School Physics notes: Electricity generation and the National Grid supply

Electricity 6. Generating electricity 'National Grid' power supply

environmental issues plus the theory, calculations and uses of transformer

Doc Brown's school physics revision notes: GCSE physics, IGCSE physics, O level physics,  ~US grades 8, 9 and 10 school science courses or equivalent for ~14-16 year old students of physics

How do power stations link up with the National Grid?  What do we use transformers for in the National Grid system?

1. The National Grid system electricity supply

2. More on transmission of electricity, need for transformers in the National Grid system

3.

4. More detailed notes on the theory, structure and uses of transformers

5. More on the uses of transformers

6. Examples of transformer calculations

See APPENDIX 1 for a summary of all electricity equations you may need.

and

Energy resources & uses, general survey & trends, comparing sources of renewables, non-renewables

1. The National Grid System electricity supply

Introduction - basic description

The power lines and transformers form the first part of the National Grid system, a country's electrical power supply.

An overall 'picture' of the National Grid system of electrical power supply

• Know that electricity is distributed from power stations to consumers along the National Grid

• You should be able to identify and label the essential parts of the National Grid.

• The National Grid consists of a vast electricity distribution network of transformers, pylons and suspended cables - insulated power lines running for long distances across the landscape.

• They are somewhat unsightly, but essential for providing bulk electrical power to towns of homes, shops and factories.

• All major power stations feed into the National Grid irrespective of their geographical location and many are needed to service millions of users in homes, transport and industry right across the country.

• You see them stretching for miles and miles across the landscape to supply you, the consumer, very conveniently with a constant (well nearly!) supply of electricity to your city, town or village across the vast majority of the country.

• Eventually the power is delivered, very conveniently, into your home as a consumer or factory etc.

• power station: energy resource to drive turbine ==> to drive generator ==> step-up transformer ==> grid system of pylons or underground cables ==> step-down transformer ==> user/consumer

• For non-renewable energy power stations: chemical/nuclear energy store (fuel) ==> thermal energy store (hot water) ==> kinetic energy store (turbine and generator) ==> electrical energy (National Grid)

• The largest power stations are usually non-renewable fossil (oil, coal, gas) or nuclear fuelled.

• The heat generated boils water to power a steam turbine which in turn drives the generator.

• The generator (a large alternator) consists of a powerful rotating electromagnet that induces a high p.d. alternating current in coils of copper wire.

• There are several copper coils all joined together in parallel to produce a single output from the generator.

• Natural gas power stations are the cheapest to build and relatively rapid start-up time.

• Nuclear power stations are the most costly to build and have the longest start-up time.

• The National Grid system of electricity supply MUST work off an alternating current (ac) for several reasons, and one important factor is that transformers only work using ac.

• With alternating current (ac), the current changes direction in a cycle e.g. 5O Hz.

• With direct current (dc) there is no reversal in current direction, it flows one way with a constant voltage.

• Oscilloscope traces comparing ac and dc current signals - showing the alternating + <=> - oscillation of the alternating current p.d. and the constant p.d. of a direct current.

• Note that some devices in the home work off a dc current - but the output from e.g. the transformer in your computer power supply, is rectified to convert it to a dc supply.

• In the UK the generator output at the power station is 25 kV.

• A step-up transformer increases the p.d. (voltage) to 400 kV in the UK for power line transmission.

• A step-down transformer decreases the p.d. (voltage) of the power line transmission to more suitable and safer levels for home and industry (typically 230 and 11000 V).

• The transformer sites are referred to as sub-stations.

• A substation will decrease the p.d. even more down to ~230-240 V that is cabled into your house - your domestic electricity supply - which operates on a frequency of 50 Hz (50 cycles per second - the rate of current reversal of the alternating current).

• Explanation of transformers is given in the last section on this page.

• See gcse physics

• Both the function of generators and transformers depend on the electromagnetic effect.

• The scheme described above is similar for most generation, except that initially for hydroelectric, tidal and wind power generation, the turbine is rotated directly by these renewable energy resources of water or wind - no fuel required.

