Pressure and depth basis worksheet for engineering students
PetroCalcHub · Original synthetic teaching exercise · 2026-10-24

This worksheet practices pressure gradients, vertical column sums and pressure references. Work each question with units before opening the answer, then compare the result with the linked hydrostatic calculator.

Uniform mud, 8,000 ft vertical height, no added surface pressure. Mud weight is the only changing input.

1. A column contributes 4,160 psi over 8,000 ft. Recover its uniform density, excluding any surface pressure.

Working: ________________________________________________


2. Reproduce the baseline Gradient (psi/ft) without rounding intermediate values.

Working: ________________________________________________


3. Compare the first and last recorded scenarios. Which input changes?

Working: ________________________________________________


4. What would be needed to apply this example to a different physical system?

Working: ________________________________________________


5. Compare hydrostatic head for 10 ppg at 8,000 ft TVD with the wrong substitution of 12,000 ft MD.

Working: ________________________________________________


6. Sum 2,000 vertical feet of 9 ppg fluid above 6,000 vertical feet of 12 ppg fluid.

Working: ________________________________________________


ANSWER KEY

1. 4,160/(0.052 × 8,000) = 10 ppg. This reverse check recovers the specified fluid density.
2. 0.052 × 10 = 0.520 psi/ft; 0.520 × 8,000 = 4,160 psi. Gradient (psi/ft) = 0.52. Fluid head is a pressure difference. A top boundary pressure would be added separately.
3. Mud weight (ppg) changes from 8 to 12. Other controls remain as stated in the premise. Gradient (psi/ft) changes from 0.416 to 0.624.
4. Replace the synthetic controls with compatible documented inputs and check the relation's assumptions. MW in ppg, gradient in psi/ft, TVD in ft, ΔP in psi. The rounded field coefficient assumes these units. Fluid head is a pressure difference. A top boundary pressure would be added separately.
5. Correct: 0.052 × 10 × 8,000 = 4,160 psi. Wrong: 6,240 psi, a 50% overstatement.
6. 0.052(9 × 2,000 + 12 × 6,000) = 4,680 psi of head. Add any top boundary separately.

Method: Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
Sources:
IADC: field-unit worksheet, pressure and depth inputs: https://iadc.org/wp-content/uploads/2015/01/IADC-SubseaStack_FIELD-012214.pdf
NIST SP 811: pressure, temperature and volume conversions: https://www.nist.gov/pml/special-publication-811/nist-guide-si-appendix-b-conversion-factors/nist-guide-si-appendix-b9
SLB: hydrostatic pressure: https://glossary.slb.com/en/terms/h/hydrostatic_pressure