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8. 1 Mixer Sizing Methods Four different sizing criteria - Velocity - Shear Stress - Yield Stress - Mixing Time.

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Presentation on theme: "8. 1 Mixer Sizing Methods Four different sizing criteria - Velocity - Shear Stress - Yield Stress - Mixing Time."— Presentation transcript:

1 8. 1 Mixer Sizing Methods Four different sizing criteria - Velocity - Shear Stress - Yield Stress - Mixing Time

2 8. 2 Velocity Mixing Duties –Circulation u specified heat or mass transfer specified... –Homogeneous suspension u depends on u settl and tank geometry Standard u values in biological treatment systems

3 8. 3 Velocity Losses (k) - Racetrack –Bends –Friction wall & bottom –Aeration –Obstacles Losses (k) - Other tanks –Tank factor (geometry) –Propeller factor –Aeration –Obstacles F req ~ u 2 k Required thrust = (Size of Mixer) = &

4 8. 4 Shear Stress Mixing Duties –Off bottom suspension & Resuspension of sediment Shear Stress calculated Shear Stress measured By experience –Erosion and transport of sediments

5 8. 5 Shear Stress Required thrust = (Size of Mixer) =  s = Requires Shear stress to resuspend –measured –calculated –Experience F ~  s

6 8. 6 Yield Stress For mixing to be possible, the fluid must move at all. If it has a finite Yield Stress, this must be overcome. Hence this is an additional mixing criterion, often decisive. Applications –Thickened sludge –Paper pulp –Drilling mud –Slurries...

7 8. 7 Yield Stress where  y is –Calculated or measured for municipal sludge, drilling mud & paper pulp –Specified by client –Measured by e.g. ITT Flygt Application Lab / known otherwise Required thrust = (Size of Mixer) = F ~  y

8 8. 8 Mixing Time Mixing Duties –Required blending time  specified or given by Throughflow; fluid leaving tank is mixed to a certain homogeneity  b. Batch; customer requires a certain maximum time  and a certain minimum homogeneity  b.

9 8. 9 Mixing Time Required thrust = (Size of Mixer) = F ~ 1 /  2  2 Specified mixing time –Given by customer –Given by process –Retention time Inflow = Q Volume = V Retention time = V/Q

10 8. 10 Quantified mixing demands VelocityF ~ u 2 Shear stressF ~  s Yield stress F ~  y TimeF ~ 1 /  2

11 8. 11 Extra study

12 8. 12 Channels - Required thrust “The velocity Solver” The required thrust is F req = A b k  is the liquid density (1000 kg/m 3 for water) k = k f + k b + k aer + k o are loss factors due to friction, bends, aerators, other obstacles. A b is the bulk flow area (projected area of cross section of main flow)  u 2 2 Extra study

13 8. 13 A racetrack example Wall friction Bends Aerators Obstacles F = k ·  · A b  u 2 2

14 8. 14 Bulk flow area Width Height AbAb

15 8. 15 Wall friction Surface roughness Length of flow loop

16 8. 16 Friction loss factor k f = L tot / (M R h ) L tot total mean length of channel R h = A b / P w hydraulic radius M  80 (Inverse) Manning number M is larger for very small channels or very smooth surfaces, and conversely smaller in the opposite cases. AbAb Wet perimeter P w = 2 H + W

17 8. 17 Bend loss factors k b 1.5 0.6 0.3 -- 1.5 2.5 1.0 0.8 0.5 1.1 1.4

18 8. 18 Bend losses k b = 1,5

19 8. 19 Aerator losses Diffusers act as flow obstacles Bubble columns increase the hydraulic losses by –causing counterflow to the bulk flow –causing velocity distributions that increase losses on the bottom and on the diffusers

20 8. 20 Aeration loss factor k aer Bottom diffuser geometry Bottom diffuser density in grid (m -2 ) h diff A  diff shape ? 1 m

21 8. 21 Aeration loss factor k aer # grids Bottom coverage (%) Air flow Q air (Nm 3 /h) Bulk flow velocity u  k aer

22 8. 22 Obstacle loss factor k o The loss force from an obstacle is F o = A o c D, And, to use A b in the F req - formula, k o = c D A o / A b. c D is typically between 1.0 and 2.0. For the pipe, say A o = 6.0  0.5 m 2, c D = 1.0 k o = c D A o / A b = 0.125  u 2 2 Projected area A o

23 8. 23 A racetrack example 50m 6m H=4m 1 grid Sanitaire diff, 20% covered area  u 2 2 2 units 4430 friction bends aerators obstacle F = k ·  · A bulk F = (0.87 + 2 · 1.5 + 0.55 + 0.125) · 1000 ·  · (6 · 4) = 4774 N 0.30 2 2

24 8. 24 Other tank shapes The same principles as in channels.... F req = A b k  u 2 2


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