GB2332227A - Optimising well numbers in oil and gas fields - Google Patents

Abstract

A process for exploiting oil and gas fields in an economically optimal manner. The process involves determining the approximate economically optimal number of wells to drill and producer:injector ratio to use, and the expected Net Present Value that will result. The process uses simple mathematical formulae derived by simplifying the oil-field economics to linear factors and discounting, and by combining discounting and a decline function into a single expression, which is then analytically optimised. The economic optimisation can be either simply to maximise Net Present Value or to maximise Net Present Value subject to a corporate limit on capital employed.

Description

OPTIMISING WELL NUMBERS IN OIL AND GAS FIELDS

In the process of extracting oil and gas from an underground accumulation, the number of wells drilled (or, equivalently, their spacing) has an important effect on profitability. Another important factor, for fields with water or gas injection, is the ratio between injector wells and producer wells.

In prior art, the usual method for deciding on both well spacing and producer: injector ratios is to base these on the results of reservoir simulation and a hydrocarbon extraction economic model. A computer reservoir simulation model is built for the oil-field in question and then run with a number of different development scenarios. The results of each scenario (in terms of hydrocarbon production profiles, activity levels etc) are then fed into an economic model (typically spreadsheet based) which calculates the net present value (NPV) expected ftom the scenario. Whichever scenario gives the best economic results is taken as the basis for the actual development of the field. The problem with this approach is that the development scheme may not be optimal. The choice of scenarios to run is based on experience, but the field may be sufficiently different from already developed fields that none the scenarios considered are close to what would be the optimal development scheme.

Another method in prior art is to create a simple mathematical model (usually based on decline curves) of the expected production profile, as a function of well numbers, amongst other factors. The model is linked in a computer program (usually a spreadsheet) to the economic model to give NPV as a function of well numbers, amongst other factors. Numerical optimisation of NPV is then carried out, to give the optimal number of wells and, if required, the optimal producer:injector ratio. (These results can then also be used as a starting point for more detailed work with a reservoir simulation model). There are two problems with this. Firstly, the resultant combined model may be rather intricate, therefore requiring considerable work to so up and modify. Secondly, because of the intricacy, it may be error-prone and difficult to audit.

F An object of this invention is provide a formula-based process for determining, for any oil or gas field, the approximate economically optimal number of wells and producer.. injector ratio for that field. The formulas are easier and quicker to use than the method of numerical optimisation, they are easier to audit and give more engineering insight.

Accordingly, in a first aspect, this invention provides a method of extracting hydrocarbons from underground accumulations, in which the number of wells to set to give the economically optimal results, the process comprising the steps of 1) taking a hydrocarbon extraction economic model and simplifying the economic model to linear factors and discounting-, 11) creating simple economic models of operating expenditure and capital expenditure as functions of the number of wells, 111) through decline analysis, modeling the expected field production profile as a function of well numbers, initial average well rates and technically recoverable reserves-,

IV) combining the economic models and the production profile model to give a single algebraic expression for the present value of the expected cash-flow, algebraically integrating this expression to give net present value (NPV), v) vi) calculating an expression for the partial derivative of NPV with respect to well numbers; V11) expressing the criteria for economic optimality in terms of the partial derivative of the T,PV function with respect to well numbersI solving the equation derived from steps (vi) and (vil) above to give an expression for the economically optimal number of wells, ix) simplifying the expression for the economically optimal number of wells by taking terms that have little effect on the output and setting them to be constants. evaluating the simplified expression for the economically optimal number of wells for the field in question.

xl) drilling substantially the number of wells so determined.

X) r In a second aspect, this invention provides a method as described above, in which, when the field is complicated, then reservoir simulation work is carried out to refine and confirm the optimal number of wells.

In a third aspect, this invention provides a method of exploiting a hydrocarbon field, comprising the steps of..

ij) vii) X) xl) 1) taking a hydrocarbon extraction economic model and simplifying the economic model to linear factors and discounting; creating simple economic models of operating expenditure and capital expenditure as functions of the number of wells; 111) through decline analysis, modeling the expected field production profile as a function of well numbers, initial average well rates and technically recoverable reserves.

iv) combining the economic models and the production profile model to give a single algebraic expression for the present value of the expected cash-flowl v) algebraically integrating this expression to give net present value (NPV), vi) calculating an expression for the partial derivative of NPV with respect to well numbers, expressing the criteria for economic optimality in terms of the partial derivative of the NPV function with respect to well numbersI v) 1 solving the equation derived from steps (vi) and (vii) above to give an expression for the economically optimal number of wells.

ix) simplifying the expression for the economically optimal number of wells by taking terms that have little effect on the output and setting them to be constants, evaluating the simplified expression for the economically optimal number of wells for the field in question; inserting the expression for the economically optimal number of wells Into the expression for NPV from step (v); algebraically manipulating the result to give a simple formula for NPV..

X111) evaluating the expression for NIPV for the field in question,

XIV) drilling wells, or not, based on the MV determined in step (xiii).

