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Dimensionless Numbers Calculator (Nu, Pr, Bi, Fo)

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Understanding the dimensionless numbers calculator — concept and principle

Heat-transfer engineers describe problems with dimensionless groups so that results from one size, fluid or speed can be applied to another. This calculator evaluates four of the most common: the Nusselt, Prandtl, Biot and Fourier numbers, from the heat transfer coefficient, a characteristic length and basic material properties.

Nusselt, Prandtl, Biot and Fourier numbersWith water-like properties: Nu = 4.17, Pr = 6.97, Bi = 4.17 and Fo = 0.00336.Four ratios that describe heat flowNuhL ÷ k50 × 0.05 ÷ 0.6= 4.17convection vs conductionPrcₚμ ÷ k4180 × 0.001 ÷ 0.6= 6.97momentum vs heat spreadBihL ÷ k_solid50 × 0.05 ÷ 0.6= 4.17inside vs surfaceFoαt ÷ L²1.4e-7 × 60 ÷ 0.05²= 0.00336how far heat soaked inBi 4.17 > 0.1: the inside is not one temperatureUse fluid k for Nu, solid k for Bi
Each number is a ratio of two effects; with the default inputs Nu = Bi = 4.17, Pr = 6.97, Fo = 0.00336.

How it is calculated

Nu and Bi have the same form but use different conductivities, so the calculator has separate inputs: k for the fluid (Nu) and k_s for the solid (Bi).

The principle

Each group is a ratio of two competing effects. A large Nu means the flow carries heat far better than still fluid would; Nu = 1 corresponds to pure conduction. Pr compares how quickly velocity and temperature disturbances spread, so it sets the relative thickness of the velocity and thermal boundary layers. Bi tells you whether a solid heats uniformly (small Bi) or develops internal gradients (large Bi). The Grashof number, Gr = gβΔT·L³/ν², which drives natural convection, needs expansion coefficient and kinematic viscosity inputs and is not computed here.

Worked example

  1. Inputs: h = 50 W/m²K, L = 0.05 m, k = 0.6 W/m·K, c_p = 4180 J/kg·K, μ = 0.001 Pa·s, α = 1.4 × 10⁻⁷ m²/s, t = 60 s (properties close to water near room temperature).
  2. Nu = 50 × 0.05 / 0.6 ≈ 4.167.
  3. Pr = 4180 × 0.001 / 0.6 ≈ 6.967.
  4. Bi = 50 × 0.05 / 0.6 ≈ 4.167 (the default k_s is also 0.6, so Bi equals Nu here).
  5. Fo = 1.4 × 10⁻⁷ × 60 / 0.05² = 0.00336.
How Nusselt Nu = hL/k changes with Fluid conductivity k_f (W/m·K)How Nusselt Nu = hL/k changes with Fluid…051015200.150.30.450.60.750.91.2Example: 4.166666667Fluid conductivity k_f (W/m·K)
Each point is the calculator re-run with only “Fluid conductivity k_f (W/m·K)” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.
Fluid conductivity k_f (W/m·K)Nusselt Nu = hL/k
0.1516.66666667
0.38.333333333
0.455.555555556
0.64.166666667
0.753.333333333
0.92.777777778
1.22.083333333

How to read the result

Typical Pr values: about 0.7 for air, about 7 for water at 20 °C, and hundreds or more for oils. If Bi < 0.1, a solid can be treated as having uniform temperature (lumped capacitance). A small Fo, as here, means heat has penetrated only a thin outer layer; one-term series solutions for transient conduction are usually accurate once Fo exceeds about 0.2.

Common mistakes

Keep in mind: These numbers only describe a problem; the heat transfer itself comes from correlations or solutions valid for specific ranges. Engineering decisions should be checked against references and by a qualified engineer.

Frequently asked questions

What characteristic length should I use?

It depends on the correlation: plate length for flat plates, diameter for tubes and cylinders, and V/A for lumped transient analysis.

How do I get h from Nu?

Rearrange: h = Nu·k / L, with Nu taken from a suitable correlation such as Dittus–Boelter for turbulent pipe flow.

Sources: Incropera, DeWitt, Bergman and Lavine, Fundamentals of Heat and Mass Transfer, Wiley · Bird, Stewart and Lightfoot, Transport Phenomena, Wiley

Checked 8 Nov 2026. Standard textbook formulas in SI units, computed in your browser. For learning and estimates — real designs must follow your local codes and be checked by a qualified engineer.

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