Question

1A:

Identify items (1) to (12) shown in the Figure 1

Solution:

1 – Screening

2 – Grit

removal

3 – Pre-treatment

4 –

Primary clarification A

5 –

Primary clarification B

6 – Aeration

tank

7 –

Secondary clarification A

8 – Secondary

clarification B

9 –

Disinfectant/chlorine contact

10 –

Primary anaerobic digester

11 –

Secondary anaerobic digester

12 –

Solids dewatering

Question

1B:

What is the main difference between primary and secondary

clarifiers?

Solution:

Primary clarifier in

most wastewater treatment plants typically follows the grit channel. It removes

around 30% of the biological oxygen demand after 5 days (BOD5),

which is a measure of biodegradable organic carbon. Around 60% of the suspended

solids (SS) are also removed; a measure of the suspended material.

After passing

through the primary clarifier, effluent will then flow through to the aeration

tank where oxygen is blown into the tank allowing bacteria to breakdown the

organic matter before it reaches the secondary clarifier. This is the second

stage of wastewater treatment and further reduces the SS levels to 30% and BOD5

to 20%. To summarise, the main difference between the primary and secondary

clarifiers is that primary clarifier’s removes sludge material that is denser.

Effluent downstream of the secondary clarifier would be cleaner than effluent

downstream of the primary clarifier.

The primary clarifier captures and

removes solids through filters by gravitation sedimentation. As sludge material

settles at the bottom of the primary clarifier it is scraped to one end, if

rectangular type, or to the middle, if circular type.

Secondary

clarification captures and removes finer particles with the use of

microorganisms to degrade the wastewater.

Question

1C:

What information do you need in

order to be able to calculate the clarifier efficiency?

Solution:

In order to calculate

the clarifier’s efficiency, samples of both the influent (upstream of

clarifier) and effluent (downstream of clarifier) must be taken. A series of

samples shall be taken over a 24 hour time period. The next step is the measure

the parameter (Biological Oxygen Demand, Chemical Oxygen Demand, Suspended

Solids, Dissolved Solids, Alkalinity, Chlorine, Phosphorus or Nitrogen) and

calculating the efficiency as a percentage by using the following equation –

Furthermore, the

efficiency of a clarifier can be altered due to factors such as:

Rate of wastewater flow

Length of time in left in collecting system

Sludge build up

Nature of solids in wastewater, if there is a

significant amount of wastewater which comes from mines or quarries,

electric power plants, nuclear industry or any another type of industrial

wastewater source

Question

2:

A settling tank with rectangular configuration has a length

to width ratio of 3 treats water at 850m3/day. And gas a retention

time of 2.4hrs and a depth of 4m. Calculate the overflow rate and the

horizontal velocity assuming that the velocity distribution through the settler

is even.

Solution:

Note: 850/24 = m3/hr

Overflow rate –

Horizontal velocity –

Question

3:

A settling tank has a horizontal flow of 8000 m3/day.

What are the dimensions of the tank for a 7:1 length to width ratio assuming a

normal loading rate of 1 m3m-2h-1 and make a

suggestion of the final shape for the overflow weir assuming that the overflow

rate stays within the acceptable loading limits of 8 m3m-1h-1

(Assume tank is 4m deep on average (typical value)).

Solution:

Average settling tank depth = 4m

Available width –

Length –

Loading time –

Minimum weir length –

Length of weir –

Required

number of weir fingers –

Note: Due to

available width being 6.9m, it has been assumed length of fingers will be 3m.

Question

4:

In the treatment of portable water, an

aluminum sulphate solution is used as a coagulant to produce an aluminum

hydroxide (sludge) floc. Assuming a settling tank with a depth of 3 meters is

used to treat the water at 10 degrees Celsius to remove the alum floc,

calculate the amount of sludge (floc) produced if 1,000 kg of alum coagulant is used

daily at 10 degrees Celsius.

Solution:

AI2(SO4)3

14H2O + 3Ca(HCO3)3 -> 2AI(OH)3

+ 3CaSO4 + 14H2O + 6CO2

1

mole of alum + 3 moles of ca bicarbonate -> 2 moles of alum hydroxide + 3

moles of ca sulphate + 14 moles of water + 6 moles of CO2

Molecular weights –

AI2(SO4)3

14H2O = 27 x 2 + (32 + 16 x 4) x 3 + 14(18) = 594g

3Ca(HCO3)3

= 340 + 2 x (1 + 12 + 3 x 16) = 486g

2AI(OH)3

= 227 + 3 x (16 + 1) = 156g

3CaSO4

= 3(40 + 32 + 4 x 16) = 408g

14H2O

= 14(2 x 1 + 16) = 252g

6CO2

= 6(12 + 2 x 16) = 264g

594g

+ 486g = 156g + 408g + 252g + 264g

1080g

= 1080g

594g

of alum produces 156g of alum hydroxide sludge therefore 1000kg of alum used

daily produces –

1000kg

= 1,000,000g

Amount of sludge (floc) produced –

References:

Reference 1 – Gerald Kiely

(1996). BOD RANGE FOR SOME TYPICAL INDUSTRIES, APPROXIMATE WASTEWATER FLOW

RATES & ADVANCED WASTERWATER TREATMENT LEVELS.

Reference

2 – Environmental Leverage. Primary clarifiers. Available: http://www.environmentalleverage.com/Primary_Clarifiers.htm. Last accessed 04th Nov 2017

Reference 3 – Clearcove.

Primary vs Secondary Sludge. Available:

http://www.clearcovesystems.com/primary-vs-secondary-sludge/.

Last accessed 05th Nov 2017