In a forth aspect, this invention provides a method of extracting hydrocarbons from underground accumulations, comprising the steps of 1) taking a hydrocarbon extraction economic model and simplifying the economic model to linear factors and discounting-, creating simple economic models of operating expenditure and capital expenditure as functions of the number of wellsl Iii) through decline analysis, modeling the expected field production profile as a function of well numbers, initial average well rates and technically recoverable reserves;

IV) combining the economic models and the production profile model to give a single algebraic expression for the present value of the expected cash-flow; algebraically integrating this expression to give net present value (NPV)-, calculating an expression for the partial derivative of NPV with respect to well numbers; VII) expressing the criteria for economic optimality in terms of the partial derivative of the NPV function with respect to well numbers, I I I vil solving the equation derived from steps (vi) and (vii) above to give an li) v) vi) X) xl) expression for the economically optimal number of wells, ix) simplifying the expression for the economically optimal number of wells by taking terms that have little effect on the output and setting them to be constants., evaluating the simplified expression for the economically optimal number of wells for the field in question. inserting the expression for the economically optimal number of wells into the expression for NPV from step (v).

X]]) algebraically manipulating the result to give a simple formula for NPV, xiii) deriving, through a material balance argument, an expression for the expected oil production per well, averaged over both injectors and producers, as a function of the producer: injector ratio.

xiv) deriving expressions for operating expenditure per well and capital expenditure per well as functions of the producer: injector ratio; t xv) inserting the expressions derived in steps (xlil) and (xlv) into the formula for N-PV determined in step (xii) to give an expression for NPV as a function of the producer: injector ratio; Xvi) calculating an expression for the partial derivative of NPV with respect to the producer: injector ratio; Xvil) expressing the criteria for economic optimality in terms of the partial derivative of the NPV function with respect to the producerinjector ratio, xvill) solving the equation derived from steps (xvi) and (xvii) to give an expression for the economically optimal producer: injector ratio; xix) evaluating the expression for the economically optimal producer: injector ratio for the field in question, and drilling producer and injector wells in substantially the optimal ratio.

A preferred embodiment of the invention will now be described, by way of example only.

First, some notation:q - initial oil production per well per year, averaged over all wells, including injectors G - gross liquid production per year for the whole field N - total number of wells, including injectors R - technical reserves Le. the amount of oil that could be recovered if the field were run for a very long time L - net revenue per tonne of oil (i.e. after all taxes and royalties, including profit tax) d - discount rate C - net capital cost per well D - net capital costs not related to numbers of wells, e.g. roads and pipelines a - decline rate E net Opex (operating expenditure) per well Note:- In this context, "net" means expressed in terms of effect on present value, after all taxes and royalties. T,h - abandonment time (x - present value of money at abandonment time.. (1 l +d)-"""

H - "hurdle" rate - NPV Net Capex (capital expenditure) ratio needed to be reached for a project to be approved In order to derive algebraic expressions for NPV and for the economically optimal number of wells, it is necessary to have an algebraic model of all the economic factors., including taxes and royalties. It is suggested that the following is an appropriate model for most tax regimes:NPV = E (L.Q1 + a.OPEXi + O.CAPEXi) / (l+d)'-' where the summation is over i=l to i=n, the year of the end of the field-life OPEXI = Opex in year i CAPEM = Capex in year 1, not including VAT Q1 = oil production in year i L = a linear factor, the---netoil price" - it represents the benefit in present value terms of one extra unit (ton, cubic metre or barrel) of oil sold cc = a linear factor, representing the effects on NPV of one extra real- terms dollar (or whatever unit of currency) of expenditure on Opex p = a linear factor, representing the effects on NPV of one extra real-terms dollar of expenditure on Capex For a given tax system, (x, 0 and L can be calculated in a conventional economics spreadsheet by increasing respectively Opex, Capex and oil production by one unit in Year 1 of the project, and seeing the effect on NPV. A more formal of describing this is to say that (x, 0 and L are the parial derivatives of NPV with repect to Year 1 Opex, Capex and oil production. The terms (x.OPEXi and.CAPEXI can be considered to be the---netOpex" and the---net Capex".

It is reasonable to model the NPV as a linear function for two reasons although, as can seen from economics spreadsheet models, the Interplay of costs, revenues and taxation is very complicated, usually all the individual steps in the spreadsheets are linear, hence combining them will usually give a linear function r if a taxation system was not linear in its effects, there would be benefits in artificially splitting or combining projects so as to take advantage of the nonlinearities; such a possibility would quickly become well known. Hence, from initial design or from progressive correction of anomalies, most tax systems have evolved to be linear. The other major assumptions about the economics are that it is desired to achieve the highest possible NPV (as opposed to the highest rate of retum, for example); capex expenditure for the field consists of a fixed element (e. g. cost of platform, road or pipeline) and a cost per well; all capex expenditure occurs in Year I; opex is proportional to the number of wells.

In analytic models, it is usually necessary to define in advance what quantity is to be minimised or maximised. (In contrast, in spreadsheet models, it is possible to run through all the various scenarios and then choose the one which gives the overall best results. For example "Putting in 15 wells rather than 10 increases the NPV by only 5%, while it increases the capex by 40%. On balance, we think that the increase in NPV is not worth it, given the overall uncertainties, so we will go for the 10 well option.") Here, NPV will be maximised, firstly because it is the simplest, and, secondly, because it is easy to modify the NPV-maximising results to cope with the situation where there is a shortage of capital, and one is only interested in returns of more than 15%, for example. The modification required is described later.

The first assumption about capex are that net capex can be expressed as Net capex = C.N + D where C is net capex per well and D is a fixed element. The second assumption is that all the capex expenditure can be considered to take place instanteously at the beginning of the project. Clearly, this is not always realistic, but, again, it is simplerl r it is unlikely that the time required to carry out the capital expenditure will affect questions such as---Whatis the optimal number of wells?" it is easy to modify the model to deal with capex spread over several years (again, the modification is described later).

