Wind, Wave, Tidal, Bio, Hydro, Geothermal

Wind power

• Advantages

–

–

–

–

low emissions

small footprint

compatible with some crops

potentially cheap source of power

• Disadvantages

–

–

–

–

–

fan

gearbox dynamo

only work in windy areas

wind is intermittent

visual & noise pollution

threat to birds

electromagnetic interference

Energy – Alternative Energies

1

Available Wind Power

• Consider a mass of air flowing through an area A.

The mass flow is given by:

𝑑𝑚

= 𝜌𝐴𝑈

𝑑𝑡

• where ρ is the air density and U is the air velocity

• The power (kinetic energy per unit time) is given by:

1 𝑑𝑚 ! 1

𝑃=

𝑈 = 𝜌𝐴𝑈 ”

2 𝑑𝑡

2

– Standard conditions (sea level, 15°C, ρ=1.225 kg/m3)

– Power is proportional to the area swept by the rotor.

Energy – Alternative Energies

2

Factors affecting usefulness of wind

• The local velocity, and therefore the power, is

affected by:

– the seasonal and daily wind variations

– local pressure and density

– the stability of the air layer (affected by

temperature, pressure variations)

– turbulence

– local terrain

Energy – Alternative Energies

3

Wind variation – seasonal, diurnal, geographical

Energy – Alternative Energies

4

Thrust

•

The propeller on a plane expels air behind it which causes a force of equal

magnitude but opposite direction on the plane, pushing it forward.

•

This force is called the thrust T:

𝑑𝑚

T=U

𝑑𝑡

•

Thrust has units of Force (kg.m.s-2) (check this !)

•

•

Effectively a propeller converts rotational energy into air flow.

A wind turbine does the opposite – it converts air flow into rotational energy.

•

So the thrust available in flowing air is given by:

𝑑𝑚

T=U

= 𝑈 – 𝜌𝐴𝑈

𝑑𝑡

Source: Wikipedia commons

Energy – Alternative Energies

5

Ideal turbine power

•

Consider the control volume shown in the figure. The air velocity drops from

U1 to U4 as it passes over the rotor area, and area increases from A1 to A4.

•

The following analysis assumes:

•

•

•

•

There is steady state flow

The air flow is homogeneous

The air is incompressible

There is no frictional drag

Stream tube boundary

Rotor

U1

U2 U3

1

•

23

U4

4

The mass flow of the air is constant we can write:

𝑑𝑚

= 𝜌𝐴! 𝑈!

𝑑𝑡

•

The net thrust will be given by:

𝑑𝑚

𝑇=

𝑈# − 𝑈$ = 𝜌𝐴! 𝑈! 𝑈# − 𝑈$

𝑑𝑡

Energy – Alternative Energies

6

Bernoulli’s equation

•

We can apply the Bernoulli equations to the flow both upstream and

downstream of the rotor, assuming it is horizontal:

1 !

1 !

𝑝# + 𝜌𝑈# = 𝑝! + 𝜌𝑈!

2

2

1 !

1 !

𝑝” + 𝜌𝑈” = 𝑝$ + 𝜌𝑈$

2

2

•

Stream tube boundary

Rotor

U1

U2 U3

where p is the pressure.

1

•

U4

23

4

We now assume that the velocity across the rotor is constant (U2=U3) and that

the pressures far from the rotor are equal (p1=p4). Solving for pressure gives:

1

𝑝! − 𝑝” = 𝜌 𝑈#! − 𝑈$!

2

•

But we can also write the thrust in terms of the pressure difference across the

rotor:

𝑇 = 𝐴! 𝑝! − 𝑝”

•

Exercise: check that this also has units of Force.

Energy – Alternative Energies

7

Bernoulli’s equation 2

•

This allows us to write the thrust as:

•

But we already have:

•

Solving yields:

•

So the velocity at the rotor is the average of the downstream

and upstream velocities.

For convenience we define the axial induction factor a:

•

•

1

1

!

!

𝑇 = 𝜌𝐴! 𝑈# − 𝑈$ = 𝜌𝐴! 𝑈# − 𝑈$ 𝑈# + 𝑈$

2

2

So we can write:

Energy – Alternative Energies

𝑇 = 𝜌𝐴! 𝑈! 𝑈# − 𝑈$

𝑈# + 𝑈$

𝑈! =

2

𝑈# − 𝑈!

𝑎=

𝑈#

𝑈! = 𝑈# 1 − 𝑎

𝑈$ = 𝑈# 1 − 2𝑎

8

Power coefficient

•

Power output from the turbine is equal to thrust times the

velocity at the turbine (U2), so substituting and simplifying gives:

𝑃%&'(

•

1

= 𝜌𝐴! 𝑈#! − 𝑈$! 𝑈!

2

1

= 𝜌𝐴! 𝑈! 𝑈# + 𝑈$ 𝑈# − 𝑈$

2

1

= 𝜌𝐴! 𝑈#” 4𝑎 1 − 𝑎 !

2

Noting that A = A2 = A3, and U = U1 is the wind speed:

𝑃%&'(

Energy – Alternative Energies

1

= 𝜌𝐴𝑈 ” 4𝑎 1 − 𝑎

2

!

9

The Betz limit

•

We can now calculate the overall performance of a rotor,

characterized by the power coefficient CP:

𝐶) =

•

+!”#$

+%&

+!”#$

=’

(

,-. )

=

‘

,-. ) $/ #0/ (

(

‘

,-. )

(

•

!

The maximum possible power can be found by differentiating this

expression and setting it equal to zero. This yields:

16

𝐶) =

= 0.593

27

•

= 4𝑎 1 − 𝑎

1

𝑤ℎ𝑒𝑛 𝑎 =

3

This is known as the Betz limit, and represents the maximum ideal

power. A consequence of this is that an ideal rotor will have a wind

speed at the rotor of 2/3 the free stream wind speed.

In analogy with the power, a thrust coefficient can also be defined,

which yields:

𝐶* =

Energy – Alternative Energies

𝑇

1

𝜌𝐴𝑈 !

2

= 4𝑎 1 − 𝑎

10

Exploring the Betz Limit

•

•

Here is a screenshot of a spreadsheet which can be downloaded from Canvas.

This allows you to explore the effects of changing parameters in the Betz equations.

Energy – Alternative Energies

11

Theoretical vs. real efficiency

•

•

Overall efficiency is a combination of the efficiency of capturing

the wind, with the mechanical and electrical efficiency ηmech of

converting this into electrical power.

So the overall efficiency is:

𝜂163’/77 = 𝜂2345 𝐶)

•

So the final output power is given by:

𝑃1&%

•

1

= 𝜌𝐴𝑈 ” 𝜂2345 𝐶)

2

The theoretical maximum cannot be reached in practice due to a

number of factors:

– Rotation of the wake behind the rotor

– Finite number of blades, and associated tip losses

– Non-zero drag.

Energy – Alternative Energies

12

Blade Velocity

•

This factor is important in bringing a real wind turbine up to the limit of

performance set by the Betz limit.

– The tangential velocity of the blade tip is:

𝑣( = 2𝜋𝑟𝑓

• Where r is the radius of the blade and f is the frequency of rotation.

– The tip velocity ratio, λ, is the ratio of this velocity to the wind velocity U:

𝑣( 2𝜋𝑟𝑓

𝜆=

=

𝑈

𝑈

– The best coefficients of performance (Cp ~ 0.5) are achieved with threebladed rotors with a tip velocity ratio λ of around 8.

Energy – Alternative Energies

13

Blade Speeds & Power

•

The mechanical power is converted into electrical energy via a gearbox

with a transmission ratio of around 1:50.

•

A wind turbine with a 60-meter rotor diameter would achieve its maximum

efficiency in a wind speed of 12 m/s with a blade rotational speed of 30

min-1, equivalent to a blade speed of 340 km/h. The largest plants reach

top speeds at the blade tips of over 450 km/h.

•

The power generated by the above example would be 1.47 MW, assuming

a Cp of 0.5.

– For comparison, the maximum output capacity of a typical traditional Dutch

windmill with a 22 m diameter blade, is about 5-6 kW.

