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Course: PGE 312, Spring 2009
School: University of Texas
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Word Count: 659

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# HOMEWORK 6 SOLUTION KEY 6-1. Assuming ideal gas, PV = nRT, nRT V= P 100 x10.732 x (60 + 459.7) = scf 14.65 = 38071.13 scf 6-8. Given, Area H SW z = 640 acres = 27878400 sq. ft. = 21 ft = 0.18 = 33% = 0.951, from example 3-10. Bg = 0.0282 zT P 0.0282 x0.951x(194 + 459.7) = cu ft/scf 3810 = 0.0046 cu ft/scf V Bg = res Vscf AH (1 - S w ) = Vscf = 15348.8 MMscf 6-10. Component H2S CO2 N2 CH4 C2H6 C3H8...

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# HOMEWORK 6 SOLUTION KEY 6-1. Assuming ideal gas, PV = nRT, nRT V= P 100 x10.732 x (60 + 459.7) = scf 14.65 = 38071.13 scf 6-8. Given, Area H SW z = 640 acres = 27878400 sq. ft. = 21 ft = 0.18 = 33% = 0.951, from example 3-10. Bg = 0.0282 zT P 0.0282 x0.951x(194 + 459.7) = cu ft/scf 3810 = 0.0046 cu ft/scf V Bg = res Vscf AH (1 - S w ) = Vscf = 15348.8 MMscf 6-10. Component H2S CO2 N2 CH4 C2H6 C3H8 Composition, mole fraction 0.1 0.05 0.021 0.703 0.062 0.037 Tc, R 672.4 547.9 227.5 343.3 549.9 666.1 Pc, psia 1300 1071 493 666.4 706.5 616 yiTc, R 67.24 27.395 4.7775 241.3399 34.0938 24.6457 yiPc, psia 130 53.55 10.353 468.4792 43.803 22.792 n-C4H10 0.027 1 734.5 527.9 19.8315 419.3234 14.2533 743.2305 = 21 T'pc = Tpc = 419.3234-21 R = 398.32 R PpcT ' pc P ' pc = Tpc + y H 2 S (1 - y H 2 S ) 743.23x 398.32 419.3234 + 0.1(1 - 0.1)21 = 702.83 psia = Ppr = P Ppc 5420 + 14 702.83 = 7.72 T T pr = Tpc = 257 + 459.7 398.32 = 1.8 z = 1.02, from fig.3-7. = Bg = 0.00502 zT resbbl / cuft P 0.00502 x1.02 x(257 + 459.7) Bg = res bbl/scf 5420 + 14 = 0.000675 res bbl/scf or 0.0038 res ft3/ scf 6-12. Given, I.D. = 6 inches = 0.5 ft V = 2.4 MMscfd T = 100 F = 559.7 R P = 1000 psia Component H2S CO2 N2 CH4 C2H6 C3H8 i-C4H10 n-C4H10 i-C5H12 n-C5H12 C6H14 C7+ Mole percent Nil 1.67 0.32 71.08 15.52 7.36 0.92 1.98 0.33 0.26 0.27 0.29 100 Tc, R 547.9 227.5 343.3 549.9 666.1 734.5 765.6 829.1 845.8 913.6 1157 Pc, psia 1071 493 666.4 706.5 616 527.9 550.6 490.4 488.6 436.9 367 yiTc, R 9.14993 0.728 244.0176 85.34448 49.02496 6.7574 15.15888 2.73603 2.19908 2.46672 3.3553 420.9384 yiPc, psia M, lb/lbmol 17.8857 1.5776 473.6771 109.6488 45.3376 4.85668 10.90188 1.61832 1.27036 1.17963 1.0643 669.018 44 28 16.04 30.07 44.1 58.1 58.1 72.1 72.1 86 100 Miyi 0.7348 0.0896 11.40123 4.666864 3.24576 0.53452 1.15038 0.23793 0.18746 0.2322 0.29 22.77075 Tpc = 420.9384 R Ppc = 669.018 psia T T pr = Tpc 559.7 420.9387 = 1.33 = Ppr = P Ppc 1000 669.018 = 1.49 = z = 0.8, from fig.3-7. Bg = 0.0282 zRT P 0.0282 x0.8 x559.7 = res cu ft/ scf 1000 = 0.01263 res cu ft/ scf Vres Vscf Bg = Actual Volumetric flow rate, Vr = BgVscf = 0.1355 x 2.4 x 106 cu ft = 30304.4 ft3/ day Calculations for Mass flow rate: From the equation of state for real gases, we have PVM m= zRT At standard conditions, 14.65 Ppr = 669.018 =0.0219 Tpr = (60 + 459.7) 421 = 1.23 from fig.3-7. z = 0.98, Mass flow rate per hour, 14.65 x 2.4 x106 x 22.771 lb/hr m= 0.98 x10.73 x(60 + 459.7) x 24(hr / day ) = 6104.4 lb/hr Superficial velocity, v = Volumetric flow rate/ cross sectional area of flow 30304.4( ft 3 / day ) 2 ( ft ) / 4 x0.52 ft / sec = 24 x 3600(sec) = 1.79 ft/s 6-15. VM = RT +B p 1 V cg = - V P Tr cg = 1 RT ( ) , on differentiating (1) V P2 1 RT = ( 2) RT ( + B) P P ........