Arps Decline Curve Analysis in R

Description

Functions for Arps decline-curve analysis. Includes exponential, hyperbolic, harmonic, and hyperbolic-to-exponential models.

Details

Index:

Author(s)

Derrick W. Turk | terminus data science, LLC <dwt@terminusdatascience.com>

References

SPEE REP#6 (https://secure.spee.org/sites/default/files/wp-files/pdf/ReferencesResources/REP06-DeclineCurves.pdf)

Examples

## Plot semi-log rate-time and Cartesian rate-cumulative
## for a hyperbolic-to-exponential decline
t <- seq(0, 10, 0.5)
q <- hyp2exp.q(5000, as.nominal(0.90), 1.5, as.nominal(0.10), t)
Np <- hyp2exp.Np(5000, as.nominal(0.90), 1.5, as.nominal(0.10), t)

old.par <- par(ask=TRUE)
plot(log(q) ~ t)
plot(q ~ Np)
par(old.par)

Arps decline classes and S3 methods

Description

Create Arps decline curve objects and compute rates, cumulative production, and nominal declines.

Usage

arps.decline(qi, Di, b=NA, Df=NA)
## S3 method for class 'arps'
arps.q(decl, t)
## S3 method for class 'arps'
arps.Np(decl, t)
## S3 method for class 'arps'
arps.D(decl, t)

Arguments

Details

Depending on whether arguments b and Df are supplied, arps.decline will select an exponential, hyperbolic, or hyperbolic-to-exponential decline and return an object appropriately. The returned object will have class "exponential", "hyperbolic", or "hyp2exp" in addition to class "arps".

Assumes consistent units of time between qi, Di, Df, and t. To convert, see the decline-rate conversion functions referenced below.

Value

arps.decline returns an object having class "arps", suitable for use as an argument to S3 methods discussed here.

q.arps returns the rate for each element of t applying decline decl, in the same units as the value of qi for decl.

Np.arps returns the cumulative production for each element of t applying decline decl, in the same units as the value of qi * t for decl.

D.arps returns the nominal decline for each element of t applying decline decl, in the same units as the value of Di for decl.

See Also

print.arps, exponential, hyperbolic, hyp2exp, as.effective, as.nominal, rescale.by.time.

Examples

## exponential decline with
## qi = 1000 Mscf/d, Di = 95% effective / year
## rate for t from 0 to 25 days
decline <- arps.decline(1000,
    as.nominal(0.95, from.period="year", to.period="day"))
arps.q(decline, seq(0, 25))

## hyperbolic decline with
## qi = 500 bopd, Di = 3.91 nominal / year, b = 1.5,
## cumulative production at t = 5 years
decline <- arps.decline(
    rescale.by.time(500, from="day", to="year", method="rate"),
    3.91, 1.5)
arps.Np(decline, 5)

EUR and time-to-limit for Arps decline curves

Description

Evaluate estimated ultimate recovery and time to economic limit for Arps decline curve objects.

Usage

arps.eur(decl, q.limit)
arps.t.el(decl, q.limit)

Arguments

Value

arps.eur returns the total production for decl at the point in time when the economic limit q.limit is reached; that is, arps.Np(decl, arps.t.el(decl, q.limit)), in the same units as q.limit.

arps.t.el returns the time until the economic limit q.limit is reached for decline decl.

See Also

arps, curtail.

Examples

## exponential decline with
## qi = 1000 Mscf/d, Di = 95% effective / year
## EUR with economic limit 10 Mscf/d
decline <- arps.decline(1000,
    as.nominal(0.95, from.period="year", to.period="day"))
arps.eur(decline, 10)

## hyperbolic decline with
## qi = 500 bopd, Di = 3.91 nominal / year, b = 1.5,
## time to reach economic limit of 3 bopd
decline <- arps.decline(rescale.by.time(500, from="day", to="year"),
    3.91, 1.5)
arps.t.el(decline, rescale.by.time(3, from="day", to="year"))

Arps declines with linear buildup period

Description

Extend Arps decline curve objects by replacing early-time declines with a buildup period in which rate is a linear function of time.

