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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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