Electrical Machines  Basic vocational knowledge (Institut für Berufliche Entwicklung, 144 p.) 
8. Transformer 

Transformers are stationary electrical machines which transmit energy from systems with certain current and voltage values into systems with generally different current and voltage values but with identical frequency.
Two separate windings are on the same iron core.
Following connection to alternating voltage U_{1} there is a standstill current I_{0}. The magnetomotive force Q = I_{0} · N_{1} generates a magnetic alternating flow (F_{1}) in the iron core.
The input and output winding of an alternating voltage are induced in accordance with the induction law. A selfinduction voltage U_{10} arises in the input winding. It is counterpositioned in accordance with Lenz’s law on applied voltage. During idling operation  because of mutual induction  there arises the output voltage U_{20} which is simultaneously the terminal voltage U_{2}.
U_{1~} ® I_{0~} ® Q_{0~} ® F_{1~} ® U_{20~}
The value of the induced voltage is derived from the following equation:
_{}
_{} 
max. flow density 
A_{Fe} 
limb crosssection 
U_{0} 
induction voltage 
f 
frequency 
N 
number of turns 
The induction voltage increases along with the number of turns, the magnetic flow density in the iron core, the iron crosssection and the frequency.
Example:
Which maximum flow density occurs in an iron core of 16 cm^{2 }crosssection when a voltage of 380 V (50 Hz) is applied to the primary coil with 980 turns?
Given: A_{Fe} = 16 cm^{2}; N_{1} = 980; U_{1} = 380 V; f = 50 Hz
Sought: _{}
Solution:
_{}
_{}
_{}
_{}
_{} » 1.09 V · s · m^{2}
_{} » 1.09 T
The iron core evidences a maximum flow density of 1.09 T.
A few field lines already close before reaching the output coil (Figure 125) so that flow F_{1} can be divided into a maximum flow F_{K} which saturates both coils and a leakage flow F_{S}.
The leakage flow may be ignored in regard to the unloaded transformer (idling). Therefore the following applies:
_{}
According to the transformer equation
_{}and
_{}.
If we relate both equation then
_{}
Shortening gives us
_{}
During idling no current flows into the output winding, thus there is no voltage decrease. Consequently the induced voltage U_{20} equal to the terminal voltage U_{2} (Cp Figure 125):
Figure 125  Transformer principle
1 Input winding/upper voltage winding/primary winding, 2 Output winding/under voltage winding/secondary winding
U_{20} = U_{2}
In the event of minimal idling current I voltage decrease in the input winding is negligibly minimal. We therefore have
U_{10} = U_{1}which results in
_{}
The voltages behave like the numbers of turns.
The interrelationship of the numbers of turns is known as the transformation ratio We have:
_{}
The rated voltages U_{1n} and U_{2n} are indicated on the rating plate of the transformer.
Example:
What secondary terminal voltage arises in a transformer where 380 V is applied to the primary winding of 980 turns and the secondary winding has 594 turns?
Given: U_{1} = 380 V; N_{1} = 980; N_{2} = 594
Sought: U_{2}
Solution:
_{}
_{}
_{}U_{2} » 230 V
Load behaviour of the transformer
If the transformer is outputloaded, current I_{2} flows into coil N_{2}. Current I_{2} generates the magnetic flow F_{2K}. According to Lenz’s Law this magnetic flow is counterpositioned to the cause (F_{1K}).
Figure 126  Loaded transformer
In this manner the magnet flow F_{1K} is weakened and induction voltage U_{10} decreases. Given uniform rated voltage, the difference increases between the two voltages U_{10} and U_{1}.
Consequently, a greater input current I_{1} flows whereby the magnetic flow F_{1K} is increased. The magnetic flow F in the iron core thus remains virtually constant:
F = F_{1K}  F_{2K} = constant
This also applies to the output voltage of the transformer.
The input current I_{1} increases as the load current I_{2} becomes greater.
Transformation ratio
Without heeded the losses of the transformer, the following applies according to the energy conservation law:
s_{1} = s_{2}U_{1} · I_{1} = U_{2} · I_{2}
If we arrange the equation so that the voltage and current values appears on respective sides, then
_{}.
The following relationships may be cited for current ratio:
_{}
Conversely the currents are proportional to the voltages or numbers of turns. A transformer converts high currents into low ones or low currents into higher ones.
Example:
A welding transformer takes up 220 (current being 10A). The output voltage is 20V. How great is the welding current?
Solution:
_{}
_{}I_{2} » 110A
A transformer idles where mains voltage U_{1} remains applied to the primary side whilst no consumer is connected to the secondary side (Z_{a}) (Figures 125/126).
Primary circuit 
U_{1} applies 

