ECE 476

POWER SYSTEM ANALYSIS Lecture 9 Transformers, Per Unit Professor Tom Overbye Department of Electrical and Computer Engineering

Announcements

  Be reading Chapter 3 HW 3 is 4.32, 4.41, 5.1, 5.14. Due September 22 in class.

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Transformer Equivalent Circuit

Using the previous relationships, we can derive an equivalent circuit model for the real transformer This model is further simplified by referring all impedances to the primary side r 2 ' 

x

2 '  2 2

r x e e r

2 '

x

2 '

2

Simplified Equivalent Circuit 3

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.

4

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.

5

Transformer Example, cont’d From the short circuit test

I sc

 100

MVA

200

kV

 P sc  2 e  500 kW  R e 

jX e

,  30

kV

500

A

 60  Hence X e  60 2  2 2  60

From the open circuit test



R c

 200

kV

2 10

kW

 4

M

 R e 

jX e



jX m

 200

kV

20

A

 10, 000 

X m

 10, 000 

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Residential Distribution Transformers

Single phase transformers are commonly used in residential distribution systems. Most distribution systems are 4 wire, with a multi-grounded, common neutral.

7

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. quantity in per unit  actual quantity base value of quantity

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Per Unit Conversion Procedure, 1

f 3.

4.

5.

1.

2.

Pick a 1 f VA base for the entire system, S B Pick a voltage base for each different voltage level, V B . Voltage bases are related by transformer turns ratios. Voltages are line to neutral.

Calculate the impedance base, Z B = (V B ) 2 /S B Calculate the current base, I B = V B /Z B 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)

9

Per Unit Solution Procedure

1.

2.

3.

Convert to per unit (p.u.) (many problems are already in per unit) Solve Convert back to actual as necessary

10

Per Unit Example

Solve for the current, load voltage and load power in the circuit shown below using per unit analysis with an S B of 100 MVA, and voltage bases of 8 kV, 80 kV and 16 kV. Original Circuit

11

Per Unit Example, cont’d

Z Left B

8

kV

2  100

MVA

 0.64



Z Middle B

80

kV

2  100

MVA Z Right B

16

kV

2  100

MVA

 2.56

 Same circuit, with values expressed in per unit.

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Per Unit Example, cont’d

I

V L   3.91



j

2.327

0.22

 0.22

    30.8

    p.u.

 

S S L G

  * 

V L

2  0.189 p.u.

Z

0.22 30.8

   

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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

V

L Actual  0.859

  30.8

 16 kV 

S

Actual L  100 MVA 

S

Actual G  0.22 30.8

 100 MVA  13.7

  I I Middle B Actual Middle   100 MVA  1250 Amps 80 kV 0.22

  30.8

 Amps  275   30.8



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Three Phase Per Unit

Procedure is very similar to 1 f except we use a 3 f 1.

VA base, and use line to line voltage bases Pick a 3 f VA base for the entire system,

S

3

B

f 2.

Pick a voltage base for each different voltage level, V B . Voltages are line to line . 3.

Calculate the impedance base

Z B

 2

V S

3 f

B

 ( 3

V

3

S

1 f

B

) 2  2

V S

1 f

B

Exactly the same impedance bases as with single phase!

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Three Phase Per Unit, cont'd

4.

I Calculate the current base, I B 3 f B 

S

3 f

B

3

V

 3

S

3 3 1 f

B V



V S

1 f

B

 I 1 f B Exactly the same current bases as with single phase!

5.

Convert actual values to per unit

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Three Phase Per Unit Example

Solve for the current, load voltage and load power in the previous circuit, assuming a 3 f power base of 300 MVA 1 f , and line to line voltage bases of 13.8 kV, 138 kV and 27.6 kV (square root of 3 larger than the 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 is exactly the same!

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3

f

Per Unit Example, cont'd

I

 3.91



j

2.327

 0.22

  V L  0.22

  30.8

    p.u.

 

S S L G

  *  2

V L

 0.189 p.u.

Z

0.22 30.8

    Again, analysis is exactly the same!

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3

f

Per Unit Example, cont'd

Differences appear when we convert back to actual values

V

L Actual 

S

Actual L 

S

Actual G  0.859

  30.8

 27.6 kV  23.8

  300 MVA   300 MVA  I Middle B  300 MVA 3 138 kV  125 0 Amps (same cur rent!) I Actual Middle  0.22

  30.

8   Amps  275   30.8



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3

f

Per Unit Example 2

Assume a 3 f load of 100+j50 MVA with V LL 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

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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

Z OriginalBase pu



Z actual



Z NewBase pu

Hence Z

OriginalBase pu



V

2

base S OriginalBase Base

/

V

2

base S NewBase Base



Z NewBase pu

Z

OriginalBase pu



S NewBase Base S OriginalBase Base



Z NewBase pu

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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?

X e

 0.0369

 100 54  0.0683 p.u.

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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)?

X e

 0.10

 100 350  0.0286 p.u.

0.0286

 230 2 100  15.1



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Three Phase Transformers

There are 4 different ways to connect 3 f

Y-Y

D

-

D transformers Usually 3 f transformers are constructed so all windings share a common core

24

3

f

Transformer Interconnections

D

-Y Y -

D

25

Y-Y Connection

Magnetic coupling with An/an, Bn/bn & Cn/cn

V An V an



a

,

V AB V ab



a

,

I A I a

 1

a

26

Y-Y Connection: 3

f

Detailed Model 27

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.

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D

-

D

Connection

Magnetic coupling with AB/ab, BC/bb & CA/ca

V AB V ab



a

,

I AB I ab

 1

I

,

a I A a

 1

a

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D

-

D

Connection: 3

f

Detailed Model

To use the per phase equivalent we need to use the delta-wye load transformation

30

D

-

D

Connection: Per Phase Model

Per phase analysis similar to Y-Y except impedances are decreased by a factor of 3. Key disadvantage is D D connections can not be grounded; not commonly used.

31

D

-Y Connection

Magnetic coupling with AB/an, BC/bn & CA/cn

32

D

-Y Connection V/I Relationships

V AB V an



a

,

V AB a



V an



V ab

Hence

V ab

 3

V AB a

For current we get  and 3

V an V an



I AB I ab a I a



a I AB I A

 3

I AB I AB

 1 3

I A a

1 3

I A

3

V An a

33

D

-Y Connection: Per Phase Model

Note: Connection introduces a 30 degree phase shift!

Common for transmission/distribution step-down since there is a neutral on the low voltage side.

Even if a = 1 there is a sqrt(3) step-up ratio

34

Y-

D

Connection: Per Phase Model

Exact opposite of the D -Y connection, now with a phase shift of -30 degrees.

35

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"

36

LTCs and Circulating Vars

slack

64 MW 14 Mvar 1 24.1 MW 12.8 Mvar

A MVA

1.000 tap

A

80%

MVA

1.00 pu 40.2 MW 1.7 Mvar 1.056 tap 24.0 MW -12.0 Mvar 2 0.98 pu 3 40.0 MW -0.0 Mvar 1.05 pu 0.0 Mvar 24 MW 12 Mvar 40 MW 0 Mvar 37

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.

38

Phase Shifter Example 3.13

500 MW 164 Mvar slack 345.00 kV 283.9 MW 39.0 Mvar 341.87 kV 283.9 MW 6.2 Mvar Phase Shifting Transformer 500 MW 100 Mvar 216.3 MW 125.0 Mvar 216.3 MW 0.0 deg 93.8 Mvar 1.05000 tap 39

ComED Control Center 40

ComED Phase Shifter Display 41

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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