Suppose if we have waste sample having BOD 4500 mg/l , how to reduce its BOD to say around 100-180 mg/l to carry out laboratory level experiment. And how to decide its dilution fraction
Use M1V1 =M2V2 relation for dilution as:
M1 = 4500 mg/I
V1 = ?
M2 = 100 mg/l
V2 = Say you want to prepare 1 litre
V1 = 100 X 1 /4500 = 0.022 litre
dilution factor = 4500/100 = 45X
good luck!
Dilution of the waste water sample may provide nutrients for microorganisms that they need to oxidize the organic matters contents. The dilution depends on the measuring device and its instructions. Finally, the results obtained could be related to the basic sample by using appropriate conversion factors (between diluted and basic sample). for your last part of question you can use Mr. Arshad Ali's formula.
Dilution is needed to prevent the depletion of entire oxygen present in in BOD bottle in less than incubation period of 5 days. For a valid BOD test APHA manual specifies some min DO of 1 mg/L to be present in the bottle after end of 5 days and DO consumed should be at least 2mg/L during incubation. Suppose saturation DO is 8 mg/l and expected BOD is 400 mg/L then, volume of sample to be added in 300 ml BOD Bottle has the following range:
minimum ml sample = [(8 mg/L 6 mg/L) x 300 mL]/400 mg/L=1.5 ml
maximum ml sample = [(8 mg/L 1 mg/L) x 300 mL]/400 mg/L=5.25 ml
Also, excess dilution reduces precision.
Mr Anushman and the others provided the correct answers already.
since the saturation of DO in water is limited. in ideal condition (distilled water and 0 oC) it can not be solved more than 14.6 mg/L. So for natural waters it decrease to below 10 mg/L. As you can not have more than this values , the waste-water should be diluted.
You have already good answers above but you may also see Metcalf & Eddy Wastewater Engineering who give guidance on dilutions depending on the BOD of the wastewater.
Very good question posed and already answers are present. Without dilution; one can not get acurate answer. You have to keep samples for 3 or 5 days and your oxygen must be present at end of that day; so dilution with nutrient water is needed. Or else all the oxygen will expire..
how to start dilution of the wastewater sample for BOD analysis or how much quantity of wastewater should take for dilution to 300 ml in the sample?
Chedly Tizaoui
Respected Sir,
Please mention the edition and page number of the book from which you have given the BOD dilution range table.
Book of Wastewater Engineering- Treatment and Reuse by METCALF & EDDY (Mc Graw Hill - 4th Edition)- pg 84 in Chapter 2.
Dilution is needed to prevent the depletion of entire oxygen present in BOD bottle over incubation period of 5 days. The dilution can be prepared as % mixture = (vol. of sample/vol. of diluted sample) x 100. The Choice of dilution depend on the type of sample. For a valid BOD test APHA manual specifies at least 1.0 mg/L DO remaining in the bottle and DO uptake should be at least 2mg/L after end of 5 days incubation for reliability. When in doubt, use more than one dilution to obtain the require range of DO. Example: when three dilutions were used (Understanding dilution and seeding procedures in BOD test : Training module # WQ – 16, New Delhi, May 1999:
http://www.mahahp.gov.in/images/16understandingdilutionandseedingproceduresinbod test):
Hello, I have one question about this, Can I use deionized water to dilute it?
Hi, I am confused with the calculation regarding the BOD5 for seeded dilutions. I could see the usage of two different formulas.
Formula 1:
BOD5 mg/l = (D1-D2)-(B1-B2)f/P
Formula 2:
BOD5 mg/l = (D1-D2)-(S)Vs/P (As per standard methods).
I am ok with formula one. Could any one of you please explain how the formula 2 is applied in calculation where 10 ml of seed is added to 300 ml BOD bottle containing buffered dilution water (seeded control) and another bottle containing 5 ml of waste water sample and 5ml of seed in a 300 ml bottle (seeded sample). The initial and final DO being 8.6 and 5.8 mg/l for seeded control and 8.7 and 4.8 mg/l for the seeded sample respectively.
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