With regards to opex, it should be noted that, whenever the gross liquid rate is constant, the simple model gives the same results as a more complicated model, in which opex contains one element proportional to the number of wells and one element proportional to the gross liquid production.

There are three important petroleum engineering assumptions technically recoverable reserves are independent of the number of wells initial well rates are independent of the number of wells the field follows exponential decline from the start of production.

Unless there are geological features like isolated fault blocks, the number of wells does not much affect the technical recovery factor. In theory, for a field producing under depletion, a single well could, over a very long time, drain the field as efficiently as twenty wells. One reason for this is that, whatever the well spacing, the area of the field that is drilled up (i.e. directly penetrated by a well-bore) is very small in relation to the total area. With vertical wells, a well spacing of 500m and a well-bore diameter of 6 inches gives a ratio (total well bore area) (area of field) of 1 X 10-7. With ten times as many wells, this would still only give a ratio of 1 x 10-6, so similar recovery factors should apply.

It can be seen ftom the formula for the steady-state productivity index (PI) of a vertical well PI = (constant x K,,.h) / [(In(r,/r,,)4-S) x B,,. p,] and the slightly more complicated formula for horizontal wells that the terms depending on well separation, r, (the effective drainage radius) only occur as a logarithmic terms, hence the effect of changes of it are usually small. So it is reasonable to assume that PI is not much affected by changes in well spacing.

Even if PIs are constant, it is does not necessarily follow that well rates are constant, because one well may reduce the average pressure seen by the other wells. However, if one scales a whole development, including the number of injectors, then the average pressures should remain the same, giving constant well rates.

The assumption that initial well rates are independent of well numbers also holds for gas fields being produced under depletion drive. Here, increasing the number of wells does give pressure interference. However, providing all the wells start production at the same time, this effect is captured by the decline rate.

The other major petroleum engineering assumption is that the field follows exponential decline from the very beginning. There may be two objections to this asssumption. Firstly, many fields show a plateau production period. Secondly, once decline starts, it may be exponential, but changes to hyperbolic decline in the later stages of field life.

In answer to the first objection, it can be argued that many of the fields showing a production plateaus did so because of staged development If this Is the case, then the effect can be captured by the modification to the model described later. However, if the plateau period is a genuine subsurface phenomenon (maybe a period of stable, field-wide dry oil production, before water break-through), then the model may require correction.

The second objection should be less of a concern. A change to hyperbolic decline usually occurs sufficiently late in field life that, with discounting, it has little effect on the economics of the field, and on the question of what is the economically optimal number of wells. All that is required for the model to work correctly is to use a modification to the reserves - namely the total production at time=infinity if expontial decline had continued for the whole life of the field.

0 10- The derivation of the formulae was done in four steps 1) By simple integration, derive a formula for NPV as a function of (amongst other things) well numbers.

2) Calculate the optimal number of wells by taking partial derivatives with respect to number of wells and abandonment time.

3) Show that the formula for the optimal number of wells can be approximated by a simpler expression.

4) Calculate the NPV for the approximation of the optimal number of wells.

Summarising assumptions to be used in the formal derivation:

1) 3) Let N be the number of wells drilled. All the wells start up at time t=0.

2) If the field were run for an infinite time, the total production would be R (the technically recoverable reserves), independent of the number of wells. The initial production rate per well is q, independent of the number of wells i.e. it is not affected by well spacing. The field oil production rate follows exponential decline Field oil production rate = initial rate x e' = N.q.c -al

5) The net oil price is a constant, L, after all taxes and deductions.

6) The net capital costs can be expressed as D + C.N.. all capital expenditure happens at time t=0. Net opex can be expressed as E.N per unit time. (The year is probably the most appropriate unit of time, but any unit can be used, providing it is the same for both E and q).

i. e.

Theorem 1 - Expressing NPV as a funútion of well numbers The NPV of a field run until abandonment can be expressed as

NPV = R.L q N Tab 'R)_ N.E.(1 - ct) - (C.N + D) 1 + R. ln(l + d) ln(l + d) 0 Proof NPV can be broken down into the component parts of the cash-flow NPV = NPV(revenue stream) + NPV(Opex) + NPV(Capex) where the NPV(Opex) and NPV(Capex) are, of course, negative.

