Engineering lesson
Explore the permeability viscosity and pressure tradeoff
Within single-phase Darcy flow, rate scales with permeability and pressure drop and inversely with viscosity and flow length. The tradeoff holds only while geometry and the linear-flow assumptions are retained.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Use proportionality to predict before solving
Doubling k doubles q if all other inputs stay fixed. Doubling viscosity halves q. Doubling both k and viscosity leaves rate unchanged. These are useful unit-independent checks on the equation.
The field calculator's linear rate is a local fluid-volume rate. A stock-tank rate requires an explicit formation-volume-factor conversion rather than a change in the Darcy coefficient.
Distinguish gradient from total pressure drop
The driving term is ΔP/L. Doubling length with fixed total pressure drop halves the gradient and the rate. Doubling both leaves the gradient unchanged.
A permeability change is not necessarily independent of stress or saturation in a real reservoir. The explorer deliberately fixes those effects to explain one relation.
Recognize where proportionality breaks
Multiphase relative permeability, non-Darcy inertia, compressibility and evolving boundaries can change the relationship. A simple sensitivity is not a model-selection test.
Use the linked Darcy–Forchheimer lesson when inertial resistance becomes relevant, and keep superficial velocity distinct from pore velocity.
Explore the example
Replayed with the same equation record and calculation engine as /calculators/darcy-s-law-for-linear-single-phase-flow. Original synthetic inputs; engine outputs are saved in the downloadable record.
Select one of 5 recorded scenarios. Intermediate states are not interpolated.
Permeability (mD) = 100. q (bbl/day): 2.8175.
- q (bbl/day)
Worked example
Darcy's Law for Linear Single-Phase Flow
Fixed controls: A = 500 ft²; DeltaP = 100 psi; mu = 2 cP; L = 1000 ft. k is varied as shown.
Relation and unit basis
q = 0.001127 k A DeltaP / (mu L)
k: Permeability (mD). A: Cross-Sectional Flow Area (ft²). DeltaP: Pressure Drop (psi). mu: Fluid Viscosity (cP). L: Flow Length (ft).
1. Define the baseline
Permeability (mD) = 100
Fixed controls: A = 500 ft²; DeltaP = 100 psi; mu = 2 cP; L = 1000 ft. k is varied as shown.
2. Select the relation
q = 0.001127 k A DeltaP / (mu L)
k: Permeability (mD). A: Cross-Sectional Flow Area (ft²). DeltaP: Pressure Drop (psi). mu: Fluid Viscosity (cP). L: Flow Length (ft).
3. Substitute the values
k = 100; A = 500; DeltaP = 100; mu = 2; L = 1000 → q = 2.8175 bbl/day
Retain intermediate precision. The displayed values are rounded only after evaluation.
4. Read the result
q (bbl/day) = 2.8175
The local fluid-volume rate is proportional to permeability at fixed linear geometry, viscosity and gradient.
Result and interpretation
q (bbl/day) = 2.8175
The local fluid-volume rate is proportional to permeability at fixed linear geometry, viscosity and gradient.
Permeability varies while all other linear-flow inputs stay fixed. Check the rate ratio against the permeability ratio.
Exact scenario data
Replayed with the same equation record and calculation engine as /calculators/darcy-s-law-for-linear-single-phase-flow. Original synthetic inputs; engine outputs are saved in the downloadable record.
| Permeability (mD) | q (bbl/day) |
|---|---|
| 25 | 0.704375 |
| 50 | 1.40875 |
| 100 | 2.8175 |
| 200 | 5.635 |
| 400 | 11.27 |
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 happens to rate if k doubles and viscosity doubles simultaneously?
Show the worked answer
It is unchanged under the same Darcy geometry and pressure gradient because the ratio k/μ is unchanged.
2. Reproduce the baseline q (bbl/day) without rounding intermediate values.
Show the worked answer
k = 100; A = 500; DeltaP = 100; mu = 2; L = 1000 → q = 2.8175 bbl/day. q (bbl/day) = 2.8175. The local fluid-volume rate is proportional to permeability at fixed linear geometry, viscosity and gradient.
3. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
Permeability (mD) changes from 25 to 400. Other controls remain as stated in the premise. q (bbl/day) changes from 0.704375 to 11.27.
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). A: Cross-Sectional Flow Area (ft²). DeltaP: Pressure Drop (psi). mu: Fluid Viscosity (cP). L: Flow Length (ft). The local fluid-volume rate is proportional to permeability at fixed linear geometry, viscosity and gradient.
Before using the result
- Hold the pressure gradient explicitly.
- Keep local and stock-tank volume bases separate.
- Use the linear model within its scope.
The local fluid-volume rate is proportional to permeability at fixed linear geometry, viscosity and gradient.
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.