Wastewater Calculations - Mathematics and Statistics for the Environment - page 727

Solution:

520,000 gal
12,600

Biosolidsretentiontime = = 41.3 days

gpd

24.13.4.4 Estimated Gas Production in Cubic Feet/Day
The rate of gas production is normally expressed as the volume of gas (ft 3 ) produced per pound of
volatile matter destroyed. The total cubic feet of gas a digester will produce per day can be calcu-
lated by

3

Gasproduction (ft /day)Volatile matterin = (lb/day) × %Volatilematterreduction

(2 4.141)

3

× Prod uction rate (ft /lb)

Key Point: Multiplying the volatile matter added to the digester per day by the percent volatile
matter reduction (in decimal percent) gives the amount of volatile matter being destroyed by the
digestion process per day.

■ EXAMPLE 24.110
Problem: The digester reduces 11,500 lb of volatile matter per day. Currently, the volatile matter
reduction achieved by the digester is 55%. The rate of gas production is 11.2 ft 3 of gas per pound of
volatile matter destroyed.

Solution:

Gas production (ft 3 /day) = 11,500 lb/day × 0.55 × 11.2 f t 3 /lb = 70,840 ft 3 /day

24.13.4.5 Volatile Matter Reduction (Percent)
Because of the changes occurring during biosolids digestion, the calculation used to determine
percent volatile matter reduction is more complicated:

(%VM − %) M 100

%%)

×

in out

%Reduction = (24.142)

[ ]

%(% VM − VM × VM out

in in

■ EXAMPLE 24.111
Problem: Using the digester data provided here, determine the percent volatile matter reduction for
the digester with raw biosolids volatile matter of 71% and digested biosolids volatile matter of 54%.

Solution:

0.71-0.54
0.71

%Volatile matter reduction = = 52%

-(0.71 × 0.54)

24.13.4.6 Percent Moisture Reduction in Digested Biosolids

( %Moisture%Moisture − ) × 100

in o ut

%Moisturereduction = (24.143)

[ ]

%Moisture − ( Moisture × %Moisture

ut ) in in o

Key Point: % Moisture = 100% - percent solids.