Question

A 2 mm diameter electric wire is insulated with 2 mm thick plastic. The plastic has a thermal conductivity of \(0.5\ \text{W.m}^{-1}\text{.K}^{-1}\). The wire is exposed to air at 320 K and outside convective heat transfer coefficient is \(25\ \text{W.m}^{-2}\text{.K}^{-1}\). Wire surface temperature is constant at 420 K and remains unaffected by the covering. Assuming steady state heat transfer conditions, the ratio of heat loss per metre of wire length with insulation to that without insulation is nearest to (Take \(\pi = 3.14\))

• A. 2.58
• B. 0.39
• C. 1.87
• D. 0.54

Hint:

Compare convection alone to conduction through the insulation plus convection in series, then check against the critical radius k/h.

Answer 1

The Correct Option is A

Step 1: Set up the two heat loss paths.
Wire radius \( r_1 = 1\,mm = 0.001\,m \). Insulation adds 2 mm, so outer radius \( r_2 = 3\,mm = 0.003\,m \).
Surface temperature stays fixed at \( T_s = 420\,K \), and air is at \( T_\infty = 320\,K \), so \( \Delta T = 100\,K \) in both cases.

Step 2: Heat loss without insulation (pure convection).
\( Q_1 = h(2\pi r_1)\Delta T = 25 \times 2 \times 3.14 \times 0.001 \times 100 = 15.7\ W/m \).

Step 3: Heat loss with insulation (conduction plus convection in series).
Conduction resistance of the plastic layer: \( R_{cond} = \dfrac{\ln(r_2/r_1)}{2\pi k} = \dfrac{\ln 3}{2 \times 3.14 \times 0.5} = \dfrac{1.099}{3.14} = 0.350\ m.K/W \).
Convection resistance at the outer surface: \( R_{conv} = \dfrac{1}{h(2\pi r_2)} = \dfrac{1}{25 \times 2 \times 3.14 \times 0.003} = 2.123\ m.K/W \).
Total resistance \( R = 0.350 + 2.123 = 2.473\ m.K/W \), so \( Q_2 = \Delta T / R = 100/2.473 = 40.4\ W/m \).

Step 4: Take the ratio asked in the question.
\( \dfrac{Q_2}{Q_1} = \dfrac{40.4}{15.7} = 2.58 \).

Final Answer:
The insulated wire actually loses more heat per metre because the outer radius sits well below the critical radius of insulation, giving ratio 2.58, option (A). \[ \boxed{Q_{ins}/Q_{bare} \approx 2.58} \]