HEAT TRANSFER 793

dharm

\M-therm\Th15-1.pm5

or

1300

1190 2 6 25

105

..LLLAAA+−

=

or^1300

2506

105

−

=

. LLAA
or 1300 LA = 105 (2 – 5.06 LA)
or 1300 LA = 210 – 531.3 LA
or LA =
210
(.)1300 5313+
= 0.1146 m or 114.6 mm. (Ans.)
∴ Thickness of insulation LB = 320 – 114.6 = 205.4 mm. (Ans.)
(ii)Heat loss per unit area, q :

Heat loss per unit area, q =

tt

LkAA

12 − = −

/

1325 1200
0.1146 /0.84
= 916.23 W/m^2. (Ans.)
If another layer of insulating material is added, the heat loss from the wall will reduce ;
consequently the temperature drop across the fire brick lining will drop and the interface tempera-
ture t 2 will rise. As the interface temperature is already fixed. Therefore, a satisfactory solution
will not be available by adding layer of insulation.
Example 15.3. An exterior wall of a house may be approximated by a 0.1 m layer of
common brick (k = 0.7 W/m°C) followed by a 0.04 m layer of gypsum plaster (k = 0.48 W/m°C).
What thickness of loosely packed rock wool insulation (k = 0.065 W/m°C) should be added to
reduce the heat loss or (gain) through the wall by 80 per cent? (AMIE Summer, 1997)
Solution. Refer Fig. 15.10.
Thickness of common brick, LA = 0.1 m
Thickness of gypsum plaster,LB = 0.04 m
Thickness of rock wool, LC = x (in m) =?
Thermal conductivities :
Common brick, kA = 0.7 W/m°C
Gypsum plaster, kB = 0.48 W/m°C
Rock wool, kC = 0.065 W/m°C
Case I. Rock wool insulation not used :

Q 1 =

At

L

k

L

k

At

A

A

B

B

() ()

.

.

.

.

∆∆

+

=

(^01) +
07
004
048
...(i)
Case II. Rock wool insulation used :
Q 2 =
At
L
k
L
k
L
k
At
A x
A
B
B
C
C
() ()
.
.
.
..
∆∆
++

(^01) ++
07
004
0 48 0 065
...(ii)
But Q 2 = (1 – 0.8)Q 1 = 0.2 Q 1 ...(Given)
∴
At
x
() At
.
.
.
..
.
()
.
.
.
.
∆∆
01
07
004
0 48 0 065
(^0201)
07
004
048
++
=×

• or^01
  07
  004
  048
  .
  .
  .
  .


• = 0.2
  01
  07
  004
  0 48 0 065
  .
  .
  .
  ..
  L ++
  NM
  O
  QP
  x
  Fig. 15.10
  A B C
  Common brick
  Gypsum plaster
  Rock wool
  LA LB LC
  = 0.1 m
  = 0.04 m = x