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1
LECTURE NOTES FOR
EE 162
ELECTROMECHANICAL ENERGY
CONVERSION & TRANSFORMERS
COURSE LECTURER: DR. PHILIP YAW OKYERE
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2
Course Outline
EE 162 ELECTRICAL MACHINES I (3 0 3)
Principles of Electromechanical Energy Conversion:
Force and torque as rate of change of energy with position. Basic Transducers: Single
Excitation, Alignment Force and Torque; Double Excitation, Alignment and Interaction
Forces and torque
Transformers:
Construction; Basic theory; Phasor Diagram; Equivalent Circuits; No-load and Short-
Circuit Tests; Voltage Regulation; Efficiency; Cooling methods; Polarity; Polyphase
transformer Connections; Per-Unit Calculation; Parallel Operation of Transformers;
Auto transformers; Tap-Changing transformers; Instrument Transformers.
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3
CHAPTER ONE Transformers
1. Introduction
The transformer transfers electrical energy from one circuit to another via the medium of
a pulsating magnetic field that links both circuits. The widespread development of ac
power systems is principally due to the transformer. It enables us to produce and transmit
power at economical voltages and to distribute it safely in factories and homes. In low-
power low-current electronic and control circuits, it is used to provide impedance
matching between a source and its load for maximum power transfer, to isolate one
circuit from another, to isolate direct current while maintaining ac continuity between
two circuits and to provide reduced ac voltages and currents for protection, metering,
instrumentation and control.
2. Principle of operation of a transformer
The transformer is a straight-forward application of Faradays Law of Electromagnetic Induction. Consider the general arrangement of a single-phase transformer shown in Fig.
1. An alternating voltage applied to coil 1, causes an alternating current to flow in the coil
and this current produces an alternating flux in the iron core. A portion of the total flux
links the second coil. The alternating flux induces a voltage in the second coil. If a load
should be connected to the coil, this voltage would drive a current through it. Energy
would then be transferred through the medium of magnetic field from coil 1 to coil 2. The
combination of the two coils is called a transformer. The coil connected to the source is
called the primary winding (or the primary) and the one connected to the load is called
the secondary winding (or the secondary).
V1
I1
E1 N1 N2 E2
I2
Load
Coil 2Coil 1
mSoft iron
Fig. 1 An elementary transformer
3. Polarity and terminal markings of a transformer Voltage E1 is induced in coil 1 and voltage E2 in coil 2. These voltages are in phase.
Suppose at any given instant when the primary terminal 1 is positive with respect to
primary terminal 2, the secondary terminal 3 is also positive with respect to secondary
terminal 4 (See Fig. 2). Then terminals 1 and 3 are said to have the same polarity. To
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4
indicate that their polarities are the same, a dot is placed beside primary terminal 1 and
secondary terminal 3. Alternatively, letters of the same suffix, A1 (for the high-voltage
winding) and a1 (for the low-voltage winding) say can be used. Current I1 entering coil 1
through the dotted terminal 1 and current I2 entering coil 2 through the dotted terminal 3
create fluxes in the same direction.
1
2
3
4
A1
A2
a1
a2
1
2
3
4
A1
A2
a1
a2
(a) Magnetic Circuit (b) Electric Circuit
Fig. 2 Polarity and terminal markings
4.0 Ideal transformer
An ideal transformer has no losses, no leakage flux and its core is infinitely permeable.
An ideal transformer is shown in Fig. 3. The mutual flux m is confined to the iron.
V1
I1
E1 N1 N2 E2
I2
m
V2
S
(I1 = 0 on no load)
Fig. 3 An ideal transformer
4.1 Properties of an ideal transformer
(a) Emf equation and voltage ratio (see Fig. 3): With the primary connected to an ac
source V1, an alternating flux m is produced in the core. Let the flux be expressed as
sinmax weberstm (1)
The induced emf e1 as indicated in the figure is given by
tNtdt
dNN
dt
de m cossin max1max111
Or
90sinmax11 tNe (2)
Hence
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5
max1max1
12
2
2
fNNE
Or
max11 44.4 fNE (3)
Similarly
max2
max2max2
2 44.42
2
2
fN
fNNE
(4)
From (3) and (4), we obtain
2
1
2
1
N
N
E
E (5)
The ratio a = N1/N2 is called the turns ratio. A step-up transformer has a < 1 and a step-
down transformer has a > 1. In an ideal transformer, the applied voltage V1 and the
induced voltage E1 must be identical. Hence we may write
11 EV (6.a)
And
max11 44.4 fNV (6.b)
Equation (6.b) indicates that for a given frequency, number of turns and voltage, the peak
flux max must remain constant.
(b) Current ratio and power equation: On no load I1 = 0. Now if a load is connected
across the secondary terminals (i.e. switch S is closed) current I2 flows through the load.
This current produces mmf N2I2 which if it acted alone would by Lenzs law, cause the mutual flux to reduce. Since when V1 is fixed the flux max is also fixed, the primary develops mmf N1I1 which is such that
2211 ININ (7.a)
Or
aII 21 (7.b)
In ideal transformer the secondary voltage
22 EV (8)
remains constant since E2 is fixed when the max is fixed. It can be deduced from above equations that for an ideal transformer
2211 IVIV (9)
That is there are no reactive and active losses in an ideal transformer.
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6
(c) Phasor diagram of an ideal transformer (Fig. 4)
E1
E2
m
E1, V1
E2, V2 I1
I2
m
(a) No load (b) Load is resistive inductive
Fig. 4 Phasor diagram of an ideal transformer
Example 1
An ideal transformer having 90 turns on the primary and 2250 turns on the secondary is
connected to a 200-V, 50-Hz source. The load across the secondary draws a current of 2
A at a power factor of 0.80 lagging. Calculate (a) the rms value of the primary current (b)
the flux linked by the secondary winding (c) Draw the phasor diagram.
Solution
AIIININa 502225090)( 112211
WbfN
Vb 01.0
905044.4
200
44.4)(
1
1max
VEN
NEc 5000)( 1
1
22
36.9(0.8)cos is and between angle phase The -122 IV
V2=E2 = 5000 V
V1=E1= 200 V I2 = 2 A
I1 = 50 A
36.9
m
Example 2
A 200-kVA, 6600-V / 400-V, 50-Hz 1-ph transformer has 80 turns on the secondary.
Calculate (a) the approximate values of the primary and secondary currents (b) the
approximate number of primary turns and (c) the maximum value of the flux.
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7
Solution
Aa 3.306600)1000200(current primary load-Full)(
A500400)1000200(current secondary load-full and
1320400)660080()( 1 Nb
)]4([0225.0508044.4
400)( max fromWbc
(d) Impedance ratio (see Fig. 5): The impedance seen by the source
ZaIEaaIaEIEIVZe2
22
2
221111
2ZaZe (10)
V1 E1 E2 Z
I2I1
a : 1
Fig. 5 impedance ratio
(e)Equivalent circuits of an ideal transformer: From (10), we can represent the
transformer in Fig. 5 by equivalent circuit shown in Fig. 6.a. We may also write
aVaEEZI 1122
Or
21 ZI
a
V (11)
and then represent the transformer by an equivalent circuit shown in Fig. 6.b
V1 a
2Z
I1 I2
a
V1Z
(a) Circuit referred to the primary side (b) circuit referred to the secondary side
Fig. 6 Equivalent circuits of ideal transformer
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8
Example 3 Calculate the voltage V and current I in the circuit of Fig. 7, knowing that the ideal
transformer has a primary to secondary turns of 1:100 (i.e. a = 1/100).
I5
V1=10V 40k
20k
V
Fig. 7 See Ex. 3
Solution:
We shall shift all impedances to the primary side
10V
5 2
4 aVV
Va
VVvoltageactualTheVIRV
AZ
VI
XXRZ
e
CLe
8008100.842
25
10
534
1
2222
5.0 Practical single-phase transformer
The windings of a practical transformer have both resistance and leakage inductance. The
core is also imperfect: it has a core loss and finite permeability. The core loss has two
components: hysteresis loss and eddy current loss.
Hysteresis loss
When ferromagnetic material is subjected to alternating magnetization, the energy put
into the magnetic field when the flux is increasing is not completely given back when the
flux dies away but a certain portion is wasted. The area of the hysteresis loop gives the
value of the energy loss taking place for each complete cycle of magnetization. The loop
area is found experimentally to vary as xBmax up to moderate values of flux density (1.0
2.0 Wb/m2)
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9
The index x is named after Steinmetz and is about 1.6 though it may be higher. In
practice the hysteresis loss, for simplicity, is often taken as proportional to 2maxB
If f = frequency of magnetization, the power wasted in magnetic hysteresis 3W/mHh fWP Or
32
max
6.1
max W/mfBfBPh (12)
The hysteresis constant depends upon the magnetic material.
Eddy-current loss
When a changing magnetic flux permeates any mass of metal, eddy currents are induced.