Start by calculating the NPV of the revenue stream:By the assumption that there is exponential decline Oil production rate = N.q.e" = N.q.e -(q. NIR)t [Since, by the definition of technical reserves o R = F N.q.e t dt N.q e-01 N.q /a _ a 0 then a = N.q/R] Oil revenue per unit time = (production rate) x (net oil price) = N.L.q.e"" AiR By the definition of NPV NPV of revenue stream = (Since (1 + d)' = e In((], d)-') = e- t Irgi+d) Tab N.L.q.e- (q N, R -ln(l.d))t q. N + In(] + d)) R R.L + _RAn(I + d) --q-.N R.L R.In(l + d) q.N fTab Revenue per unit time dt 0, (1 + d) 1 Tab N R W1 d)) = f. N.L.q.e" 'dt - q.N N.L.q - (1 - e -(q N 'R - Ink'I d),j 1 ab) R + In(l + d)) - In(l + d) Tab - q N Tab 1 R) - e e - (] - a.eq N Tab 1 R) Looking at the Opex cash-flow Opex per unit time = -N.E R1 R. In(] + d) 1 + qN.) (] - (1 + d)- Tab.e-q N Tab R) 0 NPV of opex = 1],h N.E (1 + d)' N.E (1 -(1 -d)In(l + d) dt Tat, N.E.e """''dt fl N.E. (1 ot) In(l + d) InO -d l, ' 1 ab N.E.e N-E In(] + d) In(l + d) Looking at Capex, since all the capital expenditure is assumed to occur at time t=O, NPV of Capex = - (C.N + D) By adding NPV(Revenue), N- PV(Opex) and NPV(Capex), one obtains the desired formula. QED Theorem 2 - Number of wells a -,ivinp, the hiphest NPV Parl 1 - The number of wells giving the highest NPV, N,,p,, can be calculated (Iteratively) from the expressions Nopt = R. In(l + d) q.1 where E = and (x = (1 + d Tab and Tab = R In q.N E L q E + - (Y, - + -in( 1 d) + E (x E x a.E. In L.q E- E)( 4L q - C- 1 n(l + d) + IE a EE Part 11 - If L.q >- 2.[C.In(l+d) + E] (i.e. the well is reasonably profitable at first) and 0.2 (x (equates, for d = 8%, to a field life 20 years) then the approximation of setting E = 0 gives approximate value for the number of wells, Napprox within the following bounds

1.31 N,)p, >- Nappro\ N,)pt (1-e t Proof of Part 1 Consider "V as a function of well numbers and abandonment time. For a normally behaved function NPV is a maximum c-:> cNPV - c-TN P V = 0 ON c7Tab iNPV Tab such that = 0 (3Tab can be obtained by taking the partial derivatives or, more simply, by seeing that this occurs when revenue has dropped until it equals operating expenditure i.e. N.L.q. e -q.N.TabR = N.E -so -q.N.T.t,/R = In(E/(L.q)) so Tab = R/(q. N). In(L.q/E) (N.B Also, for Tab optimal, el.N.Tab/R = EI(L. q)) For well numbers, using Theorem 1, and writing out the expression in full (not using (X, for example) iNPV - 0 N JIN Lq 11 N.

+ In(l + d) N.Lq qN + In(] + d)) R + N.Lq (1 (1 + d) - Tab e- (q.Wah 1 R q.Tab qN + ln(I + d) R R Lq qN + ln(I + d) R (1 + d) Tab e --(qNTab/R)) E(]-(l+d) -fah In(l + d) (1 + d) - Tab e -(qNTabR)) E(]-(l+d) - Tab In(] + d) (1 (1 + d) - Tab e-(""T" /R)) q R -c Setting this partial derivative equal to zero, multiplying both sides of the resultant equation by (qN/R + In(l+d))2, then using (x as shorthand for (1 +d)-Tab, and the relationships that apply when T,,1, is optimal, e -q.N.Tal)'R = E/(L.q) et equation. N.B, This equation only applies when T, ,h is optimal.

- 'I C -D c, we get the following q.N + In(] + d) L.q. 1 - a. E)_S.N + In(] + d) 2. +C -qN.L.q.]-a E) R L.q R In(] + d) R L.q +In L.q).L.q.q.N + ln(I + d) a. E = 0 E R). L.q Cancelling terms and multiplying both sides by -1 gives C+ E. (1 - et)).-N + In(l + d) - + ci.E. In Lq f q.N +In(l+d) In(l+d).(L. q-a.IE)=0 In(l + d) R E R This can be considered to a quadratic equation, with the "x" term being (q.NIR + In(l +d)) and the other terms being as followsa = C + E.(] - a)/(In(] +d) b = cc. E. 1 n(L. q/E) c = An(l+d).(L.q - (x.E) We will proceed here in Part 1 of this proof to solve the quadratic equation. In Part 11, we will show that the "b" term has little effect, and can be ignored. (It is interesting to consider how the "b" term arose. Consider the NPV of the field. If the number of wells increases, then abandonment is brought forward, so the term representing the present value of the oil lost at abandonment is increased. This decreases the overall NPV, but as can be imagined, such effects are small. This will be proved later, in Part 11).

So, ignoring the negative solution, the solution of the quadratic equation is r -4a c - b X = 2a Re -arranging this gives -C b b x ac'I + - 4a c a.E. In L.q 1 Defining c by c = b E j- 4a.c V4,(C. In(] + d) + E - et.E),(L.q a.E) and expanding x, c and a gives q.N In(l + d). (L.q - (x.E) R. + In(l + d) = E.(1 a). (i + 1 - s) In(] + d) E) 7 - In(] + d) (L q - a '(v + 1 c) -E In(l + d) + E - (x.E Re -arranging this equation gives N = R (L.q (1, E) 1 - 2 + 1c)- 1 In(l + d).

q InO +)+E - (1 + d) + E - a.] Proof of Theorem 2, Part 11 As a first step, it is useful to note that, for F-:k 0 (which is the case, providing L q E i.e. Year 1 net revenue for a well is greater than the opex for the well) 1 -t ((F-2+1) - c) I -F- hence, Napprox Nopt + 1 -c 2 E2 + 1 + 52 j = ú [Proof + 1 - +1-2 + 2E;.(F, - since J + 1 F- 1 Examining F,2 and dividing both the denominator and quotient by (Lq)2 gives 2 (In (L.q T# a. 2 (E,L.q)2 4 C. I n(l d) - (1 - a).E aE 1 L.q 1L.q I Since E: C. In(] + d) + E -< 0.5 (by assumption 1) and (x - 0.2 (by assumption 2) L.q L.q (xE 1 - L.q 1 - 0.20.5 = 0.9 Also C. In(l + d) + (1 - et).E > (1 ot). E L.q L.q L.q E > 0.8 Hence, E 2 < (In (L.q,E))2 0.2 2 (E T.J_ - 4xO.8x(E,,L.q0.9 0.014(1n(L.q/E))".

