Electricity Generation

How to do itHow to Transmit itEnergy Sources: Renewable Energy Wind Turbines Wave Power Solar PowerEnergy Sources: Non-Renewable Energy Clean Coal Nuclear Energy
How do we generate electricity? Today there are a lot of different answers to that question: Coal, Natural Gas, Nuclear, Wind, Solar and Tidal power are all viable options for electrical generation. However did you know that all of these methods except solar come down to the same thing? They extract energy in order to drive something called a generator that takes any rotating force (or torque) and turns it into electricity.
Parts of a Generator (part 1)

There are two key elements in every generator:

These elements are always arranged so that the magnets can spin past the coils of wire, or vice versa.

So what is it about magnets and coils of wire that generates electricity? Let’s quickly review a few things about magnets.

Review of Magnets

The basic thing that magnets do is attract some types of metal and also attract or repel other magnets according to their polarity. We often use an idea called a magnetic field to describe the interactions of a magnet. The direction of the magnetic field lines shows the direction that the magnetic field would push another north magnetic pole, and the density of the magnetic field lines shows the strength of the magnetic field at that location (or how strong the push would be).

From looking at the diagram below we can see that the magnetic field is strongest at the poles (where the density of lines is highest) and that the north pole would push another north AWAY from itself. Also we can see that the strength of the magnetic field gets weaker when we are far away from the magnet. We already knew those things, but it is good to review.

Image sourced from Wikimedia1

Parts of a Generator (part 2)
So we know that generating electricity involves magnets and wires. To generate electricity can we just put a loop of wire next to a magnet?
Sadly, no. it turns out that we need to CHANGE something. If we hold a magnet still near a coil, we don’t generate any electricity. If we hold a magnet still FAR from a coil, we still don’t. It’s only when we move the magnet either towards or away from the coil that we generate electricity.To figure out what’s going on here, we need to look more closely at what the magnet does to a moving wire.You can also see a simulation of this here.

Magnetic Fields and Wires

So I’ll just spit it out: when you move a wire through a magnetic field, the magnetic field pushes all the electrons to one side of the wire! Well, I should clarify: you have to orient and move the wire in a very special way.

Firstly, the wire needs to be moved perpendicular to the magnetic field. If you move it in the same direction as the magnetic field, nothing happens.

Secondly, in order to make useful electricity, we want to push the electrons down the LENGTH of the wire. In order to do this, you need to move the wire sideways. If you move the wire in the same direction as the length of the wire, the magnetic field pushes the electrons to the sides of the wire (which is not very useful).

In wire A, charges are pushed to either end of the wire, resulting in a useful voltage
In wire B, charges are pushed to the top and bottom of the wire, but this does not result in a useful voltage.
In wire C, no charges are pushed anywhere because the motion of the wire is parallel to the magnetic field.

The question of why this happens has many answers, each more mysterious than the last, but for now let’s leave it that this is a fact of nature and it Just Happens. (Teacher note: this might be a good time for a demo)

In summary, if your velocity is perpendicular to the magnetic field, there will be a force that pushes charges in a direction perpendicular to both the velocity AND the magnetic field. If the magnetic field is in the other direction, the push will be in the other direction. This is summarized by something called the “Right Hand Rule”, which is explained here.

So how do we make electricity with this effect?

That seems easy, right? We just hook up some wires to either end of this wire and run it through a magnetic field. Let’s also hook it up to a lightbulb, which represents something that USES the electricity.

Unfortunately this doesn’t work! We can see why by thinking about each of the wires individually: the two on the top and bottom have the charges pushed to the side of the wire, which doesn’t really create useful electricity so they don’t help us. The two on the side both have the positive charges pushed to the top, but that’s the problem: in order to have useful electricity we need a net flow of current AROUND the loop. Having a push upward on the lefthand wire AND a push upward on the righthand wire means these pushes are canceled out and we get no flow of current.

Each of the vertical wires has its positive charges pushed to the top of the wire. These pushes cancel out, so there is no net flow of charge and that means no electricity!

So we can see that moving a loop through a constant field won’t work. What can we try that WILL work?

Question: Which of the following will NOT light up the bulb?

Answer: C and D

Question: Which of these will light the bulb most strongly?

Answer: B

We get the best results when one side of the loop is in a field Upwards, and the other side is in a field Downwards. This way the pushes on the charges push in the same direction AROUND the loop.

Application in the Real World

Situation B above gives us a very good picture of how modern generators work! They use magnets of alternating polarity to set up alternating fields and then move coils of wire past them.

This diagram shows a generator that varies the B through each coil quite rapidly. The two parts are shown separated, but when assembled the ring of coils will be sitting on the axle of the ring of magnets below. Each magnet has the opposite polarity of its neighbors, so as an individual coil rotates around it will see a North pole followed by a South pole followed by a North etc.

Image sourced from The Norweigian University of Science and Technology 2

You can also achieve the same effect by moving the magnets instead of the coils: you still get a relative velocity between the magnets and coils, and on each transition between adjacent magnets you get a push around the coil to create electricity.

This is another type of generator. The coil is rotated in such a way that each side of the coil is moving in opposite directions in a constant magnetic field. This means that both pushes will always be in the same direction around the coil and you will generate electricity.

Image sourced from The University of New South Wales 3

This setup is very easy to build, but it has one interesting feature: the voltage generated by a generator like this is not always in the same direction. The generated voltage changes direction twice each cycle. This type of electricity is called “Alternating Current”, and it has some very useful properties that we will investigate in future lectures. It is also the kind of electricity that you find in your walls!

For some extra reading, a good animation of Electric Motors can be found here.


Electricity in our houses runs in the walls, but you may have noticed that in the countryside you occasionally see huge towers carrying electrical lines. If you are brave enough to get close to one of them you’ll see it’s marked “DANGER: HIGH VOLTAGE”. If high voltage is so dangerous, why do we need it in the first place?

Electrical Transmission Model

To answer this question, let’s think about a simple model of electricity generation and consumption.

