transformer3-perunit

1

Per-Unit System

EE341 – Energy Conversion

Ali Keyhani

Transformer

Lecture #3

2

Per-Unit SystemIn the per-unit system, the voltages, currents, powers, impedances, and other electrical quantities are expressed on a per-unit basis by the equation:

Quantity per unit = Actual value

Base value of quantity

It is customary to select two base quantities to define a given per-unit system. The ones usually selected are voltage and power.

3

Per-Unit SystemAssume:

Then compute base values for currents and impedances:

ratedb VV

ratedb SS

b

bb V

SI

b

b

b

bb S

V

I

VZ

2

4

Per-Unit SystemAnd the per-unit system is:

b

actualup V

VV ..

b

actualup I

II ..

b

actualup S

SS ..

b

actualup Z

ZZ ..

%100% .. upZZ Percent of base Z

5

Example 1An electrical lamp is rated 120 volts, 500 watts. Compute the per-unit and percent impedance of the lamp. Give the p.u. equivalent circuit.

Solution:(1) Compute lamp resistance

power factor = 1.0

8.28500

)120( 222

P

VR

R

VP

08.28Z

6

Example 1(2) Select base quantities

(3) Compute base impedance

(4) The per-unit impedance is:

VASb 500

VVb 120

8.28500

)120( 22

b

bb S

VZ

..018.28

08.28.. up

Z

ZZ

bup

7

Example 1(5) Percent impedance:

(6) Per-unit equivalent circuit:

%100%Z

..01 upZ ..01 upVS

8

Example 2An electrical lamp is rated 120 volts, 500 watts. If the voltage applied across the lamp is twice the rated value, compute the current that flows through the lamp. Use the per-unit method.

Solution:

VVb 120

..02120

240.. up

V

VV

bup

..01.. upZ up

9

Example 2The per-unit equivalent circuit is as follows:

..01 upZ ..02 upVS

..0201

02

..

.... up

Z

VI

up

upup

AV

SI

b

bb 167.4

120

500

AIII bupactual 0334.8167.402..

10

Per-unit System for 1- CircuitsOne-phase circuits

LVbLV VV

IVSSb 1

whereneutraltolineVV

currentlineII

HVbHV VV

bLV

bbLV V

SI

bHV

bbHV V

SI

11

Per-unit System for 1- Circuits

b

bLV

bLV

bLVbLV S

V

I

VZ

2)(

b

bHV

bHV

bHVbHV S

V

I

VZ

2)(

*pupu

bpu IV

S

SS

cospupub

pu IVS

PP

sinpupub

pu IVS

QQ

12

Transformation Between BasesSelection 1

Ab VV 1Ab SS 1Then

11

b

Lpu Z

ZZ

1

21

1b

bb S

VZ

Selection 2

Bb VV 2Bb SS 2Then

22

b

Lpu Z

ZZ

2

22

2b

bb S

VZ

13

Transformation Between Bases

22

2

1

21

2

11

21

2

b

b

b

b

b

b

L

b

b

L

pu

pu

V

S

S

V

Z

Z

Z

Z

Z

Z

Z

Z

1

2

2

2

112

b

b

b

bpupu S

S

V

VZZ

“1” – old“2” - new

oldb

newb

newb

oldboldpunewpu S

S

V

VZZ

,

,

2

,

,,,

14

Transformation Between BasesGenerally per-unit values given to another base can be converted to new base by by the equations:

2

11___2___ ),,(),,(

base

basebaseonpubaseonpu S

SSQPSQP

2

11___2___

base

basebaseonpubaseonpu V

VVV

12

2

22

11___2___ )(

)(),,(),,(

basebase

basebasebaseonpubaseonpu SV

SVZXRZXR

When performing calculations in a power system, every per-unit value must be converted to the same base.

15

Per-unit System for 1- TransformerConsider the equivalent circuit of transformer referred to LV side and HV side shown below:

LVV HVV LVV HVV

SS jXR

1N 2N

22 a

Xj

a

R SS

(1) Referred to LV side (2) Referred to HV side

Define 12

1 N

N

V

Va

HV

LV

S

16

Per-unit System for 1- TransformerChoose:

ratedLVb VV ,1

ratedb SS

Compute:112

1bb

LV

HVb V

aV

V

VV

b

bb S

VZ

21

1 b

bb S

VZ

22

2

2

21

21

22

21

2

1

)1(

aV

a

V

V

V

Z

Z

b

b

b

b

b

b

Normal choose rated values as base values

17

Per-unit System for 1- TransformerPer-unit impedances are:

11..

b

SSup Z

jXRZ

121

22

2

22

2..b

SS

b

SS

b

SS

up Z

jXR

aZ

ajX

aR

ZajX

aR

Z

So:2..1.. upup ZZ

Per-unit equivalent circuits of transformer referred to LV side and HV side are identical !!

