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BEKP 2443
PENGANTAR KEJURUTERAAN KUASA
Lecture 7 : Power Transformers
UNIVERSITI TEKNIKAL
MALAYSIA MELAKA UNIVERSITI TEKNIKAL MALAYSIA MELAKA
Learning Outcomes:
To review basic concepts and establish terminology & notation for :-
The ideal transformer (3.1)
Equivalent circuit for practical transformers(3.2)
3 phase transformers & phase shift (3.4)
UNIVERSITI TEKNIKAL MALAYSIA MELAKA
Transformers Overview
Transformer = a device that transfers electrical
energy from one circuit to another through inductively
coupled conductors/coil.
Operation Principle : A varying current in the first or
primary winding creates a varying magnetic flux in
the transformer's core, and thus a varying magnetic
field through the secondary winding. This varying
magnetic field induces a varying electromotive force
(EMF) or "voltage" in the secondary winding by a
process called “mutual induction”.
1st development of commercial transformer started in
1885 by William Stanley
Type of Tranformers (application)
- Unit transformer
- Substation transformer
- Distribution transformer
Power Transformers
- DC-DC converter
- Switching Mode Power Supply
- Uninterruptible Power Supply
High Frequency Transformers
- Potential Transformer
- Current Transformer
Sensing (Instrument) Transformers
Type of Transformers
UNIVERSITI TEKNIKAL MALAYSIA MELAKA
5
Step-up transformer
- provides a secondary voltage that is greater than the primary voltage.
Step-down transformer
- provides a secondary voltage that is less than the primary voltage.
Isolation transformer
- provides a secondary voltage that is equal to the primary voltage.
- to isolate the power supply electrically from the power line, which serves as a protection.
Other classification of Power Transformer
Pole-mounted transformer
Current transformer (CT) power transformer (oil immersed)
Household transformer Dry type distribution transformer
IDEAL TRANSFORMER An ideal transformer has:
– no real power losses (no winding resistance)
– magnetic core has infinite permeability (no core reluctance, similar to resistance in elec circuit))
– no leakage flux (all flux is confined in the core & links both primary & secondary windings)
– No core losses
We’ll define the “primary” side of the transformer as the side that usually takes power, and the “secondary” as the side that usually delivers power.
– primary is usually the side with the higher voltage, but may be the low voltage side on a generator step-up transformer.
Physical Model of Ideal Transformer
• Made up of
inductors/coils
• Not electrically
connected.
• An AC voltage
applied to the primary
induces an AC voltage
in the secondary.
Voltage polarity and current direction
Dot Convention/standard:
Ip Is
1. If the primary voltage is positive at the dotted end of winding with
respect to undotted end, then the secondary voltage will be positive at
the dotted end also. Voltage polarities are the same with respect to the
dots on each of the core
2. If primary current of the transformer flows into the dotted end of the
primary winding, the secondary current will flow out the dotted end of
the secondary winding
The direction of the windings determines the polarity of the voltage across the secondary winding with respect to the voltage across the primary. Phase dots are sometimes used to indicate polarities.
In phase Out of phase
Voltage polarity and current direction
Equivalent Circuit of 2 winding,
1Φ Transformer
Notation: 1=primary 2=secondary
12
2 Winding, 1Φ Transformer (Voltage
Relationships Between Primary & Secondary)
From Faradays Law & Lenz’s law, the turns ratio of a transformer is equal to the voltage ratio of the component:
2
1
2
1
N
N
V
V and we
define also 2
1
N
Nat
For example:
acac VVVN
NV 30)120(
4
11
1
22
ratio turnsampereta
13
2 winding, 1Φ Transformer (current
relationships between primary & secondary)
Assuming the transformer is 100% efficient, then the power delivered & received from winding 1 to winding 2 is SAME, thus:
Therefore:
2
*
22
*
111 SIVIVS 2
1
1
2
V
V
I
I
2
1
1
2
N
N
I
I
and
14
Example 1 Consider the source, transformer, and load shown in the circuit below. Determine the rms values of the currents and voltages (a) with the switch open and (b) with the switch closed.
