Lecture Slides for the course Fundamentals of Electrical Engineering delivered by me to First Semester Diploma in Mechanical Engineering.

1. PEE-102A
Fundamentals of Electrical Engineering
Lecture-8
Instructor:
Mohd. Umar Rehman
EES, University Polytechnic, AMU
April 7, 2021
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2. Unit-III: Electrical Machines
Principle & working of an Ideal Transformer, Transformer Construction, Transformer
losses & efficiency, applications and simple numerical examples. Construction of
Induction Motor, Principle of Induction Motor, illustrative numerical examples, Single
Phase Induction Motor, Applications. Principle of DC machines, Construction of DC
machines, EMF equation, applications.
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3. Transformer: Expectation vs Reality
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4. Transformer: Expectation vs Reality
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5. Introduction to Transformer
The transformer is probably one of the most useful electrical devices ever
invented.
It can change the magnitude of AC (voltage or current) from one value to
another. This useful property of transformer is mainly responsible for the
widespread use of AC rather than DC.
Transformers are static machines i. e. they have no moving parts, are rugged
and durable in construction, thus require very less maintenance.
They also have a very high efficiency.
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6. Principle & Working of a Transformer
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7. Principle...Contd
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8. Principle...Contd
Fig. Transformer Core Laminations
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9. Principle...Contd
A transformer essentially consists of two windings called primary and sec-
ondary (denoted by 1 and 2 in above figure).
These windings are wound on a laminated soft iron core. The windings form
two electric circuits that are separated by a magnetic circuit of iron core.
Primary winding is supplied by an alternating voltage v1, due to which an al-
ternating current i1 flows in it. This alternating current produces an alternating
flux φ as shown in the figure.
This flux links the turns N1 of the primary winding and induces an EMF e1 in it
by self induction.
For simplification let us assume that the transformer is ideal so that:
1. The resistance of each winding is negligible, i.e. no electric losses.
2. There are no magnetic losses.
3. The core of the transformer is highly permeable, and that the whole flux is setup
within the core only
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10. Principle...Contd
Then, we can say that all the flux produced by primary also links the turns N2 of
the secondary winding and an EMF e2 is induced in the secondary by mutual
induction.
If a load resistance RL is connected across the secondary winding terminals,
then an alternating current i2 will flow through it and a voltage v2 will appear
across the load terminals, and power will be delivered to the load.
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11. Remarks
(i) From Faraday’s law of EMI, we know that induced EMF is given by:
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12. Remarks
(i) From Faraday’s law of EMI, we know that induced EMF is given by:
e1 = −N1
∆φ
∆t
e2 = −N2
∆φ
∆t
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13. Remarks
(i) From Faraday’s law of EMI, we know that induced EMF is given by:
e1 = −N1
∆φ
∆t
e2 = −N2
∆φ
∆t
Hence,
e1
e2
=
N1
N2
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14. Remarks
(i) From Faraday’s law of EMI, we know that induced EMF is given by:
e1 = −N1
∆φ
∆t
e2 = −N2
∆φ
∆t
Hence,
e1
e2
=
N1
N2
But, since we are assuming negligible resistance of windings, therefore,
e1 = v1 and e2 = v2, thus we can write
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15. Remarks
(i) From Faraday’s law of EMI, we know that induced EMF is given by:
e1 = −N1
∆φ
∆t
e2 = −N2
∆φ
∆t
Hence,
e1
e2
=
N1
N2
But, since we are assuming negligible resistance of windings, therefore,
e1 = v1 and e2 = v2, thus we can write
e1
e2
=
v1
v2
=
N1
N2
= a
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16. Remarks
(i) From Faraday’s law of EMI, we know that induced EMF is given by:
e1 = −N1
∆φ
∆t
e2 = −N2
∆φ
∆t
Hence,
e1
e2
=
N1
N2
But, since we are assuming negligible resistance of windings, therefore,
e1 = v1 and e2 = v2, thus we can write
e1
e2
=
v1
v2
=
N1
N2
= a
(ii) The above ratio is known as voltage transformation ratio and is denoted by a.
