BOD / CBOD FROM A TO Z. Amy Starkey Stark County Sanitary Engineers

Transcription

1 BOD / CBOD FROM A TO Z Amy Starkey Stark County Sanitary Engineers

2 What is BOD????

3 What is BOD? It is a measure of the amount of oxygen consumed by bacteria during the decomposition of organic materials. Organic materials from the wastewater treatment facility act as a food source for the bacteria. Directly related to Dissolved Oxygen The bacteria require oxygen in the form of dissolved oxygen to decompose or eat the food source. Through a calculation, the amount of DO depletion between the initial day and final day of the analysis determines the BOD. Thus, BOD directly affects the amount of Dissolved Oxygen The greater the BOD = more rapid oxygen depletion = less oxygen available to aquatic life.

4 What is the Difference Between BOD and CBOD BOD represents the oxidation of carbons and nitrogenous compounds present in the water CBOD measures the oxidation of carbons present in water

5 TCMP

6 2-chloro-6-(trichloromethyl)pyridine prevents the oxidation of reduced forms of nitrogen such as ammonia, and organic nitrogen which exert a nitrogenous demand. Should add at the beginning of the test because nitrification will begin almost immediately if the right organisms are present (Baird and Smith, 2002).

7 BOD verses COD

8 BOD verses COD BOD represents the oxidation of carbons and nitrogenous compounds present in the water Analysis completion is done in 5-days COD Is the measure of the total amount of oxygen required to oxidize all organic material into carbon dioxide and water analysis only takes a few hours

9 BOD verses COD COD results are always higher than BOD results. Useful in determining an unknown BOD range for a sample but it can NOT replace the BOD test.

10 Methods

11 Approved Methods Standard Methods 18 th, 19 th, and 20th editions (5210B, 5-Day BOD Test)

12 Two Ways in Determining DO Iodometric Method (Winkler DO Method) Membrane Electrode Method

13 Winkler Method Azide Modification Method Preferred for most wastewaters Removes interferences caused by nitrite which is common in wastewaters. Permanganate Modification Used when ferrous iron is present Azide plus Potassium Fluoride Modification Used when 5 mg or more of ferric iron salts/l are present Alum Flocculation Modification Used when there is interference caused by suspended solids Copper Sulfate-Sulfamic Acid Flocculation Modification Used for biological flocs such as activated sludge mixtures which also have a high oxygen utilization rate

14 Sample Collection, Holding Time, and Storage

15 Grab Samples Ideally samples should be analyzed within 6-hrs of collection, however if this is not possible, then analyze samples within 48 hours of collection (40 CFR part 136). Store samples at < 6 C.

16 Composite Sampling Samples should be kept at or below < 6 C during compositing (limited to 24-hour period). start the measurement of holding time from the end of the compositing period. For example if the compositing was started at 8:30 am on Tuesday and ended at 8:30 am on Wednesday, then the 48-hour holding time would start from the end of the compositing period which would be 8:30am on Wednesday. Store samples at < 6 C

17 Quality Control & Procedure Requirements

18 BOD Quality Controls Blank Control Checks GGA Control Checks Glucose Glutamic Acid Seed Control Checks

19 Blank Control Checks Straight dilution water Used to determine cleanliness of bottles as well as the source water. It must have a DO uptake NO greater than 0.2 mg/l

20 GGA Control Checks Used to check the quality of the seeding material. Low results reflect poor seeding material. The ideal GGA range for a BOD sample is mg/l.

21 GGA Control Checks GGA Needs to be ph adjusted Initially the ph is around 4 Adjust between , like any other samples

22 Seed Controls Must have a DO uptake attributable to the seed added to each bottle between 0.6mg/L 1.0 mg/l. Most domestic wastewater, unchlorinated or undisinfected effluents will contain a sufficient population of microorganisms. Used to calculate the BOD results of samples which are seeded

