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).
I real appreciate your solutions.
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.