ECE 476 - staff.fit.ac.cystaff.fit.ac.cy/eng.ap/FALL2014/AEEE521/AEEE521_Lect8_1.pdf · 1 Announcements zStart reading Chapter 3. zHW 2 is due now. zHW 3 is 4.32, 4.41, 5.1, 5.14.Due

Lecture 8Transmission Lines, Transformers, Per Unit

Professor Tom OverbyeDepartment of Electrical and

Computer Engineering

ECE 476POWER SYSTEM ANALYSIS

1

Announcements

z Start reading Chapter 3.z HW 2 is due now.z HW 3 is 4.32, 4.41, 5.1, 5.14. Due September 22 in class. z “Energy Tour” opportunity on Oct 1 from 9am to 9pm. Visit

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2

V, I Relationships, cont’d

2 2

Define the propagation constant as

wherethe attenuation constantthe phase constant

Use the Laplace Transform to solve. Systemhas a characteristic equation

( ) ( )( ) 0

yz j

s s s

J

J D E

DE

J J J

�

� � �

3

Equation for Voltage

1 2

1 2 1 2

1 1 2 2 1 2

1 2

1 2

The general equation for V is

( )Which can be rewritten as

( ) ( )( ) ( )( )2 2

Let K and K . Then

( ) ( ) ( )2 2

cosh( ) sinh( )

x x

x x x x

x x x x

V x k e k e

e e e eV x k k k k

k k k k

e e e eV x K K

K x K x

J J

J J J J

J J J J

J J

�

� �

� �

�

� � � � �

� �

� � �

�

4

Real Hyperbolic Functions

For real x the cosh and sinh functions have the following form:

cosh( ) sinh( )sinh( ) cosh( )d x d xx xdx dx

J JJ J J J

5

Complex Hyperbolic Functions

For x = D + jE the cosh and sinh functions have the following form

cosh cosh cos sinh sinsinh sinh cos cosh sin

x jx j

D E D ED E D E

� �

6

Determining Line Voltage

R R

The voltage along the line is determined based uponthe current/voltage relationships at the terminals. Assuming we know V and I at one end (say the "receiving end" with V and I where x 0) we can �

1 2determine the constants K and K , and hence the voltage at any point on the line.

7

Determining Line Voltage, cont’d

1 2

1 2

1

1 2

2

c

( ) cosh( ) sinh( )(0) cosh(0) sinh(0)

Since cosh(0) 1 & sinh(0) 0( ) sinh( ) cosh( )

( ) cosh( ) sinh( )

where Z characteristic

R

R

R RR

R R c

V x K x K xV V K K

K VdV x zI K x K xdx

zI I z zK Iyyz

V x V x I Z x

zy

J J

J J J J

JJ J

� �

�

�

�

�

impedance

8

Determining Line Current

By similar reasoning we can determine I(x)

( ) cosh( ) sinh( )

where x is the distance along the line from thereceiving end.

Define transmission efficiency as

RR

c

out

in

VI x I x xZ

PP

J J

K

�

9

Transmission Line Example

R

6 6

Assume we have a 765 kV transmission line with a receiving end voltage of 765 kV(line to line), a receiving end power S 2000 1000 MVA and

z = 0.0201 + j0.535 = 0.535 87.8 miley = 7.75 10 = 7.75 10 90

j

j � �

�:� q

u u � .0

Then

zy 2.036 88.9 / mile

262.7 -1.1 c

mile

zy

J

q

� q�

= � q :

�

10

Transmission Line Example, cont’d

*6

3

Do per phase analysis, using single phase powerand line to neutral voltages. Then

765 441.7 0 kV3

(2000 1000) 10 1688 26.6 A3 441.7 0 10

( ) cosh( ) sinh( )441,700 0 cosh(

R

R

R R c

V

jI

V x V x I Z xJ J

� q

ª º� u �� q« »u � qu¬ ¼ � � q 2.036 88.9 )

443,440 27.7 sinh( 2.036 88.9 )x

xu � q �

�� qu u � q

11

Transmission Line Example, cont’d

12

Lossless Transmission Lines

c

c

c

For a lossless line the characteristic impedance, Z , is known as the surge impedance.

