Chapter 10_Natural Methods of Wastewater Disposal.pdf

DCE-KU Water supply and Sanitation (CIEG 313) Asst. Prof. Manish Prakash
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Natural method of wastewater disposal
After conveying the wastewater through sewers, the next step is its disposal, either after
treatment or even before treatment. The method of disposal of wastewater may be classified
as:
1. Natural methods
a) By dilution
b) By land treatment
2. Artificial methods
a) Primary treatment
b) Secondary treatment
c) Tertiary treatment
3. Combined methods
Self-purification of natural stream
The introduction of organic matter into a water body results, indirectly, in the consumption of
dissolved oxygen. This occurs as a result of the processes of the stabilization of the organic
matter undertaken by bacteria, which use the oxygen available in the liquid medium for their
respiration. As expected, the decrease in the dissolved oxygen concentration in the water
body has various implications from the environmental point of view.
The objective of this section is the study of the processes of consumption of dissolved oxygen
and of stream self-purification, in which the water body recovers itself, through purely
natural mechanisms. Both of these processes are analyzed from an ecological viewpoint and,
subsequently, more specifically through the mathematical representation of the DO profile in
the water body.
In broader terms, the process of self-purification is associated with the reestablishment of the
equilibrium of the aquatic ecosystem, after the alterations induced by the effluent discharge.
Within a more specific point of view, the conversion of organic compounds into inert
compounds, not deleterious from an ecological viewpoint, is an integral part of the process.
Action involved in self-purification
a) Dilution
b) Dispersion due to currents
c) Sedimentation
d) Oxidation
e) Reduction
f) Temperature
g) Sunlight
Because self-purification is a process that develops with time, and considering the dimension
of the river as predominantly longitudinal, the stages of ecological succession can be
associated with physically identifiable zones in the river. There are four main zones:
a) Degradation zone

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This zone is situated just below the outfall sewer when discharging its contents into the
stream. In this zone, water is dark and turbid, having the formation of sludge deposits at the
bottom. The DO is reduced to 40 % of the saturation values. There is an increase in CO2
content, and reaeration is much slower than deoxygenation. Though conditions are
unfavorable for aquatic life, fungi at higher points and bacteria at lower points breed small
worms which work over and stabilize the sewage sludge. The decomposition of solid matter
takes place in this zone and anaerobic decomposition prevails.
Fig. Zone of Pollution along a Stream

39

40
b) Active decomposition zone
This zone is just after the degradation zone and is marked by heavy pollution. Water in this
zone becomes greyish and darker than the previous zone. The DO concentration in this zone
falls down to zero. Active anaerobic organic decomposition takes place, with the evolution of
methane (CH4), hydrogen sulphide (H2S), carbon dioxide (CO2) and nitrogen (N2), bubbling
to the surface with masses of sludge forming black scum. Fish life is absent in this zone but
bacteria flora will flourish with the presence of anaerobic bacteria at upper end and aerobic
bacteria at the lower end. Protozoa and fungi will first disappear and then reappear, while
algae will mostly be absent. However, near the end of this zone, as the decomposition
slackens, reaeration sets in and DO again rises to its original level of 40 %.

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c) Recovery zone
In this zone, the process of recovery starts, from its degraded condition to its former
condition. The stabilization of organic matter takes place in this zone. Due to this, most of the
stabilized organic matter settles as sludge, BOD falls and DO content rises above the 40 %
value. Mineralization is active, with the resulting formation of products like nitrates (NO3),
sulphates (SO4), carbonates (CO3). Near the end of the zone, microscopic aquatic life
reappears, fungi decrease and algae reappears.

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d) Clean water zone
In this zone, the natural condition of stream is restored with the result that water becomes
clearer and attractive in appearance, DO rises to the saturation level, and is much higher than
BOD, and oxygen balance is attained. Thus recovery is said to be complete in this zone,
though some pathogenic organisms may be present in this zone.
Interaction Mechanisms in DO Balance
In the self-purification process there is a balance between the sources of consumption and the
sources of production of oxygen. When the consumption rate is higher than the production
rate, the oxygen concentration tends to decrease, the opposite occurring when the
consumption rate is lower than the production rate. In general, the concentrations of
constituents (such as DO) in a water body change as a result of physical processes of
advection (transportation by the water as it flows in the river channel) and dispersion
(transportation due to turbulence and molecular diffusion) and biochemical and physical

43
processes of conversion (reaction). The processes take place in the three dimensions of the
water body, although in rivers the longitudinal axis is the prevailing one.
Fig. Interaction Mechanisms in DO Balance
Fig. Axes along which Spatial and Temporal Variations of the Concentrations of the Water
Constituent Take Place (in river, the X-axis is Predominant)
Oxygen sag analysis
The oxygen sag or oxygen deficit in the stream, at any point of time during the self-
purification process is the difference between the saturation DO content and the actual DO
content at that time.
Oxygen deficit = Saturation DO - Actual DO
The normal saturation DO value for fresh water depends upon the temperature, and its value
varies from 14.62 mg/l at 0° C to 7.63 mg/l at 30° C.

