Unit  5
Transformers
A transformer is a stationary device which works on the principle of Mutual Induction. It is used to transform electric power from one circuit to another having same frequency. It can lower or higher the value of voltage in any circuit but in correspondence to change in current.
An ideal transformer has no losses i.e., its winding has no magnetic leakage and no ohmic resistance. Hence, an ideal transformer has only two purely inductive coils wound on a lossfree core.
Fig 1. Ideal Transformer
For above transformer when secondary is open and primary is having input sinusoidal voltage V1. An alternating current flow due to difference in potential. As primary coil is purely inductive so, Iµ current is drawn through it. This current is very small and logs V1 by 900.
The current Iµ produces magnetic flux φ and hence are in same phase. The flux is linked with both the windings and hence, selfinduced emf is produced E1 which is equal and opposite of V1. Similarly, E2 is induced in secondary which is mutually induced emf E2 is proportional to rate of change of flux and number of secondary windings.
The phasor is shown below.
Fig 2 Phasor for Ideal Transformer
Emf Equation of Transformer:
N1 = No of primary turns
N2 = No of secondary turns
Φm = max flux in cone (wb)
Φm = Bm x A
f = Frequency of input(ac) Hz
Average rate of change of flux
= Φm / ¼
= 4 ΦmF volts
Average emf per turn = 4F ΦmV
As Φ varies sinusoidally rms value of induced emf is given as
Form factor = r.m.s value / average value = 1.11
r.m.s value of emf/turn = 1.11 x 4f Φm
= 4.44 f Φm volt
r.m.s value of induced emf in primary winding
= (induced emf/turn) x No of primary turns
E1 = 4.44 fN1 Φm
E1 = 4.44 fN1BmA
r.m.s emf induced in secondary
E2 = 4.44 fN2 Φm
E2 = 4.44 fN2BmA
E1/E2 = N1/N2 = K
K – voltage transformation
(i). If N2 > N1 i.e., K > 1 STEP UP TRASFORMER
(II). If N1 > N2 i.e., K<1 STEP DOWN TRASFORMER
Key takeaway
FOR Ideal Transformer
V1 I1 = V2 I2 = 1/K
Hence, current is inversely proportional to the voltage transformation ratio.
Examples
Q1>. A 25 KVA transformer has 500 turns on primary and 50 on secondary. The primary is connected to 2000V, 50 Hz supply. Find full load primary and secondary currents, the secondary emf and max flux in core. Neglect leakage drop and no load primary current.
Sol: K = N2/N1 = 50/500 = 1/10
I1 = 25,000/2000 = 12.5 A
I2 = I1/K = 10 x 12.5 A = 125 A
Emf/turn on primary side = 2000/500 = 4 V
E2 = KE1
E2 = 4 x 50 = 200 v
E1 = 4.44fN1 Φm
2000 = 4.44 x 50 x 500 x Φm
Φm = 18.02 m Wb
The basic transformer and its equivalent circuit both are shown below,
Fig 3. Equivalent Transformer Circuit
Iµ  magnetising component of current
Iw = working component
R0 – Non inductive resistance
I0 – No load current
X0 = E1/I0. R0 = E1/Iw
E2/E1 = N2/N1 = K
E’2 = E2/K = E1
V’2 = V2/K
I’2 = K I2
The total equivalent circuit is again given as,
Fig 4 Transformer Equivalent Circuit
But the above circuit is exact equivalent but harder to solve so, it can be further simplified as,
Fig 5 Reduced Equivalent Circuit
Z = Z1 + Zm  (Z’2 + Z’L)
= Z1 + Zm (Z’2 + Z’L)/Zm + (Z’2 + Z’L)
Z’2 = R’2 + jX’2
Zm = impedance of exciting circuit
V1 = I1[ Z1 + Zm (Z’2 + Z’L) / Zm + (Z’2 + Z’L)]
Voltage Phasor diagrams and vector groups
Digit 1 =30° lagging (LV lags HV with 30°) because rotation is anticlockwise.
