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Heat Transfer Simulator

Biot Number Simulator — Lumped Capacitance Validity for Transient Heat

Compute the Biot number Bi = h L_c / k, the lumped capacitance regime, the cooling time constant tau = rho c_p L_c / h and the dimensionless residual T/T_0 = exp(-1) at t = tau in real time from convection coefficient, solid conductivity, characteristic length and thermal diffusivity. A temperature gradient cross-section and a Bi regime map make it obvious whether the lumped approximation is safe to use.

Parameters
Convection coefficient h
W/m^2 K
Solid conductivity k
W/m K
Characteristic length L_c
mm
Thermal diffusivity alpha
x10^-6 m^2/s
Material preset (sets k)
Scenario presets
Playback speed

Defaults represent a steel block (natural convection h = 100 W/m^2 K, k = 50 W/m K, L_c = 25 mm, alpha = 13e-6 m^2/s). rho c_p = k / alpha is computed internally. Thresholds: Bi < 0.1 (lumped, uniform), 0.1 ≤ Bi ≤ 10 (distributed, internal gradient), Bi > 10 (surface-limited). Left = same tau but forced to Bi<0.1 (uniform), right = the current Bi (gradient at high Bi).

Live results
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Biot number Bi = hL/k
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Regime
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Time constant tau
—
Elapsed time t
—
Surface temp theta_s
—
Core temp theta_c
—
Gradient theta_c-theta_s
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T/T_0 at t = tau
Cooling temperature map: uniform vs internal gradient

Left = the object forced to Bi<0.1 at the same tau: it cools as one uniform colour (lumped, no gradient). Right = the object at the current Bi: surface goes blue (cold), core stays red (hot) = a visible internal temperature gradient. Colour = red (hot) to blue (cold). The ratio of h (outer arrows) to k (interior) is Bi.

Biot regime map (Bi response)

X = Bi (log, 0.001 to 100) / Y = surface-to-core temperature gap at t = tau (gradient magnitude) / green = lumped (Bi<0.1, gradient ~ 0) / yellow = distributed (0.1<Bi<10) / red = surface-limited (Bi>10) / yellow line + dot = current Bi. The gradient rises once Bi crosses 0.1.

Theory & Key Formulas

In transient conduction the Biot number decides whether the interior of a solid can be treated as spatially uniform.

Biot number (internal vs surface resistance):

$$\mathrm{Bi} = \frac{h\,L_c}{k}$$

Lumped capacitance time constant (rho c_p = k / alpha):

$$\tau = \frac{\rho c_p\,V}{h\,A_s} = \frac{\rho c_p\,L_c}{h} = \frac{k\,L_c}{h\,\alpha}$$

Fourier number (dimensionless time):

$$\mathrm{Fo} = \frac{\alpha\,t}{L_c^{\,2}}$$

Lumped temperature response:

$$\frac{T(t)-T_\infty}{T_0-T_\infty}=\exp(-\mathrm{Bi}\cdot\mathrm{Fo})=\exp\!\left(-\frac{t}{\tau}\right)$$

$h$ is the convection coefficient [W/m^2 K], $k$ the solid conductivity [W/m K], $L_c = V/A_s$ the characteristic length [m], $\alpha = k/(\rho c_p)$ the thermal diffusivity [m^2/s]. Practical thresholds: Bi < 0.1 lumped, 0.1 ≤ Bi ≤ 10 distributed, Bi > 10 surface-limited.

