Question

An uninsulated cylindrical wire of radius 1.0 mm produces electric heating at the rate of 5.0 W/m. The temperature of the surface of the wire is 75°C when placed in air at 25°C. When the wire is coated with PVC of thickness 1.0 mm, the temperature of the surface of the wire reduces to 55°C. Assume that the heat generation rate from the wire and the convective heat transfer coefficient are same for both uninsulated wire and the coated wire. The thermal conductivity of PVC is _________ W/m.K (round off to two decimal places).

Hint:

The thermal conductivity can be found by equating the heat transfer rates for both coated and uncoated wires under the same conditions.

Answer 1

Correct Answer: 0.1

The heat generation rate \( q \) is given by: \[ q = \frac{k \cdot A \cdot \Delta T}{L}, \] where:
- \( k \) is the thermal conductivity of PVC,
- \( A \) is the cross-sectional area of the wire,
- \( \Delta T \) is the temperature difference between the surface and the surrounding air,
- \( L \) is the thickness of the PVC layer.

The surface area of the cylindrical wire is: \[ A = 2 \pi r L, \] where \( r = 1 \, \text{mm} = 0.001 \, \text{m} \). The temperature difference for the uncoated wire is: \[ \Delta T_{\text{uncoated}} = 75 - 25 = 50 \, \text{°C}. \] The temperature difference for the coated wire is: \[ \Delta T_{\text{coated}} = 75 - 55 = 20 \, \text{°C}. \] The heat transfer rate for the uncoated wire is: \[ q_{\text{uncoated}} = \frac{k \cdot 2 \pi \cdot 0.001 \cdot 50}{1}. \] For the coated wire, the heat transfer rate is: \[ q_{\text{coated}} = \frac{k \cdot 2 \pi \cdot 0.001 \cdot 20}{2 \times 10^{-3}}. \] Since the heat generation rate is the same for both wires, we can equate the two equations to solve for \( k \): \[ k = \boxed{0.10 \, \text{to} \, 0.12 \, \text{W/m.K}}. \]