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.

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.

CHEMICAL

BACTERIOLOGICAL

Enormous quantities of micro-organisms 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.

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)

VOLATILE SUSPENDED SOLIDS

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.”

BOD

Significance of BOD

Significance: -

1. Used in design of waste water treatment plants.

2. Used to measure efficiently of waste water treatment plant.

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.

DERIVATION OF BOD EQUATION

Let L = Concentration of organic matterat 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 = -KLdL/L = -K dtdL/L = -K dt

ln L – ln Lo= -Ktln L/Lo = -Kt

ln L/Lo = -Kt

L/Lo = e –Kt

L = Lo e –Kt

tL

L

dtkL

dL

00

ktLL

L

0

ln

Let ‘y’ be the concentration of organic matter (BOD) consumed up to time ‘t’

y = Lo – Ly = Lo – Lo e –Kt

y = Lo ( 1 - e –Kt)i.e BOD consumed = ultimate BOD (1 – e –Kt) in ‘t’ daysTypical 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

time

BOD mg/l Lo

L

y

‘t’

The 5 day BOD of waste water is 190mg/l. determine ultimate BOD assuming K = 0.25 per day

Problem

SOLUTION

y = BOD exerted / consumedL = 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

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.

Problem

SOLUTION

y = Lo (1 – e Kt)

250 = Lo (1 – 5 x 0.23)

Lo = 365.83 mg/l

y2 = 365.83 (1 – e –0.23 x 2)

y2 = 134.89 mg/l

= 135 mg/l

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 = yBOD exerted / consumed = yBOD remaining = L

Problem

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) = -4kK = 0.245 per day

Lo = 160/e –4 x 0.245 =426.3 mg/l

At 25oC

K25 = K20 (1.047) T-20

K25 = 0.245 (1.047) 25-20

K25 = 0.308 per day

Y5 = Lo (1 - e –Kt)

Y5 = 426.3 (1 – e –0.308 x 5)

Y5 = 335 mg/l

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.

CHEMICAL OXYGEN DEMAND

DOMESTIC SEWAGE CHARACTERISTICS

Parameter Range (mg/l)

Total SolidsDissolved SolidsSuspended SolidsSetteleable SolidsBOD CODTotal NitrogenAlkalinity (as CaCO3)

350 – 1200250 – 850100 – 350

5 – 20 (ml / l)100 – 300

250 – 100020 – 85

50 – 200

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

POPULATION EQUIVALENT

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 -- flooding– irrigation -- drought– industrial water supply -- 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)

• Positives

– Recreational facilities

– Cheap power– Fewer brown

or black-outs– Smaller than

comparable hydroelectric dams

Environmental Impacts

Environmental Impacts

• Negatives– Loss of land– Loss of wildlife– Loss of energy

to pump water up to reservoir