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Electric Ship Design Lecture 5

<Dr Ahmed El-Shenawy>

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

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Procedure for Per Unit Analysis

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

BaseV

BBBB SZVI ,,,

1

1

Base

BaseBase

V

SI

1

1

1

Base

Base

BaseI

VZ

21 :VV 32 :VV 43 :VV

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Example 5.14, p. 164-166

• 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

8.132

B

2llB

B1

1 M

k

S

VZ

190410

1382

B

2llB

B2

2 M

k

S

VZ

47610

692

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

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Step 4: All in Per Unit Quantities

+- new

B

oldB

oldp.u.new

p.u.Z

ZZZ

183.0

04.19

52.131.02

.p.u,1

MkXl

p.u.08.02 lX

1011025.51904

10010 3

B

linep.u.line,

2

jj

Z

ZZ

0913.08.13

2.13

1B

g

p.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,V63.0p.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

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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

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Symmetrical Three Phase Fault Analysis

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