PNG 410: Applied Reservoir Engineering

Home work assignment #5 Solution

April 21, 2010 Due date: April 28, 2010

NOTE: Please detail the formula, steps, and units for the calculation. Missing formula, steps, and units can lead to lost points. Problem 1: The following data are taken from an oil field that had no original gas cap and no water drive: Initial reservoir pressure = 3500 psia Initial reservoir temperature = 140 deg F Bubble point pressure of the reservoir pb= 2400 psia Oil pore volume of reservoir = 75 MM cu ft Solubility of gas in crude oil below bubble pressure = 0.42 SCF/STB/psi Formation volume factor Boi at 3500 psia = 1.333 bbl/STB Gas deviation factor z at 1500 psia and 140 deg F = 0.95 Oil produced when pressure at 1500 psia Np= 1.0 MM STB Net cumulative produced gas-oil ratio Rp at 1500 psia = 2800 SCF/STB 1.1 Calculate the initial STB of oil in reservoir

o

oi

75 MM cu ft = 75/5.615 MM bbl = 13.36 MM bblV 13.36 MM bblN = 10.02 MM STBB 1.333 bbl/STB

oV =

= =

1.2 Calculate the initial gas-oil ratio and SCF of gas in reservoir (using the definition

of gas solubility in crude oil).

soi

Solubility of gas in crude oil = 0.42 SCF/STB/psi,at bubble point pressure of 2400 psia,gas-oil ratio R = 2400 psia 0.42 SCF/STB/psia = 1,008 SCF/STBAbove bubble point pressure, gas-oil ratio rea

×

mins constant,Therefore, G = 1,008 SCF/STB 10.02 MM STB = 10.1 MMM SCF×

1.3 Calculate the SCF of gas remaining in the reservoir at 1500 psia

p p

p

N at 1500 psia = 1 MM STB, R at 1500 psia = 2800 SCF/STB,

produced gas = N 1 MM STB 2800 SCF/STB=2.8 MMM SCF

gas remaining in the reservoir = 10.1 - 2.8 = 7.3 MMM SCFpR× = ×

1.4 Calculate the SCF of free gas in the reservoir at 1500 psia

1

soi

so

p

With gas solubility of 0.42 SCF/STB/psi, and initial R 1,008 SCF/STBat 1500 psia, the R 0.42 (2400 1500) 630 SCF/STBSo the gas dissolved in oil is (N-N ) (10.02 1.0) MM STB 630 SCF/STB=

soi

so

RR

== − × − =

× = − × 5.68 MMM SCF

free gas in the reservoir = 7.3 MMM SCF - 5.68 MMM SCF = 1.62 MMM SCF

1.5 Calculate the gas volume formation factor at 1500 psia at standard condition of 14.7 psia and 60 deg F.

30.95(140 460)0.02829 0.02829 0.01075 ft /1500

rg

r

zTB SCFp

+= × = × =

1.6 Calculate the reservoir volume of the free gas at 1500 psia Vg = 1.62 MMM SCF * Bg = 17.4 MM CF

1.7 Calculate the single phase oil formation volume factor at 1500 psia

Vo at 1500 psia = total initial reservoir volume – free gas volume = 75 MM CF – 17.4 MM CF = 57.6 MM CF = 10.26 MM bbl Under this condition, the equivalent volume under standard condition in SCF is: N – Np = 10.02 – 1.00 MM STB = 9.02 MM STB Bo = Vo / (N-Np) = 1.137 bbl/STB

1.8 Calculate the total, or two phase, oil volume formation factor at 1500 psia. Bt = Bo + (Rsoi-Rso)Bg = 1.137 bbl/STB + (1008-630) SCF/STB * 0.01075 ft3/SCF / (5.615 ft3/bbl) = 1.85 bbl/STB

Problem 2: The R sand is a volumetric oil reservoir whose PVT properties are shown below. Reservoir temperature is 150 deg F. When the reservoir pressure dropped from an initial pressure of 2500 psia to an average pressure of 1600 psia, a total of 26.0 MM STB of oil had been produced. The cumulative gas-oil ratio at 1600 psia was 954 SCF/STB. No Appreciate amount of water was produced, and standard condition were 14.7 psia and 60 deg F.

2

2.1 Calculate the initial oil in place

t o

r

r

( )

From the figure, 1.29 bbl/STBB at 1600 psia = B ( )

zT0.02829 0.02829(0.82)(150 460) at 1600 psia = 0.001575 bbl/SCF5.615 p 1600

575 385 190 SC

p t ti

t p soi g

ti oi

soi so g

g

soi so

N B BRFN B R R B

B BR R B

B

R R

−= =

+ −

= =+ −

+× = =

− = − =

t

F/STBB 1.215 190 SCF/STB 0.001575 bbl/SCF

= 1.514 bbl/STB( ( ) ) 26.0(1.514 (954 575)(0.001575) 245.02 MM STB

1.514 1.29p t p soi g

t ti

N B R R BN

B B

= + ×

+ − + −= = =

− −

2.2 Calculate the SCF of free gas remaining in the reservoir at 1600 psia.

3

oi

oil

Initial volume of oil under initial reservoir condition = N B 316.08 MM bblInitially there is no gas in the reservoir, initial pore volumeat 1600 psia, the V ( ) (245.02 26.00)(1.215) 266.11p oN N B

