Transformers - University of Florida · Transformers 2. The Ideal Transformer - A primary current...
Transcript of Transformers - University of Florida · Transformers 2. The Ideal Transformer - A primary current...
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TransformersTransformersRevised October 6, 2008
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3. Transformers 1
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The Ideal Transformer
PN SN
P i S dPrimaryWinding
SecondaryWinding
Basic TransformerBasic Transformer
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3. Transformers 2
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The Ideal Transformer - A primary current produces a flux in the core.
P PP P P
N i tAN i t e
Pi t e
PN SNPN SN
PrimaryWinding
SecondaryWindingWinding Winding
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Recall Slide Set 1:
i t Applied
flux
0d t
dt
+
inducedde Ndt
dt
_
Induced (opposing) flux
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Ideal Transformer – The primary current gives rise to aninduced voltage across the primary winding.
1P P P P P
AN i t N i t e Pi t e
Note:
PN SN Pv t OpenCircuit+
PrimaryWinding
SecondaryWinding
2
P
PP Pinduced P P
di td Ne t v t Ndt dt
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dt dte
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Self-Inductance of Primary – the output (secondary) is open-circuited
2PP P
P P
di td Nv t Nd d
e
22
P P
P PP P P
dt dtdi t NL L N
e
c
similarly,
,P P PL L Ndt
ce
2 did 2SS S
S S
di td Nv t Ndt dt
e 2
2,S SS S S
di t NL L Ndt
ce
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Ideal Transformer – the flux produced by the primary winding is coupled to the secondary winding and induces a voltage across thecoupled to the secondary winding and induces a voltage across thesecondary.
1P P P P P
AN i t N i t e
Pi t
e
+ Pv t
Sv t
PN SN
S
p S PPinduced S S
N N di tde t v t Ndt dt
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S dt dte
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Mutual Inductance
p S PPS S
N N di tdv t N
S S
p SP
v t Ndt dt
N Ndi tM M
e , pM M
dt
e
2 2
,P SP S P S
N NL L M L L e ee e
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Ideal Transformer
2
PP di tN
pP P
p S PS S S
Nv t LdtN N di tv t N L
e pS S S
dte
pP Nv tTurns Ratio
v t N
S Sv t N
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Ideal Transformer – A load across the secondary winding will cause a current to flow that opposes the change in flux of the primary.
S SS S S
N i tAN i t e
Pi t Si t
e
+ Pv t
Sv t
PN SN
S
p S PPinduced S S
N N di tde t v t Ndt dt
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S dt dte
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Ideal Transformer – The flux produced by the secondary current will coupled back to the primary winding.
Pi t Si t
+ v t
v t
N N
+ Pv t
Sv t
PN SN
S SS S S
N i tAN i t e
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Ideal Transformer – The flux produced by the secondary will becoupled back to the primary winding inducing a voltage on the primary.
1S S S S S
AN i t N i t e
Pi t Si t
+ Pv t Sv tPN SN
P
SS P Sinduced P
di td N Ne t Ndt dt
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dt dte
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The relationship between the primary and secondary currents can be determined from the total mmf:
i t i t
+ t
t
N N
Pi t Si t
+ Pv t
Sv t
PN SN
f S S S
P P
S
P P
mmf N
mmf N i
i
t
t
e
e
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S S S Sf
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t t l f
For the Ideal Transformer:
:
0P P S S P S
total mmfN i t N i t
e
0
e
, Re : P P
S S
P S
S P
v t Ncallv t N
i t Ni t N
:
S SS P
total poweri t t N N
1,P P S PP S
S S P S
i t v t N N p pi t v t N N
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Pi t Si tIdeal Transformer
P S
Pv t
Sv t
PN SN
2 di t di tN N N 2P SP P S
P
di t di tN N Nv tdt dt
e e
Similarly, 2
P SP S SS
di t di tN N Nv tdt dt
e e
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dt dte e
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Ideal Transformer
Pi t Si tTransformer
Pi t Si t
Pv t
Sv t
PN SN
BLACK BOXBLACK BOX
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Ideal Transformer
Pi t Si t
Pi t Si t
Pv t
Sv t
PN SNTransformer
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Ideal Transformer
Pi t Si t
Pi t Si t
Pv t
Sv t
PN SNTransformer
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Ideal Transformer
Pi t Si t
Pv t
Sv t
PN SN
Dot Convention: When the current on the primary/secondary enters the dotted terminal the polarity of the voltage on theenters the dotted terminal, the polarity of the voltage on the secondary/primary is positive.
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i t i tIdeal Transformer
Pi t Si t
Pv t
Sv t
PN SN
i t i t Pi t Si t
Circuit
Sv t Pv tCircuitRepresentation
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Phasor Representation
1P P SV j L I j MI
V j MI j L I
22 2
1 2
S P S
P S P S
V j MI j L I
N N N NL L M
1 2, ,L L Me e e
PI SIP S
V
SV
PV
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PI SIImpedanceTransformation
SVP
inVZI
LZ S L SV Z I
PV
SV
inPI L S L SP
, ,P P P SS L S
S S S P
V N I N V Z IV N I N
2
S S S P
P PS L S
N NV Z IV N N N
P S S P
inS SP
S S
LS
V N NN NI I I
N
NN
N
Z Z
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P PNN
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Review EEL 3111 as needed for circuit analysis techniques with transformers (Chapter 13 intechniques with transformers. (Chapter 13 in Alexander and Sadiku, Fundamentals of Electric Circuits)
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Real (Non-Ideal) Transformers( )
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Revisit Flux Linkage – Note Set 1, Slide 64:
Flux Linkage:
d de N inducede Ndt dt
Flux Linkage: N
But each turn does not necessarily link the same amount of flux.