Phase Shift (Deg)  Connection  
0  Yy0  Dd0  Dz0 
30 lag  Yd1  Dy1  Yz1 
60 lag 
 Dd2  Dz2 
120 lag 
 Dd4  Dz4 
150 lag  Yd5  Dy5  Yz5 
180 lag  Yy6  Dd6  Dz6 
150 lead  Yd7  Dy7  Yz7 
120 lead 
 Dd8  Dz8 
60 lead 
 Dd10  Dz10 
30 lead  Yd11  Dy11  Yz11 




Group  O’clock  TC 
Group I  0 o’clock, 0°  delta/delta, star/star 
Group II  6 o’clock, 180°  delta/delta, star/star 
Group III  1 o’clock, 30°  star/delta, delta/star 
Group IV  11 o’clock, +30°  star/delta, delta/star 
Minus indicates LV lagging HV, plus indicates LV leading HV 
Fig 6 Clock Notation: 0
Fig 7 Clock Notation: 1
Fig 8 Clock Notation: 2
Fig 9 Clock Notation: 4
Fig 10 Clock Notation: 5
Fig 11 Clock Notation: 6
Fig 12 Clock Notation: 7
Fig 12 Clock Notation: 11
Key takeaway
Vector Groups are the IEC method of categorizing the primary and secondary winding configurations of 3phase transformers. Windings can be connected as delta, star, or interconnectedstar (zigzag). Winding polarity is also important, since reversing the connections across a set of windings affects the phaseshift between primary and secondary.
(1). Core or Iron Loss:
This includes both hysteresis loss and eddy current loss. Iron or core loss is found from open circuit test.
Hysteresis Loss (Wh) = nB1.6max f V Watt
Eddy current Loss (We) = P B2max f2 t2 Watt
(2). Copper Loss:
This is due to ohmic resistance of the transformer windings.
Total cu loss = I21 R1 + I22 R2
= I21 R01 + I22 R02
Cu loss α I2
The value of cu loss is found from shortcircuit test.
Efficiency of transformer:
Basically, efficiency is defined as
n = output/input
But for transformer there are small amount of losses so the improved way to find efficiency is
n = output/output + losses
n = output/output + cu loss + iron loss
or n = Input – losses/Input
= 1 – Losses/Input
Condition for maximum efficiency:
For n to be maximum dn/dI1 = 0
(Ww) cu loss = I21 R01 or I22 R02
Iron loss = Hysteresis loss + Eddy current loss
= Wn + We = Wi
n = Input – losses/Input
Primary Input = V1I1 Cosφ1
n = V1 I1 Cosφ1 – losses/V1 I1 cos φ1
n = V1 I1 cos φ1 – I21 R01 – Wi / V1I1 cosφ1
= 1 – I1R01/V1cosφ1 – Wi/V1I1cosφ1
differentiating w.r.t I1 both sides of above equation we have
dn/dI1 = 0 – R01/V1cosφ1 + Wi/V1I21 Cosφ1
for max value dn/dI1 = 0
R01/V1cosφ1 = Wi/V1I21 cosφ1
Wi = I21 = I21 R01
Hence,
Wi = Wcu
Iron loss = copper loss
The value of output current for maximum efficiency will be
I2 = √Wi/R02
The maximum efficiency can also be given as,
nmax = full load x √ Iron loss / F. L .cu loss
or
nmax = R’ x full load KVA x pf / R’ x full load KVA x pf + Wi + Wcu x 100
R’ = ratio of actual to full load KVA
Wi = iron loss (KW)
Wcu = copper loss (KW)
Key takeaway
n = output/output + cu loss + iron loss
The maximum efficiency can also be given as,
nmax = full load x √ Iron loss / F. L .cu loss
Example
Q1). In a 50 KVA, 2200/200 V, 1φ transformer, the iron and fullload copper losses are 400 W and 450 W respectively. Calculate n at unity power factor on (i). Full load (ii). halffull load?
Sol. (i). Total loss = 400 + 450 = 850 W
F.L output at unity power factor = 50 x 1
= 50 KVA
n = 50 / 50 + .850 = 50/50.850 = 0.98 = 98%
(ii). half full load, unity pf
= 50 KVA/2 = 25 KVA
Cu loss = 400 x (1/2)2 = 100 W
Iron loss is same = 450 W
Total loss = 100 + 450 = 550 W
n = 25/25 + 0.55 = 25/25.55 = 0.978 = 97.8 %
Q>. A 40 KVA 440/220 V, 1 φ, 50 Hz transformer has iron loss of 300 W. The cu loss is found to be 100 W when delivering half fullload current. Determine (i) n when delivering full load current at 0.8 lagging pf (ii) the percentage of fullload when the efficiency will be max.