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At the point where wastewater is discharged into the stream, it is not necessary that the DO
content of the stream will be equal to the saturation DO, there may be initial oxygen deficit
D0. At this stage, when the wastewater effluent, with an initial BOD load L0, is discharged
into the stream, the DO content of the stream starts depleting and the oxygen deficit (D)
increases. The variation of oxygen deficit (D) with the distance along the stream (and hence
time of flow from the point of pollution) is depicted by the oxygen sag curve. The major
point of interest in the oxygen sag analysis is the point of minimum DO, or point of
maximum deficit. The maximum or critical deficit, labelled as Dc occurs at the inflection
point of the oxygen sag curve.
Fig. Oxygen Sag Curve
Deoxygenation and Reoxygenation Curves
When wastewater is discharged into the stream, the DO level in the stream goes on depleting.
This depletion of DO content is known as deoxygenation. The rate of deoxygenation depends
upon the amount of organic matter remaining (Lt) to be oxidized at any time t, as well as
temperature (T) at which reaction occurs. The variation of depletion of DO content of the
stream with time is depicted by the deoxygenation curve in the absence of aeration. The
ordinates below the deoxygenation curve indicate the oxygen remaining in the natural stream
after satisfying the bio-chemical oxygen demand of oxidizable matter.
When the DO content of the stream is gradually consumed due to BOD load, atmosphere
supplies oxygen continuously to the water, through the process of re-aeration or
reoxygenation, i.e., along with deoxygenation, re-aeration is continuous process.
The rate of reoxygenation depends upon:
a) Depth of water in the stream: more for shallow depth.
b) Velocity of flow in the stream: less for stagnant water.
c) Oxygen deficit below saturation DO: since solubility rate depends on difference between
saturation concentration and existing concentration of DO.
d) Temperature of water: solubility of oxygen is lower at higher temperature and also
saturation concentration is less at higher temperature.

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Fig. Deoxygenation, Reoxygenation and Oxygen Sag Curve
Mathematical analysis of Oxygen Sag Curve: Streeter – Phelps equation
The analysis of oxygen sag curve can be easily done by superimposing the rates of
deoxygenation and reoxygenation as suggested by the Streeter – Phelps analysis. The rate of
change in the DO deficit is the sum of the two reactions as explained below:
𝑑𝐷𝑡
𝑑𝑡
= f (deoxygenation and reoxygenation)
𝑑𝐷𝑡
𝑑𝑡
= K’Lt – R’Dt ….(1)
Where,
Dt = DO deficit at any time t,
Lt = amount of first stage BOD remaining at any time t
K’ = BOD reaction rate constant or deoxygenation constant (to the base e)
R’ = Reoxygenation constant (to the base e)
t = time (in days)
𝑑𝐷𝑡
𝑑𝑡
= rate of change of DO deficit
Now, Lt =L0𝑒−𝐾′𝑡
Where, L0 = BOD remaining at time t = 0 i.e. ultimate first stage BOD
Hence,
𝑑𝐷𝑡
𝑑𝑡
= K’ L0𝑒−𝐾′𝑡
– R’Dt ….(2)
𝑑𝐷𝑡
𝑑𝑡
+ R’Dt = K’ L0𝑒−𝐾′𝑡
….(3)
This is first order first degree differential equation and solution of this equation is as under
Dt =
𝐾′𝐿0
𝑅′−𝐾′
[𝑒−𝐾′𝑡
− 𝑒−𝑅′𝑡
] + D0𝑒−𝑅′𝑡
….(4)
Changing base of natural log to 10 the equation can be expressed as:
Dt =
𝐾𝐿0
𝑅−𝐾
[10−𝐾𝑡
− 10−𝑅𝑡] + D010−𝑅𝑡
….(5)
Where,
K = BOD reaction rate constant, to the base 10
R = Reoxygenation constant to the base 10
D0 = Initial oxygen deficit at the point of waste discharge at time t = o
t = time of travel in the stream from the point of discharge = x/u
x = distance along the stream
u = stream velocity

