10. Write the expression for the self inductance of a coil wound on an iron core
3.5 REAL TRANSFORMER AND EQUIVALENT CIRCUIT
Figure 3.12 shows a real transformer on load. Both the primary and secondary have finite resistances R1 and R2 which are uniformly spread throughout the winding; these give rise to associated copper (I2R) losses.
While a major part of the total flux is confined to the core as mutual flux f linking both the primary and secondary, a small amount of flux does leak through paths which lie mostly in air and link separately the individual windings. In Fig. 3.12 with primary and secondary for simplicity assumed to be wound separately on the two legs of the core, leakage flux fl1 caused by primary mmf I1N1 links primary winding itself and fl2
caused by I2N2 links the secondary winding, thereby causing self-linkages of the two windings. It was seen in Sec. 3.2 with reference to Fig. 3.2(a) that half the primary and half the secondary is wound on each core leg. This reduces the leakage flux linking the individual windings. In fact it can be found by tracing flux paths that leakage flux is now confined to the annular space between the halves of the two windings on each leg.
In shell-type construction, the leakage will be still further reduced as LV and HV pancakes are interleaved.
Theoretically, the leakage will be eliminated if the two windings could be placed in the same physical space but this is not possible; the practical solution is to bring the two as close as possible with due consideration to insulation and constructional requirements. Shell-type construction though having low leakage is still not commonly adopted for reasons explained in Sec. 3.2. Actually some of the leakage flux will link only a part of the winding turns. It is to be
Fig. 3.12 Real transformer
As the leakage flux paths lie in air for considerable part of their path lengths, winding mmf and its self-linkage caused by leakage flux are linearly related in each winding; therefore contributing constant leakage inductances (or leakage reactances corresponding to the frequency at which the transformer is operated) of both primary and secondary windings. These leakage reactances* are distributed throughout the winding though not quite uniformly.
Both resistances and leakage reactances of the transformer windings are series effects and for low operating frequencies at which the transformers are commonly employed (power frequency operation is at 50 Hz only), these can be regarded as lumped parameters. The real transformer of Fig. 3.12 can now be represented as a semi-ideal transformer having lumped resistances R1 and R2 and leakage reactances symbolized as Xll and Xl2 in series with the corresponding windings as shown in Fig. 3.13. The semi-ideal transformer draws magnetizing current to set up the mutual flux f and to provide for power loss in the core; it, however, has no winding resistances and is devoid of any leakage. The induced emfs of the semi-ideal transformer are E1 and E2 which differ respectively from the primary and secondary terminal voltages V1 and V2 by small voltage drops in winding resistances and leakage reactances (R1, Xl1 for primary and R2, Xl2 for secondary). The ratio of transformation is
a = N N12 = E
E12 ª V
V12 (3.27)
+ l1
V1 E1 E2 V2
l2 R1 X/1
N1 N2
X/2 R2
–
+ + +
– – –
f–
Fig. 3.13 Circuit model of transformer employing semi-ideal transformer
because the resistances and leakage reactance of the primary and secondary are so small in a transformer that E1ª V1 and E2ª V2.
Equivalent Circuit
In Fig. 3.13 the current I1 flowing in the primary of the semi-ideal transformer can be visualized to comprise two components as below:
(i) Exciting current I0 whose magnetizing component Im creates mutual flux f and whose core-loss component Ii provides the loss associated with alternation of flux.
(ii) A load component I2¢ which counterbalances the secondary mmf I2N2 so that the mutual flux remains constant independent of load, determined only by E1. Thus
I1 = I0+ ¢ (3.28)I2
* The transformer windings possess inter-turn and turns-to-ground capacitances. Their effect is insignificant in the usual low-frequency operation. This effect must, however, be considered for high-frequency end of the spectrum in electronic transformers. In low-frequency transformers also, capacitance plays an important role in surge phe-nomenon caused by switching and lightning.
where I¢
The exciting current I0 can be represented by the circuit model of Fig. 3.7 so that the semi-ideal transformer of Fig. 3.13 is now reduced to the true ideal transformer. The corresponding circuit (equivalent circuit) modelling the behaviour of a real transformer is drawn in Fig. 3.14( a) wherein for ease of drawing the core is not shown for the ideal transformer.
