9. transmission & distribution sys.pdf

8/11/2019 9. Transmission & Distribution sys.pdf

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Power Transmission / Distribution- Electric power can be transmitted from one place to other with the help of conductorcable, but in doing so some of the electrical power is lost when it is transmitted to

long distance- Transmission and distribution is designed for minimum loss (less than 10%)

- At the same time it is also important to keep the costs low.

All the conductors have some resistance to the flow of electric current. This itself

show the voltage drop across the length of the conductor.

Vdrop= Ix R

"V" drop is depends on:

1. Type of cable material

2. Cross-sectional area of the cable

3. Length of the cable

Most commonly used conducting materials are:

- Copper

- Aluminum

ACSR (Aluminum Conductor Steel Reinforce)

ACSR Cable (Typical specification)

ConductorCurrent rating

(amp)

Resistance

()/km)

Weight

(kg/km)

Squirrel 76 1.374 85

Gopher 85 1.089 106

Weasel 95 0.9047 127.7

Rabbit 135 0.5404 213.6

Dog 205 0.2722 394

Aluminum Cable

Conductor

Current

rating

(amp)

Resistance

(/km)

4mm2 23 7.15

6mm2 30 4.76

10mm2 40 2.86

16mm2 51 1.78

25mm2 70 1.14

)2

(

)/(*4.18

mm

km

A

KR

K=1 for copper

k= 1.6 for aluminum

k=10 for steel 1

Ex.: A village is 1.5 km away from the powerhouse. Rabbit cable is used to

transmit 5kW power to the village in single phase. Assuming purelyresistive load, pf. =1.Find the followings:

1. Voltage drop in the cable.2. Voltage in village, if voltage at Powerhouse is 230 volt

3. What is the % of voltage drop ?

The current necessary to transmit 5kW of power, I = P/V = 5000/230 = 21.74 Amp

From the table, Rabbit conductor has a resistance of 0.5404 / km

Resistance of the whole cable = 2 * 1.5 (km) * 0.5404 (/km) = 1.62

i. Voltage drop, Vdrop= I * R = 21.74 (Amp) * 1.62 () = 35.22 Volts

ii. Voltage in the village = Voltage in PH - Vdrop= 230 - 35.22 = 194.78 Volts

iii. % of voltage drop = (35.22/230)*100 = 15.3 %

Since, Vdropis too high more than 10%, means chosen Rabbit cable is too thin. Take

bigger size cable say Dog and repeat the above calculation:

From the table, Dog conductor has a resistance of 0.2722 / km

Resistance of the whole cable = 2 * 1.5 (km) * 0.2722 (/km) = 0.82

i. Voltage drop, Vdrop = I * R = 21.74 (Amp) * 0.82 () = 17.83 Voltsii. Voltage in the village = Voltage in PH - Vdrop = 23017.83 = 212.17 Voltsiii. % of voltage drop = (17.83/230)*100 = 7.75 % 2


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Three-Phase

V

V

V

V

R Y B

Single-Phase

Three identical winding are putting out 3 voltage waveform at 1200apart.

Single winding single voltage waveform

V = Phasevoltage

I = Phase Current

Power = V.I. Cos

VP= Phase voltage

VL= Line voltage

VL= 3 . VP

IP= Phase Current

IL= Line Current

IL= 3 . IP

Power = 3. VP. IP.pf = 3.VL.IL.pf

Power = P(R)+ P(Y)+ P(B)

If three phase are balanced

If three phase are unbalanced 3

3 Phase Load (Motor)

~

~ ~

R (Red)

N (Neutral)

Y (Yellow)

B (Blue)

VR-N

VY-NVB-N

3 Phase Generator

Single Phase Loads

Ex. On a 3 phase panel board, following reading are taken:

VR= 218 V IR= 15 A

VY= 215 V IY= 16 A

VB= 220 V IB= 15 A

Find the total power output of a generator assuming loads of

unity power factor (p.f.=1)