Here’s a basic picture of a power plant connected by transmission lines to your town. In reality, the power plant would be connected to many more houses than shown in this diagram, but let’s just keep our diagram simple. (We will be focusing on those wires anyways, so the details of the houses aren’t important)Here’s a circuit diagram that shows the basic elements of this generator – wire – house system:

The power plant acts as a voltage source that sends current through the wires to your town. The town is a very complicated network of various devices that convert the electricity into any number of useful things, but for the purposes of this analysis we can treat them as just one big resistor. We don’t care about the exact details of what is happening in the town; we only care only that it needs to get power from the power plant through those wires.The key thing that we want to accomplish in electrical transmission is to get as much of the energy to your town as possible. We can’t actually get 100% of the energy to the you, because the transmission wires have some resistance so there is always some resistive heating of those wires (which costs us energy). But we want to try to keep that to a minimum.

Power Consumption in Transmission Lines

In order to show why we need high voltage for transmission, let’s figure out what would happen if we just used household voltage of 120 V. We’ll figure out the resistance of a typical transmission wire, and then see how far we can transmit with that wire without losing too much energy.The power loss in the wire is given by:


We don’t want to calculate the ∆V of the line, but we can substitute using Ohm’s Law:


This gives us

${P}_{\textnormal{wire}}={I}_{\textnormal{wire}}^2\times{R}_{\textnormal{wire}} ( *\*)$

Now, the current will be the same throughout the circuit. This means we can calculate it from the total power output of the plant.


Now we substitute back into the (*) equation to get the following result for the power loss in the wire.

${P}_{\textnormal{wire}}={I}_{\textnormal{wire}}^2\times{R}_{\textnormal{wire}}$${P}_{\textnormal{wire}}=(\dfrac{{P}_{\textnormal{plant}}}{{V}_{\textnormal{plant}}})^2\times{R}_{\textnormal{wire}} (*\**\*)$

To make use of this equation we need to choose a reasonable value for Pplant and figure out the typical resistance of the transmission wires. This will let us see how much power is wasted for a given voltage and distance.

Resistance of Transmission Lines
Transmission wires are made of layers of braided wire. Aluminum is usually used as a conductor because it has very low resistance and is relatively cheap. There are also a few strands of steel braided in to make the wire stronger.While copper is a better conductor, it’s much more expensive and much heavier. This would mean we would need to put more steel in a copper cable to hold it up, making it even more expensive.A typical wire used for long-distance transmission is called 4/0 AWG, and it has an effective diameter of 11.7 mm. We can look up the resistivity of Aluminum to figure out the resistance of a particular wire.$\textnormal{Resistivity of Aluminum}{ (\rho)}=28.2\times{10}^{-9}\Omega\textnormal{m}$Resistance is given by:$\textnormal{R}={\rho}\times\dfrac{\textnormal{L}}{\textnormal{A}}$Now we want to figure out the resistance of a transmission line per kilometer. This will let us use equation ** to figure out how far we can transmit power.The resistance for one kilometer of wire will be given by:$\textnormal{R}=\dfrac{28.2\times{10}^{-9}\Omega\textnormal{m}\times1\textnormal{km}}{\pi\times(\dfrac{11.7\textnormal{mm}}{2})^2}$$\textnormal{R}=0.263 \Omega$
Transmission Distance
Now we can figure out how far we can transmit using the standard voltage of 120V. Let’s choose that we want less than 10% of our power to be used up in the transmission line, so we choose${P}_{\textnormal{wire}}=10\%{P}_{\textnormal{plant}}$For the plant power we can use 25 MW, which is the power of a hydroelectric dam recently constructed near Squamish, BC $^{1}$.And using:${R}_{\textnormal{wire}}= 0.263 \Omega\textnormal{/km}$our expression ** becomes${P}_{\textnormal{wire}}=(\dfrac{{P}_{\textnormal{plant}}}{{V}_{\textnormal{plant}}})^2\times{R}_{\textnormal{wire}}$$25\textnormal{MW}=(\dfrac{2.5\textnormal{MW}}{120\textnormal{V}})^2\times0.263\Omega\textnormal{/km}\times\textnormal{distance}( *\* *\*)$Solving for distance we get 0.0002 km, or only 20 cm!! This would mean that in order to get 90% of the power to your house, using the biggest wire commercially available, the power plant would have to be only 20 cm away!This is an extreme example because we are using modern power plant capacities with very inappropriate voltages, but it shows a real problem: if you transmit at lower voltage, you need to have a power plant very nearby.The solution is to use a higher transmission voltage. You can see that in equation ** the power dissipated in the wire decreases as the voltage squared, so when you multiply the voltage by 10 you decrease the power (or increase the distance) by 100. This relationship is shown below.Modern transmission systems have a line loss of around 6.5% ]$^{2}$, so we can see that by going up to 300 kV we can transmit around 1000 km. This is a much more sensible transmission distance for a province as big as BC.(Note: you can do your own calculations for any % loss, wire resistance, or plant power in the following excel document: TRANSMISSION_LOSSES_CALCULATOR.xls” )
Transforming Voltage
In order to transmit electrical energy at high voltages, we need to be able to convert from one voltage to another. This is essential, as we wouldn’t want to have 300 kV power in our homes! A simple reliable devise allowing to change alternating current (AC) voltage is called a transformer.A transformer consists of two coils of wire wound around an ferromagnetic (most often iron) core. By winding a different number of turns of wire on each side, the transformer can increase the voltage at the cost of decreased current or vice versa. We discuss how a transformer works in this lecture.
This technology is easy to build, but it ONLY works for AC power. This is one of the main reasons that we use AC power today for transmission. When electricity was first being introduced to North America there were two different systems being proposed: AC power with transformers to enable long distance transmission, and DC power with many regional power plants.$^3$The competition between the two sides was fierce and acrimonious, but in the end the ability to transform AC power to higher voltages played a key part in its widespread adoption.In modern times, there are some technologies that allow transformation of DC voltages, and because they avoid some problems associated with AC transmission High-Voltage DC (HVDC) transmission is used in some parts of the world.$^4$ Because of problem associated with transmitting AC through saltwater (which is a conductive fluid) HVDC is especially suited to undersea lines that connect major power grids together.
How is wind energy created?

