Human Influence on Ecosystems

Effects of Pesticides on Ecosystems

Rachel Carson

Silent Spring

Birth of the Environmental Movement

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 NutrientUsually 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 oxygenRate ofOxygen

accumulated

Rate ofOxygen

In

Rate ofOxygen

Out

Rate ofOxygenGenerated

= - +

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 ofOxygen

accumulated

Rate ofOxygen In

(reaeration)

Rate ofOxygen Consumed

(deaeration)= - +0

Rate of deaeration = -k1C (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

D = oxygen deficit = S - C

S = oxygen saturation concentration, a function of temperature of the water, atmospheric oxygen concentration, and water chemistry

Measured in field or estimated using equations such as equation on page 187 of text

Maximum oxygen concentration increase with decreasingtemperature

Oxygen Sag Curve

Rate ofOxygen

accumulated

Rate ofOxygen In

(reaeration)

Rate ofOxygen Consumed

(deaeration)= - +0

dD/dt = k1z - k2D

z = the amount of O2 still needed by the microorganisms

= L e-k1t

L = ultimate oxygen demandSubstitute 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 0L0 = 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

ExampleA 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?

Qs =2.2 m3/sTs = 12oCBODus = 13.6 mg/lV = 0.85 m/sDs = ?

Q0 = ?T0 = ?BODu0= ?D0 = ?

Qp = 0.5 m3/sTp = 26oCBODup = 48 mg/LDOp = 1.5 mg/L

Qs =2.2 m3/sTs = 12 oCBODus = 13.6 mg/lV = 0.85 m/sDs = ?

Q0 = ?T0 = ?BODu0= ?D0 = ?

Qp = 0.5 m3/sTp = 26oCBODup = 48 mg/LDOp = 1.5 mg/L

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

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

= 14.6oC

Since the stream is saturated withDO, Ds = 0 and S = Cs =10.8oC

C0 = [(QsCs) + (QpCp)]/(Qs + Qp)

= [(2.2)(10.8) + (0.5)(1.5)]/(2.2 + 0.5)

= 9.1 mg/LD0 = S0 - C0

Since T0 = 14.6, S0 = 10.2 mg/L

D0 = 10.2 - 9.1 = 1.1 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