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FICHTNER INDIA STANDARD CALCULATIONS DOC. NO.

FOR REV. NO.

SIZING OF EARTHING GRID DATE

(IEEE METHOD) PAGE NO.

EARTHING CALCULATIONS1.0 INPUT DATA

1.1 Fault Current If kA = 31.5 kA

1.2 Duration of Fault current tc secs. = 3 secs.

for conductor sizing

1.3 Total Fault Current considered I kA = 31.5 kA

Step & Touch potential

1.4 Ratio of Soil return current to Ig / I = 0.6

Fault current (assumed)

1.5 Fault return current through Ig kA = 18.9 kA

actual Earth

1.6 Duration of shock current for ts secs. = 0.5 secs.

potential gradient calculations

1.7 Length of area to be earthed L M = 150 M

1.8 Breadth of area to be earthed B M = 65 M

1.9 Soil Resistivity of upper layer r1 = 50

1.10 Soil Resistivity of lower layer r2 = 50

1.11 Surface Soil Resistivity = 3000

1.12 Design ambient temperature Ta = 45

1.13 Depth of burial of conductor h M = 0.6 M

1.14 Area of Plot Ap = 9750 M

Wm Wm

Wm Wm

rs Wm Wm

0C 0C

M2

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2.0 CONDUCTOR SIZING2.1 Material of conductor = MS2.2 Type of conductor = ROD2.3 Type of joint = WELDING

2.4 Conductor resistivity = 15

2.5 Conductor Density SW = 7.86

2.6 Specific heat SH = 0.1142.7 Thermal co-efficient of

resistivity at temp. Tr = 0.004232.8 Reference temp. for

material constant Tr = 20

2.9 Maximum allowable temp. Tm = 620

2.10

A = If* TCAP --As per IEEE 80, eqn.33ln 1+ Tm-Ta Ko+Ta

where TCAP = 4.184*SW*SH = 3.749 --As per IEEE 80, eqn.32and Ko = = 216.407 --As per IEEE 80, page 65

Therefore, A = 658.159Considering Corrosion Allowance of 30 % --As per CBIP Tech.

Area of cross-section required is 855.606 report no. 5

Conductor Size chosen is 33.006 mm dia MS Strip

rr mW/cm3 mW/cm3

gm/cm3 gm/cm3

Cal/gm/0C Cal/gm/0C

ar

0C 0C0C 0C

Conductor cross-section, A in mm2

tc*ar*rr*104

J/cm3/0C1/ar - Tr

mm2

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Actual conductor cross-section area = 1017.88Therefore d = conductor diameter in M = 0.036 M

mm2

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3.0 TOLERABLE STEP AND TOUCH POTENTIAL

3.1 hs = 0 --depends on thickness of crushed gravel spreaded throughout the plot3.2 K -0.96721 --As per IEEE-80 page 40

3.3 Cs = 0.55 --As per Fig.-8 / page 41 of IEEE-80

3.4 = --As per IEEE 80, eqn.24 for average

human weight of 50kg


human weight of 50kgTherefore,

Volts = 1788.13 V

Volts = 570.069 V4.0 INITIAL DESIGN4.1 Length of area considered M = 150 M4.2 Breadth of area considered M = 65 M

4.3 Spacing between conductors D M = 5 MNo. of conductors on either direction

na = 31 --As per IEEE 80, page 41

nb = 14 --As per IEEE 80, page 41

4.4 Total Length of buried conductorLc M = 4115 M

4.5 Total no. of earthing pitsNp = 25 Nos.

Rod depth dr M = 3 MTherefore Rod Length Lr M = 75 M

= r-rs =

r+rs

Tol-Estep (1000+6*rs*Cs)*0.116/sqrt(ts)

Tol-Etouch (1000+1.5*rs*Cs)*0.116/sqrt(ts)

Tolerable Estep

Tolerable Etouch

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4.6 Total LengthTL=Lc+1.15Lr M = 4201.25 M -As per IEEE 80, page 114 & page 115

