TITLE.PM5

804 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th15-2.pm5

=

16964 6

0 2777 2 8881 0 7192 0 5208

.
....++ + = 3850.5 W
i.e., Rate of heat loss = 3850.5 W (Ans.)
Example 15.11. A 150 mm steam pipe has inside dimater of 120 mm and outside diam-
eter of 160 mm. It is insulated at the outside with asbestos. The steam temperature is 150°C and
the air temperature is 20°C. h (steam side) = 100 W/m^2 °C, h (air side) = 30 W/m^2 °C, k (asbestos)
= 0.8 W/m°C and k (steel) = 42 W/m°C. How thick should the asbestos be provided in order to
limit the heat losses to 2.1 kW/m^2? (N.U.)
Solution. Refer Fig. 15.21.

t = 150 Chf º

hhf

Cold fluid film

Hot fluid film

Steam pipe (A)

Insulation (B)

(Asbestos)

hcf

t = 20 Ccf º

r 1

r 2

r 3

Fig. 15.21

Given : r 1 =

120

2

= 60 mm = 0.06 m

r 2 =^160

2

= 80 mm = 0.08 m

kA = 42 W/m°C ; kB = 0.8 W/m°C

thf = 150°C ; tcf = 20°C

hhf = 100 W/m^2 °C ; hcf = 30 W/m^2 °C

Heat loss = 2.1 kW/m^2

Thickness of insulation (asbestos), (r 3 – r 2 ) :

Area for heat transfer = 2π r L (where L = length of the pipe)

∴ Heat loss = 2.1 × 2π r L kW

= 2.1 × 2π × 0.075 × L = 0.989 L kW

= 0.989 L × 10^3 watts

HFGwhere , mean radiusr ==^150 mm or 0.075 m ... GivenIKJ

2

75

Heat transfer rate in such a case 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)]