1. UNIT 3 Transformers

2. Linear Transformer
A transformer is a device that exploits the properties of two magnetically coupled coils.
A transformer is a device that changes AC electric power at one voltage level to AC electric power at another voltage level through the action of a magnetic field. It consists of two or more coils of wire wrapped around a common ferromagnetic core. These coils are (usually) not directly connected. The only connection between the coils is the common magnetic flux present within the core.

3. A transformer is a static electrical machine consisting of two or more
coils coupled through magnetic circuit.
The transformer transfers the energy from one circuit to another at different voltage levels without change in frequency.
A transformer can step up the voltage or can step down the voltage.
If the transformer steps up the voltage, it is called step up transformer.
If the transformer steps down the voltage, it is called step down transformer.
Transformers find wide applications in electric power distribution.
Laminated Iron Core
Primary
winding
Secondary

7. Example 1 Find the following
1- Self-impedance of the primary and secondary circuits
2- Impedance reflected into the primary winding
3- Input impedance seen from the primary terminals of the transformer
4- Input impedance seen from the source

8. Ideal Transformer ~ For ideal Transformer the assumptions are as:
1. The primary and secondary windings have zero resistance.
2. There is no leakage fluxes.
Losses in the core are zero.
As shown in Fig. say the
Number of turns in
Primary winding is N1, and
In secondary winding is N2
on a common magnetic core.
The primary is connected to a
Voltage source V1 and secondary
is loaded with a load ZL. ZL
Laminated Iron Core
Primary
winding
Secondary ~ I1

9. The flux ɸ causes the induced voltage at the Primary as
The induced voltage at the secondary
Thus
(1)
where, K is the transformation ratio. The above relationship clearly
exhibits that for low turns winding the voltage induced is low or for large
number of turns the voltage induced is high.
When the load in the secondary connected, the current i2 in the secondary
causes a mmf F2 = i2N2 and the primary current i1 causes the mmf F1 = i1N1,
which cancels out the mmf in the secondary.
Thus i1N1 = i2N or (2)

10. v1 i1 = v2 i2 Thus from equation(2),
Therefore the instantaneous power into primary is equal to the instantaneous power of the secondary which shows that the ideal transformer is a lossless transformer as assumed in the beginning.
From eqn(1) and (2),
Therefore, or
and

11. If secondary is open circuited, i,e., I2 = 0, Then I1 = 0 .
N1 : N2 Z2 I1 I2 V1
Thus the impedance of the secondary side can be referred to the primary side and the impedance of primary side can be referred to the secondary side. I, e.,
and As
If secondary is open circuited, i,e., I2 = 0,
Then I1 = 0 .
Ideal Transformer S
Example1: For the given Source, Transformer and Load determine the rms values of currents and voltages with the (a) switch S open, (b) switch S closed.

12. Solution: Since and , Therefore and
(a) With the switch S open the secondary current is zero. Therefore the primary current is also zero ( ) and no power is taken from the source.
(b) With the switch S closed the secondary current is
I2rms = V2rms / RL
= 22V / 10Ω = 2.2 A
Then the primary current is

13. Example-2:
In the Figure shown find the
Phasor currents and voltages.
Also find the power delivered to the load.
R1= ZL
Vs=
Vm= 1000V
Solution:
Z2 = ZL= 10 + j20
The Z2 referred (or reflected) to primary side. The impedance seen at the primary side is
Therefore the total impedance seen by the source is
The primary current
10:1

14. Now the secondary current and voltage can be computed as:
Finally the power delivered to the load is

15. Example-3: In the Figure shown below for maximum power transfer from
Rs=4Ω
Vs=
RL=400Ω
source to load, the effective resistance RL` is equal to Rs. Find the turns ratio that would result in maximum power delivered to the load.
Solution: Rs = 40 Ω, and RL = 400 Ω and RL` = (N1/ N2)2 RL
For maximum power transfer RL` = Rs
Therefore (N1 / N2)2 RL = Rs, hence (N1 / N2)2 400 = 4
Therefore (N1 / N2)2 = (4 / 400)=1/ or
= ANS.

16. rms and average relationship:
Form Factor =
Vrms = 1.11 Vav

17. E.M.F. Equation of a Transformerɸ
Let, N1 = Number of turns in primary,
N2 = Number of turns in secondary,
ɸm = Maximum flux in the core in Weber
= Bm A,
f = Frequency of the AC input in Hz.
The flux increases from zero to maximum value in one quarter of the cycle i.e., in (T/4) Sec = (1/ 4f ) Sec. ɸm t t t t t t
Therefore, the
Rate of change of flux = Change in Flux / Time
= Wb/sec
The rate of change of flux per turn is the induced emf in Volts.
Therefore the average emf induced per turn Volt
Or the rms emf induced per turn = 1.11x 4f ɸm = (4.44xf ɸm) volt
Therefore e.m.f. induced in primary, e1 = 4.44fN1ɸm = 4.44fN1BmA volt
And in secondary induced e.m.f. , e2 = 4.44fN2ɸm = 4.44fN2BmA volt

