1

Gas Condensate Reservoir

2

CharacteristicsBetween oil and gas reservoirProduce light-colored or colorless stock tank oilCondensate (distillate, oil): > 45 °APIGas-oil ratio: 5,000 ~ 10,000 SCF/bbl

3

Calculation of Average Specific Gravity of Well Fluids

1 1 3 3

1 3

4602.133,316

ow

o

wo

R R

R RM

γ γ γγ γ+ +

=+ +

1, 3

1, 3

o

R : gas-oil ratio from separator (1) and stock tank (3) : specific gravity of separator and stock tank gases: specific gravity of the stock tank oil (water = 1.00)

Rγ γ

γ

Consider a two-stage separator system

o wo,

42.43141.5 , M131.5 1.008

o

API o

γγρ γ°

= =+ −

4

Calculation of Initial Gas and Oil in Place Ex. 5.1

Initial pressure: 2740 psia Reservoir temperature: 215 deg F

Average porosity: 25% average connate water: 30%

Daily tank oil: 242 STB Oil gravity at 60°F: 48.0 °API

Daily separator gas: 3100 MCF Separator gas gravity: 0.650

Daily tank gas: 120 MCF Tank gas gravity: 1.20

Calculate the initial oil and gas in place per acre-ft for a gas-condensate reservoir.

5

Ex. 5.1 solution:

,

1 3

w

Step 1: Calculate important parameters141.5 5954 59540.788, 151.9 lb/lb-mole

48.0 131.5 8.811 48.0 8.8113,100,000 SCF 120,000 SCF12,800 SCF/STB, 496 SCF/STB

242 STB 242 STB12,810

o woo API

M

R R

γρ

γ

= = = = =+ − −

= = = =

=(0.650) 4602.(0.788) 496(1.20) 0.896133,316(0.788)12,810 496

151.9step 2: Look for z factor using pseudocritical valuesFrom one of the figures on pesudocritical values as a function of specific gravit

+ +=

+ +

pc pc

pr pr

y, we have T 423 and p =637 psia

215 460 2740T 1.60, p 4.30423 637

look up in Standing and Katz chart, we have z = 0.825 at the initial condition.step 3: Initial volume of gas in place

pc pc

R

T pT p

= °

+= = = = = =

, V= 43560 0.25 (1-0.30)Step 4: initial total lb-mole of gas in place

pV (2740 psia)(43560 0.25 (1-0.30)) n= lb-molezR'T 0.825(10.73)(215+460)

Under standard conditions each lb-mole of gas is equivalent

× ×

× ×=

to 379.4 SCFG = n lb-mole 379.4 SCF/lb-mole =1326 MCF/ac-ft×

6

step 5: partition into fraction of oil and gas. For gas, volume fraction = mole fractionfor 1 STB of gas condensate, Gas volume = 12800+496 SCF = 13296 SCF

Gas equivalent of produced condensate = 133,3 o

wo

g o

o

0.78816 133,316 691.6 /ac-ftM 151.9

13,296f 0.95, f 0.0513,296 691.61326 MCF/ac-ft 0.95 = 1260 MCF/ac-ft,

V 1326 MCF 0.95/(12800+496) = 94.8 STB/ac-ftStep 6:total daily gas-condensate

g

SCF

V

γ= × =

= = =+

= ×

= ×

p

production in MCF is:daily gas 3100 120G 3386 MCF/day

0.951 0.951total daily reservoir voidage by the gas law is:

675(14.7)(0.825)V=3,386,000 19,450 /520(2740)

cu ft day

+Δ = = =

⎛ ⎞=⎜ ⎟

⎝ ⎠

7

Differential Liberation

Keep constant V, p decreases gradually until atmospheric P.Simulate the graduate depletion of a volumetric reservoir.

p1>=pb p2 p3

GasLiquid

Gas GasLiquid

8

Composition of produced gas

9

Ex. 5.2. Calculate the volumetric depletion performance of a retrogradegas reservoir based on lab test given in the following table and thefollowing condition.

Initial p: 2960 psia; abandonment p: 500 psia; Reservoir T: 195 deg F;Connate water: 30%; porosity: 25%; standard condition: 14.7 psia and60 deg F; initial cell volume: 947.5 cu cm; molecular weight of heptaneplus in initial fluid: 114 lb.lb-mole; specific gravity of heptane plus in initialfluid: 0.755 at 60 deg F.

10

Solution: Assumption: 1. The same molecular weight and sp. Gravity for the C7content for all produced gas. 2. Liquid recovery from the gas is 25% of the butanes, 50% of the pentane, and 75% of the hexane, and 100% of heptane and heavier.

