1. SEWAGE CHARACTERISTICS

2. Composition >99.0% Water Solids 70% Organic 30% Inorganic Sewerage characteristics can be divided into three broad categories:- 1.Physical 2.Chemical 3.Bacteriological

3. 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 20 0 c, sewage will change from fresh to old in 2 ~ 6 hours.

4. Odour: Fresh domestic sewage has a slightly soapy or oil odour. Stale sewage has a pronounced odour of Hydrogen Sulphide (H 2 S). Solids:- Solids in sewage may be suspended or in solution solids are a measure of the strength of sewage.

5. 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

6. 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.

7. 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 0.45 103-105 0 C

8. 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.

9. 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 550 0 C. 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 550 0 C and are quiet resistant) VOLATILE SUSPENDED SOLIDS

10. 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 CO 2 and H 2 O. “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

11. Significance of BOD Significance: - 1.Used in design of waste water treatment plants. 2.Used to measure efficiently of waste water treatment plant.

12. 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

13. 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

14. 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 = -KL dL/L = -K dt

15. ln L – ln Lo= -Kt ln L/L o = -Kt L/L o = e –Kt L = L o e –Kt

16. Let ‘y’ be the concentration of organic matter (BOD) consumed up to time ‘t’ y =L o – L y=L o – L o e –Kt y=L o ( 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 20 o C. Value of ‘K’ is temperature dependent. K T =K 20 (1.047) T-20 K T =Value of K at temp T BOD represents amount of organic matter

17. time BOD mg/l Lo L y ‘t’

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

19. SOLUTION y=BOD exerted / consumed L=Amount of organic matter remaining at time ‘t’ L o =Ultimate BOD (Total Organic Matter) L=L o e -KT y=L o (1 - e -KT ) 190=L o (1 – e –0.25 x 5 ) L o =266.29 mg/l

20. Calculate the ultimate BOD for a sewage whose 5 day BOD at 20 o C is 250 mg/l. Assume K = 0.23 per day what will be BOD after 2 days. Problem

21. SOLUTION y=L o (1 – e Kt ) 250=L o (1 – 5 x 0.23) L o =365.83 mg/l y 2 =365.83 (1 – e –0.23 x 2 ) y 2 =134.89 mg/l = 135 mg/l

22. The BOD remaining in a sewage sample after 4 and 8 days was 160 and 60 mg/l respectively at 20 o C calculate the 5 day BOD of the sewage at 25 o C. Only if BOD = y BOD exerted / consumed =y BOD remaining =L Problem

23. SOLUTION L=L o e- Kt 160=L o e -K x 4 60=L o e -K x 8 L o =160 / e –4K 60=[160 / e –4K ] e –K8 60=160e –8K +4K 0.375=e –4K ln (0.375)=-4k K=0.245 per day

24. L o =160/e –4 x 0.245 =426.3 mg/l At 25 o C K 25 =K 20 (1.047) T-20 K 25 =0.245 (1.047) 25-20 K 25 =0.308 per day Y 5 =L o (1 - e –Kt ) Y 5 =426.3 (1 – e –0.308 x 5 ) Y 5 =335 mg/l

25. It is the amount of oxygen required to oxidize organic matter chemically (biodegradable and non-biodegradable) by using a strong chemical oxidizing agent. (K 2 Cr 2 O 7 ) in an acidic medium. For a single waste water sample the value of COD will always be greater then BOD. The oxidant (K 2 Cr 2 O 7 ) remaining is found out to find K 2 C r2 O 7 considered COD and BOD can be interrelated. CHEMICAL OXYGEN DEMAND

26. DOMESTIC SEWAGE CHARACTERISTICS ParameterRange (mg/l) Total Solids Dissolved Solids Suspended Solids Setteleable Solids BOD COD Total Nitrogen Alkalinity (as CaCO 3 ) 350 – 1200 250 – 850 100 – 350 5 – 20 (ml / l) 100 – 300 250 – 1000 20 – 85 50 – 200

27. 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 m 3 / 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

28. Water Resources Management

29. 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

30. 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

31. Factors Affecting Water Usage Population Economic considerations Cost of water Climate Type of water usage System management