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POWER SYSTEMS ILecture 3
06-88-590-68
Electrical and Computer Engineering
University of Windsor
Dr. Ali Tahmasebi
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Per Unit Calculations
l A key problem in analyzing power systems is the
large number of transformers.
– It would be very difficult to continually have to refer
impedances to the different sides of the transformers
l This problem is avoided by a normalization of all
variables.
l This normalization is known as per unit analysis.
actual quantityquantity in per unit
base value of quantity=
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Per Unit Conversion Procedure, 1
1. Pick a 1f VA base for the entire system, SB
2. Pick a voltage base for each different voltage level,
VB. Voltage bases are related by transformer turns
ratios. Voltages are line to neutral.3. Calculate the impedance base, ZB= (VB)
2/SB
4. Calculate the current base, IB = VB/ZB
5. Convert actual values to per unit
Note, per unit conversion on affects magnitudes, not
the angles. Also, per unit quantities no longer have
units (i.e., a voltage is 1.0 p.u., not 1 p.u. volts)
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Per Unit Solution Procedure
1. Convert to per unit (p.u.) (many problems are
already in per unit)
2. Solve
3. Convert back to actual as necessary
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Per Unit Example
Solve for the current, load voltage and load power
in the circuit shown below using per unit analysis
with an SB of 100 MVA, and voltage bases of
8 kV, 80 kV and 16 kV.
Original Circuit
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Per Unit Example, cont’d
2
2
2
80.64
100
8064
100
162.56
100
Left B
Middle
B
Right B
kV Z
MVA
kV Z
MVA
kV Z
VA
= = W
= = W
= = W
Same circuit, with
values expressed
in per unit.
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Per Unit Example, cont’d
L
2*
1.0 0 0.22 30.8 p.u. (not amps)3.91 2.327
V 1.0 0 0.22 30.8
p.u.
0.189 p.u.
1.0 0 0.22 30.8 30.8 p.u.
L L L L
G
I j
V S V I
Z
S
Ð °= = Ð - °+
= Ð °- Ð - °́ 2.327Ð 90°
= 0.859Ð - 30.8°
= = =
= Ð °́ Ð °=0.22Ð °
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Per Unit Example, cont’d
To convert back to actual values just multiply the
per unit values by their per unit base
LActual
ActualL
ActualG
MiddleB
ActualMiddle
0.859 30.8 16 kV 13.7 30.8 kV
0.189 0 100 MVA 18.9 0 MVA
0.22 30.8 100 MVA 22.0 30.8 MVA
100 MVAI 1250 Amps
80 kV
I 0.22 30.8 Amps 275 30.8
V
S
S
= Ð - °́ = Ð - °
= Ð °́ = Ð °
= Ð °́ = Ð °
= =
= Ð - °́ 1250 = Ð - °A
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Per Unit Equivalent of Transformers
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Three Phase Per Unit
1. Pick a 3f VA base for the entire system,
2. Pick a voltage base for each different voltage
level, VB. Voltages are line to line.
3. Calculate the impedance base
Procedure is very similar to 1f except we use a 3f
VA base, and use line to line voltage bases3
BS f
2 2 2, , ,3 1 1
( 3 )3
B LL B LN B LN B
B B B
V V V Z S S S
f f f = = =
Exactly the same impedance bases as with single phase!
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Three Phase Per Unit, cont'd
4. Calculate the current base, IB
5. Convert actual values to per unit
3 1 13 1B B
, , ,
3I I
3 3 3
B B B
B LL B LN B LN
S S S
V V V
f f f f f
= = = =
Exactly the same current bases as with single phase!
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Three Phase Per Unit Example
Solve for the current, load voltage and load power
in the previous circuit, assuming a 3f power base of
300 MVA, and line to line voltage bases of 13.8 kV,
138 kV and 27.6 kV (square root of 3 larger than the1f example voltages). Also assume the generator is
Y-connected so its line to line voltage is 13.8 kV.
Convert to per unit
as before. Note the
system (per phase) is
exactly the
same!
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3 Per Unit Example, cont'd
L
2*
1.0 00.22 30.8 p.u. (not amps)
3.91 2.327
V 1.0 0 0.22 30.8
p.u.
0.189 p.u.
1.0 0 0.22 30.8 30.8 p.u.
L L L L
G
I j
V S V I
Z
S
Ð °= = Ð - °
+
= Ð °- Ð - °́ 2.327Ð 90°
= 0.859Ð - 30.8°
= = =
= Ð °́ Ð °=0.22Ð °
Again, analysis is exactly the same in p.u.!
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3 Per Unit Example, cont'd
L
Actual
ActualL
ActualG
MiddleB
ActualMiddle
0.859 30.8 27.6 kV 23.8 30.8 kV
0.189 0 300 MVA 56.7 0 MVA
0.22 30.8 300 MVA 66.0 30.8 MVA
300 MVAI 125 (same cur 0 Amps
3138 kVI 0.22 30.
rent!)
