OCEANS AND ESTUARIES - Water Quality Engineering in Natural Systems - page 332

Assess whether consideration of longitudinal dispersion
is important in predicting the effect of wastewater on
DO levels in the river.

j x j x

k L
k

α e e

2 r c 2 a c

d 0 r

D = − (9.97)

c

− k α α

a r r a

where x c is, by definition, measured downstream of the
source.

Solution

From the data given, V = 5 cm/s = 4320 m/d, k d = k r = 0.4
d −1 , k a = 0.75 d −1 , K l = 120 m 2 /s = 1.04 × 10 7 m 2 /d, and
L 0 = 10 mg/l. Equations (9.87) and (9.88) give

EXAMPLE 9.9

After initial mixing with a wastewater discharge, a tidal
river has an ultimate BOD of 15 mg/l and a BOD
decay constant of 0.50 d −1 . The river has a mean velocity
of 3 cm/s, a reaeration constant of 0.80 d −1 , a longitudinal
dispersion coefficient of 100 m 2 /s, and a temperature of
20°C. Determine the location and magnitude of the
maximum oxygen deficit.

7

4 k K

V

4(0.4)(1.04 10 )
(4320)

×

r L

α r = 1 + = 1 + = 1.37

2 2

and

7

4 k K

V

4(0.75)(1.04 10 )
(4320)

×

a L

α a = 1 + = 1 + = 1.63

2 2

Solution

From the given data: L 0 = 15 mg/l, k d = k r = 0.50 d −1 ,
V = 3 cm/s = 2592 m/d, k a = 0.80 d −1 , K l = 100 m 2 /s =
8.64 × 10 6 m 2 /d, and T = 20°C. Using these data, the fol-
lowing parameters can be calculated:

and combining Equations (9.86) to (9.92) for x > 0 gives
the oxygen deficit 200 m downstream of the outfall as

(10)(0.4)(1.37)
0.75 0.4
1
1.37

D (200) =

−

4 k K

V

4(0.50)(8.64 10 )
(2592)

×

6

r L

(4320)(200)
2(1.04 10

α r = 1 + = 1 + = 1.89

exp

) (1 1.37)

−

2 2

×

7

4 k K

V

4(0.80)(8.64 10 )
(2592)

×

6

1
1.63

(4320)(200)
2(1.04 10 ) (1 1.63

a L

α a = 1 + = 1 + = 2.26

− exp − ) 2 2

7

×

= 1.9 mg/L V

K

2592
2(8.64 10 ) (1 1.89)

j = (1 +

α

) = + = 4.33 10 ×

− 4

1 r

2 ×

6

L

The relevant dimensionless numbers to be considered
are

V
K

2592
2(8.64 10 ) (1 1.89)

− 4

j = (1 −

α

) = − = − 1.33 10 ×

2 r

2 ×

6

L

k K
V

0.4(1.04 10 )
(4320)

×

7

V
K

2592
2(8.64 10 ) (1 2.26)

r L

= = 0.22

− 4

j = (1 +

α

) = + = 4.89 10 ×

2 2

1 a

2 ×

6

L

k K
V

0.75(1.04 10 )
(4320)

×

7

a L

V
K

2592
2(8.64 10 ) (1 2.26)

= = 0.42

j = (1 −

α

) = − = − 1.89 10 ×

− 4

2 2

2 a

6

2 ×

L

Comparing these values with the limits given in Equa-
tions (9.93) and (9.94) indicates that both advection and
dispersion

Substituting these parameters into Equations (9.96) and
(9.97) gives

should be considered, but neither

predominates.

α
α

j
j

1.89
2.26

− 1.89 10

1.33 10 4
4

×

− 4

r
a

2
2

a
r

ln ln

− ×

−

Equation (9.86) can be differentiated with respect to x
to determine the location of minimum DO, just as for
the classical Streeter-Phelps equation, and the location
of this critical point, x c , along with the critical oxygen
deficit, D c , is given by

x = = = 3040 m

c

j − j − 1.33 10 ×

−

+ 1.89 10 ×

− 4

2 r 2 a

k L
k

α e

j x

e

j x

2 r c 2 a c

d 0 r

D = −

c

− k α α

a r r a

− 4 − 4

(0.50)(15)(1.89)
0.80

e

( 1.33 10 − × )(3040)

e

( 1.89 10 − × )(3040)

α
α

j
j

r 2 a

= −

ln

− 0.50 1.89 2.26

a 2 r

x = (9.96)

c

j − j

= 4.9 mg/L

2 r 2 a