Discussion
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 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)
Answer: A
The ratio is about 2.58: the insulation more than doubles the heat loss.
Bare wire, radius 1 mm, loses heat by convection only:
\[ \begin{aligned} q_{0} &= h(2\pi r_{1})\Delta T \\ &= 25(2)(3.14)(0.001)(100) = 15.7\ \text{W/m} \end{aligned} \]
Insulated wire, outer radius 3 mm, has conduction and convection resistances in series (per metre):
\[ \begin{aligned} R &= \frac{\ln(3/1)}{2\pi k} + \frac{1}{h\,2\pi r_{2}} \\ &= \frac{1.0986}{3.14} + \frac{1}{0.471} = 0.350 + 2.123 \\ &= 2.473\ \text{K.m/W} \\ q &= \frac{100}{2.473} = 40.4\ \text{W/m} \end{aligned} \]
\[ \frac{q}{q_{0}} = \frac{40.4}{15.7} = 2.58 \]
The reason is the critical radius of insulation, \(k/h = 0.5/25 = 20\) mm. Below it, adding insulation increases the outer surface faster than it adds resistance, so the loss rises. 0.39 is the inverse ratio, which would be right only if insulation always reduced heat loss.
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