Usage

    arps.with.buildup(decl, initial.rate, time.to.peak)

Arguments

Value

arps.with.buildup returns an object having class "arps", which may be used as an argument to methods such as arps.q, arps.Np, arps.D, or print.arps.

This object implements a decline curve which behaves as decl for all time greater than time.to.peak, but implements a linear buildup of rate interpolated between initial.rate at time zero and arps.q(decl, time.to.peak) at time.to.peak.

See Also

arps.decline

Examples

## hyperbolic decline with
## qi = 500 bopd, Di = 3.91 nominal / year, b = 1.5,
## cumulative production at t = 5 years
decline <- arps.decline(
    rescale.by.time(500, from="day", to="year", method="rate"),
    3.91, 1.5)
# add buildup from initial rate of 50 bopd, over 30 days
decline.with.buildup <- arps.with.buildup(decline,
    rescale.by.time(50, from="day", to="year", method="rate"),
    rescale.by.time(30, from="day", to="year", method="time"))
# forecast 5 years and compare
forecast.time <- seq(0, 5, 0.1)
plot(arps.q(decline, forecast.time) ~ forecast.time, log="y", type="l",
    lty="dashed", col="red")
lines(arps.q(decline.with.buildup, forecast.time) ~ forecast.time,
    lty="dotted", col="blue")

Arps decline conversion from nominal to effective

Description

Convert nominal to effective Arps decline.

Usage

as.effective(D.nom,
    from.period=c("year", "month", "day"),
    to.period=c("year", "month", "day"))

Arguments

Details

The result should be applied as a tangent effective decline (see SPEE REP#6 [https://secure.spee.org/sites/default/files/wp-files/pdf/ReferencesResources/REP06-DeclineCurves.pdf]) specified in fractional (i.e. 95% = 0.95) units.

When appropriate, internally uses rescale.by.time to perform time unit conversion.

Value

Returns the tangent effective Arps decline rate in units of [1 / to.period].

See Also

as.nominal, rescale.by.time.

Examples

## 0.008 nominal daily decline to tangent effective yearly decline
as.effective(0.008, from.period="day", to.period="year")

Arps decline conversion from effective to nominal

Description

Convert effective to nominal Arps decline.

Usage

as.nominal(D.eff,
    from.period=c("year", "month", "day"),
    to.period=c("year", "month", "day"))

Arguments

Details

D.eff should be specified in fractional (i.e. 95% = 0.95) units as a tangent effective decline (see SPEE REP#6 [https://secure.spee.org/sites/default/files/wp-files/pdf/ReferencesResources/REP06-DeclineCurves.pdf]).

When appropriate, internally uses rescale.by.time to perform time unit conversion.

Value

Returns the Arps nominal decline rate in units of [1 / to.period].

See Also

as.effective, rescale.by.time.

Examples

## 95% / year effective decline to nominal daily decline
as.nominal(0.95, from.period="year", to.period="day")

Best-fitting of Arps decline curves

Description

Perform best-fits of Arps decline curves to rate or cumulative data.

Usage

best.exponential(q, t,
  lower=c( # lower bounds
    0, # qi > 0
    0), # D > 0
  upper=c( # upper bounds
    max(q) * 5, # qi < qmax * 5
    10) # = 0.99995 / [time] effective
  )

best.hyperbolic(q, t,
  lower=c( # lower bounds
    0,  # qi > 0
    0,  # Di > 0
    0), # b > 0
  upper=c( # upper bounds
    max(q) * 5, # qi < qmax * 5
    10, # = 0.99995 / [time] effective
    2)  # b <= 2.0
  )

best.hyp2exp(q, t,
  lower=c( # lower bounds
    0,  # qi > 0
    0.35,  # Di > 0
    0,  # b > 0
    0), # Df > 0
  upper=c( # upper bounds
    max(q) * 5, # qi < qmax * 5
    10, # = 0.99995 / [time] effective
    2,  # b <= 2.0
    0.35) # Df <= 0.35
  )

