Transformers
MTE 320
Spring 2006
E.F. EL-Saadany
Single-Phase Transformers
It should be mentioned here that the electric power is generated at voltage in the range of the 11-30KV.
Transmitting a given amount of power requires a fixed product of voltage and current. Thus the higher
the voltage, the lower the current can be. Lower line currents are associated with lower losses (I2Z). Transmitting the power in low voltage range for long distance is not feasible since all the power will be
lost as a voltage drop on the transmission line.
An obvious solution to this problem is to raise the voltage level to some hundred KV (ranging from
110 to 700 KV). Here comes the function of the transformer. This device is one of the most important
inventions of all times. Without a doubt, they are the most important piece of equipment along the
power transmission and distribution systems. Transformers make it possible to convert the energy
taken from generators into usable, transmittable form. Without them, it would be next to impossible to
use the energy produced by the utility companies.
Transformers are used indifferent applications, such as:
1. Step-up transformers for transmission
2. Step-down transformers for distribution
3. High voltage measurements (potential transformers)
4. High current measurement (current transformers)
5. Insulating one circuit from the other
6. Insulating DC circuits from AC circuits
1. Transformer Construction
A single-phase transformer consists basically from two or more windings coupled by magnetic core as
shown in Fig. 1. When one of the windings (primary) is connected to an AC source, a time varying
flux is produced in the core. This flux is confined within the magnetic core and it links the second
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winding (secondary). When any electric load is connected to the secondary winding, a current will
flow.
Fig. 1 Single-phase transformer construction
φ
Fig. 2 Single-phase transformer circuit
2
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Spring 2006
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2. Ideal Transformer
For ideal transformer, the following assumptions are valid:
1. No leakage flux (all the flux produced by the primary winding links the secondary)
2. No winding resistance (V1= E1 and V2= E2)
3. The core reluctance is zero ( ℜ = 0, and μ = ∞ )
4. No core losses (eddy + hysteresis)
Let the mutual flux linking both windings φ m be sinusoidal, such that: φ m = φ p sin ω t . According to
Faraday's law, the induced emf can be expressed as:
e1 =
dλ 1
e2 =
dt
dλ 2
dt
= N1
dφ m
= ωφ p N1 cos ω t
dt
= N2
dφ m
= ωφ p N 2 cos ω t
dt
The RMS values of the induced emf are:
E1 =
1
ωφ p N1 = 4.44 fφ p N1
2
E2 =
1
ωφ p N 2 = 4.44 fφ p N 2
2
The polarities of the induced emf are given by Lenz's law, that is, the emfs produces currents that tend
to oppose the flux change.
The ration between the primary and secondary induced emf is:
E1 N1 V1
=
=
= a , which is known as
E 2 N 2 V2
the transformer Turns Ratio. Since the transformer is ideal and there is no losses, then the input power
equals the output power, V1 I1 = V2 I 2 , then:
E1 N1 V1 I 2
=
=
= =a
E 2 N 2 V2 I 1
3
Transformers
MTE 320
Spring 2006
E.F. EL-Saadany
From the above equation, we can show that:
V1 = aV2 and I1 =
1
I2
a
Dividing the above two equations we obtain:
Z1 =
V1 a 2V2
=
= a 2 Z 2 = Z 2\
I1
I2
where Z 2\ is known as the secondary winding impedance referred to the primary winding. In addition,
both the current and the voltage in the secondary circuit can be referred to the primary circuit as
follows:
V1
= a , then V1 = aV2 = V2\
V2
I1 1
1
= , then I1 = I 2 = I 2\
a
I2 a
The equivalent circuit of the transformer Referred to the primary is given as shown in Fig. 3.
I 2\ =
I2
a
V2\ = aV2
E2\ = aE2
z 2\ = a 2 Z 2
Fig. 3 Transformer equivalent circuit referred to the primary side
The equivalent circuit of the transformer Referred to the secondary is given as shown in Fig. 4.
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MTE 320
Spring 2006
E.F. EL-Saadany
I1\ = aI1
V1\ =
V1
a
E1\ =
E1
a
Fig. 4 Transformer equivalent circuit referred to the secondary side
Example
A 240/120 volt, 60Hz, ideal transformer is rated at 5 KVA.
a) Calculate the turns ratio
b) Calculate the rated primary and secondary currents
c) Calculate the primary and secondary currents when the transformer delivers 3.2KW at rated
secondary voltage and 0.8 power factor lagging.
