Question

Consider a bare long copper wire of 1 mm diameter. Its surface temperature is $T_s$ and the ambient temperature is $T_a$ ($T_s > T_a$). The wire is to be coated with a 2 mm thick insulation. The convective heat transfer coefficient is 20 W m$^{-2}$ K$^{-1}$. Assume that $T_s$ and $T_a$ remain unchanged. To reduce heat loss from the wire, the maximum allowed thermal conductivity of the insulating material, in W m$^{-1}$ K$^{-1}$, rounded off to two decimal places, is

• A. 0.02
• B. 0.04
• C. 0.10
• D. 0.01

Hint:

For cylindrical systems, insulation may increase heat loss unless the thickness exceeds the critical radius $r_c = k/h$.

Answer 1

The Correct Option is A

Step 1: Understand the Heat Transfer Problem

Heat flows from the wire surface through the insulation to the ambient air. For a cylindrical geometry with insulation, we have two resistances in series:

1. Conduction through the insulation (radial)
2. Convection from the outer surface to ambient

Step 2: Set Up Heat Transfer Equations

For a long cylindrical wire, the heat transfer rate per unit length is:

$$q' = \frac{T_s - T_a}{R_{total}}$$

where the total thermal resistance per unit length is:

$$R_{total} = R_{cond} + R_{conv}$$

Conduction resistance through insulation (per unit length):

$$R_{cond} = \frac{\ln(r_2/r_1)}{2\pi k}$$

where:

• $r_1 = d_1/2 = 0.0005$ m (inner radius)
• $r_2 = d_2/2 = 0.0025$ m (outer radius)
• $k$ = thermal conductivity of insulation

Convection resistance (per unit length):

$$R_{conv} = \frac{1}{2\pi r_2 h} = \frac{1}{2\pi (0.0025)(20)}$$

$$R_{conv} = \frac{1}{0.3142} = 3.183 \text{ m}\cdot\text{K/W}$$

Step 3: Determine the Critical Condition

To reduce heat loss from the wire, we want to maximize the total resistance. However, there's a critical radius concept: adding insulation can actually increase heat loss if the outer radius is less than the critical radius.

The critical radius for a cylinder is:

$$r_{critical} = \frac{k}{h}$$

For heat loss to be reduced (insulation to be effective), we need:

$$r_2 \geq r_{critical}$$

$$0.0025 \geq \frac{k}{20}$$

$$k \leq 0.0025 \times 20 = 0.05 \text{ W/(m}\cdot\text{K)}$$

But this gives us the maximum value where insulation starts becoming counterproductive. To minimize heat loss, we want the insulation to provide significant resistance.

Step 4: Alternative Approach - Heat Loss Comparison

Without insulation, heat loss per unit length:

$$q'_{bare} = 2\pi r_1 h (T_s - T_a) = 2\pi (0.0005)(20)(T_s - T_a)$$

$$q'_{bare} = 0.0628(T_s - T_a)$$

With insulation, heat loss per unit length:

$$q'_{insulated} = \frac{T_s - T_a}{\frac{\ln(r_2/r_1)}{2\pi k} + \frac{1}{2\pi r_2 h}}$$

For the insulation to reduce heat loss: $q'{insulated} < q'{bare}$

$$\frac{1}{\frac{\ln(r_2/r_1)}{2\pi k} + \frac{1}{2\pi r_2 h}} < 2\pi r_1 h$$

$$\frac{1}{\frac{\ln(5)}{2\pi k} + \frac{1}{2\pi (0.0025)(20)}} < 0.0628$$

$$\frac{\ln(5)}{2\pi k} + 3.183 > \frac{1}{0.0628} = 15.92$$

$$\frac{\ln(5)}{2\pi k} > 15.92 - 3.183 = 12.74$$

$$\frac{1.609}{2\pi k} > 12.74$$

$$k < \frac{1.609}{2\pi \times 12.74} = \frac{1.609}{80.04} = 0.0201$$

$$k < 0.02 \text{ W/(m}\cdot\text{K)}$$

Step 5: Round to Two Decimal Places

The maximum allowed thermal conductivity is:

$$k_{max} = 0.02 \text{ W/(m}\cdot\text{K)}$$

Answer

The maximum allowed thermal conductivity of the insulating material is 0.02 W/(m·K).