10. Write the expression for the self inductance of a coil wound on an iron core
3.7 TRANSFORMER TESTING
Two chief difficulties which do not warrant the testing of large transformers by direct load test are: (i) large amount of energy has to be wasted in such a test, (ii) it is a stupendous (impossible for large transformers) task to arrange a load large enough for direct loading. Thus performance characteristics of a transformer must be computed from a knowledge of its equivalent circuit parameters which, in turns, are determined by conducting simple tests involving very little power consumption, called nonloading tests. In these tests the power consumption is simply that which is needed to supply the losses incurred. The two nonloading tests are the Open-circuit (OC) test and Short-circuit (SC) test.
In both these tests voltage, current and power are measured from which the resistance and reactance of the input impedance can be found, as seen in each test. Thus only four parameters can be determined which correspond to the approximate equivalent circuit of Fig. 3.16(a).
Before proceeding to describe OC and SC tests, a simple test will be advanced for determining similar polarity ends on the two windings of a transformer.
Polarity Test
Similar polarity ends of the two windings of a transformer are those ends that acquire simultaneously positive or negative polarity of emfs induced in them. These are indicated by the dot convention as illustrated in Sec. 3.4. Usually the ends of the LV winding are labelled with a small letter of the alphabet and are suffixed 1 and 2, while the HV winding ends are labelled by the corresponding capital letter and are suffixed 1 and 2 as shown in Fig. 3.21. The ends suffixed 2 (a2, A2) have the same polarity and so have the ends labelled 1 (a1, A1).
In determining the relative polarity of the two-windings of a transformer the two windings are connected in series across a voltmeter, while one of the windings is excited from a suitable voltage source as shown in Fig. 3.21. If the polarities of the windings are as marked on the diagram, the voltmeter should read V = V1 ~ V2. If it reads (V1 + V2), the polarity markings of one of the windings must be interchanged.
The above method of polarity testing may not be convenient in field testing of a transformer. Alternatively the polarity testing can be easily carried out by a dc battery, switch and dc voltmeter (permanent magnet type which can determine the polarity of a voltage) as shown in the simple setup of Fig. 3.21(b). As the switch on the primary side is closed, the primary current increases and so do the flux linkages of both the windings, inducing emfs in them. The positive polarity of this induced emf in the primary is at the end to which the battery pasitive is connected (as per Lenz’s law). The end of secondary which (simultaneously) acquires positive polarity (as determined by the dc voltmeter) is the similar polarity end. The reverse happens when the switch is opened, i.e. the similar polarity end of the secondary is that end which acquires negative potential.
+
V1 V1
A2 a2
A1 a1
V2
– –
+ +
–
V
S
i DC voltmeter
V
(a) (b)
Fig. 3.21 (a) Polarity test on two-winding transformer
(b)
Open-circuit (OC) or No-load Test
The purpose of this test is to determine the shunt branch parameters of the equivalent circuit of the transformer (Fig. 3.14(c)). One of the windings is connected to supply at rated voltage, while the other winding is kept open-circuited. From the point of view of convenience and availability of supply the test is usually performed from the LV side, while the HV side is kept open circuited as shown in Fig. 3.22. If the transformer is to be used at voltage other than rated, the test should be carried out at that voltage. Metering is arranged to read.
voltage = V1; current = I0 and power input = P0 (3.36)
AC supply I0 +
–
W
A
V
LV HV
Fig. 3.22 Connection diagram for open-circuit test
Figure 3.23(a) shows the equivalent circuit as seen on open-circuit and its approximate version in Fig. 3.23(b). Indeed the no-load current I0 is so small (it is usually 2-6% of the rated current) and R1 and X1 are also small, that V1 can be regarded as = E1 by neglecting the series impedance. This means that for all practical purposes the power input on no-load equals the core (iron) loss i.e.,
P0 = Pi (iron-loss) (3.37)
The shunt branch parameters can easily be determined from the three readings (Eq. (3.36)) by the following circuit computations and with reference to the no-load phasor diagram of Fig. 3.6.
