TITLE.PM5

HEAT TRANSFER 797

dharm

\M-therm\Th15-1.pm5

∴ U =^1

01006.

= 9.94 W/m^2 °C

∴ q = 9.94 × 1 × (95 – 15) = 795.2 W/m^2. (Ans.)
(ii)Temperature of the outside surface of the tank, t 2 :
We know that, q = hcf × 1 × (t 2 – tcf)
or 795.2 = 10(t 2 – 15)

or t 2 = 795 2
10

. + 15 = 94.52°C. (Ans.)

Example 15.7. The interior of a refrigerator having inside dimensions of 0.5 m × 0.5 m
base area and 1 m height, is to be maintained at 6°C. The walls of the refrigerator are constructed
of two mild steel sheets 3 mm thick (k = 46.5 W/m°C) with 50 mm of glass wool insulation (k =
0.046 W/m°C) between them. If the average heat transfer coefficients at the inner and outer
surfaces are 11.6 W/m^2 °C and 14.5 W/m^2 °C respectively, calculate :
(i)The rate at which heat must be removed from the interior to maintain the specified
temperature in the kitchen at 25°C, and
(ii)The temperature on the outer surface of the metal sheet.
Solution. Refer Fig. 15.15
Given : LA = LC = 3 mm = 0.003 m ;
LB = 50 mm = 0.05 m ;
kA = kC = 46.5 W/m°C ; kB = 0.046 W/m°C ;
h 0 = 11.6 W/m^2 °C ; hi = 14.5 W/m^2 °C ;
t 0 = 25°C ; ti = 6°C.
The total area through which heat is coming into the refrigerator
A = 0.5 × 0.5 × 2 + 0.5 × 1 × 4 = 2.5 m^2

A B C

Mild steel

sheet

Mild steel

sheet

Glass wool

Outside surface

of refrigerator Inside surfaceof refrigerator

h 0

t = 25 0 ºC

t 1

hi

t= 6i ºC

1 2 34

LA LB LC

= 3 mm = 50 mm = 3 mm

Fig. 15.15