E L.q It can be easily shown that for 1 E/(L.q) 0, the maximum value of (In(L.q/E))2. E/(L. q) is achieved when E/(L.q) 1/(e2) = 0. 1 3353, which gives (In(L.q1E))2 E/(L.q) = 0.541. [Proof - Differentiate x.(In(x))2 and set to zero]. Hence F,:5 0.014 x 0.541 = 0.00757 E: 0.087 +E2) _ C) ( 1 _ F -0.087) = 0.93 (1 Before moving on to look at a lower limit for Nopi/Napprox, 'It is useful to establish a couple of small lemmas.

Lemma A For u, v, w such that u 2 v > 0 and v > w > 0, U-W U \11 - W v Pr oof U - W - U - uv - vw - UY + uw - (U - v) W > 0 V - W v V(v - W) v.(v W) Lemma B t 17- For w constant and greater than zero, the function f (y) increasing (i.e. Y1 < Y2!:::> f (Y1) < f (Y2)).

Proof Expressing f (y) as 1 - wy / (y- 1) gives df = - W +_ W.Y W- > 0 dy y - 1 (y - 1) (y - 1) = W-w)-Y-11 / (Y- I) is strictly Moving back to the main proof, let us establish a lower bound on N,,p,N,, pp,,,, Lq - a.E + Nopt In(l + d) + E - ci.E 'V1 > Napprox - 1 By Lemma A and assumption (1) Lq _ E Lq 2 V2 -E ln(I + d) + E In(l + d) + E (x E By Lemma B 7Lq (x E 0.93 C1 + d) + In( Lq - ci.E C -- Ln ( 1 -±-c)) --- E-- (x. E Hence N(,pt/N,pprox 0.76 or equivalently Pin(lL L 0.93 1 Cin(l+d+E,. (xE Jq - 1 E_ - - 1 1L _ 7:: - C.In(l+)+E aE 0.93V2 -1 = 0.76 -2 - 1 N,pp,,, 2 1. 3 1 Npt Combining this result with the result established at the beginning of the Part 11 of the proof gives Q. E. D.

1.31 N,,p, N,pp,., t N,,p, 0 Note - In order to express the expression for the approximate optimal number of wells in the simpler form:- E CA + E a.E Number of wells that maximises NPV = RA ' Fjq - q q 7t q - 0' it suffices to rearrange the expression slightly and to note that for normal values of d (the discount rate), In(l+d);t d (e.g. In(l+0.08) = 0. 077) Theorem 3 - NPV for a development with the approximately optimal number of wells The NPV of a development with the approximately optimal number of wells, as defined in Theorem 2, is NPV = R. L -1n(l + d) -+(' - a)E D q q where (x = (1 +d)-Tab and Tab = R / (N.q). In(L.q/E) Proof Combining the results &om Theorems 1 and 2 NPV = R. ln(l + d) 1 1 L.q -c x q C. In(l + d) + (1 - (x).E 'j L.q F-, ' - q. R. In(l + d) ( L.q - a.E 1) + In(] + d) R.q Iffl + d) + (1 - a).E - D (where = R. Va _ 1 [ L f 1 - a.E - -b -jb L - FJa 1 L.q a = L.q - (x.E and b = C.In(l+d) +(1-a).E) 1 - a.E E.0 -a) c L.q In(l + d) E(]-a) C.In(l+d) D q q li - 0 = R.' ( -Ja 7b E 5 In(] + d) + (1 - (x).E D E q q aA _ b D = R. 4a _ q Va q]- q Q.E.D.

- - D q] After these formal proofs, it is useful to look at how the method can be extended to deal with more complicated situations.

A simple modification to the Capex figures makes it possible to use the formulae when the company objective is to maximise overall return on capital employed.

The formulae as they stand show the number of wells required to maximise NPV. In the real world, oil companies are faced with the problem of constrants. In particular, the companies usually have only a fixed amount of capital available, or at least limits on the amount of capital employed (gearing limits etc). The problem then is to maximise N-PV, subject to the limits on the amount of capital employed. Assuming that there are sufficient number of projects available and no other constraints intrude (such as limits on management capacity), the method to do this is to rank all the projects in order of decreasing NPV.Capex ratio, and then choose the projects in this order until one has reached the limit on the amount of capital employed.

