Lecture 9Transformers, Per Unit
Professor Tom OverbyeDepartment of Electrical and
Computer Engineering
ECE 476
POWER SYSTEM ANALYSIS
3
Transformer Equivalent Circuit
Using the previous relationships, we can derive an equivalent circuit model for the real transformer
' 2 '2 2 1 2
' 2 '2 2 1 2
This model is further simplified by referring all
impedances to the primary side
r e
e
a r r r r
x a x x x x
5
Calculation of Model Parameters
The parameters of the model are determined based upon – nameplate data: gives the rated voltages and power– open circuit test: rated voltage is applied to primary with
secondary open; measure the primary current and losses (the test may also be done applying the voltage to the secondary, calculating the values, then referring the values back to the primary side).
– short circuit test: with secondary shorted, apply voltage to primary to get rated current to flow; measure voltage and losses.
6
Transformer Example
Example: A single phase, 100 MVA, 200/80 kV transformer has the following test data:
open circuit: 20 amps, with 10 kW losses
short circuit: 30 kV, with 500 kW losses
Determine the model parameters.
7
Transformer Example, cont’d
e
2sc e
2 2e
2
e
100 30500 , R 60
200 500
P 500 kW R 2 ,
Hence X 60 2 60
2004
10
200R 10,000 10,000
20
sc e
e sc
c
e m m
MVA kVI A jX
kV A
R I
kVR M
kW
kVjX jX X
A
From the short circuit test
From the open circuit test
8
Residential Distribution Transformers
Single phase transformers are commonly used in residential distribution systems. Most distributionsystems are 4 wire, with a multi-grounded, common neutral.
9
Per Unit Calculations
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
This problem is avoided by a normalization of all variables.
This normalization is known as per unit analysis.
actual quantityquantity in per unit
base value of quantity
10
Per Unit Conversion Procedure, 1
1. Pick a 1 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 unitNote, 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)
11
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
12
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
13
Per Unit Example, cont’d
2
2
2
80.64
100
8064
100
162.56
100
LeftB
MiddleB
RightB
kVZ
MVA
kVZ
MVA
kVZ
MVA
Same circuit, withvalues expressedin per unit.
14
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.
LL L L
G
Ij
VS V I
ZS
15
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
16
Three Phase Per Unit
1. Pick a 3 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 1 except we use a 3 VA base, and use line to line voltage bases
3BS
2 2 2, , ,3 1 1
( 3 )
3B LL B LN B LN
BB B B
V V VZ
S S S
Exactly the same impedance bases as with single phase!
17
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 3B B B
B LL B LN B LN
S S S
V V V
Exactly the same current bases as with single phase!
18
Three Phase Per Unit Example
Solve for the current, load voltage and load power in the previous circuit, assuming a 3 power base of300 MVA, and line to line voltage bases of 13.8 kV,138 kV and 27.6 kV (square root of 3 larger than the 1 example voltages). Also assume the generator is Y-connected so its line to line voltage is 13.8 kV.
Convert to per unitas before. Note thesystem is exactly thesame!
19
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.
LL L L
G
Ij
VS V I
ZS
Again, analysis is exactly the same!
20
3 Per Unit Example, cont'd
LActual
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 cur0 Amps
3138 kV
I 0.22 30.
rent!)
8
V
S
S
Amps 275 30.8
Differences appear when we convert back to actual values
21
3 Per Unit Example 2
Assume a 3 load of 100+j50 MVA with VLL of 69 kV is connected to a source through the below network:
What is the supply current and complex power?
Answer: I=467 amps, S = 103.3 + j76.0 MVA
22
Per Unit Change of MVA Base
Parameters for equipment are often given using power rating of equipment as the MVA base
To analyze a system all per unit data must be on a common power base
2 2
Hence Z /
Z
base base
OriginalBase NewBasepu actual pu
OriginalBase NewBasepu puOriginalBase NewBase
BaseBase
NewBaseOriginalBase NewBaseBasepu puOriginalBase
Base
Z Z Z
V VZ
SS
SZ
S
23
Per Unit Change of Base Example
A 54 MVA transformer has a leakage reactance or 3.69%. What is the reactance on a 100 MVA base?
1000.0369 0.0683 p.u.
54eX
24
Transformer Reactance
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.
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
eX
25
Three Phase Transformers
There are 4 different ways to connect 3 transformers
Y-Y -
Usually 3 transformers are constructed so all windingsshare a common core
29
Y-Y Connection: Per Phase Model
Per phase analysis of Y-Y connections is exactly the same as analysis of a single phase transformer.
Y-Y connections are common in transmission systems.
Key advantages are the ability to ground each side and there is no phase shift is introduced.
31
- Connection: 3 Detailed Model
To use the per phase equivalent we need to usethe delta-wye load transformation
32
- Connection: Per Phase Model
Per phase analysis similar to Y-Y except impedances are decreased by a factor of 3.
Key disadvantage is - connections can not be grounded; not commonly used.
34
-Y Connection V/I Relationships
, 3 30
30 30Hence 3 and 3
For current we get
1
13 30 30
31
303
AB ABan ab an
an
AB Anab an
ABa AB
ab
A AB AB A
a A
V Va V V V
V a
V VV V
a a
II a I
I a
I I I I
a I
35
-Y Connection: Per Phase Model
Note: Connection introduces a 30 degree phase shift!
Common for transmission/distribution step-down sincethere is a neutral on the low voltage side.
Even if a = 1 there is a sqrt(3) step-up ratio
36
Y- Connection: Per Phase Model
Exact opposite of the -Y connection, now with a phase shift of -30 degrees.
37
Load Tap Changing Transformers
LTC transformers have tap ratios that can be varied to regulate bus voltages
The typical range of variation is 10% from the nominal values, usually in 33 discrete steps (0.0625% per step).
Because tap changing is a mechanical process, LTC transformers usually have a 30 second deadband to avoid repeated changes.
Unbalanced tap positions can cause "circulating vars"
38
LTCs and Circulating Vars
slack
1 1.00 pu
2 3
40.2 MW
40.0 MW
1.7 Mvar
-0.0 Mvar
1.000 tap 1.056 tap
24.1 MW 12.8 Mvar
24.0 MW-12.0 Mvar
A
MVA
1.05 pu 0.98 pu
24 MW
12 Mvar
64 MW
14 Mvar
40 MW 0 Mvar
0.0 Mvar
80%A
MVA
39
Phase Shifting Transformers
Phase shifting transformers are used to control the phase angle across the transformer
Since power flow through the transformer depends upon phase angle, this allows the transformer to regulate the power flow through the transformer
Phase shifters can be used to prevent inadvertent "loop flow" and to prevent line overloads.
40
Phase Shifter Example 3.13
slack
Phase Shifting Transformer
345.00 kV 341.87 kV
0.0 deg 216.3 MW 216.3 MW
283.9 MW 283.9 MW
1.05000 tap
39.0 Mvar 6.2 Mvar
93.8 Mvar 125.0 Mvar
500 MW
164 Mvar 500 MW 100 Mvar
43
Autotransformers
Autotransformers are transformers in which the primary and secondary windings are coupled magnetically and electrically.
This results in lower cost, and smaller size and weight.
The key disadvantage is loss of electrical isolation between the voltage levels. Hence auto-transformers are not used when a is large. For example in stepping down 7160/240 V we do not ever want 7160 on the low side!
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