Resistance of Pipe Electrode (Ref 9.2.2 of IS: 3043-1987)

Re = 100 x logρ e 4 x l

2 x π x l d

Where l = length of the rod or pipe = 300 cm. d = diameter of the rod or pipe = 3.8 cm.

ρ = resistivity of soil = 312 Ω -m assumed uniform. Re = 100 x 312 loge 4 x 300

2 x π x 300 3.8 = 16.55 loge 315.789

= 41.36 Ω.

Resistance of the Earth Strip 50 x 6 G.I Flat Old Wadi plant

5.1.1 Bulk loading silo

Rs = 100

ρ

Log e 2L2 2

π

L wt

L=Estimated Length Strip : 12500 cm Width of the Strip (t) : 50 x 6 mm Depth of the Buried Strip (W) : 60 cm

Rs = 100 x 312

Log e 2 x 125002

= 2.38 Ω

2x3.14x12500 60x5 There are 3 Nos of earth pits are used, therefore Re = Re /3 = 41.36 / 3 = 13.78 Ω

Rt = Re

x

Rs = 13.78 X 2.38 = 2.03 Ω

Re

+

Rs 13.78+2.38

Thus the Total resistance is less than 3 ohms .

50x6 G.I Flat strip is selected for vertical down conductor and horizondal earth conductor.

CLIENT : ACC -WADI REV/DATE : 0 /24.06.05 PROJECT : LIGHTNING ARRESTOR REQUIREMENT CALCULATION PRE/CHKD : LB/ MK TITLE : LIGHTNING CALCULATION SHEET : Page 33 of 48 5.1.2 Cement storage silo

Rs = 100

ρ

Log e 2L2 2

π

L wt

L=Estimated Length Strip : 14640 cm Width of the Buried Strip (t) : 50 x 6 mm Depth of the Buried Strip (W) : 60 cm

Rs = 100 x 312

Log e 2 x 146402

= 2.0 Ω

2x3.14x14640 60x5 There are 2 Nos of earth pits are used, therefore Re = Re /2 = 41.36 / 2 = 20.68 Ω

Rt = Re

x

Rs = 20.68 X 2 = 1.8 Ω

Re

+

Rs 20.68+2.

Thus the Total resistance is less than 3 ohms .

50x6 G.I Flat strip is selected for vertical down conductor and horizondal earth conductor.

5.1.3 Kiln No:3 Preheater

Rs = 100

ρ

Log e 2L2 2

π

L wt

L=Estimated Length Strip : 30160 cm Width of the Buried Strip (t) : 50 x 6 mm Depth of the Buried Strip (W) : 60 cm

Rs = 100 x 312

Log e 2 x 301602

= 1.1172 Ω

2x3.14x30160 60x5 There are 2 Nos of earth pits are used, therefore Re = Re /2 = 41.36 / 2 = 20.68 Ω

CLIENT : ACC -WADI REV/DATE : 0 /24.06.05 PROJECT : LIGHTNING ARRESTOR REQUIREMENT CALCULATION PRE/CHKD : LB/ MK TITLE : LIGHTNING CALCULATION SHEET : Page 34 of 48

Rt = Re

x

Rs = 20.68 X 1.1172 = 1.054 Ω

Re

+

Rs 20.68+1.1172. Thus the Total resistance is less than 3 ohms .

50x6 G.I Flat strip is selected for vertical down conductor and horizondal earth conductor.

5.1.4 Cement Mill Building

Rs = 100

ρ

Log e 2L2 2

π

L wt

L=Estimated Length Strip : 20405.6 cm Width of the Buried Strip (t) : 50 x 6 mm Depth of the Buried Strip (W) : 60 cm

Rs = 100 x 312

Log e 2 x 20405.62

= 1.568 Ω

2x3.14x20405.6 60x5 There are 2 Nos of earth pits are used, therefore Re = Re /2 = 41.36 / 2 = 20.68 Ω

Rt = Re

x

Rs = 20.68 X 1.568 = 1.4469 Ω

Re

+

Rs 20.68+1.568.

Thus the Total resistance is less than 3 ohms .

50x6 G.I Flat strip is selected for vertical down conductor and horizondal earth conductor.

CLIENT : ACC -WADI REV/DATE : 0 /24.06.05 PROJECT : LIGHTNING ARRESTOR REQUIREMENT CALCULATION PRE/CHKD : LB/ MK TITLE : LIGHTNING CALCULATION SHEET : Page 35 of 48 5.1.5 Cement Mill Building 4 & 5

Rs = 100

ρ

Log e 2L2 2

π

L wt

L=Estimated Length Strip : 20405.6 cm Width of the Buried Strip (t) : 50 x 6 mm Depth of the Buried Strip (W) : 60 cm

Rs = 100 x 312

Log e 2 x 20405.62

= 1.568 Ω

2x3.14x20405.6 60x5 There are 2 Nos of earth pits are used, therefore Re = Re /2 = 41.36 / 2 = 20.68 Ω

Rt = Re

x

Rs = 20.68X 1.568 = 1.4469 Ω

Re

+

Rs 20.68+1.568.

Thus the Total resistance is less than 3 ohms .

50x6 G.I Flat strip is selected for vertical down conductor and horizondal earth conductor.