Z (a real value)

If a lossless line is terminated in impedance

Z

Then so we get...

R

R

R c R

jwl ljwc c

VI

I Z V

:

13

Lossless Transmission Lines

2

( ) cosh sinh( ) cosh sinh( )( )

V(x)Define as the surge impedance load (SIL).

Since the line is lossless this implies( )( )

R R

R R

c

c

R

R

V x V x V xI x I x I xV x ZI x

Z

V x VI x I

J JJ J

� �

If P > SIL then line consumesvars; otherwise line generates vars.

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Surge Impedance Loading (SIL)

The surge impedance loading or SIL of a transmission line is the MW loading of a

transmission line at which a natural reactive power balance occurs. The following

brief article will explain the concept of SIL.

Transmission lines produce reactive power (Mvar) due to their natural capacitance.

The amount of Mvar produced is dependent on the transmission line's capacitive

reactance (XC) and the voltage (kV) at which the line is energized. In equation

form the Mvar produced is:

Transmission lines also utilize reactive power to support their magnetic fields.

The magnetic field strength is dependent on the magnitude of the current flow in

the line and the line's natural inductive reactance (XL). It follows then that the

amount of Mvar used by a transmission line is a function of the current flow and

inductive reactance. In equation form the Mvar used by a transmission line is:

A transmission line's surge impedance loading or SIL is simply the MW loading (at a

unity power factor) at which the line's Mvar usage is equal to the line's Mvar

production. In equation form we can state that the SIL occurs when:

If we take the square root of both sides of the above equation and then substitute

in the formulas for XL (=2pfL) and XC (=1/2pfC) we arrive at:

The term in the above equation is by definition the "surge impedance. The

theoretical significance of the surge impedance is that if a purely resistive load

that is equal to the surge impedance were connected to the end of a transmission

line with no resistance, a voltage surge introduced to the sending end of the line

would be absorbed completely at the receiving end. The voltage at the receiving

end would have the same magnitude as the sending end voltage and would have a

phase angle that is lagging with respect to the sending end by an amount equal to

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the time required to travel across the line from sending to receiving end.

The concept of a surge impedance is more readily applied to telecommunication

systems than to power systems. However, we can extend the concept to the

power transferred across a transmission line. The surge impedance loading or SIL

(in MW) is equal to the voltage squared (in kV) divided by the surge impedance (in

ohms). In equation form:

.

Note in this formula that the SIL is dependent only on the kV the line is energized

at and the line's surge impedance. The line length is not a factor in the SIL or

surge impedance calculations. Therefore the SIL is not a measure of a

transmission line's power transfer capability as it does not take into account the

line's length nor does it consider the strength of the local power system.

The value of the SIL to a system operator is realizing that when a line is loaded

above its SIL it acts like a shunt reactor - absorbing Mvar from the system - and

when a line is loaded below its SIL it acts like a shunt capacitor - supplying Mvar to

the system.

Figure 1 is a graphic illustration of the concept of SIL. This particular line has a SIL

of 450 MW. Therefore is the line is loaded to 450 MW (with no Mvar) flow, the Mvar

produced by the line will exactly balance the Mvar used by the line.

Figure 1

Surge Impedance Loading of a Transmission Loading

.

14

Transmission Matrix Model

Oftentimes we’re only interested in the terminal characteristics of the transmission line. Therefore we can model it as a “black box”.

VS VR

+ +

- -

IS IRTransmission

Line

S

S

VWith

IR

R

VA BIC D

ª º ª ºª º « » « »« »¬ ¼ ¬ ¼¬ ¼

15

Transmission Matrix Model, cont’d

S

S

VWith

IUse voltage/current relationships to solve for A,B,C,D

cosh sinh

cosh sinh

cosh sinh1 sinh cosh

R

R

S R c R

RS R

c

c

c

VA BIC D

V V l Z I lVI I l lZ

l Z lA B

l lC DZ

J J

J J

J J

J J

ª º ª ºª º « » « »« »¬ ¼ ¬ ¼¬ ¼

�

�

ª ºª º « » « » « »¬ ¼ « »¬ ¼

T

16

Equivalent Circuit Model

The common representation is the equivalent circuit

S

Next we’ll use the T matrix values to derive theparameters Z' and Y'.

