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New Sac manual ch 4

Test questions- multiple choice

QuestionAnswer
What is the purpose of presedimentation facilities? To reduce the solids removal load at the water treatment plant.
What size particles can be removed from water by sedimentation? Greater than 10 microns.
What property of alum attracts negatively charged particles? Positive charge.
What happens to the settling rate (settling velocity) of particles in water when the water temperature drops? Decreases
What is the largest portion of the sedimentation basin? Settling zone
Calculate the theoretical detention time for a rectangular sedimentation basin in hours. The basin is 80 feet long, 30 wide and 10 deep treating a flow of 1.8 MGD. 2.4 hrs. 80x30x10x7.48 gal/ft3 = 179,520gal/ft3. 179,520 gal/ft3 x 24 hrs = 4,308,480gal/ft3/24hrs. Divided by 1,800,000 MGD = 2.39 hrs.
What is a major means by which operators can control water treatment processes? Adjusting chemicals and chemical feed rates.
Estimate the surface loading rate in gallons per minute per square foot for a rectangular sedimentation basin 20 ft wide by 40 ft long when flow is 0.5 MGD 0.43 gpm/ft2. 0.5 MGD divided by 1440 min/day = 347 gpm divided 40x20 = 800ft2. 347 divided by 800 = 0.43 gpm/ft2
Determine the mean flow velocity in ft/min) for a rectangular sed basin with a flow of 2.0 MGD. The basin is 75 ft long, 25 wide and 14 deep. 0.53 ft/min 2.0 MGD div by 1440 min/day = 1389 gpm. multiply by 1 gal over 7.48 gal/ft3 (or divide by 7.48) = 185.68. 185.68 over (25 ft wide x 14 ft deep) = 0.53 ft/min
Determine the weir loading rate in gpm/ft of weir for a rectangular sed basin basin with flow rate of 2.0 MGD. Four effluent launders 10.5 ft long with v-notch weirs on both sides of the launder extend into basin from outlet end. 16.5 gpm/ft. 2MGD/1440 min per day = 1388 gpm. 10.5 ft per weir x 2 (per basin) = 21 ft. 21ft x 4 weirs = 84 ft. 1388 gpm/84 ft = 16.5 gpm/ft.
--- combine the caogulation, flocculation and sedimentation processes into a single basin. The solids-contact process units
Sludge produced by a solids-contact clarification unit is recycled through the process to act as a ----- coagulant aid
Changes in which process variables cause instability in solids-contact units? Flow, temperature and turbidity.
What does the volume over volume test result tell an operator about the solids-contact unit? Percentage of slurry.
If raw water at treatment plant needs a lime does of 10.5 mg/L, determine setting time on lime feeder in lbs/day. Flow is 1.5 MGD. 131 lbs/day 10.5 mg/L x 1.5 MGD x 8.34 lbs/gal = 131 lbs/day
What is the primary factor in determining the frequency of sludge removal? Rate of sludge buildup.
Low turbidity source waters may not require ---- sedimentation
Which water quality indicators should be monitored in the normal operation of the sedimentation process? Temperature and turbidity.
In solids-contact clarifiers, what should an operator do if the sludge blanket is of normal density but very close to the surface? Waste more sludge.
What does frequent clogging of the sludge discharge line indicate? Sludge concentration is too high.
How can operators measure turbidity levels at the sedimentation basin inlet and outlet to get a rough idea of process removal efficiency? Grab samples.
What is the most common reason to shut down a sedimentation basin? Annual maintenance and cleaning.
Which task should be performed immediately following dewatering of a sedimentation basin? Grease and lubricate previously submerged mechanical moving parts.
Process water samples must be --- of actual conditions in the treatment plant. representative.
How can operators determine if sludge pumping lines are clogged? Check pump's suction and discharge pressure.
Created by: sgaskell
 

 



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