RIVERS AND STREAMS - Water Quality Engineering in Natural Systems - page 82

and the corresponding time scale, T s , is given by plume will be mixed over a width of 4 m; if a 4-m-long

diffuser is used, the plume will be well mixed over a
width of 8 m. Compare the dilution achieved by the
single-port discharge and diffuser at a location 100 m
downstream of the discharge.

2

2.5
0.11

T s = = 56.8 seconds = 0 95 . minute

Since the flow velocity, V , is 1.5 m/s, the downstream
distance, L , for the wastewater to become well
mixed is

Solution

From the data given, Q o = 3 m 3 /s, c o = 10 mg/L = 0.01
kg/m 3 , and the mass flux,

M , of chromium released at

L = 0.4 VT = 0.4(1.5)(56.8) = 34 m

s

the outfall is given by

The results of this example illustrate that a 5-m-long
diffuser located at the center of the stream will
cause the wastewater to mix much more rapidly
over the stream cross section than a point discharge
at the center of the stream.

M Q c

= = (3)(0.01) = 0.03 kg/s

o o

For the single-port discharge, the plume is mixed over
an area, A m , of 4 m × 3 m = 12 m 2 . Since V = 0.5 m/s, the
average concentration of the mixed river water 100 m
downstream of the outfall is given by Equation (4.20)
as

The previous examples have illustrated why dis-
charges from diffusers generally achieve complete
cross-sectional mixing more rapidly than single-port dis-
charges. Also, at any given distance from the outfall,
discharges from diffusers are mixed over a larger portion
of the channel width than discharges from single-port
outfalls, resulting in diffusers achieving greater dilu-
tions. Consider the case where the mass flux of contami-
nant from an outfall (single port or multiport) is given
by M (MT −1 ) and the contaminant is mixed over a
width, w (L), in a river; then conservation of mass
requires that

M
A V

0.03
(12)(0.5)

c = = = 0.005 kg/m

3

= 5 mg/L

m

m

For the diffuser discharge, the plume is mixed over an
area, A m , of 8 m × 3 m = 24 m 2 , and the average concen-
tration of the mixed river water is given by

M
A V

0.03
(24)(0.5)

c = = = 0.0025 kg/m

3

= 2.5 mg/L

m

m

M c Vwd

(4.19)

=

m

Therefore, at a location 100 m downstream of the dis-
charge, the diffuser gives a dilution of 10/2.5 = 4, com-
pared with a dilution of 10/5 = 2 for the single-port
outfall.

where c m is the mean concentration of the mixture
(ML −3 ), V is the mean velocity in the mixed portion of
the river (LT −1 ), and d is the mean depth in the mixed
portion of the river (L). Rearranging Equation (4.19)
gives the mean concentration, c m , of the mixture by

one-dimensional stream models frequently assume
that the discharged contaminant is uniformly mixed
over the stream cross section in the vicinity of the dis-
charge location. Analyses described previously demon-
strate how this assertion can be assessed quantitatively.
The assumption of complete cross-sectional mixing in
the vicinity of a discharge location is justified in cases
where diffusers span the width of the stream and/or the
discharge rate of contaminated water into the stream is
comparable with the stream discharge. For single-port
discharges on the sides of wide streams, where the dis-
charge rate of contaminated water is small compared
with the river discharge, considerable distances might
be necessary for complete mixing across a river.
Consider the waste discharge shown in Figure 4.2,
where the river flow rate upstream of the wastewater
outfall is Q r (L 3 T −1 ), with a contaminant concentration,
c r (ML −3 ), and the wastewater discharge rate is Q w

M
A V

c = (4.20)

m

m

where A m is the area (L 2 ) over which the contaminant
is mixed (= wd ).

EXAMPLE 4.3

An industrial wastewater outfall discharges effluent at
a rate of 3 m 3 /s into a river. The chromium concentra-
tion in the wastewater is 10 mg/L, the average velocity
in the river is 0.5 m/s, and the average depth of the river
is 3 m. Tracer tests in the river indicate that when the
wastewater is discharged through a single port in the
middle of the river, 100 m downstream of the outfall the