About the Biot Number Simulator

🙋
My heat-transfer notes say "if Bi is below 0.1 you can use the lumped capacitance model" — but what exactly is this Biot number a ratio of?
🎓
In plain terms, Bi = h L_c / k is the ratio of the internal conduction resistance L_c / k to the external convection resistance 1 / h. When Bi is small the surface controls the heat flow, so the interior temperature stays nearly uniform. The default in this tool is a steel block (k = 50, L_c = 25 mm) cooling by natural convection (h = 100), which gives Bi = 0.05 — well inside the safe lumped regime.
🙋
The time constant shows 16.0 min. Does that mean the part cools completely in 16 minutes?
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Not quite. Tau is the time for the temperature difference to drop to 1/e ≈ 36.8% of the initial value. At t = tau, T/T_0 = exp(-1) = 0.368; at t = 3 tau the part is at about 5%, at t = 5 tau under 1%. A handy rule of thumb is "3 tau is practically done, 5 tau is fully cooled". So a 25 mm steel block in still air really needs roughly an hour to drop the last 95%.
🙋
On the regime map, why does the red "surface-limited" band appear at high Bi?
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When Bi exceeds 10 the surface resistance is less than 10% of the internal one, so the surface effectively snaps to the fluid temperature. For an oil quench (h ≈ 1000) of a copper slab (k = 400) you still get Bi = 0.25, but swap to insulating ceramic (k = 2) and Bi jumps to 50 — the surface cools fast but the centre lags badly. Set k = 2, h = 5000, L_c = 200 mm in this tool and Bi reaches 500, deep in the surface-limited band.
🙋
L_c is entered directly in mm. How do I pick the right value for my part?
🎓
Use L_c = V / A_s, volume over surface area. For an infinite plate this gives the half thickness t / 2; for a long cylinder r / 2; for a sphere r / 3. A 30 mm sphere therefore uses L_c = 5 mm (= 30 / 2 / 3), and a 30 mm plate uses L_c = 15 mm. With the same material and convection coefficient, the sphere's Bi is about a third of the plate's, so the lumped model survives longer for the sphere. Compute V / A_s for your shape and feed it directly into the slider.

Frequently Asked Questions

Bi = h L_c / k is the ratio of internal conduction resistance to surface convection resistance. When Bi < 0.1 the internal resistance is at most one-tenth of the surface resistance, which keeps the internal temperature difference below 5% of the gap between the body and the surrounding fluid. The 0.1 threshold has been the textbook practical criterion since the 1950s and is standard in ASME and other heat-transfer handbooks. In this tool, sweeping L_c across Bi = 0.1 makes the colour gradient in the solid cross-section change noticeably from nearly uniform to clearly stratified.
In the lumped model tau = rho c_p L_c / h is the convective cooling timescale, i.e. how long it takes the whole body to release heat to the fluid. The diffusion timescale L_c^2 / alpha is the time for heat to spread through the interior, and the two are linked by Bi (tau_diff / tau = Bi). When Bi = 0.05 the diffusion time is only 5% of the convective time, so the interior equilibrates instantly and lumped is fine. When Bi = 10 the diffusion time is ten times longer, so the centre lags the surface and a distributed analysis is required. For the default steel block (L_c = 25 mm) tau is about 16 min.
Above Bi = 10 the surface convection resistance is less than one-tenth of the internal resistance, so the surface effectively reaches the fluid temperature instantly. The remaining cooling problem reduces to pure conduction with the surface temperature as a boundary condition, and the convection coefficient h drops out of the math. Oil-quenched steel slabs (h about 1000 W/m^2 K) are a typical case: the surface equilibrates with the oil immediately and the interior cools at a rate determined only by alpha and L_c^2. Setting h = 5000 and L_c = 200 mm in this tool produces Bi = 20, clearly in the red surface-limited band of the regime map.
The standard definition is L_c = V / A_s, the volume to surface area ratio. For an infinite slab L_c equals the half thickness t/2, for a long cylinder L_c equals r/2, and for a sphere L_c equals r/3. This convention compresses geometry into a single number that lets Bi compare internal and surface resistances consistently. For instance a 30 mm diameter sphere and a 30 mm thick plate at the same h and k differ in Bi by about a factor of three, so the plate leaves the lumped regime first. This tool takes L_c as a direct input, so compute V / A_s for your geometry before entering it.

Real-world applications

Metal heat treatment (quench and temper): a SUS304 cylinder (k = 15 W/m K) in oil (h ≈ 1000 W/m^2 K) gives Bi = 0.67 at L_c = 10 mm (distributed analysis needed), Bi = 0.13 at L_c = 2 mm (borderline), and Bi = 0.033 at L_c = 0.5 mm (lumped is fine). Plug your part size into this tool and you immediately know whether a simple analytical model will do or you must run a transient CAE simulation. Using lumped at Bi = 1 misses the centre temperature by more than 30%, which shows up as hardness non-uniformity after quenching.

Electronics and LED thermal analysis: a silicon die (k = 130 W/m K) 1 mm x 1 mm x 0.5 mm has L_c ≈ 0.17 mm, so even in still air (h = 10) Bi = 1.3e-5 and in forced convection (h = 100) Bi = 1.3e-4. The chip is solidly in the lumped regime. An aluminium heat sink (k = 200, L_c = 10 mm) under fan cooling (h = 50) gives Bi = 0.0025, still lumped. Swap to a cheap plastic housing (k = 0.3, L_c = 20 mm) under the same convection and Bi jumps to 3.3, where internal temperature gradients in the case become important.

