Original synthetic study
One surface pressure two different equivalent mud weights
The same 500 psi surface term represents 1.9231 ppg at 5,000 ft and 0.9615 ppg at 10,000 ft. Equivalent density is inseparable from its evaluation depth.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Design the comparison around one pressure term
This synthetic study holds surface pressure at 500 psi and base mud density at 10 ppg. Depth is the only independent variable. The table records the equivalent increment separately from the total equivalent mud weight.
Holding pressure fixed is essential. If pressure increased with depth because the physical fluid changed, the experiment would no longer isolate the effect of the denominator.
Read the inverse relationship
The product of equivalent increment and TVD stays at 500/0.052 in this example. Doubling depth therefore halves the increment, while adding the unchanged 10 ppg base gives totals that approach 10 ppg at large depths.
That asymptote describes the mathematical representation of a fixed surface term. It does not predict the maximum acceptable downhole density or the strength of a formation.
Transfer the result to a worksheet
A reported surface-equivalent mud weight should name the depth, base fluid and included pressure terms. These three labels let another reader reconstruct the pressure budget.
Compare this study with the pressure-gradient conversion lesson: a real uniform density gives a fixed gradient, whereas a fixed pressure divided by changing depth gives a varying equivalent gradient.
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.
Evaluation TVD (ft) = 8000. Equivalent increment (ppg): 1.20192308; Total EMW (ppg): 11.2019231.
- Equivalent increment (ppg)
- Total EMW (ppg)
Worked example
One pressure expressed over different depths
Fixed 500 psi added pressure and 10 ppg base mud. TVD is the only changing quantity.
Relation and unit basis
ΔMW = 500/(0.052 TVD); total EMW = 10 + ΔMW
Density increments and total equivalent density are in ppg; the actual fluid remains 10 ppg.
1. Define the baseline
Evaluation TVD (ft) = 8000
Fixed 500 psi added pressure and 10 ppg base mud. TVD is the only changing quantity.
2. Select the relation
ΔMW = 500/(0.052 TVD); total EMW = 10 + ΔMW
Density increments and total equivalent density are in ppg; the actual fluid remains 10 ppg.
3. Substitute the values
500/(0.052 × 8,000) = 1.20192308; total = 11.20192308 ppg
Retain intermediate precision. The displayed values are rounded only after evaluation.
4. Read the result
Equivalent increment (ppg) = 1.20192308; Total EMW (ppg) = 11.2019231
A fixed pressure has a smaller equivalent density increment at a deeper evaluation point.
Result and interpretation
Equivalent increment (ppg) = 1.20192308; Total EMW (ppg) = 11.2019231
A fixed pressure has a smaller equivalent density increment at a deeper evaluation point.
Use the data table to verify that ΔMW × depth remains constant, then compare total EMW and the increment.
Study data and reproduction
Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
| Evaluation TVD (ft) | Equivalent increment (ppg) | Total EMW (ppg) |
|---|---|---|
| 2,500 | 3.846154 | 13.84615 |
| 5,000 | 1.923077 | 11.92308 |
| 8,000 | 1.201923 | 11.20192 |
| 10,000 | 0.9615385 | 10.96154 |
| 15,000 | 0.6410256 | 10.64103 |
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. At which depth does a 500 psi term equal a 1 ppg increment?
Show the worked answer
TVD = 500/(0.052 × 1) = 9,615.3846 ft.
2. Reproduce the baseline Equivalent increment (ppg) without rounding intermediate values.
Show the worked answer
500/(0.052 × 8,000) = 1.20192308; total = 11.20192308 ppg. Equivalent increment (ppg) = 1.20192308. A fixed pressure has a smaller equivalent density increment at a deeper evaluation point.
3. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
Evaluation TVD (ft) changes from 2500 to 15000. Other controls remain as stated in the premise. Equivalent increment (ppg) changes from 3.84615385 to 0.641025641.
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. Density increments and total equivalent density are in ppg; the actual fluid remains 10 ppg. A fixed pressure has a smaller equivalent density increment at a deeper evaluation point.
Before using the result
- Hold the 500 psi term fixed.
- Plot the increment and total with their units.
- Record the base fluid separately.
A fixed pressure has a smaller equivalent density increment at a deeper evaluation point.
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.