pdf per unit examples compress

The Per Unit System (Examples) The Per Unit System (Examples) Example 1

Example 1

A three-phase, wye-connected system is rated at 50MVA and 120kV.

A three-phase, wye-connected system is rated at 50MVA and 120kV.

Express 40MVA of three-phase apparent power as a per unit

Express 40MVA of three-phase apparent power as a per unit valuevalue referred to:

referred to:

a)

a)  The t The threehree-phas-phase syse system tem as Mas MVA VA basebase, and, and  b)

 b)  The p The per-per-phase hase systesystem MVm MVA as A as base.base.

Solution Solution

a)

a) For the three-phase system MVA as For the three-phase system MVA as base,base,

 pu S   pu

S  S  S  S 

S 

kV  kV  V 

V   MVA

 MVA S 

S 

 B  B actual  actual   pu

 pu

line line  B  B  B

 B

8 8 ..

0 50 0 50 40 40

120 120 50

50

3 3 3

3

)) ((

3 3

==

==

==

==

==

φ  φ  φ 

φ  φ  φ 

b)

b) FoFor thr the pee per phr phase ase basbasee

 pu  pu S 

S 

kV  k  kV 

V  k  V  V 

V 

 MVA  M   MVA

 M  S 

S  S  S 

 pu  pu

line line  B  B  phase

 phase  B  B

 B  B  B

 B

8 8 ..

0 67 0 67 ..

16 16

40 40 3

3 1 1

28 28 ..

69 3 69 3 120 120 3

3

67 67 ..

16 3 16

3 50 50 3

3

1 1

)) ((

)) ((

3 3 1

1

=

=

×

×

=

=

=

=

=

=

=

=

=

=

=

=

=

=

φ  φ 

φ  φ  φ 

φ 

Example 2 Example 2

Find the per unit value for X

Find the per unit value for X T1 T1, X, X T2 T2 and Xand X T T if the base values are 11kV andif the base values are 11kV and 60MVA.

60MVA.

G G

T2 T2 X

X_TT=20=20ΩΩ T1

T1

60 MVA 60 MVA 12/132kV 12/132kV

X

X_T1T1=10%=10%

30 MVA 30 MVA 132/33kV 132/33kV

X

XT2T2=10%=10%

60 MVA 60 MVA 11kV 11kV X X_GG=0.5=0.5 E

E_GG=1.5pu=1.5pu

Solution Solution

Step 1: Draw the section of

Step 1: Draw the section of the network.the network.

1 1

G G

T2 T2 X

X_TT=20=20ΩΩ T1

T1

60 MVA 60 MVA 12/132kV 12/132kV

X

X_T1T1=10%=10%

30 MVA 30 MVA 132/33kV 132/33kV

X

X_T2T2=10%=10%

60 MVA 60 MVA 11kV 11kV X X_GG=0.5=0.5 E

E_GG=1.5pu=1.5pu

Step 2: Find the base value of the voltage for each section.

Step 2: Find the base value of the voltage for each section.

kV  kV  kV 

kV  V 

V  V  V  V  V  V 

V 

kV  kV  kV 

kV  V 

V  V  V  V  V  V 

V 

kV  kV  V 

V 

 MVA  MVA S 

S 

 B  B  B

 B

 B  B  B

 B  B  B  B  B

25 25 ..

30 30 121

132 121 132 33 33

121 121 11

12 11 12 132 132 11

11 60 60

2 2 2

2 3 3 3 3

1 1 1

1 2 2 2 2 1 1

==

××

==

××

==

==

××

==

××

==

==

==

Step 3: Find the per unit values of each component Step 3: Find the per unit values of each component  Tran

 Transformsformer T1 er T1 and Tand T22

Since both transformer voltage base are the same as

Since both transformer voltage base are the same as their rated values,their rated values, their p.u reactance on a

their p.u reactance on a 60MVA are:60MVA are:

 

   

   

 

 

  

 

 

 

××  

 

   

   

 

 

  

 

 

 

××  

==

_old old 

 B  B

new new  B  B new

new  B  B

old  old   B old   B

old   pu  pu new

new  pu

 pu S S 

S  S  V 

V  V   Z  V 

 Z   Z 

 Z 

2 2

 pu  pu  X 

 X _T T ^newnew 00..1212 60

60 60 60 11

11 12 1 12 1 ..