23 BOD / CBOD Requirements ph of samples should be between a ph of Sample temperature should be adjusted to C before making dilutions After 5 days of incubation the final DO of samples must result in a DO depletion of at least 2.0 mg/l with a DO residual of no less than 1.0 mg/l. This is why it is recommended to make several dilutions of a sample. Example: The initial DO of a sample is 8.2 and after 5 days of incubation the final DO is 7.8. Then the final DO does not meet the required DO depletion of at least 2.0mg/L so a BOD result can not be calculated from this sample. Example #2: The initial DO of a sample is 8.2 and after 5 days of incubation the final DO is 0.20 mg/l. Here the final DO does not meet the required DO residual of at least 1.0 mg/l, so again the BOD result can not be calculated

24 Dilution Water

25 Dilution Water Sources Distilled Tap De-mineralized Natural Waters

26 Dilution Water Source Must be free of heavy metals, and toxic substances which inhibit micro-bacterial growth. Must also be able to maintain no more than a 0.20 mg/l DO depletion during the 5-day incubation period.

27 Reagents Added to Dilution Water Phosphate Buffer Solution Magnesium Sulfate Solution (MgSO 4 ) Calcium Chloride Solution (CaCl 2 ) Ferric Chloride Solution (FeCl 3 )

28 Purpose of Adding Trace Metals, Nutrients and Buffering Dilution Water Bacteria growth requires nutrients and trace metals. It is buffered to ensure the ph of the incubated samples remain in a range suitable for bacteria growth.

29 Why Dilute Samples Before Incubation? Because the BOD concentration in most wastewaters exceeds the concentration of DO available in an air-saturated sample.

30 Seeding

31 Why Seed? To add a population of microorganisms capable of oxidizing the biodegradable organic matter. Most domestic wastewater, unchlorinated or undisinfected effluents will contain a sufficient population of microorganisms.

32 Samples That may Require Seeding Chlorinated samples High temperature wastes Wastes with extreme ph values

33 Selecting a Seed Source Select a material to be used for seeding which will have a BOD of at least 180 mg/l. Example of seed sources according to Standard Methods 20 th Edition Raw domestic Sewage prepared as stated above Small quantities of digester supernatant, return activated sludge Commerically available seed material (Polyseed)

34 Seeding Must have a DO uptake/depletion of 2 mg/l after the 5-day incubation period, and also result in at least 1 mg/l residual DO (final DO).

35 Over Mixing the Ployseed Never let the vortex touch the stir bar Micro-organisms in the seed will be too tired to get the job done in your samples and may see low results in the seed factor.

36 Proper way to mix the Polyseed Mix on a speed of about 5, or so that the vortex is not touching the stir bar and splashing out. Mix for an hour Let bran settle out and transfer to another beaker to allow to mix for up to 5 hours on a speed setting between 1-2

37

38 Seed Calculations The DO uptake attributable to the seed (the seed factor or SF) added to each bottle is between 0.6mg/L 1.0 mg/l. The SF is calculated by using the equation below: SF = (B1 B2) x (f)

39 Seed Calculations SF = (B1 B2) x (f) Where : B1 = Initial Seed Control DO (before incubation) B2 = Final Seed Control DO (after 5-day incubation) f = ratio of seed in diluted sample to seed in seed control, or better see as f = (volume, mls of seed in diluted sample) (volume, mls of seed in seed control)

40 BOD 5, mg/l = (D 1 D 2 ) (SF) P Where: D1 = DO of diluted sample immediately after preparation, mg/l D2 = DO of diluted sample after 5-day incubation period P = decimal volumetric fraction of sample used SF = seed factor

41 General Procedure and Calculations

42 General Procedure 300ml BOD bottles are used. Filled with sample, seed, dilution water to overfilling (airtight) Samples brought to room temperature or 20 C + 1 C before making dilutions Initial DO is taken before incubation period (5-days at 20 C) Final DO is taken after incubation period

43 Unseeded BOD Calculation Where: BOD 5, mg/l = D 1 D 2 P D 1 = initial DO of sample D 2 = final DO of sample P = decimal volumetric fraction of sample used

44 EXAMPLE! 150 mls of a sample was added to a 300 ml BOD bottle and the initial DO of the sample is 8.2 and the final DO is 4.2, then what is the BOD 5 mg/l?

