Engineering lesson
Leakoff spurt loss and fracture efficiency
Leakoff and spurt are fluid-loss terms. Fracture efficiency compares retained fracture volume with injected volume under a stated fluid balance; it is not determined by a leakoff coefficient alone.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Close a defined fluid balance
A simplified ledger gives injected volume = retained fracture volume + leakoff volume when other storage and losses are excluded. Efficiency is retained volume divided by injected volume.
Wellbore storage, compressibility and other losses require additional entries when relevant. Omitting them is an assumption, not proof they are zero.
Use the same area and exposure basis
The Carter term and spurt term in the example use the same fixed effective area. If the area convention changes between them, the sum no longer represents one consistent model.
The explorer varies exposure time while injected volume is held fixed, showing how the assumed loss changes the retained fraction.
Flag impossible balances
A calculated leakoff greater than injected volume produces a negative retained volume in this simple ledger. That is an invalid scenario, not a physical negative fracture.
Do not clip efficiency to zero and hide the inconsistency. Recheck injected volume, exposure history and model inputs.
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.
Exposure (min) = 60. Leakoff volume (ft³): 1329.516; Retained volume (ft³): 1670.484; Efficiency (fraction): 0.556827999.
- Leakoff volume (ft³)
- Retained volume (ft³)
Worked example
A simplified injected-fluid ledger
Injected volume fixed at 3,000 ft³; AL = 20,000 ft², CL = 0.003 ft/√min and Sp = 0.02 ft. Other storage terms are excluded.
Relation and unit basis
Vretained = Vinjected − Vleakoff; efficiency = Vretained/Vinjected
Fixed simultaneous area exposure is assumed. This does not solve fracture propagation or pressure-dependent leakoff.
1. Define the baseline
Exposure (min) = 60
Injected volume fixed at 3,000 ft³; AL = 20,000 ft², CL = 0.003 ft/√min and Sp = 0.02 ft. Other storage terms are excluded.
2. Select the relation
Vretained = Vinjected − Vleakoff; efficiency = Vretained/Vinjected
Fixed simultaneous area exposure is assumed. This does not solve fracture propagation or pressure-dependent leakoff.
3. Substitute the values
At 60 min: leakoff = 1,329.516003 ft³; retained = 3,000 − 1,329.516003; efficiency = retained/3,000
Retain intermediate precision. The displayed values are rounded only after evaluation.
4. Read the result
Leakoff volume (ft³) = 1329.516; Retained volume (ft³) = 1670.484; Efficiency (fraction) = 0.556827999
Efficiency is conditional on this complete stated volume ledger, not CL alone.
Result and interpretation
Leakoff volume (ft³) = 1329.516; Retained volume (ft³) = 1670.484; Efficiency (fraction) = 0.556827999
Efficiency is conditional on this complete stated volume ledger, not CL alone.
The balance uses an explicit fixed injected volume. It does not solve fracture growth or pressure-dependent leakoff.
Exact scenario data
Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
| Exposure (min) | Leakoff volume (ft³) | Retained volume (ft³) | Efficiency (fraction) |
|---|---|---|---|
| 15 | 864.758 | 2,135.242 | 0.7117473 |
| 30 | 1,057.267 | 1,942.733 | 0.6475776 |
| 60 | 1,329.516 | 1,670.484 | 0.556828 |
| 120 | 1,714.534 | 1,285.466 | 0.4284886 |
| 240 | 2,259.032 | 740.968 | 0.2469893 |
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. If 1,000 ft³ is injected and 300 ft³ is lost under this simplified balance, what is efficiency?
Show the worked answer
0.70, or 70%, with 700 ft³ retained. Additional storage terms would change the interpretation.
2. Reproduce the baseline Leakoff volume (ft³) without rounding intermediate values.
Show the worked answer
At 60 min: leakoff = 1,329.516003 ft³; retained = 3,000 − 1,329.516003; efficiency = retained/3,000. Leakoff volume (ft³) = 1329.516. Efficiency is conditional on this complete stated volume ledger, not CL alone.
3. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
Exposure (min) changes from 15 to 240. Other controls remain as stated in the premise. Leakoff volume (ft³) changes from 864.758002 to 2259.03201.
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. Fixed simultaneous area exposure is assumed. This does not solve fracture propagation or pressure-dependent leakoff. Efficiency is conditional on this complete stated volume ledger, not CL alone.
Before using the result
- State all included volume terms.
- Use one area convention.
- Expose negative or above-one efficiency as inconsistent.
Efficiency is conditional on this complete stated volume ledger, not CL alone.
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.