Wind energy has been proposed as an alternative energy source, although it is currently in an early stage of large scale development$^5$. Windmills were seen in Persia as early as 500 AD and were used to grind grain and pump water rel=”noopener”> http://www.telosnet.com/wind/early.html [10 June 2009]. [/note]. It is only in modern times that humanity has attempted to extract raw power from wind in the form of electricity. The question to ask in this early developmental stage is whether or not it is possible to extract a useful amount of raw energy from the wind. To do so, we will consider the broad energetics of wind turbines to determine if harnessing wind energy could be a viable option given constraints of time, location and machinery. The first thing to consider is whether or not there is enough energy in wind to make it possible to extract a useful amount of power. It is important to note that although wind may possess a lot of kinetic energy, that is the energy resulting from the movement of masses, the rate at which this energy can be extracted limits the amount of useful power available, as power in its most rudimentary form is defined as the rate of doing useful work.

To discover whether or not a useful amount of power is available, we must first discuss where the wind comes from and how much power is available in the wind. Wind energy ultimately comes from a series of energy transformations from solar energy (radiation) to wind energy (kinetic), where about 2% of the solar energy absorbed by the Earth goes into wind[1]Aubrecht GJ. Solar Energy: Wind, Photovoltaics, and Large-Scale Installatons. In: Energy – Physical, Environmental, and Social Impact (3), edited by Erik Fahlgren. Upper Saddle River, NJ: Pearson Education Inc., 2006, chapt. 21, 461-465. . Solar radiation is absorbed by the surface of the Earth and heats it unevenly[2] Kump, L.R., Kasting, J.F., and Crane, R.G. The Atmospheric Circulation System. In: The Earth System (2), edited by Patrick Lynch. Upper Saddle River, New Jersey, USA: 2004, chapt. 4, pp. 55-82.. Different areas of the globe receive varying amounts of the incident solar intensity (W/m2) due to the angle of the Sun; the equator receives a greater percentage of solar intensity (and hence becomes hotter) than the poles[3] Kump, L.R., Kasting, J.F., and Crane, R.G. The Atmospheric Circulation System. In: The Earth System (2), edited by Patrick Lynch. Upper Saddle River, New Jersey, USA: 2004, chapt. 4, pp. 55-82.. Also, during the day the land heats up faster than the sea does, while at night the water retains heat longer than the land does[4] Kump, L.R., Kasting, J.F., and Crane, R.G. The Atmospheric Circulation System. In: The Earth System (2), edited by Patrick Lynch. Upper Saddle River, New Jersey, USA: 2004, chapt. 4, pp. 55-82.. Wind is a direct result of solar heating and the earth’s rotation as they generate changes in temperature. As the air gets warmer, it rises and cooler air must rush in to take its place, producing wind! In all, location effects how much wind energy is available (Fig. 1).

Figure 1. A wind energy map of Canada showing the average power (in W/m2 of turbine cross-section area) that can theoretically be extracted from the wind[5]Environment Canada. Canadian Atlas Level 0 (online). https://collaboration.cmc.ec.gc.ca/science/rpn/modcom/eole/CanadianAtlas0.html [20 May 2009]..

How much energy can be harnessed by wind?

The mean intercepted solar intensity at the top of the Earth’s atmosphere is 350 W/m2. Given that 2% is converted to wind, this results in 7 W/m2 going into wind energy[6]Gustavson MR. Limits to Wind Power Utilization. Science 204: 13 – 17, 1979.. This wind energy is spread out over the Earth’s atmosphere, with 35% of the energy (2.45 W/m2 of land area) dissipating in the first kilometre above Earth’s surface[7]Gustavson MR. Limits to Wind Power Utilization. Science 204: 13 – 17, 1979.. Over a period of one year, the wind energy is approximately

\textnormal{wind energy} &=& \textnormal{(intensity)(Earth’s surface area)(seconds per year)}\nonumber \\
&=& (2.45 \textnormal{ W/m}^2)(5.1 \times 10^{14} \textnormal{ m}^2)(3.2 \times 10^7 \textnormal{ s})\nonumber \\
&=& 4.0 \times 10^{22} \textnormal{ J}\nonumber

which is 200 times larger than our energy consumption on Earth, estimated to be 2 x 1020 J[8]Gustavson MR. Limits to Wind Power Utilization. Science 204: 13 – 17, 1979. Now we can calculate the energy and power harnessed from the wind. To use the basic equation for kinetic energy, $KE = \frac12mv^2$, we will need the rate at which air passes through the rotor of the wind turbine. The rotor is made up of the blades that spin on a wind turbine, and the total area of the rotor can be approximated by a circle as this is the area swept by the blades. To do this, we can imagine we are holding a hoop up in the air, and measure the mass of air traveling through the hoop in time Δt (Fig. 2).

Figure 2. At time t = 0, the mass of air is just about to pass through the hoop, but Δt later, the mass of air has passed through the hoop. The mass of this piece of air is the product of its density ρ, area A, and length vΔt.

From this, we can see that the mass is

\textnormal{mass} &=& \textnormal{density} \times \textnormal{volume} \nonumber \\
&=& \rho A v \Delta t \nonumber

where $\rho$ is the density of the air (1.2 kg/m3 for standard atmospheric pressure (1 atm) and temperature (0°C) at sea level), $v$ is the velocity of the air and $\Delta t$ is the length of time for a unit of air to pass through the loop[9]MacKay DJC. Sustainable Energy – Without the Hot Air (Online). UIT Cambridge. http://www.inference.phy.cam.ac.uk/sustainable/book/tex/ps/253.326.pdf [4 May 2009]. . The area $A$ is the area swept by the blades, not the blade area. This is because the blade moves much faster than the air and so each particle of air is affected by the blade. Therefore, the kinetic energy is found to be

K &=& \dfrac{1}{2} m v^2 \nonumber \\
&=& \dfrac{1}{2} \rho A t v^3 \nonumber

while the power of the wind passing through our hoop is

P &=& \dfrac {\dfrac{1}{2} \rho A t v^3}{t} \nonumber \\
&=& \dfrac{1}{2} \rho A v^3 \nonumber

But this is not the actual power produced by the turbine as turbines can’t extract all of the kinetic energy of the wind. Why not? If this was the case the air would stop as soon as it passed through the blades and no other wind would be able to pass through. An analysis by Betz (1919) shows that you cannot capture any more than c.60% of the wind’s energy[10]Betz’ Law http://en.wikipedia.org/wiki/Betz’_law [2012.09.27]. . In addition, there are also small losses due to friction and turbulence. So ideally you want the turbine to slow the wind down by 2/3 of its original speed (as the maximum of $P/P_0 = 0.593$ is found at $v_2 / v_1 \approx \frac13$ v2/v1 ≈1/3). For more information, click here. The power produced by one turbine is found by

