Find total temperature from static flow conditions. Switch units, solve inverses, and validate inputs instantly. Perfect for ducts, nozzles, and wind tunnel analysis work.
For adiabatic, isentropic compressible flow, stagnation temperature relates to static temperature by:
In energy form (useful when velocity and cp are known):
Notes: γ is the specific heat ratio, M is Mach number, V is flow speed, and cp is specific heat at constant pressure.
Sample values are illustrative for typical gas dynamics problems.
| Case | Mode | Inputs | Output |
|---|---|---|---|
| 1 | T₀ from T, M | T = 288.15 K, M = 0.8, γ = 1.4 | T₀ \u2248 325.033 K |
| 2 | T₀ from T, M | T = 300 K, M = 2.0, γ = 1.4 | T₀ = 540 K |
| 3 | T₀ from T, V | T = 20 \u00b0C, V = 250 m/s, cp = 1005 J/(kg\u00b7K) | T₀ \u2248 324.244 K |
| 4 | M from T₀, T | T₀ = 500 K, T = 300 K, γ = 1.33 | M \u2248 2.010 |
Stagnation temperature (T0) is the total temperature a moving gas would reach if it were brought to rest adiabatically with no external work. It is a convenient “energy level” for a flow because it stays constant across isentropic ducts and nozzles even when static temperature changes.
For an ideal gas in isentropic flow, the calculator uses T0/T = 1 + (γ−1)M2/2. With air (γ ≈ 1.40), common ratios are: M = 0.3 → T0/T ≈ 1.018, M = 0.8 → ≈ 1.128, and M = 2.0 → 1.800. These values help sanity‑check inputs.
When you know speed instead of Mach number, the energy form is practical: T0 = T + V2/(2cp). For air near room conditions, cp is often 1004–1007 J/(kg·K). At T = 293.15 K and V = 250 m/s with cp = 1005 J/(kg·K), T0 rises by about 31.1 K.
γ depends on gas composition and temperature. Air is commonly modeled as 1.40 at moderate temperatures, while CO2 is closer to 1.30. If you provide the specific gas constant R, the calculator can estimate cp from cp = γR/(γ−1), which is useful when a consistent property set is needed.
Engineering problems often require inverses: solving for T from T0 and M, estimating M from T0/T, or finding V from a measured temperature rise. This tool supports those workflows with consistent validation (for example, it enforces T0 ≥ T for real solutions).
In wind tunnels and intakes, a probe may not recover the full total temperature because of heat transfer and viscous effects. The computed T0 is the ideal isentropic value. If you use measured probe temperatures, treat them as “recovered” totals and interpret differences as losses or recovery factor effects.
Temperature inputs can be entered in K, °C, °F, or °R and are converted internally to Kelvin. Velocities can be entered in m/s, km/h, mph, or ft/s and are converted to m/s. Keeping units consistent prevents hidden errors, especially in the V2 term.
For subsonic ventilation or HVAC ducts, M is typically below 0.3, so T0 is only slightly above T. In nozzles and high‑speed jets, M can exceed 1 and T can drop significantly while T0 remains nearly constant for isentropic sections. Use the ratio T0/T as a quick indicator of compressibility intensity.
No. Static temperature is the local thermodynamic temperature of the moving gas. Stagnation temperature includes the kinetic energy effect and is higher than static temperature whenever the flow speed is nonzero.
Use Mach-number mode when compressible flow conditions are known or derived from gas dynamics. Use velocity mode when you have measured speed and a reliable cp (or R and γ) for the operating temperature range.
A common engineering value is γ = 1.40 for air near room temperature. At higher temperatures, γ can decrease. For accuracy-critical work, use property tables or a model that accounts for temperature-dependent specific heats.
In the ideal relations used here, T0/T = 1 + (γ−1)M2/2, which is always at least 1. If T0 is less than T, there is no real Mach number or velocity solution.
Yes. If you provide R and γ, the tool estimates cp using cp = γR/(γ−1). This is a useful approximation for ideal-gas calculations with constant properties.
Across a normal shock, the flow is not isentropic. Stagnation temperature is approximately conserved if there is no external heat transfer, but stagnation pressure drops due to irreversibility. Interpret results accordingly.
Accuracy depends on cp being representative and on negligible heat transfer and shaft work. For moderate speeds and small temperature changes, it works well. For very high temperatures, variable properties should be considered.
Important Note: All the Calculators listed in this site are for educational purpose only and we do not guarentee the accuracy of results. Please do consult with other sources as well.