2018 - Petroleum Engineering Exam Paper Solutions PDF

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2018 - Petroleum Engineering Exam Paper Solutions...

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Solution set and marking schedule for Reservoir Engineering Exam 2018 Prepared by Martin Blunt (1) (i)

(ii)

A wet gas produces liquid at the surface, but not in the reservoir as the pressure drops. A gas condensate produces liquid in the reservoir as the pressure drops (5 marks). Draw phase diagrams as in the notes to illustrate this (5 marks). A wet gas can be produced simply by dropping the pressure, whereas in a gas condensate it is preferable to avoid the two-phase region to prevent liquid being retained in the field and a loss of productivity at the well (4 marks). In this example the gas/oil ratio is 300 Mscf/stb (3 marks) which is in the range of a wet gas (3 marks). Start from equation:  B  Wc∆P G p = G1 − gi  +  Bg  Bg  Then: Gp Wc∆P =G + Bg − Bgi  B  1 − gi   Bg   Gp ∆P on the y axis and on the x axis. Slope = Wc and the Plot Bg − Bgi  B gi  1 −   Bg    y intercept when x=0 is the gas in place, G.

Gp (million scf)

0 100 220 300 370

P (Mpa) Bg (rb/scf) x=DP/(Bg-Bgi) y=Gp/(1-Bgi/Bg) 40 0.000456 39 0.000481 40000 1924 38 0.00055 21276.59574 1287.234043 37 0.000598 21126.76056 1263.380282 36 0.000657 19900.49751 1209.402985

From the graph below, the intercept, G is approximately 5.3×108 scf (acceptable range 5.0-5.6) and the slope Wc = 0.035 rb/Pa (acceptable range 0.03 –0. 04). (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.)

1

Gp/(1-Bgi/Bg) (million scf)

2500 y = 0.0349x + 527.35 R² = 0.9987

2000 1500 1000 500 0 0

(iii)

10000

20000 30000 DP/(Bg-Bgi) (MPa.scf/rb)

40000

50000

Recovery factor is 370/527 = 0.70 (2 marks). The gas saturation Bg  G p   = 0.34 (2 marks). For maximum recovery the water influx 1 − S g = S gi Bgi  G  is Wc∆ P = GBgi (1−Swc−Sgr)/ (1−Swc), so ∆P =4.7 MPa (2 marks). The final G B S recovery is p = 1 − gi g = 79% assuming Bg = 0.0007 (4 marks). Near the end G S gi Bg of the field life, so just drop the pressure and try to avoid excessive water production (2 marks).

(2) (i)

(ii)

(iii)

(iv)

Study the water pressure: the normal pressure (Patm + ρwgz) would be 29.4 MPa. (2 marks). Over-pressured since the measured water pressure at this depth is higher than the normal pressure (2 marks). P − P + D1 ρw g − D2 ρo g = 2635 m (8 marks); Use zowc = 2 1 ( ρw − ρo) g P − P2 + D2 ρo g − D3 ρ g g zgoc = 3 = 2521 m. (8 marks). ( ρ o − ρ g )g The oil column has a depth of 114 m to the nearest m (2 marks). This is due to capillary pressure, which retains water above the free water level – a finite capillary pressure is needed for oil to enter the rock (6 marks). In this case the true column height is 114-2 = 112 m (2 marks). This is the ratio of the rock volume that contains potentially producible oil to the total potentially oil-bearing volume of the rock: it excludes zones saturated 2

(v)

with water or of very low porosity/permeability (4 marks). It can be estimated from logs to locate water-saturated or low porosity zones, or from the analysis of cores, or from geological analogues (4 marks). The key point is the height of the oil column which must use the true contact and exclude the water-bearing zone: h = 112 m (5 marks). Then STOIIP = 13 ×106 × 112 × 0.8 × 0.19 × (1-0.26)/1.4 = 1,20 million sm3 (3 marks). This is a medium to large field (2 marks).

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