Solution:
N1 V1 240
=
=
=2
N 2 V2 120
a)
a=
b)
I1rated =
5000
= 20.83 A
240
I 2rated =
5000
= 41.67 A
120
or
I 2 N1
=
=a=2
I1 N 2
c) For the given load, I 2 =
and I1 =
3200
= 33.33 A
120 × 0.8
I 2 33.33
=
= 16.67 A
a
2
5
Transformers
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Spring 2006
E.F. EL-Saadany
3. Non-Ideal (Actual) Transformer
The assumptions made in the previous section for ideal transformer are no longer applicable when
analyzing the performance of an actual transformer.
Characteristics of non-ideal transformers
1. The primary and secondary windings have resistances
2. Not all the flux produced by one winding will link the other winding because of flux
leakage
3. The core of an actual transformer has a finite permeability.
4. There are core losses (Hysteresis and Eddy current) due to the presence of alternating flux
in the core; (iron losses).
Consider the actual transformer circuit shown in Fig. 5
m
R1
V1
I1
R2
+
+
E1
E2
-
12
11
-
N2
N1
Fig. 5 Actual transformer circuit
Where:
φm
= The mutual flux (linking flux)
φ 11
= Primary leakage flux
φ 21
= Secondary leakage flux
N1
= Primary winding number of turns
N2
= Secondary winding number of turns
R1
= Primary winding resistance
R2
= Secondary winding resistance
6
I2
V2
Load
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For the primary winding, the flux linking the winding is given as:
φ 1 = φ m + φ 11
The voltage equation for the primary loop can be written as:
V1 = R1i1 +
dλ 1
= R1i1 + N1
dt
dφ m
Thus
V1 = R1i1 + N1
+ N1
Since
N1φ 11= i1 L1 , ∴ N1
dt
∴ V1 = R1i1 + L1
dφ
dφ 1
dt
dφ
11
dt
11
dt
= L1
di1
dt
dφ m
di1
di
+ N1
= R1i1 + L1 1 + e1
dt
dt
dt
(A)
In the secondary circuit, the voltage equation may be written as follows:
V2 = − R2i2 +
dλ
2
dt
= − R2i2 + N 2
dφ 2
dt
From the flux direction, φ 2 = φ m − φ 12 ,
Thus:
V2 = − R2 i2 + N 2
dφ m
dt
∴ V2 = − R2 i2 − L2
− N2
dφ
12
dt
dφ m
di2
di
+ N2
= − R2 i2 − L2 2 + e2
dt
dt
dt
(B)
where e1 and e2 are the induced emf in the primary and the secondary windings, respectively. It can be
shown that:
e1 N1
=
=a
e2 N 2
Equations A & B can be written in frequency domain as follows:
∴ V1 = R1i1 + jω L1i1 + E1
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∴ V 2 = − R 2 i 2 − jω L 2 i 2 + E 2
To model the core losses of the transformer a parallel circuit consists of an inductor Lm and a resistor
Rc is added usually to the primary side of the transformer equivalent circuit,
where:
Lm
= Represents the core magnetization
Rc
= Represents the core losses (Hysteresis & Eddy current losses)
The core related circuit elements Lm & Rc are usually determined at the rated voltage and referred to the
primary. They are assumed constant when the transformer is operating at or near the rated conditions.
4. Equivalent Circuits
The equivalent circuit of the transformer is shown in Fig. 6.
R1
R2
L1
Rc
Lm
E2
L2
V2
Fig. 6 Equivalent circuit
The equivalent circuit of the transformer in the frequency domain is given as shown in Fig. 7.
