Power Transmission in a country is usually done through what is known as a Grid System. The Grid

System consists of extensive interconnected transmission network supplying the whole country. Its supply

from a small no. of very large and highly effective power stations. The basic network is usually 132kHZ.

For a very high industrialised nation they use 275,475,800,1250 kV. Most consumers receive supplies from

medium voltage distribution system of 3.3kV, 415V, 240V. For heavy industry consumer they may be

supplied with 11 or 33kV.

The generators produce electrical power at 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 station will step – down the

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

Generating station 11kV / 25kV

Step up Xmer25kV / 32kVSending station.

Step down Xmer 132kV / 33kVReceiving station.

Step down Xmer33kVHeavy Industry.

Step down Xmer11kVLight Industry

Step down Xmer3.3kVSubstations

Step down Xmer415V/ 240VConsumer

fig. 1, Single Line Diagram .

THE PURPOSE OF THE GRID SYSTEM.

The purpose of the grid system is to maintain a secure supply of electricity at a standard voltage and

frequency to consumers throughout the country. Having stated its purpose, we can now list several

advantages that have resulted from its introduction:

1. security of supplies;

2. standardisation of frequency and voltages;

3. economy;

4. the ability to transmit very large loads for considerable distance without loss; and

5. the ability to transfer electricity to and from different parts of the country and to step up / down the

voltages using Xmers (Transformers).

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

FUNCTION OF THE GRID SYSTEM.

In order to fill its purpose, The grid system must function in the following way. The National Grid Control

Centre in association with the various grid control centres around the country, estimates the load required in

different areas each day. This information is then used to arrange to purchase the countries power

depending on the demand. In this way stations are used to their maximum efficiency, which in turn reduces

the cost of generation. Due to the fact that the system is interconnected, bulk supply points can be fed from

other areas, should a failure of the usual supply occur.

DISADVANTAGES OF A.C TRANSMISSION:-

1. Skin effect – cable losses.

2. Heavy losses hence efficiency is reduced.

3. For high voltage higher harmonics are produced, hence it interferes with communication lines.

SYSTEM LAYOUT OF A GRID.

3- f (PHASE), 4 WIRE SYSTEM .

Vph= phase voltage

VL= Line Voltage

IL= Line Current

Iph= Phase Current

FOR STAR CONFIGURATION ( Y).

VL= 3 Vph

IL = Iph

OB =3 .

OA2

OB = OA 3

2

OC should be twice the value of OB ,

Hence OC = 2 x OA3

2

OC = OA 3

VRY = OA 3

VL = 3Vph

FOR DELTA CONFIGURATION ( )

IL = 3Iph

VL = Vph

If 3 loads are identical in every way i.e impedance and phase angle. Then the current in the 3 lines would

be identical the resultant current returning down the neutral would therefore be zero. The load in this case is

know as a balanced load. In actual practice its hard to find it exactly balanced. Hence the neutral wire is left

to carry the leftover current. The advantages of this system compared with both a single phase and 3 phase

6 wire system is like this. Suppose 3 identical loads are to be supplied with 200A each. The 2 lines for a

single phase would carry a total of 600A.. This conductor (C.S.A) would only need to be 1/3 that of single

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

conductor saves an increase in the 2nd case where in the 1st case if the proper cable selection is not used

overheating of the cable occurs, this will later result in a short circuit.

POWER DISSIPATION IN STAR AND DELTA 3 – PHASE CONNECTION.

P = VI

Pph = Vph.Iph

Pph = Vph . Iph Cos q

P3 – f = 3 Vph Iph Cosq————-1

For Star Connection.

VL =3Vp ————— 2

IL = Iph————— 3

Take 2 & 3 substitute into equation -1.

P3 – f = 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 ————— 5

Iph = IL—————- 6

3

Take 4 & 5 put it into 1.

P= 3VL. IL. Cosq

3

=3. VL . IL Cosq

3

= 3. 3. VL IL Cos q

33

:. P =3VL. IL Cosq

NEUTRAL CURRENT IN UNBALACED CIRCUIT.

Cos 60 = adj = adj

hypIB

adj = IB Cos 60

Cos 60 = adj = adj

hypIY

adj = IY Cos 60

Therefore horizontal component, HC = IR – IY Cos 60 – IB Cos 60

Sin 60 = opp = opp

hypIB

opp = IB Sin 60

Sin 60 = opp = opp

hypIY

opp = IY Sin 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 q = opp = V.C

hypH.C

q = TanV.C

H.C

>From this we can obtain the power factor.

NEUTRAL CURRENT IN UNBALACED CIRCUIT.

Cos 60 = adj = adj

hypIB

adj = IB Cos 60

Cos 60 = adj = adj

hypIY

adj = IY Cos 60

Therefore horizontal component, HC = IR – IY Cos 60 – IB Cos 60

Sin 60 = opp = opp

hypIB

opp = IB Sin 60

Sin 60 = opp = opp

hypIY

opp = IY Sin 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 q = opp = V.C

hypH.C

q = TanV.C

H.C

>From this we can obtain the power factor.