A conventional activated sludge plant treating 1000 m3-d of municipal waste water disposes off its anaerobically digested sludge on relatively impervious farmland- Use the following data-1- Raw Sewage- SS-225 mg-L - 70 Excess activated sludge returned to primary2- Primary Settling- SS-50 3- Excess Activated Sludge- 0-4gm VSS produced per gram BOD applied 80 4- Anaerobic Digester- VSS reduced 50Sludge specific gravity -15- Application on farmland- 2 m3-ha-dayTotal volatile suspended solids to be anaerobically digested kg-d VSS shall be

Wastewater discharge, Q=1000 m^3/d In Raw sewage, SS=225 mg/L (70 Volatile suspended solid, VSS=0.7× 1000× 10^3× 225 mg =157.5 kg BOD=190 mg/L BOD per day = ...

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Standard IX

Biology

Solid Waste and Different Sources

A conventiona...

Question

A conventional activated sludge plant treating $1000 m^{3} / d$ of municipal waste water disposes off its anaerobically digested sludge on relatively impervious farmland. Use the following data.

1. Raw Sewage: SS=225 mg/L ; 70 % volatile; BOD=190 mg/L
(Excess activated sludge returned to primary)

2. Primary Settling: SS=50 % removal; BOD=30 % removal

3. Excess Activated Sludge: 0.4gm VSS produced per gram BOD applied
80 % volatile of total

4. Anaerobic Digester: VSS reduced 50%; Digested sludge concentration =6%
Sludge specific gravity =1

5. Application on farmland: $2 m^{3} / h a - d a y$

Total volatile suspended solids to be anaerobically digested (kg/d VSS) shall be

144

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168

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233

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245

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Solution

The correct option is A $144$
Wastewater discharge,
$Q = 1000 m^{3} / d$

In Raw sewage,
$S S = 225 m g / L$ ( $70$ % volatile)

Volatile suspended solid,
$V S S = 0.7 \times 1000 \times 10^{3} \times 225 m g$
$= 157.5 k g$

$B O D = 190 m g / L$

$B O D$ per day $= 190 \times 1000 \times 10^{3} m g$
$= 190 k g$

Quantities shown are in per day
Total volatile suspended solids to be anaerobically digested
$= 131.95 k g / d$

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Similar questions

Q.
A conventional activated sludge plant treating

1000 m^{3} / d

of municipal wastewater disposes of its anaerobically digested sludge on relatively impervious farmland. Use the following data:

1. Raw Sewage: SS=225mg/L; 70% volatile; BOD=190 mg/L
(Excess activated sludge returned to primary)

2. Primary Settling: SS=50% removal; BOD=30% removal

3. Excess Activated Sludge: 0.4gm VSS produced per gram BOD applied
80% volatile of total

4. Anaerobic Digester: VSS reduced 50%; Digested sludge concentration =6%
Sludge specific gravity =1

5. Application on farmland:

2 m^{3} / h a - d

Area requirements (hectare) for disposal of the sludge on farmland shall be

Q. Assertion: A major part of the activated sludge is pumped into large tanks called anaerobic sludge digesters.
Reason: During anaerobic digestion, the anaerobic bacteria digest the organisms present in the primary sludge.

Q. A sewage containing 100 mg/I of suspended solids undergoes sewage treatment units. The sludge obtained contains 30% inorganic solids while rest organic solids. If 0.6

m^{3}

of gas is produced per kg of volatile matter reduced, the total gas produced in the digestion of sludge from 1.2 MLD of sewage is: (90% of solids are removed in the treatment process and efficiency of volatile matter reduction is 1).

Q. Match and find the correct option.

	I		II		III
(a)	BOD	(i)	Supernatant as effluent	(p)	Useful aerobic microbes
(b)	Flocs	(ii)	Measure of organic matter	(q)	All settled solids
(c)	Primary sludge	(iii)	Masses of bacteria associated with fungi	(r)	Methane hydrogen sulphide carbon dioxide
(d)	Biogas	(iv)	Anaerobic sludge digester	(s)	Polluting water has more BOD

Q. Which of the following is put into the anaerobic sludge digester for further sewage treatment?

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