A conventional activated sludge plant treating 1000 m3-d of municipal wastewater disposes of its anaerobically digested sludge on relatively impervious farmland- Use the following data-1- Raw Sewage- SS-225mg-L- 70Excess activated sludge returned to primary2- Primary Settling- SS-503- Excess Activated Sludge- 0-4gm VSS produced per gram BOD applied 804- Anaerobic Digester- VSS reduced 50Sludge specific gravity -15- Application on farmland- 2 m3-ha-dArea requirements hectare for disposal of the sludge on farmland shall be

In digested sludge, VSS=65.975 kg nbsp; nbsp; FSS=47.05 kg Digested sludge concentration = 6 ∴ nbsp;Weight of digested sludge =113.0250.06=1883.75 kg ...

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Standard XII

Biology

What Is Sewage and Its Effects

A conventiona...

Question

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

0.55

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2.95

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1.95

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0.95

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Solution

The correct option is D $0.95$
In digested sludge,
VSS=65.975 kg
FSS=47.05 kg

Digested sludge concentration
= 6% (by weight)

$∴ Weight of digested sludge$

$= \frac{113.025}{0.06} = 1883.75 k g$

Sludge specific gravity = 1

Volume of digested sludge

$= \frac{1883.75}{1. ρ_{ω}} = \frac{1884.75}{1 \times 1000} = 1.884 m^{3}$

Area requirement
$= \frac{Volume of digested sludge}{Application on farmland}$

$= \frac{1.884}{2} = 0.942 h e c t a r e$

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Similar questions

Q. 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

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. Study the given differences between primary sludge and activated sludge and select the incorrect ones.

	Primary Sludge	Activated sludge
(i)	It is sludge formed during primary sewage treatment.	It is sludge formed during secondary sewage treatment.
(ii)	It possesses flocs of decomposer microbes.	It does not possess flocs of decomposer microbes.
(iii)	It does not require aeration.	Formation of activated sludge requires aeration.
(iv)	A lot of decomposition occurs during formation of primary sludge.	Very little decomposition occurs during formation of activated sludge.

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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