The eddy currents cause the metal to heat. Eddy currents are also induced in a core
revolving in a stationary or constant magnetic field. If iron or metal subjected to
alternating magnetization is built up of laminations insulated from each other, the eddy
current loss is reduced.
The eddy current loss in a laminated core is given by 32
max
22 W/m/)( BtkfPe (13.a)
Where
t = thickness of the laminations
= resistivity of the material k = a constant depends on the waveform of the alternating flux
For a given thickness and waveform, (13.a) reduces to 32
max
2 / mWBfkPe (13.b)
Example 4
In a transformer core of volume 0.16 m3 the total iron loss was found to be 2170 W at 50
Hz. The hysteresis loop of the core material, taken to the same maximum flux density,
had an area of 9.0 cm2 when drawn to scales of 1 cm = 0.1 Wb/m
2 and 1 cm = 250 AT/m.
Calculate the total iron loss in the transformer core if it is energized to the same
maximum flux density, but at a frequency of 60 Hz.
Solution
Hysteresis loss = area x scale factors = 9 x 0.1 x 250 = 225 J/m3
At 50 Hz, hysteresis loss = 225 x 50 x 0.16 = 1800 W
Therefore Eddy-current loss = 2170 1800 = 370 W At 60 Hz, hysteresis loss = 1800 x (60/50) = 2160 W and
Eddy-current loss = 370 x (60/50)2 = 533 W
Therefore total iron loss = 2160 + 533 = 2693 W
Example 5
For the same maximum flux density, the total core loss in a core is 500 W at 25 Hz and
1400 W at 50 Hz. Find the hysteresis and eddy-current losses for both frequencies.
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10
Solution
Since Bmax is constant, the losses can have the following forms:
AfPh ,2BfPe and
2BfAfPPP ehc
For a frequency of 25 Hz, the core loss = 500 = A(25) + B(25)2
For a frequency of 50 Hz, the core loss = 1400 = A(50) + B(50)2
Solving the two equations, we obtain A = 12, B = 0.32 and the individual losses
Ph = 300 W, Pe = 200 W at 25 Hz and Ph = 600 W, Pe = 800 W at 50 Hz
5.1 Phasor diagram on no load
To furnish the power loss in the core (core loss), a small current must be drawn from the
source. This current Ip must be in phase with induced voltage E1. Also to create the
mutual flux m, a magnetizing current Im in phase with m and lagging 90 behind E1 must be drawn to produce the required mmf.
Im
IoIp
E2
V1=E1
Fig. 8 Phasor diagram for practical transformer on no load
We note that
(i) the no-load current Io taken by the primary is the phasor sum of Ip and Im
(ii) the difference between the value of the applied voltage V1 and that of the induced emf
E1 is only about 0.05% when the transformer is on no load so the two can be considered
to be equal
(iii) Ip is very small compared with Im. Therefore the no-load power factor is very low.
Example 4
A 1-ph transformer has 480 turns on the primary and 90 turns on the secondary. The
mean length of the flux path in the iron core is 1.8 m and the joints are equivalent to an
air gap of 0.1 mm. If the peak value of the flux density is to be 1.1 T when a voltage of
2200 V at 50 Hz is applied to the primary, find (a) the cross-sectional area of the core (b)
the secondary voltage on no load (c) the primary current and power factor on no load.
Assume the value of the magnetic field strength for 1.1 T in iron to be 400 A/m, the
corresponding iron loss to be 1.7 W/kg at 50 Hz and the density of the iron to be 7800
kg/m3.
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11
Solution
Wba 0206.05048044.42200)( maxmax
20187.01.10206.0 core of area sectional-Cross m
Vb 5.412480
902200 load noon voltageSecondary )(
Ammfc 7208.1400 coreiron for the required ofvaluePeak)(
g
o
g
gg lB
lHAmmf
5.870001.0104
1.1 gapair for the ofvaluePeak
7
A5.8075.87720 produce torequired mmf totalofvaluePeak max
AN
mmftotalofvaluePeak682.1
480
5.807current gmagnetizin of Peak value
1
AIm 19.12
682.1 sinusoidal be it to assuming valuerms Its
30337.00187.08.1iron of Volume m kg26378000337.0density volumeiron of Mass
W4477.1263 lossiron And
AI p 203.02200447
AIo 21.1203.019.1current load No22
laggingNo 168.021.1203.0II factor power load op
5.2 Mutual and leakage fluxes in a transformer (see Fig. 9) The actual flux linking a coil can be considered to have two components: the mutual flux
m linking both coils and the leakage flux l. The reluctance of the paths of the leakage flux is almost entirely due to the long air paths and is therefore practically constant.
Consequently, the value of the leakage flux is proportional to the current in the coil.
V1
I1
Ep l1 l2 Es V2
I2
m
Fig. 9 Ttransformer possessing two leakage fluxes and a mutual flux
The secondary induced voltage Es is composed of two parts:
(a) a voltage El2 induced by leakage flux l2 given by El2 = 4.44fN2l2,max (b) a voltage E2 induced by mutual flux m given by E2 = 4.44fN2max
-
12
Similarly, the primary induced voltage is composed of El1 =4.44fN1l1,max and El1 = 4.44fN1max
We can segregate the four voltages E1, E2, El1 and El2 by rearranging the transformer
circuit as shown in Fig. 10. The rearrangement of the transformer circuit makes it clear
that El1 and El2 are voltage drops across reactances. These reactances called leakage
reactances are given by 222111 and IEXIEX ll
V1
I1
E1 E2
I2
m
Es
El2 Ep
El1
l1 l2
Fig. 10 Separating out the various induced voltages
5.3 Equivalent circuit of a practical transformer
The behaviour of a practical transformer may be conveniently considered by assuming it
to be equivalent to an ideal transformer and then allowing for the imperfections of the
actual transformer by means of additional circuits or impedances inserted between the
supply and the primary winding and between the secondary winding and the load. The
complete equivalent circuit of this transformer is shown in Fig. 11. R1 and R2 are
resistances of the primary and secondary windings. X1 and X2 are the leakage reactances.
The reactance Xm is such that it takes a reactive current Im (i.e. the magnetizing current)
of the actual transformer. The core loss is accounted for by the resistor Rm which takes
the component Ip of the primary current.
R1X1
Rm
Ip
Io
Im
Xm E1 E2
I21
I2X2 R2
Z
Ideal
transformer
V1 V2
Fig. 11 Complete equivalent circuit of a practical transformer
Ideal transformer equations still apply: 2212 INNI and 2121 NNEE
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13
5.4 Equivalent circuit referred to the primary side (Fig. 12a)
R1X1
Rm
Ip
Io
Im
Xm E1
I21 X2
1
R21
Z1 V2
1
Fig. 12.a.Exact equivalent circuit referred to the primary side
Let 21 NNa (14.a)
Then
2
2
2
2
212 )( XaXNNX , 22
2
2
212 )( RaRNNR , ZaZNNZ22
21 )( (14.b)
22212 )( aVVNNV
, aII 22
(14.c)
It is worth noting that for a practical transformer, 1212 XXandRR
5.5 Equivalent circuit referred to the secondary side
R11 X1
1
Rm1
Ip1
Io1
Im1
Xm1
I2 X2 R2
Z V2a
1V
I11
Fig. 12.b.Exact equivalent circuit referred to the secondary side
etcaRRNNR ,)( 2112
121
(15.a)
etcaIIorININ pppp
12 (15.b)
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14
5.5 Approximate equivalent circuits
The exact equivalent circuit of the transformer is too exact for most practical
applications. Consequently, we can simplify it to make calculations easier. Approximate
equivalent circuits commonly used for power transformer calculations are given in Fig.
13
V1
I1
Rm
Ip
Io
Im
Xm
I21 Xe1=X1+X2
1
Z1 V2
1
Re1=R1+R21
(a) Voltage drop in the primary leakage impedance due to exciting or
no load current Io is neglected.
V1
I1 = I21 Re1
Xe1
Z1
V21
(b) Exciting current is neglected entirely. Note that primary rated or
full load current is at least 20 times larger than Io. This circuit may
be used when I1 = 1.15xfull load current
V1
I1 = I21 Xe1
Z1
V21
(c) For transformers above 500 kVA, Xe is at least 5 times
greater than Re. This circuit can be used to calculate
voltage regulation of such transformers.
Fig.13 Approximate equivalent circuits
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15
5.5 The complete phasor diagrams for loaded conditions
They are shown in Fig. 15
Fig. 15 Complete phasor diagrams for loaded conditions (drawn in two parts)
5.6 Rating of transformers To keep the transformer temperature at an acceptable level, limits are set to both the
applied voltage (this determines the iron loss at a given frequency) and the current drawn
by the load (this determines copper loss). The limits determine the rated voltage and rated
current of transformers.
