Chapter:
1. Define Sewage. Write down its components.
Sewage is defined as the type of waste water produced by the community of people mainly characterized by its physical condition, chemical and toxic constituents, and its bacteriologic status.
The sewage consists of the following two components:
- Sanitary Sewage or dry weather flow (DWF)
- Storm Sewage or wet weather flow (WWF)
The dry weather flow is the flow through the sewers that would normally be available during non-rainfall periods. The wet weather flow is the additional flow in the sewers that would occur in rainy season during rainfall.
2. Explain sources of Sanitary sewage or dry wear flow.
The dry weather flow DWF or sanitary sewage basically includes the following sources.
- Domestic sewage i.e, the sewage or wastewater derived from residential buildings, individual houses, from commercial institutional and similar public buildings such as offices, schools cinemas, hotel etc.
- Industrial sewage i.e, the sewage or wastewater obtained from manufacturing plants of industries.
- Ground water infiltrating into sewers through the pipe joints and other entry points.
- Illegal or unauthorized connection it includes entry or rain water and an unauthorized connection made by the people in the sewer.
- Private sources includes the wastewater generated from private sources like tube tube well, well etc.
3. Explain factors affecting Sanitary Sewage.
The quantity of dry weather flow or sanitary sewage is affected by the following factors.
- Rate of water supply
- Population growth
- Type of aria to be served ; and
- infiltration of ground water and ex-filtration of sewage ( addition and subtractions)
Rate of Water Supply:
Since a considerable part (generally 70 to 90 percent) of supplied water emerges as sanitary sewage, the quantity of domestic or sanitary sewage depends on the rate of water supply.
Population Growth:
Since the sewerage project is designed to serve not only the present population but also for the predicted future population within the design period also, the quantity of sanitary sewage increase with the increase in population.
Type of area to be served:
The quantity of sewage also depends upon the type of area to be served, which may be industrial or commercial or residential. The quantity of sewage produced by residential area depends upon the rate of water supply whereas the quantity of sewage produced by various industries depends on their industrial processes.
Infiltration and ex-filtration:
Infiltration is the entry of ground water into the sewers through leaking joints while ex-filtration is the leaking sewage from sewer line that percolates into the ground surrounding sewer. The infiltration increases the quantity of sanitary sewage while ex-filtration pollutes the ground water.
4. How would you determine quantity of sanitary sewage.
Generally, the rate of sanitary sewage should be equal to the quantity of water supplied but actually some additions due to private water supplies and infiltration of water and some subtractions due to leakages of water in the pipe lines and water consumption in drinking, cooking, sprinkling on roads, gardening of lawns and for various industrial purposes should be considered. Thus, the rate of sanitary sewage produced (in lpcd) may be assumed to be `70-90%` normally `80%` of the rate of water supplied.
Average quantity of sanitary sewage (DWF) = 70 to 90 % of (population `**` rate of w/s) )
This average quantity of sanitary sewage cannot be used for design purpose because practically average sewage never flows in the sewers but varies from time to time. The design of sewers should be done for the maximum possible flow.
Maximum or peak quantity of sanitary sewage is equal to the peak factor times the average flow.
The peak factor of 2 to 4 is generally adopted in Nepal.
The minimum rate of sewage flow is generally taken as the one third of the average flow.
5. Explain factors affecting source of storm sewage.
The source of storm sewage is precipitation that reaches to the sewer depend upon the following factors:
- Area of the catchment
- Characteristics of catchment area
- Slope, size, imperviousness and shape of the catchment
- Initial state of the catchment with respect to wetness.
- Nature and number of ditches present in the catchment area.
- Intensity and duration of rainfall.
- Atmospheric condition.
- Obstruction in the flow of water.
- Time required for flow to reach the sewer.
The quantity of storm water that reaches to the sewer line depends upon the following factors:
- Area to be drained off
- Topography of the area and nature of the ground surface.
- Intensity of rainfall.
                 6. A certain district of a city has a projected population of 50,000 residing over an area of 40 hectares. Find  design discharge for  sewer line, for  following data : 
- Rate of water supply = 200 lpcd 
- Average impermeability coefficient for  entire area =0.3 
- Time of concentration = 50 minutes 
The sewer line is to be designed for a flow equivalent to  wet wear flow (WWF) plus twice  dry wear flow (DWF).Use British ministry of health formula. Assume that `70%` of water supply reaches in  sewer as wastewater.(Ans: `0.648 m^3/s`) 
            
