Power Transmission CHAPTER 1. GRID SYSTEM 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 Xmer 25kV / 32kV Sending station. Step down Xmer 132kV / 33kV Receiving station. Step down Xmer 33kV Heavy Industry. Step down Xmer 11kV Light Industry Step down Xmer 3.3kV Substations Step down Xmer 415V/ 240V Consumer 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 . OA 2 OB = OA 3 2 OC should be twice the value of OB , Hence OC = 2 x OA 3 2 OC = OA 3 VRY = OA 3 VL = 3 Vph FOR DELTA CONFIGURATION ( ) IL = 3 Iph 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 = 3 Vp ————— 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 3 3 = 3 . 3 . VL. IL Cosq 3 P = 3 VL. IL Cosq For Delta Connection.
VL = Vp ————— 4 IL = 3 Iph ————— 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 3 3 :. P = 3 VL. IL Cosq NEUTRAL CURRENT IN UNBALACED CIRCUIT.
Cos 60 = adj = adj hyp IB adj = IB Cos 60 Cos 60 = adj = adj hyp IY adj = IY Cos 60 Therefore horizontal component, HC = IR – IY Cos 60 – IB Cos 60 Sin 60 = opp = opp hyp IB opp = IB Sin 60 Sin 60 = opp = opp hyp IY 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 hyp H.C q = Tan V.C H.C >From this we can obtain the power factor. NEUTRAL CURRENT IN UNBALACED CIRCUIT. Cos 60 = adj = adj hyp IB adj = IB Cos 60 Cos 60 = adj = adj hyp IY adj = IY Cos 60 Therefore horizontal component, HC = IR – IY Cos 60 – IB Cos 60 Sin 60 = opp = opp hyp IB opp = IB Sin 60 Sin 60 = opp = opp hyp IY 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 hyp H.C q = Tan V.C H.C >From this we can obtain the power factor.