Per-Unit System

Power System Analysis

Per-Unit System

In the per-unit system, the voltages, currents, powers,

impedances, and other electrical quantities are

expressed on a per-unit basis by the equation:

It is customary to select two base quantities to define

a given per-unit system. The ones usually selected

are voltage and power.

Quantity per unit =Actual value

Base value of quantity

2

Per-Unit System

Assume:

Then compute base values for currents and

impedances:

ratedb VV

ratedb SS

b

bb

V

SI

b

b

b

bb

S

V

I

VZ

2

3

Example 1

An electrical lamp is rated 120 volts, 500 watts. Compute

the per-unit and percent impedance of the lamp. Give the

p.u. equivalent circuit.

Solution:

(1) Compute lamp resistance

if power factor = 1.0 then :

8.28500

)120( 222

P

VR

R

VP

08.28Z

4

Example 1

(2) Select base quantities

(3) Compute base impedance

(4) The per-unit impedance is:

VASb 500

VVb 120

8.28500

)120( 22

b

bb

S

VZ

..018.28

08.28.. up

Z

ZZ

b

up

5

Example 1

(5) Percent impedance:

(6) Per-unit equivalent circuit:

%100% Z

..01 upZ ..01 upVS

6

Example 2

An electrical lamp is rated 120 volts, 500 watts. If the

voltage applied across the lamp is twice the rated value,

compute the current that flows through the lamp. Use the

per-unit method.

Solution:

VVb 120

..02120

240.. up

V

VV

b

up

..01.. upZ up

7

Example 2

The per-unit equivalent circuit is as follows:

..01 upZ ..02 upVS

..0201

02

..

..

.. upZ

VI

up

up

up

AV

SI

b

bb 167.4

120

500

AIII bupactual 0334.8167.402..

8

Three-Phase Systems

For a given single-line (one-line) diagram of a power

network, all component parameters are expressed in 3-

quantity whether it is the rating (capacity) expressed as

MVA or voltage as kV.

Let begin with 3- base quantity of

basebasebase IVS 3

where Vbase = line voltage, Ibase= line or phase current

(i)

9

Three-Phase Systems

Per phase base impedance,

base

base

baseI

V

Z3

(ii) This is line-to-neutral impedance

base

base

base

base

V

S

V

Z

3

3

base

basebase

MVA

kVZ

2

where kVbase and MVAbase are 3- quantities.

Combining (i) and (ii) yields,

10

Changing base impedance (Znew

)

Sometimes the parameters for two elements in the same

circuit (network) are quoted in per-unit on a different base.

The changing base impedance is given as,

2

2

base OLD base NEW

NEW OLD

base OLDbase NEW

kV MVAZ pu Z

MVAkV

11

Example 3

Determine the per-unit values of the following single-line

diagram and draw the impedance diagram.

XT1 = 0.1 p.u

50 MVA

Xg = 16%

100 MVA

275 kV/132 kV50 MVA

132 kV/66 kV

Transmission line

j 3.4

XT2 = 0.04 p.uLoad

40 MVA,

0.8 p.f. lagging

12

XT1 = 0.1 p.u

50 MVA

Xg = 16%

100 MVA

275 kV/132 kV50 MVA

132 kV/66 kV

Transmission line

j 3.4

XT2 = 0.04 p.u

Load

40 MVA,

0.8 p.f. lagging

Solution:

For S , always choose the largest rating, therefore Sbase = 100 MVA

anywhere in the power system.. For V, we have 2 transformers and

three regions with different voltage levels. So choose :

Vbase(Reg1)= 275 kV, Vbase(Reg2)=132 kV and Vbase(Reg3)=66 kV.

Region 1

Vbase=275 kV

Region 2

Vbase=132 kV

Region 3

Vbase=66 kV

13

275132

12 basebase VV 13266

23 basebase VVkVVbase 2751

Per-unit calculations:

Generator G1:(Region 1) Transformer T1:

1.0)(1 puXT

2

2

base OLD base NEW

NEW OLD

base OLDbase NEW

kV MVAZ pu Z

MVAkV

32.050

100

275

27516.0)(

MVA

MVA

kV

kV

g puX p.u.

p.u.

MVA

baseT SS 1001

14

)(

)(

2

)1(Re

2

)1(Re)()(

OLD

NEW

base

base

gbase

gbase

OLDgNEWg S

S

V

VpuXpuX

Transmission line TL:

24.174

4.3)(

)(

base

TLactual

TLZ

XpuX

Transformer T2:

MVA

MVA

T puX50

100104.0)(2

p.u.

base

basebase

MVA

kVZ

2

actual

pu

base

ZZ

Z

p.u.

