TITLE.PM5

794 ENGINEERING THERMODYNAMICS

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or 0.1428 + 0.0833 = 0.2 [0.1428 + 0.0833 + 15.385 x]
or 0.2261 = 0.2 (0.2261 + 15.385 x)
or x = 0.0588 m or 58.8 mm
Thus, the thickness of rock wool insulation should be 58.8 mm. (Ans.)
Example 15.4. A furnace wall consists of 200 mm layer of refractory bricks, 6 mm layer of
steel plate and a 100 mm layer of insulation bricks. The maximum temperature of the wall is
1150°C on the furnace side and the minimum temperature is 40°C on the outermost side of the
wall. An accurate energy balance over the furnace shows that the heat loss from the wall is
400 W/m^2. It is known that there is a thin layer of air between the layers of refractory bricks and
steel plate. Thermal conductivities for the three layers are 1.52, 45 and 0.138 W/m°C respec-
tively. Find :
(i)To how many millimetres of insulation brick is the air layer equivalent?
(ii)What is the temperature of the outer surface of the steel plate?
Solution. Refer Fig. 15.11.
Thickness of refractory bricks, LA = 200 mm = 0.2 m
Thickness of steel plate, LC = 6 mm = 0.006 m
Thickness of insulation bricks, LD = 100 mm = 0.1 m
Difference of temperature between the innermost and outermost side of the wall,
∆t = 1150 – 40 = 1110°C

A B C D

40 Cº

1150 Cº

Furnace

tso

LA LB LC LD

= 200 mm

= x mm

= 6 mm

= 100 mm

Refractory bricks

Air gap equivalent to x mm of

insulation bricks

Steel plate

Insulation bricks

Fig. 15.11

Thermal conductivities :

kA = 1.52 W/m°C ; kB = kD = 0.138 W/m°C ; kC = 45 W/m°C

Heat loss from the wall,q = 400 W/m^2

(i)The value of x = (LC) :

We know, Q = AtL

k

.∆

Σ

or

Q

A = q =

∆

Σ

t

L

k