8
V
S
S
= Ð - °́ = Ð - °
= Ð °́ = Ð °
= Ð °́ = Ð °
= =
= Ð - °́ Amps 275 30.81250 = Ð - °A
Differences appear when we convert back to actual values
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3 Per Unit Example 2
• Assume a 3f load of 100+j50 MVA with VLL of 69
kV is connected to a source through the per-phase
network below:
What is the supply current and complex power?
Answer: I=467 amps, S = 103.3 + j76.0 MVA
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Per Unit Change of MVA Base
l Parameters for equipment are often given using
power rating of equipment as the MVA base
l When only one component is considered, the
nameplate values of that component are selected as
base values
l In a larger system, all per-unit data must have the
same power base
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Per Unit Change of MVA Base
• To convert from an “old” base to a “new” base:
= , ×,
, =
, =
, ×,
,
but =
→ , = , ×
,
,
,
,
or
, = , ,
,
,
,
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Per Unit Change of Base Example
• A 30 MVA, 240/480 kV transformer has a leakage
reactance or 2.71%. What is the reactance on a 100
MVA and 230 kV base?
= 0.0271×240
230
100
30 =0.098 ..
= 0.098×
=52.032 W
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Transformer Reactance
l Transformer reactance is often specified as a
percentage, say 10%. This is a per unit value
(divide by 100) on the power base of the
transformer.l Example: A 350 MVA, 230/20 kV transformer has
leakage reactance of 10%. What is p.u. value on
100 MVA base? What is value in ohms (230 kV)?
2
1000.10 0.0286 p.u.350
2300.0286 15.1
100
e X = ´ =
´ = W
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Three Phase Transformers
• There are 4 different ways to connect 3f transformers
Y-Y -
Usually 3f transformers are constructed so all windings
share a common core
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3 Transformer Interconnections
-Y Y-
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Y-Y Connection
Magnetic coupling with An/an, Bn/bn & Cn/cn
1, , An AB A
an ab a
V V I a a
V V I a= = =
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Y-Y Connection: Per Phase Model
Per phase analysis of Y-Y connections is exactly the
same as analysis of a single phase transformer.
NOTE: In this simplified circuit, RC
is ignored.
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Y-Y Connection: Per Phase Model
Y-Y connections are common in transmission
systems.
Key advantages are the ability to ground each sideand there is no phase shift is introduced.
Main disadvantage is that the undesirable 3rd harmonic
exciting current will cause difficulties to the rest of thenetwork.
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- Connection
Magnetic coupling with AB/ab, BC/bb & CA/ca
1 1, , AB AB A
ab ab a
V I I a
V I a I a= = =
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- Connection: Per Phase Model
Per phase analysis similar to Y-Y except impedances
are decreased by a factor of 3.
To use the per phase equivalent we need to use
the delta-wye load transformation
=
Ð -30°
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- Connection
• Advantage of a Dwinding (in any combination) is
that it can trap the 3rd harmonic exciting current
• Key disadvantage is D-Dconnections can not be
grounded; not commonly used.
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-Y Connection
Magnetic coupling with AB/an, BC/bn & CA/cn
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-Y Connection V/I Relationships
, 3 30
30 30Hence 3 and 3
For current we get
1
13 30 303
130
3
AB ABan ab an
an
AB Anab an
ABa AB
ab
A AB AB A
a A
V V a V V V
V a
V V V V
a a
I I a I
I a
I I I I
a I
= = ® = Ð °
Ð ° Ð= =
= ® =
= Ð - °® = Ð °
I = Ð °
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-Y Connection: Per Phase Model
Note: Connection introduces a 30 degree phase shift!
Common for generator step-up since there is a neutralon the high voltage side which can be grounded,
reducing insulation requirement of transformer.
Even if a = 1 there is a sqrt(3) step-up ratio
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Y- Connection: Per Phase Model
Exact opposite of the D-Y connection, now with a
phase shift of -30 degrees.
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Load Tap Changing Transformers
l LTC transformers have tap ratios that can be varied
to regulate bus voltages
l The typical range of variation is ±10% from the
nominal values, usually in 33 discrete steps(0.0625% per step).
l Because tap changing is a mechanical process, LTC
transformers usually have a 30 second deadband to
avoid repeated changes.
l Unbalanced tap positions can cause "circulating
vars"
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Phase Shifting Transformers
l Phase shifting transformers are used to control the
phase angle across the transformer
l Since power flow through the transformer depends
upon phase angle, this allows the transformer toregulate the power flow through the transformer
l Phase shifters can be used to prevent inadvertent
"loop flow" and to prevent line overloads.
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Autotransformers
l Autotransformers are transformers in which the
primary and secondary windings are coupled
magnetically and electrically.
l This results in lower cost, and smaller size and weight. Also smaller leakage impedances results
in smaller series voltage drop (advantage) but also
higher short-circuit current (disadvantage).
l Another disadvantage is loss of electrical isolation between the voltage levels. This can be an
important safety consideration when a is large.