1

2 (Per Unit System)

2.1 (One Line Diagram)

!"#$%&'()'*+ ! #,$-.,/01"%+'()-., 0#12 0345 !"#$% 61",&)7.8-+#$59 ! :42%-.,7"(12 ;98<%<-+ %;'=% !"#$%

;'12 2.1 ,+

2

$>#$598=% ?+(?%@=% !"#$%

?% .,'#% !

$ 8 7A2%"+) !

&.# (?#(, -(?/0#%#%)+

2.2 !"#$%&'!'!%()%*+', per unit

)7.8 !"#$% $7?(?%@ C?+ %$+"#$% !#)<+:8 -+;' per unit =%7?,%)%7?-7?.+42%'#2+7?(?%@-.,;?-+;' per unit 3$ (?'+G

Per unit Voltage = (Actual Voltage) / (Base Voltage)Per unit Power = (Actual Power) / (Base Power)Per unit Impedance = (Actual Impedance) / (Base Impedance) Per unit Current = (Actual Current) / (Base Current)

3

&12 Base power '*+7? Base MVA3phaseBase voltage '*+ 7? line-to-line voltage

$%+$G+ Base impedance = (Base KVL-L)2/ (Base MVA3phase)Base current = (Base MVA3phase) / (√3Base KVL-L)

/, '!%7A2%"+) !=+ 75 MVA , 3.81 KV , Xd’’ = 0.6 Ohms "#$%/?"#$% !50MVA 12 3.75 KV

Solution 6,#Ac+'*+ 100MVA , 3.81 KV7A2%"+) !/?"#$% ! = 50/100 = 0.5 p.u.%$+12=$G=%7A2%"+) ! = 3.75/3.81 = 0.98 p.u.Base impedance = (3.81)2 / (100 MVA) = 0.1452 ohms

Xd’’ = 0.6/0.1452 = 4.132 p.u.

Changing the base of per unit '#2+7? per unit /c+.+42%''*+7?-+c+.+42% 1",&

new p.u. value = (old p.u. value)*(old base value) / (new base value)

".$ impedance 1",$%+G

=

given

new

new

givengivennew baseKVA

baseKVAbaseKVbaseKV

ZunitperZunitPer2

4

Example2.1The terminal voltage of a Y-connected load consisting of

three equal impedances of 20∠30° ohms is 4.4 KV line-to- line.The impedance of each of the three lines connecting the load to abus at a substation is ZL = 1.4 ∠75° ohms. Find the line-to- linevoltage at the substation bus.

Example 2.2Find the solution of example 2.1 by working in per unit on

a base of 4.4 KV , 127 A so that both voltage and current magnitudes will be 1.0 per unit.

5

Example 2.3The reactance of a generator designated Xqis given as 0.25 per unit

base on the generatorrs nameplate rating of 18 KV , 500 MVA . The basefor calculations is 20 KV , 100 MVA. Find Xqon the new base.

Example 2.4A generator (which may be represented by an emf in series with

an inductive reactance) is rated 500 MVA , 22 KV. Its Y-connected windingshave a reactance of 1.1 per unit. Find the ohmic value of the reactance of theWindings.

6

2.3 /"#%=% Synchronous Machine

,%=#>>?)% Synchronous Machine

;'12 2.2 %/ ;#=%Synchronous Machine

Synchronous impedance

Zd = R+jω(Ls+Ms) = R+jXd

Transient and sub transient effect

;'12 2.3 ;' (=% #$%/-+ Synchronous machine

7

,%=#>>?)% Synchronous machine=9 w transient

=9 w subtransient

2.4/"#%=%.,'#% !

%/ ;#=%.,'#% !

2.4.1 .,'#% !

8

Impedance Transfer

2

22 IVZ = '

/ 222

2

22

2

2

1

11 ZZa

IVa

aIaV

IVZ =====

1211' Za

Z =22

2 ' ZaZ =

9

ExampleA single phase transformer has 2000 turns on the primary winding and

500 turns on the secondary. Winding resistance are r1 = 2.0 Ω and r2 = 0.125 Ω .Leakage reactance are x1 = 8.0 Ω and x2 = 0.50 Ω . The resistance load Z2 is 12 Ω. If applied voltage at the terminals of the primary winding is 1200 V ,find V2 .

10

ExampleA single phase transformer is rated 110/440 V, 2.5 KVA. Leakage reactance

measured from the low voltage side is 0.06 ohms. Determine leakage reactance in p.u. .