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Transformers
MTE 320
R1
jX 1
Rc
Spring 2006
E.F. EL-Saadany
I 2\ = I 2 a
R2
jX m
E2
Fig. 7 Equivalent circuit in frequency domain
Where:
E1
= Primary induced voltage
E2
= Secondary induced voltage
V1
= Primary terminal voltage
V2
= Secondary terminal voltage
I1
= Primary current
I2
= Secondary current
Ie
= Excitation current
Im
= Magnetizing current
Ic
= Core loss current
Xm
=Magnetizing reactance
X1
=Primary leakage reactance
X2
=Secondary leakage reactance
Rc
= Core loss resistance
R1
= Primary winding resistance
R2
= Secondary winding resistance
9
jX 2
V2
Transformers
MTE 320
Spring 2006
E.F. EL-Saadany
4.1 Transformer equivalent circuit phasor diagram:
The load is V2 , I 2 and θ 2
V1 = R1i1 + jX 1i1 + E1
V2 = − R2 i2 − jX 2 i2 + E 2
Ie = Ic + Im
I1 = I e + I 2\
I 2\ = I 2 a
a = N1 N 2
I c // E1
I m ⊥ E1
V1
E2
Ic
I1 R1
V2
θ2
Im
Ie
jI 1 X 1
E1
jI 2 X 2
I 2 R2
I 2\
I2
I1
Fig. 8 Transformer equivalent circuit phasor diagram
4.2 Referred Transformer Equivalent Circuit
4.2.1 Referred to primary side
R2\ = a 2 R2
X 2\ = a 2 X 2
I 2\ = I 2 a
E 2\ = aE 2
V 2\ = aV 2
a = N1 N 2
E 2\ = E1
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Transformers
MTE 320
I1
jX 1
R1
Ie
Ic
V1
Spring 2006
a 2 R2
a 2 jX 2
R2\
jX 2\
I 2\
E.F. EL-Saadany
I2 a
I2
Im
jX m
Rc
V2\
aV2
V2
N1
N2
Ideal Transformer
Fig. 9 Transformer equivalent circuit referred to the primary side
V1
E1 = E 2\
Ic
I1 R1
Ie
jI 2\ X 2\
V2\
θ2
Im
jI 1 X 1
I 2\ R2\
I
\
2
I1
Fig. 10 Transformer equivalent circuit, referred to the primary side, phasor diagram
4.2.2 Referred to secondary side
R1
a2
X
X 1\ = 21
a
R1\ =
I1\ = aI1
E1
a
V
V1\ = 1
a
a = N1 N 2
E1\ =
11
Transformers
MTE 320
I1
aI1
Spring 2006
R1
a2
E.F. EL-Saadany
jX 1
a2
R2
jX 2
I2
I e\
I c\
V1
V1\
V1
N1
V1 a
Rc
a2
I m\
jX m
a2
V2
N2
Ideal Transformer
Fig. 11 Transformer equivalent circuit referred to the primary side
Example
A 25 KVA, 440/220 V, 60 Hz transformer has the following parameters:
R1 = 0.16Ω , R2 = 0.04Ω , Rc = 270Ω , X 1 = 0.32Ω , X 2 = 0.08Ω , X m = 100Ω . The transformer
delivers 20kW at 0.8 power factor lagging to a load on the low voltage side with 220 V across the load.
Find the primary terminal voltage.
Solution
Step #1
Determine the load voltage and current:
The voltage across the load is taken as a reference in this case and is equal to V2 = 220∠0 V .
For a load of 20 kW at power factor of 0.8 lag, the load current is equal to:
I2 =
P2
20,000
=
∠ − cos −1 0.8 = 113.64∠ − 36.9 A
V2 cos θ 2 220 × 0.8
Step #2
Refer the circuit to the primary side:
a=
N 1 440
=
=2
N 2 220
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Spring 2006
E.F. EL-Saadany
V2\ = aV2 = 440∠0 V
I 2\ =
I 2 113.6
=
= 56.82∠ − 36.9 A
2
a
R2\ = a 2 R2 = 0.16 Ω
X 2\ = a 2 X 2 = 0.32 Ω
Step #3
Solve the equivalent circuit
I1
R1
R2\
jX 1
jX 2\
I 2\
I2 a
Ie
Ic
V1
Im
jX m
Rc
V2\
aV2
Fig. 12 Equivalent circuit
(
E1 = V2\ + I 2\ R2\ + jX 2\
)
E1 = 440∠0 + 56.82∠ − 36.9(0.16 + j 0.32) = 458.2∠1 = 458.2 + j 9.07
The shunt branch current is:
Ic =
Im =
E1 458.3∠1
=
= 1.7 + j 0.03
Rc
270
E1
458.3∠1
=
= 0.09 − j 4.58
jX m
j100
I e = I c + I m = 1.79 − j 4.55
Thus the primary current is given as:
I1 = I e + I 2\ = 61.04∠ − 39.3 A
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Transformers
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Spring 2006
E.F. EL-Saadany
The primary voltage is:
V1 = E1 + I1 (R1 + jX 1 ) = (458.2 + j9.07) + (61.04∠ − 39.3)(0.16 + j 0.32) = 478.4∠2.2 V
4.3 Approximate Equivalent Circuits
The approximation is based on the fact that the magnetization (no load) current I e is small compared
to the full load primary input current. In practice I e = 3 − 5% I1 . Moreover, since the primary and the
secondary winding resistances and leakage reactances are very small, then the internal voltage drop is
very small such that ΔV1 = I1 (R1 + jX 1 ) ≤ 3 − 5%V1 .
