A rapid sand filter with area of filter 200m2is used to process the water for a town with a population of 10 million. The rate of filtration of the filter is 25 m3/m2/hour. The porosity of the bed and specific gravity of the material used in the bed are 0.42 and 2.72 respectively. If the bed is expanded to 0.75 m from its original undisturbed depth of 0.675 m, then the head loss in (mm) due to expanded bed will be________.Correct answer is '672-675'. Can you explain this answer?

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A rapid sand filter with area of filter 200m² is used to process the water for a town with a population of 10 million. The rate of filtration of the filter is 25 m³/m²/hour. The porosity of the bed and specific gravity of the material used in the bed are 0.42 and 2.72 respectively. If the bed is expanded to 0.75 m from its original undisturbed depth of 0.675 m, then the head loss in (mm) due to expanded bed will be________.

Correct answer is '672-675'. Can you explain this answer?

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A rapid sand filter with area of filter 200m2is used to process the wa...

Concept:

Hydraulic Head Loss and Expansion of the Filter bed during Backwash:

Backwashing is accomplished by reversing the flow in filter media to force clean water to move upward through the filter media. In order to clean the filter bed, it is required to expand the bed, so that granular media filters are no longer in contact with each other.

To hydraulically expand a porous bed, the head loss (h_Le) must be at least equal to the buoyant weight of the filter bed. Even when the bed gets expanded to depth D_e, the head loss through the expanded bed essentially remains unchanged, because the total buoyant weight of the bed is constant.

Head loss, h_Le = D (1 – n) (G – 1) …..(i)

where D = undisturbed depth of filter bed in m.

n = porosity

G = specific gravity

Also, h_Le = D_e (1 - n_e)(G – 1) ..…(ii)

where D_e = the depth of expanded bed in m.

n_e = porosity of expanded bed

∴ from equation (i) and (ii) we get:

Calculation:

⇒ n_e = 0.478

Therefore,

h_Le = D_e (1 – n_e) (G – 1)

h_Le = 0.75 (1 – 0.478) (2.72 – 1)

h_Le = 0.67338m

∴ h_Le = 673.4 mm

Alternate Solution:

Head loss in both the condition will be same i.e.

h_Le = D (1 – n) (G – 1) = De (1 – n_e)(G – 1)

h_Le = 0.675 (1 – 0.42) (2.72 – 1)

h_Le = 0.67338 m

h_Le = 673.38 mm

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A rapid sand filter with area of filter 200m2is used to process the water for a town with a population of 10 million. The rate of filtration of the filter is 25 m3/m2/hour. The porosity of the bed and specific gravity of the material used in the bed are 0.42 and 2.72 respectively. If the bed is expanded to 0.75 m from its original undisturbed depth of 0.675 m, then the head loss in (mm) due to expanded bed will be________.Correct answer is '672-675'. Can you explain this answer?

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