Upload
rohit-gupta
View
119
Download
1
Tags:
Embed Size (px)
Citation preview
Friday, April 7, 2023 Ch. 13 Transformers 1
Topics to be Discussed Introduction. Principle of Operation. Step-Up and Step-Down
Transformer. EMF Equation. Effect of Frequency. Conditions for Ideal
Transformer. Drawing the Phasor Diagram. Volt-Amperes (in Ideal
Transformer). Impedance Transformation, Practical Transformer at no
Load.1. Effect of Magnetization.2. Effect of Core Losses.
Iron Losses. Hysterisis Loss. Eddy current Loss.
Construction of a Transformer.
Laminations. Core Type Transformer. Shell Type Transformer.
How I0 changes on Loading ? Practical Transformer on
Load.1. Effect of Winding
Resistance.2. Effect of Flux Leakage.
Leakage flux in a transformer
Next
Friday, April 7, 2023 Ch. 13 Transformers 2
Equivalent Circuit. Simplified Equivalent
Circuit. Approximate Equivalent
Circuit. Voltage Regulation.
Condition for Zero Regulation.
Condition for Maximum Regulation.
Efficiency of a Transformer. Power Losses in
Transformers. Condition for Maximum
Efficiency. All-day Efficiency.
Autotransformers. Applications. Saving of copper. Advantages. Disadvantages.
Transformer Testing.(1) Open-Circuit Test.(2) Short-Circuit Test.
Next
Friday, April 7, 2023 Ch. 13 Transformers 3
Introduction A transformer is a highly efficient (about 99.5 %) static
(non-moving) device. It transfers electrical energy form one circuit to another
(usually from one ac voltage level to another), without any change in its frequency.
There exists no simple device that can accomplish such changes in dc voltages.
Transformation of voltage is necessary at different stages of the electrical network consisting of generation, transmission and distribution.
Small-size transformers are used in communication circuits, radio and TV circuits, telephone circuits, instrumentation and control systems.
Next
Friday, April 7, 2023 Ch. 13 Transformers 4
Principle of Operation It operates on the principle of mutual induction between two coils.
When two coils are inductively coupled and if current in one coil is changed uniformly, then an EMF gets induced in the other coil. This EMF can drive a current, when a closed path is provided to it.
Next
Friday, April 7, 2023 Ch. 13 Transformers 5
It consists of two inductive coils electrically separated but magnetically linked through a common magnetic circuit.
Coil in which electrical energy is fed is Primary Winding.
Coil in which other load is connected is called as Secondary Winding.
Next
Friday, April 7, 2023 Ch. 13 Transformers 6 Next
Friday, April 7, 2023 Ch. 13 Transformers 7
(a) Construction.
Next
Friday, April 7, 2023 Ch. 13 Transformers 8
N1 : Number of turns in the Primary
N2 : Number of turns in the Secondary
E1 : EMF Induced in the Primary
E2 : EMF Induced in the Secondary
(b) Symbol.
Next
Friday, April 7, 2023 Ch. 13 Transformers 9
Step-Up and Step-Down Transformer
If N1 > N2 E1 > E2 Step down
E1 < E2 Step up If N1 < N2
The transformation ratio,
2 2
1 1
N EK
N E
Click
Click
Next
Friday, April 7, 2023 Ch. 13 Transformers 10
EMF EquationDue to the sinusoidally varying voltage V1 applied to the primary voltage, the flux set up in the core,
m msin sin 2t ft The resulting induced emf in a winding of N turns,
m
m m
( sin )
cos sin ( / 2)
d de N N t
dt dtN t N t
Thus, the peak value of the induced emf, Em = ωNΦm.
Next
Friday, April 7, 2023 Ch. 13 Transformers 11
m m mm
24.44
2 2 2
E N fNE fN
mor 4.44E fN
Therefore, the rms value of the induced emf E,
This equation, known as emf equation of transformer.
Next
Friday, April 7, 2023 Ch. 13 Transformers 12
Effect of Frequency At a given flux, emf of a transformer increases with
frequency. By operating at higher frequencies, transformers
can be made physically more compact. Because a given core is able to transfer more power
without reaching saturation. Aircraft and military equipments employ 400-Hz
power supplies which reduces size and weight. Disadvantage : The core loss and conductor
resistance increases due to skin effect.
Next
Friday, April 7, 2023 Ch. 13 Transformers 13
Conditions for Ideal Transformer :1. The permeability (μ) of the core is infinite, (i.e., the
magnetic circuit has zero reluctance so that no mmf is needed to set up the flux in the core).
2. The core of the transformer has no losses.
3. The resistance of its windings is zero, hence no I2R losses in the windings.
4. Entire flux in the core links both the windings, i.e., there is no leakage flux.
Next
Friday, April 7, 2023 Ch. 13 Transformers 14
Ideal transformer
(a) The circuit.
Next
Friday, April 7, 2023 Ch. 13 Transformers 15
• We take flux Φ as reference phasor, as it is common to both the primary and secondary.
• EMF E1 and E2 lag flux Φ by 90°.
• The emf E1 in the primary exactly counter balances the applied voltage V1. Hence, E1 is called counter emf or back emf .
