Engineering lesson
Explore the square root of time in Carter leakoff
The square root of exposure time creates a declining leakoff rate for a fixed exposed area. Volume keeps increasing, but each additional minute adds less time-dependent loss.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Read volume and rate separately
For fixed AL and no spurt, VL = 2ALCL√t. Differentiating gives dVL/dt = ALCL/√t for positive t.
The idealized rate diverges at zero time, which signals the limits of this asymptotic description rather than an observable infinite pump rate.
Spurt changes the intercept
Adding ALSp increases total loss by a fixed amount. It does not alter the square-root growth of the Carter component.
The explorer retains both terms and separates them in the table. A total-volume ratio is therefore not exactly √2 when time doubles and spurt is nonzero.
Use a fixed-area interpretation
The example exposes all area at the same initial time. A growing fracture requires integrating local exposure ages over evolving area.
Do not use the final area and full job time as though every point had the same exposure history unless that simplification is explicitly intended.
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.
Compare time-dependent and total loss as time changes. The baseline includes a fixed spurt contribution.
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. Does doubling time double leakoff rate in this fixed-area model?
Show the worked answer
No. The time-dependent rate decreases by a factor 1/√2, while cumulative Carter loss increases by √2.
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
- Distinguish rate from cumulative volume.
- Treat the zero-time singularity explicitly.
- Retain local exposure-age assumptions.
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.