TITLE.PM5

796 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th15-1.pm5

Q Q

1 2 3 4

A C D

B

3 cm 8 cm 5 cm

3 cm

7 cm

Q Q

Rth–B

Rth–C

t 1 Rth–A Rth–D

t 2 t 3

t 4

= 400 Cº = 60 Cº

Fig. 15.14

Fig. 15.13. Thermal circuit for heat flow in the

composite system.

Rth–C =

L

kA

C

CC

=

×

008

65 0 007

.

. = 0.176

Rth–D = L

kA

D

DD

=

×

005

50 0 01

.
.
= 0.1
The equivalent thermal resistance for the par-
allel thermal resistances Rth–B and Rth–C is given by :
1111
089

1

()RRRth eq. th B th C. 0176.

=+=+

−−

= 6.805

∴ (Rth)eq. =

1

6 805. = 0.147

Now, the total thermal resistance is given by

(Rth)total = Rth–A + (Rth)eq. + Rth–D

= 0.02 + 0.147 + 0.1 = 0.267

∴ Q =

()

()

()

.

∆t

R

overall

th total

= 400 60−
0 267 = 1273.4 W. (Ans.)
Example 15.6. A mild steel tank of wall thickness 12 mm contains water at 95°C. The
thermal conductivity of mild steel is 50 W/m°C, and the heat transfer coefficients for the inside
and outside the tank are 2850 and 10 W/m^2 °C, respectively. If the atmospheric temperature is
15°C, calculate :
(i)The rate of heat loss per m^2 of the tank surface area ;
(ii)The temperature of the outside surface of the tank.
Solution. Refer Fig. 15.14.
Thickness of mild steel tank wall
L = 12 mm = 0.012 m
Temperature of water, thf = 95°C
Temperature of air, tcf = 15°C
Thermal conductivity of mild steel,
k = 50 W/m°C
Heat transfer coefficients :
Hot fluid (water), hhf = 2850 W/m^2 °C
Cold fluid (air), hcf = 10 W/m^2 °C
(i)Rate of heat loss per m^2 of the tank
surface area, q :
Rate of heat loss per m^2 of tank surface,
q = UA(thf – tcf)
The overall heat transfer coefficient, U is found from the relation ;
11 1 1
2850

0 012

50

1

Uh 10

L

hf khcf

=++= + +.

= 0.0003508 + 0.00024 + 0.1 = 0.1006

t 1

Water t 2

t = 95 Chf º Tank wall

Air

t = 15 Ccf º

L = 12 mm