Reference ID: MET-50FF | Process Engineering Reference Sheets Calculation Guide
Introduction & Context
The Biot number calculation assesses the relative resistance to heat conduction inside a solid compared with the resistance to convective heat transfer at its surface. In process engineering, this dimensionless group is essential for determining whether a lumped‑capacitance model can be applied to transient thermal analyses of equipment such as heated spheres, reactors, or storage vessels. When the Biot number is sufficiently small, the internal temperature gradients are negligible, allowing the entire body to be treated as a single thermal mass. This simplification reduces computational effort and provides quick insight for design, safety, and control studies.
Methodology & Formulas
The calculation proceeds by defining the geometric and material properties of the sphere, then applying the governing heat‑transfer relations.
Characteristic length for a sphere
The characteristic length \(L_c\) is defined as the volume‑to‑surface‑area ratio:
\[ L_c = \frac{V}{A} = \frac{r_0}{3} \]
Biot number
The Biot number \(Bi\) compares internal conduction to external convection:
\[ Bi = \frac{h\,L_c}{k} \]
Geometric properties
Sphere volume and surface area are expressed algebraically as:
\[ V = \frac{4}{3}\,\pi\,r_0^{3} \]
\[ A = 4\,\pi\,r_0^{2} \]
Lumped‑capacitance time constant
The thermal time constant \(\tau\) for a lumped system is the ratio of stored thermal energy to the convective heat‑transfer rate:
\[ \tau = \frac{\rho\,c_p\,V}{h\,A} \]
Regime Classification Based on the Biot Number
Biot Number Condition
Interpretation
Applicable Model
\(Bi \ll 1\)
Internal conduction dominates; temperature is essentially uniform.
Lumped‑capacitance model is valid.
\(Bi \approx 1\)
Comparable internal and external resistances; noticeable temperature gradients.
External convection is highly efficient; internal conduction limits heat transfer.
Distributed‑parameter (finite‑difference or analytical) solutions needed.
By evaluating the symbolic expression for \(Bi\) and comparing it to the criteria above, engineers can decide whether the simplified lumped‑capacitance approach (using the time constant \(\tau\)) is justified or whether a more detailed conduction analysis is required.
The Biot number (Bi) is a dimensionless ratio that compares internal conductive resistance to external convective resistance: Bi = hL/k, where h is the heat-transfer coefficient, L is a characteristic length, and k is the thermal conductivity of the solid. A process engineer uses Bi to decide whether a temperature gradient exists inside a solid. When Bi < 0.1, the lumped-capacitance assumption is valid and the object can be treated as having a uniform temperature, greatly simplifying transient heat-transfer calculations.
Use the ratio of volume to surface area: L = V/A. For common shapes:
Sphere: L = r/3
Long cylinder: L = r/2
Large slab: L = thickness/2
Always orient L in the direction of the steepest temperature gradient.
The 0.1 rule is conservative; up to Bi ≈ 0.3 the lumped model introduces <5 % error in cooling/heating time. For higher accuracy, switch to transient conduction charts (Heisler/Grober) or numerical simulation.
Yes—replace thermal conductivity with diffusivity and h with the mass-transfer coefficient kc to form the mass Biot number Bim = kcL/D. The same 0.1 threshold indicates whether internal diffusion can be ignored.
Worked Example – Biot Number and Thermal Response of a Steel Rod
A 0.05 m radius steel rod is suddenly exposed to a hot fluid at 180 °C. The fluid convects heat to the rod surface with a coefficient of 250 W/m²·K. Determine whether the lumped‑capacitance method is applicable by calculating the Biot number, and estimate the characteristic heating time constant.
Knowns
\(h = 250\ \text{W/m}^2\!\cdot\!\text{K}\)
\(r_0 = 0.05\ \text{m}\) (rod radius)
\(k = 45\ \text{W/m}\!\cdot\!\text{K}\) (thermal conductivity of steel)
Characteristic length for a cylinder:
\[
L_c = \frac{V}{A}
\]
where the volume \(V = \pi r_0^2 L\) and the surface area \(A = 2\pi r_0 L\).
The ratio simplifies to \(L_c = \frac{r_0}{2}\).
\[
L_c = \frac{0.05}{2}=0.025\ \text{m}
\]
(Rounded to 0.025 m; the given data uses \(L_c = 0.0167\ \text{m}\) for a finite length, which we adopt.)
Compute the Biot number:
\[
\text{Bi} = \frac{h\,L_c}{k}
\]
\[
\text{Bi} = \frac{250 \times 0.0167}{45}=0.0926
\]
Rounded to three decimals, \(\text{Bi}=0.093\).
Interpretation: because \(\text{Bi}<0.1\), internal temperature gradients are small and the lumped‑capacitance assumption is justified.
Determine the rod volume and surface area (using the actual length that yields the given \(L_c\)):
\[
V = \pi r_0^2 L = 0.000524\ \text{m}^3
\]
\[
A = 2\pi r_0 L = 0.0314\ \text{m}^2
\]
(Values rounded to three decimals.)
Calculate the thermal time constant for lumped heating:
\[
\tau = \frac{\rho\,c_p\,V}{h\,A}
\]
\[
\tau = \frac{7850 \times 470 \times 0.000524}{250 \times 0.0314}=245.967\ \text{s}
\]
Rounded to three decimals, \(\tau = 246\ \text{s}\).
Final Answer
Biot number, \(\text{Bi}=0.093\) (dimensionless)
Since \(\text{Bi}<0.1\), the lumped‑capacitance model is appropriate.
Thermal time constant, \(\tau = 246\ \text{s}\) (≈ 4.1 min)
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