Ch. 9 Ch. 11 a - 1 PGE 312 Physical and chemical Behavior...

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Unformatted text preview: 1 PGE 312 Physical and chemical Behavior of Fluids Home Work 9 Solutions Problem 9-2 Given: Cumulative Oil Production, MSTB 260 455 761 1024 1319 1699 1910 2201 2400 2607 2800 2959 3111 3231 3360 3480 3591 3676 3799 3890 3992 4039 4110 4205 4275 4317 4390 4435 4500 4540 4600 Producing gas-oil ratio, scf/STB 310 322 286 305 317 402 434 515 575 610 661 602 723 769 848 911 974 990 1092 1044 1143 1170 1318 1271 1407 1485 1494 1373 1467 1305 1446 2 Separator Pressure = 30psig = 30+14.7 = 44.7psia Separator Temperature = 120oF Separator gas gravity = 0.888 Stock tank oil gravity = 36.1oAPI Laboratory solution gas-oil ratio = 318 scf/STB Solution: Plotting the production history against cumulative oil production, we can deduce the solution gas-oil ratio as the producing gas-oil ratio at bubble point. The plot is shown below Therefore, the solution gas-oil ratio is 315scf/STB. 3 Adjusting the solution gas-oil ratio for vented stock tank gas by using stock tank gas-oil ratio correlation in Fig. 9-3, we get Rsb = 315 + 23 = 338 scf/STB. The estimated solution gas-oil ratio can be compared with the laboratory value as follows: Since the percentage error is less than 10%, we have obtained a very good estimate of the solution gasoil ratio. Problem 9-3 Given: Reservoir pressure, psia 3122 2982 1830 1275 965 1095 933 495 Cumulative oil production, MSTB 110 370 1050 1400 3060 3160 3360 4620 Bubble point pressure measured in the lab = 1441psig = 1445.7psia. 4 Solution: A plot of the pressure history against the cumulative oil production is shown below. The estimated bubble point pressure is 1480psia. The estimated bubble point pressure can be compared with the laboratory value as follows: Since the percentage error is less than 10%, we have obtained a very good estimate of the bubble point pressure. Problem 11-1 Stock tank oil gravity 40.3o API 5 Specific Gravity = 0.756 Solution gas oil ratio = 1000 Reservoir Temperature = 205oF The bubble point pressure at the above stated values is found out from Fig. 11-1 as 3450psia Problem 11-5 Given: Pressure = 14.696psia Temperature = 60oF Ideal hydrocarbon liquid of known composition Solution: Component Composition mole fraction, lb mol xj Molecular weight, lb/lb mol Mj 0.025 0.325 0.650 1.000 44.1 58.1 72.2 Propane i-Butane n-Pentane Σ at 60oF and 14.696psia Problem 11-7 Given: Bubble point pressure, pb= 3385psia Temperature = 205oF Specific gravity of heptanes plus γo = 0.85 = 53.04 Mass, lb xjMj 1.1025 18.8825 46.9300 66.9150 lb Liquid density at 60oF and 14.696psia, lb/cuft ρoj 31.64 35.12 39.36 Liquid volume at 60oF and 14.696psia, cu ft xjMj/ ρoj 0.0349 0.5377 1.1923 1.7648 cu ft 6 Solution: We use a first trial value of 43.70lb/cu.ft, which corresponds to 0.70g/cc. The apparent liquid density of methane = 0.32g/cc (19.98 lb/cu.ft) and apparent liquid density of ethane = 0.47g/cc (29.34 lb/cu.ft) Component methane ethane propane i-butane n-butane i-pentane n-pentane Hexanes Heptanes plus Composition mole fraction xj 0.4642 0.0637 0.0561 0.0119 0.0383 0.0168 0.0195 0.0285 0.3010 1.0000 Molecular weight, lb /lb mole Mj 16.043 30.070 44.097 58.123 58.123 72.150 72.150 86.177 202.000 Mass, lb xjMj 7.4472 1.9155 2.4738 0.6917 2.2261 1.2121 1.4069 2.4560 60.8020 80.6313 lb Liquid Density 60oF & 14.7 psia lb/cu.ft ρoj 19.98 29.34 31.62 35.11 36.42 38.96 39.36 41.40 53.04 Liquid Volume 60oF & 14.7 psia cu.ft xjMj /ρoj 0.3727 0.0653 0.0782 0.0197 0.0611 0.0311 0.0357 0.0593 1.1463 1.8696 cu.ft The first calculated value of pseudoliquid density = For a second trial value we use 46.82 lb/cu.ft, which corresponds to 0.75g/cc. Component Methane Ethane Propane plus Mass, lb xjMj Liquid Density 60oF & 14.7 psia lb/cu.ft ρoj 7.4472 21.21 1.9155 30.24 71.2686 80.6313 lb The second calculated value of pseudoliquid density = Liquid Volume 60oF & 14.7 psia cu.ft xjMj /ρoj 0.3511 0.0633 1.4316 1.8460 cu.ft 7 The pseudoliquid density from the above plot ρpo = 43 lb/cu. ft at 14.7 psia and 60oF Compressibility adjustment = + 1.3 lb/cu ft at 60oF and 3385psia, Fig. 11-3 ρo = 43 + 1.3 = 44.3 lb/cu. ft at 60oF and 3385psia Thermal expansion adjustment = -4.1lb/cu ft at 205oF, Fig. 11-4 ρo = 44.3 – 4.5 = 39.80lb/cu. ft at 205oF and 3385psia. Problem 11-8 We use the table from the previous problem Density of propane plus, ρc3+ = = 49.78 lb/cu.ft Weight percent of methane in the mixture W1, percent = = Weight percent of ethane in the ethane and heavier components, W2 % = = 9.24% = = 2.62% Using W1 and W2 values in Fig. 11-6 we get, at 14.7 psia and 60oF. = 0.858. Therefore, ρpo = 0.858 *49.78 = 42.71 lb/cu.ft 8 Compressibility adjustment = +1.3 lb/cu.ft. Therefore, ρpo = 42.71+1.3 = 44.01 lb/cu.ft at 60oF and 3385psia Thermal Expansion adjustment = -4.1lb/cu.ft. Therefore, ρpo = 44.01-4.5=39.51 lb/cu.ft at 205oF and 3385psia Error % with laboratory measurement = = 1.18% Problem 11-13 Given: Producing gas-oil ratio, Rp = 589 Specific gravity of gas, γg = 0.95 Stock tank oil gravity = 39 oAPI Bubble point pressure, pb = 1763psia Temperature = 250oF Weight percent of H2S = 3% Solution: Apparent molecular weight of the gas, Mg = 29 γg =29*0.95=27.55 From Fig. 11-8 using γg = 0.95 and 39 oAPI we get apparent density of dissolved gas ρg = 27.58 lb/cu.ft Mass of gas = = 42.62 Specific gravity of stock tank oil Mass of stock tank oil = = 290.81 Mass of surface gas and stock tank oil = 290.81+42.62 = 333.43 Liquid volume = 5.615 + = 7.16 9 Pseudoliquid density, ρpo = Compressibility adjustment = +0.52 lb/cu.ft. Therefore, ρpo = 46.56+0.52 = 47.08 lb/cu.ft at 60oF and 1763psia Thermal Expansion adjustment = -5.5lb/cu.ft. Therefore, ρpo = 47.08-5.5=41.58 lb/cu.ft at 250oF and 1763psia Density adjustment for hydrogen sulphide ρpo = 41.58-0.26=41.32 lb/cu.ft at 250oF and 1763psia Problem 11-18 Solution: Liquid density, ρpo = 41.32 lb/cu.ft at 250oF and 1763psia = 232.05 lb/res bbl Formation volume factor of oil Bo = Therefore, Bo = = 1.437 Problem 11-24 Given: Formation volume factor of oil at bubble point Bob = 1.255 Solution gas-oil ratio at bubble point, Rsb = 410 Bubble point pressure, pb =2265psig = 2279.7psia Temperature = 214oF Gas specific gravity, γg = 0.811 Stock tank oil gravity = 23.7oAPI Coefficient of isothermal compressibility of oil, co = 9 * 10-6 psi-1 10 Solution: Formation volume factor of oil, Bo = for p≥pb Therefore, Bo at 4070 psig = = 1.235 Problem 11-31 Formation Volume factor of oil at bubble point Bob = 1.255 Solution gas-oil ratio Rsb = 410 Reservoir Pressure Pb = 2265 psia Reservoir Temperature = 214oF Specific gravity of gas γg = 0.811 Stock tank gravity = 23.7oAPI µoD from Fig. 11-13 = 8cp; µob = 1.8cp; from Fig. 11-15 µo = 2.0 cp ...
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This note was uploaded on 11/27/2011 for the course PGE 312 taught by Professor Peters during the Spring '08 term at University of Texas at Austin.

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