Transcript Hydrology and Water Resources Management

SEWAGE
CHARACTERISTICS
SEWAGE CHARACTERISTICS
Composition
>99.0% Water
Solids
70% Organic
30% Inorganic
Sewerage characteristics can be divided into
three broad categories:1. Physical
2. Chemical
3. Bacteriological
PHYSICAL
Physical characteristics include:
• Temperature: The normal temperature of sewage
is slightly higher then water temperature.
Temperature above normal indicate inclusion of
hot industrial wastewaters in sewage
• Colour: Fresh sewage is light grey in colour.
While the old sewage is dark grey in colour. At a
temperature of above 200c, sewage will change
from fresh to old in 2 ~ 6 hours.
• Odour: Fresh domestic sewage has a slightly
soapy or oil odour. Stale sewage has a
pronounced odour of Hydrogen Sulphide
(H2S).
• Solids:- Solids in sewage may be suspended
or in solution solids are a measure of the
strength of sewage.
CHEMICAL
Sewage contain both organic and inorganic
chemicals. All the test representing these
organic and inorganic constituents come under
the heading of chemical characteristics. Test
like BOD, COD, NITORGON, PHOSPHOURS,
ALKALINITY
etc give
the chemical
characteristics of sewage.
BACTERIOLOGICAL
Enormous
quantities
of
microorganisms
are present in domestic
sewage. They include bacterial worms,
viruses, protozoa etc. Bacterial counts
in raw sewage may range form
500,000/ml to 50,000,000/ ml. Viruses,
protozoa, Worms etc have not enough
characteristics
that
require
measurement.
DEFINITIONS OF SOME TERMS IN
SEWAGE CHARACTERIZATION
SOLIDS
TOTAL SOLIDS:- Include both suspended and
dissolved solids. It is measured by evaporating
a known volume of sample and the weighting
the residue. Results are expressed in mg/l
SUSPENDED SOLIDS:- These are solids which
are pertained on a pre-weighed glass fiber filter
of 0.45 103-1050C
DISSOLVED SOLID:- Filtrate which
has passed thought 0.45µ filter is
evaporated in chine dish. The residue
gives the dissolved solids.
SETTLEABLE SOLIDS:It is the
fraction of the solids that will settle in
an imhoff cone in 30-60 minutes. These
are expressed as mg/l.
VOLATILE SUSPENDED SOLIDS
They give a rough measure of the organic
content or in some instances of the
concentration of BIOLOGICAL SOLIDS such
as bacteria. The determination is made by
ignition of residues on 0.45µ filter in a Muffle
furnace at 5500C. The residues following the
ignition is called non-volatile solids or ash
and is rough measure of the mineral content
of the waste water. (Note:- Most of the
inorganic and mineral content do not
volatilize at 5500C and are quiet resistant)
BOD
Bacteria placed in contact with organic matter
will utilize it as food source. In the utilization
of the organic material it will eventually be
oxidized to stable and products such as CO2
and H2O.
“The amount of oxygen required by the
bacteria to oxidize the organic matter present in
sewage to stable end products is known as
biochemical oxygen demand.”
Significance of BOD
Significance: 1.Used in design
treatment plants.
of
waste
water
2.Used to measure efficiently of waste
water treatment plant.
DERIVATION OF BOD EQUATION
Biological oxidation of organic matter by
bacteria is considered to be a first order
reaction for all practical purposes. In a first
order reaction, the rate of reaction is
proportional to the concentration of the
reactant present. So, we can say that in case of
biological oxidation of organic matter by
bacteria, the rate of oxidation is proportional to
the organic matter REMAINING.
Let
L = Concentration of organic matter at any
time ‘t’
Lo = Initial case of organic matter at t=0 i.e.
(Ultimate BOD)
Mathematically:
dL/dt  -L (-ve sign show that L is decreasing)
dL/dt = -KL
In second order reaction the rate of reaction is
proportional to square of reactions.
Where ‘K’ is const and is known as “Reaction
Rate Constant”
dL/dt
dL/L
dL/L
= -KL
= -K dt
= -K dt
L
t
dL
L L   k 0 dt
0
L
ln L   kt
L0
ln L – ln Lo= -Kt
ln L/Lo = -Kt
ln L/Lo = -Kt
L/Lo
= e –Kt
L
= Lo e –Kt
Let ‘y’ be the concentration of organic matter
(BOD) consumed up to time ‘t’
y
=
Lo – L
y
=
Lo – Lo e –Kt
y
=
Lo ( 1 - e –Kt)
i.e BOD consumed = ultimate BOD (1 – e –Kt) in ‘t’
days
Typical value of K = 0.23 per day for domestic
sewage at 20oC. Value of ‘K’ is temperature
dependent.