Y0 = Gi – jBm (3.38)
Now Y0 = I
V01 (3.39)
V12 Gi = P0
or Gi = P
V
0
12 (3.40)
It then follows that
Bm = Y02-Gi2 (3.41)
These values are referred to the side (usually LV) from which the test is conducted and could easily be referred to the other side if so desired by the inverse square of transformation ratio. The transformation ratio if not known can be determined by connecting a voltmeter on the HV side as well in the no-load test.
It is, therefore, seen that the OC test yields the values of core-loss and parameters of the shunt branch of the equivalent circuit.
Short-circuit (SC) Test This test serves the purpose of determining the series parameters of a transformer.
For convenience of supply arrangement* and voltage and current to be handled, the test is usually conducted from the HV side of the transformer, while the LV is short-circuited as shown in Fig. 3.24. The equivalent circuit as seen from the HV under short-circuit conditions is drawn in Fig. 3.25(a). Since the transformer resistances and leakage reactances are very small, the voltage VSC needed to circulate the full-load current under short-circuit is as low as 5-8% of the rated voltage. As a result the exciting current I0 (SC) under these
I0 R1 X1 I0
V1 Gi B Em 1ªV1=E G1 i
Ii Im
Bm Y0 +
–
(a) (b)
+ +
– –
Fig. 3.23 Equivalent circuit as seen on open-circuit
* Voltage needed for the SC test is typically 5% of the rated value. For a 200 kVA, 440/6600- V transformer, test on the HV side would require
6600 5
100
¥ = 330 V and 200 1000 6600
¥ = 30 A supply
while if conducted from the LV side it would need
440 5
100
¥ = 22 V and 200 1000 440
¥ = 445 A supply
Low-voltage, high-current supply needed for conducting the SC test from the LV side is much more difficult to arrange than the supply required for the same test from the HV side.
conditions is only about 0.1 to 0.5% of the full-load current (I0 at the rated voltage is 2-6% of the full-load current). Thus the shunt branch of the equivalent circuit can be altogether neglected giving the equivalent circuit of Fig. 3.25(b).
While conducting the SC test, the supply voltage is gradually raised from zero till the transformer draws full-load current. The meter readings under these conditions are:
voltage = VSC; current = ISC; power input = PSC
Isc R1 X1 X2 R2
I0(sc)
Vsc Gl Bm ª Vsc
lsc
R1 R2 X1 X2
+
–
(a)
– +
(b)
R X
Fig. 3.25 Equivalent circuit under short-circuit conditions
Since the transformer is excited at very low voltage, the iron-loss is negligible (that is why shunt branch is left out), the power input corresponds only to the copper-loss, i.e.
PSC = Pc (copper-loss) (3.42)
From the equivalent circuit for Fig. 3.25(b), the circuit parameters are computed as below:
Z = V I
SC SC
= R2+X2 (3.43)
Equivalent resistance, R = P
I
SC
( SC)2 (3.44)
Equivalent reactance, X = Z2-R2 (3.45)
These values are referred to the side (HV) from which the test is conducted. If desired, the values could be easily referred to the other side.
HV W
Isc
Low-voltage supply (variable)
A
V
LV
Fig. 3.24 Short-circuit test on transformer
It is to be observed that the SC test has given us the equivalent resistance and reactance of the transformer;
it has not yielded any information for separating* these into respective primary and secondary values.
It was observed that OC and SC tests together give the parameters of the approximate equivalent circuit of Fig. 3.16(a) which as already pointed out is quite accurate for all important computations.
EXAMPLE 3.7 The following data were obtained on a 20 kVA, 50 Hz, 2000/200 V distribution transformer:
Draw the approximate equivalent circuit of the transformer referred to the HV and LV sides respectively.
Table 3.1
Voltage (V)
Current (A)
Power (W) OC test with HV open-circuited
SC test with LV short-circuited
200 60
4 10
120 300
SOLUTION OC test (LV side)
Y0 = 4
200 = 2 ¥ 10–2 ; Gi = 120 2002
( ) = 0.3 ¥ 10–2 Bm = Y02-Gi2 = 1.98 ¥ 10–2
SC test (HV side)
Z = 60
10 = 6 W ; R = 300 102 ( ) = 3 W X = Z2-R2 = 5.2 W
Transformation ratio, N
NHL = 2000 200 = 10 Equivalent circuit referred to the HV side:
Gi (HV) = 0.3 ¥ 10–2¥ 1 102
( ) = 0.3 ¥ 10–4 Bm (HV) = 1.98 ¥ 10–2¥ 1
102
( ) = 1.98 ¥ 10–4 The equivalent circuit is drawn in Fig. 3.26(a).
Equivalent circuit referred to the LV side:
R(LV) = 3 ¥ 1 102
( ) = 0.03 W
* Resistances could be separated out by making dc measurements on the primary and secondary and duly correcting these for ac values. The reactances cannot be separated as such. Where required, these could be equally appor-tioned to the primary and secondary, i.e.
X1 = X2 (referred to anyone side) This is sufficiently accurate for a well-designed transformer.
X(LV) = 5.2 ¥ 1 102
( ) = 0.052 W The equivalent circuit is drawn in Fig. 3.26(b).
EXAMPLE 3.8 The parameters of the equivalent circuit of a 150-kVA, 2400/240-V transformer are:
R1 = 0.2 W R2 = 2 ¥ 10–3W X1 = 0.45 W X2 = 4.5 ¥ 10–3W
Ri = 10 kW Xm = 1.6 kW (as seen from 2400-V side) Calculate:
(a) Open-circuit current, power and pf when LV is excited at rated voltage
(b) The voltage at which the HV should be excited to conduct a short-circuit test (LV shorted) with full-load current flowing. What is the input power and its pf?
SOLUTION Note: Ri = 1
Gi , Xm = 1 Bm Ratio of transformation, a = 2400
240 = 10 (a) Referring the shunt parameters to LV side
Ri (LV) = 10 1000
102
¥
( ) = 100 W
Xm (LV) = 1 6 1000
102 .
( )
¥ = 16 W
I0(LV) = 240 0
100
240 0 – ∞ - j 16– ∞
= 2.4 – j 15 = 15.2–– 80.9° A
or I0 = 15.2 A, pf = cos 80.9° = 0.158 lagging
(b) LV shorted, HV excited, full-load current flowing: Shunt parameters can be ignored under this condition.
Equivalent series parameters referred to HV side:
R = 0.2 + 2 ¥ 10–3¥ (10)2 = 0.4 W X = 0.45 + 4.5 ¥ 10–3¥ (10)2 = 0.9 W
Z = 0.4 + j 0.9 = 0.985 –66° W
IH 0.03W
VH
IOH I¢L
V¢L V¢H I¢H
VL
0.3¥10–4 1.98¥10–4 1.98¥10–2
0.3¥10–2
3W 5.2 W
W W W W
0.052W
(a) Referred to HV (b) Referred to LV
lOL lL
Fig. 3.26 Equivalent circuit
Ifl (HV) = 150 1000 2400
¥ = 62.5 A
VSC (HV) = 62.5 ¥ 0.958 = 59.9 V or 60 V (say) PSC = (62.5)2¥ 0.4 = 1.56 kW
pfSC = cos 66° = 0.406 lagging Sumpner’s (Back-to-Back) Test
While OC and SC tests on a transformer yield its equivalent circuit parameters, these cannot be used for the
‘heat run’ test wherein the purpose is to determine the steady temperature rise if the transformer was fully loaded continuously; this is so because under each of these tests the power loss to which the transformer is subjected is either the core-loss or copper-loss but not both. The way out of this impasse without conducting an actual loading test is the Sumpner’s test which can only be conducted simultaneously on two identical transformers*.
