In transmission, distribution, and utilization of electrical energy: step-up or step-down voltages at a fixed frequency 50/60 Hz, at power of hundreds of watts to hundreds of megawatts
Trang 11 Lecture 4
ECE430 Power Circuits and Electromechanics
Dr Nam Nguyen-Quang Fall 2009
http://www4.hcmut.edu.vn/~nqnam/lecture.php
2 Lecture 4
Transferring electrical energy from one circuit to another through time-varying magnetic field
Applications: both power and communications fields
In transmission, distribution, and utilization of electrical energy:
step-up or step-down voltages at a fixed frequency (50/60 Hz), at power of hundreds of watts to hundreds of megawatts
In communications, transformers can be used for impedance matching, DC isolation, and changing voltage levels at power of a few watts over a very wide frequency range
This course concerns only power transformers
Transformers – Introduction
3 Lecture 4
Consider a magnetic core with two coils
wound as shown Ignore losses, stray
capacitance, and leakage flux
Magnetic permeability is infinite or zero
reluctance
Ideal transformer
v 2
v 1 + –
+ –
dt
d
N
t
dt
d N t
N t v t v
2 1 2 1
a is called turns ratio
Total mmf is given bymmf N i1 N i2 R 0
N
t
i
t
1
2
2
4 Lecture 4
It can be shown, for an ideal transformer
Ideal transformer (cont.)
Ideal
N1:N2
+
– +
–
Ideal
N1:N2
+
– +
–
a N
N i
i a N
N v
1 2 2 1 2 1 2
a N
N i
i a N
N v
1 2 2 1 2 1 2
1 2 2 0
1 i t v i t
v
i t v i t
v1 1 2 2
a v
v L
L i
i
1 2 1 2 2
2 2
5 Lecture 4
Consider an ideal transformer with resistive load accross winding 2
By Ohm’s law,
Subsituting and
Impedance-changing property of ideal transformer
RL Ideal
N1:N2
+
– +
–
L
R i
v
2
2
a v
v2 1 i 2 ai1
L
N
N R
a
i
1 2 2
1
1
The discussion can easily be extended to systems with complex load
It can be verified that
L
Z N
N I
V N
N I
2
1 2 2 2 2
1 2 1
1
6 Lecture 4
The impedance-changing property can be used for maximizing power transferring between to windings, or matching impedances
An ideal transformer is placed between power source (impedance Z o)
and load (impedance Z L) Turns ratio is so selected that
Impedance matching
Ex 3.7: Two ideal transformer (each of ratio 2:1) and one resistor R are used to maximize power transfer Find R
Load resistance 4 together with R referred to the input side is (R + 4(2)2)(2)2 For maximum power transfer,
64 4
10 R R 13 5
Trang 27 Lecture 4
Two windings mounted on a magnetic
core, minimizing leakage flux
“Primary” winding (N1turns) connected
to power supply, “secondary” winding (N2
turns) connected to load circuit
Power transformer
Assuming an ideal transformer: no leakage flux, winding resistances
are neglected, magnetic core has infinite permeability, and is lossless
Let v 1 (t) = V m1 cost is the voltage applied to the primary winding, it can
be shown that
max 1
1 2 fN
Vm or V 1 4.44fN1max
8 Lecture 4
More pictures
Control
Small power Small 3-phase
Cast resin
9 Lecture 4
Ex 3.8: Given N1, N2, core cross-sectional area, mean core length,
B-H curve, and the applied voltage Find maximum flux density, and
required magnetizing current
Example
max
1
1 4 44 fN
webers 10 32 4 200 60 44 4
max
Hence,
2 3
webers/m 864 0 005 0 10 32 4
m
B
Therefore,
The required , the peak value of
magnetizing current is (259)(0.5)/200 = 0.6475 A Hence, I rms= 0.46 A is
the magnetizing on the primary side
At/m 259 300 864
m
H
10 Lecture 4
Consider now a transformer with leakage flux and winding resistances
Equivalent directly derived from physical model is simple but somewhat useless
The equations on the secondary side is mulplied by a (= N 1 /N 2 ) and i 2is replaced
by i 2 /a, to derive a more useful equivalent circuit.
