**EXAMINATION-V **

**SOLUTION **

Consider a wastewater flow of 10 MLD with a BOD_{ 5} value of 600 mg/L. It is proposed to treat this waste using an activated sludge process (ASP). The size of the proposed aeration tank will such that the water is detained for 8 hours. Complete mixing is assumed in the aeration tank. The BSRT ( ) of the system is 10 days. In addition to the BOD_{5} , this water also contains nitrogen and phosphorus. The nitrogen concentration expressed as Total Kjeldahl Nitrogen (TKN) is 40 mg/L as N, and phosphorus concentration is not limiting. Assuming that nitrification occurs in this system, calculate the effluent TKN concentration, and the additional oxygen requirement for nitrification.

**Assumptions: **

X_{ r} = 10000 mg/L; Y_{ T} = 0.5; K_{d} = 0.05 /d; K =0.1 L/mg/d.

(K_{S} )_{ N} = 2.0 mg/L; (Y_{ T} )_{ N} = 0.2; (K_{d} )_{ N} = 0.05/d (q_{ m }) _{N} = 2 /d

• Biomass may be represented as C_{5} H_{ 7} O_{ 2} N

• Nitrogen incorporation in heterotrophic biomass must be accounted for.

• Nitrogen incorporation in autoprophic biomass may be neglected.

**(15) **

**Solution: **

; ;

Therefore,

Also,

Hence, .

Also,

Oxygen Requirement (in Kg/d) = 1.5.Q.(S o - S e ) -1.42.( D X)

= 1.5.

Formula weight of biomass, C_{5} H_{7} O_{2} N = 5(12)+7(1)+2(16)+1(14) = 113

Therefore, production of 1990 Kg/d biomass requires:

Influent Nitrogen concentration = 40 mg/L (as N), or

Therefore, Nitrogen available for nitrification = 400 - 247 = 153 Kg/d

Or, [TKN]_{o} = 15.3 mg/L (as N)

Nitrification Calculations:

;

Also,

; i.e.,

0.25 MLD of sludge is generated from the activated sludge process of a wastewater treatment plant. The solids concentration in this sludge is 13,000 mg/L. This sludge is first processed in a sludge thickener, where the solids content is increased to 4 percent (weight basis). Next the sludge is treated in the anaerobic digester for reduction in sludge solids. The digested sludge is then applied to sludge-drying beds for reduction of water content to 55%. Determine the weight and density of dried sludge that will be produced per day.

Density of Water = 1000 Kg/m^{3} ; Dry Density of Sludge Solids = 2200 Kg/m^{3}

COD reduction efficiency of sludge digester = 60 %; ;

For sludge digestion, Y_{T} = 0.06 mg/mg, and K_{d} = 0.03 /d on COD basis

** Assumptions: **

- The sludge thickener is 100 percent efficient.
- Entire effluent from the anaerobic digester is applied to sludge drying beds.
- Dried sludge is completely saturated with water.

**Answer: **

Consider 1 m^{3} of sludge after thickening

Let the sludge density be

Weight of this sludge =

Therefore, Weight of solids in 1 m 3 of sludge = (0.04).(1).(1022.305) =40.89 kg.

Solids Concentration (X o ) = 40890 mg/L

Let the discharge of thickened sludge be Q sl

Therefore, assuming 100 percent solids capture,

Q sl . (40890) = Q w . (13000) or, Q sl =

Solids Loading ( D X) =

Digester Design:

; Digester Volume (V d ) = ;

Assuming 60% treatment efficiency,

Also, , Also,

Therefore,

Anaerobic Sludge Production ( =

Therefore, Reduction in Solids =

Therefore, Solids Loading to Sludge Drying Beds = (3250-1891) = 1359 kg/d

After Drying