ExampleThree parts of a single phase electric system are designated A,B and C and are

connected to each other through transformers , as shown in figure. The transformer are rated as follows :

A-B 10,000 kVA , 13.8/138 KV , reactance 10 %B-C 10,000 kVA , 138/69 KV , reactance 8 %

If the base in circuit B is chosen as 10,000 kVA ,138 KV , find the per unit impedanceOf the 300 Ω resistive load in circuit C refered to circuit C,B and A .

11

2.4.2 .,'#% ! A. .,'#% ! (? Y-Y

B. .,'#% ! (? Y-Delta

12

C. .,'#% ! (? Delta-Y

D. .,'#% ! (? Delta-Delta

13

ExampleThree transformers , each rated 25 MVA , 38.1/3.81 KV , are connected

Y-delta with a balanced load of three 0.6 , Y-connected resistors. Choose a base of75 MVA , 66 KV for the high voltage side of the transformer and specify the base for the low voltage side. Determine the per unit resistance of the load on the base for the low voltage side. Then, determine the load resistance RL in ohms refered tothe high side and the per unit value of this resistance on the chosen base.

2.5 /"#%=% ?%

2.5.1 /"#%=% ?% $G+ ( ?)+ 80.)

IS = IR VS= VR+IR*Z

14

Phasor diagram of a short transmission line

2.5.2 /"#%=% ?%'+#% (.?% 80.- 240. )

RRRS VZIYVV +

+=

2 RRS ZIVYZV +

+= 1

2

15

2.5.3 /"#%=% ?% (? 240. )

xyzZIxyzVV OSSX •• −= cosh

xyzIxyzZVI SO

SX •• +−= coshsinh

2.6/"#%=%&.#

'()&.#=% !$/1+,7?"#$% !/)%#"#$% !/)+(w</"#%=%&.#6; ,%<A2-C,-+)7.8 6w<=% !"#$%

LL

LLoad jQP

VZ−

=2

16

Impedance diagram and Reactance diagram

w<12 1.26.+, 36stevenson

Impedance diagram

Reactance diagram

17

+?/ !?F "=G!?%)%H I*JKKL! G!?/%

7"+9#$%//"#%=% Synchronous Machine 1+, Subtransient Model<,1$G%?+"7(,+1+=% ?+'A2+@7) #?7) 12/?-.,$&.#'() ?+)#$%/

)7.8&.#& #8/"#%=% Synchronous Machine #&.# 1+, Real Power # Reactive Power <,1$G%1+ ?%, Pi equivalent circuit

PL+jQL

18

LL

LLoad jQP

VZ−

=2

)7.8 6w<

/"#%=% Synchronous Machine 1+, Transient Model #1+&.#,7?)<+:8=++

ExampleG1: 20 MVA , 6.6 KV , X” = 0.655 ohmsG2 : 10 MVA , 6.6 KV , X” = 1.31 ohmsG3: 30 MVA , 3.81 KV , X” = 0.1452 ohmsT1&T2 : Three phase transformer banks of single phase transformerseach rated 10 MVA , 3.81 KV/ 38.1 KV , x = 14.52 ohms referred tohigh voltage sideTransmission line : Reactance = 17.4 ohmsLoad A: 15 MVA , 6.6 KV , PF = 0.9 laggingLoad B: 30 MVA , 3.81 KV , PF = 0.9 lagging

19

=,7 $%(01. #A7? Base KV # Base MVA ".$ ?+- ?+.+42%-+?+

# Base KV '*+7? line-to-line voltage # Base MVA '*+ 7?7) 2. ".$ ?+=%12;?,+=%.,'#% -.,7"+97? Base KV

".$ ?++$G+ 7"+9-.,-C,$( ?+%$+ line-to-line =%.,'#% ?+7?Base MVA '*+7?$+-+10 ?+=%

3. 7? impedance =%.,'#% #7A2%"+) !12".+-.,'()/;?-+;' per unit & rating =%($%'*+c+

4. 7? impedance -+;' per unit .'*+7?121$c+12(?%$+'/7?c+=% (,%'#2+-.,'*+7?121$c+=%?+

ExampleA power house is supplying a three phase load as shown in figure

T1 : 3 phase transformer rating , 125 MVA , 11.2 KV (Y) – 230 KV (Y)Z = 0.25 + j1.25 ohms refer to low tension side

T2 : 3 phase transformer rating , 125 MVA , 6.9 KV (Y) – 230 KV (Y)Z = 0.00025 + j0.0025 ohms refer to low tension side

Using the method of transferring all the values of the parameters needed in the calculation to the same side. Determine the magnitude and phase voltage at theterminal of the power house.