\textnormal{Power} &=& \textnormal{(efficiency)(power)} \nonumber \\
P &=& \dfrac{1}{2} \eta \rho A v^3 \nonumber \\
&=& \dfrac{1}{2} \eta \rho v^3 \pi r^2 \nonumber

where $d$ is the diameter of the circle covered by the rotor. The $v^3$ term found here emphasizes the need to have a high wind speed in order to capture an useful amount of power. What we have just derived is based on a single wind turbine in constant wind conditions. In real life, however, wind conditions change. So what local conditions must be satisfied in order to make the use of wind turbines feasible? The location of the wind turbine must be carefully selected. Wind turbines only work efficiently when wind moves uniformly in the same direction. Turbulence, the unsteady flow caused by buildings, trees, and land formations, causes an inconsistent air flow which makes harnessing power inefficient and places increased stress on the rotor. The edge of a continental shelf, high ground and tundra are the best locations to build a turbine[11]Learning (online). Solacity Inc. https://www.solacity.com/SiteSelection.htm [20 May 2009]. . This is because their geography lacks any large obstructions that may create turbulence. Local wind is also an important factor, and should be, on average, at least 7 m/s at 25 m above the Earth’s surface in order to make harnessing wind from it worthwhile[12]Aubrecht GJ. Solar Energy: Wind, Photovoltaics, and Large-Scale Installatons. In: Energy – Physical, Environmental, and Social Impact (3), edited by Erik Fahlgren. Upper Saddle River, NJ: Pearson Education Inc., 2006, chapt. 21, 461-465. . One must also keep demand and dependability in mind when considering the extraction of energy from the wind. First of all, since wind is not locally predictable in the short term, the use of wind energy should be limited to only fulfill 5 – 15% of the total energy demand of the area[13]Aubrecht GJ. Solar Energy: Wind, Photovoltaics, and Large-Scale Installatons. In: Energy – Physical, Environmental, and Social Impact (3), edited by Erik Fahlgren. Upper Saddle River, NJ: Pearson Education Inc., 2006, chapt. 21, 461-465. . To overcome this problem and make wind energy more reliable, turbines need to be set up in many different locations so that the power available averages out[note value=3][/note]. In other words, on one day one of three of the locations may not have enough wind to operate, but the next day, that location may be in operation while the other two locations are not. Despite the local wind patterns, however, globally there is always a relatively constant amount of wind energy being harnessed at any one moment.

How do wind turbines work?

The machinery of a wind turbine also has its limitation on how much power can be extracted from wind. To begin, let us cement some of the terminology used in describing the structure of wind turbines, before looking at the actual mechanics of converting wind energy into electricity.

Figure 4. A turbine is composed of a foundation, a tower, a nacelle and a rotor consisting of 3 blades.

Principle components of a wind turbine unit are a foundation, a transformer, a tower, a rotor and a nacelle (Fig. 4). The wind turns the rotor, which turns the generator to produce electricity. The electricity is then transmitted to a transformer at the base of the tower before going to a substation. To maximize the power extracted, the nacelle, which connects the rotor to the tower and houses the generator, can be rotated into the direction of the wind. Rotors’ diameters range from 27 m for a 225 kW generator to 80 m for a 2500 kW generator, and depend on the desired power output, location limitations etc (Fig. 5)[note value=8][/note]. A 1 MW turbine has a rotor diameter of 54 m, a tower standing 80m tall, and works in wind speeds ranging from 3 – 25 m/s (10 – 90 km/h)[note value=7][/note]

Figure 5. The dimensions and characteristics of a typical smaller sized turbine.

The power produced by a wind turbine depends on rotor area, air density, wind speed, and wind shear. Air density increases with colder temperatures, decreased altitude, and decreased humidity. The molar mass of air (29 g/mol) is greater than the molar mass of water (18 g/mol) and so the less moisture in the air, the denser the air is. Wind shear is a difference in wind speed and direction over a short distance and is caused by mountains, coastlines and weather patterns[note value=8][/note]. Wind speed increases the farther you get away from the ground (Fig. 6)[note value=7][/note]. To maximize the power output of wind turbines, rotors are tilted slightly upwards. Why do you think this is?

Figure 6. As you get higher off the ground, the air speed increases, corresponding to a longer arrow. The rotors are tilted slightly upwards so that each part of the rotor is exposed to the same speed.

While it may be possible to use a single wind turbine for personal energy demands, entire cities and countries need huge wind farms to satisfy their energy needs. To optimize energy production in a wind farm, turbines are spread 5 – 9 rotor diameters apart in the prevailing wind direction and 3 – 5 rotor diameters apart in the perpendicular direction (Fig. 7)[note value=7][/note]

Figure 7. On a wind farm, turbines must be spaced out enough so that they do not interfere with each other. As the wind passes through the turbine it slows down, and so there is no point in putting a turbine in the region where the air is guaranteed to be slow. One common way of spacing them out is ensuring there is at least 5 rotor diameters between each turbine.

When the turbines are placed on a square grid, the power per unit land area is

\dfrac{\textnormal{power}}{\textnormal{land area}} &=& \dfrac{ \dfrac{1}{8} \eta \rho v^3 \pi d^2} {(nd)^2} \nonumber \\
&=& \dfrac{ \dfrac{1}{8} \eta \rho v^3 \pi}{n^2} \nonumber

where $n$ is the number of turbine diameters between turbines. The average power of a wind turbine farm is the product of the capacity of the farm and the fraction of the time when the wind conditions are near optimal. The capacity factor is usually around 15 – 30%[note value=7][/note].

How does wind energy compare to other energy production alternatives?

Now that it is established that wind is a possible source of power, the benefits and drawbacks need to be considered. Why use wind power in lieu of other energy sources? The harnessing of wind power does not produce hazardous wastes, use non-renewable resources or cause significant amounts of damage to the environment[note value=1][/note]. Some CO2 is produced in the manufacturing of the turbines, but as demonstrated in the problem set, it is much less than the emissions from burning an energy-equivalent amount of coal or natural gas. In addition, the use of wind power can reduce hidden costs such as those related to pollution and the consequential healthcare impacts, and in the longer term, climate change[note value=1][/note]. Also, wind turbines use less space than traditional power stations, because you can farm around them, so they can be built without extensively reducing agricultural land and locally-owned wind farms create income for communities[note value=1][/note].