Power of the individual phase

PR= VR.IR. Cos= 218*15*1= 3270 W

PY = VY.IY. Cos= 215*16*1= 3440 WPB= VB.IB. Cos= 220*15*1= 3300 W

Total power = PR+PY+PB = 3270+3440+3300=10010 W

Three phase supply system/ Load system :

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Z

IP

IP

IP

IB

IY

IRR

Y

B

Delta Connected balanced load

VP =VL

Line voltage = VL

Phase voltage, VP=

Phase Current, IP= Line Current, IL

Power drawn, P = 3.VL.IL.Cos

VL

3

Line Voltage = VL

Phase Voltage, VP= Line Voltage, VL

Phase Current, IP=

Power drawn, P = 3.VL.IL.Cos

IL

3

Three phase star (Y) connected and Delta () connected Loads.

For balanced Loads: IN= 0

VR-N = VY-N= VB-N= VPVR-Y= VY-B= V B-R= VL

IL= IP

Star Connected balanced load

Z

IR

IB

IY

IPIP

IP

R

N

Y

B

VPIN

For balanced Loads:

IR = IY= IB= ILIR-Y= IY-B= IB-R= IPVL= VP

5

VL

1 2 3 4 5 6 7 8 9 10 11 12 13 144

6

8

10

12

14

16

kW-km

% Vdrop

Single Phase(220 Volts)

(Unity power factor)

ACSR Cable Selection Chart

6


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4 8 12 16 20 24 28 32 36 40 44 48 52 564

6

8

10

12

14

16

kW-km

% Vdrop

Three Phase

(380/220 Volts)(Unity power factor)


7

30 50 70 90 110 130 150 1704

6

8

10

12

14

16

kW-km

% Vdrop

Three Phase(1 KV)

(Unity power factor)


8


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Ex.: 5 kW of power (at pf=1) has to be transmitted from the power house to a village

1.8 km away from the PH. The maximum allowable Voltage drop is 10%.

1. Using single phase transmission, what size cable will be required to transmit

this power ? How much conductor will be needed ?2. What size cable will be required to transmit this power in three phase (4-wire

system configuration) ? How much conductor will be needed ?

km-kW9km1.8*kW5productdistanceandPower dropV10%givewillconductorDogphase)single(forgraphtheFrom A

kg1418.4(kg/km)394*(km)3.6cable(Dog)theofWeight

tabletheFrom

km-kW9km1.8*kW5productdistanceandPower

dropV9.5%givewillconductorSquirrelphase)(threegraphtheFrom A

km3.62*.81conductoroflengthTotal

km7.24*.81conductoroflengthTotal

kg126(kg/km)85*(km)7.2cable(Squirrel)theofWeight

tabletheFrom

1. Single Phase

2. Three Phase

9

Example:If 30 kW, 3-phase of power is to be transmitted 2 km away from PH. What

size conductor would be required to keep the voltage drop within 10% if the

transmission is done at 1000 V line voltage ?

%5.6100*3.577

7.37V%

V37.72.178*17.3R*IVphase,perdropVoltage

2.178/km)(1.089*(km)2R,Resistance

used,isconductorGopherIf

Amp.3.173.577*3

1000*30

Iphase,perCurrent

3.5773

1000Vvoltage,line1000VatphaseperVoltage

drop

drop

ph

ph

V

Check for the Result using the graph for 3 phase; 1 kV

Transmission system.10


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6

Resistance and Reactance of a Conductor

11

dropVdrop,Voltageof(%)allowableDecide:1Step

condition)for worse5.0pf(TakepfforEstimate:2 Step

3321

**:3 sssSspacing,ConductorEffectiveFindStep

7/4.9)(saytablethefromsize,conductortheuess:4 GStep

)L*/km(ResistancetheFind:5 kmtable)(FromStep

)f.L.2.(XXReactance,InductiveFind:6 (H)LL Step

phase)per(H10*log*465(* 8)10(lengthtrans.effective)(

r

sLLH

)(ZImpedence,Find:7 2)2 LXRZStep

)*(dropVoltageFind:8 drop ZIVStep

dropvoltageof%allowableforheck:9

pro cess)repeat theandcablesizebiggerthechosemore,isdropV%(I f

CStep

For conductor spacing 0.3m

s1=0.3m

s2=0.3m

s3=0.6m

7 strand cable

Calculation of Voltage drop using Equations Conductor Size & ResistanceNominal

Area

(mm2)