17

Equivalent Circuit Parameters

'' 2

' '1 '2

' '2 2

' ' ' '' 1 14 2

' '1 '2

' ' ' '' 1 14 2

S RR R

S R R

S S R R

S R R

S R

S R

V V YV IZ

Z YV V Z I

Y YI V V I

Z Y Z YI Y V I

Z Y ZV VZ Y Z YI IY

��

§ · � �¨ ¸© ¹

� �

§ · § · � � �¨ ¸ ¨ ¸© ¹ © ¹ª º�« »ª º ª º

« »« » « »§ · § · ¬ ¼¬ ¼ « »� �¨ ¸ ¨ ¸« »© ¹ © ¹¬ ¼

18

Equivalent circuit parameters

We now need to solve for Z' and Y'. Using the Belement solving for Z' is straightforward

sinh 'Then using A we can solve for Y'

' 'A = cosh 12

' cosh 1 1 tanh2 sinh 2

C

c c

B Z l Z

Z Yl

Y l lZ l Z

J

J

J JJ

�

�

Using the definition of hyperbolic tangent and equations (10,11) we can derive the addition formula:

tanh (x± y) =tanhx± tanh y

1± tanhx tanh y(14)

5 Formulas for the double and half angle

Using equations (10,11) with x = y we immediately have:

sinh 2x = 2 sinhx coshx (15)

cosh 2x = cosh2 x+sinh2 x (16)

By plugging (8) into (16) we have the following two formulas for the squares of sine and cosine:

cosh2 x =1 + cosh 2x

2(17)

sinh2 x =cosh 2x− 1

2(18)

Notice that both (16) and (8) differ from the corresponding trig formulas by a sign, but the resultingformula for cosh2 is the same as in the trigonometric case, and the formula for sinh2 has a globalchange of sign. By substituting x with x

2 and taking the square root we have formulas for the halfangle:

coshx

2=

!1 + coshx

2(19)

sinhx

2= ±

!coshx− 1

2(20)

The first formula is the same1as the trigonometric one, and in the second one we have a globalchange of sign in the radicand. In the same way, but using (14) we have:

tanh 2x =2 tanhx

1+ tanh2 x(21)

6 Inverse Hyperbolic functions

It’s easy to check that hyperbolic sine is a monotonic increasing function on the real numbers, andfor this reason it’s invertible on all the real axis . Hyperbolic cosine has a global minimum at x = 0whose value is y = 1 and it’s decreasing from −∞ to 0 and increasing from 0 to +∞, for this reasonwe can invert it on the positive half-axis or the negative one. By convention we choose the positiveone [0,∞). Hyperbolic tangent is defined for all real numbers, it’s monotonic increasing and it hashorizontal asymptotes:

limx→∓∞

tanhx = ∓1 (22)

1You don’t need to choose the sign in front of the radical since cosh is always positive

3

19

Simplified Parameters

These values can be simplified as follows:

' sinh sinh

sinh with Z zl (recalling )

' 1 tanh tanh2 2 2

tanh 2 with Y2

2

C

c

z l zZ Z l ly l z

lZ zyl

Y l y l y lZ z l y

lY yll

J J

J JJ

J J

J

J

�

�

20

Simplified Parameters

For short lines make the following approximations:sinh' (assumes 1)

' tanh( / 2)(assumes 1)2 2 / 2

50 miles 0.998 0.02 1.001 0.01100 miles 0.993 0.09 1.004 0.04200

lZ Zl

Y Y ll

JJ

JJ

|

|

� q �� q� q �� q

sinhȖO WDQK�ȖO���LengthȖO ȖO��

miles 0.972 0.35 1.014 0.18� q �� q

21

Medium Length Line Approximations

For shorter lines we make the following approximations:sinh' (assumes 1)

