1. Problem Analysis

The question asks how the rate of heat loss ($Q$) per unit length from an electrically heated copper wire changes with the thickness ($t$) of its insulation sleeve. We need to consider the combined effects of heat generation, conduction through the insulation, and heat transfer from the outer surface to the surroundings.

2. Thermal Resistance Model

In steady state, the heat generated within the wire must be dissipated. The rate of heat loss ($Q$) depends on the temperature difference between the wire surface ($T_w$) and the ambient ($T_{\infty}$), and the total thermal resistance ($R_{total}$) per unit length.

$ Q = \frac{T_w - T_{\infty}}{R_{total}(t)} $

The total thermal resistance is the sum of the conductive resistance of the insulation ($R_{cond}$) and the convective/radiative resistance from the outer surface ($R_{conv+rad}$):

• Conductive Resistance ($R_{cond}$): For a cylindrical sleeve of inner radius $r_0$ (wire radius) and outer radius $r_1 = r_0 + t$, the resistance is: $ R_{cond}(t) = \frac{\ln((r_0+t)/r_0)}{2\pi k} $ where $k$ is the thermal conductivity of the insulation. $R_{cond}$ increases as $t$ increases.
• Convective/Radiative Resistance ($R_{conv+rad}$): This depends on the effective heat transfer coefficient ($h_{eff}$) from the outer surface area ($A = 2\pi r_1$): $ R_{conv+rad}(t) = \frac{1}{h_{eff} A} = \frac{1}{h_{eff} 2\pi (r_0+t)} $ $R_{conv+rad}$ decreases as $t$ increases (due to increasing surface area and decreasing outer surface temperature relative to wire temperature).

Therefore, the total resistance is:

$ R_{total}(t) = \frac{\ln((r_0+t)/r_0)}{2\pi k} + \frac{1}{h_{eff} 2\pi (r_0+t)} $

3. Behavior of Total Resistance

The function $R_{total}(t)$ typically has a minimum value at a certain thickness, often called the critical thickness for insulation (related to heat transfer optimization). Let's analyze the derivative of $R_{total}$ with respect to $t$ (or $r_1 = r_0+t$):

$ \frac{dR_{total}}{dr_1} = \frac{1}{2\pi k r_1} - \frac{1}{h_{eff} 2\pi r_1^2} $

Setting the derivative to zero to find the minimum:

$ \frac{1}{k r_1} = \frac{1}{h_{eff} r_1^2} \implies r_1 = \frac{k}{h_{eff}} $

This implies a minimum resistance occurs at $r_1 = r_0 + t_{min} = k/h_{eff}$, or $t_{min} = k/h_{eff} - r_0$.

• For $t < t_{min}$ (or $r_1 < k/h_{eff}$), $R_{total}(t)$ decreases as $t$ increases.
• For $t > t_{min}$ (or $r_1 > k/h_{eff}$), $R_{total}(t)$ increases as $t$ increases.

4. Heat Loss Rate ($Q$) Dependence on Thickness

Assuming the wire surface temperature $T_w$ is relatively constant (a common simplification in such problems, focusing on the resistance aspect), the heat loss rate $Q$ is inversely proportional to the total thermal resistance:

$ Q(t) \propto \frac{1}{R_{total}(t)} $

Based on the behavior of $R_{total}(t)$:

• Initially, as $t$ increases ($t < t_{min}$), $R_{total}(t)$ decreases. This causes $Q$ to increase.
• Subsequently, as $t$ increases further ($t > t_{min}$), $R_{total}(t)$ increases. This causes $Q$ to decrease.

5. Conclusion

The rate of heat loss ($Q$) first increases with an increase in insulation thickness ($t$) up to a certain point ($t_{min}$) and then decreases with further increases in thickness.