Biot Number Calculator

Find the Biot number for transient heat conduction.

Biot number 0.005

Formula: Bi = h·L_c ÷ k

Step-by-step with your numbers:
1. Values used:
2. Heat transfer coefficient = 50 W/(m²·K)
3. Characteristic length = 0.02
4. Thermal conductivity = 200 W/(m·K)
5.
6. Heat transfer coefficient x Characteristic length = 50 x 0.02 = 1
7. Biot number = (Heat transfer coefficient x Characteristic length) / Thermal conductivity = 1 / 200 = 0.005
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The Biot number tells whether an object heats/cools uniformly or with internal gradients.

How the Math Works

The Biot number (Bi) is calculated using the formula Bi = (h · L_c) ÷ k, where h represents the convective heat transfer coefficient, L_c is the characteristic length of the object, and k is the thermal conductivity of the material. This dimensionless quantity determines whether temperature gradients within a solid are negligible during transient heat transfer. When Bi < 0.1, the lumped capacitance method can be applied, simplifying transient heat conduction analysis by assuming uniform internal temperature.

Practical Applications

Engineers use the Biot number to evaluate transient heat conduction in systems like electronic components, heat exchangers, or industrial equipment. By comparing Bi to 0.1, they decide whether to use complex numerical models or simplified analytical solutions. For example, in designing a cooling system for a microprocessor, calculating Bi helps determine if the chip's temperature can be approximated as uniform during heating or cooling phases, optimizing thermal management strategies.

Day-to-Day Use

Understanding the Biot number aids in everyday thermal applications, such as selecting materials for cookware or insulation. For instance, when choosing between aluminum and stainless steel for a cooking pot, knowing the Biot number helps predict how quickly heat spreads through the material, ensuring even cooking. It also explains why thermoses with low thermal conductivity minimize heat loss, as their design inherently accounts for minimizing internal temperature gradients.

Worked example

h 50, L 0.02 m, k 200 → Bi = 0.005 (lumped).

FAQ

Why does it matter?

Low Bi lets you treat an object as a single temperature, simplifying cooling calculations.