Transformerz-PerUnit

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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 electricalquantities 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 System Assume:

Then compute base values for currents and

impedances:

rated b V V

rated bS S

b

bb

V S I

b

b

b

bb

S

V

I

V Z 2



4

Per-Unit System And the per-unit system is:

b

actualu p

V

V V ..

b

actualu p

I

I I

..

b

actualu p

S

S S

..

b

actualu p

Z

Z Z ..

%100% .. u p Z Z Percent of base Z



5

Example 1 An 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

8.28

500

)120(222

P

V R

R

V P

08.28 Z



6

Example 1(2) Select base quantities

(3) Compute base impedance

(4) The per-unit impedance is:

VAS b 500

V V b 120

8.28500

)120(22

b

bb

S

V Z

..018.28

08.28..

u p Z

Z Z

b

u p



7

Example 1(5) Percent impedance:

(6) Per-unit equivalent circuit:

%100% Z

..01 u p Z ..01 u pV S



8

Example 2 An electrical lamp is rated 120 volts, 500 watts. If

the voltage applied across the lamp is twice therated value, compute the current that flows

through the lamp. Use the per-unit method.

Solution:

V V b 120

..02120240

..u p

V V V

b

u p

..01.. u p Z u p



9

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

..01 u p Z ..02 u pV S

..0201

02

..

..

.. u p Z

V I

u p

u p

u p

AV

S I

b

bb 167.4

120

500

A I I I bu pactual

0334.8167.402..



10

Per-unit System for 1- Circuits

One-phase circuits

LV bLV V V

I V S S b 1

whereneutraltolineV V

current line I I

HV bHV V V

bLV

bbLV

V

S I

bHV

bbHV

V

S I



11


b

bLV

bLV

bLV bLV S

V

I

V Z

2)(

b

bHV

bHV

bHV bHV S

V

I

V Z

2)(

*

pu pub

pu I V S

S

S

cos pu pu

b

pu I V

S

PP

sin pu pu

b

pu I V S

QQ



12

Transformation Between BasesSelection 1

Ab V V 1 Ab S S 1

Then

1

1

b

L pu

Z

Z Z

1

2

1

1b

b

b S

V Z

Selection 2

Bb V V 2 Bb S S 2

Then

2

2

b

L pu

Z

Z Z

2

2

2

2

b

bb

S

V Z



13

Transformation Between Bases

2

2

2

1

2

1

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

1

12b

b

b

b

pu pu S

S

V

V

Z Z

“1” – old

“2” - new

old b

newb

newb

old b

old punew puS

S

V

V Z Z

,

,

2

,

,

,,



14

Transformation Between BasesGenerally per-unit values given to another base

can be converted to new base by by theequations:

2

1

1___2___ ),,(),,(base

basebaseon pubaseon pu

S

S S QPS QP

2

1

1___2___

base

basebaseon pubaseon pu

V

V V V

1

2

2

2

2

11___2___

)(

)(),,(),,(

basebase

basebasebaseon pubaseon pu

S V

S V Z X R Z X R

When performing calculations in a power system, every

per-unit value must be converted to the same base.



15

Per-unit System for 1- Transformer Consider the equivalent circuit of transformer

referred to LV side and HV side shown below:

LV V HV V LV V HV V

S S jX R

1 N

2 N

22a

X j

a

R S S

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

Define 12

1 N

N

V

V a

HV

LV

S



16

Per-unit System for 1- Transformer Choose:

rated LV b V V ,1

rated b S S

Compute: 112 1 bb

LV

HV b V

aV

V V V

b

b

b S

V Z

2

1

1

b

b

b S

V Z

2

2

2

2

2

1

2

1

2

2

2

1

2

1

)

1

(

a

V 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- Transformer Per-unit impedances are:

1

1..

b

S S u p

Z

jX R Z

12

1

22

2

22

2..

b

S S

b

S S

b

S S

u p Z

jX R

a

Z a

jX

a

R

Z

a

jX

a

R

Z

So:

2..1.. u pu p Z Z Per-unit equivalentcircuits of transformer

referred to LV side and

HV side are identical !!



