Transient Conduction & Lumped Capacitance

Transient (unsteady) conduction is a compact, formula-driven GATE ME topic (~1–2 marks). The Biot number test and the lumped-capacitance exponential answer most questions here.

Lumped-capacitance model

When internal conduction is fast compared with surface convection, a body's temperature is nearly uniform and can be "lumped." Validity test: Biot number Bi = hL_c/k < 0.1, where L_c = V/A_s (characteristic length = volume/surface area). Bi compares internal conduction resistance (L_c/k) to surface convection resistance (1/h).

Temperature history: (T − T∞)/(Tᵢ − T∞) = e^(−hA_s·t/(ρVc)) = e^(−(Bi·Fo)), where the time constant τ = ρVc/(hA_s). Fourier number Fo = αt/L_c² (dimensionless time), α = k/(ρc) = thermal diffusivity.

If Bi > 0.1, internal gradients matter — use Heisler charts or an analytical series (one-term approximation).

Exam Tricks & Tips

• 🎯 Bi < 0.1 → lumped model valid (uniform temperature); check this FIRST before using the exponential.
• 🎯 Characteristic length L_c = V/A_s — for a sphere L_c = R/3, for a long cylinder R/2, for a plate of thickness 2L it's L (half-thickness).
• 🎯 Temperature decays as e^(−t/τ), τ = ρVc/hA_s — after one time constant, ~63% of the temperature change is done.
• 🎯 Bi = hL_c/k compares internal vs surface resistance; small Bi = conduction easy relative to convection.
• 🎯 Thermal diffusivity α = k/ρc governs how fast temperature spreads (high for metals) — appears in Fo = αt/L_c².
• ❌ Common mistake: applying the lumped exponential when Bi > 0.1 — significant internal gradients then exist, so the uniform-temperature assumption (and its formula) is invalid.

Expected exam pattern

A 1-mark Biot-number check or characteristic-length NAT, and a 2-mark lumped-cooling time/temperature problem (find t to reach a temperature). The L_c = V/A_s shortcut and the Bi < 0.1 validity rule are frequently tested.

Quick recap

Lumped model valid when Bi = hL_c/k < 0.1, L_c = V/A_s. Then (T−T∞)/(Tᵢ−T∞) = e^(−hA_s t/ρVc) = e^(−t/τ), τ = ρVc/hA_s. Fourier number Fo = αt/L_c², α = k/ρc. If Bi > 0.1, use Heisler charts.

Cover the answer, recall, then check. 11 cards on transient conduction for GATE ME.

Q1. When is the lumped-capacitance model valid?
A1. When the Biot number Bi = hL_c/k < 0.1, so internal temperature gradients are negligible.

Q2. Define the Biot number and what it compares.
A2. Bi = hL_c/k — the ratio of internal conduction resistance (L_c/k) to surface convection resistance (1/h).

Q3. Define the characteristic length L_c.
A3. L_c = V/A_s (volume/surface area). Sphere: R/3; long cylinder: R/2; plate (thickness 2L): L.

Q4. Lumped temperature-history equation?
A4. (T − T∞)/(Tᵢ − T∞) = e^(−hA_s·t/(ρVc)) = e^(−t/τ).

Q5. Define the thermal time constant.
A5. τ = ρVc/(hA_s). After t = τ, about 63% of the total temperature change has occurred.

Q6. Define the Fourier number.
A6. Fo = αt/L_c² — dimensionless time, with α = thermal diffusivity.

Q7. Define thermal diffusivity.
A7. α = k/(ρc) (m²/s) — how fast temperature diffuses; high for metals, low for insulators.

Q8. Express the lumped exponent using Bi and Fo.
A8. The exponent equals −(Bi·Fo), so (T−T∞)/(Tᵢ−T∞) = e^(−Bi·Fo).

Q9. What if Bi > 0.1?
A9. Internal gradients matter; use Heisler charts or a one-term analytical series instead of the lumped model.

Q10. Characteristic length of a sphere of radius R?
A10. L_c = V/A_s = (4/3πR³)/(4πR²) = R/3.

Q11. Physical meaning of small Biot number?
A11. Conduction inside the body is easy relative to surface convection, so temperature stays nearly uniform throughout.

How fast does a hot billet cool, or a thermocouple respond? That is transient conduction. The lumped-capacitance model gives a beautifully simple exponential answer when the body is "thermally small," and the Biot number tells you when you may use it. GATE tests exactly this judgement.

Core concept: during transient heating/cooling, if internal conduction is fast compared with surface convection (small Biot number), the whole body stays at nearly uniform temperature and cools exponentially.