best.exponential.curtailed(q, t,
  lower=c( # lower bounds
    0, # qi > 0
    0, # D > 0
    0  # t.curtail > 0
  ),
  upper=c( # upper bounds
    max(q) * 5, # qi < qmax * 5
    10, # = 0.99995 / [time] effective
    t[length(t)])
  )

best.hyperbolic.curtailed(q, t,
  lower=c( # lower bounds
    0,  # qi > 0
    0,  # Di > 0
    0,  # b > 0
    0   # t.curtail > 0
  ),
  upper=c( # upper bounds
    max(q) * 5, # qi < qmax * 5
    10, # = 0.99995 / [time] effective
    2,  # b <= 2.0
    t[length(t)])
  )

best.hyp2exp.curtailed(q, t,
  lower=c( # lower bounds
    0,  # qi > 0
    0.35,  # Di > 0
    0,  # b > 0
    0,  # Df > 0
    0   # t.curtail > 0
  ),
  upper=c( # upper bounds
    max(q) * 5, # qi < qmax * 5
    10, # = 0.99995 / [time] effective
    2,  # b <= 2.0
    0.35, # Df <= 0.35
    t[length(t)])
  )

best.fit(q, t)

best.curtailed.fit(q, t)

best.exponential.from.Np(Np, t,
  lower=c( # lower bounds
    0, # qi > 0
    0), # D > 0
  upper=c( # upper bounds
    max(c(Np[1], diff(Np)) / diff(c(0, t))) * 5, # qi < max(rate) * 5
    10) # = 0.99995 / [time] effective)
  )

best.exponential.from.interval(volume, t, t.begin=0.0,
  lower=c( # lower bounds
    0, # qi > 0
    0), # D > 0
  upper=c( # upper bounds
    max(volume / diff(c(t.begin, t))) * 5, # qi < max(rate) * 5
    10) # = 0.99995 / [time] effective)
  )

best.hyperbolic.from.Np(Np, t,
  lower=c( # lower bounds
    0,  # qi > 0
    0,  # Di > 0
    0), # b > 0
  upper=c( # upper bounds
    max(c(Np[1], diff(Np)) / diff(c(0, t))) * 5, # qi < max(rate) * 5
    10, # = 0.99995 / [time] effective
    2)  # b <= 2.0
  )

best.hyperbolic.from.interval(volume, t, t.begin=0.0,
  lower=c( # lower bounds
    0,  # qi > 0
    0,  # Di > 0
    0), # b > 0
  upper=c( # upper bounds
    max(volume / diff(c(t.begin, t))) * 5, # qi < max(rate) * 5
    10, # = 0.99995 / [time] effective
    2)  # b <= 2.0
  )

best.hyp2exp.from.Np(Np, t,
  lower=c( # lower bounds
    0,  # qi > 0
    0.35,  # Di > 0
    0,  # b > 0
    0), # Df > 0
  upper=c( # upper bounds
    max(c(Np[1], diff(Np)) / diff(c(0, t))) * 5, # qi < max(rate) * 5
    10, # = 0.99995 / [time] effective
    5,  # b <= 2.0
    0.35) # Df <= 0.35
  )

best.hyp2exp.from.interval(volume, t, t.begin=0.0,
  lower=c( # lower bounds
    0,  # qi > 0
    0.35,  # Di > 0
    0,  # b > 0
    0), # Df > 0
  upper=c( # upper bounds
    max(volume / diff(c(t.begin, t))) * 5, # qi < max(rate) * 5
    10, # = 0.99995 / [time] effective
    5,  # b <= 2.0
    0.35) # Df <= 0.35
  )

best.exponential.curtailed.from.Np(Np, t,
  lower=c( # lower bounds
    0, # qi > 0
    0, # D > 0
    0  # t.curtail > 0
  ),
  upper=c( # upper bounds
    max(c(Np[1], diff(Np)) / diff(c(0, t))) * 5, # qi < max(rate) * 5
    10, # = 0.99995 / [time] effective
    t[length(t)])
  )

