Heat Transfer · Chapter 2 of 10 · Beginner
Steady Conduction and Thermal Resistance
Treat heat flow like an electric current: temperature difference drives it, and each layer and film is a resistance. Add the resistances and the heat rate follows in one line.
Readiness check
From Chapter 1. Tick only what you can do closed-notes.
- State Fourier's law q = kA ΔT/L.
- State Newton's law of cooling q = hA ΔT.
- Add resistors in series and combine them in parallel.
- Look up a thermal conductivity and a convection coefficient.
- Keep area and thickness in SI units.
The core idea
Heat flow is a current driven by a temperature difference through a chain of resistances.
q = ΔT / RthRcond = L/(kA)Rconv = 1/(hA)The electrical analogy is exact: temperature difference is the voltage, heat rate is the current, and every wall layer or surface film is a resistor. Resistances in the same heat path add in series; the largest one controls the flow. The whole network collapses into one overall coefficient U.
The skills, taught in order
One analogy carries this whole chapter: heat flow obeys Ohm's law. Learn the resistance of each element, string them together, and read off the heat rate and the overall coefficient.
2.1 Fourier's law as a resistance
Rewrite steady conduction q = kA ΔT/L as q = ΔT/R with R = L/(kA). A thick, low-conductivity, small-area layer is a large resistance. The form q = ΔT/R is what lets layers be chained.
Micro-example. A 0.10 m brick wall (k = 0.7) of 1 m² has R = L/(kA) = 0.10/(0.7 × 1) = 0.143 K/W.
2.2 The electrical analogy
Heat conduction maps exactly onto a DC circuit. Once you see the map, every circuit trick (series, parallel, dividers) carries over.
| Electrical | Thermal |
|---|---|
| Voltage V | Temperature difference ΔT |
| Current I | Heat rate q |
| Resistance R (Ω) | Thermal resistance Rth (K/W) |
| I = V/R | q = ΔT/Rth |
2.3 Convection as a resistance
A surface losing heat to a fluid is also a resistance: Rconv = 1/(hA). Every place a solid meets a fluid adds one of these films in series. A low h (still air) is a large resistance, often larger than the wall it sits on.
2.4 Composite walls
Layers stacked in the heat path are resistances in series, so they simply add: Rtotal = ΣR. Compute q from the total, then find any interface temperature by walking the drops: ΔT across an element is q·R for that element. The element with the largest R carries the largest temperature drop.
2.5 Radial conduction
For a pipe or cylindrical shell the area grows with radius, so the resistance is R = ln(r₂/r₁)/(2πkL) per length L, not L/(kA). Insulation on a pipe is a radial resistance added in series with the outside film.
| Element | Thermal resistance |
|---|---|
| Plane wall | L/(kA) |
| Cylindrical shell (length L) | ln(r₂/r₁)/(2πkL) |
| Convection film | 1/(hA) |
2.6 The overall heat-transfer coefficient U
Engineers package the whole network into one number: q = UA ΔT, where 1/(UA) = ΣR. U (W/m²·K) is the conductance per unit area, handy for comparing walls and for the heat-exchanger work of Chapter 9.
2.7 The critical radius of insulation
On a thin pipe or wire, adding insulation increases the conduction resistance but also enlarges the outer surface, which lowers the convection resistance. Below the critical radius rcr = k/h, the second effect wins and a little insulation can actually raise the heat loss. Above it, insulation helps as expected.
Engineering connection: the workhorse method for insulation, building envelopes, and the U-values that Chapter 9 uses to size heat exchangers.
Worked example 1: a composite wall
A wall (per square metre) is inside air at 20 °C, then 0.10 m of brick (k = 0.7), then 0.05 m of foam (k = 0.030), then outside air at −5 °C. The inside film gives h = 8 W/m²·K and the outside film h = 20 W/m²·K. Find the heat loss per m², the overall U, and the brick-foam interface temperature.
- ProblemFind q/A, U, and the brick-foam interface temperature for the wall in Figure 1.
- Given / findInside air 20 °C, hi = 8; brick L = 0.10, k = 0.7; foam L = 0.05, k = 0.030; outside air −5 °C, ho = 20. Per 1 m². Find q/A, U, Tinterface.
- AssumptionsSteady one-dimensional conduction, constant properties, perfect contact between layers.
- ModelFour resistances in series (per m²): inside film, brick, foam, outside film. Add them, then q = ΔT/ΣR and U = 1/ΣR. Interface temperature comes from the drops upstream of it.
- EquationsΣR = 1/hi + Lb/kb + Lf/kf + 1/ho q/A = ΔT/ΣR, U = 1/ΣR
- SolveΣR = 0.125 + 0.143 + 1.667 + 0.050 = 1.985 m²·K/W. q/A = 25/1.985 = 12.6 W/m². U = 1/1.985 = 0.50 W/m²·K. From the inside: 20 − 12.6(0.125 + 0.143) = 16.6 °C at the brick-foam interface.
- CheckThe foam alone is 1.667 of the 1.985 total, 84%, so it should carry 84% of the 25 °C drop, about 21 °C; indeed the temperature falls from 16.6 °C to about −4.4 °C across the foam. Units of ΣR are (m²·K/W), so q/A comes out in W/m².
- ConclusionThe thin foam layer does almost all the insulating; thickening the brick would barely help. Finding the controlling resistance, not just the total, is what turns a number into a design decision.
Worked example 2: an insulated pipe
A pipe whose outer surface sits at 90 °C has radius 25 mm. It is wrapped to 50 mm radius with insulation (k = 0.040 W/m·K), and the outside air at 20 °C gives h = 10 W/m²·K. Find the heat loss per metre of pipe, and compare it with the bare pipe.
