 # PHY 2066 Kennedy King College Alternative Energies Questions

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## Full Question

Alternative Energies
Wind, Wave, Tidal, Bio, Hydro, Geothermal
Wind power

low emissions
small footprint
compatible with some crops
potentially cheap source of power

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:

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:

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

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

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

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

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

UniWave200 King Island Project – Wave Swell

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 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

– 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.

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

– in general the materials burn more cleanly than fossil fuels, as they
are purer.
– make portable fuels
– 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
– 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
– 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
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.
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

Nem Watch

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
• 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
1

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:
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
Peak Wavelengths
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
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
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
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
• 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
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.
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:
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
• 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
• 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
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
• 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
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
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:
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
Name:
Student Number:
Signature:
• 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.
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)
PHYS2066
7 of 7
RMIT – SCHOOL OF SCIENCE
PHYS2066/2129 – Energy and the Earth’s Environment
Practice Skills & Capability Assessment – SOLUTIONS
TIME ALLOWED:
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
Name:
Student Number:
Signature:
• 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.
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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