0 0

2 2 1

1

    ==

 

 

 

  

 

 

 

××  

 

   

 

 

  

 

 

 

××  

==

:: Looking at the LV side of T1:: Looking at the LV side of T1 OR

OR

 pu  pu  X 

 X ^newnew

T  T 

12 12 ..

0 0 60 60 60 60 121

121 132 1 132 1 ..

0 0

2 2

1

1

        ==

 

 

 

 

 

 

××

 

 

 

 

 

 

 

 

 

 

××

==

:: Looking at the HV side of T1 :: Looking at the HV side of T1

 pu  pu  X 

 X _T T ^newnew 00..2424 30

30 60 60 121

121 132 1 132 1 ..

0 0

2 2 2

2

    ==

 

 

 

  

 

 

 

××  

 

   

 

 

  

 

 

 

××  

==

:: Looking at the HV side of T2:: Looking at the HV side of T2 OR

OR

 pu  pu  X 

 X _T T ^newnew 00..2424 30

30 60 60 25

25 ..

30 30

33 1 33

1 ..

0 0

2 2 1

1

    ==

 

 

 

  

 

 

 

××  

 

   

 

 

  

 

 

 

××  

==

:: Looking at the LV side of T2:: Looking at the LV side of T2

NOTE: The per unit value for the transformer impedance is the same whether it is seen NOTE: The per unit value for the transformer impedance is the same whether it is seen

in the HV or LV side of the transformer- one of the advantages of per unit system.

in the HV or LV side of the transformer- one of the advantages of per unit system.

Line Impedance Line Impedance

Since the impedance given in actual value, we have to find the base value Since the impedance given in actual value, we have to find the base value for the i

for the impedance.mpedance.

( ( )) ( ( ))

 pu  X   pu

 X   X   X   X 

 X 

 M   M  k  k  S 

S  V   X  V 

 X 

base base T  T 

actual  actual  T  T   pu

 pu T  T 

 B  B  B  B base

base T  T 

082 082 ..

0 244 0 244 20 20

244 60 244

60 121 121

*

*

)) ((

)) ((

)) ((

2 2 2

2 2 2 ))

((

=

=

=

=

=

=

Ω Ω

=

=

=

=

=

=

Example 3 Example 3  The on

 The one-line diae-line diagram ogram of three-pf three-phase pohase power syswer system is shtem is shown beown below. Selelow. Selectct a

a common base of common base of   100 MVA and 22 kV on the generator side100 MVA and 22 kV on the generator side. Draw. Draw an

an imimpepedadancnce e didiagagraramm wwitith h alall l imimpepeddanance ce ininclclududining g ththe e loloadad impedance marked in per-unit.

impedance marked in per-unit.

Solution Solution

Step 1: Draw the zone in the circuit diagram Step 1: Draw the zone in the circuit diagram

3 3

G

G M M

1

1 22

3 3

4 4 T

T₁₁ TT₂₂

T

T₃₃ TT₄₄

Line 1 Line 1 220 kV 220 kV

Line 2 Line 2 110 kV

110 kV LoadLoad

90 MVA 90 MVA 22 kV 22 kV X = 18%

X = 18%

50 MVA 50 MVA 22/220 kV 22/220 kV X = 10%

X = 10%

40 MVA 40 MVA 22/110 kV 22/110 kV X = 6.4%

X = 6.4%

40 MVA 40 MVA 110/11 kV 110/11 kV X = 8.0%

X = 8.0%

40 MVA 40 MVA 220/11 kV 220/11 kV X = 6.0%

X = 6.0%

66.5 MVA 66.5 MVA 10.45 kV 10.45 kV X = 18.5%

X = 18.5%

57 MVA 57 MVA 0.6 pf lag 0.6 pf lag 10.45 kV 10.45 kV X = 48.4

X = 48.4ΩΩ

X = 64.43 X = 64.43ΩΩ

Note that the line impedance has only the resistive value.

