Janitariengineering

ws
Source Raw water Intake
ofwater usage

Disposalofsewage a treatment
pipes ofsewage

Sanitary engineering is a branch public health engineering that deals with

of
The preservation and maintenance
of public health of individual community
bypreventing
communicable diseases

It is foul in nature it is a part of stormwater from streets and terraces openyards

as well which also carries A good lot organic matter which is washed into it
of
Sewage needs to be removed and disposed
off for the following reasons

1 It causes insanitary conditions

2 It may breed flies bacteria mosquitos causing health hazards

34 It may percolate and contaminate Subsurface water

4 Waste from Industries is

very
complex in nature and may contain toxic matter harmful

to everyone

Domestic Waste liquid waste collected from residences buildings and institutions

Industrial waste liquid waste from industries

storm water Run off from streets and open yards

 Sewage Combination domeshe Industrial and storm water
of
Water
Sullage from kitchen bath sinks or similar application from buildings

which does not contain human or animal excreta

Refuse Dry Rubbish from streets or house sweeping

Garbage semisolid and solid waste from food products such as vegetables

waste meats

Sewerage A system
of pipes laid for carrying sewage is called sewerage

Conservancy System
Methods water carrier
of Disposal system

There are 2 methods

of disposal
of sewage

1 Conservancy System

2 water carrier system

Conservancy system

1. Conservancy is an old system and dry system

2. Night soil, garbage and sullage collected separately in vessels and pits

3. Removed periodically once in 24 hrs

4. Night soil is burried in trenches and used as manure after 2-3 yrs

5. Garbage - twice a day, and dumped in open area or in sanitary field

6. Sullage or storm water collected separately and disposed in closed or open gutters
1. Collection, conveyance and disposal is carried out with the help of water.

2. W/c’s are used which are flushed that leads to design and maintained sewers.

3. Kitchens and baths are led to sewers

4. Sewers are underground closed pipes, those are led to suitable longitudinal slope so that the flow

takes place by gravity.

5. Sewage so collected is disposed off after the treatment.

6. Garbage is disposed similar to the conservancy system.

Classification of water carriage system

1. Separate -a) Domestic waste water, industrial waste water, drainage from slaughter house, stable

and laundries b) for carrying rainwater and overflow from public baths.

2. Combined - One sewer carries both rain water and foul sewage which is to be treated before

disposal.

3. Partially separate - One sewer only but stormwater to be collected in open drains after overflow.

Waste water flow

A (Dry Flow DWF)

1. Flow due to domestic industrial and waste water from public facilities.

2. Groundwater infiltration

3. Factors affecting DWF:

• Rate of water supply

• Population growth
• Type of area served

• Infiltration of ground water

4. The discharge can be calculated by

i Babbit's formula Omax Soav

po2

2 Harman's formula Omax a

pos

B (Storm water flows WWF)

1. Amount of water collected from ground surface, pavements, house roofs, also called as runoff or

stormwater.

2. Factors affecting WWF O

5
• Catchment area

• Ground slope C Runoff coefficient

• Permeability A catchment area

• Rainfall duration i mm hr
Rainfall intensity
• Climatic condition

28 Ai Ri

Discharge in Sewer line

Q1. A Certain district of a city has projected population of 50,000 residing over an area of 40 hectors.

Find the design discharge for the sewer line, for the following data
Rate of water supply 200 lpcd , avg impermeability constant 0.3, time of concentration 50 mins

The sewer is to be designed for the flow equivalent to the wet weather flow and twice the dry weather

flow.

Assume 75% of water supply reaches the waste water.

Lol 1 DWT 75
of total discharge

i Sewage flow DWF 0 75 200 501000

75 106 eld 24 60 60

86.8 lls

WWF A 03 40
ii
5
iii t somin Ri i a a 40 5 20
224

Ri i 25.4 40 14.51
50 20

0.3 14 40
6
C constant

0.48 m s i mm hr

iv d sewer A hectares

2DWF WWF 0 653 m S Q m s

Q2. The following data is available for various types of areas and corresponding impermeability

factors. Determine the average coefficient of runoff. If the total area of the district is 20 hectares,

determine the maximum storm water flow for the rainfall intensity of 50mm/hr.

