Electricity Generation

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.

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

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.

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.

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.

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.

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:

${P}_{\textnormal{wire}}={I}_{\textnormal{wire}}\times\Delta{V}_{\textnormal{wire}}$

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

$\Delta{V}_{\textnormal{wire}}={I}_{\textnormal{wire}}\times{R}_{\textnormal{wire}}$

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.

${P}_{\textnormal{plant}}={I}_{\textnormal{plant}}\times{V}_{\textnormal{plant}}$${I}_{\textnormal{plant}}=\dfrac{{P}_{\textnormal{plant}}}{{V}_{\textnormal{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.

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$

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” )

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.

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

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

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

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

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

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

while the power of the wind passing through our hoop is

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

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

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

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.

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.

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]

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?

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]

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

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

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].

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

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].

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 GW_ey [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.