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Sunday, 16 December 2018

Characteristics of Sewage


Decomposition of sewage:

Physical characteristics:
  • Turbidity 
  • Colour 
  • Odour 
  • Temperature

Colour:         
Indicates the freshness of sewage
Yellowish, grey or light brown - fresh sewage
Black or dark brown - stale and septic sewage

Temperature:
Has an effect on
  • biological activity 
  • solubility of gases in sewage

Chemical characteristics:


Organic matter: 
  • Carbohydrates such as cellulose, fibre, starch, sugar. etc..
  • Fats and oils
  • Wastes from animals, urea fatty acids, hydrocarbons.

Determination of solids:
Total solid &(in mg/7):
  • Determined by evaporating a known volume of sewage sample, and weighing the dry residue left.
  • The mass of the residue divided by the volume of the sample evaporated ,will represent total solids in mg/l  

Suspended solids: 
  • Those solids retained by a filter of 1 
  • glass fibre felter appratus is used


Dissolved and Colloidal solids:
  • Difference between the total solids and suspended solids

Volatile and fixed solids:
Total suspended solids = Volatile + fixed
  • Volatile solids:

  1. Suspended solids are burnt and ignited at about 550°C in an electric muffle furnace for about 15 to 20 min.
  2. Loss of weight due to ignition represents volatile solids.

  • The difference between the suspended solids and volatile solids gives fixed solids.

Settleable solids:
  • Sewage is allowed to stand in this imhoff cone for a period of 2 hours
  • The quantity of solids settled in tile bottom of the cone can be directly read as settle able solids.

Dissolved oxygen (D.O):
  • To know the extent of pollution of sewage 
  • At least 4 ppm of DO should be ensured while discharging sewage into river stream.

Types of Organic matter:
  • Biologically active (means bacteria can stabilize this organic matter)
  • Biologically inactive (means bacteria can not act on this organic matter)

Oxygen Demand: 
  • It is the amount of of oxygen consumed by readily oxidizable organic matter for oxidation.

Chemical Oxygen Demand(C.O.D) :
  • It is chemical oxidation with Potassium Permanganate or Potassium dichromate in an acid solution.

Theoretical oxygen Demand: 
  • It is the oxygen demand that can be worked out theoretically.

Biochemical Oxygen Demand (B.O.D): It is the amount of oxygen required for biological decomposition of biodegradable organic matter under aerobic conditions at a specified temperature and for a specified duration.

  • Standard B.O.D. is reported at 20°C for 5 days period.
  • B.O.D. -5 days at 20o- 68% of the total demand
  • C.O.D represent both biologically active and inactive organic matter
  • Where as D.O.D. gives biologically active organic matter only
  • Therefore C.O.D. > B.O.D
Test method:
  • Initial D.O measured in the beginning for the sample diluted with water.
  • Sample Incubated for 5 days at 20°C
  • Final D.O of the sample measured after 5 days.
  • B.O.D. (Initial D.O - Final D.O) x Dilution Factor

Dilution factor: Number of times sewage diluted with distilled water
Eg: For 2 % solution, Dilution
Factor =100/2=50
  • Rate of change of Organic matter with time is directly proportional to the organic matter present in the sewage at the time

Where Lt=Amount of organic matter present at time "t"
K = Rate constant, KD = deoxygenation constant

Carbonaceous Demand: The first demand that occurs during the first 20 days due to
the Oxidation of Organic matter is called Carbonaceous demand (or) I stage demand.
  • The term B.O.D. usually mean I stage demand.
  • Represented by "OAB” in the above figure

Nitrogenous demand:
  • The latter demand that occurs due to biological Oxidation of Ammonia.  
  • Represented by “AC” in the above figure.

B.O.D/C.O.D. Ratio:
  • If BODu/COD lies between 0.92 and I, then the waste water can be considered to fully biodegradable.
  • If BOD5 /COD) vary between 0.63 and 0.68 then the waste water can be considered to b fully biodegradable wastes.

Relative Stability:
  • Ratio of amount of oxygen available in the effluent (D.O) to the total oxygen required to satisfy the first stage BOD demand.

Population Equivalent:
  • Industrial waste waters arc generally compared with per capita normal domestic waste water so as to charge industries properly.

Population equivalent =
Total standard BOD5 of industrial sewage per day/Standard BOD5 of domestic sewage per person per day
  • Average standard BOD5 of domestic sewage is about 0.08 kg/day/person.


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