802 ENGINEERING THERMODYNAMICS

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\M-therm\Th15-2.pm5

∴ Q =

2

11 11

1

21 3 2

3

πLt t

hr k

rr

k

rr

hr

hf cf

hf ABcf

()

.

ln ( / ) ln ( / )

.

−

++ +

L

N

M

M

M

M

O

Q

P

P

P

P

∴ Q =

2

11

1

21 32

3

πLt t

hr

rr

k

rr

khr

hf cf

hf ABcf

()

.

ln ( / ) ln( / )

/

−

+++

L

N

M

M

O

Q

P

P

...(15.34)

If there are ‘n’ concentric cylinders, then

Q =

2

11 1

(^11)
1
1
πLt t
hr k
rr
hr
hf cf
hf n n
nn
nn
cf n
()
.
ln { / }
() .()
−
++
L
N
M
M
O
Q
P
= P

• 
  

  ∑ +
  ...(15.35)
  If inside the outside heat transfer coefficients are not considered then the above equation
  can be written as
  Q =
  2
  1
  11
  1
  1
  πLt t
  k
  rr
  n
  n n
  nn
  nn
  []
  ln [ / ]
  ()
  ()
  − +

  
  

  =
  ∑ +
  ...(15.36)
  Example 15.9. A thick walled tube of stainless steel with 20 mm inner diameter and
  40 mm outer diameter is covered with a 30
  mm layer of asbestos insulation (k = 0.2 W/
  m°C). If the inside wall temperature of the pipe
  is maintained at 600°C and the outside
  insulation at 1000°C, calculate the heat loss
  per metre of length. (AMIE Summer, 2000)
  Solution. Refer Fig. 15.19,
  Given, r 1 =
  20
  2 = 10 mm = 0.01 m
  r 2 =^40
  2
  = 20 mm = 0.02 m
  r 3 = 20 + 30 = 50 mm = 0.05 m
  t 1 = 600°C, t 3 = 1000°C, kB = 0.2 W/m°C
  Heat transfer per metre of length,
  Q/L :
  Q =
  (^213)
  21 32
  πLt t
  rr
  k
  rr
  ABk
  ()
  ln ( / ) ln ( / )
  −

  
  


• 
  

  Since the thermal conductivity of satinless steel is not given therefore, neglecting the resist-
  ance offered by stainless steel to heat transfer across the tube, we have
  Q
  L
  tt
  rr
  kB
  =^2 − =2 600 1000005 002−
  02
  13
  32
  ππ()
  ln ( / )
  ()
  ln (. /. )

  
  

. = – 548.57 W/m. (Ans.)
Negative sign indicates that the heat transfer takes place radially inward.

r 1

r 2

r 3

t = 600 C 1 º A B

t 1 t^2 t = 1000 C 3 º

Asbestos

Stainless steel

Fig. 15.19