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
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