R-rated Fuses

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Jager

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I have seen these fuses on 4160V motors at a plant we're working on. The info I've found on them says that they are backup current limiting fuses for medium voltage motors. I have also found that they are supposed to melt in 15-35s at a current value of 100 times the "R" number.

How would one go about sizing these things? This is not really part of my job, and I'm sure the motor manufacturer could tell me, but I am curious.

For example, I have a 400 HP motor with a motor contactor and a 6R fuse with a disconnect. FLA is ~52 A. Why choose a fuse rated for six times that, and one that would burn for 15-35s at that? I don't see anything in the code for this kind of thing.

Thanks for helping out this new guy.
 
Motors draw alot of current when they first start. 400 HP = 298.4 KW. 298.4 KW / 4,160 volts = 71.7 amps. A motor can draw up to 6 times that current when starting. 71.7 X 6 = 430.2 amps. Then 430.2 X 125% = 537.75 amps. The fuse would have to be able to handle 537.75 amps for a short time while the motor is starting. Once the motor reaches full speed it would only need to be able to handle 71.7 X 125% = 89.6 amps. If something is wrong the fuse will burn instead of the feeders or motor. Normally the motor and feeders will heat up and then cool down once the motor reaches full speed.
 
Okay, I looked up that web site. It says these fuses will start to burn at 100 times their R value. You said you had 6R fuses installed. 6 X 100 = 600 amps. My previous calculation came to 537.75 amp current draw when starting that 400 HP motor at 4,160V. So if that motor refused to start the fuse would burn up in 15 to 35 seconds preventing a fire or serious equipment damage. This info would be found in "electrical theory" and/or "motor circuits" type text books. The NEC assumes that you have already studied this material. Hope I was able to help.
 
I guess I never said these were three phase motors, but they are, which I think is pretty typical for 4160V.

So my calculations are:
298.4kW/4160V/1.732 = 41.4A. (FLA)
41.4 * 6 = 248.5A. (starting current)
248.5*1.25 = 310A. (Code multiplier)
100*6R = 600A is just under 200% of that. Hence my question.

Originally I had said the fuses were rated for six times the FLA when I meant twelve. Why make the fuse twice the size it needs to be?

Anyways, I guess this must be specific to the motor. Maybe it has an extra high starting current. Thanks for the help though.
 
Many medium voltage fuses are to provide short circuit protection and not overload protection.
That's why an E rated medium voltage fuse will reach its "handle value" in approx 10,000 seconds or so.
 
Ok, so this fuse is to protect the motor less from overcurrent protection, which is what the contactors are for, and more from short circuits? That makes sense, as I guess the 4160V motors are expensive enough to warrant the extra protection.

So, this fuse can carry 12x the motors FLA for 15-35s. That sounds like a reasonable thing to do, as the motor could probably handle that kind of current for a short period before it started to get seriously hot.

Thanks Ron.

-Another Ron
 
For low or medium voltage motors the fuse is used for short circuit protection only so it could be sized at 13 times the FLC before you get into the motor damage curves (NEC 430.52). The motor overload circuit (overloads on low voltage motors and CT's on medium voltage motors) provide the overload protection for the motor (125% of FLC).
 
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