Engineering lesson
Initial versus final circulating pressure on a worksheet
Initial circulating pressure adds stabilized SIDPP to a recorded slow-rate friction pressure. Final circulating pressure scales that friction reference by kill mud weight divided by original mud weight, under the worksheet assumptions.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Give each pressure term a job
SIDPP represents the pressure shortfall left by the original static column in the simplified balance. The slow-circulating-rate pressure is a friction reference at a specified pump rate and flow path. Adding them gives an initial schedule value.
For 500 psi SIDPP and a 900 psi slow-rate reference, ICP is 1,400 psi. That addition does not mean that all pump pressure is transmitted as an extra bottomhole pressure.
The final estimate changes the fluid reference
With original mud 12 ppg and kill mud 12.9615 ppg, the worksheet ratio gives approximately 972.1 psi FCP. This is a scaling approximation for the chosen circulation condition.
The estimate depends on preserving the reference pump rate, path and relevant rheological behavior. Changing these conditions while retaining 900 psi as the reference describes a different system.
A schedule needs context
Keep the original recorded value, pump speed and mud basis alongside the derived pressures. These values explain the pressure chain but do not prescribe choke movements or override observed well behavior.
Initial and final schedule pressures also do not replace shoe-pressure or equipment checks. A friction estimate and a separate surface pressure constraint answer different questions.
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.
Slow-rate reference (psi) = 900. ICP (psi): 1400; FCP (psi): 972.115385.
- ICP (psi)
- FCP (psi)
Worked example
Initial and final worksheet schedule values
SIDPP 500 psi, original mud 12 ppg, TVD 10,000 ft. Kill density = 12 + 500/(0.052 × 10,000) = 12.96153846 ppg.
Relation and unit basis
ICP = SCR + SIDPP; FCP = SCR × KWM/OMW
SCR must be recorded at the same intended pump rate and path. This example is schedule arithmetic, not an operating procedure.
1. Define the baseline
Slow-rate reference (psi) = 900
SIDPP 500 psi, original mud 12 ppg, TVD 10,000 ft. Kill density = 12 + 500/(0.052 × 10,000) = 12.96153846 ppg.
2. Select the relation
ICP = SCR + SIDPP; FCP = SCR × KWM/OMW
SCR must be recorded at the same intended pump rate and path. This example is schedule arithmetic, not an operating procedure.
3. Substitute the values
ICP = 900 + 500 = 1,400; FCP = 900 × 12.96153846/12 = 972.115385 psi
Retain intermediate precision. The displayed values are rounded only after evaluation.
4. Read the result
ICP (psi) = 1400; FCP (psi) = 972.115385
ICP contains an additive pressure imbalance; FCP uses the mud-density ratio.
Result and interpretation
ICP (psi) = 1400; FCP (psi) = 972.115385
ICP contains an additive pressure imbalance; FCP uses the mud-density ratio.
Vary the recorded slow-rate pressure. The different slopes of ICP and FCP expose the additive and proportional parts of the worksheet.
Exact scenario data
Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
| Slow-rate reference (psi) | ICP (psi) | FCP (psi) |
|---|---|---|
| 500 | 1,000 | 540.0641 |
| 700 | 1,200 | 756.0897 |
| 900 | 1,400 | 972.1154 |
| 1,000 | 1,500 | 1,080.128 |
| 1,200 | 1,700 | 1,296.154 |
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 a 1,000 psi slow-rate reference with the same mud ratio and 500 psi SIDPP, what is ICP?
Show the worked answer
1,500 psi. FCP is approximately 1,080.1 psi from 1,000 × 12.9615/12.
2. Reproduce the baseline ICP (psi) without rounding intermediate values.
Show the worked answer
ICP = 900 + 500 = 1,400; FCP = 900 × 12.96153846/12 = 972.115385 psi. ICP (psi) = 1400. ICP contains an additive pressure imbalance; FCP uses the mud-density ratio.
3. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
Slow-rate reference (psi) changes from 500 to 1200. Other controls remain as stated in the premise. ICP (psi) changes from 1000 to 1700.
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. SCR must be recorded at the same intended pump rate and path. This example is schedule arithmetic, not an operating procedure. ICP contains an additive pressure imbalance; FCP uses the mud-density ratio.
Before using the result
- Match the reference pump rate.
- Keep original and kill mud density distinct.
- Use schedule values within their stated worksheet assumptions.
ICP contains an additive pressure imbalance; FCP uses the mud-density ratio.
Continue learning
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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.