Question
A 400 KVA ,11 KV/415 volt transformer is to be installed at the premises of
a proposed industries. The fault level at the 11 KV sides of the consumers
Premises is 250 MVA the earth resistivity at the site was measured and
Found to be 150 ohm meter find the number of earth electrode required
And size of earthing conductor on HT and LT side .assume Duration of earth
Fault as 3 second.
The percentage of impedance of the transformer is 5%
MALAYALAM NOTE CLICK HERE
Answer
1. Transformer Capacity = 400 KVA
2. fault level at 11 KV = 250 MVA
3. Soil Resistivity = 150 ohm meter
4. Duration of Earth Fault = 3 second
5. percentage impedance of the transformer = 5 %
6. Assume base MVA = 100
7. earth plate size = 1.2 × 1.2 × 0.12 mtr (Standerd)
8. Area of plate Both side =2.88 m2
9. Current Density assumed as = 118 A / Sq mm for Copper
Formula applicable =
Power = √(3 )×V I
I =P/(√3×V)
For rectifying current density =〖7.57 ×10 〗^3/√(ϱ×t) (Standerd value)
Calculation
Fault level at 11 KV =250 MVA
There for percentage impedance at 11 KV side assuming 100 MVA as Base=
= (Base MVA)/(11 KV side fault level)
= (100 )/250 × 100 = 40 %
Transformer 400 KVA Impedance = 5 %
There for
Impedance of the transformer at 100 MVA Base =
= (transformer percentage impedence )/(transformer capacity in MVA) × 100=
(5 )/(.400) × 100 = 1250 %
There for
Total impedance of LT side of the transformer = 40 + 1250 = 1290 %
LT side
Fault level at LT side = (Base MVA×100)/(Total Impedence) × 100
Fault level at LT side = 100/1290 × 100 = 7.75 MVA
Fault current LT side = (Fault level in Volt Ampiar)/(√3×415 volt)
Fault current LT side = (7.75 ×〖10〗^6)/(√3×415 volt) = 10781.84 A
Current density Assumed as = 118 A / Sq mm For Copper and there for Size of
the earth Conductor Required =
=(Fault current )/(Current density of the Copper) = 10781.84/118 = 91.37 mm2
Standerd size of the copper 25× 6 copper strip of 150 Sq mm
There for we can use 25 × 6 mm copper strip for earthing
HT Side
Fault current at 11 KV side =(Fault current at 11 KV side )/(√3×11)
Fault current at 11 KV side =(250 ×1000)/(√3×11) = 131121.59 A (13122 A)
Permissible current density = 〖7.57 ×10 〗^3/√(ϱ×t)
Permissible current density = 〖7.57 ×10 〗^3/√(150×3 ) = 356.8 Ambiar/m2
Total area of plate electrode required for the fault level current =
= (Fault level current )/(current density) = (13122 )/356.8 = 36.77 m2
Hence number of plate earth required (Plate) =
= (Total area of plate electrode required for the fault level current )/(Area of the plate Both side)
= (36.77 )/2.88 =12.77
Number of Earth Plate= 12.77 = 13 plat
MALAYALAM NOTE CLICK HERE
A 400 KVA ,11 KV/415 volt transformer is to be installed at the premises of
a proposed industries. The fault level at the 11 KV sides of the consumers
Premises is 250 MVA the earth resistivity at the site was measured and
Found to be 150 ohm meter find the number of earth electrode required
And size of earthing conductor on HT and LT side .assume Duration of earth
Fault as 3 second.
The percentage of impedance of the transformer is 5%
MALAYALAM NOTE CLICK HERE
Answer
1. Transformer Capacity = 400 KVA
2. fault level at 11 KV = 250 MVA
3. Soil Resistivity = 150 ohm meter
4. Duration of Earth Fault = 3 second
5. percentage impedance of the transformer = 5 %
6. Assume base MVA = 100
7. earth plate size = 1.2 × 1.2 × 0.12 mtr (Standerd)
8. Area of plate Both side =2.88 m2
9. Current Density assumed as = 118 A / Sq mm for Copper
Formula applicable =
Power = √(3 )×V I
I =P/(√3×V)
For rectifying current density =〖7.57 ×10 〗^3/√(ϱ×t) (Standerd value)
Calculation
Fault level at 11 KV =250 MVA
There for percentage impedance at 11 KV side assuming 100 MVA as Base=
= (Base MVA)/(11 KV side fault level)
= (100 )/250 × 100 = 40 %
Transformer 400 KVA Impedance = 5 %
There for
Impedance of the transformer at 100 MVA Base =
= (transformer percentage impedence )/(transformer capacity in MVA) × 100=
(5 )/(.400) × 100 = 1250 %
There for
Total impedance of LT side of the transformer = 40 + 1250 = 1290 %
LT side
Fault level at LT side = (Base MVA×100)/(Total Impedence) × 100
Fault level at LT side = 100/1290 × 100 = 7.75 MVA
Fault current LT side = (Fault level in Volt Ampiar)/(√3×415 volt)
Fault current LT side = (7.75 ×〖10〗^6)/(√3×415 volt) = 10781.84 A
Current density Assumed as = 118 A / Sq mm For Copper and there for Size of
the earth Conductor Required =
=(Fault current )/(Current density of the Copper) = 10781.84/118 = 91.37 mm2
Standerd size of the copper 25× 6 copper strip of 150 Sq mm
There for we can use 25 × 6 mm copper strip for earthing
HT Side
Fault current at 11 KV side =(Fault current at 11 KV side )/(√3×11)
Fault current at 11 KV side =(250 ×1000)/(√3×11) = 131121.59 A (13122 A)
Permissible current density = 〖7.57 ×10 〗^3/√(ϱ×t)
Permissible current density = 〖7.57 ×10 〗^3/√(150×3 ) = 356.8 Ambiar/m2
Total area of plate electrode required for the fault level current =
= (Fault level current )/(current density) = (13122 )/356.8 = 36.77 m2
Hence number of plate earth required (Plate) =
= (Total area of plate electrode required for the fault level current )/(Area of the plate Both side)
= (36.77 )/2.88 =12.77
Number of Earth Plate= 12.77 = 13 plat
MALAYALAM NOTE CLICK HERE
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