So why, in light of these positive elements, is there so much resistance against wind turbines? Arguments against include the following fears of damages from collapsing turbines, noise, a less attractive skyline, an unreliable power source, unnecessarily high bird fatality, and significantly modifying the Earth’s wind patterns. First of all, the noise of a typical turbine is 45 dB at 250 m away[14]Clarke S. Electricity Generation Using Small Wind Turbines At Your Home Or Farm (Online). Ontario Ministry of Agriculture, Foods and Rural Affairs. https://www.omafra.gov.on.ca/english/engineer/facts/03-047.htm#noise [25 May 2009].. This level is lower than the background noise at an office or a home[note value=11][/note]. Secondly, the reliability of wind energy increases depending on location and how many farms are operating in a variety of sites within the area. In regards to bird deaths, in the US less than 40,000 are said to die from turbine blades while hundreds of millions are said to die from domestic cats[15]Marris E and Fairless D. Wind Farms’ Deadly Reputation Hard to Shift. Nature 447: 126, 2007.!  Finally, in regards to modifying the Earth’s climate, it is plausible that one would see local climate change surrounding areas with a high concentration of wind farms, but the large-scale climatic effects will likely be negligible[16]Keith D. Wind Power and Climate Change (online). University of Calgary. https://www.ucalgary.ca/~keith/WindAndClimateNote.html [20 May 2009].. In addition, since wind turbines will be replacing coal-fired power plants, if anything, we anticipate a considerable reduction in CO2 emissions.

While currently wind turbines are the accepted machinery used to extract wind energy, technology can change quite easily as people discover new and more efficient ways to harness power. Currently, wind cells are being investigated by Accio Energy, which takes a completely different approach to the method of extracting wind energy than the tradition wind turbine[17]Accio Energy. About Accio Energy (online). https://accioenergy.com/about.html [12 June 2009]..

Above: Making full use of the wind: Wolfe Island Wind Farm, Lake Ontario

Energy required to disturb the surface of water

Before we ask how much power we can extract from waves, we have to estimate how much energy there is in a wave. Consider a moving disturbance in the surface of a body of water that looks something like this:

The energy required to create this disturbance is just the work done in moving the water that was in the trough of the wave up to the crest of the wave. Thus a body of water height h, width w and length L has been moved vertically up a distance h (centre-of-mass to centre-of-mass).


If the density of water is ρ then the mass m involved in the move is ρwhL. As the work done W to raise this mass a height h is mgh,

$W = \rho g w h^2 L$

Power in a wave-train

In the open ocean such disturbances usually occur repetitively, with a frequency f and a spacing λ, i.e. the speed of the waves, v = f λ. If the disturbances are packed together, λ = 2w.

We won’t worry about the unnatural shape of the waves for now. Rectangles are easier to deal with than real wave shapes.

If we have some kind of energy absorber at the end of this wave train capable of using waver power to generate, say, electricity, the rate at which waver energy crosses this absorber is

$P = \rho g w h^2 L f = \rho g (\dfrac{1}{2} \lambda ) h^2 L f = \dfrac{1}{2} \rho g h^2 L v$

Consider a typical ocean wave train: h = 1 m, λ = 10 m, f = 0.1 Hz (period = 10 s), i.e. v = 1 m/s.

Assume the energy absorber is L = 10 m long, typical for such an installation:

$P = \dfrac{1}{2} (1000 \textnormal{ kg/m}^3)(10 \textnormal{m/s}^2)(1 \textnormal{ m})^2 (10 \textnormal{ m})(1 \textnormal{ m/s}) = 50,000 \textnormal{ W , i.e. 50 kW}$

Although a crude approximation, this result is very close to that obtained with a much more sophisticated water-wave model. It shows that a significant amounts of power are potentially available in water waves. However, one must remember that 50 kW from this 10-metre size device would only service the energy needs of a few single-family dwelling. And we haven’t said anything yet about how this power can be extracted[fn]Wikipedia. Wave Power (online). http://en.wikipedia.org/wiki/Wave_power [27 May 2010].[/fn].

For real current wave data, suitable for use in student projects, see the U.S. NOAA National Data Buoy Center. Zoom in on a buoy and click to see current conditions[fn] National Oceanic and Atmospheric Administration. National Data Buoy Centre (online). http://www.ndbc.noaa.gov/ [27 May 2010]. [/fn].

We all have seen solar panels. They are mounted on traffic signs, light houses (see title photograph), and in our calculators. They are great green sources of energy. So why isn’t every roof in [Vancouver] covered with them? This would reduce our reliance on fossil fuels.

Let’s learn something about solar panels and do some simple calculations.

To create electric potential energy (or colloquially speaking electricity) we have to separate positive and negative charges. In batteries, the work required to separate positive and negative charges is done by an electrochemical reaction. In silicon-based solar cells, the work is done by the incoming solar radiation in a quantum process: a visible photon from the sun has enough energy to separate an electron from a silicon atom leaving behind a site that is positively charged, called  a “hole” (Remember the duality of light, we can treat light as electromagnetic wave or as a stream of particles called photons). Electrons are attracted to the positive site and holes to the negative site, so once we connect a solar panel to some load (for example a light bulb, a motor or a heater) we have a current that can do work on such external load.

Let’s calculate how much power we can obtain from a solar panel in Vancouver. The solar constant (the amount of solar radiation per m2 at the top of the atmosphere) is about 1400 W/ m2. We have on average 12 hours of sunlight a day, which should give us 1400 W/ m2 times 12 hours = 16.8 kWh/ m2 per day. But according to Natural Resources Canada[fn]http://pv.nrcan.gc.ca/index.php?m=r[/fn] we can only expect on average 5.2 kWh/ m2 on a surface perpendicular to the direction from the Sun. This is due to atmospheric absorption and cloud cover.

There is another problem. As the Sun’s position changes during the day (differently in each day of the year) it is expensive to keep the panel perpendicular to the direction from the Sun. To do this requires directional alignment in two axes so the panels cannot be fixed to the roof.