Note that the line impedance has only the resistive value.

Therefore the complex power conjugate value is the same since

Therefore the complex power conjugate value is the same since ^θθ is equal to 0.is equal to 0.

G

G MM

1

1 22

3 3

4 4 T

T11 TT22

T

T₃₃ TT44

Line 1 Line 1 220 kV 220 kV

Line 2 Line 2 110 kV

110 kV LoadLoad

90 MVA 90 MVA22 kV22 kV X = 18%

X = 18%

50 MVA 50 MVA 22/220 kV 22/220 kV X = 10%

X = 10%

40 MVA 40 MVA 22/110 kV 22/110 kV X = 6.4%

X = 6.4%

40 MVA 40 MVA 110/11 kV 110/11 kV X = 8.0%

X = 8.0%

40 MVA 40 MVA 220/11 kV 220/11 kV X = 6.0%

X = 6.0%

66.5 MVA 66.5 MVA 10.45 kV 10.45 kV X = 18.5%

X = 18.5%

57 MVA 57 MVA 0.6 pf lag 0.6 pf lag 10.45 kV 10.45 kV X = 48.4

X = 48.4ΩΩ

X = 64.43 X = 64.43ΩΩ

Step 2: Find the base voltage for each zone.

Step 2: Find the base voltage for each zone.

kV  kV  k 

k  V 

V  V  V  V  V  V 

V 

kV  kV  k 

k  V 

V  V  V  V  V  V 

V 

kV  kV  k 

k  V 

V  V  V  V  V  V 

V 

kV  kV  V 

V 

b b b

b

b b b

b

b b b

b b b

11 11 220

220 220 220 11 11

110 110 22

22 22 22 110 110

220 220 22

22 22 22 220 220 22

22

2 2 2 2 4 4 4 4

1 1 1 1 3 3 3 3

1 1 1 1 2 2 2 2 1 1

==

××

==

××

==

==

××

==

××

==

==

××

==

××

==

==

Step 3: Find the per unit value Step 3: Find the per unit value Generator and T

Generator and Transformransformerer

Since generator & transformer voltage base are the same as their

Since generator & transformer voltage base are the same as their ratedrated values, their p.u reactance on a 100 MVA

values, their p.u reactance on a 100 MVA

Motor Motor

 pu  pu  Z 

 Z 

 pu  pu  Z 

 Z 

 pu  pu  Z 

 Z 

 pu  pu  Z 

 Z 

 pu  pu  Z 

 Z 

S  S  S   Z  S   Z   Z 

 Z 

S  S  S  S  V 

V  V   Z  V 

 Z   Z 

 Z 

new new new new T  T 

new new T  T 

new new T  T 

new new G G

old  old   B  B

new new  B old   B old   pu  pu new

new  pu  pu

old  old   B  B

new new  B  B new

new  B  B

old  old   B old   B old   pu  pu new

new  pu  pu

2 2 ..

0 100 0 08 100 08 ..

0 0

16 16 ..

0 40 0

40 100 064 100 064 ..

0 0

15 15 ..

0 40 0

40 100 06 100 06 ..

0 0

2 2 ..

0 50 0

50 100 10 100 10 ..

0 0

2 2 ..

0 90 0

90 100 18 100 18 ..