Food and pharmaceutical thermal processing: tofu (k ≈ 0.5, L_c = 20 mm) cooked in 100 C water (h = 300) sits at Bi = 12, deep in the surface-limited regime — reaching pasteurisation temperature in the centre takes far longer than the surface heating time. This is why retort foods need surprisingly long cook cycles. Thin ham (L_c = 2 mm) gives Bi = 1.2 (distributed), thinly sliced bacon (L_c = 0.5 mm) Bi = 0.3 (borderline). HACCP pasteurisation tables are built directly on this kind of conduction analysis.

Earth sciences and building heat flow: bedrock (k = 2.5, alpha = 1e-6) heated by sunlight, with surface convection h = 10 at night, gives Bi = 0.4 for L_c = 10 cm — distributed analysis, with the surface and the rock 10 cm below the surface differing by 30–50% of the diurnal swing, exactly as field probes report. In buildings, 100 mm of insulation (k = 0.04) with h = 8 on the room side yields Bi = 20, surface-limited; the temperature gradient inside the insulation linearly bridges the outside and inside temperatures (the classical steady-state limit).

Common pitfalls and notes

The most common misconception is "small Bi means the part cools quickly". Bi only measures internal uniformity, not cooling speed. The cooling speed is set by the time constant tau = rho c_p L_c / h: raising h shrinks tau (faster cooling) but raises Bi (less uniform). In this tool, taking h from 100 to 1000 cuts tau from 16 min to 1.6 min — ten times faster — yet Bi jumps from 0.05 to 0.5, crossing from lumped into distributed. Treat them as two separate questions.

The next mistake is "k in the Bi formula is the fluid conductivity". It is not. k is the solid conductivity and h is the surface convection coefficient — two completely different material properties. If you accidentally type the water value (k = 0.6) instead of the solid value, Bi comes out wildly off. The slider is labelled "Solid conductivity k" precisely so you enter the steel, copper or insulation value, not the surrounding fluid.

Finally, beware of treating Bi = 0.1 as a hard pass/fail line. At exactly Bi = 0.1 the centre and surface differ by roughly 5%, which may not be tight enough for precision heat treatment where 1% non-uniformity matters. The cross-section colour map in this tool already shows faint stratification at Bi = 0.1. A safer mental model is Bi = 0.01–0.05 "lumped is safe", 0.05–0.2 "lumped only if 5% error is acceptable", and Bi > 0.2 "strongly prefer a distributed analysis".

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How to Use

  1. Enter convection coefficient h (W/m²·K) using slH slider: range 10–10,000 for air/water/oil cooling
  2. Set thermal conductivity k (W/m·K) using slK slider: 0.6 for water, 15 for brick, 50 for steel, 200 for aluminum
  3. Define characteristic length L (mm) via slL_mm slider: use (Volume/Surface Area) for lumped node approximation
  4. Input thermal diffusivity α (mm²/s) with slAlpha slider: pre-calculated as k/(ρ·cp) for your material
  5. Simulator computes Biot number Bi = hL/k, validates lumped capacitance regime (Bi < 0.1), calculates time constant τ and temperature ratio T/T₀ at t = τ

Worked Example

Steel ball (ρ=7850 kg/m³, cp=490 J/kg·K, k=50 W/m·K, r=25 mm diameter, L=12.5 mm) cooled in oil (h=200 W/m²·K). Calculate: Bi = (200 × 0.0125)/50 = 0.05 (valid lumped capacitance). Thermal diffusivity α = 50/(7850 × 490) = 1.30 × 10⁻⁵ m²/s = 13.0 mm²/s. Time constant τ = L²/α = (12.5)²/13.0 = 12.0 seconds. At t = τ, T/T₀ = e⁻¹ = 0.368 (36.8% of initial excess temperature remains). Fourier number at one time constant: Fo = α·t/L² = 1.0.

Practical Notes

  1. Regime validity: Bi < 0.1 enables spatial uniformity assumption; Bi > 0.1 requires finite-difference or analytical series solutions (error > 5% in lumped analysis)
  2. Increase h (spray cooling, turbulent flow) → higher Bi → exit lumped regime; reduce L (smaller part) or increase k (copper vs. plastic) to restore validity
  3. Time constant τ scales with L²: doubling part thickness increases cooling time 4×; critical in injection mold design and quench hardening cycles
  4. Temperature profile exponential decay T(t) = T∞ + (T₀−T∞)exp(−t/τ) holds only if Bi < 0.1; larger Bi values show internal gradients requiring spatial discretization