TITLE.PM5

HEAT TRANSFER 803

dharm

\M-therm\Th15-2.pm5

Example 15.10. Hot air at a temperature of 65°C is flowing through a steel pipe of 120 mm
diameter. The pipe is covered with two layers of different insulating materials of thickness 60 mm
and 40 mm, and their corresponding thermal conductivities are 0.24 and 0.4 W/m°C. The inside
and outside heat transfer coefficients are 60 and 12 W/m°C. The atmosphere is at 20°C. Find the
rate of heat loss from 60 m length of pipe.
Solution. Refer Fig. 15.20.

Hot air

thf hhf

Steel pipe

hcf

A Insulation layers

B

Atmospheric air t = 20 Ccf º

r 1

r 2

r 3

60

mm

40

mm

Fig. 15.20

Given : r 1 =^120

2

= 60 mm = 0.06 m

r 2 = 60 + 60 = 120 mm = 0.12 m

r 3 = 60 + 60 + 40 = 160 mm = 0.16 m

kA = 0.24 W/m°C ; kB = 0.4 W/m°C

hhf = 60 W/m^2 °C ; hcf = 12 W/m^2 °C

thf = 65°C ; tcf = 20°C

Length of pipe, L = 60 m

Rate of heat loss, Q :

Rate of heat loss is given by

Q =

2

11

1

21 3 2

3

πLt t

hr

rr

k

rr

khr

hf cf

hf ABcf

()

.

ln ( / ) ln ( / )

.

−

++ +

L

N

M

M

O

Q

P

P

[Eqn. (15.34)]

=^2606520

1

60 0 06

012 006

024

016 012

04

1

12 0 16

π× −

×

+++

×

L

N

M

O

Q

P

()

.

ln (. /. )

.

ln (. /. )

..