For the projects chosen, let H (the "hurdle" rate) be the lower limit to the NPV:Capex ratio i.e. the NPV-Capex ratio for the last project chosenJo determine, for a new project, the optimal number of wells, it is useful to consider the project as consisting of (Minimalist development olytion) + Series of increments to the mininalist development option. (Note that thesplit is purely conceptual; all the increments start at time All the increments can be analysed as if they were separate projects. They pass the screening criteria if their NPV:Capex ratio is greater than H 0 i e. 8NPV / 6Capex H So, while the NMCapex ratio for additional wells is greater than H, the wells are worth adding to the development scheme. The optimal total number of wells is reached when the limit is reached, so the criteria for determining the total number of wells is 6NI?V / 6Capex = H This can be converted into a more workable criteria as follows 6NI?V / 6Capex = H <::> 6(NPV-H.Capex) / 6Capex = 0 ≤> [6(NPV-H.Capex) / 5N1 x [6N / 6Capexl = 0,4=-> 6(NPV-H.Capex) / 6N = 0 amount of money to drill a new well).

where N is the number of wells since 8N / 6Capex:# 0 (it never costs an infinite Hence, the new problem (find the number of wells that maximises corporate NPV subject to a limit on capital employed) can be converted into the old (find the number of wells that maximises NPV) by using an artifical NPV, defined to be NW= NPV - H.Capex = NPV - R(C.N + D) and artificial C' and D' defined to be C=(] +FI)xC U=(1 +H)xD Then E Optimal number of wells, N = RA q q NPV with this number of wells= NPV'+ R(C.N D) = (the -H. D and +H.D terms cancel each other out).

r:+:-E - (- 7 (1 a E CA + E - a.E l q t ld+: E - j. E q 2 - + RGN - D t If a field is very large, so that the development is spread over several years, with different areas coming on stream in different years, then the situation can be modelled as follows.

Let NPV(N) be the NPV that would have been achieved if the development had been had carried out in the space of one year. Let M be the number of years that will be required. (Assume M to be independent of N, the number of wells). The new problem is to find N to maximise M NPY NPV(N) (, + d)-('--') = (1 + d) - (1 + d)-('-' MV(N) M d Clearly, the value of N that maximises NPV(N) also maximises "V' and the above equation gives NPV'. So the problem is solved.

In moving on to the problem of the economically optimal producer: i nj ector ratio for a water-flood, two separate cases will be consideredIf there are limits to bottom-hole flowing pressure for the producers, Pprod, the bottomhole injection pressure, Pinj, and there is a target pressure at which to maintain the average reservoir pressure, Pe. In this case, it will be shown that Optimal Producer: injector ratio = R(Pinj-Pe) / (PI.Bo. (Pe-Pprod) If there is flexibility about the average reservoir pressure, Pe, and the only constraints are the limits to bottom-hole flowing pressure for the producers, Pprod, the bottom-hole injection pressure, Pinj. In this case, Producer: Injector ratio giving the highest production = (11/PI.Bo) Economically optimal ratio = (Xi.11/(Xp.PI.B) where 11 is the average injectivity index (in reservoir volume), PI.Bo is the average productivity index (in reservoir volume), X1 is the annual net cost of running an injector and Xp is the average net cost of running a producer. 11 is assumed that PI does not change significantly with increased water cut.

In almost every water-flood, there will be a limit to Pprod, dictated by the need to keep the pressure in the near well-bore to above bubble-point. There may also be a higher t limit, e.g. the limit imposed by the need to keep ESP inlet pressures above (or at least not much below) the bubble-point, to avoid pump reliability problems. There will also be limits to bottom-hole injection pressures - perhaps dictated by pump limits, or by a desire not to exceed the fracture propagation pressure. In most cases, it will worthwhile to operate the wells at these limits otherwise, one could achieve higher production or get by with fewer injection wells.

Case (a) In some fields, where there is a possibility of causing crossflow, for example, it may be desirable to aim for a target average reservoir pressure, Pe. In this scenario (Case A), the pressure is stable when the injection rate (in reservoir volumes) equals the production rate (in reservoir volumes). Now, if m is the number of injectors and n is the number of producers, Reservoir injection rate = mll(Pinj-Pe) Reservoir production rate = nPI.Bo(Pprod-Pinj) So, mll(Pinj-Pe) = nPI.Bo(Pe-Pprod) Hence, if stability is to be achieved and the wells are all to operate at capacity, Producer injector ratio = n/m = II(Pinj-Pe) / [P1.1Bo.(PePprod)]

Case (b) technical optimum In case B, there is some flexibility about the choice of average reservoir pressure at which to operate the field. The field may be large in area, with relatively low permeability and thick shale barriers above and below. In such a case, increasing the field pressure to above the aquifer pressure will have few consequences in terms of oil pushed out into the aquifer, since the flow rates are small, and could be reduced further by arranging for there to be an outer ring of injectors. Hence, Pe can be chosen to give the optimal production rates. From the equation mll(Pinj-Pe) = nPI.Bo(Pe-Pprod) above, it can be seen that in the long term, assuming cross-flows and aquifer influxes are negligible, that Pe is determined by the injection-production pattern. Rearranging the equation gives