(1) ........(2) ........(3) ........(4) = 1 10.73 x559.7 ( ) psia -1 2 10.73x559.7 200 ( - 2.12) 200 = 0.00538 psia-1 For an ideal case, 1 cg ,ideal = p 1 = psia-1 200 = 0.005 difference psia-1 Percentage = (0.00538 - 0.005) x100% 0.005 = 7.6% 6-18. Known values, Tpc = 350.17 R Ppc = 666.9 psia T pr = T Tpc (170 + 459.7) 350.17 = 1.7983 = Ppr = P Ppc 3335 666.9 =5 = z = 0.91, c pr = from fig.3-7. 1 1 z - Ppr z Ppr Tpr 1 1 - ( 0.333)Tpr =1.8 5 0.91 = 0.153 Co-efficient of thermal compressibility, c cg = pr Ppr = 0.153 psia-1 666.9 = 2.29x10-4 psia-1 = 6-25. Component H2S CO2 N2 CH4 C2H6 C3H8 n-C4H10 Composition, mole fraction 0.1 0.05 0.021 0.703 0.062 0.037 0.027 1 Tc, R 672.4 547.9 227.5 343.3 549.9 666.1 734.5 Pc, psia 1300 1071 493 666.4 706.5 616 527.9 yiTc, R 67.24 27.395 4.7775 241.3399 34.0938 24.6457 19.8315 419.3234 yiPc, psia M, lb/lbmol 130 53.55 10.353 468.4792 43.803 22.792 14.2533 743.2305 34 44 28 16.04 30.07 44.1 58.1 Miyi 3.4 2.2 0.588 11.27612 1.86434 1.6317 1.5687 22.52886 Specific gravity, = 22.53/29 = 0.78 Correction added = 0.0002 + 0.0003 + 0.00022 cp, from fig. 6-8 = 0.00072 cp Uncorrected viscosity at 1atm, 1g = 0.013 cp, from fig. 6-8 Corrected viscosity at 1atm, 1g = 0.013 + 0.00072 cp = 0.01372 cp (257 + 459.7) 419.3234 = 1.7 (5420 + 14) Ppr = 743.2305 = 7.74 Viscosity ratio, g/ 1g = 2.2, from fig. 6-9 Co-efficient of viscosity, g = 2.2 x 0.01372 cp = 0.03018 cp Tpr = SP1. Given, Pinitial = 6500 psia T = 180 F = 0.75 = 0.25 Sw = 0.16 A = 3 miles x 3 miles H = 580 ft From the specific gravity we can find Tpc and Ppc. Tpc = 390 R Ppc = 657 psia 180 + 459.7 Tr = 390 = 1.64 6500 Pr = 657 = 9.89 z = 1.12, from fig. 3-7. Bg = 0.0282 zT P 1.12 x(180 + 459.7) = 0.0282 res cu ft/ scf 6500 = 0.00311 res cu ft/ scf Vres G Therefore, Initial gas in place, V G = res Bg Bg = = 3x3x5280 x5280 x580 x0.25 x0.84 scf 0.00311 = 9.83x1012 scf Thus, initial Oil in Place = 9.83x1012 scf SP2. Component Composition, mole fraction Tc, R Pc, psia yiTc, R yiPc, psia CH4 C2H6 C3H8 n-C4H10 0.89 0.6 0.03 0.02 343.3 549.9 666.1 734.5 666.4 706.5 616 527.9 305.537 32.994 19.983 14.69 373.204 593.096 42.39 18.48 10.558 664.524 a. Tpc = 373 R Ppc = 664.524 psia P P 1 P = -( i )( )G p + i z zi G zi G scf, Cummulative Pressure, psia Production 0 8000 50x106 7500 100x106 150x106 200x10 6 Ppr 12.038692 11.286274 10.383372 9.6309539 8.4270847 zi 1.26 1.22 1.16 1.105 1.04 Pi/zi 6349.206 6147.541 5948.276 5791.855 5384.615 6900 6400 5600 Plot of Pi/zi vs cumulative production(scf) The equation of the straight line is found to be, y = -5E-06x + 6381.3 therefore, slope = -5E-06 Also, P 1 slope = -( i )( ) zi G Therefore, P 1 G = -( i )( ) zi slope intercept = thus, Pi zi = 6381.3 psia G= -6381.3 -5E - 06 = 1.276x109 scf b. At 500 psia, Tpr = 1.714 Ppr = 0.752 z = 0.95 Using the equation of the line found in (a) we get, 500 = -5 xE - 05G p + 6381.6 0.95 Therefore, Recoverable oil, Gp = 1.171x109 scf c. Bg = 0.0282 zT P 0.0282 x1.26 x(180 + 459.7) = 8000 = 0.00284 res cu ft/scf Vres G AH (1 - S w ) = 1.276 x109 scf 0.00284 Bg = Area = 539261.9 ft2 = 12.35 acres
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Notes 19 Notes Notes 14 Notes 13 Notes 12 Notes 11 Notes 10 Notes 2 Notes 1