Oxygen Demand of Water (exp:10) - assignment - Q-A

Determination of Biochemical Oxygen Demand of Water
(1) Q: A sample of sewage is mixed with water( no seeding done) in the ratio of 1: 20 (i.e., 1 ml of sewage diluted to 20 ml by adding water) for BOD test. The initial DO is 8.5 mg/l and final DO after 5 days, is 3.1 mg/l. calculate BOD5 of the sewage.
Ans: The five-day BOD of a diluted sample is given by
BOD5 = [DOI - DOf ] X D.F. - - - - - - - - - - - - (1)
Where
D.F. = Dilution factor = (vol. of wastewater + dilution water) / (vol. of
wastewater)
= (1+19)/ 1
= 20
Here, DOI =initial DO = 8.5 mg/l
DOf =final DO = 3.1 mg/l
By using equation (1)
BOD5 = (8.5 – 3.1) X 20
= 5.4 X 20
= 108 mg/l
Therefore, BOD5 of the sewage is 108 mg/l
(2) Q: A test bottle containing just seeded dilution water has its DO level drop by 0.8 mg/l in a 5-day test. A 300 ml BOD bottle filled with 30 ml of wastewater and the rest with seeded dilution water experiences a drop of 7.3 mg/l in the same period (5-day). Calculate the BOD5 of the wastewater.
Ans: The oxygen demand of the wastewater (BODw) is determined by the following equation
BODm Vm = BODw Vw + BODd Vd - - - - - - - - ( 1 )
D.F. = Dilution factor = (vol. of wastewater + dilution water) / (vol.
of wastewater)
= (30 + 270)/ 30
= 10
Where,
BODm = the BOD of the mixture of wastewater = 7.3 mg/l
BODd = the BOD of the dilution water alone = 0.8 mg/l
BODw = the BOD of the of wastewater
Vm = the volume of wastewater and dilution water = 300 ml
Vw = the volume of wastewater = 30 ml
Vd = the volume of dilution water = 270 ml
By using equation (1)
BODm Vm = BODw Vw + BODd Vd
7.3 X 300 = BODw X 30 + 0.8 X 270
2190 = BODw X 30 + 216
BODw = (2190 – 216)/ 30
BODw = 1974/30
BODw = 65.8 mg/L
Therefore, BOD5 of the wastewater is 65.8 mg/l
(3) Q: In a BOD test on a diluted wastewater sample ( 1: 20 dilution, but not needed), the initial DO is 8.4 mg/L, and final DO after 5 days is 4.2 mg/L. If the reaction rate constant is 0.22/day, calculate
(a) 5- day BOD of the wastewater,
(b) Ultimate carbonaceous BOD of the wastewater
(c) Remaining Oxygen demand after 5-days
(a) Ans: determination of 5- day BOD of the wastewater,
The five day BOD of a diluted sample is given by
BOD5 = [DOI - DOf ] X D.F. - - - - - - - - - - - - (1)
Where
D.F. = Dilution factor = (vol. of wastewater + dilution water ) / (vol.
of wastewater)
= (1+19)/ 1
= 20
here, DO0 = initial DO = 8.4 mg/l
DO5 = final DO = 4.2 mg/l
By using equation (1)
BOD5 = (8.4 – 4..2) X 20
= 4.2 X 20
= 84 mg/l
5- day BOD of the wastewater is 84 mg/L
(b) Determination of ultimate carbonaceous BOD of the wastewater. Lo
BOD5 = L0 (1 – e-kt )
84 = L0 (1 – e-0.22 X 5 )
L0 = 125.91= 126 mg/L
ultimate carbonaceous BOD of the wastewater.Lo = 126 mg/L
(c) remaining Oxygen demand after 5-days, L5
L0 = BOD5 + L5
L5 = L0 – BOD5
L5 = 126- 84
L5 = 42 mg/L
remaining Oxygen demand after 5-days, L5 = 42 mg/L

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