(b) The phasor diagram.
Next
Friday, April 7, 2023 Ch. 13 Transformers 16
Flux,
V1 = -E1
E1
E2
O90
Drawing the Phasor Diagram
Click
Click
Next
Friday, April 7, 2023 Ch. 13 Transformers 17
Volt-Amperes (in Ideal Transformer) The current I1 in the primary is just sufficient to provide mmf
I1N1 to overcome the demagnetizing effect of the secondary mmf I2N2. Hence,
2 11 1 2 2
1 2
1or
I NI N I N
I N K
Note that the current is transformed in the reverse ratio of the voltage. If V2 > V1, then I2 < I1. Also, we have
1 1 2 2E I E I
Hence, in an ideal transformer the input VA and output VA are identical.
Next
Friday, April 7, 2023 Ch. 13 Transformers 18
Impedance Transformation
1 1 2 2 1 2 2eq L
1 1 2 2 2 1 2
( ) 1 1
( )
V V V I V I VZ Z
I I V I V I I K K
2
eq Lor /Z Z K
The concept of impedance transformation is used for impedance matching.
Next
Friday, April 7, 2023 Ch. 13 Transformers 19
Example 1 A single-phase, 50-Hz transformer has 30
primary turns and 350 secondary turns. The net cross-sectional area of the core is 250 cm2. If the primary winding is connected to a 230-V, 50-Hz supply, calculate
(a) the peak value of flux density in the core,
(b) the voltage induced in the secondary winding, and
(c) the primary current when the secondary current is 100 A. (Neglect losses.)
Next
Friday, April 7, 2023 Ch. 13 Transformers 20
Solution :
(a) The peak value of the flux,
1m
1
2300.034534 Wb
4.44 4.44 50 30
E
fN
mm 4
0.034534
250 10B
A
1.3814 T
(b) The voltage induced in the secondary,
22 1
1
350230 2683.33 V .
30
NE E
N 2 683 kV
(c) The primary current,
21 2
1
350100 1166.67 A .
30
NI I
N
1 167 kA
Next
Click
Click
Click
Friday, April 7, 2023 Ch. 13 Transformers 21
Example 2Determine the load current IL in the ac circuit shown.
Next
Friday, April 7, 2023 Ch. 13 Transformers 22
Transforming the load impedance into the primary
p 2
30 00.872 35.53° A
20 20 2 (2 10)j j
I
L p2 2 0.872 35.53 I I 1.74 35.53° A
Solution :
Next
Click
Click
Friday, April 7, 2023 Ch. 13 Transformers 23
Practical Transformer at no Load There are following two reasons why the no-load
current (also called exciting current) I0 is not zero :
1. Effect of Magnetization : No magnetic material can have infinite permeability. A finite mmf is needed to establish magnetic flux in
the core. An in-phase magnetizing current Im in the
primary is needed. Im is purely reactive (current Im lags voltage V1 by
90°). This effect is modelled by putting X0 in parallel with
the ideal transformer.
Next
Friday, April 7, 2023 Ch. 13 Transformers 24
2. Effect of Core Losses : There exist hysteresis and eddy current losses for
the energy loss in the core. The source must supply enough power to the
primary to meet the core losses. These iron losses can be represented by putting a
resistance R0 in parallel.
The core-loss current Iw flowing through R0 is in phase with the applied voltage V1,
Next
Friday, April 7, 2023 Ch. 13 Transformers 25
(a) The circuit.
(b) The equivalent circuit.
Next
Friday, April 7, 2023 Ch. 13 Transformers 26
• The R0-X0 circuit is called exciting circuit.
2 2 10 w m 0 m w
1 w 1 0 0
; tan ( / );
and Input power Iron loss
cos
I I I I I
V I V I
Next
Friday, April 7, 2023 Ch. 13 Transformers 27
Modified phasor diagram
O Flux,
E2 = V2
E1
V1 = -E1
Iw
Im
I00Click
Click
Next
Friday, April 7, 2023 Ch. 13 Transformers 28
Iron Losses
The core losses occur in iron core, hence these are also called iron losses.
There are two reasons for these losses:
1. Hysteresis Loss.
2. Eddy current loss.
Next
Friday, April 7, 2023 Ch. 13 Transformers 29
Hysterisis Loss
Next
Friday, April 7, 2023 Ch. 13 Transformers 30
When alternating current flows through the windings, the core material undergoes cyclic process of magnetization and demagnetization.
h h mnP K B f V
Kh = hysteresis coefficient whose value depends upon the material(Kh = 0.025 for cast steel, Kh = 0.001 for silicon steel)Bm = maximum flux density (in tesla)n = a constant, depending upon the material f = frequency (in hertz)V = volume of the core material (in m3)
Next
Friday, April 7, 2023 Ch. 13 Transformers 31
Eddy current Loss2 2 2
e e mP K B f t V
where Ke = a constant dependent upon the material t = thickness of laminations (in metre)
i h eP P P Next
Friday, April 7, 2023 Ch. 13 Transformers 32
Construction of a Transformer There are two basic parts of a transformer :
Magnetic Core Winding or Coils
Limb
Yoke
Magnetic core Winding or coil
Next
Friday, April 7, 2023 Ch. 13 Transformers 33
Laminations The core of a
transformer is usually laminated to reduce the eddy currents.
