Transcript Document

ENERGY CONVERSION ONE
(Course 25741)
Chapter Two
TRANSFORMERS
…continued
Three Phase Transformers
• Almost all major generation & Distribution Systems in the world
are three phase ac systems
• Three phase transformers play an important role in these
systems
• Transformer for 3 phase cct.s is either:
(a) constructed from 3 single phase transformers, or
(b) another approach is to employ a common core for
the three sets of windings of the three phases
• The construction of a single three phase transformer is the
preferred today, it is lighter, smaller, cheaper and slightly more
efficient
• There is an advantage that each unit in the bank could be
replaced individually in the event of a fault, however this does
not outweigh the other advantages of combined 3 ph. unit
Three Phase Transformers
• How the core of compact three phase is built
• φa+φb+φc=0 and central leg can be removed
Three Phase Transformers
• The two constructions
Three Phase Transformers
• 3 phase transformer connections
• The windings of primary and secondary (in any
construction) can be connected in either a wye (Y) or
delta (Δ)
• This provides a total of 4 possible connections for 3
phase transformer (if Neutral is not grounded):
(a) Wye-wye Y-Y
(b) Wye-delta Y-Δ
(c) Delta-wye Δ-Y
(d) Delta-Delta Δ-Δ
Three Phase Transformers
• To analyze a 3-phase transformer, each single
transformer in the bank should be analyzed
• Any single phase in bank behaves exactly like 1 phase
transformer just studied
• impedance, V.R., efficiency, & similar calculations for 3
ph. are done on per phase basis, using the same
technique already used in single phase Transformer
• The applications, advantages and disadvantages of
each type of three phase connections will be
discussed next
Three Phase Transformers
• WYE-WYE connection
• In Y-Y connection, primary voltage on each phase is
VφP=VLP/√3
• Primary phase voltage is related to secondary phase voltage by
turns ratio of transformer
• Phase voltage of secondary is related to Line voltage of
secondary by VLS=√3 VφS
• Overall the voltage ratio of transformer is:
•
VLP

VLS
3VP
3VS
a
Y-Y
Three Phase Transformers
• Two serious concerns on Y-Y connection
1- if loads on transformer cct. are unbalanced,
voltages on phases of transformer severely
unbalanced, also source is loaded in an
unbalanced form
2- Third harmonic voltages can be large (there
is no path for passage of third harmonic
current)
• Both concerns on unbalance load condition & large 3rd
Harmonic voltages can be rectified as follows:
Three Phase Transformers
• Solidly grounding the neutrals of windings
specially primary winding, this connection provide a path for 3rd
harmonic current flow, produced and do not let build up of large
3rd voltages . Also provides a return path for any current
imbalances in load
• Adding a third winding (tertiary) connected in Δ
(a) 3rd harmonic components of voltage in Δ will add up, causing
a circulating current flow within winding
(b) tertiary winding should be large enough to handle circulating
currents (normally 1/3 of power rating of two main windings)
One of these corrective techniques should be employed with YY, however normally very few transformer with this type of
connection is employed (others can do the same job)
Three Phase Transformers
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WYE-DELTA CONNECTION
VLP=√3 VφP, while :
VLS= VφS
Voltage ratio of each phase : VφP/ VφS=a
VLP/ VLS= √3 VφP/ VφS= √3 a  Y-Δ
Y-Δ doesn’t have shortcomings of Y-Y regarding
generation of third harmonic voltage since the Δ
provide a circulating path for 3rd Harmonic
• Y-Δ is more stable w.r.t. unbalanced loads, since Δ
partially redistributes any imbalance that occurs
• This configuration causes secondary voltage to be
shifted 30◦ relative to primary voltage
• If secondary of this transformer should be paralleled
with secondary of another transformer without phase
shift, there would be a problem
Three Phase Transformers
• WYE-DELTA CONNECTION
Three Phase Transformers
Y-Δ Connection
• The phase angles of secondaries must be
equal if they are to be paralleled, it means that
direction of phase shifts also should be the
same
• In figure shown here, secondary lags primary if
abc phase sequence applied,
• However secondary leads primary when acb
phase sequence applied
Three Phase Transformers
Δ-Y Connection
• DELTA-WYE CONNECTION
• In Δ-Y primary line voltage is equal to primary
phase voltage VLP=VφP, in secondary VLS=√3VφS
• Line to line voltage ratio ;
• VLP/ VLS = VφP/ [√3 VφS ]=a/√3
 Δ-Y
• This connection has the same advantages &
phase shifts as Y- Δ
• And Secondary voltage lags primary voltage by
30◦ with abc phase sequence
Three Phase Transformers
Δ- Δ Connection
• DELTA-DELTA CONNECTION
• In Δ-Δ connection VLP= VφP and VLS= VφS
• Voltage ratio : VLP/VLS= VφP / VφS =a  Δ-Δ
• This configuration has no phase shift and there
is no concern about unbalanced loads or
harmonics
Three Phase Transformers
Δ- Δ Connection
THREE PHASE TRANSFORMERS
PER UNIT
• In 3 phase, similarly a base is selected
• If Sbase is for a three phase system, the per phase basis is :
S1φ,base= Sbase/3
• base phase current, and impedance are:
• Iφ,base= S1φ,base/ Vφ,base= Sbase /(3Vφ,base)
• Zbase=(Vφ,base)²/ S1φ,base
• Zbase=3(Vφ,base)²/ Sbase
• Relation between line base voltage, and phase base voltage
depends on connection of windings, if connected in Δ ;
VL,base=Vφ,base
and
• if connected in Wye: VL,base= √3Vφ,base
• Base line current in 3 phase transformer:
• IL,base= Sbase/ (√3 VL,base)
THREE PHASE TRANSFORMERS
PER UNIT
• A 50 kVA 13800/208 V Δ-Y distribution transformer
has a resistance of 1 percent & a reactance of 7
percent per unit
(a) what is transformer’s phase impedance referred to
H.V. side?
