Question 2) The unsteady state partial differential diffusivity equation could also be written as follows: +...

 

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Question 2) The unsteady state partial differential diffusivity equation could also be written as follows: + 2 1 1 r dr at 0.000264k n = The units are k [mD], r[ft], P[psia],c[1], t[hrs], and [cp]. Assuming the initial reservoir pressure is P; at all points before the production initiate and an infinite acting reservoir (reco) and constant flow rate in a cylindrical reservoir the following analytical solution would be obtained: P(r,t) = P + [70.6QBo kh -9480cr2 E( kt The units are k [mD],r[ft], P[psia],c[+], t[hrs], Q[ST], and [cp]. Day Using the above equation, the reservoir pressure could be calculated at different times for each point in the reservoir. E is simply a mathematical function that is represented by the following graph. Therefore, by -9480cr2 -9480cr2 calculating the term and locating it on the x-axis of the graph, E( kt kt could be read from the y-axis. 10 8 6 43 2 1 EXPONENTIAL INTEGRAL VALUES 9-C E;(-x) = -du x For x <0.02 U E(-x) In(x)+0.577 -.02 -.04 -.06 -.08 -10 1.0 1 E;(-x) 0.8 0.6 0.4 0.3 0.2 0.1 .08 .06 .04 .03 .02 .01 -0.5 -1.0 -1.5 -2.0 -2.5 -3.0 -3.5 E; (-x) The E-function. (After Craft, Hawkins, and Terry, 1991.) Adopted from (Reservoir engineering handbook, second edition, Tarek Ahmed) An oil well is producing at a constant flow rate of 300STB/Day under unsteady state conditions and has the following properties: Bo bbl 1 psi , k = 60 mD, h = 15 ft, = 15%, w = 0.25 ft = 1.25 STB'Mo = 1.5 cp, c = 12 10-6. The initial pressure of the reservoir is P = 4000 psia. A) After 1 hr of production, what is the reservoir pressure at r = 5ft? B) After 24 hr of production, what is the reservoir pressure at r = 5ft? 2 Question 2) The unsteady state partial differential diffusivity equation could also be written as follows: + 2 1 1 r dr at 0.000264k n = The units are k [mD], r[ft], P[psia],c[1], t[hrs], and [cp]. Assuming the initial reservoir pressure is P; at all points before the production initiate and an infinite acting reservoir (reco) and constant flow rate in a cylindrical reservoir the following analytical solution would be obtained: P(r,t) = P + [70.6QBo kh -9480cr2 E( kt The units are k [mD],r[ft], P[psia],c[+], t[hrs], Q[ST], and [cp]. Day Using the above equation, the reservoir pressure could be calculated at different times for each point in the reservoir. E is simply a mathematical function that is represented by the following graph. Therefore, by -9480cr2 -9480cr2 calculating the term and locating it on the x-axis of the graph, E( kt kt could be read from the y-axis. 10 8 6 43 2 1 EXPONENTIAL INTEGRAL VALUES 9-C E;(-x) = -du x For x <0.02 U E(-x) In(x)+0.577 -.02 -.04 -.06 -.08 -10 1.0 1 E;(-x) 0.8 0.6 0.4 0.3 0.2 0.1 .08 .06 .04 .03 .02 .01 -0.5 -1.0 -1.5 -2.0 -2.5 -3.0 -3.5 E; (-x) The E-function. (After Craft, Hawkins, and Terry, 1991.) Adopted from (Reservoir engineering handbook, second edition, Tarek Ahmed) An oil well is producing at a constant flow rate of 300STB/Day under unsteady state conditions and has the following properties: Bo bbl 1 psi , k = 60 mD, h = 15 ft, = 15%, w = 0.25 ft = 1.25 STB'Mo = 1.5 cp, c = 12 10-6. The initial pressure of the reservoir is P = 4000 psia. A) After 1 hr of production, what is the reservoir pressure at r = 5ft? B) After 24 hr of production, what is the reservoir pressure at r = 5ft? 2