Engineering lesson
Carter leakoff coefficient time and area bases
Carter leakoff volume scales with exposed area and the square root of exposure time, plus any specified spurt term. The area must use the same face-count convention as the equation.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Read the implemented definition
The site's relation is VL = AL[2CL√t + Sp], with CL in ft/min^0.5, t in minutes, Sp in feet and AL in ft². The result is ft³.
Whether AL represents one face, both faces or a defined effective leakoff area must be established before adding a separate factor of two.
Exposure time is local
A growing fracture exposes new area at different times. Applying the full job duration to all final area is a fixed-area approximation, not the full moving-front integral.
The worked case explicitly assumes a fixed area exposed simultaneously. This makes the square-root-time relation directly testable.
Keep spurt and time-dependent loss separate
Spurt contributes ALSp independently of the square-root term. When time doubles, the time-dependent component grows by √2, while spurt stays fixed.
Use consistent time units throughout. A coefficient in ft/√min cannot be multiplied by the square root of seconds without conversion.
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 time (min) = 60. Carter component (ft³): 929.516003; Total with spurt (ft³): 1329.516; Carter ratio to 60 min: 1.
- Carter component (ft³)
- Total with spurt (ft³)
Worked example
Fixed-area Carter loss with spurt separated
Effective area 20,000 ft², CL = 0.003 ft/√min, Sp = 0.02 ft. All area is exposed at once.
Relation and unit basis
VCarter = 2ALCL√t; Vtotal = VCarter + ALSp
Exposure in minutes; no extra face factor is added to the declared effective area.
1. Define the baseline
Exposure time (min) = 60
Effective area 20,000 ft², CL = 0.003 ft/√min, Sp = 0.02 ft. All area is exposed at once.
2. Select the relation
VCarter = 2ALCL√t; Vtotal = VCarter + ALSp
Exposure in minutes; no extra face factor is added to the declared effective area.
3. Substitute the values
At 60 min: 2 × 20,000 × 0.003 × √60 = 929.516003 ft³; spurt = 20,000 × 0.02 = 400 ft³; total = 1,329.516003 ft³
Retain intermediate precision. The displayed values are rounded only after evaluation.
4. Read the result
Carter component (ft³) = 929.516003; Total with spurt (ft³) = 1329.516; Carter ratio to 60 min = 1
Only the Carter component follows the pure square-root ratio when spurt is nonzero.
Result and interpretation
Carter component (ft³) = 929.516003; Total with spurt (ft³) = 1329.516; Carter ratio to 60 min = 1
Only the Carter component follows the pure square-root ratio when spurt is nonzero.
Exposure time varies for a fixed effective area. The output separates time-dependent loss and spurt.
Exact scenario data
Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
| Exposure time (min) | Carter component (ft³) | Total with spurt (ft³) | Carter ratio to 60 min |
|---|---|---|---|
| 15 | 464.758 | 864.758 | 0.5 |
| 30 | 657.2671 | 1,057.267 | 0.7071068 |
| 60 | 929.516 | 1,329.516 | 1 |
| 120 | 1,314.534 | 1,714.534 | 1.414214 |
| 240 | 1,859.032 | 2,259.032 | 2 |
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 the time-dependent term if exposure time is multiplied by four?
Show the worked answer
It doubles. The spurt term does not change.
2. Reproduce the baseline Carter component (ft³) without rounding intermediate values.
Show the worked answer
At 60 min: 2 × 20,000 × 0.003 × √60 = 929.516003 ft³; spurt = 20,000 × 0.02 = 400 ft³; total = 1,329.516003 ft³. Carter component (ft³) = 929.516003. Only the Carter component follows the pure square-root ratio when spurt is nonzero.
3. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
Exposure time (min) changes from 15 to 240. Other controls remain as stated in the premise. Carter component (ft³) changes from 464.758002 to 1859.03201.
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. Exposure in minutes; no extra face factor is added to the declared effective area. Only the Carter component follows the pure square-root ratio when spurt is nonzero.
Before using the result
- State the face-count area convention.
- Match CL and time units.
- Distinguish fixed area from growing exposure.
Only the Carter component follows the pure square-root ratio when spurt is nonzero.
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.