Using values obtained above:
,.. [(25,0OO Ib/d) - (10,000 Ib/d) - 1.42(621 lb/d)]
o/o Stab^on = ± 25,000 lb/d ' ~x 100 = 56.5%
Related Calculations. This disadvantages and advantages of the anaerobic treat-
ment of sludge, as compared to aerobic treatment, are related to the slow growth rate of
the methanogenic (methane producing) bacteria. Slow growth rates require a relatively
long retention time in the digester for adequate waste stabilization to occur. With
methanogenic bacteria, most of the organic portion of the sludge is converted to methane
gas, which is combustible and therefore a useful end product. If sufficient quantities of
methane gas are produced, the methane gas can be used to operate duel-fuel engines to
produce electricity and to provide building heat.
DESIGN OFA CHLORINATION SYSTEM
FOR WASTE WATER DISINFECTION
Chlorine is to be used for disinfection of a municipal wastewater. Estimate the chlorine
residual that must be maintained to achieve a coliform count equal to or less than 200/100
ml in an effluent from an activated sludge facility assuming that the effluent requiring dis-
infection contains a coliform count of 1O^7 AOO ml. The average wastewater flow requiring
disinfection is 0.5 Mgd (1,892.5 m^3 /d) with a peaking factor of 2.8. Using the estimated
residual, determine the capacity of the chlorinator. Per regulations, the chlorine contact
time must not be less than 15 minutes at peak flow.
Calculation Procedure:- Find the required residual for the allowed residence time
The reduction of coliform organisms in treated effluent is defined by the following equa-
tion:
-J-=(1yv +0.23QO-^3
owhere Nt = Number of coliform organisms at time t
NQ = Number of coliform organisms at time ^ 0
Ct = Total chlorine residual at time t (mg/L)
t = Residence time (min)
Using values for coliform count from above:
irf =0+0.23Q)-32.Ox 10-^5 = (1+0.23Q)-^3
5.Ox 104 = (1+0.23Q)^3