R (nPI.Bo + mll),Pe = mll.Pinj + n.PI.Bo.Pprod hence Pe = [mll.Pinj + n.PI. Bo.Pprodl / [n.PI.Bo + m.111 The oil production achieved is 011 production= n.PI.Bo.(Pe-Pprod).(1-WC) where WC is the water-cut Substituting the above expression for Pe gives 011 production = n.PI.Bo.( [m1I.Pinj + n.PI.Bo.Pprodl / [n.PI.Bo + mAll - Pprod). (IWC) [n.PI.Bo.(]-WC) / (n.PI.Bo + m.11)l x [m.11.Pinj + n.PI.Bo. Pprod n.PI.Bo.Pprod m.II.Pprod] = [n.PI.Bo.(1-WC) / (n.PI.Bo + m.ll)] x [m.ll.(Pinj-Pprod)l Multiplying both above and below the dividing sign by (m+n) - the total number of wells - gives 011 production = (Pinj-Pprod). (n+m).(1-WC).(n.PI.Bo.m.11) / [(n+m).(n.PI-Bo + m.11)l For a fixed total number of wells, fixed P1,11, Pprod, Pinj and fixed water-cut (at any one moment in time), the split of producers to injectors that gives the highest production is the one that gives the highest value of (n.PI.Bo.m.11) / [(n+m).(n.PI.Bo + mAI)l or, equivalently, the one that gives the lowest value of the inverse of the above expression i.e. the lowest value to (n+m).(n.PI.Bo + mAl) / (n.PI. Bo.m.II), which we will call cx Simplifying (x gives (n+m)/(m.11) + (n+ m)/(n.PI.Bo) Let r be the ratio of injectors.total number of wells i-e. r = m/(n+m). Since (I-r) n/(n+m), (x can be written as 1/(r.11) + 1/[(1-r). PI.Bol a is at a minimum when its derivative, da/dr, is zero N ow d(x/dr = - 1 /(. 11) + 1 /[(1 -r)2. pl.BO] r If -1/(r 2.11) + 1 /[(1 -r)2. PI.Bol = 0 then ( 1 -r)2/r2 = 11/P1.13o and ( 1 -r)/r = (1 1/P1. Bo) Since (I-r)/r = [n/(n+m)l / [m/(n+m)l = n/m we have shown that production is maximised, for a fixed number of wells when Producer: injector ratio = (11/PI.Bo) If injector wells and producer wells cost the same to construct and to operate, then this will also be the economic optimum.

Case (b) Economic optimum Using the formula for field NPV, we will show that the economically optimal ratio of producers to injectors is (Xi. IIJ(Xp.PI.Bo), where Xi and Xp are the annual costs of running an injector and a producer.

It has been shown above that, given a producer/injector mix that gives an overall initial oil production rate of q per well (i.e. q = (Field Production Rate)/(Number of Wells then the NPV of the development, using the best possible well density is NPV = Reserves x oil price x (1 - [(net opex per well + annuity value of net capex per well)/(Year 1 net revenue per well)])^2 - Net non-well-related capex. or NPV = R.L x (IA[(b+cq +CIn(l+d))/ql)12 - D splitting the Opex term, E, into a well-related element, b, and a gross fluid element, c.

Since R- L and D can be considered to be fixed, the NPV will be at its greatest, with respect to the choice of producer: injector ratio, whenever (b-cq +C.in(l--d))/q - call this 'P is minimised. Expressing b as [r.bi + (I-r).b.] and C.in(l+d) as [r.C, + (1-r).Cp] where b, is the well-related net opex for each injector 0 and Cj is the well-related annuity equivalent of the net capex for each injector and r is the proportion of injectors to total well stock, and (1 -r) is the proportion of producers to the total well stock T = [r,(bi+Q + (I-r).(bp+Cp) + c.q] lq Defining Xi = (bi+Ci) and Xp = (bp+Cp) and expressing q as (P]-Pprod).(1- r).PI.Bo.r.11 / [(I-r).M.Bo + r.111 gives T = [r.(bi+Ci) + (I-r).(bp+Cp)] X [(1-r).PI.Bo + r.111 / [(Pi-Pprod).(]- r).PI.Bo.r.11] + c Since PI, Pprod, P1.13o, 11 and c are fixed, this is minimised when [r.(bi+Q + (I-r).(bp+Cp)] x [(1r).PI.Bo + rAll / [(]-r).r] is minimised - call this 0 Now p = X1.11 / X + Xp.11 + XI.M.BO + Xp.PI.Bo..X where X is the producer. injector ratio, (I-r)/r So NPV will be at a maximum whenever 0 is at a minimurn i.e. when the derivative do/dy, is zero.

Now, when do/dX = -(X1AI)/X2 + Xp.PI.Bo = 0 X = (Xl.II/(Xp-PI.Bo)) i.e. The NPV is maximised when the producer: injector ratio is (Xi.11/(Xp. PI Bo)).

Q. E. D.

t The final stage is to evaluate the formulae. As an example, consider a field with the following technical and economical parametersTechnically recoverable reserves = 400,000,000 bbis 11 = 4 x P1.13o Net fixed capex = ú10,000,000 Net capex per producer = ú3,000,000 Net capex per injector = ú2,000,000 Net opex per producer per year ú200,000 Net opex per injector per year 100,000 Discount rate = 8% Net oil price = ú4 / bbl The oil company aims to maximise NPV subject to a limit on the total amount of capex, and has a cut-off of 15% for the Net NPV:Net Capex ratio.