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This is Streeter-Phelps oxygen sag equation.
Fig. Oxygen sag curve of Streeter-Phelps equation
Note: Deoxygenation and reoxygenation occurs simultaneously. After critical point, the rate
of re-aeration is greater than the deoxygenation and after some distance the DO will reach to
original level and stream will not have any effect due to addition of wastewater. At time t=0
at x = 0.
Determination of Critical D0 deficit (Dc) and distance Xc
The value of Dc can be obtained by putting
𝑑𝐷𝑡
𝑑𝑡
= 0 in equation 3,
Hence,
Dc =
𝐾′
𝑅′L0𝑒−𝐾′𝑡𝑐….(6)
Dc =
𝐾
𝑅
L010−𝐾 𝑡𝑐….(7)
Where, tc is time required to reach the critical point.
The value of tc can be obtained by differentiation equation 4 or 5 with respect to t and setting
𝑑𝐷𝑡
𝑑𝑡
= 0. Therefore,
tc =
1
𝑅′−𝐾′ log𝑒
𝑅′
𝐾′ [1 −
𝐷0(𝑅′−𝐾′)
𝐾′𝐿0
]….(8)
or
tc =
1
𝑅−𝐾
log10
𝑅
𝐾
[1 −
𝐷0(𝑅−𝐾)
𝐾𝐿0
]….(9)
The distance Xc is given by Xc = tc .u
Where, u = velocity of flow in the stream
The deoxygenation constant K, is obtained by laboratory test or field tests, and varies with
temperature as given below:
KT = K20 𝜃𝑇−20
….(10)
K = 0.1 to 0.3 for municipal sewage, base 10, (0.23 to 0.70 for base e)
The reoxygenation constant R also varies with the temperature and can be expressed as:
RT = R20 1.024𝑇−20
Where, R’/R = 2.303
R= 0.10 to 0.15 for lakes and sluggish stream
= 0.15 to 0.20 for low velocity large stream
= 0.20 to 0.30 for normal velocity large stream

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= 0.30 to 0.50 for Swift stream
= > 0.50 for rapids and water falls
The ratio of R/K (or R’/K’) is called the self-purification constant fs and it is equal to 0.50 to
5.0.
Substituting R/K = fs in equation 9,
tc =
1
𝐾(𝑓𝑠−1)
log10 [𝑓
𝑠 {1 − (𝑓
𝑠 − 1)
𝐷0
𝐿0
}]….(11)
Also from equation 7, we get
Dc =
𝐿0
𝑓𝑠
10−𝐾𝑡𝑐
Taking log on both sides,
log10 𝐷𝑐 = log10
𝐿0
𝑓𝑠
- Ktc….(12)
Substituting value of tc from equation 11, we get
log10 𝐷𝑐 = log10
𝐿0
𝑓𝑠
-
1
(𝑓𝑠−1)
log10 [𝑓
𝑠 {1 − (𝑓
𝑠 − 1)
𝐷0
𝐿0
}]
or, (𝑓𝑠 − 1) [log10
𝐿0
𝑓𝑠
− log10 𝐷𝑐] = log10 [𝑓
𝑠 {1 − (𝑓
𝑠 − 1)
𝐷0
𝐿0
}]
or, log10 [
𝐿0
𝑓𝑠𝐷𝑐
]
(𝑓𝑠−1)
= log10 [𝑓
𝑠 {1 − (𝑓
𝑠 − 1)
𝐷0
𝐿0
}]
[
𝑳𝟎
𝒇𝒔𝑫𝒄
]
(𝒇𝒔−𝟏)
= [𝒇𝒔 {𝟏 − (𝒇𝒔 − 𝟏)
𝑫𝟎
𝑳𝟎
}]….(12)
Table: Saturation Concentration for Oxygen in Clean Water (mg/L)
Dilution into sea
The saturation concentration of dissolved oxygen in water decreases with increasing salt
content. Due to this reason, the saturation concentration in sea water is approximately 80 %
of that in water. In addition to this deficiency, the temperature of sea water is lower than the

