Environmental Engineering Reference
In-Depth Information
TABLE 7.3. Trophic Status of Lakes
Water Quality
Oligotrophic Mesotrophic
Eutrophic
Source
Total P ( µ g/L)
<10
10-20
>20
USEPA (1974)
<10
10-30
>30
Nürnberg (1996)
Chlorophyll a ( µ g/L)
<4
4-10
>10
USEPA (1974)
0.8-3.4
3-7.4
6.7-31
Ryding and Rast (1989)
<4
4-10
>10
Novotny and Olem (1994)
3.5-9
Nürnberg (1996)
<3.5
>9
<3.5
3.5-9
9-25
Smith (1998)
0.3-3
2-15
>10
Wetzel (2001)
Secchi disk depth (m)
>4
2-4
<2
USEPA (1974) and Nürnberg (1996)
Hypolimnetic oxygen (% saturation)
>80
10-80
<10
USEPA (1974)
Phytoplankton production (g org C/m 2 ·d
7-25
75-250
350-700
Mason (1991)
Trophic status index (TSI)
<40
35-45
>45
Carlson (1977)
EXAMPLE 7.7
solids in the water column, k e (L), can be estimated
using the relation (Beeton, 1958)
Measurements in a lake indicate a Secchi depth of
3.0 m, a chlorophyll a concentration of 6 µ g/L, and a TP
concentration of 17 µ g/L. Estimate the trophic state of
the lake.
a
(7.15)
k
=
e
SD
where SD is the Secchi depth (L), and a is a constant
with typical values in the range of 1.7-1.9. The depth of
the euphotic zone can be estimated in terms of the
Secchi depth by combining Equations (7.14) and (7.15),
which gives
Solution
From the given data: SD = 3.0 m, Chl a = 6 µ g/L, and
TP = 17 µ g/L. According to the guidelines given in
Table 7.3, the lake is mesotrophic with respect to Secchi
depth, mesotrophic with respect to chlorophyll a , and
mesotrophic with respect to TP. Using the given data to
determine the TSI yields
I d
I
( )
SD
d
= −
ln
(7.16)
a
s
and taking I ( d )/ I s = 0.01 and a = 1.8, the depth of the
euphotic zone, d e , can be estimated as
60 14 43
.
ln(
SD
Chl
)
=
60 14 43
.
ln( . )
3 0
=
44
TSI
=
30 56 9 81
.
+
.
ln(
a
=
30
.
56 9 81
+
.
ln( )
6
=
48
d e
=
2 .
×
SD
(7.17)
4 14 14 43
.
+
.
ln(
TP
)
=
4 14 14 43
.
+
.
ln(
17
)
=
45
which indicates that the depth of the euphotic zone is
around two to three times the Secchi depth.
The average of these results is (44 + 48 + 45)/3 = 46,
which according to Table 7.3, the lake is eutrophic
(barely). The weight of the evidence here would indi-
cate that the lake is mesotrophic, although the indica-
tors are not unanimous in this regard.
EXAMPLE 7.8
Estimate the minimum depth of the euphotic zone in an
oligotrophic lake.
Beer's law (Eq. 7.8) describes the light penetration
into a water body by the relation
Solution
According to Table 7.3, the minimum Secchi depth in an
oligotrophic water body is 4 m. Combining this value of
SD with Equation (7.17) gives
I d
( ) =
I e k d
(7.14)
e
s
where I ( d ) is the solar radiation (PL −2 ) penetrating to a
depth d (L) below the water surface, I s is the solar radia-
tion at the water surface (PL −2 ), and k e is the light extinc-
tion coefficient (L −1 ). In cases where the Secchi depth is
used as a measure of the concentration of suspended
d e
=
2 6
.
×
SD
=
2 6
.
×
( )
4
=
10 4
.
m
Hence, the minimum depth of the euphotic zone in an
oligotrophic lake is 10.4 m.
 
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