11

Solution:

scp

g sc

43,560 0.25 (1 0.30) 7623 cu ft/ac-ftFor the increment produced from 2960 to 2500 psia,

175.3 cu cmV=7623 1410 cu ft/ac-ft at 2500 psia and 195 F,947.5 cu cm

T V VVGB p R

HC

sc

V

p V pzT z T

= × × − =

Δ × = °

Δ ΔΔΔ = = =

p

379.4 379.4 2500 1410 = 240.1 M SCF1000 1000 0.794 10.73 655

For column (3) (gross gas production), G p

p VzRT

G

Δ × ×= =

× × ×= Δ∑

Step 1: Calculate the increments of gross production in M SCF per ac-ft of net, bulk reservoir rock. Enter in column (2).

12

Solution: Step 2: Calculate the M SCF of residue gas and the barrels of liquid obtained from each increment of gross gas production. Enter in column (4) and (6). Assume that 0.25 C4, 0.50 C5, and 0.75 C6, and all C7

+ is recovered as stock tank liquid. For example, in the 240.1 M SCF produced from 2960 to 2500 psia, the mole fraction recovered as liquid is:

0.25 0.028 0.50 0.019 0.75 0.016 0.034 0.007 0.0095 0.0120 0.034 0.0625 mole fractionAs the mole fraction also equals the volume fraction in gas, the M SCF recovered as liquid from 240.1 M SC

LnΔ = × + × + × += + + + =

F is:0.007 240.1 0.0095 240.1 0.012 240.1 0.034 240.1

= 1.681+2.281+2.881+8.163 15.006 M SCFThe gas volume can be converted to gallons of liquid using the gal/M SCF figures for each carbon,

LGΔ = × + × + × + ×=

4

5 6

7

for C , the conversion is 32.04 (average of iso and n-butane) gal/M SCF;for C , the conversion is 36.32 gal/M SCF; for C , the conversion is 41.03 gal/M SCF;

114 lb/lb-mofor C , the conversion factor = le 47.71 gal/M SCF0.3794M SCF/lb-mole 8.337lb/gal 0.755

Therefore, the total liquid recovered from 240.1 M SCF:1.681 32.04+2.281 36.32+2.881 41.03+8.163 47.71 =53.9+82.8+118.2+389.5=644.4 gal =15.3 bbl

=× ×

× × × ×.

Volume of residue gas after recovery = 240.1 - 15.006 = 225.1 M SCFThese two number enter cols. (4) and (6). Cumulative values are calculated and enter in cols.(5) and (7).

13

Step 3: Calculate the gas-oil ratio for each increment of gross production in units of residue gas per barrel of liquid. Enter in column (8). For example, the gas-oil ratio of the increment produced from 2960 to 2500 psia is 225.1*1000/15.3 = 14,700 SCF/STB.

Step 4: Calculate the cumulative recovery percentages of gross gas, residue gas, and liquid. Enter in Cols (9), (10), (11).

379.4pV 379.4 2960 7623The initial gas in place is: 1580 M SCF/ac-ft1000 zRT 1000 0.771 10.73 655

Of this, the liquid mole fraction is 0.088 and the total liquid recovery is 3.808 gal/M SCF of gross ga

× ×= =

× × ×s,

which are calculated from the intial composition in the same amnner as shown in step 2. ThenG=(1-0.088) 1580=1441 M SCF residue gas/ac-ft

3.808 1580N= 143.2 bbl/ac-ft.42

At 2500 psia, the

Gross gas rec

××

=

100 240.1overy factor = 15.2%1580

100 225.1Residue gas recovery factor = 15.6%1441

100 15.3Liquid recovery factor = 10.7%143.2

×=

×=

×=

14

Solution: Assumption: 1. The same molecular weight and sp. Gravity for the C7content for all produced gas. 2. Liquid recovery from the gas is 25% of the butanes, 50% of the pentane, and 75% of the hexane, and 100% of heptane and heavier.

15P decrease

Recovery factor of retrograde liquid = 71.6/143.2*100% = 50%

Recovery factor of gas = 1200/1441*100% = 80.4%

16

Calculation for the whole reservoir: IF the reservoir has produced 12.05 MMM SCF of gross well fluid when the average reservoir pressure declined from 2960 psia initial to 2500 psia. According to table 4.5 the recovery at 2500 psia under volumetric depletion is 15.2% of the initial gross gas in place, and therefore the initial gross gas in place for the whole reservoir is:

9

9

12.05 10 79.28 MMM SCF0.152

Because the recovery factor is 80.4 at an abandonment pressure of 500 psia, the initial recoverable gross gas or initial reserve is:Initial reserve = 79.28 10 0.804 63.7

G ×= =

× × = 4 MMM SCFSince 12.05 MMM SCF has already been recovered, the reserve at 2500 psia is 63.74-12.05=51.69 MMM SCF

17

How accurate is this kind of calculation?

