Lumped Capacitance Method
Theory: describing a body by a single temperature
Overview
I want a rough idea of how many minutes a part takes to cool after coming out of a furnace. It doesn't seem worth building an FEM model…
That's exactly what the lumped capacitance method is for. Ignore temperature differences inside the body and treat it as having one temperature. The partial differential equation becomes an ordinary one and the answer is a single exponential. Whether that's allowed depends on the Biot number — the ratio of how hard it is to get heat out through the surface compared with how easily it moves inside.
Energy balance and time constant
| Elapsed time | Remaining temperature difference | Change completed |
|---|---|---|
| τ | 36.8 % | 63.2 % |
| 2τ | 13.5 % | 86.5 % |
| 3τ | 5.0 % | 95.0 % |
| 4.6τ | 1.0 % | 99.0 % |
Biot number and characteristic length
The Biot number is the ratio of the internal resistance $L_c/k$ to the surface resistance $1/h$. If the internal resistance is much smaller, temperature inside evens out quickly and the body can be treated as uniform. The characteristic length $L_c$ is usually volume divided by surface area; for a sphere:
Newton's law of cooling
The idea that a body cools at a rate proportional to its temperature difference from the surroundings is usually traced to a paper Newton published anonymously in 1701. He observed how heated iron cooled, hoping to estimate temperatures far above red heat from cooling times. The lumped capacitance method simply combines that law with the body's heat capacity, and more than three centuries later it is still the entry point to thermal CAE. That the proportionality fails at high temperatures, where radiation and natural convection make the heat-transfer coefficient strongly temperature-dependent, was debated even then.
Worked examples
Example 1: thermocouple response time
A spherical thermocouple bead 0.7 mm in diameter ($\rho = 8500$ kg/m³, $c = 400$ J/(kg·K), $k = 20$ W/(m·K)) sits in a gas stream with $h = 400$ W/(m²·K). Then $\tau = \rho c d/(6h) \approx 0.99$ s and Bi ≈ 0.002, so the lumped model is valid. After a step change it takes about 3 s to follow 95% of the change and about 4.6 s for 99%.
If the fluid temperature changes on a one-second timescale, this sensor will show it late and smaller than it is. The calculation is the starting point for choosing thermocouple size and for correcting measured temperatures back to true ones.
Example 2: steel sphere cooling versus the exact solution
Consider a 50 mm steel sphere ($k = 50$ W/(m·K), $\rho = 7850$ kg/m³, $c = 460$ J/(kg·K)) cooled in still air, a strong air blast and water. The time for the mean temperature difference to halve is compared between the lumped model and the exact series solution for a sphere, together with centre and surface temperatures at that moment (as fractions of the initial difference).
| Cooling | Bi (V/A) | Half-time: lumped / exact | Error | Centre / surface |
|---|---|---|---|---|
| Air, h = 20 | 0.003 | 1043 s / 1045 s | −0.2 % | 0.501 / 0.499 |
| Air blast, h = 200 | 0.033 | 104 s / 106 s | −2.0 % | 0.515 / 0.490 |
| Water, h = 2000 | 0.33 | 10.4 s / 12.4 s | −16 % | 0.645 / 0.411 |
In water, the time is underestimated by 16% and the centre and surface are far apart.
Right. Bi = 0.1 is a rule of thumb; the error depends on geometry, time and the temperature of interest, and is not guaranteed to be within 5% in every case. When Bi is large, as in water quenching, the surface cools first while the centre stays hot. That very temperature difference is what causes quench cracking and residual stress, and the lumped model can't see it at all. Switch to the Heisler chart calculator or a numerical analysis then.
Using it in practice
From estimate to numerical analysis
- Estimate $V/A$ and the heat-transfer coefficient, and compute the Biot number.
- If it is below 0.1, get the timescale from the lumped model and use it to size test conditions or process times.
- If it is 0.1 or more, or if the internal temperature difference itself matters (thermal stress, quenching, solder quality), move to Heisler charts or numerical analysis.
- In the numerical model, use the lumped time constant to choose the time step and total simulated time.
Checking with FEM
- When you model a small thermocouple or thin plate transiently, compare it with the lumped exponential under the same conditions. If they disagree at small Bi, suspect how the heat-transfer coefficient was applied (surface selection, units) or the heat capacity (density × specific heat) input.
- Mixing kilograms and tonnes shifts the time constant by a factor of 1000. In FEM unit systems such as mm–t–s the specific-heat units change too, so take extra care.
- At high temperatures where radiation matters, the effective coefficient varies with temperature, so a single exponential will not match. With radiation, solve the ODE numerically for the comparison.
Experiments that never notice a slow thermometer
Putting a thermocouple inside a thick protective sheath can stretch its time constant from seconds to tens of seconds. In tests with sudden fluid-temperature changes, the smoothed reading is easily misread as a genuinely slow temperature change. When simulation and test disagree, try filtering the simulated temperature with the sensor's time constant before changing the model — sometimes they then agree immediately. Modelling the response of the measuring device is the reliable way to validate.
Common mistakes
Mistakes and fixes
| Mistake | Effect | Fix |
|---|---|---|
| Mixing Biot-number length definitions | For a sphere, radius-based Bi is three times Bi based on V/A | Match the criterion to its definition: for convection over the whole sphere, Bi(V/A) < 0.1 corresponds to Bi(radius) < 0.3 |
| Using it without checking Bi | Large timing errors, e.g. with water cooling | Compute Bi first |
| Using it when internal gradients matter | Causes of thermal stress and cracking are missed | Use Heisler charts or numerical analysis |
| Assuming constant h where radiation dominates | High-temperature cooling is wrong | Solve the ODE numerically with temperature-dependent $h$ |
| Comparing without the sensor's time constant | Misreading the simulation–test mismatch | Filter the simulation with the sensor response before comparing |
I'd like to try changing the numbers myself.
Check the Biot criterion with the Biot number simulator, thermocouple lag with the thermocouple response simulator, and an everyday cooling curve with the coffee cooling (Newton's law) simulator. For large Bi use the Heisler chart calculator; short events where only the surface layer changes are covered in transient conduction in a semi-infinite solid, and dimensionless time in the article on the Fourier number.
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