Question

A 3 mm diameter and 5 m long electric wire is tightly wrapped with a 2 mm thick plastic cover whose thermal conductivity is k= 0.15 W/m°C. Electrical measurements indicates that a current of 10A and voltage drop of 8V along the wire. If the insulated wire is exposed to a medium at T∞= 30C with h= 12 W/m2°C, determine the temperature at the interface of the wire and the plastic cover in steady operation. Also, determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature.

Answer 1

According to the problem the diameter of the wire , r1

The outer diameter after wrapping , r2 = 7mm

The thickness of the plastic cover , t

Length of the wire = I

The Thermal conductivity of cover = = k

Current pass through the wire = I

Due to current passing voltage dropped, is V

The temperature = T

Heat transfer coefficient = h

The rate at which the heat is generated in the wire,

q= VI = 8 * 10 =80 Watt

Now for calculating the resistance for each medium .

x1 = 1/hA2 = 1/2 x 2π x 0.0035 x 5 = 0.76 K/ W

x2 = ln r2/r1/2πkl = ln 0.0035/0.0015 /2π 0.15 x 5 = 0.179 K/W

Therefore the total resistance ,

R = x1+x2 = 0.76 +0.179 = 0.939 K/W

Now to calculate the temperature at the interface of the wire,

q= R(T - t)

=> 80 = 0.939(T -30- 273)

T = 378.12 K

Critical radius is r(critical)

r(critical) = K/h = 0.15/12 = 0.0125 m or 12.5 mm