Unsteady conduction: lumped capacitance and Heisler charts

Why it matters

Most real thermal events are transient: a motor warming up after start, a power transistor during a current pulse, a quenched gear, a thermocouple following a changing gas temperature. Transient conduction tells you how fast a body heats or cools and how long a sensor takes to respond. The lumped model gives the answer in one line when it is valid; the Biot number tells you whether it is.

Key ideas

Unsteady (transient) conduction. Temperature changes with time as well as position. It ends in a new steady state, which is not the same as thermal equilibrium unless the body is isolated.

Biot number. Bi = h L_c / k compares the internal conduction resistance (L_c/k) with the external convection resistance (1/h). A small Bi means the inside of the body equalises much faster than heat can leave the surface, so the temperature is nearly uniform inside at every instant.

• For the lumped check use L_c = V/A_s: for a plane wall of thickness 2L cooled on both faces L_c = L (half-thickness); for a long cylinder L_c = r₀/2; for a sphere L_c = r₀/3.
• Do not confuse Bi with the Nusselt number: Bi uses the solid's k, Nu uses the fluid's k.

Lumped capacitance model (Bi < 0.1). An energy balance on the whole body, −hA_s(T − T∞) = ρVc dT/dt, gives an exponential approach to T∞ with time constant τ = ρVc/(hA_s). After one τ the body has covered 63.2 % of the change; after 4.6 τ it is within 1 %. The criterion Bi < 0.1 keeps the error in the lumped result to about 5 %.

Fourier number. Fo = αt/L_c² is dimensionless time: it compares the rate of heat conduction through the body with the rate of energy storage in it. α = k/(ρc) is the thermal diffusivity (m²/s): metals have high α and respond quickly; insulators, concrete and soil have low α. The lumped exponent can be written hA_s t/(ρVc) = Bi·Fo.

When Bi > 0.1: spatial effects. The centre lags the surface. For an infinite plane wall, a long cylinder and a sphere, with uniform initial temperature, constant h and T∞, constant properties and no generation, the exact series solution is plotted as the Heisler charts (centre temperature θ₀/θᵢ against Fo for curves of 1/Bi, plus position-correction and heat-ratio Q/Q₀ charts). Here Bi and Fo use the half-thickness L or the outer radius r₀, not V/A_s. For Fo > 0.2 the first term of the series alone is accurate to about 2 %; the one-term constants (λ₁, C₁) come from tables in your data book.

• Short bodies (finite cylinder, rectangular bar) are handled by multiplying the one-dimensional solutions (product solution).

Semi-infinite solid. For early times or very thick bodies (ground, thick concrete, a brake disc in a short stop), the heat has not yet reached the far side. With a sudden change of surface temperature, the solution uses the error function of η = x/(2√(αt)). The penetration depth grows as √(αt), not linearly with time.

Formulas

Bi = h L_c / k — h (W/m²·K), L_c (m), k of the solid (W/m·K); lumped valid if Bi < 0.1. L_c = V / A_s — wall: half-thickness L; long cylinder: r₀/2; sphere: r₀/3 (m). α = k / (ρ c) — thermal diffusivity (m²/s); ρ (kg/m³), c (J/kg·K). Fo = α t / L_c² — t (s). θ/θᵢ = (T − T∞)/(Tᵢ − T∞) = exp(−t/τ) = exp(−Bi·Fo) — lumped body. τ = ρ V c / (h A_s) = ρ c L_c / h — time constant (s). t = τ · ln[(Tᵢ − T∞)/(T − T∞)] — time to reach T. Q(t) = ρ V c (Tᵢ − T∞) [1 − exp(−t/τ)] — heat lost (J) up to time t. θ₀/θᵢ ≈ C₁ exp(−λ₁² Fo) — one-term centre solution, Fo > 0.2 (C₁, λ₁ from tables for the given Bi). (T − T_s)/(Tᵢ − T_s) = erf[x / (2√(α t))] — semi-infinite solid, surface suddenly at T_s.

