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. 

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. 

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`

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`

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`

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`

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.

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.

Solution:

Assuming 80% of supplied water contributes to sewage and peck factor to be 2.5.

The sanitary sewage discharge is,

`Q_(DWF)=(0.8**2.5**225**50000)/(1000**86400)`

`=0.2604\ m^3/s`

Here,

`t=10+30=40 min`

The intensity of rainfall is, `i=1020/(40+20)=17` mm/hr

Now, the storm sewage discharge is,

`Q_(WWF)=(0.3**17**20)/360`

`=0.233\ m^3/s`

Now, the design discharge for the combined system is,

`Q=0.233+0.2604`

`=0.4934\ m^3/s`

Solution:

Assuming 80% of supplied water contributes to sewage and peck factor to be 2.5.

The sanitary sewage discharge is,

`Q_(DWF)=(0.8**2.5**225**50000)/(1000**86400)`

`=0.2604\ m^3/s`

Here,

`t=10+30=40 min`

The intensity of rainfall is, `i=1020/(40+20)=17` mm/hr

Now, the storm sewage discharge is,

`Q_(WWF)=(0.3**17**20)/360`

`=0.233\ m^3/s`

Now, the design discharge for the combined system is,

`Q=0.233+0.2604`

`=0.4934\ m^3/s`

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`

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`

Solution:

Average Impermeability factor is,

`C=(0.85**0.48+0.2**0.22+0.1**0.3)`

`=0.482`

Now, the runoff from the area is,

`Q=(CiA)/360`

`=(0.482**55**6)/360`

`=0.442 \ m^3/s`

Solution:

Average Impermeability factor is,

`C=(0.85**0.48+0.2**0.22+0.1**0.3)`

`=0.482`

Now, the runoff from the area is,

`Q=(CiA)/360`

`=(0.482**55**6)/360`

`=0.442 \ m^3/s`

The source of storm sewage is precipitation that reaches to the sewer depend upon the following factors:

1. Area of the catchment
2. Characteristics of catchment area
3. Slope, size, imperviousness and shape of the catchment
4. Initial state of the catchment with respect to wetness.
5. Nature and number of ditches present in the catchment area.
6. Intensity and duration of rainfall.
7. Atmospheric condition.
8. Obstruction in the flow of water.
9. 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.

The source of storm sewage is precipitation that reaches to the sewer depend upon the following factors:

1. Area of the catchment
2. Characteristics of catchment area
3. Slope, size, imperviousness and shape of the catchment
4. Initial state of the catchment with respect to wetness.
5. Nature and number of ditches present in the catchment area.
6. Intensity and duration of rainfall.
7. Atmospheric condition.
8. Obstruction in the flow of water.
9. 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.

Solution:

For impervious area, `C=1`

For area X,

`t=20 min`

`t_c=30 min`

Since `t

The contribution area is,

`=sum A**t/t_c`

`=20**20/30`

`=13.33 ha`

Now,

`i=760/(30)`

`=25.33` mm/hr

Thus, `Q=(CiA)/360`

`=(1**13.33**25.33)/360`

`=0.939 m^3/s`

For area Y,

`t=30 min`

`t_c=30 min`

In this case, contributing area is equal to entire area i.e, 20 ha.

Thus, `i=1020/(50)=20.4` mm/hr

Thus, `Q=(1**20.4**20)/360=1.134 m^3/s`

Again for area Z,

`t=35 min`

`t_c=30 min`

since `t>t_c`,  contributing area is equal to entire area i.e, 20 ha.

Thus, 

`i=1020/(35+20)`

`=18.55` mm/hr

Now, 

`Q=(1**18.55*20)/360`

`=1.03\ m^3/s`

Solution:

For impervious area, `C=1`

For area X,

`t=20 min`

`t_c=30 min`

Since `t

The contribution area is,

`=sum A**t/t_c`

`=20**20/30`

`=13.33 ha`

Now,

`i=760/(30)`

`=25.33` mm/hr

Thus, `Q=(CiA)/360`

`=(1**13.33**25.33)/360`

`=0.939 m^3/s`

For area Y,

`t=30 min`

`t_c=30 min`

In this case, contributing area is equal to entire area i.e, 20 ha.

Thus, `i=1020/(50)=20.4` mm/hr

Thus, `Q=(1**20.4**20)/360=1.134 m^3/s`

Again for area Z,

`t=35 min`

`t_c=30 min`

since `t>t_c`,  contributing area is equal to entire area i.e, 20 ha.

Thus, 

`i=1020/(35+20)`

`=18.55` mm/hr

Now, 

`Q=(1**18.55*20)/360`

`=1.03\ m^3/s`

Solution:

Average impermeability factor is,

`C=(C_1A1+C_2A_2)/A`

`=(0.40**0.85+0.40**0.2+0.2**0.1)`

`=0.44`

Now,

The maximum probable runoff is given by,

`Q=(CIA)/360`

`=(0.44**50**5)/360`

`=0.3056\ m^3/s`

Solution:

Average impermeability factor is,

`C=(C_1A1+C_2A_2)/A`

`=(0.40**0.85+0.40**0.2+0.2**0.1)`

`=0.44`

Now,

The maximum probable runoff is given by,

`Q=(CIA)/360`

`=(0.44**50**5)/360`

`=0.3056\ m^3/s`