TITLE.PM5

HEAT TRANSFER 799

dharm

\M-therm\Th15-1.pm5

The rate of heat transfer per unit area of the furnace wall,

q = hhf (thf – t 1 )

= 25(1250 – 1100) = 3750 W/m^2

Also, q = ()

()

∆toverall

Rthtotal

or q =

()

()conv ()conv

tt

RRRRR

hf cf

th hf th A th B th C th cf

−

− −−−−+++ −

or 3750 =

()1250 25

11

−

++++

h

L

k

L

k

L

hf kh

A

A

B

B

C

Ccf

or 3750 =

1225

1

25

025

165

01 015

92

1

12

++++.

.

..

kB.

=^1225

0 04 01515 01 0 0163 0 0833

1225

..+++ +. .. .0 2911^01.

=

+

kkBB

or 3750 0 289

F. + 01.

HG

I

kBKJ

= 1225 or

01 1225

3750

.

kB

= – 0.2911 = 0.0355

∴ kB = k =

01

0 0355

.

.

= 2.817 W/m^2 °C. (Ans.)

(ii)The overall transfer coefficient, U :

The overall heat transfer coefficient, U =

1

()Rthtotal

Now, (Rth)total =

1

25

1

12

++ ++0.25

1.65

0.1

2.817

0.15

9.2

= 0.04 + 0.1515 + 0.0355 + 0.0163 + 0.0833 = 0.3266°C m^2 /W

∴ U =

1

()Rthtotal =

1

0 3266.

= 3.06 W/m^2 °C. (Ans.)

(iii)All surface temperature ; t 1 , t 2 , t 3 , t 4 :

q = qA = qB = qC

or 3750 =

(–)

/

()

/

()

/

tt

Lk

tt

Lk

tt

AA BB LkCC

(^12) = 23 − = 34 −
or 3750 =
()
./.
1110
025 165
−t (^2) or t
2 = 1100 – 3750 ×
025
165
.
.
= 531.8°C
Similarly, 3750 =
(. )
./.
531 8
01 2817
−t (^3) or t
3 = 531.8 – 3750 ×
01
2817
.
.
= 398.6°C
and 3750 =
(. )
(. / .)
398 6
015 92
−t (^4) or t
4 = 398.6 – 3750 ×
05
92
.
.
= 337.5°C
[Check using outside convection, q =
(. )
/
(. )
/
337 5 25
1
337 5 25
112
− = −
hcf = 3750 W/m
(^2) ]

15.2.7. Heat conduction through hollow and composite cylinders

15.2.7.1. Heat conduction through a hollow cylinder
Refer Fig. 15.17. Consider a hollow cylinder made of material having constant thermal
conductivity and insulated at both ends.