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17. Remarks...contd
(iii) The above equation expressed in terms of RMS values becomes
V1
V2
=
N1
N2
= a
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18. Remarks...contd
(iii) The above equation expressed in terms of RMS values becomes
V1
V2
=
N1
N2
= a
(iv) If N1 > N2, a > 1, V2 < V1 → step-down transformer.
If N1 < N2, a < 1, V2 > V1 → step-up transformer.
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19. Remarks...contd
(iii) The above equation expressed in terms of RMS values becomes
V1
V2
=
N1
N2
= a
(iv) If N1 > N2, a > 1, V2 < V1 → step-down transformer.
If N1 < N2, a < 1, V2 > V1 → step-up transformer.
(v) The same transformer can be used as both step-up or step-down depending
upon the connections of source and load
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20. Remarks...contd
(iii) The above equation expressed in terms of RMS values becomes
V1
V2
=
N1
N2
= a
(iv) If N1 > N2, a > 1, V2 < V1 → step-down transformer.
If N1 < N2, a < 1, V2 > V1 → step-up transformer.
(v) The same transformer can be used as both step-up or step-down depending
upon the connections of source and load
(vi) There is no change in frequency in secondary and primary voltage.
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21. Remarks...contd
(iii) The above equation expressed in terms of RMS values becomes
V1
V2
=
N1
N2
= a
(iv) If N1 > N2, a > 1, V2 < V1 → step-down transformer.
If N1 < N2, a < 1, V2 > V1 → step-up transformer.
(v) The same transformer can be used as both step-up or step-down depending
upon the connections of source and load
(vi) There is no change in frequency in secondary and primary voltage.
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22. Remarks...contd
(vii) Secondary power (in volt-amperes) is equal to the primary power (in volt-
amperes), i.e.
V1I1 = V2I2
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23. Remarks...contd
(vii) Secondary power (in volt-amperes) is equal to the primary power (in volt-
amperes), i.e.
V1I1 = V2I2
Hence,
V1
V2
=
I2
I1
= a
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24. Remarks...contd
(vii) Secondary power (in volt-amperes) is equal to the primary power (in volt-
amperes), i.e.
V1I1 = V2I2
Hence,
V1
V2
=
I2
I1
= a
or,
I1
I2
=
1
a
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25. Remarks...contd
(vii) Secondary power (in volt-amperes) is equal to the primary power (in volt-
amperes), i.e.
V1I1 = V2I2
Hence,
V1
V2
=
I2
I1
= a
or,
I1
I2
=
1
a
(viii) A transformer cannot work on DC, as there is no change in current, no change
in flux, thus there will be no electromagnetic induction, and no EMF will be
induced in the primary winding. Further, the resistance of primary winding is
very small and a heavy current will flow through it leading to burning out and
damage.
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26. Transformer Construction
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27. Transformer Construction
Transformers are designed and constructed in such a way so that their character-
istics are as close as possible to an ideal transformer. To achieve this following
features are incorporated in transformer design:
1. High quality silicon steel laminations are used to make the core to minimize the
iron losses.
2. Instead of placing primary on one limb and secondary on the other, it is a
usual practice to wind one-half of each winding on one limb. This ensures tight
coupling between the two windings and maximum flux linkage.
3. Good quality conductor is used for windings, so that resistance is small and
copper loss is minimized, leading to less heating and high efficiency.
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28. Transformer Construction...Contd
Depending upon the manner in which the primary and secondary are wound on the
core, transformers are of two types:
(i) core-type transformer, and
(ii) shell-type transformer
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29. Core Type Transformer
In core type transformer, L-shaped laminations are stacked and joined together
through bolts to form the core, the windings are split into halves and placed on
each vertical side of the core. The low voltage (LV) winding is placed next to the
core and the high voltage (HV) winding is placed around the LV winding to reduce
the insulation. Such windings are called cylindrical windings.
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30. Shell type Transformer
In shell type transformer, E-shaped laminations form the core, and the windings are
arranged on the central limb. The other two limbs act as a low-reluctance flux path.
HV and LV winding subsections are placed alternately to form a sandwich. There-
fore, such winding is also known as sandwich winding. The core type is generally
more suitable for high voltage while the shell-type is generally more suitable for low
voltage applications.
