PGE 312_Spring10_Solution_HW9

# Specific gravity of gas γ g 095 stock tank oil

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??? Specific gravity of gas , γ g = 0.95 Stock tank oil gravity = 39 o API Bubble point pressure, p b = 1763psia Temperature = 250 o F

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7 Weight percent of H 2 S = 3% Solution: Apparent molecular weight of the gas, M g = 29 γ g =29*0.95=27.55 ?? ?? ???? From Fig. 11- 8 using γ g = 0.95 and 39 o API we get apparent density of dissolved gas ρ g = 27.58 lb/cu.ft Mass of gas = 589 scf STB ?? ???? 380.7 27.55 ?? ?? ???? = 42.62 ????? ??? Specific gravity of stock tank oil 𝛾 ??? = 141.5 131.5 + 39 = 0.830 Mass of stock tank oil = 5.615 cu.ft STB 0.83 ?? ?𝑖? / ?? ?? ?𝑖? ?? ????? / ?? ?? ????? 62.4 ?? ????? ?? ?? ????? = 290.81 ?? ?𝑖? ??? Mass of surface gas and stock tank oil = 290.81+42.62 = 333.43 ?? ??? Liquid volume = 5.615 ?? ?? ??? + 42.62 ?? ??? ??? 27.58 ?? ?? . ?? = 7.16 ?? ?? ??? Pseudoliquid den sity, ρ po = 333.43 7.16 = 46.56 ?? ?? . ?? Compressibility adjustment = +0.52 lb/cu.ft. Therefore, ρ po = 46.56+0.52 = 47.08 lb/cu.ft at 60 o F and 1763psia Thermal Expansion adjustment = -5.5lb/cu.ft. Therefore, ρ po = 47.08-5.5=41.58 lb/cu.ft at 250 o F and 1763psia Density adjustment for hydrogen sulphide ρ po = 41.58-0.26= 41.32 lb/cu.ft at 250 o F and 1763psia Problem 11-18 Solution: Liquid density, ρ po = 41.32 lb/cu.ft at 250 o F and 1763psia = 232.05 lb/res bbl Formation volume factor of oil B o = ???? ?? 1.0 ??? + ?????? ??? ??????? ???? 1,0 ??? ???? ?? 1.0 ???????𝑖? ?????? Therefore, B o = 290.81 ?? ?𝑖? / ??? +42.62 ?? ??? / ??? 232.05 ?? / ??? ??? = 1.437 ??? ??𝒍 ??𝑩
8 Problem 11-22 Given: Problem 11-2 Bubble point pressure, p b = 2050psia Solution: The solution gas-oil ratio for pressures below the bubble point pressure can be obtained from the results of problem 11-4. Using this information in Fig.11-9 we can determine the formation volume factor of oil at pressure below the bubble point pressure. Pressure, p psia Solution gas oil Ratio R s ??? ??𝑩 Formation volume factor, B o res bbl/STB 1950 540 1.395 1800 490 1.364 1650 440 1.335 1500 395 1.305 1350 355 1.285 1200 310 1.257 1050 265 1.230 900 220 1.208 750 175 1.182 600 132 1.158 450 94 1.142 300 62 1.130 150 - - Plotting the above values for more insight, notice that we observe the trend discussed in chapter 10(page 260) for black oils (for pressures lesser than bubble point pressure). 1 1.1 1.2 1.3 1.4 1.5 0 100 200 300 400 500 600 Formation volume factor of oil, B o resbbl/STB Solution Gas oil Ratio, R s scf/STB B o vs R s ; p<p b

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9 Problem 11-24 Given: Formation volume factor of oil at bubble point B ob = 1.255 ??? ??? ??? Solution gas-oil ratio at bubble point, R sb = 410 ??? ??? Bubble point pressure, p b =2265psig = 2279.7psia Temperature = 214 o F Gas specific gravity, γ g = 0.811 Stock tank oil gravity = 23.7 o API Coefficient of isothermal compressibility of oil, c o = 9 * 10 -6 psi -1 Solution: Formation volume factor of oil, B o = ? ?? ? ? ? ( ? ? −? ) for p≥p b Therefore, Bo at 4070 psig = 1.255 ∗ ? 9 10 6 (2279.7 4084.7) = 1.235 ??? ??𝒍 ??𝑩 Problem 11-27 Given: Problem 11-2 Solution gas-oil ratio = 589 ??? ??? Specific Gravity = 0.95 API Gravity = 39.0 o API Temperature = 250 o F Bubble point pressure, p b = 2050psia Solution: Using the given information, from Fig. 11-11 the coefficient of isothermal compressibility of oil at 3500psia, c o = 14 x 10 -6 psi -1
10 Problem 11-32 Given: Bubble point pressure, p b =1250psia Reservoir temperature = 142 o F Producing gas-oil ratio, R p = 370 ???

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