The power rating of transformer Srated = rated voltage x rated current can be expressed in
VA, kVA or MVA depending on the size of the transformer. The rated kVA (i.e. the rated
power), frequency and voltage are always shown on the name plate. In large
transformers, the corresponding currents are also shown. We note that
Rated kVA = V1r x I1fl x10-3
= V2r x I2fl x10-3
= V2n xI2fl x10-3
=E2n x I2fl x10-3
Where
V1r = rated primary voltage
I1fl = rated primary current = primary full load current
V2r = rated secondary voltage = V2n (no load secondary voltage corresponding to the rated
primary voltage) = E2n (no load induced secondary voltage corresponding to primary
rated voltage)
I2fl = rated secondary current = secondary full load current
5.7 The turns ratio
It is given by
21 EEa
rVE 11 on no load because ZI 1o is very small. Since V E 22 on no load
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16
rrVVa
21 (16.a)
And aII flfl 21 (16.b)
Example 7
A transformer is rated 10 kVA, 2400 / 240 V, 60 Hz. The parameters for the approximate
equivalent circuit of Fig. 14.a are Rm = 80 k, Xm = 35 k, Re1 = 8.4 and Xe1 = 13.7 . Determine the voltage to be applied to the primary to obtain the rated current in the
secondary when the secondary terminal voltage is 240 V. What is the input power factor?
The load power factor is 0.8 lagging.
Solution
AIaII flfl 17.4240010000122
The power factor angle 9.368.0cos 1 If we choose the load current I2 as the reference phasor, then 9.362402V
volts144019209.362400 ja
22 VV
The voltage across the equivalent leakage impedance is
volts5735017.47.134.8 jjj 2e1IZ
The primary voltage required
voltsjjj 4.37246214971955 573514401920 2e121 IZVV
mAjk
j
Rm7125.184375.24
80
14971955
1p
VI
mAjkj
j
jX m8571.557714.42
35
14971955
1m
VI
0558.00427.00187.00244.0017.4 jjj mp21 IIII Aj 50.0237.40371.02371.4
9.3750.04.37 and between angle Phase IV 11 lagging79.09.37cosfactor power Input
Example 8
A 1-ph transformer operates from a 230-V supply. It has an equivalent resistance of 0.1 and an equivalent leakage reactance of 0.5 referred to the primary. The secondary is connected to a coil having a resistance of 200 and a reactance of 100 . Calculate the secondary terminal voltage. The secondary winding has four times as many turns as the
primary.
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17
Solution
Refer to approximate equivalent circuit of Fig. 14.b
25.65.12100200,100200,4
1 2
2
1 jjajN
Na ZZ
75.66.1225.65.125.01.0impedanceTotal jjj ZZe1
25.65.1275.66.12
0230
75.66.12
0230j
jand
j
22 VI
voltsj
j8719.224
2941.14
9754.13230
75.66.12
25.65.12230
2V
voltsa
VV 8998719.224422
5.8 Definition of per-unit impedances The leakage impedances Z1 and Z2 on the primary and secondary side are expressed in
per unit as follows:
2
1
1
1
1
1
1
11
r
rated
r
fl
base V
SZ
V
IZ
Z
ZZ (17.a)
2
2
2
2
2
2
2
22
r
rated
r
fl
base V
SZ
V
IZ
Z
ZZ (17.b)
rated
rbase
rated
rbase
S
VZand
S
VZ
22
2
21
1
where
(17.c)
The impedances are said to be expressed in per unit with reference to the bases V1r, Srated
in the case of Z1 and V2r, Srated in the case of Z2. The total impedance of the transformer in
per unit . 21 ZZ
Example 5: A single-phase transformer that is rated 3000 kVA, 69 kV / 4.16 kV, 60 Hz
has an impedance of 8 percent. Calculate the total impedance of the transformer referred
to (a) the primary side (b) the secondary side
Solution
1587
103000
1069)(
3
622
1
rated
rbase
S
VZa
127158708.0 111 baseee ZZZ
7685.5103000
1016.4)(
3
6222
2
rated
r
baseS
VZb
46.07685.508.0 222 baseee ZZZ
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18
Alternatively
46.0127
69
16.42
1
2
1
22 ee Z
N
NZ
5.9 Voltage regulation
With the primary voltage maintained constant, the secondary terminal voltage at no load
differs from the secondary voltage under load. The voltage regulation or simply
regulation of a transformer is the change in secondary voltage which occurs when the
rated kVA output at a specified power factor is reduced to zero, with the primary voltage
maintained constant. It is usually expressed as a percentage (called percentage regulation)
or a fraction of the rated no-load terminal voltage (in per unit).
The equivalent circuit given in Fig. 14.b is used to calculate voltage regulation. The
circuit may be either referred to the primary or the secondary side. The circuit in general
form can thus be represented as shown in Fig. 16.
If the circuit is referred to the primary side, then
1121 ,, eeee XXandRRVVVE
eReX
Load VE
Fig. 16 Circuit for computing voltage regulation
Voltage regulation puV
VV
E
VE
1
21
(18.a)
If the circuit is referred to the secondary side, then
2221
2 ,, eeeen XXandRRVVa
VVE
Voltage regulation puV
VV
n
n
2
22 (18.b)
It can be shown thatn
n
V
VV
V
VV
2
22
1
21
(19)
And nVV 22 1 (20)
In general, let the load current be I lagging behind the load voltage V by . Then taking
the load voltage as the reference phasor, we can write
-
19
sincossincos
sincos01
eeee
ee
IRIXjIXIRV
jIIjXRjV
E
And hence
22 sincossincos eeee IRIXIXIRVE E
The second term under the root is usually negligible except at low leading power factors.
Considering the first term only gives
sincos ee IXIRVE
Or sincos ee IXIRVE (21)
The angle is negative when current is leading and positive when current is lagging.
The voltage regulation is maximum when = where
e
e
R
X1tan
0cossin
ee IXIRVE
d
dFrom
Hence ee
ee
e
ee
e
ee ZI
Z
XRI
Z
XIX
Z
RIRVE
22
max (22.a)
Where22
eee XRZ (22.b)
From (21)
sincos
E
IX
I
I
E
IR
I
I
E
VE
fl
fl
fl
fl Or
sincos ........ upupupup XIRIE
VE
(23)
Where
etc,unitperiniresistanceunitperincurrent .... EIRRIII upflup
We note the following:
(i) Usually the quantities will be referred to the secondary side (ii) With (23), it is not necessary to refer quantities from primary to secondary side, for
per-unit values of primary and secondary impedances can be added directly
(iii)The equations are correct at any current or kVA and at rated current or kVA.
(iv) At rated current or kVA, Ip.u.=1. (v) It is supposed that E is the rated voltage. When E is not the rated voltage, voltage
regulation can still be calculated using (21).
-
20
Example 9
A 100-kVA 1-ph transformer has 400 turns on the primary and 80 turns on the secondary.
The primary and secondary resistances are 0.3 and 0.01 respectively, and the corresponding leakage reactances are 1.1 and 0.035 respectively. The supply voltage is 2200 V. Calculate (a) the equivalent impedance referred to the primary circuit and (b)
the voltage regulation and the secondary terminal voltage for full load having a power
factor of (i) 0.8 lagging and (ii) 0.8 leading (c) the maximum voltage regulation
Solution
55.0)80400(01.03.0)()( 222
2111 RNNRRa e
975.1)80400(035.01.1 21eX
05.2975.155.0 21
22
1eZ
6.0sin8.0cos)()( ib
AI fl 45.452200101003
1
pu0336.02200
6.0975.18.055.045.45
VV n 440400
8022002
VVV n 2.4250336.01440122
puii 0154.02200
6.0975.18.055.045.45)(
VV 8.4460154.014402
.0424.02200
05.245.45regulationvoltagemaximum)(
1
11
max puV
ZIc
efl
Example 10
The primary and secondary windings of a 30-kVA, 6000-V / 230-V transformer have
resistances of 10 and 0.016 respectively. The total reactance of the transformer referred to the primary is 23 . Calculate the percentage regulation of the transformer when supplying full load current at a power factor of 0.8 lagging.
Solution
89.20)2306000(016.010)( 222
2111 RNNRRe ]
231eX
AI fl 5600010303
1
pu0254.06000
6.0238.089.205
%54.2regulationPercent
-
21
5.10Transformer output
The transformer supply voltage and frequency are substantially constant; the heating
therefore depends on the current taken by the load. Since the secondary voltage of the
transformer is also substantially constant it means that the heating also depends on the
load kVA. The transformer output is therefore usually quoted in kVA. The transformer
load in kVA is given by
kVA10 322 IVS (24)
Where
V2 = actual load voltage and
I2 = actual load current.]
When S is given and V2 is unknown the load current can be estimated using the
approximate equation
kVA10 322 IVS n (25)
5.11 Efficiency
The losses which occur in a transformer on load are composed of
(i) Copper losses in primary and secondary windings, namely
22
212
12221
21 ee RIRIRIRI
(ii) Iron losses in the core due to hysteresis and eddy currents. The iron losses depend on
the peak value of the mutual flux m and frequency. It is therefore independent of load current if voltage and frequency are constant.
