Quiz 4 Flashcards

(33 cards)

1
Q

Why add oxygen to lakes and reservoirs (2)

A

Deposition of carbonaceous materials, decomposition of algae, and chemical reactions reduce oxygen demand in lakes and reservoirs

Adding oxygen meets the increased oxygen demand that this creates in stratified eutrophic lakes

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2
Q

Whole lake oxygen depletion (2)

A

Occurs in winter in eutrophic lakes because wind and diffusion inputs can’t penetrate the ice cover

Results in loss of suitable habitat for cold water fish (salmonids) as oxygen goes less than 3-5mg/L

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3
Q

What is a fish kill called and why? (2)

A

In the summer it is called a summerkill or a “thermocline squeeze” and this is because, fish are squeezed in the epilimnion during stratification as oxygen is too low at depth.

In the winter it is called a winterkill. These occur usually in mid-late February or at spring circulation.

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4
Q

What causes a winterkill? (2)

A

Mid to late February it is caused by fish running out of oxygen as they can only go ~90 days without

In the spring it is caused by epilimnion layer mixing into lower oxygen depleted layers, causing dilution of oxygen in upper layers, as well as release of nutrients and chemicals from the sediment layer

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5
Q

What negative effect can spring mixing cause? (3)

A

During the winter, anoxic conditions cause 10X more toxic compounds and nutrients to be released from the sediment (lid is removed because of lack of chemical reactions)

This causes winterkills in spring because those toxins are circulated throughout the lake

Therefore: lack of oxygen causes more nutrients causes more lack of oxygen and it is a positive reinforcement cycle

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6
Q

What are the toxic compounds that can be released from the sediment layer? (3)

A

NH3, H2S - very toxic

Mn and Fe ions - less toxic

P and NH3 - algal nutrients

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7
Q

Internal loading

A

Increased nutrient loading caused by the removal of the oxic layer at the bottom of a eutrophic lake

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8
Q

Results of internal loading (4)

A

Grows more algae from nutrient input

Further depletes oxygen in the hypolimnion

Causes fish kills and releases even more nutrients from their bodies

= runaway process that leads to an alternate stable state

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9
Q

Two categories of artificial circulation (2)

A

Destratification - aerates through removing stratification

Hypolimnetic aeration - aerates only the hypolimnion, preserving stratification

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10
Q

Destratification (3)

A

Type of artificial circulation that breaks up lake layers

Only a cosmetic procedure, unless underlying source of excess nutrient loading has been identified and reduced

Uses compressed air or mechanical mixers

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11
Q

Compressed air destratification (3)

A

Type of destratification system

Oxygen added to lake from rising bubbles (from diffusers at bottom of lake), but majority is on the surface when rising bubbles come in contact with the atmosphere and transfer the air to the surface water

Winter aeration creates an open ice area when the system is operating ( need fence)

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12
Q

Mechanical surface aerators (4)

A

Type of destratification system

Adds oxygen by:
Propeller turbulence causing internal motion
Water spray contact with atmosphere at surface

Requires an underwater electrical cable and voltage of electricity to operate propellor = major limitation

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13
Q

What is a major limitation of mechanical aerators?

A

They require electricity and long cables, making them expensive and difficult to implement in rural/wilderness areas

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14
Q

Solar and wind powered mechanical aerators

A

Not usable as mechanical destratification systems they don’t have enough power due to lack of sun and wind

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15
Q

Garton pumps (2)

A

Type of destratification system that uses a propellor and electricity

Very common in drinking water reservoirs in Australia

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16
Q

Electricity need (3)

A

Mechanical aerators are the preferred method for many lake destratification projects

Therefore we need electricity

In B.C., electricity is 60 hertz and single or three phase

17
Q

Single phase voltage (5)

A

115 or 230 volts

As voltage doubles, the amperage decreases

On a “virgin line” the max power you can operate is two 230 volt, 7.5hp (5.6kW) electric motors

Because of the rise and falls in voltage, power is not delivered at a constant rate and single phase motors larger than 10kW are rare

Single phase motors need additional circuits for starting because single phase supply does not produce a revolving magnetic field

18
Q

Watts equation

A

Watts (ie. power) = volts* amps

Ex. 0.746kW electric motor running on 115 volt single phase electricity, requires 746W/115v= 6.5 amps

If 230 volt single phase electricity with same kW electric motor, 746W/230v= 3.2 amps

19
Q

Relationship between volts and copper wire (6)

A

Higher volts means longer, thinner copper wire

115 volts = 49m distance
230 volts = 120m distance

Why? Because increasing voltage decreases the amps and decreases resistance

Therefore, always use highest available voltage to decrease amperage, weight, and cost of cable

Unwritten rule: on a “virgin line” can exceed these distances by 50%

Can also use multiple units, as long as you don’t exceed 11.2kW/230v limit

20
Q

Three phase power (4)

A

In three phase power, the currents reach their peak values sequentially, first one, then second one; then third one.