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Flux Linkage
Taken from: J.D. Kraus & D.A. Fleisch, Electromagnetics with Applications, Fifth Edition, McGraw-Hill, 1999, page 98.
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Flux Linkage inducedd de t Ndt dt
induced
dt dte t dt
1inducede t dt
N
Average flux per turn
inducedde t NdtHence we really mean:
dt
How does this flux effect the secondary?
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Flux Linkage – When a voltage is applied across the primary of a transformer the average flux in the winding will be:g g
1P
P
v t dtN
P
How does this flux effect the secondary?
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Not all the primary flux is coupled to the secondary
Leakage Flux (ideally small)
I Mutual Flux
L k Fl
PI
Leakage Flux (ideally small)
Mutual Leakage PrimaryP
M LP
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Similarly,
Leakage Flux (ideally small)
IMutual Flux
L k Fl
SI
Leakage Flux (ideally small)
Mutual Leakage SecondaryS
M LS
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Re-express Faraday’s Law as:
Pd PP P
dv t Ndt
d d
M LPP P
d dN Ndt dt
e t e t
P LPe t e t
Si il l Sdv t N Similarly: S
S S
M LS
v t Ndt
d dN N
M LSS SN N
dt dte t e t
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S LSe t e t
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The primary voltage due to the mutual flux is:
d MP P
de t Ndt
The secondary voltage due to the mutual flux is:
d MS S
de t Ndt
Hence:
P S P PM Pe t e t e t v td N
P S P PM P
P S S S SN dt N e t N v t
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Magnetization Current in a Real Transformer
Pi t
N NPN SN
Primary SecondaryWinding Winding
Note again that when a primary source is connected current flowsNote again that when a primary source is connected, current flows in the primary even the secondary is open circuited. This is the current required to produce the flux in the (real) ferromagnetic core.
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Magnetization Current in a Real Transformer
The applied current consists of two components:
Magnetization current; the current required to produce the flux in the core, and
The core loss current to account for eddy current and hysteresis loss.
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Eddy Current Loss – A changing flux will induce a current in the magnetic core just as is does for a wire.g j
Eddy currents flowing inEddy currents flowing in the core.
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Since the eddy current loss is proportional to the path length, making the core using the laminated, layered structure shown minimizes these losses.
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Magnetization Current: Ignoring leakage, recall from Slide 28 that the average flux in the core is:
1 v t dt If the primary voltage is:
PP
v t dtN
If the primary voltage is:
cosP Mv t V t
then
P M
V
Note that the flux (and hence the magnetization current) lags the applied
sinM
P
V tN
) g pp
voltage by 90o.
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Example – Suppose the magnetization curve for a real core can be
2
Ni 7
modeled as
2
0
210 5 0 5 10
2
Ni ( )
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22If the sinusoidal flux variation
0
0.1 sin t( )
is ‘small’, the magnetization current is nearly sinusoidal as well.
10 0 102
2
1010 Ni ( ) 1010 Ni ( ) t
0.10 1
0
0.1
Ni 0.1 sin t( )( ) Ni
10 0 100.1
0.1
1010 t
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1010 t
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22
Once the peak flux reaches the
0
2
1.0 sin t( )
saturation point in the core, the magnetization current is clearly on longer sinusiodal. Note also that
10 0 102
2
geven a small additional increase in the flux will cause a large corresponding increase in the 1010 Ni ( ) t
22
corresponding increase in the magnetization current. (See next slide as well).
0Ni 1.0 sin t( )( ) N
Ni
0 10 0 102
2Ni
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22
0
2.0 sin t( )
10 0 102
2
1010 Ni ( )200
1010 Ni ( ) t
0
129.984
Ni 2.0 sin t( )( )
200129.984
( )( )
10 0 10
1010 t
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Hysteresis and Eddy Current Loss – The other component of the no-load current is current required to supply power for the q pp y physteresis and eddy current losses. These are core losses.
As before assume that the flux in the core is sinusoidal ByAs before, assume that the flux in the core is sinusoidal. By Faraday’s law, the eddy currents in the core will be proportional to
deddy
didt
The eddy currents will be largest when the core flux is passing through zero.
Since core losses are resistive in nature, the eddy current will be in-phase with the applied primary voltage.
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magnetization hysteresis eddyexcitation iii
4
magnetization hysteresis eddyexcitation
2
Ni 1.1 sin t( )( ) 1 cos t( )
0
( )( ) ( )
Ni 1.1 sin t( )( )
cos t( )
2
10 5 0 5 104
t
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Linear Circuit Model for a Real Transformer
H d d l li lHow do we model a linear real transformer?
Copper lossesEddy current lossesHysteresis lossesHysteresis lossesLeakage flux
N t th ti t ti ff t thNot the magnetic saturation effects – these are non linear.
