Unsteady conduction: lumped capacitance and Heisler charts

Why it matters

Engines warm up, brake discs heat during a stop and cool afterwards, gears are quenched after heat treatment, and thermocouples take time to respond. All of these are transient conduction problems. Knowing when a body can be treated as having one uniform temperature, and what to do when it cannot, lets you estimate warm-up times, quench times and sensor lag quickly.

Key ideas

Transient (unsteady) conduction. Temperatures inside a body change with time, typically after a sudden change in the surrounding temperature or heat input. Energy stored in the body changes at the rate ρ c V dT/dt.

Biot number. Bi = h L_c/k compares the internal conduction resistance (L_c/k) with the external convection resistance (1/h). For the lumped criterion the characteristic length is L_c = V/A_s:

• plane wall of thickness 2L cooled on both sides: L_c = L (half-thickness);
• long cylinder of radius r₀: L_c = r₀/2;
• sphere of radius r₀: L_c = r₀/3. (Heisler charts and one-term solutions use L or r₀ instead; always check which definition a formula expects.)

Lumped capacitance model. If Bi < 0.1, the temperature inside the body is uniform to within about 5% at any instant, and an energy balance on the whole body gives an exponential decay toward the fluid temperature. The time constant τ = ρ c V/(h A_s) is the time to cover 63.2% of the total temperature change; after 4–5 τ the body has essentially reached the fluid temperature. Small, highly conductive bodies in gases (thermocouple beads, thin sheets) are almost always lumped.

Fourier number. Fo = α t/L_c² is dimensionless time; α = k/(ρc) is the thermal diffusivity (m²/s), which measures how quickly temperature changes propagate. The lumped solution can be written θ/θ_i = exp(−Bi·Fo).

When Bi > 0.1: Heisler charts and one-term solutions. For a plane wall, long cylinder or sphere suddenly exposed to convection, the exact series solution is dominated by its first term once Fo > about 0.2. Heisler charts plot (a) the centre temperature against Fo for various 1/Bi, (b) the ratio of temperature at another position to the centre temperature, and (c) the fraction of the maximum possible heat transfer Q/Q₀ (Gröber charts). The eigenvalue and coefficient for the one-term solution come from tables in your data book.

Semi-infinite solid. A thick body heated at one face behaves as semi-infinite for short times, before the temperature change reaches the far side. With a sudden change in surface temperature, (T − T_s)/(T_i − T_s) = erf(x/(2√(αt))). Useful for brake surfaces during a single stop and for soil and road surfaces.

Multidimensional bodies. A short cylinder or a rectangular bar can be treated as the product of one-dimensional solutions (product solution), provided Bi conditions apply in each direction.

Formulas

Bi = h L_c / k — h: W/m²·K; L_c = V/A_s for lumped analysis, m; k: of the solid, W/m·K.

(T − T_∞)/(T_i − T_∞) = exp(−t/τ) — lumped capacitance; valid for Bi < 0.1.

τ = ρ c V / (h A_s) — time constant, s; ρ: kg/m³; c: J/kg·K; V: m³; A_s: m².

t = τ ln[(T_i − T_∞)/(T − T_∞)] — time to reach temperature T.

Q = ρ c V (T_i − T) — energy lost (or gained) by the body up to time t, J.

α = k / (ρ c); Fo = α t / L_c²; exp(−t/τ) = exp(−Bi·Fo).

τ = ρ c D / (6 h) — sphere of diameter D (V/A_s = D/6), e.g. a thermocouple bead.

(T − T_s)/(T_i − T_s) = erf(x / (2√(αt))) — semi-infinite solid, sudden surface temperature change.

q″_s = k (T_s − T_i)/√(π α t) — surface heat flux for that case, W/m².

Worked examples

Example 1 (standard, lumped cooling). A steel sphere (k = 50 W/m·K, ρ = 7800 kg/m³, c = 500 J/kg·K) of diameter 0.1 m at 300 °C is placed in air at 50 °C with h = 25 W/m²·K. Find the time to cool to 100 °C and the heat removed.

1. L_c = V/A_s = r₀/3 = 0.05/3 = 0.01667 m.
2. Bi = h L_c/k = 25 × 0.01667/50 = 0.0083 < 0.1, so lumped analysis is valid.
3. τ = ρ c L_c / h = 7800 × 500 × 0.01667/25 = 2600 s.
4. θ/θ_i = (100 − 50)/(300 − 50) = 0.2.
5. t = τ ln(1/0.2) = 2600 × 1.609 = 4185 s ≈ 70 min.
6. Heat removed: V = (4/3)π(0.05)³ = 5.236 × 10⁻⁴ m³; Q = ρ c V (T_i − T) = 7800 × 500 × 5.236 × 10⁻⁴ × 200 = 408 kJ.

Example 2 (GATE level, thermocouple response). A thermocouple junction is a sphere of diameter 1 mm (ρ = 8500 kg/m³, c = 400 J/kg·K, k = 20 W/m·K). It is moved from air at 25 °C into a gas stream at 200 °C with h = 200 W/m²·K. Find the time constant, the time to read 199 °C (99% of the change), and the reading after 5 s.

1. Bi = h (D/6)/k = 200 × (0.001/6)/20 = 0.0017, so lumped.
2. τ = ρ c D/(6h) = 8500 × 400 × 0.001/(6 × 200) = 2.83 s.
3. For 99% of the step, θ/θ_i = 0.01: t = 2.83 × ln 100 = 2.83 × 4.605 = 13.0 s.
4. After 5 s: T = 200 − 175 × exp(−5/2.833) = 200 − 175 × 0.1712 = 170.0 °C.
5. To halve the response time, halve the bead diameter (τ ∝ D) or double h.

Common mistakes

• Using the radius as L_c in the lumped Biot criterion. For a sphere V/A_s = r₀/3; for a long cylinder r₀/2. With the wrong L_c a body can be wrongly judged non-lumped (or lumped).
• Using k of the fluid instead of the solid in Bi (that would be the Nusselt number).
• Forgetting that τ depends on h; the same thermocouple has a different time constant in still air and in a fast stream.
• Applying the lumped formula when Bi > 0.1; use Heisler charts or the one-term solution.
• Using one-term Heisler results at very short times (Fo < 0.2).
• Writing the ratio as (T − T_i)/(T_∞ − T_i); that is the fraction of change completed, which equals 1 − exp(−t/τ), not exp(−t/τ).

For GATE ME

Expect lumped-capacitance numericals (time to cool or heat, time constant of a thermocouple, temperature after a given time, energy transferred), Biot-number checks with the right characteristic length, and conceptual MCQs on Bi, Fo, thermal diffusivity and when Heisler charts are needed. Practise writing τ = ρcV/(hA) for each shape quickly.

Quick check

1. What is V/A_s for a sphere of radius r₀?
2. A body has Bi = 0.05. Can it be treated as lumped?
3. After one time constant, what fraction of the total temperature change has occurred?
4. Doubling the bead diameter does what to a thermocouple's time constant?

Answers: 1. r₀/3. 2. Yes, Bi < 0.1. 3. About 63.2%. 4. Doubles it.