WATER QUALITY - Water Quality Engineering in Natural Systems - page 14

river Thames has a travel time to the sea of less than 5
days, so there was no need to consider oxygen demand
at longer times. An additional consideration in choosing
the 5-day duration was that nitrification is seldom sig-
nificant for the first 5 days, and so the 5-day BOD is
typically a measurement of carbonaceous BOD only.

ria can oxidize organic molecules without the use of
dissolved oxygen, but the end products include com-
pounds such as hydrogen sulfide (H 2 S), ammonia (NH 4 ),
and methane (CH4), which are toxic to many aquatic
organisms.
Waste discharges from nonpoint (diffuse) sources
rarely cause significant reduction of dissolved oxygen in
receiving streams. Exceptions to this include runoff with
high concentrations of biodegradable organics from
concentrated animal feeding operation s (CAFOs) and
spring runoff from fields with manure spread on still-
frozen soils. river water with BOD 5 less than 5 mg/l
can be regarded as unpolluted (Davie, 2008), and river
water with a BOD 5 greater than 10 mg/l is grossly pol-
luted. Water for salmon or trout should have a BOD 5
value below 3 mg/l, for coarse fish (i.e., other than sal-
monids) less than 6 mg/l, and drinking water sources
may have a value up to 7 mg/l.

EXAMPLE 2.2

The results of a BOD test on secondary-treated sewage
give a 5-day BOD of 25 mg/l and a rate constant of
0.2 d −1 . (a) Estimate the ultimate carbonaceous BOD
and the time required for 90% of the carbonaceous
BOD to be exerted. (b) If the ultimate nitrogenous BOD
is 20% of the ultimate carbonaceous BOD, estimate the
oxygen requirement per cubic meter of wastewater.

Solution

(a) From the data given, BOD 5 = 25 mg/l and
k 1 = 0.2 d −1 ; hence, the ultimate carbonaceous BOD,
L 0 , is given by Equation (2.13) as

Chemical Oxygen Demand. The chemical oxygen
demand (COD) is the amount of oxygen consumed
when the substance in water is oxidized by a strong
chemical oxidant. The COD is measured by refluxing a
water sample in a mixture of chromic and sulfuric acid
for a period of 2 hours. This oxidation procedure almost
always results in a larger oxygen consumption than the
standard BOD test, since many organic substances that
are not immediately available as food to aquatic
microbes (e.g., cellulose) are readily oxidized by a
boiling mixture of chromic and sulfuric acid. Domestic
wastewaters typically have a BOD 5 /COD ratio in the
range of 0.4-0.5 (Metcalf & Eddy, Inc., 1989). Compari-
son of BOD 5 and COD results can help identify the
occurrence of toxic conditions in a waste stream or indi-
cate the presence of biologically resistant (refractory)
wastes. For example, a BOD 5 /COD ratio approaching 1
indicates a highly biodegradable waste; a ratio approach-
ing zero suggests a poorly biodegradable material.

BOD 25

5

L = = = 39 5 . mg/L

0

1 − e

− 5 k

1 − e

− 5 0 2 ( . )

1

letting t * be the time for the BOD to reach 90%
of its ultimate value, Equation (2.12) gives

*

− 0 2 . t

0 9 39 5 . ( . ) = 39 5 1 . ( − e )

which gives

t * = 11 6 days .

(b) Since the ultimate nitrogenous BOD is 20% of the
ultimate carbonaceous BOD, the ultimate BOD is
given by

Ultimate BOD = 1 2 . L = 1 2 39 5 . ( . ) = 47 4 . mg/L

0

2.3.3 Suspended Solids

Suspended solids (SS) is the amount of suspended
matter in water. SS is typically measured by filtering a
known volume of water through a 1.2- µ m microfiber
filter, drying the filter at 105°C, and calculating the SS
value by dividing the mass of solids retained on the filter
by the volume of water filtered. The concentration of
particles in the water that passes through the 1.2- µ m
filter is called the total dissolved solids (TDS). Particles
in the size range of 0.001-1.2 µ m are classified as col-
loidal solids . The suspended solids value is normally
expressed in mg/l, and SS concentrations are usually
quite high in surface runoff. A high level of SS produces
a turbid receiving water, blocks sunlight needed by

and for 1 m 3 = 1000 l of wastewater, the ultimate
mass of oxygen consumed by biochemical reactions
is

Mass of oxygen = 47 4 1000 . ( ) = 47 400 , mg/m

3

= 47 4 . g/m

3

If the wastewater is discharged into a surface water,
this oxygen will be taken from the ambient water.

If the dissolved oxygen concentration falls below about
1.5 mg/l, the rate of aerobic biological oxidation is
reduced (Clark, 1997). In cases where adequate amounts
of dissolved oxygen are not available, anaerobic bacte-