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A ) Calculation For Earth Electrodes

I ) Design Data:- a) Fault level – at 11KV – 500MVA, 1Sec ie 26240Amps

- at 3.3KV – 150MVA, 1Sec ie 26240Amp - at 415V – 31MVA, 1Sec ie 43130Amp

b) Soil Resistivity ( ρ ) - 100ohm Mtere c) Earthing Material – Galvanized iron d) Duration of fault level ( t ) – 1Sec

II ) Size of conductor

As = 21.6 x I x √t circular mills = 21.6 x 43130 x √1 = 931608 circular mills ( 1 Circular mill = 0.0005067 Sqmm ) = 472.0 Sqmm Adding 25% extra As = 1.25 x 472 = 590 Sq mm Main earth conductor size chosen = 75 x 8 mm III) Calculation of no of earth electrodes : Dissipation of fault current = 7.57 x 10³ / ( √ ρ x √ t ) Amp / M² Where p = 100ohm – meter and t = 1 Sec Dissipation of fault current = 7.57 x 10³ / ( √100 x √1 ) = 757 Amp / M² Area of Earth electrode = 43130 / 757 = 56.97 M² Adding 10% extra = 56.97 x 1.1 = 62.67 M² Area of 1 Pipe electrode = π x d x l d = 3.8 cm l = 250 cm A = π x 3.8 x 250 = 0.2985 M² No of Pipe electrodes = 62.67 / 0.2985 = 210 Nos Since it is difficult to distribute such a high no of electrodes in the plant area, we will use plate electrodes for earthing of the system B ) Calculation for number of earth electrodes for receiving station. a) Limiting values of Touch Potential ( Et ) and step Potential ( Es ) to be maintained as , Et = ( 165 + 0.25 ρ ) / √ t and Es = ( 165 + ρ) / √ t Where ρ = resistivity of ground in ohm meter = 47.55 ohm –m ( average value ) t = time in sec to clear the fault = 1sec Et = ( 165 + 0.25 x 47.55 ) / √ 1 = 176.89 Volts Es = ( 165 + 47.55 ) / √ 1 = 212.55 Volts b ) Maximum value of attainable Step and Touch potential is given by, Et = ( Kt x Ki x p x I ) / L and Es = ( Ks x Ki x p x I ) / L Where Kt = 1/2π ln ( S² / 16 d D ) + 1/π ln ( 3/4 x 5/6 x ……x (2n-3)/(2n-2) Ks = 100/π ( 1/2D + 1 / (S +D) + 1/2S + 1/3S + …. + n terms ) S = Average distance between two parallel conductors in cm along X-axis and along Y – axis calculated as

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S’ for Y- axis average space = (Total space length along Y-axis ) / ( No of Spaces ) = 20 / 10 = 2.00 M = 200 cm S’ for X- axis average space = (Total space length along X axis ) / ( No of Spaces ) = 30 / 12 = 2.50 M = 250 cm d = equivalent diameter of earth strip in cms = 75 x 8 = π x d² / 4 d = √ ( 75 x 8 x 4 ) / ( 100 x 3.14 ) = 2.76 cm D = depth of burial of earth strip in cms = 50 cms ( IS ) n = no of parallel conductors along X-axis / Y – axis therefore , n = 13 in the X-axis and n = 11 in Y-axis Ki = 0.635 + 0.172 n ρ = Soil resistivity in ohm-m = 47.55 ohm M I = max fault current for 11KV system = 500 / (√3 x 11) = 26.24KA = 26240 Amps L = length of buried conductor in meters = 20 x 11 + 30 x 13 = 610 Ki = 0.655 + 0.172x13 = 2.891 for grid along X – axis And Ki = 0.655 + 0.172x11 = 2.547 for grid along Y– axis To find out Kt we have to find out , Kt1 = for Y axis configuration and Kt2 = for X – axis configuration Kt1 for Y – axis configuration, Kt1 = 1/2π ln ( S² / 16 d D ) + 1/π ln { ( 3/4 x 5/6 x ……x (2n-3)/(2n-2)} = 1/2π ln ( 200² / 16 x 2.76 x 50 ) + 1/π ln { ( 3/4 x 5/6 x ……x (2x11-3)/(2x11-2)} = 1/2π ln ( 18.12 ) + 1/π ln ( 0.3524) = 1/2π x 2.89 + 1/π ( - 1.043 ) = 0.46 – 0.33 = 0.13 Kt2 for X – axis configuration, Kt2 = 1/2π ln ( S² / 16 d D ) + 1/π ln { ( 3/4 x 5/6 x ……x (2n-3)/(2n-2)} = 1/2π ln ( 250² / 16 x 2.76 x 50 ) + 1/π ln { ( 3/4 x 5/6 x ……x (2x13-3)/(2x13-2)} = 1/2π ln ( 28.306 ) + 1/π ln ( 0.3224 ) = 1/2π ( 3.343 ) + 1/π ( -1.1319 ) = 0.17 Kt = Kt1 x Kt2 = 0.13 x 0.17 = 0.022 Et1 for Y – axis configuration,

