Fuse selection for transformer inrush current

[Jeffduk]

Hi

new to this forum, but it looks very helpful.

my question is this: I am trying to select an REC Cut-out fuse (BS88) that will not rupture whilst powering up a 125kVA transformer. The inrush rating given by the manufacturer is 5 times full load current. i understand that whilst looking at a fuse characteristic curve i can see whether the fuse is clear of this inrush current.

The problem i have is that i am unsure on how to calculate the I2t (energy let through) created by the transformer on inrush. the equations i have found are 5xfull load = 1600A squared = 2.56MA x 0.1 secs = 256kA.

this high value is posing a problem, which is why i seek clarification on how to calculate this value as i have a feeling i am wrong. I am sure there must be another method as the above equation is normally used for a short circuit fault, but with current inrush the current starts high and then normallises after a few cycles.

thanks for any help you can give.

 

[DanDel]

Typical IEEE specifications for inrush currents are 10x the FLA for transformers below 2500kVA, and 12x for above, all at a maximum of 0.1s.
I'm not sure where the '5xfull load = 1600A squared' part of your calcuation comes from. What is the primary (or energizing side) voltage of your 125kVA transformer?

 

[Jeffduk]

The primary voltage is 400V 3 phase

the 5x full load current refers to the manufacturers rating of a maximum inrush current of 5 x the full load capable of the transformer.

full load is approximately 320A therefore 5 x is 1600A.

you then square this number and multiply by 0.1 secs.

my question is really how do you calculate the I squared t let through on the transformer before the current normalises.

 

[jbartos]

Suggestion: The transformer manufacturer may be asked to provide an inrush current dot on I-t chart. The transformer protecting fuse I-t characteristic should pass this point on the higher current and the longer time side on I-t coordination chart. However, the fuse characteristic should be below the transformer damage curve, namely, on the lower current and shorter time side on the I-t coordination chart.

 

[powerlnman]

I believe you have forgotten (1.732) when calculating the rated current of the 3 phase transformer 125kva/(400*1.732)=180A

 

[jghrist]

Normally the fuse minimum melt time-current characteristic is compared to a multiple of full load at different times. Typically 2x for 100 sec, 3x for 10 sec, 6x for 1 sec, 12x for 0.1 sec, 25x for 0.01 sec.

It is not usual practice to compare the inrush energy let-through I²t to the fuse min melt I²t, but if you want to, you will have to integrate the inrush over time because it is not a constant current. Multiplying the current squared times the inrush time assumes that the current is constant for that period of time and then goes to zero. This is not the case. Also, energy let-through units are amp²·sec not kA or MA.

 

[DanDel]

Jeffduk, hold on a minute. Let me try to tie together the issues brought up by myself and my esteemed colleagues that should be corrected in your analysis.

125kVA 3-phase transformer, full load amps is 125000/(400V x 1.73) = 181A
181A x 5 = 903A at 0.1s for the inrush point, which is plotted on a log-log TCC graph.

The term 'I squared t' refers to a value of energy, not current, based on the square of the current multiplied by the amount of time that current flows.

For your application, simply plot the inrush point against the fuse manufacturer's melting curve(which is an I2t curve) to tell if the fuse will allow the inrush current.

 

[busbar]

Aside — On log-log TCC plots, I²t is conveniently a line with a slope of –2.

 

[Jeffduk]

Thanks for all the help chaps, it would appear i have made a mistake in that the transformer is not 3-phase but is connected phase to phase, so is 400V but the 1.73 is not needed in the calc.


JGhrist seems to be on track i think with the fact that i will need to integrate the area below the curve to get the total i2t.

Also it is the area of the fuse where thermal damage is caused to the fuse when too much energy is let through it. i dont think the i2t value would be high enough to melt the fuse but need to be sure it wont damage it at all.

 

[tulum]

Jeffduk,

Why would you even want to compare the I^2t curves?

Follow the adivce given by Jbartos and Dandel... plot the fuses melting curve vs. the transformers damage curve and insure it is below and to the left of the xfmrs damage curve at the aforementioned inrush points... within reason; at all other points the fuse should protect the transformer (above and to the right).

 

[CastleHydro]

S&C have this very useful calculator on their website that provides fuse coordination including damage curves for transformers.

http://www.sandc.com/fsa/fsaa1/(u1nkdt555bngesawsoj2wkmb)/FSAAInit.aspx

You will probably need to copy and paste this link so that it works in your browser.

Might be helpful.
John

 

[rbulsara]

I am with Dandel and tulum.

Jeffduk is perhaps, confusing I^2.t with kA interrupting rating of a fuse which are to different things.

All you need is to see the point of intersectio of the max. inrush current and 0.1 sec line on TCC lies below (to the right) of the fuse curve.

In more practical terms, for this size/voltage transformer a fuse rated 1.25 times FLA will suffice.

 

[jbartos]

Suggestion: Reference:
IEEE Std 141-1993 Red Book - IEEE Recommended Practice for Electric Power Distribution for Industrial Power Plants,
Chapter 5 Application and Coordination of Protective Devices