These laminations may be different sections of E,I,T,F.
They are stacked finally to get the complete core of the transformer.
Next
Friday, April 7, 2023 Ch. 13 Transformers 34
Two Types of Transformers (1) Core Type Transformer :
Next
Friday, April 7, 2023 Ch. 13 Transformers 35
The windings surround a considerable part of the core.
Both the windings are divided into two parts and half of each winding is placed on each limb, side by side.
This is done to reduce the leakage of the magnetic flux.
Next
Friday, April 7, 2023 Ch. 13 Transformers 36
Practically, the windings are placed as follows.
To minimize the cost of insulation, the low voltage (LV) winding is placed adjacent to the core and high voltage (HV) winding is placed around the LV winding
Next
Friday, April 7, 2023 Ch. 13 Transformers 37
(2) Shell Type Transformer The core surrounds a considerable part of the windings. It has three limbs. Both the windings are placed on the central limb. The flux divides equally in the central limb and returns
through the outer two legs.
Next
Friday, April 7, 2023 Ch. 13 Transformers 38 Next
Friday, April 7, 2023 Ch. 13 Transformers 39
How I0 changes on Loading ?
Φ Φ’
V1 E2E1 V2
N1 N2
I0 + I1’ I2
Click
Click
Next
Friday, April 7, 2023 Ch. 13 Transformers 40 Next
Friday, April 7, 2023 Ch. 13 Transformers 41
Before connecting the load, there exists a flux Φ requiring current I0 in the primary.
On connecting the load, a current I2 flows in the secondary.
The magnitude and phase of I2 with respect to V2 depends upon the nature of the load.
The current I2 sets up a flux Φ’, which opposes the main
flux Φ. Hence, it is called demagnetizing flux. This momentarily weakens Φ, and back emf E1 gets
reduced. As a result, V1 - E1 increases and more current is drawn
from the supply. This again increases E1 to balance the applied voltage
V1. Next
Friday, April 7, 2023 Ch. 13 Transformers 42
In this process, the primary current increases by I1’. This current is known as primary balancing
current, or load component of primary current. Under such a condition, the secondary ampere-turns
must be counterbalanced by the primary ampere-turns.
'1 1 2 2N I N I
21 2 2
1
' NI I KI
N
and
'1 0 1 I I I
Next
Friday, April 7, 2023 Ch. 13 Transformers 43
Volt Ampere Rating of a Transformer Output power depends on cos2 ( power factor of
secondary). As pf can change depending on the load, the rating is
not specified in watts or kilowatts. But is indicated as a product of voltage and current
called VA RATING.
2211 IVIV
100022 IV
1000
IV rtransforme a of rating kVA 11
1load) (full1
1000 ratingkVA
VI
2load) (full2
1000 ratingkVA
VI
Fore ideal transformer :
Next
Friday, April 7, 2023 Ch. 13 Transformers 44
Transformers rating in kVA ?
• Transformers are rated in VA, because the manufacturer does not know the power factor of the load which you are going to connect.
• So the customer should not exceed the VA rating of the transformer.
• In case of motors, the manufacturer knows exactly the power factor at full load.
• That is why motors are rated in kW.
Next
Friday, April 7, 2023 Ch. 13 Transformers 45
Example 3 A single-phase, 230-V/110-V, 50-Hz transformer takes an
input of 350 volt amperes at no load while working at rated voltage. The core loss is 110 W. Find
(a) the no-load power factor,
(b) the loss component of no-load current, and
(c) the magnetizing component of no-load current.
Next
Friday, April 7, 2023 Ch. 13 Transformers 46
Solution : (a) Given : 1 0 350 VAV I
01
3501.52 A
230
VAI
V
The core loss = Input power at no load, 1 0 0cosiP V I
01 0
110 Wcos
350VAiP
pfV I
0.314
(b) The loss component of no-load current,
w 0 0cos 1.52 0.314I I 0.478 A
(c) The magnetizing component of no-load current,
2 2 2 2m 0 w (1.52) (0.478)I I I 1.44 A
Next
Click
Click
Click
Click
Click
Friday, April 7, 2023 Ch. 13 Transformers 47
Example 4 A 100-kVA, 4000-V/200-V, 50-Hz, single-phase
transformer has 100 secondary turns. Determine
(a) the primary and secondary currents,
(b) the number of primary turns, and
(c) the maximum value of the flux.
Solution : The kVA rating = V1I1 = V2I2 = 100 kVA.
11
22
kVA rating 100000
4000
kVA rating 100000
200
IV
IV
25 A
500 A
Next
Click
Friday, April 7, 2023 Ch. 13 Transformers 48
1 1
2 2
11 2
2
Since
4000100
200
N V
N V
VN N
V
2000
(b)
(c)2 2
2
2
4.44
200
4.44 4.44 50 100
m
m
E f N
E
fN
9.01 mWb
Next
Friday, April 7, 2023 Ch. 13 Transformers 49
Flux,
E1
E2= V2
OI0
0
I1I1
I2
V1 = -E1
1
Phasor Diagram for Resistive Load
Click
Click
Next
Friday, April 7, 2023 Ch. 13 Transformers 50
Flux,
E1
E2 = V2
OI0
0
I1I1
I2
V1 = -E1
1
Phasor Diagram for Inductive Load
2
Click
Click
Next
Friday, April 7, 2023 Ch. 13 Transformers 51
Flux,
E1
E2 = V2
OI0
0
I1
I1
I2
V1 = -E1
1
Phasor Diagram for Capacitive Load
2
Click
Click
Next
Friday, April 7, 2023 Ch. 13 Transformers 52
(a) Resistive. (b) Inductive. (c) Capacitive.
Phasor Diagrams for Different Types of Loads
Next
Friday, April 7, 2023 Ch. 13 Transformers 53
Is it ever possible that the load connected to the secondary is capacitive but the overall power factor is inductive ?
Ans. : Yes. See the phasor diagram for capacitive load.
Is it ever possible that the load connected to the secondary is inductive but the overall power factor is capacitive?
Ans. : No. Not possible.
Next
Click
Click
Friday, April 7, 2023 Ch. 13 Transformers 54
Example 5 A single-phase, 440-V/110-V, 50-Hz
transformer takes a no-load current of 5 A at 0.2 power factor lagging. If the secondary supplies a current of 120 A at a power factor of 0.8 lagging to a load, determine the primary current and the primary power factor. Also, draw the phasor diagram.
Next
Friday, April 7, 2023 Ch. 13 Transformers 55
Solution :1 1
0 2cos 0.2 78.46 and cos 0.8 36.87
2
1
110 1
440 4
VK
V
' '1 2 1(1/ 4) 120 30 A; 30 36.87 AI K I I
'1 1 0 30 36.87 5 78.46 I I I 33.9 42.49° A
Primary power factor,
1cos cos 42.49pf 0.737(lagging)
Next
Click
Click
Click
Friday, April 7, 2023 Ch. 13 Transformers 56 Next
Friday, April 7, 2023 Ch. 13 Transformers 57
Practical Transformer on Load We now consider the deviations from the last two
ideality conditions :
1. The resistance of its windings is zero.
2. There is no leakage flux.
The effects of these deviations become more prominent when a practical transformer is put on load.
Next
Friday, April 7, 2023 Ch. 13 Transformers 58
(1) Effect of Winding Resistance Current flow through the windings causes a power
loss called I2R loss or copper loss. This effect is accounted for by including a resistance
R1 in the primary and resistance R2 in the secondary
Next
Friday, April 7, 2023 Ch. 13 Transformers 59
(2) Effect of Flux Leakage The difference between the total flux linking with the
primary and the useful mutual flux Φu linking with both the windings is called the primary leakage flux, ΦL1.
Similarly, ΦL2 represents the secondary leakage flux.
Flux leakage results in energy being alternately stored in and discharged from the magnetic fields with each cycle of the power supply.
It is not directly a power loss, but causes the secondary voltage to fail to be directly proportional to the primary voltage, particularly under heavy loads.
Next
Friday, April 7, 2023 Ch. 13 Transformers 60
Leakage flux in a transformer
(a) Its definition. (b) Its effect accounted for.
• The useful mutual flux Φu is responsible for the transformer action.
• The leakage flux ΦL1 induces an emf EL1 in the primary winding.
Next
Friday, April 7, 2023 Ch. 13 Transformers 61
Similarly, flux ΦL2 induces an emf EL2 in the secondary.
Hence, we include reactances X1 and X2 in the primary and secondary windings, in the equivalent circuit.
The paths of leakage fluxes ΦL1 and ΦL2 are almost entirely due to the long air paths and are therefore practically constant.
The reluctance of the paths being very high, X1 and X2 are relatively small even on full load.
However, the useful flux Φu remains almost independent of the load.
Next
Friday, April 7, 2023 Ch. 13 Transformers 62 Next
Friday, April 7, 2023 Ch. 13 Transformers 63
Equivalent Circuit of a Transformer It is merely a representation of the following KVL equations :
1 1 1 1 1 1 1 1 1 1( )I R jI X I R jX V E E
2 2 2 2 2 2 2 2 2 2( )I R jI X I R jX E V V
Next
Friday, April 7, 2023 Ch. 13 Transformers 64
m
E2
E1
-E1
I1R1
I1X1V1
I1
I0
I1'
I2
V2
I2R2
I2X2
0
1
Pha
sor
Dia
gram
for
Pra
ctic
al T
rans
form
er o
n R
esis
tive
Lo
ad
1 1 1 1 1
2 2 2 2
( )
( )
I R jX
V I R jX
2
V E
E
O
I1Z1
I2Z2
Next
Friday, April 7, 2023 Ch. 13 Transformers 65
Pra
ctic
al T
rans
form
er o
n In
du
ctiv
e L
oad
Next
Friday, April 7, 2023 Ch. 13 Transformers 66
Pra
ctic
al T
rans
form
er o
n C
apac
itiv
e L
oad
Next
Friday, April 7, 2023 Ch. 13 Transformers 67
Simplified Equivalent Circuit The no-load current I0 is only about 3-5 % percent of
the full-load current. The exciting circuit R0-X0 in is shifted to the left of
impedance R1-X1.
