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Ally KanyondoParticipant
Dear All,
The mentioned software are Good but the best one is SOLUTIONS ELECTRICAL SOFTWARE. Currently im using this software.
IT is capable of doing the following:-
CABLE CALCULATIONS
CIERCUIT DESIGNER
INSTALLATION SIMULATOR
AIR CONDITIONING SIZES
HEATING CALCULATIONS
HEATING ELEMENTS
SHORT CIRCUIT CALCULATIONS
TEST INSPECTION FORMS
ISTALLATION OPTIMIZER
ELECTRICAL ESTIMATOR
PROTECTION AND COORDINATION
BOARD TEMPERATURE
POWER FACTOR CORRECTION
TRANSFORMER REGULATION
MAX DEMAND FOR DOMESTIC
MAX DEMAND FOR INDUSTRIAL.
I advise you to try on this software.
For more informations you can visit the following web: http://www.solutionselectricalsoftware.com
Kind regards
Ally 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
Ally 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 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. 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 = 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.
Ally 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.
Ally 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.
Ally 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 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.
2011/07/14 at 12:13 pm in reply to: Why and Where to use distribution voltages 33kV,11kV or 0.4kV #12315Ally KanyondoParticipantDear Ghali,
Thank you very much for briefly explanation.
You have provided me the way forward.
I thought there will be a formula for that but it seems only the utillity company can set the levels.
Kind Regards.
Ally KanyondoParticipantDear Laurent
I have been trying to access it several times but im told im already registered. What can i do now
Ally KanyondoParticipantHI Jenny just go to 4shared website and type this”tansmission and distribution” then search in the link titled books.
you will get the exactly book. i just did that n got it.
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