## Wastewater Denitrification-II

SOLUTION

1.                  10 MLD of wastewater is to be denitrified, and glucose (C6H12O6) is to be added for this purpose to the denitrification tank.  Nitrate concentration influent to the denitrification tank is 40 mg/L (as N).  qc = 3 days, q = 2 hours.  Calculate the effluent nitrate concentration and glucose requirement in Kg/d.  (YT)DN = 0.8; (Ks)DN = 0.1 mg/L; (Kd)DN = 0.04 /d; (qm)DN = 0.5/d.  All bio-kinetic constants are in terms of  Neglect nitrogen incorporation into bio-mass.

Hint:

·        Glucose consumption is due to both energy requirement for denitrifying microorganisms and bio-mass growth

·        Bio-mass Formula:

·        Balanced equation describing nitrate reduction to nitrogen gas with glucose as electron donor has to be formulated and used.

SOLUTION:

BSRT (qc) = 3 days;

Therefore,

Also,    ;

or,        ;                          Also,

Also,

Therefore,        X = 992 mg/L

V = q.Q =

Sludge produced per day =

Bio-mass Formula is ;          Formula Weight = 113

i.e.,                   113 g bio-mass has 5 moles of carbon

Therefore,        275 g of bio-mass has  moles of carbon

275 Kg of bio-mass has 12194 moles of carbon

Effluent Nitrate =

Nitrate conversion = 400 – 14 = 386 Kg/d

Equation for nitrate conversion:

i.e.,       336 Kg of nitrate-N requires 30000 moles of carbon

386 Kg of nitrate-N requires  of carbon

i.e., total carbon requirement per day = 12194 + 34464 = 46658 moles

6 moles of carbon = 180 g glucose

46658 moles of carbon =

Glucose requirement is 1399 Kg/d