Sol

Type Area Impermeability factor c

1 Roofs 15 0.9

2 Pavements 20 08

3 Lawns 40 0.15

4 unpaved Area 1St 0.2

5 woods 10 0.05

i Can CA CzAz GAs CUAU CSAs A

A

0.15 0.9 020 08 040 015 Co Is to 2 0 1 0 05

0 39
Cang

a Cang 3

039 20 1.083 m s
 360

Q3. The drainage area of 1 sector of a town is 20 hectares. The classification of the surface is given

as below. Use time of concentration as 30 mins, find maximum runoff, use following formula to

calculate. i 9

of area Type of surface Runoff coefficient

25 Hard pavement 0.85

25 Roof Surface 0.8

IS unpaved street 0.3

25 Gardens 0.15

10 woods 01

i Can CA C2Az GAs CUAU CSAs A

A

0.25 0.85 0 25 408 0.15 0.3

0.25 0 15 01 0.1

0.505

i i 10 900
30 60

ii D 0 505 10 20 0 28
Gf
m s
360
Q4. For a small town having projected population 30,000 residing over an area of 20 hectares. Find

the design discharge of the combined sewer system. If the rate of water supply is 150 lpcd, time of

concentration 30 mins, runoff coefficient 0.4, and assumed 75% of water reaches waste water.

Lol A 20 p 30,000

t 30 C O4

DWF 0 75 150 30,000

3375000 l d 24 60 60

39.062 lls 0.0390 m s

m s

WWF I 04 2032 20 0.451

9
m s
360

I 25.4 a 9 40 b 20
b

20 32 25 4 40
30 20
 Design Sewers
Hydraulic
of

Formulae

i cheay's formula
V CES a CTmi

27 Mannings formula
1243g

3 Kutter's formula
for C

C 23 0 0 55 R Hydraulic radius

s Bedstop

I 23 0.0 55 R
I fetedperimeter

Q. Determine The velocity of flow in the circular sewer of dia = 150cm, laid on a slope of 1:750

while flowing full. Consider n=0.013, consider chezy’s and mannings formula.

5 D 150cm 0013 n
agN manning
 c 23 23
0.095s 9.39 8 3

7
I 23 0.0 5 1 423 0 013
0.95 5037

101 08 66 94
1 51

A D2 P TD V CARS

R Fp 0.37m 1122135112

Q. Determine the velocity of flow in Sewer running half full. The sewer is laid at 1 in 550 slope.

The dia of sewer is 150cm also determine the discharge following through it. Assume n=0.012

in mannings formula.

27 Mannings formula
1243g

0.75 2 8m s
o

dia 150cm 1 Sm R
slope
wetted
0.75
Q. Design a combined sewer section, Area to be served 150 hectares, population 50,000, max.

Permissible velocity 3.2, duration is 25mins, rate of water supply 270 lpcd, impermeability

factor 0.45, assume peak factor as 1.5.

1 DWF 15 270 50,000 20250000

24 60 60 lls
27 WWF 1000 m S

0.234m S

O.US 22.57 150 WWF 75 Of DWF 0 176

360
i 25.4 a
ttb
4.23 m S

22.57 mm hr

3 D 2 5
DWF WWF 0.41 m

4 d AV 0.41 A 3.2

5 A

D 0.406M
Q. A town has a population of 1L with per capita supply of 200 lpcd. Design the sewers running

0.7 times the maximum discharge, take n= 0.01. S= 1/500, peak factor = 3, consider 80% of the

total supply to be converted to sewage.

