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POWER APPARATUS MODELING AND PER UNIT SYSTEMS System Modeling Generator , Transformer and Load Modeling Per Unit (PU) Analysis 17-Jan-11 EE 308 Power Systems 18

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POWER APPARATUS MODELING AND PER UNIT

SYSTEMS

System Modeling

Generator, Transformer and Load Modeling

Per Unit (PU) Analysis

17-Jan-11 EE 308 Power Systems 18

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System Modeling

Systems are represented on a per-phase basis

• A 1-φ representation is used for a balanced system

– the system is modeled as one phase of a Y-connected network 

Symmetrical components are used for unbalanced

systems

• unbalance systems may be caused by: generation,

network components, loads, or unusual operating

conditions such as faults

The per-unit (PU) system of measurements is used

Review of basic network component models

• Generators, Transformers, Loads, and Transmission

lines

17-Jan-11 EE 308 Power Systems 19

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Generator Models

17-Jan-11 EE 308 Power Systems 20

Generator may be modeled in three different ways:

• Power Injection Model - the real, P, and reactive, Q, power of the generator is specified at the node that the generator is

connected

–  either the voltage or injected current is specified at the

connected node, allowing the other quantity to be determined

• Thevenin Model - induced AC voltage, E  , behind the

synchronous reactance, X  d  

• Norton Model - injected AC current, I G , in parallel with thesynchronous reactance, X  

d  

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Transformer Model

Transformer equivalent circuit, with secondary impedancesreferred to the primary side

17-Jan-11 EE 308 Power Systems 21

Figure source: wikipedia

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Load Models

Models are selected based on both the type of analysis and

the load characteristics. Three main static load models are: Constant impedance, Z

load 

–  Load is made up of  R, L, and C elements connected to a network node

and the ground (or neutral point of the system)

Constant current, I load –  The load has a constant current magnitude I , and a constant power factor,

independent of the nodal voltage

–  Also considered as a current injection into the network 

Constant power (PQ) , Sload 

–  The load has a constant real, P, and reactive, Q, power component

independent of nodal voltage or current injection

–  Also considered as a negative power injection into the network 

17-Jan-11 EE 308 Power Systems 22

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Per Unit (PU) Analysis (1/4)

17-Jan-11 EE 308 Power Systems 23

P.U. : ratio of the actual quantity to its base values

1-Ø system

Base

Base

Base

Base

S MVA

V kV 

3-Ø system

AV 

SI 

Base

Base

Base

1000

Base

Base

Base

Base

Base

S

2

1000

2

Base

PU Actual

Base

SZ Z 

Base

Base

Base

Base

Base

S

2

3

1000

AV 

SI 

Base

Base

Base

3

1000

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Per Unit (PU) Analysis (2/4)

17-Jan-11 EE 308 Power Systems 24

Changing the base of PU quantities

2

2

Old Old  ActualNew PU Base

PU  New New

Base Base

New Old 

Base BaseOld 

PU  Old  NewBase Base

Z  Z Z Z 

Z Z 

S V Z 

S V 

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Per Unit (PU) Analysis (3/4)

17-Jan-11 EE 308 Power Systems 25

1. Pick for the whole system

2. Pick arbitrarily (according to line-to-line voltage).

Relate all the others by transformer ratio.

3. Calculate for different zones.

4. Express all quantities in P.U.

5. Draw impedance diagram and solve for P.U. quantities.

6. Convert back to actual quantities if needed.

BaseS

BaseV 

BaseZ 

Steps for a PU analysis

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Per Unit (PU) Analysis (4/4)

Divide circuit into zones by transformers.

Specify two base values out of ; for example,

and

Specify voltage base in the ratio of zone voltage (L-L).

17-Jan-11 EE 308 Power Systems 26

BaseSBaseV 

BBBB SZ V I  ,,,

Source

Zone 1 Zone 2 Zone 3 Zone 4

1BaseV 2BaseV 

3BaseV 4BaseV 

21 :V V  32 :V V  43 :V V 

1

1

Base

Base

Base

SI 

V  1

1

1

Base

Base

Base

V Z 

How to Choose Base Values ?