5.0 GRID RESISTANCE

5.1 = 0.235 W --As per IEEE 80, eqn.40

6.0 CALCULATION OF MESH AND STEP(ATTAINABLE) VOLTAGES

6.2 Es = 519.5815 V --As per IEEE 80, eqn.73

TL Tolerable 1788.1326.1 Em = 506.3446 V --As per IEEE 80, eqn.71

TL Tolerable 570.0695where,Km

= 0.375934 --As per IEEE 80, eqn.68

where,Kii for grids with ground rods / earth pits predominantly along the perimeter 1ho(reference depth of burial) = 1 M (assumed)

Spacing between conductors 'D' in M = 5 M= 20.8326667

Kh = = 1.26491106

Ks = 0.385761 --As per IEEE 80, eqn.74

Ki = 0.656+.172nx 5.988 --As per IEEE 80, eqn.69where, nx = max(na,nb)

Grid resistance Rg=r é 1 + 1 1+ 1 ë TL Ö20Ap 1+hÖ20/Ap

= r*Ig*Ks*Ki

= r*Ig*Km*Ki

=1 ( ln D 2 + (D+2h)2 - h ) + Kii ln ( 8 )) 2p 16hd 8Dd 4d Kh p(2n-1)

n = Öna*nbÖ(1+h/ho)

=1 1 + 1 + 1 (1-0.5n-2)

p 2h D+h D

(

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1.2 Duration of Fault current tc secs. = 0.5 secs.







actual Earth











Wm Wm

Wm Wm

rs Wm Wm

0C 0C

M2

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2.10






rr mW/cm3 mW/cm3

gm/cm3 gm/cm3

Cal/gm/0C Cal/gm/0C

ar

0C 0C0C 0C


tc*ar*rr*104

J/cm3/0C1/ar - Tr

mm2

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Actual conductor cross-section area = 34Therefore d = conductor diameter in M = 0.01 M

mm2

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3.1 hs = 0.102 --depends on thickness of crushed gravel spreaded throughout the plot3.2 K -0.72414 --As per IEEE-80 page 40






Volts = 2686.582 V






4.5 Total no. of earthing pitsNp = 0 Nos.

Rod depth dr M = 3 MTherefore Rod Length Lr M = 0 M

= r-rs =

r+rs



Tolerable Estep

Tolerable Etouch

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4.6 Total LengthTL=Lc+Lr M = 1540 M -As per IEEE 80, page 114 & page 115

5.0 GRID RESISTANCE

5.1 = 2.776 W --As per IEEE 80, eqn.40

Ground Potential Rise 5296.02351 Volt


6.2 Es Actual = 457.29519 V --As per IEEE 80, eqn.73

TL Tolerable 2686.58156.1 Em Actual = 1001.6143 V --As per IEEE 80, eqn.71

TL Tolerable 838.16902 NOT OK, NEED TO ADD RODS AROUD PERIPHERYwhere, SEE NEXT SHEETKm

= 0.88955859 --As per IEEE 80, eqn.68

where,Kii for grids with ground rods / earth pits predominantly along the perimeter 0.5700634 Without Ground Rods, Kii is not 1, use formula

ho(reference depth of burial) = 1 M (assumed)

Spacing between conductors 'D' in M = 7 M= 11

Kh = = 1.22474487

Ks = 0.40613523 --As per IEEE 80, eqn.74


= r*Ig*Ks*Ki

= r*Ig*Km*Ki



=1 1 + 1 + 1 (1-0.5n-2)

p 2h D+h D

(

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Ki = 0.644+.148n 2.272 --As per IEEE 80, eqn.69

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1.2 Duration of Fault current tc secs. = 0.5 secs.