18. Real Transformers
The equivalent circuit of a real transformer is shown as follows:
Leakage
Inductance
Coil Resistances
Leakage
Inductance
L R1
R L2
Magnetizing
Inductance Lm Rc
Core Loss Resistance
R1 and R2 are the resistance of the wires of the primary and secondary windings respectively. L1 and L2 are the leakage inductance caused due to Leakage flux in the primary and secondary coils. The Lm is magnetizing inductance of the core and Rc is of the core accounts for power dissipated in the core due to hysteresis and eddy currents.

19. L R1
α2R2 α2L2 Lm Rc
α = ( N1 / N2)
L R1
α2R2 α2L2 Lm Rc
α = ( N1 / N2)
(L1+ α2L2) (R1+ α2R2) Lm Rc
α = ( N1 / N2)

20. Transformer Regulation and Efficiency
Because of the non-idealities such as L1, L2, R1 and R2, the voltage delivered to the load side varies with the load current. Usually, this is an undesirable effect. The regulation of a transformer is defined as
Where, Vno-Load is the rms voltage across the load terminals for an open circuit load and VLoad is the rms voltage across the actual load.
Ideally, the percentage regulation is to be zero.
Where PLoad is the power delivered to the load, PLoss is the power dissipated in the transformer, and Pin is the power delivered by the source to the transformer primary terminals.

21. Table: Circuit values of a 60 Hz, 20KVA, 2400V / 240V Transformer
Element Name
Symbol
Ideal
Real
Primary Resistance R1
3 Ω
Secondary Resistance R2
0.03 Ω
Primary leakage reactance
X1 = ωL1
6.5 Ω
Secondary leakage Reactance
X2 = ωL2
0.07 Ω
Magnetizing reactance
Xm = ωLm ∞
15KΩ
Core loss resistance Rc
100K Ω

22. Example:Find the percentage regulation and power efficiency for the transformer
of the Table on the last page for a rated load having a lagging power factor of 0.8.
Solution: VLOAD = V rms
For the rated load of 20KVA, the load current is
I2 = 20KVA / 240V = 83.33A rms
The given load power factor is
Cosθ =0.8, hence θ =
Thus the phasor load current is
I2 = A rms
The primary current is
Now the secondary voltage
V2 =VLOAD + (R2 + jX2) I2
= ( j0.07)x
= x
= = e j29.93
= ( Cos j x Sin29.930)
= x( jx0. 499) = j 3.166
V2 = j V rms

23. Therefore the primary voltage

24. The power loss in the transformer
Power delivered to the load is given by
PLoad = VLoad x I2 x Cosθ = 20KVAx0.8 =16KW
Input Power
PIN = PLoad + PLoss = 16KW KW = KW

25. Power efficiency =
For no load I2 = I1 = 0, then V1 = VS = 2508
Therefore VNo-Load = V2 = V1(N2 / N1) = 2508/10 = 250.8V rms
Hence, the percentage regulation is =

26. Condition for maximum efficiency
R01 is sum of primary resistances, and secondary resistances transferred to the primary side,
R02 is sum of secondary resistances, and the primary resistances transferred to the secondary side,
I1 is the primary current and I2 is the secondary current,
Then the
Copper loss PCu = I12R or PCu = I22R02
And the Iron loss PIron = Hysteresis loss + Eddy current loss= Ph + Pe
Input power in primary Pi = V1I1Cosɸ1

27. Hence, the above equation becomes
For ɳ to be maximum,
Hence, the above equation becomes
Or or
Therefore for maximum efficiency of the transformer
Copper Loss = Iron Loss
Thus, the output current corresponding to maximum efficiency is
It is this value of the load current for which the efficiency of the transformer is Maximum.

28. Efficiency
Max. Efficiency
Total Loss
Losses
% Efficiency
Cu Loss
Iron Loss
Load Power

29. (b) At half load and 0.8 pf, Efficiency = ?
Example-1: A 200/400V, 10KVA, 50 Hz, single phase transformer has at full load, a copper loss of 120W. (a) If it has an efficiency of 98% at full load unity power factor, determine the iron loss. (b) What would be the efficiency of the transformer at half load 0.8 power factor lagging?
Solution:
(a)Output at full load = V I x power factor = V I Cosɸ = 10 x 1 = 10 KW
η = ( Output / Input ) =0.98
Therefore Input = Output / η = 10 / 0.98 = KW
Total full load losses = Input – Output = – 10 = KW 204 W
Total Losses Copper Loss + Iron Loss =120W + Iron Loss = 204 W
Therefore, Iron Loss = 204 – 120 = 84 W
(b) At half load and 0.8 pf, Efficiency = ?
Copper Loss = (1/2)2 x PCUFL =(1/2)2 x 120 = 30W, Iron Loss = 84W Hence the total loss = = 114W = KW, Thus Output = (10/2)x0.8 = 4KW Therefore Input = Output + Losses= 4.114KW, η =Output / Input = 4/4.114 =0.97 = 97 %

