__SOLUTION__

1.
10 MLD of wastewater is to be denitrified, and glucose
(C_{6}H_{12}O_{6}) is to be added for this purpose to
the denitrification tank. Nitrate
concentration influent to the denitrification tank is 40 mg/L (as N). q_{c}
= 3 days, q = 2 hours. Calculate the effluent nitrate concentration
and glucose requirement in Kg/d. (Y_{T})_{DN}
= 0.8; (K_{s})_{DN} = 0.1 mg/L; (K_{d})_{DN}
= 0.04 /d; (q_{m})_{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
(q_{c}) = 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

Influent
nitrate loading = _{}

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