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Chapter 3: TRANSFORMERS and PER UNIT
Calculations
A transformer is a device that converts ac electrical energy at one voltage level to ac
electrical energy at another voltage level.
Transformers are divided into two main categories: single-phase transformers and threephase transformers.
Transformers are used primarily to change voltage levels in the range from a few watts to
mega-watts.
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Background:
1.2 The Magnetic Field
Magnetic fields are the fundamental mechanism by which energy is converted from one form to
another in motors, generators and transformers.
First, we are going to look at the basic principle – A current-carrying wire produces a magnetic field
in the area around it.
Production of a Magnetic Field

Ampere’s Law – the basic law governing the production of a magnetic field by a current:
 H dl  I
net
where H is the magnetic field intensity produced by the current Inet and dl is a differential element of
length along the path of integration. H is measured in Ampere-turns per meter.
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Background:

Consider a current currying conductor is wrapped around a ferromagnetic core;

I
CSA
N turns
mean path length, lc

Applying Ampere’s law, the total amount of magnetic field induced will be proportional to the
amount of current flowing through the conductor wound with N turns around the ferromagnetic
material as shown. Since the core is made of ferromagnetic material, it is assume that a majority
of the magnetic field will be confined to the core.

The path of integration in Ampere’s law is the mean path length of the core, lc. The current
passing within the path of integration Inet is then Ni, since the coil of wires cuts the path of
integration N times while carrying the current i. Hence Ampere’s Law becomes,
Hlc  Ni
H 
Ni
lc
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Background:

In this sense, H (Ampere turns per meter) is known as the effort required to induce a magnetic
field. The strength of the magnetic field flux produced in the core also depends on the material of
the core. Thus,
B  H
B = magnetic flux density (webers per square meter, Tesla (T))
µ= magnetic permeability of material (Henrys per meter)
H = magnetic field intensity (ampere-turns per meter)

The constant  may be further expanded to include relative permeability which can be defined as
below:
r 

o
where: o – permeability of free space (ex. air)

Hence the permeability value is a combination of the relative permeability and the permeability
of free space. The value of relative permeability is dependent upon the type of material used. The
higher the amount permeability, the higher the amount of flux induced in the core. Relative
permeability is a convenient way to compare the magnetizability of materials.
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Background:

Also, because the permeability of iron is so much higher than that of air, the majority of the flux
in an iron core remains inside the core instead of travelling through the surrounding air, which has
lower permeability. The small leakage flux that does leave the iron core is important in
determining the flux linkages between coils and the self-inductances of coils in transformers and
motors.

In a core such as in the figure,
B = H =
Ni
lc
Now, to measure the total flux flowing in the ferromagnetic core, consideration has to be made in
terms of its cross sectional area (CSA). Therefore,
   BdA
A
Where: A – cross sectional area throughout the core
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Background:
Assuming that the flux density in the ferromagnetic core is constant throughout hence constant
A, the equation simplifies to be:
  BA
Taking into account past derivation of B,

