Conventional Filtration

1. Conventional Filtration CE 547

2. Filtration is a unit operation of separating solids from liquids. Types of Filters (to create pressure differential to force the water through the filter): • Gravity filters • Pressure filters • Vacuum filters In terms of media • Perforated plates • Septum of woven materials • Granular materials (such as sand) Sand Filters • Slow sand filters (1.0 – 10 m3/m2.d) • Rapid sand filter (100 – 200 m3/m2.d)

5. Medium Specification for Granular Filters 1.Medium is the most important component of granular filters • Small grain sizes tend to have higher head losses • Large grain sizes may not be effective in filtering • The most effective grain sizes are found from previous experience 2.Effective size of medium is specified in terms of : • Effective size • Uniformity coefficient

7. What is Effective Size? It is the size of sieve opening that passes the 10% finer of the medium sample (the 10th percentile size, P10) What is Uniformity Coefficient? It is the ratio of the size of the sieve opening that passes the 60% finer of the medium sample (P60) to the size of the sieve opening that passes the 10% finer of the medium sample (P10). It is P60 to P10. For Slow Sand Filters • P10 = 0.25 – 0.35 mm • P60/P10 = 2 – 3 For Rapid Sand Filters • P10 = 0.45 mm and higher • P60/P10 = 1.5 and higher

8. Example 7.1

10. Linear Momentum Equation Applied to Filters (Fig 7.10) Movement of water through a filter bed is similar to the moment of water in parallel pipes, except that the motion is not straight but tortuous. If momentum equation applied on water flow in the downward direction of the element, then:

12. F2 = unbalanced force in downward (z) direction (inertia force) p = hydrostatic pressure A = cross-sectional area of the cylindrical element of the fluid Fg = weight of water in the element Fsh = shear force acting on the fluid along the surface areas of the grains dV = volume of element dl = differential length of the element As = surface area of the grains k = factor which converts As into an area such that ( kAsdl = dV )  = porosity of the bed a2 = acceleration of fluid element in downward (z) direction  = fluid mass density  = fluid element velocity in the (z) direction t = time

13. Since Ki = proportionality constant V = average water velocity

14. Since Ps = drop in pressure due to shear force  = liquid viscosity l = length of pipe D = diameter of pipe

15. Ks = proportionality constant rH = hydraulic radius = (area of flow / wetted perimeter) then Since Fg is constant, it can be included in Ki and Ks and removed from the equation. Then: This is a good linear momentum equation which can be applied to any filter

16. If particles are spherical d = diameter of particle

17. Thus rH = (area of flow / wetted perimeter) = (volume of flow / wetted area) if N = number of grains vp = volume of each grain Thus Since V can be expressed as = (vs / ) vs = superficial velocity

18. Head Loss in Grain Filters There are two categories • head loss in clean filters • head loss due to the deposited materials A. Clean-filter Head Loss Sp = surface area of a particle N = number of grains in the bed Volume of bed grains = S0l(1-) S0 = empty bed or surficial area of the bed  = porosity of bed l = length of bed

19. if vp is volume of a grain, then:

20. Now we have

21. Substitute (2), (3), and (4) in (1)

22. fp is a form of friction factor. Since:

23. Since particles are not spherical and  = shape factor dp = sieve diameter After backwash, the grain particles are allowed to settle. That means, the particles will deposit layer by layer, and hence particles will be of different sizes.

24. The bed is said to be stratified, and the head loss will be the sum of head losses of each layer. Then: If xi is the fraction of the di particles in the ith layer, then Assume  is the same throughout the bed, then

25. Example 7.2

29. Head Losses Due to Deposited Materials If q = deposited materials per unit volume of the bed, then: If hL0 = clean-bed head loss, then C = concentration of solids introduced into bed l = length of the filter bed

30. How to determine (a) and (b) in: It is clear that the equation represents a straight line if ln (hd) was plotted against ln (q).

31. This means that only two data points are needed in order to determine (a) and (b). If we have:

32. Examples 7.3 - 7.5

42. Backwashing Head Loss in Granular Filters During backwashing, the filter bed expands. For this to happen, a force must be applied: • le = expanded depth of the bed • l’ = the difference between the level at the trough and the limit of bed expansion • hLb = backwashing head loss • w = specific weight of water

43. weight of suspended solids is: e = expanded porosity of bed p = specific weight of particles weight of column water

44. Weights of suspended grains and column water are acting downward against the backwashing force. Thus: Solving for backwashing head loss:

45. If • vb = backwashing velocity • v = settling velocity of grains Then, and according to the following empirical equation:

46. for stratified bed if xi = fraction of particles in layer I, then: So,

47. Assuming expanded mass = unexpanded mass; then: Then, the fraction bed expression is:

48. Study Example 7.6