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Human Influence on Ecosystems. Rachel Carson. Effects of Pesticides on Ecosystems. Birth of the Environmental Movement. Silent Spring. Biomagnification. As you move up through the food chain the concentration of the pollutant increases. DDT the classical Example.

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Presentation Transcript
slide2

Rachel Carson

Effects of Pesticides on Ecosystems

Birth of the Environmental Movement

Silent Spring

slide3

Biomagnification

As you move up through

the food chain the

concentration of the

pollutant increases

DDT the classical Example

PCB’s another example

slide4

Effect of Nutrients on a Lake Ecosystem

Limiting Nutrient

Usually Nitrogen or Phosphorus in Aquatic ecosystems

homeostasis

slide5

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

slide6

Lake Stratification

External Nutrient Input

Algae Blooms

Eutrophication

slide7

Rate of

Oxygen

accumulated

Rate of

Oxygen

In

Rate of

Oxygen

Out

Rate of

Oxygen

Generated

=

-

+

Effect of Organic Wastes on Streams

A stream may be thought of as a plug flow reactor

Mass balance on dissolved oxygen

slide8

Rate of

Oxygen

accumulated

Rate of

Oxygen In

(reaeration)

Rate of

Oxygen Consumed

(deaeration)

=

-

0

+

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 deaeration = -k1C (first order)

k1 = deaeration constant, function of type of waste,

temperature, etc. Units, day-1

slide9

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

slide12

Rate of

Oxygen

accumulated

Rate of

Oxygen In

(reaeration)

Rate of

Oxygen Consumed

(deaeration)

=

-

0

+

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

slide13

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

slide14

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 = ?

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?

slide15

Qp = 0.5 m3/s

Tp = 26oC

BODup = 48 mg/L

DOp = 1.5 mg/L

Qs =2.2 m3/s

Ts = 12 oC

BODus = 13.6 mg/l

V = 0.85 m/s

Ds = ?

Q0 = ?

T0 = ?

BODu0= ?

D0 = ?

T0 = [(QsTs) + (QpTp)]/(Qs + Qp)

= [(2.2)(12) + (0.5)(26)]/(2.2 + 0.5)

= 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

= 9.1 mg/L

Since T0 = 14.6, S0 = 10.2 mg/L

D0 = 10.2 - 9.1 = 1.1 mg/L

slide16

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

slide17

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