Equivalent circuit of transformer with linear core
RL
N1:N2
+
– +
–
a2R2
aM
2/a
R1 L1– aM a2L2– aM
– +
– +
L1– aM is termed the leakage inductance of winding 1, a2L2– aM is termed
“referred” (to side 1) leakage inductance of winding 2 aM is the magnetizing
inductance, and its associated current is called magnetizing current
11 Lecture 4
There are losses in the magnetic core due to hysteresis and eddy current
These losses are very difficult to calculate analytically The sum of these losses
represents the total loss in the magnetic circuit of the transformer, and depends
only upon the value of Bm They are called core or iron losses A resistance can
be placed in parallel with the magnetizing inductance aM to account for them
Equivalent circuit of transformer with linear core (cont.)
RL Ideal
N1:N2
+
– +
–
i1
i2
+
–
av2
R1 L1– aM
Rc1 (aM)1
a2R2 a2L2– aM
Real load RLand its associated voltage and current can be retained by
referring them back to the secondary side, using an ideal transformer
12 Lecture 4
For steady-state operation, impedances and phasors can be used in the equivalent circuit
Transformer under sinusoidal steady-state conditions
ZL Ideal
N1:N2
+
– +
–
+
–
R1 jxl1
m1
a2R2 ja2xl2
where
1
I
2
I
2
V
1
a
I2
2 2 2
2
2 2
1 1 1
l l m l
x a aM L a
x a M L
X aM x aM L
Magnetizing reactance referred to winding 1 Leakage reactance of winding 2 Leakage reactance of winding 2 referred to side 1
Trang 313 Lecture 4
All quantities can be referred to winding 1
Transformer under steady-state (cont.)
a2ZL +
–
+
–
R1 jxl1
Rc1 jXm1
a2R2 ja2xl2
1
I
1
a
I2
ZL +
–
+
–
R1/a2 jxl1/a2
Rc1/a2 jXm1/a2
R2 jxl2
1
I a
a
2
I
Or they can be referred to winding 2
14 Lecture 4
Magnetizing branch makes computation somewhat difficult, hence this
branch is moved to the terminals of winding 1, yielding an approximate
equivalent circuit, with no serious numerical error introduced.
Approximate equivalent circuit
R1
1
I
+
–
m1 1
+
–
jxl1 a2R2 ja2xl2
2
V a a
I2
R1eq
1
I
+
–
Rc1 jXm1
1
+
–
jx1eq
2
V a a
I2
2 2 1 1
2 2 1 1
l l eq eq
x a x x
R a R R
15 Lecture 4
Parameters in equivalent circuit can be determined by two simple
tests: open-circuit test and short-circuit test.
In power transformers, the windings are called high-voltage (HV) and
low-voltage (LV) windings
Open- and short-circuit tests of transformers
A
V
W
oc
V
Open-circuit test
oc
V
R
oc
I
Equivalent circuit
16 Lecture 4
The test is performed with all instrumentation on the LV side with the
HV side being open-circuited Rated voltage is applied to LV side V oc,
I oc , and P ocare measured with the meters
Open-circuit test
oc
oc c
P
V R
2
c
oc R
R
V
I Ioc IR IX
Hence, IX Ioc2 IR2
X
oc m
I
V
R c and X mare the values referred to the LV side
oc
V
R
oc
I
17 Lecture 4
All the instrumentation is on the HV side Rated current is supplied to
HV side V sc , I sc , and P scare measured with the meters
Short-circuit test
R eq and X eqare referred to the HV side
A
V
W
sc
V
HV LV
Req Xeq
sc
V
sc
I
2
sc
sc
eq
I
P
sc
sc eq
I
V
eq eq
18 Lecture 4
Ex 3.9: Given OC and SC tests’ readings Find equivalent circuit parameters referred to the HV side
From OC test
Example
From SC test
50
2202
c
968
220
R
I
0 227 0 974 A
12 2
X
974 0
220
m
X
0 2076 17
60
2
eq
R
0 882 17
15
eq
Z
0 8822 0 20762 0 8576
eq
X
Trang 419 Lecture 4
Efficiency is defined as the ratio of output power to input power
Efficiency and voltage regulation
Voltage regulation is defined as
% 100
%
i c out out out
out in
out
P P P
P losses
P
P P
P
Losses are copper loss Pc and iron losses Pi.
Alternatively, if input power is known,
% 100
in i c in
P P P P
% 100 regulation
voltage
%
load load load
V V V
20 Lecture 4
Problem 3.22 and 3.23
In-class quiz