Let
Pi = the iron losses (fixed loss) in kW and
Pc = the copper loss with full-load S kVA in kW
Then the total loss at any load xS kVA at power factor cos is Pi + x2 Pc and the
efficiency is
c
ici xPx
PS
S
PxPxS
xS
input
output
cos
cos
cos
cos2
For a given power factor, the efficiency is maximum when the expression in brackets is a
minimum. Hence for a maximum efficiency, we have
cic
i
c
i
c
i PxPorPx
PorxP
x
P
dx
dFrom
P
Px 2
200 (26)
i.e. efficiency is maximum when the copper loss, ic PPx 2 , the fixed loss or iron losses.
The efficiency of a transformer is calculated using this form of efficiency equation:
ci
ci
PxPxS
PxP
outputlosses
losses2
2
cos11
(27)
-
22
Example 13
The primary and secondary windings of a 500-kVA transformer have resistances of 0.42
and 0.0011 respectively. The primary and secondary voltages are 6600 V and 400 V respectively and the iron loss is 2.9 kW. Calculate the efficiency on full load at a power
factor of 0.8.
Solution
AI fl 1250400
1000500currentsecondary load-Full 2
AI currentprimary load-Full fl 8.756600
10005001
W17200011.01250load fullon losscopper Secondary 2
W241542.08.75load fullon losscopper Primary 2
kW135.4W413524151720load fullon losscopper Total cP ,
kW035.79.2135.4load fullon loss Total
kW4008.0500pf 0.8at load fullon power Output
%27.989827.0035.7400
035.71
load full on Efficiency
Example 12
Find the output, at which the efficiency of the transformer of example 11 is maximum
and calculate its value assuming the power factor of the load to be 0.8.
Solution
837.0135.4
9.2
c
i
P
Px
kVA5.418500837.0efficiency maximumat output Therefore
kW8.59.22loss total,efficiency maximumAt
kW8.3348.05.418pf0.8at power output and
%30.98983.08.3348.5
8.51efficiency maximum Therefore,
Example 13
A 400-kVA transformer has an iron loss of 2 kW and the maximum efficiency at 0.8 pf
occurs when the load is 240 kW. Calculate (a) the maximum efficiency at unity power
factor and (b) the efficiency on full load at 0.71 power factor
Solution
kW422efficiency maximumat loss Total(a)
kVA3008.0
240 efficiency maximumat kVA Output
kW3001300pfunity at efficiency maximumat power Output
-
23
%68.989868.03004
41pfunity at efficiency Maximum
75.0400
300max is efficiency heat which tkVA load full of fraction x The)( b
kW56.375.0
2losscopper load Full
2
2
ic
x
PP
pf at efficiency load Full %08.989808.071.040056.32
56.32171.0
5.12 All-day efficiency
Power transformers operate at substantially constant load. They are designed such that
the maximum efficiency occurs at the normal operating load. Generally their efficiency
varies little as the load varies from 50 to 130 % of its rated kVA.
Distribution transformers on the other hand supplies load which varies widely over a 24-h
period. The efficiency of these transformers is better assessed on energy basis. The
output and losses are calculated in kW hours over a 24-hour day. The all-day efficiency is
defined as
kWhinoutputkWhinlosses
kWhinlossesdayall
1 (28)
Example 14
A 200-kVA 1-ph transformer has full load copper loss of 3.02 kW and iron loss of 1.6
kW. The transformer is in circuit continuously. For a total of 8 hours, it delivers a load of
160 kW at 0.8 pf. For a total of 6 hours, it delivers a load of 80 kW at unity power factor.
For the remainder of the 24-h cycle, it is on no load. What is the all-day efficiency?
Solution
kWlossTotal
kWlossiron
kWlosscopper
62.4
60.1
02.3load),(fullkVA200kVApf,0.8kW,160At
kWlossTotal
kWlossiron
kWlosscopper
08.2
60.1
48.002.3200
80,kVA80kVAupf,kW,80At
2
kW loss Total 6.1load), noon losscopper no is (there load noOn
kWhlosstotalkWhoutputtotal
kWhlosskWh output
kWhlosskWh output
kWhlosskWh output
5.651760h,24In
16106.10100h 10For
5.12608.2480680h 6For
37862.412808160h 8For
-
24
%41.969641.017605.65
5.651efficiencydayAll
5.13 Open-circuit and short-circuit tests on a transformer
These two tests enable the efficiency and voltage regulation to be calculated without
actually loading the transformer.
(a) Short-circuit test: This test is used to determine the leakage impedance. During this
test, one winding is short-circuited and a reduced voltage Vsc applied to the other to cause
rated current to flow. The test circuit and the equivalent circuit are shown in Fig. 18.a and
Fig. 18.b respectively. The magnetizing branch is neglected because its current under
this condition is less than 1 % of the total.
Voltage Vsc, current Isc and power Psc measured by the instruments are used to make the
following calculations.
)()()()( 21211211
cRZXcI
PRb
I
VZa eee
sc
sce
sc
sce (29)
We note that the following:
(i) Isc need not be the rated current since the equivalent circuit is linear. However, it is desirable that it should be near to the rated value so that stray losses (they are due to
eddy currents set up in large section conductors, tank and metallic supports by
leakage fluxes) are normal.
(ii) The supply could be fed to either winding. It is often convenient on the higher voltage transformers to supply the high-voltage winding, thus using a smaller current.
Vsc which will be about 3-15 % of the rated value may also be more suitable for test
facilities.
(iii)In laboratory experiments using small transformers, the instruments positions shown
minimize measurement errors.
(iv) The circuit parameters obtained with (27) are referred to the side which the test voltage is applied.
)()()()( 21211211
cRZXcI
PRb
I
VZa eee
sc
sce
sc
sce
Fig. 18 Short-circuit test
(b) Open-circuit test or no load test: During this test, one winding is open-circuited and
rated voltage at rated frequency is applied to the other. Quite often the low-voltage side is
supplied to reduce the test voltage required for safety reasons. As with the short-circuit
-
25
test, the equivalent circuit parameters will be referred to the side to which the test voltage
is applied. The test circuit and equivalent circuit are shown in Fig. 19.a and b.
The following calculations can be made
2
11122
1
)()()()()(V
Vae
I
VXd
I
VRcIIIb
V
PIa
m
m
p
mpomo
p (30)
Psc and Po represent the full load copper loss and the core loss (or iron losses)
respectively. They can be used directly to calculate efficiency.
.
2
11122
1
)()()()()(V
Vae
I
VXd
I
VRcIIIb
V
PIa
m
m
p
mpomo
p
Fig. 19 Open-circuit or no-load test
Example 15
The circuit shown below was used in a test on a 3-kVA transformer. A variable voltage
supply of fixed frequency was connected to terminals A and B and two tests were
performed:
(a) The voltage was raised to normal rated voltage and the meter reading were then 200 V, 24 W, 1.2 A
(b) The terminals C and D were short-circuited and the voltage was raised until the transformer full-load current was flowing. The meter readings were then 6.4 V, 28 W,
15 A
From the results of the tests, obtain:
(i) the no-load current and its power factor (ii) the iron losses of the transformer at normal frequency and voltage (iii)the full-load copper loss
(iv) the transformer resistances, Re1 and Rm and reactances Xe1 and Xm (v) the efficiency of the transformer at full load at a power factor of 0.8
-
26
A
B
V
A C
D Circuit diagram for Example 13
Solution
A1.2I i o current load-No )(
W 24PP (ii) oi lossesIron
W 28PP (iii) scc losscopper load Full
:nscalculatiost circuit te-Open (iv)
AIIIAV
PI pom
op 19.112.02.112.0
200
24 2222
1
16819.1
20067.1
12.0
200 11
m
m
p
mI
VXk
I
VR
:nscalculatiost circuit te-Short
12.015
2843.0
15
4.6211 2
sc
sce
sc
sce
I
PR
I
VZ
41.012.043.0 2221211 eee RZX
%88.979788.02824)8.03000(
28241)
puEfficiencyv
Example 16
A 10-kVA 1-ph transformer has a voltage ratio 1100 / 250 V. On no load and at normal
voltage (1100 V) and frequency the input current is 0.75 A at a pf of 0.2 lagging. With
the secondary short-circuited, full-load currents flow when the primary applied voltage is
77 V, the power input being 240 W. Calculate
(a) the transformer equivalent resistance and reactance referred to the secondary side (b) the maximum value of the voltage regulation at full load and the load power factor at
this regulation
(c) the percentage of full-load current at which the transformer has maximum efficiency
-
27
Solution
(a) Referring the circuit to the secondary side,
V5.171100
250771
a
V
A40250
00010
2
n
rated
V
SI
438.040
5.172eZ , 15.0
40
24022e
R and 41.015.0438.0 222eX
(b) The maximum voltage drop at full load = 5.1722 efl ZI V
The maximum regulation occurs when or when
lagging342.0438.0
15.0coscos
2
2 e
e
Z
R
(c) Maximum efficiency occurs when oc PPx 2 . Therefore
%9.82or829.015.040
2.075.011002
x
5.14 Construction of transformers
Power transformers are designed so that their characteristics approach those of an ideal
transformer
(a) To attain high permeability and consequently a small magnetizing current, the core is
made of iron and the core forms a closed magnetic circuit
(b) To keep the hysteresis loss down, high-grade grain-oriented steel having a narrow
hysteresis loop is used
(c) To keep the eddy current loss down, the core is laminated and 3 to 4% silicon is added
to increase the resistivity of the steel. The laminations (about 0.4mm thick) are insulated
from each other by a thin layer of insulation, thus overall cross-sectional area is slightly
greater than the actual cross-sectional area of the iron.
