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Biot Number Calculator

Biot number equals heat transfer coefficient times characteristic length divided by thermal conductivity

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Biot Number

The Biot number compares the internal thermal resistance within a body to the external convective thermal resistance at its surface. When Bi < 0.1, the lumped capacitance method applies. When Bi >> 1, significant internal temperature gradients exist.

Bi = h × L / kₛ

How It Works

The Biot number (Bi) is a dimensionless quantity used in heat transfer analysis. It compares the internal thermal resistance within a body to the external convective thermal resistance at its surface. The formula is Bi = hL / kₛ, where h is the heat transfer coefficient, L is a characteristic length, and kₛ is the thermal conductivity of the solid. When Bi < 0.1, the temperature gradient inside the body is negligible, allowing the simpler lumped capacitance method. When Bi >> 1, significant internal temperature gradients exist and a detailed conduction analysis is required.

Example Problem

A sphere with a diameter of 0.1 m has a heat transfer coefficient of h = 50 W/m²·K and thermal conductivity kₛ = 200 W/m·K. For a sphere, the characteristic length is L = V/A = r/3 = 0.0167 m. What is the Biot number?

  1. Identify the known values: h = 50 W/(m²·K), radius r = 0.05 m, and kₛ = 200 W/(m·K).
  2. Compute the characteristic length for a sphere: L = V/A = r/3 = 0.05/3 = 0.0167 m.
  3. Determine what we are solving for: the Biot number to check if lumped capacitance is valid.
  4. Write the Biot number equation: Bi = h × L / kₛ.
  5. Substitute the known values: Bi = 50 × 0.0167 / 200.
  6. Compute the numerator: 50 × 0.0167 ≈ 0.833.
  7. Divide to get the result: Bi ≈ 0.833 / 200 = 0.00417. Since Bi < 0.1, the lumped capacitance method is valid.

Because Bi < 0.1, the lumped capacitance model applies and the temperature within the sphere can be assumed uniform during transient cooling or heating.

When to Use Each Variable

  • Solve for Biot Number — when you know the heat transfer coefficient, characteristic length, and thermal conductivity and need to determine whether lumped capacitance analysis is valid, e.g., checking if a steel billet cools uniformly.
  • Solve for Heat Transfer Coefficient (h) — when you know the Biot number, geometry, and material and need to find the required convection conditions, e.g., designing a forced-air cooling system to achieve a target Bi.
  • Solve for Characteristic Length (L) — when you know Bi, h, and k and need to find the maximum object size that permits lumped analysis, e.g., determining the largest part that can be treated as uniform temperature during quenching.
  • Solve for Thermal Conductivity (k) — when you know Bi, h, and L and need to identify the minimum material conductivity for uniform cooling, e.g., selecting a material for a heat sink.

Key Concepts

The Biot number is the ratio of internal conduction resistance to external convection resistance. When Bi is less than 0.1, the object's internal temperature gradients are negligible and the simpler lumped capacitance method gives accurate transient heat transfer results. When Bi is much greater than 1, significant temperature differences exist within the body and a full spatial conduction analysis is required. The characteristic length is typically the body's volume divided by its surface area.

Applications

  • Metal quenching: determining whether a heated part cools uniformly or develops internal thermal stresses
  • Food processing: validating pasteurization models by checking if small food items heat uniformly
  • Electronics cooling: verifying that small components reach steady state quickly enough for lumped thermal analysis in circuit board design
  • Casting and forging: deciding whether to use simplified or detailed thermal models for solidification simulations

Common Mistakes

  • Using the wrong characteristic length — for a sphere it is radius/3 (V/A), not the diameter or radius; using the wrong dimension changes Bi by a factor of 2-6 and may invalidate the lumped capacitance assumption
  • Applying lumped capacitance when Bi exceeds 0.1 — this threshold exists because above it, internal temperature gradients become significant enough to affect accuracy by more than 5%
  • Confusing thermal conductivity of the solid with that of the surrounding fluid — Bi uses the solid's conductivity in the denominator, not the fluid's thermal conductivity

Frequently Asked Questions

When can you assume uniform temperature inside an object?

