Heat Transfer · Chapter 1 of 10 · Beginner
Introduction: The Three Modes of Heat Transfer
Thermodynamics tells you how much energy moves; heat transfer tells you how fast. Three modes carry it: conduction, convection, and radiation, each with its own rate law.
Readiness check
From Thermodynamics and Thermal Physics. Tick only what you can do closed-notes.
- Write an energy balance: in minus out plus generated equals stored.
- Use Q = mcΔT for stored thermal energy.
- Keep watts (a rate) distinct from joules (an amount).
- Convert °C to kelvin, and know radiation demands kelvin.
- Judge when steady state is a fair assumption.
The core idea
Heat always flows from hot to cold, by three modes, each with a rate law in watts.
q = kA ΔT/Lq = hA ΔTq = εσA(Ts⁴ − Tsurr⁴)Conduction carries heat through matter, convection carries it to a moving fluid, and radiation carries it as electromagnetic waves across any gap, even a vacuum. Real surfaces usually lose heat by two or three modes at once, and an energy balance keeps the books.
The skills, taught in order
Six ideas turn "it gets hot" into a number in watts. Learn the three rate laws, the energy balance that links them, and the habit of finding the mode that dominates.
1.1 What heat transfer adds to thermodynamics
Thermodynamics counts totals: how much energy a process moves, and the equilibrium it ends at. Heat transfer asks the next two questions: how fast does the energy move (a rate, in watts), and what temperature distribution does that produce. Those rates set the size of every heat sink, radiator, and layer of insulation.
1.2 Conduction
Conduction is heat diffusing through matter, down the temperature gradient. Fourier's law for steady one-dimensional flow is q = kA ΔT/L, where k is the thermal conductivity (W/m·K), A the area, and L the thickness. Metals conduct hundreds of times better than insulators, which is the whole reason both exist.
| Material | k (W/m·K, near room temperature) |
|---|---|
| Copper | ≈ 400 |
| Aluminium | ≈ 237 |
| Carbon steel | ≈ 50 |
| Stainless steel | ≈ 15 |
| Glass | ≈ 1.0 |
| Water | ≈ 0.6 |
| Foam or fibre insulation | ≈ 0.04 |
| Still air | ≈ 0.026 |
Micro-example. A 3 mm glass pane (k ≈ 0.8) with 20 K across it passes q/A = kΔT/L = 0.8 × 20 / 0.003 ≈ 5300 W/m², which is why a window's heat loss is set by the air films, not the glass.
1.3 Convection
Convection is heat carried away by a moving fluid. Newton's law of cooling lumps the messy fluid mechanics into one coefficient: q = hA(Ts − T∞), where h is the convection coefficient (W/m²·K). Finding h for a given flow is most of the rest of this course.
| Situation | Typical h (W/m²·K) |
|---|---|
| Free convection, gases | 2 to 25 |
| Free convection, liquids | 50 to 1000 |
| Forced convection, gases | 25 to 250 |
| Forced convection, liquids | 100 to 20 000 |
| Boiling or condensation | 2500 to 100 000 |
Micro-example. A chip surface 35 K above the air sheds hA ΔT per unit area: 35 × 10 ≈ 350 W/m² in still air, but 35 × 80 ≈ 2800 W/m² under a fan. The fan, not the chip, changed h.
1.4 Radiation
Every surface radiates, with no medium required. The net exchange with large surroundings is q = εσA(Ts⁴ − Tsurr⁴), where σ = 5.67 × 10⁻⁸ W/m²·K⁴ is the Stefan-Boltzmann constant and ε is the emissivity (0 to 1). The fourth-power law means radiation is negligible when cool but dominant when hot, and the temperatures must be absolute (kelvin).
1.5 The energy balance
Every heat-transfer problem is an accounting statement: energy in minus energy out plus energy generated equals the change in stored energy. At steady state the stored term is zero, so in equals out. On a surface (which stores nothing) the modes crossing it must sum to zero.
1.6 Finding the dominant mode
Before computing anything, ask which mode carries most of the heat. Compare the would-be rates or, better, the thermal resistances of Chapter 2. A surface in still air loses comparable heat by free convection and radiation; the same surface in a strong wind is convection-dominated. Naming the controlling mode first keeps the analysis honest and short.
Engineering connection: the foundation of cooling, insulation, and thermal management in every later chapter and in Energy Systems. Builds directly on Thermodynamics.
Worked example 1: how a hot surface loses heat
The outer face of an electronics enclosure is at 70 °C, has area 0.05 m², and sits in a room at 25 °C with still air (free-convection coefficient h = 10 W/m²·K). Its painted surface has emissivity ε = 0.9. Find the heat lost by convection and by radiation, and say which matters.
- ProblemFind the convective and radiative heat loss from the 70 °C surface of Figure 1, and compare them.
- Given / findTs = 70 °C, T∞ = Tsurr = 25 °C, A = 0.05 m², h = 10 W/m²·K, ε = 0.9. Find qconv, qrad, and the total.
- AssumptionsSteady state, uniform surface temperature, large surroundings at the air temperature, σ = 5.67 × 10⁻⁸ W/m²·K⁴.
- ModelTwo parallel paths off the surface: Newton's law of cooling for convection and the Stefan-Boltzmann net exchange for radiation. Radiation needs kelvin: 343 K and 298 K.
- Equationsqconv = hA(Ts − T∞) qrad = εσA(Ts⁴ − Tsurr⁴)
- Solveqconv = 10 × 0.05 × (70 − 25) = 22.5 W. qrad = 0.9 × 5.67×10⁻⁸ × 0.05 × (343.15⁴ − 298.15⁴) = 15.2 W. Total = 37.7 W, of which radiation is 40%.
- CheckBoth terms are positive (the surface is hotter than its surroundings) and the same order of magnitude, which is exactly what free convection plus radiation gives at moderate temperatures. Units: (W/m²·K)(m²)(K) = W, and (W/m²·K⁴)(m²)(K⁴) = W.
- ConclusionIgnoring radiation here would underestimate the cooling by 40% and overpredict the chip temperature. At moderate temperatures with still air, radiation is not optional. A fan would raise h tenfold and make convection dominate instead.
Worked example 2: sizing insulation
A cold-room wall of area 12 m² must hold its heat gain to 60 W when the temperature difference across the foam insulation (k = 0.030 W/m·K) is 25 °C. How thick must the foam be?
- ProblemFind the foam thickness L that limits the heat gain to 60 W for the wall in Figure 2.
- Given / findq = 60 W, A = 12 m², ΔT = 25 °C, k = 0.030 W/m·K. Find L.
- AssumptionsSteady one-dimensional conduction through the foam, the foam the controlling resistance (surface films neglected), constant k.
- ModelFourier's law for a slab, solved for thickness.
- Equationsq = kA ΔT/L L = kA ΔT/q
- SolveL = (0.030 × 12 × 25) / 60 = 9 / 60 = 0.15 m = 15 cm.
- CheckUnits: (W/m·K)(m²)(K)/(W) = m. Scale: thick foam for a small heat leak is right, since foam's k is tiny. Halving the allowed heat gain to 30 W would double the thickness to 30 cm, the inverse relation Fourier predicts.
- ConclusionFifteen centimetres of foam meets the target. Conduction set the answer through one material constant and two geometry choices, which is the everyday arithmetic of insulation and the launching point for the resistance networks of Chapter 2.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Heat and temperature confused | "It is hot, so a lot of heat flows" | "Is this a temperature or a rate of energy?" | Temperature (K) is a level; heat rate (W) depends on the temperature difference and the path. A warm, well-insulated wall passes little heat. |
| Radiation forgotten | Free-convection problems solved with q = hA ΔT alone | "Is the surface warm and the air still?" | Then radiation rivals convection. Add q = εσA(T⁴) whenever h is small. |
| Celsius inside the T⁴ | Radiation answers wildly wrong | "Are my radiation temperatures in kelvin?" | The Stefan-Boltzmann law needs absolute temperature. Convert before raising to the fourth power. |
| Area or thickness dropped | Conduction reported in W/m² called the heat rate | "Did I multiply by area and divide by thickness?" | q = kA ΔT/L is a rate in watts; q/A = kΔT/L is a flux in W/m². Keep them separate. |
Practice ladder
A 2 m² window of glass 4 mm thick (k = 1.0 W/m·K) has 18 °C across the pane itself. Find the conduction heat rate through the glass.
Show answer
q = kA ΔT/L = 1.0 × 2 × 18 / 0.004 = 9000 W. The glass barely resists heat; real windows rely on the air films and gas gaps, the resistances of Chapter 2.
Repeat Worked Example 1 with a fan that raises h to 60 W/m²·K. Find the new convective loss and the new radiation share of the total.
Show answer
qconv = 60 × 0.05 × 45 = 135 W; radiation is still 15.2 W, so the total is 150.2 W and radiation is now only 10%. Forced convection swamps radiation, which is why radiation is most often dropped for cooled electronics but not for still-air ones.
A 1500 W electric heater element reaches steady state losing all its power from a 0.02 m² surface to a 20 °C room. If convection gives h = 30 W/m²·K, estimate the surface temperature, ignoring radiation. Then comment on whether ignoring radiation was safe.
Show answer
1500 = 30 × 0.02 × (Ts − 20) gives Ts − 20 = 2500 K, so Ts ≈ 2520 °C, absurdly hot. That signals the model is wrong: at such temperatures radiation (T⁴) carries most of the heat, so it cannot be ignored. Real heater elements glow precisely because radiation dominates.
Pick a real warm object (a laptop charger, a radiator, a mug of coffee). Estimate its surface temperature, area, and surroundings, then compute its convective and radiative losses and name the dominant mode.
What good work looks like
Stated temperatures in kelvin for radiation, both rate laws applied with reasonable h and ε, the two losses compared, and a one-sentence verdict on which mode controls and why.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Write a one-page "Modes Card": the three rate laws with units, the k and h tables in your own words, and one worked surface where you compute all the active modes and declare the dominant one.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. State the three rate laws with their units.
Conduction q = kA ΔT/L; convection q = hA ΔT; radiation q = εσA(Ts⁴ − Tsurr⁴). All give watts.
2. What does heat transfer add to thermodynamics?
Rates (how fast, in watts) and temperature distributions, where thermodynamics gave only totals and end states.
3. When can radiation not be ignored?
Whenever the convection coefficient is small (still air) or the surface is hot, because radiation grows as T⁴.
4. Write the general energy balance and its steady-state form.
In − out + generated = stored; at steady state stored = 0, so in = out.
5. Why must radiation temperatures be in kelvin?
The law uses absolute temperature raised to the fourth power; Celsius has an arbitrary zero and breaks T⁴.
Textbook mapping
| Item | Mapping |
|---|---|
| Primary source | Lienhard and Lienhard, A Heat Transfer Textbook (6th ed), Chapter 1 (modes of heat transfer) |
| Cross-reference | Incropera, Fundamentals of Heat and Mass Transfer, Ch. 1 · Çengel and Ghajar, Ch. 1 |
| Core topics | 1.1 Rates versus totals · 1.2 Conduction · 1.3 Convection · 1.4 Radiation · 1.5 Energy balance · 1.6 Dominant mode |
| Engineering connection | The basis of all cooling, insulation, and thermal management; builds on Thermodynamics. |
| Read next | Chapter 2: Steady Conduction and Thermal Resistance. |