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sewage temperature, whereas the specific gravity is higher. Due to these reasons, when
sewage is discharged into the sea water, the lighter and warmer sewage will rise up to the
surface, resulting in the spreading of the sewage at the top surface of sea in a thin film or
sleek. Moreover, sea water contains a large amount of dissolved matter which chemically
react with the sewage solids, resulting in the precipitation of some of the sewage solids,
giving a milky appearance to the sea water and resulting in formation of sludge banks. These
sludge banks and thin milky layer formed at the top of sea water produce offensive hydrogen
sulphide gas by reacting with the sulphate rich water of the sea. The various chemical actions
and the prevailing dissolved matter in the sea water reduce its capacity to absorb more
quantity of sewage. However, since the sea contains large volume of water, most of these
deficiencies can be overcome if the sewage is discharge deep into the sea.
 Sewage should be discharged deep into the sea, preferably 1 to 1.5 Km away from the
shore.
 The minimum depth of water at the outfall point should be 3-5 m.
 The sewage should be disposed off only during the low tides. This is accomplished by
holding the wastewater into large sized tanks constructed near the shore.
Fig. Disposal into Sea
Disposal by land treatment
When the wastewater, either raw or partly treated, is applied or spread on the surface of land,
the method is called disposal by land treatment. The three principal processes of land
treatment of wastewater are:
1. Broad irrigation or sewage farming
2. Rapid infiltration
3. Overland runoff

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Fig. Three Principal Processes of Land Treatment

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Numerical 1 The domestic sewage of a town is to be discharged into a stream after treatment.
Determine the maximum permissible effluent BOD and the percentage purification required
in the treatment plant, given the following particulars:
Population of town = 50,000
DWF of sewage = 150 litres per capita per day
BOD contribution per capita = 0.075 kg per day
Minimum flow of stream = 0.20 m3
/s
BOD of stream = 3 mg/l
Max. BOD of stream on downstream = 5 mg/l
Solution:
DWF of sewage = 150 X 50000 litres per day = 0.0868 m3
/s
Stream discharge = 0.20 m3
/s
The BOD (Ym) of the mixture at the downstream is
Ym =
𝑌1𝑄1+ 𝑌2𝑄2
𝑄1 + 𝑄2
Where Ym = 5 mg/L (given), Y1 is the BOD of the effluent having discharge Q1 and Y2 is the
BOD of the stream having discharge Q2.
5 =
𝑌1 𝑋 0.0868 +3 𝑋 0.2
0.0868+0.2
Y1 = 9.608 mg/L
Now BOD per capita per day = 0.075 kg per day = 75000 mg/day
Sewage DWF = 150 litres/day
Initial BOD of effluent =
75000
150
= 500 mg/L
Hence percentage purification required =
500−9.608
500
X 100 = 98.1 %
Numerical 2 A city discharges sewage at the rate of 1200 litres per second, into a stream
whose minimum flow is 5000 litres per second, the temperature of both being 20° C. The 5
day BOD at 20° C for sewage is 160 mg/l and that of river water is 2 mg/l. The DO content
of sewage is zero while that of stream is 90 % of the saturation DO. Find out the degree of
treatment required if the minimum DO to be maintained in the stream is 4 mg/l. Assume
deoxygenation coefficient as 0.1 and reoxygenation coefficient as 0.3. Given saturation DO at
20° C as 9.17 mg/l.
Solution:
DO of stream = 90 % of saturation DO = 0.9 X 9.17 = 8.253 mg/l
DO of the mix =
8.253 𝑋 5000+0 𝑋 1200
5000+1200
= 6.656 mg/l
Initial DO deficit = D0 = 9.17 – 6.565 = 2.514 mg/l
Minimum DO to be maintained in the stream = 4 mg/l
Maximum permissible saturation deficit, Dc = 9.17 – 4 = 5.17 mg/l
Now, the first stage BOD of mixture of sewage and stream is
[
𝑳𝟎
𝒇𝒔𝑫𝒄
]
(𝒇𝒔−𝟏)
= [𝒇𝒔 {𝟏 − (𝒇𝒔 − 𝟏)
𝑫𝟎
𝑳𝟎
}]
Where, fs = R/K =0.3/0.1 = 3
D0 = 2.514 mg/l and Dc = 5.17 mg/l