•

Modern glass and carbon fibre reinforced blades (up to 80 m diameter) with

aerodynamic profiles can achieve ~ 10 MW output.

Energy – Alternative Energies

14

Examples to work through

1. A wind turbine with 50m diameter blades is operating in a wind speed of

60 km/h. After the wind has gone beyond the turbine to where the

pressure is again atmospheric, the average wind speed is 35 km/h. The

frequency of rotation is observed to be 1 rotation per second (1 Hz).

Assume the density of air is 1.225 kg.m-3 and there is 100% efficiency in

converting the mechanical energy in the blades into electrical energy.

a.

b.

c.

d.

e.

Energy – Alternative Energies

Calculate the wind velocity at the turbine in SI units.

Calculate the axial induction factor.

Calculate the turbine power.

Calculate the power coefficient.

Calculate the ratio of blade tip velocity to frequency, and the actual blade

tip velocity.

15

Wind Power in Australia and Victoria

• Current (end 2019):

– Australia

• 101 major wind farms

• 6.3 GW

• 8.3% of Australia’s electricity, 35.4% of renewable energy

– Victoria:

• 29 major wind farms, 834 turbines

• 1.9 GW

• New Victorian wind farms:

– Approved or under construction: 13 windfarms, 2.6 GW

•

•

•

https://www.energy.vic.gov.au/renewable-energy/wind-energy

http://en.wikipedia.org/wiki/Wind_power_in_Australia

http://en.wikipedia.org/wiki/List_of_largest_power_stations_in_the_world

Energy – Alternative Energies

16

Installed Wind Energy Capacity – Worldwide

Energy – Alternative Energies

Source: Global Wind Energy Council

http://www.gwec.net/global-figures/graphs/

17

Difficulties

• Intermittency

– Wind is inherently

unpredictable, so

wind is not a good

solution for base

power.

• Cost, including

– Storm & lightning

damage

– Transmission to

users

Energy – Alternative Energies

18

Threat to

birds ?

Source: A. Manville, US

Fish and Wildlife Service,

Nature.com

Energy – Alternative Energies

19

Wave Power

• The power of waves is simply related to the fact that the water is

given gravitational potential energy by the wind.

– In the first approximation we consider uniform waves on the surface of a a

body of deep water. In this case only gravity is important.

• We consider the simplistic model shown in the figure. We

approximate the sine wave by a square wave. We assume that a

volume V of liquid moves up and down as the wave propagates.

– The wave has a lateral length L (going into the page)

l

𝑉 =

Amplitude

•

•

•

L = Length of the wave (not

the wavelength)

V =Volume

h = Amplitude = height

Energy – Alternative Energies

h

20

ℎ𝜆

2

𝐿

Wave Power

•

Let the length of a wavefront be L, and the wavelength λ. The potential

energy U in the wave is given by:

𝜆

𝜆 !

𝑈 = 𝑚𝑔ℎ = 𝑉 𝜌𝑔ℎ = 𝐿ℎ 𝜌𝑔ℎ = ℎ 𝜌𝑔𝐿

2

2

•

If we assume that all this potential energy can be converted into kinetic

energy, then the power available per unit length is:

𝑈

𝜆 ℎ! 𝜌𝑔𝐿 𝜆 ℎ! 𝜌𝑔 1 !

𝑃8 =

=

=

= 𝑣ℎ 𝜌𝑔

𝐿𝑇 2 𝑇𝐿

𝑇 2

2

– where the velocity v = fλ = λ/T

•

•

The total power will depend on the length of the wavefront being

intercepted by the power generator:

1 !

𝑃 = 𝑃8 𝐿 = 𝑣ℎ 𝜌𝑔𝐿

2

Finally there will be a mechanical efficiency of conversion, so final

output power will be:

𝑃1&%

Energy – Alternative Energies

1

= 𝜂 – 𝑃 = 𝜂𝑣ℎ! 𝜌𝑔𝐿

2

21

Extracting Wave Power 1

• A typical ocean wave has l~100m, T~8s, and Dh~1.5m.

The density for seawater is 1030 kg/m3.

– For this situation the power carried forward is ~140 kW/m.

• There is actually more energy in a wave than calculated above – there is also

motion of water below the surface which could be harnessed.

• In order to extract the power, the vertical wave

energy must be converted into usable

mechanical/electrical energy.

– There are many possible schemes, only a few of which

have been implemented:

•

Salter’s duck

– consists of a line of vanes fixed to

a central shaft. The wave motion is

converted to an oscillatory motion

which drives a rotary pump which

powers a generator.

Energy – Alternative Energies

22

Extracting Wave Power 2

• Oscillating air column

– this method uses the vertical

wave motion to compress a

column of air, which is driven

through a one way turbine. As the

wave falls, the air is sucked back

in, again running the turbine.

Experimental units in large waves

can produce 75 kW.

•

Tapered channel (Toftestallen, Norway)

– in this method, a site was chosen where water is focused up a

natural channel. A reservoir was built 2 m above the average sea

level. The waves crash above the wall into the reservoir, and are

released back to the ocean using a standard hydro-power generator.

Energy – Alternative Energies

23

Australian Case Studies – CETO

•

Uses the wave effect on

submerged buoys to generate

water flow through a pipe, which

can drive a turbine

•

•

Projects in Fremantle

Technology still being developed

•

•

http://www.carnegiewave.com

Energy – Alternative Energies

24

Australian Case Studies – OceanLinx

• Prototype plant in Port Kembla, NSW was

connected to the grid on 29 March 2010

– Has ceased operation (technology sold)

Energy – Alternative Energies

25

Australian Case Studies – Uniwave200

• Wave power project near King Island

Energy – Alternative Energies

26

Examples to work through

2. A 1 MW wave power station is installed in an area with a wave speed of

1.5 m.s-1 and a wave amplitude of 2.5 m. Assume the density of seawater

is 1030 kg.m-3 and g = 9.8 m.s-2:

a. Calculate the power available per unit length of the station (in kW).

b. Calculate the total power (in MW) available if the power station is 50 m

long.

c. Calculate the % efficiency of the power station.

Energy – Alternative Energies

27

Tidal Power

• Tides are caused by the gravity of the

Moon (and the Sun)

– Tides are complex but predictable

• There a 2 high and 2 low tides per day

• Away from the coast, tidal changes are

of order 1 m, and very slow.

– However, near shallow coastlines, and near

islands, tidal variations can be many m.

– Simple tidal water mills have been used in

some areas of the world for centuries (at

least since roman times).

Tidal Mill

Île-de-Bréhat,

France

Source:

Wikipedia commons

Energy – Alternative Energies

High Tide

due to

gravity

Low

tide

Low

tide

Earth

High tide

due to

inertia

28

Tidal Power

• The largest tidal power station

in the world are:

– Sihwa Lake, South Korea, 254

MW

– La Rance, France, 240 MW.

• These are based around

estuaries, which provide a

concentration of tidal energy,

with tides up to 20 m.

• These photos are from the

Bay of Fundy in Canada (at

low tide)

Energy – Alternative Energies

29

Extracting estuarine tidal power

•

A barrier is built across an estuary. Water is allowed into the

barrier as the tide rises, then the gates are closed.

– As the tide recedes, the height difference can be used to generate

power.

•

Consider a storage basin of area A, at height h above low tide.

The centre of gravity is at height h/2. So the maximum amount

of energy available each tidal cycle is:

ℎ

ℎ 𝜌𝑔𝐴ℎ!

𝐸 = 𝑚𝑔 = 𝜌𝐴ℎ 𝑔 =

2

2

2

•

This maximum power is influenced by diurnal and seasonal

variations, how quickly the water can be released, efficiency of

power generation etc.

Energy – Alternative Energies

30

Tidal Power

•

Advantages

– Once you’ve built the storage basin (dam), tidal power is free

– It produces no greenhouse gases or other waste

– It needs no fuel

– It produces electricity reliably

– Not expensive to maintain

– Tides are totally predictable.

•

Disadvantages

– Very expensive to build

– Affects a very wide area

– The environment is changed for many miles upstream and

downstream, affecting ecosystems.

– Only provides power for around 10 hours each day, when the tide is

moving in or out

– There are very few suitable sites for such tidal power stations.