18

Per-unit Eq. Ckt for 1- Transformer

LVV HVV

SS jXR

1N 2N

Fig 1. Eq Ckt referred to LV side

12

1 N

N

V

Va

HV

LV

S

1bZ

1bV 2bV

Fig 2. Per-unit Eq Ckt referred to LV side Fig 3.

puSZ ,

1:1

1bV 2bVpuSZ ,

1bV 2bV

bS

19

Per-unit Eq. Ckt for 1- Transformer

LVV HVV

1N2N

Fig 4. Eq Ckt referred to HV side

12

1 N

N

V

Va

HV

LV

S

2bZ

2bV

Fig 5. Per-unit Eq Ckt referred to HV side Fig 6.

puSZ ,

1:1

1bV 2bVpuSZ ,

1bV 2bV

1bV

22 a

Xj

a

R SS

bS

20

Voltage Regulation

%100

loadfull

loadfullloadno

V

VVVR

Voltage regulation is defined as:

%100,

,,

loadfullpu

loadfullpuloadnopu

V

VVVR

In per-unit system:

Vfull-load: Desired load voltage at full load. It may be equal to, above, or below rated voltage

Vno-load: The no load voltage when the primary voltage is the desired voltage in order the secondary voltage be at its desired value at full load

21

Voltage Regulation ExampleA single-phase transformer rated 200-kVA, 200/400-V, and 10% short circuit reactance. Compute the VR when the transformer is fully loaded at unity PF and rated voltage 400-V.

Solution:

Fig 7. Per-unit equivalent circuit

PV SV1.0j

loadS

VVb 4002

kVASb 200

puS puload 01,

pujX puS 1.0,

SX

22

Voltage Regulation ExampleRated voltage:

puV puS 00.1,

pu

jj

XIVV

o

puSpupuSpuP

7.5001.1

1.011.000.100.1

,,,

puV

SI

puS

puloadpuload 00.1

00.1

00.1**

,

,,

23

Voltage Regulation Example

puVV opuPloadnopu 7.5001.1,,

puVV puSloadfullpu 00.1,,

Secondary side:

Voltage regulation:

%1.0%1000.1

0.1001.1

%100,

,,

loadfullpu

loadfullpuloadnopu

V

VVVR

24

Select Vbase in generator circuit and Sb=100MVA, compute p.u. equivalent circuit.

Problem 1

G

100j

20 kV 22kV/220kV80MVA14%

220kV/20kV50MVA10%

50MVA0.8 PF lagging

25

Per-unit System for 3- CircuitsThree-phase circuits

LVLbLV VV ,

IVSSSb 33 13 where 3/)(lineLneutraltoline VVV

Lcurrentline III

HVLbHV VV ,LLb IVS 3

bHVbHVbLVbLVb IVIVS 33

26

Per-unit System for 3- Circuits

b

bLV

b

bLVbLV

LV

LVbLV S

V

S

VV

I

VZ

2)(3

3

b

bHVbHV S

VZ

2)(

**

3

3

3pupu

bb

LL

bpu IV

IV

IV

S

SS

bLV

bbLV

V

SI

3

bHV

bbHV

V

SI

3

27

Three 25-kVA, 34500/277-V transformers connected in -Y. Short-circuit test on high voltage side:

Determine the per-unit equivalent circuit of the transformer.

Per-unit System for 3- Transformer

VV SCLine 2010,

AI SCLine 26.1,

WP SC 912,3

28

(a) Using Y-equivalent


3

34500 277

SS jXR

VASb 250003

2010SCV

26.1SCI

00.92126.1

47.1160SCZ

VVSC 47.11603

2010

29

So


86.90048.191921 2222

SSCS RZX

WP 3043

912 48.191

26.1

30422

SCS I

PR

86.90048.191 jZSC

VASb 25000 VV HVb 58.199183

34500,

99.1586925000

58.19918 2

,HVbZ

pujj

Z YpuSC 0568.0012.099.15869

86.90048.191,,

30

(b) Using -equivalent


34500 277

,SCZ

VASb 250002010SCV

3

26.1SCI

79.2764727.0

2010,SCZ

VVSC 2010 AISC 727.03

26.1

31

So


30.270418.57579.2764 222,

2

,, SSCS RZX

WP 3043

912 18.575

727.0

30422,

SCS I

PR

86.90048.191 jZSC

VASb 25000 VV HVb 34500,

4761025000

345002,HVbZ

pujj

Z puSC 0568.0012.047610

30.170418.575,,

32

Related Materials in Textbook(1) Section 2.6 and 2.7, page 83~90, Chapman

book(2) Section 2.10, page 113~116, Chapman book

transformer3-perunit

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