VrmsV 110)(1 Solution Voltage applied to the primary,
VrmsVN
NrmsV 22)110(
5
1)()( 1
1
22
(a) With the switch open, the secondary current is zero. Hence, the primary current is also zero. Why?. Because no power transfer between the two “circuit” (remember : S=VI*).
(b) With the switch closed:
AR
rmsVrmsI
L
2.210
22)()( 2
2 ArmsIN
NrmsI 44.0)2.2(
5
1)()( 2
1
21
Transformer Rating
The rating of a transformer is stated as Volt Ampere (VA) that it can transform without overheating.
The transformer rating can be calculated as either V1I1 or V2I2 where I2 is the full load secondary current.
E.g. transformer is rated at 20kVA, 480/120V, 60Hz : Means: “This transformer is capable of transferring a normal/nominal
of up to 20kVA of complex power between primary & secondary winding. Its primary & secondary winding is capable of having a nominal/normal voltage of up to 480V & 120V respectively. The whole transformer can only be placed at a system with frequency of 60Hz.”
Impedance Transformation/Reflection
2
2
2
I
VZ
2
11
2
12
1
I
V
Z
NN
NN
2
2
2
2
2
1
1
12
I
VZaZ
N
NZ t
The phasor current and voltage in the secondary are related to the load impedance by
Then,
The impedance seen by the source (referred to primary & secondary) are:
2
11
2
1
2
2
21
I
V
ta
ZZ
N
NZ
Secondary impedance referred to primary:
Primary impedance referred to secondary :
Example of Reflection of impedances & equivalent circuit
UNIVERSITI TEKNIKAL MALAYSIA MELAKA
2
2
2
2
2
1
1
1'
2I
VZaZ
N
NZ t
2
11
2
1
2
2
2"
1I
V
ta
ZZ
N
NZ
Secondary impedance referred to primary:
Primary impedance referred to secondary :
Example 2 (Glover, pg100)
A single phase , 2 winding transformer is rated at 20kVA, 480/120V, 60Hz. A source connected to the 480V winding supplies an impedance load connected to the secondary winding. When the secondary voltage is at 118V, the load absorbs 15kVA at 0.8pf lagging.(assume ideal transformer & select 118 V as reference) Calculate:
a) Voltage applied across the primary winding (V1)
b) The actual load impedance (Z2)
c) The load impedance referred to the primary winding (Z’2)
d) The real & reactive power supplied at the primary winding
UNIVERSITI TEKNIKAL MALAYSIA MELAKA
Solution Step1 : draw & label equivalent diagram (picture tells a 1000 words!!) &
write down all information given!.
a :
b:
c :
d:
UNIVERSITI TEKNIKAL MALAYSIA MELAKA
60Hzf 3.
kVA )8.0(cos15lag 0.8pf @ 15S .2
4120
480 .1
:given info
1
22
2
1
SkVA
N
Nat
oo
tVaV 0472)0118(421
o
oo
o
I
VZ
Z
VI
V
SI
IIVS
87.3693.0
A87.3612787.36127*
kVA 87.3615)0118(
2
22
2
22
2
22
*
2
0*
222
oo
tZaZ 87.3688.14)87.3693.0(42
2
'
2
Var 9000Q
W12000P
jQP900012000kVA 87.361512
jSS o
Badariah Bais KKKF163 Introduction to EE Sem II 2006/07 20
Example 3
Consider the circuit shown below. Find the phasor currents and voltages at both primary & secondary winding. Also, find the real & reactive power delivered to the load.
Solution Step1: Impedance at the secondary: )2010( jZL
Step 2: Load Impedance reflected/referred at the primary:
)20001000()2010(
1
1022
2
12'jjZ
N
NZaZ LLtL
Step 3: Total impedance
(all referred at primary) 452828)20002000(200010001000
'
1 jjZRZ LT
Badariah Bais KKKF163 Introduction to EE Sem II 2006/07 21
Step 4: Primary current and voltage (refer equivalent circuit referred to primary):
452828TZ
AZ
VI
T
S
453536.0
452828
010001
)20001000)(453536.0(11 jZIV'
L
V 43.186.790)43.632236)(453536.0(
Step 5: Secondary current and voltage (refer original circuit): AII
45536.3)453536.0(
1
10
N
N1
2
12
VVV
43.1806.79)43.186.790(
10
1
N
N1
1
22
)20001000('
jZL
Step 6: Real & Reactive Power delivered to the load:
222
*
2221
jQP
4.63279)45536.3)(43.1806.79(
S
IVSS
original circuit New equivalent circuit with load referred to primary
Rule of thumb to find primary & secondary winding
current & voltages
If given the turns ratio & you know the voltage or current of winding 1, you can always find V&I for winding 2 by only using the turns ratio relationship.
