# Re: sizing load ,cable& breaker

#12326

ally said:

Dear Ghali

Let me add to the above question,

Assume the ratings are as follows:-

1. Transformer rating is 1000kVA, 11kV/0.4kv

2. Type of earthing system is TN-C

3. Cable length is 35m.

Help us to calculte the remaining items(short circuits).

Your are a real MENTOR most of in the forum.

Regards.

Dear Mr. Ally ;

Thanks a lot for your complement.

Please find hereafter the manual Calculation of the important Short Circuit Currents ” …max  &  …min “.

To calculate the Short Circuit Currents, we should calculate the Impedances ” R  &  X ” for : Medium Voltage – MV, Transformer – Tr, and the Cables between the Output of Tr and the Input of MDB that we do as follow :

– Medium Voltage :

We have the following information :

– The value of the Short Circuit Power on ” 11 kV ” is : ” Scs ≈ 500 MVA “

– The Voltages : 11 / 0.4 kV

– RMV / ZMV = 0.1

So :

ZMV = U2 / Scs   →   ( 1.05 x 400 )2 / 500 x 106   →   ZMV = 0.352 mΩ

Rem. : as we should calculate the values of ” R  &  X ” from LV side of transformer, we can use directly ” 400V “, or we use ” 11kV ” but in this case and after calculation we should multiply the result by the ” SQR of the Transformer Ratio – ( 11/0.4 )2  “.

RMV = 0.1 x ZMV    →                                                                      RMV = 0.035 mΩ

XMV2 = ( ZMV )2 – ( RMV )2    →    ( 0.352 )2 – ( 0.035 )2    →         XMV = 0.35 mΩ

– Transformer :

As the Transformer's power is ” Sn = 1000 kVA   →   In = 1445 A ” and I haven't the other information, I will assume that :

– Usc = 6 % ( percentage of Short Circuit Voltage )

– Wtr = 15 kW ( full load Copper Loses )

So :

ZTR = Usc x U2 / Sn   →   0.06 x ( 400 )2 / 1000 x 103   →               ZTR = 9.6 mΩ

RTR = Wtr / 3 x In2   →   15000 / 3 x ( 1445 )2   →                          RTR = 2.39 mΩ

XTR2 = ( ZTR )2 – ( RTR )2    →   ( 9.6 )2 – ( 2.39 )2    →                XTR = 9.29 mΩ

Rem. : if the values of ” Usc  &  Wtr ” are different, the calculation should be re-done.

– Cables :

As the previous cables have been defined accordingly to the Loads' Current ” 1175 A “, but now :

– They should be defined accordingly to the transformer's nominal current ” 1445 A “

– The number of cables per phase will be more than 4

– These cables will be installed on 2 Cable Trays.

So, The Reduction Factors are :  0.81   &   0.61

By doing the same previous calculation, we obtain the cables' sections in the same working conditions that are :

– For each Phase        : ” 5 Single Core Cables of  300 mm2

Regarding the following points :

– All loads are ” 3 Phases “.

– Assuming that we haven't the 3rd harmonic.

– The used Earthing System is ” TN-C “.

We can use the following cables :

– For  PEN                   : ” 1 Single Core Cables of  300 mm2

Calculation of ” X ” for Cables :

As the Installation's Method is ” Touching – Flat “, the ” X  per 1m = 0.13 mΩ/m “, so :

Xca( Ph ) = 0.13 x 35 / 5    →                                                                Xca( Ph ) = 0.91 mΩ

Xca( PEN ) = 0.13 x 35    →                                                                   Xca( PEN ) = 4.55 mΩ

Calculation of ” R ” for Cables :

As the Resistivity ” ρ “of Copper is different between the ” Cold ( 20 °C )  or  Hot ( 70 °C ) ” states of cables, we should calculate ” R ” in these 2 states, that we do as follow :

In Cold state ” 20 °C ” :

Rca(20)( Ph ) = ρ x L / S = 18.51 x 35 / 5 x 300    →                              Rca(20)( Ph ) = 0.432 mΩ

Rca(20)( PEN ) = ρ x L / S = 18.51 x 35 / 300    →                                 Rca(20)( PEN ) = 2.159 mΩ

In Hot state ” 70 °C ” :

Rca(70)( Ph ) = 1.23 x ρ x L / S = 1.23 x 18.51 x 35 / 5 x 300    →          Rca(70)( Ph ) = 0.531 mΩ

Rca(70)( PEN ) = 1.23 x ρ x L / S = 1.23 x 18.51 x 35 / 300    →             Rca(70)( PEN ) = 2.656 mΩ

Calculation of Short Circuit Currents :

To calculate the ” Loop Impedances ” of Short Circuits, it will be better to integrate all results in a table like the following :

 Component X : mΩ R : mΩ 1 Medium Voltage : 500MVA – 11/0.4kV 0.35 0.035 2 Transformer : 1000kVA – 6% – 15kW 9.29 2.39 3 Cables : 20 °C 70 °C Phase : 5 x 300mm2 0.91 0.432 0.531 PEN    : 1 x 300mm2 4.55 2.159 2.656

Calculation of the maximum Short Circuit Current ” Isc3max ” :

We assume in this case that we have a short circuit between ” 3 Phases ” at the input of MDB but the cable are in the ” Cold ” state :

Isc3max = Cmax x Uo / 1.73 x Zsc      ,      where :

– Cmax = 1.05

– Uo = 400V

Zsc2 = ( 0.35 + 9.29 + 0.91 )2 + ( 0.035 + 2.39 + 0.432 )2     →                         Zsc = 10.93 mΩ

So :

Isc3max = 1.05 x 400 / 1.73 x 10.93 10-3      →   Isc3max = 22.21 x 103 A      →    22.21 kA

The Breaking Capacity of the Circuit Breaker should be  ” BC  ≥  23 kA “

Rem. : The ” Isc3max ” at the Transformer's Output is ” 24.42 kA “.

Calculation of the minimum Short Circuit Currents ” Isc1min   &  If ” :

As we have ” PEN ” that means the ” Isc1min  &  If ” have the same value, so, we will calculate only one.

We assume in this case that we have a short circuit between ” 1 Phase & PEN ” at the input of MDB but the cable are in the ” Hot ” state :

Isc1min = If = Cmin x Vo / Zsc      ,      where :

– Cmin = 0.95

– Vo = 231 V

Zsc2 = ( 0.35 + 9.29 + 0.91 + 4.55 )2 + ( 0.035 + 2.39 + 0.531 + 2.656 )2   →     Zsc = 16.109 mΩ

So :

Isc1min = If = 0.95 x 231 / 16.109 x 10-3     →   Isc1min = If = 13.62 x 103 A      →   13620 A

The Magnetic Protection should be adjusted at ” Im = 10600 A “ or smaller accordingly to the protection unit adjustments.