Sol i DWF population lped peakfactor

60 106 lped 0 694 m 2

To
D
2 I 08 DWF

0 556
x Id

3 1 050 0.7 1 650

0.7 12 1 050

1.4 1 Cosd

1.4 1 0050
0.4 Cosa

a A
32 20 It 0 113.570
113.574 1 98
p DO R AIP

A 2 1.98 sin 2413.57 A 05802

5 0 X 2 1.98 sin 2413.57

0.556 05802 0 140

of X
got

0.917 m
 Types of Sewers

1. Main

It receives sewage from from many tributaries

2. Branch

It receives sewage from small areas.

3. Lateral

It receives sewage from houses

4. Separate

From household and industries

5. House

Pipe carrying sewage from building to the street sewer

6. Depressed

Section of sewer constructed Lower than the adjacent section to pass beneath the obstacle or

obstruction.

7. Intercepting

Laid transversely to the main sewer to intercept the DWF.

8. Combined sewer

Carrying both sewage and storm water

9. Storm water sewer

Carrying only storm water

Materials for Sewer

1. The material for sewer needs to be tough, sustainable, and non-corroding.

2. Sewers can be of asbestos, RCC, vitrified clay, brick, CI (Cast iron) Pipes, steel pipes, plastic

pipes and DI (Ductile iron) pipes.

The forces acting on Sewers

1. Internal pressure

2. Temperature stresses

3. Forces due to external loads

Backfill load Superimposed load Flexural stresses

Internal pressure is
Hoop stress 2T
g thickness

Temperature stress
in temp
he coeffof
change

Modulus
of thermal
elasticity expansion

Bd t.GE
3 External 1
Backfill load

Fᵗʰ
We Cr BE soil

vertical load per

ftp.ient units
unitlength acting
due to backfill
BC
pipeshouldbe able to sustain
backfillwed

Q. A concrete pipe of 1.5 m diameter is installed in a trench of width 2.75m and depth 4.5m. The

thickness of pipe is 80mm. Determine the load on the pipe due to soil fill which consists of clay for

which unit weight is 1900N/m^3. value for Ku’ = 0.13.

Sol i Bc D at

1s 2 80 10 3

1 66 m

2 1.03 2BC
Trench condition

ku H bd 137 1 03
37 c e 1 26 0 903
2 Ksu 2 0 13

Ku friction factor

4 WC Cr Bd
0 903 1900 2 75

12983 78 Mm

Q. A concrete pipe of diameter 900mm, is installed in a trench of width 2.5m and depth 6.5m, the

thickness of pipe is 50mm, ku’ = 0.15, and y=1920N/m^3. Determine the load on the pipe.
 1 Bc D 2t

900 10 3 2 50 10 3

2 5.5 2.2 in b w IBC BBC

25
trench condition

2 5 22
ku H bd
34 C e g g
2 Ksu 2 0 15

4 WC Cr Bd

1 61 1920 2 52
19325 94 NIM

Sewer Oppurtenances

Devices used at suitable intervals for inspection, maintenance and cleaning

1. Inlet

2. Catch basin

3. Clean outs

4. Manholes

5. Deep manholes
6. Drop manholes

7. Lamp holes

8. Flushing tanks

9. Grease and oil tanks

10. Inverted siphon

11. Stream regulator

1. Inlet

It is to admit the storm water and surface wash to convey to the storm water drain. Not necessary for

separate type of sewers. It so placed that no cross roads are flooded.

Foot path
road
9 1
I Inlet

I
concrete wall

To stormwater drain

2. Catch basin

Special type of inlet in which basin in provided which allows grit, sand and debris flowing with storm

water to settle out. Hood is an outlet provided for removal for odour. Settled matter is removed

periodically.
Road

1 1 1 19 I
 hoof
Sand got

3. Clean outs

Inclined pipe for clearing out the lateral sewers, the cover at the top is removed and water is forced

to clean out the pipe. For removing large obstacles flexible rods are inserted.

mm
clearout pipe

sewer I

4. And 5. Manholes

Masonry or RCC structures constructed on the sewer for providing access to the sewer line for

inspection. Also helps in joining the sewers, changing the direction or alignments. Provided at every

junction, bend, change of gradient or change of diameter. 3 types of manholes are:

