ELEN 30093
Module 7
Transformer Connections and PU Computations
Objectives
After successful completion of this module, you should be able to:
u
Determine the different connections of a three phase transformer.
u
Differentiate balanced and unbalanced load.
u
Determine the importance of Per Unit (PU) computations.
Generation of Three-phase Supply
A1
Stator
B2
C2
S
N
B1
C1
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A2
vA = Vm sin ωt = ⇒ VA = |Vph|∠0◦
vB = Vm sin(ωt − 120◦) = ⇒ VB = |Vph|∠ − 120◦
vC = Vm sin(ωt − 240◦) = ⇒ VB = |Vph|∠ − 240◦
Meaning of Phase Sequence
ABC Sequence
ACB Sequence
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Phase sequence can be changed by reversing the direction of
rotation of rotor of the alternator.
Meaning of Balanced and
Unbalanced Supply
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(a)
(b)
(c)
(d)
If |VA| = |VB| = |VC| = |Vph| and if the phase
difference between VA and VB , VB and VC , VC and VA
is equal to 120◦ as shown in (a) then, the supply is
said to be balanced or symmetrical.
Phasor diagrams (b), (c), and (d) represent unbalanced
supply.
Meaning of Balanced/Unbalanced Load and
System
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If the three impedances, which may be Y or ∆ connected
are equal, then, the three-phase load is said to be
balanced.
If load and supply are both balanced, then three-phase
system is said to be balanced.
Under normal working conditions, a three-phase system
can be taken to be balanced.
Relation between Line and Phase Voltages
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ABC Sequence
If supply is balanced, the n,line voltage magnitudes will be
|VAB| = |VBC| = |VCA| = s q r t 3|Vph| = |V|.
When phase sequence is ABC, VAB leads VAN by 30◦, VBC
leads VBN by 30◦, and VCA leads VCN by 30◦.
Relation between Line and Phase Voltages
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If supply is balanced, the line voltage magnitudes will be
|VAB| = |VBC| = |VCA| = s q r t 3|Vph| = |V|.
When phase sequence is ACB, VAB lags VAN by 30◦, VBC
lags VBN by 30◦, and VCA lags VCN by 30◦.
Line and Phase Currents in Three-phase Circuits
Line
Current
Star Point
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Current flowing through
each impedance is
called phase current
Current flowing through
each suppy line is
called line current
Load can be
Y or delta
connected
as shown
above
Relation between Line and Phase Currents
• In Y-connected load, the line current is equal to the phase current.
• From the phasor diagram given in the previous slide, it is clear that,
𝐼! = 2 𝐼!" cos 30 = 3 𝐼!" = 3
#
$
= 𝐼" = 𝐼% = 𝐼
i.e in balanced delta connected three phase load the line current is,
3 𝐼&' = 3
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#
$
Zero Neutral Shift Voltage in Balanced System
A
Neutral Shift
Voltage
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C
B
Balanced Supply
Balanced Load
It can be shown that in a balanced system the neutral shift
voltage is zero so that VAN = VAN' , VBN = VBN' , and
VCN = VCN' .
Power in Balanced System
A
Balanced
B
Three-phase
Supply
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Total power consumed by the load is
P = 3P() = 3 𝑉&' 𝐼&' cos ∅ = 3 3 𝑉&' 𝐼&' cos ∅
But 3 𝑉&' = 𝑉 and 𝐼&' = 𝐼 in a Y-connected load and delta connected can
always be replaced by equivalent Y. So,
P = 3 𝑉 𝐼 cos ∅
Three Phase Transformers
WYE-DELTA CONNECTION
u VLP=√3 VφP, while :
VLS= VφS
u Voltage ratio of each phase : VφP/ VφS=a
u VLP/ VLS= √3 VφP/ VφS= √3 a
ç Y-Δ
u 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
u Y-Δ is more stable with regards to unbalanced loads, since Δ partially
redistributes any imbalance that occurs
u This configuration causes secondary voltage to be shifted 30◦ relative
to primary voltage
Three Phase Transformers
u
WYE-DELTA CONNECTION
Three Phase Transformers
Δ-Y Connection
DELTA-WYE CONNECTION
u
In Δ-Y primary line voltage is equal to primary phase voltage VLP=VφP, in secondary VLS=√3VφS
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Line to line voltage ratio ;
VLP/ VLS = VφP/ [√3 VφS ]=a/√3
ç Δ-Y
u
This connection has the same advantages & phase shifts as Y- Δ
u
And Secondary voltage lags primary voltage by 30◦ with abc phase sequence
Three Phase Transformers
Δ- Δ Connection
DELTA-DELTA CONNECTION
u
In Δ-Δ connection VLP= VφP and VLS= VφS
u
Voltage ratio :
u
This configuration has no phase shift and there is no concern about unbalanced loads or
harmonics.
VLP/VLS= VφP / VφS =a ç Δ-Δ
Three Phase Transformers
Δ- Δ Connection
PER UNIT SYSTEM
For the analysis of electrical machines or electrical machine system, different values
are required, thus, per unit system provides the value for voltage, current, power,
impendence, and admittance.
The per-unit value of any quantity is defined as the ratio of actual value in any unit to
the base or reference value in the same unit.
𝐴𝑐𝑡𝑢𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 𝑖𝑛 𝑎𝑛𝑦 𝑢𝑛𝑖𝑡
𝑃𝑒𝑟 𝑈𝑛𝑖𝑡 𝑉𝑎𝑙𝑢𝑒 =
𝐵𝑎𝑠𝑒 𝑜𝑟 𝑟𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑣𝑎𝑙𝑢𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑢𝑛𝑖𝑡
Advantages of Per Unit System
1. The parameters of the rotating electrical machines and the transformer lie roughly
in the same range of numerical values, irrespective of their ratings if expressed in a
per-unit system of ratings
2. It relieves the analyst of the need to refer circuit quantities to one or the other side
of the transformer, making the calculations easy.
THREE PHASE TRANSFORMERS
PER UNIT
In 3 phase, similarly a base is selected
u If Sbase is for a three phase system, the per phase basis is :
S1φ,base= Sbase/3
u 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
u Relation between line base voltage, and phase base voltage
depends on connection of windings, if connected in Δ ;
VL,base=Vφ,base
and
u 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 and
to the L.V side?
(b) Calculate this transformer’s voltage regulation at full load and
0.8 PF lagging using the calculated high-side impedance
(c) Calculate this transformer’s voltage regulation under the same
conditions, using the per-unit system
THREE PHASE TRANSFORMERS
PER UNIT
SOLUTION
u
(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
u
Base of Low voltage=208 V, Sbase=50 kVA
Zbase=3(208)²/50000=2.59584Ω
Zeq=Zeq,pu Zbase =(0.01+j0.07 pu)(2.59584)=0.0259584 + j 0.1817088 Ω
THREE PHASE TRANSFORMERS
PER UNIT
(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
u
THREE PHASE TRANSFORMERS
PER UNIT
(c)
Rated secondary phase voltage: 208 V/√3=120V
Referred to H.V. è V’φS=a VφS=13800 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
(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+0.008j0.006+0.042+j0.056 =1.05+j0.05=1.051/_2.73◦
V.R.= (1.051-1.0)/1.0 x100% = 5.1%
Objectives
After successful completion of this module, you should be able to:
u
Determine the different connections of a three phase transformer.
u
Differentiate balanced and unbalanced load.
u
Determine the importance of Per Unit (PU) computations.