Transforming the impedances from the secondary to the primary side.
Next
Friday, April 7, 2023 Ch. 13 Transformers 68
Equivalent resistance and reactance referred to the primary side
2 2e1 1 2 e1 1 2( / ) and ( / )R R R K X X X K
Next
Friday, April 7, 2023 Ch. 13 Transformers 69
Approximate Equivalent Circuit
As referred to primary side.
As referred to secondary side.
Next
Friday, April 7, 2023 Ch. 13 Transformers 70
Example 5 A single-phase, 50-kVA, 4400-V/220-V, 50-Hz
transformer has R1 = 3.45 Ω, R2 = 0.009 Ω, X1 = 5.2 Ω and X2 = 0.015 Ω. Calculate
(a) the Re as referred to the primary,
(b) the Re as referred to the secondary,
(c) the Xe as referred to the primary,
(d) the Xe as referred to the secondary,
(e) the Ze as referred to the primary,
(f) the Ze as referred to the secondary, and
(g) the total copper loss.
Next
Friday, April 7, 2023 Ch. 13 Transformers 71
Solution : Full-load primary current,
11
kVA 5000011.36 A
4400I
V
Full-load secondary current, 22
kVA 50000227.27 A
220I
V
2
1
220 10.05
4400 20
VK
V
(a) 2 2e1 1 2( / ) 3.45 [0.009 /(0.05) ]R R R K 7.05 Ω
(b)
(c)
2 2e2 1 2 (0.05) 3.45 0.009R K R R 0.0176 Ω
2 2e1 1 2( / ) 5.2 [0.015 /(0.05) ]X X X K 11.2 Ω
(d) 2 2e2 1 2 (0.05) 5.2 0.015X K X X 0.028 Ω
Next
Click
Click
Click
Click
Click
Click
Friday, April 7, 2023 Ch. 13 Transformers 72
(e) 2 2 2 2e1 e1 e1 (7.05) (11.2)Z R X 13.23 Ω
(f)
(g) Total copper loss
2 2 2 2e2 e2 e2 (0.0176) (0.028)Z R X 0.0331 Ω
2 2 2 21 1 2 2 (11.36) 3.45 (227) 0.009I R I R 909 W
Alternatively, by considering equivalent resistances, total copper loss
2 21 e1 (11.36) 7.05I R 909.8 W
2 22 e2 (227.27) 0.0176I R 909 W
Next
Click
Click
Click
Friday, April 7, 2023 Ch. 13 Transformers 73
Voltage Regulation
V2(0) = secondary terminal voltage at no load,
and V2 = secondary terminal voltage at full load.
The voltage regulation of a transformer is defined as the change in its secondary terminal voltage from no load to full load, the primary voltage being assumed constant.
The voltage drop V2(0) - V2 is called the inherent regulation.
2(0) 2
2(0)
( ) Per unit V V
iV
regulation down
2(0) 2
2(0)
% 100V V
V
regulation down
Next
Friday, April 7, 2023 Ch. 13 Transformers 74
2(0) 2
2
( ) Per unit V V
iiV
regulation up
2(0) 2
2
% 100V V
V
regulation up
Normally, when nothing is specified, ‘regulation’ means ‘regulation down’.
Next
Friday, April 7, 2023 Ch. 13 Transformers 75
Exact voltage drop =
2(0) 2 OC OA OG OA AG AF+FGV V
Next
Friday, April 7, 2023 Ch. 13 Transformers 76
In case of leading power factor,
2 e2 2 e2
Approximate voltage drop, AF AE EF AE BD
cos sinI R I X
2 e2 2 e2
Approximate voltage drop, AF AE EF AE BD
cos sinI R I X
In general,
2 e2 2 e2Approximate voltage drop cos sinI R I X
Next
Friday, April 7, 2023 Ch. 13 Transformers 77
2 2 2 2
2(0)
cos sin% Regulation 100
cos sin
e e
r x
I R I X
V
V V
Condition for Zero Regulation :
Possible only if the load has leading power factor.
2 2 2 2cos sin 0e eI R I X 2
2
tan e
e
R
X
Use + sign for lagging power factor and – sign for leading power factor.
Next
Friday, April 7, 2023 Ch. 13 Transformers 78
2 e2 sinI X 2 e2 cosI R
Note that for leading power factor, if the magnitude of the phase angle is high, we may have
• The regulation then becomes negative.
• It means that on increasing the load, the terminal voltage increases.
