๐Ÿ“ Pitot Tube Calculator

Compute flow velocity from pitot tube differential pressure. Includes Bernoulli incompressible and isentropic compressibility corrections.

โšก Fortran 90 Engine Double Precision (IEEE 754) โœ“ ISO / ASME Validated
Pitot Tube Calculator Fluid Mechanics
๐Ÿ“Š Solver Telemetry โ— ACTIVE
๐Ÿ‘๏ธ Views 113
โšก Solves 89
๐Ÿ’พ Downloads 525 ๐Ÿ“ฆ Fortran Code 4.5 KB
๐Ÿ“… Released Jun 2026
โฑ๏ธ Latency < 1 ms
โšก TOOLS & REPORTS:
๐Ÿ’พ Download Fortran 90

๐Ÿ”ฌ Pitot-Static Probe Schematic

๐Ÿ“ Configuration

๐Ÿ“ Pressure Readings
๐Ÿ’ง Fluid Properties
๐Ÿ”ง Pipe (optional)
Key Equations:

Bernoulli: V = โˆš(2ฮ”P/ฯ)
Isentropic: Pt/Ps = (1+(ฮณโˆ’1)/2 Mยฒ)ฮณ/(ฮณโˆ’1)
V = Mยทa; a = โˆš(ฮณRT)
Correction = Vcomp/Vincomp

๐Ÿ“Š Results

Configure inputs and click Compute to view results.

๐Ÿ“˜ Methodology

Bernoulli vs Compressible

For M < 0.3, Bernoulli's incompressible equation gives less than 2% error. Above M = 0.3, the isentropic pressure-Mach relation must be used to avoid significant velocity underestimation.

Correction Factor

The ratio Vcomp/Vincomp quantifies how much the incompressible assumption over- or under-estimates the true velocity. The chart shows this divergence as ฮ”P increases.

Assumptions

  • Steady, one-dimensional flow at probe location.
  • Perfect gas with constant ฮณ.
  • No probe interference or alignment error.
  • Subsonic isentropic relation (no normal shock for M > 1).

๐Ÿ“˜ Calculation Methodology: Pitot-Static Tube Velocity Measurement

Mathematical Model & Theory

A Pitot-static tube measures stagnation pressure $P_0$ and static pressure $P_\infty$. Applying Bernoulli's equation yields local flow velocity $V$:

$$V = C \sqrt{\frac{2(P_0 - P_\infty)}{\rho}}$$
$$\text{Compressible: } V = \sqrt{\frac{2\gamma R T}{\gamma-1}\left[\left(\frac{P_0}{P_\infty}\right)^{\frac{\gamma-1}{\gamma}} - 1\right]}$$

Assumptions

  • Steady, frictionless streamline along probe stagnation point.
  • Instrument coefficient $C pprox 0.98 - 1.00$.

Academic References

  1. ISO 3966: Measurement of fluid flow in closed conduits.
  2. White, F. M.: Fluid Mechanics, Ch. 6.

Worked Engineering Example

Problem Statement:
A pitot tube in air ($\rho = 1.20\text{ kg/m}^3$) reads $\Delta P = 450\text{ Pa}$ with $C = 0.995$. Calculate airspeed.

Step-by-step Solution:
1. $V = 0.995 \times \sqrt{2 \times 450 / 1.20} = 0.995 \times \sqrt{750} \approx 27.25\text{ m/s}$.
2. $V = 27.25 \times 3.6 = 98.1\text{ km/h}$.
Final Result:
Airspeed is $V = \mathbf{27.25\text{ m/s}}$ ($98.1\text{ km/h}$).
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