Lecture 9Transformers, Per Unit

Professor Tom OverbyeDepartment of Electrical and

Computer Engineering

ECE 476

POWER SYSTEM ANALYSIS

2

Announcements

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

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

4

Simplified Equivalent Circuit

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

26

3 Transformer Interconnections

-Y Y-

27

Y-Y Connection

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

1, ,An AB A

an ab a

V V Ia a

V V I a

28

Y-Y Connection: 3 Detailed Model

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.

30

- Connection

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

1 1, ,AB AB A

ab ab a

V I Ia

V I a I a

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.

33

-Y Connection

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

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

41

ComED Control Center

42

ComED Phase Shifter Display

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!