AGSM 337/BAEN 465 Sedimentation, Flow Equalization Page 1 of 7 Definition of Sedimentation Gravitational accumulation of solids (particles) at the bottom of a fluid (air or water) Essentially settling of solid particles Where is Sedimentation Used? Removal of solids from drinking water Most water sources, especially surface water, have some solids that must be removed prior to consumption Removal of solids from waste waters Settling of solids in a septic tank Settling of solids in a primary lagoon Solids removal at a treatment plant Settling of solids from air emissions Dust deposition near a feed lot Removal of solids from runoff water Sedimentation basins at construction sites Solids deposition in stock tanks Types of Sedimentation Discreet Settling individual particles settle independently, low solids concentration Flocculant Concentration is great enough to cause individual particles to stick together and form flocs. Flocs settle unhindered. Hindered Particle concentration is great enough to inhibit water movement between particles during settling, water must move through spaces between particles Compression Particles settle by compressing mass below Influencing Factors Particle factors size, density, shape, to some extent electrical charge Size bigger particles settle faster than smaller Density denser particles settle faster than less dense Shape spherical particles settle faster than large, flat particles Charge some particles (clays) carry an electrical charge which interacts with polarity of water Fluid factors flow velocity, density, viscosity Flow velocity particles settle faster when there is no fluid movement, velocity causes turbulence that keeps particles suspended (greater velocity allows suspension of larger, denser particles) Fluid density determines amount of buoyancy (weight of displaced water), denser fluids provided greater buoyancy and slower settling Fluid viscosity (fluid thickness) greater drag from thicker fluid
AGSM 337/BAEN 465 Sedimentation, Flow Equalization Page 2 of 7 Determining Sedimentation Rate Stoke s Law applies to spherical particles settling in a very quiet fluid (usually won t apply to gasses) v p 2 (ρ p ρw )d g (1) 18μ where: v p = particle settling velocity (m/s or ft/s), ρ p = particle density (kg/m 3 or lb m /ft 3 ) ρ w = fluid density (kg/m 3 or lb m /ft 3 ) d = particle diameter (m or ft), g = gravitational acceleration (9.81 m/s 2 or 32.2 ft/s 2 ) μ= fluid viscosity (kg/m s or lb m /ft s) Typical values for ρ and μ Particle densities for mineral sediments (mineral fraction of soil) ρ p = 2650 kg/m 3 Water density and viscosity at 20 ºC ρ w = 998 kg/m 3 (62.3 lb m /ft 3 ) μ = 1.01 x 10-3 kg/m s 6.74 x 10-4 lb m /ft s) (Note: values will vary with temperature see attached table) Example Determine the settling velocity for a 0.5mm diameter particle with a density of 2000 kg/m 3 in 20 ºC water. ρ p = 2000 kg/m 3 ρ w = 998 kg/m 3 d = 0.5mm = 0.0005 m g = 9.81 m/s 2 μ = 0.00101 kg/m s v p 3 2 2 2000 998kg/m 0. 0005m 9.81m/s 18 0. 00101kg/m s 014m/s.
AGSM 337/BAEN 465 Sedimentation, Flow Equalization Page 3 of 7 Application to Settling Tanks and Basins The surface area of a settling tank or basin needed to trap particles of a given size and density can be determined using Stoke s Law. The critical settling velocity is set equal to the settling velocity of the smallest particle to be trapped. The overflow rate (OFR) for a settling tank is equal to the flow rate into the tank divided by the surface area. This is similar to a vertical velocity. Setting the overflow rate equal to the critical settling velocity will allow time for capture of the smallest particles of interest: Q OFR v c A (2) where OFR = the overflow rate (m/s or ft/s) v c = the critical settling velocity (m/s or ft/s) Q = the flow rate into the tank (m 3 /s or cfs) A = the surface area of the tank (m 2 or ft 2 ) Example Particles in a wastewater flow have a density of 2000 kg/m 3 and range in size from 0.2 mm to 1.5 mm. The flow rate into a settling basin is 2,000,000 m 3 /d. What surface area is needed for a settling basin if it is to capture particles 0.5 mm and larger? The temperature is 20 ºC. We need v p for 0.5mm particles which we will use as the critical velocity. From the previous example, we know that v p = 0.14 m/s for 0.5mm diameter particles having a density of 2000 kg/m 3. This is the critical velocity needed which is set equal to the OFR: v c 2, 000, 000m 014m/s. d 3 1 A d 24 h h 3600 s Solving for the surface area, we get A = 165 m 2. Note that we don t get any information from this about the depth of the basin, only the surface area.