• For renewable energy power stations: kinetic energy store (water/wind) ==> kinetic energy store (turbine and generator) ==> electrical energy (National Grid)

• The only kind of power generation that does not require a turbine and generator is the solar panel.

• For solar power: nuclear energy store (the Sun) ==> electromagnetic radiation (visible light) ==> electrical energy (National Grid)

• OR if for charging a battery:  nuclear energy store (the Sun) ==> electromagnetic radiation (visible light) ==> electrical energy ==> chemical energy store (battery).

2. More detail on the transmission of electricity, need for transformers in the National Grid system

• For a given power increasing the voltage reduces the current required and this reduces the energy losses in the cables.

• You should know why transformers are an essential part of the National Grid.

• You also need to know why the long distance power lines use a very high p.d. and relatively low current.

• So that you can transmit (transfer) the very large quantities of electrical energy per unit time needed, you need to use either a very high current or a very high voltage or both

• Since P = IV, to deliver lots of power you need to increase current or p.d. or both.

• The theoretical four possible choices are (i) low current and low voltage, (ii) high current and low voltage, (iii) low current and high voltage or (iv)  high current and high voltage ...

• (i) obviously couldn't deliver what is needed.

• (ii) and (iv) increase heat losses, which is a function of the current flowing through a resistor - you always get some conversion of electrical energy into heat - increasing the thermal energy store of the cable and surroundings.

• but (iii) is the actual choice.

• So why is 'low current and high voltage' the desired choice for electrical power line transmission?

• The greater the current flowing through a wire, the greater the heat generated, which in the context of power lines means more waste heat energy the higher the current, which is why (ii) and (iv) are not employed.

• Since P =  E/t = I2R, the power loss is a function of I2 for a fixed resistance - the National Grid cables.

• This is a good numerical argument for minimising the current I.

• However, since power = current x voltage, to deliver a particular power rating, you must still increase one of the two variables and decrease the other.

• Therefore by using a very high voltage (eg 400 000 V, 400 kV) and relatively low current you maximise power transmission for the minimal heat loss of wasted electrical energy.

• So for a given power transmission increasing the p.d. and reducing the current makes the National Grid system as efficient it can be with the minimum of electrical energy lost to the thermal energy store of the surroundings.

• See the equations in the transformer section.

• However, use of these extremely high voltages (1667 x your domestic voltage of 240 V), means health and safety issues arise and you need lots of big ceramic insulators on pylons and transformers and lots of barbed wire to deter people from climbing up pylons!

• You should know and understand the uses of step-up and step-down transformers in the National Grid.

• With cables with a p.d. of 400 kV, not only requires transformers, you have to have, rather unsightly, tall pylons to support them and well insulated too! In some places the cables run underground, but this is more expensive (see later comparison discussion).

• Now, (i) since the national power transmission uses 400 kV, you can hardly use this in the home,

• and (ii) generators themselves cannot deliver 400 kV, you need a way of increasing (for efficiency), and then decreasing (for safety), the voltage in power lines.

• A transformer is a means of changing an input voltage in one circuit, into another output voltage in a separate circuit.

• At the power station end is a step-up transformer to increase the voltage for power line transmission.

• At the user end is a step-down transformer, to reduce the voltage that is a safe level for factories, domestic homes, street lighting etc.

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The comparison table below addresses the issues.

 COMPARISON Installation - cost Installation - ease of Environmental impact - visual impact Environmental impact - the land itself Maintenance and reliability Overhead power lines lower, simple erection of pylons and linking cables carrying the high voltage current relatively easy, simple erection job considerable, miles of pylons and cables stretching across the countryside - controversial in designated areas of outstanding natural beauty slight - foundations of pylons much more needed and much less reliable - weather damage eg from frost, snow, corrosion of structure Underground cables much higher, costly trenches and insulated cables much more work, all that digging and filling! minimal, not really seen at all considerable, but temporary - digging trenches is disturbing the land, but no lasting damage much less and more reliable, not affected by weather BUT not as easy to trace and access if fault develops

Variation of electricity demand - meeting industrial and consumer needs

• The demand for electricity varies through the day e.g. there are peak times in the morning and evening and low demand through the night.

