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A Recurring Problem in Estimating Prospect Reserves—Determining Reasonable “Low-side” Values (P99% and P90%)

During the period 1935-1969, a field-size distribution of discoveries in the Texas Gulf Coast (Figure C-l) indicates that the smallest field sizes (= P99%) contained between 3,000 and 4,000 recoverable bbl. During the next 19 years (1970-1988), the P99% discovery size was only a little smaller—about 2,000 bbl (Figure C-2). Similarly, in the northern part of the Midland Basin/Red River Arch province, the P99% field size was about 1,000 bbl as of 1945, and about 650 bbl as of 1957. In the Silurian (Niagaran) pinnacle-reef play of northern Michigan, P99% for discovered fields was about 5,000 bbl in the period 1968-1974 (inclusive), and about 1,000 bbl for the period 1975-1988. And yet all these very small, money-losing events qualify as “discoveries”—mobile hydrocarbons were sensed, the wells were completed, and oil was produced, presumably for sale. The implications are depressing but profound: there is always a possibility that your discovery is going to turn out to be a very small accumulation!

Most experienced exploration risk analysts know that the most common cause of overestimation of prospect reserves is that the “low-side” estimates (P99% and P90%) are too large. But how to arrive at appropriate low-side estimates? This section addresses that problem using six different examples.

Example 1

In our short courses, we use a simple prospect exercise called the RMAG prospect (for the Rocky Mountain Association of Geologists, where it was first used). The RMAG exercise provided good examples whereby the low-side values for Area, Average Net Pay, and HC-recovery Factor were all quite low because the existing data demonstrated that such marginal values were entirely possible. Thus the P90% value for Area was about 60 acres (24 ha), and the P99% value was about 20 acres (8 ha). The P90% value for Average Net Pay thickness was about 6 feet (2 m) and P99% was 4 feet. HC-recovery Factor was given (P90% = 100 bbl/af [barrels/acre-foot]; P10% = 300 bbl/af, consequentially P99% was about 60 bbl/af, which approaches an effective reservoir minimum). The resulting reserves distribution had P90% reserves of about 96,000 bbl and P99% reserves of 27,000 bbl.

Example 2

But consider a second situation, where consistent and reliable regional log and rock data indicate that average net reservoir thickness is never less than 16 feet (5 m) and ranges systematically and regionally up to 80 feet (24 m). Even taking into account the geometric adjustment factor, P99% for Average Net Pay probably shouldn't be less than about 12 feet; P90% should be around 16 feet. Consider a prospect in this setting, where HC recoveries are consistently poor, ranging from 60 bbl/af to 150 bbl/af. Here, the productive area necessary to provide around 10,000 bbl—about the minimum amount of recoverable oil necessary to sustain flow—would of necessity be around 14 acres (5.7 ha) if Average Net Pay was 12 feet and HC-recovery was 60 bbl/af. However, only 6.5 acres would be required to contain 10,000 bbl if Average Net Pay was 16 feet and HC-recovery was 100 bbl/af.

Example 3

You must also take into account the effects of geometries consistent with the anticipated trap type. For example, suppose your prospect is a fault trap as shown in Figure C-3, minimum reservoir quality is projected to be 60 bbl/af (P99%), and average net pay thickness in such poor reservoir rock must be at least 40 feet (12 m).

Given that the dip-rate of the reservoir-zone is about 300 feet (90 m) per mile, this geometry imposes minimum value upon possible area of the low-side case (Figure C-4).

In order to arrive at the requisite 40 feet of average net pay in such trap geometry, the height of the hydrocarbon column must be about 65 feet. This forces the downdip position of the oil/water contact to be positioned about 1,000 feet south of the trapping fault, measured along the north-south axis of the structure, as shown by Figure C-4. Because the structural geometry requires an elongated east-west closure, the length of the minimum closure measured along the fault must be about 1.85 miles, or 9,800 feet. The area of this minimum closure (shaded in Figure C-3) is about 112 acres (9,800 ft. χ 1,000 ft. + 2 χ 43,560 ft2/acre). The product of the three P99% values is 269,000 bbl (112 acres χ 40 ft. ave. net pay χ 60 bbl/af). Obviously, the corresponding P90% values for Average Net Pay, Area, and Reserves would have to be much larger.