(b) Calculate this transformer’s voltage regulation at
full load and 0.8 PF lagging using the calculated highside impedance
(c) Calculate this transformer’s voltage regulation
under the same conditions, using the per-unit system
THREE PHASE TRANSFORMERS
PER UNIT
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SOLUTION
(a) Base of High voltage=13800 V, Sbase=50 kVA
Zbase=3(Vφ,base)²/Sbase=3(13800)²/50000=11426Ω
The per unit impedance of transformer is:
Zeq=0.01+j 0.07 pu
Zeq=Zeq,pu Zbase =(0.01+j0.07 pu)(11426)=
114.2 + j 800 Ω
• (b) to determine V.R. of 3 phase Transformer bank, V.R. of any
single transformer can be determined
• V.R. =(VφP-a VφS)/ (aVφS) x 100%
• Rated phase voltage on primary 13800 V, rated phase current
on primary: Iφ=S/(3 Vφ) =50000/(3x13800)=1.208 A
THREE PHASE TRANSFORMERS
PER UNIT
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Example …
Rated secondary phase voltage: 208 V/√3=120V
Referred to H.V.  V’φS=a VφS13800 V
At rated voltage & current of secondary:
VφP=a VφS+Req Iφ + j Xeq Iφ =
13800/_0◦ +(114.2)(1.208/_-36.87)+(j800)(1.208)/_-36.87)=
13800+138/_-36.87+966.4/_53.13= 13800+110.4j82.8+579.8+j773.1= 14490+j690.3= 14506/_2.73◦ V
V.R. = (VφP-a VφS )/ (a VφS ) x 100%=
(14506-13800)/13800 x 100% = 5.1%
THREE PHASE TRANSFORMERS
PER UNIT
• Example …
• (c) V.R. using per unit system
• output voltage 1/_0◦ & current 1/_-36.87◦ pu
VP=1/_0◦ +(0.01) (1/_-36.87◦)+(j0.07)(1/_36.87◦)=1+0.008-j0.006+0.042+ j0.056
=1.05+j0.05=1.051/_2.73◦
• V.R.= (1.051-1.0)/1.0 x100% = 5.1%
Apparent Power Rating of a
Transformer
• Apparent power rating & Voltage rating set current flow of
windings
• Current flow important as it controls I²R losses
in turn control heating of coils
Heating is critical, since overheating the coils reduce insulation
life
• Actual VA rating of a transformer may be more than a single
value:
In real Transformer: (a) may be a VA rating for transformer by
itself, (b) another (higher) rating for transformer with forced
cooling
• If a transformer’s voltage reduced for any reason (i.e. operating
with lower frequency than normal) then transformer VA rating
must reduced by an equal amount, otherwise current exceed
permissible level & cause overheating
Inrush Current
• This is caused by applied voltage level at energization
of transformer, or due to residual flux in the
transformer core
• Suppose that voltage is : v(t)=VM sin(ωt+θ) V
• The maximum flux reached during first half-cycle of
applied voltage depends on θ
• If θ=90◦ or :
v(t)=VM cos(ωt)
& no residual flux in core  max. flux would be :
φmax=Vmax/(ωNP)
• However if θ=0 the max. flux would be
φmax=2Vmax/(ωNP) and is twice the steady-state flux
Inrush Current
• With this high maximum
flux if the magnetization
curve examined it shows
passage of enormous
magnetizing current,
(looks like short circuit in
part of cycle)
• In these cases that θ is
not 90◦ this excess current
exist, therefore power
system & transformer
must be able to withstand
these currents
Transformer Nameplate
• Example: