CC-SC Current Calculation Moose-Zebra

Calculation of Continious Current Rating : (Considering Zebra) 1 Connected Load

=

2 Current Requirement

:

100 MVA

On 230 kV Side

=

251.03 A

Considering 20 % Overloading

=

302.00 A

On 110 kV Side

=

524.88 A

Considering 20 % Overloading

=

630.00 A

Hence Maximun current requirement is 630 A on LV Side .

Selected Conductor Zebra : Stranding and Wire Diameter (Alm +Steel)

=

Sectional Area (Al.)

=

Current Carring capacity at 75 Deg

=

737 A

Curring carring capacity for twin Zebra

=

1474 A

54+7/3.18

428.9 Sq mm

Hence selected Twin Zebara is sufficient to catter the total load.

Calculation of Continious Current Rating : (Considering Moose) 1 Connected Load

=

2 Current Requirement

:

100 MVA

On 230 kV Side

=

251.03 A

Considering 20 % Overloading

=

302.00 A

On 110 kV Side

=

524.88 A

Considering 20 % Overloading

=

630.00 A

Hence Maximun current requirement is 630 A on LV Side .

Selected Conductor Zebra : Stranding and Wire Diameter (Alm +Steel)

=

Sectional Area (Al.)

=

Current Carring capacity at 75 Deg

=

836 A

Curring carring capacity for twin Moose

=

1672 A

54+7/3.53

528.5 Sq mm

Hence selected Twin Moose is sufficient to catter the total load.

Calculation of Short Circuit Current Rating : (Considering Zebra) INPUT PARAMETERS

CONDUCTOR PARAMETERS

No. of Sub Conductors

ACSR-ZEBRA Cross Sectional Area of conductor (mm2) 0 Nc 2 Max Allowable Conductor temperature( C)

Symmetrical initial short circuit current (kA)

Ik"

40

Specific Thermal Capacity (J / (kg. C))

X/R

15

Duration of short circuit (s)

Tk

3

Specific mass (kg / m ) Specific conductivity at 20 °C (1/ Ω.m)

Conductor temperature (Initial) ( °C)

θb

75

f

50

Type of Conductor 

X over R Ratio (X/R)

System Frequency (Hz)

 A

0

3

0

Temperature coefficient of Resistance (1/ C) Rated Short-Time (sec)

428.8821

θc

200

C

910

ρ

2700

κ20

3.48E+07

α 20

0.0036

Tkr 

1

Reference Document : (IEC 865 PART-I) CALCULATION OF THERMAL WITHSTAND SHORT TIME CURRENT CAPACITY 1

Calculation of Peak Short Circuit Current Factor (k) (IEC-60909, Sec.4.3, Page:101) k= =

2

1 .0 2 + 0. 98 * E XP ( - 3 / 15 ) 1.822

Calculation of Thermal Equivalent Short Time Current Ith = Ik" x ( m + n )

0.5

n=

1

= Ith =

(Figure-12(a), IEC-865-I, considered as 1 (maximum value))

[ 1 / ( 2 x 50 x 3 x ln ( 1.82235613801642 - 1) ) ] x [ EXP (4 x 50 x 3 x ln ( 1.82235613801642 - 1) - 1]

= 0.017043 40 x ( 0.0170431742612367 + 1 ) ^ (0.5) 40.33942 kA

3

Calculation of Rated short -time withstand current density " Sthr  " of conductor -

  K



= = 0.5

Sthr  = K / ( Tkr  ) Sthr  = 4

20



C       In

  20

1    (  20 c 1    (  20 b



20 0 C )



20 0 C )

SQRT(((34800000x910x2700)/0.0036)*LN((1+0.0036x (200- 20))/(1+0.0036x (75- 20)))) 8.70E+07 A*s0.5 /m2 = 87031063.6441945 / ( 1 ) ^ 0.5 8.70E+07 A /m2

