Sewage disposal
The liquid wastes from industrial and domestic sources must eventually
be disposed of in some manner , whether by reuse , by discharge to
surface water , by injection or percolation to ground water , or by
evaporation to the atmosphere in nearly all cases the water must first be
treated to remove the look of the contaminants either as a Matter of
engineering necessity or to meet the either as a matter of engineering
necessity or to meet the requirements of environmental regulation .
Effect of stream Discharge
The introduction of excessive quantities of pollutants can upset the
natural balance in a variety of ways . changes in PH or the concentration
of some organic and inorganic species may be . toxic to specific life
forms . Excessive quantities of organic material may cases bacterial
growth and depletion of the dissolved oxygen resources of the stream.
As the concentration of pollutants is reduced by dilution ,precipitation
aeration , bacterial oxidation , or other natural processes the normal
cycle and distribution of life forms will tend to be reestablished . the self
– purification of natural water results from a variety of physical ,
chemical , and biological phenomena . water quality standers are based
on the maintenance of minimum dissolved oxygen concentration ≥ 40%
DOset or DOmin ≥ ( 4-5)
1
mg/l.
X(distance from the point of disposed )
The oxygen sag care is govern by streeter – phelps equations thes
governed equations either interms of the distance from the point of
disposed or interms of time in which t=
Where v is the average flow velocity of the river .
The following equation qives the values of consumed dissolved oxygen
at any distance or at any time starting from the point of disposal
DOx
=
DOa
*
In terms of x
Xc =
Dot = DOa .
-
2
In terms of t are most
suitable for using
Tc =
Doc =
DO min = DO set - DOc
Where
DOx : the concentration of consumed dissolved oxygen at distance
from point of disposal
x
DOa : initial deficit of dissolved oxygen ( ot x = 0 . or at t= 0 )
DO sat.: dissolved oxygen concentration at saturation at saturation level .
See table on page 835 to find DO sat for the
Rivers with different values of temperature .
DO mix : concentration of dissolved oxygen for mixture of river water
and wastewater .
Xc = critical distance from the point of disposed in which maximum
dissolved oxygen consumption is happened
Tc= the time from disposed of sewage in which maximum dissolved
oxygen consumption is happened .
Doc: Max. consumed dissolved oxygen concentration .
DOmin : minimum remaining dissolved oxygen concentration .
( at BOD max . or at Doc).
River
sewage
BOD mix=
River
sewage
3
DOM mix =
K1 = de oxygenation constant , d -1 ( is usually used with sewage )
K2= re aeration constant , d-1
( is usually used with river water )
The re aeration constant is determined by using the following formulas
K2= (
or K2 = 2.2
Where V = average stream ( river) velocity
H = average depth
Dm =molecular diffusion coefficient
= 2.037 * 10-5
= 2.037 *10-5 cm2/s at 20 c0
K2 varies with temperature in accord with
K2 (T) = K2( 20c0 ) (
EX: river of velocity 10 km/d and its flow rate is 5 m3/s A treatment plant
discharge sewage of flow rate of 0.5 m3/s
To the river the BOD5 concentration of sewage is 400 mg/l .find the
distance from the point of discharge in which BOD5 concentration of
sewage is 400 mg/l . find the distance from the point of discharge in
which BOD5 concentration at the end of such distance is become 10
mg/l . the river has not organic matters before sewage discharging , and
the value of K1 for the minture of river water and waste water is 0.35 d-1
at 20 co .
Sal:
4
BOD mix =
BOD5 = L ( 1-
= 36.36 mg/ l
) 36.36 = L (
L = 44 mg/l
BOD t = L ( 10 = 44 (1)
T= 0.735 day
V=
x= v.t = 10km/d * 0.735d = 7.35 km
Ex : A stream with a flow of 0.75 m3/s of BOD5 = 8.3 mg/l DO = 9.17
mg/l receive an effluent of 0.35 m3/s of
BOD5= 300 mg/l and DO= 3 mg/l determine the DO at a point of 5 km
downstream if the velocity of flow is
0.2 m/s Assume temperature is 20 c0 . K1 for effluent is 0.10 d-1 and K2
for the river is 0.4 d-1
Sal : BOD mix =
101 mg /l
=
DO mix =
=
7.2 mg/l
From table p.835 , for 20 c0 , DO sat = 9.17 mg/l
DOa = 9.17 – 7.20 = 1.97 mg/l
101 = Lo (1DOx = DOa
DOx = 1.97
Lo = 256 . 69 mg/l
+
(+
(256.9) * (
DOa
-(
DO=
DOx = 8.66 mg/l
)
DOx=
DO remaining = 9.17 – 8.66 = 0.51 mg /l
8.66 mg/l
DO sat.
DO min
DO rem.
5
5000 m
0.51 mg/l
x
EX: Waste water effluent of 0.54 m3/s , with BOD5 = 50 mg/l, DO =
3mg/l and temperature of 23 co is to be disposed to river of 2.8 m3 / s ,
and BOD5 = 4 mg/l DO = 8.2 mg/L , and temperature 17 co K1 for waste
0.1 day -1 , at 20 co K2 for river 0.3 /day -1 , at 20 co the flow velocity is
0.18 m/s , calculate the (DO min) and its distance down stream , and its
time after disposal .
Sol :
BOD mix =
= 11.66 mg/l , DO mix =
= 7.3 mg/l
= 18 co
Temp. mixed =
K1 ( 18) = 0.1 ( 1.047 )18-20 = 0.09 day-1
K1( T) = K1 ( 20 ) ( 1.047)
K2 ( 18) = 0.31 ( 1.025 )18-20 = 0.295 day-1
K2( T) = K2 ( 20 ) ( 1.025)
.
BOD5 = Lo (
mg/L.
)
11.66 = Lo(
) :. Lo = 29.63
Lo(18) = Lo(20) (0.02 Tmix + 0.6 ) = 28 .44 mg/l
*from table at 18 co
mg/l.
Tc =
ln [
Tc =
DOsat. = 9.5 mg/l
( 1- DOa
ln [
:. DOa = 9.5 – 7.3 = 2.2
)
( 1- 2.2
) ] = 4.84 day .
Xc = V.tc = 0.18 * 3600 * 24 *4.84 = 75271.68 =75.27 km.
Doc =
Lo
=
( 28.44) e-0.09 * 4.84 = 5.61 mg/l
DO min = 9.5 – 5.61 = 3.89 mg/l ≥ 0.4 * 9.5 = 3.8 O.K
6