SEWAGE CHARACTERISTICS

SEWAGE CHARACTERISTICS

Composition>99.0% WaterSolids70% Organic 30% Inorganic Sewerage characteristics can be divided into three broad categories:-Physical Chemical Bacteriological

PHYSICALPhysical 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

BACTERIOLOGICALEnormous 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 CHARACTERIZATIONSOLIDS 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 0.45 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 0.45 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 BODSignificance: -

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 ConstantdL/dt = -KLdL/L = -K dtdL/L = -K dt

ln L ln Lo= -Ktln L/Lo = -Ktln L/Lo = -KtL/Lo = e KtL = Lo e Kt

Let y be the concentration of organic matter (BOD) consumed up to time ty =Lo Ly=Lo Lo e Kty=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-20KT=Value of K at temp TBOD represents amount of organic matter

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 tLo=Ultimate BOD (Total Organic Matter)L=Lo e-KTy=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/ly2=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 =LProblem

SOLUTION

L=Lo e-Kt160=Lo e-K x 4 60=Lo e-K x 8Lo=160 / e 4K 60=[160 / e 4K ] eK860=160e 8K +4K0.375=e 4Kln (0.375)=-4kK=0.245 per day

Lo=160/e 4 x 0.245=426.3 mg/l

At 25oCK25=K20 (1.047) T-20K25=0.245 (1.047) 25-20K25=0.308 per dayY5=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 3505 20 (ml / l)100 300250 100020 8550 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/lQ=10,000 m3 / dayTotal BOD load by tannery =BOD x Q=500 x 10000 / 1000=5000 kg BOD/dayPopulation equivalent =(5000 / 80) x 1000=62,500 persons POPULATION EQUIVALENT

Water Resources Management

DefinitionsHydrology the scientific study of the properties, distribution and effects of water on the earths surface, in the soil and underlying rocks and in the atmosphereWater resources management -- control and utilization of water for beneficial uses or to avoid adverse impactsdrinking water supply -- floodingirrigation-- droughtindustrial water supply-- subsidence

Determining Water DemandAverage daily water consumptionAbility to meet continuing demand over critical periodsEstimate stored water requirement quantitiesPeak demand rates (~ 2.2 x daily average flow for metered dwellings)Size of plumbing and piping, pressure lossesEstimate storage requirements during periods of peak water demand

Factors Affecting Water UsagePopulationEconomic considerationsCost of water ClimateType of water usageSystem management

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