If only given : turn ratio, impedances for winding 1 & 2, & source voltage (Vs) of winding 1, then…
To calculate V & I at winding 1
1. Reflect all impedance from winding 2 to 1
2. use Ohms law to calculate the V & I of winding A using the total impedance
(Total impedance = impedance at winding A plus the ones reflected at winding A)
To calculate V & I at winding 2
1. Convert V &I at winding A to winding B using only the turns ratio relationship
To calculate power
Power at winding A = power at winding B
UNIVERSITI TEKNIKAL MALAYSIA MELAKA
Real/Practical Transformer
Real transformers
– have losses
– have leakage flux
– have finite permeability of
magnetic core
Real power losses
– resistance in windings (i2 R)
– core losses due to eddy
currents and hysteresis
Transformer Core losses
1. Eddy currents in the core
arises because of changing
flux in core. Eddy currents
are reduced by laminating
the core.
2. Hysteresis losses are
proportional to area of
BH curve and the
frequency. These losses
are reduced by using
material with a
“thin” BH curve
Equivalent Circuit of real transformer
Equivalent Circuit of real transformer
Phenomena physics represented by equivalent circuit
Proportional to voltage square
Core loss current is a current proportional to the voltage applied to the core that is in phase with the aplied voltage, so it is modeled by a resistance Rc or Gc
connected with the primary winding
- Core (eddy current) losses
- Copper losses Proportional to current square
Copper loss are resistive losses in the primary and secondary windings and are modeled by resistor R1 and R2 connected to the primary and secondary winding so respectively
- Hysterisis current losses Proportional to excitation frequency
The high excitation frequency is due to magnetic saturation in the transformer core
- Leakage flux Flux component which links winding 1 but does not link winding 2
Magnetization flux is flux produced by magnetization current
- Magnetization flux Represented by magnetization inductance
Equivalent circuit of single phase 2 winding transformer & parameter definition
R1 = primary winding resistance I2R losses (real power loss)
R2 = secondary winding resistance I2R losses (real power loss)
X1= primary winding leakage reactance I2X loss (reactive power loss)
X2= secondary winding leakage reactance I2X loss (reactive power loss)
UNIVERSITI TEKNIKAL MALAYSIA MELAKA
Equivalent circuit of single phase 2 winding transformer
Im= magnetizing
Ic= core loss current
Ie = excitation current
UNIVERSITI TEKNIKAL MALAYSIA MELAKA
)usceptanceinductor(sshunt 2
1
11
2
11
N
RBwhere
EjBIN
NII
cm
mm
1)( EjBGIII mcmce
Equivalent Circuit of Transformer
Another version of equivalent circuits: All losses take into account & secondary impedances all referred to primary
windings
UNIVERSITI TEKNIKAL MALAYSIA MELAKA
Another version of equivalent circuits: All losses take into account & secondary impedances all referred to
primary windings
'
2
'
2
'
21
'
22
2
2
12
2
11
IZVE
IZN
NV
N
NE
2
2
2
12
2
2
1'
2
'
2
'
2
'
2
XN
NjR
N
NZ
jXRZ
Another version of equivalent circuits: neglect core losses (ignore shunt branch) & secondary impedances all referred
to primary windings
UNIVERSITI TEKNIKAL MALAYSIA MELAKA
Another version of equivalent circuits: neglect core losses (ignore shunt branch) & ignore resistive winding losses with
secondary impedances all referred to primary windings
UNIVERSITI TEKNIKAL MALAYSIA MELAKA
Equivalent Circuit (referred to primary)- only if (R1, X1 << Rc, Xm)
Approximate equivalent circuit referred to the primary.
