Lecture 7

Power Transformer

Reading: Chapters 3.1; 3.3 ; 3.4

Homework 2 is due on Feb. 8th

Dr. Lei Wu

Department of Electrical and Computer Engineering

EE 333

POWER SYSTEMS ENGINEERING

Outline

� Ideal Transformer

� Phase-Shifting Transformer

� Three-phase Transformers

� Per Unit Calculation

2

1 1

2 2

V N

V Nα= = 1 2

2 1

1I N

I N α= =

1 2S S=2

' 1

2LL

NZ Z

N

=

'L LZ Z=1 2S S=1

2

jVe

Vφ= 1

2

jI

Ie φ=

Y/Y and ∆/∆

3 30 3 30

AB ab

AN an

AN an

V NV

V V

V NV

=

∠ = ∠=

� �

AN anV NV=

No phase shift between corresponding quantities on the low and high voltage windings 3

∆/Y and Y/∆

3 30 180

3 30 180 240

303

CB cn

CB BC BN

CN

CN cn

V NV

V V V

V

NV V

=

= − = ∠ +

= ∠ + −

= ∠

� �

� � �

�

3 30AN ab anV NV NV= = ∠ �

High voltage side leads low voltage side by 30o

Y for the high voltage side- Neutral on the high voltage side reduce insulation requirements

- High voltage gain of∆ for the low voltage side

- third order harmonic magnetic currents remain inside

3N

4

How to Describe Three Phase Transformers

5

� Distinguish the two statements

� A balanced 3Φ 208-V source -- line-to-line voltage

� A balanced 3Φ source with 120-V in each phase -- line-to-

neutral voltage

Three Phase Transformer Example

� Consider a bank of three single-phase two-winding transformers

whose high-voltage terminals are connected to a three-phase

13.8kv feeder. The low-voltage terminals are connected to a

three-phase substation load rated 2.3KV. Determine the

required ratings of both windings of each transformer with for different connection ways.

�

� 13.8 Y/ 2.3 ∆, the single-phase transformer is

� 13.8 ∆/ 2.3 Y, the single-phase transformer is

� 13.8 ∆/ 2.3 ∆, the single-phase transformer is 13.8/2.3

� 13.8 Y/ 2.3 Y, the single-phase transformer is

, is 13.8 for HV side, and 2.3KV for LV sideB LLV KV

13.8/ 2.3

3

2.313.8/

3

13.8 2.3/

3 3 6

Per Unit Calculation

� 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 -

known as per unit analysis.

� Transformer equivalent circuit can be simplified. Windings are

eliminated, voltages, currents, and external impedances expressed in

per unit values are the same when referring from one side to the other.

� Avoid serious calculation errors.

� Easy checking – per unit values lie within a narrow numerical range

actual quantityper unit quantity

base value of quantity=

7

Per Unit Conversion Procedure - 1φφφφ

� Pick up two independent base values.

� 1φ VA base for the entire system, SB

� A voltage base for each different voltage level, VB. Voltage

bases are related by transformer turns ratios. Voltages are line

to neutral.

� Calculate the impedance base, ZB= (VB)2/SB

� Calculate the current base, IB = VB/ZB =SB /VB

� Convert actual values to per unit using the base values.

SB is the same for the system, VB are related to turns ratiosPer unit conversion affects magnitudes, not the angles. Per unit quantities no longer have units (i.e., a voltage is 1.0 p.u., not 1 p.u. volts)

8

Per Unit Analysis for 1f Two-Winding Transformer

� Pick up two independent base values.

� 1φ VA base for the entire system, SB

� A voltage base for each different voltage level, VB1 and VB2 ,

which satisfy that

� Thus,

1 2 1 2B BV V N N=

( )( )

1 2 21 21, 2,

1 1 2 2 2pu pu

B B B

N N VV VV V

V N N V V= = = =

( ) ( )( )[ ]

2 21 2 1 2

, ,2 2 21 1 1 2 21 2 2 2

L LL LL pu L pu

B B B B BB B

N N Z N N ZZ ZZ Z

Z V S V SN N V S

′′ = = = = =

9