design of sewage level.pdf

8/13/2019 Design of Sewage level.pdf

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Quiz No. 3Wastewater Collection

SOLUTION

1. The present population of a certain colony in a city is 10000. The present (Year 2003) per capita average water supply to the colony is 180 lpcd (liters per capita per day). The population of the colony is expected to be 45000 at the end of the design period of 20 years (Year 2023). Per capita average water

supply at the end of the design period is expected to be 240 lpcd. The entire wastewater generated in the colony is collected and discharged through aconnecting sewer to one of the trunk sewers of the city. Determine an acceptable slope and diameter of this sewer. Make appropriate assumptionswherever required and state them explicitly.

d/D v/V q/Q

Mannings Formula

1.0 1.000 1.000

V = Velocity in m/sR = Hydraulic Radius, mS = Slope in m/1000m Assume, n = 0.013Assume circular sewer

0.9 1.124 1.0660.8 1.140 0.968

0.7 1.120 0.8380.6 1.072 0.6710.5 1.000 0.5000.4 0.902 0.3370.3 0.776 0.1960.2 0.615 0.0880.1 0.401 0.021

Solution: Present Population (in Year 2003) = 10,000Present Average Water Supply (in Year 2003) = 10,000.(180) = 1.8 million liters per dayAssuming that 80 percent water becomes wastewater,Present Average Wastewater Generation = 1.8.(0.8) = 1.44 million liters per dayAssuming a Peaking Factor of 3,Present Peak Wastewater Generation = 3.(1.44) = 4.32 million liters per day

Present Peak Wastewater Generation =

Design Population (in Year 2023) = 45,000


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Design Average Water Supply (in Year 2023) = 45,000.(240) = 10.8 million liters/dayAssuming 80 percent water becomes wastewater,Design Average Wastewater Generation = 10.8.(0.8) = 8.64 million liters per dayAssuming a Peaking Factor of 3,Design Peak Wastewater Generation = 3.(8.64) = 25.92 million liters per day

Design Peak Wastewater Generation =For a pipe flowing full:

At q = 0.3 m 3/s, the pipe is flowing 0.8 full

When, . Therefore,

Assuming, n = 0.013,Solving, D = 0.628 m, say, 0.700 m or 700 mm

Corresponding to D = 700 mm,

Therefore, the new value of

Corresponding , and

;Therefore, v Design = Velocity at design peak flow = (1.076).(1.1) = 1.184 m/s(which is within the recommended range of 0.8 m/s to 3 m/s)

At present peak flow, ; correspondingTherefore, v

pres ent = velocity at present peak flow = (0.65).(1.076) = 0.699 m/s

(which is greater than the recommended minimum value of 0.6 m/s)


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Hence use 700 mm circular concrete pipe at a slope 0f 2 m per 1000 m for accommodating the above flow.

2. Consider the sewer system shown in the figure below. A main sewer (MS5) with manholes, M5.4 (starting manhole) to M 5.1 discharges into manholeM5 of the trunk sewer. Branch sewers discharge into manhole M5.1 M5.4 of the main sewer (MS5) as shown in the figure. Invert levels of the branchsewers discharging into manhole M5.4 M5.1 are 97.223, 98.775, 98.311, and 96.334 m respectively. Invert level of the trunk sewer leaving manholeM5 is 93.700 m. Based on this information explain how values in the last three columns of the design table for main sewer (MS5) have been calculated. All

branch sewers have 400 mm diameter. The ground level in the area under question is between 101 and 102 m.

Table 1. Tabular Calculations for Design of main Sewer M51 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Manhole

Length,m

PeakFlow,(2003)m3/s

Peak Flow,(2023)m3/s

Diameter,mm

SlopeDischarge,

(2023) Litre/s

Velocity,(2023)

m/s

PresentPeak Flow,Year 2020Litre/s

Self-

ClensingVelocity(2000)

m/s

TotalFall,m

Invert

Elevation,m

From To Full Actual Full Actual Upper Lower

M5.4 M5.3 88 0.05 0.15 500 0.002 0.169 0.15 0.860 0.972 0.05 0.688 0.176 96.773 96.597M5.3 M5.2 77 0.09 0.28 700 0.002 0.414 0.28 1.076 1.152 0.09 0.829 0.154 96.397 96.243M5.2 M5.1 101 0.12 0.33 700 0.002 0.414 0.33 1.076 1.184 0.12 0.861 0.202 96.213 96.011M5.1 M5 122 0.15 0.45 800 0.002 0.591 0.45 1.176 1.235 0.15 0.929 0.244 95.734 95.490

Solution:Elevation of the centerline of branch sewer at manhole M5.4 = 97.223 + 0.2 = 97.423Elevation of centerline of the upper end of the main sewer from M5.4 to M5.3

= 97.423 0.4 = 97.023Invert level of the upper end of the main sewer for M5.4 to M5.3


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= 97.023 0.25 = 96.773Fall from M5.4 to M5.3 = 88.(0.002) = 0.176Invert level of the lower end of the main sewer for M5.4 to M5.3

= 96.773 0.176 = 96.597 Invert level of the upper end of the main sewer for M5.3 to M5.2

= 96.597 0.2 = 96.397

Considering the above invert level, minimum invert level of the branch sewer at M5.3 should be 96.397 + 0.35 + 0.4 0.2 = 96.957, while the actual invert level is98.775.

Hence okay.Fall from M5.3 to M5.2 = 77.(0.002) = 0.154Invert level of the lower end of the main sewer for M5.3 to M5.2

= 96.397 0.154 = 96.243 Invert level of the upper end of the main sewer for M5.2 to M5.1

= 96.243 0.03 = 96.213

Considering the above invert level, minimum invert level of the branch sewer at M5.2 should be 96.213 + 0.35 + 0.4 0.2 = 96.763, while the actual invert level is98.311.Hence okay.

Fall from M5.2 to M5.1 = 101.(0.002) = 0.202Invert level of the lower end of the main sewer for M5.2 to M5.1

= 96.213 0.202 = 96.011Invert level of the upper end of the main sewer for M5.1 to M5

= 96.011 0.1 = 95.911 Considering the above invert level, minimum invert level of the branch sewer at M5.1 should be 95.911 + 0.4 + 0.4 0.2 = 96.511, while the actual invert level is96.334.

Hence not okay.Hence, actual invert level of the upper end of the main sewer for M5.1 to M5

= 96.334 + 0.2 0.4 0.4 = 95.734Fall from M5.1 to M5 = 122.(0.002) = 0.244Invert level of the lower end of the main sewer for M5.1 to M5

= 95.734 0.244 = 95.490

design of sewage level.pdf

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