30. Three Phase Transformers
Symbolic Representation of 3-Phase Transformer
Construction of 3-Phase Transformer

31. Three Phase Transformer Connections
The windings of primary and secondary can be connected in either a wye (Y) or delta (Δ)
This provides a total of 4 possible connections for 3-phase transformer (if Neutral is not grounded):
(a) Wye-wye Y-Y
(b) Wye-delta Y-Δ
(c) Delta-wye Δ-Y
(d) Delta-Delta Δ-Δ

32. Analysis of 3-Phase Transformers:
To analyze a 3-phase transformer, each single transformer in the bank should be analyzed.
Any single phase in bank behaves exactly like
1-phase transformer just studied.
Impedance, V.R., efficiency, & similar calculations for 3 ph. are done on per phase basis, using the same technique already used in single phase Transformer.
The applications, advantages and disadvantages of each type of three phase connections will be discussed next.

33. WYE-WYE Connection of the 3-Phase Transformers:
In Y-Y connection, primary voltage on each phase is VφP = VLP/√3
Primary phase voltage is related to secondary phase voltage by turns ratio of transformer
Phase voltage of secondary is related to Line voltage of secondary by VLS=√3 VφS
Overall the voltage ratio of transformer is:

34. RP YP BP RS YS BS RP YP BP RS YS BS

35. Two serious concerns on Y-Y connection
1- If loads on transformer cct. are unbalanced,
voltages on phases of transformer severely
unbalanced, also source is loaded in an
unbalanced form.
2- Third harmonic voltages can be large (there
is no path for passage of third harmonic
current)
Both concerns on unbalance load condition & large 3rd Harmonic voltages can be rectified as follows:

36. Solidly grounding the neutrals of windings
specially primary winding, this connection provide a path for 3rd harmonic current flow, produced and do not let build up of large 3rd voltages. Also provides a return path for any current imbalances in load.
Adding a third winding (tertiary) connected in Δ
(a) 3rd harmonic components of voltage in Δ will add up, causing a circulating current flow within winding.
(b) tertiary winding should be large enough to handle circulating currents (normally 1/3 of power rating of two main windings).
One of these corrective techniques should be employed with Y-Y, however normally very few transformer with this type of connection is employed (others can do the same job).

37. WYE-DELTA Connection 3-Phase Transformers:
VLP=√3 VφP, while : VLS= VφS
Voltage ratio of each phase : VφP/ VφS = a
VLP/ VLS= √3 VφP/ VφS= √3 a  Y-Δ
Y-Δ doesn’t have shortcomings of Y-Y regarding generation of third harmonic voltage since the Δ provide a circulating path for 3rd Harmonic.
Y-Δ is more stable w.r.t. unbalanced loads, since Δ partially redistributes any imbalance that occurs.
This configuration causes secondary voltage to be shifted 30◦ relative to primary voltage.
If secondary of this transformer should be paralleled with secondary of another transformer without phase shift, there would be a problem.

38. RS YS BS RP YP BP RS YS BS RP YP BP

39. Y-Δ Connection of three Phase Transformers:
The phase angles of secondaries must be equal if they are to be paralleled, it means that direction of phase shifts also should be the same.
In figure shown here, secondary lags primary if abc phase sequence applied.
However secondary leads primary when acb phase sequence applied.

40. Δ-Y Connection of the 3-Phase Transformers:
In Δ-Y primary line voltage is equal to primary phase voltage VLP=VφP, in secondary VLS=√3VφS
Line to line voltage ratio :
VLP/ VLS = VφP / [√3 VφS ] = a/√3  Δ-Y
This connection has the same advantages & phase shifts as Y- Δ
And Secondary voltage lags primary voltage by 30◦ with abc phase sequence.

41. RP BP YP RS YS BS YP RP BP RS YS BS

42. Δ-Δ Connection of the 3-Phase Transformers:
In Δ-Δ connection
VLP= VφP and VLS= VφS
Voltage ratio :
VLP/VLS= VφP / VφS =a  Δ-Δ
This configuration has no phase shift and there is no concern about unbalanced loads or harmonics.

43. RP RS YP YS BP BS RP YP BP RS YS BS

44. Example: The primary line voltage of a transformer is 33KV and the
transformation ratio is α = 100 : 1. Find the VLP, VϕP, VSP, VϕS ,
if the transformer is connected in
(i) Y : Y mode
(ii) Y : Δ mode,
(iii) Δ : Y mode,
(iv) Δ : Y mode.

45. Electric Substation: