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A population 50,000 resides in a town covering an area of 75 hectares. Calculate design discharge for combined system of sewers. Given,
- Coefficient of runoff =0.7
- Time of entry = 5 minutes
- Time of flow = 20 minutes
- Water supply rate = 120 lpcd. [PU-2008]
A population 50,000 resides in a town covering an area of 75 hectares. Calculate design discharge for combined system of sewers. Given,
- Coefficient of runoff =0.7
- Time of entry = 5 minutes
- Time of flow = 20 minutes
- Water supply rate = 120 lpcd. [PU-2008]
Solution:
Here, time of concentration is, `t_c=20+5=25` min
Assuming `80%` of supplied water contributes to sewage and peck factor to be `2.5`.
Now, the sanitary discharge is,
`Q_(DWF)=(0.8**2.5**120**50000)/(1000**86400)`
`=0.139\ m^3/s`
Similarly,
`i=(1020)/(25+20)=22.67` mm/hr
Thus, the storm discharge is,
`Q_(WWF)=(0.7**22.67**75)/360`
`=3.306\ m^3/s`
Now, the design discharge for the combined system is,
`Q=3.306+0.139`
`=3.445\ m^3/s`
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Quantity of Waste Water
Design of sewers Tutorial
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Similar Question
Assuming the surface on which rainfalls is thickly built residential district as follows :
| % of total surface | Types of surface | Impermeability |
| 48 | Roofs and pavements | 0.85 |
| 22 | Lawns and gardens | 0.2 |
| 30 | Ground surface | 0.1 |
Calculate the average impermeability factor. If the total area draining to the sewer is `6` hectares and maximum intensity of rainfall is `55` mm/hr. Estimate the runoff from the area.[TU 2006]
A population 50,000 resides in a town covering an area of 75 hectares. Calculate design discharge for combined system of sewers. Given,
- Coefficient of runoff =0.7
- Time of entry = 5 minutes
- Time of flow = 20 minutes
- Water supply rate = 120 lpcd. [PU-2008]
A certain district of a city with a population of `50,000` over an area of `20` hectares is given Find the design discharge of the combined sewer line for the following data : [TU 2004]
- Rate of water supply =`225` lpcd
- Average impermeability coefficient for the entire district=`0.3`
- Time of entry =`10` minutes
- Time of flow = `30` minutes.
