Lesson video

Hello, my name's Dr.

George, and this lesson is called The National Grid.

It's all about how electricity is distributed around the country and how it reaches our homes.

It's part of the unit Mains electricity.

The outcome for this lesson is I can describe the National Grid and explain why transmission lines use a very high transmission voltage.

These are the keywords for the lesson.

I'm not going to read them now because I'll introduce them as we go along, but this slide is here in case you want to come back anytime and check the meanings.

This lesson has three parts.

They're called National Grid, transmission lines, and transformers.

Now, most of the mains electricity that we use in the UK is generated either by power stations or by wind turbines.

But often the best places to put these are well away from the towns and cities where most of the electricity is needed.

So we have the National Grid, which is a network of high-voltage cables that connects places where electricity is generated to places where it's needed.

And we call these cables transmission lines.

National Grid stretches all across the UK, and has a few connections across to Europe.

Transmission lines carry voltages of up to 400 kilovolts, with currents of over 1000 amps.

Now, we normally think of air as an electrical insulator, but if the voltage across a small air gap is large enough, then current can flow across the gap.

That's what we see when there's lightning or if you've ever seen a demonstration of a Van de Graaff generator and seen sparks go through the air.

So these transmission lines have to be kept well away from the pylons, which are made of metal and connected to the earth.

And so the transmission lines hang from ceramic insulators, ceramic as in literally a type of china porcelain is often used.

These are excellent insulators.

So take a look at that next time you see a pylon.

So what is the voltage of some of the transmission lines in the National Grid? And when I ask a question, I'll wait five seconds, but you may well need longer, in which case press pause and press play when you're ready.

And the answer is 400,000 volts, or 400 kilovolts.

230 volts is what we use in the home.

Power stations often generate electricity that has a voltage of about 25 kilovolts, 25,000 volts, and wind turbines generate electricity with a voltage of about 1000 volts.

So we need to increase that voltage to get to those high voltages for transmission lines.

And to do that, we use devices called step-up transformers.

So they increase these voltages up to around 400,000 volts.

I'll be talking more later about how transformers work.

At the other end of the transmission lines, there are step-down transformers, and what these do is reduce the voltage to the right size for the user, which could be homes or it could be factories, which usually use a higher voltage than homes.

Voltages supplied to factories vary.

Some use 33 kilovolts, some 11 kilovolts, or 450 volts.

Electric trains, by the way, use 33 kilovolts.

So what do step-down transformers do in the National Grid? Step-down transformers decrease voltages.

Step-up transformers increase voltages.

Now can you use the words listed here to label the diagram showing parts of the National Grid? You might find it helpful to make a simplified sketch of this and then you can write the labels onto it.

Press pause while you're working on this, and press play when you're ready to check your answers.

And here's the diagram with the labels added to it.

Power station generating electricity at 25 kilovolts.

Step-up transformer increases that to 400 kilovolts.

Cables transmit the electricity across the country, then to a step-down transformer.

And here we have another step-down transformer.

And the voltage comes down to 230 volts for homes.

Now for the second part of the lesson about transmission lines.

Each transmission line can carry around 500 megawatts of power.

That's 500 million watts, or 500 million joules of energy per second.

And that means the currents through the cables are huge.

You may remember this relationship that power of an electrical device is current times potential difference.

For a 400-kilovolt transmission line, here's the equation with the power and potential difference substituted in.

We can rearrange to find the current by dividing both sides by the potential difference.

And we get a current of 1250 amps.

That size of current is enough to make transmission lines hot.

They're often at about 90 degrees Celsius.

And we need that much power.

We need that much energy per second to be transmitted so that homes and factories can get the electricity needed.

You might have wondered whether it might be easier just not to have those step-up and step-down transformers and just use a voltage of 230 volts throughout the National Grid, but we can't send that much power at that voltage.

Going back to the equation, P = I x V.

If the transmission line were at 230 volts, but we still need 500 million watts of power, we find that the current would be 2 million amps.

And a current that high would heat the transmission lines so quickly that they would melt.

So is this true or false? Transmission lines need a very high voltage in order to transfer the amount of energy we need from electricity.

And when you've chosen your answer, I'd also like you to think about why you chose it.

How do you know? So press pause while you think about that, and press play when you're ready.

And the statement is true.

And here's why.

As we've seen, P = I x V, so if the voltage is very big, the current can be smaller.

If the current is too high, it will melt the transmission lines.

These transmission lines have a diameter of about five centimetres, far thicker than any wire that you will have used in an experiment.

And that means it has a very low resistance.

The resistance of a transmission line is only about 3 ohms per 100 kilometres.

