Engineering lesson
Stretched exponential Duong and Arps choosing a decline model
Arps, stretched exponential and Duong forms describe different empirical curve shapes and time behavior. Model choice needs the flow regime, fitting interval and prediction checks, not just the smallest in-sample residual.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Match definitions before comparing parameters
Arps uses an initial rate, nominal decline and b. Stretched exponential introduces a time scale and exponent. Duong's time-dependent form has its own coefficients and domain. Their parameters are not interchangeable.
The Duong calculator on this site uses its documented daily time basis. Copying a coefficient fitted with another time definition changes the model.
Evaluate prediction outside the fitting interval
A flexible curve can fit a short production segment well while giving an implausible tail. Hold out later observations when the goal is forecasting.
The numerical panel uses matched Arps branches to demonstrate how shape alone changes a tail. It is not presented as a fitted three-family comparison or an accuracy ranking of Duong and stretched exponential.
Use physical and operational context
Boundary-dominated assumptions, transient behavior, shut-ins and changing controls affect interpretation. A named empirical equation does not identify those conditions automatically.
Choose a finite horizon and document any terminal decline rule for each model. Compare both rate and cumulative on the same basis.
Explore the example
Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
Select one of 5 recorded scenarios. Intermediate states are not interpolated.
Elapsed time (day) = 365. Exponential rate (STB/day): 694.196651; Hyperbolic b=0.5 rate (STB/day): 715.150919; Harmonic rate (STB/day): 732.600733.
- Exponential rate (STB/day)
- Hyperbolic b=0.5 rate (STB/day)
- Harmonic rate (STB/day)
Worked example
Matched initial rate and nominal decline
qi = 1,000 STB/day and Di = 0.001/day for all curves. Only b changes among the three model columns.
Relation and unit basis
q = qi exp(−Dit) for b = 0; q = qi(1 + bDit)^(−1/b) otherwise
Time in days, rate in STB/day. b = 1 is harmonic. These are synthetic model curves, not field fits.
1. Define the baseline
Elapsed time (day) = 365
qi = 1,000 STB/day and Di = 0.001/day for all curves. Only b changes among the three model columns.
2. Select the relation
q = qi exp(−Dit) for b = 0; q = qi(1 + bDit)^(−1/b) otherwise
Time in days, rate in STB/day. b = 1 is harmonic. These are synthetic model curves, not field fits.
3. Substitute the values
At day 365: exponential = 1,000 exp(−0.365); hyperbolic = 1,000(1 + 0.5 × 0.365)^−2; harmonic = 1,000/1.365
Retain intermediate precision. The displayed values are rounded only after evaluation.
4. Read the result
Exponential rate (STB/day) = 694.196651; Hyperbolic b=0.5 rate (STB/day) = 715.150919; Harmonic rate (STB/day) = 732.600733
Higher b sustains more late-time rate under these matched initial conditions.
Result and interpretation
Exponential rate (STB/day) = 694.196651; Hyperbolic b=0.5 rate (STB/day) = 715.150919; Harmonic rate (STB/day) = 732.600733
Higher b sustains more late-time rate under these matched initial conditions.
This panel demonstrates controlled Arps shape sensitivity. Use the linked separate tools for the other empirical families with their own documented inputs.
Exact scenario data
Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
| Elapsed time (day) | Exponential rate (STB/day) | Hyperbolic b=0.5 rate (STB/day) | Harmonic rate (STB/day) |
|---|---|---|---|
| 0 | 1,000 | 1,000 | 1,000 |
| 90 | 913.9312 | 915.73 | 917.4312 |
| 365 | 694.1967 | 715.1509 | 732.6007 |
| 730 | 481.909 | 536.7038 | 578.0347 |
| 1,825 | 161.2176 | 273.3991 | 353.9823 |
Scroll the table to see all columns. The highlighted row is the worked baseline. Downloads retain full numerical precision.
Test your understanding
Work through the question before revealing its answer. Use the stated assumptions and include units in every numerical step.
1. Can a Duong coefficient fitted on months be entered unchanged into a daily-basis formula?
Show the worked answer
No. The time definition is part of the coefficient meaning and requires a consistent transformation or refit.
2. Reproduce the baseline Exponential rate (STB/day) without rounding intermediate values.
Show the worked answer
At day 365: exponential = 1,000 exp(−0.365); hyperbolic = 1,000(1 + 0.5 × 0.365)^−2; harmonic = 1,000/1.365. Exponential rate (STB/day) = 694.196651. Higher b sustains more late-time rate under these matched initial conditions.
3. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
Elapsed time (day) changes from 0 to 1825. Other controls remain as stated in the premise. Exponential rate (STB/day) changes from 1000 to 161.217644.
4. What would be needed to apply this example to a different physical system?
Show the worked answer
Replace the synthetic controls with compatible documented inputs and check the relation's assumptions. Time in days, rate in STB/day. b = 1 is harmonic. These are synthetic model curves, not field fits. Higher b sustains more late-time rate under these matched initial conditions.
Before using the result
- Keep each model's time definition.
- Use held-out observations.
- Avoid ranking families from an unmatched example.
Higher b sustains more late-time rate under these matched initial conditions.
Continue learning
Start with the broader workflow guide
Sources and method
The explanation and synthetic examples are original PetroCalcHub teaching material. The references below support the definitions and model context. They do not supply field validation for the synthetic scenarios.