Next
Friday, April 7, 2023 Ch. 13 Transformers 79
Condition for Maximum Regulation
2 2 2 2( cos sin ) 0e e
dI R I X
d
2 2 2 2( sin cos ) 0e eI R I X
2
2
tan e
e
X
R
Maximum regulation can occur only for inductive load. The voltage drop is maximum when
Next
Friday, April 7, 2023 Ch. 13 Transformers 80
Example 6
Solution :
Click
Next
Friday, April 7, 2023 Ch. 13 Transformers 81
Example 7
Solution :
2the load voltage, 240 6V 234V
Click
Next
Friday, April 7, 2023 Ch. 13 Transformers 82
Example 8 A single-phase, 40-kVA, 6600-V/250-V,
transformer has primary and secondary resistances R1 = 10 Ω and R2 = 0.02 Ω, respectively. The equivalent leakage reactance as referred to the primary is 35 Ω. Find the full-load regulation for the load power factor of
(a) unity,
(b) 0.8 lagging, and
(c) 0.8 leading.
Next
Friday, April 7, 2023 Ch. 13 Transformers 83
2 2e2 1 2 (0.0379) 10 0.02 0.0343R K R R
2 2e2 e1and (0.0379) 35 0.0502X K X
(a) For power factor, cos = 1; sin = 0. Hence,
2 2 2 2
2(0)
cos sin% Regulation 100
160 0.0343 1 0100
250
e eI R I X
V
2.195 %
Solution : Given : R1 = 10 Ω; R2 = 0.02 Ω; Xe1 = 35 Ω
2
250 the turns-ratio, 0.0379
660040000
the full-load current, 160 A250
K
I
Next
Click
Click
Click
Friday, April 7, 2023 Ch. 13 Transformers 84
(b) For power factor, cos = 0.8 (lagging, positive);
(c) For power factor, cos = 0.8 (leading, negative);
sin 0.6 2 e2 2 e2
2(0)
cos sin% Regulation 100
160 0.0343 0.8 160 0.0502 0.6100
250
I R I X
V
0.172 %
2sin 1 cos 0.6
2 e2 2 e2
2(0)
cos sin% Regulation 100
160 0.0343 0.8 160 0.0502 0.6100
250
I R I X
V
3.68 %
Next
Click
Click
Friday, April 7, 2023 Ch. 13 Transformers 85
Efficiency of a Transformer Like any other machine, the efficiency of a transformer is defined as
o
o l
Power output Power output
Power input Power output + Power losses
P
P P
• Large-size transformers are designed to be more efficient ( > 98 %)
• But, the efficiency of small transformers (used in power adapters for charging mobile phones) is not more than 85 %.
Power lost
Outputpower
Inputpower
Next
Friday, April 7, 2023 Ch. 13 Transformers 86
Power Losses in Transformers(i) Copper losses or I2R losses :
The copper losses are variable with current. Assuming the voltage to remain constant, the current is proportional to the VA. Therefore, the copper losses for a given load (and hence for given VA) is given as
2
c c(FL)FL
VA
VAP P
2 2 2 2c 1 1 2 2 1 e1 2 e2P I R I R I R I R
Next
Friday, April 7, 2023 Ch. 13 Transformers 87
(ii) Iron losses or core losses :
Due to hysteresis and eddy-currents. Pi = Ph + Pe.
Since the flux Φm does not vary more than about 2 % between no load and full load, it is usual to assume the core losses constant at all loads.
In general, the efficiency,
o o 2 2 22
o l o c 2 2 2 2 2 i
cos
P cosi e
P P V I
P P P P V I I R P
Next
Friday, April 7, 2023 Ch. 13 Transformers 88
Condition for Maximum EfficiencyAssuming the operation at a constant voltage and a constant power
factor, for what load (i.e., what value of I2) the efficiency becomes maximum ?
Let us first divide the numerator and denominator by I2, to get
The efficiency will be maximum when the denominator of the above equation is minimum,
i2 2 2 e2 2 e2 2
2 2
22 e2 c i
( cos / ) 0 or 0
or or
i
i
d PV I R P I R
dI I
I R P P P
2 2
2 2 2 e2 i 2
cos
cos /
V
V I R P I
Next
Friday, April 7, 2023 Ch. 13 Transformers 89
Copper loss = Iron loss
Condition :
Next
Friday, April 7, 2023 Ch. 13 Transformers 90
All-day Efficiency The efficiency defined above is called commercial
efficiency. In a distribution transformer, the primary remains
energized all the time. But the load on the secondary is intermittent and variable during the day.
The core losses occur throughout the day, but the copper losses occur only when the transformer is loaded.
Such transformers, therefore, are designed to have minimum core losses. This gives them better all-day efficiency, defined below.
all-day
Output energy (in kW h) in a cycle of 24 hours
Total input energy (in kW h)
Next
Friday, April 7, 2023 Ch. 13 Transformers 91
Example 9 For a single-phase, 50-Hz, 150-kVA transformer,
the required no-load voltage ratio is 5000-V/250-V and the full-load copper losses are 1800 W and core losses are 1500 W. Find (a) the number of turns in each winding for a maximum core flux of 0.06 Wb, (b) the efficiency at half rated kVA, and unity power factor,(c) the efficiency at full load, and 0.8 power factor lagging, and (d) the kVA load for maximum efficiency.