actual Earth











Wm Wm

Wm Wm

rs Wm Wm

0C 0C

M2

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2.10






rr mW/cm3 mW/cm3

gm/cm3 gm/cm3

Cal/gm/0C Cal/gm/0C

ar

0C 0C0C 0C


tc*ar*rr*104

J/cm3/0C1/ar - Tr

mm2

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Actual conductor cross-section area = 34Therefore d = conductor diameter in M = 0.01 M

mm2

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3.1 hs = 0.102 --depends on thickness of crushed gravel spreaded throughout the plot3.2 K -0.72414 --As per IEEE-80 page 40






Volts = 2686.582 V






4.5 Total no. of earthing pits Lp 280 MNp = 20 Nos.

Rod depth dr M = 7.5 MTherefore Rod Length Lr M = 150 M

= r-rs =

r+rs



Tolerable Estep

Tolerable Etouch

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4.6 Total LengthTL=Lc+Lr M = 1690 M -As per IEEE 80, page 114 & page 115

5.0 GRID RESISTANCE

5.1 = 2.753 W --As per IEEE 80, eqn.40

Ground Potential Rise 5252.03673 Volt


6.2 Es Actual = 549.11079 V --As per IEEE 80, eqn.73

TL Tolerable 2686.58156.1 Em Actual = 749.05861 V --As per IEEE 80, eqn.71

TL Tolerable 838.16902where, 1786.36434367398 0.41932Km

= 0.77168339 --As per IEEE 80, eqn.68

where,Kii for grids with ground rods / earth pits predominantly along the perimeter 1 Without Rods, Kii is set to unity = 1

ho(reference depth of burial) = 1 M (assumed)

Spacing between conductors 'D' in M = 7 M= 11

Kh = = 1.22474487

Ks = 0.40613523 --As per IEEE 80, eqn.74


= r*Ig*Ks*Ki

= r*Ig*Km*Ki



=1 1 + 1 + 1 (1-0.5n-2)

p 2h D+h D

(

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Ki = 0.644+.148n 2.272 --As per IEEE 80, eqn.69

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Schwarz's Formula

Rg = R1*R2 - R12 -0.102282 WR1 + R2 - 2*R12

where,

R1 = 0.198113 W

R2 = 0.695697 W

R12 = 0.190316 W

Soil Resistivity 5050

---- for rods penetrating 50 to 2 layers---- for rods and grid conductors 50 in the same layer---- for uniform soil

5050

where, H is the upper layer thickness 2 m

Total Length of buried conductor 4115 m

L = Length of plot plan 150 m

B = Breadth of plot plan 65 mh' = sqrt(d1*h) or [0.5*d1 if h = 0] 0.146969

where,

h = Depth of burial of conductor 0.6 m

d1 = dia of the grid conductors in m 0.036 m

A = Area of Plot 9750 m²

0.006076K1 = -0.04*L / W + 1.41 ---- for h = 0 1.084615

-0.05*L / W + 1.2-0.05*L / W + 1.13

K2 = 0.15*L / W + 5.5 ---- for h = 0 4.9107690.10*L / W + 4.68-0.05*L / W + 4.4

n = no. of ground rods 25

dr = Rod depth 3 m

d2 = dia of the ground rods in m 0.036 m

[r/(p*LC)]*[ln(2*LC/h') + K1*(LC/ÖA) - K2]

[ra/(2*n*p*dr)]*[ln(8*dr/d2) - 1 + 2*K1*(dr/ÖA)*(Ön - 1)²]

[ra/(p*LC)]*[ln(2*LC/dr) + K1*(LC/ÖA) - K2 + 1]

r = W-mra = Apparent soil resistivity as seen by a ground rod in W-m W-m

dr*(r1*r2) / [r2*H + r1(dr - H)] W-m

dr*(r1*r2) / [r2*(H-h) + r1(dr + h - H)] W-m

r1 = r2 r1 = W-mr2 = W-m

LC =

h/ÖA =

---- for h / ÖA = 0.1---- for h / ÖA = 0.16667

---- for h / ÖA = 0.1---- for h / ÖA = 0.16667

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