A copper wire of radius 0.5 mm is insulated uniformly with plastic (k = 0.5 W/m K) sheathing 1mm thick. The wire is exposed to atmosphere at $30^0 C$ and the outside surface coefficient is $8 W/m^2 K$. Find the maximum safe current carried by the wire so that no part of the insulated plastic is above $75^0 C$. Also calculate critical thickness of insulation.

For copper: Thermal conductivity = 400 W/m K, specific electrical resistance = $2×10^{-8} ohm-m$.

Given,

$r1 = 0.5mm = 0.5 × 10^{-3}m$ (Radius of wire)

$r2 = 1.5mm = 1.5 × 10^{-3}m$ (Outer radius of insulation)

$Kins = 0.5W/mK$ (Thermal conductivity of insulation material)

$t3 = 30℃ = 303K$ (Atmospheric temperature)

$h = 8W/m2K$ (Convective heat transfer coefficient for surrounding atmosphere)

$t1 = 75℃ = 348K$ (Temperature of inside surface of insulation)

$K = 400W/mK$ (Thermal conductivity of wire material)

$ρ = 2 × 10^{-8}Ωm$ (Specific electrical resistance)

$L = 1m$ (Length of wire)(Assuming)

Find:

(1) I (Safe current carried by wire)

(2) rc (Critical thickness of insulation)

Solution:

Calculation for thermal resistance

Thermal resistance due to insulation

R1 = ln⁡(r2r1)2πKinsL = ln⁡(1.50.5)2π × 0.5 × 1

R1 = 0.349 K/W

Thermal resistance due to convective layer

R2 = 1h.As = 18 × 2π × 1.5 × 10-3 × 1

R2 = 13.262 K/W

Total thermal resistance

R = R1 + R2 = 0.349 + 13.262

R = 13.611 K/W

Heat transfer rate

Q = dtR = (t1 - t3)R = (348 - 303)13.611

Q = 3.306 W

Condition for safe current to flow through wire

Q = I2Re = I2.ρ.LAc

3.306 = I2 × 2 × 10-8 × 1π4 × (0.5 × 10-3)2

I = 5.697 A

Critical radius of insulation

rc = Kinsh = 0.58

rc = 0.0625m = 62.5mm

∴Critical thickness = 62.5 - 0.5 = 62mm