1. Shallow (0.75-9) 2. Normal (1.5m) 3. Deep (more than 1.5m)

Foreseen

2 steps
 I 1 slab
FRC
1
I
L Brasher
I main sewer
I
4usection

I Maingewer

6. Drop manhole

Special type of manhole to provide the connection between high level branch sewer to low level main

sewer. Can be installed when branch sewer is 0.5-0.6m above main sewer.

iii tier
steps
g

am
Inspection
or Branch sewer

vertical
pipe
Mider

7. Lamp holes

Special opening on the ground for the sewer line for lowering a lamp inside a sewer. Located where

there is a change in gradient, direction and where the construction of manhole is difficult. Use for

inspection and flushing.

 8. Flushing tank

Device that holds the water and then throws it into the sewers for flushing purpose. Can be operated

manually or automatically. When the sewers are led on the flat terrain then it will not produce the self

cleansing velocity. In such situations flushing tank are used or are used at the dead ends.

9. Grease and oil chambers

Special built chambers to exclude grease or oil from sewage before they enter the sewer line.

Located near the garages, hotel kitchens, or repair workshops. Forms sludge layers on inside of

pipes and is explosive. If grease or oil enter the sewage it may stick to the walls of pipe and will

become hard in the presence of cold water and obstruct the flow. May cause explosion and

treatment will be difficult. Prevent oxygen to penetrate so the aerobic bacteria will not survive and

decomposition will be difficult.

Hever

areasp Grees J
outlet

10. Inverted siphon

When the sewer line lies below the hydraulic gradient line it is called as inverted siphon or depressed
sewer. It is to carry sewer under the obstructions like roadways, railways or stream rivers. The self

cleansing velocity in depressed sewers has to be 1m/s min.

Depressed sewer Inverted siphon

Q. A concrete pipe of 1.5 m diameter is installed in a trench of width 2.75m and depth 4.5m. The

thickness of pipe is 80mm. Determine the load on the pipe due to soil fill which consists of clay for

which unit weight is 1900N/m^3. value for Ku’ = 0.13.

Sol i Bc D at

1s 2 80 10 3

1 66 m

2 1.03 2BC
Trench condition

Ku H bd 0 13 1 03
37 c 0 903
2 Ksu 2 0 13

Ku friction factor
 4 WC Cr Bd
0 903 1900 2 75

12983 78 Mm

Q. A concrete pipe of diameter 900mm, is installed in a trench of width 2.5m and depth 6.5m, the

thickness of pipe is 50mm, ku’ = 0.15, and y=1920N/m^3. Determine the load on the pipe.

1 Bc D at

900 10 3 2 50 10 3

2 5.5 2.2 in b w IBC BBC

25
trench condition

2 s 22
bd

if g g
34 C UGH 1 e
2 0 is

4 WC Cr Bd

1 61 1920 2 52
19325 94 NIM
 Inverted siphon numerical

I A sewer line an
carrying average discharge
of 200 lls had to cross a stream Design
a 3 barrel siphon for this purpose if
the length of
siphon measured army the

centre line is 90m The invert level of

the inlet and outlet ends of sewer

are 152.5 151 78 m The max

min Hours are 2501 40 of

avg
Assume that minor losses are 0.07 m

self is 1m n 0.013
cleansing velocity s

1 Imax 200 250 0.5 m s

and Imax Vs A
0.5 I xd
I 0 79 m 0.8m

i 0.99 m s
Vact
 and Vact R 8112

S 1 in 706.12

90
hf S L
2

hf 0 13 m

i
he he minor loceel
0 13 0.07
0 2m C 0 72

152 5 151 78 0 12
Imin 0 08 m s

0min Vs
DII
0 08 1 d

i d 0 319 0.32m

Vact 0 9947 m s

Vael 12 351 2

0.9997_ 0 08 g a

S 1 in 206 15

h t S L 490
2
0 436 m

0.07
HL 0.436
0.506 0 72 m
 darg 0.2 0.08
0.12 m s

darg

sewage pumping
Priematic Ejectors

centrifugal
Reciprocating

Whenever a branch sewer is connected to main sewer for treatment it may happens that the main

sewer is above the branch sewer for which pumping has to be done.

