Problem 51: A 150-degF gas reservoir, originally 4,000
psia...known water drive...10 Mbbl of water and 2 MMMCF of gas has been produced...the percentage recovered to date is closest to:
This problem is wordy (I've cut most for the blog post) but is just plug -and-chug. Using 13 RES 2:
Bi = 5.04(150+460)0.9/(4000) = 0.692
Bf = 5.04(150+460)0.85/(2500) = 1.045
Delta B = 0.354 bbl/MCF
G = [GpBf – We + BwWp]/(delta B)
= ((2000000*1.0453)-1000000+(10000*1))*(1/0.354)
G = [(2 MMMCF(1.0453 bbl/MCF) – 1 MMbbl + (10 Mbbl*1 bbl/STB)]*(1/0.354 bbl/MCF = 3.1 MMMCF
2/3.1 = 64%
For this one, after Bi and Bf are found, why can't the RF=1-(Bi/Bf) equation be used? That gives me 33%?
ReplyDeleteIf you look on 13 RES 1, you can see the equation is prefaced by the statement it must not have water drive (which this problem has).
DeleteThat would be a great word question here, to give the problem and then ask, what would be the difference in recovery at this point with and without water drive. Quite a bit, eh? Pray for water drive :-).