Next
Friday, April 7, 2023 Ch. 13 Transformers 92
Solution :
2 2 m
22
m
4.44
25018.8(say, )
4.44 4.44 50 0.06
E fN
EN
f
19 turns
11 2
2
5000and 19
250
EN N
E 380 turns
0.5 (kVA) (power factor) 0.5 150 1 75 kWoP 2 2(0.5) (full-load copper loss) (0.5) 1800 W 0.45 kWcP
Iron losses (fixed), Pi = 1500 W = 1.5 kW
(a) Using the emf equation, we have
(b) At half rated-kVA, the current is half the full-load current, and hence the output power too reduces by 0.5. Thus,
Next
Click
Click
Click
Friday, April 7, 2023 Ch. 13 Transformers 93
o
o c i
75100 100
75 0.45 1.5
P
P P P
97.47 %
(c) At full load and 0.8 power factor,
(kVA) (power factor) 150 0.8 120 kWoP 1800 W 1.8 kW; and 1500 W 1.5kWc iP P
o
o c i
120100 100
120 1.8 1.5
P
P P P
97.3 %
(d) Let x be the fraction of full-load kVA at which the efficiency becomes maximum
2c i or 1800 1500 1500 /1800 0.913P P x x
(Full-load kVA) 150 0.913x 137 kVA
Therefore, the load kVA under the condition of maximum efficiency,
Next
Click
Click
Click
Friday, April 7, 2023 Ch. 13 Transformers 94
Example 10 For a single-phase, 200-kVA, distribution transformer
has full-load copper losses of 3.02 kW and iron losses of 1.6 kW. It has following load distribution over a 24-hour day :
(i) 80 kW at unity power factor, for 6 hours.
(ii) 160 kW at 0.8 power factor (lagging), for 8 hours.
(iii) No load, for the remaining 10 hours.
Determine its all-day efficiency.
Next
Friday, April 7, 2023 Ch. 13 Transformers 95
Solution :
(i) For 80 kW load at unity power factor (for 6 hours) :
Output energy 80 6 480 kW h
o 80kVA 80kVA
1
P
pf
2 2
c(FL)FL
kVA 80(3.02) 0.4832 kW
kVA 200cP P
Iron losses, Pi = 1.6 kW
Total losses, Pl = Pc + Pi = 0.4832 kW + 1.6 kW = 2.0832 kW
Total energy losses in 6 hours 2.0832 6 12.50 kW h Next
Click
Click
Friday, April 7, 2023 Ch. 13 Transformers 96
Output energy 160 8 1280 kW h
FL
160200kVA kVA
0.8oP
kVApf
c c(FL)Copper losses, 3.02 kWP P
Total energy losses in 8 hours 4.62 8 36.96 kW h
(ii) For 160-kW load at 0.8 power factor (for 8 hours) :
Iron losses, Pi = 1.6 kWTotal losses, Pl = Pc + Pi = 3.02 kW + 1.6 kW = 4.62 kW
(iii) For the no-load period of 10 hours :
Output energy Po = 0
Copper losses, Pc = 0
Iron losses, Pi = 1.6 kW
Total losses, Pl = Pc + Pi = 0 + 1.6 = 1.6 kW Next
Click
Click
Friday, April 7, 2023 Ch. 13 Transformers 97
Total energy losses in 10 hours 1.6 10 16 kW h
Thus, for 24-hour period :
Total output energy, Wo = 480 + 1280 = 1760 kW h
Total energy losses, Wl = 12.50 + 36.96 +16 = 65.46 kW h
oall-day
o l
All-day efficiency,
1760100 100
1760 65.46
W
W W
96.41%
Next
Click
Friday, April 7, 2023 Ch. 13 Transformers 98
Autotransformer It is a special transformer that is useful in power
systems, motor starters, variable ac sources, etc. An autotransformer is a transformer which has a part
of its winding common to the primary and secondary circuits.
Next
Friday, April 7, 2023 Ch. 13 Transformers 99
Applications Boosting or buckling of supply voltage by a
small amount. Starting of ac machines, where the voltage is
raised in two or more steps. Continuously varying ac supply as in variacs.
(a) Step-down (b) Step-up
Next
Friday, April 7, 2023 Ch. 13 Transformers 100
The turns-ratio is given as
2 2
1 1
N VK
N V
• The portion YZ of the winding is called common winding.
• The portion XY is called series winding. • In variacs (variable autotransformers), point Y is made
a sliding contact so as to give a variable output voltage.
Consider the Step-down autotransformer :
The volt-amperes on the two sides must be the same,
1 1 2 2V I V I 2 2 2 1 2 2 1( )V I V I V I I
• The part V2I1 is conductively transferred through the winding XY.
• The remaining part is inductively transferred through the winding YZ.
Next
Friday, April 7, 2023 Ch. 13 Transformers 101
Saving of copper in an autotransformer
• For the same voltage ratio and capacity (volt-ampere rating), an autotransformer needs much less copper compared to a two-winding transformer.
