What is the Biot Number?

The Biot number (Bi) is a dimensionless quantity used in heat transfer analysis that compares the rate of heat conduction within a body to the rate of heat convection at its surface. Named after French physicist Jean-Baptiste Biot, this number helps engineers determine whether the temperature within an object can be assumed uniform during transient (time-dependent) heat transfer.

The Biot number is defined as Bi = hLc/k, where h is the heat transfer coefficient at the surface, Lc is the characteristic length (typically volume/surface area), and k is the thermal conductivity of the material. A small Biot number (Bi < 0.1) indicates that heat conduction within the object is much faster than heat convection away from its surface, meaning temperature gradients inside the object are negligible.

When Bi < 0.1, the "lumped capacitance" or "lumped system analysis" method can be used, greatly simplifying transient heat transfer calculations. In this regime, the entire object can be treated as having a single, uniform temperature that changes with time. For Bi > 0.1, temperature varies significantly within the object, requiring more complex analytical or numerical methods to solve the heat equation.

How to Calculate Biot Number

The Formula

Bi = (h × Lc) / k

Bi = Biot number (dimensionless)

h = Heat transfer coefficient (W/m²·K)

Lc = Characteristic length (m), typically V/As

k = Thermal conductivity of material (W/m·K)

Interpretation Guidelines

Bi < 0.1 (Lumped System)

Internal conduction resistance is negligible. Temperature is uniform throughout the object. Lumped capacitance analysis is valid.

0.1 < Bi < 10 (Intermediate)

Both internal and external resistances are significant. Temperature gradients exist within the object. Requires analytical solutions (Heisler charts) or numerical methods.

Bi > 10 (Non-Lumped System)

Surface convection resistance is negligible compared to internal conduction. Surface temperature equals fluid temperature. Internal temperature distribution governs the problem.

Example Calculation

A steel sphere (k = 50 W/m·K) with diameter 2 cm is cooled in air with h = 25 W/m²·K:

Given:

Thermal conductivity: k = 50 W/m·K

Heat transfer coefficient: h = 25 W/m²·K

Diameter: D = 0.02 m

Step 1:

Calculate characteristic length (for sphere: Lc = V/As = r/3):

Lc = r/3 = 0.01/3 = 0.00333 m

Step 2:

Calculate Biot number:

Bi = (h × Lc) / k = (25 × 0.00333) / 50 = 0.00167

Result:

Bi = 0.00167

Since Bi = 0.00167 < 0.1, the lumped capacitance method is valid! The steel sphere can be treated as having uniform temperature during cooling, greatly simplifying the transient heat transfer analysis.

Frequently Asked Questions

What is characteristic length (Lc)?

Characteristic length is the ratio of volume to surface area (V/As). For simple shapes: sphere = r/3, cylinder = r/2, flat plate = thickness/2, cube = side/6. It represents how 'thick' the object is relative to its surface.

Why is Bi < 0.1 the criterion?

When Bi < 0.1, the maximum temperature difference within the object is less than 5% of the overall temperature difference between object and surroundings. This makes the uniform temperature assumption acceptably accurate for most engineering applications.

What if my Biot number is exactly 0.1?

Bi = 0.1 is a rule-of-thumb boundary, not a strict cutoff. For Bi between 0.05 and 0.15, lumped analysis introduces some error but may still be acceptable depending on required accuracy. For critical applications, use more exact methods when Bi > 0.05.

How does Biot number differ from Nusselt number?

Both are dimensionless heat transfer numbers, but Biot number (hLc/k) uses the solid's thermal conductivity and determines if internal gradients matter. Nusselt number (hLc/kfluid) uses fluid conductivity and relates convection to conduction in the fluid boundary layer.

Can Biot number change during cooling?

Yes! The heat transfer coefficient h often depends on temperature, flow regime, or phase change. For example, h decreases significantly during film boiling. Recalculate Bi if conditions change substantially during the process.

What materials typically allow lumped analysis?

Metals (high k) in gases (low h) commonly satisfy Bi < 0.1: steel in air, aluminum fins, copper spheres cooling naturally. Poor conductors (plastics, ceramics) or liquids with high h typically require distributed analysis.

Is there a similar number for mass transfer?

Yes! The Biot number for mass transfer uses mass transfer coefficient and diffusivity: Bim = hmLc/D. It determines if concentration can be assumed uniform. The criteria and interpretation are analogous to thermal Biot number.

What's lumped capacitance analysis?

When Bi < 0.1, the temperature response follows: T(t) = T∞ + (T0 - T∞)exp(-hAs/ρVc × t). The entire object cools exponentially as one 'lump' with time constant τ = ρVc/hAs. Much simpler than solving the heat equation spatially.

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