0 0

3 3 2 2 1 1

2 2

  ==

   

  

 

==  

  ==

   

 

 

  

 

 

 

==  

  ==

   

 

 

  

 

 

 

==  

  ==

   

 

 

  

 

 

 

==  

  ==

   

 

 

  

 

 

 

==  

 

   

   

 

 

  

 

 

 

==  

 

   

   

 

 

  

 

 

 

   

   

   

 

 

  

 

 

 

==  

 pu  pu

S  S  S  S  V 

V  V   Z  V 

 Z   Z 

 Z   Z   Z 

kV  kV  V 

V  kV 

kV  V 

V 

 MVA  MVA S 

S   MVA

 MVA S 

S 

old  old   B  B

new new  B  B new

new  B  B

old  old   B old   B

old  motor  motor  new

new motor  motor  old  old  motor  motor 

old  old   B  B new

new  B  B

old  old   B  B new

new  B  B

25 25 ..

0 0

5 5 ..

66 66 100 100 11

11 45 45 ..

10 10 18

18 ..

0 0

18 18 ..

0 0

45 45 ..

10 10 11

11

5 5 ..

66 66 100

100

2 2 2 2

==

 

   

 

 

 

  

 

 

××

 

   

 

 

 

  

 

 

××

==

  ××

   

   

 

 

  

 

 

 

××  

==

==

==

==

==

==

Line Impedance Line Impedance Line 1

Line 1

( ( )) _{( ( ))}

 pu  Z   pu

 Z   Z   Z   Z 

 Z 

 M   M  k  k  S 

S  V   Z  V 

 Z 

 MVA  MVA S 

S 

kV  kV  V 

V 

line line  B  B

actual  actual   pu

 pu line line

 B  B  B  B line

line  B  B  B  B  B  B

1 1 ..

0 484 0 484 4 4 ..

48 48

484 100 484

100 220 220 100

100 220 220

1 1 ))

((

1 1

2 2 2

2 2 2 1

1 2 2

=

=

=

=

=

=

Ω Ω

=

=

=

=

=

=

=

=

=

= Line 2Line 2

( ( )) ( ( ))

 pu  Z   pu

 Z   Z   Z   Z 

 Z 

 M   M  k  k  S 

S  V   Z  V 

 Z 

 MVA  MVA S 

S 

kV  kV  V 

V 

line line  B  B

actual  actual   pu

 pu line line

 B  B  B  B line

line  B  B  B  B  B  B

54 54 ..

0 121 0 121 43 43 ..

65 65

121 100 121

100 110 110 100

100 110 110

2 2 ))

((

2 2

2 2 2

2 3 3 2

2 3 3

==

==

==

Ω Ω

==

==

==

==

==

Load Load

Step 4: Draw the per unit

Step 4: Draw the per unit impedance diagram.impedance diagram.

5 5

( ( )) (

( ) ) ( ) ( )

 pu  pu  j

 j  j  j  Z 

 Z   Z   Z   Z 

 Z 

 M   M  k  k  S 

S  V   Z  V 

 Z 

 j  M   j

 M  k  k  S 

S  V   Z  V 

 Z 

 MVA  MVA S 

S 

lagging  lagging   pf  

 pf   kV  kV  V 

V   MVA  MVA S 

S 

base base load  load 

act  act  load  load   pu

 pu load  load 

 B  B  B  B base

base load  load 

load  load   L  L act 

act  load  load 

load  load 

 L  L

267 267 ..

1 1 95 95 ..

0 21 0

21 ..

1 1

5327 5327 ..

1 1 1495 1495 ..

1 1

21 21 ..

1 100 1

100 11 11

5327 5327 ..

1 1 1495 1495 ..

1 13 1 13 ..

53 53 57

57

45 45 ..

10 10

*

* 13 13 ..

53 53 57 57

13 13 ..

53 53 6 6 ..

0 0 cos cos

6 6 ..

0 0 45

45 ..

