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The Biot number compares internal conduction resistance in a body with external convective resistance.
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Formula
Bi dimensionless; h in W/(m2·K); Lc in m; k in W/(m·K)
| Symbol | Meaning | Unit |
|---|---|---|
| Biot number | dimensionless | |
| Convection heat-transfer coefficient | W/(m2·K) | |
| Characteristic length, often volume divided by surface area | m | |
| Solid thermal conductivity | W/(m·K) |
\mathrm{Bi}=\frac{hL_c}{k}Bi≈hLc/k is a resistance ratio, not a temperature or material constant. Small Biot numbers indicate that internal temperature equalization is faster than external transfer, so a lumped-capacitance model may be suitable. Bi<0.1 is a common practical rule of thumb, not a law of nature. Geometry and Lc must match the body and heat-flow problem.
Bi≈hLc/k is a resistance ratio, not a temperature or material constant. Small Biot numbers indicate that internal temperature equalization is faster than external transfer, so a lumped-capacitance model may be suitable. Bi<0.1 is a common practical rule of thumb, not a law of nature. Geometry and Lc must match the body and heat-flow problem.
The Biot number compares internal conduction resistance in a body with external convective resistance.
Base equation for the target quantity
Preserve signs, boundary conditions and units when rearranging.
Work through the numerical example for “Biot number for transient conduction” and check its unit and magnitude.
For h=25 W/(m2·K), Lc=0.01 m and k=50 W/(m·K), Bi=0.005, suggesting small internal gradients.
Substituting a related quantity that shares the same symbol.
Use variable definitions and check: Bi dimensionless; h in W/(m2·K); Lc in m; k in W/(m·K).
Formula cluster
The Biot number compares internal conduction resistance in a body with external convective resistance.
For h=25 W/(m2·K), Lc=0.01 m and k=50 W/(m·K), Bi=0.005, suggesting small internal gradients.
Check whether a body can be approximated as having uniform temperature during transient cooling.
Here is how to work through a typical Biot number for transient conduction (Bi = hLc/k) task step by step:
Task
Work through the numerical example for “Biot number for transient conduction” and check its unit and magnitude.
Solution path
For h=25 W/(m2·K), Lc=0.01 m and k=50 W/(m·K), Bi=0.005, suggesting small internal gradients.
A small Biot number means internal conduction resistance is relatively small compared with external convection resistance. The solid may then be approximated as having uniform temperature while cooling overall. The common Bi<0.1 limit is a practical modeling rule, not a sharp transition, and should match desired accuracy. Biot compares conduction within a solid with convection at its boundary and is dimensionless. Characteristic length is tied to geometry, often V/A for lumped analysis. A small Bi supports a uniform-temperature approximation but does not guarantee every other transient assumption.
Lc must follow from geometry and the heat-transfer problem. Lumped capacitance often uses body volume divided by heat-transfer area. Analytical solutions may use another characteristic length, such as half-thickness. Do not change Lc between Biot and Fourier numbers without checking the associated solution. Biot compares conduction within a solid with convection at its boundary and is dimensionless. Characteristic length is tied to geometry, often V/A for lumped analysis. A small Bi supports a uniform-temperature approximation but does not guarantee every other transient assumption.
No. 0.1 is a frequently used criterion suggesting internal gradients are small enough for many geometries. Actual error depends on geometry, boundary condition and required accuracy. For critical calculations, compare the lumped result with a spatially resolved solution or validated approximation. Biot compares conduction within a solid with convection at its boundary and is dimensionless. Characteristic length is tied to geometry, often V/A for lumped analysis. A small Bi supports a uniform-temperature approximation but does not guarantee every other transient assumption.
Write the quantity and unit from the variable list next to every value: Bi dimensionless; h in W/(m2·K); Lc in m; k in W/(m·K). Substitute only after converting to a consistent unit system, and state sign or reference direction explicitly. The example For h=25 W/(m2·K), Lc=0.01 m and k=50 W/(m·K), Bi=0.005, suggesting small internal gradients. is a calculation check: result, unit and order of magnitude must fit the described situation. Round after intermediate calculations so small contributions are not lost prematurely. Biot compares conduction within a solid with convection at its boundary and is dimensionless. Characteristic length is tied to geometry, often V/A for lumped analysis. A small Bi supports a uniform-temperature approximation but does not guarantee every other transient assumption.
The background states this starting condition: Bi≈hLc/k is a resistance ratio, not a temperature or material constant. Small Biot numbers indicate that internal temperature. That limits when the result is a valid model prediction. Compare the problem’s boundary conditions with this assumption before substituting values; real components, material properties or loads can introduce additional effects. The cited reference supports the relation used here but does not remove the need to check whether the specific case fits the model. Biot compares conduction within a solid with convection at its boundary and is dimensionless. Characteristic length is tied to geometry, often V/A for lumped analysis. A small Bi supports a uniform-temperature approximation but does not guarantee every other transient assumption.
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Curated exam set for "Biot number for transient conduction":
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Which formula describes Biot number for transient conduction?
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How do you rearrange Biot number for transient conduction for Base equation for the target quantity?
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Which common mistake happens with Biot number for transient conduction?
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