45 BOD 5, mg/l = D 1 D 2 D 1 = 8.2 D 2 = 4.2 P = 150 mls 300 mls P = 0.5 P BOD 5, mg/l = ( ) 0.5 BOD 5 mg/l = 8 mg/l

46 SEEDED BOD CALCULATION BOD 5, mg/l = (D 1 D 2 ) (SF) P Where D 1 = initial DO of sample D 2 = final DO of sample P = decimal volumetric fraction of sample used SF = the DO uptake attributable to the seed

47 EXAMPLE! 15 mls of seed was added to a 300 ml BOD bottle and labeled as the seed control. The initial DO was 8.2 mg/l and the final DO is 5.0 mg/l. What is the seed factor, SF if 4 mls of seed was added to the samples? Using the calculated SF value, what would be the BOD 5 mg/l if 150 mls of sample was added to a 300 ml BOD bottle along with 4 mls of seed and the initial DO was 8.2 mg/l and the final DO is 4.2 mg/l?

48 First Calculate the SF Value SF = (B1 B2) x (f) B1 = 8.2 B2 = 5.0 f = 4 mls 15 mls SF = ( ) x 4 mls SF = mls

49 Second calculate the BOD 5 mg/l of the sample: BOD 5, mg/l = (D 1 D 2 ) (SF) P D 1 = 8.2 D 2 = 4.2 P = 0.5 SF = BOD 5, mg/l = ( ) BOD 5, mg/l = 6.3 mg/l

50 Interferences

51 Common Interferences Samples with caustic alkalinity (ph > 8.5) Neutralize to a ph between Samples with acidity (ph < 6.0) Neutralize to a ph between Samples Supersaturated with DO (> 9 mg/l ) Residual Chlorine

52 Adjusting ph With samples of high alkalinity or acidity, use sulfuric or sodium hydroxide solutions of a concentration of which does not dilute the sample by more than 0.5% Standard Methods Recommends a concentration of 1N acid and alkali solutions to neutralize caustic or acidic waste samples.

53 Samples Supersaturated with DO Samples containing more than 9 mg/l DO at 20 C Cold water samples Where photosynthesis has occurred

54 Reducing Super-saturation Bring cold samples to room temperature or 20 C in a partially filled bottle and agitate the bottle by shaking vigorously or aerating with clean filtered compressed air.

55 Removing Residual Chlorine Residual chlorine can cause erroneous results, where no depletion may occur. In some samples chlorine will dissipate within 1 to 2 hours of standing in the light. For samples in which chlorine residual does not dissipate in a reasonably short time, destroy the residual chlorine by adding sodium sulfite (Na 2 SO 3 ).

56 Removing Residual Chlorine De-chlorinating with Sodium Sulfite (Na 2 SO 3 ) To the 100 ml sample: Add 1 ml of a 1:50 sulfuric acid solution Add 1 ml of potassium iodide solution 2 ml starch indicator Titrate against 0.025N Na 2 SO 3 X 1 = X Where, X 1 = known amount of titrant calculated to neutralize a 100 ml sample X 2 = amount of titrant needed to neutralize 150 ml sample

57 To calculate the amount of Na 2 SO 3 solution needed to neutralize a 75mL, 150mL and 250mL sample X 1 = X 2 X 1 = X 3 X 1 = X (X 1 )*(75) = (X 2 )*(100) (X 1 )*(150) = (X 3 )*(100) (X 1 )*(250) = (X 4 )*(100) (X 1 )*(75) = X 2 (X 1 )*(150) = X 3 (X 1 )*(250) = X Where: X 1 = known amount of titrant calculated to neutralize a 100mL sample X 2 = amount of titrant needed to neutralize a 75mL sample X 3 = amount of titrant needed to neutralize a 150mL sample X 4 = amount of titrant needed to neutralize a 250mL sample

58 EXAMPLE #1! If only using 150 mls of sample for the BOD test, and it is known that 3 mls of 0.025N Na 2 SO 3 will remove the residual chlorine from 100 mls of the same sample, then the equation for how much sodium sulfite the analyst would need is: X 1 = X

59 X 1 = 3 mls X 2 =? 3.0 = X (3.0)*(150) = (X 2 )*(100) (3.0)*(150) = X X 2 = 4.5 mls 0.025N Na 2 SO 3

60 EXAMPLE #2! The residual chlorine in 100 mls of a sample was removed by titrating with 0.025N Na 2 SO 3 solution. 3 mls of the titrant was used. Calculate the amount of titrant necessary to remove the residual chlorine from 75, 150, and 250 mls of the same sample.