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31. Losses in Transformer
The simplifying assumptions on page 1 are not valid for a real transformer. A real
transformer has certain power losses that are as follows:
1. Copper losses
2. Iron losses
3. Other losses
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32. Losses in Transformer
The simplifying assumptions on page 1 are not valid for a real transformer. A real
transformer has certain power losses that are as follows:
1. Copper losses
2. Iron losses
3. Other losses
These losses appear in the form of heat and produce:
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33. Losses in Transformer
The simplifying assumptions on page 1 are not valid for a real transformer. A real
transformer has certain power losses that are as follows:
1. Copper losses
2. Iron losses
3. Other losses
These losses appear in the form of heat and produce:
(i) an increase in temperature,
(ii) a drop in efficiency.
Let us discuss briefly each type one by one.
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34. 1. Copper Losses, Pc
These losses occur in both the primary and secondary windings due to their
resistances.
Also known as I2R loss.
Depends upon the load current, which is variable, hence also called variable
loss.
Total Copper loss is given by:
Pc = I2
1R1 +I2
2R2
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35. 2. Iron Losses, Pi
These consist of hysteresis and eddy current losses and occur in the trans-
former core due to the alternating flux.
Hysteresis loss is due to reversal of magnetization in the transformer core. This
loss depends upon the frequency and value of flux density in the core.
The iron core is also electrically conductive, therefore the alternating magnetic
flux induces an EMF and then circulating loops of current in it, called eddy
currents, due to electromagnetic induction. These eddy currents cause power
loss and heating of the core.
Since, frequency and flux density are constant for a transformer, these losses
are also called fixed losses.
Total Iron Loss is given by:
Pi = Ph +Pe
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36. Other Losses
These include stray loss and dielectric loss and are usually negligible.
Stray loss occurs due to leakage flux that produces eddy currents and conse-
quent heating in the conductors, tanks.
Dielectric loss occurs in transformer insulation and oil, and is considerable only
in high voltage transformers.
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37. Transformer Efficiency
Efficiency of any machine is defined as ‘the ratio of its output power to its input
power’, which can be expressed either as a per unit (p.u.) or, more commonly,
a percentage (%) value, denoted by the symbol η
η =
Output Power
Input Power
As a machine’s output power can never match its input power, its efficiency will
always be less than 100%
The above expression of efficiency can be re-written as:
η =
Output Power
Output Power + Losses
=
Po
Po +Pc +Pi
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38. Transformer Efficiency...Contd
It can be proved mathematically that, a transformer’s maximum efficiency oc-
curs when the copper losses and iron losses are equal to each other, i. e.
Pc = Pi
Distribution transformers are connected to the load all the time, and another
efficiency in terms of energy is specified for them, known as all-day efficiency.
It is given by
ηall-day =
Energy Output in 24 h
Energy Input in 24 h
=
Energy Output in 24 h
Energy Output in 24 h + Total losses in 24 h
All-day efficiency is also known as commercial efficiency.
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39. Numerical Example 1
A transformer with an output voltage of 4.6 kV is supplied at 230 V. If the sec-
ondary winding has 2000 turns, calculate the number of primary turns.
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40. Numerical Example 1
A transformer with an output voltage of 4.6 kV is supplied at 230 V. If the sec-
ondary winding has 2000 turns, calculate the number of primary turns.
Solution:
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41. Numerical Example 1
A transformer with an output voltage of 4.6 kV is supplied at 230 V. If the sec-
ondary winding has 2000 turns, calculate the number of primary turns.
Solution: V1 = 230 V, V2 = 4.6 kV = 4600 V, N2 = 2000, N1 =?
We know that,
V1
V2
=
N1
N2
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42. Numerical Example 1
A transformer with an output voltage of 4.6 kV is supplied at 230 V. If the sec-
ondary winding has 2000 turns, calculate the number of primary turns.
Solution: V1 = 230 V, V2 = 4.6 kV = 4600 V, N2 = 2000, N1 =?
We know that,
V1
V2
=
N1
N2
230
4600
=
N1
2000
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43. Numerical Example 1
A transformer with an output voltage of 4.6 kV is supplied at 230 V. If the sec-
ondary winding has 2000 turns, calculate the number of primary turns.
Solution: V1 = 230 V, V2 = 4.6 kV = 4600 V, N2 = 2000, N1 =?
