Thermal Resistance and Equivalent Conductivity
Thermal resistance is R_th = Delta T/H; for planar steady conduction R_th=L/(kA), series resistances add, and parallel conductances add under common node temperatures.
Why this shows up in the exam
Layered walls · Series and parallel rod combinations · Effective conductivity of composite cylinders in axial flow
Learn the idea
Composite conductors combine like resistance networks when each branch has a well-defined steady heat current. Temperature difference plays the role of voltage and heat current the role of electric current, so layers in series add resistance while parallel paths add conductance.
🧠 Memory hook: Series adds thermal resistance; parallel adds thermal conductance.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- R_th = L/(kA) — thermal resistance of a uniform planar element
- R_series = sum_i R_i — series elements carrying the same heat current
- 1/R_parallel = sum_i 1/R_i — parallel paths sharing the same temperature difference
- k_eq = L_total/(A R_total) — equivalent conductivity only after total geometry is defined
How to approach it
- 1Draw thermal nodes and branches
- 2Assign one resistance to each path
- 3Reduce the network and recover temperatures or equivalent k
Common slip-ups that cost marks
- •Averaging conductivities without geometry
- •Calling paths parallel when their endpoint temperatures differ
- •Using planar L/kA resistance for radial flow
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A wire 1 m long and cross-sectional area 2 mm^2 extends by 1 mm under a 200 N load. Find Young modulus.
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