As illustrated Figure 1  the power of radiation delivered to a given surface depends on the incident angle. Notice that the power contained in a 1 m2 column of radiation impacts 1 m2 of surface if it falls vertically but 2 m2 of surface if it is at an angle of 30 degrees. So at 30 degrees we are getting only ½ of power per m2! The angular dependence of the power delivered is:

$P_{\theta} = P_{90} \sin(\theta)$

where $ P_{90}$ is the power per unit area of radiation impacting the surface from the direction perpendicular to the surface, $\theta$ is the angle of the surface from the horizontal and $P_{\theta}$ is the power per unit area of the same radiation impacting the surface from the direction at the angle $\theta$ from the horizontal.

Figure 1. The radiation of the same intensity illuminating a surface at different angles

Notice that the angle is measured from the direction perpendicular to the direction of incoming radiation as described in[fn value=1][/fn]. So if we mount our solar panel at the angle equal to our latitude on a south facing roof we will only get on average 3.7 kWh/ m2 of energy per day impacting our panel.

Why should we tilt the panel at the angle equal to our latitude? Lets look at Figure 2. Twice a year at solstice the axis of rotation of the Earth is perpendicular to the direction of the incoming solar radiation (as shown). The panel tilted from the horizontal plane at the angle equal to the latitude of it’s position is at noon perpendicular to the direction of the incoming solar radiation.  Over the year the angle between the axis of the Earth and the direction of the incoming solar radiation changes between -23.5° and +23.5°. Also the angle between the solar panel and the direction of the incoming solar radiation will change during the day. But such a position of the panel gives us a best yearly average of the power of the the solar radiation impacting the panel. How much do we loose if the tilt angle of the panel is different.  Between the tilt angle equal to latitude and tilt angle equal to latitude -15° almost nothing.  At  tilt angle equal to latitude +15° about 10%.

Figure 2. An illustration explaining why we set up solar panels at an angle from horizontal roughly equal to latitude.

A two person family living in a small detached house uses about 40kWh per day for cooking, hot water, heating, lights, TV, laundry, computers, microwaves and so on at a cost of about \$850 a year. It would seem that a 10m2 solar panel should cover their needs. Unfortunately not! Only about 10% of energy hitting the panel is converted to AC electrical energy. This is because of the solar panel efficiency (15-25%), DC/AC conversion efficiency and resistive loses. So from a nominally 1kW panel (the panel which gives us 1 kW electrical power at noon in full sunshine, when oriented perpendicular to incoming solar radiation) we can expect on average only about 3kWh per day (1 gives us so called Photovoltaic potential in kWh/kW, which is how many kWh per month or year can we expect at a particular location from a 1kW panel). And such a panel has an area of 7-10 m2 depending on the technology! At the moment the complete 1kW solar system costs about \$7500. On a typical house and garage there is usually about 50 m2 of a south facing roof space to be covered by solar panels. So we could install 5-7 kW panels and get 15 – 20 kWh per day, almost half of the expected consumption. This installation would cost \$40-50K and would recoup the initial cost in 50-60 years; this is unfortunately not a good investment.

And there is an other problem: energy storage. Solar panels give us a peak production on a midsummer day. But peak consumption occurs in midwinter evenings. If the house is connected to the grid we can sell energy at peak production and buy it at peak consumption and all our previous analysis is valid. But if our house is located on a small island we have to think about energy storage. We would have to store about ¼ year’s energy = about 4 000 kWh! A big car battery (mass 55 kg) stores about 200Ah at 12 V = 2.4 kWh. We would need about 1700 such batteries!

On the other hand a solar panel and a modest battery bank would be a perfect solution for a summer cottage used only on weekends. It would be a good exercise to try to calculate the expected energy consumption, the necessary size of solar panels, and the capacity of the batteries.

It sounds pessimistic at the moment but the technology is improving. Printed “roll on the roof” solar cells are coming [fn]http://www.nanosolar.com/[/fn]. In a few years we might be able to save some money and more importantly the enviroment by covering our roofs with solar cells.

Reasonable quality coal of the type used in power stations has an energy content (higher heating value, or HHV) of around 30 MJ/kg. This number varies considerably depending on the source of coal. The chemical formula for coal is roughly (CH)n. Let’s calculate how much coal we have to burn in a typical large power station to produce 1 GWe for a year. For this we have to assume a conversion efficiency η for thermal to electrical energy. Really good modern steam generators have η ≈ 0.4, so to generate 1 GWe we’ll need a a thermal power of 1/0.4 = 2.5 GWth.

The total amount of thermal energy required to generate 2.5 GWth for one year is

(2.5)(3600 s/h)(24 h/d)(365 d/y)(109 W) = 7.9 x 1016 J.

The mass of coal that needs to be burnt to produce this energy is

(7.9 x 1016 J)/(30 x 106 J/kg) = 2.6 x 109 kg = 2.6 Mtonnes.

To convert this mass of coal to the mass of CO2 produced on burning, consider the chemical reaction:

4CH + 5O2 → 4CO2 + 2H2O

As, always when burning fossil fuels, every carbon atom in the fuel ends up in a CO2 molecule. The molecular mass of CH is 13; that of CO2 is 44. Thus 13 tonnes of coal produce 44 tonnes of CO2.

In other words, the 2.6 Mtonnes of coal burnt each year in a 1 GWe power station produces (2.6)(44/14) = 8.9 Mtonnes of CO2.

As we are using some rough numbers here and also ignoring CO2 emissions caused by mining and transportation of coal, let’s call our result 10 Mtonnes of CO2 per GWe per year. Its hard to know what to do with 10 million tonnes of anything, let alone a gas (which at STP[fn]Hyper Physics.  Ideal Gas Law (online).  http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/idegas.html  [9 June 2010].[/fn] would fill 5 cubic km). However, this is the CO2 production from only one large coal-fired plant in one year. Two such plants are being opened in China every week[fn]BBC News.  China Building More Power Plants (online).  http://news.bbc.co.uk/2/hi/asia-pacific/6769743.stm  [9 June 2010].[/fn]. Although ideas and plans abound[fn]Wikipedia.  Carbon Capture and Storage (online).  http://en.wikipedia.org/wiki/Carbon_capture_and_storage  [9 June 2010].[/fn], no plant yet disposes of its CO2 anywhere other than in the atmosphere.