KT =
K20 (1.047) T-20
KT =
Value of K at temp T
BOD represents amount of organic matter
L
BOD
mg/l
Lo
y
‘t’
time
Problem
The 5 day BOD of waste water is 190mg/l.
determine ultimate BOD assuming K = 0.25 per
day
SOLUTION
y =
BOD exerted / consumed
L =
Amount of organic matter
remaining at time ‘t’
Lo =
Ultimate BOD (Total Organic
Matter)
L =
Lo e-KT
y =
Lo (1 - e-KT)
190
=
Lo (1 – e –0.25 x 5)
Lo =
266.29 mg/l
Problem
Calculate the ultimate BOD for a sewage
whose 5 day BOD at 20oC is 250 mg/l.
Assume K = 0.23 per day what will be
BOD after 2 days.
SOLUTION
y
250
Lo
=
=
=
Lo (1 – e Kt)
Lo (1 – 5 x 0.23)
365.83 mg/l
y2
y2
=
365.83 (1 – e –0.23 x 2)
=
134.89 mg/l
= 135 mg/l
Problem
The BOD remaining in a sewage sample after 4
and 8 days was 160 and 60 mg/l respectively at
20oC calculate the 5 day BOD of the sewage at
25oC.
Only if BOD = y
BOD exerted / consumed
BOD remaining
=
=
y
L
SOLUTION
L =
Lo e-Kt
160 =
Lo e-K x 4
60 =
Lo e-K x 8
Lo =
160 / e –4K
60 =
[160 / e –4K ] e–K8
60 =
160 e –8K +4K
0.375
=
e –4K
ln (0.375) =
-4k
K =
0.245 per day
Lo
At 25oC
K25 =
K25 =
K25 =
Y5 =
Y5 =
Y5 =
=
160/e –4 x 0.245
K20 (1.047) T-20
0.245 (1.047) 25-20
0.308 per day
Lo
(1 - e –Kt)
426.3 (1 – e –0.308 x 5)
335 mg/l
=426.3 mg/l
CHEMICAL OXYGEN
DEMAND
It is the amount of oxygen required to oxidize
organic matter chemically (biodegradable and
non-biodegradable) by using a strong chemical
oxidizing agent. (K2Cr2O7) in an acidic
medium. For a single waste water sample the
value of COD will always be greater then BOD.
The oxidant (K2Cr2O7) remaining is found out
to find K2Cr2O7 considered COD and BOD can
be interrelated.
DOMESTIC SEWAGE
CHARACTERISTICS
Parameter
Range (mg/l)
Total Solids
350 – 1200
Dissolved Solids
250 – 850
Suspended Solids
100 – 350
Setteleable Solids
5 – 20 (ml / l)
BOD
100 – 300
COD
250 – 1000
Total Nitrogen
20 – 85
Alkalinity (as CaCO3)
50 – 200
POPULATION EQUIVALENT
1 person excrete 80gm BOD/day. Population
equivalent of an industry is the number of
persons which may produce the same amount
of BOD per day.
Let BOD of tannery is
=500 mg/l
Q
=10,000 m3 / day
Total BOD load by tannery =BOD x Q
=500 x 10000 / 1000
=5000 kg BOD/day
Population equivalent
=(5000 / 80) x 1000
=62,500 persons
Water Resources
Management
Definitions
• Hydrology – the scientific study of the
properties, distribution and effects of water on
the earth’s surface, in the soil and underlying
rocks and in the atmosphere
• Water resources management -- control and
utilization of water for beneficial uses or to
avoid adverse impacts
– drinking water supply
– irrigation
– industrial water supply
-- flooding
-- drought
-- subsidence
Determining Water Demand
• Average daily water consumption
• Ability to meet continuing demand over critical
periods
• Estimate stored water requirement quantities
• Peak demand rates (~ 2.2 x daily average flow
for metered dwellings)
• Size of plumbing and piping, pressure losses
• Estimate storage requirements during periods of
peak water demand
Factors Affecting Water Usage
•
•
•
•
•
•
•
•
Population
Economic considerations
Cost of water
Meterage (reduction by ~40%)
Climate
Type of water usage
System management
Conservation practices
How can the demand be met?
• Locate close to the water supply
• Store and transport water
– NYC: over 100 miles
– LA: about 350 miles
California aqueduct
From: Water, National Geographic Special Edition; Vol. 184 (5), © 1993
Examples of Water Resource
Projects
Ludington Pumped Storage
Hydroelectric Plant
• Water is pumped from Lake Michigan into an
upper reservoir during “off-peak” hours when
demand for electricity is low
• Water is stored until electrical demand increases,
then released through turbines
• Max. of 17.5 billion gal water may be
transferred at rate of >33 million gal/min
• 1,300 ft long penstocks (24-28 ft in diam.)
• Upper Reservoir: 842 acres (2.2 mi long x 0.8
mi wide; cap. 27 billion gallons)
Environmental Impacts
• Positives
– Recreational
facilities
– Cheap power
– Fewer brown
or black-outs
– Smaller than
comparable
hydroelectric
dams
Environmental Impacts
• Negatives
– Loss of land
– Loss of wildlife
– Loss of energy
to pump water
up to reservoir