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Linear Circuit Model for a Real Transformer
How do we model a real transformer?
Copper losses – Add a resistance (RPand R ) in the primary and secondaryand RS) in the primary and secondary circuit, respectively.
EEL 3211© 2008, H. Zmuda
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Linear Circuit Model for a Real Transformer
Copper Losses Model
PR SR
I I V
V
N NPI SI
SV
PV
PN SN
IDEAL
EEL 3211© 2008, H. Zmuda
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Linear Circuit Model for a Real Transformer
How do we model a real transformer?
Leakage flux
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Leakage Flux
Recall from Slide 31 that,
LPLP P
de t N LSLS S
de t N
Since much (most) of the leakage flux path is through air, and since i h l hi h i h h h f h
LP P dt LS S dt
air has a constant reluctance which is much greater than that of the core, the primary/secondary leakage flux is proportional to the primary/secondary coil current, respectively, orp y y p y
P PLP P P
N i N i ce Permeance of
fl thS S
LS S SN i N i
e
ce
flux pathReluctance of flux path
EEL 3211© 2008, H. Zmuda
3. Transformers 48
e
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Leakage Flux
2LP P Pd di die t N N L c
2
LP P P P
LS S S
e t N N Ldt dt dt
d di dit N N L
c
c 2LS S SLS S S Se t N N L
dt dt dt c
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Recall Slide 31,
MLPdt t N dt N
MP P
M
LPP LP Pv t e t N
dte t N
dtdN
MP Pe t N
dt
which suggests an inductor as a model.
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SLS S
die t L PLP P
die t L
PjXPR SjXRFlux Leakage Model
LS Se t Ldt
LP Pe t Ldt
I I
N N
PP SSR
P PX LPI SI
SV
PV
PN SN
IDEAL
d d LPLP P
MP P P
de t Ndt
dv t e t Ndt
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How do we model these effects?
•Magnetization current•Hysteresis (core) losses
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Recall (Slide 37) that the magnetization current (in the unsaturated region) is proportional to the voltage applied to the core and lags the applied voltage by 90o,
cosP Mv t V t sinM
P
V tN
P PN i e
1
d 1 PPP
Pv t d tt dv NN dt
Hence it can be modeled by a reactance (an inductor) connected across the primary voltage source
EEL 3211© 2008, H. Zmuda
3. Transformers 53
across the primary voltage source.
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Magnetization Current
PjXPR SjXSR
g
I I
N N
PP SS
PI SISV
PV
PN SNMjX
IDEAL
LP Mdv t e t N de t N LPP LP P
MP Pv t e t N
dte t N
dt
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The core loss current ihysteresis + eddy is a current proportional to the y yvoltage applied to the core that is in-phase with the applied voltage, hence it can be modeled by a resistance RC connected across the primary voltage source.across the primary voltage source.
Remember, the core effects are really nonlinear, and the circuit d l h d d id l li i timodel we have produced provides only a linear approximation
in this complicated situation.
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Core Losses
PjXPR SjXSR(Eddy + hysteresis)
I I
N N
PP SS
PI SISV
PV
PN SNCRMjX
IDEAL
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PjXPR SjXSRWhy here ...
N NRjX PN SNCRMjX
PjXPR SjXSRIDEAL
PN SNCRMjX P S
IDEAL
CMj
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3. Transformers 57
IDEAL... and not here?
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Because the voltage actually applied to the core is equal to g y pp qthe applied voltage less the internal voltage drops of the windings.
PjXPR SjXSR
PN SNCRMjX P S
IDEAL
CRMjX
IDEAL
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Simplifications of the Modelp
Recall from Slide 8:
pP Nv tTurns Ratio
N
So that
S Sv t N
pP NV aV N
Also recall from Slide 22:
S SV N
2
22p
in L LS
NZ Z a Z
N
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Simplifications of the Model
jXR jXRPjXPR SjXSR
Original Model
PN SNCRMjXPI SISVPV
PjXPR 2Sja X2
Sa R
RjX 1 I VIV
CRMjX
SIa SaV
PI
PV
EEL 3211© 2008, H. Zmuda
3. Transformers 60
Model referred to its primary side
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Simplifications of the Model
PjXPR SjXSROriginal Model
N NRjXI I VV
g
PN SNCRMjXPI SISV
PV
2PXj
a2PR
a SjXSR
CRMXj I VaIPV
2a2ja SI
SV
PaIP
a
EEL 3211© 2008, H. Zmuda
3. Transformers 61
Model referred to its secondary side
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Approximations to the Model
Series Impedance
PjXPR SjXSR
PN SNRjXPI SI VV
PN SNCRMjXPI SI
SV
PV
Shunt (Excitation)Impedance
For large transformers (several kilovoltamperes or more), the shunt impedance is much greater than the series impedance, and some approximations can be made. For example...
