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).
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:
\[
T_{{interface}} = 14.8^\circ C
\]
