# Step-by-Step Solution Let's break down the problem: - **Wire diameter**: \( d_{\text{wire}} = 2\,\text{mm} = 0.002\,\text{m} \) - **Insulation thickness**: \( t_{\text{ins}} = 2\,\text{mm} = 0.002\,\text{m} \) - **Thermal conductivity of insulation**: \( k_{\text{ins}} = 0.13\,\text{W/m-K} \) - **Thermal contact resistance**: \( R_c = 6 \times 10^{-4}\,\text{m}^2\text{-K/W} \) - **Ambient temperature**: \( T_\infty = 25^\circ\text{C} \) - **Convective heat transfer coefficient**: \( h = 7\,\text{W/m}^2\text{-K} \) - **Max insulation surface temp**: \( T_{\text{ins}} \leq 55^\circ\text{C} \) - **Find**: 1. **Maximum power dissipated per meter** (\( \dot{Q}_{\text{max}} \), W/m) 2. **Corresponding wire surface temperature** (\( T_{\text{wire}} \), in \( ^\circ\text{C} \)) --- ## 1. **Thermal Resistances** ### a. **Geometry** - **Wire radius**: \( r_1 = d_{\text{wire}}/2 = 0.001\,\text{m} \) - **Outer insulation radius**: \( r_2 = r_1 + t_{\text{ins}} = 0.001 + 0.002 = 0.003\,\text{m} \) ### b. **Thermal Resistance Network** Heat escapes from the wire core through: 1. **Contact resistance** (\( R_c \)) at the interface (per unit length) 2. **Cylindrical insulation** (\( R_{\text{ins}} \)) 3. **Convection** (\( R_{\text{conv}} \)) to air #### i. **Thermal Contact Resistance** (per length) \[ R_{c,\,\text{per}\,m} = \frac{R_c}{2\pi r_1} \] #### ii. **Insulation (Cylinder) Resistance** (per length) \[ R_{\text{ins}} = \frac{\ln(r_2/r_1)}{2\pi k_{\text{ins}} L} \] For \( L = 1\,\text{m} \). #### iii. **Convection Resistance** (per length) \[ R_{\text{conv}} = \frac{1}{2\pi r_2 h L} \] For \( L = 1\,\text{m} \). --- ## 2. **Calculate Each Resistance** ### a. **Contact Resistance per meter** \[ R_{c,\,\text{per}\,m} = \frac{6 \times 10^{-4}}{2\pi \times 0.001} \ \text{K/W} = \frac{6 \times 10^{-4}}{0.006283} \ \text{K/W} \approx 0.0955\,\text{K/W} \] ### b. **Insulation Resistance** \[ R_{\text{ins}} = \frac{\ln(0.003/0.001)}{2\pi \times 0.13 \times 1} \] \[ \ln(3) \approx 1.0986 \] \[ R_{\text{ins}} = \frac{1.0986}{2\pi \times 0.13} = \frac{1.0986}{0.8168} \approx 1.345\,\text{K/W} \] ### c. **Convection Resistance** \[ R_{\text{conv}} = \frac{1}{2\pi \times 0.003 \times 7 \times 1} = \frac{1}{0.131946} \approx 7.577\,\text{K/W} \] --- ## 3. **Total Thermal Resistance** \[ R_{\text{total}} = R_{c} + R_{\text{ins}} + R_{\text{conv}} = 0.0955 + 1.345 + 7.577 = 9.0175\,\text{K/W} \] --- ## 4. **Maximum Power Dissipation (\( \dot{Q}_{\text{max}} \))** The temperature drop from the insulation surface to the air is: \[ T_{\text{ins}} - T_\infty = \dot{Q} \cdot R_{\text{conv}} \] But to find **maximum power**, use **total resistance from insulation surface (max allowed at \( T_{\text{ins}} \)) to air**: \[ \dot{Q}_{\text{max}} = \frac{T_{\text{ins,\,max}} - T_\infty}{R_{\text{conv}}} = \frac{55 - 25}{7.577} = \frac{30}{7.577} \approx 3.96\,\text{W/m} \] But this is **heat transfer from the insulation surface to air**. The same heat must come from the wire, and all resistances matter. Instead, the correct approach is: \[ \dot{Q}_{\text{max}} = \frac{T_{\text{ins,\,max}} - T_\infty}{R_{\text{conv}}} \] But, as the insulation surface cannot exceed \( 55^\circ\text{C} \), and the wire will be hotter due to the other resistances. But power generated by the wire is the same as what exits at the insulation surface, so: \[ \dot{Q}_{\text{max}} = \frac{T_{\text{ins,\,max}} - T_\infty}{R_{\text{conv}}} = 3.96\,\text{W/m} \] --- ## 5. **Wire Surface Temperature** The temperature at the wire surface: \[ T_{\text{wire}} = T_{\text{ins}} + \dot{Q}_{\text{max}} \times (R_{\text{ins}} + R_c) \] \[ = 55 + 3.96 \times (1.345 + 0.0955) \] \[ = 55 + 3.96 \times 1.4405 \] \[ = 55 + 5.705 \] \[ = 60.7^\circ\text{C} \] --- ## **Final Answers** ### 1. **Maximum Electric Power Dissipated per Unit Length** \[ \boxed{3.96\,\text{W/m}} \] ### 2. **Corresponding Surface Temperature of the Wire** \[ \boxed{60.7^\circ\text{C}} \] --- ## **Summary Table** | Quantity | Value | Units | |-----------------------------|--------------|------------| | Maximum power per length | 3.96 | W/m | | Wire surface temperature | 60.7 | °C | --- ### **Illustration of Heat Transfer Path** ``` [Wire Surface] -- R_c -- [Inner Insulation Surface] -- R_ins -- [Outer Insulation Surface (max 55°C)] -- R_conv -- [Air (25°C)] ``` --- **Let me know if you want a diagram or need clarification on any step!**