ࡱ> <>; |`bjbjVV .d<<%iii}}}8aD}K"z44 43K5K5K5K5K5K5KM`Pn5KQi42z4445K-K8884<"i3K843K88wHr+J`f}5IKK0KIP5xP(+J+JPi?J448444445K5K\7Z444K4444P444444444 : Assignment 4: Settling and Flotation Settling tanks 1. A settling tank of 3m depth treats water at 10C for the removal of alum floc with a relative density of 1.002 and diameter of 1 mm. a. Calculate theoretical settling velocity, loading rate and retention time required. b. What would the effect of a temperature of 25C be? c. What would the effect be of a removal efficiency of only 50% on the loading rate and retention time? Solution When T = 10 C, the dynamic viscosity of water is 1.31  EMBED Equation.3  10-6 m2/s So, the theoretical settling velocity is:  EMBED Equation.3  loading rate:  EMBED Equation.3  retention time:  EMBED Equation.3 , say 1h b) when the temperature is 25C, the dynamic viscosity of water decreases, therefore, the settling velocity and loading rate increase and the retention time decreases. The effect of water density will be minimal, so  EMBED Equation.3  c) Assume that efficiency in linearly related to loading rate. With a removal efficiency of only 50%, the loading rate will be halved and the retention time will be doubled. 2. A rectangular settling tank with a length: width of 3 treats water at 850 m3/d. The retention time is 2.4h and depth 4m. What would the overflow rate be and the horizontal velocity assuming an even velocity distribution through the settler. Solution  EMBED Equation.3 ; the overflow rate:  EMBED Equation.3   EMBED Equation.3  so, the horizontal velocity  EMBED Equation.3 or 3.3 m/h 3. Design a horizontal flow settling tank for a flow of 8000 m3/d. Determine the dimensions for a 7:1 length to width ratio. Assume a normal loading rate of 1 m3m-2h-1. Suggest a shape for the final overflow weir if the overflow rate is to stay within acceptable loading limits of 8 m3m-1h-1. Solution  EMBED Equation.3  Assume tank is 4m deep on average (typical value).  EMBED Equation.3  Retention time at 1 m/h loading rate and 4m depth will be 4h  EMBED Equation.3  L = 7W = 48.3m Minimum weir length =  EMBED Equation.3  Weir design Weir loading rate, WLR = 8 m3/m.h = 192 m3/m.d Total weir length needed = Q/WLR = 8000/192 = 41.7 m Available width = 6.9 m Further weir length needed = 41.7 m 6.9 m = 34.8 m Assume length of weir fingers = 2.5 m each Then required number of fingers = 34.8/(2.5 x 2) = 6.95 say 7 Width of finger = 6.9 m/ (7 fingers x 2 + 1) = 0.46 m  Flotation Make a semi-quantitative design for a flotation saturation chamber, given that the mass concentration of nitrogen and oxygen in water from air at atmospheric pressures equals about 15 mg/L and 10 mg/L, respectively. The gas requirements for flotation amount to about 10 mg/L (excess). Suitable air compressors can deliver air at 700 kPa. (Tip: You will need to have a high-pressure air saturation tank on a recycle stream.) Solution In reality, it is a simple mass balance. The recycle stream needs to contain enough dissolved air so that if mixed with the main stream, the total dissolved air will be 30 mg/L, that is the saturated level at atmospheric pressure (20 mg/L) plus the required excess of 10 mg/L. The water leaving the saturation tank will (assume) be saturated with air at the higher pressure maintained in this tank. The mass of air in the recycle water Cr at 800kPa(absolute): according to Henrys law values as provided:  EMBED Equation.3  Assume the flocculated water has no air deficit, so Cfl = 20mg/L  EMBED Equation.3  So, this DAF plant operates on a 7.7% recycle.   (  compressed air Note: Another question that could be asked about this is to size the compressor. Since we only need 10 mg/L or 10g/m3, we need to move 10g air/m3, about 8L at sea level (for a density of 1.25 g/L). Again for safety, assume 10L/m3 is required. For a flow of, say 1MGD, we have 158 m3/h and we will need 1580 L/h = 45 ft3/h of air (volume at ambient). Again, we need to have leeway, as compressors are generally not used continuously, so we would use the size of a compressor somewhat larger. A compressor of 1 or 2 cfm (cubic feet/min) would probably be chosen and operated on a pressure control switch from the saturator. Note that this is still a very small compressor 0.5 to 1 kW. (This could also be calculated from the ideal gas law, of course). A DAF plant, operates on a 6% recycle and a saturator pressure of 700kPa. The flocculated water has a particle concentration of 8000 particles/mL and a floc volume concentration of 10 ppm. Calculate the bubble volume, air mass and number concentrations in the contact zone. Flocculated water is saturated with air. Compare ratios of bubbles to particles in numbers and volumes at 20C Repeat at 5C Calculate expected limiting loading rates. Make the necessary assumptions. Solution at 20C (1) mass of air in DAF tank: according to Fig 7.34, the mass of air in the recycle water would be approximately 170mg/l for 800kPa ( 700kPa gauge pressure plus 101kPa atmospheric). Next, a mass balance for air is made for the contact zone of the DAF tank using Eq. 7.62.   EMBED Equation.3  (2) bubble volume concentration:b is calculated from the density at 20C, air = 1.19kg/m3  EMBED Equation.3  (3) bubble number concentration. Nb is calculated from Eq. 7.64 using a mean bubble diameter of 60(m  EMBED Equation.3  (4) ratios of concentration  EMBED Equation.3   EMBED Equation.3  at 5C (1) mass of air in DAF tank: according to Fig 7.34, the mass of air in the recycle water would be approximately 240mg/l for 800kPa ( 700kPa gauge pressure plus 101kPa atmospheric). Next, a mass balance for air is made for the contact zone of the DAF tank using Eq. 7.62.  EMBED Equation.3  (2) bubble volume concentration:b is calculated from Eq. 7.63: at 5C, the air = 1.27kg/m3  EMBED Equation.3  (3) bubble number concentration. Nb is calculated from Eq. 7.64 using a mean bubble diameter of 20(m  EMBED Equation.3  (4) ratios of concentration  EMBED Equation.3   EMBED Equation.3  c) At 20 C, the number of attached air bubbles to the floc particle is 7.4 (from part a) There are 8 million floc particles per liter occupying 10 ppm by volume. Therefore the volume of each particle is (10/106)/8x106 l or 1.25 x 10-12 l = 1.25 x 103 (m3. The diameter of each, if spherical, on average would therefore be dp = (6/( x 1.25 x 103)-3 = 13.6(m Assume the particle density is 1010kg/m3 According to Eq.7.66 on P7.58, the floc particle-bubble aggregate density is:  EMBED Equation.3   EMBED Equation.3  the rise velocity is:  EMBED Equation.3  so, the limiting loading rates at 20 C is 42 m/h. This is a high rate for flotation. Appreciate though, that it would be very difficult from a steric orientation point of view to attach more than 4 or 6 bubbles to a floc. Think of the three-dimensional arrangement of bubbles around a smaller object.     PAGE  PAGE 1 0.46 m 2.5 m %&'(789Wkl     ɻq`OD5Djh h OJQJUh h OJQJ h h CJOJQJ^JaJ h h CJ OJQJ^JaJ&h h 5CJOJQJ\^JaJ&h h 5CJOJQJ\^JaJ h]Yh CJOJQJ^JaJ hMGh CJOJQJ^JaJh CJOJQJ^JaJh h CJOJQJ^Jh OJQJ\hyh CJ OJQJ\aJ h CJ OJQJ\aJ &'(78 V $ O g u ] ^ U ^gd gd gd $a$gd       ! 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