Per Unit CalculationsA key problem in analyzing power systems is the large number of transformers. – It would be very difficult to continually refer

impedances to the different sides of the transformers

This problem is avoided by a normalization of all variables.

This normalization is known as per unit analysis.

actual quantityquantity in per unit base value of quantity

Per Unit Conversion Procedure, 11. Pick a 1 VA base for the entire system, SB

2. Pick a voltage base for each different voltage level, VB. Voltage bases are related by transformer turns ratios. Voltages are line to neutral.

3. Calculate the impedance base, ZB= (VB)2/SB

4. Calculate the current base, IB = VB/ZB 5. Convert actual values to per unit

Note, per unit conversion affects magnitudes, not the angles. Also, per unit quantities no longer have units (i.e., a voltage is 1.0 p.u., not 1 p.u. volts)

Three Phase Per Unit

1. Pick a 3 VA base for the entire system, 2. Pick a voltage base for each different

voltage level, VB. Voltages are line to line. 3. Calculate the impedance base

Procedure is very similar to 1 except we use a 3 VA base, and use line to line voltage bases

3BS

2 2 2, , ,3 1 1

( 3 )3

B LL B LN B LNB

B B B

V V VZ

S S S

Exactly the same impedance bases as with single phase!

Three Phase Per Unit, cont'd4. Calculate the current base, IB

5. Convert actual values to per unit

3 1 13 1B B

, , ,

3I I3 3 3B B B

B LL B LN B LN

S S SV V V

Exactly the same current bases as with single phase!

Load Representation

Transmission Circuit Calculations

Short Transmission lineIn the case of a short transmission line the capacitance and

conductance to earth may be neglected. Leaving only the series resistance and inductance to be taken

into consideration. The current entering the line at the sending-end termination

is equal to the current leaving at the receiving-end, and this same current flows through all the line sections.

The R and L parameters may therefore be regarded as ' lumped ' .

The equivalent circuit diagram and the vector diagram for a short line are shown below in which:

Equivalent circuit for a short transmission line

Vector diagram for a short transmission line .

The currents IS and IR will be equal in magnitude but not in phase.

Since there is a phase-shift of voltage along the line. R is obtained from a knowledge of the line length ,the size of

conductor and the specifics resistance of the conductor material ,

while XL is calculated from the conductor spacing and radius using the formula derived in Chapter 3.

Referring to the equivalent circuit :

Hence, if the receiving-end conditions are known the necessary sending-end voltage may be calculated .

(6.1 )S RI I a

( ) (6.1 )

S R L R

R R

V V R jX I bV Z I

It will be noted that previous are phasor equations , a more approximate method involving scalar quantities is as follows: Referring to the vector diagram,

cos sinSX R R R R L RV V I R I X

cos sinSY R L R R RV I X I R

2

2 1 2

=[ ( cos sin )

+( cos sin ) ] S R R R R L R

R L R R R

V V I R I X

I X I R

However (IR XL) and (IR R) are very much less than VR and the small voltage is in quadrate with the much larger VSX ,

The voltage regulation of the line is given by the rise in voltage when full loads is removed , or :

cos sinS SX R R R R L RV V V I R I X

( cos sin )% S R R L R

RR R

V V R Xage voltage regulation I

V V

Example A three-phase line delivers 3 MW at 11 KV for a distance of 15 Km . Line loss is 10 % of power delivered , load power factor is 0.8 lagging . frequency is 50 Hz , 1.7 m equilateral spacing of conductors . Calculate the sending-end voltage and regulation .

Solution

11,000Receiving-end phase voltage = 6.3603 RV

3

3

Line current = phase current ( assuming a star connection ) 3,000 10 = 197 A

3 11 10 0.8

Assuming that the conductors are manufactured from copper having a resistance of 0.0137 ohms per meter for a cross-sectional area of 1 mm2 , the conductor cross-section is 80 mm2 corresponding to a radius of 5 mm .

2

3

Total line loss =3 (in three conductors)10 = 3,000 10

100

I R

3

2

300 10 3 197

2.58 ohms

R

71Inductance = (1 4 log ) 10 /2 c

dL H metre

r

37 3

length

1 1.7 10 = 314 (1 4 log ) 10 15 102 5

=5.75 ohms

L

c

X L

cos sin = 6,350 + ( 197 2.58 0.8) + ( 197 5.75 0.6) = 6,350 + 1057 = 7,407 V per phase = 12,780 V line

S R R R R L RV V I R I X

( cos sin )Regulation = =

1,057 = 16.7 %6,350

S RR L RR

R R

V VR XI

V V