61 X 1 = 3.0 mls X 2 = amount of titrant needed to neutralize a 75mL sample X 3 = amount of titrant needed to neutralize a 150mL sample X 4 = amount of titrant needed to neutralize a 250mL sample 3.0 = X = X = X (3.0)*(75) = (X 2 )*(100) (3.0)*(150) = (X 3 )*(100) (3.0)*(250) = (X 4 )*(100) (3.0)*(75) = X 2 (3.0)*(150) = X 3 (3.0)*(250) = X X 2 = 2.25 mls titrant X 3 = 4.5 mls titrant X 4 = 7.5 mls titrant

62 Determining Dilutions to Make on a Unknown BOD Sample

63 Determining Dilutions of an Unknown A PT Demand Standard has a BOD concentration range of mg/l. What dilutions are needed to find the result with in this range?

64 Determining Dilutions of an Unknown Where: BOD5, mg/l = D 1 D 2 (%X) DO 1 = Initial DO DO 2 = Final DO X = % volume of sample to be used

65 Step #1: Finding the dilutions on the lower end of the range (15.0 mg/l) BOD mg/l = 15.0 mg/l DO 1 = 8.0 DO 2 = 6.0 (assume a depletion of 2.0) 15.0 mg/l = (%X) (15.0 mg/l) x (%X) = 2.0 (%X) = mg/l (%X) = ( ) x 100 = 13.33% (round off to 13.3 %) = 40 mls

66 Step #2: Finding the dilutions on the lower end of the range (15.0 mg/l) BOD mg/l = 15 mg/l DO 1 = 8.0 DO 2 = mg/l = (%X) (15 mg/l) x (%X) = 7.0 (%X) = mg/l (%X) = (0.4666) x 100 = % (round off to 47 %) = 141 mls or round off to 140 mls

67 Finding the dilutions on the lower end of the range (15.0 mg/l) Step #1 yielded 13.3% (or 40 mls sample) Step #2 yielded 47% (or 141 mls sample)

68 Step #1: Finding the dilutions on the higher end of the range (250.0 mg/l) BOD mg/l = mg/l DO 1 = 8.0 DO 2 = 6.0 (assume a depletion of 2.0) mg/l = (%X) (250.0 mg/l) x (%X) = 2.0 (%X) = mg/l (%X) = (0.008) x 100 = 0.8% = 2.4 mls (round off to 2.0 mls)

69 Step #2: Finding the dilutions on the higher end of the range (250.0 mg/l) BOD mg/l = 250 mg/l DO 1 = 8.0 DO 2 = mg/l = (%X) (250 mg/l) x (%X) = 7.0 (%X) = mg/l (%X) = (0.028) x 100 = 2.8% (round off to 3 %) = 9 mls

70 Finding the dilutions on the higher end of the range (250.0 mg/l) Step #1 yielded 0.8% (or 2.0 mls sample) Step #2 yielded 3% (or 9 mls sample)

71 Recommended Dilutions from Standard Method s 20 th Edition % for strong industrial wastes % for raw and settled wastewaters % for biologically treated effluents % for polluted river waters

72 Questions??

73 References American Public Health Association; American Water Works Association; Water Environment Federation (1998). Standard Methods for the Examination of Water and Waste Water. 20th Edition, 4500-O Oxygen (Dissolved) American Public Health Association; American Water Works Association; Water Environment Federation (1998). Standard Methods for the Examination of Water and Waste Water. 20th Edition, 5210B 5-Day BOD Test Baird, R.B., & Smith, R.K.P. (2002). Third Century of Biochemical Oxygen Dmand. Alexandria, VA: Water Environment Federation.