• Peak times are associated with cooking and transport needs and demands will increase in the winter when more energy is used for heating.

• Power companies know the demand patterns and can adjust to society's needs.

• Power stations do not run at their maximum output, there must be spare capacity most of the time, so if there is suddenly a huge increase in demand, it can be taken care of.

• There might be an unplanned shut-down of a power station due to unforeseen circumstances.

• There are smaller power stations on standby that can be quickly brought into use.

• There are also pumped-storage systems that are very useful to meet electricity demands at peak times.

• Check out your practical work you did or teacher demonstrations you observed, all of this is part of good revision for your module examination context questions and helps with 'how science works'.

• modelling the National Grid.

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4. More detailed notes on the theory and structure of transformers

A transformer is a device that can change the potential difference (p.d.) of an alternating current (a.c.)

It is another example of electromagnetic induction - the effect of magnetic fields inducing an ac current in a coil.

Transformers are nearly 100% efficient - important in the context of the National Grid electricity supply.

A basic transformer consists of two coils of insulated wire, a primary coil and a secondary coil independently wound on an iron core - see the diagram further down.

The wire is usually made of copper with a thin coating of insulation material.

Iron is used as because it is easily temporarily magnetised.

Both coils MUST consist of complete, but separate circuits - no electrical connection between them.

When an a.c. p.d. is applied across the primary coil, the iron core magnetises and demagnetises quickly due to the nature of the alternating current.

Therefore an alternating current (a.c.) in the primary coil (p) of a transformer automatically produces a alternating (changing) magnetic field in the iron core and hence in the secondary coil (s) by induction.

This induces an alternating potential difference across the ends of the secondary coil (Vs).

If the secondary coil is part of a complete circuit, an induced current will flow in the secondary coil.

Remember, the current MUST BE ac to get an alternating magnetic field that 'cuts' through the iron core', otherwise continuous induction will NOT take place.

Using a  d.c. current is no good - you don't get a constantly changing magnetic field if the current only flows one way!

A step-up transformer increases the p.d. and has a greater number of coils in the secondary coil than the primary.

A step-down transformer reduces the p.d. and has a smaller number of coils in the secondary coil than the primary.

If you know the input p.d. and the number of turns on each coil, you can calculate the output p.d. using the equation below.

The ratio of the potential differences across the input primary coil and the output secondary coil of a transformer, Vp and Vs, depends on the ratio of the number of turns of wire on each coil, np and ns. and is given by a simple ratio equation. The data is 'pictured' on the diagram above.

 Vp np ---- = ---- Vs ns

vp / vs = np / ns   The p.d. and coils transformer equation

the ratios are the same i.e. input p.d. / output p.d = turns on primary coil / turns on secondary coil

and the potential differences, Vp and Vs in volts, V

In a step-up transformer Vs > Vp

In a step-down transformer Vs < Vp

See the diagram above for a visual appreciation of the equation.

If transformers were 100% efficient, the electrical power output would equal the electrical power input.

Vs × Is = Vp × Ip    The transformer power equation

(remember power P = I x V, P in watts, W   and   I current in amps, A)

where    Vs × Is     equals the power output from the secondary coil (never 100% efficient)

and       Vp × Ip     equals the power input to the primary coil

However, no transformer is ever 100% efficient, so the power output never equals the power input, because there are always energy losses eg as heat energy - remember that the two coils of wire in a transformer are still acting as resistances.

However, you expected to do calculations based on an efficiency of 100% - see last section below.

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5. More on the uses of transformers

Battery charging

Transformers are used to control the voltage that enters the battery during the charging process in order to prevent any damages that can occur to the internal battery components.

So, transformers are used in many devices in the home to convert the mains electricity p.d. of ~230 V down to a smaller safer voltage e.g. charging up your computer battery.

The laptop adapter for my website consists of the mains supply

plug and what is basically does what a transformer in the case, but its NOT a simple transformer, technically it is what is known as a 'switched-mode power supply.

Inside the adaptor 100-240 V AC (50/60 Hz)1.5 A mains supply and converts it to 19 V 3.16 A DC output.

