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 bbl V 13.36 MM bbl N = 10.02 MM STB B 1.333 bbl/STB o V = = =
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/STB Above bubble point pressure, gas-oil ratio rea
×
mins constant,
soi so
p
With gas solubility of 0.42 SCF/STB/psi, and initial R 1, 008 SCF/STB at 1500 psia, the R 0.42 (2400 1500) 630 SCF/STB
So the gas dissolved in oil is (N-N ) (10.02 1.0) MM STB 630 SCF/STB=
soi so R R = = − × − = × = − × 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.
3 0.95(140 460) 0.02829 0.02829 0.01075 ft / 1500 r g r zT B SCF p + = × = × =
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.
2.1 Calculate the initial oil in place
t o
r r
( )
From the figure, 1.29 bbl/STB B at 1600 psia = B ( ) zT 0.02829 0.02829(0.82)(150 460) at 1600 psia = 0.001575 bbl/SCF 5.615 p 1600 575 385 190 SC p t ti t p soi g ti oi soi so g g soi so N B B RF N B R R B B B R R B B R R − = = + − = = + − + × = = − = − = t F/STB B 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 1.29 p t p soi g t ti N B R R B N B B = + × + − + − = = = − −
oi
oil
Initial volume of oil under initial reservoir condition = N B 316.08 MM bbl Initially there is no gas in the reservoir, initial pore volume
at 1600 psia, the V (N Np)Bo (245.02 26.00)(1.215) 266.11
× =
= − = − =
freeGas oil, initial oil
6 10 freeGas 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 so G G G N R N R N N R = − = − = = × = × = − − = × − − − × × × × − × × 6 615 954 (245.02-26) 10× − × ×5.615 385×
2.3 Calculate the average gas saturation in the reservoir at 1600 psia.
oi pore N B 316.08 MM bbl V 385.46 MM bbl (1 ) (1-0.18) 50.7 0.1315 385.46 wi freeGas g pore S V S V × = = = − = = =
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 1.29
R 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 B N 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.
t g o soi s ( ) 1.29 bbl/STB B ( ) 0.02829 0.02829 0.82(150 460) at 2000 psia, B 0.00126 bbl/STB 5.615 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 B RF N B R R B B B R R B zT p − = = + − = = + − + = = = = = o t p,2000 ,2000 ,2000 (220 ,2000 510 SCF/STB B 1.27 bbl/STB + (575-510)SCF/STB 0.00126 bbl/STB=1.434 bbl/STB 1.434 1.29 so N 245.02 (1) 1.434 ( 575)(0.00126) ( ) p pb soi p pb avg p R N R N N R R = = × − = × + − × + − = 0,2000) ,2000 avg , , pb , , (2) 575+510
Here R between 2200 and 2000 psia is 542.5 SCF/STB 2 1.30 1.25 N 245.02 9.424 MM STB 1.30 insert back p t bp ti o bp oi t bp o bp N B B B B N RF N N B B = − − − = × = × = × = × = ,2000 p,2000 ,2000 p,2000 ,2000 p,2000 p
into the Equation (2),
9.424 575 ( 9.424)(542.5)
we have R ,
1.434 1.29 solve together with N 245.02
1.434 ( 575)(0.00126) we have N 25.05 MM STB, R 554.73 SCF/STB p p p N N R × + − = − = × + − = =
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.
1
Above the bubble point, B 1.256 bbl/STB B 1.265 bbl/STB, 2110 1700 410 psia c o o w wi f ti e p t wi o ti oi o e c S c S c NB p N B S B B B B p + + ⎡ ⎤ Δ = ⎢ − ⎥ ⎣ ⎦ = = = = = Δ = − = 5 5 6 6 5 -e 5 1.265 1.256 1.748 10 bbl/STB 1.256 410 1.748 10 0.75 3.1 10 0.25 4.0 10 c 2.385 10 psia 1 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 f ti e p wi t B B B p c S c S c S c S c S c B p N S B N − − − − − − − − = = = × × Δ + + × × + × × + × = = = × − + + ⎡ ⎤ Δ ⎢ − ⎥ × × × ⎣ ⎦ = = = 00970 1 6
