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02_chap_wang.qxd 05/05/2004 12:40 pm Page 84
84 Lawrence K. Wang et al.
µ = air viscosity = 1.21 × 10 −5 lb /(s ft)
f m
g = 32.174 ft/s 2
Determine the following design variables:
1. Cake fabric-filter resistance coefficient, K
2
2. Filtration cycle time, t
3. Blower horsepower
4. Fabric-filter area
5. Solids removal rate
Solution
1. Using Eq. (7),
k µ f S 1− ε
2
K = ( . × − 3
32 10 )
2 g ρ p ε 3
×
6
121 10 )( .
32 10 )
938)
= ( . × − 3 5 ( . × − 5 1 83 10 ) 2 ( .
32 174 323
.
= 585 in. H O 1 ( b m ft 2 )(ft min )
2
Operating data in the literature (3) show that for an installation of this type, using
Orlon fabric filters, K = 45. This is obtained via Eq. (6) for an inlet dust loading of
2
3
0.8 gr/ft .
2. Assume that the filtration should operate so that the pressure drop increases by up to
about 3 in. H O. The filtration cycle time can then be estimated by rearranging Eq. (6)
2
(use K = 45):
2
∆ P
t = 2
2
Kv L
2
( in. H O3 2 )(7000 grains ) b 1
= 2
[ 45 in. H O (1b ft 2 )(ft min )]( ft min3 ) (0 8. grains ft 3 )
2
= 65 min
Therefore, it would be necessary to shake the system about once an hour.
3. Considering that the residual fabric-filter resistance is also about 3 in. H O and there
2
are other gas flow pressure losses, assume an overall ∆P of 7 in. H O. The size of the
2
blower can be estimated (7) using 60% blower efficiency:
−4
Blower horserpower(HP) = (3 × 10 )(∆P)(Q)
∆P = 7 in. H O
2
3
Q = 8000 ft /min
−4
HP = (3 × 10 )(7)(8000) = 17
4. The size of the filter area required is
3
air-to cloth ratio = 3 ft min = 3 ft min
-
ft 2
3
8000 ft min
filter area = = 2670 ft 2
3 ft min