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3. Transformers 62
pp p
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Approximations to the Model
PjXPR 2Sja X2
Sa RPjP SjS
1
CRMjX 1SI
a
SaV
PIPV
Model referred to its primary side
PjXPR 2Sja X2
Sa RPI PjP SjaSa
RjX 1
P
CRMjX 1
SIa SaV
PV
Convince yourself of this
EEL 3211© 2008, H. Zmuda
3. Transformers 63Approximation
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Further Approximations to the Model2
P SX ja X2P SR a R P SjP S
CRMjX Neglect this entirely
2P SX ja X2
P SR a R P SjP S
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Determination of Values of Components in the ModelComponents in the Model
(EEL 4201, Lab 6)
Only two tests required:Open Circuit TestShort Circuit TestShort Circuit Test
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Open Circuit Test
TransformerAmmeter W i t
Measures average power
A
TransformerUnder Test
Ammeter Wattmeter Pi t
+ Pv tV v tOCV OCI OCP
Voltmeter
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Open Circuit Test – Since the magnetizing impedance is generally much greater that the series primary impedance, the series g p y p ,impedance can be neglected.
PjXPR This has no effect
SjXSRRjX
neglect this
CRMjXthis
1 1
IDEAL
1 11 1 1 1
E P PZ R jX
jX R jX REEL 3211© 2008, H. Zmuda
3. Transformers 67
M C M CjX R jX R
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Open Circuit Test
1 1 1 1,1 1E EE C M
Z Y jZ R X
E C M
C M
OC OC
R jXI P
, cosOC OCE
OC OC OC
I PY PFV V I
1cosOC OCE
I IY PF The power factor is always lagging for a cosE
OC OC
Y PFV V
always lagging for a real transformer
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1 1 1
Open Circuit Test
1 1 1E
E C M
Y jZ R X
2 21 1
EYR X
1
C MR X
1 1tan tan1CM
M
RXX
M
CR
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1 1cos tan C
M
RPFX
1tan cos
M
C M
X
R X PF
2 2
1 1EY
R X
2 21 1
EC MR X
1
1 1tan cos MM XX PF
2 1
1 1 1tX PF
EEL 3211© 2008, H. Zmuda
3. Transformers 70
2 1tan cosMX PF
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1 1 1Y 2 1
2 1
1tan cosE
M
YX PF
2 1
2 1
1 tan cos1tan cosM
PFX PF
1sec cos1
M
PF
1
1
tan cos
cos cos1 1
MX PF
PF
1 1
cos cos1 1cos cos sin cosM
PFX PF PF
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1 1Y 1sin cosE
M
YX PF
Z
1sin cosE
M
ZX
PF
1tan cosC MR X PF
Z Z
11 1
tan cossin cos cos cos
E EZ ZPF
PF PF
EZPF
EEL 3211© 2008, H. Zmuda
3. Transformers 72
PF
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1,E E
M C
Z ZX R
1,
sin cosM C PFPF
CRMjX1
1 1EZ
M CjX R
PF RC |XC|
0 ( ) Z0 (open) ZE
1 ZE (open)
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3. Transformers 73
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Short Circuit Test
A
TransformerUnder Test
Pi tPave
+ Pv t
V v t SCV SCI SCP Si t
Begin with a small voltage for and increase it untilis at its rated value.
v t Si t
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Short Circuit Test
jXR jXRPjXPR SjXSR
PN SNCRMjXPI SISVPV
PjXPR 2Sja X2
Sa RPjP SjS
CRMjXPIPV
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Short Circuit Test
PjXPR 2Sja X2
Sa R
Neglect thisCRMjXPIPV
Neglect thisCMjPPV
Since the shunt magnetizing impedance is generally large d ith th i i d R + jX th h tcompared with the series impedance RP + jXP, the shunt
impedance can be neglected..
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Short Circuit Test
jXR 2j X2RPjXPR 2Sja X2
Sa R
PIPV
2
P SjX ja X2P SR a R P SjX ja XP SR a R
2 2in P S P SZ R a R j X a X
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Transformer Voltage Regulation and Efficiency
Since any real transformer has an internal impedance (resistance), the output voltage varies with the load even if the input voltage remains constantremains constant.
To quantify this effect, it is customary to define the voltage l i (VR) f h fregulation (VR) of the transformer as:
V V, ,
,
100%Secondary no load Secondary full load
Secondary full load
V VVR
V
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Transformer Voltage Regulation and Efficiency
, , 100%Secondary no load Secondary full loadV VVR
V
,Secondary full load
Primary
V
VV ,
ySecondary no load
Primary
Va
V
,100%
PrimarySecondary full load
S d f ll l d
VV
aVRV
,Secondary full loadV
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Transformer Voltage Regulation and Efficiency
100%outPP
100%
in
out
PP
P P
Losses: Copper (I2R)
out lossP P
pp ( )Hysteresis (RC)Eddy Current
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Transformer Voltage Regulation and Efficiency
100%outPP P
cos 100%
out loss
S S
P PV I
100%cosS S Copper Hysteresis EddyV I P P P
core losses
cos 100%cos
S S
core lossesV I
V I P P
cosS S Copper CoreV I P P
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Transformer Phasor Diagrams – A useful visualization tool
Recall our earlier model:
PXjPR2Pj
a2P
a SjXSR
2CR
a2MXj
a SI SVPaIPV
aa S S
a
Model referred to its secondary side
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Transformer Phasor Diagrams – A useful visualization tool
Recall our earlier model:
PXjPR2Pj
a2P
a SjXSR
2CR
a2MXj
a SI SVPaIPV
aa S S
a
The excitation branch will have little effect on voltage regulation.
