Engineering lesson
Darcy versus Fanning friction factors in drilling calculations
For matching definitions, the Darcy friction factor is four times the Fanning factor. Use a factor with the pressure-loss equation that defines it.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
The factor's name is part of the formula
Darcy–Weisbach uses Δp = fD(L/D)ρv²/2. The corresponding Fanning expression contains 4fF in that position. Setting fD = 4fF makes the predicted loss identical.
Copying a numerical factor between conventions without changing the equation creates a factor-of-four pressure error, even when the Reynolds number and roughness are correct.
Use the correlation within its flow regime
The linked Haaland calculator returns a Darcy factor for turbulent flow. It is not a laminar annular correlation. The worked convention example fixes fD = 0.02 and fF = 0.005 so the algebra can be checked directly.
For a circular Newtonian laminar pipe, fD = 64/Re and fF = 16/Re. Noncircular and non-Newtonian flows require their own definitions and geometry treatment.
Audit before diagnosing the fluid model
If two pressure models differ by almost exactly four, compare factor definitions before changing roughness or viscosity. A convention mismatch can masquerade as a physical calibration problem.
Keep density and velocity on the same local basis. The convention conversion does not fix a separate error in flow area or gas volume.
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.
Velocity (m/s) = 2. Matched Darcy loss (kPa): 40; Matched Fanning loss (kPa): 40; Wrong convention loss (kPa): 10.
- Matched Darcy loss (kPa)
- Matched Fanning loss (kPa)
- Wrong convention loss (kPa)
Worked example
A matched factor convention
Liquid density 1,000 kg/m³, L = 100 m, D = 0.1 m; Darcy fD = 0.02 and Fanning fF = 0.005.
Relation and unit basis
ΔpD = fD(L/D)ρv²/2; ΔpF = 4fF(L/D)ρv²/2
The deliberately wrong calculation inserts fF into the Darcy expression without the factor four.
1. Define the baseline
Velocity (m/s) = 2
Liquid density 1,000 kg/m³, L = 100 m, D = 0.1 m; Darcy fD = 0.02 and Fanning fF = 0.005.
2. Select the relation
ΔpD = fD(L/D)ρv²/2; ΔpF = 4fF(L/D)ρv²/2
The deliberately wrong calculation inserts fF into the Darcy expression without the factor four.
3. Substitute the values
0.02 × (100/0.1) × 1,000 × 2²/2 = 40,000 Pa = 40 kPa; 4 × 0.005 gives the same result
Retain intermediate precision. The displayed values are rounded only after evaluation.
4. Read the result
Matched Darcy loss (kPa) = 40; Matched Fanning loss (kPa) = 40; Wrong convention loss (kPa) = 10
The two matched conventions agree; the mismatched one underpredicts by four.
Result and interpretation
Matched Darcy loss (kPa) = 40; Matched Fanning loss (kPa) = 40; Wrong convention loss (kPa) = 10
The two matched conventions agree; the mismatched one underpredicts by four.
The matched Darcy and Fanning calculations coincide; the deliberately mismatched column shows the fourfold error.
Exact scenario data
Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
| Velocity (m/s) | Matched Darcy loss (kPa) | Matched Fanning loss (kPa) | Wrong convention loss (kPa) |
|---|---|---|---|
| 0.5 | 2.5 | 2.5 | 0.625 |
| 1 | 10 | 10 | 2.5 |
| 2 | 40 | 40 | 10 |
| 3 | 90 | 90 | 22.5 |
| 4 | 160 | 160 | 40 |
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. A paper reports Fanning f = 0.006. Which number belongs in Darcy–Weisbach?
Show the worked answer
Darcy f = 0.024 for the same state and definitions.
2. Reproduce the baseline Matched Darcy loss (kPa) without rounding intermediate values.
Show the worked answer
0.02 × (100/0.1) × 1,000 × 2²/2 = 40,000 Pa = 40 kPa; 4 × 0.005 gives the same result. Matched Darcy loss (kPa) = 40. The two matched conventions agree; the mismatched one underpredicts by four.
3. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
Velocity (m/s) changes from 0.5 to 4. Other controls remain as stated in the premise. Matched Darcy loss (kPa) changes from 2.5 to 160.
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. The deliberately wrong calculation inserts fF into the Darcy expression without the factor four. The two matched conventions agree; the mismatched one underpredicts by four.
Before using the result
- Identify Darcy or Fanning in the source.
- Check the equation's prefactor.
- Keep the correlation's geometry and regime.
The two matched conventions agree; the mismatched one underpredicts by four.
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.