Ejectors can be used for small of quantity of sewage to be lifted up or if it is impossible to construct

an assembly for the pump in the area.

I to lift the
Design a
ejector
than a low level area
sewage having
of 7800
a
population persons supplied w
water of 180 I d Assume the
supply
velocity of o.am s velocity
sewage
of compressed air is 6m s
Ejector works for 10hr of
day height
eje for is 2 5m Assume that 701 of

water is converted to
supply sewage

3
0.015m
1 5
Imax 20 0 03125

Sewage discharge 0.2 Omax

Is 0.021875 m 9

of Qs 0
Capacity ejector

7 87 5 m hr

Area
IF
2.002 m
Design of air pipe

Is A
XVA
Dair 0.068 m
3
3 631 10
Aregirt m s

of sewer line
Design

Is As Vs
i 7.875 d 0.9

i
dg 3.33 m

i are 8 709 m
as
Q. A town having population of 60,000 is supplied with a per capita demand of 180 L/day. Aa

separate sewer from this town enters a pumping station through a low level sewer at RL 120m which

is to be pumped to a high level sewer of RL 129m. Assuming that 80% of water reaches the sewer

determine 1. Size of sump well 2. BHP of pump motor. 3. Size of rising main.

Assume following data: length of rising main 120m. Peak flow =2 times avg. velocity of flow in rising

main 1m/s. Min. Time of pumps running continuously is 15 mins. Head losses in bends 0.4 m.

Efficiency of pumps 65%. Efficiency of driving units . Depth of sump well is 3m.

1 Dang 0.8 600 O 10 3 o 1m s

max Peak Doug 3 0.1 0.3

Omax AV

0.3 x x 1

A 0.3 mm

D 0.618 m 4062

Vact 0.994 ms

2 Design sump well

of

Vol of sewage collected in is mins

V1 Imax 15 60 270 m

V2 Area of rising main L rising main

 0.62 120 36.22

Total V V2 306.228 m

Let us provide 2 sumps of 3062228 153.114m

Lump 2 23

Area Sump 153.114

of 3
depth

3 Design pump
of

0.170m S
capacity of pump Yee

H his hf t hminor

9 04
1
his 129 120 9m

0093m
5s
µ at 0.093 0.4 9.493m

BHP 00H 44138 up

9.81 75 xp nd
Q. Per capita demand of a township is 150 lpcd having total population 60,000. The sewage

generated from this town is required to be lifted up for 7m of static head and 70m distance. Consider

the loss of head in bends and valves as 0.4m. Velocity in rising main as1 m/s. Determine 1. Size of

sump well. 2. HP required for the pump. 3. Diameter of the rising main. If min. Time of pumping is 20

min. Efficiency of pump as 70%. Efficiency of driving units as 75%. F =0.03. Depth of sump well is

3.5m. Assume 75% of sewage reaching the sewer.

Dang 0.75 150 10 3 60,000 0.078 ms

24 60 60

Omax 3x daug 0.234

Omax AV

0.234 x x I

0.546 m

Vactual 0.98 m s

Vol collected in 20 mins

V1 Imax 20 60 280.8 m

V2 Area of rising main L

of rising main

0.55 70 16.63 e 2b

Total V V2 297 43 m

Let 148.71 m
us provide 2 sumps of 297243
 2
ins
3 Design
of pump
b
IIIb
8.622
I 17 244
Capacity
of pump 0.124m S
Yee

H his hf t hminor

7 0.4 7 453
1

9242 0.053
EI
BHP 23.47 HP
a Egypn

ET
Oxygen Demand

1. Bod - Biochemical oxygen demand

2. Cod - Chemical oxygen demand

3. Tod - Total oxygen demand

4. Theoretical oxygen demand

Oxygen is demanded in waste water for oxidation of both inorganic and organic matter. There are 4

types of demand as mentioned above.