And transmission lines are part of a huge circuit that connects with all of the appliances we use in our homes, our lighting, and everything else that uses mains electricity.

Most of the voltage is used to push electricity through these appliances, but a little is used to push electricity along the transmission lines themselves against their resistance, because they do have a little bit of resistance.

We can use the resistance of a transmission line and the current it carries to work out how quickly it dissipates energy to the surroundings by heating.

Energy dissipated each second equals the power dissipated.

Voltage pushes the current through the transmission line.

So we can write this equation as well, V = I x R.

That's true for any electrical component or any piece of wire or cable.

Combining these two equations, we can say that if P = I x V, and V = I x R, then P = I x I x R.

And we could shorten that to I squared R.

So a typical transmission line, as we've seen, has a resistance of 3 ohms per 100 kilometres and a current through it of 1000 amps.

So we can now work out the energy dissipated by this 100 kilometres of transmission line each second using P = I squared R.

And substituting in those values, we find that it's 3 million watts, or 3 megawatts.

So every second, 3 million joules of energy are transferred by heating and spread out into the surroundings.

And this works out to be nearly 1% of the power transmitted along the transmission line.

And across the whole National Grid, up to 10% of power is dissipated by heating.

So this is simply wasted energy.

We're not trying to heat the environment by using transmission lines, but this energy loss is unavoidable.

And now I'm going to show you a worked example and then ask you a similar question.

How much power is dissipated by 300 kilometres of a transmission line if the resistance is 3 ohms per 100 kilometres and the current is 1200 amps? So we can start by working out the total resistance.

300 kilometres at 3 ohms per 100 kilometres has a total resistance of 9 ohms. And then using the equation power dissipated is current squared times resistance, substituting in the numbers we get about 13 million watts, 13 megawatts.

Okay, now here's one for you to try.

How much power is transferred to the surroundings by 250 kilometres of a transmission line? Again, the resistance is 3 ohms per 100 kilometres, but this time, the current is 950 amps.

Press pause while you write out your working, and press play when you've finished.

And here's the solution.

First, we work out the resistance of 250 kilometres if 100 kilometres has resistance 3 ohms. So it's 2.

5 times as long.

2.

5 x 3 is 7.

5 ohms. And we use the same equation.

So write down the equation, then substitute in the values, and we get this answer, but we can't claim to know that many significant figures based on the quantities we used.

So we could round that to say two significant figures, 6.

8 million watts, or 6.

8 megawatts.

Well done if you got that right.

And now a couple more questions for you.

Imagine that engineers are going to build a new part to the National Grid.

Question one: how much power would be dissipated if they used 400 kilovolts to send a current of 800 amps through each of three transmission lines that are 80 kilometres long? And how much power would be dissipated if they used 275 kilovolts to send a current of 1280 amps through the same three transmission lines? And the resistance of each transmission line again is 3 ohms per 100 kilometres.

Show your working in your answers.

Press pause while you work these out, and press play when you're ready.

I'll show you the worked answers now.

For the first question, we can find the resistance of one transmission line.

We know that resistance is 3 ohms per 100 kilometres, but we only have 80 kilometres, so it's 0.

8 as long.

0.

8 x 3 is 2.

4 ohms. Then we use P = I squared R to find the rate of energy dissipated.

Substitute in the values, and we get this answer.

Don't round it yet because we haven't finished.

So we have three transmission lines, each with the same current.

So three times that power, and we could round that to 4.

6 megawatts, 4.

6 million watts.

And the steps in the working are the same for the second question.

We just have different values of the quantities.

And this time we get 11.

8 megawatts.

So this is almost three times the rate of energy loss by dissipation.

So we can see that a lower transmission voltage dissipates much more power by heating because we have a higher current.

Well done if you got these right.

And now for the third part of the lesson, transformers.

We're going to look at how these work.

They have no moving parts.

They're almost 100% efficient.

That means there's very little wasted energy.

And this image shows step-up transformers at a power station.

One transformer would be made of two coils of insulated wire, both wrapped around a soft iron core.

This one we could call the primary coil, and this one the secondary coil.

And here's the core.

So no current flows directly from one coil to the other.

And no current flows between a coil and the core because the wires are insulated.

Transformers only work with alternating current, AC.

They don't work with direct current.

When a potential difference is put across the primary coil, a current flows in the coil, and it's then an electromagnet because an electromagnet is a current-carrying coil, it has a magnetic field around it.

But that makes the soft iron core become magnetised as well.

And when that core becomes magnetised, it's like pushing a magnet into the secondary coil.

And if you do that, the generator effect causes a p.

d.

to be induced across the coil.

So when the primary coil becomes magnetised, the potential difference is induced across the secondary coil, and a current can flow if it's connected to a circuit.

