Original synthetic study
Doubling exposure time does not double Carter leakoff
For fixed area and zero spurt, doubling exposure time multiplies Carter leakoff volume by √2 rather than two. A nonzero spurt term makes the total-volume ratio smaller than √2.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Design a ratio experiment
The study uses one effective area and one CL. It compares t with multiples of a baseline exposure time, retaining the same unit convention.
The time-dependent ratio is √(t/t0), independent of area and CL. This provides an analytical check on every generated row.
Add spurt as a controlled second ledger
The dataset reports both a zero-spurt component and total loss with a fixed Sp. At increasing time, the fixed spurt becomes a smaller fraction of total loss.
Comparing total volumes without separating that term can obscure the expected square-root scaling.
Keep the physical scope narrow
All area is assumed exposed at once. The experiment does not simulate fracture propagation, changing pressure or a changing leakoff coefficient.
Use the download to reproduce the ratios. A field study would additionally need exposure history, area definition and fluid-loss measurements.
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.
The baseline ratio is one at t0; four times t0 should give twice the Carter component.
Study data and reproduction
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. Why is total loss not exactly doubled at four times t0 when spurt is nonzero?
Show the worked answer
Only the time-dependent component doubles. The fixed spurt contribution remains unchanged.
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
- Use one fixed area.
- Separate spurt from Carter loss.
- Compare against the analytical square-root ratio.
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.