1. Bod - Biochemical oxygen demand

It is oxygen required for the microorganisms to carry out biological decomposition of dissolved solids

or organic matter in the waste water under aerobic conditions at standard temperature.

1 BOD oxygenconsumed dilution ratio

2 Dilutionratio Volof diluted sample

Vol of undiluted sewage sample

3 It to 1 to kt

where It Amount
of firststage of BOD
to oxygen equivalent
of organic matter present in sewage

K BOD constant

time I constant given

4 1420 0 12
day
5 KT 1 20 0 20

Q. The domestic sewage of a town was tested for total solids and following results were obtained.

Weight of sample of sewage = 1000 g

Weight of solids after evaporation of liquid = 0.952 g

Weight of dry residue after ignition = 0.516 g

Determine total solids, fixed solids, volatile solids.

to convert toppm

Sol. 952 ppm

1 Total solids Tds 0,6932 106 ST
 2 Fixed solids 16 106 516 ppm SF
8
3 volatile solids St SF 436 ppm 952 516

Q. In order to conduct a 5 day bod test, the sample of waste water was diluted with specially

prepared dilution water with a dilution factor of 150. The contents of dissolved oxygen in the

beginning and end of the test were found to be, 11ppm and 7ppm. Compute 5 day BOD.

Sol

BOD oxygen demand dilution factor

11 7 150

600 ppm

Q. 5 ml of raw sewage was diluted by a specially prepared water in 300ml capacity BOD bottle. The

DO(dissolved oxygen) concentration of the diluted sample at the beginning of the test was 9mg/l.

And 6mg/l after 5 days incubation at 20 degree c. Find the BOD of raw sewage.

Sol

1 Dilution ratio Vol diluted sample

of undiluted 3 60
Vol sample
of

loss DO 9 6 3mg L
of

BOD 3 60 180 mg L
Q. Determine the bod of a sewage having 5 day BOD at 20 degree as 160ppm. Assume the

deoxygenated constant as 0.12 per day.

14
1 It to 1 10 Ls co 1 to kt

Lo will be more than lt.

Lt 160 ppm

K 0 12 day
t S days

Q. A sample of waste water has 4 days of 20 degree c BOD value as 75% of the final. Find the rate of

constant per day.

See t 4

It 0.7s to

Ans K 0.15
day

Q. Find the rate of constant at a temperate of 30 degrees if its value at 20 degrees is 0.12 per day.

Sol
20
KT 1 20 0 0 1.056 for 20 30

0 1.047 for 40 200

30 20
1430 1 20 1.0563

0.21
Q. The bod of a sewage incubated for 1 day at 30 degrees has been found out 100 mg/l. What will be

5 day BOD at 20 degree. Assume k=0.12 per day At 20 degrees.

Sol It to 1 to kt

4 at 30 100mg
1 0.12 20
i
30 20
1430 1 20 1.056 0 206

to 264
Is at 20 198.25

Q. For a waste water sample 5 day BOD at 20 degree c is 200mg/l and is 67% of the ultimate. What

will be 4 day BOD at 30 degree c.

Sol
10 200 298.51 5 200 s 67 10
g

1 30 1 70 1.056 0.206

y 253 73
at 30
DO-Oxygen Demand
 saturation eace

DO oxygen generation

Deoxygeneration

time

De critical point

fig oxygen Sag

to
5

Oxygen Demand

1. When the pollution load is discharged into her stream, DO content goes on decreasing called

deoxygenation.

