Question

An electric cable of Aluminium conductor (k = 230 W/mK) is to be insulated with rubber (k = 0.18 W/mK). The cable is to be located in the air (h = 6 W/m²K). The critical thickness of insulation will be
1. 40 mm
2. 10 mm
3. 30 mm
4. 50 mm

Answer 1

Correct Answer - Option 2 : 10 mm

Concept:

Critical radius:

The insulation radius at which resistance to heat flow is minimum and consequently heat flow rate is maximum is called “critical radius”.

The critical radius of insulation for a cylindrical body:

\({r_{cr,cylinder}} = \frac{k_{ins}}{h}\)

The critical radius of insulation for a spherical shell:

\({r_{cr,sphere}} = \frac{{2k_{ins}}}{h}\)

Critical thickness:

The thickness up to which heat flow increases and after which heat flow decreases is termed as Critical thickness.

(r_cr)_thickness = r_cr - r.

Calculation:

Given:

K_ins = 0.18 W/mk, h = 6 W/m²K

\({r_{cr}} = \frac{k}{h} = \frac{{0.18}}{6} \) = 0.03 m = 30 mm

The critical radius is 30 mm.

NOTE: Critical thickness is always less than the critical radius.

Out of the given option, only 10 mm is less than the critical radius therefore most probable answer is option 2.