10 10 57

57

)) ((

)) ((

)) ((

2 2 2

2 4 4 ))

((

2 2

3 3

2 2 ))

((

3 3

1 1 3

3

+ +

= Ω =

Ω

Ω Ω +

= +

=

=

=

Ω Ω

=

=

=

=

=

=

Ω Ω +

+

=

°° =

∠

= ∠

=

=

=

°°

∠

∠

=

=

°°

=

=

=

=

=

=

=

=

=

=

−

−

φ  φ  φ 

φ  φ  φ 

θ  θ 

G

G MM

X

XGG= j0.2p.u= j0.2p.u

X

XT1T1==jj00..22pp..uu XXT2T2= j0.15p.u= j0.15p.u

X

XT3T3= j0.16p.u= j0.16p.u XXT4T4= j0.20p.u= j0.20p.u Z

ZL1L1= j0.10p.u= j0.10p.u

Z

ZL1L1= j0.54p.u= j0.54p.u

Z

Zmotor motor = j0.25p.u= j0.25p.u

Z

ZLoadLoad= 0.95+j1.267= 0.95+j1.267

Example 4 Example 4

Using base values of 30kVA and 240V in the generator side, draw the per Using base values of 30kVA and 240V in the generator side, draw the per uni

unit t circcircuit, and uit, and detdetermermine ine the per the per uniunit t imimpedpedancances es and the and the per unitper unit source voltage. Then calculate the load current both in per unit and in source voltage. Then calculate the load current both in per unit and in amperes. Transformer winding resistance and shunt admittance branches amperes. Transformer winding resistance and shunt admittance branches are neglected.

are neglected.

G G

T2 T2 X

X_lineline=2=2ΩΩ T1

T1

30 kVA 30 kVA 240/480V 240/480V X

X_T1T1=0.1pu=0.1pu

20 kVA 20 kVA 460/115V 460/115V X

XT2T2=0.1pu=0.1pu 30 kVA

30 kVA V

V_GG=220=220∟0∟0ºVºV

Z

Z_loadload=0.9+j0.2=0.9+j0.2ΩΩ

Solution Solution

Step 1: Draw the zone in

Step 1: Draw the zone in the network.the network.

G G

T2 T2 X

Xlineline=2=2ΩΩ T1

T1

30 kVA 30 kVA 240/480V 240/480V X

XT1T1=0.1pu=0.1pu

20 kVA 20 kVA 460/115V 460/115V X

XT2T2=0.1pu=0.1pu 30 kVA

30 kVA V

VGG=220=220∟0∟0ºVºV

Z

Z_loadload=0.9+j0.2=0.9+j0.2ΩΩ

Step 2: Draw the per unit circuit Step 2: Draw the per unit circuit

G G

V VS(pu)S(pu)

X

X_T1(pu)T1(pu) ZZline(pu)line(pu) XXT2(pu)T2(pu)

Z Z_{load(pu)load(pu)} II_loadload(pu)(pu)

II_SS(pu)(pu)

Step 3: Find the base values Step 3: Find the base values S

SBB for the entire network is for the entire network is 30kVA. Find the base voltages and 30kVA. Find the base voltages and impedancesimpedances of each zone.

of each zone.

Ω Ω

==

==

==

××

==

Ω Ω

==

==

==

××

==

××

==

Ω Ω

==

==

==

==

48 48 ..

0 30 0

30 120 120 120

120 480

460 480 460 115 115

68 68 ..

7 30 7

30 480 480 480

480 240

240 240 240 480 480

92 92 ..

1 30 1

30 240 240 240

240

2 2 3

3 3

3

2 2 2

2 1

1 1

1 2 2 2 2

2 2 1

1 1

1

2 2

k   Z  k 

 Z  V 

V  V 

V 

k   Z  k 

 Z  V 

V  V 

V  V  V  V  V  V 

V 

k   Z  k 

 Z  V 

V  V 

V 

S  S  V   Z  V 

 Z 

 B  B  B

 B

 B  B  B

 B  B

 B

 B  B  B

 B

 B  B  B  B  B

 B

Find base current in zone 3 [!:to calculate the load current later]

