TITLE.PM5

HEAT TRANSFER 795

dharm

\M-therm\Th15-1.pm5

or 400 =

1110

L

k

L

k

L

k

L

k

A

A

B

B

C

C

D

D

+++

or 400 = 02 1110

152

1000

0138

0 006

45

01

0138

.

.

(/ )

.

..

.

+++x

=

1110

01316 0 0072 0 00013 0 7246

1110

..... .+++0 8563 0 0072

=

xx+

or 0.8563 + 0.0072 x =^1110

400

= 2.775

or x =

2 775 0 8563

0 0072

..

.

−

= 266.5 mm. (Ans.)

(ii)Temperature of the outer surface of the steel plate tso :

q = 400 = ()

/

t

Lk

so

DD

− 40

or 400 = ()

(./. )

tso− 40

010138

= 1.38(tso – 40)

or tso =^400
138.

• 40 = 329.8°C. (Ans.)
  Example 15.5. Find the heat flow rate
  through the composite wall as shown in
  Fig. 15.12. Assume one dimensional flow.
  kA = 150 W/m°C,
  kB = 30 W/m°C,
  kC = 65 W/m°C and
  kD = 50 W/m°C.
  (M.U. Winter, 1997)
  Solution. The thermal circuit for heat
  flow in the given composite system (shown in
  Fig. 15.12) has been illustrated in Fig. 15.13.
  Thickness :
  LA = 3 cm = 0.03 m ; LB = LC = 8 cm = 0.08 m ; LD = 5 cm = 0.05 m
  Areas :
  AA = 0.1 × 0.1 = 0.01 m^2 ; AB = 0.1 × 0.03 = 0.003 m^2
  AC = 0.1 × 0.07 = 0.007 m^2 ; AD = 0.1 × 0.1 = 0.01 m^2
  Heat flow rate, Q :
  The thermal resistances are given by

Rth–A =

L

kA

A

AA

=

×

003

150 0 01

.

. = 0.02

Rth–B = L

kA

B

BB

=

×

008

30 0 003

.

.

= 0.89

A

B D

400 Cº C

60 Cº

10 cm

3 cm

5 cm

8 cm

3 cm

7 cm

Fig. 15.12