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: Note the fixed driving temperature difference.
Both cases share the same $\Delta T = 420 - 320 = 100\ K$, so the ratio of heat loss is just the ratio of the two conductances.

Step 2: Find the bare wire conductance.
Bare wire radius $r_1 = 0.001\ m$. Conductance per metre is $UA_1 = h \times 2\pi r_1 = 25 \times 2 \times 3.14 \times 0.001 = 0.157\ W/(m.K)$.

Step 3: Find the critical radius of insulation, for context.
$r_{crit} = k/h = 0.5/25 = 0.02\ m = 20\ mm$. Since the insulated outer radius $r_2 = 3\ mm$ is far below $r_{crit}$, adding insulation should raise heat loss, not lower it.

Step 4: Combine conduction and convection conductances in series.
$1/UA_2 = \dfrac{\ln(r_2/r_1)}{2\pi k} + \dfrac{1}{h \times 2\pi r_2} = 0.350 + 2.123 = 2.473\ m.K/W$. So $UA_2 = 1/2.473 = 0.4044\ W/(m.K)$.

Step 5: Take the conductance ratio.
$\dfrac{UA_2}{UA_1} = \dfrac{0.4044}{0.157} = 2.58$.

Final Answer:
Since $r_2$ sits well inside the critical radius, insulation increases heat loss by a factor of about 2.58, so option (A) fits. \[ \boxed{2.58} \]