Size

notation

Resistance

(/km)

22 7/2.06 1.227

30 7/2.44 0.875

35 7/2.59 0.777

40 7/2.79 0.669

60 7/3.40 0.450

70 7/3.66 0.389

75 7/3.79 0.365

90 7/4.18 0.300

100 7/4.29 0.270

125 7/4.90 0.217

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A factory requiring 50KW, 3-phase, 380V, 50Hz, power to be supplied by a MHPplant. The PH is 550 m away from the factory. The worse pf is estimated to be of

pf=0.6. The maximum voltage that can a generator produce is 420 V.

Find the size of the conductor mm2, If aluminum conductor is to be used andcorresponding Power loss in the conductor.

9.5%~V40380420 dropVAllowable

6.0pf

m0.3spacingConductorAssuming

cable)](7/4.9tablethe[from125mmConductorofsizethessuming 2A

119.055.0*217.0*/ LengthkmR

4810

810

10*6.410*00735.0

378.0log*465(*550

10*log*465(*

r

sLengthLH

378.06.0*3.0*3.0** 33 321 ssss

mstrandr 00735.035.73*2

4.93*radius

s3s1

s2

13

/phase144.010*6.4*50**2***2 4 LfXL

phaseper0.1860.1190.144 2222 RXZ L

V35.2178.114*186.0*phase)( IZperVdrop

V98.363*35.213*)()( phaseVlineV dropdrop

wattRIPloss 1568119.0*78.114*phaseper22

VVVV drop 38398.36420supplysiteat

Amp78.114

6.0*)3/420(*3

50000

**3

,

pfV

PICurrent

wattsPTotal loss 47041568*3 14


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Resistance and Reactance of a Conductor

15

Transmission & Distribution SystemLine ConfigurationSingle Phase 2-wire systemPower consume by single phase load

cos**IVP

Voltage drop

IxrLVdrop

*)sin*cos*(**2

Voltage drop on each phase

IxrLVdrop

*)sin*cos*(*

Power consume by each phase load

cos**3

IVP

P phph

Three Phase 4-wire system

Voltage drop on each phase

IxrLVdrop

*)sin*cos*(**3

Power consume by each phase load

cos**3

phph IVP

P

Three Phase 3-wire system

16


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17

Example:A 21 mm2 ACSR line with a spacing of 0.30 m is used to bring power 500 m fromthe power house to a load of 3 kW with a power factor of 0.95, resistance (r) = 1.41

ohm/km and reactance (x) = 0.32 ohm/km. What is the percent voltage drop withfollowing transmission line systems;

a. Single phase 2-wire systemb. Three phase 4 wire system (per phase)

p.f. = 0.95 = Cos 180 Sin 180=0.31

VIxrLVdrop

75.1973.13*)31.0*32.095.0*41.1(*5.0*2*)sin*cos*(**2

AmppfV

PI 73.13

95.0*230

1000*3

*

%58.8100*230

75.19100*%

V

VV

drop

drop

a. Single ph ase 2-wire system

b. Three phase 4- wire system (per phase)

VIxrLVdrop

29.358.4*)31.0*32.095.0*41.1(*5.0*)sin*cos*(*

AmppfV

PI 58.4

95.0*230*3

1000*3

*

%43.1100*230

29.3100*%

V

VV drop

drop

A 10 kW power at a pf=0.8 is being transmitted over a transmission line whoseresistance is given as (Z=0.3 +J*0.1). Find the voltage drop. What would be the

voltage drop if pf is unity

AV

PowerTrueTCCurrentTrue 45.45

220

10000)(

Apf

TCACCurrentApparent 81.56

8.0

45.45)(

AACRCCurrentactive 346.0*81.56sin*)(Re )34*45.45( jI

17.96V

)1.03.0(*)3445.45(

)1.0*3.0mod(*)34*45.45mod(

).mod(*)mod(

2222

jj

ZIVdrop

14.36V

)1.03.0(*)45.45(

)1.0*3.0mod(*)45.45mod(

).mod(*)mod(

222

j

ZIVdrop

Vdropwhen pf=0.8

Vdropwhen pf=1

18


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Volts = 230 VConductor spacing = 0.3 m

