RIVERS AND STREAMS - Water Quality Engineering in Natural Systems - page 111

along the 4-km stretch of river, k d = 0.5 d −1 , S L = 3 mg/L,
and Equation (4.125) gives the contribution of the dis-
tributed load to the Do deficit as

where Δ D S ( x ) is given by Equation (4.116), and Δ D BoD ( x )
is given by Equation (4.125). The form of Equation
(4.127) explicitly gives the contributions of distributed
oxygen sources and sinks on the oxygen deficit and
provides a convenient form for assessing the relative
impact of the various oxygen sources and sinks on the
Do in rivers.

k S
k k

k x
V

d L a

∆ D

BOD ( )

x = 1 − exp −

r a

k S

d L

−

( k − k k

k

)

a r r

EXAMPLE 4.19

x
V

k x
V

−

r a

exp − exp −

A wastewater treatment plant discharges effluent into a
river that has a mean velocity of 3 cm/s. Analysis of the
waste discharge and distributed BoD input over a
500-m river stretch indicates that the Do 500 m down-
stream of the waste outfall is predicted to be 5 mg/L. In
this analysis, it is assumed that the reaeration constant
is equal to 0.7 d −1 , and the average temperature of the
stream is 20°C. Photosynthesis within the river stretch
is expected to generate 3 mg/L·d, plant respiration
1 mg/L·d, and sediment oxygen demand 0.5 mg/L·d.
Estimate the expected fluctuation in Do 500 m down-
stream of the outfall.

0.5(3)
(0.6)(0.7)

0.7 4000
2592

×

∆ D BOD (4000) = 1 − exp −

.5(3)
(0.7 0.6)(0.6)

0 0.6 4000

2592

×

− exp −

−

0.7 4000
25

×

− exp −

92

= 0.9 mg/L

Therefore, the distributed BoD load increases the Do
deficit 4 km downstream by 0.9/6.7(100) = 13%. In the
absence of the distributed BoD load, the Do concen-
tration 4 km downstream of the wastewater discharge
would be 2.4 mg/L, and with a distributed BoD load,
the Do concentration is 2.4 mg/L − 0.9 mg/L = 1.5 mg/L.
In either case, whether the distributed BoD load is
present or not, the estimated Do level would be devas-
tating to the indigenous aquatic life, which typically
requires at least 5 mg/L of Do.
The asymptotic oxygen deficit caused by the distrib-
uted BoD source is given by Equation (4.126) as

Solution

At 20°C, the saturation concentration in the stream is
9.1 mg/L. From the data given, Δ D SF + Δ D BoD = 9.1
mg/L − 5 mg/L = 4.1 mg/L, k a = 0.7 d −1 , S p = 3 mg/L·d,
S r = −1 mg/L·d, and S b = −0.5 mg/L·d. Since the average
velocity in the river is 3 cm/s (= 0.03 m/s = 2592 m/d),
the time, T , to travel 500 m is given by

500
0.03

T = = 16,667 seconds = 4.6 hours

k S
k k

(0.5)(3)
(0.6)(0.7)

Since the travel time is less than the photoperiod, the
most critical condition for Do 500 m downstream of the
outfall is when the waste travels for 4.6 hours in dark-
ness, photosynthesis is nonexistent, and both respiration
and sediment oxygen demand are exerted. In this case,
the dissolved oxygen deficit is given by Equations
(4.127) and (4.116) as

d L

lim
x

∆ D ( ) x = − = − = 3.5 mg/L

BOD

→∞

r a

Therefore, at best, the Do concentration will be
9.1 mg/L − 3.5 mg/L = 5.6 mg/L. This will be “too close
for comfort” for the aquatic life, and consideration
should be given to reducing the magnitude of the dis-
tributed BoD.

D x ( ) = [ D x ( ) + ∆ D ( )] x + ∆ D x ( )

SF BOD S

Composite Sources and Sinks of Oxygen. In cases where
the river BoD originates from both a wastewater source
and distributed sources of BoD along the channel,
where the sources of oxygen are atmospheric reaeration
and photosynthesis, and where the sinks of oxygen are
the demand of the discharged wastewater plus respira-
tion and benthic oxygen demand along the channel, the
oxygen deficit, D ( x ), along the channel, is given by

= [ D x ( ) + ∆ D ( )] x

SF BOD

S + S + S x

V

p r b

− 1− exp − k

a

k

a

− − −

1 0 5
0.7

.

0.7 500
2592

D (500) = [4.1] 1 − exp −

= 4 4 . mg/L

which corresponds to a dissolved oxygen concentration
of 9.1 mg/L − 4.4 mg/L = 4.7 mg/L.

D x D x ( ) = ( ) + ∆ D x ( ) + ∆ D ( ) x (4.127)

SP S BOD