Sulphite:

The Determination of sulfite usually is made only on boiler waters.Generally speaking sulfite is not present in natural water. in boiler feed water sodium sulphite is fed to remove dissolved oxygen.

Reagents

1. 1N Hydrochloric acid 1+1

2. Starch indicator solution

3. Standard potassium iodide-iodate solution.

 

Procedure:

Place 10ml of 1+1 HCL in a 250 ml flask. rapidly add 100 ml of sample, submerging the pipette tip below the acid surface to minimize sir exposure. After adding 1 ml.starch indicator solution add 5ml.KI Solution(5% solution) titrate with standard KIO3 Solution, to the first appearance of persistent blue color. Determine the blank titration by taking 100 ml Distilled water.

Calculation:

Sulfite,as SO3 mg/litre = (A-B) x 100

 

Where,

        A= ml.of titrant for sample

        B= ml.of titrant for blank

As mg./litre, the effect on P will be equal to one half of the content in mg./litre of phosphate and this shall be subtracted from P reading before using the above table.the alkalinity to phenolphthalein after addition of BaCL2 is not affected by the presence of phosphate.

 

Procedure:

To 100 ml.of sample, add a crystal of sodium sulfate and 10 ml.of 10% BaCl2 Solution.

Mix well for 2 minutes. Titrate using phenolphthalein as indicator.

 

Calculation:

Ml.of 0.02 H2SO4 X 1000/ Ml. of sample = Mg./litre of P(BaCl2)value.    

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For a completely biodegradable wastewater such as glucose, approximately ten percent of the original organics remain as non-biodegradable cellular residues after biological oxidation. Hence, the cellular residues are not measured by the BOD test. Therefore:

BODu = 0.9ThOD  ---1

where BODu = ultimate BOD

            ThOD = theoretical oxygen demand For domestic sewage and some biodegradable industrial wastes, the relationship between BOD5 and BODu is:

BOD5 = 0.77BODu  ---2

where BOD5 = 5 day BOD

            BODu = ultimate BOD

For most wastewaters:

ThOD = COD ---3

where ThOD = theoretical oxygen demand

COD = chemical oxygen demand since the COD test oxidizes all organics except for those which are totally resistant to dichromate oxidation.

Stoichiometrically, the COD/TOC ratio should be approximately the molecular ratio of oxygen to carbon:

COD/TOC =32/12= 2.66 ---4

The ratio will actually range from zero, when organic material is resistant to dichromate oxidation, to as much as 6.0 when inorganic reducing agents are present.

For raw domestic sewage and some biodegradable industrial wastes, the following ratio of BOD5/TOC occurs:

BOD5/TOC =32/12(0.90)(0.77) = 1.85 ---5

where BOD5 = 5 day BOD

TOC = Total Organic Carbon

0.90 = BOD5/BODu as per ---1
0.77 = BODu/ThOD as per ---2
              As a wastewater is oxidized through a wastewater treatment plant, the BOD5/TOC ratio will drop. A treatment plant effluent may have a BOD5 /TOC ratio of as low as 0.5 since the effluent wastewater is so much less biodegradable. (It has already been largely degraded).

              The BOD5 to COD ratio for domestic waste and certain biodegradable industrial wastes can be computed as follows:

BOD5 = 0.7 COD ---6

where BOD5 = 5 day BOD

              COD = chemical oxygen demand

                 0.7 = 1.85 /2.66

               1.85 = BOD5/TOC as per --5
               2.66 = COD/TOC as per   --4
                 This ratio can also vary widely depending on the state of biodegradation of the wastewater. The author has found this ratio as low as 0.1 after several days of oxidation. If the BOD of a biodegradable wastewater equals zero, the wastewater will be completely biodegraded. There is some controversy about whether this ever occurs. Many authors will say that the Non-biodegradable Residue (NBDR) is as high as 0.10 as explained in the first paragraph of this section. The author has found that in activated sludge systems with hydraulic detention times in the range of 14 days, there is no accumulation of volatile suspended solids, which indicates that all organics are ultimately degraded under certain anoxic conditions.

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a AOBr AOCl AOI AOX b B0 B5 BOD BOD5 BODS BODSEED c CBOD COD CL CMWD D0 D5 DBP DO DOC DOX ECD ECNI EOX EPA f f ’ FAAS FIA FI–CL GC HAA HAA5 HAA9 HBOD HTC HX IC ICP–AES LOD MS MW NDIR NOM NDOC NPOC NPOX NVOC OMWD P POC POX R RSD TAME TC THM ThOD TIC TOC TOBr TOCl TOI TOX US USD UV VOC Vsample

Volume of titrating solution employed with the blank Adsorbable organic bromine Adsorbable organic chlorine Adsorbable organic iodine Adsorbable organic halide Volume of titrating solution employed with the sample Dissolved oxygen in the dilution water before incubation Dissolved oxygen in the dilution water after 5 days of incubation at 20 Deg C Biochemical oxygen demand Biochemical oxygen demand after an incubation period of 5 days BOD supplied by sensors Formulated uniform dehydrated microbial consortium for BOD estimation Normality of the titrating solution Carbonaceous BOD Chemical oxygen demand Chemiluminescence Closed microwave-assisted digestion Dissolved oxygen in the diluted sample after preparation Dissolved oxygen in the diluted sample after 5 days of incubation at 20 Deg C Disinfection by-products Dissolved oxygen Dissolved organic carbon Dissolved organic halide Electron capture detection Electron capture negative ionization Extractable organic halide Environmental Protection Agency ratio of the % seed in the diluted water to the % seed in the dilution water titrating solution correction factor Flame atomic absorption spectrometry Flow injection analysis Flow-injection chemiluminescence Gas chromatography Haloacetic acids Five species of HAAs currently regulated Nine species of HAAs Headspace biochemical oxygen demand High-temperature combustion Hydrogen halide Inorganic carbon Inductively coupled plasma–atomic emission spectrometry Iimit of detection Mass spectrometry Microwave radiation Nondispersive infrared spectrometry Natural organic matter Nondissolved organic carbon Nonpurgeable organic carbon Nonpurgeable organic halide Nonvolatile organic carbon Open microwave-assisted digestion Decimal volumetric fraction of sample used Purgeable organic carbon Purgeable organic halide Regression coefficient Relative standard deviation Tert-amyl methyl ether Total carbon Trihalomethanes Theoretical chemical oxygen demand Total inorganic carbon Total organic carbon Total organic bromine Total organic chlorine Total organic iodine Total organic halide Ultrasound radiation Ultrasound-assisted digestion Ultraviolet Volatile organic carbon Volume of the water sample analyzed

