Engineering lesson
Why well radius and drainage radius affect productivity differently
Well radius and drainage radius enter the radial resistance logarithm through their ratio. Their percentage changes can have equal algebraic effects, although their physical meanings and feasible changes differ.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Inspect the logarithm rather than the raw distance
Doubling drainage radius adds ln 2 to ln(re/rw). Doubling well radius subtracts ln 2. This differs from a linear pipe-area relationship.
A radial pressure profile also depends on the observation radius. Keep r, rw and re distinct so an interior pressure is not mistaken for the outer boundary.
Separate geometric sensitivity from completion benefit
Increasing effective well access may also change skin and flow geometry. A radius-only comparison deliberately excludes those changes.
A larger drainage radius can represent a different boundary or connected volume. It is not an independent design knob that creates pressure support without changing the reservoir description.
Use a limiting check
At r = rw, the ideal radial logarithmic pressure increment is zero in the no-skin profile. At larger r it grows logarithmically for the stated inward flow.
The explorer varies a radius while other model terms are fixed. The numerical result is a sensitivity of that model, not an estimate of an actual drainage boundary.
Explore the example
Replayed with the same equation record and calculation engine as /calculators/steady-state-radial-liquid-flow-rate. Original synthetic inputs; engine outputs are saved in the downloadable record.
Select one of 5 recorded scenarios. Intermediate states are not interpolated.
Drainage Radius (ft) = 1000. q_o (STB/day): 735.844574.
- q_o (STB/day)
Worked example
Steady-State Radial Liquid Flow Rate
Fixed controls: k = 100 mD; h = 50 ft; p_e = 3000 psi; p_wf = 2500 psi; mu_o = 2 cP; B_o = 1.2 bbl/STB; r_w = 0.328 ft; s = 2 dimensionless. r_e is varied as shown.
Relation and unit basis
q_o = 0.00708 k h (p_e - p_wf) / (mu_o B_o (ln(r_e / r_w) + s))
k: Permeability (mD). h: Net Pay Thickness (ft). p_e: External Boundary Pressure (psi). p_wf: Flowing Bottom-Hole Pressure (psi). mu_o: Oil Viscosity (cP). B_o: Oil Formation Volume Factor (bbl/STB). r_e: Drainage Radius (ft). r_w: Wellbore Radius (ft). s: Skin Factor (dimensionless).
1. Define the baseline
Drainage Radius (ft) = 1000
Fixed controls: k = 100 mD; h = 50 ft; p_e = 3000 psi; p_wf = 2500 psi; mu_o = 2 cP; B_o = 1.2 bbl/STB; r_w = 0.328 ft; s = 2 dimensionless. r_e is varied as shown.
2. Select the relation
q_o = 0.00708 k h (p_e - p_wf) / (mu_o B_o (ln(r_e / r_w) + s))
k: Permeability (mD). h: Net Pay Thickness (ft). p_e: External Boundary Pressure (psi). p_wf: Flowing Bottom-Hole Pressure (psi). mu_o: Oil Viscosity (cP). B_o: Oil Formation Volume Factor (bbl/STB). r_e: Drainage Radius (ft). r_w: Wellbore Radius (ft). s: Skin Factor (dimensionless).
3. Substitute the values
k = 100; h = 50; p_e = 3000; p_wf = 2500; mu_o = 2; B_o = 1.2; r_e = 1000; r_w = 0.328; s = 2 → q_o = 735.844574 STB/day
Retain intermediate precision. The displayed values are rounded only after evaluation.
4. Read the result
q_o (STB/day) = 735.844574
Increasing drainage radius increases logarithmic resistance under a maintained outer pressure.
Result and interpretation
q_o (STB/day) = 735.844574
Increasing drainage radius increases logarithmic resistance under a maintained outer pressure.
Compare radius ratios and the resulting logarithmic resistance, with all fluid and rock properties held fixed.
Exact scenario data
Replayed with the same equation record and calculation engine as /calculators/steady-state-radial-liquid-flow-rate. Original synthetic inputs; engine outputs are saved in the downloadable record.
| Drainage Radius (ft) | q_o (STB/day) |
|---|---|
| 250 | 853.9633 |
| 500 | 790.516 |
| 1,000 | 735.8446 |
| 2,000 | 688.2461 |
| 4,000 | 646.4313 |
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. What does doubling re do to ln(re/rw)?
Show the worked answer
It adds ln 2 ≈ 0.693147. It does not double the logarithm or pressure resistance.
2. Reproduce the baseline q_o (STB/day) without rounding intermediate values.
Show the worked answer
k = 100; h = 50; p_e = 3000; p_wf = 2500; mu_o = 2; B_o = 1.2; r_e = 1000; r_w = 0.328; s = 2 → q_o = 735.844574 STB/day. q_o (STB/day) = 735.844574. Increasing drainage radius increases logarithmic resistance under a maintained outer pressure.
3. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
Drainage Radius (ft) changes from 250 to 4000. Other controls remain as stated in the premise. q_o (STB/day) changes from 853.963293 to 646.431318.
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. k: Permeability (mD). h: Net Pay Thickness (ft). p_e: External Boundary Pressure (psi). p_wf: Flowing Bottom-Hole Pressure (psi). mu_o: Oil Viscosity (cP). B_o: Oil Formation Volume Factor (bbl/STB). r_e: Drainage Radius (ft). r_w: Wellbore Radius (ft). s: Skin Factor (dimensionless). Increasing drainage radius increases logarithmic resistance under a maintained outer pressure.
Before using the result
- Distinguish observation and boundary radii.
- Use a dimensionless radius ratio.
- Keep skin separate from a radius-only sensitivity.
Increasing drainage radius increases logarithmic resistance under a maintained outer pressure.
Continue learning
Start with the broader workflow guide · Related method 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.