Thermal Resistance Networks

Recast conduction as a ΔT–Q relationship and build series networks like electrical circuits.

sophomoremodelingconduction
⏱️ About 17 min

Instead of re-deriving Fourier’s law each time, what if you could solve heat-transfer problems like circuit problems?

💡
The big idea: Thermal resistance turns conduction into a simple proportionality: heat rate equals temperature drop divided by resistance, and series resistances add.
🎯 By the end, you'll be able to
  • Define conduction resistance for a plane wall
  • Use Q = ΔT/R to compute heat rate
  • Combine series thermal resistances correctly
  • Check equivalence to Fourier’s-law form
📎 Helpful to know first

Conduction resistance for a plane wall

For steady 1‑D conduction through a homogeneous plane wall, define a thermal resistance:

This lets you write a compact “Ohm’s-law” form for heat transfer.

\[ R_{\text{cond}} = \frac{L}{kA} \]
Thermal resistance for 1‑D conduction through a plane wall.
\[ Q = \frac{\Delta T}{R_{\text{cond}}} \]
Resistance form: heat rate equals temperature drop divided by resistance.
✨ Electrical analogy (be careful with symbols)

The conduction form Q = ΔT/R mirrors I = ΔV/R in circuits.

Here, temperature difference plays the role of voltage difference, and heat rate plays the role of current.

Series resistances add

If heat must pass through multiple layers in sequence (same heat rate through each at steady state), the resistances add:

R_total = R1 + R2 + ...

\[ R_{\text{total}} = \sum_i R_i \quad \Rightarrow \quad Q = \frac{T_1 - T_{n+1}}{R_{\text{total}}} \]
Series thermal resistances for layers in sequence.
📝 Worked example: Using the same glass window as Lesson 1 (k = 0.8 W/(m·K), L = 0.008 m, A = 1.50 m²), with surface temperatures 22°C and 2°C, compute the heat rate using the resistance method.
  1. Compute resistance: R = L/(kA) = 0.008 / (0.8 × 1.50).
  2. Denominator: 0.8 × 1.50 = 1.20.
  3. R = 0.008 / 1.20 = 0.0066667 K/W.
  4. ΔT = 22 − 2 = 20 K.
  5. Q = ΔT/R = 20 / 0.0066667 = 3000 W.
✓ 3000 W
✏️ Practice: A carbon-steel plate (k = 50 W/(m·K)) is 5.0 mm thick with area 0.50 m². What is its conduction resistance R?
K/W
Solution
  1. R = L/(kA).
  2. L = 0.005 m, kA = 50 × 0.50 = 25.
  3. R = 0.005 / 25 = 0.0002 K/W.

Check your understanding

1. For a plane wall, which expression is the correct conduction resistance?
For 1‑D steady conduction through a plane wall, R_cond = L/(kA).
2. If two layers are in series, what is the total resistance?
Series resistances add because the same heat rate passes through each layer sequentially.
3. In the thermal–electrical analogy, heat rate Q corresponds to:
Temperature difference ↔ voltage difference, heat rate ↔ current, thermal resistance ↔ electrical resistance.
🔎 Related lessons
Composite Walls & Cylinders · soon
✅ Key takeaways
  • Define conduction resistance for a plane wall as R = L/(kA).
  • Use the compact form Q = ΔT/R to compute steady conduction heat rates.
  • Series layers add resistances: R_total = ΣR_i.
  • The resistance method is algebraically identical to the Fourier’s-law slab formula.
➡️ Now we’ll extend series networks to multi-layer walls and introduce the cylindrical-shell resistance used for pipes and insulation.
Want to test yourself on this? Try the Chemical Aptitude test →