24.17410100

)10132(6

23

baseZ

base

MVA

T SS 502

2

2

baseOLD base NEW

NEW OLD

baseOLDbase NEW

kV MVAZ pu Z

MVAkV

08.0)(2 puXT

0195.0)( puXTL

(Region 2)

15

)(

)(

)3(Re

)3(Re

2

2

22 )()(

OLD

NEW

g

g

OLDNEW

base

base

base

base

TTS

S

V

VpuXpuX

Inductive load:

o

actual LoadZ 87.3612.87

8.0106631040

31066

)(

3

6

3

)2.16.1(87.36256.43

87.3612.87)( jorpuZ o

o

L

p.u.

PFV

S

CosV

SI

.3.3

I

V

Z3

In 3-phase systems :

and

56.4310100

)1066(6

23

baseZ

(Region 3)

16

17

Now, we have all the

impedance values in per-

unit with a common base

and we can now combine

all the impedances and

determine the overall

impedance.

17

Load

G

j 0.32 p.u.

j 0.1 p.u. j 0.0195 p.u.

Transformer

T1

Transformer

T2

Transmission Line

TL

j 0.08 p.u.

1.6 p.u..

j 1.2 p.u.

Generator

17

Example 4

Eg=275 kV T1 T2Transmission line

Load

18

In the previous example , assume that the generator

voltage is Eg=275 kV. Find the values of Ig , Iline , Iload ,Vloadand Pload.

Ig ILineILoad

19

From the previous example we know per-phase, per-unit

equivalent circuit of this power system:

Load

j 0.32 p.u.

j 0.1 p.u. j 0.0195 p.u.

Transformer

T1

Transformer

T2

Transmission Line

TL

j 0.08 p.u.

1.6 p.u..

j 1.2 p.u.

Generator

pupuEg 01275

275)(

puV

EpuE

gbase

g

g 01275

275)(

)1(Re

puj

jpuZeq

4735.27195.16.1

)2.108.00195.01.032.0(6.1)(

pupuZ

puEpuI

eq

g474256.0

4735.2

01

)(

)()(

I(pu)

˚

˚

20

)(94.209102753

10100

.3 3

6

)1(Re

)1(Re AmpV

SI

gbase

basegbase

)(473.8994.209)474256.0()( )1(Re AmpIpuII gbaseg

)(39.437101323

10100

.3 3

6

)2(Re

)2(Re AmpV

SI

gbase

basegbase

)(471.18639.437)474256.0()( )2(Re AmpIpuII gbaseLine

)(77.87410663

10100

.3 3

6

)3(Re

)3(Re AmpV

SI

gbase

basegbase

)(472.37277.874)474256.0()( )3(Re AmpIpuII gbaseLoad

˚

˚

˚

21

1.1043.32

)87.3612.87()472.372(

kV

LoadLoadLoad ZIV

o

actual LoadZ 87.3612.87

8.0106631040

31066

)(

3

6

3

From the previous example, we know :

Moreover:

)(65.98.0)1043.32(2.372 3 MWCosVIP LoadLoadLoad

Power Factor

˚

˚ ˚

Example 5

A power system consists of one synchronous generator and

one synchronous motor connected by two transformers and

a transmission line. Create a per-phase, per-unit equivalent

circuit of this power system using a base apparent power of

100 MVA and a base line voltage of the generator G1 of 13.8

kV. Given that:

G1 ratings: 100 MVA, 13.8 kV, R = 0.1 pu, Xs = 0.9 pu;

T1 ratings: 100 MVA, 13.8/110 kV, R = 0.01 pu, Xs = 0.05 pu;

T2 ratings: 50 MVA, 120/14.4 kV, R = 0.01 pu, Xs = 0.05 pu;

M ratings: 50 MVA, 13.8 kV, R = 0.1 pu, Xs = 1.1 pu;

L1 impedance: R = 15 , X = 75 .

22

The system base apparent power is Sbase = 100 MVA everywhere

in the power system. The base voltage in the three regions will

vary as the voltage ratios of the transformers that delineate the

regions. These base voltages are:

23

The corresponding base impedances in each region are:

24

The impedances of G1 and T1 are specified in per-unit on a base

of 13.8 kV and 100 MVA, which is the same as the system base

in Region 1. Therefore, the per-unit resistances and reactances of

these components on the system base are unchanged:

There is a transmission line in Region 2 of the power system. The

impedance of the line is specified in ohms, and the base impedance

in that region is 121 . Therefore, the per-unit resistance and

reactance of the transmission line are:

25

The impedance of T2 is specified in per-unit on a base of 14.4 kV

and 50 MVA in Region 3. Therefore, the per-unit resistances and

reactances of this component on the system base are:

The impedance of M2 is specified in per-unit on a base of 13.8 kV

and 50 MVA in Region 3. Therefore, the per-unit resistances and

reactances of this component on the system base are:

26

Therefore, the per-phase, per-unit equivalent circuit of this

power system is shown:

27