Geoscience Reference
In-Depth Information
Penetration is
P
t
= 1.0 - η = 0.005
Now calculate the bag failure parameter (ϕ) (a dimensionless number):
ϕ =
Q
/(
LD
2
(
T
+ 460)
0.5
) = 50,000/(2)(6)
2
(60 + 460)
0.5
= 30.45
Calculate penetration correction
P
tc
, which determines penetration from bag failure:
P
tc
= 0.582(∆
p
)
0.5
/ϕ = (0.582)(6)
0.5
/30.45 = 0.0468
Calculate the penetration and efficiency after the two bags failed. Use the earlier results to cal-
culate
P
t
1
:
P
t
1
=
P
t
2
+
P
tc
= 0.005 + 0.0468 = 0.0518
η
*
= 1 - 0.0518 = 0.948
Calculate the new outlet loading after the bag failures. Relate inlet loading and new outlet loading
to the revised efficiency or penetration:
New outlet loading = (Inlet loading)
P
t
1
= (4.0)(0.0518) = 0.207 grains/acf
■
EX AMPLE 17.15
Problem:
A plant emits 50,000 acfm of gas containing a dust loading of 2.0 grains/ft
3
. A particulate
control device is employed for particle capture and the dust captured from the unit is worth $0.01/lb
of dust. We are required to determine at what collection efficiency is the cost of power equal to the
value of the recovered material. Also determine the pressure drop in inches of H
2
O at this condition
(USEPA, 1984b, p. 122).
Given:
Overall fan efficiency = 55%
Electric power cost = $0.06/kWh
For this control device, assume that the collection efficiency is related to the system pressure drop
∆
p
through the following equation:
η = ∆
p
/(∆
p
+ 5.0)
where
η = Fractional collection efficiency.
∆
p
= Pressure drop (lb/ft
2
).
Solution
: Express the value of the dust collected in terms of collection efficiency η:
Amount of dust collected = (
Q
)(Inlet loading)(η)
Note that there are 7000 grains per pound.
=
(
)
×
(
)
×
(
)
×
(
)
η
3
3
Valueof dustcollected
50,000 ft /min2
grains/ft
/7000 lb/grain
0.01 $/lb
= 0 143
.
η
$/min
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