Producing GOR and Gas-Injection Material Balance: 24-Pet-A6 Well Logging and Formation Evaluation, May 2015, Q2

Problem 2: Producing GOR and Gas-Injection Material Balance (15 marks)

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Given. $N_p$ (0–7 MMstb) with its producing GOR $R_p$ (scf/stb) and cumulative injected gas $G_{inj}$ (MMscf, injection starts at $N_p=5$), tabulated above.

Find. Cumulative produced gas $G_p$, net cumulative produced gas ($G_p-G_{inj}$), and a single plot of $R_p$, $G_p$, $G_p-G_{inj}$ and $G_{inj}$, all vs. $N_p$.

Approach. $R_p$ is the instantaneous producing GOR at each $N_p$, so cumulative produced gas is its running integral, $G_p=\int R_p\,dN_p$, evaluated between table rows with the trapezoidal rule (average $R_p$ of the two endpoints times the $N_p$ increment); net produced gas simply removes whatever has been re-injected.

Cumulative produced gas (trapezoidal integration of $R_p$). Between consecutive rows, $\Delta G_p=\bar{R}_p\,\Delta N_p$ with $\bar{R}_p$ the average of the two endpoint GORs and $\Delta N_p=1$ MMstb throughout: e.g. $0\to1$: $\bar{R}_p=(300+280)/2=290$, $\Delta G_p=290$ MMscf; $4\to5$: $\bar{R}_p=(560+850)/2=705$, $\Delta G_p=705$ MMscf; and so on. Running the sum over all seven intervals gives $\boxed{G_p=0,\,290,\,570,\,880,\,1330,\,2035,\,3020,\,4290\text{ MMscf}}$ at $N_p=0,1,\dots,7$ MMstb respectively.
Net cumulative produced gas. $G_{p,\text{net}}=G_p-G_{inj}$. Injection is zero through $N_p=5$, so $G_{p,\text{net}}=G_p$ there; at $N_p=6$: $3020-520=2500$; at $N_p=7$: $4290-930=\boxed{3360\text{ MMscf}}$.

Fig. 1: Producing GOR and the three cumulative-gas curves vs. cumulative oil production (dual axis: GOR left, gas volumes right).

$N_p$ (MMstb)	$R_p$ (scf/stb)	$G_p$ (MMscf)	$G_{inj}$ (MMscf)	$G_p-G_{inj}$ (MMscf)
0	300	0	0	0
1	280	290	0	290
2	280	570	0	570
3	340	880	0	880
4	560	1330	0	1330
5	850	2035	0	2035
6	1,120	3020	520	2500
7	1,420	4290	930	3360

Topic: Producing-GOR integration and gas-injection bookkeeping in a material-balance study.

Key relations: $G_p=\int_0^{N_p} R_p\,dN_p'\approx\sum \bar{R}_p\,\Delta N_p$ (trapezoidal); net produced gas $=G_p-G_{inj}$.

Why this works: $R_p=dG_p/dN_p$ by definition (instantaneous produced-gas-to-oil ratio), so recovering $G_p$ from a table of $R_p$ values is exactly a numerical integration; only the NET gas actually leaves the reservoir once injection begins, which is the quantity the reservoir's own material balance (and gas-cap $E_g$ term) must use, not the gross produced gas.

Common pitfall: using $R_p\times N_p$ directly (a rectangle rule at the endpoint GOR) instead of integrating the whole history — this badly overstates early-life gas production once $R_p$ has risen sharply late in life; forgetting to subtract injected gas from produced gas before using it in a gas material balance.

Source: Craft & Hawkins, Applied Petroleum Reservoir Engineering, 3rd ed., Ch. 4; Ahmed, Reservoir Engineering Handbook, 5th ed., Ch. 4 (material balance with gas injection).

Question 2 of 8: Producing GOR and Gas-Injection Material Balance

Problem 2: Producing GOR and Gas-Injection Material Balance (15 marks)