HEAT TRANSFER 847

dharm

\M-therm\Th15-4.pm5

=

π

π

π

×××× −

+ ××

××

L −

NM

O

QP

0 022 1 5 67 10− 813 1373

1

062

0 022

018

1

082

1

..(^844 )

.

.

..

l

l

=

−12214.3

1.64

= – 7447.7 W/m

∴ Average rate of heat absorption during the heating process

Qav. = 7940.5 7447.7

2

+ = 7694.1 W/m

Time required for heating, th is obtained from the equation

mcp ∆T = Qav. × th

∴ th = [/ (. )π^4 0 022 1 7845]^0 (540 420)^1000

7694.1

×^2 ×× ××−×.67

= 31.16 s. (Ans.)
Example 15.28. Calculate the heat transfer rate per m^2 area by radiation between the surfaces
of two long cylinders having radii 100 mm and 50 mm respectively. The smaller cylinder being in the
larger cylinder. The axes of the cylinders are parallel to each other and separated by a distance of
20 mm. The surfaces of inner and outer cylinders are maintained at 127°C and 27°C respectively. The
emissivity of both the surfaces is 0.5.
Assume the medium between the two cylinders is non-absorbing. (P.U.)
Solution. Given : r 1 = 50 mm = 0.05 m ; r 2 = 100 mm = 0.1 m, T 1 = 127 + 273 = 400 K,
T 2 = 27 + 273 = 300 K, ε 1 = ε 2 = 0.5
The heat transfer between two concentric or eccentric cylinders is given by

()Q AT T()

F

A

A

(^12) net
114 24
1
112
2
2
1
2
111
= −
F −
HG
I
KJ
++F −
HG
I
− KJ
σ
ε
ε
ε
ε
Here F 12 − =^1 and
A
A
rL
rL
r
r
1
2
1
2
1
2
2
2

π
π
Substituting the values, we have
()
.
Q 12 net
44
167400
100
300
100
10 1 10
0
005
0

× GFH IKJ −FHG IKJ
L
N
M
M
O
Q
P
P
F −
HG
I
KJ+
F −
HG
I
KJ×

5. .5
   0.5
   .5
   .5 .1
   992.25
   2.5 = 396.9 W/m
   (^2). (Ans.)
   Example 15.29. Three thin walled infinitely long hollow cylinders of radii 5 cm, 10 cm and
   15 cm are arranged concentrically as shown in Fig. 15.54. T 1 = 1000 K and T 3 = 300 K.
   Assuming ε 1 = ε 2 = ε 3 = 0.05 and vacuum in the spaces between the cylinders, calculate the
   steady state temperature of cylinder surface 2 and heat flow per m^2 area of cylinder 1. (P.U.)
   Solution. Given : r 1 = 5 cm ; r 2 = 10 cm ; r 3 = 15 cm ; T 1 = 1000 K ; T 3 = 300 K
   ε 1 = ε 2 = ε 3 = 0.05.
   For steady state heat flow,
   Q 12 = Q 23
   or AT T
   F
   A
   A
   AT T
   F
   A
   A
   11
   4
   2
   4
   1
   112
   2
   2
   1
   2
   22
   4
   3
   4
   2
   223
   3
   3
   2
   3
   111 111
   σ
   ε
   ε
   ε
   ε
   σ
   ε
   ε
   ε
   ε
   ()− ()
   F −
   HG
   I
   KJ
   ++−
   F
   HG
   I
   KJ
   = −
   F −
   HG
   I
   KJ
   ++−
   F
   HG
   I
   − − KJ