' tanh( / 2)(assumes 1)2 2 / 2

50 miles 0.998 0.02 1.001 0.01100 miles 0.993 0.09 1.004 0.0

lZ Zl

Y Y ll

JJ

JJ

|

|

� q �� q� q ��

sinhȖO WDQK�ȖO���LengthȖO ȖO��

4200 miles 0.972 0.35 1.014 0.18

q� q �� q

22

Three Line Models

(longer than 200 miles)

tanhsinh ' 2use ' ,2 2

2 (between 50 and 200 miles)

use and 2

(less than 50 miles)use (i.e., assume Y is zero)

ll Y YZ Z ll

YZ

Z

JJ

JJ

Long Line Model

Medium Line Model

Short Line Model

23

Power Transfer in Short Lines

Often we'd like to know the maximum power that could be transferred through a short transmission line

V1 V2

+ +

- -

I1 I1TransmissionLine with

Impedance ZS12 S21

1

** 1 2

12 1 1 1

1 1 2 2 2

21 1 2

12 12

with , Z

Z Z

V VS V I VZ

V V V V Z Z

V V VSZ Z

T T T

T T T

�§ · ¨ ¸© ¹

� � �

� � � �

24

Power Transfer in Lossless Lines

21 1 2

12 12 12

12 12

1 212 12

If we assume a line is lossless with impedance jX and are just interested in real power transfer then:

90 90

Since - cos(90 ) sin , we get

sin

Hence the maximu

V V VP jQZ Z

V VPX

T

T T

T

� � q � � q �

q �

1 212

m power transfer is

Max V VPX

25

Limits Affecting Max. Power Transfer

z Thermal limits– limit is due to heating of conductor and hence depends

heavily on ambient conditions.– For many lines, sagging is the limiting constraint.– Newer conductors limit can limit sag. For example, in

2004 ORNL working with 3M announced lines with a core consisting of ceramic Nextel fibers. These lines can operate at 200 degrees C.

– Trees grow, and will eventually hit lines if they are planted under the line.

26

Other Limits Affecting Power Transfer

z Angle limits– while the maximum power transfer occurs when line

angle difference is 90 degrees, actual limit is substantially less due to multiple lines in the system

z Voltage stability limits– as power transfers increases, reactive losses increase as

I2X. As reactive power increases the voltage falls, resulting in a potentially cascading voltage collapse.

27

Transformers Overview

z Power systems are characterized by many different voltage levels, ranging from 765 kV down to 240/120 volts.

z Transformers are used to transfer power between different voltage levels.

z The ability to inexpensively change voltage levels is a key advantage of ac systems over dc systems.

z In this section we’ll development models for the transformer and discuss various ways of connecting three phase transformers.

28

Transmission to Distribution Transfomer

29

Transmission Level Transformer

30

Ideal Transformer

z First we review the voltage/current relationships for an ideal transformer– no real power losses– magnetic core has infinite permeability– no leakage flux

z We’ll define the “primary” side of the transformer as the side that usually takes power, and the secondary as the side that usually delivers power.– primary is usually the side with the higher voltage, but

may be the low voltage side on a generator step-up transformer.

31

Ideal Transformer Relationships

1 1 2 2

1 21 1 2 2

1 2 1 1

1 2 2 2

Assume we have flux in magnetic material. Then

= turns ratio

m

m m

m m

m

N Nd d d dv N v Ndt dt dt dt

d v v v N adt N N v N

IO I O I

O I O I

I

o

32

Current Relationships

'1 1 2 2

'1 1 2 2

'1 1 2 2

'1 1 2 2

To get the current relationships use ampere's law

mmf

lengthlength

Assuming uniform flux density in the corelengtharea

d N i N i

H N i N iB N i N i

N i N i

P

IP

* �

u �u

�

u �

u

³ H L��

33

Current/Voltage Relationships

'1 1 2 2

1 2 1 2'

1 2 12

1 2

1 2

If is infinite then 0 . Hence1or

Then010

N i N ii N i N

N i N ai

av vi i

a

P �

�

ª ºª º ª º« » « » « »« »¬ ¼ ¬ ¼¬ ¼

34

Impedance Transformation Example

Example: Calculate the primary voltage and current for an impedance load on the secondary

2121

010

a vvvi Za

ª º ª ºª º « » « » « » « »¬ ¼ « »¬ ¼¬ ¼

21 2 1

21

1

1 vv av ia Z

v a Zi

35

Real Transformers

z Real transformers– have losses– have leakage flux– have finite permeability of magnetic core