            
           
                
            
            
                   
        
Solution:
Given,
Population = 50,000
Area = 40 ha.
Peak factor = 2
Rate of water supply = 200 lpcd
Avg. impermeability factor = 0.3
Time of concentration = 50 min
Now,
The discharge of DWF is,
`Q_(DWF)=(2**0.7**50000**200)/(1000**86400)`
`=0.1620 m^2/s`
Again,
Intensity of rainfall is given by,
`i=1020/(t+20)=1020/70=14.5714` mm/hr
Now, the storm water flow rate is,
`Q_(WWF)=(CiA)/360`
`=(0.3**14.5714**40)/360`
`=0.4857 m^3/s`
Thus, the combined design discharge is,
`Q=Q_(DWF)+Q_(WWF)`
`=0.1620 + 0.4857`
`=0.6477 m^3/s`
                 7. Assuming  surface on which rain falls in a thickly build up residential district as follows : 
- `40 %` area consists of roofs and pavements (`C_1 = 0.80`) 
- `60 %` of area consists of lawns and gardens (`C_2 = 0.2`)  
Calculate  coefficient of runoff. If  area of  district is 2 hectares and  maximum intensity of rainfall is taken as 50 mm/hr, what is  runoff of  district?  (Ans: `0.44, 0.122` `m^3`/sec)  
            
            
           
                
            
            
                   
        
Solution:Average runoff coefficient is,
`C=(C_1A_1+C_2A_2)/(A_1+A_2)`
`=(0.4**0.8+0.2**0.6)/1`
`=0.44`
Again,
Runoff is given by,
`Q_(WWF)=(CiA)/360`
`=(0.44**50**2)/360`
`=0.122\ m^3/s`
                 8. The surface of  town on which  rainfalls is classified as follows : 
| Types of surface | % area | Runoff coefficient | 
| Roof | 25 | 0.8 | 
| Pavements and yards | 25 | 0.85 | 
| Macadamized roads | 15 | 0.32 | 
| Gravel roads | 10 | 0.2 | 
| Unpaved streets | 20 | 0.15 | 
| Lawns and gardens | 5 | 0.2 | 
What is runoff of catchment if,
- Total area of catchment is 12 hectares 
- Time of concentration for  area is 15 minutes.
Solution:
Average impermeability factor is,
`C=(sum(C_1A_1))/A`
`=(0.8**0.25+0.85**0.25+0.32**0.15+0.2**0.1+0.15**0.2+0.2**0.05)`
`=0.5205`
Now, the intensity of rainfall is given by,
`I=760/(15+10)`
`=30.4` mm/hr
Thus, the runoff of the catchment is given by,
`Q=(CIA)/360`
`=(0.5205**30.4**12)/360`
`=0.527\ m^3/s`
                 9. The catchment area of  city is 500 hectares. Assuming that  surface on which rainfalls is classified as follows :  : 
| S.NO | Types of surface | % area | C | 
| 1 | Forest and wooden area | 10 | 0.15 | 
| 2 | open ground + unpaved street | 10 | 0.20 | 
| 3 | parks + lawns + gardens | 15 | 0.15 | 
| 4 | Gravel roads | 20 | 0.25 | 
| 5 | Asphalt pavements | 20 | 0.85 | 
| 6 | Water tight roof surfaces | 25 | 0.9 | 
Calculate run off coefficient of area. Also calculate storm water, if area to be served is 500 hectares, time of entry is 15 minutes and time of flow is 10 minutes.
[Ans: `0.5025`, `15.32 m^3/s`]
Solution:
Average runoff coefficient is given by,
`C=(C_1A_1+C_2A_2+C_3A_3+C_4A_4+C_5A_5+C_6A_6)/(A_1+A_2+A_3+A_4+A_5+A_6)`
`=(0.15**0.1+0.20**0.1+0.15**0.15+0.25**0.20+0.85**0.20+0.9**0.25)/1`
`=0.5025`
Now,
Time of concentration is,
`t=15+10=25\ mi n`
Now,
Storm water flow rate is,
`Q=(CiA)/360`
where,
`i=1020/(25+20)=22.67` mm/hr
Thus,
`Q=(0.5025**22.67**500)/360`
`=15.82 m^3/s`
                 10. Determine  quantity of sewage in `m^3`/sec for  combined system from  following data : 
- `50 %` of  area is roofs and pavements (`C_1 = 0.8`) 
- `50 %` of  area is vacant plots (`C_2 =0.1`) 
 
- Area of  community ` 5` hectares 
 
- Population density = `700` persons/hectare 
 
- Water supply rate = `150` lpcd. 
 
