Per Unit System

The Per Unit System

• Allows engineers to analyze a single phase network where:– All P and Q quantities are three phase– Voltage magnitudes are represented as a fractional

part of their standard or “base” value– All phase angles are represented in the same units as

normally used

3

Advantages

1. Per-unit representation results in a more meaningful and correlated data. It gives relative magnitude information. 2. There will be less chance of missing up between single - and three-phase powers or between line and phase voltages. 3. The p.u. system is very useful in simulating machine systems on analog, digital, and hybrid computers for steady-state and dynamic analysis. 4. Manufacturers usually specify the impedance of a piece of apparatus in p.u. (or per cent) on the base of the name plate rating of power (P) and voltage (V). Hence, it can be used directly if the bases chosen are the same as the name plate rating. 5. The p.u. value of the various apparatus lie in a narrow range, though the actual values vary widely. 6. The p.u. equivalent impedance (Zsc) of any transformer is the same referred to either primary or secondary side. For complicated systems involving many transformers or different turns ratio, this advantage is a significant one in that a possible cause of serious mistakes is removed. 7. Though the type of transformer in 3-phase system, determine the ratio of voltage bases, the p.u. impedance is the same irrespective of the type of 3-phase transformer. (Yç D , D ç Y, D ç D , or Yç Y) 8. Per-unit method allows the same basic arithmetic operation resulting in per-phase end values, without having to worry about the factor '100' which occurs in per cent system.

4

Conversion Procedure

-Specify the MVA base. Typically this will be related to the rating of a generator, transformer, or transmission line. Just choose the one that will result in the least amount of computation. This base will remain constant throughout the system.

-At any location in the circuit, specify a voltage base. This will typically be the nominal voltage for that particular location.

-Determine the voltage base for all other areas in the circuit by adjusting by the turns ratio every time a transformer is encountered.

-Having specified the voltage and MVA base throughout the system, current and impedance bases may be determined as:

-For each value, the per unit quantity is the actual value divided by the base value.

-For 3phase circuits, the following relationships must also be included:

5

Set Up the Per Unit System

• Each region of the power system is uniquely defined by a standard voltage determined by the transformer windings, this sets base voltage

• The entire system is given a base power to which everything in the power flow is referred

Procedure for Per Unit Analysis

1. Pick for the system.

2. Pick according to line-to-line voltage.

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

How to Choose Base Values ?

• Divide circuit into zones by transformers.• Specify two base values out of ;

for example, and • Specify voltage base in the ratio of zone line to

line voltage.

Source

Zone 1 Zone 2 Zone 3 Zone 4

BaseS

1BaseV2BaseV

3BaseV4BaseV

BaseVBBBB SZVI ,,,

1

1

Base

BaseBase V

SI

1

1

1

Base

BaseBase I

VZ

21 :VV 32 :VV 43 :VV

Example - 1

• Given a one line diagram,

Find , , , , and .gI loadV loadP

~5 MVA

13.2 Δ – 132 Y kV

10 MVA

138 Y - 69 Δ kV

10010line jZgI

p.u.1.01 lX p.u.08.02 lX

kVVg 2.13

300loadZ

loadIline-tI

Step 1, 2, and 3: Base Values

Zone 1 Zone 2 Zone 3

MVAS 10B

kVV 8.131B kVV 138

2B kVV 69

3B

04.1910

8.13 2

B

2llB

B1

1 M

k

S

VZ

190410

138 2

B

2llB

B2

2 M

k

S

VZ

47610

69 2

B

2llB

B3

3 M

k

S

VZ

4.4188.133

10

3 l-lB

3B

B

1

1

1

k

M

V

SI 84.41

1383

10

3 l-lB

3B

B

2

2

2

k

M

V

SI 67.83

693

10

3 l-lB

3B

B

3

3

3

k

M

V

SI

~5 MVA

13.2 Δ – 132 Y kV

10 MVA

138 Y - 69 Δ kV

10010line jZgI

p.u.1.01 lX p.u.08.02 lX

kVVg 2.13

300loadZ

Step 4: All in Per Unit Quantities

+- new

B

oldB

oldp.u.new

p.u. Z

ZZZ

183.0

04.19

52.131.0 2

.p.u,1

MkX l

p.u.08.02 lX

1011025.51904

10010 3

B

linep.u.line,

2

jj

Z

ZZ

0913.08.13

2.13

1B

gp.u.g, kV

kV

V

VV

63.0476

300

3B

loadp.u.load,

Z

ZZ

Step 5: One Phase Diagram & Solve

+-

183.0.p.u,1 lX 08.02 lX 1011025.5 3p.u.line, jZ

0913.0p.u.g,V 63.0p.u.load, Z

4.2635.1

4.26709.0

096.0

p.u.total,

p.u.g,p.u.load, Z

VI

4.2635.1p.u.load,p.u.line,-tp.u.g, III

4.268505.0p.u.load,p.u.load,p.u.load, ZIV

148.1*p.u.load,p.u.load,p.u.load, IVS

Step 6: Convert back to actual quantities

Zone 1 Zone 2 Zone 3

~5 MVA

13.2 Δ – 132 Y kV

10 MVA

138 Y - 69 Δ kV

10010line jZgI

p.u.1.01 lX p.u.08.02 lX

kVVg 2.13

300loadZ

1Bp.u.g,g III 2Bp.u.line,-tline-t III 3Bp.u.load,load III

3Bp.u.load,load VVV

Bp.u.load,load SSS

4.2635.1p.u.load,p.u.line,-tp.u.g, III

4.268505.0p.u.load,V

148.1p.u.load, S

Advantage of per unit calculation

• Simplify calculation by eliminating transformers.• Helps to spot data errors

– p.u. is more uniform compare to actual impedance value of different sizes of equipment.

• Helps to detect abnormality in the system– Operator at control center can spot over/under

voltage/current rating easily.