Subtopic: Steady state heat transfer in conduction, convection and radiation
Question Statement:
A 2 mm diameter electric wire is insulated with 2 mm thick plastic. The plastic has a thermal conductivity of 0.5 W·m⁻¹·K⁻¹. The wire is exposed to air at 320 K and the outside convective heat transfer coefficient is 25 W·m⁻²·K⁻¹. Wire surface temperature is constant at 420 K and remains unaffected by the covering. Assuming steady state heat transfer conditions, the ratio of heat loss per metre of wire length with insulation to that without insulation is nearest to (Take π = 3.14)
Options:
(A) 2.58
(B) 0.39
(C) 1.87
(D) 0.54
Official Answer Key:
A
Step-by-Step Solution & Explanation
Case 1 — bare wire (convection from the wire surface only):
Q₁ = h(πd L)ΔT
Per metre: Q₁ = 25 × 3.14 × 0.002 × (420 − 320) = 15.70 W·m⁻¹
Case 2 — insulated wire. Inner radius r₁ = 1 mm; outer radius r₂ = 1 + 2 = 3 mm. Two resistances in series:
Conduction through the plastic:
R_cond = ln(r₂/r₁)/(2πk) = ln(3)/(2 × 3.14 × 0.5) = 1.0986/3.14 = 0.3499 K·m·W⁻¹
Convection at the outer surface:
R_conv = 1/(h · 2πr₂) = 1/(25 × 2 × 3.14 × 0.003) = 1/0.471 = 2.1231 K·m·W⁻¹
Total: ΣR = 2.4730 K·m·W⁻¹
Q₂ = ΔT/ΣR = 100/2.4730 = 40.44 W·m⁻¹
Ratio:
Q₂/Q₁ = 40.44/15.70 = 2.58
Concept — critical radius of insulation:
r_c = k/h = 0.5/25 = 0.02 m = 20 mm
Since the insulated outer radius (3 mm) is far below the critical radius, adding insulation increases heat loss — the added outer surface area outweighs the added conduction resistance. This is why the ratio exceeds 1, and is exactly the principle exploited when insulating small electrical wires for heat dissipation.
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