× =

= − = − =

freeGas oil, initial oil

6 10freeGas g

freeGas

6 6

MM bbl

V V V 316.8 266.1 50.7 MM bbl

G 50.7 10 / B 3.219 10 SCF

Alternatively, G ( )

= 245.02 10 5.615 575 26 10 5.p dissolved soi p p p soG G G N R N R N N R

= − = − =

= × = ×

= − − = × − − − ×

× × × − × × 6615 954 (245.02-26) 10 5.615 385× − × × ×

2.3 Calculate the average gas saturation in the reservoir at 1600 psia.

oipore

N B 316.08 MM bblV 385.46 MM bbl(1 ) (1-0.18)

50.7 0.1315385.46

wi

freeGasg

pore

SV

SV

×= = =

−

= = =

2.4 Calculate the barrels of oil that would have been recovered at 1600 psia if all the produced gas had been returned to the reservoir.

p

p

If all the gas has been returned to the reservoir,1.514 1.29R 0, then RF 0.368

(0 ) 1.514 575 0.001575

so N 245.02 0.368 90.21 MM STB,

compared to 26.0 MM STB that has been pro

p t ti

t soi g

N B BN B R B

N RF

− −= = = = =

+ − − ×

= × = × =

duced without injecting the gas back to reservoir. 2.5 Assuming no free gas flow, calculate the recovery expected by depletion drive performance down to 2000 psia.

4

t

g

o soi s

( )

1.29 bbl/STBB ( )

0.02829 0.02829 0.82(150 460)at 2000 psia, B 0.00126 bbl/STB5.615 5.615 2000

From the PVT figure, B 1.27 bbl/STB, R 575 SCF/STB, R

p t ti

t p soi g

ti

o soi so g

r

r

N B BRFN B R R B

BB R R B

zTp

−= =

+ −

=

= + −

+= = =

= = o

t

p,2000,2000

,2000 (220,2000

510 SCF/STBB 1.27 bbl/STB + (575-510)SCF/STB 0.00126 bbl/STB=1.434 bbl/STB

1.434 1.29so N 245.02 (1)1.434 ( 575)(0.00126)

( )p

pb soi p pb avgp

R

N R N N RR

=

= ×−

= ×+ −

× + −= 0,2000)

,2000

avg

, ,pb

, ,

(2)

575+510Here R between 2200 and 2000 psia is 542.5 SCF/STB2

1.30 1.25N 245.02 9.424 MM STB1.30

insert back

p

t bp ti o bp oi

t bp o bp

N

B B B BN RF N N

B B

=

− − −= × = × = × = × =

,2000p,2000

,2000

p,2000,2000

p,2000 p

into the Equation (2),9.424 575 ( 9.424)(542.5)

we have R ,

1.434 1.29solve together with N 245.021.434 ( 575)(0.00126)

we have N 25.05 MM STB, R 554.73 SCF/STB

p

p

p

NN

R

× + −=

−= ×

+ −

= =

5

Problem 3: The following table provides fluid property data for an initially undersaturated oil reservoir. The initial connate water saturation was 25%. Initial reservoir T and p were 97 deg F and 2110 psia, respectively. The bubble-pint pressure was 1700 psia. Average compressibility factors between the initial and bubble-point pressures were 4.0(10)-6 psia-

1 and 3.1(10)-6 psia-1 for the formation and water, respectively. The critical gas saturation at which gas flow starts to form is estimated to be 10%. Determine the recovery factor at the bubble point pressure. Assume it is a volumetric reservoir. Pressure (psia) Oil formation

volume factor (bbl/STB)

Solution gas-oil ratio (SCF/STB)

Gas formation volume factor (ft3/SCF)

2110 1.256 540 1700 1.265 540 0.007412 1500 1.241 490 0.008423 1300 1.214 440 0.009826 1100 1.191 387 0.011792 900 1.161 334 0.014711 700 1.147 278 0.019316 500 1.117 220 0.027794 Solution:

t

t

o

We need to use the following equation for recovery factor calculation.

1Above the bubble point, B

1.256 bbl/STBB 1.265 bbl/STB,

2110 1700 410 psia

c

o o w wi fti p te

wi

o

ti oi

o

e

c S c S cNB p N B

SB

B BB

p

+ +⎡ ⎤Δ =⎢ ⎥−⎣ ⎦

=

= == =

Δ = − =

5

5 6 65 -

e

5

1.265 1.256 1.748 10 bbl/STB1.256 410

1.748 10 0.75 3.1 10 0.25 4.0 10c 2.385 10 psia1 0.75

1 1.256 2.385 10 410 0.1.265

o oi

oi e

o o w wi f

wi

o o w wi fti e

pwi

t

B BB pc S c S c

Sc S c S c

B pNS

B N

−

− − −−

−

− −= = = ×

×Δ

+ + × × + × × + ×= = = ×

−

+ +⎡ ⎤Δ⎢ ⎥− × × ×⎣ ⎦ = = = 00970

1

6