5.1.6 Cement Mill Control Room Rs = 100

ρ

Log e 2L2 2

π

L wt

L=Estimated Length Strip : 10394.8 cm Width of the Buried Strip (t) : 50 x 6 mm Depth of the Buried Strip (W) : 60 cm

Rs = 100 x 312

Log e 2 x 10394.82

= 2.799 Ω

2x3.14x10394.8 60x5 Only one earth pit is used, therefore Re = Re /1 = 41.36 / 1 = 41.36 Ω

CLIENT : ACC -WADI REV/DATE : 0 /24.06.05 PROJECT : LIGHTNING ARRESTOR REQUIREMENT CALCULATION PRE/CHKD : LB/ MK TITLE : LIGHTNING CALCULATION SHEET : Page 36 of 48

Rt = Re

x

Rs = 41.36 X 2.799 = 2.6 Ω

Re

+

Rs 41.36+2.799.

Thus the Total resistance is less than 3 ohms .

50x6 G.I Flat strip is selected for vertical down conductor and horizondal earth conductor.

5.1.7 Packing Plant

Rs = 100

ρ

Log e 2L2 2

π

L wt

L=Estimated Length Strip : 17730 cm Width of the Buried Strip (t) : 50 x 6 mm Depth of the Buried Strip (W) : 60 cm

Rs = 100 x 312

Log e 2 x 177302

= 1.77 Ω

2x3.14x17730 60x5 There are 2 Nos of earth pits are used, therefore Re = Re /2 = 41.36 / 2 = 20.68 Ω

Rt = Re

x

Rs = 20.68 X 1.77 = 1.68 Ω

Re

+

Rs 20.68+1.77.

Thus the Total resistance is less than 3 ohms .

50x6 G.I Flat strip is selected for vertical down conductor and horizondal earth conductor.

5.1.8 Cement silo Expansion Rs = 100

ρ

Log _e

2L2 2

π

L wt

L=Estimated Length Strip : 32033 cm Width of the Buried Strip (t) : 50 x 6 mm Depth of the Buried Strip (W) : 60 cm

CLIENT : ACC -WADI REV/DATE : 0 /24.06.05 PROJECT : LIGHTNING ARRESTOR REQUIREMENT CALCULATION PRE/CHKD : LB/ MK TITLE : LIGHTNING CALCULATION SHEET : Page 37 of 48

Rs = 100 x 312

Log e 2 x 320332

= 1.07 Ω

2x3.14x32033 60x5 There are 2 Nos of earth pits are used, therefore Re = Re /2 = 41.36 / 2 = 20.68 Ω

Rt = Re

x

Rs = 20.68 X 1.07 = 1.0 Ω

Re

+

Rs 20.68+1.07.

Thus the Total resistance is less than 3 ohms .

50x6 G.I Flat strip is selected for vertical down conductor and horizondal earth conductor.

5.1.9 Kiln No:1& 2 Preheater Rs = 100

ρ

Log _e

2L2 2

π

L wt

L=Estimated Length Strip : 25672 cm Width of the Buried Strip (t) : 50 x 6 mm Depth of the Buried Strip (W) : 60 cm

Rs = 100 x 312

Log e 2 x 256722

= 1.285 Ω

2x3.14x25672 60x5 Only one earth pit is used, therefore

Re = Re /1 = 41.36 / 1 = 41.36 Ω

Rt = Re

x

Rs = 41.36X1.285 = 1.240 Ω

Re

+

Rs 41.36+1.285.

Thus the Total resistance is less than 3 ohms .

50x6 G.I Flat strip is selected for vertical down conductor and horizondal earth conductor.

CLIENT : ACC -WADI REV/DATE : 0 /24.06.05 PROJECT : LIGHTNING ARRESTOR REQUIREMENT CALCULATION PRE/CHKD : LB/ MK TITLE : LIGHTNING CALCULATION SHEET : Page 38 of 48 5.1.10 Raw Mill 4 & 5

Rs = 100

ρ

Log e 2L2 2

π

L wt

L=Estimated Length Strip : 56730cm Width of the Buried Strip (t) : 50 x 6 mm Depth of the Buried Strip (W) : 60 cm

Rs = 100 x 312

Log e 2 x 567302

= 0.64 Ω

2x3.14x56730 60x5 There are 2 Nos of earth pits are used, therefore Re = Re /2 = 41.36 / 2= 20.68 Ω

Rt = Re

x

Rs = 20.68X0.64 = 0.62 Ω

Re

+

Rs 20.68+0.64.

Thus the Total resistance is less than 3 ohms .

50x6 G.I Flat strip is selected for vertical down conductor and horizondal earth conductor.

5.1.11 Coal Mill

Rs = 100

ρ

Log e 2L2 2

π

L wt

L=Estimated Length Strip : 52000cm Width of the Buried Strip (t) : 50 x 6 mm Depth of the Buried Strip (W) : 60 cm

Rs = 100 x 312

Log e 2 x 520002

= 0.64 Ω

2x3.14x52000 60x5

There are 2 Nos of earth pits are used, therefore Re = Re /2 = 41.36 / 2= 20.68 Ω

CLIENT : ACC -WADI REV/DATE : 0 /24.06.05 PROJECT : LIGHTNING ARRESTOR REQUIREMENT CALCULATION PRE/CHKD : LB/ MK TITLE : LIGHTNING CALCULATION SHEET : Page 39 of 48

Rt = Re

x

Rs = 20.68X0.64 = 0.62 Ω

Re

+

Rs 20.68+0.64.

Thus the Total resistance is less than 3 ohms .

50x6 G.I Flat strip is selected for vertical down conductor and horizondal earth conductor.

5.1 Power plant

In document Lightning Calculations (Page 32-39)