The impedance (R2 + jXl2) on the secondary side of the ideal transformer can now be referred to its primary side resulting in the equivalent circuit of Fig. 3.14(b) wherein
Xl2¢ = N
The load voltage and current referred to the primary side are
¢
Therefore there is no need to show the ideal transformer reducing the transformer equivalent circuit to the T-circuit of Fig. 3.14(c) as referred to side 1. The transformer equivalent circuit can similarly be referred to side 2 by transforming all impedances (resistances and reactances), voltages and currents to side 2. It may be noted here that admittances (conductances and susceptances) are transformed in the inverse ratio squared in contrast to impedances (resistances and reactances) which as already shown in Sec. 3.4 transform in direct ratio squared. The equivalent circuit of Fig. 3.14(c) referred to side 2 is given in Fig. 3.14(d) wherein
¢
With the understanding that all quantities have been referred to a particular side, a superscript dash can be dropped with a corresponding equivalent circuit as drawn in Fig. 3.14(d).
In the equivalent circuit of Fig. 3.14(c) if Gi is taken as constant, the core-loss is assumed to vary as E21 or f2max f 2 (Eq. (3.6)). It is a fairly accurate representation as core-loss comprises hysteresis and eddy-current
+
Fig. 3.14 Evolution of transformer equivalent circuit
loss expressed as (Khf1.6max f + Kef2max f 2). The magnetizing current for linear B-H curve varies proportional to fmaxμE
f1 . If inductance (Lm) corresponding to susceptance Bm is assumed constant.
Im = E f L1 m
2p = Bm E1
It therefore is a good model except for the fact that the saturation effect has been neglected in which case Bm would be a nonlinear function of E1/f. It is an acceptable practice to find the shunt parameters Gi, Bm at the rated voltage and frequency and assume these as constant for small variations in voltage and frequency.
The passive lumped T-circuit representation of a transformer discussed above is adequate for most power and radio frequency transformers. In transformers operating at higher frequencies, the interwinding capacitances are often significant and must be included in the equivalent circuit. The circuit modification to include this parameter is discussed in Sec. 3.12.
The equivalent circuit given here is valid for a sinusoidal steady-state analysis. In carrying out transient analysis all reactances must be converted to equivalent inductances.
The equivalent circuit developed above can also be arrived at by following the classical theory of magnetically coupled circuits, section 3.22. The above treatment is, however, more instructive and gives a clearer insight into the physical processes involved.
Phasor Diagram of Exact Equivalent Circuit of Transformer [Fig. 3.14(a)]
The KVL equations for the primary and secondary circuits are V1 = E I R1+ 1 1+ j I X1 1 V2 = E2-I R2 2- j I X2 1 The ideal transformer relationships are
E
E12 = a ; I¢ I22
= 1 a The nodal equation for the primary side currents is
I1 = I2¢ + ¢I0 =I2¢ +
(
Ii+Im)
where Ii is in phase with E1
Im is 90° lagging E1
The exact phasor diagram from these equations is drawn in Fig. 3.15.
Alternative Phasor Diagram Alternatively if we use e = -d dt
l ; direction of E1 and E2 will reverse in Fig. 3.13 and these will lag the flux phasor by 90°. The direction of secondary current is now into the dotted terminal. So for mmf balance
¢ =
-I2 I2. The KVL equation on the primary side is
V1 = (-E1)+ ¢ +I R2 2 j I X2 2¢
The corresponding phasor diagram is drawn in Fig. 3.16. Note that the polarity of V2 reverses in Fig. 3.13 but it is of no consequence.
Ii
q2
I0 Im
q1
v2 I X2 2
I R2 2
I R1 1 E1 E2
I X1 1
I¢2
I1
I0 I2
V1
O
Fig. 3.15
I1
I¢2 Im
q1
q2 V2
E2
I X2 2
I R2 2
I2 –E1
O
I X1 1
V1
I R1 1
Ii
f–
Fig. 3.16
EXAMPLE 3.4 Consider the transformer of Example 3.2 (Fig. 3.8) with load impedance as specified in Example 3.3. Neglecting voltage drops (resistive and leakage reactive drops), calculate the primary current and its pf. Compare with the current as calculated in Example 3.3.
SOLUTION As per Eqs (3.28) and (3.29)
I1 = I0+ ¢ and I2 I¢ I22 = N
N21 As calculated in Example 3.3
¢
I2 = 10 ––30° A
Further as calculated in Example 3.2
Compared to the primary current computed in Example 3.3 (ignoring the exciting current) the magnitude of the current increases slightly but its pf reduces slightly when the exciting current is taken into account.
In large size transformers the magnitude of the magnetizing current is 5% or less than the full-load current and so its effect on primary current under loaded conditions may even be altogether ignored without any significant loss in accuracy.
This is a usual approximation made in power system computations involving transformers.
