Question

Calculate BOD5 of all 3 samples.

	Source	Volume of BOD (mL)	Volume of WW sample added (mL)	Initial DO (mg/L) dddddd	Final DO (mg/L) ddddd	Dilution factor (P) ddddddd	Seed Control ddddddddd	Amount of Seed added (mL) dddddd	BOD5 (mg/L) dddddddddd
Sample 1	MWW	300	5	6.90	3.60	0.0167	WAS	1
Sample 2	MWW	300	5	6.60	4.23	0.0167	WAS	1
Seed Control		300	5	6.81	6.56	0.0167	WAS	1

Answer 1

where:

IDO = initial D.O. of dil

uted sample, mg/L

FDO = final D.O. of diluted sample, mg/L

VS = volume of sample, ml

VB = volume of bottle, ml

BOD = biochemical oxygen demand, mg/L

When seed is added, seed control tests must be done to determine the amount of oxygen depleted by the seed. Calculation of BOD includes a correction factor for seed as expressed by Equation 2.

where:

IDO = initial D.O. of diluted sample, mg/L

FDO = final D.O. of diluted sample, 0 mg/L

VS = volume of sample, ml

VB = volume of bottle, ml

Cavg= correction factor for oxygen depletion resulting from presence of seed

SV =  Volume of Seed added to Sample in ml.

BOD = biochemical oxygen demand, mg/L

So by using eq .2 Final BOD5 is calculate for sample 1 as:

(6.9-3.6) - (0.0167*1)   

> ------------------------------------ = 196.6 mg/l

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