1. Real power losses– resistance in windings (i2 R)– core losses due to eddy currents and hysteresis

36

Transformer Core losses

Eddy currents arise because of changing flux in core.Eddy currents are reduced by laminating the core

Hysteresis losses are proportional to area of BH curveand the frequency

These losses are reduced by using material with a thin BH curve

37

Effect of Leakage Flux

2

22

1 1 1

2 2 2

'1 1 1 2 2

11 1 1 1 1

''

2 2 2 2

Not all flux is within the transformer core

Assuming a linear magnetic medium we get

v

v

l m

l m

l l l l

ml

ml

NN

L i L i

ddir i L Ndt dtdi dr i L Ndt dt

O O IO O I

O O

I

I

� �

� �

� �

� �

38

Effect of Finite Core Permeability

m

1 1 2 2 m

m 21 2

1 1

2 m1 2 m

1 1

Finite core permeability means a non-zero mmf is required to maintain in the core

NThis value is usually modeled as a magnetizing current

where im

i N i

Ni iN NNi i iN N

II

I

I

� �

� �

� �

39

Transformer Equivalent Circuit

Using the previous relationships, we can derive an equivalent circuit model for the real transformer

' 2 '2 2 1 2' 2 '2 2 1 2

This model is further simplified by referring allimpedances to the primary side

r e

e

a r r r r

x a x x x x

�

�

40

Simplified Equivalent Circuit

41

Calculation of Model Parameters

z The parameters of the model are determined based upon – nameplate data: gives the rated voltages and power– open circuit test: rated voltage is applied to primary with

secondary open; measure the primary current and losses (the test may also be done applying the voltage to the secondary, calculating the values, then referring the values back to the primary side).

– short circuit test: with secondary shorted, apply voltage to primary to get rated current to flow; measure voltage and losses.

42

Transformer Example

Example: A single phase, 100 MVA, 200/80 kV transformer has the following test data:

open circuit: 20 amps, with 10 kW lossesshort circuit: 30 kV, with 500 kW losses

Determine the model parameters.

43

Transformer Example, cont’d

e

2sc e

2 2e

2

e

100 30500 , R 60200 500

P 500 kW R 2 ,

Hence X 60 2 60

200 410

200R 10,000 10,00020

sc e

e sc

c

e m m

MVA kVI A jXkV A

R I

kVR MkW

kVjX jX XA

� :

o :

� :

:

� � : :

From the short circuit test

From the open circuit test

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2

Open Circuit And

Short Circuit Test

On Transformer

These two transformer tests are performed

to !nd the parameters of equivalent circuit

of transformer and losses of the

transformer. Open circuit test and short

circuit test on transformer are very

economical and convenient because they

are performed without actually loading of

the transformer.

Open Circuit Or No

Load Test On

Transformer

Open circuit test or no load test on a

transformer is performed to determine 'no

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load loss (core loss)' and 'no load current I0'.

The circuit diagram for open circuit test is

shown in the !gure below.

Usually high voltage (HV) winding is kept

open and the low voltage (LV) winding is

connected to its normal supply. A wattmeter

(W), ammeter (A) and voltmeter (V) are

connected to the LV winding as shown in

the !gure. Now, applied voltage is slowly

increased from zero to normal rated value

of the LV side with the help of a variac.

When the applied voltage reaches to the

rated value of the LV winding, readings from

all the three instruments are taken.

The ammeter reading gives the no load

current I0. As I0 itself is very small, the

voltage drops due to this current can be

neglected.

The input power is indicated by the

wattmeter (W). But, as the other side of

transformer is open circuited, there is no

output power. Hence, this input power only

consists of core losses and copper losses.

But as described above, short circuit current

is so small that these copper losses can be

neglected. Hence, now the input power is

almost equal to the core losses. Thus, the

wattmeter reading gives the core losses of

transformer. Voltage is applied to the HV

side and increased from the zero until the

ammeter reading equals the rated current.

All the readings are taken at this rated

current.

The ammeter reading gives primary

equivalent of full load current (Isc).

The voltage applied for full load current is

very small as compared to rated voltage.