Deep explanation

Beginner — the Biot number

The Biot number compares internal conduction resistance to surface convection resistance:
Bi = hL_c/k, where L_c = V/A_s (characteristic length = volume/surface area).

• Bi < 0.1: lumped analysis valid (uniform temperature inside).
• Bi > 0.1: significant internal gradients — need charts (Heisler) or series solutions.

Intermediate — lumped-capacitance solution

When Bi < 0.1, an energy balance (ρVc dT/dt = −hA_s(T − T_∞)) integrates to:
(T − T_∞)/(T₀ − T_∞) = e^(−t/τ), with time constant τ = ρVc/(hA_s).
The body's temperature approaches ambient exponentially; after one τ it has covered 63% of the total change, after ~5τ it is essentially there. Small τ (small mass, high h) means fast response — desirable for a thermocouple.

Advanced — Fourier number and charts

The exponent can be written using the Fourier number Fo = αt/L_c² (dimensionless time, α = k/ρc thermal diffusivity) and Biot number: exponent = Bi·Fo. For Bi > 0.1, one-term or Heisler-chart solutions give centre and surface temperatures as functions of Fo and Bi for slabs, cylinders and spheres. Thermal diffusivity α governs how fast a temperature disturbance penetrates — metals (high α) equilibrate quickly, insulators slowly.

Worked example

A steel ball (ρ = 7800 kg/m³, c = 470 J/kgK, k = 40 W/mK) of 20 mm diameter at 400°C is quenched in oil at 30°C with h = 200 W/m²K. Check lumped validity and find the time constant.
L_c = V/A_s = (πd³/6)/(πd²) = d/6 = 0.020/6 = 3.33×10⁻³ m.
Bi = hL_c/k = 200 × 3.33×10⁻³/40 = 0.667/40 = 0.0167 < 0.1 → lumped valid.
τ = ρVc/(hA_s) = ρ c L_c/h = 7800 × 470 × 3.33×10⁻³/200 = 12,210/200 = 61.1 s. (Cooling to 63% of the drop takes ~61 s.)

GATE relevance

Biot-number validity check and the lumped exponential cooling law (with time constant τ = ρVc/hA_s) are frequent, high-scoring Heat Transfer questions. The Fourier-number formulation and Heisler-chart applicability (Bi > 0.1) appear as concept items.

Exam tricks & shortcuts

• Always check Bi = hL_c/k < 0.1 before using lumped analysis; L_c = V/A_s.
• For a sphere L_c = d/6; cylinder d/4; slab (both sides) = half-thickness.
• Response is exponential with τ = ρVc/hA_s; small τ = fast (good sensor).
• Mnemonic: "Biot below a tenth, lump it with confidence."

Key formulas

• Biot number: Bi = hL_c/k (L_c = V/A); lumped valid if Bi < 0.1.
• Lumped response: (T − T∞)/(T_i − T∞) = e^(−t/τ), τ = ρVc/(hA).
• Time constant: τ = ρVc/hA; Fourier number Fo = αt/L_c².
• Thermal diffusivity: α = k/(ρc).

The lumped-capacitance model (uniform internal temperature) applies when Biot number is small (Bi < 0.1).

Worked Example

Problem: A 20 mm diameter steel ball (ρ = 7800 kg/m³, c = 480 J/kg·K, k = 50 W/m·K) initially at 500°C is quenched in oil at 30°C with a convection coefficient h = 400 W/m²·K. (a) Check whether the lumped-capacitance method is valid. (b) Find the time for the ball to cool to 100°C.

Solution:
Characteristic length for a sphere (volume/surface area):
L_c = V/A = (πd³/6)/(πd²) = d/6 = 0.020/6 = 0.003333 m.

(a) Biot number:
Bi = h·L_c/k = (400 × 0.003333)/50 = 1.333/50 = 0.0267.
Since Bi < 0.1, internal temperature gradients are negligible and the lumped model is valid.

(b) Time constant of the lumped system:
τ = ρ·c·L_c/h = (7800 × 480 × 0.003333)/400 = 12480/400 = 31.2 s.

Apply the exponential cooling law θ/θᵢ = e^(−t/τ):
(T − T∞)/(Tᵢ − T∞) = (100 − 30)/(500 − 30) = 70/470 = 0.149.
Solve for t: t = −τ·ln(0.149) = −31.2 × (−1.905) ≈ 59.4 s.

Answer: (a) Bi = 0.027 < 0.1, so lumped analysis is valid; (b) t ≈ 59.4 s.