Engineering lesson
What an inflow curve cannot tell you about a producing system
An inflow curve describes reservoir delivery versus flowing bottomhole pressure. The producing-system rate also needs an outflow relation and compatible boundary conditions.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
An intersection supplies the operating condition
A well's feasible steady point is where inflow and outflow agree on rate and the same bottomhole pressure basis. An inflow maximum alone does not include tubing or surface constraints.
The teaching example uses a transparent synthetic outflow relation with a linear inflow. It is a geometry-of-intersection exercise, not a calibrated tubing model.
Keep pressure locations consistent
Wellhead pressure is not flowing bottomhole pressure. Converting between them can require elevation, friction, gas expansion and temperature behavior.
A comparison that puts wellhead pressure on an IPR axis without an outflow model does not describe a valid operating point.
Use the right scope for the next calculation
The linked Vogel calculator evaluates inflow under its assumptions. The site's multiphase profile workflows can support separate pressure-path studies but do not automatically form a fully coupled nodal solver.
Document how the curves were constructed and whether their property states remain consistent at the intersection.
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.
Outflow base pressure (psi) = 1500. Intersection rate (STB/day): 1500; Matched bottom pressure (psi): 2250.
- Intersection rate (STB/day)
Worked example
A transparent synthetic inflow–outflow intersection
Inflow q = 2(3,000 − pwf) STB/day. Teaching outflow pwf = pbase + 0.5q psi; pbase varies.
Relation and unit basis
q = (6,000 − 2pbase)/2; pwf = 3,000 − q/2
The outflow is an explicitly invented linear teaching relation, not a tubing correlation.
1. Define the baseline
Outflow base pressure (psi) = 1500
Inflow q = 2(3,000 − pwf) STB/day. Teaching outflow pwf = pbase + 0.5q psi; pbase varies.
2. Select the relation
q = (6,000 − 2pbase)/2; pwf = 3,000 − q/2
The outflow is an explicitly invented linear teaching relation, not a tubing correlation.
3. Substitute the values
At pbase = 1,500: q = 2[3,000 − (1,500 + 0.5q)]; 2q = 3,000; q = 1,500 STB/day and pwf = 2,250 psi
Retain intermediate precision. The displayed values are rounded only after evaluation.
4. Read the result
Intersection rate (STB/day) = 1500; Matched bottom pressure (psi) = 2250
Only a compatible intersection supplies the synthetic operating point.
Result and interpretation
Intersection rate (STB/day) = 1500; Matched bottom pressure (psi) = 2250
Only a compatible intersection supplies the synthetic operating point.
The imposed surface-system burden changes a synthetic outflow curve; the inflow model remains fixed.
Exact scenario data
Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
| Outflow base pressure (psi) | Intersection rate (STB/day) | Matched bottom pressure (psi) |
|---|---|---|
| 500 | 2,500 | 1,750 |
| 1,000 | 2,000 | 2,000 |
| 1,500 | 1,500 | 2,250 |
| 2,000 | 1,000 | 2,500 |
| 2,500 | 500 | 2,750 |
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. Does qmax from an IPR equal the actual producing rate?
Show the worked answer
Generally no. It is an inflow-model intercept; actual rate depends on the compatible outflow and boundary conditions.
2. Reproduce the baseline Intersection rate (STB/day) without rounding intermediate values.
Show the worked answer
At pbase = 1,500: q = 2[3,000 − (1,500 + 0.5q)]; 2q = 3,000; q = 1,500 STB/day and pwf = 2,250 psi. Intersection rate (STB/day) = 1500. Only a compatible intersection supplies the synthetic operating point.
3. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
Outflow base pressure (psi) changes from 500 to 2500. Other controls remain as stated in the premise. Intersection rate (STB/day) changes from 2500 to 500.
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. The outflow is an explicitly invented linear teaching relation, not a tubing correlation. Only a compatible intersection supplies the synthetic operating point.
Before using the result
- Use one bottomhole-pressure datum.
- Construct outflow explicitly.
- Do not claim an uncoupled tool solves the whole system.
Only a compatible intersection supplies the synthetic operating point.
Continue learning
Start with the broader workflow guide · Related method 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.