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Filter Loading and Backwash Rates Well Yields and Chlorine Dosage in Waterworks Operation Math for Water Technology MTH 082 Lecture Chapter 4 & 8- Applied Math for Water Plant Operators Loading Rate Calculations (pg 62-66); Well yield (163-177) Mathematics Chapter 19-20 &23- Basic science Concepts and Applications (pg 183-190; 197-203) Objectives Reading assignment: Chapter 4 & 8- Applied Math for Water Plant Operators Loading Rate Calculations (pg 62-66); Well yield (163-177) Mathematics Chapter 19-20 &23- Basic science Concepts and Applications (pg 183-190; 197-203) 1. 2. 3. 4. Filter overviews Filter loading and backwash rates Well yield Chlorine dosage for a new well Conventional Treatment • Conventional Treatment – common treatment steps used to remove turbidity from the initial source water. Chlorination Ozone UV 1. Coagulation Raw Water Pretreatment 2. Flocculation 3. Sedimentation 5. Clear Well 4. Filtration Disinfection Distri Fast bution Slow Rapid or flash Mixing Alum, polymer Sludge Washwater Clearwell Backwash pumps High Rate Filter Dual or Multi Media Filter -4 times faster then rapid sand -granular activated carbon -anthracite coal -garnet sand and gravel -backwash ready (70 hr longevity) -excellent water quality High Rate Filters Anthracite Coal Fine sand Garnet sand High Rate Filter Dualmedia Filter Monomedia Filter Anthracite Coal Fine sand Garnet sand Monomedia Coarse sand Dual-media and multimedia filters 91% 1. Require an extremely deep bed 2. Can operate at three or four times the rate of sand filters 3. Cannot reduce turbidity 4. Do not require backwashing 5% eq ui tr no o D C an no t op an C re ... ... er a re d te uc e at ... t.. . ex an ui re eq R 5% 0% When mixed media filters composed of garnet, sand, and crushed anthracite coal are used, which of the following describes their placement in the filter bed? 92% a. .. ite co ,a ... nt hr ac on nd Sa A n to ar ne G 0% to p to p, co ite nt hr ac 0% ... a. .. 8% A 1. Anthracite coal on top, garnet in the middle, and sand on the bottom 2. Garnet on top, anthracite coal in the middle, and sand on the bottom 3. Sand on top, anthracite coal in the middle, and garnet on the bottom 4. Anthracite coal on top, sand in the middle, and garnet on the bottom In a filter using gravel, anthracite, and sand, the anthracite should be? 88% The top layer of media Beneath the gravel Between the sand and the gravel Mixed with the sand 12% th e. .. ... ith w d ix e M ee n et w B ea th en B th e th e sa er la y to p e 0% gr ... ... 0% Th 1. 2. 3. 4. Pilot Filters Anthracite Coal Fine sand Garnet sand The main action of a mixed media filter is: 87% Straining Disinfecting Coagulating None of the above ab ... th e gu l N on e of oa C D is i nf ec at in g tin g in in g 9% 4% 0% St ra 1. 2. 3. 