For additional perspective, suppose that the required P99% value for average net pay is 20 feet, rather than 40 feet; still using 60 bbl/af, this places the oil/water contact 500 feet south of the fault, and reduces the rTL?nimum area (P99%) by half, to 56 acres. Additionally, prospec-tors must remember that the reserves P99% value is the approximate product of the P90% values for area, average net pay, and HC-recovery factor. For elongated structural traps, P99% reserves accordingly may be substantially greater than the rninimum 10,000 to 50,000 bbl required to support sustained flow.

Example 4

Consider a prospect located on a large, gentle domai anticline (Figure C-5), in which the reservoir dip away from the apex of the closure is very gentle, say 30 feet per mile. Further, let us suppose that regional reservoir data indicate that P90% = 120 bbl/af and P99% = 60 bbl/af are appropriate estimates for this prospect. Also, assume that appropriate estimates for net reservoir thickness are 15 feet (P90%) and 10 feet (P99%). The structural geometry leads to the fol-lowing possible P99% outcome, keyed to a 10,000-bbl minimum reserves case (Figure C-5):

We now adjust net reservoir thickness to obtain average net pay: P99% = 8 feet and P90% = 12 feet. Now, in order to achieve the requisite 8 feet of P99% average net pay thickness, the thickness of the oil column at the apex of the structure must be around 15 feet, which then requires a linear distance of about 1.0 mile for the length of the elongated axis of the saturated area. The corresponding area of closure is: πr2 = 3.1416 × (2,640 ft.)2 ÷ 43,560 ft.2/acre = 503 acres = P99%. Obviously, P90% area would have to be much larger, perhaps approaching 1,000 acres.

Example 5

Suppose you are exploring an undrilled frontier trend containing perhaps 100 structural closures, some large, some intermediate, some small. Given the usual exploration pattern, your prospectors can identify closures too small to contain the 25 million barrels (MM bbl) they think is the minimum required to develop; but what your prospectors cannot consistently do in such a trend is to identify those large closures that contain very small accumulations (because of severe underfilling, or very poor reservoir quality or thickness, or because of faulty geophysical data or interpretations). In any case, if you drill only prospects you think are large enough to contain at least 25 MM bbl, your low-side areas (P99%) could well be larger than 20 or 40 acres, and P90% areas could certainly be several hundred acres or even more. As a result, P99% reserves could well be larger than 10-20,000 bbl—it could properly be 50-100,000 bbl, or maybe more. Correspondingly, P90% reserves might be from about 200,000 up to around one million barrels, or more. But only because you are not going to drill any but the largest closures.

Example 6

Consider a prospect in the Gulf of Mexico that features a fairly well-defined amplitude anomaly that is congruent with structural contours, and a “flat spot” that may indicate a gas-water contact, also consistent with the structural picture. Your geotechnical staff think there is ± 20% possible variation in the median Area parameter of 400 acres because of imperfect resolution. So, in this case, P90% = 320 acres and P90% = 480 acres.

What's the message? There's no “cookbook”—you cannot blindly assign a low-side value to the reserves parameters. You must understand the range of geologic values, the structural patterns, and the exploration maturity of the trend. Then you must integrate these considerations and think probabilistically. But, in any case, it's not a bad idea to start with very small assumptions for P90% and P99% values for Area, Average Net Pay, and HC-recovery Factor—and then adjust them upward as the geotechnical evidence justifies.

Final admonition:—Remember that the most common field size discovered in west Texas in the 1930s and 1940s (during the “flush” period of discovery, when many very large fields were being found) was a one-well field—around 10,000 to 20,000 bbl. During all the subsequent decades, even up to the present time, the most common field size has remained about the same—the one-well field. Sure, such reserves sizes have gradually decreased, but only a little. So, when constructing the reserves distribution for your prospect, ask yourself, “Do I really think that there's no chance of my prospect turning out to be a mediocre little one-well field?”





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