Lumped Thermal Response

Estimate an object’s transient temperature and check whether internal gradients can be neglected.

How this calculation works

The lumped-capacitance method treats an object as having one temperature while it exchanges heat with a constant-temperature environment. Enter the object's mass, specific heat, volume, exposed area and conductivity, along with the convection coefficient, initial temperature, ambient temperature and elapsed time. Thermal capacity divided by convection conductance gives the time constant. The temperature difference from ambient decays exponentially, reaching about 63.2% of its total change after one time constant. Volume divided by area supplies the characteristic length for the Biot-number check. A value below 0.1 is the model's screening criterion for small internal gradients; larger values trigger a warning. The time to 95% response is also reported. Internal heating, phase change, radiation and contact heat transfer are not included.

Inputs and units

Object mass (kg)
Specific heat (J/(kg·K))
Object volume (m³)
Exposed surface area (m²)
Object conductivity (W/(m·K))
Convection coefficient (W/(m²·K))
Initial object temperature (°C): Must be above absolute zero
Constant ambient temperature (°C): Must be above absolute zero
Elapsed time (s)

Method and formula

Lc = V/A; Bi = hLc/k; τ = mcp/(hA); T(t) = Tair + (Ti − Tair) exp(−t/τ); t95 = −τ ln(0.05).

Worked example

Example inputs

Object mass: 10 kg
Specific heat: 500 J/(kg·K)
Object volume: 0.001 m³
Exposed surface area: 1 m²
Object conductivity: 50 W/(m·K)
Convection coefficient: 20 W/(m²·K)
Initial object temperature: 100 °C
Constant ambient temperature: 20 °C
Elapsed time: 250 s

Calculation steps

Calculate characteristic length: Lc = 0.001/1 = 0.001 m.
Check Biot number: Bi = 20 × 0.001/50 = 0.0004, dimensionless and below 0.1.
Calculate time constant: τ = 10 × 500/(20 × 1) = 250 s.
Evaluate temperature after 250 s: T = 20 + (100 − 20) × exp(−250/250) = 49.430355 °C.
Calculate time to 95% of the full change: t95 = −250 × ln(0.05) = 748.933068 s.

Example results

Thermal time constant: 250 s
Biot number (Lc = V/A): 0.0004 1
Predicted object temperature: 49.43035529 °C
Time to 95% of total temperature change: 748.9330684 s

Assumptions

Initially uniform object temperature, constant ambient and properties
Small internal temperature gradients; Bi below 0.1 is the screening criterion

Limitations

No internal generation, phase change, radiation or contact heat transfer
Reported temperature is unreliable when lumped-capacitance assumptions fail

Sources

MIT Unified Engineering: Transient Heat Transfer: Section 18.3, lumped convective response and Biot number
MIT 3.185 Problem Set 4 Solutions: Biot-number screening for uniform-temperature analysis