Question

An electrical wire of 2 mm diameter and 5 m length is insulated with a plastic layer of thickness 2 mm and thermal conductivity \( k = 0.1 \) W/(m·K). It is exposed to ambient air at 30°C. For a current of 5 A, the potential drop across the wire is 2 V. The air-side heat transfer coefficient is 20 W/(m²·K). Neglecting the thermal resistance of the wire, the steady-state temperature at the wire-insulation interface __________°C (rounded off to 1 decimal place).

GIVEN: 
Kinematic viscosity: \( \nu = 1.0 \times 10^{-6} \, {m}^2/{s} \) 
Prandtl number: \( {Pr} = 7.01 \) 
Velocity boundary layer thickness: \[ \delta_H = \frac{4.91 x}{\sqrt{x \nu}} \]

Hint:

For heat transfer through insulation, use the formula for a cylindrical shell and solve for the interface temperature.

Answer 1

The power dissipated in the wire due to the current is given by: \[ P = I^2 R \] Where:
\( I = 5 \, {A} \) is the current,
\( R \) is the resistance of the wire.
The resistance \( R \) is related to the resistivity of the material, but for this case, it can be derived from the potential drop and current: \[ R = \frac{V}{I} = \frac{2 \, {V}}{5 \, {A}} = 0.4 \, \Omega \] Now calculate the power: \[ P = 5^2 \times 0.4 = 25 \times 0.4 = 10 \, {W} \] Now, using the heat transfer formula for the cylindrical insulation: \[ Q = \frac{2 \pi k L (T_{{interface}} - T_{{ambient}})}{\ln\left(\frac{r_{{outer}}}{r_{{inner}}}\right)} \] Where:
\( k = 0.1 \, {W/(m·K)} \) is the thermal conductivity of the insulation,
\( L = 5 \, {m} \) is the length of the wire,
\( r_{{outer}} = 3 \, {mm} = 0.003 \, {m} \),
\( r_{{inner}} = 1 \, {mm} = 0.001 \, {m} \),
\( T_{{ambient}} = 30^\circ C \).
Substitute the values and solve for \( T_{{interface}} \), we get: \[ T_{{interface}} = 14.8^\circ C \]