 NiA
lc
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Magnetic Circuit of a single phase transformer
Windings wrapped around laminations
Picture Source: http://www.sayedsaad.com
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Magnetic Circuit of a single phase transformer
Hollow-core construction
Picture Source: http://www.sayedsaad.com
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Magnetic Circuit of a single phase transformer
Shell-type core construction
Picture Source: http://www.sayedsaad.com
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Magnetic Circuit of a single phase transformer
Exploded view of shell-type transformer construction
Picture Source:: http://www.sayedsaad.com
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Types of transformers:
i)
Step up/Unit transformers – Usually located at the output of a generator. Its function is to step up
the voltage level so that transmission of power is possible.
ii) Step down/Substation transformers – Located at main distribution or secondary level
transmission substations. Its function is to lower the voltage levels for distribution 1st level
purposes.
iii) Distribution Transformers – located at small distribution substation. It lowers the voltage levels
for 2nd level distribution purposes.
iv) Special Purpose Transformers - E.g. Potential Transformer (PT) , Current Transformer (CT)
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Step up transformer in a power station
Generator Side: 15 kV
Transmission Side: 132 kV
Power: orders of 100 MVA
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850 MVA step down transformer
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25 MVA Distribution transfomer
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Potential Transformer (PT) is a special purpose transformer for monitoring single-phase
and three-phase voltages for metering and protection purposes.
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Current transformer (CT) is a special purpose transformer to measure high AC currents at
low/medium/high voltage levels for metering and protection purposes.
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Keban Dam Substation 380 kV Current Transformer (CT)
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Auto transformer is a single/three-phase transformer with a changable turns ratio. They are
used in laboratories to obtain variable AC voltage and in substations to adjust voltage levels
against changing load conditions.
Single-phase variac
İnternal structure of three-phase auto transformer
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Power in an Ideal Transformer
1.
The power supplied to the transformer by the primary circuit:
Pin = Vp Ip cos θp
Where θp = the angle between the primary voltage and the primary current. The power supplied by
the transformer secondary circuit to its loads is given by:
Pout = Vs Is cos θs
Where θs = the angle between the secondary voltage and the secondary current.
2.
The primary and secondary windings of an ideal transformer have the SAME power factor –
because voltage and current angles are unaffected θp - θs = θ
3.
How does power going into the primary circuit compare to the power coming out?
Pout = Vs Is cos θ
Also, Vs = Vp/a and Is = a Ip
So,
Pout 
Vp
a
aI cos 
p
Pout = Vp Ip cos θ = Pin
The same idea can be applied for reactive power Q and apparent power S.
Output power = Input power
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Equivalent circuit of a practical single-phase transformer
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Approximate Equivalent circuits of a Transformer
(a)
(b)
(c)
(d)
Referred to the primary side
Referred to the secondary side
With no excitation branch, referred to the primary side
With no excitation branch, referred to the secondary side
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Three phase Transformers
Transformers for 3-phase circuits can be constructed in two ways:
- connect 3 single phase transformers
- Three sets of windings wrapped around a common core.
Three-Phase Transformer Connections
The primaries and secondaries of any three-phase transformer can be independently connected in either a
wye (Y) or a delta (∆).
The important point to note when analyzing any 3-phase transformer is to look at a single transformer in
the bank. Any single phase transformer in the bank behaves exactly like the single-phase
transformers already studied.
The impedance, voltage regulation, efficiency, and similar calculations for three phase transformers are
done on a per-phase basis, using same techniques as single-phase transformers.
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The Buchholz Relay working principle of is very simple.
Buchholz Relay function is based on very simple
mechanical phenomenon. It is mechanically actuated.
Whenever there will be a minor internal fault in the
transformer such as an insulation faults between turns,
break down of core of transformer, core heating, the
transformer insulating oil will be decomposed in different
hydrocarbon gases, CO2 and CO. The gases produced
due to decomposition of transformer insulating oil will
accumulate in the upper part the Buchholz Container
which causes fall of oil level in it.
Fall of oil level means lowering the position of float and
thereby tilting the mercury switch. The contacts of this
mercury switch are closed and an alarm circuit energized.
Sometime due to oil leakage on the main tank air bubbles
may be accumulated in the upper part the Buchholz
Container which may also cause fall of oil level in it and
alarm circuit will be energized. By collecting the
accumulated gases from the gas release pockets on the
top of the relay and by analyzing them one can predict the
type of fault in the transformer.
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Magnetic Circuit of a three phase transformer
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Magnetic Circuit of a three phase transformer
Core Assembly of 10 MVA, 110 KV Transformer
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Picture Source:
http://www.standardtransformers.com/d_3.htm
Photo:http://www.infolytica.com/en/coolstuff/ex0104/TransformerModel.png
Photo: http://www.mttcorp.us/trafo3_6b.jpg
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A simple concept that all students must remember is that, for a Delta configuration,
V P  V
L
I P 
IL
3
S P 
S
3
For Wye configuration,
VP 
VL
3
I P  I
L
S P 
S
3
Calculating 3 phase transformer turns ratio
The basic concept of calculating the turns ratio for a single phase transformer is utilised where,
a
VP
VS
Therefore to cater for 3 phase transformer, suitable conversion into per phase is needed to relate the turns
ratio of the transformer with the line voltages.
Three-phase delta/star transformer wiring diagram showing earthing arrangements
Figure: http://www.tlc-direct.co.uk/Figures/5.1b.gif
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Wye-Delta connection
Voltage Relationships for Delta Connection
Delta-Wye connection
Voltage Relationships for Wye Connection
Figures: http://www.elec-toolbox.com/usefulinfo/xfmr-3ph.htm
Source: http://www.sick-fm.de/m/3-trafo.html
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Per-Unit System
It is frequent to express the electrical quantities (voltage, current, power, impedance) in
terms of appropriately chosen base values. All the usual computations are carried out in
per unit (pu) values instead of conventional units of volts, amps, watts, etc...
In a per-unit system the electrical quantitiy is expressed on a per-unit basis by the
following equation:
Actual value
Quantity per unit =
Base value of quantity
The advantges of pu system are:
 Narrow numerical range
 Easy to refer circuit parameters from one side to another side if there is a transformer in
the power system.
 Very useful in simulation studies of electrical machines and power systems on computers
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If subscript «b» denotes base value then we can compute base values for the currents and
impedances using the following equations for a single-phase system:
S
Ib  b
Vb
Vb Vb2
Zb 

Ib
Sb
Ib: base current
Sb: base power
Vb: base voltage
Zb: base impedance
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Following after the calculations of base values the following p.u. values can be computed:
V p .u .
V
 actual
Vb
I p.u . 
I actual
Ib
S actual
Sb
Z p.u . 
Z actual
Zb
S p.u. 
Z %  Z p.u .  100%
Percent of base Z
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Example 1:
A 500 W single-phase electrical load (@ unity pf) operates at 120 volts rms. Compute the perunit and percent impedance of the load. Draw the p.u. equivalent circuit.
Solution:
First we compute the equivalent resistance of the load since load power factor is 1.0
V2
V2
(120) 2
P
R