Energy – Alternative Energies

31

Other Tidal Power – Offshore Turbines

•

Marine current turbines are, much like submerged wind turbines.

– They are installed in the sea at places with high tidal current velocities, to

take out energy from the huge volumes of flowing water.

•

Prototypes consist of twin axial flow rotors of 15-20m in diameter, each

driving a generator via a gearbox.

– The twin power units of each system are mounted on wing-like extensions

either side of a tubular steel monopile some 3m in diameter which is set into

a hole drilled into the seabed from a jack-up barge.

•

SeaGen (UK)

– Operational unit generated power July 18 2008, generating 150kW

(eventually 1.2 MW). Decommissioned in 2017.

Energy – Alternative Energies

32

Alternative extraction methods

• Many systems have been

proposed.

• In all cases, marine fouling is a

major problem.

Energy – Alternative Energies

http://www.emec.org.uk/marine-energy/tidal-devices/

33

Tidal Power in Australia

•

Little development so far Cost per kWh is very

expensive compared to the

alternatives.

•

Australian Tidal Energy

is a consortium formed to

explore tidal power in

Australia

http://austen.org.au/

Energy – Alternative Energies

34

Examples to work through

3. An estuarine tidal station has a storage dam 50 m x 70 m in area, and a

height above low tide of 12 m. Assume the density of seawater is 1030

kg.m-3 and g = 9.8 m.s-2:

a. Calculate the energy available per high tide.

b. Calculate the total power (in kW) available (averaged over 24 hours) if it is

running at 60% efficiency.

Energy – Alternative Energies

35

Hydroelectric power

•

•

Traditional power source – a water mill

on a small stream can produce a few

kW.

Modern schemes use large dams to trap

rainwater at a fixed height, then channel

the water in a controlled fashion through

turbines that generate electricity (the

largest schemes are 20 GW).

Cross section of a hydro scheme

Source: Wikipedia commons

Advantages

• the cheapest of all power sources – about 2 cents per kWh.

Disadvantages

• need lots of water – to power a 60 W light bulb for 1 day, you would

need to drop ~10,000 litres of water from the top of building 14.

• full development would occupy 40% more land than currently, 42

million more acres = 66,000 sq miles.

• substantially changed ecosystems upstream and downstream aridity, temperature, etc.

Energy – Alternative Energies

36

Mechanism of hydroelectric power

•

•

Hydroelectric power simply makes use of the potential energy of having

a mass at an elevated height.

Consider a body of water which is flowing a rate of Q m3/s, has a

density of ρ, and is a relative height h above some reference point. The

potential energy per unit time (ie the input Power available Pin) is:

𝑃9: = 𝜌𝑔ℎ𝑄

•

The efficiency of a system is determined by:

– pipe friction (slows the water speed), 5-30% depending on size of

system

– turbine losses (5-10%)

– electricity generation (5%)

– transmission and distribution losses (~10%).

•

Overall efficiencies h = 0.5-0.7 (50-75%). So output power is:

𝑃1&% = 𝜂 – 𝑃9: = 𝜂𝜌𝑔ℎ𝑄

Energy – Alternative Energies

37

Mechanism of hydroelectric power

•

Worldwide hydropower generates ~17% of

the world’s electricity

•

Largest hydro schemes are:

•

Three Gorges Dam, China (22 GW)

–

the largest single electricity production

facility of any kind

•

Itaipu Dam, Brazil/Paraguay (14 GW)

•

In Australia about 8% of energy is

generated by hydro electricity

–

Three Gorges Dam – Source: Wikipedia commons

The Snowy mountains scheme generates

3.8 GW

Energy – Alternative Energies

Murray-1 Power Station – Snowy Mountains Scheme

Source: Wikipedia commons

38

Examples to work through

4. A hydroelectric dam releases water through its turbines at a rate of 15

m3.s-1. If the water drops through a height of 120 m and is converted to

electricity with an overall efficiency of 75%, calculate the output power.

Energy – Alternative Energies

39

Hydro pumped storage

• During low demand/off peak periods (eg overnight), where

electricity is cheap or wasted, water can be pumped into

an elevated storage reservoir.

• When additional generating capacity is needed, water is

released from the upper reservoir, running turbine

generators to return power to the grid.

• Such a setup is a very efficient storage device (much

more efficient than batteries).

– An example of negative net energy – uses more kWh to pump the

water up at night than the kWh that the falling water produces

during the day, but provides a convenient and efficient way of

helping to cope with peak demand.

https://www.energy.vic.gov.au/renewable-energy/hydroelectricity

Energy – Alternative Energies

40

Biomass and Biofuels

•

This refers to replacing fossil fuels with fuels derived from vegetation

•

Gaseous biofuels

– methane (CH4) and CO2 from anaerobic digestion of plant and

animal wastes

– CO and H2 from gasification of plants, wood and wastes

•

Liquid biofuels

– vegetable oils from crop seeds, and derivatives

– ethanol, methanol from distillation and fermentation

•

Sold biofuels

– Wood (from plantations, waste, timber yards etc)

– charcoal from pyrolysis

– solid fuels derived from waste products (compressed manure etc)

Energy – Alternative Energies

41

Biomass and Biofuels 2

•

•

Advantages

– in general the materials burn more cleanly than fossil fuels, as they

are purer.

– make portable fuels

Disadvantages

– producing ethanol and other biofuels causes air/water/soil pollution

– planting, harvesting and conversion to ethanol/biofuels all cost

energy.

– Examples

• corn ethanol – provides 25% more energy than it costs to produce

• soy biodiesel – provides 93% more energy than it costs to produce

– all of these processes lead to the production of greenhouse gases.

Only a percentage (25%-50%) will be absorbed by growing the next

crop.

– a major disadvantage is that crops used for biofuels cannot be used

for food.

• This has led to significant food price rises worldwide, and food

shortages in some parts of the world.

Energy – Alternative Energies

42

Geothermal Energy

•

•

•

•

At tectonic plate boundaries, geothermal plants can tap the heat

of the earth’s interior

Uses hot springs, geysers and dry rock to convert (added) water

to steam, which can power turbines

Advantages

– no significant pollution

– no fuel is needed

– environmental impact low (small footprint)

– once you’ve built a geothermal power station, the energy is

almost free

Disadvantages

– location dependent, very limited

– source decays over time, becomes non-renewable

– hazardous gases and minerals may come up from

underground, and can be difficult to safely dispose of.

Energy – Alternative Energies

43

Geothermal Energy

• Individual power plants operate at capacities ranging

between 100kW and 100MW

• Over 8 GW of electricity from geothermal plants

worldwide.

From www.our-energy.com

Energy – Alternative Energies

44

Distribution of Major Geothermal Energy Reserves

Photo: www.geothermal.marin.org/

Energy – Alternative Energies

From www.geni.org

45

Geothermal Energy

•

Geothermal energy has been utilized for centuries where the circulating

fluid is used directly:

– building and hot water heating, melting snow on pavements/roads, warming

of greenhouses (eg Iceland)

– cooking, laundry, washing (eg NZ)

•

Heat Pumps – modern heat pumps harness geothermal energy on a

small scale

– A circulating fluid absorbs or releases heat to the soil

– A heat pump removes the heat, which is ducted to the building

• eg home heating in Iceland

From www.radiantfloorheating.com.au

Energy – Alternative Energies

46

Geothermal Power Generation

•

Large scale harnessing requires large sources, which come in

three main types:

•

Dry hyper-thermal field (vapor-dominated)

– produces dry saturated, or slightly superheated steam at P > Patm

– needs a dry steam power plant

•

Wet hyper-thermal field (water-dominated)

– produces pressurized water > 100°C

– needs a flash steam power plant

•

Semi-thermal field

– produces water up to 100°C from drilling depths of 1-2 km

– needs a Binary Cycle power plant

Energy – Alternative Energies

47

Dry Steam

•

•

First used in 1904

In this type of plant steam moves from geothermal wells directly

to a turbine which then produces electricity.