A starting point for comparison: Chemical Energy

All the energy transfer in chemical reactions comes from or goes into the rearrangement of electrons in the atom. The amount of energy involved is therefore controlled by Coulomb’s Law relating the magnitude of the repulsive force F between two charges to the amounts of charge $q_1$ and $q_2$, and the distance between them $r$.

$F = \dfrac{k q_1 q_2}{r^2}$                      1)

Here $k$ is the universal constant, 9.0 × 109 Nm 2/C2 in SI units. If $q_1$ and $q_2$ are of opposite sign, the force is attractive. From this expression we can find the potential energy $U$ of two charges a distance $r$ apart, with respect to the state where they are infinitely far apart.

$U = \dfrac{k q_1 q_2}{r}$                           2)

If $q_1$ and $q_2$ are of opposite sign, this potential energy is negative; it takes positive work to pull to the two charges apart. This much is standard textbook work.

Consider a hydrogen atom, consisting of a proton and an electron in its lowest orbital state. Both particles have a charge of magnitude 1.6 × 10-19 C, and the mean distance between them is the so-called Bohr radius 5.3 × 10-11 m. The potential energy of this state can be found from the above formula, and is -4.35 × 10-18 J. Because the electron also has kinetic energy, it only takes half of 4.35 × 10-18 J to rip it out of the hydrogen atom. Most chemical reactions involve electronic changes far less dramatic than ionizing a hydrogen atom, and a typical energy is of order 10-19 J.

For example burning carbon in oxygen produces 393.5 kJ/mol or 6.54 × 10-19 J per atom[fn]National Institute of Standards and Technology.  Carbon Dioxide (online).  http://webbook.nist.gov/cgi/cbook.cgi?ID=C124389&Units=SI&Mask=1 [4 June 2010].[/fn].

Energy in the Atomic Nucleus

Now consider what happens when a uranium nucleus splits in half, or, as we say, undergoes fission.

The first thing we have to appreciate is that atomic nuclei are made up of a collection of protons and neutrons (collectively called nucleons) held together by the strong nuclear force. This force is very different from the Coulomb force in that it is only attractive and much stronger, but only over short distances (just a few nucleon radii, after which it disappears to zero), and is altogether too complicated to be described by a simple formula like Coulomb’s Law. Hence, atomic nuclei are a fine balance between the attractive strong nuclear force and the electrostatic repulsion between the positively-charged protons. When nuclei become too big, the short-range nuclear force can no longer hold everything together against the electrostatic force. This is why the heaviest stable nucleus is 209Bi, with 83 protons and 126 neutrons. Heavier common nuclei like 238U are stable enough (half-life 4.46 Gyr) that a large fraction is left over from the formation of the solar system 4.5 or 4.6 Gyr ago[fn]U.S. Geological Survey.  Age of the Earth (online).  http://pubs.usgs.gov/gip/geotime/age.html [4 June 2010].[/fn], but heavier nuclei than this only last for a few thousand years, days, or mere fractions of a second.

Food for thought: why don’t we see nuclei made up of only neutrons, that experience only the attractive strong force and not the replusive electrostatic force?

Now consider a nucleus that is teetering on the edge of stability, 235U (half-life 700 Myr, 0.7% of natural uranium, which is mostly 238U). If we add one more neutron, it becomes just too big to hold together, and splits in two. This process is called fission and it releases a large amount of energy in the form of the kinetic energy of the two mutually repelling fragments. This energy we can estimate from Coulomb’s Law. For now, assume the two fragments have equal numbers of protons (this is usually not the case, but good for our rough estimate). Uranium is element 92, so that means 46 protons each. To estimate the mean distance between the two fragments, lets take it to be the radius of the original nucleus[fn]K. S. Krane, Introductory Nuclear Physics, (Wiley, New York, 1988).[/fn]: 6 × 10-15 m.


From equation 2 we find that the potential energy is approximately:

$U = \dfrac{(9.0\times10^9 \textnormal{Nm}^2/\textnormal{C}^2)(46 \times 1.6 \times 10^{-19} \textnormal{C})^2}{6 \times 10^{-15}\textnormal{m}} = 8\times 10^{-11}\textnormal{J}$

This is an over-estimate by a factor of about 2.5, (a) because the nucleus has to deform considerably before it breaks, and so the effective distance between the two fission fragments is much greater than the original radius, and (b) because real nuclei seldom split into two equal parts (so $q_1q_2$ is less than if it did); but we’re in the right ball-park. The bottom line is that the energy released in the fission of one uranium nucleus is of the order of 100 000 000 times more than that released in burning a carbon atom in air (a few times 10-11 J compared to a few times 10-19 J). And the result of a fission event, however nasty the fission products (and they do tend to be highly radioactive) does not involve putting yet another carbon atom into the atmosphere. Thus fission as a potential source of useful energy is worth looking at very closely.

The Chain Reaction

How do we keep uranium nuclei fissioning to provide a reliable source of power? It turns out that every time a uranium nucleus splits, it releases two or three free neutrons. If the lump of uranium is large enough, these neutrons can cause another nucleus to fission, which causes more neutrons to be released, etc. etc. Plainly if one is doubling the energy output for each generation of neutrons, and the time for a neutron to find its target nucleus in a lump of uranium is very short (these neutrons travel at around 107 m/s) and the neutrons are not being lost to other reactions or the outside world, then we have a problem on our hands. These conditions occur in a few kg of fairly pure 235U (238U absorbs neutrons harmlessly). This is a bomb.

On the other hand it is possible to slow the neutrons down before they find another uranium nucleus and yet still keep the reaction going. This slowing occurs in a moderator, a material that allows neutrons to rattle around and lose their kinetic energy without absorbing them. Common materials used are water (heavy water is better than ordinary “light” water) and carbon. In addition, if one ensures that by losing enough neutrons to absorption or escape each uranium fission causes precisely one other fission, nor more or less, then the system calms down and produces power at a constant, controllable rate.  This is a reactor.

Some reactors can work with natural uranium if the moderator is good enough (i.e. absorbs very few neutrons). The Canadian CANDUs are like this: they use heavy water as a moderator. Most reactors use slightly enriched uranium (~3% 235U) and ordinary water as a moderator.

