Meaning of short circuit MVA

[abrahamJP]

Hi,
Given datas: 11kv,350MVA Fault level &
11kv is feeding to 11kv/400v,500kva with%z=5%; Transformer;
-------------------------------
Now to calculate Impedances;
primary impedance-->(11000^2)/350-->(equation1)
secondary side impedance-->(5)*400/(100*500)-->(equation2)
--------------------------------
Is this correct or whether we need to use in (equation1) 400
instead of 11000 for short circuit calculations?
I want to calculate for a fault point at MDB at 400V. Please help..

regards

Abraham

 

[mosesnbklyn]

The 350MVAsc is the rating of a family of MV circuit breakers; if the system is designed properly the max fault from the utility is 350MVA / (11kV x 1.732) ~18.4kA; on these systems 15kA is a better estimate.

use this 11kV fault current thru the transformer and you will get your solution.

Even with an infinite bus its only 500/(.4 x 1.732)=14.5kA (below standard 22kA ratings).

hope this helps.



M. Nissen,P.E.
Senior Electrical Engineer
Waldron Engineering & Construction

 

[mosesnbklyn]

correction:
*Even with an infinite bus its only 500 /(0.4 x 1.732 x 0.05 )=14.5kA (below standard 22kA ratings).

M. Nissen,P.E.
Senior Electrical Engineer
Waldron Engineering & Construction

 

[odlanor]

% Hi,
% Given datas: 11kv,350MVA Fault level &
% 11kv is feeding to 11kv/400v,500kva with%z=5%; Transformer;
% -------------------------------
% Now to calculate Impedances;
%%%%%%%%%%%%%%%%%%%%%%%%%% MVA method %%%%%%%%%%%%%%%%%

V = 400/1000;%kV
Zpu = 5/100;%pu
N = 500/1000;%MVA
MVA1 = 350; %MVA

%%%%%%%%%%%%%%%%%%%%
MVA2 = N / Zpu; %MVA
MVA12 = (MVA1 * MVA2) / (MVA1 + MVA2); %MVA
ICC = MVA12 / (1.732*V)
%fault 400V side ==> 14.03 kA

 

[7anoter4]

First of all I agree with odlanor, of course.
I’ll try to sketch a “hand made” calculation.
This is a simple classical one-line diagram. If we’ll take 0.4 kV as reference voltage then
Zsystem=0.4^2/350=0.000457=0.457E-3
Zxfrm=0.4^2/0.5*5%=15.972E-3
The transformer rated current will be 500/0.4/sqrt(3)=721.7 A then for underground cable 2 cables of 3*300 mm^2 copper conductor will be suitable[in usual conditions: RHO=100 ,30 dgr.C only 2 cable running touching, load factor 1-See VDE]
Let’s say the cable length 100 m and according to BS7671 Table 4E2B r=0.140 mV/A/m x=0.120 that means
r=0.140/sqrt(3)=0.08083 ohm/km ; x=0.120/sqrt(3)=0.06928 ohm/km
Zcable=sqrt(0.008083^2+0.006928^2)/2=10.6458E-3/2=5.32E-3 ohm.
Total Ztotal=Zsystem+Zxfrm+Zcable=0.46E-3+15.972E-3+5.32E-3=0.0218 ohms
I”k3=0.4/Ztotal=0.4/0.0218=10.6 kA
A more accurate calculation could be if you would separate R and X of all equipment [System, Transformer and cable) and then Ztotal=sqrt (Sum(x)^2+Sum(r)^2).

 

[7anoter4]

Sorry, I”k3=0.4/sqrt(3)/Ztotal=10.6 kA