4. Backwash • Suspended particles entrapped by filter media. • Accumulation occurs: – head loss within the filter to reach excessively high levels (6 to 8 feet of hydraulic head). - Particles pass through the filter, water turbidities reach unacceptable levels - Rule Backwash at 0.1 NTU - SWTR Allows 0.3 NTU. Head Loss • Clean filter =0 psi= one foot of head loss on a new filter • As filter clogs more negative pressure • Pressure builds in a linear fashion -2.5 to -4.0 psi = 6 to 10 ft of head loss • More clogged greater the head loss • Remember 1 ft of water column = 0.433 psi • 2.31 ft of water for 1 psi change Filter Backwash • Some plants use head loss, some use time • Some plants use operator knowledge and turbidity • Each operator has their own scheme!! • Parents and diapers-- The most critical criterion for determining when a mixed media filter should be backwashed is: Filter effluent quality Flow rate Head loss Visual inspection of the filter surface 39% Vi su a H ea li ns d pe lo s s ct ... 4% ra te ow Fl lte r ef flu en ... 4% Fi 1. 2. 3. 4. 52% Filtration Rate Calculation Unit Filter Run= (Total gal filtered gal) Filter Area (sq ft) Downward Filtration Rate= (flow gpm) Area (sq ft) Units will be gpm ft2! Backwash Rate = (flow gpm) Area (sq ft) Units will be gpm Upward The total water filtered during a filter run (between backwashes) is 2,950,000 gal. If the filter is 15 ft by 20 ft, What is the unit filter run volume (UFRV)? Given Formula L= 15 ft, W=20 ft; Rate 2,950,000 gal A=L X W UFVR= (Total gallons filtered g) Area (sq ft) Solve: A= 20 ft X 15 ft = 300 ft2 UFVR= (2,950,000 gal) 300 (sq ft) UFVR = 9833 g/ft2 25% ft2 0. 98 33 g/ 33 1. 10 98 7 g/ ft2 g/ ft2 0% g/ ft2 0% 3 9.833 g/ft2 101.7 g/ft2 9833 g/ft2 0.9833 g/ft2 9. 83 1. 2. 3. 4. 75% The total water filtered during a filter run (between backwashes) is 4.8 MG. If the filter is 20 ft by 30 ft, What is the unit filter run volume (UFRV)? Given Formula L= 20 ft, W=30 ft; Rate 4.8 MG A=L X W UFVR= (Total gallons filtered g) Area (sq ft) Solve: A= 20 ft X 30 ft = 600 ft2 UFVR= (4,800,000 gal) 600 (sq ft) UFVR = 8000 g/ft2 8 0. 00 80 0 g/ ft2 g/ ft2 80 00 g/ ft 0% g/ ft2 0% 2 0% 00 0 96000 g/ft2 8000 g/ft2 800 g/ft2 0.008 g/ft2 96 1. 2. 3. 4. 100% A filter 20 ft by 25 ft receives a flow of 1940 gpm. What is the filtration rate in gpm/ft2? Given Formula L= 20 ft, W=25 ft; Rate 1940 gpm A=L X W Filtration Rate= (flow gpm) Area (sq ft) Solve: A= 20 ft X 25 ft = 500 ft2 Filtration Rate= (1940 gpm) 500 (sq ft) Filtration Rate = 3.9 gpm/ft2 11% d/ ft3 gp 0. 25 gp m 25 0. 9 3. 0% /ft 2 m /ft 2 6% gp /ft 3 gp d 3.9 gpd/ft3 3.9 gpm/ft2 0.25 gpm/ft2 0.25 gpd/ft3 3. 9 1. 2. 3. 4. 83% A filter 20 ft by 35 ft receives a flow of 1530 gpm. What is the filtration rate in gpm/ft2? Given Formula L= 20 ft, W=35 ft; Rate 1530 gpm A=L X W Filtration Rate= (flow gpm) Area (sq ft) Solve: A= 20 ft X 35 ft = 700 ft2 Filtration Rate= (1530 gpm) 700 (sq ft) Filtration Rate = 2.2 gpm/ft2 11% d/ ft3 gp 0. 45 gp m 0. 45 gp 2 2. 0% /ft 2 m /ft 3 0% gp m /ft 2 2.2 gpm/ft2 2.2 gpm/ft3 0.45 gpm/ft2 0.45 gpd/ft3 2. 2 1. 2. 3. 4. 89% A filter 25 ft by 30 ft receives a flow of 3.3 MGD. What is the filtration rate in gpm/ft2? L= 25 ft, W=30 ft;Rate 3.3 MG 1,000,000 gal 1 Day D 1MG 1440 min A=L X W Filtration Rate= (flow gpm) Area (sq ft) Solve: 100% ft2 32 1 gp gp d/ ft3 m /ft 2 0% 3. gp m 0. 