2. Even tho DO goes on decreasing the atmosphere supplies oxygen continuously through

reservation or re-oxygenation. The rate of re-oxygenation depends on 1. Depth of water 2.

velocity of flow 3. Oxygen deficit 4. Time and 5. Temperature

 daft f deoxygenation and deoxygenation

K It R d
where
K BOD coefficient

R reoxygenation constant Re

It 1st stage BOD

kt
Lt to e

kt
K 10 e R Dt
daft

Ridt K'Zoe Kt
1ft
Dt e k't e't Doe
R't

kt Rt
Dt 10 é Doe

kt
09 Lt Lo 1 10
Q. The domestic sewage of a town is to be discharged into a stream after treatment. Determine the

maximum permissible effluent BOD and percentage of purification required in the treatment plant. For

the given following particulars :

1. Population = 50,000

2. DWF = 150 lpcd

3. BOD contribution per capita = 0.075 kg/day.

4. Min. Flow of stream = 0.02 m^3

5. BOD of stream = 3mg/l

6. Max. BOD on downstream is 5mg/l

1 DWF 10 0.0868 m s
529 5 8
2 BOD of mixture
of dis

5 Yt Lt TIO t 9202
Qi 02

5 Y 0 0868 3 02
0.0868 0.2

y 9 65 mg L BOD
of eff

37 BOD per capita per day 0 075 Kg day

 Actual BOD 0 07 106 500 mg l
0

t 500 9.608 100 98.07

held
purification soo

91 mix 4
I distream

Q. A stream saturated with DO has a flow of 1.2 m^3/s, BOD of 4mg/L and rate of constant as 0.3

per day. It receives an effluent discharge of 0.25m^3/s having BOD =20mg/L and DO=5mg/L and

rate of constant as 0.13 per day. The average velocity of flow is 0.18 m/s. Calculate DO deficit at

0.20km and 40km downstream. Assume that temperature is 20 degrees throughout and BOD is

measured at 5 days. Take saturation DO at 20 degrees as 9.17 mg/L

Sol

up
ds.gg
of
In
eggment 20km
norm
in

882

1 L 4 mg L

BOD 2 Q 1.2 m S
 3 12 20 mgt
4 02 0.25 m s

5 Do stream 9.17 msll

6 Do effluent Smgl
7 t for BOD Sdays 9 R 0.3 day
8 initial Do 0 13 day 9174 1 104 Velocitymix 0.18ms

1 LS 401 411.2 025 6.76mg l

2202 tt
kt
2 Ls Lo 1 10
013 5
6.76 Lo 1 10 8.71 to

3 Do mix DO di t Do e Q2
Qi 02

9.17 1.2 s 0.25

1.2 0.25

8 45 mg L

4 Initial Do deficit DO

9.17 8 45

0.72 mall
A For 20 km
 57 Avgvelocity 0.18ms

for 20 km with this velocity how many days

would be required

V 0 18 20.000
x 2
1.28 days

Rt
57 Dt 15 10 Do 15

013 128 128 1.28

0.13 871 10 1503 0.72 450.3
0 3 0.13

2.089 mg L

B For 40 km

57 Avgvelocity 0.18ms

for 20 km with this velocity how many days

would be required

V 0 18 40.000 222222.12
x 24 60 60

2.57 days
 Rt
57 Dt 15 10 Do 15

2 2 257
15013 1503 0.72 4503
3

2.079 mg 2

Q. Using following data find the DO at the end of 1 days and 2 days.

River wastewater

Flow m s 25 Q 2 02

Do mall 9.1 Docs 0 Dose

S day BOD 2 11 200 12

K 0.1

R 0.3

saturated DO 9.1 mgle

14 Do mix DO di t Do e Q2
Qi 02

91 25 O 2
25 2

8.426 mg l
2 Is Lid 1202
dit 2

2 25 200 2
2,1
16 67

kt
Ls Lo 1 10

16 67 10 1 1050 10 24.375

Do 91 8 426 0 675 mgt

A For 1 Day
Rt
At 15kt 10 Do 15

1
15 1503 0.673 1503
Of 24,37s

3.91 mgt

B For 2 Days
2 2
15 1503 0.673 1503
Of 24,37s

4 79 mgt
3 Deficit after 1 day
9 17 3 91 5 26 Ans

4 2 damp
Deficit after
9 17 4 79 4 38 Ans