Wet Scrubbers for Emission Control - Mathematics and Statistics for the Environment - page 538

Note: The values of α and β can be in either metric or British units and can be found in USEPA
(1984c, p. 9-15).

Scrubber efficiency is also expressed as the number of transfer units (USEPA, 1984a, p. 9-17):

N t = α( P T ) β = ln[l/(l - η)] (18.23)

where
N t = Number of transfer units.
η = Fractional collection efficiency.
α and β = Characteristic parameters for the type of particulates being collected.

Unlike the cut power and Johnstone theories, the contact power theory cannot predict efficiency
from a given particle size distribution. The contact power theory gives a relationship, which is inde-
pendent of the size of the scrubber. With this observation, a small pilot scrubber could first be used
to determine the pressure drop needed for the required collection efficiency. The full-scale scrubber
design could then be scaled up from the pilot information. Consider the following example.

■ EXAMPLE 18.4
Problem: A wet scrubber is to be used to control particulate emissions from a foundry cupola. Stack
test results reveal that the particulate emissions must be reduced by 85% to meet emission stan-
dards. If a 100-acfm pilot unit is operated with a water flow rate of 0.5 gal/min at a water pressure
of 80 psi, what pressure drop (∆ p ) would be needed across a 10,000-acfm scrubber unit (USEPA,
1984c, p. 9-15; USEPA, 1984a, p. 9-18)?

Solution: From the table in USEPA (1984c, p. 9-15), read the α and β parameters for foundry cupola
dust.

α = 1.35
β = 0.621

Calculate the number of transfer units N t using Equation 18.20:

η = 1 - exp(- N t )

N t = ln[l/(l - η)] = ln[l/(l - 0.85)] = 1.896

Now calculate the total contacting power ( P T ) using Equation 18.21:

N t = α( P T ) β
1.896 = 1.35( P T ) 0.621
1.404 = ( P T ) 0.621
ln 1.404 = 0.621(ln P T )
0.3393 = 0.621(ln P T )
0.5464 = ln P T
P T = 1.73 hp/1000 acfm

Calculate the pressure drop (∆ p ) using Equation 18.19:

P T = 0.1575∆p + 0.583 p L ( Q L / Q G )

1.73 = 0.1575∆ p + 0.583(80)(0.5/100)

∆ p = 9.5 in. H 2 O