Steel manufacturing

Steel manufacturing plants use high/low voltage transformers to provide a range of voltages for the manufacturing process - regulation of the current flowing. High currents are required for the melting and welding of steel, and lower current are required during the cooling process.

Electrochemicals

In chemical engineering and manufacturing processes, electrolysis is normally fuelled by the functioning of transformers to control both the p.d. and current flowing in e.g. electrolysis processes. Metals used for electroplating include copper, zinc, and aluminum and during the process, transformers provide a regulated electrical current to effect the chemical plating reaction.

National Grid system

Transformers enable the National Grid system to efficiently deliver electricity to all parts of a country.

The National Grid makes extensive use of step-up and step-down transformers.

Transformers come in all sizes in terms of physical size and power output e.g. from laptop computer charger to large step-up transformers at power stations of the National Grid electricity supply.

For example, the images above illustrate the local electricity supply to some farms in a rural area.

The small transformer converts the higher voltage supply to a lower one appropriate for domestic supply.

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6. Examples of transformer calculations

Everything you need to know is on the transformer diagram above and take note of the abbreviations below!

To save repetition in the questions PLEASE note the following abbreviations which I will use for problem solving:

P = power (W) = I (A) x V (V),  P = IV = I2R

Vp = p.d. across the input primary coil

Vs = induced p.d. generated across the output secondary coil

np = number of wire turns on the primary input coil

ns = number of wire turns on the secondary output coil

Ip = input current flowing through the primary coil

Is = induced output current flowing out through the secondary coil

The ratio of the potential differences across the coil equals the ratio of the turns on each coil.

 Vp np ---- = ---- Vs ns

Vp / Vs = np / ns    The p.d. and coils transformer equation

Vpns = Vsnp

VsIs = VpIp  The transformer power equation

Since P = IV, it means power input = power output,

assuming 100 % efficiency (never this in reality)

Rearrangement: Vs / Vp = Ip / Is

Q1 A transformer has 200 turns on the primary coil and 10 turns on the secondary coil.

If the output p.d. required is 12.0 V from the secondary coil, what p.d. must be put across the primary coil?

Vp / Vs = np / ns

Vpns = Vsnp

Vp = Vsnp / ns

Vp = 12 x 200/10 = 240 V

Q2 The p.d. across the primary coil of a transformer was 12000 V and the current flowing through it was 20 A.

If the current flowing through the secondary coil was 1000 A, what is the p.d. across the secondary coil?

VsIs = VpIp  (assuming 100 % efficiency)

Vs = VpIp / Is  = 12000 x 20/1000 = 240 V

Q3 A power station step-up transformer has 500 turns of wire on the primary coil and 8000 turns of wire on the secondary coil.

(a) If the generator output is 25 000 V, what is the output p.d. in kV for the transmission lines of the National Grid?

Vp / Vs = np / ns

Vpns = Vsnp

p.d. output = Vs = Vpns / np = (25 000 x 8000)/500 = 400 000 V = 400 kV

(b) What is the power input, in MW, from the generator to the transformer, if the input current to the primary coil is 20 A? (assume 100% efficiency)

Pinput  = IpVp = 20 x 25 000 = 500 000 W = 500 kW = 0.50 MW

(Note: In reality, a power station will usually have several generators and step-up transformers running at the same time.)

(c)  If the power output from the transformer to the transmission line is 0.48 MW, what is the % efficiency of the energy transfer and what has become of the lost energy?

% efficiency = 100 x useful power output / total power input = 100 x 0.48/0.50 = 96%

There are always thermal energy losses from the transformer to the surrounding thermal energy store.

(d) How much energy is wasted from the transformer every minute?

The power wastage is equal to 0.50 - 0.48 = 0.02 MW

0.02 MW = 20 kW = 20 000 W

This equals a heat loss of 20 000 J/s

Therefore in one minute 20 000 x 60 = 1 200 000 J is lost (1.2 MJ/min)

Q4 The p.d. across the primary coil of a transformer is 240 V and carries a current of 5.0 A.

If the p.d. across the secondary coil is 12.0 V, what current is flowing in secondary coil?

