Original synthetic study
How much a small washout changes a synthetic cement estimate
This synthetic study keeps casing and interval length fixed and enlarges only the hole. A small diameter change can produce a much larger percentage change in annular slurry demand.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Define a reproducible geometric experiment
The baseline uses an 8.5 in hole, 5 in casing and 1,000 ft measured length. Every comparison uses the site's same rounded annular coefficient, the same 20% volume excess and the same 1.32 ft³/sk yield.
The independent variable is hole diameter. No caliper data are implied, and no separate loss model is added. This makes the observed difference attributable to the chosen geometry.
Compute a ratio before rounding sacks
The annular-volume ratio is (Dh,new² − Do²)/(Dh,base² − Do²). Length and the common geometric coefficient cancel. The ratio can therefore be checked independently of the unit conversion.
Compare this ratio with the final sack ratio. The latter can differ slightly because each batch is rounded upward. The downloadable table keeps theoretical and whole-sack demand separate.
Use the experiment to prioritize measurements
The result explains why a diameter assumption deserves attention in a cement estimate. It does not tell an operator which excess to choose or how much loss will occur.
For a nonuniform washout, sum measured sections or integrate a caliper-derived area profile. One average diameter can hide the effect of short large cavities because area depends on the square.
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.
Hole diameter (in) = 8.5. Geometry volume (bbl): 45.9; Theoretical sacks (sk): 234.28125; Whole sacks (sk): 235.
- Geometry volume (bbl)
Worked example
Hole diameter propagated to volume and sacks
5 in casing, 1,000 ft measured length, 20% volume excess, yield 1.32 ft³/sk; hole diameter varies.
Relation and unit basis
Vplan = Va × 1.20; N = Vplan × (42 × 231/1728)/1.32
Geometry uses the rounded library field coefficient. The barrel conversion uses 42 US gallons and 231 in³/US gal.
1. Define the baseline
Hole diameter (in) = 8.5
5 in casing, 1,000 ft measured length, 20% volume excess, yield 1.32 ft³/sk; hole diameter varies.
2. Select the relation
Vplan = Va × 1.20; N = Vplan × (42 × 231/1728)/1.32
Geometry uses the rounded library field coefficient. The barrel conversion uses 42 US gallons and 231 in³/US gal.
3. Substitute the values
Va = 45.9 bbl; planned = 55.08 bbl; 55.08 × 5.614583333/1.32 = 234.28125 sk; round to 235
Retain intermediate precision. The displayed values are rounded only after evaluation.
4. Read the result
Geometry volume (bbl) = 45.9; Theoretical sacks (sk) = 234.28125; Whole sacks (sk) = 235
Diameter-squared geometry and final integer rounding are different sources of the response.
Result and interpretation
Geometry volume (bbl) = 45.9; Theoretical sacks (sk) = 234.28125; Whole sacks (sk) = 235
Diameter-squared geometry and final integer rounding are different sources of the response.
The nominal hole is the control. Reproduce a volume ratio from the diameter-squared terms before comparing sacks.
Study data and reproduction
Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
| Hole diameter (in) | Geometry volume (bbl) | Theoretical sacks (sk) | Whole sacks (sk) |
|---|---|---|---|
| 8.5 | 45.9 | 234.2812 | 235 |
| 8.75 | 50.08929 | 255.6641 | 256 |
| 9 | 54.4 | 277.6667 | 278 |
| 9.5 | 63.38571 | 323.5312 | 324 |
| 10 | 72.85714 | 371.875 | 372 |
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 is the annular-area ratio at 9 in versus 8.5 in with 5 in casing?
Show the worked answer
(81 − 25)/(72.25 − 25) = 1.185185, an 18.5185% increase although diameter rose only 5.8824%.
2. Reproduce the baseline Geometry volume (bbl) without rounding intermediate values.
Show the worked answer
Va = 45.9 bbl; planned = 55.08 bbl; 55.08 × 5.614583333/1.32 = 234.28125 sk; round to 235. Geometry volume (bbl) = 45.9. Diameter-squared geometry and final integer rounding are different sources of the response.
3. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
Hole diameter (in) changes from 8.5 to 10. Other controls remain as stated in the premise. Geometry volume (bbl) changes from 45.9 to 72.8571429.
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. Geometry uses the rounded library field coefficient. The barrel conversion uses 42 US gallons and 231 in³/US gal. Diameter-squared geometry and final integer rounding are different sources of the response.
Before using the result
- Use identical coefficients and recipe inputs.
- Compare volume before integer rounding.
- Do not interpret synthetic geometry as measured washout.
Diameter-squared geometry and final integer rounding are different sources of the response.
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.