Engineering lesson
Excess cement volume versus hole enlargement
A percentage increase in hole diameter is not the same percentage increase in annular volume. Annular area depends on the difference of squared diameters.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Diameter enlargement has a nonlinear effect
For fixed 5 in casing, the area term is Dh² − 25. Raising an 8.5 in hole diameter by 5% increases this area term by more than 5%, because the casing area is unchanged while the outer boundary expands.
A volume excess instead multiplies an already calculated volume by 1 + e. These are different ways of specifying demand and should not be interchanged.
Caliper geometry and contingency need separate labels
If a caliper-derived diameter is already used, an additional excess is a separate allowance. Calling both entries washout can count the same uncertainty twice.
The example uses a uniform enlarged diameter solely to isolate geometry. Real washouts vary along the hole and require interval or caliper integration rather than one diameter inferred from a percentage.
Translate the effect through the batch
The geometric increase carries through planned slurry volume and theoretical sacks. Whole-sack rounding adds discrete steps to an otherwise smooth relationship.
Record whether an excess is applied to open hole only or to the whole job. The geometric sensitivity itself does not establish an appropriate contingency.
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.
Compare hole diameter, annular volume and whole sacks. The same casing and measured length are held throughout.
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. Does a 10% volume excess mean the hole diameter is 10% larger?
Show the worked answer
No. The diameter must be recovered from the squared-area relationship with the casing diameter retained.
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
- Distinguish diameter enlargement from volume excess.
- Do not count caliper enlargement twice.
- State which intervals receive excess.
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.