Home › Electrical Engineering Forum › General Discussion › sizing load ,cable& breaker
 This topic has 9 replies, 3 voices, and was last updated 11 years, 10 months ago by Ally Kanyondo.

AuthorPosts

2011/07/11 at 10:25 pm #10545FreeknowParticipant
Helllo
Whta is the right rule to calculate the total load?
Cable size?
Breaker size?
I have abuilding its area around 1670 msq 1st floor the same for 2nd floor. We A/C unit 70 unit of 4 tone.
Thanks in advance
2011/07/14 at 4:19 pm #12318Spir Georges GHALIParticipantDear ;
I think that you posted another topic for Cables' Section Calculation where I explained the procedure to define the suitable sections of Cables to electrify the 70 AC, the cables are :
– For each Phase : ” 4 x Single Core Cable of 300mm² “
– For Neutral : ” 2 x Single Core Cable of 300mm² “
About the Circuit Breaker size :
we do as follow to define the specifications of the CB
– The Nominal Current for CB :
In = 1250A
– The Thermal Protection :
Ir = 1200A ( as the cables can carry ” 1246A ” with the installation conditions )
– The Magnetic Protection :
Im : we should calculate : the minimum Short Circuit Current at the end of cable ” Isc1min “, and the Earth Fault at the end of cable, then the value of ” Im ” will be smaller by ” 20% ” than the smallest one.
– The Breaking Capacity :
BC: we should calculate the maximum Short Circuit Current at the beginning of cable ” Isc3max “, then the value of ” BC ” will be equal or bigger than the calculated value.
But as I don't know : the Power of Transformer, the Earting system, and the Length of cable, I can't calculate the 2 last specifications.
Regards.
2011/07/15 at 7:36 pm #12321Ally KanyondoParticipantDear 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 TNC
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.
2011/07/18 at 9:04 am #12326Spir Georges GHALIParticipantally 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 TNC
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 = U^{2} / Scs → ( 1.05 x 400 )^{2} / 500 x 10^{6} → 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
XMV^{2} = ( 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 U^{2} / Sn → 0.06 x ( 400 )^{2} / 1000 x 10^{3} → ZTR = 9.6 mΩ
RTR = Wtr / 3 x In^{2} → 15000 / 3 x ( 1445 )^{2} → RTR = 2.39 mΩ
XTR^{2} = ( 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 redone.
– 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 mm^{2} “
Regarding the following points :
– All loads are ” 3 Phases “.
– Assuming that we haven't the 3^{rd} harmonic.
– The used Earthing System is ” TNC “.
We can use the following cables :
– For PEN : ” 1 Single Core Cables of 300 mm^{2} “
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 300mm^{2}
0.91
0.432
0.531
PEN : 1 x 300mm^{2}
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
Zsc^{2} = ( 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 10^{3} 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
Zsc^{2} = ( 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 10^{3} A → 13620 A
The Magnetic Protection should be adjusted at ” Im = 10600 A “ or smaller accordingly to the protection unit adjustments.
2011/07/18 at 8:17 pm #12331Ally KanyondoParticipantDear Mr. Ghali
Thank you very much for your solution. Now i can calculate the short circuits at different levels
by using your knowledge as you gave us above.
Be blessed and hope to get and leran a lot from you.
Kind Regards.
2011/07/19 at 4:11 pm #12335Spir Georges GHALIParticipantDear Mr. Ally ;
Certainly you can.
Please find below another important points :
– The ” Loop Impedance “ for a fault at any point in the network should be calculated from the source to this point, that means : MV, Tr, and all cables till the point of fault.
– If the used Earthing System is different than ” TNC “, in this case, we should calculate ” Isc1min & If ” specialy if the ” N & PE ” have different sections that means ” Isc1min ≠ If “.
– If you use ” IT system ” : it's better to define first the ” Fault's Loop ” then calculate the Loop Impedance, as there's many method to realize the ” IT “.
I's at your disposal for any inquiry.
Regards.
2011/07/20 at 7:29 am #12337Ally KanyondoParticipantDear Mr Ghali.
