TITLE.PM5

HEAT TRANSFER 811

dharm

\M-therm\Th15-2.pm5

Solid

sphere

Insulation

t hO

1

k

r 1

r– r 21

r 2

tair

Fig. 15.27
Example 15.13. A small electric heating application
uses wire of 2 mm diameter with 0.8 mm thick insulation (k
= 0.12 W/m°C). The heat transfer coefficient (ho) on the in-
sulated surface is 35 W/m^2 °C. Determine the critical thick-
ness of insulation in this case and the percentage change in
the heat transfer rate if the critical thickness is used, assum-
ing the temperature difference between the surface of the
wire and surrounding air remains unchanged.
Solution. Refer Fig. 15.28.

Given : r 1 =

2

2 = 1 mm = 0.001 m

r 2 = 1 + 0.8 = 1.8 mm = 0.0018 m

k = 0.12 W/m°C, ho = 35 W/m^2 °C

Critical thickness of insulation :

The critical radius of insulation is given by

rc =

k

ho

=^012

35

. = 3.43 × 10–3 m or 3.43 mm.

∴ Critical thickness of insulation

= rc – r 1 = 3.43 – 1 = 2.43 mm. (Ans.)

Percentage change in heat transfer rate :

Case I : The heat flow through an insulated wire is given by

Q 1 =

2

1

2

0 0018 0 001

012

1

35 0 0018

2

20 77

1

21

2

πππLt t 11

rr

khr

air Lt t Lt t

o

() air air

ln ( / )

.

()

ln (. /. )

..

()

.

−

+

=

−

+

×

=

−

...(i)

Fig. 15.28

tair

Wire

Insulator

hO

1 2

t 1

k

0.8

r (^1) mm
r 2