Frequency = 50 Hz

Decide the % VdropFind the kW-m, as per system and load configuration shown in above tableWith known value of kW-m draw vertical line to meet the line of % dropDraw horizontal line to fine area of conductor (left for aluminum and right copper)

19

Volts = 230 VConductor spacing = 0.3 m

Frequency = 50 Hz

Decide the % VdropFind the kW-m, as per system and load configuration shown in above tableWith known value of kW-m draw vertical line to meet the line of % drop

Draw horizontal line to fine area of conductor (left for aluminum and right copper)

20


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11

Volts = 230 V

Conductor spacing = 0.3 mFrequency = 50 Hz

Decide the % VdropFind the kW-m, as per system and load configuration shown in above tableWith known value of kW-m draw vertical line to meet the line of % dropDraw horizontal line to fine area of conductor (left for aluminum and right copper)

21

Typical Load distribution system for MHP

Uniformly distributed Load

End Concentration Load

Unevenly distributed Load

22


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12

Single Phase 2 wire configuration

3)()( 10*cos*

1000*V

PV

PI kWkVA 3)()( 10*cos*

*)sin*cos*(**2

VPxrLV kWLdrop

Generalized Equation

5

2

)(10*

cos*

**)sin*cos*(*2

V

PLxrY

kW

L

3

2

)(10*

cos****2

V

PLrZ kW

Line

`Configuration

Balance Load of P

concentrated at the

end of the line

Balance Load of P

uniformly distributed

along the line

Voltage drop factor

for a 50% unbalanced

load that totals P

% Vdrop PLoss(kW) % Vdrop PLoss(kW)Single-Phase

(2-wire)Y Z Y/2 Z/3 1

Three-Phase

(3-wire, delta)Y/2 Z/2 Y/4 Z/6 1.1

Three-Phase

(4-wire, wye)Y/6 Z/6 Y/12 Z/18 1.5

Comparison table, % voltage drop and power loss for different load configuration

Comparison of different line configuration and load condition

23

System

configuration

Load condition/

configuration

Modification of (kW-m) value (k' )

For Single

Phase

If the load concentratedat the end

If the load is evenlydistribute

If the load is unevenly

distribute

For ThreePhase, delta

system

For perfectly balance

load system

First obtain the k'

(kW-m) value as

assuming single

phase for

corresponding type

of load d is t r ibu t ion

system as above

and modify it for

three phase

corresponding to

load balance

For 50% balance load

system

For ThreePhase, wye

system

For perfectly balanceload system

For 50% balance load

system

(km)(kW) L*Pk'km)-(kW k

2

L*Pk'km)-(kW

(km)

(kW)k

(km)x(kW)x L*Pk'km)-(kWk

2

'km)-(kW

kk

8.1

'km)-(kW

kk

6

'km)-(kW

kk

4

'

km)-(kW

k

k

To find the equivalent kW-km value using single phase configuration

24


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20 kW, 3-phase, 50 Hz power with pf=0.6 is to be transmitted at a distance of

360m. Find size of conductor so that voltage drop should be less than 6%.

m)-(kW2.736.0*20L*Pk'km)-(kW (km)(kW) kFind (kW-m) assuming single phasesystem (concentrated load at end)

km)-(kW2.16

2.7

6

k'km)-(kW kmodification of k' (kW-m) value For 3-phase, 4-

wire, wye configuration perfectly balance loadsystem

- Aluminum conductor of ~ 40mm2to be used to keep Vdropless than 6%

- Copper conductor of ~25mm2to be used to keep Vdropless than 6%25

Generator TypesThere are two types of AC Generator (Alternator)

1. Synchronous Generator

- Single Phase / Three Phase- Brush / Brushless type

2. Induction (Asynchronous) Generator

- Single Phase / Three Phase

Induction Generator are

simpler and more reliable

machine

It can withstand 200%

overspeed

less maintenance required.