(d) To keep the leakage fluxes low, the windings are arranged as shown in Fig. 18 in the
case of 1-ph transformers. Two main forms of magnetic circuits are used: core- and shell-
type arrangements. The core-type construction has the primary and secondary windings
distributed over both core legs in order to reduce the amount of copper (i.e. length of
turns is reduced).
The transformer windings are carefully insulated from each other and from the core.
Winding resistances R1 and R2 are also minimized to minimize copper losses.
-
28
L L
H H
L LH H
H L L H H L L H
windingsconcentric
on with constructi type-Core (a)
windingssandwiched
on with constructi type-Shell (b)
LL
Fig. 20 Common Single-phase transformer construction
5.15 Polarity tests
The four terminals of a single-phase transformer may be mounted so that it has either
additive or subtractive polarity as shown in Fig. 21
A2 a1
A1 a2
A2
a1
a2
A1
polarity additive (a) polarity esubtractiv (b)
Fig. 21Two standard ways of mounting transformer terminals
If it becomes necessary to determine whether a transformer has additive or subtractive
polarity then either of the following polarity tests may be used:
(a) Using a low-voltage ac source: Referring to Fig. 22,
(i) Connect the HV winding to ac source
(ii) Connect a jumper J between any two adjacent HV and LV terminals (1, 3)
(iii) Connect a voltmeter (V2) between the other two adjacent HV and LV terminals (2, 4)
(iv) Connect a voltmeter (V1) across the HV winding.
Reading of voltmeter V2 = E1 E2 = V1 E2 if 1 and 3 possess the same polarity = E1 + E2 = V1 + E2 if they do not
Therefore if V2 is lower than V1 then the polarity is subtractive and if V2 is higher than V1
then the polarity is additive
1 3
2 4
V1V2
J
Fig. 22 Polarity test using ac source
-
29
(b) Using dc source: Referring to Fig. 23,
(i) Connect a dc source to LV winding
(ii) Mark the terminal connected to the positive side of the source a2
(iii) Connect a dc voltmeter across the HV winding
(iv) Close the switch and observe the movement of the pointer of the voltmeter
(v) If the pointer moves upscale, the transformer terminal connected to the (+)
terminal of the voltmeter is marked A2 and the other marked A1.
a2
V+
-
Fig. 23 Polarity test using dc source
5.15 Parallel Operation of single-phase transformers
Parallel connection of several transformers is widely used in electrical systems for the
following reasons:
(a) In many cases, the amount of power to be transformed is greater than that which can be built into one transformer
(b) Frequently, the growth of load requires that the installed transformers supply an output greater than their total kVA capacity. Additional transformers are then
installed to run in parallel with the existing transformer.
(c) It is sometimes found desirable to supply a load through two or more units in order to reduce the cost of the spare unit required to ensure continuity of service in case of
damage.
The following conditions must be fulfilled when operating two or more single-phase
transformers in parallel:
(a) The polarity should be the same. The polarity can be either right or wrong. A wrong polarity results in a severe short circuit. Terminals of the same markings are
connected together to ensure correct polarity. See Fig. 24. If the polarity markings
are either incorrect or not present, the polarity of the incoming transformer can be
checked by connecting a voltmeter across the paralleling switch
(b) The voltage ratio should be the same. This is to avoid no-load circulating current and also over-loading on one transformer when the paralleled transformers are loaded.
(c) The per-unit impedances should be equal in magnitude and have the same angle. When they are equal in magnitude, the transformers share kVA loads in proportion to
their respective ratings. If both their magnitudes and angles are the same they will not
only share kVA loads in proportion to their respective ratings but also the combined
load kVA will be the algebraic sum of the kVA carried by each transformer. If the
angles are different, the resultant kVA capacity of the paralleled group will be slightly
smaller than the sum of their individual ratings if none should be overloaded. It is not
very necessary that the angles should be the same
-
30
A2 a2
A1 a1
Tx1
Tx2
A2
A1
a2
a1
Load
Fig. 24 Connection ensuring correct polarity
5.16 Load sharing of parallel-connected transformers The equivalent circuit of two transformers in parallel feeding a common load ZL is shown
in Fig. 25. The voltage ratios are supposed to be equal and the magnetizing branch is
neglected. The circuit is referred to the secondary side but it may also be referred to the
primary side.
a
V1
1I
2I
1Z
2Z
I
VLZ
Fig. 25 Equivalent circuit of two transformers in parallel
Z1 is the impedance of transformer 1 referred to the secondary side and Z2 is the
impedance of transformer 2 referred to the secondary side.
The voltage drops across the impedances are the same. Therefore
)( 21
212211
ZZ
ZZIZIZI
Or
IZZ
ZIandI
ZZ
ZI
21
12
21
21
Multiplying by the terminal voltage V gives
21
21
ZZ
ZSS
(31.a)
21
12
ZZ
ZSS
(31.b)
-
31
Where
loading 1r transforme11 VIS (32.a)
loading 2r transformeS 22 VI (32.b)
load combinedVIS (32.c)
Equations (31) hold for per-unit impedances provided that all are expressed with
reference to a common base power. The following equation can be used to obtain a new
per unit value with reference to a new base power.
old
base
new
baseold
pu
new
puS
SZZ (33)
Example 17
A 500-kVA transformer (Transformer 1) is connected in parallel with a 250-kVA
transformer (Transformer 2). The secondary voltage of each is 400 V on no load. Find
how they share a load of 750 kVA at power factor of 0.8 lagging if
pujZ
andpupujZc
pupujZ
andpupujZb
pupujZ
andpupujZa
62.6005095.00444.0025.0
69.7805099.005.001.0)(
69.7805099.005.001.0
69.7805099.005.001.0)(
44.6904272.004.0015.0
69.7805099.005.001.0)(
2
1
2
1
2
1
Solution
If all the impedances are referred to a base power of 500 kVA, then only the impedances
of transformer 2 will change.
old
pu
old
puold
base
new
baseold
pu ZZS
SZ 2
250
500Znewpu
Case (a)
69.7805099.005.001.01 jZ and
44.6908544.008.003.004.0015.022 jjZ
Further
90.721360.013.004.008.003.005.001.021 jjjZZ kVAS 9.367508.0cos750loadkVATotal 1
kVAZZ
ZSS
36.40471
9.72136.0
44.6908544.09.36750
21
21
lagging 762036.40cos offactor power at 471 .kVA
-
32
Similarly
11.31281
9.72136.0
69.7805099.09.367502S
lagging 8560 offactor power at 281 .kVA
Remark: Transformer 1 with larger per-unit impedance is under-loaded whereas
transformer 2 with lower per-unit impedance is overloaded.
Case (b)
pujjZ 69.7810198.01.002.005.001.022 pujZZ 69.7815297.015.003.021
kVAS
9.36500
69.7815297.0
69.7810198.09.367501
lagging 80 offactor power at 500 .kVA
kVA
9.36250
69.7815297.0
69.7805099.09.36750S2
laggging 80 offactor power at kVA 250 .
Remark: Load shared in proportion to transformer ratings. Arithmetic sum of loadings is
equal to the combined load. A shorter approach can be used on recognizing that
pupu ZZ 21
Case (c)
pujjZ 62.601019.00888.005.00444.0025.022 62.661512.01388.006.021 jZZ
9.42505
62.661512.0
62.601019.09.367501S
lagging 7320 offactor power at 505 .kVA
82.24253
62.661512.0
69.7805099.09.367502S
lagging 910 offactor power at kVA 253 .
Remark: Transformers are slightly overloaded when the combined load is equal to the
sum of individual kVAs
-
33
6.0 Three-phase transformers
These are required to transform 3-phase power. The three-phase transformer may be
either of the following:
(a) A three-phase transformer bank: This consists of three identical single-phase
transformers having their windings externally connected for three-phase working. The
single-phase transformers retain all their basic single-phase properties such as current
ratio, voltage ratio and the flux in the core. The kVA capacity of the bank is the sum of
their individual ratings. See Fig.26.
(b) A three-phase transformer unit: This is a single unit of special construction for three-
phase working. Modern large transformers are usually of the three-phase three-legged
core type shown in Fig.27. A leg carries the primary and secondary windings of a phase.
The windings are internally connected. For a given total capacity, 3-phase units are much
cheaper in capital cost, lighter, smaller and more efficient.
Tx1
Tx2
Tx3
1L
2L
3L
1L
2L
3L
2A
1a1A
2a
2A
2A
1A
1A
2a
2a
1a
1a
ers transformphase-single of connectionStar -Delta 9 Fig.