V1 = E1 + I1 (R1 + jX 1 )
Since I 1 = I e + I 2\ , then:
V1 = E1 + I e (R1 + jX 1 ) + I 2\ (R1 + jX 1 )
4.3.1 First approximation
If I e is very small compared to I1 and R1 & X 1 are also very small, then I e (R1 + jX 1 ) is very small
and neglecting it will have negligible effect on both E1 and V2 . The approximate equivalent circuit
can thus be given in the form:
Rc
R1
jX m
E1 = E2\
V1
I 2\ = I 2 a
jX 1
R2\
jX 2\
V2\ = aV2
Fig. 13 Approximate equivalent circuit
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MTE 320
Spring 2006
I 2\ = I 2 a
jX m
Rc
jX eq
E1 = E2\
V1
Req
E.F. EL-Saadany
V2\ = aV2
Fig. 14 Approximate equivalent circuit
Where:
Req = R1 + R2\ = R1 + a 2 R2
X eq = X 1 + X 2\ = X 1 + a 2 X 2
The phasor diagram of the approximated equivalent circuit is given as:
V1
Ic
V2\
θ2
Im
Ie
\
2
jI 2\ X eq1
I Req1
I
\
2
I1
Fig. 15 Approximate equivalent circuit phasor diagram
4.3.2 Second Approximation
If the current I e can be neglected, then the circuit is reduced to the second approximation as
shown in the figure below:
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MTE 320
Spring 2006
I 2\ = I 2 a
Req
E.F. EL-Saadany
jX eq
V1
V2\ = aV2
Fig. 16 Second approximation equivalent circuit
Req and X eq are the equivalent resistance and reactance referred to the primary side. The phasor
diagram of this circuit will be as shown below:
V1
V2\
θ2
\
2
jI 2\ X eq1
I Req1
I = I1
\
2
I1
Fig. 17 Second approximation equivalent circuit phasor diagram
5. Voltage Regulation
Distribution and power transformers are often used to supply loads that are designed to operate at
essentially constant voltage. The amount of the secondary current drawn by the load depends on the
load magnitude. As this current change, the load voltage will change consequently. This change is due
to the voltage drop on the transformer internal impedance. A measure of how much the voltage will
change as the load is varied is called “voltage regulation”
Definition
The voltage regulation is defined as the change in the magnitude of the secondary voltage as the
current changes from full load to no load with the primary voltage held constant.
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Transformers
MTE 320
Voltage regulation = ε =
ε=
Spring 2006
V2 nd no −load − V2 nd
V2 nd
E.F. EL-Saadany
full −load
full −load
V1 − V2\ V1\ − V2
=
V2\
V2
6. Efficiency
The percentage efficiency of the transformer is defined as the ratio of the power output to the power
input.
η=
η=
Poutput
Pinput
× 100
Poutput
Poutput + losses
× 100
where the losses are the core and copper losses.
∑ losses = P
∑ losses = P
core
+ Pcopper
core
+ I12 Req1
∑ losses = P
NL
+ Psc− fl
6.1 Efficiency at any load “x”
If the transformer is loaded with x% of its full load, then the copper losses at this loading level will
be x 2 Psc .
For any loading percentage x, I x2 Req1 = x 2 I 2fl Req1 . The output power at any percentage x of the full load
is:
Poutputx = xPoutput fl = xV2 I fl cosθ 2
The efficiency for any loading condition is:
ηx =
Poutput
Poutput + losses
=
Pinput − ∑ losses
Pinput
17
= 1−
PNL + x 2 Psc
xPoutput + x 2 Psc + PNL
Transformers
MTE 320
Spring 2006
E.F. EL-Saadany
6.2 Maximum Efficiency
For maximum efficiency,
∂η
= 0 . This will lead to PNL = x 2 Psc . In other words, for the maximum
dx
efficiency will occur at the loading level where the no-load losses is equal to the copper losses.
Consequently, the loading level at which the maximum efficiency occurs is given by:
x=
PNL
Psc
7. Transformer Parameters Determination
The transformer parameters are R1 , R2 , X 1 , X 2 , Rc , and X m . These parameters can be determined
experimentally by two tests; the open circuit and the short circuit tests.
7.1 Open circuit test
This test gives information regarding the losses in the transformer core. It can be used to
determine Rc and X m .
If we connect the circuit as shown in Fig. 18.