32 gp 1 0% /ft 2 m /ft 3 0% 3. 1. 2. 3. 4. A= 25 ft X 30 ft = 750 Filtration Rate= (2292 gpm) 750 (sq ft) 3.1 gpm/ft3 2 Filtration Rate = 3.1 gpm/ft 0.32 gpm/ft2 3.1 gpm/ft2 0.32 gpd/ft3 0. Given Formula A filter 25 ft by 10 ft has a backwash rate of 3400 gpm. What is the filter backwash rate in gpm/ft2? Given L= 25 ft, W=10 ft; Rate 3400 gpm Formula A=L X W Backwash Rate = (flow gpm) Area (sq ft) Solve: 100% 0% gp d /ft 3 /ft 2 6 13 .6 13 4 07 0. gp m gp m /ft 3 gp m .1 0% /ft 2 0% 13 1. 2. 3. 4. A= 25 ft X 10 ft = 250 ft2 Filter Backwash Rate= (3400 gpm) 250 (sq ft) 2 Filter Backwash Rate = 13.6 gpm/ft 13.1 gpm/ft3 0.074 gpm/ft2 13.6 gpm/ft2 136 gpd/ft3 A filter 20 ft by 15 ft has a backwash rate of 4.5 MGD. What is the filter backwash rate in gpm/ft2? Given L= 20 ft, W=15 ft; ft;Rate 4.5 MG 1,000,000 gal 1 Day Formula D 1MG 1440min A=L X W Backwash Rate = (flow gpm) Area (sq ft) Solve: 94% 6% ft3 gp d/ 01 0. gp m .4 10 0. 01 gp m /ft 2 0% /ft 2 0% /ft 3 gp m .4 10 1. 2. 3. 4. A= 20 ft X 15 ft = 300 ft2 Filter Backwash Rate= (3125 gpm) 300 (sq ft) 3 10.4 gpm/ft Filter Backwash Rate = 10.4 gpm/ft2 0.01 gpm/ft2 10.4 gpm/ft2 0.01 gpd/ft3 Well Problems • Drawdown ft = pumping water level – static water level ft • Well yield = Flow gallons duration of Test, min • Specific yield, gpm/ft = (Well yield gpm) (Drawdown ft) • Well casing disinfection lbs= (dose mg/L Cl2)(water in well casing MG)(8.34 lb/gal) Chlorine lbs = chlorine lbs % available chlorine 100 Before the pump is started the water level is measured at 140 ft. The pump is then started. If the pumping water level is determined to be 167 ft, what is the drawdown in ft? Given Static WL= 140 ft, Pumped WL=167 ft Formula Drawdown ft = pumping water level – static water level ft Drawdown = 167 ft- 140 ft Solve: Drawdown = 27 ft ft 0 27 -2 7 ft 0% ft 0% ft 0% 7 307 ft -27 ft 27 ft 0 ft 30 1. 2. 3. 4. 100% During a five minute test for well yield, a total of 740 gallons are removed from the well. What is the well yield in gpm? Given total = 740 gal, time = 5 minutes Formula Well yield = Flow gallons Duration of Test, min 96% 4% gp m 0 gp 00 37 14 8 gp m m 0% gp m 0% 67 Solve: Well yield = 740 gallons = 148 gpm 5 min 1. 67 gpm 2. 148 gpm 3. 3700 gpm 4. 0 gpm How many lbs of calcium hypochlorite (65% available chlorine) is required to disinfect a well if the casing is 18 inches in diameter and 220 ft long, with water level at 100 ft from the top of the well? The desired dose is 50 mg/L? Given Cl= 65/100 D=18 in=1.5 ft Well 220-100 =120 ft Formula 220 ft - 100 ft = 120 ft water in well (0.785)(D2)(H) = ft3 (0.785)(1.5 ft)(1.5 ft) (120 ft)(7.48 gal/ft3)= 1585 gal (50 mg/L)(.001585 MG)(8.34 lb/gal) = 1.01lbs Solve: 65/100 71% 24% 6% lb s 65 0. 2 .0 1 lb s lb s 0% lb s 2 lbs 1 lbs .02 lbs 0.65 lbs 2 1. 2. 3. 4. Today’s objective: Filter Loading Rates, Filter Backwash Rates, Well yield and chlorine dosage of new wells been met? Strongly Agree Agree Neutral Disagree Strongly Disagree 0% ag re e 0% St ro ng ly D Di s is a gr ee tra l 0% eu A gr ee 0% N ly Ag re e 0% St ro ng 1. 2. 3. 4. 5.