ece421 3-gentranspu - utkweb.eecs.utk.edu/~kaisun/ece421/ece421_3-gentranspu_2.pdf•y- and y-...
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cos d dE V X I
3 3 cosaP V I
cos cos sina q dI ab de I I
sin q qV X I
sinq
q
VI
X
cosd
d
E VI
X
23 3 cos sin 3 sin 3 sin 2
2d q
q dd d q
X XE VP V I I V
X X X
3 3 sind
E VP
X Approximately, the second item can be ignored:
Consider a simple case ignoring Ra and Xl
d d q qE V jX I jX I
q d d
d q q
E V jX IV jX I
q axisd axis
Vq
Vd
Iq
Id
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Power Transformer
F=IN=/P
Ideal Transformer Real TransformerWinding resistance 0 0
Leakage flux 0 0 (windings do not link the same flux)
Permeabilityof the core
Infinite (flux is produced even with magnetizing current IP=0)
Finite(magnetizing current IP0 is needed to
produce flux)
Core losses0
0 (including hysteresis losses and eddy
current losses due to time varying flux)
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Real Transformers• Modeling the current under no-load
conditions (due to finite core permeability and core losses)When I2=I’2=0 I1=I0 0 I0=Im +Ic
Im is the magnetizing current to set up the core flux – In phase with flux and lagging E1 by 90o, modeled by E1/(jXm1)
Ic supplies the eddy-current and hysteresis losses in the core– A power component, so it is in phase with E1, modeled by E1/(Rc1)
• Modeling flux leakages– Primary and secondary flux leakage reactances: X1 and X2
• Modeling winding resistances– Primary and secondary winding resistances: R1 and R2
Ideal Transformer
1 2 1
2 2 2
E I N aE I N
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22 2 2 2
2 2
1 12 2
2 2
Z R jX a Z
N NR j XN N
Referred to the primary
side.
Exact Equivalent Circuit Ideal Transformer
1 2 1
2 2 2
E I N aE I N
22 2 2
1
/ NI I a IN
a2 Z
a V
1/a I
1/a2Z
1/aV
aI
12 2 2
2
NV aV VN
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Approximate Equivalent Circuits• Since V1E1, Z1 can be combined with Z’2 to become an equivalent Ze1
Referred to the primary side:
1 2 1 1 2e eV V R jX I
22 1
1 1 2 1 22
eNR R a R R RN
2
2 11 1 2 1 2
2e
NX X a X X XN
Referred to the secondary side:
1 2 2 2 2e eV V R jX I 2
2 22 1 2 1 2
1
/eNR R a R R RN
2
2 22 1 2 1 2
1
/eNX X a X X XN
V2
• If power transformers are designed with very high permeability core and very small core loss, the shunt branch can be ignored. Then, I1=I’2 and I’1=I2.
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Determination of Equivalent Circuit Parameters
• Open-circuit (no-load) test– Neglect (R1+jX1)I0 (since |R1+jX1|<<|Rc1//jXm1|)– Measure input voltage V1, current I0, power P0 (core/iron loss)
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10
| |c
VRP
1
1c
c
VIR
2 20| | | |m cI I I
11
| || |m
m
VXI
P0
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• Short-circuit test– Apply a low voltage VSC to create rated current ISC
– Neglect the shunt branch due to the low core flux
1| || || |
sce
sc
VZI
1 2| |sc
esc
PRI
2 21 1 1| |e e eX Z R
ISC
VSC
PSC
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Transformer Performance
• Efficiency: 95% - 99%• Given |V2,rated|, |S|=3|V2,rated||I2,rated| (full-load rated VA) and PF• Actual load is |I2|=n|I2,rated| where n<1 is the fraction of the full load power
Pout=n|S|PF
Pc, rated=3Re2|I2,rated|2 : full-load copper loss (current dependent)
Pc=3Re2|I2|2 =3Re2|I2,rated|2 n2
Pi, rated: core/iron loss at rated voltage (mainly voltage dependent, almost constant)
• Maximum efficiency (constant PF) occurs when
output powerinput power
2
0 0d dd I dn
,
,
when i rated
c rated
Pn
P
Learn Example 3.4
2, , , ,
| | | || | | | /c rated i rated c rated i rated
n S PF S PFn S PF n P P S PF n P P n
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Voltage Regulation• % change in terminal voltage from no-load to full load (rated)
– Generator
– Transformer:Utilizing the equivalent circuit referred to one side:
VR= 100nl rated
rated
V VV
VR= 100rated
rated
E VV
2, 2,
2,
VR= 100nl rated
rated
V VV
1 2
2
VR= 100V V
V
1 2
2
VR= 100V V
V
1 2 1 1 2e eV V R jX I
1 2 2 2 2e eV V R jX I
+
V
-
I
V=Vnl when I=0
V=Vrated when I=Irated
Source (generator,
transformer, etc.)