Sol. Full load efficiency at 0.8 pf
= 40 x 0.8/ (40 x 0.8) + losses
Full load cu loss = (440/220)2 x 100
= 400 W
Iron loss = 400 + 300
= 700 W
n = 40 x 0.8/ (40 x 0.8) + 0.7 = 97.8 %
(ii). KVA for maximum / F.L KVA = √ iron loss / F.L cu loss
= √300/400 = 0.866
(1). Open circuit test
(2). Short circuit test
(1). OPEN CIRCUIT TEST:
Basically, these tests are performed to find the basic parameters of transformers. The open circuit test is performed to determine noload loss (core loss) and noload current I0 which is helpful in finding X0 and R0.
Fig. 13 Open Circuit Test
In this test the high voltage winding is left open and the other is connected to supply.
A wattmeter W, voltmeter V and an Ammeter A are connected to L.V. winding.
Due to applied voltage in primary, flux is set up in core, and hence normal iron loss occurs. This is recorded in Wattmeter. The current I0 is measured by Ammeter A.
Wattmeter reading represents core loss under noload condition.
The wattmeter reading will be given as
W = V1 I0 cos φ0
Iµ = I0 Sin φ0
Iw = I0 cos φ0
X0 = V1/Iµ, R0 = V1/Iw
At no load I0 ≈ Iµ
I0 = V1 Y0
Y0 – Exciting Admittance
W = V21 G0
G0 – Exciting conductance
B0 = √ Y20 – G20
B0 – Exciting susceptance
(2). SHORT CIRCUIT TEST:
This test is used to determine
(i). Equivalent Impedance (Z01, Z02), leakage reactance’s (X01 or X02), Total resistance (R01, R02).
(ii). cu losses are also calculated at full load.
(iii). the regulation of transformer can be determined as Z01, Z02 can be calculated.
Fig 14 Short Circuit Test
Here low voltage is short circuited. And the supply is through high voltage (primary here).
As applied voltage is small, hence the flux φ induced is also small.
As core losses are very small, the wattmeter reading is the culoss at full load for whole transformer.
The equivalent circuit is shown below,
Fig 15 Equivalent Circuit
Z01 = Vsc/I1
Vsc = voltage to circulate rated load current
W = I21 R01
R01 = W / I21
X01 = √ Z201 – R201
R’2 = R01 – R1
Polarity test
The polarity test of a transformer can be done by measuring only voltage at no load. The basic connections are shown in below circuit.
Fig 16: Polarity Test
From above circuit input is applied for one winding. There is another voltage ‘V’ measured between one winding from each terminal. If voltage V’ > V, polarity is additive. If voltage V’ < V, polarity is subtractive.
BacktoBack Test:
Sumpner's test or backtoback test on transformer is another method for determining transformer efficiency, voltage regulation and heating under loaded conditions. Sumpner's test or backtoback test is done with two identical transformers. Both transformers are connected to supply such that one transformer is loaded on another. Primaries of the two identical transformers are connected in parallel across a supply. Secondaries are connected in series such that emf's of them are opposite to each other. Another low voltage supply is connected in series with secondaries to get the readings, as shown in the circuit diagram shown below.
Fig 17: BacktoBack Test
In above diagram, T1 and T2 are identical transformers. The secondaries are of opposite polarity the emf's cancel each other, as transformers are identical. In this case, as per superposition theorem, no current flows through secondary. The current drawn from V1 is 2I0, where I0 is equal to no load current of each transformer. Thus, input power measured by wattmeter W1 is equal to iron losses of both transformers. So, we can write iron loss per transformer
Pi = W1/2.
Now, a small voltage V2 is injected into secondary with the help of a low voltage transformer. The voltage V2 is adjusted so that, the rated current I2 flows through the secondary. In this case, both primaries and secondaries carry rated current. Thus, short circuit test is simulated and wattmeter W2 shows total full load copper losses of both transformers.
i.e., copper loss per transformer PCu = W2/2.