 28.8
R
P
500
Z  28.80
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Then we select the following base quantities for the question: (usually given in problems)
S b  500VA
Vb  120V
The base impedance is calculated as:
Vb2 (120) 2
Zb 

 28.8
Sb
500
Finally we compute the per-unit impedance of the load:
Z p.u. 
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Z 28.80

 10 p.u.
Zb
28.8
Example 2:
For the following single-phase system, obtain pu equivalent circuit and find the generator’s
voltage VG both in pu and actual. Choose base values as Sbase=1kVA and Vbase=200V
I=5A
ZG=j10 ohm
+
+
VG
LOAD
VL
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Solution:
I(pu)=1.0 pu
Base Values:
Base S = 1000 VA (given)
Base V = 200 V (given)
Base I = 1000/200=5 A
Base Z = 200/5=40 ohm
ZG(pu)=j0.25 pu
+
LOAD
VG(pu)
+
VL(pu)=1.0 pu
VL(pu)=200/200=1.0 pu
I(pu)=5/5=1.0 pu
ZG(pu)=j10/40=j0.25 pu
pu equivalent circuit
KVL => VG(pu)=VL(pu)+(j0.25)(I(pu))
VG(pu) = 1.0 + (j0.25)(1.0)
VG(pu) = 1.03 /__14° pu
 VG(actual)=Vbase*VG(pu)=(200)*(1.03)=206/__14° Volts
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Example 2: Pu Calculation of a Three-Phase Transformer
Obtain the actual parameters of the three-phase two winding transformer in SI units and
draw its per-phase equivalent circuit based on actual parameters.
The following typical parameters could be provided by the manufacturer:
Nominal power = 300 kVA total for three phases
Nominal frequency = 60 Hz
Winding 1: connected in wye, nominal voltage = 25 kV RMS line-to-line
resistance 0.01 pu, leakage reactance = 0.02 pu
Winding 2: connected in delta, nominal voltage = 600 V RMS line-to-line
resistance 0.01 pu, leakage reactance = 0.02 pu
Magnetizing losses at nominal voltage in % of nominal current:
Magnetizing Resistance =100 pu, magnetizing reactance = 100 pu
Source: http://www.mathworks.com/help/physmod/sps/powersys/ref/per-unit-and-international-systems-of-units.html
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Solution:
(Y connection)
Source: http://www.mathworks.com/help/physmod/sps/powersys/ref/per-unit-and-international-systems-of-units.html
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Source: http://www.mathworks.com/help/physmod/sps/powersys/ref/per-unit-and-international-systems-of-units.html
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X1=0.02*2083=41.66 Ω
X2=0.02*3.60=0.072 Ω
 Xm=100*2083=208.3 kΩ
Source: http://www.mathworks.com/help/physmod/sps/powersys/ref/per-unit-and-international-systems-of-units.html
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j41.66 Ω
j0.072 Ω
208.3 kΩ
0.036 Ω
j208.3 kΩ
20.83 Ω
Per-phase equivalent circuit of the three-phase transformer
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Change of Base
If the base is changed (usually the case) the per unit impedance changes and takes a new value.
2
Z new
V
S
 Z old old 2 new
Vnew S old
Example 3:
A 54 MVA transformer has a leakage reactance of 3.69% (on its own MVA base). What is the
reactance on a 100 MVA base? (Assume that voltage values do not change).
2
X new
V
S
100MVA
 X old old 2 new  0.0369 *1*
 0.0683 pu
54MVA
Vnew S old
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In summary: Why prefer to use Per Unit (pu) System Instead of the Standard SI Units?
 When values are expressed in pu, the comparison of electrical quantities with their "normal" values is
straightforward. For example, a transient voltage reaching a maximum of 1.42 pu indicates immediately that
this voltage exceeds the nominal value by 42%.
 The values of impedances expressed in pu stay fairly constant whatever the power and voltage ratings.
 For example, for all transformers in the 3 kVA to 300 kVA power range, the leakage reactance varies
approximately between 0.01 pu and 0.03 pu, whereas the winding resistances vary between 0.01 pu and
0.005 pu, whatever the nominal voltage.
 For transformers in the 300 kVA to 300 MVA range, the leakage reactance varies approximately between
0.03 pu and 0.12 pu, whereas the winding resistances vary between 0.005 pu and 0.002 pu.
 It means that if you do not know the parameters for a 10 kVA transformer, you are not making a major error
by assuming an average value of 0.02 pu for leakage reactances and 0.0075 pu for winding resistances.
Source: http://www.mathworks.com/help/physmod/sps/powersys/ref/per-unit-and-international-systems-of-units.html
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End of Chapter 3
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