M multiplier in short circuit calcs

Status
Not open for further replies.
Grouch1980

Grouch1980

Senior Member
Location
New York, NY
So i've been reading some prior threads on short circuit calcs... there's an f-factor, which represents the impedance of the cable or bus between your 2 point on a cable (or between the primary and secondary of a transformer)... and then you have the M multiplier, which is 1 / (1+f), and that M multiplier is used to bring down your fault current from point A to point B. I know the SPD handbook by cooper bussmann has some good examples. My question is... what is this M multiplier and how did it come about? I under the f-factor. i don't understand why it then goes into the 1 / (1+f) formula...
 
I wanna start off by saying I was never told what it means, so this is just a guess. Also I this is my first post so I don't know how to make equations look nice.

First to get an understanding of 'f' lets take the formula for a 3 phase fault as an example.
f = (sqrt(3) * L * Isc) / (N * C * E)

C is just 1 / (impedance of a single conductor per ft), so N * C is the 1 / (impedance of the feeder/ft)
1 / (impedance of the feeder/ft) = 1/(Zf/L) = L/Zf, where Zf is the impedance of the feeder between points 1 and 2.

E / (sqrt(3) * Isc) = E_LN / Isc is the upstream impedance limiting the AFC at the first point.
so E / (sqrt(3) * Isc) = Zu

so to clean this up we have
f = ((sqrt(3) * Isc) / E) * L / (N * C) = (1 / Zu) * L / (L / Zf)
f = Zf/Zu, so its like the ratio of the downstream impedance to the upstream impedance.

Now to answer this question,
M = 1 / (1 + f) = 1 / (1 + Zf/Zu)
M = 1 / (Zu/Zu + Zf/Zu)
M = 1 / ((Zu + Zf) / Zu)
M = Zu / (Zu + Zf)

Zu + Zf is just the total impedance from upstream all the way to your new point 2.

So Zu / (Zu + Zf), aka 'M' is the ratio of the upstream impedance to the total impedance.

Someone please correct me if I'm wrong lol.
 
I think it makes sense actually. and at the very end, you multiply the numerator of 'M' (Zu) times the short circuit current at point 1, which gives you voltage, and then divide by the total impedance (up to point 2), and that gives you the short circuit current at point 2. Thanks for the patience of showing all that math.
 
mathematically this makes sense to me, however something's not clear when i try to picture it... see my post above; in order to get such a high short circuit current at point 2, or wherever you're trying to calculate the fault current, multiplying Zu times the Isc at point 1 has to give you a very high voltage value... and when you divide by the total impedance (Zu + Zf), this has to give you a high Isc at point 2. but when you have a short circuit condition, doesn't that bring the voltage values close to 0? How can the voltage be so high?
 
The voltage is close to 0 at the fault, but the reason a current exists is because the sources upstream continue to have a voltage. So the voltage you get by multiplying Isc by (Zu + Zf) should be equal to V_LN (if the fault occurs at point 2).
 
Status
Not open for further replies.
Top