Heat Transfer · Chapter 4 of 10 · Intermediate
Transient Conduction
Things heat up and cool down. The Biot number decides whether a body warms as one lump or with a slow interior, and the time constant tells you how long it takes.
Readiness check
From Chapters 1 and 2. Tick only what you can do closed-notes.
- Write an energy balance on a body that is storing heat.
- Use mc to convert heat to a temperature change.
- Handle the exponential function and natural logs.
- Compare a conduction resistance with a convection resistance.
- Keep density, specific heat, and conductivity straight.
The core idea
If a body conducts heat internally far faster than it loses it from the surface, its whole temperature falls together, exponentially.
Bi = hLc/kθ/θi = e−t/ττ = ρVc/(hAs)The Biot number compares the internal conduction resistance to the surface convection resistance. When Bi < 0.1, the inside is nearly uniform (the lumped model) and the temperature decays as a single exponential with time constant τ. When Bi is larger, the centre lags the surface and you need the distributed solutions.
The skills, taught in order
Transient conduction starts with one yes-or-no question, the Biot number, and then either a one-line exponential or the distributed solutions. Six skills cover the decision and the lumped answer.
4.1 Why transient matters
Steady state is the exception. Engines warm up, parts are quenched, sensors must respond, electronics cycle on and off. Transient conduction predicts how long these changes take and what temperatures occur on the way.
4.2 The Biot number
Bi = hLc/k compares the body's internal conduction resistance (Lc/k) to its surface convection resistance (1/h), using the characteristic length Lc = V/As. Small Bi means the inside equalises far faster than the surface loses heat, so the body is nearly isothermal.
4.3 Lumped capacitance
When Bi < 0.1, an energy balance on the whole body gives a single exponential: θ/θi = e−t/τ, where θ = T − T∞. One temperature describes the part at every instant, and the cooling or heating curve is a pure decay.
Micro-example. A body at Bi = 0.05 is safely lumped; one at Bi = 0.5 is not, and its centre will lag its surface noticeably.
4.4 The time constant
The decay rate is set by τ = ρVc/(hAs) = mc/(hAs): thermal mass over surface conductance. In one τ the body covers 63% of its temperature change; the milestones below are worth memorising.
| Elapsed time | Fraction of change completed |
|---|---|
| 1 τ | 63% |
| 2 τ | 86% |
| 3 τ | 95% |
| 5 τ | 99% (practically settled) |
4.5 Thermal diffusivity and the Fourier number
How fast a temperature change penetrates a solid is set by the thermal diffusivity α = k/(ρc) (m²/s). The dimensionless time is the Fourier number Fo = αt/L². A useful estimate: a disturbance takes roughly t ≈ L²/α to reach a depth L.
4.6 When the body is not lumped
If Bi > 0.1, the surface responds while the centre lags, and the temperature is a function of position and time. The one-term approximation (and the classic Heisler charts) give the answer in terms of Bi and Fo for slabs, cylinders, and spheres. Worked Example 2 shows why a thick wall needs this treatment.
Engineering connection: quenching and heat treatment, sensor response time, thermal start-up, and the cycling of electronics. The h comes from Chapters 6 and 7.
Worked example 1: cooling a copper sphere
A 6 mm copper sphere (k = 400, ρ = 8900 kg/m³, c = 385 J/kg·K) at 200 °C is cooled in 25 °C air with h = 120 W/m²·K. Confirm the lumped model applies, find the time constant, and find how long it takes to reach 50 °C.
- ProblemCheck the lumped assumption, find τ, and find the time to reach 50 °C for the sphere in Figure 1.
- Given / findD = 0.006 m, k = 400, ρ = 8900, c = 385, Ti = 200 °C, T∞ = 25 °C, h = 120. Find Bi, τ, and t to 50 °C.
- AssumptionsUniform h over the surface, constant properties, negligible radiation.
- ModelCompute Lc = V/As = D/6, check Bi < 0.1, then τ = ρcLc/h and invert the exponential for time.
- EquationsBi = hLc/k τ = ρVc/(hAs) = ρcLc/h t = −τ ln(θ/θi)
- SolveLc = 0.006/6 = 0.001 m. Bi = 120 × 0.001 / 400 = 3×10⁻⁴, far below 0.1, so lumped is valid. τ = 8900 × 385 × 0.001 / 120 = 28.6 s. θ/θi = (50 − 25)/(200 − 25) = 0.143, so t = −28.6 ln(0.143) = 55.6 s.
- CheckReaching 0.143 of the start is just past two time constants (where 0.135 remains), and 55.6 s ≈ 1.94 τ, consistent. The tiny Bi is exactly what makes copper parts respond as one lump.
- ConclusionThe whole sphere cools together because copper conducts internally far faster than the air removes heat. Cooling time scales with thermal mass over surface conductance, so a bigger sphere or weaker airflow would take proportionally longer.
Worked example 2: a thick wall is not a lump
A 0.12 m thick clay wall (k = 1.0, ρ = 1900 kg/m³, c = 800 J/kg·K) is suddenly exposed on one face to hot gas with h = 30 W/m²·K. Decide whether the lumped model applies, and estimate how long heat takes to penetrate to the far face.
- ProblemDecide if the wall in Figure 2 is lumped, and estimate the time for heat to reach the far face.
