Environmental Engineering Reference
In-Depth Information
TABLE 9.11. Longitudinal Dispersion Coefficients in Select
Estuaries
8
V
Q
S
S
=
1.8 10
8.64 10
×
×
10
35
=
T
=
1
1
149
days
f
5
f
0
River Flow
(m 3 /s)
Dispersion
Coefficient (m 2 /s)
Estuary
Alternatively, if Equation (9.102) is used to estimate the
flushing time, T f , then
Cooper River (SC)
283.2
900
Savannah River
(gA, SC)
198.2
300-600
8
V
V T
1.8 10
7.2 10
×
×
T
=
=
(0.5175)
=
129
days
Hudson River (NY)
141.6
600
f
5
Delaware River
(DE, NJ)
70.8
150
tp
Therefore, the estimated flushing time of the estuary is
in the range of 129-149 days.
Cape Fear River
(NC)
28.3
60-300
Hudson ship
channel (TX)
25.5
800
Potomac River
(MD, VA)
15.5
30-300
9.3.5.2  Net  Flow.  In estuaries that are dominated
by tidal flows, the cross-sectionally averaged flow rate
can sometimes be closely approximated as varying
sinusoidally, with tidal inflows occurring on the
half-sinusoidal flood tide and outflows on the half-
sinusoidal ebb tide ; when the tides reverse, the flow
rate is zero, and this condition is called slack tide . The
flow rate, Q (l 3 T −1 ), exiting the estuary, can then be
expressed as
lower Raritan
River (NJ)
4.3
150
Compton Creek
(NJ)
0.3
30
Wappinger and
Fishkill Creeks
(NY)
0.1
15-30
East River (NY)
0.0
300
San Francisco Bay
Southern bay
18-180
π
t
Northern bay
46-1800
Q
sin
0
≤ ≤
t T
e
e
Thames River (UK)
(low river low)
53-84
T
e
Q
=
(9.105)
Thames River (UK)
(High river flow)
338
π
(
t T
T
)
e
Q
sin
π
+
T
≤ ≤
t T T
+
f
e
e
f
Sources of data : USEPA (1984), and Fischer et al. (1979).
f
where Q e and Q f are the maximum flow rates on the
ebb and flood tides, respectively (l3T−1), 3 T −1 ), t is time (T),
and T e and T f are the durations of the ebb and flood
tides, respectively (T). Based on the sinusoidal
approximation given by Equation (9.105), the net
outflow, V out , from the estuary during each tidal cycle is
given by
EXAMPLE 9.10
An estuary has a surface area of 12 km 2 , an average
depth of 15 m, an average salinity of 10 ppt, a tidal range
of 6 cm, and a tidal period of 12.42 hours. The average
flow rate of the river discharging into the estuary is
10 m 3 /s, and the surrounding coastal waters have an
average salinity of 35 ppt. Estimate the flushing time of
the estuary.
π
t
T
π
(
t T
T
)
T
T
e
f
e
V
=
Q
sin
dt Q
+
sin
π
+
dt
out
e
f
0
T
e
e
f
Solution
(9.106)
From the given data: A e = 12 km 2 = 1.2 × 10 7 m 2 ,
H = 15 m, S = 10 ppt , Δ H = 6 cm = 0.06 m, T = 12.42
hours = 0.5175 d −1 , Q f = 10 m 3 /s = 8.64 × 10 5 m 3 /d, and
S 0 = 35 ppt. The volume of the estuary, V , and the
volume of the tidal prism, V tp , can be estimated as
which simplifies to the useful relation
2
π
(
)
(9.107)
V
=
QT QT
out
e e
f
f
This useful relationship should be used with caution
when V out is small, since this result usually corresponds
to large outflow volumes being approximately equal
to large inflow volumes, in which cases the numerical
differences might have zero or very few significant
digits.
7
8
3
V A H
=
=
( .
1 2 10
×
)(
15
)
=
1 8 10
.
×
m
e
V
=
A
×
H
=
( .
1 2 10
×
7
)( .
0 06
)
=
7 2 10
.
×
5
m
3
tp
e
If Equation (9.101) is used to estimate the flushing time,
T f , then
 
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