Checking for Thermal withstand Short-time rating of Conductor  Calculated STC rating = =

87031063.6441945 x ( 1 / 3 ) ^ 0.5 *2 *428.882060927179 *40 /40.3394233823189 /1000 /1000000 42.74 KA

LHS

RHS

Sth = Ith / ( A x Nc )

Sthr  x ( Tkr  / Tk ) ^ 0.5

= =

40.3394233823189 x 1000 / (428.882060927179 x 10 ^ -6 4.70E+07 A /m2

Since LHS < RHS, Selected Conductor has Thermal withstand Short-Time rating as required. Hence Twin Zebar conductor is suiable for 40KA for 3 Sec

= =

 

87031063.6441945 x ( 1 / 3 ) ^ 0.5 5.02E+07 A /m2

Calculation of Short Circuit Current Rating : (Considering Moose) INPUT PARAMETERS

CONDUCTOR PARAMETERS

No. of Sub Conductors

ACSR-MOOSE Cross Sectional Area of conductor (mm2) 0 Nc 2 Max Allowable Conductor temperature( C)

Symmetrical initial short circuit current (kA)

Ik"

40

Specific Thermal Capacity (J / (kg. C))

X/R

15

Duration of short circuit (s)

Tk

3

Specific mass (kg / m ) Specific conductivity at 20 °C (1/ Ω.m)

Conductor temperature (Initial) ( °C)

θb

75

f

50

Type of Conductor 

X over R Ratio (X/R)

System Frequency (Hz)

 A

0

3

0

Temperature coefficient of Resistance (1/ C) Rated Short-Time (sec)

528.4855

θc

200

C

910

ρ

2700

κ20

3.48E+07

α 20

0.0036

Tkr 

1

Reference Document : (IEC 865 PART-I) CALCULATION OF THERMAL WITHSTAND SHORT TIME CURRENT CAPACITY 1

Calculation of Peak Short Circuit Current Factor (k) (IEC-60909, Sec.4.3, Page:101) k=

1 .0 2 + 0. 98 * E XP ( - 3 / 15 )

= 2

Calculation of Thermal Equivalent Short Time Current Ith = Ik" x ( m + n )

0.5

n=

1

= = Ith =

3

(Figure-12(a), IEC-865-I, considered as 1 (maximum value))

[ 1 / ( 2 x 50 x 3 x ln ( 1.82235613801642 - 1) ) ] x [ EXP (4 x 50 x 3 x ln ( 1.82235613801642 - 1) - 1] 0.017043

40 x ( 0.0170431742612367 + 1 ) ^ (0.5) 40.33942 kA

Calculation of Rated short -time withstand current density " Sthr  " of conductor -

  K



= = 0.5

Sthr  = K / ( Tkr  ) Sthr  = 4

1.822

20



C       In

1    (  20 c

  20

1    (  20 b



20 0 C )



20 0 C )

SQRT(((34800000x910x2700)/0.0036)*LN((1+0.0036x (200- 20))/(1+0.0036x (75- 20)))) 8.70E+07 A*s0.5 /m2 = 87031063.6441945 / ( 1 ) ^ 0.5 8.70E+07 A /m2

Checking for Thermal withstand Short-time rating of Conductor  Calculated STC rating =

87031063.6441945 x ( 1 / 3 ) ^ 0.5 *2 *528.48547061108 *40 /40.3394233823189 /1000 /1000000 =

52.66 KA

LHS

RHS

Sth = Ith / ( A x Nc )

Sthr  x ( Tkr  / Tk ) ^ 0.5

= =

40.3394233823189 x 1000 / (528.48547061108 x 10 ^ -6 3.82E+07 A /m2

Since LHS < RHS, Selected Conductor has Thermal withstand Short-Time rating as required. Hence Twin Moose conductor is suiable for 40KA for 3 Sec

= =

 

87031063.6441945 x ( 1 / 3 ) ^ 0.5 5.02E+07 A /m2