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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?
Sub-index for this page
1.
The
National Grid system electricity supply
2.
More on transmission of
electricity, need for transformers in the National Grid system
3.
Comparing advantages/disadvantages of overhead power lines
& underground cables
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.
See also
Energy resources
and comparison of methods of generating electricity
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
-
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).
-
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
12.
Generator effect, applications e.g. generators
generating electricity and microphone
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.
TOP OF PAGE
3. Comparing the advantages and disadvantages of overhead power lines and underground cables
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.
TOP OF PAGE
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 /
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.
TOP OF PAGE
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
The use of transformers in the
National Grid system has already been described on this
page.
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.
TOP OF PAGE
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 / 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
TOP OF PAGE
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
Ohm's Law)
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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TOP OF PAGE
What next?
Electricity and
magnetism revision
notes index
1.
Usefulness of electricity, safety, energy transfer, cost & power calculations, P = IV = I2R,
E = Pt, E=IVt
2.
Electrical circuits and how to draw them, circuit symbols, parallel
circuits, series circuits explained
3. Ohm's Law, experimental investigations of
resistance, I-V graphs, calculations V = IR, Q = It, E = QV
4. Circuit devices and how are they used? (e.g.
thermistor and LDR), relevant graphs gcse physics revision
5. More on series and parallel circuits,
circuit diagrams, measurements and calculations
gcse physics
6. The 'National Grid' power supply, environmental
issues, use of transformers
gcse
physics revision notes
7.
Comparison of methods of generating electricity
gcse
physics revision notes (energy 6)
8. Static electricity and electric fields, uses
and dangers of static electricity gcse
physics revision notes
9.
Magnetism
- magnetic materials - temporary (induced) and permanent magnets - uses gcse
physics
10.
Electromagnetism, solenoid coils, uses of electromagnets gcse
physics revision notes
11. Motor effect of an electric current,
electric motor, loudspeaker, Fleming's left-hand rule, F = BIL
12.
Generator effect, applications e.g. generators
generating electricity and microphone
gcse
physics
ALL My GCSE
Physics Revision Notes
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 How
to explain how the effect of an alternating current in
one coil in inducing a current in another is used
in transformers explain how the ratio of the potential differences across the two
coils depends on the ratio of the number of turns
on each how to calculate the current drawn
from the input supply to provide a particular power output
how to apply the equation linking the potential differences and number
of turns in the two coils of a transformer to the
currents and the power transfer involved, and
relate these to the advantages of power
transmission at high voltages. GCSE
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