Environmental Engineering Reference
In-Depth Information
Solution:
Proportioning factor = 3825 mL ÷ (9 samples × 4.25 MGD) = 100
Volu me
8a.m.
= 3.88 × 100 = 388 mL
Volu me
9a.m.
= 4.10 × 100 = 410 mL
Volu me
10a.m.
= 5.05 × 100 = 505 mL
Volu me
11a.m.
= 5.25 × 100 = 525 mL
Volu me
12noon
= 3.80 × 100 = 380 mL
Volu me
1p.m.
= 3.65 × 100 = 365 mL
Volu me
2p.m.
= 3.20 × 100 = 320 mL
Volu me
3p.m.
= 3.45 × 100 = 345 mL
Volu me
4p.m.
= 4.10 × 100 = 410 mL
BIOCHEMICAL OXYGEN DEMAND CALCULATIONS
Biochemical oxygen demand (BOD
5
) measures the amount of organic matter that
can be biologically oxidized under controlled conditions (5 days at 20°C in the dark).
Several criteria determine which BOD
5
dilutions should be used for calculating test
results. Consult a laboratory testing reference manual (such as
Standard Methods
)
for this information. Two basic calculations are used for BOD
5
. The irst is used for
samples that have not been seeded, and the second must be used whenever BOD
5
samples are seeded. Both methods are introduced and examples are provided below.
Bod
5
(u
nseeded
)
[
]
× 300
DO
(mg/L)
−
DO
(mg/
L)
mL
start
final
BOD unseeded)
=
(16.5)
5
Sample volume(mL)
■
Example 16.6
Problem:
A BOD
5
test has been completed. Bottle 1 of the test had dissolved oxygen
(DO) of 7.1 mg/L at the start of the test. After 5 days, bottle 1 had a DO of 2.9 mg/L.
Bottle 1 contained 120 mg/L of sample. Determine the unseeded BOD
5
.
Solution:
(
)
× 300
7.1 mg/L
−
2.9 mg/L
mL
BOD unseeded)
=
=
10.5 mg/L
5
1
20 mL
Bod
5
(s
eeded
)
If the BOD
5
sample has been exposed to conditions that could reduce the number of
healthy, active organisms, the sample must be seeded with organisms. Seeding requires
the use of a correction factor to remove the BOD
5
contribution of the seed material:
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