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Example 5.14, p. 164-166

17-Jan-11 EE 308 Power Systems 27

Given a one line diagram of 

the 3-φ system,

gI  loadV  loadPloadI Tr I Find , , , , and .

“Power system analysis” by

A.R. Bergen and V. Vittal

~ 5 MVA

13.2 Δ – 132 Y kV

10 MVA

138 Y - 69 Δ kV

line 10 100Z j g

1 0.1 p.u.T X 

2 0.08 p.u.T X 

( ) 13.2g L LV kV 

load 300Z 

Tr I 

Load I 

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Example 5.14 (Cont’d)

17-Jan-11 EE 308 Power Systems 28

Zone 1 Zone 2 Zone 3

3 10BS MVA 

1B 13.8L LV kV 

B2

138L LV kV 

3B 69V kV 

1

1

22L L

B

B

B

13.819.04

10

V  kV Z 

S MVA

2

2

22

B

B

B

1381904

10

L LV  kV Z 

S MVA

3

3

22

B

B

B

69476

10

L LV  kV Z 

S MVA

Step 1, 2, and 3: Base Values

~5 MVA

13.2 Δ – 132 Y kV

10 MVA

138 Y - 69 Δ kV

line 10 100Z j gI 

1 0.1 p.u.T X  2 0.08 p.u.

T X 

( ) 13.2g L LV kV 

load 300Z 

Tr I 

Load I 

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Example 5.14 (Cont’d)

17-Jan-11 EE 308 Power Systems 29

Step 4: All per unit quantities

+

-

2

1,p.u. 2

13.2100.1 0.183

5 13.8T 

kV MVAX 

MVA kV 

2 0.08p.u.T X 

2

3lineline,p.u.

B

10 1005.25 10 1 10

1904

Z  jZ j

1

gg,p.u.

B

13.2 0.9565 013.8

V  kV V V kV 

3

loadload,p.u.

B

300 0.63476

Z Z Z 

2

2

new old  

base basenew old  pu pu old  new

base base

S V Z Z S V 

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Example 5.14 (Cont’d)

17-Jan-11 EE 308 Power Systems 30

Step 5: One phase diagram & solve

+

-

1,p.u.0.183T X 

2 0.08T X  1011025.5 3

p.u.line, jZ 

g,p.u. 0.9565 0V 

63.0p.u.load, Z 

g,p.u.

load,p.u.

total,p.u.

0.9565 01.35 26.4

0.709 26.4

V I 

g,p.u. Tr,p.u. load,p.u. 1.35 26.4I I I 

load,p.u. load,p.u. load,p.u. 0.8505 26.4V I Z 

*

load,p.u. load,p.u. load,p.u. 1.1474S V I 

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Example 5.14 (Cont’d)

17-Jan-11 EE 308 Power Systems 31

Zone 1 Zone 2 Zone 3

~line 10 100Z j

gI 

kV V g 2.13

load 300Z 

1g g,p.u. BI I I 

2,p.u. BTr Tr  I I I 

3load load,p.u. BI I I 

3load load,p.u. BV V V 

load load,p.u. BS S S

g,p.u. Tr,p.u. load,p.u. 1.35 26.4I I I 

load,p.u. 0.8505 26.4V 

load,p.u. 1.1474S

Step 6: Convert back to actual quantities

5 MVA

13.2 Δ – 132 Y kV

1 0.1 p.u.T X 

10 MVA

138 Y - 69 Δ kV

2 0.08 p.u.T X 

1

1

3

B L L

B

BS

I  V 

 

2B

13.2 418.4 41.84132

6

3

10 10 3418.4

3 13.8 10

3B

138

41.84 83.6769I 

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Advantages of P.U. System

P.U. representation results in a more meaningful data. It 

gives a clear idea of relative magnitudes of various

quantities.

It is more uniform compare to actual impedance value of 

different sizes of equipment 

It is very useful in simulating power systems for steady-

state and dynamic analysis.

The P.U. equivalent impedance, voltages, and currents of 

any transformers are the same referred to either primary

or the secondary side.

–  Different voltage levels disappear across the entire system.

–  The system reduces to a system of simple impedances

–  P.U. impedance is the same irrespective of the type of 3-φ

transformer

17-Jan-11 EE 308 Power Systems 32