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Copyright © 2014 John Wiley & Sons, Inc. All rights reserved. Chapter 2-3 Transformers Third Edition P. C. Sen Principles of Electric Machines and Power Electronics

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Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

Chapter 2-3Transformers

Third Edition

P. C. Sen

Principles of Electric Machines

and

Power Electronics

Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

Auto transformer

Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.Fig_2-15

Per unit system

𝑆𝑏 = 𝑆𝑟𝑎𝑡𝑒

𝑉𝑏1 = 𝑉𝑟𝑎𝑡𝑒1 𝑉𝑏2 = 𝑉𝑟𝑎𝑡𝑒2

𝐼𝑏1 =𝑆𝑏

𝑉𝑏𝑎𝑠𝑒1=

𝑆𝑟𝑎𝑡𝑒𝑉𝑟𝑎𝑡𝑒1

= 𝐼𝑟𝑎𝑡𝑒1

𝐼𝑏2 =𝑆𝑏

𝑉𝑏𝑎𝑠𝑒2=

𝑆𝑟𝑎𝑡𝑒𝑉𝑟𝑎𝑡𝑒2

= 𝐼𝑟𝑎𝑡𝑒2

𝑍𝑏1 =𝑉𝑏1𝐼𝑏1

=𝑉𝑟𝑎𝑡𝑒1𝐼𝑟𝑎𝑡𝑒1

𝑍𝑏2 =𝑉𝑏2𝐼𝑏2

=𝑉𝑟𝑎𝑡𝑒2𝐼𝑟𝑎𝑡𝑒2

Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.Fig_1-1

Chapter 2 Main contents (continued)

•Main contentsBalanced three phase system

Three-phase transformer

Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.Fig_2-17

Balanced three-phase power system

• Y-connection with positive phase sequence

• 120 degree shift between any two lines

• Neutral connection

cnV

c

anV

bnV

a

n

bo

o

120 2 cos( )

120 2 cos( 120 )

120 2 cos( 120 )

a

b

c

v t

v t

v t

Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

a1 a2

b1

b2

c1

c2

a-b-c-a-b-c c-b-a-c-b-a

a b c c b a

Phase sequence

Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.Fig_2-17

Three-phase Y-source

• Y-connection: line to line variables

Line-to line voltages

Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

𝑉𝑎𝑏 = 𝑉𝑎𝑛 − 𝑉𝑏𝑛

= 2𝑉𝑟𝑚𝑠 cos𝜔𝑡 − 2𝑉𝑟𝑚𝑠 cos(𝜔𝑡 − 120°)

= 2𝑉𝑟𝑚𝑠 (cos𝜔𝑡 − cos(𝜔𝑡 − 120°))

= 2𝑉𝑟𝑚𝑠 (−2sin𝜔𝑡+(𝜔𝑡−120°)

2sin

𝜔𝑡−(𝜔𝑡−120°)

2)

= 2𝑉𝑟𝑚𝑠 (−2sin(𝜔𝑡 − 60°) sin 60°)

= 2𝑉𝑟𝑚𝑠 (2 cos(𝜔𝑡 + 30°)3

2)

= 2𝑉𝑟𝑚𝑠 3 cos(𝜔𝑡 + 30°)

Deriving line to line voltages

Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.Fig_2-17

Three-phase Y-source

•Phase voltage: voltage per source

•Line voltage: voltage between output terminals

•Phase current: current per source

•Line current: current of one transmission line between source and load

•Y-source: phase current = line current

YZ

cnV

c

anV

bnV

a

n

b

A

C

N

B

YZYZ

lZ

lZ

lZcCI

aAI

bBI

nI

0303

3 | || | cos 3 | || | cos

P P

P L

j

L L P

P P L L P

V I

I I

V V e

P V I V I

Three phase load

Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.Fig_2-17

Three-phase: Δ -source

0303

3 | || | cos 3 | || | cos

P P

j

L P

L L P

P P P L

V I

I I e

V V

P V I V I

= 𝑉𝑝𝑘 cos𝜔𝑡 + 𝑉𝑝𝑘 cos 𝜔𝑡 − 120° + 𝑉𝑝𝑘 cos 𝜔𝑡 + 120°

= 𝑉𝑝𝑘(cos𝜔𝑡 + cos 𝜔𝑡 − 120° + cos 𝜔𝑡 + 120° )

phase voltage = line voltage

= 𝑉𝑝𝑘(cos𝜔𝑡 + 2cos(2𝜔𝑡

2) cos

−240

2)

