TITLE.PM5

HEAT TRANSFER 805

dharm

\M-therm\Th15-2.pm5

0.989 L × 10^3 =

2 150 20

1

100 0 06

008 006

42

008

08

1

30

3

3

πL

r

r

()

.

ln(. /. ) ln( /. )

.

−

×

+++

×

L

N

M

O

Q

P

0.989 × 10^3 =

816 81

0 16666 0 00685^008

08

1

30

3

3

.

..ln ( /. )

.

++ +

L

N

M

O

Q

P

r

r

or ln ( /. )

.

.

.

r

r

3

3 3

008

08

1

30

816 81

0 989 10

+=

×

• (0.16666 + 0.00685) = 0.6524

or 1.25 ln (r 3 /0.08) +

1

30 r 3

• 0.6524 = 0
  Solving by hit and trial, we get
  r 3 ~− 0.105 m or 105 mm
  ∴ Thickness of insulation = r 3 – r 2 = 105 – 80 = 25 mm. (Ans.)

15.2.8. Heat conduction through hollow and composite spheres

15.2.8.1. Heat conduction through hollow sphere
Refer Fig. 15.22. Consider a hollow
sphere made of material having constant ther-
mal conductivity.
Let r 1 , r 2 = Inner and outer radii,
t 1 , t 2 = Temperature of inner and
outer surfaces, and
k = Constant thermal conductiv-
ity of the material with the
given temperature range.
Consider a small element of thickness
dr at any radius r.
Area through which the heat is trans-
mitted, A = 4πr^2

∴ Q = – k. 4πr^2. dt
dr
Rearranging and integrating the above
equation, we obtain

Q dr

r r

r

1 2

2

z = – 4πk^ tdt

t

1

2

z

or Q

r

r

−+ r

−+

L

N

M

M

O

Q

P

P

21

21

1

2

= – 4πk t

t

L t

N

M

O

Q

P

1

2

or – Q

11

rr 21

−

F

HG

I

KJ

= – 4πk(t 2 – t 1 )

or

Qr r

rr

() 21

12

−

= 4πk (t 1 – t 2 )

or Q =

4

4

12 1 2

21

12

21

12

π

π

kr r t t

rr

tt

rr

kr r

()

() ()

()

−

−

= −

L −

N

M

O

Q

P

...(15.37)

Fig. 15.22. Steady state conduction

through a hollow sphere.

Q t 1 t 2

R=th 4k rrr– r2 1

π 12

Q

dr

Element

Hollow sphere

t 2

Q (Heat flows radially

outwards, t > t ) 12

r 2

rr (^1) t 1