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In Extended Aeration, This process takes raw sewage directly into an aerated mix tank for 8 h or more to provide bacteria with optimum conditions to consume the BOD present in the wastewater. The effluent from this mix tank goes to a sedimentation tank where the flocculated colonies of organisms are settled to produce a clear overflow.

Extended Aeration Process Scheme.

A portion of the settled microbial floc is returned to the head works and a portion sent to sludge disposal. The clear effluent is then directed to final treatment such as disinfection, perhaps passing through a final polishing filter. This method of
treatment is particularly suited to plants that have a low concentration of settleable solids in the raw sewage. It minimizes the number of unit operations involved in smaller plants.

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Solids, liquids, and gases may dissolve in water to form solutions. The amount of solute present may vary below certain limits, so-called solubility. The strength of a solution can be expressed in two ways:

(1) weight (lb) of active solute per 100 pounds (i.e. %) and

(2) weight of active solute per unit volume (gallons or liters) of water.

Either expression can be computed to the other if the density or specific gravity is known. If the solution is dilute (less than 1%), the specific gravity can be assumed to be 1.0; i.e. 1 L of solution is equal to 1 kg and 1 gal of solution equals 8.34 lb.
In water chemistry, molarity is defined as the number of gram-molecular weights or moles of substance present in a liter of the solution. If solutions have equal molarity, it means that they have an equal number of molecules of dissolved substance per unit volume. The weight of substance in the solution can be determined as follows:
The molarity (M) of a solution can be expressed as:

M (mol/L) = moles of solute (mole) / 1.0 L of solution

For uniform purpose,

the Normality (N) is used for preparation of laboratory solutions. The
normality can be written as:
    N (eq/L or meq/L) = equivalent of solute (eq or meq) / 1.0 L of solution

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Integrity Test is very essential to evaluate whether the module/element’s present state is good or it is going to end of life. Integrity Test is really means of fiber repair procedure.

The test is very useful for OUT to IN Flow style element Like DOW and Hyflux

Requirement:

You will need the following to complete the fiber repair

procedure:

• Concentrate tube plug

• Air supply apparatus

• Oil-free compressed air [recommended pressure: 3.0 bars (44 psi)]

• Loctite 406 glue

• Repair pins

• Personal protection equipment (gloves, safety glasses, etc.)

• Knife or diagonal cutters

Figure 1. Integrity test Feed Valve (OFF)Con. Valve (OFF)Air Inlet Valve (ON) PI Transparent TubePermeate Valve (ON)Observe whether air bubbles continuously appear

Procedure:

The bubble test using transparent tubes has been selected to illustrate how it is done.

1. Take the module out of the filtration mode.

2. Drain the module from the feed side.

3. Close feed and concentrate valve and keep the permeate valve open.

4. Pressurize the drained side of the module with oil-free compressed air from the air inlet valve, and slowly raise the air pressure to 1.5 bars (21 psi). Some displaced water will flow out the permeate side.

5. If large continuous air bubbles appear in the transparent tube then the module has broken fibers. Smaller and infrequent bubbles are the result of air diffusion through the pores of the ultrafiltration membrane. If leaks are confirmed, move to STEP 2.

Step-2

1. Drain the module from the feed side.

2. Remove the top end cap. Keep the bottom end cap on.

3. Isolate the module to perform repair by closing valves or sealing remaining openings.

4. Place the positioning block into the concentrate outlet tube. Then put in the fastening nut. Next, put in the cylinder block and screw in the fastening bolt. [Positioning block, fastening nut, cylinder block and fastening bolt are part of the concentrate tube plug assembly.]

5. Connect the air supply line.

6. Supply oil-free compressed air to the module and slowly raise pressure to 1.5 bars (or 21 psi). 7. Provide a water stream to cover the permeate end of the fibers to help locate any leaks.

8. As the air pressure rises, continuous air bubbles will appear at the location of a broken fiber. Mark the broken fiber with a pin.

9. Continue applying air and water until all broken fibers are located and marked.

10. Depressurize the module using the air release valve.

11. If the broken fibers are near the concentrate tube plug then remove the concentrate tube plug to provide room for repair.

12. Take a new repair pin and place a drop of glue on the end of the pin. Use it to replace the fiber- marking pin immediately. Push the pin firmly into the leaking fiber. Let the repair cure for 5 minutes before trimming the extruding portion of the pin with a knife or diagonal cutters. Repeat for all broken fibers.

13. Repeat steps 4 - 8 to make sure all broken fibers are repaired.

14. To complete the repair, depressurize the membrane, remove the concentrate tube plug, and reassemble the top end cap, remove seals, and realign valves for operation.

You have successfully repaired a Ultrafiltration module!

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