Find base current in zone 3 [!:to calculate the load current later]

 A k   A

k  V 

V  S   I  S 

 I 

 B  B  B  B  B

 B 250250

120 120 30 30

3 3 3

3 == == ==

Step 4: Find the per unit values Step 4: Find the per unit values

7 7

 pu  pu  j

 j  j  j  ZB

 ZB  Zload   Zload 

 pu  pu S 

S  S  S  V 

V  V   X  V 

 X   X 

 X 

or  or 

 pu  pu S 

S  S  S  V 

V  V   X  V 

 X   X 

 X 

 pu  Z   pu

 Z   X   X   X 

 X 

 pu  pu  X 

 X   X 

 X 

 pu V   pu

V  V  V  V 

V 

old  old   B  B

new new  B  B new

new  B  B

old  old   B old   B

old  T  T  new

new T  T 

old  old   B  B

new new  B  B new

new  B  B

old  old   B old   B

old  T  T  new

new T  T 

 B  B line line  pu

 pu line line

old  old  T  T  new

new T  T 

 B  B  s  s  pu

 pu  s  s

4167 4167 ..

0 0 875 875 ..

1 48 1

48 ..

0 0

2 2 ..

0 0 9 9 ..

0 0 3 Z 3

Z

ratio ratio voltage voltage using

using n

n calculatio calculatio ::

::

1378 1378 ..

0 20 0 20 30 30 120

120 115 1 115 1 ..

0 0

value value  base

 base using using n

n calculatio calculatio ::

::

1378 1378 ..

0 20 0 20 30 30 480

480 460 1 460 1 ..

0 0

2604 2604 ..

0 68 0 68 ..

7 7

2 2 1 1 ..

0 0

0 0 9167 9167 ..

0 240 0

240 0 0 220 220

load(pu) load(pu)

2 2

2 2 2

2 2

2

2 2

2 2 2

2 2

2

2 2 ))

((

1 1 1

1

1 1 ))

((

++

++ ==

==

==

  ==

   

 

 

  

 

 

 

××  

 

   

 

 

  

 

 

 

××  

==

 

   

   

 

 

  

 

 

 

××  

 

   

   

 

 

  

 

 

 

××  

==

  ==

   

 

 

  

 

 

 

××  

 

   

 

 

  

 

 

 

××  

==

 

   

   

 

 

  

 

 

 

××  

 

   

   

 

 

  

 

 

 

××  

==

==

==

==

==

==

°°

∠

∠

°° ==

∠

== ∠

==

  

Step 5: Find the load current Step 5: Find the load current

 A  A  I 

 I   I 

 I   I 

 I 

 pu  pu

 j  j  j

 j

 Z   Z   X 

 X   X 

 X   X 

 X   j  j

V  V   Z   Z  V   I  V 

 I   I 

 I 

 B  B  pu  pu load  load  load 

load 

 pu  pu load  load  T 

T  line

line T 

T 

 pu  pu  s  s  pu  pu total  total   pu  pu  s  s  pu

 pu  s  s  pu  pu load  load 

°°

−−

∠

∠

==

××

°°

−−

∠

∠

==

××

==

°°

−−

∠

∠

==

°°

∠

∠

°°

∠

== ∠

++

++

++

++

°°

∠

== ∠

++

++

== ++

==

==

01 01 ..

26 26 9

9 ..

109 109

250 250 01

01 ..

26 26 4395

4395 ..

0 0

01 01 ..

26 26 4395

4395 ..

0 0

01 01 ..

26 26 086 086 ..

2 2

0 0 9167 9167 ..

0 0

)) 4167 4167 ..

0 0 875 875 ..

1 1 ((

)) 1378 1378 ..

0 0 2604 2604 ..

0 0 1 1 ..

0 0 ((

0 0 9167 9167 ..

0 0

)) ((

3 3 )) ((

)) ((

2 2 1

1

)) ((

)) ((

)) ((

)) ((

)) ((