best.exponential.curtailed.from.interval(volume, t, t.begin=0.0,
  lower=c( # lower bounds
    0, # qi > 0
    0, # D > 0
    0  # t.curtail > 0
  ),
  upper=c( # upper bounds
    max(volume / diff(c(t.begin, t))) * 5, # qi < max(rate) * 5
    10, # = 0.99995 / [time] effective
    t[length(t)])
  )

best.hyperbolic.curtailed.from.Np(Np, t,
  lower=c( # lower bounds
    0,  # qi > 0
    0,  # Di > 0
    0,  # b > 0
    0   # t.curtail > 0
  ),
  upper=c( # upper bounds
    max(c(Np[1], diff(Np)) / diff(c(0, t))) * 5, # qi < max(rate) * 5
    10, # = 0.99995 / [time] effective
    5,  # b <= 2.0
    t[length(t)])
  )

best.hyperbolic.curtailed.from.interval(volume, t, t.begin=0.0,
  lower=c( # lower bounds
    0,  # qi > 0
    0,  # Di > 0
    0,  # b > 0
    0   # t.curtail > 0
  ),
  upper=c( # upper bounds
    max(volume / diff(c(t.begin, t))) * 5, # qi < max(rate) * 5
    10, # = 0.99995 / [time] effective
    5,  # b <= 2.0
    t[length(t)])
  )

best.hyp2exp.curtailed.from.Np(Np, t,
  lower=c( # lower bounds
    0,  # qi > 0
    0.35,  # Di > 0
    0,  # b > 0
    0,  # Df > 0
    0
  ),
  upper=c( # upper bounds
    max(c(Np[1], diff(Np)) / diff(c(0, t))) * 5, # qi < max(rate) * 5
    10, # = 0.99995 / [time] effective
    5,  # b <= 2.0
    0.35, # Df <= 0.35
    t[length(t)])
  )

best.hyp2exp.curtailed.from.interval(volume, t, t.begin=0.0,
  lower=c( # lower bounds
    0,  # qi > 0
    0.35,  # Di > 0
    0,  # b > 0
    0,  # Df > 0
    0
  ),
  upper=c( # upper bounds
    max(volume / diff(c(t.begin, t))) * 5, # qi < max(rate) * 5
    10, # = 0.99995 / [time] effective
    5,  # b <= 2.0
    0.35, # Df <= 0.35
    t[length(t)])
  )

best.fit.from.Np(Np, t)

best.fit.from.interval(volume, t, t.begin=0.0)

best.curtailed.fit.from.Np(Np, t)

best.curtailed.fit.from.interval(volume, t, t.begin=0.0)

best.exponential.with.buildup(q, t,
  lower=c( # lower bounds
    0,  # qi > 0
    0), # D > 0
  upper=c( # upper bounds
    max(q) * 5, # qi < qmax * 5
    10),        # = 0.99995 / [time] effective
  initial.rate=q[1], time.to.peak=t[which.max(q)])

best.hyperbolic.with.buildup(q, t,
  lower=c( # lower bounds
    0,  # qi > 0
    0,  # Di > 0
    0), # b > 0
  upper=c( # upper bounds
    max(q) * 5, # qi < qmax * 5
    10, # = 0.99995 / [time] effective
    2), # b <= 2.0
  initial.rate=q[1], time.to.peak=t[which.max(q)])

best.hyp2exp.with.buildup(q, t,
  lower=c( # lower bounds
    0,  # qi > 0
    0.35,  # Di > 0
    0,  # b > 0
    0), # Df > 0
  upper=c( # upper bounds
    max(q) * 5, # qi < qmax * 5
    10, # = 0.99995 / [time] effective
    2,  # b <= 2.0
    0.35), # Df <= 0.35
  initial.rate=q[1], time.to.peak=t[which.max(q)])

best.fit.with.buildup(q, t)