Key takeaway
(i) The power required to carry out these tests is very small as compared to the fullload output of the transformer. In case of open circuit lest, the power required is equal to the iron loss whereas, for a shortcircuit test, the power required is equal to fullload copper loss.
(ii) These tests enable us to determine the efficiency of the transformer accurately at any load and p.f. without actually loading the transformer.
(iii) The shortcircuit test enables us to determine R01 and X01 (or R02 and X02). We can thus find the total voltage drop in the transformer as referred to primary or secondary. This, permits us to calculate voltage regulation of the transformer.
Examples
Q1>. The primary and secondary windings of a 30 KVA, 5000/330 V, 1 φ transformer having resistance of 15 Ω and 0.02 Ω. The reactance referred to primary is 34 Ω. Calculate primary voltage required to circulate fullload current when the secondary is S.C. Also calculate the pf?
Sol. K = 330/5000 = 33/500
X01 = 34 Ω
R01 = R1 + R2/K2 = 15 + 0.02(500/33)2 = 19.59 Ω
Z01 = √ R201 – X201
= √ 19.592 + 342
Z01 = 39.23 Ω
F. L I1 = 30,000/5000 = 6A
Vsc = I1 Z01 = 6 x 39.23 = 235.4 V
S.C. power factor = R01 / Z01 = 19.59/39.23 = 0.5
Q.2). In noload test of 1 φ transformer, the test data are
Primary voltage = 200 V,
Resistance of primary = 0.6 Ω
Secondary voltage = 100 V
Primary current = 0.6 A
Power input = 30 W
Find (a) Turns ratio (b) Magnetising component of noload current (c) Working component (d) Iron loss
Sol>. (a). N1/N2 = 200/100 = 1
(b). W = V1 I0 cos φ0
cos φ0 = 30/200 x 0.6 = 0.25, sin φ0 = 0.97
Iµ = I0 sin φ0 = 0.6 x 0.97 = 0.58 A
©. Iw = I0 Cos φ0 = 0.6 x 0.25 = 0.15 A
(d). cu loss = I20 R1 = (0.6)2 x 0.6 = 0.216 W
Iron loss = 30 – 0.216 = 29.78 W
Q.3 Obtain the secondary voltage when delivered 10 KW at 0.8 PF. logging. The primary volt be 220v. the 300/600 v 50H2, 1 Transformer has following test results.
O.C test:200v, 0.8A 70WL. V side
S.C test: 12V,10A 80WHV side
Soln:  O.C TEST
W=COS
70=200*0.8*Cos
Cos= 0.436
Sin=0.899
= COS=0.8*0.436=0.35A
I = sin = 0.8*0.899=0.72A
===571.4
===277.8𝛺
S.C TEST:
= ==1.2A
K==2
Z01===0.3
I22=W
RO2==0.7
R01===0.175𝛺
X01=
=
X01 = 0.244
output KVA ==12.5
I2= = 20.83A
Z02=1.2A R02=0.7𝛺
X02= =0.975𝛺
Total transformer drop refereed to secondary
= I2 (RO2COS + X02Sin)
= 20.83(0.7*0.8+0.975*0.6)
= 23.85v
= 600  23.85=576.15v
The core loss of a transformer depends on the frequency and the maximum flux density when the volume and the thickness of the core laminations are known. The core loss comprises of two losses:
a) Hysteresis Loss Wh=P f
b) Eddy Current Loss We= Q f2 where Q is constant
Total core loss Wi= Wh+ We
= P f+ Q f2
To find P and Q we need to carry out two different experiments at two different frequencies but with same maximum flux density. After having values of P and Q hysteresis and eddy current losses can easily be calculated.
As the supply is in 3phase so we need this transformer. Its alternating voltage and currents differ in phase by 120 degrees.
Fig:17 Three phase Transformer
Fig18: Line and phase voltage
Construction:
a) Core Type:
The construction of a core type three phase transformer is as shown in the figure. The core consists of three legs or limbs. The core is made up of thin laminated sheets to reduce eddy current losses. Each limb has primary and secondary windings in cylindrical shape (former wound) arranged concentrically. The construction is shown in the figure.