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Transformer Phasor Diagrams – A useful visualization tool
Recall our earlier model:
PR PXjX jX j2 P
eq SRR Ra 2 P
eq SjX jX jaPaI
2CR
a2MXj
a SI SVPV
a
aa S S
a
Now this has no effect – neglect it
PS S S
V I R jI X V
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S eq S eq Sj Va
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Transformer Phasor Diagrams – A useful visualization toolfor this equation:q
PS eq S eq S
V I R jI X Va
Case 1: Lagging Power Factor
a
SV
SI
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Transformer Phasor Diagrams – A useful visualization toolfor this equation:q
PS eq S eq S
V I R jI X Va
Case 1: Lagging Power Factor
a
SV
S eqI R
SI
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Transformer Phasor Diagrams – A useful visualization toolfor this equation:q
PS eq S eq S
V I R jI X Va
Case 1: Lagging Power Factor
a
SVS
S eqI RS eqjI X
SI
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Transformer Phasor Diagrams – A useful visualization toolfor this equation:q
PS eq S eq S
V I R jI X Va
Case 1: Lagging Power Factor
aPV
a
SV
a
S
S eqI RS eqjI X
SI
0P
SP
V VV aV VR
For lagging loads:
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0SS
V VRa V gg g
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Transformer Phasor Diagrams – A useful visualization toolfor this equation:q
PS eq S eq S
V I R jI X Va
Case 2: Unity Power Factor
a
SVSSI
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Transformer Phasor Diagrams – A useful visualization toolfor this equation:q
PS eq S eq S
V I R jI X Va
Case 2: Unity Power Factor
aPV
a
SV S eqjI Xa
S
SI S eqI R
0P
SP
S
V VV aV VR
For unity PF:Note that VR is smaller here than i h i
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SSa V
yin the previous case.
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Transformer Phasor Diagrams – A useful visualization toolfor this equation:q
Case 3: Leading Power FactorPV
S eqjI X
P
a
SVSI
q
S
S eqI R
V
0P
SP
S
V VV aV VRa V
For leading loads: Note that VR is
could be negative.
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Sa V g
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Example
A 50 kVA 2400 240 V 60 H di ib i f hVP : VS
A 50 kVA, 2400:240 V, 60 Hz distribution transformer has a leakage impedance of 0.72 + j 0.92 Ohms in the high-voltage winding and 0.0070 + j 0.0090 Ohms in the low-voltage winding. g j g gAt the rated frequency, the impedance of the shunt exciting branch (the impedance of magnetization branch RC and jXM in parallel) is 6 32 + j 43 7 Ohms when viewed from the low-voltage side Draw6.32 + j 43.7 Ohms when viewed from the low-voltage side. Draw the equivalent circuit referred to
) th hi h lt id da) the high-voltage side and
b) the low-voltage side.
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PjXPR SjXSR10 :1Solution
PN SNCRMjX0.72 0.92j 0.007 0.009j
P SCRMjX6.32 43.7j
PjXPR 2Sja X2
Sa R
Solution – Part (a)
PjXPR Sja XSa R
0.72 0.92j 0 7 0 9jCRMjX
0.72 0.92j 0.7 0.9j
632 4370j
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PjXPR SjXSR10 :1Solution
PN SNCRMjX0.72 0.92j 0.007 0.009j
632 4370j
CMj
2PXj
a2PR
a
632 4370j
SjXSR
Solution – Part (b)
aa
0.0072 0.0092j 0 007 0 009j
SjS
CRMjX0.0072 0.0092j 0.007 0.009j
6.32 43.7j
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Example – Continued Solution
If 2400 V rms is applied to the high voltage side of the transformer, calculate the current in the magnetizing impedance in parts (a) and (b) respectivelyparts (a) and (b), respectively.
Solution (a):632 0 72 4370 0 92Z 632 0.72 4370 0.92632.72 4370.92
TZ j jj
4416.5 81.76
2400 0.543 81.764416 5 81 76
MI amps rms
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4416.5 81.76
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Example – Continued Solution
If 2400 V rms is applied to the high voltage side of the transformer, calculate the current in the magnetizing impedance in parts (a) and (b) respectivelyparts (a) and (b), respectively.
Solution (b):
6.3272 43.7092 TZ j44.171 81.76
240
240 5.43 81.7644.171 81.76
MI amps rms
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Example – Continued
Consider the equivalent “T” circuit found with the impedances referred to the high-voltage side (below).
(a)Draw the “cantilever” equivalent circuit with the shunt branch(a)Draw the cantilever equivalent circuit with the shunt branch at the high-voltage terminal.
(b) With the low-voltage terminal an open-circuited and 2400 V li d h hi h l i l l l h l happlied to the high-voltage terminal, calculate the voltage at the
low-voltage terminal as predicted by each equivalent circuit.
jXR 2j X2RPjXPR 2Sja X2
Sa R
0 72 0 92j 0 7 0 9jCRMjX
0.72 0.92j 0.7 0.9j
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j
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Example – Continued Solution
(a)Draw the “cantilever” equivalent circuit with the shunt branch at the high-voltage terminal.
