Engineering lesson
Buckley Leverett breakthrough pore volumes explained
Buckley–Leverett breakthrough pore volumes follow from a fractional-flow shock construction. The shock speed is a tangent slope, not simply the fractional-flow value at an arbitrarily selected saturation.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Begin with the fractional-flow function
For an incompressible one-dimensional displacement, water fractional flow depends on water and oil mobilities. Relative permeability and viscosity therefore determine the curve.
The teaching case uses fw = Sw²/[Sw² + (1 − Sw)²] with zero initial water saturation. These are synthetic endpoint assumptions, not a calibrated core model.
Locate the shock with a tangent
The shock saturation satisfies fw'(Sf) = [fw(Sf) − fw(Swi)]/(Sf − Swi). Breakthrough injected pore volumes are the reciprocal of the shock slope on this nondimensional convention.
For the symmetric synthetic curve, Sf = 1/√2. The initial saturation and endpoint definitions must remain part of the calculation.
Do not confuse the ideal front with field sweep
The classical construction excludes capillary diffusion, gravity segregation, compressibility and multidimensional heterogeneity. A real waterflood can break through differently.
Use the linked calculator for its specified inputs and compare the shock condition explicitly. A result labelled pore volumes is dimensionless, not a barrel quantity until multiplied by the relevant pore 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.
Water saturation = 0.707106781. Water fractional flow: 0.853553391; Shock tangent value: 0.853553391.
- Water fractional flow
- Shock tangent value
Worked example
A synthetic shock construction
Symmetric Corey-style fractional flow fw = S²/[S² + (1 − S)²], Swi = 0. Shock saturation Sf = 1/√2.
Relation and unit basis
shock slope = fw(Sf)/Sf; breakthrough PV = 1/shock slope
S is an idealized saturation fraction with zero residual endpoints; gravity, capillary diffusion and compressibility are excluded.
1. Define the baseline
Water saturation = 0.707106781
Symmetric Corey-style fractional flow fw = S²/[S² + (1 − S)²], Swi = 0. Shock saturation Sf = 1/√2.
2. Select the relation
shock slope = fw(Sf)/Sf; breakthrough PV = 1/shock slope
S is an idealized saturation fraction with zero residual endpoints; gravity, capillary diffusion and compressibility are excluded.
3. Substitute the values
Sf = 0.707106781; fw(Sf) = 0.853553391; slope = 1.20710678; breakthrough = 0.828427125 PV
Retain intermediate precision. The displayed values are rounded only after evaluation.
4. Read the result
Water fractional flow = 0.853553391; Shock tangent value = 0.853553391
The tangent touches at the shock state. Its reciprocal slope gives breakthrough pore volumes on this convention.
Result and interpretation
Water fractional flow = 0.853553391; Shock tangent value = 0.853553391
The tangent touches at the shock state. Its reciprocal slope gives breakthrough pore volumes on this convention.
The table evaluates the fractional-flow curve around the analytically defined shock. It does not solve a full reservoir simulation.
Exact scenario data
Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
| Water saturation | Water fractional flow | Shock tangent value |
|---|---|---|
| 0.2 | 0.05882353 | 0.2414214 |
| 0.4 | 0.3076923 | 0.4828427 |
| 0.6 | 0.6923077 | 0.7242641 |
| 0.7071068 | 0.8535534 | 0.8535534 |
| 0.9 | 0.9878049 | 1.086396 |
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. How do you convert 0.8 injected pore volumes to a physical injected volume?
Show the worked answer
Multiply 0.8 by the pore volume used to nondimensionalize the displacement, on a compatible reservoir-volume basis.
2. Reproduce the baseline Water fractional flow without rounding intermediate values.
Show the worked answer
Sf = 0.707106781; fw(Sf) = 0.853553391; slope = 1.20710678; breakthrough = 0.828427125 PV. Water fractional flow = 0.853553391. The tangent touches at the shock state. Its reciprocal slope gives breakthrough pore volumes on this convention.
3. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
Water saturation changes from 0.2 to 0.9. Other controls remain as stated in the premise. Water fractional flow changes from 0.0588235294 to 0.987804878.
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. S is an idealized saturation fraction with zero residual endpoints; gravity, capillary diffusion and compressibility are excluded. The tangent touches at the shock state. Its reciprocal slope gives breakthrough pore volumes on this convention.
Before using the result
- State initial and endpoint saturations.
- Use a shock tangent, not an arbitrary slope.
- Keep ideal displacement separate from field sweep.
The tangent touches at the shock state. Its reciprocal slope gives breakthrough pore volumes on this convention.
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.