(R1, X1 << Rc, Xm)
- Move the excitation branch close to V1
Equivalent Circuit (referred to secondary) Approximate equivalent circuit referred to the secondary.
(R1, X1 << Rc, Xm)
- Move the excitation branch close to V1
Simplified circuit referred to one side.
-Neglect the excitation branch
(very high permeability core and very small core loss)
Method to determine equivalent Circuit parameters for real transformer
PARAMETERS
to find the shunt admittance (Y)
Data given : I1/ Io and P1/ Po, you must know that V1 = rated voltage
no load
Io = open circuit current
Open-circuit test (no load)
No load
to find the series impedances (Z)
Data given : V1/ Vsc & P1/ Psc, you must know that I1/ Isc = rated current
Short-circuit test
Transformer Performance - Efficiency of a transformer in percent is given by %100
x
powerinput
poweroutput
- Voltage Regulation – the change in the magnitude of the secondary terminal voltage from no-load to full-load.
100egulation2
22x
V
VVR
nl
- Primary no-load.
- Secondary no-load.
100egulation'
2
'
21x
V
VVR
100egulation2
2
'
1x
V
VVR
Refer to primary
Refer to secondary
Example 4
Data obtained from short-circuit and open-circuit tests of a 20 kVA, 480/120 V 60 Hz transformer are as follows. Determine the parameter of the equivalent circuit.(solution: refer Glover, P 105~106)
Result of Short-circuit test: V1 = 35V & P1 = 300W Result of Open-circuit test: I2 = 12A, & P2 = 200 W Questions: 1. From the short circuit test, find the equivalent series impedances (neglect shunt
admittances) 2. From open circuit test, determine the shunt admittances (Ym) referred to
winding 1. neglect the series impedances.
solution
Part a: step 1 draw equivalent circuit of SC test & write down all info given:
UNIVERSITI TEKNIKAL MALAYSIA MELAKA
OR
82.01728.0XR
82.0RZX
XR84.042
35
I
VZ
173.042
300
I
PRRIP
42480
20
V
SII
eqeq
2
eq
2
eqeq
2
eq
2
eq
1
1eq
22
sc
eqeq
2
sc
rated 1,
ratedSC1
jjZ
AV
kVA
eq
Given : V1 = 35V P1 = 300W
solution Part b: step 1 draw equivalent circuit of OC test & write down all info given:
UNIVERSITI TEKNIKAL MALAYSIA MELAKA
OR
o
m
m
mm
m
mcom
c
cmco
c
ot
jjY
ZYjYremember
I
VXAIII
AR
VIjIII
P
VR
AIN
NIIVa
02.820063.000619.0000868.0BG
00619.0B
000868.0G
X
1B ,
R
1G ,
1 where,BG :
6.161,97.2
42.01152
480,
1152200
480
312480
120,480120
120
480VV
mc
m
c
m
m
c
cmc
122
1
22
1
2
1
2121
Given : I2 = 12A P2 = 200W
Example 5
A single phase transformer has 2000 turns on the primary winding and 500 turns on the secondary. Winding resistances are R1 = 2Ω and R2= 0.125 Ω; leakage reactances are X1 = 8Ω and x2 = 0.5Ω. The resistance load on the secondary is 12Ω.
(a)If the applied voltage at the terminals of the primary is 1000V, determine V2 at the load terminal of the transformer, neglecting magnetizing current.
(b) Compute the percent voltage regulation
UNIVERSITI TEKNIKAL MALAYSIA MELAKA
solution
UNIVERSITI TEKNIKAL MALAYSIA MELAKA
3 phase transformer
connections & phase shift
Three-phase Transformer
a. Three one-phase transformers are composed to be a three-phase transformer bank
b. Three-phase transformer wrapped around single three-legged core
Three-Phase Transformer
Connection technique
Connect the end of a winding with the
begining of another winding
Delta
Wye (star)
Connect the end of each winding at a
common point
a
b
c
b
a
c
Three-Phase Transformer
Three-Phase Transformer Connections.
End of Lecture 7
UNIVERSITI TEKNIKAL MALAYSIA MELAKA