Unsteady conduction and lumped capacitance analysis

Transient conduction: Biot and Fourier numbers, the lumped capacitance model and time constant, and when to use charts or the semi-infinite solid instead.

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Why it matters

Start-up and shutdown of equipment, quenching of metal parts, heating of catalyst pellets, sterilisation of canned food and the response time of a thermocouple are all transient conduction problems. The lumped capacitance model answers "how long will it take?" with one exponential, and the Biot number tells you whether you are allowed to use it.

Key ideas

Transient conduction. When a body's surroundings change suddenly, its temperature changes with time and, in general, with position. The full description is the heat diffusion equation ∂T/∂t = α·∇²T (no generation, constant properties), where α = k/(ρc) is the thermal diffusivity: high α means the solid responds quickly.

Two resistances. Heat leaving a body must first be conducted to the surface (internal resistance ~ L_c/k) and then convected away (external resistance ~ 1/h). Their ratio is the Biot number Bi = h·L_c/k.

  • Bi ≪ 1: internal resistance negligible, the body is practically isothermal at each instant — lumped.
  • Bi ≫ 1: surface temperature quickly becomes the fluid temperature; large internal gradients.

Characteristic length. For the lumped criterion use L_c = V/A_s: L for a plane wall of thickness 2L cooled on both sides, r/2 for a long cylinder, r/3 for a sphere, a/6 for a cube of side a. (Heisler charts use the half-thickness or the full radius instead — read the definition each time.)

Lumped capacitance. If Bi < 0.1 (error under about 5 %), an energy balance on the whole body, −ρ·c·V·dT/dt = h·A_s·(T − T∞), gives an exponential approach to the fluid temperature with time constant τ = ρ·c·V/(h·A_s). After one τ the excess is 36.8 % of the initial value; after 3τ it is 5 %; after 4.6τ, 1 %. The analogy is an RC circuit: R = 1/(hA_s), C = ρcV.

Fourier number. Fo = α·t/L_c² is dimensionless time. The lumped solution can be written θ/θ_i = exp(−Bi·Fo).

When Bi > 0.1. Use the one-term series solutions or Heisler/Gröber charts for plane walls, long cylinders and spheres (valid for Fo > about 0.2), or the semi-infinite solid solution for short times or thick bodies: for a sudden change in surface temperature, (T − T_s)/(T_i − T_s) = erf[x/(2√(αt))]. The penetration depth grows as √(αt).

Assumptions of lumped analysis. Constant ρ, c and h; no generation; uniform initial temperature; a step change in fluid temperature. Radiation can be included only if linearised into h.

Formulas

Bi = h·L_c/k h (W/m²·K), L_c = V/A_s (m), k of the solid (W/m·K). Lumped analysis valid when Bi < 0.1.

(T − T∞)/(T_i − T∞) = exp(−t/τ) T body temperature (°C), T_i initial, T∞ fluid, t time (s).

τ = ρ·c·V/(h·A_s) = ρ·c·L_c/h Time constant (s); ρ density (kg/m³), c specific heat (J/kg·K), V volume (m³), A_s surface area (m²).

t = τ·ln[(T_i − T∞)/(T − T∞)] Time to reach T.

Q = ρ·c·V·(T_i − T) Heat transferred up to time t (J).

Fo = α·t/L_c², α = k/(ρ·c) Fourier number; thermal diffusivity (m²/s). Lumped: θ/θ_i = exp(−Bi·Fo).

(T − T_s)/(T_i − T_s) = erf[x/(2·√(α·t))] Semi-infinite solid after a sudden change of surface temperature to T_s; x depth (m).

Worked examples

Example 1 (standard). Steel balls 10 mm in diameter (ρ = 7800 kg/m³, c = 460 J/kg·K, k = 40 W/m·K) at 800 °C are quenched in oil at 50 °C with h = 200 W/m²·K. How long until they reach 150 °C, and how much heat does each ball release?

Given: r = 0.005 m, T_i = 800 °C, T∞ = 50 °C, T = 150 °C.

