Generator short circuit.

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weam

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Hay… I have 3 generator's feed MSB , etch generator have S=925 KVA, X’d=27.1%, I need to calculate 3 phase short circuit in generator and in MSB for 3 generator's , the length between generator and MSB is 100m , and cable is A=8*(3*95+50)mm2.
Pleas help
 
I

Ingenieur

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what is system voltage?

calc base X = v^2 / S
calc act X = pu X x base
sc i per gen = v / act X
times 3 for total
calc wire Z
at MSB sc i = v / ((act X)/3+ wire Z)
 
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Julius Right

Julius Right

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Electrical Engineer Power Station Physical Design Retired
In my opinion you are speaking about 400 V 50 HZ system. Since the generator current is 1335 A then the short-circuit current has to be 1335/0.271=4.927 kA
that means 3*4.927 =14.78 kA [three phase or phase to neutral] at MSB.
The 8 cables[parallel] reactance is 1.7 mohm and the generator 46.88 mohm then the cable reactance may be neglected.
See BS7671 Table 4E2B for cable reactance.
 
I

Ingenieur

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using 440 vac
X base = 400^/925000 • 0.173 Ohm
act X = 0.271 x 0.173 = 0.0469 per ph converted to 3 ph = 0.0812
400/0.0812 = 4926 A
14778 total

cable Z ~ 0.006 Ohm (based on NEC estimation ~4 x 1000 kcmil steel cnd)
at MSB
400/(0.0812/3 + 0.0066) = 11881

using cable X only 0.0076
11538 A
 
Julius Right

Julius Right

Senior Member
Occupation
Electrical Engineer Power Station Physical Design Retired
At first 95 mm^2=95/.5067=187.48 MCM –it is more than 3/0 and less than 4/0.
Let’s say the cable run in steel conduit[for 300 ft.?] so the reactance will be :
0.171 ohm/km=0.0171 ohm per 100 m and divided by 8[8 parallel cables]=0.002138 ohm/feeder.
The generator reactance per phase it is 0.4^2/0.925*27.1/100=0.046875676 ohm.
so the short-circuit current will be:
400/sqrt(3)/(0.04688+0.002138)=4711 A per one generator.
British Standard ,closer to the actual position ,states 0.130 ohm/phase for this cable then the total feeder reactance will be 1.7 milliohm.
 
G

GoldDigger

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FWIW, for any time period greater than a few cycles, the available energy into a short circuit from the generator will be limited to the available kinetic energy of the rotating mass. The prime mover of the generator may not be able to supply anywhere near the short circuit current for a significant length of time for, say, arc flash calculations.
 
I

Ingenieur

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yep
should have used the same number of cables
interpolating between 2/0 and 3/0 for the 95 and 1 and 2 for the 50
X = 0.0013
Z = 0.0023
not adjusted for 50 vs 60 Hz

moot <<< gen X
X 14.1 ka
Z 13.6 ka

adjust for 50 Hz
14.2 ka
13.7 ka
 
mivey

mivey

Senior Member
Ingenieur said:
looks like euro data sheets give much more info
direct and quadrature axis values
sub and transient valued

http://www.kohlerpower.com.sg/onlinecatalog/pdf/KX2200C.pdf

how are breakers rated?
transient, sub, sym steady-state???
is the subtransient value factored into the rating?
is the tc so short as to not be a factor? I^2 t ? ~ 1 cycle
Click to expand...
My understanding has been that moulded case breakers should be rated using the subtransient fault current.
 
I

Ingenieur

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mivey said:
My understanding has been that moulded case breakers should be rated using the subtransient fault current.
Click to expand...
with a genset that seems to be the case

obtaining the direct quadrature value from the util might be hard
I guess you could calc/est a number based on system x/r

isc = S/(sqrt3 v puZ) is a sym value which is typically used for sizing
xfmr L will dampen the subtransient I guess
 
mivey

mivey

Senior Member
Ingenieur said:
with a genset that seems to be the case

obtaining the direct quadrature value from the util might be hard
I guess you could calc/est a number based on system x/r

isc = S/(sqrt3 v puZ) is a sym value which is typically used for sizing
xfmr L will dampen the subtransient I guess
Click to expand...
The OP has the generator data.
 
I

Ingenieur

Senior Member
Location
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mivey said:
That does not make sense.
Click to expand...

that could be interpreted 2 ways
I was not clear
you do not understand

from link
isc" = V(sqrt3 X"d)
isc' = V/X'd
let
V = 1 pu
X"d = 0.10 pu
X'd = 0.17 pu

isc" = 5.77 V pu
isc' = 5.88 V pu
moot

some texts say mult isc" x 1.5 (1.6) to account for the dc component
and 1.2 for isc'
 
mivey

mivey

Senior Member
Ingenieur said:
that could be interpreted 2 ways
I was not clear
you do not understand

from link
isc" = V(sqrt3 X"d)
isc' = V/X'd
let
V = 1 pu
X"d = 0.10 pu
X'd = 0.17 pu

isc" = 5.77 V pu
isc' = 5.88 V pu
moot

some texts say mult isc" x 1.5 (1.6) to account for the dc component
and 1.2 for isc'
Click to expand...
Not clear.

Where is your link stating E/X'd? Where is it using a different voltage?

The DC component should be higher at isc'' because it will have less asymmetric decay than isc' (which is taken to be later in the fault). A conservative value to size equipment is to use 2X for the DC offset at isc''.
 
I

Ingenieur

Senior Member
Location
Earth
mivey said:
Not clear.

Where is your link stating E/X'd? Where is it using a different voltage?

The DC component should be higher at isc'' because it will have less asymmetric decay than isc' (which is taken to be later in the fault). A conservative value to size equipment is to use 2X for the DC offset at isc''.
Click to expand...

to you
link was previously posted
 
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