I_{0} flows (idling current) 
Secondary circuit 
Z_{a} = ¥ 

I_{2} = 0 

U_{2} = U_{20} 
Idling current
The applied voltage U drives the idling current I_{0}. This is needed to establish the magnetic field Iµ. This lags behind the voltage U_{1}.
Figure 127  Indicator image for
idling operation
1 Iron loss current I_{Fe}
The phase position of the idling current I_{0} to voltage U_{1} can be determined according to the circuitry of Figure 128.
Figure 128  Circuitry to determine
idling losses
1 Rated voltage
_{}
The value of idling current I_{0} is between 2 and 5 per cent of idling current in big transformers and up to 15 per cent in smaller transformers.
Noload curve
The idling curve I = f (U_{1}) in Figure 129 indicates that noload current I_{0} increases proportionally to the input voltage U_{1}. Noload current increases markedly over and beyond the input rated speed U_{1n}. It can, moreover, even attain values greater than the rated current.
Figure 129  Idling curve of a
transformer I_{0} = f (U_{1})
Transformers shall not be driven by voltages greater than the rated voltage.
Idling losses (iron losses)
The active power derived from the circuit in Figure 128 can only be transformed into heat in the input winding and iron core as no current flows into the secondary winding during idling. The active power P_{0}, which is converted into heat in the iron core, is made up of eddy current and hyteresis loss.
The following example shows that the iron losses almost always arise during idling.
Example:
The following idling values were measured in a transformer:
U_{1n} = 220 V; I_{0} = 0.5 A; P_{0} = 40 W; R_{1} = 3.
What percentage of winding losses are contained in idling power?
Solution:
P_{0} = P_{VFe} + P_{VW}
_{}P_{VW} = 0.75 W
_{}P_{VFe} = P_{0}  P_{VW} = 40 W  0.75 W = 39.25 W
Thus, the power loss determined during idling is an iron loss.
Iron losses are determined during noload operation and are independent of load.
Shortcircuit curves
Secondary current I_{2} increases if load resistance is decreased. Where Z_{a} = 0 the transformer has been shortcircuited.
Primary circuit 
U_{1} is applied 

I_{K} flows 
Secondary circuit 
Z_{a} = 0 

U_{2} = 0 
Shortcircuit voltage
The shortcircuited transformer can be replaced by resistor Z_{1} which corresponds to the transformer internal resistor.
Figure 130  Shortcircuited
transformer
1 Shortcircuit current I_{K}
Figure 131 depicts the commensurate duplicate circuit diagram.
Figure 131  Duplicate circuit
diagram for short circuit run
1 Ohmic winding resistance, 2 Scattered reactance (is made up of the scatter flow of the input and output coils), 3 Inner resistance of the transformer (impedance)
During a shortcircuit attempt (Figure 132) the input voltage given a shortcircuited output winding is increased until primary and secondary nominal currents flow. The voltage applied to the input side is then the shortcircuit voltage U_{K}.
Figure 132  Circuitry to determine
shortcircuit losses
1 Short circuit voltage
The shortcircuit voltage is the overall voltage decrease of a transformer during rated loading.
The relative shortcircuit voltage U_{K} in % is determined by the following equation:
_{}
The relative shortcircuit voltage is, on average, 2 to 10% of input rated voltage (U_{1n}) in mains transformers.
Shortcircuit losses (winding losses)
In the shortcircuit experiment (Figure 132) a power meter indicates shortcircuit losses as the primary and secondary rated currents generate winding losses. The iron core is only slightly magnetised by the applied shortcircuit voltage (U_{K} U_{1}).
The winding losses can be metered during the shortcircuit experiment. They are dependent on the load current (P_{VW} = I_{2} R).
In contrast to operational idling, during loading the secondary circuit is closed through an external resistance Z_{a} (Figure 126). Secondary current I_{2} flows. According to the energy conservation law the transformer must also take up commensurate primary power, thus a primary current I_{1} also flows.
Primary circuit 
U_{1} is applied 