Turbine

Generator

Electrical output

Projection well

Injection well

Dry rocks

Energy – Alternative Energies

48

Flash Steam Power Generation

•

•

•

•

•

Most common type used today

Needs water-dominated fields

with fluid T>182°C

Fluid pumped under high

pressure to power generation

equipment where it is sprayed

into low pressure tank

Pressure reduction causes

some fluid to vapourize

quickly (called the flash)

Flash vapour drives turbine.

Flash Tank

Generator

Turbine

Electrical output

Projection well

Injection well

Dry rocks

Energy – Alternative Energies

49

Binary Cycle Power Generation

•

•

•

•

Used where geothermal fluid is

at moderate temperatures

Hot geothermal fluid and a

binary fluid pass through heat

exchanger

Binary fluid has much lower

boiling point than water, thus

heat from the geothermal fluid

causes binary fluid to flash

Flash steam drives

turbines/generators.

Turbine

Generator

Working

Fluid

Heat

Exchanger

Projection well

Electrical output

Injection well

Dry rocks

Energy – Alternative Energies

50

Hot rock geothermal in Australia

• Geothermal energy is present in

Australia, but it is a long way

below the surface.

• Modern drilling techniques make

this accessible via “hot rock”

extraction techniques.

• Main disadvantage is the

distance to market for electricity

(significant transmission losses).

•

•

There has been little development in

Australia.

Australian company Geodynamics

closed several test wells in the Cooper

Basin after several years of testing due

to lack of financial viability.

Energy – Alternative Energies

51

Electricity Production in Australia – Recap

2019 Australian Energy Update

https://www.energy.gov.au/publications/australian-energy-update-2019

Energy – Introduction

Renewable Energy in Australia – Recap

2019 Australian Energy Update

https://www.energy.gov.au/publications/australian-energy-update-2019

Energy – Introduction

Examples to work through – Solutions

1. A wind turbine with 50m diameter blades is operating in a wind speed of

60 km/h. After the wind has gone beyond the turbine to where the

pressure is again atmospheric, the average wind speed is 35 km/h. The

frequency of rotation is observed to be 1 rotation per second (1 Hz).

Assume the density of air is 1.225 kg.m-3 and there is 100% efficiency in

converting the mechanical energy in the blades into electrical energy.

a.

Calculate the wind velocity at the turbine in SI units.

First convert the speeds into SI units:

U1 = 60 kph = 16.67 m.s-1 and U4 = 35 kph = 9.72 m.s-1.

1

𝑈! = 𝑈# + 𝑈$ = 13.2 𝑚. 𝑠 0#

2

b.

Calculate the axial induction factor.

𝑈# − 𝑈!

𝑎=

= 0.21

𝑈#

Energy – Alternative Energies

54

Examples to work through – Solutions p2

1. A wind turbine with 50m diameter blades is operating in a wind speed of

60 km/h. After the wind has gone beyond the turbine to where the

pressure is again atmospheric, the average wind speed is 35 km/h. The

frequency of rotation is observed to be 1 rotation per second (1 Hz).

Assume the density of air is 1.225 kg.m-3 and there is 100% efficiency in

converting the mechanical energy in the blades into electrical energy.

c.

Calculate the turbine power.

𝑃%&'(

1

= 𝜌𝐴𝑈 ” 4𝑎 1 − 𝑎

2

!

= 2.92 𝑀𝑊

d. Calculate the power coefficient.

The inherent power of the wind striking the turbine is:

So the efficiency is:

Or:

𝑃%&'(

𝑃%&'(

2.92 𝑀𝑊

𝐶) =

=

=

= 0.52

1

𝑃9:

𝜌𝐴𝑈 ” 5.57 𝑀𝑊

2

𝐶) = 4𝑎 1 − 𝑎

!

= 0.52

The efficiency of the system is 52%, compared to 59.3% set by the Betz limit.

Energy – Alternative Energies

55

Examples to work through – Solutions p3

1. A wind turbine with 50m diameter blades is operating in a wind speed of

60 km/h. After the wind has gone beyond the turbine to where the

pressure is again atmospheric, the average wind speed is 35 km/h. The

frequency of rotation is observed to be 1 rotation per second (1 Hz).

Assume the density of air is 1.2 kg.m-3 and there is 100% efficiency in

converting the mechanical energy in the blades into electrical energy.

e.

Calculate the ratio of blade tip velocity to frequency, and the actual blade

tip velocity.

The tip velocity ratio is:

2𝜋𝑟𝑓 2𝜋 – 25 – 1

𝜆=

=

= 9.42

𝑈

16.67

This is close to the ideal value of about 8.

So the blade tip velocity will be:

𝑣( = 𝜆𝑈 = 9.4 – 16.67 = 157 𝑚. 𝑠 0# = 565 𝑘𝑚. ℎ0#

Energy – Alternative Energies

56

Examples to work through – Solutions

2. A 1 MW wave power station is installed in an area with a wave speed of

1.5 m.s-1 and a wave amplitude of 2.5 m. Assume the density of seawater

is 1030 kg.m-3 and g = 9.8 m.s-2:

a. Calculate the power available per unit length of the station (in kW).

𝑃! =

1

2

”

𝑣ℎ 𝜌𝑔 =

1

2

”

, 1.5 , 2.5 , 1030 , 9.8 = 47315 𝑊 = 47.3 𝑘𝑊

b. Calculate the total power (in MW) available if the power station is 50 m

long.

𝑃 = 𝑃! 𝐿 = 47315 , 50 = 2365781 𝑊 = 2.37 𝑀𝑊

c. Calculate the % efficiency of the power station.

𝜂 =

Energy – Alternative Energies

#!”#

#$%

=

$

“.&’

= 0.42 = 42%

57

Examples to work through – Solutions

3. An estuarine tidal station has a storage dam 50 m x 70 m in area, and a

height above low tide of 12 m. Assume the density of seawater is 1030

kg.m-3 and g = 9.8 m.s-2:

a. Calculate the energy available per high tide.

𝐸 =

𝜌𝑔𝐴ℎ

”

”

=

1030 , 9.8 , 50 , 70 , 12

(

= 2.54×10 𝐽 = 2.54 𝐺𝐽

2

2

b. Calculate the total power (in kW) available (averaged over 24 hours) if it is

running at 60% efficiency.

There are 2 tides per day, so the average power available over 24 hours is:

P =

Energy – Alternative Energies

𝜂,2,𝐸

24 , 3600

(

=

0.6 , 2 , 2.54×10

24 , 3600

= 35329𝑊 = 35.3 𝑘𝑊

58

Examples to work through – Solutions

4. A hydroelectric dam releases water through its turbines at a rate of 15

m3.s-1. If the water drops through a height of 120 m and is converted to

electricity with an overall efficiency of 75%, calculate the output power.

𝑃1&% = 𝜂𝜌𝑔ℎ𝑄 = 0.75 – 1000 – 9.8 – 120 – 15

= 13230000 𝑊 = 13.2 𝑀𝑊

Energy – Alternative Energies

59

Atmospheric Physics

Modelling the Atmosphere

• Most light at visible wavelengths (400nm – 700 nm) is

transmitted by the atmosphere (except if there is heavy

cloud cover).

• As much of the Sun’s radiation is transmitted, the Earth’s

surface receives a lot of energy. This causes it to heat up, and

it re-radiates that heat.

• However, the Earth re-radiates over a different wavelength

range – in the infrared.

• Infrared radiation is strongly absorbed by some gases in the

atmosphere.

• This is how the atmosphere provides insulation to the

Earth’s surface.

• In order to understand this we must re-visit Blackbody

Radiation.

1

Blackbody Radiation (recap)

•

•

Any body at a temperature T radiates heat to its surroundings with a

characteristic spectrum. This behaviour is called blackbody radiation.

