Name ______(1 point)
CEE 454 SSSWS: Prelim 2 Nov. 15, 2007 - 8:40 AM -9:55 AM Open book May the time you spent preparing for this exam pay off. Read each problem carefully before beginning to work on the answer! Make sure you answer each part of each question (7 points each).
1) Is the required baffle spacing for a desired maximum velocity gradient a function of the depth of the flocculator tank? Explain why or why not. Baffle spacing is not a function of flocculator tank depth. The spacing is based on the maximum velocity gradient generated by the U turn. That velocity gradient is not affected by tank depth unless the tank depth is small relative to the baffle spacing.
2) Why might it be beneficial to use obstacles between baffles for flocculators designed for low flow rates, but not for high flow rates? Specifically, which dimensionless geometric ratio determines whether obstacles would be useful? At low flow rates the baffles are tall and thin and the energy dissipation from the U turn is expended long before reaching the next U turn. The dimensionless parameter is b/h (baffle spacing/tank depth).
3) What transport mechanism is responsible for mixing aluminum sulfate with the raw water at the molecular scale? Diffusion
4) You have designed a hydraulic flocculator to work with water at 20°C. If the water temperature dropped to 1°C will the velocity gradients increase, decrease, or stay the same? Explain why and write the relevant equation. Please note that viscosity increases as temperature decreases. You may assume most of the energy loss is due to minor losses. e G = The viscosity will increase, but if the energy dissipation is due to minor n losses it will remain unchanged. Thus the velocity gradients will decrease.
5) In sweep flocculation which type of material occupies more volume in the flocculator, particles that were present in the source water or aluminum hydroxide precipitate? The aluminum hydroxide precipitate occupies much more volume than the source water particles.
1 6) It is possible that the floc volume fraction increases as the flocs grow in size. If the floc volume fraction is lower for small flocs than what I assumed in the floc model,
then how will the Gmax vs G design curve change? Describe the change to the curve and explain how this change will affect baffle spacing and the number of baffles of a given spacing in a flocculator.
400 G max design G max at high flow 300 G max at low flow )
s G average at high flow / 1
( 200
G
100
0 0 5000 10000 15000 20000
G The G for each collision will increase that will cause the first part of the curve to stretch in the x direction. The baffle spacing will not be changed by this, but the number of baffles with narrow spacing will increase.
2 Design Challenges (57 points)
1) (15 points) A rapid mix pipe is being designed to have a G of 1000. The plant flow rate is 400 L/min (0.0067 m3/s), the pipe diameter is 10 cm, and the pipe length is 3 m. What is the minimum amount of head loss required to get the required G? e G = n gh e = l q ghq Gq = l n p d2 L q = 4Q gh p d2 L Gq = l n 4Q
2 4Qn hl = ( Gq ) gp d2 L 骣 m3 m 2 4琪 0.00667 10-6 2 桫 s s hl =(1000) = 2.9 cm 骣 m 2 琪9.8p ( 0.1m) ( 3 m) 桫 s2
2) (15 points) What is the baffle spacing for a flow of 400 L/min, a baffle width of 0.5 m, a value of Gmax of 50/s, a tank depth of 2 m, and a tank width of 0.5 m. The loss coefficient for the U-turn is approximately 3. Make reasonable estimates for any other parameters you need. 1 3 1 骣 1 2 骣Q4 骣 K 4 b = 琪 琪 琪 桫Gmax 桫 w 桫2n 1 3 4 骣 m3 4 骣 1 琪0.00667 琪 骣s 2 3 b=琪 s 琪 = 19.4 cm 琪 琪 2 桫50琪 0.5m 骣 -6 m 琪 琪2琪 10 桫 桫 桫 s
3 3) (15 points) What is the average value of G if the baffle spacing is 10 cm, for the conditions given in problem 2? To make the calculation simpler, you may neglect major losses. 1 骣 h G= V琪 Kminor +f 2nq 桫 4Rh K G= V minor 2nq wbh q = Q Q V = wb 3 骣Q 2 Kminor G = 琪 桫wb2n h 3 骣骣 m3 2 琪琪0.00667 桫 s 3 G=琪 = 42 / s 琪 2 (0.5m)( 0.10 m) 骣 -6 m 琪 2琪 10( 2m) 桫 桫 s
4 4) (12 points)The flocs accumulate in the bottom of the sedimentation tank. I’d like to know how fast the bottom of the sedimentation tank will fill with flocs. To make this calculation assume that the flocs don’t compress significantly. The floc volume fraction is approximately 0.001. Provide an estimate of how many days it will take for the bottom of the sedimentation tank to fill with flocs. You may assume that there is 1 m between the sludge drain and the bottom of the lamella and that the upflow velocity is 100 m/day. You may neglect the influence of the inactive area at the end of the sedimentation tank that is due to the sloping lamella. launder Find the volume of the bottom of the tank 1 volume= h wL 2 sludge tank
Sludgevolume= Qtf floc
Q= Vup Lw
Sludgevolume= Vup L tank wtf floc nowsetthevolumesequal 1 h wL= V L wtf 2 sludge tank up tank floc h t = sludge Sludge 2V f up floc drain 1m t= = 5 days 骣 m 2琪 100 0.001 桫 day
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