Worked examples

Example 1 (standard, quenching). A steel ball of diameter 10 mm (k = 40 W/m·K, ρ = 7800 kg/m³, c = 460 J/kg·K) at 900 °C is quenched in oil at 30 °C with h = 300 W/m²·K. Find the time to reach 100 °C.

1. L_c = r₀/3 = 0.005/3 = 1.667 × 10⁻³ m.
2. Bi = h L_c / k = 300 × 1.667 × 10⁻³ / 40 = 0.0125 < 0.1, so the lumped model holds.
3. τ = ρ c L_c / h = 7800 × 460 × 1.667 × 10⁻³ / 300 = 19.93 s.
4. t = τ ln[(Tᵢ − T∞)/(T − T∞)] = 19.93 × ln(870/70) = 19.93 × 2.520 = 50.2 s.

t ≈ 50 s

Example 2 (GATE level, thermocouple response). A spherical thermocouple bead of diameter 1 mm (k = 35 W/m·K, ρ = 8500 kg/m³, c = 320 J/kg·K) is placed suddenly in a gas stream with h = 210 W/m²·K. Find its time constant and the time to show 99 % of the step change.

1. L_c = D/6 = 1.667 × 10⁻⁴ m; Bi = 210 × 1.667 × 10⁻⁴ / 35 = 0.001, so lumped.
2. τ = ρ c D / (6h) = 8500 × 320 × 0.001 / (6 × 210) = 2.159 s.
3. 99 % response: θ/θᵢ = 0.01, so t = τ ln(100) = 2.159 × 4.605 = 9.94 s.
4. Halving the bead diameter halves τ; doubling h (higher gas velocity) also halves τ.

τ ≈ 2.16 s, t₉₉ ≈ 9.9 s

Example 3 (semi-infinite solid). A thick concrete floor (α = 7 × 10⁻⁷ m²/s) initially at 20 °C has its surface suddenly raised to 80 °C. Find the temperature 50 mm below the surface after 1 h.

1. η = x/(2√(αt)) = 0.05 / (2√(7 × 10⁻⁷ × 3600)) = 0.05 / (2 × 0.0502) = 0.498.
2. erf(0.498) = 0.519 (from erf tables).
3. T = T_s + (Tᵢ − T_s) erf(η) = 80 + (20 − 80) × 0.519 = 48.9 °C.

T ≈ 49 °C

Common mistakes

• Using L_c = radius (or diameter) in the lumped Biot check instead of V/A_s; for a sphere it is r₀/3.
• Using the fluid's k in the Biot number. Bi uses the solid's conductivity.
• Applying the lumped exponential when Bi > 0.1, or using V/A_s in the Heisler charts, which are built on L or r₀.
• Writing the temperature ratio upside down: θ/θᵢ starts at 1 and decays to 0.
• Forgetting to convert hours to seconds, or mm to m, inside exponentials.
• Reading a Heisler chart for Fo < 0.2, where the one-term approximation and many chart curves become inaccurate.

For GATE ME

Questions here are mostly lumped-capacitance numericals: check Bi, find the time constant, the time to reach a temperature, or the temperature after a given time; thermocouple response time and the effect of bead diameter or h on it; and conceptual questions on the meaning of Bi and Fo. Practise writing τ = ρcV/(hA) for spheres, cylinders and plates quickly, and ratio problems where properties cancel.

Quick check

1. What is the lumped characteristic length of a sphere of radius r₀?
2. A body has τ = 10 s. What fraction of the temperature step remains after 10 s?
3. Does a larger bead make a thermocouple faster or slower?
4. What does a Biot number of 5 tell you?
5. In the semi-infinite solid, how does the penetration depth scale with time?

Answers: 1. r₀/3. 2. e⁻¹ ≈ 0.368 (36.8 %). 3. Slower (τ ∝ D). 4. Internal resistance dominates; large temperature gradients inside, so the lumped model is invalid. 5. As √(αt).