I_{1} > I_{0} 
Secondary circuit 
Z_{a} < ¥ 

I_{2} > 0 

U_{2} ¹ U_{20} 
Voltage curve U_{2} = f (I_{2})
As the curve in Figure 133 shows, terminal voltage U_{2} decreases during loading.
Figure 133  Voltage behaviour
during loaded operation U_{2} = f (I)_{2}
1 U_{K} small, 2 U_{K} big
Figure 134 depicts the duplicate circuit diagram for the loaded transformer.
Figure 134  Duplicate circuit for
the loaded transformer with a transformation ratio = 1:1
The duplicate circuit diagram corresponds to a transformer with a transformation ratio _{}
_{}
_{}
As rated current flows the shortcircuit voltage U_{K} decreases at the internal transformer resistance Z_{i} as a result of which the terminal voltage U_{2} declines by the power decrease of the shortcircuit voltage U_{K}.
Transformers with considerable shortcircuit voltage U_{K} have powerful internal resistors, that is to say pronounced voltage changes as load alters.
U_{K} = 2...10% 
minimal voltage losses 

voltagerigid behaviour 
U_{K} = 20...50% 
considerable voltage losses 

voltageflexible behaviour 
Example:
A 220/42 V transformer has a shortcircuit voltage of 10%.
How great is the voltage change between idling and rated current loading?
Solution:
_{}
Output voltages at differing loads
Given differing loads with ohmic, inductive or capacitive external resistance gives rise to the dependence of output voltage on load current as shown in Figure 135.
Figure 135  Secondary terminal
voltage depending on the degree and nature of loading
1 Idling, 2 Rated load
Given capacitive load, the output voltage may even be greater than noload voltage.
The output voltage of a transformer depends on the
 degree of load current I_{2}
 the magnitude of relative shortcircuit voltage
 the nature of the load (ohmic, inductive or capacitive).
Efficiency represents the ratio of power output to power input.
_{}
Efficiency is determined by the separate loss indication.
P_{in} = P_{out} + P_{VFe} + P_{VW}
P_{out} 
rated power of the transformer given 

