TITLE.PM5

798 ENGINEERING THERMODYNAMICS

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\M-therm\Th15-1.pm5

(i)The rate of removal of heat, Q :

Q =

At t

h

L

k

L

k

L

kh

i

o

A

A

B

B

C

Ci

() 0

11

−

++++

=

25 25 6

1

11 6

0 003

46 5

005

0 046

0 003

46 5

1

14 5

.( )

.

.

.

.

.

.

..

−

++++

= 38.2 W. (Ans.)

(ii)The temperature at the outer surface of the metal sheet, t 1 :

Q = ho A(25 – t 1 )

or 38.2 = 11.6 × 2.5 (25 – t 1 )

or t 1 = 25 –

38 2

11 6 2 5

.
..×
= 23.68°C. (Ans.)
Example 15.8. A furnace wall is made up of three layers of thicknesses 250 mm, 100 mm
and 150 mm with thermal conductivities of 1.65, k and 9.2 W/m°C respectively. The inside is
exposed to gases at 1250°C with a convection coefficient of 25 W/m^2 °C and the inside surface is
at 1100°C, the outside surface is exposed air at 25°C with convection coefficient of 12 W/m^2 °C.
Determine :
(i)The unknown thermal conductivity ‘k’ ;
(ii)The overall heat transfer coefficient ;
(iii)All surface temperatures.
Solution. LA = 250 mm 0.25 m ; LB = 100 mm = 0.1 m ;
LC = 150 mm = 0.15 m ; kA = 1.65 W/m°C ;
kC = 9.2 W/m°C ; thf = 1250°C ; t 1 = 1100°C
hhf = 25 W/m^2 °C ; hcf = 12 W/m^2 °C
(i)Thermal conductivity, k (= kB) :

Gases Air

t = 1250 Chf º A B C

h = 25 W/m Chf^2 º

t = 25 Ccf º

t 4

t 3

t 2

t = 1100 C 1 º

1 2 3 4

LA LB LC

= 250 mm

= 100 mm

150 mm

(a) Composite system.

thf t 1 t 2 t 3 t 4 tcf

1250 Cº 1100 Cº Rth–A Rth–B Rth–C 25 Cº

(R )thconv.-hf (R )thconv.-cf

(b) Thermal circuit.

Fig. 15.16