the per unit system€¦ · lecture 08 power engineering - egill benedikt hreinsson 1 the per unit...

12-Oct-11

1Lecture 08 Power Engineering - Egill Benedikt Hreinsson

The Per Unit System

Introducing new units of measurements in order to simplify the analysis of power systems

12-Oct-11


The Per-unit System

• Conventional units of measurement for voltage, current, power and impedance are for instance: kV, kA, MW, (MVA, MVar) and ohms

• We can always (!) introduce new units of measurement or base quantities for (1)voltage, (2) current, (3) power and (4) impedance, e.g.:

100 /132 kA=0.757 kA

132 / 0.757 174.4

bb

b

bb

b

SI

V

VZ

I

= =

= = = Ω

100 MVA

132 kVb

b

S

V

=

=

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Consistency of Units

• The new units must be consistent, i.e. the base quantities must satisfy:

b b b

b b b

S V I

V Z I

= ⋅

= ⋅

• It is sufficient to define 2 of the above 4 quantities and the other 2 quantities can be calculated.

• The quantities in p.u. are dimensionless• The base quantities are real numbers (not complex)

• An actual quantity such as voltage V can be expressed as: pu

b

VVV

=

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Per Unit Quantities

• Actual voltages, currents, power and impedance, as complex phasors can be expressed in the new units:

pu pu pub b

S P jQS P jQS S

+= + = =

Vj

pub b

V eVVV V

δ

= =Ij

pub b

I eIII I

δ

= =

pu pu pub b

Z R jXZ R jXZ Z

+= + = =

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Example of Per Unit Voltage

• For instance, if a certain actual measured voltage, is V = 234 kV and the base quantity or base voltage is |Vb| = 220 kV, the voltage in units (p.u.) is:

234 1.0636220pu

b

VVV

= = =

…or about 106.4 % voltage...

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6Lecture 08 Power Engineering - Egill Benedikt HreinssonPrimary Objectives and Advantages of the P.U. System

• Quantities, such as voltage, are expressed in relation to their “normal” value. The voltages should normally lie in the range:

0.95 pu < |V | < 1.10 puThis helps when many standard voltages and levels in power systems are present, such as 0,4 kV (400 V), 11kV, 33 kV, 66kV, 132 kV, 220 kV, 400 kV

• The p.u. system removes the need for considering the transformation ratio of transformers. It removes all transformers from the system!!

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Voltage and P.U. In TransformersConsider an ideal transformer (IT) with one impedance on the left side and define 2 sets of base quantities, one on each side of the IT . Define new units and examine the voltage on each side:

Assume that : 1 2bbS S=

2

11

2

b

b

V N aNV

= =and 1

2

'Va

V=Since

and:

11

1

''pub

VV

V = 22

2pu

b

VV

V = 1 2' pu puV V=

Z1 I2I1

V1 V'1 V2

An ideal transformer (IT)

1 1 1 1, , ,b b b bS V I Z 2 2 2 2, , ,b b b bS V I Z

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Current and P.U. In Transformers

Examine the current: 2

1

I aI

=

1

1

111

1p

b

b bu

IIIV

I S⋅

= =2

2

222

2p

b

b bu

IIIV

I S⋅

= =

1

2b

bVV

a= 1 2bbS S=

1 2pu puI I=

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Impedance and P.U. In TransformersExamine the impedance in p.u. quantities:

1 111 2

1 1

bpu

b b

Z SZZZ V

⋅= =

2 222 1b bV a V⋅ =

1 2b bS S=

1 2pu puZ Z=

1 222 122

2 2

bpu pu

b b

Z SZZ ZZ a V

⋅= = =

⋅2

2 1a Z Z⋅ =

1

2

b

b

Va

V=

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P.U. And Transformers• Define a different p.u. base system on each side of

transformers in the system• Define the same S-base (power)• Define voltage-bases according to the turns ratios• Then the ideal transformer can be

removed!!(i.e. the following formulas apply to the transformer

sides:)

1 2pu puZ Z=1 2pu puI I=1 2' pu puV V=

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The p.u. Transformer Circuit Model (d)

