Practice and answer pack
Pressure and depth basis worksheet for engineering students
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.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Build a pressure-basis ledger
For every problem write the location, vertical depth, fluid density and top boundary pressure. Leave a separate column for MD even when it is not used, so the choice of depth is explicit.
The worked case begins with a 10 ppg uniform column. Its pressure gradient is 0.520 psi/ft. The additional exercises deliberately change one item at a time so a dimensional error can be traced to its source.
Use the reverse calculation as a check
After calculating pressure, divide the fluid-head contribution by 0.052 and TVD to recover density. A surface boundary must first be subtracted; otherwise the reverse calculation returns an equivalent total rather than actual mud density.
For multiple fluids, use the weighted vertical integral. A simple arithmetic average is valid only for equal vertical heights. Keep each interval in the ledger rather than hiding it inside an unlabelled average.
Explain the answer, including its scope
An answer key should state why TVD was used, whether the result is a difference or a state pressure, and what was assumed about density. Numerical agreement alone cannot reveal a mistaken pressure reference.
Print the page for an exercise sheet or download the case data. These problems address static arithmetic; a circulating or well-control procedure requires additional information that the worksheet does not provide.
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.
Mud weight (ppg) = 10. Gradient (psi/ft): 0.52; Fluid head (psi): 4160.
- Gradient (psi/ft)
Worked example
Mud weight, gradient and fluid head
Uniform mud, 8,000 ft vertical height, no added surface pressure. Mud weight is the only changing input.
Relation and unit basis
gradient = 0.052 MW; ΔP = gradient × TVD
MW in ppg, gradient in psi/ft, TVD in ft, ΔP in psi. The rounded field coefficient assumes these units.
1. Define the baseline
Mud weight (ppg) = 10
Uniform mud, 8,000 ft vertical height, no added surface pressure. Mud weight is the only changing input.
2. Select the relation
gradient = 0.052 MW; ΔP = gradient × TVD
MW in ppg, gradient in psi/ft, TVD in ft, ΔP in psi. The rounded field coefficient assumes these units.
3. Substitute the values
0.052 × 10 = 0.520 psi/ft; 0.520 × 8,000 = 4,160 psi
Retain intermediate precision. The displayed values are rounded only after evaluation.
4. Read the result
Gradient (psi/ft) = 0.52; Fluid head (psi) = 4160
Fluid head is a pressure difference. A top boundary pressure would be added separately.
Result and interpretation
Gradient (psi/ft) = 0.52; Fluid head (psi) = 4160
Fluid head is a pressure difference. A top boundary pressure would be added separately.
Use the baseline worked example, then solve the additional questions without changing more than one basis at a time.
Exact scenario data
Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
| Mud weight (ppg) | Gradient (psi/ft) | Fluid head (psi) |
|---|---|---|
| 8 | 0.416 | 3,328 |
| 9 | 0.468 | 3,744 |
| 10 | 0.52 | 4,160 |
| 11 | 0.572 | 4,576 |
| 12 | 0.624 | 4,992 |
Scroll the table to see all columns. The highlighted row is the worked baseline. Downloads retain full numerical precision.
Practice problems and answer key
Work through the question before revealing its answer. Use the stated assumptions and include units in every numerical step.
1. A column contributes 4,160 psi over 8,000 ft. Recover its uniform density, excluding any surface pressure.
Show the worked answer
4,160/(0.052 × 8,000) = 10 ppg. This reverse check recovers the specified fluid density.
2. Reproduce the baseline Gradient (psi/ft) without rounding intermediate values.
Show the worked answer
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. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
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. 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. 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. Compare hydrostatic head for 10 ppg at 8,000 ft TVD with the wrong substitution of 12,000 ft MD.
Show the worked answer
Correct: 0.052 × 10 × 8,000 = 4,160 psi. Wrong: 6,240 psi, a 50% overstatement.
6. Sum 2,000 vertical feet of 9 ppg fluid above 6,000 vertical feet of 12 ppg fluid.
Show the worked answer
0.052(9 × 2,000 + 12 × 6,000) = 4,680 psi of head. Add any top boundary separately.
Before using the result
- Show units at each substitution.
- Identify the vertical height and pressure zero.
- Check the result by reversing the calculation.
Fluid head is a pressure difference. A top boundary pressure would be added separately.
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.