- Given / findL = 0.12 m, k = 1.0, ρ = 1900, c = 800, h = 30. Find Bi and a penetration time.
- AssumptionsOne face heated, constant properties, the far face initially undisturbed (semi-infinite early on).
- ModelUse the full wall thickness as the conduction length. Compute Bi to test lumping, then α = k/(ρc) and the diffusion-time estimate t ≈ L²/α.
- EquationsBi = hL/k α = k/(ρc) t ≈ L²/α
- SolveBi = 30 × 0.12 / 1.0 = 3.6, far above 0.1, so the wall is not lumped. α = 1.0/(1900 × 800) = 6.6×10⁻⁷ m²/s. Penetration time t ≈ 0.12² / 6.6×10⁻⁷ = 21 900 s ≈ 6 hours.
- CheckThe large Bi and the hours-long diffusion time are consistent: a thick, low-conductivity wall cannot equalise quickly. This is the opposite regime to the copper sphere, decided entirely by Bi.
- ConclusionA lumped exponential would badly mispredict this wall, since the centre lags the surface by hours. When Bi exceeds 0.1, reach for the one-term or chart solutions, not τ. Thermal mass that responds slowly is also why masonry buildings even out day-night swings.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Lumped used without checking Bi | Exponential applied to a thick block | "Is Bi below 0.1?" | Always compute Bi = hLc/k first; above 0.1 the centre lags and τ is meaningless. |
| Wrong characteristic length | Bi off by a large factor | "Did I use Lc = V/As?" | For lumping, Lc = volume over surface area (D/6 for a sphere, not the radius). |
| Diffusivity confused with conductivity | Response time computed from k alone | "Did I divide k by ρc?" | Speed of response is α = k/(ρc); a high-k metal can still be slow if it is massive. |
| "Settled" after one τ | Process assumed complete too early | "How many time constants have passed?" | One τ is only 63%; allow about 5τ for the change to finish. |
Practice ladder
A thermocouple bead behaves as a 1 mm sphere (ρ = 8500, c = 400, k = 20) in a gas stream with h = 250 W/m²·K. Find Bi and the time constant.
Show answer
Lc = 0.001/6 = 1.67×10⁻⁴ m. Bi = 250 × 1.67×10⁻⁴ / 20 = 2.1×10⁻³ (lumped). τ = 8500 × 400 × 1.67×10⁻⁴ / 250 = 2.3 s. The sensor needs about 5τ ≈ 11 s to fully register a step, so it lags fast fluctuations.
For the Worked Example 1 sphere, how long to reach 30 °C instead of 50 °C? Why does the last few degrees take so long?
Show answer
θ/θi = (30 − 25)/175 = 0.0286, so t = −28.6 ln(0.0286) = 102 s, nearly double the time to 50 °C. The exponential flattens as it approaches the fluid temperature, so the final approach is always the slowest part.
A 4 cm solid steel cylinder (k = 50, ρ = 7850, c = 470) is quenched in oil with h = 500 W/m²·K. Compute Bi using Lc = r/2 and decide whether lumped capacitance is acceptable.
Show answer
r = 0.02 m, Lc = r/2 = 0.01 m. Bi = 500 × 0.01 / 50 = 0.10, right at the boundary. Lumped is marginal: usable for a rough estimate but the centre will lag the surface, so a one-term solution is safer for the centre temperature.
Time a real cooling or heating event (coffee cooling, a part out of an oven, a fridge item warming). Estimate the body's Bi and τ, then compare your model's predicted curve with what you observe.
What good work looks like
A Bi check that justifies (or rejects) the lumped model, a computed τ, and measured temperatures at a few times compared with the e−t/τ prediction, with the dominant error source named.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Write a one-page transient note for a real part: the Bi check, the model chosen, τ if lumped, and the predicted time to a target temperature, with a measured or sanity-checked comparison.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Define the Biot number and its threshold.
Bi = hLc/k, the ratio of internal conduction resistance to surface convection resistance; below 0.1 the body is lumped.
2. Write the lumped cooling law and the time constant.
θ/θi = e−t/τ with τ = ρVc/(hAs) = mc/(hAs).
3. What does one time constant correspond to?
63% of the total temperature change; about 5τ to reach 99%.
4. Define thermal diffusivity and the Fourier number.
α = k/(ρc); Fo = αt/L²; a disturbance reaches depth L in roughly L²/α.
5. What do you do when Bi exceeds 0.1?
Use distributed solutions (one-term approximation or Heisler charts) in terms of Bi and Fo; the lumped τ no longer applies.
Textbook mapping
| Item | Mapping |
|---|---|
| Primary source | Lienhard and Lienhard, A Heat Transfer Textbook (6th ed), Chapter 5 (transient and multidimensional conduction) |
| Cross-reference | Incropera, Ch. 5 (transient conduction) · Çengel and Ghajar, Ch. 4 |
| Core topics | 4.1 Why transient · 4.2 Biot number · 4.3 Lumped capacitance · 4.4 Time constant · 4.5 Diffusivity and Fourier number · 4.6 Distributed response |
| Engineering connection | Quenching, sensor response, thermal start-up; the h comes from Chapters 6 and 7. |
| Read next | Chapter 5: Convection and Boundary Layers. |