AGSM 337/BAEN 465 Sedimentation, Flow Equalization Page 4 of 7 Flow Equalization (BAEN 465) Because the flow rate and strength (BOD concentration) of a wastewater stream are continually changing, it is difficult to maintain efficient treatment process operation. Flow equalization is used to dampen the variations in flow and waste strength so the wastewater can be treated at an approximately constant flow rate. A flow equalization tank (basin) allows wastewater to accumulate when flow is above average and provides additional wastewater to downstream processes when inflow is below average to maintain a constant flow rate at the average value. Assuming the density of the wastewater remains constant, a volumetric balance can be made around the equalization tank: dv dt Q in Q out (3) or for a finite time interval (Δt) V Qin t Q t (4) out where Q is volumetric flow rate. In addition to maintaining a constant flow to downstream processes, an equalization tank will also dampen variations in the BOD 5 concentration in the flow to these processes. The best way to approach designing an equalization tank is a spreadsheet solution using a table of hourly flow rates and BOD 5 concentrations which comprise the daily inflow cycle for the WWTP. Calculate the average flow rate then rearrange the table so the first row is the time where the inflow first exceeds the average flow. The rest of the times follow in sequence to complete the daily cycle. Calculate the volume accumulating in the basin as the difference between inflow and outflow, which is the average flow rate, throughout the day. The accumulated volume should increase to a maximum and then decrease to zero at the end of the daily cycle. The maximum is the volume needed for the equalization tank. Additional volume may be added to allow for equipment failures, unexpected flow variation, and solids accumulation, e.g., the flow equalization volume requirement might be increased by 25%. Equalization also will provide a dampening effect on fluctuations in the BOD 5 loading to downstream processes. Similar to the volume balance used above, a mass balance on BOD 5 can be used to determine the fluctuation in the BOD 5 concentration in the basin, and hence to the WWTP, throughout the day. The mass of BOD 5 (M BOD-in ) entering the basin is given by the product of the inflow (Q in ), the BOD 5 concentration (S 0 ) and the time interval (Δt): M BOD -in QinS0t (5) The average BOD 5 concentration (S avg ) is determined by adding the total amount of BOD 5 contained in the basin at the end of the previous period (V tank-prev S prev ) to the BOD 5 that enters during the current period and dividing by the total volume in the basin at the end of the previous period plus the volume entering during the current period:
AGSM 337/BAEN 465 Sedimentation, Flow Equalization Page 5 of 7 S avg QinS0t Vtank-prev Sprev (6) Q t V in tank-prev The mass of BOD 5 flowing out of the basin during the time period is the product of the average flow rate, the average concentration and the time interval: M BOD Q S t (7) -out out avg Equalization can significantly reduce the fluctuation in the BOD 5 mass loading to downstream processes. Example (also see Example 11-2 in the text) The daily flow cycle for a municipal WWTP is given in the table below using 4 h time increments: Flow BOD 5 Time (m 3 /s) (mg/l) 0000 0.345 138 0400 0.141 89 0800 0.305 131 1200 0.498 251 1600 0.562 255 2000 0.550 146 Avg 0.400 The average flow rate is found to be 0.400 m 3 /s. Rearrange the table so the first row contains the first period where the flow rate is greater than the average (1200 in this example). We assume that the equalization tank is empty at the beginning of this time period. For each row, calculate the difference between the flow in and flow out (ΔV) then add that to the amount in the tank at the beginning of that time period (Σ(V)) from the previous time period. The maximum value in the Σ(ΔV) column will be the minimum size for the equalization tank. The tank volume may be increased to allow for equipment outages or other contingencies. Notice that the Σ(ΔV) column will decrease to 0 after 24 h (last row). Flow BOD 5 V in V out ΔV Σ(ΔV) Time (m 3 /s) (mg/l) (m 3 ) (m 3 ) (m 3 ) (m 3 ) 1200 0.498 251 7171 5762 1409 1409 1600 0.562 255 8093 5762 2330 3739 2000 0.550 146 7920 5762 2158 5897 0000 0.345 138 4968 5762-794 5102 0400 0.141 89 2030 5762-3732 1370 0800 0.305 131 4392 5762-1370 0 To determine the effect of equalization on the BOD 5 concentration in the tank, mass balances are completed for each row. The mass of BOD 5 flowing into the basin for a time period is given by equation 5, and the average concentration in the tank by equation 6. The average concentration
AGSM 337/BAEN 465 Sedimentation, Flow Equalization Page 6 of 7 for the time period is given by equation 7. For the first time period, the average concentration is the same as the inflow concentration since the tank is empty at the beginning of this time period. Flow BOD 5 V in V out ΔV Σ(ΔV) M BOD-in BOD tank Time (m 3 /s) (mg/l) (m 3 ) (m 3 ) (m 3 ) (m 3 ) (kg) (mg/l) 1200 0.498 251 7171 5762 1409 1409 1800 251 1600 0.562 255 8093 5762 2330 3739 2064 254 2000 0.550 146 7920 5762 2158 5897 1156 180 0000 0.345 138 4968 5762-794 5102 686 161 0400 0.141 89 2030 5762-3732 1370 181 140 0800 0.305 131 4392 5762-1370 0 575 133 The BOD tank concentration is the concentration flowing to downstream processes. Notice that the range is smaller (133-254 mg/l) than the range of concentrations entering the equalization tank (89-255 mg/l). Note: In this case the maximum BOD tank concentration occurred during the same time period as the maximum BOD 5 coming into the tank. This will not always be true, particularly when hourly data are available, since the highest BOD 5 concentrations will not necessarily occur at the same time as the highest flow rates. Properties of water (taken from http://www.thermexcel.com/english/tables/eau_atm.htm): Temp ( C) Density (kg/m 3 ) Viscosity (kg/m s) Temp ( C) Density (kg/m 3 ) Viscosity (kg/m s) 0.00 999.82 0.001792 21.00 998.08 0.000979 1.00 999.89 0.001731 22.00 997.86 0.000955 2.00 999.94 0.001674 23.00 997.62 0.000933 3.00 999.98 0.001620 24.00 997.38 0.000911 4.00 1000.00 0.001569 25.00 997.13 0.000891 5.00 1000.00 0.001520 26.00 996.86 0.000871 6.00 999.99 0.001473 27.00 996.59 0.000852 7.00 999.96 0.001429 28.00 996.31 0.000833 8.00 999.91 0.001386 29.00 996.02 0.000815 9.00 999.85 0.001346 30.00 995.71 0.000798 10.00 999.77 0.001308 31.00 995.41 0.000781 11.00 999.68 0.001271 32.00 995.09 0.000765 12.00 999.58 0.001236 33.00 994.76 0.000749 13.00 999.46 0.001202 34.00 994.43 0.000734 14.00 999.33 0.001170 35.00 994.08 0.000720 15.00 999.19 0.001139 36.00 993.73 0.000705 16.00 999.03 0.001109 37.00 993.37 0.000692 17.00 998.86 0.001081 38.00 993.00 0.000678 18.00 998.68 0.001054 39.00 992.63 0.000666 19.00 998.49 0.001028 40.00 992.25 0.000653 20.00 998.29 0.001003
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