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Transient Conduction

Steady state says where a body ends up and nothing about how long it takes. One dimensionless number decides which of two entirely different methods applies.

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Transient conduction asks how long a body takes to reach a new temperature, and the Biot number decides the method: below 0.1 the body is effectively isothermal and cools as a single exponential; above it, internal gradients must be solved for.

Why the transient is the question

Steady-state analysis gives the final distribution and nothing about the journey — yet the journey is usually what matters: quench times, cooking times, how long a steel column survives a fire, how fast a sensor responds.

The Biot number decides the method

Two resistances act in series: internal conduction L/kA and external convection 1/hA. Their ratio is Bi = hL/k.

Biot numberPhysical situationMethod
Bi < 0.1Internal gradient under about 5%Lumped capacitance — one exponential
0.1 < Bi < 100Genuine internal gradientsSeries solution, or Heisler charts
Bi > 100Surface effectively at fluid temperatureConstant surface temperature solution

The lumped model

Below Bi = 0.1 the body is treated as one temperature and the equation is first-order with τ = ρVc/hA — thermal mass over surface conductance, structurally identical to RC.

Every intuition from a charging capacitor transfers: 63% of the way in one time constant, 99.3% in five. Since τ ∝ V/A, response time grows roughly with linear size — which is why a fine thermocouple bead responds so much faster than a thick probe.

When gradients matter

Above Bi = 0.1 the surface responds while the interior lags, and the governing equation becomes partial — temperature varying with position and time. Solved historically with Heisler charts, now numerically.

That gradient is also a stress: the cooled surface tries to contract around an interior that has not, producing surface tension. Slow cooling avoids cracking a casting, while toughened glass is quenched deliberately so the surface ends in useful compression.

The semi-infinite solid

If a body is large enough that the far side never notices, it is treated as semi-infinite and the disturbance penetrates as √(αt) — slowing as it deepens.

That is why ground temperature is stable at two metres while the surface swings 30 °C annually, and why a brief flash scorches a surface without heating what lies behind it.

Diffusivity is not conductivity

Thermal diffusivity α = k/ρc governs how fast a temperature change propagates; conductivity governs how much heat flows in steady state. They answer different questions.

A material with high conductivity and high heat capacity conducts well and responds slowly, because every joule passing through must first warm the material it passes through.

The numbers you will be asked for

Lumped response

(T − T_∞)/(T₀ − T_∞) = e^(−t/τ)

Thermal time constant

τ = ρVc / hA

Biot number

Bi = h·L_c / k, with L_c = V/A

Fourier number

Fo = αt / L²

dimensionless time

Thermal diffusivity

α = k / ρc

Penetration depth

δ ≈ 3.6 √(αt)

Watch it work

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Check yourself

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What does the Biot number decide?
Why does a fine thermocouple bead respond faster than a thick probe?
Why does quenching crack a thick component?
Ground temperature at two metres depth is stable all year while the surface swings 30 °C. Why?

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