Engineering lesson
Calculate hydrostatic pressure across two fluid columns
Add the hydrostatic contribution of each fluid using that fluid's density and vertical height. A two-column calculation is a sum of pressure increments, not an average of densities without length weighting.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Locate the interface on a vertical datum
Suppose 2,000 vertical feet of 9 ppg fluid lie above 6,000 vertical feet of 12 ppg fluid. The interface divides the pressure integral into two intervals. The top fluid contributes 936 psi and the lower fluid contributes 3,744 psi.
Neither interval is the measured length of an inclined segment. Map interface locations to TVD before assigning vertical heights. The pressure at the interface becomes the boundary pressure for the lower interval.
A weighted average is an equivalent description
The length-weighted density is (9 × 2,000 + 12 × 6,000)/8,000 = 11.25 ppg. Applying that equivalent density over 8,000 ft reproduces 4,680 psi. The arithmetic mean 10.5 ppg would describe equal vertical heights and gives the wrong answer here.
The section ledger is usually easier to audit because it retains the physical interface. More than two fluids can be handled with the same sum when each density is suitably representative of its interval.
Add boundary pressure separately
The sum gives pressure gained below the top of the column. If the top is held at 200 psig, bottom pressure becomes 4,880 psig in this static example. Adding 200 psi does not change the hydrostatic gradient of either fluid.
Friction, gas expansion, temperature gradients and changing interfaces are outside this static calculation. Their presence calls for a model with additional state information.
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.
Upper vertical height (ft) = 2000. Upper head (psi): 936; Lower head (psi): 3744; Total head (psi): 4680.
- Upper head (psi)
- Lower head (psi)
- Total head (psi)
Worked example
Two static fluid columns
Total vertical height 8,000 ft; upper fluid 9 ppg, lower fluid 12 ppg; zero top gauge pressure.
Relation and unit basis
ΔP = 0.052[MW1 h1 + MW2(8,000 − h1)]
Each height is vertical and nonoverlapping. The fluid interface, not measured path length, divides the sum.
1. Define the baseline
Upper vertical height (ft) = 2000
Total vertical height 8,000 ft; upper fluid 9 ppg, lower fluid 12 ppg; zero top gauge pressure.
2. Select the relation
ΔP = 0.052[MW1 h1 + MW2(8,000 − h1)]
Each height is vertical and nonoverlapping. The fluid interface, not measured path length, divides the sum.
3. Substitute the values
0.052 × 9 × 2,000 + 0.052 × 12 × 6,000 = 936 + 3,744 = 4,680 psi
Retain intermediate precision. The displayed values are rounded only after evaluation.
4. Read the result
Upper head (psi) = 936; Lower head (psi) = 3744; Total head (psi) = 4680
Moving the lighter-fluid interface deeper reduces total static head at fixed total TVD.
Result and interpretation
Upper head (psi) = 936; Lower head (psi) = 3744; Total head (psi) = 4680
Moving the lighter-fluid interface deeper reduces total static head at fixed total TVD.
Move the interface while keeping total TVD fixed. The response shows why the location of a lighter fluid matters.
Exact scenario data
Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
| Upper vertical height (ft) | Upper head (psi) | Lower head (psi) | Total head (psi) |
|---|---|---|---|
| 0 | 0 | 4,992 | 4,992 |
| 1,000 | 468 | 4,368 | 4,836 |
| 2,000 | 936 | 3,744 | 4,680 |
| 4,000 | 1,872 | 2,496 | 4,368 |
| 8,000 | 3,744 | 0 | 3,744 |
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 if both vertical intervals contain 12 ppg fluid?
Show the worked answer
The full 8,000 ft column gives 0.052 × 12 × 8,000 = 4,992 psi, independent of an artificial interface.
2. Reproduce the baseline Upper head (psi) without rounding intermediate values.
Show the worked answer
0.052 × 9 × 2,000 + 0.052 × 12 × 6,000 = 936 + 3,744 = 4,680 psi. Upper head (psi) = 936. Moving the lighter-fluid interface deeper reduces total static head at fixed total TVD.
3. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
Upper vertical height (ft) changes from 0 to 8000. Other controls remain as stated in the premise. Upper head (psi) changes from 0 to 3744.
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. Each height is vertical and nonoverlapping. The fluid interface, not measured path length, divides the sum. Moving the lighter-fluid interface deeper reduces total static head at fixed total TVD.
Before using the result
- Use nonoverlapping vertical intervals.
- Carry interface pressure into the next interval.
- Keep top boundary pressure outside the fluid sum.
Moving the lighter-fluid interface deeper reduces total static head at fixed total TVD.
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.