  1. L_c = V/A_s = r/3 = 0.001667 m.
  2. Bi = h·L_c/k = 200 × 0.001667/40 = 0.0083 < 0.1, so lumped analysis is valid.
  3. τ = ρ·c·L_c/h = 7800 × 460 × 0.001667/200 = 29.9 s.
  4. t = τ·ln[(T_i − T∞)/(T − T∞)] = 29.9 × ln(750/100) = 29.9 × 2.015 = 60.2 s.
  5. V = (4/3)π(0.005)³ = 5.236 × 10⁻⁷ m³; Q = ρ·c·V·(T_i − T) = 7800 × 460 × 5.236 × 10⁻⁷ × 650 = 925.7 J.

Answer: t ≈ 60 s; Q ≈ 0.93 kJ per ball.

Example 2 (GATE level). A spherical thermocouple junction (ρ = 8500 kg/m³, c = 320 J/kg·K, k = 20 W/m·K) is placed in a gas stream with h = 400 W/m²·K. What junction diameter gives a time constant of 1 s, and how long does it take to register 99 % of a step change in gas temperature?

Given: τ = 1 s.

  1. For a sphere L_c = d/6, so τ = ρ·c·d/(6h) → d = 6·h·τ/(ρ·c).
  2. d = 6 × 400 × 1/(8500 × 320) = 2400/2 720 000 = 8.82 × 10⁻⁴ m = 0.88 mm.
  3. Check Bi = h·(d/6)/k = 400 × 1.47 × 10⁻⁴/20 = 0.0029 < 0.1, so lumped is valid.
  4. 99 % response: e^(−t/τ) = 0.01 → t = τ·ln(100) = 4.61 s.

Answer: d ≈ 0.88 mm; about 4.6 s for 99 % response.

Common mistakes

  • Using L_c = r (or d) instead of V/A_s = r/3 when checking the lumped criterion for a sphere.
  • Using the fluid conductivity in Bi (that would be the Nusselt number); Bi uses the solid's k.
  • Inverting the temperature ratio in the logarithm and getting a negative time.
  • Computing τ with V/A_s in the wrong units (mm instead of m).
  • Using lumped analysis when Bi > 0.1, or Heisler charts without checking which length they define Bi with.

For GATE CH

Expect Biot-number checks, time to cool or heat a sphere, cylinder or plate by the lumped method, thermocouple time-constant and response problems, and conceptual questions comparing Bi and Nu or asking what α measures. Occasionally a semi-infinite solid question appears (ground temperature, penetration depth). Practise the logarithm form t = τ·ln(θ_i/θ) and always check Bi first.

Quick check

  1. What is L_c = V/A_s for a long cylinder of radius r?
  2. After a time equal to 2τ, what fraction of the initial temperature excess remains?
  3. Bi = 0.5. Can you use the lumped model?
  4. Which property appears in Bi but not in Nu?
  5. Units of thermal diffusivity?

Answers: 1. r/2. 2. e⁻² ≈ 0.135. 3. No. 4. The conductivity of the solid (Nu uses the fluid's k). 5. m²/s.

Interview questions

All Heat Transfer interview questions

Try answering each one aloud before you open it.

  1. 1.What is unsteady conduction in heat transfer?Concept

    Unsteady conduction, also known as transient conduction, refers to the process where the temperature within an object changes with time. Unlike steady-state conduction, where temperatures are constant over time, unsteady conduction involves time-dependent temperature fields. This occurs when a body is subjected to a sudden change in its thermal environment, such as a change in surrounding temperature or heat generation within the body.

  2. 2.Explain the lumped capacitance method in heat transfer.Concept

    The lumped capacitance method is an approach used to simplify the analysis of transient heat conduction problems. It assumes that the temperature within a solid body is uniform at any given time, meaning the temperature gradient within the body is negligible. This method is applicable when the Biot number (Bi = hL_c/k) is less than 0.1, indicating that the thermal resistance within the body is much smaller than the thermal resistance at the surface.