2P SjX ja X2
P SR a R
1.42 1.82j CRMjX
1.42 1.82j
632 4370j
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Example – Continued Solution
(b) With the low-voltage terminal an open-circuit and 2400 V applied to the high-voltage terminal, calculate the voltage at the low-voltage terminal as predicted by each equivalent circuitlow-voltage terminal as predicted by each equivalent circuit.
PjXPR 2Sja X2
Sa R
RjX0.72 0.92j 0.7 0.9j
2400
V
CRMjX632 4370j
2400
2V
Open
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Example – Continued (b) For the “T”:
Z
Solution
2 2400M
M P
ZVZ Z
632 4370 2400632 4370 0.72 0.92
jj j
j j
632 j 4370632 72 j 4370 92
2400 2.399 103 0.316i632.72 j 4370.92
632 j 4370632 72 j 4370 92
2400 2.3994 103632.72 j 4370.92
180
arg632 j 4370
632 72 j 4370 922400
7.536 10 3
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Example – Continued (b) For the “T”: Solution
Reflected back to the secondary side gives,
2 239 94V
V V
2 239.9410SV V
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Example – Continued (b) For the “Cantilever”: Solution
2P SjX ja X2
P SR a R
CRMjX1.42 1.82j
2400
V
CMj
632 4370j
2V
O632 4370j Open
2 2400 240SV V V 2 2400 240
240 239.94 100% 0.025%240
SV V V
error
Well within
engineering accuracy
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240 y
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Example – Continued
Suppose that the 2400 V feeder has an internal impedance of 0.30 + j1.60 Ohms. Find the voltage at the secondary terminals of the transformer when the load connected to the secondary draws ratedtransformer when the load connected to the secondary draws rated current from the transformer and the power factor of the load is 0.80 lagging. Neglect the voltage drops in the transformer and the f d d b h i ifeeder caused by the exciting current.
LISolution
1.42 1.82j V
2400
0.30 1.60j
Load
CRMjX
632 4370j
2V
2400 Neglect
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Example – ContinuedLI
Solution
1.42 1.82j 2400
0.30 1.60j
L d
L
2V
2400
Load
This is the high-voltage end:
50 kVA50-kVA 20.83 A2400LI
1cos 0.8 36.87lagging
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Example – Continued
20 83 36 87 AI Solution
20.83 36.87 ALI V
1.72 3.42j 2V
2400 Load
1 2 3 42 20 83 36 8 2
1.72 3.42 20.83 36.87
2400 2400 1.72 3.42 20.83 36.87LV ZI j V
V V j
2 j
2400 1 72 j 3 42( ) 20 8 e
j 36.87
180
2 329 103
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2400 1.72 j 3.42( ) 20.8 e 2.329 10
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Example – Continued Solution
At the load2 2329V
At the load,
V2 232.910L
VV
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Example – Continued
With instrumentation on the high-voltage side and the output short-circuited, the short-circuit test readings for this example are 48 V, 20 8 A and 617 W An open circuit test with the low-voltage side20.8 A, and 617 W. An open circuit test with the low-voltage side energized gives instrument readings on that side of 240 V, 5.41 A, and 186 W. Determine the efficiency and the voltage regulation at f ll l d 0 80 f l ifull load, 0.80 power factor lagging.
Transformer Pi tPave
AUnder Test
P
V I P + Pv t
V v t SCV SCI SCP Si t
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Example – Continued Solution
From Slide 77 2P SjX ja X2
P SR a R
2 2i P S P SZ R a R j X a X in P S P SZ R a R j X a X
48 2.3120.8in
VZ
I
22
617 1.4220 8eq eqP I R R
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20.8
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Example – Continued Solution
2 22 2 2in P S P SZ R a R X a X
2 222 2P S in P SX a X Z R a R
2 2
2 22 31 1 42
eq in eqX Z R 2 22.31 1.42
1.82
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Example – Continued Solution
Operation at full load, 0.80 power factor, lagging, corresponds to a primary current of
50,000-kVA 20.8 A2400P
P
VIIV
and an output power ofP
cos 50,000 VA 0.8 40,000 avP VI W
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Example – Continued Solution
The loss in the windings is2 220 8 1 42 617 WP I R
The core loss is determined from the open-circuit test,
20.8 1.42 617 Wwindings P eqP I R
The total loss is 186 W givencoreP
617 186 803 Wwinding coreP P
The power supplied is thus
40,000 803 40,803 WlP P EEL 3211© 2008, H. Zmuda
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40,000 803 40,803 Wave lossP P
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Example – Continued Solution
The efficiency is
P 100%out
in
PP
100%out
out loss
PP P
40,000 100% 98.03%40 803
out loss
40,803
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Example – Continued Voltage Regulation: Solution
Secondary no load Secondary full loadV V, ,
,
100%Secondary no load Secondary full load
Secondary full load
VRV
2P SjX ja X2
P SR a R
LI
2400full loadV
Load
0 8PF laggingEEL 3211© 2008, H. Zmuda
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0.8PF lagging
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Example – Continued Solution
Voltage Regulation:
50 000VI 36.950,0002,400
j jL
VII e eV
20.83 0.8 0.616.66 12.5
jj
j
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Example – Continued Solution
The required input voltage is
in eq L full loadV Z I V
1.42 1.82 16.66 12.5 2400in eq L full load
j j 2446 13j V
2P SjX ja X2
P SR a R LIP SjX ja XP SR a R
LI
2400full loadV
LoadinV
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Example – Continued Solution
The required input voltage is
1.42 1.82 16.66 12.5 2400in eq L full loadV Z I V
j j
1.42 1.82 16.66 12.5 24002446 13
j jj V
2446inV
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Example – Continued Solution
Under loaded conditions, the load voltage is 2400 volts, by design. If the load were removed the load voltage would jump to 2446 volts Thusvolts. Thus
, , 100%Secondary no load Secondary full loadV VVR
, ,
,
100%
2446 2400
Secondary no load Secondary full load
Secondary full load
VRV
2446 2400 100%2400
1 92%
1.92%
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The Per–Unit System – An Industry-Standard Normalization
Computations relating to machines and transformers are often carried out in per-unit form. This is where all quantities expressed as fractions of appropriately chosen base values. s c o s o pp op e y c ose base values.