Because the current in the primary coil is AC, it keeps switching direction, the magnetic field around the primary coil keeps changing.

And so the secondary coil experiences a changing magnetic field.

So that continues to induce potential difference across the secondary coil.

And it turns out that the more turns the secondary coil has, the greater the induced p.

d.

across that coil.

So what sort of current does a transformer need to work? And the answer is AC, alternating current, as I said earlier.

Now, as transformers are almost 100% efficient, they dissipate very little energy by heating.

And when each coil is connected to a circuit, current flows through each of the coils.

And when this happens, the power supplied to the primary circuit is almost equal in size to the power output of the secondary circuit.

So the primary circuit could be a connection to a power station, and the secondary circuit could be a connection to consumers who are using the electricity.

The power into a transformer is equal to the power out because of its near 100% efficiency.

So we could write this equation showing that the power in equals the power out.

And here we have Ip being the current in the primary coil, and Vp is the p.

d.

across it.

And we've got Is and Vs for the secondary coil.

P and the s are not full-size letters, they're not representing quantities.

They're labels on the I and the V to show which coil they're referring to.

And now I'll show you how to answer a question and then give you one to try.

A transformer increases voltage from 25 kilovolts to 400 kilovolts.

If the current in the primary coil is 2000 amps, what is the current in the secondary coil? We're going to have to assume that this transformer is 100% efficient so that the power in equals the power out.

So we can use this equation.

And we know three of these quantities.

We know the current in the primary coil.

And if this transformer is increasing voltage, then it must be the lower voltage that's across the primary coil.

So we use those quantities on the left.

We substitute the secondary voltage, 400,000 volts on the right.

Then we need to rearrange to make the current in the secondary the subject.

So what we've done here is multiply the two quantities on the left and divide both sides by 400,000.

And we find that the current in the secondary is 125 amps.

We get the same power out as in, but in a different way with a higher voltage and lower current.

And now this question for you.

A transformer decreases voltage from 275 kilovolts to 11 kilovolts.

If the current through the secondary coil is 400 amps, what is the current in the primary coil? Press pause while you write out your solution.

So again, let's write the equation that we're going to use and substitute in the values that we know.

This time we know both the voltages, but we know the current through the secondary coil.

So substituting those values into the right places, we can multiply the two numbers on the right hand side, then divide both sides by 275,000 and we get 16 amps.

This is a step-down transformer.

It decreases the voltage from 275 to 11 kilovolts and it increases the current from 16 amps to 400 amps.

And now a set of questions for you.

When you answer these, show all your working out.

So press pause while you do this, and when you're ready, press play and I'll show you the worked solutions.

So let's start, question one.

A step-down transformer reduces the voltage from 230 volts to 12 volts.

If the current in the secondary coil is 1.

5 amps, what is the current in the primary coil? And again, we are using the equation power in equals power out.

And substituting in the values we know, we know that the primary has 230 volts across it and then that is reduced to 12 volts across the secondary.

And we know the current in the secondary.

We need to divide both sides by 230.

And our rounded answer is 0.

078 amps.

And it's sensible to round to two significant figures because we only know the secondary voltage and the secondary current, two significant figures each.

Question two: a step-up transformer at a wind turbine increases the voltage from 1000 volts to 275 kilovolts.

If the current from the wind turbine is 2000 amps, what is the current in the secondary coil? And we need to use voltage with the same units both times, so we'll write 275 kilovolts as 275,000 volts.

And we can substitute those values in.

We're stepping up.

So it must be the primary that has the lower voltage, the secondary has the higher voltage.

And we can divide both sides of the equation by 275,000.

And we get a current in the primary of 7.

3 amps to two significant figures.

And question three: a power station produces a p.

d.

across a transformer of 25 kilovolts, and a current through the primary coil of 1800 amps.

What current is applied to the National Grid by the transformer if it increases the p.

d.

to 400 kilovolts? So we have the power station on one side of this transformer and we have the National Grid on the other side.

Again, we substitute these values into the equation in the right places.

We can divide both sides by 400,000 and we get a current of 110 amps to two significant figures.

So well done if you're getting these right.

And now we're at the end of the lesson.

So here's a summary of what we've done.

The National Grid is a network of transmission lines that connect power stations and wind turbines to places where electricity is used.

Step-up transformers increase the voltage from power stations to very high voltages for transmission.

Step-down transformers reduce the transmission voltage to voltages that are safer to use.

Transformers are almost 100% efficient.

So, I primary times V primary equals I secondary times V secondary.

Higher transmission voltages reduce dissipation of energy by heating.

So well done for working through this lesson.

I hope you found it interesting and I hope to see you again in a future lesson.

Bye for now.