl
Vvi = initial volume of vapor phase, L3, res bblVw = volume of water, L3, res bbiVwi = initial volume of water, L3, res bblVwv = volumeof water vapor phase,L3, res bblVwvi = initial volume of water vapor phase, L3, res bblf:J.Vf = change in reservoir formation (rock) volume, L3,
res bblf:J.Vw = change in water volume, L3, res bbl
w = width of linear reservoir, L, ftWe = cumulative water influx volume, L3, res bblWei = initial "encroachable" volume of water in aquifer,
L3, res bblWi = initial volume of water in aquifer, L3, res bblWp = cumulative water production, L3, STByw = fraction of total vapor-phase volume that is waterYwi = initialfractionof total vaporphasevolumethaiis
water
:>VOLUMES AND MATERIAL-BALANCE CALCULATIONS
Z = gas compressibility factor, dimensionlessZa = gas compressibility factor at abandonment, dimen-
sionless
Zi = gas compressibility factor evaluated at initial condi-tions, dimensionless
zsc = gas compressibility factor at standard conditions,dimensionless
Zz = two-phase gas compressibility factor, dimensionlessZZi= two-phase gas compressibility factor evaluated at ini-
tial conditions, dimensionless'10 = specific gravity of condensate (water= 1.0 g/cm3)'Iw = specific gravity of reservoir gas at reservoir condi-
tions (air= 1.0)'II = specific gravity of high-pressure separator gas
(air= 1.0)'Y2= specific gravity of low-pressure separator gas
($= 1.0)'13 = specific gravity of stock-tank gas (air=1.0)
e = angle encompassed by aquifer, degreesA = dummy integration variable for Eq. 10.57Il = viscosity,miLt, cp<p= porosity, fraction
.scriptaq = aquifer
-erscript- = average
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