6.1 Winding arrangement
The three windings, primary or secondary, can be connected in three different ways:
(a) Star connection: For this connection %583 voltagelinevoltagephase of the
line voltage. This enables the insulation of the winding to be reduced to a minimum for a
given supply voltage. Line current = phase current. It is the most economical connection
for a high-voltage winding.
(b) Delta connection: Phase voltage = Line voltage. Therefore winding must be
insulated for the full Line voltage. More turns are also required. With very high voltages
a saving of 10 % may be achieved by using star-connection rather than delta connection
on account of insulation. The saving is small, however, at voltages below 11 kV. For
delta connection 3currentlinecurrentphase so the winding cross-sectional area is
-
34
58 % of that required for the star connection. Therefore it is the most economical for
low-voltage winding.
(c) Zigzag (or interconnected star) connection: It is a modification of the star
connection. Each phase winding is divided into two sections and placed on two different
legs. The two sections are then connected in phase opposition. The zigzag connection is
restricted to the low-voltage winding. 15 % more turns are required for a given phase
terminal voltage compared with a normal star.
The three different winding arrangements give rise to several possible connection
combinations: star-star, star-delta, star-zigzag, delta-star, delta-delta, etc
1L 2L 3L
2A2B
1B
2C
1C
2a
1a
2b
1b
2c
1c
1L
2L
3L
1A
Fig. 27 Delta-Star connection of three-phase, three-legged core-type transformer
6.2 Phase groupings
The voltage induced in a primary phase winding is in phase with its corresponding
secondary phase winding. However, the phase angles of the primary and secondary line
voltages may differ depending on the type of winding connection. Three-phase
transformers are given symbol which indicates the type of connection used for the high-
voltage and the low-voltage winding and the phase displacement between the high-
voltage line phasor and its corresponding low-voltage line phasor. The symbol is in the
form Xxn where X (the capital letter) and x (the small letter) indicate the type of
-
35
connection for the high-voltage and low-voltage windings respectively and n is a clock
hour number which indicates the phase displacement. In this method of indicating the
phase shift, the high-voltage line phasor is represented by the minute hand of a clock
always set at 12 (or the zero hour) and the corresponding low-voltage line phasor by the
hour hand. Thus when n = 11 it means the clock reads 11 Oclock and the low-voltage line phasor leads by 30
o. The groups into which three-phase transformers are classified
are as follows:
Group Phase displacement Winding connections
1 zero Yyo Ddo Dzo
2 180o Yy6 Dd6 Dz6
3 30o lag Dy1 Yd1 Yz1
4 30o lead Dy11 Yd11 Yz11
6.3 Three-phase transformer connections More common transformer connections are
(a) Delta-Delta connection (Fig 28). : This connection is economical for large low
voltage transformer. This connection is not often used because there is no neutral point
and a four-wire supply cannot be given. It is used in a 3-phase transformer bank but
rarely in 3-phase transformer unit. It is possible to use this arrangement to provide 3-
phase power with one transformer removed. This connection, known as open-delta or
vee connection, can supply up to 57.7 % of the load capacity of the delta-delta
connection.
1L
2L
3L
LV phV
phI
LI
1A 2A
1B
2B1C
2C
2L
1L
3L
1a 2a
1b
2b1c
2c
phL
phL
II
VV
3
Fig. 28 Delta-delta connection
(b) Delta-Star (Fig. 29): It is commonly used to step up alternator voltage to
transmission line voltage. Another common application is in distribution service where
as a step-down transformer, the windings are not the most economical. The secondary
star point can be earthed and a four-wire supply given.
(c) Star-Delta: There is no secondary neutral and four-wire supplies cannot be given.
The main use is as a step-down transformer at the load end of transmission line.
-
36
2A
1B
2B1C
2C
1A
1L
3L
2L
2c
1c
1b
2b
1a 2a
phV
phI
LI 3L
1L
2L
LV
3
Lph
VV Lph II
Fig. 29 Delta-star connection
(d) Star-Star: This is economical for high-voltage transformer. However, the primary
and secondary phase voltages contain pronounced third harmonic voltages, which cause
the neutral point to oscillate at three times the line frequency. The effect of this
oscillating neutral is to cause fluctuation in the line to star-point voltage. Also if the
secondary load is unbalanced, the neutral point will be displaced and the line-to-neutral
voltages will become unequal. To stabilize the neutral, the neutral of the primary and the
neutral of the source are connected together usually by way of ground. Another way is to
provide a third delta-connected winding called tertiary winding. Although it can be used
to supply additional power, the tertiary winding generally has no external connection
Example 18
Three single-phase step-up transformers rated 40MVA, 13.2 kV / 80 kV are connected in
delta-star on a 13.2 kV transmission line. If they feed a 90MVA load, calculate the
following:
(a) the secondary line voltage
(b) the currents in the transformer windings
(c) the incoming and outgoing transmission line currents.
Assume transformers are ideal.
Solution
2.13
801)(
1
2
aV
Va
ph
ph
voltage)lineion(transmiss2.13But 1 kVVph
kVVph 80Therefore
kV138380sidesecondary on the voltageline theand
-
37
2phI
2phV
1phV
1phI
kV2.13
MVA b 303
90formereach transby carried load The)(
AkV
MVAI ph1 2272
2.13
30ndingprimary wi in theCurrent
AkV
MVAI ph2 375
80
30windingsecondary in theCurrent
AIc ph 3932322723 line incominginCurrent)( 1
AII Lph 375line outgoingin Current 2
Example 19
Three single-phase transformers have their primaries joined in delta to a 6600 V, three-
phase, three-wire supply. Their secondaries are connected to give a three-phase, four-wire
output at 415 V across lines. The total load on the transformers is a balanced load of 150
kW at 0.8 pf lag. If the voltage per turn on the primaries is 4, find
(a) the number of turns on the primary winding and the secondary winding
(b) the currents and voltages in all windings and lines, including the neutral wire on the
secondary side
(c) kVA load on each transformer
Assume transformers are ideal
Solution
turnsturnvolt
Va
ph1650
4
6600phaseperturnsPrimary)(
1
turnsturnvolt
Vph60
43
415phaseperturnsSecondary
2
-
38
kV6.61phV
1phI
1LI2LI
V415
Load2phV
kW150
laggingpf8.0
AV
PowerIIb
L
phL 2618.04153
10150
cos3Secondary)(
3
22
balanced is load because0 AI N
AN
INI
ph
ph 5.91650
26160
1
22
1
AIL 4.1635.9 .
Example 20
A three-phase 415 V load takes a line current of 800 A from a 3300 / 415 V delta/star
transformer. The 3300 V system is supplied from an 11000/3300 star/star transformer.
Draw the circuit diagram and assuming no losses, find both line and phase values of
voltages and currents in each part of the circuit. What will be the turns ratios of both
transformers?
Solution
110001 V
1phV 1phI2phI 2ph
V
32 VV
32 II
3phI 3phV4phI
4phV
AI 8004
V
V
415
4
1IArTransformeBrTransforme
Solution
Voltages:
VV
VVV ph 2403
415
3;415 444
VVVVV ph 3300;3300 333
VV
VVVV ph 19053
3300
3;3300 2232
-
39
VV
VV ph 63503
11000
3;11000 111
Turn ratios:
75.13240
3300
4
3
ph
ph
V
VArTransforme
33.31905
6350
2
1
ph
ph
V
VBrTransforme
Currents:
AIIAI ph 800;800 444
AAratioturns
II
ph
ph 2.5875.13
80043
AIII ph 8.1002.5833 332
AII ph 8.10022
ABratioturns
II
ph
ph 2.3033.3
8.10021
AII ph 2.3011
Check:
kVAVAIVkVAInput 5755750002.301100033 11
kVAVAIVkVAOutput 57557500080041533 44
6.4 Parallel operation of three-phase transformers (Fig. 30)
Three-phase transformers operating in parallel should have
(a) the same line voltage ratios
(b) the same per-unit impedances, i.e., they are equal in magnitude and in phase
(c) the same phase displacement between primary and secondary line voltages
(d) the same phase sequence
The last two conditions which are absolutely essential ensure that the secondary line
voltages of the transformers are in phase. When these conditions are not fulfilled a
potential difference appears across the paralleling switches S1 and S2.
From the view point of phase sequence and phase displacement, three-phase transformers
which can operate in parallel are:
-
40
(a) transformers of the same group. This case the terminals with the same letter must be
connected to the same line as shown in Fig. 30.
(b) transformers having -30o phase displacement (Group 3 transformers) and those having
+30o phase displacement (group 4 transformers). In this case two of the high voltage
connections and the corresponding low voltage connections are interchanged as shown in
Fig. 31.
Faulty internal connections in the transformer tank can cause the phase sequence of a
transformer to be reversed. Voltage across paralleling switches should therefore be
monitored before the switches are closed.
7. Cooling methods Cooling of transformer windings and core is provided to prevent rapid deterioration of
the insulating materials. There are several methods of transformer cooling. Each method
is described by a standard designation (or nomenclature) consisting of letter symbols.