cos j = 1 (resistive load) 
P_{VW} 
metered winding losses in shortcircuit trial 
P_{VFe} 
metered iron losses in noload trial 
_{}
The following equation applies for determining efficiency for partially inductive or capacitive load:
_{}
Example:
Given:
S_{n} = 100 kVA
P_{VFe} = 570 W
P_{VW} = 2.1 kW
cos j = 0.85
Sought: h
Solution:
_{}
For economical reasons the transmission of electric power these days is not undertaken by singlephase systems but by threephase systems. Thereby, threephase alternating voltage has to be transformed into another, like frequency and number of phases. The transformation is possible by means of three identical singlephase transformers.
The resultant voltages must not only possess the same value but shall also evidence a mutual phase displacement of 120 degrees.
Consequently, the mains connection of the singlephase transformers must ensure a delta or star circuit despite the spatially separate installation of electric primary and secondary winding connections.
Figure 136  Transformation through
three singlephase transformers
In view of their size, big transformers of this kind come as socalled threephase (transformer) bank. They are generally added by a fourth singlephase transformer. This latter unit constitutes the reserve and can be switched on if another transformer fails.
Material and space requirements are usually too great for medium and small power units for this kind of transformation. The constructional fusion into a unit leads to substantial material economies.
Core transformers are most frequently constructed.
An input and an output coil each have been positioned on the common limb. Following threephase mains connection the three input coils along with threephase consumers, can be linked up into a star (Y) or delta (D) connection.
Figure 137  Threephase transformer
in Y_{y} circuitry
1 Upper voltage winding
2 Under voltage winding
Circuitry of windings
 The primary and secondary circuits of the threephase transformer each consist of three strands. These three strands can form a delta connection if the terminals x, y and z are connected to v, w and u.
Figure 138  Delta connection
In the delta connection the conductor voltage U equals the phase voltage U. Strand current is made up thus:
_{}
 Where the terminals x, y and z are interconnected we obtain a star circuit.
Figure 139  Star connection
As opposed to the delta circuit, phase voltage is _{} phase and conductor current values are identical.
 A special kind of star connection is the zigzag connection which, however, is only very rarely employed.
Phase position of upper and undervoltages
 The delta and star connection of the upper and under voltages yields the following combinations:
Yy 
Yd 
Dy 
Dd 
The designation Yy indicates that the upper and undervoltage windings have been starconnected. Yd denotes uppervoltage winding as star and undervoltage winding as delta.
Figure 140 indicates that these designations are not final.
Figure 140  Connection options of a
starstar circuit
(1) Upper voltage windings, (2) Undervoltage windings
Circuits 1 and 2 and 3 and 4 are identical; however, both groups differ as regards the phase position of under to upper voltage. The upper and undervoltage windings of circuits 1 and 2 feature opposing winding senses. As a result, in line with the transformer principle, there is no phase displacement between upper and lower voltage.
Figure 141  Voltage indicator for
the voltages 1 and 2 of Figure 140
The windings of circuits 3 and 4 possess the same winding sense. For this reason there is a phase displacement of 180 degrees between upper and undervoltage, that is to say the voltages are counterdirected to each other.
Figure 142  Voltage indicator for
circuits 3 and 4 of Figure 40
Consequently any comprehensive vector group designation must not only indicate winding circuits but also data pertaining to the phase position of the voltages.
The example of the star delta connection shows how to determine the phase position from the circuit diagram.
Figure 143  Stardelta connection
The circuit diagram is added by the phase voltages (I, II, III, 1, 2, 3) whose indicators are always directed towards the terminals. The uppervoltage indicators (I, II, III) are inserted into a twelvesegment circle which serves as construction aid (Figure 144). The position of the indicator can be varied ad lib; however, amongst themselves they should heed a mutual phase displacement of 120 degrees and the winding circuit (star). The position of the under voltage indicator is determined by the uppervoltage indicator. The circuit diagram shows that the undervoltage indicators are counterdirected to the uppervoltage indicators (indicator 1 counter to indicator 1 etc.). Where the indicators 1, 2 and 3 are inscribed into the twelvesegment circle heeding the (delta) undervoltage winding circuit, the position of the undervoltage terminals u, v and w are stipulated. The phase position of likenamed conductor voltages, for example between the uppervoltage terminals U, V and the undervoltage terminals u and v can now be derived from the indicator figure. In our example the undervoltage lags behind the upper voltage by 150 degrees.
Figure 144  Indicator of a
stardelta connection
A phase displacement of the undervoltage against the upper voltage of 30 degrees in each case, from zero; 30; 60 etc. up to 360 (0) degrees can be attained through varying linkage of the delta and star connections. However, in practice, one sticks to those connections where the displacement is 0; 150; 180 and 330 degrees. Thereby angle indication does not ensue directly but by means of a socalled index. This is derived from the division of the phase angle by 30 degrees.
Vector group designation
Vector group = circuit + index
Example:
Yy0 
Y star connection of the uppervoltage winding OS 

y star connection of the undervoltage winding US 
0 30 degrees = 
0 degrees phase displacement 
Dy5 
D delta connection OS 

y star connection US 
5 30 degrees = 
150 degrees phase displacement 
The index indicates by how many times of 30 degrees the undervoltage lags behind the upper voltage
Standardized vector groups
Survey 21 focuses attention on the most common of the 12 vector groups.
Survey 21  Standardized vector groups of threephase transformers
Vector group circuit 
Circuit diagram 
Indicator image 
Transformation ratio 
Dy5 