Fig 19: Core type 3phase transformer
b) Shell Type:
In a shell type three phase transformer, three phases are more independent than they are in core type. Each phase has its individual magnetic circuit. The construction of shell type three phase transformer is illustrated in the figure at right. The construction is similar to that of three single phase shell type transformers kept on the top of each other.
Fig 20: Shell Type 3phase Transformer
Star – star or Y/Y connection: 
The below figure shows transformers connected in Y/Y on both primary and secondary sides. The ratio of line voltage and transformation ratio of line voltage and transformation ratio of both primary and secondary of each transformer in same. Hence the phase voltage VP = . VLine (line voltage)
This connection works only if the load is balanced.
Fig 21: Three phase transformer YY connection
Delta – Delta or Δ – Δ connection: 
Fig 22: ΔΔ connection
Wye / Delta or Y/Δ connections: 
Fig 23: Y/Δ connection
Delta/Wye or Δ/Y connection:
Fig 24: ΔY connection
Comparative features:
Key takeaway
Que 1. A 3Ø, 50Hz transformer has delta connected primary and starconnected secondary, the line voltage being 20,000 V and 500 V respectively. The secondary has a star connected balanced load at 0.8 lagging p.f. The line current of primary side is 5A. Determine the current in each coil of the primary and secondary line. Also find the output of the transformer?
→ Phase voltage on primary side = 20,000 V
Phase voltage on secondary = V
K =
Primary phase current =
Secondary phase current =
=
= 200 A
Output = VLILcosØ
=
= 138.56 kW
Que 2. A 100 kVA, 3Ø, 50 Hz, 3500/500 V transformer is Δ – connected on the h. V side and Yconnected on the L.V. side. The resistance of h.V. winding is 3.5 Ω / phase and that of L.V is 0.02 Ω per phase. Calculate the iron losses of the transformer at normal voltage and frequency if its full load efficiently be 95.1 % at 0.8 p.f.?
→ Full load output = 100 × 0.8 = 80 kW
Input = kW
Total loss = Input – Output
= 83.2 – 80
= 3200 kW
K =
K =
=
RO2 = R2 + k2R1
=
= 0.044 Ωs
Fullload secondary phase current I2
=
= 115.4 A
Total Cu Loss = 3I22RO2
= 3 × 0.044 × (115.4)2
= 1760 W
Iron Loss = Total loss – Cu Loss
= 3200 – 1760
= 1440 W
Parallel operation of singlephase transformer
If the present transformers are not capable of supplying any amount of additional load which is required, then the parallel connections are used to fulfil the requirement. The basic circuit for parallel transformer is shown below.
The condition required for parallel connections: 
Fig.25 Parallel Connection of 1ØTransformer
1) Ideal Case:
Here both the transformers are considered to have same voltage ratio and impedance voltages triangles identical in size and shape.
Fig.26 Phasor for ideal Case
Fig 27: Ideal Case
I = IA + IB
V2 = E  IAZA
IAZA = IBZB
IA =
IB =
E – No load secondary voltage
V2 – Terminal voltage
IA, IB – Currents of each Transformer
2) Equal Voltage Ratio: 
Let Noload voltage of each secondary be same E = EA = EB. The change in the secondary is shown below.
Fig 28: Equal Voltage Ratio
All values are considered referred to the secondary.
ZA, ZB = impedances of transformer
IA, IB = currents of transformer
V2 = common terminal voltage
IAZA = IBZB = IZAB
ZAB = ZA  ZB
ZA = IZABZA,
IA =
V2IA =
V2IB =
Let V2I × 103 = S (Combined Load KVA)
SA =
SB =
3) Unequal Voltage Ratios: 
Here the no load secondary voltages are unequal.
Fig.29: Unequal Voltage Ratios
EA, EB = noload secondary emp
Z1 = load impedance across secondary.
Here circulating current exist as the voltage are unequal (IC).
IC =
EA = IAZA + V2
EB = IBZB + V2
V2 = IZL = (IA + IB) ZL (1)
EA = IAZA + (IA + IB) ZL (2)
EB = IBZB + (IA + IB) ZL (3)
EA – EB = IAZA – IBZB
IA =
If ZA and ZB are smaller than ZL.
