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CHAPTER 1. GRID SYSTEM Essay

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.

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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.

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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.

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CHAPTER 1. GRID SYSTEM Essay
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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 in
2018-12-27 03:50:47
CHAPTER 1. GRID SYSTEM Essay
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