They are
(a) Letters for medium: air A, gas G, synthetic oil L, mineral oil O, solid insulation S,
water W
(b) Letters for circulation: natural N, forced F
Up to four letter symbols are used for each method for which the transformer is assigned
a rating: some big transformers are designed to have a variable rating, depending on the
method of cooling used. The order of the symbols is
S2
S1
Transformer 2
A2 a2
B2 b2
C2 c2
Transformer 1
A2 a2
B2 b2
C2 c2
C B A a b c
Fig. 30 Parallel operation of 3-phase transformers
-
41
(i) the medium and
(ii) the circulation of the coolant in contact with the windings; and
(iii) the medium and
(iv) the circulation of the coolant in an external heat exchanger system.
Common methods of cooling transformers are
(a) Air Cooling (Dry type transformers)
(i) AN: the ambient air as coolant and natural circulation by convection. The metallic
housing is fitted with ventilating louvers. Ratings: up to 50 kVA (low-power
transformers). Using high-temperature insulating materials (glass and silicone resins)
make ratings up to 1.5 MVA possible. The low-power transformers are used inside
buildings where the air is clean and high-power ones are for special conditions such as
those in mines.
(ii) AF: Forced air circulation is used to raise transformer loadings
(b) Oil-immersed, Oil cooling: Oil is a much better insulator than air. Consequently, it is
invariably used on high-voltage transformers
(i) ONAN: natural oil circulation and natural air flow over the tank. It is very common
for transformers rated up to 5 MVA. With radiators, it is possible to build units up to 40
MVA. (Note that limit of output is determined by tank size and cost).
(ii) ONAF: Cooling fans blow air over the radiators to enable a much bigger output from
a transformer of a given size. With this method of cooling ratings up to about 75 MVA
can be built.
S2
S1
Transformer 2
A2 a2
B2 b2
C2 c2
Transformer 1
A2 a2
B2 b2
C2 c2
C B A a b c
Fig. 31 Parallel operation of group 3 & 4 transformers
-
42
(iii) OFAF: Pumps are used to circulate the oil and cooling fans to blow air over
radiators. This is the usual method for transformers of 30 MVA and upward. Both OFAF
and ONAN may be used on a unit with ONAN up to 0.5 p.u. rating. Change over is
initiated automatically by temperature-sensing elements. Three-phase type OFAF step-up
transformer rated 1300 MVA installed at a nuclear power generating station is one of the
largest units ever built.
(c) Oil-immersed, Water cooling
(i) ONWF: Copper cooling coils are mounted in the tank above the level of the
transformer core, but below the oil surface.
(ii) OFWF: Oil is circulated by pump from the top of the transformer tank to an external
oil/water heat exchanger. Oil returns when cold to the bottom of the tank. Its advantages
over ONWF include
-The transformer is smaller and the tank does not have to contain the cooling coils.
-Leakage of water into oil is improbable if oil pressure is greater than that of water
This method is used for large installations. It is commonly used in generating stations,
particularly hydro stations where ample supply of water is available.
8.0 Tap-changing Transformers Most power transformers have tappings on coils brought out to terminals so that the
number of turns on one winding can be changed. The turns ratios are changed in order
(a) to maintain the secondary voltage at their rated value under the varying conditions of
load and power factor. The secondary terminal voltage may vary with changes in load
over an undesirably large range, because of changes in the impedance drop in the
transmission lines and transformers
(b) to control the flow of reactive power between two interconnected power systems or
between component parts of the same system, at the same time permitting the voltages at
specified points to be maintained at desired values.
Tappings on power transformers permit voltage adjustment within %5 . Low- and medium-power transformers usually have three taps per phase: +5%, 0, and -5%
variations in the turns ratio. Higher power ratings usually have five: +5%, +2.5%, 0, -
2.5% and -5% variations in the turns ratio. The principal tapping 0 is that to which the rating of the winding is related. A positive tapping includes more, and a negative less
turns than those of the principal tapping.
Tappings are usually fitted on the higher voltage winding to obtain tappings within fine
limits. Consider say, 11kV / 433 V, 600 kVA delta-star distribution transformer having
volts / turn = 10. On the low-voltage side, 25310433 N . Adjustment can then be
in steps of 4%. If 5 % and 10 % are required, we shall use 4 % and 8 % or 12 %.
On the high-voltage side 11001011000 N and it is possible to make adjustment in
steps of 0.09 %.
-
43
8.1 Changing the taps of transformers
Tap-changing may be either on-load or off-load.tap-changing:
(a) Off-load tap changing: the changes are made when the transformer is disconnected
from the primary circuit. The most common off-load tap changing transformer has
tappings inside the tank and connected to an internal switch which is operated by an
external switch handle (usually by rotary movement of a handwheel). A three-phase star
connected winding with taps made at the neutral is shown in Fig. 32.
Fig. 32 Three-phase star connected winding with taps at the neutral point
Off-load tap changing is simple and inexpensive and it is commonly used with
distribution transformers where occasional adjustments are required..
(b)On-load tap changing: Daily and short-time adjustment is generally by means of on-
load tap-changing gear. Tap changing is done without breaking the circuit. Momentary
connection must be made simultaneously to two adjacent taps during the transition, and
the short-circuit current between them must be limited by some form of impedance
known as transition impedance. Centre-tapped iron-core inductors (the reactor method) or
resistors (the resistor method) are used for this purpose.
(i) Reactor method: This method has now almost entirely been superseded by the resistor
method. It is manufactured and used only in the USA.
(ii) Resistor method: In modern designs the transition impedance is almost invariably
obtained by means of a pair of resistors. An arrangement of such a tap changer for one
phase is shown in Fig. 33. In this figure, the diverter switch and the even tap selector are
shown in the position when the T2 tap is brought in circuit.
To move to the next tap, T3, the odd tap selector should first be moved to that tap (see the
dashed lines in Fig. 33), and the diverter switch may then be rotated clockwise. The
ensuing sequence of events is as follows:
-
44
selector tapodd
1TS1T
3T
5T
7T
2T
4T
6T
8T
2TS
selector even tap
1R 2R
1
2 3
4
I
I
Fig. 33.a. Resistor tap changer
- contacts 3 and 4 break, contacts 1 and 4 make, Fig. 33.b
1R 2R
1
2 3
4
I
i-I2
1iI
2
1i
Fig 33.b
-
45
- contacts 1 and 4 break, and contacts 1 and 2 make, Fig. 33.c
1R 2R
1
2 3
4
I
Fig 33.c
The tap selectors may be moved from tap to tap only when their circuits are de-energized.
The resistors are short-time rated and it is essential to minimize their time of duty. For the
same reason, means must be adopted to ensure that it cannot be inadvertently left in the
bridging position.
On-load tap changer control gear can vary from simple push-button initiation to a
complex automatic control of as many as four transformers in parallel.
9.0 Autotransformers
An autotransformer has a single tapped winding which serves both primary and
secondary functions as shown in Fig. 34. The circuit diagrams are shown in Fig. 35.
1V
1I A
B
1N
2N
2I C
B
2V Load
1V 1N
A
B
2N
2I
2V Load
ormerautotransfdown -Step 34.a Fig. ormerautotransf up-Step 34.b Fig.
C
B
1I
9.1 Autotransformer equations If we neglect losses, leakage flux and magnetizing current then
2
1
1
2
1
2
I
I
V
V
N
Nn
9.2 Advantages and disadvantages of autotransformer over two-winding
transformer
The main advantage gained in the use of autotransformer is the saving of copper. For a
two-winding transformer and an autotransformer which can perform the same duty (they
-
46
1V
1I
S
C
2I C
B
2VLoad
1V
A
B
1I
S
C
2VLoad
C
BB
ormerautotransfofdiagramsCircuit35 Fig.
dingcommon win C winding;series S
should have the same voltage per turn and therefore the same flux. We can also assume
the same mean length per turn)
H
L
V
V
rtransforme winding-two in copper of Volume
ormerautotransf in copper of Volume1
Or
ormerautotransf an using
by effected copper of saving rtransforme windingtwo the in copper of Volume
V
V
H
L
sidevoltage high on voltageV
sidevoltage low the onvoltageV
Where
H
L
In practice, voltage ratios HL VV less than about 31 show little economic benefit over
two-winding transformer because of other factors such as cost of insulation.
The main disadvantage is that the primary and secondary circuits are not isolated from
each other.
Example 21
An autotransformer is required to step up a voltage from 220 to 250 V. The total number
of turns is 2000. Determine (a) the position of the tapping point (b) the approximate
value of the current in each part of the winding when the output is 10 kVA and (c) the
economy in copper over the two winding transformer having the same peak flux and the
same mean length per turn.
-
47
Solution:
1V
1I
2V
2I
1N
2N
17602000250
220
250
220
250
220)( 21
2
1
2
1 NNorN
N
V
Va
Position is 240 turns from one end.