- ProblemFind the heat loss per metre for the insulated pipe in Figure 2, and compare with the bare pipe.
- Given / findPipe surface 90 °C at r₁ = 25 mm; insulation to r₂ = 50 mm, k = 0.040; air 20 °C, h = 10. Find q′ (per metre).
- AssumptionsSteady radial conduction, constant k, the pipe-wall resistance negligible (its surface is the 90 °C boundary), uniform outside film.
- ModelPer unit length, the insulation is a log-law resistance in series with the outside convection film: R′ = ln(r₂/r₁)/(2πk) + 1/(h·2πr₂).
- EquationsR′ins = ln(r₂/r₁)/(2πk) R′conv = 1/(h·2πr₂) q′ = ΔT/(R′ins + R′conv)
- SolveR′ins = ln(2)/(2π·0.040) = 2.76 K·m/W. R′conv = 1/(10·2π·0.05) = 0.32 K·m/W. q′ = (90 − 20)/3.08 = 22.8 W/m. Bare pipe (just the film at r₁): R′ = 1/(10·2π·0.025) = 0.64, giving q′ = 110 W/m.
- CheckThe insulation resistance (2.76) is nearly nine times the film (0.32), so it controls. Insulation cut the loss from 110 to 22.8 W/m, a factor of about 4.8, which is the point of lagging hot pipes. Units: K·m/W in R′, so q′ is W/m.
- ConclusionFor a pipe this size the insulation is well past its critical radius (rcr = k/h = 0.004 m = 4 mm), so every millimetre of foam reduces the loss. Radial geometry changes the resistance formula but not the resistance-network method.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Surface films forgotten | Only wall layers added; q too high | "Does the wall touch a fluid on either face?" | Add 1/(hA) at every solid-fluid boundary; in still air the film can dominate. |
| Series and parallel confused | Parallel paths added like series | "Does the heat have one path or several?" | One path: add R in series. Several paths (a stud beside insulation): combine in parallel. |
| Plane formula on a pipe | L/(kA) used for radial conduction | "Is the area constant or growing with radius?" | Use ln(r₂/r₁)/(2πkL) for cylinders; the area is not constant. |
| Insulation always helps | Thin wire insulated to cut loss, loss rises | "Is the radius above rcr = k/h?" | Below the critical radius the larger surface wins; check rcr for small pipes and wires. |
Practice ladder
A 0.2 m concrete wall (k = 1.4) of 6 m² has 15 °C across it. Find its conduction resistance and the heat rate, ignoring films.
Show answer
R = L/(kA) = 0.2/(1.4 × 6) = 0.0238 K/W. q = ΔT/R = 15/0.0238 = 630 W. Bare concrete resists little; the films and any insulation would dominate.
Add a 4 cm foam layer (k = 0.035) to the Level 1 wall and recompute the heat rate. By what factor does it fall?
Show answer
Rfoam = 0.04/(0.035 × 6) = 0.190 K/W. New ΣR = 0.0238 + 0.190 = 0.214 K/W, so q = 15/0.214 = 70 W, a ninefold drop. The foam, not the concrete, now sets the loss.
A double-glazed window is two 4 mm glass panes (k = 1.0) with a 12 mm still-air gap (k = 0.026), all of area 1.5 m², with 18 °C across the assembly (ignore the room and outdoor films). Find the heat rate and identify the controlling layer.
Show answer
Glass each: 0.004/(1.0 × 1.5) = 0.00267 K/W. Air gap: 0.012/(0.026 × 1.5) = 0.308 K/W. ΣR = 0.00267 + 0.308 + 0.00267 = 0.313 K/W. q = 18/0.313 = 57.5 W. The trapped air gap, not the glass, does essentially all the insulating, which is the entire point of double glazing.
Build a resistance model of a real wall, window, or insulated container near you. Estimate each layer's R, identify the controlling resistance, and propose the one change that would most reduce the heat loss.
What good work looks like
A labelled series network with every layer and film, the dominant resistance named, q and U computed, and a change aimed at the controlling resistance rather than a minor one.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Produce a one-page "Resistance Network Card" for one real assembly: the sketch, every series resistance with its formula and value, the total, U, q, and the single change that would help most.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Write the resistance of a plane wall, a cylinder, and a convection film.
L/(kA); ln(r₂/r₁)/(2πkL); 1/(hA).
2. State the heat-flow form of Ohm's law and the rule for series layers.
q = ΔT/Rth; resistances in one path add, Rtotal = ΣR.
3. How do you find an interface temperature?
Start from a known temperature and subtract q·R for each element up to that interface.
4. Define the overall heat-transfer coefficient.
q = UA ΔT with 1/(UA) = ΣR; U is the conductance per unit area.
5. What is the critical radius of insulation, and why does it matter?
rcr = k/h. Below it, added insulation enlarges the surface enough to raise the loss, so small pipes and wires can behave counterintuitively.
Textbook mapping
| Item | Mapping |
|---|---|
| Primary source | Lienhard and Lienhard, A Heat Transfer Textbook (6th ed), Chapter 2 (conduction, thermal resistance, overall U) |
| Cross-reference | Incropera, Ch. 3 (one-dimensional steady conduction) · Çengel and Ghajar, Ch. 3 |
| Core topics | 2.1 Conduction resistance · 2.2 Electrical analogy · 2.3 Convection resistance · 2.4 Composite walls · 2.5 Radial conduction · 2.6 Overall U · 2.7 Critical radius |
| Engineering connection | Insulation, building envelopes, and the U-values used in Chapter 9. |
| Read next | Chapter 3: Fins and Extended Surfaces. |