solution2017

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1 Solution set and marking schedule for Reservoir Engineering Exam 2017 Prepared by Martin Blunt (1) (i) Prepare a figure as in the notes (10 marks). Then define the different cases: dry gas (mainly methane) produces no liquid in the reservoir or at surface; wet gas (methane with some C2-C8) produces no liquid in the reservoir but at surface; gas condensate (C1-C10) produces liquid in the reservoir as the pressure drops; near-critical oil (C1-C12) has an initial temperature near the critical temperature; black oil (C1-C20) produces gas as the pressure drops in the reservoir (2 marks for each explanation). (ii) Start from equation: gg gi p B PWc B BGG +      −= 1 Then: gig g gi p BB PWcG B B G − +=       −1 Plot       − g gi p B B G 1 on the y axis and gig BB P −  on the x axis. Slope = Wc and the y intercept when x=0 is the gas in place, G. Gp (million scf) P (Mpa) Bg (rb/scf) x=DP/(Bg-Bgi) y=Gp/(1-Bgi/Bg) 0 40 0.000456 150 39 0.000481 40000 2886 450 38 0.00055 21276.59574 2632.978723 630 37 0.000598 21126.76056 2653.098592 800 36 0.000657 19900.49751 2614.925373 From the graph below, the intercept, G is approximately 2.4×109 scf (acceptable range 2.5-2.2) and the slope Wc = 0.013 rb/Pa (acceptable range 0.011 –0. 015). (2 marks for method, 4 marks for table, 4 marks for graph and 8 marks for values, including correct units. Lose 2 marks for any values quoted to 3 or more significant figures.) 2 (iii) Recovery factor is 800/2360.5 = 0.34 (2 marks). The gas saturation       −= G G B BSS p gi g gig 1 = 0.75 (2 marks). For maximum recovery the water influx is WcP = GBgi (−Swc−Sgr)/(−Swc), so P =47 MPa (3 marks). This is greater than the reservoir pressure, so water influx will never fill the field – just keep dropping the pressure (5…