Substitute IA in equation (3) and solve
IA =
IB =
Neglecting ZA ZB in comparison with ZL (ZA + ZB)
We get above equations.
I = IA + IB
I =
V2 = EA  IAZA and V2 = EB – IBZB
Parallel operation of 3Ø Transformer
The conditions are same as in case of 1Ø transformer.
Key takeaway
Example
Que 1. What should be the kVA rating of each transformer in VV bank when 3Ø balanced load is 50 kVA. If a third transformer is connected for operation find % increase in rating?
→ kVA for each transformer has to be 15%
kVA per transformer =
= 28.75
ΔΔ bank rating = 28.75 × 3
= 86.75
Increase =
= 72.5 %
Que 2. A balanced 3Ø load of 120 kW at 1100V 0.866 lag p.f is supplied from 1500V, 3Ø main through 1Ø transformer connected – (i) ΔΔ (ii) YY
Find current in the windings of each transformer and p.F at which they operate in each case.
→ (i) ΔΔ
VLILcosØ = 120,000
× 1100 × IL × 0.866 = 120,000
IL = 72.73 A
Secondary line current = 72.73 (= IL)
Secondary phase current = = 41.99 A
K =
Primary phase current = = 20.99 A
(ii) YY connection
× 1100 × I × 0.866 = 120,000
I = 72.73 A
Secondary phase current = 72.73 A
Primary phase current = = 41.99 A
Transformer p.f = 86.6 % of 0.866
= 0.75 (lag)
Que 3. Two transformers are required for a Scott connection operating from 410 V 3Ø supply for supplying two 1Ø furnaces at 200 V on the twophase side. If total output is 130 kVA, calculate secondary to primary turn ratio and the winding of each transformer?
→ For main Transformer
Primary voltage = 410 V
Secondary voltage = 200 V
Secondary current = = 325 A
Primary current = 325 ×
= 153.54 A
For teaser Transformer
Primary volts = = 355 V
Secondary volts = 200 V
It is a transformer in which only one winding is common it both primary and secondary because it has one windy it uses loss copper and is cheaper.
It can be step up and step down autotransformer. the circuit for both is shown below.
Fig 30a: Step up auto transformer
Fig30 b: Step Down Autotransformer
CBPrimary winding
BA Secondary windings
Negating iron loss and noload current
===K
The current in AB () is vector difference () >
An auto transformer when compared to as ordinary 2 winding transformer having same output, given higher efficiency and is smaller in size with superior voltage regulation.
FOR AUTO TRANSFORMER:
Weight of copper in AC (N1N2) I1
Weight of copper in BA (N2(I2I1)
Total weight of copper (N1N2) I1+N2(I2I1) auto transformer
For 2 winding transformer:
Weight of on primary N1I1
Weight of on primary N2I2
Total weight of N1I1+N2I2
=
= = 1=1k
= K, =
Weight in auto transformer =(1k) *(weight of in ordinary transformer)
Saving = WoWa
= W0(1k) W0
Saving = Kwo
Power transformer inductively =(1k) input hence, saving will increase at K approaches unity.
Example
Q. An auto transformer suppliers load of 4KW at 100v at unity pf. IF the applied primary voltage is 220v. calculating power transferred to load (a) Inductively (b) conductively.
Sol: Power transferred inductively = Input(1k)
Power transferred conductively = K* Input
K= =
Input= Output =4KW
Inductively transferred power =4()
=3.82KW
Conductively transferred power = *4
= 0.182 kw
Magnetizing current is the term used to denote the total current that flows into the primary of a transformer when the transformer is energized at a specific voltage and frequency, with the secondaries open circuited. Although known as magnetizing current it is actually the combination of the current required to magnetize the core (I1) and the current required to supply the losses in the core (I2).
Fig 31 Magnetising Current
We know from BiotSavart’s law that a current carrying conductor produces magnetic field. Magnetic field strength H is proportional to the current which produces the field. From Ampere’s Circuital law, it can be proved that H is proportional to current I. If a current carrying coil produces magnetic flux which traverses an average length of l in complete flux path, then
Hl = NI
In a magnetic circuit, this field is represented by magnetomotive force. It is analogous to the electromotive force in electrical circuit. This field is responsible to “set up” certain flux, which in turn gives rise to certain flux density B. Note that, here H is cause and B is its effect. The amount of flux which can be setup in a material is determined by an inherent property of the material, called as permeability, denoted by µ.
