HEAT TRANSFER 807

dharm

\M-therm\Th15-2.pm5

If there are n concentric spheres then the above equation can be written as follows :

Q =

4

11

1

2

1

1 1 2 1

π()

. ...

()

() ()

tt

hr

rr

krr hr

hf cf

hf

nn

n nn n

nn

cf n

−

+

R −

S

|

T|

U

V

|

W|

+

L

N

M

M

O

Q

P

P

+

= +

=

∑ +

...(15.39)

If inside and outside heat transfer coefficients are considered, then the above equation can
be written as follows :

Q =

(^411)
1
1
1
π()
..
()
()
()
tt
rr
krr
n
n
nn
nn
nn n
−
L −
N
M
M
O
Q
P
P

• 
  

  =

  
  


• 
  

  ∑ +
  ...(15.40)
  Example 15.12. A spherical shaped vessel of 1.4 m diameter is 90 mm thick. Find the rate
  of heat leakage, if the temperature difference between the inner and outer surfaces is 220°C.
  Thermal conductivity of the material of the sphere is 0.083 W/m°C.
  Solution. Refer Fig. 15.24.
  Spherical shaped
  vessel
  k r^2
  t 1 t 2
  90 mm
  r 1
  Fig. 15.24
  Given : r 2 =
  14
  2
  .
  = 0.7 m ;
  r 1 = 0.7 –
  90
  1000
  = 0.61 m ;
  t 1 – t 2 = 220°C ; k = 0.083 W/m°C
  The rate of heat transfer/leakage is given by
  Q =
  ()
  ()
  tt
  rr
  kr r
  12
  21
  (^412)
  −
  L −
  N
  M
  O
  Q
  π P
  ...[Eqn. (15.37)]

  
  

  220
  07 061
  4 0 083 0 61 0 7
  (.. )
  ...
  −
  ×××
  L
  N
  M
  O
  Q
  π P
  = 1088.67 W
  i.e., Rate of heat leakage = 1088.67 W. (Ans.)