The peak wavelength

is given by:

Blackbody Radiation – Solar Temperatures

200

lmT = 2.9×10-3 m.K

F = sT4

•

This is called the

Stefan-Boltzmann law

where:

150

-2

The integral of the

curve gives the total

flux:

Flux (TW.m )

•

‘4000’

‘5000’

‘5778’

‘6000’

‘7000’

100

50

0

s = 5.67×10-8 Wm-2K-4

500

1000

1500

2000

Wavelength (nm)

2500

3000

2

Infrared Radiation

Peak Wavelengths

Blackbody Radiation – Earth Temperatures

35

254 K = -19 °C (Earth without atmosphere)

273 K = 0 °C (freezing)

287 K = 14 °C (average Earth temperature)

305 K = 32 °C (skin temperature)

-19°C – 11422 nm

30

0°C – 10627 nm

25

-2

Flux (MW.m )

14°C – 10109 nm

32°C – 9512 nm

20

15

10

5

0

5

3

10

15

Wavelength (nm)

20

25

3

30 x10

Infrared images

Wilson’s Promontory

Bushfires Mar 14 2009.

Source: http://visibleearth.nasa.gov

Human body

Source: Wikipedia commons

4

Thermal

Radiation &

Reflected

Sunlight

Credit: NASA/Goddard Space Flight Center Scientific Visualization Studio

• Left – sunlight reflected back to space by the ocean, land,

aerosols, and clouds.

• Right – heat (or thermal radiation) emitted to space from

Earth’s surface and atmosphere

Source: https://science.nasa.gov/ems/13_radiationbudget

5

Emissivity

•

For any real body, the radiation emitted at any wavelength will

always be less than or equal to, but never greater than, the radiation

from a black body of the same temperature.

• The black body curves therefore represent the maximum that

might be expected to emanate from a body at a particular

temperature.

•

For a real body, the actual radiation emitted is determined by a

quantity known as the emissivity.

• This is a number, e , which lies between 0 and 1. A white body

has a value close to 0 whereas a pure black body has a value of

exactly 1.

• e is not a constant for the material, but can depend on the

wavelength of the radiation – some materials are more like black

bodies at some wavelengths than at other wavelengths.

6

Examples to work through

1. Assume the average person radiates as a perfect black body with a

surface temp of ~32 °C and an area of ~1.5 square metres.

a. At what wavelength does the person radiate in (micrometres)?

b. Calculate the power being radiated?

c. If the emissivity is 0.5 (eg due to wearing clothes) calculate the

power being radiated?

7

Solar Insolation

• Just above the atmosphere of the earth, the power

coming from the sun is S = 1360 W/m2.

• This is called solar insolation, S. Given this we can

calculate the amount of energy the sun radiates:

Sun

Incident

Sunlight

Earth

• Information we will need:

𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑆𝑢𝑛 𝑡𝑜 𝐸𝑎𝑟𝑡ℎ: 𝐷 = 1.5×10!! 𝑚

𝑅𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝐸𝑎𝑟𝑡ℎ: 𝑅” = 6.37×10# 𝑚

𝑅𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑆𝑢𝑛: 𝑅$ = 6.96×10% 𝑚

8

Radiation from the Sun

• What is the total amount of power radiating from the

surface of the sun?

• If we know the flux at the Earth, then we just need to multiply by

the surface area at the radius of the earth to determine the total

Power of the Sun.

𝑃 = 𝑆 ? 𝐴 = 𝑆 ? 4𝜋𝐷 & = 1360 ? 4𝜋 1.5×10!!

&

= 3.85×10 𝑊

Earth

Sun

D=1.5×1011 m

9

Radiation from the Sun (continued)

Assuming the sun radiates as an ideal Blackbody,

what is the temperature of the surface of the sun?

∴𝑇=

!

𝑃

=

𝜎𝐴

!

3.85×10

& = 5778 𝐾

𝜎 ? 4𝜋𝑅’

RS

Here the relevant

area is the surface

area of the sun.

10

Examples to work through

2. The surface of the sun is at a temperature of about 5778 K.

a.

b.

c.

At what wavelength is the maximum amount of energy being given off?

What is the flux produced by the sun?

Calculate the total power striking the earth from the sun?

Hint – in this case it is the cross-sectional area of the Earth that is important,

not the surface area. From the point of view of the Sun, the Earth looks like a

disc, not a sphere.

Sun

Incident

Sunlight

Earth

11

Simple Model A

no atmosphere, uniform surface temperature

• For the moment let’s assume there is no atmosphere

and that all the incident energy from the sun is

absorbed.

•

If the temperature of the earth remains constant, it must

re-radiate the same amount of energy.

•

•

The re-radiation will be in the infrared wavelength range.

Remember that the sunlight is striking only one side of the

Earth, whereas heat leaves from the whole surface.

Re-radiated infrared

Incident Sunlight

12

Simple Model A (continued)

Q7:

What is the resultant temperature of the earth, TE? We have:

𝑃() = 𝑆 ? 𝐴* = 𝑆 ? 𝜋𝑅”&

Where:

𝑃,-. = 𝐹,-. ? 𝐴 = 𝜎𝑇”+ ? 4𝜋𝑅”&

Ac is the cross sectional area of the Earth, A is the surface area of the Earth

S is the Solar insolation (flux), Pin is the incoming power, Pout is the radiated power from the Earth

Fout is the radiated flux from the Earth’s

Setting Pin = Pout gives:

Re-radiated infrared

𝑆 ? 𝜋𝑅”& = 𝜎𝑇”+ ? 4𝜋𝑅”&

Incident Sunlight

And solving for TE gives:

𝑇” =

!

𝑆 ? 𝜋𝑅”&

& =

𝜎 ? 4𝜋𝑅”

!

𝑆

= 278 𝐾

4𝜎

13

Simple Model B

Simple Greenhouse Effect

• This is a bit cold, so obviously we need the blanket of the

atmosphere to keep us warmer.

• As well as protecting us from harmful ultraviolet radiation and

cosmic rays.

• Consider the model shown below where now we’ve added

an atmosphere. It’s a rather crude atmosphere at just one

uniform temperature TA.

Incident Sunlight

14

Simple Model B (continued)

• As a very simple start, we’ll assume that all the incident

sunlight can travel through the atmosphere and is totally

absorbed by the earth.

• Actually the amount absorbed depends on where it hits – for

example the sea is a very good absorber of sunlight, but the

polar ice caps reflect a lot of the sunlight.

• The earth radiates off the heat in the infrared (red

arrows) which is absorbed by the atmosphere.

• The atmosphere then heats up,

so it re-radiates, also in the

infrared, not only out into space

but also back towards Earth

(orange arrows).

Incident Sunlight

15

Simple Model B (continued)

•

•

Assuming temperatures aren’t changing, then there must be a heat

balance for the atmosphere as well as the Earth.

Below are two simultaneous equations, one for the heat balance of the

Earth and one for the atmosphere.

• At this steady state condition, the outgoing infrared must still match

the incident sunlight in terms of power.

𝐹” ? 4𝜋𝑅”& = 𝑆 ? 𝜋𝑅”& + 𝐹/ ? 4𝜋𝑅”&

Heat balance for

the Earth’s surface

Heat balance for

the atmosphere

𝐹” ? 4𝜋𝑅”& = 𝐹/,-. ? 4𝜋𝑅”& + 𝐹/() ? 4𝜋𝑅”& = 2𝐹/ ? 4𝜋𝑅”&

Here we assume the atmosphere radiates 50% inwards and 50% outwards,

so FA in = FA out = FA. Simplifying these two equations yields:

Eliminating FA gives:

4𝐹” = 𝑆 + 4𝐹/

𝐹”

4𝐹” = 𝑆 + 4

= 𝑆 + 2𝐹”

2

𝑆

∴ 𝐹” =

2

𝐹” = 2𝐹/

16

Simple Model B (continued)

•

•

𝑆

𝐹” = 𝜎𝑇 =

2

+

So, this yields:

Solving for TE:

∴ 𝑇” =

𝑆

B𝑢𝑡 𝐹” = = 2𝐹/

2

𝑆

+

∴ 𝐹/ = 𝜎𝑇/ =

4

! 𝑆

∴ 𝑇/ =

= 278 𝐾

4𝜎

•

Finally, this gives:

• TE = 331 K

• TA = 278K

!

𝑆

=

2𝜎

!

1360

= 331 𝐾

0%

2 ? 5.67×10

Incident Sunlight

17

Preliminary models – summary

• With some very simple physics and some

rather gross modelling, we have calculated

a temperature of the Earth which is quite

realistic.

• We are well on the way to making a simple

model of the effect of the Earth’s

atmosphere.

