Engineering lesson
Water resistivity temperature basis in a saturation workflow
Water resistivity must represent the water at the temperature basis required by the saturation interpretation. A surface-temperature Rw cannot automatically be used as a formation-temperature value.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Temperature changes the electrical input
Brine resistivity depends on temperature and composition. A laboratory measurement therefore needs its temperature, salinity and method retained before it is transferred into a log interpretation.
The explorer varies an already supplied formation-basis Rw. It does not apply a universal temperature correction or infer salinity from one reading.
Read the Archie sensitivity
With other inputs fixed, Sw is proportional to Rw^(1/n). For n = 2, doubling Rw multiplies calculated saturation by √2.
This sensitivity shows why a temperature mismatch can matter. It does not determine the correct Rw; that needs a suitable measured or corrected water-property input.
Keep all resistivities on compatible units
Rw and Rt must use matching resistivity units, commonly Ω·m. Conductivity is the reciprocal of resistivity and cannot be substituted without conversion.
Archive the original measurement and correction method. If several water populations or salinity gradients exist, one reservoir-wide Rw may be inadequate.
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.
Formation Water Resistivity (ohm m) = 0.05. S_w (fraction): 0.25.
- S_w (fraction)
Worked example
Archie Water Saturation from Resistivity Logs
Fixed controls: a = 1 dimensionless; phi = 0.2 fraction; m = 2 dimensionless; R_t = 20 ohm m; n = 2 dimensionless. R_w 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
Formation Water Resistivity (ohm m) = 0.05
Fixed controls: a = 1 dimensionless; phi = 0.2 fraction; m = 2 dimensionless; R_t = 20 ohm m; n = 2 dimensionless. R_w 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
These are supplied formation-basis Rw values; no temperature correction is inferred.
Result and interpretation
S_w (fraction) = 0.25
These are supplied formation-basis Rw values; no temperature correction is inferred.
The values are assumed formation-basis resistivities. No hidden temperature correction is made.
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.
| Formation Water Resistivity (ohm m) | S_w (fraction) |
|---|---|
| 0.025 | 0.1767767 |
| 0.04 | 0.2236068 |
| 0.05 | 0.25 |
| 0.075 | 0.3061862 |
| 0.1 | 0.3535534 |
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. For n = 2, what does a 21% relative Rw increase do to Sw?
Show the worked answer
It multiplies Sw by √1.21 = 1.10, a 10% relative increase if other inputs stay 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. These are supplied formation-basis Rw values; no temperature correction is inferred.
3. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
Formation Water Resistivity (ohm m) changes from 0.025 to 0.1. Other controls remain as stated in the premise. S_w (fraction) changes from 0.176776695 to 0.353553391.
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). These are supplied formation-basis Rw values; no temperature correction is inferred.
Before using the result
- Record measurement and formation temperatures.
- Use a documented water-property correction.
- Match resistivity rather than conductivity units.
These are supplied formation-basis Rw values; no temperature correction is inferred.
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.