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Lumped Capacitance Time Constant

τ=ρVchA\tau = \frac{\rho V c}{h A}

Worked example: 100 cm3 aluminium block, h 30 over 0.02 m2 → tau 405 s — press Try an example to run it live, then adjust anything.

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Lumped Capacitance Time Constant explained

ρVcphAτ

ρVc is the heat a body stores per kelvin — its thermal capacitance — and hA is the conductance draining it. Their ratio is a time constant in exactly the sense an electrical engineer means, and the analogy is complete: the body is a capacitor, the surface film is a resistor, and the temperature decays exponentially. In one τ the gap to ambient closes by 63%, in three τ by 95%, in five τ by over 99%.

Worked example: an aluminium block, ρ = 2700 kg/m³, V = 100 cm³, c = 900 J/(kg·K), cooling in air at h = 30 W/(m²·K) over A = 0.02 m². τ = 2700 × 0.0001 × 900/(30 × 0.02) = 243/0.6 = 405 s, so it is essentially at room temperature in about twenty minutes. This is the number that sizes thermocouple response — a fine bead responds in milliseconds, a 6 mm thermowell in a stagnant pocket can lag a minute, and a control loop tuned without knowing which one you have will hunt forever. Trap: τ assumes the lumped regime, so check Bi < 0.1 before trusting it, and remember h is not constant during a violent transient — free convection off a hot block starts strong and weakens as the block cools.

Lumped Capacitance Time Constant formula

τ=ρVchA\tau = \frac{\rho V c}{h A}
Where
  • τ\tau= Thermal time constant (s)
  • ρ\rho= Density (kg/m³)
  • VV= Body volume (L)
  • cpc_p= Specific heat (J/(kg·K))
  • hh= Film coefficient (W/(m²·K))
  • AA= Surface area (m²)

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