Engineering lesson
Explore how hole diameter affects cement demand
Hole diameter can strongly change cement demand because the open-hole area grows with diameter squared while casing displacement stays fixed.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Hold the other geometric inputs still
The explorer keeps casing OD, measured length, yield and the selected excess unchanged. Moving hole diameter therefore isolates one geometric cause of changing sacks.
The response becomes particularly sensitive when the gap is small: a modest increase in outer diameter can represent a large fractional increase in the remaining annular area.
Read smooth volume and stepped sacks separately
Theoretical volume changes continuously with hole diameter. Whole-sack demand changes in steps because the final batch is rounded upward. Those steps are a material-count effect rather than a discontinuity in the hole geometry.
Use the unrounded volume for sensitivity analysis. A small diameter change may produce no visible sack change in one interval and cross a rounding threshold in another.
Keep uncertainty separate from recommendation
The diameter choices are synthetic scenarios. They do not predict a washout distribution or select a field contingency. A caliper measurement should retain its interval resolution and uncertainty.
After a geometry change, revisit slurry volume and water demand together. The explorer leaves recipe properties fixed so that a geometry effect is not confused with a design change.
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.
Select a hole-size scenario and compare the volume and sack columns. The baseline is the nominal 8.5 in hole.
Exact scenario data
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. Why can sacks stay unchanged after a very small volume increase?
Show the worked answer
Both theoretical requirements may round upward to the same whole sack. Volume still changed, even if the batch count did not.
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
- Keep recipe inputs fixed during geometry sensitivity.
- Inspect unrounded volume.
- Treat scenario diameters as assumptions.
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.