best.exponential.from.Np.with.buildup(Np, t,
  lower=c( # lower bounds
    0, # qi > 0
    0), # D > 0
  upper=c( # upper bounds
    max(c(Np[1], diff(Np)) / diff(c(0, t))) * 5, # qi < max(rate) * 5
    10), # = 0.99995 / [time] effective
  initial.rate=Np[1] / t[1],
  time.to.peak=(t[which.max(diff(Np))] + t[which.max(diff(Np)) + 1]) / 2.0)

best.exponential.from.interval.with.buildup(volume, t, t.begin=0.0,
  lower=c( # lower bounds
    0, # qi > 0
    0), # D > 0
  upper=c( # upper bounds
    max(volume / diff(c(t.begin, t))) * 5, # qi < max(rate) * 5
    10), # = 0.99995 / [time] effective
  initial.rate=volume[1] / (t[1] - t.begin),
  time.to.peak=(t - diff(c(t.begin, t)) / 2)[which.max(volume)])

best.hyperbolic.from.Np.with.buildup(Np, t,
  lower=c( # lower bounds
    0,  # qi > 0
    0,  # Di > 0
    0), # b > 0
  upper=c( # upper bounds
    max(c(Np[1], diff(Np)) / diff(c(0, t))) * 5, # qi < max(rate) * 5
    10, # = 0.99995 / [time] effective
    2), # b <= 2.0
  initial.rate=Np[1] / t[1],
  time.to.peak=(t[which.max(diff(Np))] + t[which.max(diff(Np)) + 1]) / 2.0)

best.hyperbolic.from.interval.with.buildup(volume, t, t.begin=0.0,
  lower=c( # lower bounds
    0,  # qi > 0
    0,  # Di > 0
    0), # b > 0
  upper=c( # upper bounds
    max(volume / diff(c(t.begin, t))) * 5, # qi < max(rate) * 5
    10, # = 0.99995 / [time] effective
    2), # b <= 2.0
  initial.rate=volume[1] / (t[1] - t.begin),
  time.to.peak=(t - diff(c(t.begin, t)) / 2)[which.max(volume)])

best.hyp2exp.from.Np.with.buildup(Np, t,
  lower=c( # lower bounds
    0,  # qi > 0
    0.35,  # Di > 0
    0,  # b > 0
    0), # Df > 0
  upper=c( # upper bounds
    max(c(Np[1], diff(Np)) / diff(c(0, t))) * 5, # qi < max(rate) * 5
    10, # = 0.99995 / [time] effective
    5,  # b <= 2.0
    0.35), # Df <= 0.35
  initial.rate=Np[1] / t[1],
  time.to.peak=(t[which.max(diff(Np))] + t[which.max(diff(Np)) + 1]) / 2.0)

best.hyp2exp.from.interval.with.buildup(volume, t, t.begin=0.0,
  lower=c( # lower bounds
    0,  # qi > 0
    0.35,  # Di > 0
    0,  # b > 0
    0), # Df > 0
  upper=c( # upper bounds
    max(volume / diff(c(t.begin, t))) * 5, # qi < max(rate) * 5
    10, # = 0.99995 / [time] effective
    5,  # b <= 2.0
    0.35), # Df <= 0.35
  initial.rate=volume[1] / (t[1] - t.begin),
  time.to.peak=(t - diff(c(t.begin, t)) / 2)[which.max(volume)])

best.fit.from.Np.with.buildup(Np, t)

best.fit.from.interval.with.buildup(volume, t, t.begin=0.0)

Arguments

Details

Best-fitting is carried out by minimizing the sum of squared error in the rate or cumulative forecast, using nlminb as the optimizer.

Appropriate bounds are applied to decline-curve parameters by default, but may be altered using the lower and upper arguments to each specific function.

Value

best.exponential, best.hyperbolic, and best.hyp2exp return objects of the appropriate class (as from arps.decline) representing best fits of the appropriate type against q and t, in the same units as q and t.

best.fit returns the best overall fit, considering results from each function above.

best.exponential.from.Np, best.hyperbolic.from.Np, and best.hyp2exp.from.Np return objects of the appropriate class (as from arps.decline) representing best fits of the appropriate type against Np and t, in the same units as Np and t.

best.fit.from.Np returns the best overall fit, considering results from each function above.

best.exponential.from.interval, best.hyperbolic.from.interval, and best.hyp2exp.from.interval return objects of the appropriate class (as from arps.decline) representing best fits of the appropriate type against volume and t, in the same units as volume and t.