In a per unit system (without the ideal transformer) as a T-link

1V

−

+

1I 1ljX

mjb−cg

−

+

1R

Álag:

LZ

2R′2I′

cI mI

2ljX ′

2 2V aV′ =

−

+

1 2E aE=

0I

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In a per unit system (without the ideal transformer),as an L-link

The p.u. Transformer Circuit Model (e) - L circuit

1V

−

+1I

mjb−cg

−

+

Álag:

LZ

heild 1 2R R R′= +2

2I

Ia

′ =

cI mI

heild 1 2l ljX jX jX ′= +

2aV

0I

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The p.u. Transformer Circuit Model (f) - Impedance

In a per unit system (without the ideal transformer), as a impedance, R+jX

1V

−

+1 2I I ′=

−

+

Álag:

LZ

heildRheildjX

2 2V aV′ =

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The p.u. Transformer Circuit Model (g) - Reactance

In a per unit system (without ideal transformer), as a reactance, jX

1V

−

+1 2I I ′=

−

+

Álag:

LZ

heildjX

2 2V aV′ =

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Change of Base• We can change between base 1

and base 2. Assume in base 1:

• Similarly for base 2:

• Divide the 2nd equation into the 1st

• If the voltage bases are the same we get:

12

1 1

111

b

b bpu

ZZZS

Z V

⋅= =

22

2 2

222

b

b bpu

ZZZS

Z V

⋅= =

22 1

22

2

22b

bu

bp

bpu

V

SZ

S

VZ=

12

22pu pu

b

bZ

SZ

S=

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Summary of Transformer Circuit Models• The ideal transformer can be removed in p.u.• Impedances can be referred to either side• An accurate model accounts for

– winding resistances (real power losses)– hysteresis and eddy current real power losses– magnetization current (reactance)– leakage reactances

• The leakage reactance is the most important circuit model element

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17Lecture 08 Power Engineering - Egill Benedikt HreinssonA Phasor Diagrams for Voltages and Currents in the Transformer Circuits

1V

1 1jI X

1 1I R

2 2ljaI X2 2aI R

1 2E aE=2 2V aV′ =

cI

mI0I

φ

1I

2Ia

Lθ1θ

12δ

1V

−

+

1I 1ljX

mjb−cg

−

+

1R

Álag:

LZ

2R′2I ′

cI mI

2ljX ′

2 2V aV′ =

−

+

1 2E aE=

0I

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3 Phase Transformers

a

b

c

a

b

c

c

ab

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3 Phase Transformers• Can be either: A bank

of 3 identical single phase units with 3 separate cores …or….

• ...one unit with a single 3 phase core

aφ cφbφ

2φ

φ

2φ

corecoil corecoil

φ

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Principal Transformer Connections

• Y-Y connection

• Y-D (or D-Y) connection

a

b

c

a

b

c

c

ab

a

b

c

ab

c

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21Lecture 08 Power Engineering - Egill Benedikt Hreinsson3 Phase Transformer Symbols and Circuits

a

b

c

ab

c

a

b

c

2

4

6

a

b

c

1

3

5

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22Lecture 08 Power Engineering - Egill Benedikt HreinssonPlus or Minus 30° Phase Shift in Y-D Transformers

• +30 degrees phase shift

• -30 degrees phase shift

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Turns ratio for a 3 phase transformer

• The turns ratio for a 1 phase transformer is as follows:

• However the turns ratio for a 3 phase transformer is as follows where we assume symmetrical voltages both on the primary and secondary:

11

2f

V aV

=

1 1 13

2 2 2

a b cf

a b c

V V V aV V V

= = =

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Example of a Dy11 transformer

Phase shift +30 degreesDy11: The primary is always “12 o’clock” but the secondary is here at 11 o’clock

1 1

2 2

3 3

4 4

5 56 6

7 7

8 8

9 9

10 10

11 1112 12

A a

b

c

B

Forvaf (háspenna) Bakvaf (lágspenna)

C

A a

c

bB

C

Connection diagramPrimary (low voltage) Secondary (high voltage)

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Common transformer connections