VsIs = VpIp  (assuming 100 % efficiency)

Is = VpIp / Vs  = 240 x 5.0 / 12.0 = 100 A

Q5 A transformer has 20 turns of wire on the secondary coil and the p.d. across it is 3.0 V.

How many turns must there be on the primary coil if the primary coil p.d. is 230 V?

Vp / Vs = np / ns

Vpns = Vsnp

np = Vpns / Vs = 230 x 20/3.0 = ~1533 turns of wire.

Q6 The output p.d. across the secondary coil of a transformer is 6.0 V and a current flowing through it of 0.50 A.

What p.d. must be applied across the primary coil to give a current of 0.0125 A to flow through it?

VsIs = VpIp  (assuming 100 % efficiency)

Vp = VsIs / Ip = 6.0 x 0.50/0.0125 = 240 V

Q7 The secondary coil of a transformer has a p.d. output of 120 V.

If the primary coil has 300 turns, how many turns must there be on the secondary coil if the primary coil input p.d. is 3.0 V?

Vp / Vs = np / ns

Vpns = Vsnp

ns = Vsnp / Vp = 120 x 300/6.0 = 6000 turns

Q8 The p.d. across the secondary coil of a transformer is 12.0 V and 0.30 A flows through it.

If the p.d. across the primary coil is 240 V, what current is flowing through the primary coil?

VsIs = VpIp  (assuming 100 % efficiency)

Ip = VsIs / Vp = 12 x 0.3 / 240 = 0.015 A

Q9 A laptop charger works off the 230 V AC mains supply.

Inside the adapter, the transformer produces an a d.c. supply current of 3.0 A at a p.d. of 19 V.

(a) What current did the charger draw from the mains supply?

power input (mains) = power output (adapter)

P = IV, power output = 3.0 x 19 = 57 W

power input = 57 W, therefore since current drawn I = P / V, I = 57 / 230 = 0.25 A (2 sf)

(b) What assumptions has been made? and how can you tell from your everyday experience that the assumption is incorrect?

The calculation assumes there is no energy loss in the transformer process.

The adapter feels 'warm' due to electrical energy ==> thermal energy by the resistances of the wire causing a rise in temperature - this loss increases the thermal energy store of the surrounding air.

Q10

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APPENDIX 1: Important definitions, descriptions and units

 Note: You may/may not (but don't worry!), have come across all of these terms, it depends on how far your studies have got. In your course, you might not need every formula - that's up to you to find out. V the potential difference (p.d., commonly called 'voltage') is the driving potential that moves the electrical charge around a circuit - usually electrons. Potential difference is the work done in moving a unit of charge. It indicates how much energy is transferred per unit charge when a charge moves between two points in a circuit e.g. between the terminals of a battery. The p.d. across any part of a circuit is measured in volts, V. I the current is rate of flow of electrical charge in coulombs/second (C/s), measured in amperes (amps, A). The quantity of electric charge transferred in a give time = current flow in amps x time elapsed in seconds Formula connection: Q = It,  I = Q/t,  t = Q/I, Q = electrical charge moved in coulombs (C), time t (s) R the resistance in a circuit, measured in ohms (Ω). A resistance slows down the flow of electrical charge - it opposes the flow of electrical charge. Formula connection: V = IR,   I = V/R,   R = V/I  (This is the formula for ) P is the power delivered by a circuit = the rate of energy transfer (J/s) and is measured in watts (W). Formula connection: P = IV,  I = P/V,  V = P/I   also  P = I2R   (see also P = E/t below) E = QV,  the energy transferred by the quantity of electric charge by a potential difference of V volts. energy transferred (joules) = quantity of electric charge (coulombs) x potential difference (volts) Q = E/V,  V = E/Q,   E = energy transfer in joules (J), Q = electrical charge moved (C), V = p.d. (V) E = Pt,  P = E/t,  t = E/P,  where P = power (W), E = energy transferred (J), t = time taken (s) Energy transferred in joules = power in watts x time in seconds Formula connection: Since E = Pt and P = IV, energy transferred E = IVt

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