Transcript Chapter 6 - UniMAP Portal

BASIC ELECTRICAL TECHNOLOGY
DET 211/3
Chapter 6: Single Phase
Transformer
(Continued)
Parameter determination of the transformer
• Open circuit test
– Provides magnetizing reactance and core
resistance
– Obtain components are connected in parallel
loss
Experiment Setup:
In the open circuit test, transformer rated voltage is
applied to the primary voltage side of the transformer
with the secondary side left open. Measurements of
power, current, and voltage are made on the primary
side.
Since the secondary side is open, the input current IOC
is equal to the excitation current through the shunt
excitation branch. Because this current is very small,
about 5% of rated value, the voltage drop across the
secondary winding and the winding copper losses are
neglected.
Admittance
I oc
Yoc 
Voc
Open circuit Power Factor
PF  cos  
Poc
Voc I oc
Open circuit Power Factor Angle
  cos 1
Poc
Voc I oc
Angle of current always lags angle of voltage by 
Yoc 
I oc
1
1
    GC  jBM 
j
Voc
RC
XM
Short circuit test
– Provides combined leakage reactance and winding
resistance
– Obtain components are connected in series
Experiment Setup:
In the short circuit test, the secondary side is short
circuited and the primary side is connected to a
variable, low voltage source. Measurements of power,
current, and voltage are made on the primary side.
The applied voltage is adjusted until rated short circuit
currents flows in the windings. This voltage is
generally much smaller than the rated voltage.
Impedances referred to the primary side
Z sc 
Vsc
I sc
Power Factor of the current
Psc
PF  cos  
Vsc I sc
Angle Power Factor
  cos 1
Psc
Vsc I sc
Therefore
Z sc 
Vsc 0 0
I sc     0

Vsc
 0
I sc

 
Z sc  Req  jX eq  R p  a 2 Rs  j X p  a 2 X s

Exercise
The equivalent circuit impedances of a 20-kVA, 8000/240-V, 60Hz transformer are to be determined. The open circuit test and
the short circuit test were performed on the primary side of the
transformer and the following data were taken:
Open- circuit test
(on primary)
Voc = 8000 V
Short- circuit test
(on primary)
Vsc = 489 V
Ioc = 0.214 A
Isc = 2.5 A
Poc = 400 W
Psc = 240 W
Find the impedances of the approximate equivalent circuit
referred to the primary side and sketch that circuit
Per unit System
The per unit value of any quantity is defined as
Actual Quantity
Per Unit, pu 
Base value of quantity
Quantity – may be power, voltage, current or impedance
Two major advantages in using a per unit
system:
1. It eliminates the need for conversion of the voltages,
currents, and impedances across every transformer in
the circuit; thus, there is less chance of computational
errors.
2. The need to transform from three phase to single phase
equivalents circuits, and vise versa, is avoided with the
per unit quantities; hence, there is less confusion in
handling and manipulating the various parameters in
three phase system.
Per Unit (pu) in Single Phase System
Pbase ,Qbase , Sbase  VbaseI base
Vbase
Z base 
I base
I base
Ybase 
Vbase
( Vbase )2
Z base 
Sbase
Voltage Regulation (VR)
The voltage regulation of a transformer 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 held fixed.
VR 
VS ,nl  VS , fl
VS , fl
Req
X 100%
At no load, VS  VP a
Vp
VR  a
 VS , fl
VS , fl
Xeq
+
+
Vp/a
Vs
-
X 100%
Is
-
Phasor Diagram
Vp/a
jIsXeq
Vs
IsReq
Is
Lagging power factor
Vp/a
jIsXeq
Is
Vs
IsReq
Unity power factor
Vp/a
jIsXeq
Is
IsReq
Vs
Leading power factor
Efficiency
The efficiency of a transformer is defined as the ratio
of the power output (Pout) to the power input (Pin).
Pout

X 100%
Pin
Pout

X 100%
Pout   P losses
Vs I s cos 

X 100%
Vs I s cos   Pcu  Pcore
Pcore = Peddy current +
Physteresis
And
Pcu=Pcopper losses
Copper losses are resistive losses in the primary and secondary winding of
the transformer core. They are modeled by placing a resistor Rp in the
primary circuit of the transformer and resistor Rs in the secondary circuit.
Core loss is resistive loss in the primary winding of the transformer core. It
can be modeled by placing a resistor Rc in the primary circuit of the
transformer.
Assignment 5
A 15-kVA, 2400/240-V transformer is to be tested to
determine its excitation branch components, its series
impedances and its voltage regulation. The following
test data have been taken from the primary side of the
transformer:
Open- circuit test
Short- circuit test
Voc = 2400 V
Vsc = 48 V
Ioc = 0.25 A
Isc = 6.0 A
Poc = 50 W
Psc = 200 W
Assignment 5
The data have been taken by using the connections of
open-circuit test and short-circuit test:
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 and 0.8 leading power factor
d) What is the efficiency of the transformer at full load
with a power factor of 0.8 lagging?