Geoscience Reference
In-Depth Information
6. Calculate the specific amount to add to the composite container:
Required volume (mL) = FlowT
T
× Proportioning factor
(24.171)
where
T
is the time the sample was collected.
7. Mix the individual sample thoroughly, measure the required volume, and add to composite
storage container.
8. Keep the composite sample refrigerated throughout the collection period.
■
EXAMPLE 24.138
Problem:
The effluent testing will require 3825 mL of sample. The average daily flow is 4.25 MGD.
Using the flows given below, calculate the amount of sample to be added at each of the times shown:
8 a.m.
3.88 MGD
9 a.m.
4.10 MGD
10 a.m.
5.05 MGD
11 a.m.
5.25 MGD
12 noon
3.80 MGD
1 p.m.
3.65 MGD
2 p.m.
3.20 MGD
3 p.m.
3.45 MGD
4 p.m.
4.10 MGD
Solution:
3825 mL
9samples
Proportioning factor (PF)
=
=
100
× 4.25 MGD
Volu me
8a.m.
= 3.88 × 100 = 388 (400) mL
Volu me
9a.m.
= 4.10 × 100 = 410 (410) mL
Volu me
10a.m.
= 5.05 × 100 = 505 (500) mL
Volu me
11a.m.
= 5.25 × 100 = 525 (530) mL
Volu me
12noon
= 3.80 × 100 = 380 (380) mL
Volu me
1p.m.
= 3.65 × 100 = 365 (370) mL
Volu me
2p.m.
= 3.20 × 100 = 320 (320) mL
Volu me
3p.m.
= 3.45 × 100 = 345 (350) mL
Volu me
4p.m.
= 4.10 × 100 = 410 (410) mL
24.15.3 b
ioChemiCal
o
xygen
d
emand
C
alCulations
Biochemical oxygen demand
(BOD
5
) measures the amount of organic matter that can be biologi-
cally 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 refer-
ence manual (such as
Standard Methods
) for this information. Two basic calculations are used for
BOD
5
. The first is used for samples that have not been seeded, and the second must be used when-
ever BOD
5
samples are seeded. Both methods are introduced and examples are provided below.
24.15.3.1
BOD
5
(Unseeded)
(
)
DO
(mg/L)
− DO
final
(mg/L)
×
300 mL
start
BOD unseeded)
=
(24.172)
5
Sample volume(mL)
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