For these functions, t is taken to represent the time at the end of each producing interval; the beginning time for the first interval may be specified as t.begin if it is non-zero.

best.fit.from.interval returns the best overall fit, considering results from each function above.

best.exponential.curtailed, best.hyperbolic.curtailed, best.hyp2exp.curtailed, best.curtailed.fit, best.exponential.curtailed.from.Np, best.hyperbolic.curtailed.from.Np, best.hyp2exp.curtailed.from.Np, best.curtailed.fit.from.Np, best.exponential.curtailed.from.interval, best.hyperbolic.curtailed.from.interval, best.hyp2exp.curtailed.from.interval, and best.curtailed.fit.from.interval work as the corresponding functions above, but may return curtailed declines (as from curtail).

best.exponential.with.buildup, best.hyperbolic.with.buildup, best.hyp2exp.with.buildup, best.fit.with.buildup, best.exponential.from.Np.with.buildup, best.hyperbolic.from.Np.with.buildup, best.hyp2exp.from.Np.with.buildup, best.fit.from.Np.with.buildup, best.exponential.from.interval.with.buildup, best.hyperbolic.from.interval.with.buildup, best.hyp2exp.from.interval.with.buildup, and best.fit.from.interval.with.buildup work as the corresponding functions above, but will return a fit including a linear buildup portion (as from arps.with.buildup).

See Also

arps, curtailed, arps.with.buildup, nlminb

Examples

fitme.hyp2exp.t <- seq(0, 5, 1 / 12) # 5 years
fitme.hyp2exp.q <- hyp2exp.q(
    1000, # Bbl/d
    as.nominal(0.70), # / year
    1.9,
    as.nominal(0.15), # / year
    fitme.hyp2exp.t
) * rnorm(n=length(fitme.hyp2exp.t), mean=1, sd=0.1) # perturb

hyp2exp.fit <- best.hyp2exp(fitme.hyp2exp.q, fitme.hyp2exp.t)
cat(paste("SSE:", hyp2exp.fit$sse))
dev.new()
plot(fitme.hyp2exp.q ~ fitme.hyp2exp.t, main="Hyperbolic-to-Exponential Fit",
     col="blue", log="y", xlab="Time", ylab="Rate")
lines(arps.q(hyp2exp.fit$decline, fitme.hyp2exp.t) ~ fitme.hyp2exp.t,
      col="red")
legend("topright", pch=c(1, NA), lty=c(NA, 1), col=c("blue", "red"), legend=c("Actual", "Fit"))

Arps decline curves with initial curtailment

Description

Create decline curve objects and compute rates, cumulative production, and nominal declines for Arps decline curves with an initial period of curtailment to constant rate. The resulting decline curve models will impose a constant rate of production equal to the initial rate of the underlying model until the specified end-of-curtailment time.

Usage

    curtail(decl, t.curtail)
    curtailed.q(decl, t.curtail, t)
    curtailed.Np(decl, t.curtail, t)
    curtailed.D(decl, t.curtail, t)

Arguments

Details

Assumes consistent units as described for underlying Arps decline types.

Value

curtail returns an object having class "arps", suitable for use as an argument to S3 methods listed in arps.

curtailed.q returns the rate for each element of t, under a decline described by decl and curtailed until time t.curtail.

curtailed.Np returns the cumulative production for each element of t, under a decline described by decl and curtailed until time t.curtail.

curtailed.D returns the nominal instantaneous decline for each element of t, under a decline described by decl and curtailed until time t.curtail.

See Also

arps, print.arps, exponential, hyperbolic, hyp2exp.

