Electrical Machines

EE-260

Instructor: Dr Alina Mirza

Department of Electrical Engineering, Military College of Signals
National University of Sciences & Technology (NUST)
Determining Transformer Model’s Parameters
The parameter of the transformer model i.e., the
values of resistances and inductances can be
experimentally determined using two simple
tests: The open circuit test and the short circuit
test.

1.The Open Circuit Test: One of the winding is open circuited, and a full load rated
voltage is applied to the other side.
This measurement is normally done on the low voltage side of transformer, since lower voltages
are easier to work with)
 (a). The Open Circuit Test
Rp & Xp are too small as compared to Rc and Xm. Approximately all the voltage
drops across the excitation impedance

Convenient to consider admittance

of the excitation branch and then
compute Rc and Xm.

Conductance of the core- loss resistor

The total excitation admittance is:

Susceptance of the magnetizing inductor

1 1
YE  j
Rc XM
 (a). The Open Circuit Test (cont…
Magnitude of YE referred to primary The admittance is
side
I OC
YE   
VOC
The P.F. for real transformer is always I OC
lagging. Therefore: I lags V by θ
YE    cos 1 PF
VOC
Y E  A- jB

Finally RC and XM can be obtained

 (b) Short Circuit Test
• In this test the low voltage terminals of the transformer are short circuited.
• Apply a variable voltage source to high voltage side. Fairly low voltages are applied
on this side and are adjusted such that current in short circuited winding is equal to
rated current.
• Caution: Make sure to keep the applied voltage at a safe level, otherwise you could
burn the transformer’s winding.

• (this measurement is normally done on the high voltage side of transformer, since
currents will be lower on that side, and lower currents are easier to work with)
 (b) Short Circuit Test (cont…
• The input voltage, current, and power are measured.
• Negligible current flows through the excitation branch as the input voltage is
very low.
• Ignoring the excitation current, voltage drop can be attributed to series
elements in the circuit.

Magnitude and angle of the series impedances The series impedance ZSE is :
referred to the primary side of the transformer is

VSC 0 VSC
Z SE    
VSC I SC    I SC
Z SE 
I SC

PF  cos  
PSC
   cos 1 sc
P Z SE  Req  jX eq
VSC I SC I SCVSC
Z SE  ( RP  a 2 Rs )  j ( X P  a 2 X S )
 Tests : Find Values of Components

Determining the Values of Components in the Transformer Model

Open Circuit Test Short Circuit Test

Secondary
Ampere Ampere
Meter
Short
Wattmeter Meter Wattmeter
IP(t) IP(t) Circuit
A • • A • •

Volt
VP(t) Open V
Volt
VP(t) A
V(t) V Meter
V(t) Meter
Circuit

Figure 2-19 Connection for Transformer Figure 2-20 Connection for Transformer
transformer open-circuit test transformer short-circuit test

Apply rated primary (input) Voltage Apply Low voltages at input

IO No-Load current flows primary
Full load IP & IS current flows
Wattmeter reads …..? Core losses
Wattmeter reads I2R (Copper)
Example 2-2:
The equivalent circuit impedance of a 20 kVA, 8000/240 V. 60 HZ transformer
are to be determined. The open circuit test was performed on the secondary side
of the transformer, and short circuit test was performed on the primary side of
the transformer, and the following data were taken. Find the impedance of the
approximate equivalent circuit referred to primary side, and sketch the circuit.

Open-circuit test (on Short-circuit test (on primary)

secondary)
Voc= 240 V Vsc= 489 V

Ioc=7.133 A Isc= 2.5 A

Poc= 400 W Psc= 240 W

1. Transformer Voltage Regulation and
Efficiency
Due to series impedance in the transformer,
the output voltage of the transformer varies
with the load even if input voltage remains
constant.

Full-load voltage regulation is a quantity that compares the output voltage of the
transformer at no load with the output voltage at full load
VS ,nl  VS , fl
VR   100%
VS , fl
Since at no load: VS  VP / a • Low voltage regulation means smaller
winding impedance.

• For an ideal transformer, VR = 0 %

How to determine Voltage Regulation?