  3. 3.Why is the Biot number important in the lumped capacitance analysis?Application

    The Biot number is a dimensionless parameter that compares the internal thermal resistance of a body to the external thermal resistance. In lumped capacitance analysis, a Biot number less than 0.1 indicates that the temperature within the body can be assumed uniform, as the internal resistance is negligible compared to the external resistance. This simplifies the analysis by allowing the use of a single temperature to describe the entire body.

  4. 4.What happens if the Biot number is greater than 0.1 in a heat transfer problem?Application

    If the Biot number is greater than 0.1, the assumption of uniform temperature within the body is invalid. This means that there are significant temperature gradients within the body, and the lumped capacitance method cannot be used. Instead, more complex methods, such as solving the heat conduction equation with spatial variations, are required to accurately model the temperature distribution.

  5. 5.How does the thermal diffusivity of a material affect unsteady conduction?Application

    Thermal diffusivity is a measure of how quickly heat can spread through a material. It is defined as the ratio of thermal conductivity to the product of density and specific heat capacity (α = k/ρc_p). A higher thermal diffusivity means that the material can reach thermal equilibrium faster, leading to quicker changes in temperature during unsteady conduction. Materials with low thermal diffusivity will take longer to respond to changes in thermal conditions.

  6. 6.Explain how you would determine if the lumped capacitance method is applicable to a given problem.Application

    Compute the Biot number Bi = h·L_c/k using the solid's conductivity and the characteristic length L_c = V/A_s (r/3 for a sphere, r/2 for a long cylinder, half-thickness for a plate cooled on both faces). If Bi < 0.1 the internal temperature differences are small compared with the surface-to-fluid difference and the lumped model is accurate to within about 5 %. Also check that h, ρ and c are roughly constant and that the fluid temperature change is a step; otherwise use the one-term/Heisler solutions or a numerical model.

  7. 7.What is the significance of the time constant in transient heat conduction?Concept

    The time constant in transient heat conduction is a measure of the time it takes for a system to respond to changes in thermal conditions. It is defined as τ = ρc_pV/hA, where ρ is the density, c_p is the specific heat capacity, V is the volume, h is the convective heat transfer coefficient, and A is the surface area. A smaller time constant indicates a faster response to thermal changes, while a larger time constant indicates a slower response.

  8. 8.Calculate the time required for a copper sphere of diameter 0.05 m to cool from 100 °C to 50 °C in air at 25 °C. Take h = 10 W/m²·K, ρ = 8950 kg/m³, c_p = 385 J/kg·K and k = 401 W/m·K.Numerical

    L_c = V/A_s = r/3 = 0.025/3 = 0.00833 m, so Bi = 10 × 0.00833/401 = 2.1 × 10⁻⁴ ≪ 0.1 and lumped analysis applies. Time constant τ = ρ·c·L_c/h = 8950 × 385 × 0.00833/10 = 2871 s. Then t = τ·ln[(T_i − T∞)/(T − T∞)] = 2871 × ln(75/25) = 2871 × 1.0986 ≈ 3155 s, i.e. about 53 minutes.

  9. 9.A steel rod (k = 50 W/m·K) is suddenly exposed to a high-temperature environment. Explain how you would model the temperature distribution over time.Application

    To model the temperature distribution in the steel rod over time, first determine if the lumped capacitance method is applicable by calculating the Biot number. If Bi < 0.1, use the lumped capacitance method. If not, solve the transient heat conduction equation, which involves partial differential equations considering both time and spatial variables. Use appropriate boundary and initial conditions, such as the initial temperature of the rod and the temperature of the surrounding environment, to solve the equations.

  10. 10.What assumptions are made in the lumped capacitance method, and how do they affect the accuracy of the analysis?Concept

    The lumped capacitance method assumes that the temperature within the body is uniform at any given time, meaning there are no temperature gradients within the body. This assumption is valid when the Biot number is less than 0.1. The method also assumes constant material properties and boundary conditions. These assumptions simplify the analysis but can lead to inaccuracies if the Biot number is not sufficiently small or if material properties change significantly with temperature.

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