Human Influence on Ecosystems
Effects of Pesticides on Ecosystems
Birth of the Environmental Movement
Rachel Carson
Silent Spring
Biomagnification
As you move up through
the food chain the
concentration of the
pollutant increases
DDT the classical Example
PCB’s another example
Effect of Nutrients on a Lake Ecosystem
Limiting Nutrient
Usually Nitrogen or Phosphorus in Aquatic ecosystems
homeostasis
Note algae must live
near surface so they
can get light, but the
decomposers live at
the bottom.
Mixing between layers
is necessary
Aerobic decomposers need oxygen supplied by algae.
Algae need nutrients supplied by decomposers
Lake Stratification
External Nutrient Input
Algae Blooms
Eutrophication
Effect of Organic Wastes on Streams
A stream may be thought of as a plug flow reactor
Mass balance on dissolved oxygen
Rate of
Oxygen
=
accumulated
Rate of
Oxygen
In
-
Rate of
Oxygen
Out
+
Rate of
Oxygen
Generated
Oxygen is introduced into the stream by reaeration or reoxygenation
There is no oxygen leaving the stream since we will assume it is
unsaturated
Dissolved oxygen may be produced in water by algae during
Photosynthesis, but is a swift stream the algae don’t have time to
grow and there is no oxygen produced in this way.
Oxygen may be used by microorganisms respiration. This is called
deaeration or deoxygenation.
So the new mass balance is:
Rate of
Rate of
Oxygen
= Oxygen In accumulated
(reaeration)
Rate of deaeration = -k1C
0
+
Rate of
Oxygen Consumed
(deaeration)
(first order)
k1 = deaeration constant, function of type of waste,
temperature, etc. Units, day-1
Rate of reaeration = k2 D
(first order)
k2 = reaeration constant based on characteristics of the stream
and the weather
Measured in field or
estimated using equations
such as equation on page 187
of text
D = oxygen deficit = S - C
S = oxygen saturation concentration, a function of temperature of
the water, atmospheric oxygen concentration, and water
chemistry
Maximum oxygen concentration increase with decreasing
temperature
Oxygen Sag Curve
Rate of
Rate of
Oxygen
= Oxygen In accumulated
(reaeration)
0
+
Rate of
Oxygen Consumed
(deaeration)
dD/dt = k1z - k2D
z = the amount of O2 still needed by the
microorganisms
= L e-k1t
L = ultimate oxygen demand
Substitute and integrate:
D = [(k1 L0)/(k2 – k1)] (e-k1t – e-k2t) = D0e-k2t
D0 = the oxygen deficit at the point in the stream we call 0
L0 = the oxygen demand at the point in the stream we call 0
To find the point on the stream where the deficit, D, is the
greatest we will set dD/dt = 0 and solve the equation for t:
tC = [1/(k2 – k1)] ln[(k2/k1)(1 – (D0 x (k2- k1))/(k1L))]
tC then is the time downstream where the oxygen
concentration is at its lowest
Example
A stream flows at 2.2 m3/sec with a velocity of 0.85 m/s and temperature of
12oC. The stream is saturated with dissolved oxygen, has an ultimate oxygen
demand of 13.6 mg/L, and has a reaeration constant of 0.4 day-1. Treated waste
from a municipal waste treatment plant is discharged to the stream. The waste
stream has an ultimate BOD of 48 mg/L, a flow rate of 0.5 m3/s, a dissolved
oxygen concentration of 1.5 mg/L, and a temperature of 26oC. After the waste
is mixed with the stream the deaeration constant is 0.2 day –1. What is the
dissolved oxygen concentration 48.3 km downstream? What is the minimum
dissolved oxygen concentration, and where does it occur?
Qp = 0.5 m3/s
Tp = 26oC
BODup = 48 mg/L
DOp = 1.5 mg/L
Qs =2.2 m3/s
Ts = 12oC
BODus = 13.6 mg/l
V = 0.85 m/s
Ds = ?
Q0 = ?
T0 = ?
BODu0= ?
D0 = ?
Qp = 0.5 m3/s
Tp = 26oC
BODup = 48 mg/L
DOp = 1.5 mg/L
T0 = [(QsTs) + (QpTp)]/(Qs + Qp)
= [(2.2)(12) + (0.5)(26)]/(2.2 + 0.5)
m3/s
Qs =2.2
Ts = 12 oC
BODus = 13.6 mg/l
V = 0.85 m/s
Ds = ?
Q0 = ?
T0 = ?
BODu0= ?
D0 = ?
= 14.6oC
Since the stream is saturated with
DO, Ds = 0 and S = Cs =10.8oC
C0 = [(QsCs) + (QpCp)]/(Qs + Qp)
= [(2.2)(10.8) + (0.5)(1.5)]/(2.2 + 0.5)
D0 = S0 - C0
Since T0 = 14.6, S0 = 10.2 mg/L
D0 = 10.2 - 9.1 = 1.1 mg/L
= 9.1 mg/L
L0 = [(Ls)(Qs) + (Lp)(Qp)]/(Qs + Qp)
= [(13.6)(2.2) + (48)(0.5)]/(2.2 + 0.5)
= 20 mg/L
Deficit at 48.3 km:
48.3 x 103 m/0.85 m/s = 56.8 x 103 sec = 0.66 days
D = (k1L0)/(k2 –k1) [(e-k1 t - e-k2 t)] + D0 e-k2 t
= (0.2)(20)/(0.4 – 0.2)[e-0.2(0.66) – e-0.4(0.66)) + 1.1 (e-0.4 (0.66))
= 3.0 m/L
C = S – D = 10.2 – 3.0 = 7.2 mg/L
tC = [1/(k2 – k1)] ln[(k2/k1)(1 – (D0 x (k2- k1))/(k1L))]
= [1/(0.4 – 0.2)] ln[(0.4/0.2)(1 – (1.1(0.4-0.2)/(0.2)(20)]
= 3.18 days
At 3.18 Days
D = (k1L0)/(k2 –k1) [(e-k1 t - e-k2 t)] + D0 e-k2 t
= (0.2)(20)/(0.4 –0.2) [(e-0.2(3.18) – e-0.4(3.18))] + 1.1 e-0.4(3.18)
= 5.29 mg/L
C = S – D = 10.2 – 5.29 = 4.91 mg/L