All the usual computations are then carried out in these per unit values instead of the familiar volts amperes ohms etcvalues instead of the familiar volts, amperes, ohms, etc.
There are a number of advantages to the system. One is that the parameter values of machines and transformers typically fall in a reasonably narrow numerical range when expressed in a per-unit system based upon their rating. The correctness of their values is y p gthus subject to a rapid approximate check.
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The Per–Unit System
A second advantage is that when transformer equivalent-circuit parameters are converted to their per-unit values, the ideal transformer turns ratio becomes 1:1 and hence the ideal transformer s o e u s o beco es : d e ce e de s o ecan be eliminated.
This greatly simplifies analyses For complicated systems involvingThis greatly simplifies analyses. For complicated systems involving many transformers of different turns ratios, this simplification is a significant one in that a possible cause of serious mistakes is removed.
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The Per–Unit System Quantifies such as voltage V, current I, power P, reactive power Q, voltamperes VA, resistance R, reactance p , p Q, p , ,X, impedance Z, conductance G, susceptance B, and admittance Ycan be translated to and from per-unit form as follows:
actual valuePer unit valuebase value of quantity
Once Vbase and Ibase have been chosen, then
f q y
P Q VA V I, ,base base base base base
base
P Q VA V IVR X Z
Clearly only two independent base values can be chosen arbitrarily.
, ,base base basebase
R X ZI
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The Per–Unit System
Typically VAbase and Vbase are chosen, then the remaining quantities are uniquely established as per the previous slide.
The value of VAbase must be the same over the entire system of analysis.
When a transformer is encountered, the values of Vbase differ on each side and should be chosen to have the same ratio as the turns ratio of the transformer.
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The Per–Unit System
Usually the rated or nominal voltages of the respective sides are chosen as base valueschosen as base values.
If the base voltages of the primary and secondary are chosen to b i h i f h f h id l f h ibe in the ratio of the turns of the ideal transformer, the per-unit ideal transformer will have a unity turns ratio and hence can be eliminated.
If these rules are followed, the procedure for performing system analyses in per-unit can be summarized as follows:analyses in per-unit can be summarized as follows:
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The Per–Unit System We will illustrate via example, but the procedure is:p
1. Select a VA base and a base voltage at some point in the system.
2. Convert all quantities to per unit on the chosen VA base and with a voltage base that transforms as the turns ratio of any transformer which is encountered as one moves through the systemwhich is encountered as one moves through the system.
3. Perform a standard electrical analysis with all quantities in per unit.
4. When the analysis is completed, all quantities can be converted back to real units (e.g., volts, amperes, watts, etc.) by multiplying their per-unit values by their corresponding base values.
5. The procedure is best illustrated with an example.
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5. The procedure is best illustrated with an example.
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The Per–Unit System – Example
The equivalent circuit for a 100-MVA, 7.97-kV:79.7-kV transformer is shown below. Convert this equivalent circuit to per unit using the transformer rating as baseunit using the transformer rating as base.
7.97-kV : 79.7-kV30.72 10 0.040j 3.75 0.085j
114j
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The Per–Unit System – Example Solution
The base quantities for the transformer are:
Low-voltage side: VA = 100-MVA V = 7 97-kVLow-voltage side: VAbase = 100-MVA, Vbase = 7.97-kV2
0.635basebase base
VR X
High-Voltage side: VAbase = 100-MVA, Vbase = 79.7-kV
base basebaseVA
g g base base
2
63.5basebase base
VR X base basebaseVA
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The Per–Unit System – Example – Compute the per unit values:
S l ti0.04 0.0630.635PX
Solution
3.75 0.059163.5SX
114 1800.635MX
47.6 10 0.00120 635PR
0.635
0.085 .001363 5SR
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63.5
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The Per–Unit System – Example – Solution
Compute the per unit turns ratio:
7.97-kV 79.7-kV: 1:17 97 kV 79 7 kVpuTurns Ratio
There is no need to keep the transformer as far as analysis is
7.97-kV 79.7-kV
p yconcerned.