In conducting the Sumpner’s test the primaries of the two transformers are connected in parallel across the rated voltage supply (V1), while the two secondaries are connected in phase opposition as shown in Fig. 3.27. For the secondaries to be in phase opposition, the voltage across T2T4 must be zero otherwise it will be double the rated secondary voltage in which case the polarity of one of the secondaries must be reversed.
Current at low voltage (V2) is injected into the secondary circuit at T2T4. The supply (1) and supply (2) are from the same mains.
W1
–
A1
2l0
T1 T3
T2 T4
lfl V2 A2
W2 AC supply (1)
+
Low Voltage supply (2) V1
Fig. 3.27 Sumpner’s test on two identical single-phase transformers
* In very large sizes two identical transformers may not be available as these are custom-built.
As per the superposition theorem, if V2 source is assumed shorted, the two transformers appear in open-circuit to source V1 as their secondaries are in phase opposition and therefore no current can flow in them.
The current drawn from source V1 is thus 2I0 (twice the no-load current of each transformer) and power is 2P0 (= 2Pi, twice the core-loss of each transformer). When the ac supply (1) terminals are shorted, the transformers are series-connected across V2 supply (2) and are short-circuited on the side of primaries. Therefore, the impedance seen at V2 is 2Z and when V2 is adjusted to circulate full-load current (Ifl), the power fed in is 2Pc (twice the full-load copper-loss of each transformer). Thus in the Sumpner’s test while the transformers are not supplying any load, full iron-loss occurs in their cores and full copper-loss occurs in their windings; net power input to the transformers being (2P0 + 2Pc). The heat run test could, therefore, be conducted on the two transformers, while only losses are supplied.
In Fig. 3.27 the auxiliary voltage source is included in the circuit of secondaries; the test could also be conducted by including the auxiliary source in the circuit of primaries.
The procedure to connect a 3-phase transformer for the back-to-back test will be explained in Sec. 3.12.
EXAMPLE 3.9 Two transformers of 20 kVA each with turn-ratios respectively of 250 : 1000 and 250 : 1025 are connected in back-to-back test; the two primaries being fed from a 250 V supply and secondaries being connected in phase opposition. A booster transformer connected on primary side to the same 250 V supply is used to inject voltage in the circuit of secondaries such as to circulate a current of 20 A. The core losses of each transformer are 350 W and each transformer has a reactance 2.5 times its resistance.
Calculate the possible readings of the wattmeter connected to measure the input to the primaries.
SOLUTION Using the principle of superposition the currents on the primary side are first found, caused by the circulating current in the secondaries with the primary voltage source shorted; the voltage injected on the secondary side being intact. The primary currents necessary to balance the secondary circulating current are shown in Fig. 3.28; these being in phase with each other. The difference of these currents is 2 A which flows in the lines connecting the primaries to the main (refer to figure). This current has a power factor of
cos tan–1 2.5 = 0.371
Therefore, the power exchanged with the mains by this current is
250 ¥ 2 ¥ 0.371 = 185.5 W
This power will be drawn from or fed into the mains depending on the polarity of the injected voltage.
Mains
250 V 2A 250:1000
20 1000 250
¥ = 80 A
250: 1025
20 1025 250
¥ = 82 A
20 A
Fig. 3.28
Consider now the currents owing to the voltage source connected to the primaries with the secondary injected voltage source shorted. The primaries now draw the magnetizing currents from the mains with associated core-loss of both the transformers equal to 2 ¥ 350 = 700 W. The currents in the secondaries because of the small voltage unbalance (transformers have a slightly different turn ratio) would be small with very little associated loss.*
Hence power drawn from the mains is
700 ± 185.5 = 885.5 W or 514.5 W