B = µH
When external magnetic field H is applied to a material, all the domains align in a particular direction, setting up net flux in the material. Due to domain alignment B (i.e., the magnitude of B) increases. However, after a certain value of B, the slope of B – H curve starts reducing as shown in Figure below
Fig 32: Magnetization curve
BH Curve:
We cannot measure B and H directly. Further if we have transformer, we only have terminal measurements are with us. Hence, it is required to process the signals to get values of B and H. From Faraday’s law, V = N dφ/dt. Also, B is directly proportional to flux φ.
Fig 33: BH curve
Fig 34: Circuit for BH curve
Key takeaway
Magnetizing current is the term used to denote the total current that flows into the primary of a transformer when the transformer is energized at a specific voltage and frequency, with the secondaries open circuited. Although known as magnetizing current it is actually the combination of the current required to magnetize the core (I1) and the current required to supply the losses in the core (I2).
Harmonics in transformers cause an increase in the iron and copper losses. Voltage distortion increase losses due to hysteresis and eddy currents and causes overstressing of the insulation material used. The primary effect of power line harmonics in transformer is, thus the additional heat generated. Other problems include possible resonance between the transformer inductance and the system capacitance, thermal fatigue due to temperature cycling and possible core vibrations.
Scott connection or TT connection
For making 3Ø to 3Ø transformation from two transformers we use this connection.
Fig.35: Scott Connection
The voltage diagram is shown below
Fig.36: Voltage and Phase diagram
Three to six phase conversions:
A greater number of phases results in the reduction of harmonics in the alternating current. Therefore, 6phase is preferred to 3phase for rectification. There are several methods of connecting transformers for obtaining 3to6phase transformation. A simple arrangement known as the 6phase star connection is shown in Fig.
Fig 37: 3phase delta to six phase star connection
Fig 38: Phasor of primary and secondary voltages
The secondaries are provided with centre taps, which are connected together to form the neutral of the 6phase side. The primaries are shown connected in delta but they may also be connected in wye. The 6phase forkedwye or doublezigzag connection is shown in Figure below. In this arrangement, each secondary winding is divided into three equal parts and the portions that are shown parallel to each other. A 6phase transformation may be affected by means of three singlephase transformers.
Fig 39: Six phased forkedwye connection
Key takeaway
OnLoad Tap changing:
For uninterrupted supply we use onload tap changing transformer. Such a transformer is known as a tapchanging under load transformer. While tapping, two essential conditions are to be fulfilled.
The tap changing employing a centre tapped reactor R show in the figure.
Fig 40: On load Tap changing
Here S is the diverter switch, and 1, 2, 3 are selector switch. The transformer is in operation with switches 1 and S closed. To change to tap 2, switch S is opened, and 2 is closed. Switch 1 is then opened, and S closed to complete the tap change. It is to be noted that the diverter switch operates on load, and no current flows in the selector switches during tap changing. It is to be noted that the diverter switch operates on load, and no current flows in the selector switches during tap changing. During the tap change, only half of the reactance which limits the current is connected in the circuit.
NoLoad Tap Changing:
The cheapest method of changing the turn ratio of a transformer is the use of Off Load Tap Changer. As the name indicates, it is required to deenergize the transformer before changing the tap. A simple Off Load Tap Changer is shown in Figure below
Fig 41: NoLoad Tap Changing
It has eight studs marked one to eight. The winding is tapped at eight points. The face plate carrying the suitable studs can be mounted at a convenient place on the transformer such as upper yoke or located near the tapped positions on the windings. The movable contact arm A may be rotated by handwheel mounted externally on the tank.
If the winding is tapped at 2% intervals, then as the rotatable arm A is moved over to studs 1, 2; 2, 3; . . . 6, 7; 7, 8 the winding in circuit reduces progressively by it from 100% with arm at studs (1, 2) to 88% at studs (7, 8). The stop F which fixes the final position of the arm A prevents further anticlockwise rotation so that stud 1 and 8 cannot be bridged by the arm.
