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24-Pet-A6 Well Logging and Formation Evaluation · December 2019

Question 2 of 5

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Reference texts: Craft, B.C. and Hawkins, M. (rev. Terry & Rogers), Applied Petroleum Reservoir Engineering, 3rd ed.; Ahmed, T., Reservoir Engineering Handbook, 5th ed.; Lyons, W.C. (ed.), Standard Handbook of Petroleum and Natural Gas Engineering, 3rd ed.

Question 2 (15 Marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. An undersaturated-to-saturated reservoir producing under solution-gas drive plus a small ("pot" / unsteady, single-tank) aquifer.

PVT and production history
$P$ (psia)$B_o$$R_s$$B_g$$N_p$ (MMSTB)$G_p$ (MMscf)
3,000 ($p_i$)1.316650–00
2,500 ($p_b$)1.3246500.000820.09259.8
1,5001.2525100.001350.850490.0
1,300 (current)1.2314500.001601.100970.0

Find. OOIP ($N$) and cumulative water influx ($W_e$) at the current pressure (1,300 psi).

Approach. Cast the material balance in Havlena–Odeh straight-line form, $F=N\,E_o+W_e$, and combine it with the Pot (small, instantaneous-equilibrium) aquifer model $W_e=k\,\Delta p$ to get a straight line in $F/E_o$ vs. $\Delta p/E_o$, whose intercept is $N$ and slope is the aquifer constant $k$.

  1. Underlying-drive functions. $$F=N_pB_o+(G_p-N_pR_s)B_g,\qquad E_o=(B_o-B_{oi})+(R_{si}-R_s)B_g$$ with $B_{oi}=1.316$, $R_{si}=650$ (both at $p_i=3{,}000$ psi, above $p_b$).
  2. Evaluate $F$ and $E_o$ at each pressure.
    Havlena–Odeh terms
    $P$$E_o$ (rb/STB)$F$ (res bbl)$\Delta p=p_i-p$$F/E_o$$\Delta p/E_o$
    2,5000.0080121,80850015,226,00062,500
    1,5000.12501,140,4751,5009,123,80012,000
    1,3000.23502,114,1001,7008,996,1707,234
    The 2,500 psi (bubble-point) row is excluded from the regression: $E_o$ there is only 0.008, so any small data error is amplified enormously in $F/E_o$ (its own value, 15.2 million, is wildly out of line with the other two rows) – the standard, well-documented Havlena–Odeh instability near $p_b$.
  3. Fit the straight line through the 1,500 and 1,300 psi points. With the Pot aquifer model $W_e=k\,\Delta p$, the material balance becomes $F/E_o=N+k(\Delta p/E_o)$, a straight line in $(\Delta p/E_o,\,F/E_o)$: $$k=\frac{8{,}996{,}170-9{,}123{,}800}{7{,}234-12{,}000}\approx 26.8\ \text{res bbl/psi}$$ $$N=\left(\frac{F}{E_o}\right)_{1{,}300}-k\left(\frac{\Delta p}{E_o}\right)_{1{,}300}=8{,}996{,}170-26.8(7{,}234)$$ $$\boxed{N\approx 8.80\text{ MMSTB (OOIP)}}$$
  4. Cumulative water influx at the current pressure. $$W_e(1{,}300\text{ psi})=k\,\Delta p=26.8\times1{,}700$$ $$\boxed{W_e\approx 45{,}525\text{ res bbl}}$$
Question 2 – final results
QuantityValue
Pot-aquifer constant, $k$26.8 res bbl/psi
OOIP, $N$8.80 MMSTB
Cumulative water influx at 1,300 psi45,525 res bbl