Arps Exponential Decline Curve (Well Rate over Time)

An exponential decline means the well loses the same fraction of its rate in every unit of time, so the rate falls on a straight line when plotted on a log scale. This is the simplest of Arps' three decline curves, the limit $b = 0$ of the hyperbolic curve. It fits a well in boundary-dominated flow with a fixed bottom-hole pressure, and it is the tail on which long hyperbolic forecasts are usually finished. The same curve describes radioactive decay and any other first-order loss; this page keeps the oil and gas names.

The decline rate is the nominal one. $D_i$ is the continuous rate that sits in the exponent. It is not the effective rate that operators often quote, the fraction lost over one year, which is smaller. An effective decline of 30 per cent a year means the rate is multiplied by 0.70 over the year, and 0.70 is $e$ to the power of minus 0.357, so $D_i = -\ln(0.70)$ = 0.3566749 per year. Time and the decline rate must be in the same unit.

Worked examples. A well at 1 000 barrels a day with $D_i$ = 0.357 per year is at 700 after one year and 490 after two (1 000 × 0.7²). Run backward, a fall from 1 000 to 700 in a year gives $D_i = \ln(q_i/q)/t$ = 0.3566749 per year. The page's own example sets $D_i t$ = 1: 1 000 bbl/d at 0.5 per year for 2 years leaves $q_i e^{-1}$, which is 367.88 bbl/d ($e^{-1}$ = 0.3678794). A rate that halves every 2 years has $D_i$ = ln 2 / 2 per year.

Source and limits. Arps (1945), Analysis of decline curves, gave the exponential, hyperbolic and harmonic curves. Fetkovich (1980) prints the rate-time equations: $q/q_i = 1/(1 + b D_i t)^{1/b}$, which is $e^{-D_i t}$ for $b = 0$ and $1/(1 + D_i t)$ for $b = 1$. The choice of curve matters for the forecast: in Fetkovich's worked case, at a limit of 20 barrels a month, the life is 285 months on the exponential curve against 1 480 months on the harmonic. If the real decline rate slows with time, use the hyperbolic page instead.