REVERSE OSMOSIS

Principle: 

Osmosis is a natural process that when a pure water and saline water are separated by a membrane which can permit only water molecules to pass through. Pure water will flow into the saline water due to a tendency called osmosis. This natural tendency can be reversed by applying an opposite pressure to the saline water so that water molecules from saline water tend flow out as pure water. This is called reverse osmosis.

Process:

·         It consists of several horizontal compartments separated by semi-permeable membranes.

·          A pressure of 15-40 kg/cm² is applied to the saline water to force its pure water to pass through the semi-permeable membrane leaving behind the dissolved salts.

·         The pure water from different compartments are collected and the concentrated saline water left out is discarded.

·         The membrane is generally a thin film of cellulose acetate or polymethyl methacrylate.

Advantages:

1. This method is economical, compact and very simple.

2. Life time of the membrane is high about 2 years.

3. Easy to replace the membrane within two minutes.

4. Ionic, non ionic, colloidal and high molecular weight organic matters are easily removed.

Chemical Oxygen Demand (COD)

Definition

COD is defined as the amount of oxygen used while oxidizing the total organic load of the sample with a strong chemical oxidant like K₂Cr₂O₇ in acid medium. It is represented in mg/dm³ or ppm.

Importance of COD:

1. It is satisfactory, quantitative method for measuring total organic load.

2. Rapidly measurable parameter and needs about 3 hours for completion.

3. In general COD>BOD since both biodegradable and non biodegradable organic load are completely oxidized.

Experiment to determine Biological oxygen demand

Principle: A known volume of sewage water is diluted with a known amount of dilution water (containing nutrient for bacterial growth). The solution is kept in incubator for a period of 5 days at 20⁰ C. The DO of diluted water sample (D₁) and the solution taken out of the incubator (D₂) are determined. From these values of BOD is calculated.

Procedure:

A)   Preparation of two separate diluted water samples:

Pipette out a known volume of sewage sample (A ml) and dilute it to a known volume (B ml) by adding dilution water containing nutrient and Free O₂. Fill it in two separate BOD bottles.

B)   Blank titration:

In bottle 1, find DO immediately as follows.

·         Add 2ml of MnSO₄ and 3ml of alkaline KI. Stopper the bottle and mix well for 10- 15min.

·         Wait for 2 min. Then add 1ml 1 ml of conc. H₂SO₄ to dissolve MnO₂ ppt.

·         Pipette out known volume (V ml) of this solution and titrated against std. Na₂S₂O₃ using starch as indicator.

·         Note the volume of Na₂S₂O₃ consumed (V₁ml).

C)   Back titration:

·         Bottle 2 is incubated for 5 days at 20^o C.

·         After 5 days, add 2ml of MnSO₄ and 3ml of alkaline KI. Stopper the bottle and mix well for 10- 15min.

·         Wait for 2 min .Then add 1ml 1 ml of conc. H₂SO₄ to dissolve MnO₂ ppt.

·         Pipette out same known volume (V ml) of this solution and  titrated against same std. Na₂S₂O₃ using starch as indicator.

·         Note the volume of Na₂S₂O₃ consumed (V₂ml)

D) Calculation:

1)   DO in blank titration:

Volume of water sample taken= V ml

         Volume of sodium thiosulphate consumed = V₁ ml

         Normality of sodium thiosulphate used = N₁

         Normality of oxygen content in water =    V₁ x N₁

                                                                     V

                                                               = N₂

          Hence oxygen content in blank solution = N₂ x Equivalent weight of oxygen

                                             = N₂ x 8 x 1000 mg/dm³ = D₁ mg/dm³

2)   DO in back titration:

           Volume of water sample taken= V ml

           Volume of sodium thiosulphate consumed = V₂ ml

           Normality of sodium thiosulphate used = N₁

           Normality of oxygen content in water =    V₂ x N₁

                                                                         V

                                                               = N₃

           Hence oxygen content in back solution = N₃ x Equivalent weight of oxygen

                                               = N₃ x 8 x 1000 mg/dm³ = D₂ mg/dm³

                 Therefore BOD = (D1-D2) x B   mg/dm³

                                                      A