Circuit breakers are typically a bit more sensitive to spikes and trip off. That will in turn break the circuit, kill fan, reset, spike again, break circuit, kill fan, reset, ad nauseum. I don't think the fan would ever run.

 

Could get a manual reset type and use an led to tell you when the fan is on.
http://www.wiringproducts.com/contents/en-us/d29.html
These go from 50A to 150A

 

Ok. So, this means that by design, these are not efficient. The in-rush current is what the problem is.

The fuse rating and wiring is factored in for in-rush current per component ratings and design in standard applications. If you've got a 100A of draw for ANY amount of time, and, a standard draw of ~40A nominal, then, the AWG, fuse holder type as well as fuses should be rated way above nominal load.

What type of wire: Al or Cu (aluminum or copper)?

What is the total distance of all wire used in feet?

<IF> the total length of all wiring falls within 7-10 feet, step up to 8 AWG and, only use copper. Make sure all splice joints are fine and, no corrosion that will increase resistance on that CKT.

You always must rate the wire HIGHER than maximum current draw. The fuse holders shall also be rated for 120A like the generator fuse - if it's drawing that much current. The fuse and holder has to be rated higher than the current draw to prevent degrading it. Sooner or later, the fuse is going to blow upon start up.

CM

 

it is copper wire, roughly 7ft i would say......i ordered another fuse holder and a 100 amp fuse last night......im thinkin about ditchin this fan

 

That's the best thing to do. Even that 100A fuse is not adequate and, for a simple 12 VDC motor, this is ridiculous. I almost was tempted to mention putting in capacitors as a workaround to which, is even MORE ridiculous and, that's why I didn't mention it. IT's not the right thing to do.

The right thing to do is to rid the fan and find a better / more modern fan system. (if truly necessary)

CM

 

Late to the discussion, but here I am:

You can expect inrush currents of approximately 10x full load current from any electrical motor that instantaneously receives full operating voltage. Its duration should be short, unless you've got excessive power cabling impedance, so it shouldn't blow normal or slow-blow fuses that are rated properly. If the fan pulls 40A at full load, then its inrush is probably more like 350-450A -- but, again, the duration of that current spike should be very short if your power cable impedance is sufficiently low.

10AWG stranded copper wire has a nominal resistance of .0012 ohms/foot, so seven feet of that with 400A flowing instantaneously is going to drop 3.36VDC. If you've got the same seven feet of cable on the return run, you're dropping another 3.36VDC, and the motor is only seeing 7VDC at start-up. That will certainly extend your inrush period and be very hard on the motor.

If you increase the conductor diameter to 6AWG (.000465 ohms/foot), over the same seven feet, each one-way conductor will drop just 0.186VDC, and over the 14' in the circuit just 0.372VDC. -- and your fans will get past the inrush period much, much faster. That would be a very good thing to ensure a long life for the motor. Instead of seeing around 7VDC at start-up, it'll see more like 13.4VDC.

Make sure your ground connections are every bit as stout. Were it mine to do, I'd run the grounds right back to the battery, and use a high quality battery terminal connector of the sort audio enthusiasts prefer.

Leave just enough pigtail hanging out of the fan for future maintenance (aim for three or four cuts/re-splices) and no more. Make sure those connections are nice and meaty so the splices don't dissipate a lot of power.

Inrush current itself is not what's melting your fuse holders unless your fans are cycling like crazy. Chances are good, if you're using the inline/spring tension AGC variety, they're just not up to the task and are presenting resistance in the circuit. Go with an AGU type fuse and holder to eliminate the high connection impedance and you should be good to go. Since it'll be under the hood in a corrosion-promoting environment, a nice thin coating of dielectric grease applied to prevent air from reaching the contact surfaces of the conductors is a good way to prevent corrosion (the resistance it causes). No-Ox-Id battery terminal grease would work, too, if you've got that on hand. Don't go nuts with it -- a very thin void-free coating that covers 100% of the exposed metal is what you're after.

Ensure that your control circuit isn't cycling the fan excessively. You likely want a ten or 15 degree coolant temperature hysteresis band, e.g. on at 185, off at 170. The more the fan cycles, the shorter the lifespan of the motor and the greater the power dissipation in the power cabling/controls.

Oh, a final thought: It's pretty certain that the engineers who originally designed the circuit for the Lincoln ended their concern at the point of "won't catch fire", just as the folks who write the NEC and local wiring codes do. There's often a huge difference between "won't catch fire" and optimal (or even satisfactory) operation.

Hope this helps!

 

Good points there. This is why I initially recommended a MIN of 6 AWG on this CKT. We don't know what the cycle time actually is. If it is cycling ON / OFF in the heat, in-rush current is going to be a problem.

He's also witnessing 100A draw that tapers off to 40 but, what we don't know is, over what period of time. That's another factor.

We also don't know what the actual loss is. It may be > than what is outlined above.
Good post.

CM

 

alright, i guess im gonna try out the 100 amp fuse and try to use some bigger wire and see if that helps for now, thanks for the inputs guys

 

Don't! For a 40A full load current, you don't want to go any more than 60A. If the fan occasionally eats 60A fuses, go for slow-blow 60A fuses.

Imagine what happens when that motor shorts a few windings so it pulls 85A at full load and you've got a 100A fuse in there. Think: FIRE and melted bits of crap flying into the moving parts on the front of the engine.

If 6AWG, properly installed, and a type AGU fuse and holder don't work, you've got a severely hosed fan. In which case the plumbing will be useful when you get another that's not severely hosed.

 

If he's got the proper hysteresis and good plumbing, it won't be a significant problem unless it's spraying enough electromagnetic noise to annoy control electronics. 14 feet of 6AWG has a lot of heat dissipating surface area, and at 400A the peak power dissipation is only going to be in the neighborhood of 75W per conductor and for just a few milliseconds. The conductors and the AGU fuse/holder will handle that cycling at a 1Hz rate all day and all night -- the alternator might not like it, and rest of the vehicle might not like it, but the power feed would hold up just fine even if the inrush weren't limited by the counter-EMF on the second and subsequent starts due to the fan flywheeling.

As long as he's running a good, properly sized battery and the fan is plumbed directly to it, he should have no problems.

Myself, when I get around to installing the Perma-Cool fans, I'm going to add an additional control circuit so the fans won't run unless the coil primary is pulsing at more than 25Hz (375RPM), and with a few seconds of start-up delay to ensure that the electrical system is stable before the fan load hits. And with a manual on-off-auto override for those times when I think I'm smarter after the fact than I was when I designed the thing, of course. That way I can avoid running the fans and the starter at the same time, or, worse, load-dumping the fans and the starter at the same time.

It'll probably be next year before I get around to doing that, but when I do I'll post a schematic for anyone who wants to build his own ultra-paranoid fan control.

I gotta wonder where that 100A number is coming from. Unless there's some kind of current limiting device within the fan, I'd guess that the 100A number is coming from the poorly informed who are measuring the current with a clamp-on ammeter. 100A "peak" is about what I'd expect to see by doing precisely that.

A pretty good ballpark figure can be easily enough calculated: Measure the winding resistance (with all other components isolated from the circuit) with a properly zeroed ohmmeter, and divide that measurement into the nominal battery voltage of 13.8. If the winding resistance is 0.14 ohms, then the peak inrush current will be around 100A. But if the winding resistance is more like .03 ohms the peak inrush current is going to be closer to 400A.

Feel like doing that, dakotas ram?

Thanks. That means a lot coming from you.