EXAMPLE 3.5 A 20-kVA, 50-Hz, 2000/200-V distribution transformer has a leakage impedance of 0.42 + j 0.52 W in the high-voltage (HV) winding and 0.004 + j 0.05 W in the low-voltage (LV) winding. When seen from the LV side, the shunt branch admittance Y0 is (0.002 – j 0.015) (at rated voltage and frequency).
Draw the equivalent circuit referred to (a) HV side and (b) LV side, indicating all impedances on the circuit.
SOLUTION The HV side will be referred as 1 and LV side as 2.
Transformation ratio, a = N
N12 = 2000
200 = 10 (ratio of rated voltages; see Eq. (3.27))
(a) Equivalent circuit referred to HV side (side 1) Z2¢ = (10)2 (0.004 + j 0.005) = 0.4 + j 0.5 W Y0¢ = 1
102
( ) (0.002 – j 0.015) (Notice that in transforming admittance is divided by a2)
The equivalent circuit is drawn in Fig. 3.17(a).
(b) Equivalent circuit referred to LV side (side 2).
Z1¢ = 1 102
( ) (0.42 + j 0.52) = 0.0042 + j 0.0052 The equivalent circuit is drawn in Fig. 3.17(b).
Approximate Equivalent Circuit
In constant frequency (50 Hz) power transformers, approximate forms of the exact T-circuit equivalent of the transformer are commonly used. With reference to Fig. 3.14(c), it is immediately observed that since winding resistances and leakage reactances are very small, V1ªE1 even under conditions of load. Therefore, the exciting current drawn by the magnetizing branch (Gi || Bm) would not be affected significantly by shifting it to the input terminals, i.e. it is now excited by V1 instead of E1 as shown in Fig. 3.18(a). It may also be observed that with this approximation, the current through R1, Xl1 is now I¢2 rather than I1 = I0+ ¢ . Since II2 0
is very small (less than 5% of full-load current), this approximation changes the voltage drop insignificantly.
Thus it is basically a good approximation. The winding resistances and reactances being in series can now
0.42 + 0.52j W 0.4 + 0.5j W 10:1
be combined into equivalent resistance and reactance of the transformer as seen from the appropriate side (in this case side 1). Remembering that all quantities in the equivalent circuit are referred either to the primary or secondary dash in the referred quantities and suffixes l, 1 and 2 in equivalent resistance, reactance and impedance can be dropped as in Fig. 3.18(b).
Here Req (equivalent resistance) = R1 + R2 Xeq (equivalent reactance) = Xl1 + Xl2 Zeq (equivalent impedance) = Req + j Xeq
In computing voltages from the approximate equivalent circuit, the parallel magnetizing branch has no role to play and can, therefore, be ignored as in Fig. 3.18(b).
The approximate equivalent circuit offers excellent computational ease without any significant loss in the accuracy of results. Further, the equivalent resistance and reactance as used in the approximate equivalent circuit offer an added advantage in that these can be readily measured experimentally (Sec. 3.7), while separation of Xl1 and Xl2 experimentally is an intricate task and is rarely attempted.
The approximate equivalent circuit of Fig. 3.18(b) in which the transformer is represented as a series impedance is found to be quite accurate for power system modelling [7]. In fact in some system studies, a transformer may be represented as a mere series reactance as in Fig. 3.18(c).
This is a good approximation for large transformers which always have a negligible equivalent resistance compared to the equivalent reactance.
The suffix ‘eq’ need not be carried all the time so that R and X from now onwards will be understood to be equivalent resistance and reactance of the transformer referred to one side of the transformer.
Phasor Diagram
For the approximate equivalent circuit of Fig. 3.18(b), (suffix ‘eq’ is being dropped now),
V2 = V1-I Z (3.32a)
or V1 = V2+ (R + jX ) (3.32b) I
The phasor diagram corresponding to this equation is drawn in Fig. 3.19(a) for the lagging power factor (phase angle* f between V2 and I ) and in Fig. 3.19(b) for the leading power factor (pf ). It is immediately observed from these phasor diagrams that for the phase angle indicated
V2 < V1 for lagging pf
* Phase angle f should not be confused with flux though the same symbol has been used.
lZ V1
O d
f
f
B f D
E C F
IR I 90° IX V2
(a) Lagging pf
A
I
V1
IX
V2 f d IR
(b) leading pf O
Fig. 3.19
In the plasor diagrams of Figs. 3.19(a) and (b) (these are not drawn to scale), the angle d is such that V1
leads V2. This is an indicator of the fact that real power flows from side 1 to side 2 of the transformer (this is proved in Section 8.9). This angle is quite small and is related to the value of the equivalent reactance:
resistance of the transformer being negligible.
