1. (40 points) The properties of a dry gas reservoir are as follow: -1 9 = 0.000

Question

1. (40 points) The properties of a dry gas reservoir are as follow: -1 9 = 0.000025 psi Cv = 0.000007 psi-1 Swe - 30% Reservoir temperature is 150 °F Gas production date are listed in following table Cumulative gas production Average reservoir Gas z-factor at (MMSCF) pressure (psia) reservoir pressure and temperature 0) 4000 4000 2600 0.77 (a) Estimate the original gas in place by assuming that the reservoir is volumetric (b) Estimate the original gas in place by taking into consideration the water and (c) Gas z-factor is 0.9 at pressure of 1000 psia and reservoir temperature. Taking with zero water and rock compressibility. rock compressibility given above. into consideration the water and rock compressibility, and given that the reservoir abandonment pressure is 1000 psia, estimate recoverable gas volume in SCF and recovery factor.

Explanation / Answer

a) Original Gas in place = G

G = (pi/zi/pi/zi-p/z)Gp

Here Gp(Gas Production,MMSCF) = 4000

z= Gas formation volume factor = 0.77

zi = Initial formation volume factor = 0.75

Pi = Initial pressure = 4000

P = average reservoir pressure = 2600

so G = (4000/0.75/4000/0.75-2600/0.77)4000

       = (5333.33/5333.33-3376.62)4000

        = (5333.33/1956.71)4000

        = 2.72*4000

       = 10880 MMSCF

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