= 𝑉𝑝𝑘(cos𝜔𝑡 + 2cos(𝜔𝑡)(−1

2))

= 𝑉𝑝𝑘(cos𝜔𝑡 − cos(𝜔𝑡))

= 0

Where 𝐼𝐿𝑖𝑠 𝑙𝑖𝑛𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡, 𝐼𝑃𝑖𝑠 𝑝ℎ𝑎𝑠𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡

= 𝑉𝑎𝑏 + 𝑉𝑏𝑐 + 𝑉𝑐𝑎

Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.Fig_2-17

Δ-Y transformation of source

0( 30 )3

aba Y

VV

• Use Y source to replace Δ source

• Maintain same Vl-l and Il

𝐼𝑎 = 3𝐼𝑏𝑎𝑒𝑗(−30°)

𝑉𝑎𝑏 = 3𝑉𝑎𝑛𝑒𝑗30°

𝐼𝑎𝑏 =𝐼𝑏𝑎

3𝑒𝑗(30°)

Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.Fig_2-17

Three-phase load

• phase current = line current

• phase voltage=line to line voltage

Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.Fig_2-17

Δ-Y transformation of load

𝑍∆ =𝑉𝐴𝐵𝐼𝐴𝐵

= 3𝑒𝑗(−30°)𝑉𝐴𝐵𝐼𝑎𝐴

𝑉𝐴𝑁 =𝑉𝐴𝐵

3𝑒𝑗(−30°)

𝑍𝑌 =𝑉𝐴𝑁𝐼𝑎𝐴

=𝑒𝑗(−30°)

3

𝑉𝐴𝐵𝐼𝑎𝐴

𝑍𝑌𝑍∆

=1

3

• With same Vl-l and Il

• Find the relation between Y load and Δ load

Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.Fig_2-17

Balanced three-phase power system

• Per-phase analysis of balanced three-phase systemBalanced voltage sources

Balanced line and load impedances

Balanced resulting currents

For Y-source and Y-load (Y- Y), we have

aVYZ

lZ

n N

an bn cn 0V V V

a b c 0nI I I I

YZ

cnV

c

anV

bnV

a

n

b

A

C

N

B

YZYZ

lZ

lZ

lZcCI

aAI

bBI

nI

Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.Fig_2-17

Balanced three-phase power system

• How to analyze other types of three-phase source-power

connection (Page 14, reference materials)

Y- Δ

Δ-Y

Δ- Δ

•Solution: transform to Y-Y connection and apply per-phase

analysis

•During transformation, keep two variables unchanged

Line to line voltage

Line current

As a result: power is conservative

Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.Fig_2-17

Power of balanced three-phase loads

•Instantaneous power is constant, and equal to average

power

Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

Simulation: Power of balanced three-phase loads and

single phase load

1. Show phase voltage and phase current

2. Show line voltage

3. Show neutral voltage add to zero

4. Show single phase power/three phase power

Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.Fig_2-17

Example (Reference, Page 24)

A balanced 120 V-rms three-phase source with positive phase sequence. ZY1 = (30 + j40) Ω, ZΔ2 = (60 - j45)Ω. Zline = (2 + j4)Ω. Determine

(a.) the per phase line current

(b.) the total real and reactive power drawn from the source

(c.) the per phase line-to-neutral voltage across the parallel load

(d.) the per phase load current in the wye and delta loads

(e.) the total real and reactive three-phase power delivered to each load and the line.