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The Per–Unit System – Example – in per unit values Solution
1:10.0012 0.063j 0.0591 0.0013j
180j
0 0012 0 063j 0 0591 0 0013j0.0012 0.063j 0.0591 0.0013j
180j
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The Per–Unit System – Example
Consider again our earlier example of a 50-kVA, 2400:240-V transformer. We found earlier that the current on the low voltage side is 5 43 amps (Slide 96) and that its equivalentvoltage side is 5.43 amps (Slide 96) and that its equivalent impedance referred to the high voltage side is 1.42 + j1.82 Ohms (Slide 98). Using the transformer rating as the base,
i i h l d hi h l id ( ) hexpress in per unit on the low- and high-voltage sides (a) the exciting current and (b) the equivalent impedance.
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The Per–Unit System – Example Solution
The base values are
2400 V 240 VV V
h i
, ,
, ,
2400 V 240 V20.8 A 208 A
base P base S
base P base S
V VI I
these give
2400 240, ,
2400 240115.2 1.15220.8 208base P base SZ V
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The Per–Unit System – Example Solution
Using the values found in Slides 95 and 96,
0 543,
0.543 0.02620.8
5 43
M PI per unit
,5.43 0.026208
M SI per unit
As expected, the per-unit values are the same referred to either side corresponding to a unity turns ratio for the ideal transformer This is a direct consequence of the choice oftransformer. This is a direct consequence of the choice of base voltages in the ratio of the transformer turns ratio and the choice of a constant volt-ampere base.
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The Per–Unit System – Example Solution
From Slide 100:
1 42 1 82 j,
1.42 1.82 0.0123 0.0158115.2
0 0142 0 0182
eq PjZ j per unit
j,
0.0142 0.0182 0.0123 0.01581.152 eq S
jZ j per unit
Again the values are the same, with the effect of the turns ratio accounted for by choice of base valuesratio accounted for by choice of base values.
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The Per–Unit System – Example
A 15-kVA 120:460-V transformer has an equivalent series impedance of 0.018 + j0.042 per unit. Calculate the equivalent series impedance in ohms (a) referred to the low-equivalent series impedance in ohms (a) referred to the low-voltage side and (b) referred to the high-voltage side.
120 0.96b LZ
Solution
, 0.9615,000 /120
460 14 1
base LZ
Z
, 14.115,000 / 460base HZ
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The Per–Unit System – Example Solution
, 0.96 0.018 0.042eq LZ j
0.0173 0.040314 1 0 018 0 042
q
jZ j
, 14.1 0.018 0.042
0.254 0.592eq HZ j
j
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Transformer Ratings
Transformers have four major ratings:Transformers have four major ratings:
apparent powerlvoltage
currentfrequencyq y
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Transformer Ratings
The voltage rating serves two purposes; the first is to insure that breakdown does not occur between the transformer windings. The second is prevent the core from becoming saturated which as wesecond is prevent the core from becoming saturated which, as we have seen (see results of our earlier example below), can result in very large magnetization currents.
22
200129.984
0
2
2.0 sin t( ) 0Ni 2.0 sin t( )( )
10 0 102
1010 Ni ( ) t10 0 10
200129.984
1010 t
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1010 t
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Transformer Ratings
If a steady-state voltage v(t) is applied to the transformer’s primary winding,
hi i h fl
max sinv t V t
this gives the flux as
1P
d tv t N t v t dt
max
PP
v t N t v t dtdt N
Vt t
max cosP
t tN
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Transformer Ratings
From max cosVt t
N
we see that there are two ways to drive the core into saturation. O i i h l ( l ) A h i d
PN
One is to increase the voltage (our example). Another is to reduce the operating frequency.
Accordingly, a (European) transformer rated at 50 Hz may be operated at a 20% high voltage at 60 Hz assuming that this does not cause insulation breakdownnot cause insulation breakdown.
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Transformer Ratings
The purpose of the apparent power rating it that, together with the voltage rating, it sets the current flow through the windings. The current flow is important because it controls the i2R lossesThe current flow is important because it controls the i R losses which in turn controls the heating of the coils. Heating significantly reduces insulation lifetime.
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The Problem of Current Inrush – a transient effect
Suppose that the moment the transformer is connected to the power line the voltage across the primary winding is,
The maximum flux value reached during the first half-cycle of the
li d l d d h l f F 90
max sinv t V t
applied voltage depends on the value of . For = 90o,
max maxsin 90 cosv t V t V t and
max maxsin maxV Vt t
N N
or just the maximum value at steady state.
maxP PN N
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The Problem of Current Inrush – a transient effect
Now suppose that = 0o, then
and the maximum flux during the first half-cycle is
max sinv t V t
and the maximum flux during the first half-cycle is1
2 21T
f
max0
1 sinP
t V tdtN
max0
cosP
Vt tN
maxmax
2
P
VN
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PN
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The Problem of Current Inrush – a transient effect
Now the maximum flux is twice its steady-state value. Recall again our earlier example showing what happens if, near saturation the flux is further increased by even a small amountsaturation, the flux is further increased by even a small amount.
A doubling of flux could result (as a transient effect) is a hugei h f h h f i fi l d i I finrush of current when the transformer is first plugged in. In fact, for this fist part of a cycle the primary really looks like a short.
The primary winding must be designed to handle this effect.
There is lots of engineering involved in designing magneticThere is lots of engineering involved in designing magnetic circuits! This holds even truer for machines – our next topic.
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Autotransformers (Text Section 3.9) and Three Phase Transformers (Text Section 3.10) are left as a reading exercise.( ) g
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