Page 1003 - Industrial Power Engineering and Applications Handbook

311948 Industrial Power Engineering and Applications Handbook
36.9 where
x
... w, = - 0.85 x 1411.49 [3784 - 36a4] A, = 2223.12 square inches
1012
= 91.92 Wlft p = 1 (near sea level)
2 By convection e=95-48
(i) From the outer surface = 47°C

Po-5 ' I9=5 h = 59 inches
W, = 0.0022 A, ho,25 Wlft
W, = 0.0022 x 2223.12 x -
1
X47l.25
A,, = 1411.49 square inches 59.25
p = 1 (near sea level) = 217.17 Wlft
e= i05-95=io'c :. Total heat dissipated
h = 37.46 inches = 411.76 + 217.17
101.25 = 628.93 Wtft
.'. W, = 0.0022 x 141 1.49 x ~
37.46°.25 compared to the total heat generated of 798.87 Wlft.
= 22.32 Wlft
Notes
(ii) From the inner surface'
If we assume the emissivity factor to be 0.85 the shortfall
Asi = 1379.09 square inches in the heat dissipation is nearly made up. However, it
would be a matter of laboratory testing to establish the
h = 37.46 - 2 x 0.43 = 36.6 inches suitability of the enclosure section chosen. Otherwise the
101.25
W, = 0.0022 x 1379.09 x - thickness of the enclosure may be slightly increased and
the calculations repeated.
36.6°.25 Add a suitable margin in the conductor's cross-section to
= 21.93 W/ft account for the opening considered to dissipate the inside
heat. Similarly, add a suitable margin in the enclosure's
I.. Total heat dissipated cross-section to account for the inspection windows.
In fact, it is the solar effect that is causing the maximum
= 91.92 + 22.32 + 21.93 heat. The factors considered for the solar effect are also
= 136.17 Wlft highly conservative. Nevertheless, a canopy over the
outdoor part is advisable in the above case. This will
as against the total heat generated of 149.22 wattlft. The ensure the same size of enclosure for the outdoor as well
conductor chosen appears to be very close to the required as the indoor parts and also eliminate the requirement for
size. Its diameter or thickness may be slightly modified to a thicker enclosure or a forced cooling arrangement. Now
achieve thermal equilibrium and a more appropriate size. there will be no direct solar radiation over the bus system
and the total solar effect can be eliminated, except for
3 By enclosure substituting the indoor ambient temperature of 48°C with
(a) By radiation the maximum outdoor temperature for the outdoor part of
the bus system.
36.9 If we can overcome the solar effect, the size of the conductor
W, =-.e.AS[T,4-T:]W/ft
10'2 and the enclosure can be reduced to economize their
where costs. Another exercise with reduced dimensions will be
necessary for this until the most economical sections are
e = 0.65 (This is on a conservative side. For dull, non- established.
magnetic matt finish painted surfaces, which such Once the required sizes of the conductor and the enclosure
enclosures generally are, some manufacturers are established, one can choose the nearest economical
consider this factor up to 0.85) sizes of the extruded sections available in the market, or
A, = 2223.1 2 square inches have them specially manufactured if possible.
TI = 95 + 273
= 368°C
T2 = 48 + 273 A design engineer will be the best judge of the best
= 321°C
and the most economical design and extruded sections.
36.9 The above example is only for the outdoor part of the
:. W, = - x 0.65 x 2223.12 [3684 - 321"]
10l2 bus system. The indoor part, in any case, would be cooler
= 411.76 Wlft than the outdoor one and will also provide a heat sink to
the hotter enclosure and the conductor constructed
(b) By convection (W,)
outdoors. No separate exercise is therefore carried out
for the indoor part of the bus system, for the sake of
PO5 .
W, = 0.0022 A, ho,25 Wlft brevity. For a realistic design that would be essential.
The above example provides a basic approach to the
'This dissipation of heat will not be applicable when the circular dcsign of an IPB system. With some permutations and
section is made of circular extruded sections which have no combinations, a more realistic and economical design
surface openings. can be achieved. A computer is necessary for this exercise.