To maximise the artificial NPV, the optimal producer: Injector ratio is given by (XI'.11/(Xp'.PI.Bo)) where M' = (]+hurdle) x Net capex x In (1 + discount rate) + Opex = 1. 15 x 2,000,000 x In(l. 08) x 100,000 = ú277,000 and, similarly, Xp' = ú466,000 So, optimal producer: injector ratio = (Xi'. 11/(Xp'.PI.Bo)) = (4 x 277,000 / 466,000) L54 In practice, a non-integral ratio would usually give a strange sweep pattern and so reduce reserves. Hence, the value of the calculation of the optimal producer injector ratio is to focus attention on the choice between a 1. 1 ratio or a 2:1 ratio - the calculation suggests that a 3:1 or a 1:2 ratio would be unlikely to be optimal. Let us assume here that the decision was taken to go for a seven-spot pattern, with producer: injector ratio of 2: 1, and the resultant parameters are Average initial production per well, q = 200, 000 bbl Average capex per well, C = ú2,700,000 Average artificial capex per well, C' = (I+H) x C = 1. 15 x 2,700,000 = ú3, 100,000 Average opex per well, E ú 170,000 0 -2 7- Feeding these into the formulae and calculating well numbers and abandonment time iteratively until there is convergence gives Optimal number of wells = 12.5 NPV = ú45,000,000 0 28 -

Claims (1)

1) A method of extracting hydrocarbons from underground accumulations, in which the number of wells to set to give the economically optimal results, the process comprising the steps of 1) 11) taking a hydrocarbon extraction economic model and simplifying the economic model to linear factors and discounting; creating simple economic models of operating expenditure and capital expenditure as functions of the number of wells; Ili) through decline analysis, modeling the expected field production profile as a function of well numbers, initial average well rates and technically recoverable reserves-, IV) combining the economic models and the production profile model to give a single algebraic expression for the present value of the expected cash-flow, V) algebraically integrating this expression to give net present value (NPV).

VI) calculating an expression for the partial derivative of NPV with respect to well numbers, vii) expressing the critefia for economic optimality in terms of the partial derivative of the NPV function with respect to well numbers-, Vill solving the equation derived from steps (v) and (vii) above to give an expression for the economically optimal number of wellsl 1x) X) xl) simplifying the expression for the economically optimal number of wells by taking terms that have little effect on the output and setting them to be constants, evaluating the simplified expression for the economically optimal number of wells for the field in question., drilling substantially the number of wells so determined.

A method as claimed in claim 1, in which, when the field is complicated, then reservoir simulation work is carried out to refine and confirm the optimal number of wells.

3) A method of exploiting a hydrocarbon field, comprising the steps of r i) ii) taking a hydrocarbon extraction economic model and simplifying the economic model to linear factors and discounting, creating simple economic models of operating expenditure and capital expenditure as functions of the number of wells-, 111) through decline analysis, modeling the expected field production profile as a function of well numbers, initial average well rates and technically recoverable reserves; iv) combining the economic models and the production profile model to give a single algebraic expression for the present value of the expected cash-flow; algebraically integrating this expression to give net present value (NIPV).1 calculating an expression for the partial derivative of NIPV with respect to well numbers. expressing the criteria for economic optimality in terms of the partial derivative of the NIPV function with respect to well numbers.

viii) solving the equation derived ftom steps (vi) and (vil) above to give an expression for the economically optimal number of wellsI ix) simplifying the expression for the economically optimal number of wells by v) vi) vii) taking terms that have little effect on the output and setting them to be constants, X) evaluating the simplified expression for the economically optimal number of wells for the field in question; xl) inserting the expression for the economically optimal number of wells into the expression for NPV from step (v); xii) algebraically manipulating the result to give a simple formula for NPV1 xiii) evaluating the expression for NPV for the field in question., xiv) drilling wells, or not, based on the NPV determined in step (xiii).

4) A method of extracting hydrocarbons from underground accumulations, comprising the steps of 1) taking a hydrocarbon extraction economic model and simplifying the economic model to linear factors and discounting., li) 1 i 1) creating simple economic models of operating expenditure and capital expenditure as functions of the number of wells; through decline analysis, modeling the expected field production profile as a function of well numbers, initial average well rates and technically recoverable reserves, iv) combining the economic models and the production profile model to give a single algebraic expression for the present value of the expected cash-flow; v) algebraically integrating this expression to give net present value (NPV), vO calculating an expression for the partial derivative of NPV with respect to well numbersI V11) expressing the criteria for economic optimality in terms of the partial derivative of the NPV function with respect to well numbers; viii) solving the equation derived from steps (vi) and (vil) above to give an expression for the economically optimal number of wells.

lx) simplifying the expression for the economically optimal number of wells by taking terms that have little effect on the output and setting them to be constants, evaluating the simplified expression for the economically optimal number of wells for the field in question; inserting the expression for the economically optimal number of wells into the expression for NPV from step (v); algebraically manipulating the result to give a simple formula for NPV, xiii) deriving, through a material balance argument, an expression for the expected oil production per well, averaged over both injectors and producers, as a function of the producer.. injector ratio, xiv) deriving expressions for operating expenditure per well and capital expenditure per well as functions of the producer: injector ratio, Xv) inserting the expressions derived in steps (xlli) and (xiv) into the formula for NPV determined in step (xii) to give an expression for NPV as a function of the producer: Injector ratiol XVI) calculating an expression for the partial derivative of NPV with respect to the producer: injector ratio, X) xi) xii) XVII) expressing the criteria for economic optimality in terms of the partial derivative of the NPV function with respect to the producer: injector ratio, xvili) solving the equation derived from steps (xvi) and (xvil) to give an expression for the economically optimal producer.. injector ratio, xlx) evaluating the expression for the economically optimal producer: injector ratio for the field in question, and drilling producer and injector wells in substantially the optimal ratio.

5) A method as claimed in any preceeding claim in which the hydrocarbon comprises oil.

6) A method as claimed in any preceeding claim in which the hydrocarbon comprises gas.