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[
𝐿0
3 𝑋 5.17
]
(3−1)
= [3 {1 − (3 − 1)
2.514
𝐿0
}]
Solving this by trial and error, we get L0 = 23.86 mg/l
Y5 = L0(1 - 10−𝐾𝑡
) = 23.86(1 - 10−0.1𝑋5
) = 16.31 mg/l
This is permissible BOD of the mix.
16.31 =
2 𝑋 5000+𝑌𝑒 𝑋 1200
5000+1200
Ye = 75.94 mg/l
Hence permissible BOD of effluent = Ye = 75.94 mg/l
Actual BOD of effluent = 160 mg/l
Degree of treatment required =
160−75.94
160
X 100 = 52.5 %
Numerical 3 A river having a flow of 5 m3
/s, 5 day BOD 1 mg/l and saturated with oxygen
receives at a point A, a sewage effluent discharge of 2 m3
/s with 5 day BOD 20 mg/l and DO
2 mg/l.
A tributary meets the stream at a point B, 20 km on the downstream, having a flow of 2 m3
/s,
5 day BOD 2 mg/l and DO 9 mg/l.
As the river further advances by a distance of 25 km, at point C on downstream, another
effluent having flow of 2 m3
/s, 5 day BOD 16 mg/l and DO 3 mg/l joins the main stream.
Compute the DO deficiency at point P 20 km downstream of C. Assume saturation DO as
9.17 mg/l for a constant temperature of 20° C, K = 0.12 per day and R = 0.36 per day.
Assume uniform velocity of flow equal to 0.3 m/s.
Solution:
(a) At junction point A
(Y5)A =
5 𝑋 1+2 𝑋 20
5+2
= 6.43 mg/l
But (Y5)A = L0(1- 10−𝐾𝑡
)
6.43 = L0(1- 10−0.12 𝑋 5
)
L0 = 8.59 mg/l
(DO)A =
5 𝑋 9.17+2 𝑋 2
5+2
= 7.12 mg/l
(D0)A = 9.17 – 7.12 = 2.05 mg/l

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(b) At a point just upstream of B
u = 0.3 m/s and XB = 20 km = 20000 m
tB =
20000
0.3 𝑋 24 𝑋 36000
= 0.772 days
DO deficit prior to point B is given by Streeter-Phelps Equation:
D0.772 =
𝐾𝐿0
𝑅−𝐾
[10−𝐾𝑡
− 10−𝑅𝑡] + D010−𝑅𝑡
=
0.12 𝑋 8.59
0.36−0.12
[10−0.12 𝑋 0.772
− 10−0.36 𝑋 0.772] + 2.05(10)−0.36 𝑋 0.772
= 2.29 mg/l
DO of river = 9.17 – 2.29 = 6.88 mg/l
BOD remaining after t = 0.772 days
𝐿𝑡 = 𝐿010−𝑘𝑡
𝐿0.772 = 8.59 ∗ 10−0.12∗0.772
= 6.94 𝑚𝑔/𝑙
20° C BOD5 = 6.94(1 − 10−0.12∗5) = 5.20 𝑚𝑔/𝑙
(c) Just after junction point B
(DO)B =
6.88 (5+2)+ (9 𝑋 2)
(5+2)+ 2
= 7.35 mg/l
(D0)B = 9.17 – 7.35 = 1.82 mg/l
Also, (Y5)B =
5.20 (5+2)+ (2 𝑋 2)
(5+2)+ 2
= 4.49 mg/l
But (Y5)B = L0(1- 10−𝐾𝑡
)
4.49 = L0(1- 10−0.12 𝑋 5
)
(L0)B = 6.00 mg/l
(d) At a point just before junction point C
Distance XBC = 25 km; u = 0.3 m/s
tBC =
25000
0.3 𝑋 24 𝑋 3600
= 0.965 days
D0.965=
0.12 𝑋 6.00
0.36−0.12
[10−0.12 𝑋 0.965
− 10−0.36 𝑋 0.965] + 1.82(10)−0.36 𝑋 0.965
= 1.77 mg/l
DO of river = 9.17 – 1.77 = 7.4 mg/l
BOD remaining after t = 0.772 days
𝐿𝑡 = 𝐿010−𝑘𝑡
𝐿0.965 = 6.00 ∗ 10−0.12∗0.965
= 4.60 𝑚𝑔/𝑙
20° C BOD5 = 4.60(1 − 10−0.12∗5) = 3.44 𝑚𝑔/𝑙
(e) Just after junction point C
(DO)C =
7.2 (5+2+2)+ (3 𝑋 2)
(5+2+2)+ 2
= 6.44 mg/l
(D0)C = 9.17 – 6.44 = 2.73 mg/l
Also (Y5)C =
3.44 (5+2+2)+ (16 𝑋 2)
(5+2+2)+ 2
= 5.72 mg/l
(Y5)C = L0(1- 10−𝐾𝑡
)
5.72 = L0(1- 10−0.12 𝑋 5
)
(L0)C = 7.64 mg/l

