Geoscience Reference
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bottom with a web of drain tile placed on 20- to 30-ft centers. Sidewalls and partitions between bed
sections are usually made of wooden planks or concrete and extend about 14 in. above the sand sur-
face. Typically, three calculations are used to monitor sand drying bed performance: total biosolids
applied, solids loading rate, and biosolids withdrawal to drying beds.
24.14.4.1 Total Biosolids Applied
The total gallons of biosolids applied to sand drying beds may be calculated using the dimensions
of the bed and depth of biosolids applied, as shown by Equation 24.159:
Volume (gal) = Length (ft) × Width (ft) × Depth (ft) × 7.48 (gal) l /f (ft)
3
(24.159)
■
EXAMPLE 24.127
Problem:
A drying bed is 220 ft long and 20 ft wide. If biosolids are applied to a depth of 4 in., how
many gallons of biosolids are applied to the drying bed?
Solution:
3
Volume
=
Length (ft)width (ft)depth (ft)7.
×
×
× 48 gal/ft
3
=
220 ft
×
20 ft
×
0.33 ft
×
7.48 gal/f
t
=
10,861 gal
24.14.4.2 Solids Loading Rate
The solids loading rate may be expressed as lb/yr/ft
2
. The loading rate is dependent on biosolids
applied per application (lb), percent solids concentration, cycle length, and square feet of sand bed
area. The equation for solids loading rate is given below:
× (
Biosolidsa
pplied(lb)
Days of application
365
days/yr)
×
(
%Solids/100)
2
(24.160)
Solids loadingrate(lb/yr/ft )
=
Length (ft)W
×
idth (ft)
■
EXAMPLE 24.128
Problem:
A biosolids bed is 210 ft long and 25 ft wide. A total of 172,500 lb of biosolids is applied
during each application of the sand drying bed. The biosolids have a solids content of 5%. If the
drying and removal cycle requires 21 days, what is the solids loading rate in lb/yr/ft
2
?
Solution:
Biosolidsa
pplied(lb)
Days of application
×
365 d
ays
%solids
100
×
yr
2
=
Solids loadingrate(lb/yr/ft )
Length (ft)width (
×
ft)
172,500 lb
21 days
365 days
yr
5
100
×
×
=
210 ft
×
25 ft
37.5 lb/yr/ft
2
=
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