AIIVb 40250
10101010)(
3
23
22
AIIV 45.45220
10101010
3
1
3
11
Therefore current in series winding is 40 A and current in common
winding A45.54045.45
%8888.0250
220)( orpu
V
VcopperinSavingc
H
L of copper used in the two-
winding transformer
9.3 Two-winding transformer connected as an autotransformer
A two-winding transformer can be changed into an autotransformer by connecting the
primary and secondary windings in series. The following rules apply whenever a two-
winding transformer is connected as autotransformer:
(a) the current in any winding should not exceed its current rating
(b) the voltage across any winding should not exceed its voltage rating
(c) rated current in one winding gives rise to rated current in the other
(d) rated voltage across one winding gives rise to rated voltage across the other
(e) if current in one winding flows from say A2 to A1, then current in the other winding
must flow from a1 to a2 and vice versa
(f) the voltages add when terminals of opposite polarity (A1 and a2 or A2 and a1) are
connected together by a jumper. The voltages subtract when A1 and a1 (or A2 and a2) are
connected together.
-
48
Example 22
A two-winding single-phase transformer rated 15 kVA, 600 V / 120 V, 60 Hz. We wish
to reconnect it as an autotransformer in three different ways to obtain three different
voltage ratios:
(a) 600 V primary to 480 V secondary (b) 600 V primary to 720 V secondary (c) 120 V primary to 480 V secondary Calculate the maximum load the transformer can carry in each case.
Solution
(a) The secondary voltage 120 V must be subtracted from the primary voltage to obtain
the 480 V.
%8888.0250
220)( orpu
V
VcopperinSavingc
H
L
(b) The secondary voltage 120 V must be added to the 600 V to obtain 720 V
%8888.0250
220)( orpu
V
VcopperinSavingc
H
L
-
49
(c) The 120 V becomes the primary of the autotransformer and the 120 V is subtracted
from the 600 V to obtain its secondary
%8888.0250
220)( orpu
V
VcopperinSavingc
H
L
9.4 Applications of autotransformers
They are mainly used for
(a) variac
(b) interconnecting power systems that are operating at roughly the same voltage (eg. 132
kV, 275 kV, 400 kV) and
(c) starting squirrel-cage induction motors.
10.0 Instrument transformers
They are used in ac circuits to serve these purposes:
(a) to make possible the measurement of high voltages with low-voltage instruments or
large currents with low current ammeters
(b) to insulate high voltage circuits being monitored from measuring circuit in order to
protect the measuring apparatus and operator
(c) to energize relays for the operation of protective and automatic control devices
The load on the secondary of an instrument transformer is called its burden and is
expressed in volt-amperes (VA). There are two types of instrument transformers: the
voltage (or potential) transformer and the current transformer
10.1 The voltage or potential transformers (VTs or PTs) The construction is similar to a power transformer. The primary is connected directly to
the power circuit either between two phases or between a phase and ground and the
secondary is connected to instruments and coils of relays. Sufficient insulation is
provided between the primary and the secondary to withstand the full line voltage as well
as the very high impulse voltage. Voltage transformers are designed to step down the
primary voltage to a nominal or rated voltage of 110 V so that standard instruments and
relays can be used. They introduce errors of two kinds into measurement being made: the
ratio errors (the ratio between input and output voltages is not constant under all
conditions of load) and the phase angle errors (the phase shift between input and output
voltages is not zero). These errors are due to the exciting current and the equivalent
-
50
series impedance of the transformer and they are kept low by using high quality iron
(high permeability and low loss) and operating it at low flux densities so that the exciting
current is very small. The resistance and reactance of the windings are also made very
low.
Fig. 36 shows the circuit for a potential transformer. One terminal of the secondary
winding is always earthed. The windings though insulated from each other, are
connected invisibly together by distributed capacitance between them. By earthing one
of the secondary terminals, the highest voltage between the secondary lines and earth can
never rise above that of the secondary voltage.
V
V1500
V110
PT
circuit ac H.V.
Fig. 36 Potential transformer installed on H.V. circuit
10.2 Current transformers (CTs)
Their primary consists of small number of turns connected in series with the power
circuit load. The secondary consists of a larger number of turns and it is connected to
ammeter, current coils of other instruments or current coils of relays. Current
transformers have the ratio of primary to secondary current approximately constant. The
nominal or rated secondary currents are usually 5 A or 1 A, irrespective of the primary
current rating. The transformer ratio is usually stated to include the secondary current
rating. Current transformers also introduce two errors in measurement: the ratio error
(the ratio between primary and secondary currents is not constant) and phase angle error
(the phase angle between the primary and secondary currents is not zero. The basic cause
of ratio and phase angle errors is the exciting current. To keep the exciting current small,
a high quality iron operating at very low flux densities is used as in PTs. In Cts the
secondary leakage impedance and impedances of the secondary leads and instruments
should also be very low; for any increase in these impedances increases the core flux and
therefore the exciting current. The tansformer is connected in the power circuit as shown
in Fig. 37 As in the case of PT (and for the same reasons) one of the secondary terminals
is always earthed
Current transformer secondary circuit must not be opened while current is flowing in the
primary. Without opposing ampere-turns the line current, which may be 100 to 200 times
the normal exciting current, becomes the exciting current. The iron core becomes.
-
51
A
AorA 51
Load
currentLoad
supplyac
AorA 51ammeter
CT
Fig. 37 Current transformer intalled in H.V. circuit
saturated and very high voltage spikes (several thousand volts) are induced across the
open-circuited secondary. These voltages are dangerous to life and to the transformer
insulation. The core when it becomes saturated can also cause excessive heating of the
core and windings. Therefore when it is desired to remove a load from the secondary
circuit, the secondary winding must first be short circuited.
When the line current exceeds 100 A we can sometimes use a toroidal or bar-primary
(N1=1) transformer shown in Fig. 38. It consists of a laminated ring-shaped core which
carries the secondary winding. The primary is composed of a single conductor that
simply passes through the centre of the ring as shown in the figure. Toroidal CTs are
simple and inexpensive and are widely used in HV and MV indoor installations.
AAI 32
AI 6001
11
N
primaryBar
Fig. 38 Toroidal or bar-primary transformer having a ratio 1000A / 5A connected to
measure a current in a line
-
52
Current transformers are also commonly used for the measurements of large currents
even when the circuit voltage is not dangerously high. This avoids bringing heavy leads
to the instrument panels. Whereas instrument CTs have to remain accurate up to 12 %
rated current, protection CTs must retain proportionality up to 20 times normal full load
Example 23
A potential transformer rated 14400 V / 115 V and a current transformer rated 75 A / 5 A
are used to measure the voltage and current in a transmission line. If the voltmeter
indicates 111 V and the ammeter reads 3 A, calculate the voltage and current in the line.
Solution
VV is line the on voltage The 13900115
14400111
AI is line the in currentThe 455
753
Example 24
The toroidal current transformer of Fig. 38 has a ratio of 1000 A / 5 A. The line
conductor carries a current of 600 A.
(a)Calculate the voltage across the secondary winding if the ammeter has an impedance
of 0.15 (b) Calculate the voltage drop the transformer produces on the line conductor
(c) If the primary conductor is looped four times through the toroidal opening, calculate
the new current ratio
Solution
AI secondarythe in Current (a) 36001000
52
V burden the across drop Voltage 45.015.03
mVVI
IVor
I
I
N
N
V
V b 25.245.0
1000
5)( 2
1
21
1
2
2
1
2
1
41)( 1122 xIxINIc
AIthatimpliesThis 2501
AAratiocurrentnewtheTherefore 5250
-
53
Exercises One
(1) A single-phase transformer has a primary winding with 1,500 turns a nd a secondary
winding with 80 turns. If the primary winding is connected to a 2300-V, 50-Hz supply,
calculate (a) the secondary voltage (b) the maximum value of the core flux. Neglect the
primary impedance
(2) A single-phase 2,300/230-V, 500-kVA, 50-Hz transformer is tested with the
secondary open-circuited. The following test results were obtained: V1 = 2,300 V, Io =
10.5 A, and Po = 2,300 W. Calculate (a) the power factor (b) the core-loss current Ip (c)
the magnetizing current Im.
(3) A 150-kVA, 2,400/240-V, 50-Hz single-phase transformer has the following
resistances and reactances: R1 =0.225 , Xl1 = 0.525 , R2 = 0.00220 and Xl2 = 0.0445 . Calculate the transformer equivalent values (a) referred to the primary (b) referred to the secondary.
(4) A 150-kVA, 2,400/240-V, 50-Hz single-phase transformer has the following
resistances and reactances: R1 =0.225 Xl1 = 0.525 , R2 = 0.00220 , Xl2 = 0.00445 , Rm = 10 k and Xm = 1.5 k . The transformer is supplying full-rated load at 0.85 lagging power factor and rated secondary terminal voltage. Calculate (a) I2 (b) Ip (c) Im
(d) Io (e) I1 (f) V1. Use the exact equivalent circuit referred to the primary side.
(5) The results of open- and short-circuit tests carried out on a 230/115-V, 60-Hz single phase transformer are
Cal