Thanks for the additoinal informations.
i am also Concerned about the Lightining protection.
I want to know how i can calculate the maximum covereage of lighting arrestors in high rise buildings and in those builidings in which the
lightning protection device is provided at the highest point of the building like church towers.(as there is a maximum area any lightning
protection device can cover).
Thanks again in advance.
Hope to hearing from you soon.
Best Regards.
2011/07/20 at 2:44 pm #12339Ally KanyondoParticipantSpir Georges GHALI said:
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 TNC
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. Ghali ;
I was reading your solution but i am little confused on how you get those data highligted in red colour. Please let me know
where did you get those data below.
Kind regards
– 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 = U^{2} / Scs → ( 1.05 x 400 )^{2} / 500 x 10^{6} → 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
XMV^{2} = ( 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 U^{2} / Sn → 0.06 x ( 400 )^{2} / 1000 x 10^{3} → ZTR = 9.6 mΩ
RTR = Wtr / 3 x In^{2} → 15000 / 3 x ( 1445 )^{2} → RTR = 2.39 mΩ
XTR^{2} = ( 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 redone.
– 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 mm^{2} “
Regarding the following points :
– All loads are ” 3 Phases “.
– Assuming that we haven't the 3^{rd} harmonic.
– The used Earthing System is ” TNC “.
We can use the following cables :
– For PEN : ” 1 Single Core Cables of 300 mm^{2} “
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 300mm^{2}
0.91
0.432
0.531
PEN : 1 x 300mm^{2}
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
Zsc^{2} = ( 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 10^{3} 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
Zsc^{2} = ( 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 10^{3} A → 13620 A
The Magnetic Protection should be adjusted at ” Im = 10600 A “ or smaller accordingly to the protection unit adjustments.
2011/07/21 at 11:11 am #12341Spir Georges GHALIParticipantDear Mr. Ally ;
Regarding the highlighted points, please find below the explanation :
– Ssc : 500 MVA : The IEC defined the maximum value of Short Circuit Power for each level, exp. for ” 11kV & 20kV ” this value is till now ” 500MVA “, but if in your country the Ssc is smaller, you can either use this value or the defined value in your country. Some persons prefer do not calculate the RMV & ZMV, and they consider that the Ssc value is too big ( infinitif value )( making the calcultaion more easier ).
– RMV / ZMV : The value of this ratio have been also defined by IEC as the value of ” XMV ” is too bigger than ” RMV “, noting that in the past this ratio was ” 0.2 ” but as the MV networks is composed now of thousands of KM it's better to taken into consideration ” 0.1 “
– The value ” 1.05 & Cmax ( 1.05 ) : it's the voltage's flactuation of the medium voltage network.
– The value of ” X ” : as this value is depending on the Cable's Length, the IEC defined this value accordingly to the kind of cable and the method of installation, that means if the kind of cable is different than ” Touching – Flat ” the X will be also different. for exp. for Multi cores cable ” X = 0.08 mΩ ” wahtever the installation method.
– The value ” Cmin ” : as there's a Voltage Drop when a short circuit happen that will be in the most of cases around 5%, and we want to calculate the minimum value of the short circuit current, we should take it into consideration. For me, when I calculate the minimum short circuit current for the final circuits in the network I prefer to take it into consideration even smaller than ” 0.95 ” ( 0.800.85 ).
Regards.
2011/07/21 at 1:22 pm #12342Ally KanyondoParticipantDear Mr Ghali
Thanks very much for your clarifications. I am also begging you to pass through my article in the general discussion entitled
“LIGHTNING PROTECTION” or
Tell me or send me softcopy of any good electrical books about protection.
My email address is: allygk@gmail.com
Be much blessed for helping us young and emerging engineers.
Kind regards

AuthorPosts
 You must be logged in to reply to this topic.