RIVERS AND STREAMS - Water Quality Engineering in Natural Systems - page 108

where D SP ( x ) is the oxygen deficit predicted by the con-
ventional Streeter-Phelps equation (Eq. 4.71), which
assumes that the BoD in the stream originates from the
wastewater discharge, and the only source of oxygen is
from atmospheric reaeration, and Δ D S ( x ) is the addi-
tional oxygen deficit caused by photosynthesis, respira-
tion, and benthic oxygen demand, and is given by
(Thomann and Mueller, 1987)

S = S

20 (1.065)

T − 20

(4.111)

b b T

where S b and S b 20 are the values of S b at temperature
T and 20°C, respectively. At temperatures below 10°C,
S b declines faster than indicated by Equation (4.111),
and in the range of 5-0°C, S b approaches zero (Chapra,
1997). The constant 1.065 in Equation (4.111) is some-
times taken to be in the range of 1.05-1.06, not a specific
value such as 1.065 (novotny, 2003). There has been
some debate about whether S b should be taken as a
constant at a given temperature, since it is certainly a
function of the organic content of the sediments and the
oxygen concentration in the overlying water, both of
which vary with distance from the source. The SoD can
be assumed independent of the oxygen concentration,
c o , in the overlying water as long as c o > 3 mg/L, for
c o ≤ 3 mg/L the dependence of S b on c o should be taken
into account, since S b decreases to zero as c o goes to
zero. This can be taken into account by using the follow-
ing relation (Thomann and Mueller, 1987)

S + S + S

x
V

p r b

∆ D x ( ) = − 1 − exp − k

(4.116)

S a

k

a

Equations (4.115) and (4.116) leads to a critical point,
x c , with maximum oxygen deficit given by

k D k ( − k )

a o a r

V k

k

+ ( S + S + S )( k − k )

p r b a r

a

ln − k ≠ k

a r

k − k k k L

a r r d r o

x =

c

V
k

k
k

D
L

S + S + S

c

d 0 p r b

− + k = k

O 0

S = S (4.112)

a r

k L

b b

c + k

d a 0 a 0

O so

(4.117)

where k so (ML −3 ) is the half-saturation constant that has
been found to be in the range of 0.7-1.4 mg/L, and S 0
is the SoD when there is not any oxygen limitation.
Most of the SoD is attributed to biochemical oxida-
tion of evolving methane from the lower anaerobic sedi-
ment in the upper oxygenated interstitial sediment
layer. Generally, if the stream velocity is high (>30 cm/s),
most of the fine sediments remain in suspension, and
the bed is composed mostly of sand and gravel, which
have a very low organic content, and the bed will exhibit
very low SoD.

and a corresponding critical oxygen deficit, D c , given by

S + S + S

k
k L

x
V

−

p r b

d c

D = exp − k (4.118)

c o r

k

a a

EXAMPLE 4.16

An outfall discharges wastewater into a slow-moving
river that has a mean velocity of 3 cm/s and an average
depth of 3 m. After initial mixing, the Do concentration
in the river is 9.5 mg/L, the saturation concentration of
oxygen is 10.1 mg/L, the ultimate BoD of the mixed
river water is 20 mg/L, the rate constant for dissolved
BoD is 0.48 d −1 , the reaeration rate constant is 0.72 d −1 ,
and removal of BoD by sedimentation is negligible.
During the night, algal respiration exerts an oxygen
demand of 2 g/m 2 ·d, and sludge deposits downstream of
the outfall exert a benthic oxygen demand of 4 g/m 2 ·d.
Estimate the minimum Do and the critical location in
the river.

Combined Processes. Incorporating photosynthetic,
respiratory, and benthic oxygen fluxes into the oxygen
sag model yields the following modified form of the
Streeter-Phelps formulation

∂
∂

c
x

(4.113)

V = − k L k c + ( − c ) + ( S + S + S )

d a s p r b

dL
dt

(4.114)

= − k L

r

where the addition of the ( S p + S r + S b ) term in Equa-
tion (4.113) is the only difference with the conventional
Streeter-Phelps formulation. Equations (4.113) and
(4.114) yield a solution that can be expressed in terms
of the oxygen deficit, D (= c s − c ), as

Solution

From the data given, the initial oxygen deficit, D 0 ,
is 10.1 − 9.5 = 0.6 mg/L, k d = k r = 0.48 d −1 , k a = 0.72 d −1 ,
S r

= − ⋅ . Since the average
depth, d , of the river is 3 m, the volumetric oxygen

*

= − 2 g /m d

2

⋅ , and S b

*

4 g/m d

2

D x D x ( ) = ( ) + ∆ D x ( ) (4.115)

SP S