Name Plate Rating
The voltage ratio is specified as V1 (rated)/V2 (rated). It means that when voltage V1 (rated) is applied to the primary, the secondary voltage on fullload at specified pf is V2 (rated). The ratio V1 (rated)/V2 (rated) is not exactly equal to N1/N2, because of voltage drops in the primary and secondary. These drops being small are neglected and it is assumed that for all practical purposes
V V12
( )
( )
rated rated = N
N12 (3.33)
The rating of the transformer is specified in units of VA/kVA/MVA depending upon its size.
kVA(rated) = V(rated)¥I(full-load)
1000 (3.34)
where V and I are referred to one particular side. The effect of the excitation current is of course ignored.
The transformer name plate also specifics the equivalent impedance, but not in actual ohm. It is expressed as the percentage voltage drop (see Sec. 3.8) expressed as
I Z
V
( )
( )
full-load
rated ¥ 100%
where all quantities must be referred to anyone side.
EXAMPLE 3.6 The distribution transformer described in Example 3.5 is employed to step down the voltage at the load-end of a feeder having an impedance of 0.25 + j 1.4 W. The sending-end voltage of the feeder is 2 kV. Find the voltage at the load-end of the transformer when the load is drawing rated transformer current at 0.8 pf lagging. The voltage drops due to exciting current may be ignored.
SOLUTION The approximate equivalent circuit referred to the HV side, with the value of transformer impedance from Fig. 3.17(a), is drawn in Fig. 3.20. The feeder being the HV side of the transformer, its impedance is not modified.
Rated load current (HV side) = 20 2 = 10 A
Z = (0.25 + j 1.4) + (0.82 + j 1.02) = 1.07 + j 2.42 = R + jX
One way is to compute V2 from the phasor Eq. (3.32). However, the voltage drops being small, a quick, approximate but quite accurate solution can be obtained from the phasor diagram of Fig. 3.19(a) without the necessity of carrying out complex number calculations.
From Fig. 3.19(a)
OE = (OA)2-(AE)2 From the geometry of the phasor diagram
AE = AF – FE
= IX cos f – IR sin f
= 10(2.42 ¥ 0.8 – 1.07 ¥ 0.6) = 12.94 V
Now OE = (2000)2-( . )12 94 2 = 1999.96 V It is therefore seen that
OE ª OA = V1 (to a high degree of accuracy; error is 2 in 105) V2 can then be calculated as
V2 = OE – BE ª V1 – BE
Now BE = BD + DE
= I(R cos f + X sin f)
= 10(1.07 ¥ 0.8 + 2.42 ¥ 0.6) = 23.08 V
Thus V2 = 2000 – 23.08 = 1976.92 V
Load voltage referred to LV side = 1976 92 10
. = 197.692 V
Remark It is noticed that to a high degree of accuracy, the voltage drop in transformer impedance can be approximated as
V1 – V2 = I(R cos f + X sin f); lagging pf (3.35a) It will soon be shown that
V1 – V2 = I(R cos f – X sin f); leading pf (3.35b) 3.6 TRANSFORMER LOSSES
The transformer has no moving parts so that its efficiency is much higher than that of rotating machines. The various losses in a transformer are enumerated as follows:
Feeder
0.25 + 1.4j 0.82 + 1.02j I = 10 A 0.8 pf lag Transformer
– +
Load +
– V2 V1= 2000V
Fig. 3.20
Core-loss
These are hysteresis and eddy-current losses resulting from alternations of magnetic flux in the core.
Their nature and the remedies to reduce these have already been discussed at length in Sec. 2.6. It may be emphasized here that the core-loss is constant for a transformer operated at constant voltage and frequency as are all power frequency equipment.
Copper-loss (I2R-loss)
This loss occurs in winding resistances when the transformer carries the load current; varies as the square of the loading expressed as a ratio of the full-load.
Load (stray)-loss
It largely results from leakage fields inducing eddy-currents in the tank wall, and conductors.
Dielectric-loss
The seat of this loss is in the insulating materials, particularly in oil and solid insulations.
The major losses are by far the first two: Pi, the constant core (iron)-loss and Pc, the variable copper-loss.
Therefore, only these two losses will be considered in further discussions.
It will be seen in Sec. 3.7 that transformer losses and the parameters of its equivalent circuit can be easily determined by two simple tests without actually loading it.