Question
Download Solution PDFA village has a population of 200 with an average rate of water demand of 100 litres per capita per day. A rapid sand filter having an average filtration rate of 100 liter/hour/m2 is to be designed for water treatment. The area of rapid sand filter required is:
(Take the ratio of maximum demand to average demand as 1.5)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
Given Data:
Village population = 200
The average rate of water demand = 100 litres per capita per day
average filtration rate = 100 liter/hour/m2 = 100 x 24 liter/day/m2
Ratio of maximum demand to average demand = 1.5
Design dicharge = 1.5 x Village population x average rate of water demand
= 1.5 x 200 x 100 = 30000 liter/day/m2
Area of rapid sand filter = \(\frac{Design \ discharge}{Average \ filtration \ rate}\)
Area of rapid sand filter = \(\frac{30000}{100\ \times 24}\)
Area of rapid sand filter = 12.5 m2
Additional InformationFor Rapid Sand Filter
- Numbers of filters (N) = 1.22 \(\sqrt (\)Q), where Q is in MLD
- Total cross-sectional area of perforations = 0.2% of filtered area
- Cross-sectional area of on lateral is 2 or 4 times the total cross-sectional area of perforation.
Last updated on May 28, 2025
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