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Thus D0 and L0, just after the junction point C are known.
(f) At point P, 20 km from C
Distance XCP = 20 km; u =0.3 m/s
tCP =
20000
0.3 𝑋 24 𝑋 3600
= 0.772 days
DO deficit at point P is given by Streeter Phelps equation:
D0.772=
0.12 𝑋 7.64
0.36−0.12
[10−0.12 𝑋 0.772
− 10−0.36 𝑋 0.772] + 2.73(10)−0.36 𝑋 0.772
= 2.51 mg/l
Numerical 4 A city discharges 1.25 m3
/s of wastewater into a stream whose minimum rate of
flow is 8 m3
/s. The velocity of stream is 0.8 m/s. The temperature of the wastewater is 20° C
and that of the stream is 15° C. The 20° C BOD5 of the wastewater is 250 mg/l, and that of
the stream is 2.0 mg/l. The wastewater contains no dissolved oxygen but the stream is 90 %
saturated upstream of the discharge. At 20° C, K' is estimated to be 0.3 per day and R' is 0.9
per day. Determine the critical-oxygen deficit and its location. Also, estimate the 20° C BOD5
of a sample taken at the critical point. Use temperature coefficients of 1.135 for K' and 1.024
for R'. Also plot the dissolved-oxygen sag curve. Given saturation concentration at 15° C =
10.15 mg/l.
Solution:
a) DO of the stream
Saturation concentration = 10.15 mg/l
DO of stream = 0.9 X 10.15 = 9.14 mg/l
b) Characteristics of the mixture
(T)mix =
1.25 𝑋 20+8 𝑋 15
1.25+8
= 15.68° C
(BOD5)mix =
1.25 𝑋 250+8 𝑋 2
1.25+8
= 35.51 mg/l
(L0)mix =
35.51
1− 𝑒−0.3 𝑋 5
= 45.71 mg/l
(DO)mix =
1.25 𝑋 0+8 𝑋 9.14
12.5+8
= 7.90 mg/l
c) Initial dissolved oxygen deficit at T = 15.68° C
DO saturation concentration at 15.68° C = 10.01 mg/l
Initial deficit D0 = 10.01 – 7.90 = 2.11 mg/l
d) Rate constant at 15.68° C
K' = 0.3 (1.135)15.68−20
= 0.174/d
R' = 0.9 (1.024)15.68−20
= 0.812/d
e) Determination of tc and Xc
tc =
1
𝑅′−𝐾′
log𝑒
𝑅′
𝐾′
[1 −
𝐷0(𝑅′−𝐾′)
𝐾′𝐿0
]
=
1
0.812−0.174
log𝑒
0.812
0.174
[1 −
2.11(0.812−0.174
0.174 𝑋 45.71
] = 2.124 days

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Xc = u tc = 0.8 X 2.124 X 24 X 3600 X 10−3
= 146.8 km
f) Determination of Dc and DO at Xc
Dc =
𝐾′
𝑅′
L0𝑒−𝐾′𝑡𝑐 =
0.174
0.812
X 45.71 X 𝑒−0.174 𝑋 2.124
= 6.77 mg/l
DO at Xc = 10.01 – 6.77 = 3.24 mg/l
g) Determination of BOD5 of a simple taken at Xc
Lt = L0𝑒−𝐾′𝑡𝑐 = 45.71𝑒−0.174 𝑋 2.124
= 31.59 mg/l
20° C BOD5 = 31.59[1 - 𝑒−0.3 𝑋 5
] = 24.54 mg/l
h) Computation of additional points on the oxygen sag curve
Dt =
𝐾′𝐿0
𝑅′−𝐾′
[𝑒−𝐾′𝑡
− 𝑒−𝑅′𝑡
] + D0𝑒−𝑅′𝑡
Dt =
0.174 𝑋 45.71
0.812−0.174
[𝑒−0.174𝑡
− 𝑒−0.812𝑡] + 2.11𝑒−0.812𝑡
Dt = 12.466[𝑒−0.174𝑡
− 𝑒−0.812𝑡] + 2.11𝑒−0.812𝑡
(DO)t = Saturation concentration of DO – Dt = 10.01 - Dt
Distance x, km 0 50 100 146.8 200 250
Time t, day 0 0.723 1.447 2.124 2.849 3.617
DO, mg/l 7.9 4.71 3.52 3.24 3.46 3.92
Fig. Oxygen Sag Curve
0
2
4
6
8
10
12
0 50 100 150 200 250 300
DO
,mg/l
Distance, km
DO

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