Transcript Transformers HNC

Transformers
HNC/D
Engineering Science
Transformers
No not them
Transformers – the electrical ones
Transformers come in many shapes and
sizes but basically operate the same way.
Three-phase polemounted step-down
transformer.
Potential-transformer
Laminated core transformer showing
edge of laminations at top of unit
Transformers
A transformer is a device that converts alternating voltage at one
level to an alternating voltage at a higher or lower level.
The above diagram shows a simple double wound transformer. The coil on
the left is called the primary and the coil or winding on the right is called the
secondary. The bit in the middle is called the core.
Double Wound Transformers
Most double wound transformers look slightly different in that the two windings
primary and secondary will be on the same leg of the core. This helps reduce
losses.
Three-phase Transformers
In a three-phase transformer, the transformer is slightly different.
There are a series of three windings on the primary and secondary.
The windings on a transformer are sandwiched or wound on top of each other
to help the distribution of magnetic flux.
Three Phase Supply
In a three phase supply there are effectively three supplies all rising
and falling at the same time.
The secondary windings
can differentiate between
the varying magnetic
fields because each
produces a particular
phase and these are at
maximum and minimum at
different times.
Instrument Transformers
In large power circuits it is not possible to measure either current or
voltage directly.
This is because the meter used would have to be very large for the
levels of voltage and current.
When measuring instruments placed in circuits is not appropriate,
because of the large voltage and current, then voltage transformers
and current transformers are used.
Air Cooled Transformers
The simplest kind of cooling available is air.
There are two types of air cooling, these are natural air (AN) and
forced air (AF)
The air cooled transformer will be labelled with one (AN) or the other
(AF) or possibly both.
It is important to check for damage before installing the transformer.
This type of transformer must be installed indoors and surrounded
by an earthed metal case or enclosure.
If the transformer has been stood for some time it is necessary to
carry out an insulation resistance test.
Oil Cooled Transformer
With oil cooled transformers we have to consider the cooling
of the transformer and the cooling of the oil itself.
The labelling on oil-cooled transformers has four letters,
possibly ONAN (oil-natural air-natural)
The first two letters describe the removal of the heat from the
transformer and the second two denote the removal of the
heat from the oil.
When installing oil-cooled transformers there needs to be a
means of containing the oil and a means of containing the oil
should it leak.
Transformer on no-load
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An ideal transformer is one where
all the magnetic flux is confined to
the core and links both the primary
and secondary coils
Calculations for transformers
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U = voltage
N = number of turns
I = current
p = primary, s = secondary
Calculations
Example 1
An idea transformer has a primary with 900 turns. How
many turns will there need to be on the secondary coil if a
240 V a.c input is to give rise to a no-load output of 80 V ?
Example 2
An idea transformer has 800 primary turns and 200
secondary turns. What will be the secondary voltage when
the primary is supplied from a 240 V a.c supply ?
Transformer – resistive load
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If there is a resistive load connected to
the secondary.
A secondary current flows and hence
power is dissipated.
The secondary coil current produces its
own alternating flux in the core.
If the power losses is negligible the power
supplied to the primary coil must equal
the secondary coil when there is a current
in the primary coil

I1V1 = I2V2
Calculations
Example 1
An ideal transformer is used to light a 12V, 15W lamp from
the 240V mains supply. Determine the turns ratio required
and the current taken from the mains supply.
Example 2
An ideal transformer has 20 primary turns and 80 secondary
turns and supplies a secondary current of 1.5A. What is the
value of the ampere-turns of the transformer and the
primary current ?
Losses
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Losses are inherent to the nature of the equipment
that is used and fall into two key types in
transformers:
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Copper losses are created by the current flowing
through the cables. All conductors have resistance
and when current flows then heat is dissipated.
This is a loss. Copper losses are never affected by
frequency.
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Solution = cooling
Losses
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Eddy current, and hysteresis losses are called ‘iron
losses’. This type of loss is a magnetic loss.
The eddy current, particularly within a material such as
iron, can cause quite a large increase in temperature, and
consequent power loss.
Solution = laminated cores (these reduce the losses)
Hysteresis is a magnetic phenomenon which causes
residual magnetism to hang around after the field has
collapsed, again causing heat, this happens more with
changes in frequency
Solution = silicon steel as the core material is better
Power Rating
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Power rating in VA = V2I2FL
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V2 = the secondary voltage
I2FL = the secondary current on full load
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Full Load = Maximum power output
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(determined by the heat generated by internal losses
dissipated)
Power Rating
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Power Loss = I2R
The loss is termed the copper loss
This is the current which can be obtained
before the transformer overheats
When a transformer is said to be
operating at half load, it means half its
rated power output
Therefore the secondary current is half
the full load value
Calculations
Example 1
A 2.75 KVA single phase transformer has a primary with 600
turns and a secondary with 100 turns. If the alternating
primary voltage is 3300V what will be the minimum
resistance which can be connected as the load ?
Efficiency
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Power loss that occurs with a transformer
are :
Copper Loss – generated by current
through primary and secondary winding
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Power Loss = I12R1 + I22R2
I1 = current through primary of resistance R1
I2 = current through primary of resistance R2
Iron Loss – is the loss resulting from
hysteresis and eddy currents.
Efficiency
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Efficiency = output power x 100
input power
Output Power = V2I2 cosФ (this
equals the input power minus
losses)
Calculations
Example 1
A transformer delivers a power of 5 kW when on-load. What
will be its efficiency if the copper loss is then 100W and the
iron loss 75W ?
Example 2
A 40 kVA single phase transformer has an iron loss of 400W
and a full load copper loss of 600W. What is the efficiency of
the transformer for a power factor of 0.8 at
(a) full load, (b) half load ?