Treatment Plant HydraulicsLecture No. 101. General2. Frictional Pipe Headlosses3. Minor Headloss4. Headlosses from Control Structures (Weirs)5. Inlet and Outlet HydraulicsA. InletB. Outlet6. ExamplesTreatment Plant HydraulicsLecture No. 101. General- Hydraulic Profile. A hydraulic profile is a complete tracing of he water surface, HGL or hydraulic grade line, through a treatment plant.- Purpose:- To ensure that the hydraulic gradient is adequate for flow through the treatment works.- To establish the energy, head, needed for pumps.- To ensure that the plant facilities will not be flooded or backed up during periods of peak flow. - The hydraulic profile is typically a regulatory requirement.- Most conventional water treatment plants require 16-17ft of headloss across the plant. This means that a difference of 16-17ft must exist between the water level at the head of the plant and the high water level in the clearwell, which is at the tailend of he processes. An advanced plant with preozonation, postozonation and GAC may require25 ft of headloss. A flat and level site is NOT the best choice for the average plant. Theideal plant will have a 3-5% slope.- The hydraulic profile involves a careful consideration of frictional losses in pipes, minor losses and losses in control structures.- Typical headlosses:2. Frictional Pipe Headlosses- Any accurate headloss equation may be used.- The Darcy-Weisbach equation is preferred because it considers the fluid being transported.h = f Where:h = headloss, ftf = Darcy-Weisbach friction factor, dimensionless (.01-.07, usually .02-.03)Treatment Unit Headloss RangeftBar Screen .5-1Grit Chamber 1.5-4Primary Sedimentation 1.5-3Aeration Tank .7-2Trickling Filter 10-20Secondary Sedimentation 1.5-3Filtration 10-16Carbon adsorption 10-20Chlorine contact tank .7-6Treatment Plant Hydraulics, Page No. 2L = pipe length, ftD = pipe diameter, ftv = mean velocity, fpsg = acceleration due to gravity, ft2/s3. Minor Headloss- A minor headloss is involves any pipe appurtenance which is NOT a straight run of pipe.- Typical minor losses:- valves- expanders, reducers- turns, tees, bends- entrances, exits- Empirical equations, most useful:hminor = KminorWhere:hminor = headloss, ftKminor = coefficient, dimensionlessv = mean velocity, fpsg = acceleration due to gravity, ft2/s- Typical K values:- K=.5 Perpendicular square entrance- K=.15 Gate valve, full open- K = 10 Globe valve- K=1 Exit from pipe to reservoir4. Headlosses from Control Structures (Weirs)- The Francis Formula may be used for rectangular weirs:Q = 1.84(L-.1nh)h3/2where:Q = discharge, m3/s or cfs 1.84 = constant for metric system3.33 = constant of for US Customary systemL = length of crest of weir, m or ftn = number of end contractionsh = head on weir crest, ftFor 90- triangular weirs:Treatment Plant Hydraulics, Page No. 3Q = .55 h5/2Q = discharge, m3/s or cfs .55 = constant for metric system2.48 = constant of for US Customary systemh = head on weir crest, ft5. Inlet and Outlet HydraulicsA. InletPLANAABBSection A-Ad2d2Center LineSection B-BQ1Q2Q3Q4- Q4 - .90 Q1, In order to achieve equal flow through the orifices, the head through the orifices should be >> than the flume losses. - Orifice discharge formula:Q = .6AWhere:Q = discharge, cfs .6 = constant of for US Customary systemh = head on weir crest, ftg = acceleration due to gravity, ft2/s- The velocity in the influent pipe should be sufficient to main the suspended solids in suspension but < 4fps to avoid excessive headloss.B. OutletTreatment Plant Hydraulics, Page No. 4PLANAABBSection A-ACenter LineSection B-BWeirPlateH0V-notch weird- If the weir is suppressed:Q = 3.33LH3/2- If a 90- V-notch weirQ = 2.54LH5/2Where:Q = discharge, cfs 3.33, 2.54 = constant of for US Customary systemh = head on weir crest, ftL= Weir length, ft- The depth of flow in the effluent channel is give by the lateral spillway equation for a level channel:H0 = Where:H0 = upstream depth, ftd= downstream depth, ft dbar = mean depth, ftQ = discharge, cfs g = acceleration due to gravity, ft2/sb = channel width, ftf = Darcy-Weisbach friction factor, dimensionless (.01-.07, usually .02-.03), .03-.12 for concreteL = channel length, ftTreatment Plant Hydraulics, Page No. 5Rbar = mean hydraulic radius, ft- Solution procedure:- Perform a trial calculation for H0 ignoring friction. The friction term is the third term in equation, ) Estimate friction at .06 to .16 of the water surface drawdown. The water surface drawdown is H0-d, the depths at either end of the channel.- dbar and Rbar can then be calculated- recalculate H0 considering friction- Add 3-4” from the weir crest to the maximum water surface at H0 - d is set by the elevation of the water surface in the effluent box, a downstream condition:dmin = yc (critical depth)For a rectangular channel:yc = 1/3Where:yc = critical depth, ftq = Q/b, ft2/sQ = discharge, cfs g = acceleration due to gravity, ft2/sb = channel width, ft6. ExamplesA. Given: Circular primary clarifier. ADF=1.5 MGD. Qpeak=2.6ADF. OR(ADF) = 800 gpd/ft2. t(ADF)=2hrs. OR(peak)=75.0gal/(hr.ft2). t(peak)=.5hr. dmin=8.0Find:1.) diameter and depth of tank2.) H over V-notch weirs, use 90- triangular weirs, 8” OC.3.) Weir depth, 1” of Freeboard4.) depth of water in the influent box, b=15”5.) H06.) depth of effluent channel, triangular weirs with a 4” freefall1.) diameter and depth of tankA(ADF) = Q/OR = 1.5x106 gal/day x ft2/day/800galA(ADF) = 1,875ft2depth(ADF) = ORt = 800 gpd/ft2. x 2hr x x 1day/24hrdepth(ADF) = 8.91’, say 9.0’Area(peak) = 1.5x106 gal/day x 2.6(peaking factor) x 1ft2.hr/75gal x 1day/24hrArea(peak) = 2,166.7ft2depth(peak) = 75gal/ft2.hr x x .5hr depth(peak) = 5.01’Peak controls area, ADF depth controlsD2 = 2,166.7ft2D=52.52’, day 55’Use diameter=55’, depth = 9.0’2.) H over V-notch weirs, use 90- triangular weirs, 8” OC.weir length = -D = -(55’)weir length = 172.79’number of weirs = 172.79’ x 1/8” x 12”/ftnumber of weirs = 259Q = 1.5x106 gal/day x 2.6(peaking factor) x x 1day/24hr x 1min/60s x 1hr/60minTreatment Plant Hydraulics, Page No. 6Q = 6.035cfsQ/weir = 6.035cfs/259weirsQ/weir = .0233cfs/weirQ = 2.48 h5/2Q = 2.48 h5/2 = .0233cfs/weirh = .155’ = 1.85”3.) Weir depth, 1” of FreeboardWeir Depth = height required by weir + freeboard = 1.85” + 1”Weir Depth = 2.85”4.) depth of water in the influent box, b=15”dmin = yc (critical depth)For a rectangular channel:yc = 1/3Where:yc = critical depth, ftq = Q/b,