• The cross-sectional area of a conductor is proportional to the current carried by it, and its length is proportional to the number of turns. Therefore,
Weight of copper NI kNI
For a two-winding transformer :
1 1
2 2
1 1 2 2
Weight of copper in primary
Weight of copper in secondary
Total weight of copper ( )
kN I
kN I
k N I N I
Next
Friday, April 7, 2023 Ch. 13 Transformers 102
For a step-up autotransformer :
1 2 1
2 2 1
1 2 1 2 2 1
1 2 1 2 2
Weight of copper in portion XY ( )
Weight of copper in portion YZ ( )
Total weight of copper ( ) ( )
[( 2 ) ]
k N N I
kN I I
k N N I kN I I
k N N I N I
Therefore, the ratio of copper- weights for the two cases is
2 1 2
1 2 11 2 1 2 2
1 1 2 2 1 2
2 1
1 2[( 2 ) ] [1 2 ]
1( )
N I NN I Nk N N I N I K K K
Kk N I N I K KI N
I N
Next
Friday, April 7, 2023 Ch. 13 Transformers 103
• Evidently, the saving is large if K is close to unity. • A unity transformation ratio means that no copper is
needed at all for the autotransformer. • The winding can be removed all together. • The volt-amperes are conductively transformed
directly to the load !
Next
Friday, April 7, 2023 Ch. 13 Transformers 104
Advantages of autotransformers
A saving in cost since less copper is needed. Less volume, hence less weight. A higher efficiency, resulting from lower I2R
losses. A continuously variable output voltage is
achievable if a sliding contact is used. A smaller percentage voltage regulation. Higher VA Rating.
Next
Friday, April 7, 2023 Ch. 13 Transformers 105
Disadvantages of autotransformersThe primary and secondary windings are not electrically separate, hence if an open-circuit occurs in the secondary winding the full primary voltage appears across the secondary.
Low impedance hence high short circuit currents for short circuits on secondary side.
No electrical separation between primary and secondary which is risky in case of high voltage levels.
Economical only when the voltage ratio is less than 2.
Next
Friday, April 7, 2023 Ch. 13 Transformers 106
Transformer Testing There are two simple tests to determine the equivalent-
circuit parameters and its efficiency and regulation: Open-circuit test (OC Test) Short-circuit test (SC Test)
Advantage of these tests is without actually loading the transformers, we can determine the Losses and Regulation, for full-load.
Next
Friday, April 7, 2023 Ch. 13 Transformers 107
(1) Open-Circuit Test This test determines the no-load current and the
parameters of the exciting circuit of the transformer. Generally, the low voltage (LV) side is supplied rated
voltage through a variac. The high voltage (HV) side is left open.
Open Circuit
Low voltage
Next
Friday, April 7, 2023 Ch. 13 Transformers 108
• The I2R loss on no load is negligibly small compared with the core loss.
• Hence the wattmeter reading, Wo, can be assumed to give the core loss of the transformer.
Calculations :
2i o 0 o
1
2 2ow m 0 w
1
1 10 0
w m
; ;
; ;
;
VP W I I K
V
WI I I I
V
V VR X
I I
Next
Friday, April 7, 2023 Ch. 13 Transformers 109
(2) Short-Circuit Test This test determines the equivalent resistance and leakage
reactance. Generally, the LV side of the transformer is short-circuited
through a suitable ammeter A2. A low voltage is applied to the primary (HV) side. This voltage is adjusted with the help of a variac so as to
circulate full-load currents in the primary and secondary circuits.
Short Circuit
Next
Friday, April 7, 2023 Ch. 13 Transformers 110
• The reading of ammeter A1, Isc, gives the full-load current in the primary winding.
• Since the applied voltage (and hence the flux) is small, the core loss is negligibly small.
• Hence, the wattmeter reading, Wsc, gives the copper loss (Pc).
Calculations :
2 2sc sce1 e1 e1 e1 e12
sc sc
; ;W V
R Z X Z RI I
Next
Friday, April 7, 2023 Ch. 13 Transformers 111
Example 11 A single-phase, 50-Hz, 12-kVA, 200-V/400-V transformer
gives the following test results :
(i) Open-circuit test (with HV winding open) : 200 V, 1.3 A, 120 W
(ii) Short-circuit test (with LV winding short-circuited) : 22 V, 30 A, 200 W
Calculate :
(a) the magnetizing current and the core-loss current, and
(b) the parameters of equivalent circuit as referred to the low voltage winding.
Next
Friday, April 7, 2023 Ch. 13 Transformers 112
Solution :(a) The wattmeter reading, 120 W, in the open-circuit test gives the core losses. Therefore, the core-loss current is given as
(b) The parameters of the exciting circuit are given by the open-circuit test, as
FL
12 kVAand 30 A
400 VI
This confirms that the short-circuit test has been done at the rated full-load
ow
1
120 W
200 V
WI
V 0.6 A
2 2 2 2m 0 w (1.3) (0.6)I I I 1.15 A
1 10 0
w m
200 V 200 V and
0.6 A 1.15 A
V VR X
I I 333 Ω 174 Ω
2
1
200 V 1Now,
400 V 2
VK
V
Click
Next
Click
Friday, April 7, 2023 Ch. 13 Transformers 113
sc sce1 e12 2
sc sc
200 W 22 V0.222 and 0.733
(30 A) 30 A
W VR Z
I I
The equivalent resistance and reactance as referred to the secondary side (low voltage winding),
22
e2 e1
10.222
2R K R
0.055 Ω
22
e2 e1
1and 0.699
2X K X
0.175 Ω
Next
Click