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 Vo = 75 MM cu ft = 75/5.615 MM bbl = 13.36 MM bbl
N=
Vo 13.36 MM bbl = = 10.02 MM STB Boi 1.333 bbl/STB
1.2 Calculate the initial gas-oil ratio and SCF of gas in reservoir (using the definition of gas solubility in crude oil). Solubility of gas in crude oil = 0.42 SCF/STB/psi,
at bubble point pressure of 2400 psia, gas-oil ratio R soi = 2400 psia × 0.42 SCF/STB/psia = 1,008 SCF/STB Above bubble point pressure, gas-oil ratio reamins 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 N p at 1500 psia = 1 MM STB, R p at 1500 psia = 2800 SCF/STB, produced gas = N p × R p = 1 MM STB × 2800 SCF/STB=2.8 MMM SCF gas remaining in the reservoir = 10.1 - 2.8 = 7.3 MMM SCF
1.4 Calculate the SCF of free gas in the reservoir at 1500 psia
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With gas solubility of 0.42 SCF/STB/psi, and initial R soi = 1, 008 SCF/STB at 1500 psia, the R so = Rsoi − 0.42 × (2400 − 1500) = 630 SCF/STB So the gas dissolved in oil is (N-N p ) × Rso = (10.02 − 1.0) MM STB × 630 SCF/STB=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. zT 0.95(140 + 460) Bg = 0.02829 × r = 0.02829 × = 0.01075 ft 3 / SCF pr 1500 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.
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2.1 Calculate the initial oil in place RF =
Np N
=
Bt − Bti Bt + ( R p − Rsoi ) Bg
From the figure, Bti = Boi = 1.29 bbl/STB Bt at 1600 psia = Bo + ( Rsoi − Rso ) Bg Bg at 1600 psia =
0.02829 zTr 0.02829(0.82)(150 + 460) × = = 0.001575 bbl/SCF 5.615 pr 1600
Rsoi − Rso = 575 − 385 = 190 SCF/STB
Bt = 1.215 + 190 SCF/STB × 0.001575 bbl/SCF = 1.514 bbl/STB N p ( Bt + ( R p − Rsoi ) Bg ) 26.0(1.514 + (954 − 575)(0.001575) N= = = 245.02 MM STB 1.514 − 1.29 Bt − Bti 2.2 Calculate the SCF of free gas remaining in the reservoir at 1600 psia.
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Initial volume of oil under initial reservoir condition = N × Boi = 316.08 MM bbl Initially there is no gas in the reservoir, initial pore volume at 1600 psia, the Voil = ( N − N p ) Bo = (245.02 − 26.00)(1.215) = 266.11 MM bbl VfreeGas = Voil, initial − Voil = 316.8 − 266.1 = 50.7 MM bbl G freeGas = 50.7 × 106 / Bg = 3.219 ×1010 SCF Alternatively, G freeGas = G − G p − Gdissolved = N × Rsoi − N p R p − ( N − N p ) × Rso = 245.02 ×106 × 5.615 × 575 − 26 ×106 × 5.615 × 954 − (245.02-26) ×106 × 5.615 × 385
2.3 Calculate the average gas saturation in the reservoir at 1600 psia. N × Boi 316.08 MM bbl = = 385.46 MM bbl Vpore = (1 − S wi ) (1-0.18)
Sg =
V freeGas V pore
=
50.7 = 0.1315 385.46
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. If all the gas has been returned to the reservoir, R p = 0, then RF =
Np N
=
Bt − Bti 1.514 − 1.29 = = 0.368 Bt + (0 − Rsoi ) Bg 1.514 − 575 × 0.001575
so N p = N × RF = 245.02 × 0.368 = 90.21 MM STB, compared to 26.0 MM STB that has been produced 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.
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RF =
Np N
=
Bt − Bti Bt + ( R p − Rsoi ) Bg
Bti = 1.29 bbl/STB Bt = Bo + ( Rsoi − Rso ) Bg at 2000 psia, Bg =
0.02829 zTr 0.02829 0.82(150 + 460) = = 0.00126 bbl/STB 5.615 pr 5.615 2000
From the PVT figure, Bo = 1.27 bbl/STB, R soi = 575 SCF/STB, R so = 510 SCF/STB Bt = 1.27 bbl/STB + (575-510)SCF/STB × 0.00126 bbl/STB=1.434 bbl/STB so N p,2000 = 245.02 × R p ,2000 =
1.434 − 1.29 1.434 + ( R p ,2000 − 575)(0.00126)
N pb × Rsoi + ( N p ,2000 − N pb ) Ravg (2200,2000) N p ,2000
(1) (2)
575+510 = 542.5 SCF/STB 2 Bt ,bp − Bti Bo ,bp − Boi 1.30 − 1.25 = N× = 245.02 × = 9.424 MM STB N pb = N × RF = N × Bt ,bp Bo ,bp 1.30
Here R avg between 2200 and 2000 psia is
insert back into the Equation (2), 9.424 × 575 + ( N p ,2000 − 9.424)(542.5) we have R p,2000 = , N p ,2000 solve together with N p,2000 = 245.02 ×
1.434 − 1.29 1.434 + ( R p ,2000 − 575)(0.00126)
we have N p,2000 = 25.05 MM STB, R p = 554.73 SCF/STB
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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 psia1 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)
2110 1700 1500 1300 1100 900 700 500
Oil formation volume factor (bbl/STB) 1.256 1.265 1.241 1.214 1.191 1.161 1.147 1.117
Solution gas-oil ratio (SCF/STB)
540 540 490 440 387 334 278 220
Gas formation volume factor (ft3/SCF)
0.007412 0.008423 0.009826 0.011792 0.014711 0.019316 0.027794
Solution: We need to use the following equation for recovery factor calculation. ⎡c S + c S + cf ⎤ NBti ⎢ o o w wi ⎥ Δ p e = N p Bt 1 − S wi ⎣ ⎦ Above the bubble point, Bt = Bo Bti = Boi = 1.256 bbl/STB Bt = Bo = 1.265 bbl/STB, Δ p e = 2110 − 1700 = 410 psia co = ce =
Bo − Boi 1.265 − 1.256 = = 1.748 × 10−5 bbl/STB 1.256 × 410 Boi Δ p e co So + cw S wi + c f 1 − S wi
=
1.748 ×10−5 × 0.75 + 3.1×10−6 × 0.25 + 4.0 ×10−6 = 2.385 × 10−5 psia -1 0.75
⎡c S + c S + cf ⎤ Bti ⎢ o o w wi ⎥ Δ pe N 1 − S wi 1.256 × 2.385 × 10−5 × 410 ⎣ ⎦ = p = = 0.00970 1.265 Bt N
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