You can assume uniform temperature when the Biot number is less than 0.1 (Bi < 0.1). This means the internal conduction resistance is so small relative to the surface convection resistance that any temperature gradients inside the body are negligible — typically less than 5% error compared to an exact spatial solution.

What does a Biot number less than 0.1 mean physically?

A Biot number below 0.1 means the solid conducts heat internally much faster than heat is removed (or added) at its surface. The entire body stays at nearly the same temperature during heating or cooling, which allows the simplified lumped capacitance analysis where the object is treated as a single thermal node.

What does a high Biot number mean?

A high Biot number (Bi >> 1) indicates that internal conduction resistance is much larger than external convection resistance. This means large temperature gradients develop inside the body, and the lumped capacitance approach is not valid.

How is characteristic length determined?

The characteristic length is typically the body's volume divided by its surface area (L = V/A). For a sphere this equals r/3, for a long cylinder r/2, and for a flat plate half the thickness. Using V/A provides a consistent basis for comparing different geometries.

What does a Biot number of 1 signify?

A Biot number of 1 means that the internal conduction resistance and external convection resistance are equal. Neither dominates, so a full conduction-convection analysis is needed to accurately predict the temperature distribution.

Can the Biot number be negative?

No. The Biot number is always a non-negative value because it is a ratio of positive thermal resistances. All three quantities in the formula (h, L, and kₛ) are positive physical properties.

How is the Biot number used in quenching and heat treatment?

In quenching, the Biot number determines whether the part develops internal thermal stresses. When Bi is large (oil or water quench on thick parts), the surface cools much faster than the center, creating steep temperature gradients that can cause warping or cracking. Engineers use Bi to choose between quench media and to size parts for uniform cooling.

Biot Number Formula

The Biot number quantifies the ratio of internal conduction resistance to external convection resistance during transient heat transfer:

Bi = h × L / kₛ

Where:

  • Bi — Biot number (dimensionless)
  • h — heat transfer coefficient, measured in W/(m²·K)
  • L — characteristic length, typically V/A, measured in meters (m)
  • kₛ — thermal conductivity of the solid, measured in W/(m·K)

When Bi < 0.1, the internal temperature gradient is negligible and the lumped capacitance model is valid. When Bi >> 1, significant temperature differences exist within the body and a full spatial conduction analysis is required.

Worked Examples

Metallurgy

Can a quenched steel billet be modeled with lumped capacitance?

A cylindrical steel billet (k = 45 W/(m·K)) with characteristic length L = 0.03 m is quenched in oil with h = 300 W/(m²·K).

  • Bi = h × L / k = 300 × 0.03 / 45
  • Bi = 9 / 45
  • Bi = 0.2

Since Bi > 0.1, lumped capacitance is not valid. A spatial conduction model is needed to capture the internal temperature gradient during quenching.

Food Processing

Does a small fruit heat uniformly during pasteurization?

A blueberry (k = 0.5 W/(m·K)) approximated as a sphere with characteristic length L = 0.003 m is heated in hot water with h = 150 W/(m²·K).

  • Bi = h × L / k = 150 × 0.003 / 0.5
  • Bi = 0.45 / 0.5
  • Bi = 0.9

Bi = 0.9 is well above 0.1, so significant temperature gradients exist inside the fruit. A spatial heat transfer model is needed for accurate pasteurization time predictions.

Electronics Cooling

Can a silicon chip be treated as isothermal during thermal analysis?

A silicon chip (k = 148 W/(m·K)) has a characteristic length of L = 0.001 m and is cooled by forced air with h = 50 W/(m²·K).

  • Bi = h × L / k = 50 × 0.001 / 148
  • Bi = 0.05 / 148
  • Bi = 0.000338

With Bi << 0.1, the chip can be modeled as isothermal. Lumped capacitance analysis gives accurate transient temperature predictions for junction temperature estimation.

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