- Peak factor = `3` 
 
- Time of concentration =`10` minutes. 
 
           
                
            
            
                   
        
Solution:
Average impermeability factor is,
`C=(C_1**A_1+C_2**A_2)/(A_1+A_2)`
`=(0.8**0.5+0.1**0.5)/1`
`=0.45`
Now, the sanitary discharge is given by,
`Q_(DWF)=(3**700*5**150**0.80)/(1000**86400)`
`=0.0146\ m^3/s`
Again,
The intensity of rainfall is,
`i=760/(10+10)=38` mm/hr
Now, the storm discharge is,
`Q_(WWF)=(CiA)/360`
`=(0.45**38**5)/360`
`=0.2375\ m^3/s`
Now, the quantity of sewage for the combined system is given by,
`Q=Q_(DWF)+Q_(WWF)`
`=0.0146+0.2375`
`=0.2521\ m^3/s`
All Chapters
Introduction
Quantity of Waste Water
Design of sewers Tutorial
Treatment of WasteWater
Topics
Sources of sanitary sewages
factors affecting sanitory sewage
determination of quantity of sanitary sewages
variation in the quantity of sanitary sewage
determination of quantity of sanitary sewages
factors affecting storm sewages
time of concentration
time area graph and many more illustrative examples
Define Sewage. Write down its components.
Explain sources of Sanitary sewage or dry wear flow.
Explain factors affecting Sanitary Sewage.
How would you determine quantity of sanitary sewage.
Explain factors affecting source of storm sewage.
A certain district of a city has a projected population of 50,000 residing over an area of 40 hectares. Find design discharge for sewer line, for following data :
- Rate of water supply = 200 lpcd
- Average impermeability coefficient for entire area =0.3
- Time of concentration = 50 minutes
The sewer line is to be designed for a flow equivalent to wet wear flow (WWF) plus twice dry wear flow (DWF).Use British ministry of health formula. Assume that `70%` of water supply reaches in sewer as wastewater.(Ans: `0.648 m^3/s`)
Assuming surface on which rain falls in a thickly build up residential district as follows :
- `40 %` area consists of roofs and pavements (`C_1 = 0.80`)
- `60 %` of area consists of lawns and gardens (`C_2 = 0.2`)
Calculate coefficient of runoff. If area of district is 2 hectares and maximum intensity of rainfall is taken as 50 mm/hr, what is runoff of district? (Ans: `0.44, 0.122` `m^3`/sec)
The surface of town on which rainfalls is classified as follows :
| Types of surface | % area | Runoff coefficient | 
| Roof | 25 | 0.8 | 
| Pavements and yards | 25 | 0.85 | 
| Macadamized roads | 15 | 0.32 | 
| Gravel roads | 10 | 0.2 | 
| Unpaved streets | 20 | 0.15 | 
| Lawns and gardens | 5 | 0.2 | 
What is runoff of catchment if,
- Total area of catchment is 12 hectares 
- Time of concentration for  area is 15 minutes.
The catchment area of city is 500 hectares. Assuming that surface on which rainfalls is classified as follows : :
| S.NO | Types of surface | % area | C | 
| 1 | Forest and wooden area | 10 | 0.15 | 
| 2 | open ground + unpaved street | 10 | 0.20 | 
| 3 | parks + lawns + gardens | 15 | 0.15 | 
| 4 | Gravel roads | 20 | 0.25 | 
| 5 | Asphalt pavements | 20 | 0.85 | 
| 6 | Water tight roof surfaces | 25 | 0.9 | 
Calculate run off coefficient of area. Also calculate storm water, if area to be served is 500 hectares, time of entry is 15 minutes and time of flow is 10 minutes.
[Ans: `0.5025`, `15.32 m^3/s`]
Determine quantity of sewage in `m^3`/sec for combined system from following data :
- `50 %` of area is roofs and pavements (`C_1 = 0.8`)
- `50 %` of  area is vacant plots (`C_2 =0.1`) 
- Area of  community ` 5` hectares 
- Population density = `700` persons/hectare 
- Water supply rate = `150` lpcd. 
- Peak factor = `3` 
- Time of concentration =`10` minutes. 
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