18

Albedo

•

•

Actually about 30-35% of incident sunlight gets

reflected off the atmosphere back to space without

reaching the Earth.

•

i.e the Earth’s albedo is 30-35%. This varies with

the amount of cloud cover etc. Let’s assume a

value of 31%:

So Seff = (1-0.31) S = 0.69 S = 938 W/m2

Q8: Exercise: Modify the previous analysis to take this

into account.

A:

• TE = 302 K

• TA= 254 K

19

Simple Greenhouse Model C

• To improve the model we also need to consider other factors:

1. As stated above, some of the sunlight will be reflected back

from the clouds without ever striking the earth

• Call this the albedo, a1.

2. A fraction, a2, will go straight to Earth, and therefore the

fraction, (1-a1-a2 ), of the sunlight will be absorbed by the

atmosphere.

•

So the cloud emissivity to sunlight is (1-a1-a2 ).

3. Some of the infrared emitted by the Earth will go through the

atmosphere instead of being absorbed (eg on a clear night).

•

•

The fraction of infra-red emitted by the earth which is

absorbed by the atmosphere we’ll call a3. We will assume

none is reflected.

The amount of greenhouse gases in the atmosphere directly

affects the value of a3.

4. We will still assume that the atmosphere radiates half its

energy inwards and half outwards (as in model B).

20

Model C

• On the next slide is a simple schematic of a model which

takes some of these features into account.

• We’ve flattened out the earth to make it easier to view.

• But remember that the sunlight falls on a fraction of the earth (half),

whereas the infrared is emitted from the whole earth.

• You should confirm these equations:

Heat balance for

the surface of the

Earth

4𝐹” = 𝛼& ? 𝑆 + 𝛼1 ? 4𝐹/

Heat balance for

the atmosphere

𝛼1 ? 2 ? 4𝐹/ = 𝛼1 ? 4𝐹” + 1 − 𝛼! − 𝛼& ? 𝑆

• To start with, assume that a1, a2 and a3 are 0.31, 0.65

and 0.87 respectively.

• Note that the following relationship must always be

satisfied: a1+ a2 ≤ 1.

21

A Flat(tened) Earth

22

Questions

•

What is the temperature of the earth in this model?

•

What is the amount of infrared being given off into

space from the atmosphere?

•

What is the total amount of infrared radiation being

given off into space from the Earth?

23

Solving the equations (derivation not examinable)

• Eliminating FA we can solve for TE:

2 4𝐹” − 𝛼& ? 𝑆 = 8𝐹” – −2𝛼& ? 𝑆 = 𝛼1 ? 4𝐹” + 1 − 𝛼! − 𝛼& ? 𝑆

1 − 𝛼! + 𝛼& ? 𝑆

∴ 𝐹” =

= 𝜎𝑇”+ ⇒ 𝑇” =

8 − 4𝛼1

!

𝑆 1 − 𝛼! + 𝛼&

4𝜎 2 − 𝛼1

• Substituting back we can solve for TA:

4𝐹” = 4𝜎𝑇”+ = 𝛼& ? 𝑆 + 𝛼1 ? 4𝐹/

∴ 𝑇/ =

!

4𝜎𝑇”+ − 𝛼& ? 𝑆

=

4𝜎𝛼1

!

+

4𝜎𝑇

− 𝛼& ? 𝑆

”

+

∴ 𝐹/ = 𝜎𝑇/ =

4𝜎𝛼1

𝑇”+

𝛼& 𝑆

−

𝛼1 4𝜎𝛼1

• With a1, a2 and a3 equal to 0.31, 0.65 and 0.87, this gives:

TE = 290 K and TA = 247 K.

24

Model C as an excel simulation

Note: Models A and B can also be calculated with the

spreadsheet by using appropriate values of a1, a2 and a3.

25

Things to think about

• Primary effect: increase greenhouse gases leads to increased in

temperature

• How might other activities affect temperatures using this model?

•

• What is the effect of the heat created directly from the burning of

fossil fuels and other activities?

• Consider how bush fires, which generate smoke and dust which

goes into the atmosphere, are likely to affect the temperatures.

•

Dust particles absorb both sunlight and infrared. What is the

likely influence on the coefficients a1, a2 and a3 ?

•

In the Excel spreadsheet try changing these parameters & try to

predict what the results will be.

If the temperature of the Earth and atmosphere did heat up, what is

the likely effect on the cloud cover?

•

•

Will a change in cloud cover cause an increase or a decrease in

temperatures? i.e. will it produce positive or negative feedback in the

cycle?

Consider the effect of reduced polar icecaps on the model. How

might you alter the above model to simulate this factor?

26

Nuclear Winter

• It has been suggested that if there were a nuclear war,

enormous quantities of soot and dust would be pumped

up into the upper atmosphere

• no sunlight could penetrate down to earth, so creating

a severely cold climate – a nuclear winter.

• Change Model B so that the atmosphere now absorbs all

the sunlight (ie a2=0), with none getting to the Earth. The

atmosphere still absorbs all the IR radiation coming from

the Earth.

Q1: What would be the temperature of the earth under

these conditions? (ans: TE=244K, TA=253K)

27

Effect of Water Vapour

•

•

In the simple model to date we’ve ignored the heat transfer

associated with the movement of water vapour.

In the schematic below, the effect of water being evaporated off the

oceans is added.

• The water vapour condenses in the atmosphere, so releasing its

latent heat. This is called the adiabatic lapse rate.

• We won’t include this in our model.

28

Rising Sea Level

• A major concern for Pacific Island states and many other

countries is the potential rise in sea level which would

accompany any global warming.

• Why would sea levels rise?

1. Melting of ice which is on land (Antarctic and Greenland

ice sheets, glaciers) would increase the sea level.

•

Note that the melting of floating ice (eg Arctic ice) would not

lead to a sea level rise (consider what happens to the level

of a glass of water with ice after the ice melts).

2. the thermal expansion of water with increasing

temperatures is also a factor to be considered

•

eg between 4°C and 12°C, the density of water decreases

from 1.000 g/ml to 0.9998 g/ml.

29

Examples to work through

3. The Greenland ice sheet has a volume of 2.85×10! 𝑘𝑚” . Assuming the

density of ice is 920 kg.m-3:

a.

b.

c.

d.

Calculate the volume of ice in m3.

Calculate the volume of water that would be created if the ice melted?

Calculate the surface area of the Earth (𝑅! = 6.37×10″ 𝑚)

71% of the Earth’s surface is covered in water. If the Greenland ice sheet

melted, calculate how much the water level would rise, other things being

equal.

30

Examples to work through

4. The average depth of the ocean is 3,688 m.

a.

b.

c.

If the average temperature of the ocean increased by 3 °C, which is

accompanied by a 0.06% reduction in density, and assuming that the %

coverage of the Earth’s surface remains constant, what would be the sea

level rise?

How does this compare to melting of Greenland’s ice sheet?

Comment on the assumptions in this question

31

Comparison between Earth, Mars and Venus

Mars, Earth, Venus.

Altitude (km)

100

A little greenhouse

effect is good.

Arrows show

surface temperature

without the

greenhouse effect..

50

100

800

400

Temperature (Kelvin)

32

Modelling Venus

• A few facts:

• Its orbit is 108,200,000 km (0.72 AU) from the Sun and it

has a diameter of 12,103.6 km

• Its albedo (reflectivity of sunlight) is 0.59

• The pressure of Venus’ atmosphere at the surface is 90

atmospheres (about the same as the pressure at a depth

of 1 km in Earth’s oceans).

• The atmosphere is composed mostly of carbon dioxide.

• There are several layers of clouds many kilometres thick

composed of sulphuric acid.

• Venus’ surface temperature is over 740 K (hot

enough to melt lead!)

33

Venus’ Atmosphere

• The absorbing thickness of the atmosphere is so large

that you might need to model it using several layers.

• The top one would be relatively cool. Suppose you had

two layers and the infrared from one layer was

completely absorbed by the adjacent layer. In other

words, extend model B to two atmospheric layers.

• Compare this with what you would get if there were just

one layer.