This is in contrast to the single phase that has two wires reaching their peak at the same time.

Therefore, they have much higher voltage than the single phase (240, 480, 600 and higher)

Three phase motors are self starting because they produce a magnetic field

21
Q

When should you switch to three phase? (2)

A

When you require more than 11.2kW of power

Bottom line: you cannot install a destratification system from a single phase power line if your energy requirements exceed 2 x 7.5hp (2 x 5.6kW) of energy (two 230 volt motors)

22
Q

Phase converters (2)

A

Phase converters are meant to create three phase electricity out of single phase electricity

Generally don’t work

23
Q

Fossil fuel electricity (2)

A

Can’t justify, causes climate change

Diesel, propane, liquified natural gas all options, but worth it?

24
Q

Possible energy sources for power generation (4)

A
For destratification:
Electricity 
Wind power
Solar power
Fossil fuels
25
Operational costs of electric motors (2)
If use low hp (min single phase - 7.5hp = 5.6kW) 5600 watts x 1 hour = 5.6kWh 5.6kWh x 24 = 134.4kWh per day @ $0.08 per kWh = $10.75 per day If operate for 5 months, can cost $1612 If use high hp (three phase - 50hp = 37kW 37000 watts x 1 hr = 37kWh @ $0.08 per kWh = $71.04 per day If operate for 5 months, can cost $10,656
26
Formula for sizing destratification systems
Based on water flow induced by rising bubble plume Qw(x)=35.6C(x+0.8)sqrt(-Voln(1-x/h+10.3))/ub) ``` Qw(x) = water flow (m3/s) x = height above orifice (m) C = 2Vo + 0.05 Vo = air flow in m3/s at 1atm h = depth of orifice (m) ub = 25Vo + 0.7m/s ```
27
Field method for sizing destratification systems (3)
Two types: 1. Use at least 1 SCFM air per 1,000,000 cubic feet of lake volume 2. Use at least 20 SCFM air per 1,000,000 square feet of lake surface area Because lake morphometry, weather, and wind can affect this, may want to take both and average the value
28
What does SCFM stand for? (4)
Cubic feet per minute = CFM = is for feet per minute at actual conditions “Standard” cubic foot of air per minute = feet per minute of air at standard conditions Standard is always the same, so it’s considered a mass flow rate Air density at standard conditions = 0.0752lb/ft3
29
Why do we state air flow in SCFM units?
Because it is easier to do calculations, and makes it easier to compare conditions between lakes as it is standardized
30
Calculation for converting CFM to SCFM
SCFM = CFM*(Pactual/14.7psi)*(528degreesR/Tactual) Eg. Air flow of 25 CFM at a pressure of 30psig, and 90degreesF (same as 550degreesR) (25)*(44.7/14.7)*(528/550)=73SCFM 32degrees F = 491.67R
31
Compressed air lines and pressure losses (6)
As compressed air is forced through air lines, it loses pressure due to friction Therefore, ensure that pressure losses are kept to a minimum (less than 10%) by using the correct diameter line Remember that as pipe diameter doubles, the cross section area quadruples (a=pie*r2) 5 dia. hose has 4x area of a 2.5 dia. hose As SCFM airflow and length needed increases thickness of air line increases Compressed air also has to overcome hydrostatic pressure (in fresh water 10m = 1atm of pressure) => make sure you know the hydrostatic pressure of the depth you wish to release air at
32
Types of air compressors (4)
Never use a reciprocating air compressor, as they are not intended for continuous use Always use a rotary vane air compressor for lower power installations <10hp (<7.5kW) Or rotary screw air compressor for high power installations >10hp (15kW) Or an air blower if you don’t need high depth (<6m)
33
Cleaning compressed air (4)
Most compressors are oil lubricated and carry over a small amount of oil mist This is removed by filters on a valve Otherwise could reduce oxygen transfer and put oil film on the water If compressed air is used to generate oxygen on site, a third filter is used with activated carbon