Engineering lesson
Annular pressure loss through several hole sections
Calculate frictional pressure loss for each interval using its own geometry, length, flow and rheology. Add compatible losses along the same path; do not average diameters before solving.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Serial intervals share rate, not gradient
In a steady incompressible path with no branches, the same Q passes through successive annuli. A narrow interval can require a much larger pressure gradient than a wider interval at that rate.
The worked example uses the site's Herschel–Bulkley annular solver for two concentric intervals. The common rheology is retained while outer diameter and interval length change.
Sum losses after solving each geometry
Multiply each computed gradient by that interval's length, then add the pressure drops. A total measured path length is appropriate for friction, while hydrostatic terms use the vertical rise or fall.
For variable fluid properties, evaluate local states as well. A constant-rheology ledger assumes its measured parameters represent every interval in the example.
Translate only the intended loss into ECD
Annular loss can be represented as a density increment at a specified TVD. Pipe and bit losses are different terms and should not automatically be added to annular ECD.
The educational model also needs an applicability check: solving the laminar HB relation does not demonstrate that the selected flow is laminar.
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.
Second interval length (m) = 500. First loss (Pa): 226381.448; Second loss (Pa): 145796.177; Total loss (Pa): 372177.625.
- First loss (Pa)
- Second loss (Pa)
- Total loss (Pa)
Worked example
Two separately solved HB annular intervals
First interval D/d = 0.2/0.1 m, L = 500 m; second D/d = 0.25/0.1 m with variable length. Q 0.001 m³/s, K 0.2 Pa·s^0.5, n 0.5, τy 10 Pa, ρ 1,200 kg/m³.
Relation and unit basis
Δptotal = G1 L1 + G2 L2
Both gradients are separately solved by the supported HB annulus engine. Pressure units are Pa.
1. Define the baseline
Second interval length (m) = 500
First interval D/d = 0.2/0.1 m, L = 500 m; second D/d = 0.25/0.1 m with variable length. Q 0.001 m³/s, K 0.2 Pa·s^0.5, n 0.5, τy 10 Pa, ρ 1,200 kg/m³.
2. Select the relation
Δptotal = G1 L1 + G2 L2
Both gradients are separately solved by the supported HB annulus engine. Pressure units are Pa.
3. Substitute the values
First loss at 500 m = 226381.448; second at 500 m = 145796.177; sum = 372177.625 Pa
Retain intermediate precision. The displayed values are rounded only after evaluation.
4. Read the result
First loss (Pa) = 226381.448; Second loss (Pa) = 145796.177; Total loss (Pa) = 372177.625
Serial sections share the same Q but not the same gradient. This is a laminar HB calculation with its stated regime assumptions.
Result and interpretation
First loss (Pa) = 226381.448; Second loss (Pa) = 145796.177; Total loss (Pa) = 372177.625
Serial sections share the same Q but not the same gradient. This is a laminar HB calculation with its stated regime assumptions.
Change the second interval's length. Its geometry is solved separately before losses are added.
Exact scenario data
Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
| Second interval length (m) | First loss (Pa) | Second loss (Pa) | Total loss (Pa) |
|---|---|---|---|
| 0 | 226,381.4 | 0 | 226,381.4 |
| 250 | 226,381.4 | 72,898.09 | 299,279.5 |
| 500 | 226,381.4 | 145,796.2 | 372,177.6 |
| 750 | 226,381.4 | 218,694.3 | 445,075.7 |
| 1,000 | 226,381.4 | 291,592.4 | 517,973.8 |
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. If the second interval has zero length, what should the total loss become?
Show the worked answer
Exactly the first interval's loss. This is a useful ledger boundary check.
2. Reproduce the baseline First loss (Pa) without rounding intermediate values.
Show the worked answer
First loss at 500 m = 226381.448; second at 500 m = 145796.177; sum = 372177.625 Pa. First loss (Pa) = 226381.448. Serial sections share the same Q but not the same gradient. This is a laminar HB calculation with its stated regime assumptions.
3. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
Second interval length (m) changes from 0 to 1000. Other controls remain as stated in the premise. First loss (Pa) changes from 226381.448 to 226381.448.
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. Both gradients are separately solved by the supported HB annulus engine. Pressure units are Pa. Serial sections share the same Q but not the same gradient. This is a laminar HB calculation with its stated regime assumptions.
Before using the result
- Solve each distinct cross section.
- Use measured path length for friction.
- Check the rheology and flow-regime scope.
Serial sections share the same Q but not the same gradient. This is a laminar HB calculation with its stated regime assumptions.
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.