Engineering lesson
Effective versus total porosity in saturation calculations
Total porosity counts the selected total pore volume; effective porosity describes the connected or interpreted effective pore space on the chosen method. A saturation model must use the porosity basis for which its parameters were established.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Keep the pore-volume definition attached
Clay-bound water and isolated pores can contribute differently to total and effective interpretations. The distinction depends on the measurement and model, so effective porosity is not obtained by a universal fixed subtraction.
Using a smaller porosity in Archie increases calculated Sw when all other inputs remain fixed. That sensitivity does not establish which porosity is physically appropriate.
Calibration and input basis belong together
Archie parameters a, m and n are calibrated for a stated rock and porosity interpretation. Combining total-porosity calibration with an unrelated effective-porosity input changes the model.
The worked comparison holds resistivities and exponents fixed to expose this input effect. It is not a recommendation to switch porosity definitions until saturation appears plausible.
Audit the full saturation ledger
A saturation fraction refers to the pore volume used as its denominator. Multiplying effective porosity by a total-pore saturation can create an inconsistent fluid volume.
Record the log or core method, clay correction and calibration interval. A saturation number without its pore-volume basis is incomplete.
Explore the example
Replayed with the same equation record and calculation engine as /calculators/archie-water-saturation-from-resistivity-logs. Original synthetic inputs; engine outputs are saved in the downloadable record.
Select one of 5 recorded scenarios. Intermediate states are not interpolated.
Porosity (fraction) = 0.2. S_w (fraction): 0.25.
- S_w (fraction)
Worked example
Archie Water Saturation from Resistivity Logs
Fixed controls: a = 1 dimensionless; m = 2 dimensionless; R_w = 0.05 ohm m; R_t = 20 ohm m; n = 2 dimensionless. phi is varied as shown.
Relation and unit basis
S_w = ((a / phi^m) * (R_w / R_t))^(1 / n)
a: Tortuosity Factor (dimensionless). phi: Porosity (fraction). m: Cementation Factor (dimensionless). R_w: Formation Water Resistivity (ohm m). R_t: True Formation Resistivity (ohm m). n: Saturation Exponent (dimensionless).
1. Define the baseline
Porosity (fraction) = 0.2
Fixed controls: a = 1 dimensionless; m = 2 dimensionless; R_w = 0.05 ohm m; R_t = 20 ohm m; n = 2 dimensionless. phi is varied as shown.
2. Select the relation
S_w = ((a / phi^m) * (R_w / R_t))^(1 / n)
a: Tortuosity Factor (dimensionless). phi: Porosity (fraction). m: Cementation Factor (dimensionless). R_w: Formation Water Resistivity (ohm m). R_t: True Formation Resistivity (ohm m). n: Saturation Exponent (dimensionless).
3. Substitute the values
a = 1; phi = 0.2; m = 2; R_w = 0.05; R_t = 20; n = 2 → S_w = 0.25 fraction
Retain intermediate precision. The displayed values are rounded only after evaluation.
4. Read the result
S_w (fraction) = 0.25
Changing porosity with fixed exponents is a sensitivity; it does not determine the correct porosity definition.
Result and interpretation
S_w (fraction) = 0.25
Changing porosity with fixed exponents is a sensitivity; it does not determine the correct porosity definition.
Only the supplied porosity changes; the interaction diagnoses sensitivity rather than selecting a porosity model.
Exact scenario data
Replayed with the same equation record and calculation engine as /calculators/archie-water-saturation-from-resistivity-logs. Original synthetic inputs; engine outputs are saved in the downloadable record.
| Porosity (fraction) | S_w (fraction) |
|---|---|
| 0.1 | 0.5 |
| 0.15 | 0.3333333 |
| 0.16 | 0.3125 |
| 0.2 | 0.25 |
| 0.25 | 0.2 |
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. If φ falls from 0.20 to 0.16 with m = n = 2, how does Archie Sw change?
Show the worked answer
It increases by a factor 0.20/0.16 = 1.25, with a, Rw and Rt fixed.
2. Reproduce the baseline S_w (fraction) without rounding intermediate values.
Show the worked answer
a = 1; phi = 0.2; m = 2; R_w = 0.05; R_t = 20; n = 2 → S_w = 0.25 fraction. S_w (fraction) = 0.25. Changing porosity with fixed exponents is a sensitivity; it does not determine the correct porosity definition.
3. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
Porosity (fraction) changes from 0.1 to 0.25. Other controls remain as stated in the premise. S_w (fraction) changes from 0.5 to 0.2.
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. a: Tortuosity Factor (dimensionless). phi: Porosity (fraction). m: Cementation Factor (dimensionless). R_w: Formation Water Resistivity (ohm m). R_t: True Formation Resistivity (ohm m). n: Saturation Exponent (dimensionless). Changing porosity with fixed exponents is a sensitivity; it does not determine the correct porosity definition.
Before using the result
- State total or effective porosity.
- Match calibration to the input basis.
- Keep saturation and pore-volume denominators consistent.
Changing porosity with fixed exponents is a sensitivity; it does not determine the correct porosity definition.
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.