_{} 
Yd5 


_{} 
Yz5 


_{} 
Yy0 


_{} 
Figure 145  Utilisation of
transformer vector groups in power supplies
1 Power station generator, 2 Machine transformer, 3 Network transformer, 4 Distribution transformer, 5 Substation transformer
Block and machine transformers
 Block transformers along with a generator make up one unit. They establish the connection between the generator and the highvoltage side.
 Machine transformers operate in the same manner as block transformers whereby, initially, several generators work together on a bus bar.
 Preferred vector group for both transformers is Yd5.
Mains transformers
 Mains transformers function as a link between transmission networks of differing voltage planes, e.g. between the 380 kV and 220 kV mains.
 Network transformers are preferred in the Yy0 vector group.
Distribution and urban network transformers
 Distribution transformers link the transmission network to the consumer system.
 Urban network transformers are transformers whose undervoltage is less than 1 kV. Particular significance accrues to supplying the asymmetrically loaded urban network.
 The vector group Dy5 is suitable for urban network and distribution transformers.
Basic information on parallel operation
The extension of existing electrotechnical installations makes it necessary to parallel connect further transformers to the existing ones.
Excessive transmission ratings may also necessitate multiple operation of several transformers.
Parallel operation signifies the upper and undervoltage interswitching of several transformers.
Conditions for parallel operation
In order to prevent the transformers being preloaded or subject to unequal load distribution amongst themselves because of compensating currents, the following conditions must prevail:
 the vector group must have the same index figure same transformation ratio
 same short circuit voltages U. They shall not deviate by more than 10 per cent from one another
 rated power ratios should not be greater than 3 : 1.
The surveys 22 and 23 feature the index figures of several mains and distribution transformers.
Survey 22  Characteristic values of distribution transformers (threephase oil transformer)
Rated power 
kVa 
100 
160 
250 
400 
630 
1000 
1600 
Rated uppervoltage 
kV 
(6; 10; 15; 20) 
(6; 10; 15; 20; 30) 
(6; 10; 15; 20; 30)  
Adjustment range 
% 
± 4 
± 5 
± 5  
Rated undervoltage 
kV 
(0.231; 0.4; 0.525) 
(0.4; 0.525) 
(0.4; 0.525; 6.3)  
Idling losses 
W 
380 
550 
700 
1000 
1450 
2200 
3200 
Short circuit losses 
W 
2200 
2900 
4400 
5900 
7800 
11000 
16000 
Short circuit voltage 
% 
3.8 

6 
6 
6 
6 
6 
Dimensions length a 1 
mm 
1110 
1260 
1810 
1980 
2110 
2300 
2650 
width b 
mm 
640 
800 
800 
1100 
1100 
1000 
1000 
height h_{1} 
mm 
1420 
1590 
1870 
1950 
2215 
2490 
2700 
Oil filling 
kg 
215 
300 
470 
620 
755 
1150 
1550 
total weight 
kg 
790 
1070 
1520 
2020 
2620 
4000 
5750 
Vector group 
 
(Yy0; Yz5) 
(Yy0; Dy5) 
(Yy0; Dy5) 
Survey 23  Characteristic values of drytype transformers
Rated power 
kVa 
63 
100 
160 
250 
400 
630 
1000 
Rated uppervoltage 
kV 
(2; 3; 5; 6; 10; ) 
(2; 3; 5; 6; 10; )  
rated undervoltage 
kV 
(0,231; 0,4; 0,525) 
(0,4; 0,525)  
Rated frequency 
Hz 
(50) 
(50)  
Idling losses 
W 
580 
750 
900 
1200 
1750 
2500 
2900 
Shortcircuit losses 
W 
1330 
1700 
2570 
3200 
5250 
6500 
10400 
Shortcircuit voltage 
% 
(3, 8(4)) 
(6) 
Figure 146 serves as an example of the dimensional size of a threephase oil transformer.
Figure 146  Oil transformer for the
250  16000 kVA range
1 Oil level, 2 Thermopockets, 3 Oil removal device, 4 Converter, 5 Eye bolt, 6 Earthing screw, 7, 8 Oil opening, 9 Buchholz relay, 10 Stop valves (from 1000 kVA onwards), 11 Shoulder hooks, 12 Hoisting points, 13 Variable dimensions
Questions for revision and control
1. Describe the basic construction and range of a
transformer.
2. How can iron and winding losses be determined in a
transformer?
3. How can shortcircuit voltage be determined?
4. Which
factors cause a voltage drop in a transformer?
5. What is the significance of
the index figure in vector group data? Which index figures are cited?
6.
Which are the most common vector groups and for which purposes are they
used?
7. Name the parallel switching
conditions.