Page 150 - A Practical Companion to Reservoir Stimulation

PRACTICAL COMPANION TO RESERVOIR STIMULATION

EXAMPLE 5-4
presented in Example 5-3 would suggest the lower possible
Optimum Horizontal Well Length bottomhole pressure (i.e., largest Ap) that would still avoid
two-phase or other operational problems.
For the well in Example 5-3, it is shown that in spite of the
pressure-dependent skin effect no optimum bottomhole flow- Solution (Ref. Sections 8-2.8 and 19-2)
ing pressure can be identified. This is because any horizontal The economic analysis presented here uses the Net Present
well length with that type of inflow performance relationship Value (NPV) concept. In terms of incrementals, it can be
does not have an inflection point leading to a maximum flow written as
rate. However, it is evident that while flow rates are mono-
tonically increasing, the values for different well lengths are
getting nearer at higher pressures. This means that an optimi- (5-12)
zation, based on incremental revenue and incremental costs,
is necessary to identify the appropriate well length. This can where (A$)n is the incremental revenue in year II, i is the time
be done for any bottomhole pressure, although the results value of money (at least inflation rate, preferably the rate of
return) and ACost is the incremental cost associated with the
investment.
AP L 4 Aq A$ NPV The basis for the analysis is a 500-ft horizontal well, and
(Psi) (fi) [ 1000) (1 000) the incremental costs are $440/ft. Table J-5 contains the
300 500 266 incremental NPV for various pressure drops and well lengths.
750 283 17 6205 255 145 The price of oil was taken as $ I8/STB, and the time value of
money as 15%. An example calculation for the results in Table
1000 296 30 10950 450 230 J-5 is shown below. (Ap = 600 psi, and a comparison between
1250 307 41 14965 61 5 285 a 500-ft well and a 1000-ft well is done.)
1500 31 7 51 1861 5 764 324 First, from Eq. J-9, qjo0 = 342 STB/d, and from Eq. J-10,
q,ooo = 366 STB/d. Thus, Aq (the daily rate difference) = 24
600 500 342 STB/d, which can be roughly translated to A Np (incremental
750 356 14 51 10 21 0 100 annual production) = 8760 STB. (Even if it is not entirely
1000 366 24 8760 360 140 steady-state flow for both wells, this difference is a reasonable
1250 374 32 11680 480 150 approximation without respect to the type of flow.)
Using $I8/STB and i = 0.15, the incremental net revenue
1500 382 40 14600 600 160 discounted to time 0 (for 3 yr) is then
900 500 378 3
-
750 389 11 401 5 166 56 (A$),l - (8760) (18) + (8760) (18)
11 = I 1.15 ( 1.15)2
1000 397 19 6935 284 64
1250 404 26 9490 389 59 + (8760) (18) = $360 x lo'. (5-13)
1500 409 31 11315 465 25 ( 1.15)2
However,
1200 500 399
ACost = (1000 - 500) ft ($440/ft)
750 408 9 3285 135 25
1000 415 16 5840 239 19 = $220 x lo', (5-14)
1250 42 1 22 8030 330 0 resulting in
1500 425 26 9490 389 -5 1
NPV = 360 x lo3 - 220 x 10' = $140 x 10'. (J-15)
1500 500 41 3 Figure 5-7 is a plot of the incremental NPV vs. horizontal
750 42 1 8 2920 120 10 well length for a number of pressure drops. Since Ap = 1500
1000 427 14 51 10 21 0 -1 0 psi is more desirable than any other Ap (see Fig. J-5), then it
1250 431 18 6570 270 -60 is obvious that the intended length of 500 ft is very near the
1500 435 22 8030 330 -110 best. At 750 ft, the incremental NPV is only $10,000, and it
becomes negative for wells of greater length. Thus, a decision
Table J-+Incremental NPV for various pressure drops and to drill a short, 500-ft well is justified by this economic
well lengths for Example J-4. analysis.

J-10