11
 Voltage Regulation
• Voltage Regulation depends on both
– Magnitude of series impedances
– Phase angle of the current

• How to measure ?
– Phasor diagram

12
 Power Factor
• Cosine angle between fundamental current and voltage

13
 2. The Transformer Phasor Diagram
IS -
VReq VjXeq
L Lagging
VP =ReqIS =XeqIS VS o Power Factor
a a Load
d

Phase relation
VP
between IS & VReq ?
a
 VRe q  VjXeq  VS
Phase relation
between IS & VXeq ? VS is assumed to be at
VP/a angle 0o ( VS-). All
other voltages & currents
are compared to this
reference
VReq
VjXeq
VS
IS VP/a > VS
Vp/a  Vsfl
% VR   100  Positive
?
Vsfl 14
Fig 2-27(a) Vector Dig 2. The Transformer Phasor Diagram
Unity Power Factor
IS (cont...
-
VReq VjXeq
L
VP =ReqIS =XeqIS o Unity
VS Power Factor
a a
d Load

VP
a
 VRe q  VjXeq  VS

VP/a

VP/a > VS
Vp/a  Vsfl
% VR   100  ?
VS
Fida Muhammad (Air University) Vsfl
Positive 15
Fig 2-27(a) Vector Dig
2. The Transformer Phasor Diagram
Leading PF
I 
(cont…
S

VReq VjXeq L
VP o Leading
=ReqIS =XeqIS VS
a Power Factor
a Load
d

VP
a
 VRe q  VjXeq  VS

VP/a Vp/a  Vsfl

% VR   100  ?
Vsfl

VP/a <VS
VReq

Negative
VS 16
 3. Transformer Efficiency
The efficiency is given by:
Pout
  100%
Pin
Pout
  100%
Pout  Plosses

From the transformer equivalent circuit efficiency can be easily calculated

The power losses comprise At any given load efficiency can be

– Copper losses calculated as:
– Hystersis losses Pout  Vs I s Cos s
– Eddy current losses
Vs I s Cos s
 100
The efficiency for a power transformer is Pcu  Pcore  Vs I s Cos
between 0.9 to 0.99. 17
Example 2-5:
A 15 kVA, 2300/230-V transformer is to be tested to determine its excitation
branch components, its series impedance and its voltage regulation.

The data have been taken by using the

connections shown in the following
Figures.
Open-ckt test Short-ckt test
Voc=2300 V Vsc=47 V

Ioc=0.21 A Isc=6.0 A

Poc= 50 W Psc= 160 W

1
Example 2-5:
a. Find the equivalent circuit of this transformer referred to the
high-voltage side.
b. Find the equivalent circuit of this transformer referred to the low-
voltage side.
c. Calculate the full-load voltage regulation at 0.8 lagging power
factor, 1.0 power factor, and at 0.8 leading power factor.
d. Plot the voltage regulation as load is increased from no load to
full load at power factors of 0.8 lagging, 1.0, and 0.8 leading.
e. What is the efficiency of the transformer at full load with a power
factor of 0.8 lagging?
Example 2-5: (Solution)
a)
From the OCT data

2
Example 2-5: (Solution)

For equivalent circuit referred to primary side the Excitation branch elements
are found as
a = 2300/230
= 10

The short-circuit impedance angle

The series elements referred

to the primary
The equivalent series impedance
Example 2-5: (Solution)

The transformer equivalent circuit referred

its primary side
Example 2-5: (Solution)
b) To find the equivalent circuit referred to the low-voltage side, the impedance is
divided by a2.

The transformer equivalent circuit referred its secondary side

Example 2-5: (Solution)
c) VR ? 15 kVA, 2300/230-V transformer
Full-load current on the secondary side

From the circuit

Example 2-5: (Solution)
c.

Tetha=cos^-1(PF)

The Voltage regulation is

7
Example 2-5: (Solution)
Phasor Diagram

8
Example 2-5: (Solution)
c.
At PF = 1.0, current Is = 65.2 L 0° A

9
Example 2-5: (Solution)
c.
At PF = 1.0, current Is = 65.2 L 0° A

9
Example 2-5: (Solution)

10
Example 2-5: (Solution)

10
Example 2-5: (Solution)
Plot of voltage regulation versus load
d. (Matlab)

11
Example 2-5: (Solution)
e.

To find the efficiency of the transformer. first calculate its losses.

The copper losses are:

The core losses are

The output power at 0.8 leading PF