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A plastic sleeve of outer radius \(r_o=1\,\text{mm}\) covers a wire of radius \(r=0.5\,\text{mm}\) carrying electric current. The thermal conductivity of the plastic is \(0.15\,\text{W/m-K}\). The heat transfer coefficient on the outer surface of the sleeve exposed to air is \(25\,\text{W/m}^2\text{-K}\). Due to the addition of the plastic cover, the heat transfer from the wire to the ambient will

1. increase
2. remain the same
3. decrease
4. be zero

Correct answer: increase

Solution

For cylindrical insulation, the critical radius is \(r_c=k/h\). Here \(r_c=0.15/25=0.006\,\text{m}=6\,\text{mm}\), which is greater than the sleeve outer radius of 1 mm. Since the added radius is below the critical radius, heat transfer increases.

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