HW #3_Solutions - Applied Reservoir Engineering HOMEWORK #3...

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Unformatted text preview: Applied Reservoir Engineering HOMEWORK #3 Given the flow rate and bottom pressure. W e calculate the value of (PrAPWf2 ) , then calculate the value of log(PrZ-ow2), and the value of log(q). The average reservoir pressure is Pr = 1952 psi Ex: (PH—PW? ) 1700 = 19522 — 17002 = 920304 l0g(Pr2—ow2) : log(920304) 2 596393131 l0g(q) = log(2625) = 3.419129 PrZ-owz Loglql o 920304 596393131 3.419129 1560304 6.193209222 3.618571 2120304 6326398133 3.734079 Then plot the value of IoglPrZ—ow) versus loglq) to find the linear line y = ax + b Xe; W 1593:? 8:"? i: = 9.67 3,WW..A.M,.,,W.V,W“Wasw1,04% Iog(Pr2-ow2) vs. log(q) 6‘35 MW...W...............W.WWWWWWWW ,. WMWWWWMWW..,.....WWW....M.,.......,..w,WWWWWW...“ 5 3 ..,2...-W...WW...WWW..-W...._...x.:;$925.:.;.9.;a.......... WW.-- 6 i 2 5 ;.WMWN.WWWWWWM.m..m...mmW..m.mm.,_wmwm WWWWWWWMW A 6.2 « w E M. .. . .. . W... W. . .. .. , ,, .....,.U.M.WW v W... 6.1 6.05 2 log(Pr2-P 5.95 l From the graph above we have the equation y = 1.1507X + 2.0294 Thus, we have the value of: y = ax + b m x 1.1507 b = 2.0294 The value of n is calculated: n = 3— : ~3— = 0.8690362 . 1‘1507 m 1.1507 0.8690362 2.0294 0.0172337 ~1 The value of the intercept b= 2.0294 = log(C7) —l 102'0294= C77 -> C = 0.0172337 S _ 0.037SK3H ” "f2 Then we calculate the value of Tave. skin, Fr Bythese formulas F I 0394362 0’ g 4.27‘7' F {3, 3.: 43 if i” ffilififi c. 0.00400? 0' > 4.277 ” Tave : 250:1“) + 460 = 635 ° R _ 0.0375(0.74)(10000) _ ' 635*088 S 0.5 0.004362 Fr 3 W = 0.00341 We come up with these values specifi gravity 0.74 Tr 250 Tt 100 10000 12000 0.88 1800 3000 0.0172337 0.87 635 0455 0.00341 Find the ow From the equation _~ PxAIfZ--95PI:2 2 2 n 0— ————-—-—~—-—-= P —P Q” S(F7‘TZ)2(65~1) C( T Wf ) 1 fl 1 owzweSPtz c %(FTTZ)2(es—1) ow2——e0‘5(1800)2 1 F (0.0034*635*0.88)2(60505—1) 12000 10000 Using excel to find the relationship between assumed PM and calculated PM We have 3 Relative assumed l Calculated Deviation 2900 2946 -0.0160331 2910 2946 0.0123405 2920 2945 0.008674 2930 2945 “00050333 Relative deviation = 2940 2944 0.0014183 1"” assumed 2950 2944 000217152 2960 2943 0.0057363 2970 2942 000927635 2980 2942 001279193 2990 2941 001628331 3000 2943. 0.01975074 ow assumedewf calculated Plot the assumed ow versus relative deviation, we have the graph: VMM...M.W.-Mi4M._..0,W«.WMWM.~.MflVw"WWMWW.mnmwm‘wmwmmwmnwWW...mm,.4w.Mc.ww.wmmmm..m.u..a.ww..wMMflwMWWW? Relative deviation vs. assumed ow i 0.025 0.02 0.015 0.01 0 Relative deviation -0 01 ammmwmwm , .0 v WMWMW.aaaMmewmwmmwWMiWV.WWMMMWMWWW -002 wwwwMW.”WWWWWW,WWW.WW,MWwM.MmeWWWWWW.WWMWWWWWWWWw assumed ow rm«mmmmwmmamacaw»meWWW».0 0 MW, WWW WWWMWM M. News“ 0. 4.. W, 2 mm, W“ m (MWWWWMMMir/mmwmm,r/Wmlmwwumdw“WWW”WWW.“WWWW/VWWWWW/m. From the graph plotted between relative deviation versus assumed bottom pressure we can obtained the value of PM is psia ...
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This note was uploaded on 02/04/2012 for the course PE 4533 taught by Professor Civan during the Fall '11 term at The University of Oklahoma.

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HW #3_Solutions - Applied Reservoir Engineering HOMEWORK #3...

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