1)A) Consider a Cross-Section of the Filter Occupied by 4 Spheres of Radius = D. the Total
CEE 6560
Answer Key
Prob. Set No. 8
Nov. 11, 2009
1)a) Consider a cross-section of the filter occupied by 4 spheres of radius = d. The total area of the cross-section is A = 4d2. The area occupied by the 4 spheres is (see figure below):
Void volume = A-As
This implies that diameter of the particle has nothing to do with the porosity. To verify this use the same cross-sectional area but this time fill the area with twice as many spheres with 1/2 the diameter (see figure below).
Then A =
And
Same answer as with 4 spheres. Diameter doesn't influence porosity but it does influence the size of the pores.
b) First do the clean filter head loss calculations. For each layer use the geometric mean of particle diameter as the effective diameter for each layer:
dp1= (0.6 x 0.75)0.5 = 0.671 mm (2.201 x 10-3 ft)
dp2= (0.75 x 0.85)0.5 = 0.798 mm (2.62 x 10-3 ft)
V = Q/A = =0.688 ft/min = 0.011 ft/sec
Use the Carmen-Kozeny equation for each filter layer.
(2.548 ft)
similarly for layer two:
Total head loss = sum of head loss through each layer.
hftotal = 1.051 m (3.447 ft)
Now determine the solids loading to the filter (on volumetric basis)
Volume of suspended solids per liter of suspension = (50 mg/L)/(1.15gm/cm3) = 4.348x 10-8 m3/liter.
Q = (5 gpm/ft2)(21.53 ft2) = 1.55 gal/day =586.74 m3/day (to each filter).
Rate of void volume reduction for filter bed = Q x 4.348 x 10-8m3/liter = 0.026 m3/day.
Initial void volume of filter bed (top layer) = (2 m2)(depth of layer)() = (2 m2)(0.4m) () = 0.28 m3
Void volume of top layer after 1 day = 0.28 m3- 0.026 m3 = 0.254 m3
Porosity of top layer after 1 day = 0.254 m3/ 0.4 m x 2m2 = 0.3175
Porosity of 2nd layer doesn't change, so we need to calculate the new head loss in only in layer 1. Again using the Carmen-Kozeny equation, assuming effective diameter of porous media remains constant we get.
hf1 = 1.138 m (3.734 ft)
hf2 = 0.274 m (0.899 ft)
Total head loss = hf1 + hf2 = 1.412 m (4.633 ft)
c) i. The three filters left on line will split the flow that previously was handled by the filter that is being backwashed. So the flow to each filter is 1.33 (5 gpm/ft2). Assume the porosity of the 3 on-line filters is the same as calculated above for 1 day. Head loss calculations proceed as above except that the flow rate (or V in the Carmen-Kozeny equation) is increased by a factor of 1.33. Since the velocity term is linear in the Carmen -Kozeny eqn. we just need to multiply the previously calculated head loss term by 1.33.
Then:
hf1 = 1.518 m (4.979 ft). hf2 = 0.365 m (1.199 ft).
Total head loss = 1.883 m (6.18 ft)
ii) After the backwashed filter is returned to line the flow is split between 3 dirty and 1 clean filter. The dirty filters have the “1 day” porosity. The clean filter has the initial or clean porosity. If the total flow to the filters is maintained at a constant 2.347 x 103 m3/day through a common influent manifold the flow will distribute itself so that the head loss is equal in all filters. For a constant porosity head loss is a linear function of velocity or flowrate. The head loss calculation from the Carmen-Kozeny equation can be written in a simplified form:
For the dirty filters the constant is 207.93 sec/m2
For the clean filter the constant is 154.77 sec/m2
solve for h = 1.3 m
(Check: 3x 540.17 + 725.75 = 2347 m3/day)
2)
D = 1.969 ft (0.6 m) De = 1.2 D, initial porosity = 0.48, effective size = 0.8 mm, uniformity coefficient = 1.7
To calculate the minimum (incipient) fluidization velocity use the empirical relationship:
Now need to solve the coefficients of the empirical relationship for fluidization velocity :
ne =3.59Ref -0.1
ne = 2.91
Pressure drop calculation:
3) For a 400 ft2 filter total backwash flow = (15.95 gpm/ft2)(400 ft2) = 14.22 ft3/sec = 0.403 m3/sec
Number of gutters is selected by trial and error. For example select 3 gutters. Spaced evenly across the filter surface area this would give 5 ft 2in spacing between each gutter plus 2 ft 7 in. between the gutters and the wall of the filter bed. This meets the criterion that the horizontal travel distance for backwash water to reach a gutter is less than 3 ft. The bottom of the gutters should be located 12" above the top of the expanded bed.
Depth of the gutters can be determined from:
Select W = 18 " = 1.5 ft
For 3 gutters Q = (14.22 ft3/sec)/3 = 4. 74 ft3/sec
Du = 1.42 ft (17.04 in)
Total depth = 17.04 + 2 = 19.48 in (round up to 20 in). (need 2” – 3” freeboard).
Gutter size = 20” x 18”
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