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= Kt x Ki x ρ x I / L = 0.22 x 2.547 x 47.55 x 26240 / 650 =107.56V Et2 for X – axis configuration, = Kt x Ki x ρ x I / L = 0.22 x 2.891 x 47.55 x 26240 / 650 =122.08V Step Potential Es = Ks x Ki x ρ x I / L Where Ks1 for X-axis configuration, Ks1 = 1/π ( 1/2D + 1/(S+D) + 1/2S + 1/3S + ….. + 1/nS ) = 1/π ( 1/2x50 + 1/(250+50) + 1/2x250 + 1/3x250 + ….. + 1/13x250 ) = 1/π ( 0.01 + 0.003 + 0.0087 ) = 0.0069 Ks2 for Y-axis configuration, Ks1 = 1/π ( 1/2D + 1/(S+D) + 1/2S + 1/3S + ….. + 1/nS ) = 1/π ( 1/2x50 + 1/(200+50) + 1/2x200 + 1/3x200 + ….. + 1/9x200 ) = 1/π ( 0.01 + 0.003 + 0.00044 ) = 0.0066 therefore Ks = Ks1 x Ks2 = 0.0069 x 0.0066 = 0.0000455 Es1 for X – axis = ( 0.0000455 x 2.891 x 47.55 x 26240 ) / 650 = 0.25 Volts Similarly Es2 for Y axis = 0.22 Volts Therefore to restrict step and touch potential within tolerance limit the no of pipes electrodes required are, = (13+9 ) x 2 = 44 As distribution of 44 nos of pipe electrodes in substation area of 20M x 30M is not practicable Therefore equivalent plate electrodes should be given, Area of 44 Nos of pipe electrodes = π x d x l x 44 = 13.13 Sq M Area of one Plate electrode = 1.2 x 1.2 x 2 = 2.88 Sq M No of plate electrodes = 13.13 / 2.88 = 4.55 = 6Nos 6 Nos of Plate electrodes shall be given for main receiving station. Therefore remaining 210 – 44 = 166 electrodes shall be distributed over the plant area. Area of 166 pipe electrodes = 49.54 Sq M Therefore no of plate electrodes required = 17.20 = 18nos A ) Calculation of combined resistance for receiving station, i) Resistance of Plate electrode = R = ρ / 4 √ ( π / A ) = 47.55/4 √ ( π / 2.88) = 11.88 x 1.044 = 12.4 therefore combined resistance of 6nos plates electrodes = 12.4 / 6 = 2.066 ohm

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ii) Resistance of earth strip R = 100 ρ / 2 π l x ln ( 2 l² / wt ) Where ρ = 47.55 ohm-meter l = length of strip buried = 610 x 100 cms w = depth of buried strip = 50cms t = width of strip = 75mm = 7.5cm

therefore R = 100x47.55 / 2 π x 610 x 100 ln ( 2 x (780x100)l² / 50 x 7.5 ) = 0.21 ohm Combined resistance of plate and strip electrode, 1/ R comb = 1/ 2.066 + 1/0.21 = ( 2.276 + 0.21 ) / ( 2.066 x 0.21 ) = 1 / Rc Rcomb = 0.19ohm B) Calculation of combined Resistance of plant Area i) Resistance of plate electrode = 12.4ohm therefore combined resistance of 18nos of plate electrodes = 12.4 / 18 = 0.69 ohm ii) Resistance of strip conductor R = 100 ρ / 2 π l x ln ( 2 l² / wt ) = 100 x 100 / 2 π x 660 x 100 x ln ( 2 x ( 660 x 100) ² / 50 x 7.5 ) = 0.409 ohm therefore combined resistance of plate and strip Rcomb = ( 0.69 + 0.409 ) / 0.69 x 0.409 = 0.26 ohm Distribution of electrodes over plant area, 1) Main receiving station - -------------------- --------- 6 nos Lightning arrestor = ------------------1 no Neutral earthing = --------------------3 nos Equipment earthing = ----------------2 nos 2) Load Center – 1 -------------------------------- 2 nos 3) Load center –2 / load center – 6 ------ ------------ 4 nos ( Raw mill / cement mill ) 4) Packing Plant - ------------------------------------- 1 nos 5) Blending Silo and Compressor House - ---------- 1 nos 6) Load Center- 3 ------------------------------------ -2 nos 7) Kiln feed / Heat exchanger --- -----------------------3 nos 8 ) Load Center- 4 - ------------------------------------2 nos 9) Cooler Department / coal mill – ----------------------2 nos 10) Coal crusher = ----------------------------------------01no Total = -----------------------------------------------------24 nos

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