Relationship between BOD, COD, TOC and ThOD

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.

Article by Sivanandan

T.P.Sivanandan

4 Comments

  1. Liza says:

    Good day!
    The relationship between BOD and COD that your site gave is very useful!
    I would like to know your reference book about this relationship, that is, the name of the book, its author, and its copyright date.
    I would greatly appreciate your help.
    Hoping to hear from you very soon.

    With much hope and gratitude,
    Liza

  2. Anonymous says:

    I think that a typo error is present at the paragraph where it is written :

    0.90 = BOD5/BODu as per —1
    0.77 = BODu/ThOD as per —2

  3. seyed khademi says:

    i,m manager of water & wastewater treatment and i want to tell that your site is very useful, you have been very helpful for me.
    best regards

    seyed mohammad seyed khademi

  4. Kanhaiya says:

    Stoichiometrically COD/TOC = 32 * no. O2 used for complete oxidation/(12 * no of carbon present in the compound). It is not just the ratio of molar mass of O2/ Molar mass of carbon
    e.g for methane CH4 + 2O2 = CO2 + 2H2O
    COD = (2*32/16) * W (g), TOC = (1*12/16)* W(g)
    COD/TOC = 2*32/(1*12) = 5.33

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