Key takeaway
OnLoad Tap changing:
For uninterrupted supply we use onload tap changing transformer.
NoLoad Tap Changing:
The cheapest method of changing the turn ratio of a transformer is the use of Off Load Tap Changer. As the name indicates, it is required to deenergize the transformer before changing the tap.
As the name suggests this transformer can be made by adding one additional winding. The main use of this type of transformer is for interconnections with the transmission line at various voltages, for measuring voltages of an H.V side transformer, synchronous condensers can be connected to this winding to inject reactive power in system for voltage control. Below is shown the figure which explains how the tertiary winding can be used for stabilization.
Fig 42: Tertiary winding for stabilization
The unbalanced load cab be divided into three phase sets i.e., positive, negative and zero sequence component. The starstar connection unbalanced load provides high reactance between line and neutral. The zerosequence current component on secondary side cannot be balanced by primary current and hence magnetic flux is developed on the secondary side. Then the delta connected tertiary allows the circulation of zerosequence current in it, as shown in the above figure. The equivalent circuit of the three winding transformer is shown below. There are three terminals primary (1), secondary (2), and tertiary (3) and one common terminal, neglecting the exciting current.
Fig 43: Single phase equivalent circuit of three winding transformer
The impedance can be calculated by the short circuit test. If is the short circuit impedance of winding 1 and 2 with winding 3 open, then we can write
= +
The above three equations can also be written as
Key takeaway
The core loss, magnetizing loss and turn ratio can be determined by open circuit test on any one of the three windings
The main source of heat generation in transformer is its copper loss or I2R loss. If this heat is not dissipated properly, the temperature of the transformer will rise continually which may cause damages in paper insulation and liquid insulation medium of transformer. So, it is essential to control the temperature with in permissible limit to ensure the long life of transformer by reducing thermal degradation of its insulation system.
1) ONAN (Oil Natural Air Natural):
This is the simplest transformer cooling system. In this the natural convectional flow of hot oil is utilized for cooling. In convectional circulation of oil, the hot oil flows to the upper portion of the transformer tank and the vacant place are occupied by cold oil. This hot oil which comes to upper side, will dissipate heat in the atmosphere by natural conduction, convection and radiation in air and will become cold. In this way the oil in the transformer tank continually circulates when the transformer put into load.
As the rate of dissipation of heat in air depends upon dissipating surface of the oil tank, it is essential to increase the effective surface area of the tank. So additional dissipating surface in the form of tubes or radiators connected to the transformer tank. This is known as radiator bank of transformer.
2) ONAF (Oil Natural Air Forced):
The Heat dissipation can be increased, if dissipating surface is increased but it can be made further faster by applying forced air flow on that dissipating surface. Fans blowing air on cooling surface is employed. Forced air takes away the heat from the surface of radiator and provides better cooling than natural air. As the heat dissipation rate is faster and more in ONAF transformer cooling method than ONAN cooling system.
3) OFAF (Oil Forced Air Natural):
In OFAF cooling system the oil is forced to circulate within the closed loop of transformer tank by means of oil pumps. The main advantage of this system is that it is compact system and for same cooling capacity OFAF occupies much less space than farmer two systems of transformer cooling. Actually, in oil natural cooling system, the heat comes out from conducting part of the transformer is displaced from its position, in slower rate due to convectional flow of oil but in forced oil cooling system the heat is displaced from its origin as soon as it comes out in the oil, hence rate of cooling becomes faster.
4) OFWF (Oil Forced Water Forced):
In OFWF cooling system of transformer, the hot oil is sent to water heat exchanger by means of oil pump and there the oil is cooled by applying sowers of cold water on the heat exchanger’s oil pipes.
5) ODAF (Oil Directed Air Forced):
The forced circulation of oil directed to flow through predetermined paths in transformer winding. The cool oil entering the transformer tank from cooler or radiator is passed through the winding where gaps for oil flow or predecided oil flowing paths between insulated conductor are provided for ensuring faster rate of heat transfer. This cooling method is generally used in very high rating transformer.
6) ODWF (Oil Directed Water Forced):
This is just like ODAF only difference is that here the hot oil is cooled in cooler by means of forced water instead of air. Both of these transformer cooling methods are called forced directed oil cooling of transformer.
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