(No brushes that need

maintenance) It is cheap and easily

available.

Both, Synchronous and Asynchronous

generators are available in single and

three phase. Induction generator and

single phase generator are generally

used for small scheme. For larger

scheme 3 phase synchronous generators

are used.

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Generator Parts

Prime mover: mechanical work which turns the rotor, may be a water turbine, gasturbine, diesel engine...

Armature windings: the conductor in which the output voltage is induced Field windings: the conductors used to produce the electromagnetic field (needs a

DC power supply)

Stator: stationary housing of the generator

Rotor: rotates inside the stator, moved by a prime mover (water turbine, dieselengine etc)

Sliding contacts (slip-rings and brushes): used to conduct the field or armaturecurrent to and from the rotor

Generator

Rotating rectifierExciter rotor

Exciter stator

Fieldwinding

Phase

windings

Rotates

Stationary

Idc

27

V

Load (A)

220 V

Load curve at fixed RPM

V

RPM

220 V

1500

No load curve

V

Load (A)

220 V

Max (A)

Load curve at fixed RPM

(with compounding)

RPM =120 * f

P(1+s)

Where;

s= slip (010%)

p = No. of poles (2,4...)

RPM =120 * f

PWhere;

P = No. of poles

When;

P = 4, f = 50 ;

RPM = 1500

When;

P = 4, f = 50 ; s = 0-5%

RPM = 1500 - 1575

Operating Characteristic of Synchronous Generator

Operating Characteristic of Induction Generator

V

RPM

220 V

1500

No load curve

I.G.

S.G.

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Synchronous Generator Induction Generator

DC Exciter required for fieldexcitation

Magnetizing and excitation power is provided from thepower system towhich it is connected

Synchronization is required for

parallel operation with grid

Synchronization is not required as the machine is

brought above synchronous speed and the breaker is

closed

Complicated construction Simple and rugged construction

Cost is higher Cost is less than synchronous generator

Operation cost is higher Operating cost is less than synchronous generator

Suitable for base load and peak

load duties

Suitable for base load operation

Higher efficiency Slightly lower efficiency than synchronous generator

Comparison between Synchronous and Induction Generator

Generally, Induction Generator are cheaper for small scheme. But, they are not

good for supplying power to motors.

29

Selection of Generator typeSize of scheme Up to 10 kW 10 to 25 kW More than 25 kW

Type of Generator Induction

or

Synchronous

Synchronous

or

Induction

Synchronous

Phase Single

or

Three

Three Three

Factors to be considered for selecting the size of the Generator.

A

Max. ambient Temp.(0C) 20 30 40 50

Temp. factor 1.10 1.06 1.00 0.92

B

Altitude 1000 2000 3000 4000

Altitude factor 1.00 0.93 0.86 0.8

C ELC Correction factor 0.83

D Power factor

When load resistive (light bulbs) only 1.00

When loads are light bulbs + Tube-lights

etc. (Resistive + Inductive)

0.80

Generator KVA =Power output in kW

A x B x C x D 30


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Controlling & protection Panel board (Volt meter, Current meter, Circuit

breaker, Emergency cut-out) Under & over voltage protection

Under & over current protection

MCBs / MCCBs must be rated at less than 110% of

generator rating.

