A town has a population of 1,50,000 persons with per capita water supply of 200 litres/day. Assuming 85% of water usage is appears to be sewage, design a sewer running 0.7 times full at maximum discharge. Take a constant value of N = 0.013 at all depths of flow. The sewer is to be laid at a slope of 1 in 500. Take a peak factor of 3.
Anna .... a town has a population of 100000 persons with a per capita water supply of 200 Lpd . Design a sewer running 0.7 times full. take n=0.013 and slope 1 in 500 and peak factor of 3. assume 85% of water supply turns into sewer. .....intha sum ku answer venum na Anna .. pls Anna
Step 1: Determine the total water supply to the town per day: Total water supply = population * per capita water supply Total water supply = 100000 * 200 = 20,000,000 Lpd Step 2: Determine the amount of water that will turn into sewer: Water that turns into sewer = 85% of total water supply Water that turns into sewer = 0.85 * 20,000,000 = 17,000,000 Lpd Step 3: Determine the peak flow rate using the peak factor: Peak flow rate = Peak factor * Average daily flow rate Peak flow rate = 3 * (Water that turns into sewer / 24) Peak flow rate = 3 * (17,000,000 / 24) Peak flow rate = 1,275,000 Lph (liters per hour) Step 4: Determine the diameter of the sewer pipe using the Manning's formula: Q = (1.486/n) * A * R^(2/3) * S^(1/2) where: Q = flow rate (m3/s) n = Manning's roughness coefficient A = cross-sectional area of the pipe (m2) R = hydraulic radius of the pipe (m) S = slope of the pipe (m/m) Since the sewer is running 0.7 times full, the cross-sectional area of the pipe will be: A = Q / (0.7 * V) where V is the velocity of the flow in the pipe. We can assume a velocity of 0.75 m/s. A = 1.275 / (0.7 * 0.75) = 2.427 m2 Assuming a circular pipe, the diameter can be calculated as: d = sqrt(4A/π) = sqrt(42.427/π) = 1.77 m Therefore, a sewer pipe with a diameter of approximately 1.77 meters should be used to carry the peak flow rate of wastewater in the town.
A town has a population of 1,50,000 persons with per capita water supply
of 200 litres/day. Assuming 85% of water usage is appears to be sewage,
design a sewer running 0.7 times full at maximum discharge. Take a
constant value of N = 0.013 at all depths of flow. The sewer is to be laid at
a slope of 1 in 500. Take a peak factor of 3.
Anna .... a town has a population of 100000 persons with a per capita water supply of 200 Lpd . Design a sewer running 0.7 times full. take n=0.013 and slope 1 in 500 and peak factor of 3. assume 85% of water supply turns into sewer. .....intha sum ku answer venum na Anna .. pls Anna
Step 1: Determine the total water supply to the town per day:
Total water supply = population * per capita water supply
Total water supply = 100000 * 200 = 20,000,000 Lpd
Step 2: Determine the amount of water that will turn into sewer:
Water that turns into sewer = 85% of total water supply
Water that turns into sewer = 0.85 * 20,000,000 = 17,000,000 Lpd
Step 3: Determine the peak flow rate using the peak factor:
Peak flow rate = Peak factor * Average daily flow rate
Peak flow rate = 3 * (Water that turns into sewer / 24)
Peak flow rate = 3 * (17,000,000 / 24)
Peak flow rate = 1,275,000 Lph (liters per hour)
Step 4: Determine the diameter of the sewer pipe using the Manning's formula:
Q = (1.486/n) * A * R^(2/3) * S^(1/2)
where:
Q = flow rate (m3/s)
n = Manning's roughness coefficient
A = cross-sectional area of the pipe (m2)
R = hydraulic radius of the pipe (m)
S = slope of the pipe (m/m)
Since the sewer is running 0.7 times full, the cross-sectional area of the pipe will be:
A = Q / (0.7 * V)
where V is the velocity of the flow in the pipe. We can assume a velocity of 0.75 m/s.
A = 1.275 / (0.7 * 0.75) = 2.427 m2
Assuming a circular pipe, the diameter can be calculated as:
d = sqrt(4A/π) = sqrt(42.427/π) = 1.77 m
Therefore, a sewer pipe with a diameter of approximately 1.77 meters should be used to carry the peak flow rate of wastewater in the town.
Why are you multiplying with 1/36 in rational formula?
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