Original synthetic study
Same flow rate different hole sizes a synthetic hydraulics study
This controlled study sends the same 600 gpm through several synthetic hole sizes around one 5 in string. Only annular area changes, so the velocity difference follows directly from continuity.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Define the controls before interpreting a trend
The pipe OD, actual volumetric rate and fluid-volume basis are fixed. Each row changes only hole diameter. The experiment uses the site's published annular-velocity relationship and reports both ft/s and ft/min.
No pressure-loss correlation is attached to the velocity result. Pressure response also depends on rheology, hydraulic geometry and regime, so a velocity ratio alone is insufficient to predict ECD.
Read the reciprocal-area response
For two diameters, v2/v1 = (Dh1² − Dp²)/(Dh2² − Dp²). The rate and conversion coefficient cancel. This independent ratio is a useful check on the tabulated values.
The curve is steep near a small annular gap. It flattens as hole diameter grows, because each additional inch adds increasingly large outer area.
What the experiment supports
The study demonstrates why one pump rate does not imply one annular velocity throughout a changing well geometry. It does not identify a critical cleaning velocity or predict cuttings transport.
Use the downloadable rows to reproduce the ratio calculation. A physical cleaning study would additionally need inclination, solids characteristics, rheology and local flow observations.
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. Average velocity (ft/s): 5.18726009; Average velocity (ft/min): 311.235605.
- Average velocity (ft/s)
Worked example
Continuity through changing hole sizes
Actual rate 600 US gpm, pipe OD 5 in, hole diameter varies. All local volume assumptions remain fixed.
Relation and unit basis
v = Q/[2.448(Dh² − Dp²)] ft/s; ft/min = 60v
This average speed is not a local velocity field or a cuttings-transport criterion.
1. Define the baseline
Hole diameter (in) = 8.5
Actual rate 600 US gpm, pipe OD 5 in, hole diameter varies. All local volume assumptions remain fixed.
2. Select the relation
v = Q/[2.448(Dh² − Dp²)] ft/s; ft/min = 60v
This average speed is not a local velocity field or a cuttings-transport criterion.
3. Substitute the values
600/[2.448 × (8.5² − 5²)] = 5.18726009 ft/s; multiply by 60 for ft/min
Retain intermediate precision. The displayed values are rounded only after evaluation.
4. Read the result
Average velocity (ft/s) = 5.18726009; Average velocity (ft/min) = 311.235605
The wider local flow area lowers average speed at fixed actual Q.
Result and interpretation
Average velocity (ft/s) = 5.18726009; Average velocity (ft/min) = 311.235605
The wider local flow area lowers average speed at fixed actual Q.
The original dataset is a geometry sweep, with no measured hole-cleaning outcomes or accuracy ranking.
Study data and reproduction
Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
| Hole diameter (in) | Average velocity (ft/s) | Average velocity (ft/min) |
|---|---|---|
| 6.5 | 14.20858 | 852.5149 |
| 7.5 | 7.843137 | 470.5882 |
| 8.5 | 5.18726 | 311.2356 |
| 10 | 3.267974 | 196.0784 |
| 12.25 | 1.959804 | 117.5883 |
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 doubling hole diameter not equivalent to halving annular speed?
Show the worked answer
The open area depends on the difference of squared hole and pipe diameters. It does not increase linearly with hole diameter.
2. Reproduce the baseline Average velocity (ft/s) without rounding intermediate values.
Show the worked answer
600/[2.448 × (8.5² − 5²)] = 5.18726009 ft/s; multiply by 60 for ft/min. Average velocity (ft/s) = 5.18726009. The wider local flow area lowers average speed at fixed actual Q.
3. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
Hole diameter (in) changes from 6.5 to 12.25. Other controls remain as stated in the premise. Average velocity (ft/s) changes from 14.208582 to 1.95980441.
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. This average speed is not a local velocity field or a cuttings-transport criterion. The wider local flow area lowers average speed at fixed actual Q.
Before using the result
- Hold actual Q fixed.
- Use the same pipe OD throughout.
- Do not infer transport performance from speed alone.
The wider local flow area lowers average speed at fixed actual Q.
Continue learning
Start with the broader workflow guide · Related method 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.