CHAPTER 1. GRID SYSTEM Essay Example For Students

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 / 25kVStep up Xmer25kV / 32kVSending station. Step down Xmer 132kV / 33kVReceiving station. Step down Xmer33kVHeavy Industry. Step down Xmer11kVLight IndustryStep down Xmer3.3kVSubstationsStep down Xmer415V/ 240VConsumerfig. 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; and5. 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 expensiveFUNCTION 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 voltageVL= Line VoltageIL= Line CurrentIph= Phase CurrentFOR STAR CONFIGURATION ( Y). VL= 3 VphIL = Iph OB =3 . OA2 OB = OA 3 2OC should be twice the value of OB ,Hence OC = 2 x OA3 2 OC = OA 3VRY = OA 3 VL = 3VphFOR DELTA CONFIGURATION ( )IL = 3IphVL = VphIf 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 n ot used overheating of the cable occurs, this will later result in a short circuit.POWER DISSIPATION IN STAR AND DELTA 3 PHASE CONNECTION. .ufcea00b610b3632ad1bf052ec8e1a6c1 , .ufcea00b610b3632ad1bf052ec8e1a6c1 .postImageUrl , .ufcea00b610b3632ad1bf052ec8e1a6c1 .centered-text-area { min-height: 80px; position: relative; } .ufcea00b610b3632ad1bf052ec8e1a6c1 , .ufcea00b610b3632ad1bf052ec8e1a6c1:hover , .ufcea00b610b3632ad1bf052ec8e1a6c1:visited , .ufcea00b610b3632ad1bf052ec8e1a6c1:active { border:0!important; } .ufcea00b610b3632ad1bf052ec8e1a6c1 .clearfix:after { content: ""; display: table; clear: both; } .ufcea00b610b3632ad1bf052ec8e1a6c1 { display: block; transition: background-color 250ms; webkit-transition: background-color 250ms; width: 100%; opacity: 1; transition: opacity 250ms; webkit-transition: opacity 250ms; background-color: #95A5A6; } .ufcea00b610b3632ad1bf052ec8e1a6c1:active , .ufcea00b610b3632ad1bf052ec8e1a6c1:hover { opacity: 1; transition: opacity 250ms; webkit-transition: opacity 250ms; background-color: #2C3E50; } .ufcea00b610b3632ad1bf052ec8e1a6c1 .centered-text-area { width: 100%; position: relative ; } .ufcea00b610b3632ad1bf052ec8e1a6c1 .ctaText { border-bottom: 0 solid #fff; color: #2980B9; font-size: 16px; font-weight: bold; margin: 0; padding: 0; text-decoration: underline; } .ufcea00b610b3632ad1bf052ec8e1a6c1 .postTitle { color: #FFFFFF; font-size: 16px; font-weight: 600; margin: 0; padding: 0; width: 100%; } .ufcea00b610b3632ad1bf052ec8e1a6c1 .ctaButton { background-color: #7F8C8D!important; color: #2980B9; border: none; border-radius: 3px; box-shadow: none; font-size: 14px; font-weight: bold; line-height: 26px; moz-border-radius: 3px; text-align: center; text-decoration: none; text-shadow: none; width: 80px; min-height: 80px; background: url(https://artscolumbia.org/wp-content/plugins/intelly-related-posts/assets/images/simple-arrow.png)no-repeat; position: absolute; right: 0; top: 0; } .ufcea00b610b3632ad1bf052ec8e1a6c1:hover .ctaButton { background-color: #34495E!important; } .ufcea00b610b3632ad1bf052ec8e1a6c1 .centered-text { display: table; height: 80px; padding-left : 18px; top: 0; } .ufcea00b610b3632ad1bf052ec8e1a6c1 .ufcea00b610b3632ad1bf052ec8e1a6c1-content { display: table-cell; margin: 0; padding: 0; padding-right: 108px; position: relative; vertical-align: middle; width: 100%; } .ufcea00b610b3632ad1bf052ec8e1a6c1:after { content: ""; display: block; clear: both; } READ: Racial profiling EssayP = VIPph = Vph.IphPph = Vph . Iph Cos qP3 f = 3 Vph Iph Cosq-1For Star Connection. VL =3Vp 2IL = Iph 3Take 2 3 substitute into equation -1. P3 f = 3. VL . IL Cos q3 = 3. 3. VL IL Cos q33 = 3 .3.VL. IL Cosq3 P=3VL. IL CosqFor Delta Connection. VL = Vp 4IL =3Iph 5Iph = IL- 6 3Take 4 5 put it into 1. P= 3VL. IL. Cosq 3=3. VL . IL Cosq 3= 3. 3. VL IL Cos q33:. P =3VL. IL CosqNEUTRAL CURRENT IN UNBALACED CIRCUIT. Cos 60 = adj = adj hypIBadj = IB Cos 60 Cos 60 = adj = adjhypIY adj = IY Cos 60 Therefore horizontal component, HC = IR IY Cos 60 IB Cos 60Sin 60 = opp = opp hypIBopp = IB Sin 60Sin 60 = opp = opp hypIYopp = IY Sin 60Therefor vertical components, V.C = IB Sin 60- IY Sin 60To find Neutral Current,IN = H.C+ V.CIN=H.C+ V.CTan q = opp = V.ChypH.Cq = TanV.CH.CFrom this we can obtain the power factor. NEUTRAL CURRENT IN UNBALACED CIRCUIT. Cos 60 = adj = adj hypIBadj = IB Cos 60 Cos 60 = adj = adjhypIY adj = IY Cos 60 Therefore horizontal component, HC = IR IY Cos 60 IB Cos 60Sin 60 = opp = opp hypIBopp = IB Sin 60Sin 60 = opp = opp hypIYopp = IY Sin 60Therefor vertical components, V.C = IB Sin 60- IY Sin 60To find Neutral Current,IN = H.C+ V.CIN=H.C+ V.CTan q = opp = V.ChypH.Cq = TanV.CH.CFrom this we can obtain the power factor.