We know that,
V1
V2
=
N1
N2
230
4600
=
N1
2000
N1 = 100 turns
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44. Numerical Example 2
The transformation ratio of a transformer is 0.2. If the secondary winding has
1000 turns and voltage is 50 V, find (a) voltage ratio, (b) primary side voltage,
(c) number of turns in the primary winding, & (d) Current ratio
Solution:
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45. Numerical Example 2
The transformation ratio of a transformer is 0.2. If the secondary winding has
1000 turns and voltage is 50 V, find (a) voltage ratio, (b) primary side voltage,
(c) number of turns in the primary winding, & (d) Current ratio
Solution: (a) Voltage ratio = transformation ratio = 0.2
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46. Numerical Example 2
The transformation ratio of a transformer is 0.2. If the secondary winding has
1000 turns and voltage is 50 V, find (a) voltage ratio, (b) primary side voltage,
(c) number of turns in the primary winding, & (d) Current ratio
Solution: (a) Voltage ratio = transformation ratio = 0.2
(b) V1 =?
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47. Numerical Example 2
The transformation ratio of a transformer is 0.2. If the secondary winding has
1000 turns and voltage is 50 V, find (a) voltage ratio, (b) primary side voltage,
(c) number of turns in the primary winding, & (d) Current ratio
Solution: (a) Voltage ratio = transformation ratio = 0.2
(b) V1 =?
V1
V2
= 0.2 = a
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48. Numerical Example 2
The transformation ratio of a transformer is 0.2. If the secondary winding has
1000 turns and voltage is 50 V, find (a) voltage ratio, (b) primary side voltage,
(c) number of turns in the primary winding, & (d) Current ratio
Solution: (a) Voltage ratio = transformation ratio = 0.2
(b) V1 =?
V1
V2
= 0.2 = a
V1
50
= 0.2
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49. Numerical Example 2
The transformation ratio of a transformer is 0.2. If the secondary winding has
1000 turns and voltage is 50 V, find (a) voltage ratio, (b) primary side voltage,
(c) number of turns in the primary winding, & (d) Current ratio
Solution: (a) Voltage ratio = transformation ratio = 0.2
(b) V1 =?
V1
V2
= 0.2 = a
V1
50
= 0.2
V1 = 0.2×50 = 10 V
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50. Numerical Example 2...Contd
(c) N1 =?
N1
N2
= 0.2
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51. Numerical Example 2...Contd
(c) N1 =?
N1
N2
= 0.2
N1
1000
= 0.2
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52. Numerical Example 2...Contd
(c) N1 =?
N1
N2
= 0.2
N1
1000
= 0.2
N1 = 0.2×1000 = 200 turns
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53. Numerical Example 2...Contd
(c) N1 =?
N1
N2
= 0.2
N1
1000
= 0.2
N1 = 0.2×1000 = 200 turns
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54. Numerical Example 2...Contd
(c) N1 =?
N1
N2
= 0.2
N1
1000
= 0.2
N1 = 0.2×1000 = 200 turns
(d)
I1
I2
=
1
a
=
1
0.2
= 5
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55. Numerical Example 3
In a 25 kVA, 2000/200 V transformer the iron and full load copper losses are
350 W and 400 W respectively. Find the efficiency at unity p.f. & full load
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56. Numerical Example 3
In a 25 kVA, 2000/200 V transformer the iron and full load copper losses are
350 W and 400 W respectively. Find the efficiency at unity p.f. & full load
Solution:
Efficiency, η =
Output Power
Output Power + Losses
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57. Numerical Example 3
In a 25 kVA, 2000/200 V transformer the iron and full load copper losses are
350 W and 400 W respectively. Find the efficiency at unity p.f. & full load
Solution:
Efficiency, η =
Output Power
Output Power + Losses
=
25×1000×1
25×1000×1+350+400
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58. Numerical Example 3
In a 25 kVA, 2000/200 V transformer the iron and full load copper losses are
350 W and 400 W respectively. Find the efficiency at unity p.f. & full load
Solution:
Efficiency, η =
Output Power
Output Power + Losses
=
25×1000×1
25×1000×1+350+400
= 0.97 = 97%
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