Examples

## qi = 1000 Mscf/d, Di = 95% effective / year, b = 1.2, t from 0 to 25 days,
## curtailed until 5 days
curtailed.q(arps.decline(
        1000,
        as.nominal(0.95, from.period="year", to.period="day"),
        1.2
    ),
    5, seq(0, 25))

## hyperbolic decline with
## qi = 500 bopd, Di = 3.91 nominal / year, b = 1.5,
## curtailed for 1 month
## cumulative production at t = 5 years
decline <- curtail(
    arps.decline(rescale.by.time(500, from="day", to="year"),
                 3.91, 1.5),
    (1 / 12)
)
arps.Np(decline, 5)

Arps exponential declines

Description

Compute rates and cumulative production values for Arps exponential decline curves.

Usage

exponential.q(qi, D, t)
exponential.Np(qi, D, t)

Arguments

Details

Assumes consistent units of time between qi, D, and t. To convert, see the decline-rate conversion functions referenced below.

Value

exponential.q returns the rate for each element of t, in the same units as qi.

exponential.Np returns the cumulative production for each element of t, in the same units as qi * t.

See Also

as.effective, as.nominal, rescale.by.time.

Examples

## qi = 1000 Mscf/d, Di = 95% effective / year, t from 0 to 25 days
exponential.q(1000, as.nominal(0.95, from.period="year", to.period="day"), seq(0, 25))

## qi = 500 bopd, Di = 3.91 nominal / year, t = 5 years
exponential.Np(rescale.by.time(500, from.period="day", to.period="year"), 3.91, 5)

Format methods for Arps decline objects

Description

Get human-readable representation of Arps decline-curve objects.

Usage

## S3 method for class 'arps'
format(x, ...)

Arguments

Value

A character representation of x.

See Also

format, print.arps, arps.decline.

Examples

## exponential decline with
## qi = 1000 Mscf/d, Di = 95% effective / year
## rate for t from 0 to 25 days
decline <- arps.decline(1000,
    as.nominal(0.95, from.period="year", to.period="day"))
format(decline, digits=2)

Arps harmonic declines

Description

Compute rates, cumulative production values, and instantaneous nominal declines for Arps harmonic decline curves (i.e. hyperbolic with b = 1).

Usage

harmonic.q(qi, Di, t)
harmonic.Np(qi, Di, t)
harmonic.D(Di, t)

Arguments

Details

Assumes consistent units of time between qi, D, and t. To convert, see the decline-rate conversion functions referenced below.

Value

harmonic.q returns the rate for each element of t, in the same units as qi.

harmonic.Np returns the cumulative production for each element of t, in the same units as qi * t.

harmonic.D returns the nominal instantaneous decline for each element of t. This can be converted to effective decline and rescaled in time by use of as.effective and rescale.by.time.

See Also

as.effective, as.nominal, rescale.by.time.

Examples

## qi = 1000 Mscf/d, Di = 95% effective / year, t from 0 to 25 days
harmonic.q(1000, as.nominal(0.95, from.period="year", to.period="day"), seq(0, 25))

## qi = 500 bopd, Di = 3.91 nominal / year, t = 5 years
harmonic.Np(rescale.by.time(500, from.period="day", to.period="year"), 3.91, 5)

## Di = 85% effective / year, t = 6 months
harmonic.D(as.nominal(0.85), 0.5)

Arps hyperbolic-to-exponential declines

Description

Compute rates, cumulative production values, instantaneous nominal declines, and transition times for Arps hyperbolic-to-exponential decline curves.

Usage

hyp2exp.q(qi, Di, b, Df, t)
hyp2exp.Np(qi, Di, b, Df, t)
hyp2exp.D(Di, b, Df, t)
hyp2exp.transition(Di, b, Df)

Arguments

Details

Assumes consistent units of time between qi, Di, Df, and t. To convert, see the decline-rate conversion functions referenced below.

When appropriate, internally uses harmonic.q and harmonic.Np to avoid singularities in calculations for b near 1.