• To find out more about atmospheric modelling by the

CSIRO, try http://www.cmar.csiro.au/

34

Greenhouse Gases

• The table shows the estimated present warming effect

due to the presence of the main two greenhouse gases,

H2O and CO2.

Gas

Concentration

(ppmv)

Warming

Effect (°C)

H2O vapour

5000

20.6

CO2

358

7.2

• How does a3 vary with the H2O and CO2 levels?

• We can only take our model so far. It is possible that

increases in CO2 levels might lead to ever increasing

temperatures due to positive feedback.

35

Feedback

• Amongst the positive feedback mechanisms are:

• Melting of the snow and ice, lowering the reflectivity;

• More water vapour in the air leads to more absorption of

thermal radiation from the Earth;

• Higher sea water temperatures leads to less CO2

absorption; there is also a faster decay or organic material,

which can increase CO2 and CH3 levels.

• Mitigating (negative feedback) factors include:

• Increasing temperatures can produce increased algal

growth in the oceans which will consume CO2 in

photosynthesis;

• Humid air can carry more heat up into the atmosphere, and so

helps take heat away from the surface of the Earth (the

adiabatic lapse rate).

36

Summary

• These ideas should give you an idea of how the

modelling could be extended to include more effects.

• One should also consider how the atmosphere changes

with altitude, longitude and latitude, the effect of rotation

of the earth etc etc.

• Real models take all these into account.

37

CO2 and Temperature – State of Play

https://www.esrl.noaa.gov/gmd/ccgg/trends/mlo.html

38

Recent temperature increases

https://data.giss.nasa.gov/gistemp/graphs/

39

CO2 through History

https://climate.nasa.gov/vital-signs/carbon-dioxide/

40

Conclusions

•

•

•

•

•

•

•

What is energy?

Energy requirements, consumption and usage

Fossil fuels

Introduction to Energy Concepts

• Relevant physics needed for understanding later topics:

• Force and Motion, Work and Energy, Temperature, Heat and

Thermodynamics, Fluids, Basic Electricity

Energy Efficiency and Storage

Other Methods of Energy Production

• Solar energy, Hydro, Geo and Bio energy, Wind, Wave, Tidal,

Nuclear energy

Introduction to Atmospheric Physics

• basic physical models for understanding the greenhouse effect.

41

Examples to work through – Solutions

1. Assume the average person radiates as a perfect black body with a

surface temp of ~32 °C and an area of ~1.5 square metres.

a. At what wavelength does the person radiate (in micrometres)?

𝜆234

2.9×1001 2.9×1001

=

=

= 9.5 𝜇𝑚

𝑇

305

b. Calculate the power being radiated?

𝑃 = 𝐴𝜎𝑇 + = 1.5 ? 5.67×100% ? 305+ = 736 𝑊

c.

If the emissivity is 0.5 (eg due to wearing clothes) calculate the

power being radiated?

𝑃 = 𝜀𝐴𝜎𝑇 + = 0.5 ? 1.5 ? 5.67×100% ? 305+ = 368 𝑊

42

Examples to work through – Solutions

2. The surface of the sun is at a temperature of about 5778 K.

a.

At what wavelength is the maximum amount of energy being given off?

𝜆234

b.

2.9×1001

=

= 502 𝑛𝑚

𝑇

What is the flux produced by the sun?

𝐹 = 5.67×100% ? 5778+ = 6.32×105 𝑊. 𝑚 0&

c.

Calculate the total power striking the earth from the sun?

𝑃 = 𝑆 ? 𝐴 = 1360 ? 𝜋𝑅”& = 1.73×10!5 𝑊

43

Examples to work through – Solutions

3. The Greenland ice sheet has a volume of 2.85×10! 𝑘𝑚” . Assuming the

density of ice is 920 kg.m-3:

a.

Calculate the volume of ice in m3.

𝑉7 = 𝑉7 ? 10001 = 2.85×10!8 𝑚 1

b.

Calculate the volume of water that would be created if the ice melted?

Ice is pure water (no salt etc), so will melt to make pure water whose density

is 1000 kg.m-3, so the ice would shrink by a factor determined by the density

ratio:

920

𝑉6 = 𝑉7

= 2.62×10!8 𝑚 1

1000

c.

Calculate the surface area of the Earth (𝑅! = 6.37×10″ 𝑚)

𝐴 = 4𝜋𝑅”& = 5.10×10!+ 𝑚 &

d.

71% of the Earth’s surface is covered in water. If the Greenland ice sheet

melted, calculate how much the water level would rise, other things being

equal.

𝐴 = 0.71 ? 5.10×10!+ = 3.62×10!+ 𝑚 &

𝑉 2.62×10!8

𝐻= =

= 7.24 𝑚

!+

𝐴 3.62×10

44

Examples to work through – Solutions

4. The average depth of the ocean is 3,688 m.

a.

If the average temperature of the ocean increased by 3 °C, which is

accompanied by a 0.06% reduction in density, and assuming that the %

coverage of the Earth’s surface remains constant, what would be the sea

level rise?

0.06

Δ𝐻 =

? 3688 = 2.2 𝑚

100

b.

How does this compare to melting of Greenland’s ice sheet?

This is less than the Greenland ice sheet but still a considerable rise.

c.

Comment on the assumptions in this question

The temperature of the ocean is a function of depth, so a more detailed

model would need to be used to do this calculate properly.

45

RMIT – SCHOOL OF SCIENCE

PHYS2066/2129 – Energy and the Earth’s Environment

Practice Skills & Capability Assessment

TIME ALLOWED:

1.5 hours for the test, plus 30 minutes to allow for downloading, scanning

and uploading to canvas.

Academic Honour Code declaration

The work I am submitting for this Assessment Task is entirely my own work. Nothing I submit is

from another source – either from another student or person, or from another resource, in whole or in

part. All written work is in my own words.

I have not communicated with any other student during the period of this Assessment Task. I

have not discussed the content of this Assessment with anyone else during the period of this

Assessment Task.

Name:

Student Number:

Signature:

INSTRUCTIONS – READ CAREFULLY

• Test can be printed, worked through and scanned, or you can do the test on blank paper and

scan, making sure to carefully label the questions. If the latter, you must write and sign the

Honour code on your question sheet.

• ANSWER ALL QUESTIONS – All questions are of equal value.

• Show ALL working. The use of UNITS in final answers is ESSENTIAL.

• In answering the questions please ensure there is adequate balance between any mathematics

you may use and a description of what you are saying in English.

• Calculators may be used.

Useful Physical Constants

s = 5.67×10-8 W/K4.m2

Wien’s constant = 2.90×10-3 m.K

k = 1.38×10-23 J/K

cp = 4179 Jkg-1K-1 (specific heat for water)

g = 9.8 m.s-2

S = 1360 W.m-2

PHYS2066/2129 – Test 2 – practice test A

e = -1.6×10-19 C

r = 1000 kg/m3 (fresh water)

r = 1030 kg/m3 (seawater)

r = 1.225 kg/m3 (air at sea level)

RE = 6.37×106 m

TK=TC+273

1 of 7

Question 1

A wind turbine with an 11 m radius rotor has an axial induction factor of 0.15. If the wind is

blowing at 25 km/h, and the turbine outputs 20 kW then:

(a) Calculate the power coefficient for the turbine

(b) Calculate the wind velocity in m/s.

(c) Calculate the mechanical efficiency of the turbine (as a %).

A wave power station produces 10 MW of electricity.

(d) if the overall efficiency is 45%, calculate the amount of power (in MW) in the waves.

(e) if the average wave speed and amplitude are 8 m.s-1 and 2.5 m respectively, calculate the

length of the power station.

PHYS2066

2 of 7

Question 2

A wall consists of 3 layers with the following properties:

Weatherboard: k = 0.15 W.m-1.K-1 L = 2 cm

Insulation:

k = 0.05 W.m-1.K-1 L = 40 mm

Plaster:

k = 0.5 W.m-1.K-1

L = 15 mm

(a) Calculate the thermal resistance Rtot for the wall

(b) Calculate the heat transfer coefficient U value for the wall

(c) The heat flux through the wall is measured to be 5.2 𝑊. 𝑚!” . Calculate the temperature

difference between the inside and outside under these conditions.