Sources of error: spatial variation:

Uneven distribution of pressure and properties

varying stage of depletion

varying gas-oil ratio

varying gas composition

18

Material Balance for Retrograde Gas Reservoirs

( )1

w wi fg gi gi e p g w p

wi

c S cG B B GB p W G B B W

S+⎡ ⎤

− + Δ + = +⎢ ⎥−⎣ ⎦

( )g gi e p g w pG B B W G B B W− + = +

Compressibility term negligible compared to the gas expansion

(For single phase gas reservoir in general)

Here the Bg needs to be calculated using the two phase z factor, not a single phase z factor.

( )g gi p gG B B G B− =

19

Two Phase z factor379.4( ) '

res res

p res

p VzG G R T

×=

−

z factor for the gas phase remaining in reservoir

z factor of produced gas

20

Gas Production History

21

Differences Between Production curve and prediction

Possible cause of differences:

-Sampling error

- N2 as part of the gas

- possibility of retrograde liquid flow

Res

ervo

ir P

ress

ure

(psi

a)

Here z is two phase gas deviation factor

22

Lean Gas CyclingLean gas: the dry gas produced on surface after the condensate liquid being removed from the total produced (wet) gas. Lean gas cycling: dry gas is returned to the reservoir through injection wells. Advantage: maintain reservoir pressure and retards retrograde condensation. Advantage: drives the wet gas toward producing wells.

23

Lean Gas Cycling

Disadvantage: deferred income from the sales of gasesDisadvantage: required additional expenditures (more wells sometimes)Disadvantage: liquid recovery may still be considerably lower than 100%.

24

Three separate recovery factors1. Microscopic displacement efficiency: 70% ~ 90%2. Due to flow rate of production and injection wells,

there are areas not swept by dry gas – 50% - 90%3. Stratified reservoirs: wet gas remain in less

permeable regions while the lean gas tends to sweep through more permeable regions.

If 80% in each stage, what is the overall efficiency? = 0.8*0.8*0.8 = 51.2%

25

To Cycle or Not to Cycle

Not attractive, because retrograde loss rarely exceeds 50%Whether or not to cycle is balancing between the gain and the loss

26

Other Types of Gases for Pressure Maintenance

1. Wet hydrocarbon gas2. N2

3. CO2

4. Combustion flue gas

What to consider:Composition and properties of injection gasPossibility of changing dew point values Which gas works best depends on the properties of gas and also reservoir condition

27

Gas Condensate Reservoir Under Active Water Drive

Under active water drive, pressure drop very little - there is little or no retrograde condensation- Gas-oil ratio of the production remains constant- Formation volume factor would remain the same- behave the same as non-retrograde gas reservoirRecovery factor depends on: - Initial connate water saturation: Swi

- Residual gas condensation: Sgr

- The fraction of initial reservoir invaded by water: F

28

Fractional Recovery from Retrograde Gas Reservoirs under Active Water Drive

(Pa = Pi, Bgi = Bga)

gi

a

If under an active water drive there is essential no decline in reservoir pressure,B

(1 ) (1 )G-GRecovery factor = (1 ) 1

ga

i wi gr gi wi gr

i wi gi wi

B

V S S B F S S FG V S B S

φφ

=

− − − −= =

− −

What determines F:

well location, well spacing, permeability stratification, water coning in bottom-drive reservoirs

Fraction of the reservoir invaded by water drive

29

Swi

30

From Table 4.8:We know that

With the same Swi and Sgr, recovery factor increases with increasing F.With the same Sgr and F, recovery factor decreases with increasing Swi.With the same Swi and F, recovery factor increases with increasing Sgr.

31

32

33

Unit Recovery from Single Phase Gas Reservoirs under Active Water Drive (Pa =

Pi, Bgi = Bga)

gi

gr wi grwi

gi ga gi

wi gr

ga

If under an active water drive there is essential no decline in reservoir pressure,B

S 1 S S1 Sunit recovery = 43,560 ( - )=43,560 ( )B B B

1 S S100(

B100(G-G )Recovery factor =

gaB

G

φ φ

=

− −−× × × ×

− −

= i wi gr

wi wi

gi

)100(1 S S )

%1 S 1 SB

− −=

− −