Value

hyp2exp.q returns the rate for each element of t, in the same units as qi.

hyp2exp.Np returns the cumulative production for each element of t, in the same units as qi * t.

hyp2exp.D returns the nominal instantaneous decline for each element of t. This can be converted to effective decline and rescaled in time by use of as.effective and rescale.by.time.

hyp2exp.transition returns the transition time (from hyperbolic to exponential decline), in the same units as 1 / Di.

See Also

as.effective, as.nominal, rescale.by.time.

Examples

## qi = 1000 Mscf/d, Di = 95% effective / year, b = 1.2,
## Df = 15% effective / year, t from 0 to 25 years
## result in Mscf/d
hyp2exp.q(1000, as.nominal(0.95), 1.2, as.nominal(0.15), seq(0, 25))

## qi = 500 bopd, Di = 3.91 nominal / year, b = 0.5,
## Df = 0.3 nominal / year, t = 20 years
hyp2exp.Np(rescale.by.time(500, from.period="day", to.period="year"),
    3.91, 0.5, 0.3, 20)

## Di = 85% effective / year, b = 1.5, Df = 15% effective / year,
## t = 6 and 48 months
hyp2exp.D(as.nominal(0.85), 1.5, as.nominal(0.15), c(0.5, 4))

## Di = 85% effective / year, b = 1.5, Df = 5% effective / year
hyp2exp.transition(as.nominal(0.85), 1.5, as.nominal(0.05))

Arps hyperbolic declines

Description

Compute rates, cumulative production values, and instantaneous nominal declines for Arps hyperbolic decline curves.

Usage

hyperbolic.q(qi, Di, b, t)
hyperbolic.Np(qi, Di, b, t)
hyperbolic.D(Di, b, t)

Arguments

Details

Assumes consistent units of time between qi, D, and t. To convert, see the decline-rate conversion functions referenced below.

When appropriate, internally uses harmonic.q and harmonic.Np to avoid singularities in calculations for b near 1.

Value

hyperbolic.q returns the rate for each element of t, in the same units as qi.

hyperbolic.Np returns the cumulative production for each element of t, in the same units as qi * t.

hyperbolic.D returns the nominal instantaneous decline for each element of t. This can be converted to effective decline and rescaled in time by use of as.effective and rescale.by.time.

See Also

as.effective, as.nominal, rescale.by.time.

Examples

## qi = 1000 Mscf/d, Di = 95% effective / year, b = 1.2, t from 0 to 25 days
hyperbolic.q(1000, as.nominal(0.95, from.period="year", to.period="day"),
    1.2, seq(0, 25))

## qi = 500 bopd, Di = 3.91 nominal / year, b = 0.5, t = 5 years
hyperbolic.Np(rescale.by.time(500, from.period="day", to.period="year"),
    3.91, 0.5, 5)

## Di = 85% effective / year, b = 1.5, t = 6 months
hyperbolic.D(as.nominal(0.85), 1.5, 0.5)

Print methods for Arps decline objects

Description

Print human-readable representation of Arps decline-curve objects using format.arps.

Usage

## S3 method for class 'arps'
print(x, ...)

Arguments

Value

Invisibly (see invisible) returns x.

See Also

print, format.arps, arps.decline.

Examples

## exponential decline with
## qi = 1000 Mscf/d, Di = 95% effective / year
## rate for t from 0 to 25 days
decline <- arps.decline(1000,
    as.nominal(0.95, from.period="year", to.period="day"))
print(decline)

Time unit conversion for DCA

Description

Scales rates, declines, and time periods from one time unit to another.

Usage

rescale.by.time(value,
    from.period=c("year", "month", "day"),
    to.period=c("year", "month", "day"),
    method=c("decline", "rate", "time"))

Arguments

Value

Returns value scaled from from.period to to.period according to its type as specified by method.

See Also

as.nominal, as.effective.

Examples

## 3 MMscf/D to MMscf/year
rescale.by.time(3, from.period="day", to.period="year", method="rate")

## Nominal decline of 3.2/year to nominal decline per month
rescale.by.time(3.2, from.period="year", to.period="month", method="decline")

## 5 years in days
rescale.by.time(5, from.period="year", to.period="month", method="time")