# Ch. 8 a - 1 PGE 312 – Physical and Chemical Behavior of...

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Unformatted text preview: 1 PGE 312 – Physical and Chemical Behavior of Fluids Homework 7 – Solutions Problem 8-4 Given: API Gravity = 47.3 oAPI Solution: The API gravity is given by equation 8-2 o API = Therefore, The specific gravity of the stock tank liquid is 0.7914 Problem 8-6 Given: Density = 56.4 Solution: o API = Liquid specific gravity, Therefore, oAPI = Problem 8-9 Volume of oil at reservoir conditions = 86.3 res bbl/d Volume of oil in stock tank = 57.9 STB/d Total volume of gas at evolved at surface = 43150+7240 = 50390 scf/d 2 Formation volume factor, Bo = Therefore, Bo = = 1.491 Solution gas oil ratio Rs = Therefore, Rs = = 870.29 Since the Solution-gas oil ratio (Rs) is constant at pressures above the bubble point pressure, the value calculated applies also at bubble point. Problem 8-10 Pressure p psig 5000 4500 4000 3500 3400 3300 3200 3100 3054 2700 2400 2100 1800 1500 1200 900 600 Oil Formation Volume gas–oil factor Bo res bbl/STB 1.498 1.507 1.517 1.527 1.530 1.532 1.534 1.537 1.538 1.484 1.441 1.401 1.361 1.323 1.287 1.252 1.215 Solution gas-oil ratio Rs scf/STB Gas Formation volume factor Bg res bbl/scf 941 941 941 941 941 941 941 941 941 819 732 646 562 481 400 321 240 0.000866 0.000974 0.001090 0.001252 0.001475 0.001795 0.002285 0.003108 0.004760 3 The shapes of the Bo vs p and Rs vs p plots match the shapes in Fig. 8-1 and Fig. 8-2, respectively. Above the bubble point pressure, Bo decreases with an increase in pressure as the single phase oil is supercompressed. The solution gas oil-ratio on the other hand is unaffected by increase in pressure because above the bubble point, the volume of gas dissolved in the oil is constant. 4 For all pressures below the bubble point, both the formation volume factor and the solution gas-oil ratio decrease with decrease in pressure due to the evolution of gas from the oil. Problem 8-11 The solution gas-oil ratio at bubble point Rsb = 941 scf/STB (from table) Total formation volume factor Bt = Bo + (Rsb – Rs)* Bg ---------- (1) At 2400 psig Bo = 1.441 , Rs = 732 , Bg = 0.001090 Therefore, Bt = 1.441+(941-732)*0.001090 = 1.669 Problem 8-12 Since the pressure 3500 psig is above the bubble point pressure of 3054 psig, the volume of gas evolved in the reservoir pore space is 0. Therefore, Rs = Rsb. Thus, the total formation volume factor is equal to the formation volume factor of oil Bo. Bt = Bo = 1.527 Problem 8-13 Given: Temperature: 300oF Pressure psig 5000 4500 4000 3500 3000 2500 2200 1972 = pb Oil volume cc 219.80 220.55 221.33 222.17 223.07 223.99 224.57 225.05 5 Solution: The coefficient of isothermal compressibility of oil as a function of pressure at constant reservoir temperature at pressures above the bubble point is given by equation 8 -8 on page 232 in the textbook. Therefore, plotting ln V vs. p we can obtain as the negative slope of the curve at specific pressure, p. The dotted lines show the lines for slope calculation at different pressure. Pressure psig (p) 5000 4500 4000 3500 3000 2500 2200 1972 = pb Oil volume cc (V) 219.80 220.55 221.33 222.17 223.07 223.99 224.57 225.05 Oil volume cuft 0.007763 0.007790 0.007817 0.007847 0.007879 0.007911 0.007932 0.007949 Ln V -4.8583 -4.8549 -4.8514 -4.8476 -4.8436 -4.8395 -4.8369 -4.8347 -(Slope) 6.00E-6 6.00E-6 6.00E-6 6.50E-6 7.50E-6 1.25E-5 6 Plotting the coefficient of compressibility of the oil as a function of the pressure above the bubble point we observe that the trend is pretty close to the shape in Fig. 8 -5 on page 232 in the textbook. Problem 8-16 Given: Pressure between 3500 – 4000 (Pressure range > Bubble point pressure Pb= 3054 psig) The coefficient of isothermal compressibility at pressure greater than bubble point pressure of the oil is given in equation 8-9. Integrating the above equation, assuming co remains constant with pressure we get 7 results in --------- (1) At p1 = 3500 psig, Bo1 = 1.527 values into equation (1) we get and p2 = 4000psig, Bo2 = 1.517 . Therefore, . Substituting these = 13.14 x 10-6 psi-1 Problem 8-17 The coefficient of isothermal compressibility at pressure lesser than bubble point pressure of the oil is given in equation 8-24 in book. --------- (1) At 2400 psig Bo = 1.441, Bg = 0.001090, 8 The value of is the slope of the tangent in the above graph and is = 0.06/400 = 1.5x10 -4 The value of is the slope of the tangent in the above graph and is = Therefore, substituting the values into equation (1) we get = 113.0x10-6 psi-1 = 0.2875 ...
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