An uninsulated cylindrical wire of radius 1.0 mm produces electric heating at the rate of 5.0 W/m. The temperature of the surface of the wire is 75 °C when placed in air at 25 °C. When the wire is coated with PVC of thickness 1.0 mm, the temperature of the surface of the wire reduces to 55 °C. Assume that the heat generation rate from the wire and the convective heat transfer coefficient are same for both uninsulated wire and the coated wire. The thermal conductivity of PVC is _____________ W/m.K (round off to two decimal places).

The correct answer is 0.11.

To find the thermal conductivity of the PVC coating, we analyze the heat transfer process from the cylindrical wire under two conditions: first, when the wire is uninsulated, and second, when it is coated with PVC insulation.

Step 1: Analyze the uninsulated wire to find the convective heat transfer coefficient (h)
Initially, the bare wire of radius r₁ generates heat at a rate per unit length of Q/L = 5.0 W/m.
The surface temperature of the wire is T_s1 = 75 °C, and it is placed in surrounding air at temperature T_∞ = 25 °C.
Under steady-state conditions, the rate of heat generation is equal to the convective heat transfer rate from the wire surface to the air:

$$\frac{Q}{L}=h\cdot A_{s1}\cdot (T_{s1}-T_{\infty })$$

Where the surface area per unit length is A_s1 = 2��·r₁.
Given:
r₁ = 1.0 mm = 1.0 × 10^-3 m
T_s1 = 75 °C
T_∞ = 25 °C
Q/L = 5.0 W/m

Substituting these values into the heat transfer equation:

$$5.0=h\cdot (2\pi \cdot 1.0\times {10}^{-3})\cdot (75-25)$$

Solving for h:

$$5.0=h\cdot (2\pi \cdot {10}^{-3})\cdot 50$$

$$5.0=h\cdot \left(0.1\pi \right)$$

$$h=\frac{5.0}{0.1\pi }=\frac{50}{\pi }\approx 15.915\text{ W/}{\text{m}}^2\text{·K}$$

Step 2: Analyze the PVC coated wire
When the wire is coated with PVC of thickness t = 1.0 mm, the new outer radius (r₂) becomes:
r₂ = r₁ + t = 1.0 mm + 1.0 mm = 2.0 mm = 2.0 × 10^-3 m.
The surface temperature of the wire reduces to T_s2 = 55 °C.
The total thermal resistance per unit length (R'_th) is the sum of the conductive thermal resistance of the PVC insulation layer and the convective thermal resistance at the outer surface of the insulation:

$$R_{\text{th}}^{\text{'}}=\frac{\ln (r_2/r_1)}{2\pi k}+\frac{1}{h\cdot \left(2\pi r_2\right)}$$

Where k is the thermal conductivity of PVC. The heat transfer rate per unit length is given by:

$$\frac{Q}{L}=\frac{T_{s2}-T_{\infty }}{R_{\text{th}}^{\text{'}}}$$

Step 3: Calculate the thermal conductivity (k) of PVC
Substituting the known values into the equation:

$$5.0=\frac{55-25}{\frac{\ln (2/1)}{2\pi k}+\frac{1}{h\cdot (2\pi \cdot 2.0\times {10}^{-3})}}$$

$$5.0=\frac{30}{\frac{\ln \left(2\right)}{2\pi k}+\frac{1}{(50/\pi )\cdot (4.0\times {10}^{-3}\pi )}}$$

We simplify the second term in the denominator:

$$\frac{1}{(50/\pi )\cdot 4.0\times {10}^{-3}\pi }=\frac{1}{50\cdot 4.0\times {10}^{-3}}=\frac{1}{0.2}=5$$

Now we substitute this back into the heat transfer equation:

$$5.0=\frac{30}{\frac{\ln \left(2\right)}{2\pi k}+5}$$

Rearranging the terms:

$$\frac{\ln \left(2\right)}{2\pi k}+5=\frac{30}{5.0}$$

$$\frac{\ln \left(2\right)}{2\pi k}+5=6$$

$$\frac{\ln \left(2\right)}{2\pi k}=1$$

Solving for k:

$$k=\frac{\ln \left(2\right)}{2\pi }$$

Using the approximation ln(2) ≈ 0.69315:

$$k\approx \frac{0.69315}{2\cdot 3.14159}\approx \frac{0.69315}{6.28318}\approx 0.1103\text{ W/m·K}$$

Rounding to two decimal places, we get 0.11 W/m·K.