φm
Fig. 18 Open circuit test connection
The equivalent circuit of the transformer will be as shown in Figs. 19 and 20. Measuring the voltage,
current and power, then:
18
Transformers
MTE 320
R1
jX m
E1 = E2\
V1
I 2\ = 0
Rc
Spring 2006
jX 1
E.F. EL-Saadany
R2\
jX 2\
V2\ = aV2
Fig. 19 Open circuit test connection equivalent circuit
I 2\ = 0
V1
\
jX m V2 = aV2
Rc
Fig. 20 Open circuit test connection equivalent circuit
The measured values are: PNL , I NL and V1 . From the above circuit:
V12
I
and Yo = NL
V1
PNL
RC =
Yo =
1
1
−j
Rc
Xm
∴Xm =
1
⎛ 1 ⎞
Yo2 − ⎜⎜ 2 ⎟⎟
⎝ Rc ⎠
7.2 Short circuit test
This test provides the values of the total leakage impedances and the value of the losses in the winding
at full load.
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Transformers
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Spring 2006
E.F. EL-Saadany
In short circuit test, we short-circuit the low voltage winding and we put the input voltage on the high
voltage winding. Increase the input voltage in steps until we reach the full load current (about 20-30%
of input voltage) and measure the voltage, current and power. The circuit connection and the
equivalent circuits will be as shown in Figs. 21 and 22, respectively.
LV
HV
φm
I1
W
A
V1
V
+
+
E1
V2
-
N1
N2
Fig. 21 Short circuit test connection
I 2\ = I 2 a
R1
jX 1
R2\
jX 2\
V1
Fig. 22 Short circuit test connection equivalent circuit
Req =
Psc
I12
Z eq =
V1
I1
X eq = Z eq2 − Req2
If the transformer is designed to have equal losses on the primary and secondary circuits, then:
R1 = R2\ =
Req
X 1 = X 2\ =
2
X eq
2
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Transformers
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Spring 2006
E.F. EL-Saadany
Example
A 50 KVA, 2400/240, 60 Hz, single-phase transformer has a short circuit and open circuit tests
performed on the high-voltage and low-voltage sides respectively, and the following results were
obtained:
Voltage (V)
Current (A)
Power (W)
Open circuit test
240
5.4
186
Short circuit test
48
20.8
620
a) Determine the approximated equivalent circuit referred to the primary side.
b) Determine the voltage regulation and the efficiency at rated load, 0.8 power factor lagging
and rated voltage at the secondary terminals.
Solution
a) The parameters of the transformer referred to the primary side are: R1 , R2\ , X 1 , X 2\ , Rc and X m .
From the short circuit test
Z eq =
Vsc
48
=
= 2.3Ω
I sc 20.8
Req =
Psc
620
=
= 1.43Ω
2
I sc (20.8)2
X eq = Z eq2 − Req2 =
(2.3)2 − (1.43)2
= 1.8Ω
From the open circuit test
Since the open circuit test was performed on the low-voltage side, then the determined
parameters are going to referred to secondary side.
Yo\ =
I NL 5.4
=
= 0.0225 S
Voc 240
⎛ PNL ⎞
⎟⎟ = 81.8 Lagging
⎝ Voc I NL ⎠
θ oc = cos −1 ⎜⎜
Yo\ = 0.0225 ∠ − 81.8 ° S
Yo\ = Gc − jBm = 0.0225 ∠ − 81.8 ° = (3.23 − j 22.3) × 10 −3 S
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Transformers
MTE 320
Rc\ =
Spring 2006
E.F. EL-Saadany
1
= 309.6Ω
Gc
X m\ =
1
= 44.8Ω
Bm
Rc\ =
1
= 309.6Ω
Gc
Rc\ =
Rc
,
a2
∴ Rc = Rc\ × a 2 = 309.6 × (10) = 30.96 kΩ
2
∴ X m = X m\ × a 2 = 44.8 × (10) = 4.48 kΩ
2
b) At the rated secondary conditions and 0.8 power factor lagging,
V2\ = aV2 = 10 × 240∠0 = 2400∠0V
I 2\ =
50,000
∠ − cos −1 0.8 = 20.83∠ − 36.9 A
2400
V1 = V2\ + I 2\ (Req + jX eq ) = 2400∠0 + (20.83∠ − 36.9)(1.43 + j1.8) = 2446.6∠0.3V
The percentage voltage regulation is:
ε=
2446.6 − 2400
× 100 = 1.9 %
2400
The transformer efficiency is:
η=
Output
× 100
Input
Output power
= 50,000 * 0.8 = 40,000 Watt
Input power
= Output power + Losses
Losses
= Core losses + Copper losses
= 186 + 620 = 806 Watt
∴η =
40,000
×100 = 98 %
40,000 + 806
22