Generator neutral and all metal casing should be

earthed

One Lightning arrestor must be mounted on eachphase on the first pole outside the PH

One arrestor on each phase per km of transmission,

and additional as required so that no consumer is

more than .5 km away from the arrestor31

Voltage Control kVAR Control pf Control

Excitation Control

Generator Control

By pass

Jet deflector

Speed Control

Main Discharge

Niddle

Guide

Wicket

Level Control kW Control

Flow Control

ELC IGC

Load Control

Turbine Control

Hydro Plant Control

32


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Load

Controller

Generator Consumers

Load

Electronic Switch

Ballast

Load

Fuses

A simple Schematic diagram of Load Controller

The purpose of the ELC is to maintain constant speed by accurately adjusting the secondary

load (ballast load).

The ELC sense the generator frequencyWhen the consumers Load decreases, Generator tends to run fast resulting increase in

frequency.

When the frequency increases, the electronic switch move to on position that transfer extra

load to the ballast load. The reverse happens when consumer load increases.

The ballast load must be rated 20% higher than the maximum generator output

ELC is suitable to use with Synchronous Generators

Since it sense only frequency, an AVR is needed with ELC to keep voltage constant.

Electronic Load Controller (ELC)

33

ELC

Main Load

Ballast Load

Main Load draws

full power

No power to

ballast load

Alternator

Supply

20kW

240V

ELC

Main Load

Ballast Load

12 kW used by

Main Load

8 kW chopped to

ballast load

Alternator

Supply

20kW

240V

ELC

Main Load

Ballast Load

No current drawn

to Main Load

20 kW to ballast

load

Alternator

Supply

20kW240V

Load

Controller

Generator Consumers

Load

Electronic Switch

Ballast

Load

Fuses

Working of ELC

34


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Induction Generator Controller (IGC)

The IGC sense the generator voltage

When the voltage is too high, it divert the extra load to the ballast load.

When the voltage is too low, it divert the additional load to the consumers load.

The ballast load of the IGC is same as the maximum output of the plant.

The main function of a Load controller is to maintain constant voltage by keeping

constant load to the generator.

Automatic Voltage Regulator (AVR)

In a simple Synchronous generator, voltage falls as load current increase and vice versa.

AVR is an electronic unit which regulates the voltage within the limit.

AVR regulate the generator voltage by varying the field current.

35

Typical Generator Specification for MHP application

Turbine type: (Ex.: Pelton wheel)

Turbine power : ............ kW

Generator power rating : ............ KVA

Duty : Maximum continuous rated

Operating Speed : .......... RPM

Over speed : 180% continuously

Voltage: 415 V

Phase : 3

Frequency : 50 Hz

Power factor : 0.8

AVR : Voltage regulation 2.5% with under and over voltage andfrequency protection.

Ambient temperature: ......... 0CAltitude : .. ....... msl

Induction Motor as Generator (IMG)

Induction Motor are most commonly used machine. They are simple cheap and plentiful.

It can be used as generator by putting suitable size capacitor across phase winding.

Induction motor are generally used as generator for smaller size machine.

36


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TransformerA transformer is a static electrical

device, which transforms electricalenergy from one voltage level to

another voltage level.

This permits electrical energy to begenerated at relatively low voltages

and transmitted at high voltages

and low currents, thus reducing

voltage drop and line losses.

Need of Transformer After the power distance product gets to about

54 kW-km, even the largest size conductor

(say Dog) cable used becomes insufficient in

even three phase to give the required 10% of

the voltage drop or better.

4 8 12 16 20 24 28 32 36 40 44 48 52 5646

8

10

12

14

16

kW-km

% Vdrop


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11 March 2011

Classification of Transformer

Class -A: Dry type : Core and winding are not contained ininsulating liquid. Heat losses ard dissipated to the ambient air,

hence require large surface area and low current density.

Class -O: Oil immersed type: Core and winding are contained in

mineral oil which is simultaneously work as coolant and insulating.

Transformer Specification

(i) Type Phase

(ii) Installation outdoor

(iii) Rated capacity kVA

(iv) Rated H.V. (Secondary)kV

(v) Rated L.V. (Primary) kV

(vi) Cooling O/A

(vii) Rated frequency 50 Hz

(viii) Primary connection Delta

(ix) Secondary connection Star

(xi) Efficiency not less than 98%

40

9. transmission & distribution sys.pdf

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