Chapter 1 grid program essay

Power Indication in a country is usually performed through what is known as a Main grid System. The Grid

System contains extensive connected with each other transmission network supplying the whole country. It is supply

from a small number of huge and successful power areas. The basic network is usually 132kHZ.

For a very high industrialised nation they use 275, 475, 800, 1250 kV. Many consumers get supplies via

channel voltage division system of three or more. 3kV, 415V, 240V. For heavy industry consumer they could be

provided with 11 or 33kV.

The generators develop electrical power by 11kV / 25kV and it is stepped up by using a step- up

Transformer (Xmer) to a value of 132kV before it is transmitted. The receiver place will stage down the

voltage to a value of 33kV by various distributions centres.

Generating station 11kV / 25kV

Boost Xmer25kV as well as 32kVSending station.

Step straight down Xmer 132kV / 33kVReceiving station.

Stage down Xmer33kVHeavy Industry.

Step down Xmer11kVLight Industry

Step down Xmer3. 3kVSubstations

Step down Xmer415V/ 240VConsumer

fig. 1, One Line Picture.

THE PURPOSE OF THE GRID SYSTEM.

The purpose of the grid method is to maintain a secure way to obtain electricity in a standard voltage and

frequency to consumers through the entire country. Having stated it is purpose, we could now list several

advantages which have resulted from the introduction:

1 . security of supplies

2 . standardisation of frequency and voltages

three or more. economy

5. the ability to send very large tons for substantial distance with no loss, and

5. to be able to transfer electric power to and from various areas of the country and step up as well as down the

voltages employing Xmers (Transformers).

6. Convenient way to convert A. C to D. C but the reverce is expensive

FUNCTION WITH THE GRID SYSTEM.

In order to load its goal, The grid system must function in the following approach. The National Grid Control

Middle in association with the many grid control centres about the country, quotes the load required in

different areas each day. This information can then be used to organise to purchase the countries electric power

with regards to the demand. This way stations are used to their maximum efficiency, which reduces

the cost of era. Due to the fact that the machine is connected with each other, bulk source points can be fed coming from

other areas, should a failure of the typical supply arise.

DRAWBACKS OF A. C TRANSMISSION: –

1 . Skin effect wire losses.

installment payments on your Heavy failures hence effectiveness is lowered.

3. To get high voltage bigger harmonics will be produced, hence it interferes with communication lines.

SYSTEM STRUCTURE OF A MAIN GRID.

3- farreneheit (PHASE), four WIRE PROGRAM.

Vph= phase volts

VL= Collection Voltage

IL= Line Current

Iph= Phase Current

PERTAINING TO STAR SETUP ( Y).

VL= several Vph

ELLE = Iph

OB =3.

OA2

OB = OA several

2

OCCITAN should be twice the value of OB

Hence OCCITAN = 2 x OA3

2

OC = OA several

VRY = OA three or more

VL = 3Vph

FOR DELTA CONFIGURATION ( )

ARIANNE = 3Iph

VL = Vph

In the event 3 a lot are the same in every approach i. elizabeth impedance and phase position. Then the current in the several lines will

end up being identical the resultant current returning throughout the neutral would therefore always be zero. Force in this case is usually

find out as a balanced load. In actual practice its difficult to find it specifically balanced. Therefore the fairly neutral wire is left

to carry the leftover current. The advantages of the system in contrast to both a single phase and 3 stage

6 wire method is like this. Assume 3 identical loads have to be supplied with 200A each. The 2 lines for any

sole phase could carry an overall total of 600A.. This caudillo (C. S. A) will only need to become 1/3 regarding single

phase system but becoming 6 lines it would nevertheless be the 50mA current of conductor material. Hence the

caudillo saves a rise in the 2nd circumstance where in the 1st circumstance if the right cable assortment is certainly not used

overheating from the cable arises, this will later on result in a short circuit.

POWER WASTE IN STAR AND DELTA 3 STAGE CONNECTION.

S = VI

Pph sama dengan Vph. Iph

Pph sama dengan Vph. Iph Cos queen

P3 farreneheit = several Vph Iph Cosq-1

Intended for Star Connection.

VL =3Vp 2

ARIANNE = Iph 3

Consider 2 & 3 alternative into formula -1.

P3 f sama dengan 3. VL. IL Cos q

3

= 3. 3. VL IL Cos q

33

= 3. 3. VL. IL Cosq

3

P=3VL. IL Cosq

For Delta Connection.

VL = Vp 4

IL =3Iph five

Iph sama dengan IL- 6

3

Have 4 & 5 said into 1 )

P= 3VL. IL. Cosq

3

=3. VL. IL Cosq

3

= a few. 3. VL IL Cos q

33

:. P =3VL. IL Cosq

NEUTRAL CURRENT IN UNBALACED CIRCUIT.

Cos 60 sama dengan adj sama dengan adj

hypIB

adj = IB Cos 60

Cos 62 = adj = adj

hypIY

adj = IY Cos 62

For that reason horizontal part, HC sama dengan IR IY Cos 62 IB Cos 60

Trouble 60 = opp sama dengan opp

hypIB

opp = IB Sin 60

Trouble 60 sama dengan opp = opp

hypIY

opp = IY Desprovisto 60

Therefor vertical components, V. C = IB Sin 60- IY Sin 60

To find Neutral Current

IN = H. C+ V. C

IN=H. C+ V. C

Tan queen = opp = V. C

hypH. C

queen = TanV. C

They would. C

&gt, From this we are able to obtain the electric power factor.

SIMPLE CURRENT IN UNBALACED OUTLET.

Cos sixty = adj = adj

hypIB

adj = IB Cos 62

Cos 60 sama dengan adj = adj

hypIY

adj = IY Cos 60

Therefore side to side component, HC = VENTOSEAR IY Cos 60 IB Cos sixty

Sin 60 = opp = opp

hypIB

opp = IB Sin sixty

Sin 62 = opp = opp

hypIY

opp = IY Sin 60

Therefor top to bottom components, V. C = IB Bad thing 60- IY Sin sixty

To find Fairly neutral Current

IN = L. C+ V. C

IN=H. C+ Sixth is v. C

Tan q sama dengan opp sama dengan V. C

hypH. C

q sama dengan TanV. C

H. C

&gt, From this we can obtain the power component.