Engineering lesson
Fracture conductivity versus dimensionless conductivity
Fracture conductivity is permeability times fracture width. Dimensionless conductivity divides that quantity by a reservoir permeability–length scale, so the same fracture can have different dimensionless significance in different reservoirs.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Define the dimensional quantity
With kf in mD and width w in ft, kfw has units mD·ft. A dimensionless ratio such as Fcd = kfw/(kxf) requires compatible reservoir k and fracture half-length xf units.
The linked bilinear-flow calculator estimates a dimensional conductivity from its test relation. It does not by itself determine every input needed for the dimensionless ratio.
Keep the normalization visible
Doubling reservoir permeability at fixed kfw and xf halves Fcd. Doubling fracture half-length also halves the ratio on this definition.
Those changes do not mean the propped fracture's intrinsic conductivity changed. They change the reference flow resistance used to judge its significance.
Retain test and geometry assumptions
A conductivity inferred from a flow-test slope depends on identifying the intended regime and using the documented coefficient and units.
The teaching panel uses a specified kfw rather than claiming a test inversion. It isolates normalization, while the calculator link provides the separate dimensional estimate.
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.
Reservoir permeability (mD) = 1. Dimensionless conductivity: 1; Fracture conductivity (mD·ft): 100.
- Dimensionless conductivity
Worked example
Normalize a fixed fracture conductivity
Specified kfw = 100 mD·ft and fracture half-length xf = 100 ft. Reservoir k varies.
Relation and unit basis
Fcd = kfw/(k xf)
This normalization is separate from the linked bilinear-test estimate of dimensional conductivity.
1. Define the baseline
Reservoir permeability (mD) = 1
Specified kfw = 100 mD·ft and fracture half-length xf = 100 ft. Reservoir k varies.
2. Select the relation
Fcd = kfw/(k xf)
This normalization is separate from the linked bilinear-test estimate of dimensional conductivity.
3. Substitute the values
100 mD·ft/(1 mD × 100 ft) = 1
Retain intermediate precision. The displayed values are rounded only after evaluation.
4. Read the result
Dimensionless conductivity = 1; Fracture conductivity (mD·ft) = 100
The same dimensional fracture conductivity can have different significance relative to reservoir resistance.
Result and interpretation
Dimensionless conductivity = 1; Fracture conductivity (mD·ft) = 100
The same dimensional fracture conductivity can have different significance relative to reservoir resistance.
Reservoir permeability varies at fixed fracture conductivity and half-length.
Exact scenario data
Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
| Reservoir permeability (mD) | Dimensionless conductivity | Fracture conductivity (mD·ft) |
|---|---|---|
| 0.1 | 10 | 100 |
| 0.5 | 2 | 100 |
| 1 | 1 | 100 |
| 5 | 0.2 | 100 |
| 10 | 0.1 | 100 |
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 Fcd results from kfw = 100 mD·ft, k = 1 mD and xf = 100 ft?
Show the worked answer
1, a dimensionless ratio. All length units cancel on the stated definition.
2. Reproduce the baseline Dimensionless conductivity without rounding intermediate values.
Show the worked answer
100 mD·ft/(1 mD × 100 ft) = 1. Dimensionless conductivity = 1. The same dimensional fracture conductivity can have different significance relative to reservoir resistance.
3. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
Reservoir permeability (mD) changes from 0.1 to 10. Other controls remain as stated in the premise. Dimensionless conductivity changes from 10 to 0.1.
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. This normalization is separate from the linked bilinear-test estimate of dimensional conductivity. The same dimensional fracture conductivity can have different significance relative to reservoir resistance.
Before using the result
- Label kfw units.
- State the half-length convention.
- Keep inferred conductivity separate from its normalization.
The same dimensional fracture conductivity can have different significance relative to reservoir resistance.
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.