Hence, core loss due to small applied

voltage can be neglected. Thus, the

wattmeter reading can be taken as copper

loss in the transformer.

Therefore, W = Isc2Req....... (where Req is the

equivalent resistance of transformer)

Zeq = Vsc/Isc.

Therefore, equivalent reactance of

transformer can be calculated from the

formula Zeq2 = Req

2 + Xeq2.

These, values are referred to the HV side of

the transformer.

Hence, it is seen that the short circuit test

gives copper losses of transformer and

approximate equivalent resistance and

reactance of the transformer.

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the transformer.

Sometimes, a high resistance voltmeter is

connected across the HV winding. Though, a

voltmeter is connected, HV winding can be

treated as open circuit as the current

through the voltmeter is negligibly small.

This helps in to !nd voltage transformation

ration (K).

The two components of no load current can

be given as,

Iμ = I0sinΦ0 and Iw = I0cosΦ0.

cosΦ0 (no load power factor) = W / (V1I0). ...

(W = wattmeter reading)

From this, shunt parameters of equivalent

circuit parameters of equivalent circuit of

transformer (X0 and R0) can be calculated as

X0 = V1/Iμ and R0 = V1/Iw.

(These values are referring to LV side of the

transformer.)

Hence, it is seen that open circuit test gives

core losses of transformer and shunt

parameters of the equivalent circuit.

Short Circuit Or

Impedance Test On

Transformer

The connection diagram for short circuit

test or impedance test on transformer is as

shown in the !gure below. The LV side of

transformer is short circuited and

wattmeter (W), voltmere (V) and ammeter

(A) are connected on the HV side of the

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Emai l

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AC Machines

DC Machines

Transformers

In KVA?

From the above transformer tests, it can be

seen that Cu loss of a transformer depends

on current, and iron loss depends on

voltage. Thus, total transformer loss

depends on volt-ampere (VA). It does not

depend on the phase angle between voltage

and current, i.e. transformer loss is

independent of load power factor. This is

the reason that transformers are rated in

kVA.

44

Residential Distribution Transformers

Single phase transformers are commonly used in residential distribution systems. Most distributionsystems are 4 wire, with a multi-grounded, commonneutral.

45

Per Unit Calculations

z A key problem in analyzing power systems is the large number of transformers. – It would be very difficult to continually have to refer

impedances to the different sides of the transformersz This problem is avoided by a normalization of all

variables.z This normalization is known as per unit analysis.

actual quantityquantity in per unit base value of quantity

46

Per Unit Conversion Procedure, 1I

1. Pick a 1I 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 on 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)

47

Per Unit Solution Procedure

1. Convert to per unit (p.u.) (many problems are already in per unit)

2. Solve3. Convert back to actual as necessary

48

Per Unit Example

Solve for the current, load voltage and load power in the circuit shown below using per unit analysis with an SB of 100 MVA, and voltage bases of 8 kV, 80 kV and 16 kV.

Original Circuit

49

Per Unit Example, cont’d

2

2

2

8 0.64100

80 6410016 2.56

100

LeftB

MiddleB

RightB

kVZMVAkVZMVAkVZMVA

:

:

:

Same circuit, withvalues expressedin per unit.

50

Per Unit Example, cont’d

L

2*

1.0 0 0.22 30.8 p.u. (not amps)3.91 2.327

V 1.0 0 0.22 30.8p.u.

0.189 p.u.

1.0 0 0.22 30.8 30.8 p.u.

LL L L

G

Ij

VS V IZ

S

� q �� q

� � q � �� qu ��������q ������� ����q�

� qu � q ����� q�

51

Per Unit Example, cont’d

To convert back to actual values just multiply theper unit values by their per unit base

LActual

ActualLActualG

MiddleB

ActualMiddle

0.859 30.8 16 kV 13.7 30.8 kV

0.189 0 100 MVA 18.9 0 MVA

0.22 30.8 100 MVA 22.0 30.8 MVA100 MVAI 1250 Amps

80 kVI 0.22 30.8 Amps 275 30.8

V

S

S

�� qu �� q

� qu � q

� qu � q

�� qu����� �� q�$