18

Per-unit Eq. Ckt for 1- Transformer

LV V HV V

S S jX R

1 N

2 N

Fig 1. Eq Ckt referred to LV side

12

1 N

N

V

V a

HV

LV

S

1b Z

1bV 2bV

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

puS Z ,

1:1

1bV 2bV

puS Z ,

1bV 2bV

bS



19

Per-unit Eq. Ckt for 1- Transformer

LV V HV V

1 N

2 N

Fig 4. Eq Ckt referred to HV side

12

1 N

N

V

V a

HV

LV

S

2b Z

2bV

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

puS Z ,

1:1

1bV 2bV

puS Z ,

1bV 2bV

1bV

22

a

X j

a

R S S

bS



20

Voltage Regulation

%100

load full

load fullload no

V

V V VR

Voltage regulation is defined as:

%100

,

,,

load full pu

load full puload no pu

V

V V VR

In per-unit system:

V full-load : Desired load voltage at full load. It may be equalto, above, or below rated voltage

V no-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 Example A single-phase transformer rated 200-kVA, 200/400-V, and

10% short circuit reactance. Compute the VR when thetransformer is fully loaded at unity PF and rated voltage

400-V.

Solution:

Fig 7. Per-unit equivalent circuit

PV S V 1.0 j

load S

V V b 4002

kVAS b 200

puS puload 01,

pu j X puS 1.0,

S X



22

Voltage Regulation ExampleRated voltage:

puV puS 00.1,

pu

j j

X I V V

o

puS pu puS puP

7.5001.1

1.011.000.100.1

,,,

puV

S I

puS

puload

puload

00.100.1

00.1*

*

,

,

,



23

Voltage Regulation Example

puV V o

puPload no pu 7.5001.1,,

puV V puS load full pu 00.1,,

Secondary side:

Voltage regulation:

%1.0%1000.1

0.1001.1

%100,

,,

load full pu

load full puload no pu

V

V V

VR



24

Select V base in generator circuit and S b=100MVA,compute p.u. equivalent circuit.

Problem 1

G

100 j

20 kV 22kV/220kV

80MVA

14%

220kV/20kV

50MVA

10%

50MVA

0.8 PF

lagging



25


Three-phase circuits

LV LbLV V V ,

I V S S S b 33 13

where3 / )(line Lneutraltoline V V V

Lcurrent line I I I

HV LbHV V V ,

L Lb I V S 3

bHV bHV bLV bLV b I V I V S 33



26


b

bLV

b

bLV bLV

LV

LV

bLV S

V

S

V V

I

V

Z

2)(3

3

b

bHV bHV

S

V Z

2)(

**

3

3

3 pu pu

bb

L L

b

pu I V I V

I V

S

S S

bLV

bbLV

V

S I

3

bHV

bbHV

V

S I 3



27

Three 25-kVA, 34500/277-V transformers

connected in -Y. Short-circuit test on high voltageside:

Determine the per-unit equivalent circuit of the

transformer.

Per-unit System for 3- Transformer

V V SC Line 2010,

A I SC Line 26.1,

W P SC 912,3



28

(a) Using Y-equivalent


3

34500277

S S jX R

VAS b 25000

3

2010SC V

26.1SC I

00.92126.1

47.1160SC Z

V V SC 47.11603

2010



29

So


86.90048.1919212222

S SC S R Z X

W P 3043

912

48.19126.1

304

22SC

S I

P

R

86.90048.191 j Z SC

VAS b 25000 V V HV b 58.199183

34500,

99.1586925000

58.199182

, HV b Z

pu j j

Z Y puSC 0568.0012.099.15869

86.90048.191,,



30

(b) Using -equivalent


34500 277

,SC Z

VAS b 250002010SC

V

3

26.1SC I

79.2764727.0

2010,SC Z

V V SC 2010 A I SC 727.03

26.1



31

So


30.270418.57579.2764222

,

2

,, S SC S R Z X

W P 3043

912

18.575727.0

304

22,SC

S I

P

R

86.90048.191 j Z SC

VAS b 25000 V V HV b 34500,

4761025000

345002

, HV b Z

pu j j

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

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