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Transformer connections

Dy1 30o

Dd2 60o

Dd6 180o

Dy7 210o

Yd5 150o

Dd4 120o

Note: A fixed phase shift on each side of the transformer by a whole multiple of 30o does not affect the power flow but is a constant on top a variable phase shift, for instance:

1 2 1 212

sin( )( )

V VP

Xδ δ

δ⋅ −

=

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27Lecture 08 Power Engineering - Egill Benedikt HreinssonSystem stability with many generators, transformers and lines

Note: The power system behaves dynamically in a similar a similar manner as a system of weights (in a gravitational field) which are connected by elastic ties.These ties are the transformers an transmission lines. If a disturbance occurs or an oscillation is created somewhere it will create a perturbation throughout the systems. (The system can be described as a set of differential equations)

GeneratorsTransformers

Transmissionlines

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3 power system zones. An example one line diagram with standard voltages and phase shifts

220kV

T1 T2 T3

T4

132kV400kV

0o

+30o+90o

Notes: generators and loads not shown

What are the needed turns ratios and phase shifts for T1, T2, T3 and T4 ?

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Example 12Gefið er 3 fasa, 50 Hz raforkukerfi samkvæmt eftirfarandi mynd:

Kerfishlutarnir hafa eftirtalda eiginleika: Rafalinn G1 hefur

málspennu 38 kV, málraun 170 MVA og innra spanviðnám rafalans er 1.15 p.u. miðað við eigin málraun. (Sleppa má raunviðnámi rafala). Spennirinn T1 hefur málraun 130 MVA, umsetningu 38/220 kV og lekaspanviðnám 11% miðað við eigin málraun. (Sleppa má raunviðnámi spennis) Línan L1 hefur spankennt seríuviðnám sem nemur 39 ohm/fasa. (Sleppa má raunviðnámi háspennulínu) Álagið S er 100 MW með cosφ = 0.84 og er spankennt. Finnið spennuna á teini nr 1 ef spennan á teini nr 3 er 220 kV. Finnið einnig strauminn (í Amper) í hverjum fasa háspennulínunnar. Reiknið þessar stærðir út í einingakerfi miðað við aflgrunninn 150 MVA, en gefið svör í hefðbundnum einingum.

L1 T1 G1

2 1

S

3

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Example 13Til athugunar er raforkukerfi samkvæmt eftirfarandi mynd:

L1T1G1

S2 31

4

L2

S4

S3

T2

Fyrir rafalann gildir: Málraun 180 MVA, Mállspenna: 33kV; Samfasaspanviðnám: xd = 1.10 p.u. Fyrir spenninn T1 gildir: Umsetning 33kV/220kV. Skammhlaupsviðnám: 7%; Málraun 150 MVA. Fyrir spenninn T2 gildir: Umsetning 220/22 kV; Skammhlaupsspanviðnám: 5%. Málraun 100 MVA. Línurnar L1 og L2 eru eins. Fyrir hvora um sig gildir: Lengd 140 km; raðviðnám= 0.04+j0.35 ohm/fasa/km. Álagið S4 er hreint samviðnámsálag (spennuháð), sem við 22 kV er 60 MW og 45 Mvar. Nota skal einingakerfi með aflgrunni 100 MVA og athuga eftirfarandi 2 rekstrartilfelli þessa kerfis. a) Álagið S3 er spennuháð álag (hreint samviðnám), 70 MW með cosφ = 0.8. Hver þarf segulmögnunarspenna (E) rafalans að vera til að halda 220

kV stífri spennu á teini nr 2. Hvaða gildi tekur álagið S4 í þessu tilfelli og hver verður spennan á teini nr 4? Hver verður aflframleiðsla rafalans bæði raunafl og launafl í þessu tilfelli?.

b) Gert er ráð fyrir að á teini nr 2 sé 220 kV spenna og 33 kV spenna á teini nr 1. Rafalinn framleiðir 100 MW raunafl. Finnnið straum frá rafalanum (IG) og launaflsframleiðslu (QG) rafalans.

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Example 12 - solution

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Example 12 – solution (2)

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Example 13 solution (1)

12-Oct-11



the per unit system€¦ · lecture 08 power engineering - egill benedikt hreinsson 1 the per unit...

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