Thermal Resistance: Series & Parallel
Heat flow does circuit arithmetic.
Lesson at a glanceShow the key points
- R = x/KA; i = (T₁ − T₂)/R (28.3), like Ohm's law.
- Series: same heat current; R = R₁ + R₂ + …; junction temperature from T₁ − T = R₁i.
- Parallel: same temperature difference; 1/R = 1/R₁ + 1/R₂ + ….
Heat flow through a rod obeys the same arithmetic as electric current through a wire, so composite walls can be worked out like circuits.
Thermal resistance
Write the heat current as i and the quantity x/KA as the thermal resistance R. Then equation 28.1 becomes
Find the thermal resistance, along its length, of a steel rod 50 cm long with cross-section 2.0 cm² (K = 46 W m⁻¹ K⁻¹).
The same heat current flows through each rod. With junction temperature T: T₁ − T = R₁i and T − T₂ = R₂i. Adding: i = (T₁ − T₂)/(R₁ + R₂). R = R₁ + R₂ + …
Each rod has the same temperature difference and its own current; i = i₁ + i₂ = (T₁ − T₂)(1/R₁ + 1/R₂). So 1/R = 1/R₁ + 1/R₂ + …
A copper rod and a steel rod, each 10 cm long and 1.0 cm² in cross-section, are joined end to end. The free ends are at 100°C and 0°C. Find the heat current and the junction temperature (K: 385 and 46).
- 1R = x/KA; i = (T₁ − T₂)/R, like Ohm's law.
- 2Series: same current, R = R₁ + R₂; the higher resistance takes the larger temperature drop.
- 3Parallel: same temperature difference, 1/R = 1/R₁ + 1/R₂.
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