3. The total available driving force (i.e. head) on a dual-media filter is 21.0

Question

3.

The total available driving force (i.e. head) on a dual-media filter is 21.0 ft. The clean bed losses when filtering at a rate of 8.0 gpm/ft2 are as follows:

Anthracite coal = 1.1 ft
Sand medium = 1.5 ft. Underdrain = 0.5 ft.
Filter rate controller = 0.42 ft.

Calculate:

The net available head for filtration if the rate is reduced to 3.0 gpm/ft2. (Hint: assume that laminar flow conditions prevail only in the sand and coal media).

The expected run length at the new filtration rate if the headloss in the filter bed increases at the rate of 0.08 ft. per hour.

PLEASE HELP WILL AWARD TOP RATING AND EXTRA POINTS TO WHOEVER CAN ANSWER!!!

03. The total available driving force (i.e. head) on a dual-media filter is 21.0 ft. The clean bed losses when filtering at a rate of 8.0 gpm/ft are as follows: Anthracite coal 1.1 ft Sand medium 1.5 ft Underdrain 0.5 ft Filter rate controller 0.42 ft. Calculate: a. The net available head for filtration if the rate is reduced to 3.0 gpm/ft (Hint: assume that laminar flow conditions prevail only in the sand and coal media) b. The expected run length at the new filtration rate if the headloss in the filter bed increases at the rate of 0.08 ft. per hour.

Explanation / Answer

LET THE theckness of filter media is L then we know that

Q= KiA

OR K= Q/(i*A) =constant

area A will be constant in both case

Q/i = constant or Q1/i1=Q2/i2

i=  h/L

in first case  h= 1.1+1.5+0.5+0.42=3.52 ft

i1 = (21-3.52)/L =17.48/l

8/(17.48/L ) = 3/(21-h/L)

21- h = avalable head=6.555 ft

time= h /0.08 = 14.445/0.08=180.56 hr

thanks

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