Question

On an electrical wire of 1 mm diameter and 1 mm thick plastic sheeting is provided to increase current carrying capacity. The thermal conductivity of plastic is $0.1W/mK$ and heat transfer coefficient of surrounding air is $25W/m^{2}K.$ On further addition of insulation material, the current carrying capacity will

• A. Increases
• B. Increases and then decreases
• C. Decreases
• D. Decreases and then increases

Answer 1

The correct option is B Increases and then decreases

$\text{Diameter of wire}\left(d\right)=1mm$
$(k{)}_{insulation}=0.1W/m-K,h_{outer}=25W/m^2$
$\text{Critical radius of insulation}=\frac{(k{)}_{insulation}}{h_{outer}}{r}_c=\frac{0.1}{25}=4\times {10}^{-3}=4mm$

$r_1=0.5mm{r}_2=0.5+1=1.5mm<4mm$
Further insulation will increase the heat loss upto 4 mm radius (critical radius). Insulation after the critical radius will decrease the heat loss.