Solution2016

Pagina 1

1 Solution set and marking schedule for Reservoir Engineering Exam 2016 Prepared by Martin Blunt (1) (i) This is a field whose initial temperature and pressure are close to the critical point of the hydrocarbon mixture in the subsurface, meaning that there is little distinction between oil and gas: illustrate this with a phase diagram that shows this (7 marks). Since the gas is in equilibrium with oil, the gas must be near the dew point: the initial conditions are where the oil and gas curves cross (5 marks). Therefore the gas will be a gas condensate (5 marks) and as the pressure drops oil will drop out in the gas cap (3 marks). (ii) Start from equation: gg gi p B PWc B BGG ∆+      −= 1 Then: gi g g gi p BB PWc G B B G − ∆+=       −1 Plot       − g gi p B B G 1 on the y axis and gi g 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 45 0.0035 30 44 0.004 2000 240 56 43 0.0047 1666.666667 219.3333333 80 42 0.0058 1304.347826 201.7391304 95 41 0.0072 1081.081081 184.8648649 From the graph below, the intercept, G is approximately 1.2×10 8 scf (acceptable range 1.1-1.3) and the slope Wc = 0.058 rb/Pa (acceptable range 0.05 –0. 065). (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 95/120 = 0.79 (2 marks). The gas saturation       −= G G B BSS p gi g gi g 1 = 0.37 (2 marks). For maximum recovery the water influx is Wc ∆P = GB gi (1− Swc −Sgr )/ (1− Swc ) = 26,000 rb, so ∆P =4.4 MPa (3 marks). Need to estimate Bg sensibly – from the table the value is close to 0.0078 (2 marks). Then…