RIVERS AND STREAMS - Water Quality Engineering in Natural Systems - page 109

demand rates for respiration and benthic consumption,
S r and S b , respectively, are

S
k

x
V

d

∆ D x ( ) = − 1 − exp − k

(4.120)

S a

a

*

S
d

2
3

r

S = = − = − 0.667 g/m d

3

⋅ = − 0.667 mg/L d ⋅

This result indicates that the oxygen deficit generated
by a distributed source/sink asymptotically (as x → ∞)
approaches − S d / k a since

r

and

S
k

*

S
d

4
3

d

lim
x

∆ D x ( ) = − (4.121)

b

S = = − = − 1.33 g/m d

3

⋅ = − 1.33 mg/L d ⋅

S

b →∞

a

For distributed oxygen sinks ( S d < 0) with low reaera-
tion ( k a small), which is typical of slowly moving streams,
the asymptotic oxygen deficit is large and is approached
after a long travel time. In contrast, with high reaeration
( k a large), which is typical of fast moving streams, the
asymptotic oxygen deficit is small and is approached
after a short travel time.

Since L 0 = 20 mg/L and V = 3 cm/s = 2592 m/d, Equa-
tion (4.117) gives the location, x c , of the critical oxygen
deficit as

k D k

0 (

− k )

a a r

V k

k

+ ( S + S + S )( k − k )

a p r b a r

x = ln −

c

k − k k k L

a r r d r o

EXAMPLE 4.17

(0.72)(0.6)(0.72 0.48)
( 0.

−

Photosynthesis, benthic oxygen demand, and respira-
tion collectively produce a net oxygen demand of
1.997 mg/L·d in a stream that moves with an average
velocity of 5 cm/s and has an aeration rate coefficient of
0.72 d −1 . What is the ultimate oxygen deficit caused by
these processes? If these processes were to suddenly
begin at a certain location, how far downstream from
this location would the induced deficit be 95% of its
ultimate value?

2592
0.72 0.48

0.72
0.48

+ − 667 1.33)(0.72 0.48)

(0.48)(0.48)(20)

− −

= ln −

−

= 4951 m

and Equation (4.118) gives the critical oxygen deficit,
D c , as

k
k L

k x
V

S + S + S

d r c p r b

Solution

D = exp −

−

c o

k

a a

0.48
0.72 (20)

0.48 4951
2592

×

− −

0.667 1.33
0.72

−

From the given data: S d = −1.997 mg/L·d, V = 5 cm/s =
4320 m/d, and k a = 0.72 d −1 . The ultimate oxygen deficit
is given by Equation (4.121) as

= exp −

= 8.1 mg/L

S
k a

= − −

1.997
0.72

Hence, the minimum Do level in the stream is
10.1 − 8.1 = 2.2 mg/L and occurs 4951 m downstream of
the outfall location.

d

D

∞ = −

= 2.77 mg/L

S

If x 95 is the distance to attain 95% of the ultimate deficit,
then Equation (4.120) requires that

Photosynthesis, respiration, and benthic oxygen
demands collectively add up to a distributed source of
oxygen, S d (ML −3 T −1 ), where

S
k

S
k

x
V

= −

d d

0.95 − 1 − exp − k

a

a a

x

95

0.95 = 1 − exp − 0.72 4320

S = S + S + S (4.119)

d p r b

Therefore, the solution for the more general case in
which the oxygen deficit is caused by any distributed
source (or sink) of oxygen along a stream can be derived
directly from Equation (4.116) and expressed as

which yields x 95 = 17,970 m. Hence, the deficit will be
at 95% of its ultimate value of 2.77 mg/L at about
17.97 km downstream from where the oxygen sinks are
initiated.