Question tank 7 – Secondary clarification A 8

Question1A:Identify items (1) to (12) shown in the Figure 1 Solution:1 – Screening2 – Gritremoval3 – Pre-treatment4 –Primary clarification A5 –Primary clarification B6 – Aerationtank 7 –Secondary clarification A8 – Secondaryclarification B9 –Disinfectant/chlorine contact10 –Primary anaerobic digester 11 –Secondary anaerobic digester 12 –Solids dewateringQuestion1B: What is the main difference between primary and secondaryclarifiers?       Solution:Primary clarifier inmost wastewater treatment plants typically follows the grit channel. It removesaround 30% of the biological oxygen demand after 5 days (BOD5),which is a measure of biodegradable organic carbon.

Around 60% of the suspendedsolids (SS) are also removed; a measure of the suspended material.After passingthrough the primary clarifier, effluent will then flow through to the aerationtank where oxygen is blown into the tank allowing bacteria to breakdown theorganic matter before it reaches the secondary clarifier. This is the secondstage of wastewater treatment and further reduces the SS levels to 30% and BOD5to 20%. To summarise, the main difference between the primary and secondaryclarifiers is that primary clarifier’s removes sludge material that is denser.Effluent downstream of the secondary clarifier would be cleaner than effluentdownstream of the primary clarifier.              The primary clarifier captures andremoves solids through filters by gravitation sedimentation. As sludge materialsettles at the bottom of the primary clarifier it is scraped to one end, ifrectangular type, or to the middle, if circular type. Secondaryclarification captures and removes finer particles with the use ofmicroorganisms to degrade the wastewater.

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Question1C:What information do you need inorder to be able to calculate the clarifier efficiency? Solution:In order to calculatethe clarifier’s efficiency, samples of both the influent (upstream ofclarifier) and effluent (downstream of clarifier) must be taken. A series ofsamples shall be taken over a 24 hour time period. The next step is the measurethe parameter (Biological Oxygen Demand, Chemical Oxygen Demand, SuspendedSolids, Dissolved Solids, Alkalinity, Chlorine, Phosphorus or Nitrogen) andcalculating the efficiency as a percentage by using the following equation –   Furthermore, theefficiency 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          Question2: A settling tank with rectangular configuration has a lengthto width ratio of 3 treats water at 850m3/day. And gas a retentiontime of 2.

4hrs and a depth of 4m. Calculate the overflow rate and thehorizontal velocity assuming that the velocity distribution through the settleris even.  Solution:          Note: 850/24 = m3/hr Overflow rate –  Horizontal velocity –   Question3: 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 anormal loading rate of 1 m3m-2h-1 and make asuggestion of the final shape for the overflow weir assuming that the overflowrate 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 –Requirednumber of weir fingers – Note: Due toavailable width being 6.

9m, it has been assumed length of fingers will be 3m.Question4: In the treatment of portable water, analuminum sulphate solution is used as a coagulant to produce an aluminumhydroxide (sludge) floc. Assuming a settling tank with a depth of 3 meters isused 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 useddaily at 10 degrees Celsius.  Solution:AI2(SO4)314H2O + 3Ca(HCO3)3 -> 2AI(OH)3+ 3CaSO4 + 14H2O + 6CO2 1mole of alum + 3 moles of ca bicarbonate -> 2 moles of alum hydroxide + 3moles of ca sulphate + 14 moles of water + 6 moles of CO2 Molecular weights –AI2(SO4)314H2O = 27 x 2 + (32 + 16 x 4) x 3 + 14(18) = 594g3Ca(HCO3)3= 340 + 2 x (1 + 12 + 3 x 16) = 486g2AI(OH)3=  227 + 3 x (16 + 1) = 156g3CaSO4= 3(40 + 32 + 4 x 16) = 408g14H2O= 14(2 x 1 + 16) = 252g6CO2= 6(12 + 2 x 16) = 264g 594g+ 486g = 156g + 408g + 252g + 264g1080g= 1080g 594gof alum produces 156g of alum hydroxide sludge therefore 1000kg of alum useddaily produces – 1000kg= 1,000,000g Amount of sludge (floc) produced -References: Reference 1 – Gerald Kiely(1996). BOD RANGE FOR SOME TYPICAL INDUSTRIES, APPROXIMATE WASTEWATER FLOWRATES & ADVANCED WASTERWATER TREATMENT LEVELS.  Reference2 – Environmental Leverage. Primary clarifiers.

Available: Last accessed 04th Nov 2017Reference 3 – Clearcove.Primary vs Secondary Sludge.

Available: accessed 05th Nov 2017


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