The choice is basically between isotopically enriching the uranium or isotopically enriching water. Both processes are expensive, in energy and money.


How much natural uranium is needed to produce a GW of electricity (the output of a typical big power station) for one year (i.e. a GWey of energy)?

First of all it is necessary to convert 1 GWey to joules: (109 W)(3600 s/h)(24 h/d)(365 d/y) = 3.15 ×1016 J

Each 235U atom produces 3.2 ×10-11 J of energy.

In principle (3.15 ×1016 J)/(3.2 ×10-11 J per atom) = 9.84 ×1026 atoms can produce the required energy.

Each atom has a mass of (235)(1.66 ×10-27 kg) = 3.90 ×10-25 kg, so the total mass of 235U is 384 kg.

However, this is not the whole story because (a) natural uranium is only 0.7% 235U by mass, (b) the thermal energy produced by fission is only convertible to electricity with at most 40% efficiency, and (c) reactors can only “burn” a small fraction of the 235U fuel before the build-up of neutron-absorbing products reduces the neutron population below the level of a self-sustaining chain reaction.

Let us take these points in turn:

(a) The amount of natural uranium containing 384 kg of 235U is 384 kg/0.007 = 54 900 kg ≈ 55 tonnes.

(b) If we want 1 GWey rather than 1 GWthy we’ll need about 55 tonnes/0.4 = 140 tonnes

(c) Operators of CANDU natural uranium power reactors quote a “fuel burnup” of 180 MWeh/kg of uranium[fn]Rouben, B.  Canteach.  CANDU Fuel-Management Course (online).  http://canteach.candu.org/library/20031101.pdf  [4 June 2010].[/fn]. Hence the mass of uranium required for one GWey is:

(3.15 ×1016 J/GWey)/(3.6 ×109 J/MWeh)/(180 MWeh/kg) = 8.8 ×106 kg = 8800 tonnes.

The fuel burnup for reactors using enriched uranium is about double this[fn]World Nuclear Association.  The Economics of Nuclear Power (online).  http://www.world-nuclear.org/info/inf02.html [4 June 2010].[/fn], and so the uranium needed per 1 GWey is about 4000 tonnes.

Nuclear Waste and the Environment

How much nuclear waste does each 1000 tonnes of spent uranium produce? As only a tiny fraction of the original mass ends up as energy (via E=mc2), the answer is pretty much 1000 tonnes. This material contains many different isotopes of widely varying radioactivity and chemical toxicity. Short-lived isotopes (minutes, hours, days) decay away very quickly. Very long-lived isotopes (thousand, millions of years) have low radioactivity. The most dangerous isotopes are those that are biologically active (they mimic common atoms in the human body) and have half-lives of the same order as the human lifespan,  e.g. 90Sr (29 years, replaces the calcium in our bones). The total radioactivity of all reactor waste products will decay to the level of the original uranium ore in a few thousand years[fn] B. Comby, The Solutions for Nuclear Waste.  International Journal of Environmental Studies, Vol. 62, No. 6, December 2005, 725–736 http://www.efn.org.au/NucWaste-Comby.pdf [/fn].

Greenhouse gas emissions from nuclear power are not zero, as is sometimes assumed. Mining, refining, transport and construction all contribute CO2 to the atmosphere. The total CO2 production is estimated to be about 100 000 tonnes per GWey [fn]S. Andseta et al., Candu Reactors and Greenhouse Gas Emissions,  Proceedings of the 19th Annual Conference, Canadian Nuclear Society, Toronto, Ontario, Canada, October 18-21, 1998. http://www.computare.org/Support%20documents/Publications/Life%20Cycle.htm [/fn]. A GWe coal-fired power station produces around 10 million tonnes of CO2 each year[fn]Clean coal /article/clean-coal[/fn]. Unless we take active and expensive steps to remove it from the atmosphere, most of the CO2 will stay there forcing the climate forever; this is a useful fact to bear in mind when worrying about the long-term storage of nuclear waste.

In addition, any discussion of the danger of nuclear power should be set against the approximately 6000 miners who die in China’s coal mines each year[fn]China Labour Bulletin.  Deconstructing Deadly Details from China’s Coal Mine Safety Statistics (online). http://www.clb.org.hk/en/node/19316 [4 June 2010].[/fn], and the unfolding disaster in the Gulf of Mexico.


1, 12, 13 Aubrecht GJ. Solar Energy: Wind, Photovoltaics, and Large-Scale Installatons. In: Energy – Physical, Environmental, and Social Impact (3), edited by Erik Fahlgren. Upper Saddle River, NJ: Pearson Education Inc., 2006, chapt. 21, 461-465.
2, 3, 4 Kump, L.R., Kasting, J.F., and Crane, R.G. The Atmospheric Circulation System. In: The Earth System (2), edited by Patrick Lynch. Upper Saddle River, New Jersey, USA: 2004, chapt. 4, pp. 55-82.
5 Environment Canada. Canadian Atlas Level 0 (online). https://collaboration.cmc.ec.gc.ca/science/rpn/modcom/eole/CanadianAtlas0.html [20 May 2009].
6, 7, 8 Gustavson MR. Limits to Wind Power Utilization. Science 204: 13 – 17, 1979.
9 MacKay DJC. Sustainable Energy – Without the Hot Air (Online). UIT Cambridge. http://www.inference.phy.cam.ac.uk/sustainable/book/tex/ps/253.326.pdf [4 May 2009].
10 Betz’ Law http://en.wikipedia.org/wiki/Betz’_law [2012.09.27].
11 Learning (online). Solacity Inc. https://www.solacity.com/SiteSelection.htm [20 May 2009].
14 Clarke S. Electricity Generation Using Small Wind Turbines At Your Home Or Farm (Online). Ontario Ministry of Agriculture, Foods and Rural Affairs. https://www.omafra.gov.on.ca/english/engineer/facts/03-047.htm#noise [25 May 2009].
15 Marris E and Fairless D. Wind Farms’ Deadly Reputation Hard to Shift. Nature 447: 126, 2007.
16 Keith D. Wind Power and Climate Change (online). University of Calgary. https://www.ucalgary.ca/~keith/WindAndClimateNote.html [20 May 2009].
17 Accio Energy. About Accio Energy (online). https://accioenergy.com/about.html [12 June 2009].