( )a s aE V R jX I
Secondary
Primary
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Three‐Phase Transformer Connections• A bank of three single-phase transformers connected in Y or arrangements• Four possible combinations: Y-Y, △-△, Y-△ and △-Y
– Y-connection: lower insulation costs, with neutral for grounding, 3rd harmonics problem (3rd harmonic voltages/currents are all in phase, i.e. van3=vbn3=vcn3=Vmcos3t)
– -connection: more insulation costs, no neutral, providing a path for 3rd harmonics (all triple harmonics are trapped in the loop), able to operate with only two phases (V-connection)
– Y-Y and - : HV/LV ratio is same for line and phase voltages; Y-Y is rarely used due to the 3rd harmonics problem.
– Y- : commonly used as voltage step-down transformers– -Y : commonly used as voltage step-up transformers
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3rd harmonics problem with three‐phase transformers• △ configuration provides a closed path for 3rd harmonics, or in
other words, all triple harmonics are trapped in the △ loop.– The 120o phase difference between the fundamental harmonic of I1 an I2 and
I3 becomes 360o (in phase) for the 3rd harmonic– Then by KCL, Ia(3)=I1(3)-I3(3)=0; also Ib(3)=Ic(3)=0
Distorted waveform
Fundamental
3rd harmonic3
9
, 3, 5, 7 9
, 3, 5, 7, 9
, 3, 5, 7 9
, 5, 7
, 5, 7
, 5, 7
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Y‐△ and △ ‐Y Connections• Y-△ and Y-△ connections result in a 30o phase shift between the primary and
secondary line-to-line voltages• According to the American Standards Association (ASA), the windings are
arranged such that the HV side line voltage leads the LV side line voltage by 30o
– E.g. Y- (HV-LV) Connection with the ratio of turns a= NH/NX>1
,
,
H P An H
X P ab X
V V N aV V N
,
,
3 30H L AB
H P An
V VV V
, ,X L X PV V
,
,
3 30H L AB
X L ab
V V aV V
HV Side (indicated by “H”) in Y connection:
LV Side (indicated by “X”) in △ connection:
=VH,P
=VX,P =VX,L
=VH,L
(Complex ratio)
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Per‐Phase Model for Y‐△ or △ ‐Y Connection
• Neglect the shunt branch• Replace the △ connection by an equivalent Y connection• Work with only one phase (equivalent impedances are line-to-
neutral values ZeY=Ze/3)• All voltages are line-to-neutral voltages
E.g. for Y-△ Connection, V1 is the phase voltage of the Y side and V2 is the phase (line-to-neutral) voltage of the △ side
V1
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Autotransformers• A conventional two-winding transformer can be changed into an autotransformer by
connecting its two coils in series. • The connection may use a sliding contact to providing variable output voltage.• An autotransformer has kVA rating increased but loses insulation between primary
and secondary windings
20MVA (115/69kV) McGraw-Edison Substation Auto-Transformer (Y-Y) (Source:
http://www.tucsontransformer.com)(Source: EPRI Power System Dynamic Tutorial)
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Autotransformer Model
1 2 1
2 1 2
V I N aV I N
1 1
2 1 2 22 2
(1 )H L L LN NV V V V V V V a VN N
IL
11 2 1 1
2
(1 )L HNI I I I I a IN
21 2 1 1 1
1
2 2
(1 )
1 (1 )
auto
w w conducted
NS V V I V IN
S S Sa
Power rating advantage
Equivalent Circuit (if the equivalent impedance is referred to the HV side)
Transformed power (thru
EM induction)
Conducted power (V2I1)
• Apparent power:
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+
600V
-
-
120V
+
+
480V
-
H1
H2
X2
X1
I2=125A
I2-I1=100A
I1=25A
+
600V
-
+
120V
-
+
720V
-
H1
H2
X1
X2
I2=125A
I2+I1=150A
I1=25A
+
120V
-
-
600V
+
-
480V
+
X1
X2
I1=25A
I2-I1=100A
I2=125A
H2
H1
|Si|=600150=90kVA |Si|=120100=12kVA|So|=720125=90kVA |So|=48025 =12kVA
The maximum apparent power |Smax|=max(|E1|,|E2|)(|I1|+|I2|)= (|E1|+|E2|) max(|I1|, |I2|)
600V 120V
H1
H2 X2
X1
|Si|=600100=60kVA |So|=480125=60kVA
Conventional transformer connected as an autotransformer (Example 11‐2 from ECE325 – Wildi’s book)