A flywheel rotates with a kinetic energy of 100 kJ at 2500 rpm and has a mass of 25 kg.

(d) Calculate the moment of inertia of the flywheel

(e) Calculate the energy density of the flywheel in kJ/kg

PHYS2066

3 of 7

Question 3

Electricity generation methods fall into several categories. Provide 2 examples of each of these

types and explain why:

(a) Variable and intermittent power sources

(b) Variable and predictable power sources

The Sun outputs 3.85 × 10″# 𝑊 of power, and Mercury has a radius of 2.44 × 10# 𝑚 and is at a

distance of 5.79 × 10$% 𝑚 from the Sun.

(c) Calculate the solar insolation on Mercury.

(d) Assuming simple atmospheric model B, calculate the temperature of Mercury’s atmosphere

(e) Assuming simple atmospheric model B, calculate the temperature of Mercury’s surface

PHYS2066

4 of 7

Question 4

A solar updraft tower has a height of 1 km.

(a) If the ambient temperature is 15 °C and the temperature of the air at the input to the tower is

65°C, what is the maximum velocity of air in the tower?

(b) If the area of the turbine is 10 m2 and the density of air is 1.225 kg/m3, calculate the total

power available in MW.

(c) A flat plate collector has ε = 0.9, t = 0.85, h = 10 Wm-2K-1 and A = 4.6 m2. For a solar

insolation of S = 600 W/m2, if the air temperature is 20°C, the temperature in the collector is

40°, and the mass flow is 0.01 kg/s, calculate ΔT.

(d) Explain how a silicon solar cell works. Use diagrams and/or equations as appropriate.

(e) In the space provided discuss the relative advantages and disadvantages converting solar

radiation to electricity using photovoltaic solar cells and solar concentrators.

PHYS2066

5 of 7

Question 5

Consider the following nuclear reaction:

(a)

(b)

(c)

(d)

“‘(

&’𝑈

)

→ “‘)

&%𝑇ℎ + “𝐻𝑒

Describe this reaction in 1-2 sentences and state what type of reaction this is.

For each uranium atom a total of 4.25 MeV of energy is released. What is the mass

equivalent of this energy?

If the mass of )”𝐻𝑒 is 4.00260u and the mass of “‘)

&%𝑇ℎ is 234.05079u, what is the

mass of the Uranium in atomic mass units?

A solar concentration plant consists of 100 parabolic reflectors of diameter 5 m with a focal

length of 2 m. If the solar insolation is 360 W.m-2:

(d) Calculate the flux at the focus of each reflector (in MW.m-2)

(e) Calculate the total amount of power the solar plant collects (in kW)

PHYS2066

6 of 7

Spare page if extra room needed – clearly label the question(s)

Add additional pages if necessary

PHYS2066

7 of 7

RMIT – SCHOOL OF SCIENCE

PHYS2066/2129 – Energy and the Earth’s Environment

Practice Skills & Capability Assessment – SOLUTIONS

TIME ALLOWED:

1.5 hours for the test, plus 30 minutes to allow for downloading, scanning

and uploading to canvas.

Academic Honour Code declaration

The work I am submitting for this Assessment Task is entirely my own work. Nothing I submit is

from another source – either from another student or person, or from another resource, in whole or in

part. All written work is in my own words.

I have not communicated with any other student during the period of this Assessment Task. I

have not discussed the content of this Assessment with anyone else during the period of this

Assessment Task.

Name:

Student Number:

Signature:

INSTRUCTIONS – READ CAREFULLY

• Test can be printed, worked through and scanned, or you can do the test on blank paper and

scan, making sure to carefully label the questions. If the latter, you must write and sign the

Honour code on your question sheet.

• ANSWER ALL QUESTIONS – All questions are of equal value.

• Show ALL working. The use of UNITS in final answers is ESSENTIAL.

• In answering the questions please ensure there is adequate balance between any mathematics

you may use and a description of what you are saying in English.

• Calculators may be used.

Useful Physical Constants

s = 5.67×10-8 W/K4.m2

Wien’s constant = 2.90×10-3 m.K

k = 1.38×10-23 J/K

cp = 4179 Jkg-1K-1 (specific heat for water)

g = 9.8 m.s-2

S = 1360 W.m-2

PHYS2066/2129 – Test 2 – practice test A

e = -1.6×10-19 C

r = 1000 kg/m3 (fresh water)

r = 1030 kg/m3 (seawater)

r = 1.225 kg/m3 (air at sea level)

RE = 6.37×106 m

TK=TC+273

1 of 8

Question 1

A wind turbine with an 11 m radius rotor has an axial induction factor of 0.15. If the wind is

blowing at 25 km/h, and the turbine outputs 20 kW then:

(a) Calculate the power coefficient for the turbine

(b) Calculate the wind velocity in m/s.

(c) Calculate the mechanical efficiency of the turbine (as a %).

A wave power station produces 10 MW of electricity.

(d) if the overall efficiency is 45%, calculate the amount of power (in MW) in the waves.

(e) if the average wave speed and amplitude are 8 m.s-1 and 2.5 m respectively, calculate the

length of the power station.

Solution

(a) 𝐶! = 4𝑎(1 − 𝑎)” = 4 ∙ 0.15 ∙ (1 − 0.15)” = 0.434

“#∙%&&&

(b) 𝑣 = ‘(&& = 6.94 𝑚. 𝑠 )%

(c) The mechanical efficiency is given by:

1

𝑃 = 𝜌𝐴𝑈 ‘ 9𝜂*+,- ∙ 𝐶! ;

2

2𝑃

2 ∙ 20 × 10’

∴ 𝜂*+,- =

=

= 0.592 = 59.2%

𝐶! 𝜌𝐴𝑈 ‘ 0.43 ∙ 1.225 ∙ 𝜋 ∙ 11″ ∙ 6.9’

(d) If the station produces 10 MW of power at an efficiency of 45% then the power in the

waves must be:

𝑃./0!/0 10 × 10(

𝑃=

=

= 2.22 × 101 𝑊 = 22.2 𝑀𝑊

𝜂

0.45

(e) The total power is:

1

𝑃 = 𝑃2 𝐿 = 𝑈𝐴” 𝜌𝑔𝐿

2

2𝑃

2 ∙ 2.22 × 101

∴𝐿=

=

= 88.1 𝑚

𝑈𝐴” 𝜌𝑔 8 ∙ 2.5″ ∙ 1030 ∙ 9.8

PHYS2066

2 of 8

Question 2

A wall consists of 3 layers with the following properties:

Weatherboard: k = 0.15 W.m-1.K-1 L = 2 cm

Insulation:

k = 0.05 W.m-1.K-1 L = 40 mm

Plaster:

k = 0.5 W.m-1.K-1

L = 15 mm

(a) Calculate the thermal resistance Rtot for the wall

(b) Calculate the heat transfer coefficient U value for the wall

(c) The heat flux through the wall is measured to be 5.2 𝑊. 𝑚)” . Calculate the temperature

difference between the inside and outside under these conditions.

A flywheel rotates with a kinetic energy of 100 kJ at 2500 rpm and has a mass of 25 kg.

(d) Calculate the moment of inertia of the flywheel

(e) Calculate the energy density of the flywheel in kJ/kg

Solution

(a)

Remembering to convert the L values to m, this becomes:

𝐿3 𝐿4 𝐿5

𝑅0.0 = 𝑅3 + 𝑅4 + 𝑅5 =

+ +

= 0.963 𝑚” . 𝐾. 𝑊 )%

𝑘3 𝑘4 𝑘5

(b)

𝑈=6

(c)

%

!”!

= 1.04 𝑊. 𝑚)” . 𝐾 )%

Remembering that U has the same dimensions as k/L:

𝑇7 − 𝑇8

𝐽

5.2

𝐽=𝑘

= 𝑈 ∙ (𝑇7 − 𝑇8 ) ∴ Δ𝑇 = (𝑇7 − 𝑇8 ) = =

= 5.01 𝐾

𝐿

𝑈 1.04

%

“9

“9

“∙%×%&$

(d)

𝐾 = ” 𝐼𝜔” ∴ 𝐼 = :# = (”

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