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Re: (ET) 12 v marine batteries



hPower is what matters. Power is what runs the lift motor and power is what lifts the snowblower.

Power=Current*Voltage. Or P=I*E.

So for a given amount of power you can double the current, half the voltage, and off you go. Or you can double the voltage and halve the current. Wires are rated to carry current, not so much voltage so reducing the current and boosting voltage means less load on the wires for a given amount of power.

It also means the motor can develop way more power than it should, and if loaded will dissipate more power as heat. So you need to limit the amount of power the motor can produce. One way is with a fuse, if you half the size of the fuse it will blow when the motor exceeds the same amount of power as with lower voltage and higher current.

I think the lift motor fuse was 30a. At 18v that's 30*18v that's 540 watts of power before it goes. Dropping the fuse to 15 and doubling the voltage gives you 15*36=540 so the world is good.

If we didn't derate the fuse the motor would produce over 1kw of power, which is insane, and will either break things or (if the motor is stalled and all power is being dissipated in the motor) will put a 1kw heat load in the armature of the motor. Melted armature and such.

The one theoretical problem would be in the lift interrupt switch. That's designed to shut off the motor if the load gets too high. Because it operates on current it will open a lot later in the power curve, possibly damaging the motor. And with the increased voltage there is a chance it could arc over on opening and weld shut. Thus again the smaller fuse. And a higher voltage can arc over on the commutator, but probably not 36v.

Fun stuff. Really the motor was designed for 12v@30a peak or 360 watts but that's for decades of duty cycles. Bumping it up a bit will not cause the world to end IMO.

C


On 1/30/2019 9:23 PM, David Roden wrote:
On 30 Jan 2019 at 18:44, Chris Zach wrote:

remember, twice the voltage means half the amps

I'm not an EE, just a hobbyist, but that's not what I learned.  Ohm's law
says that I = E/R, so for a given load, current is directly proportional to
voltage.

The situation is more complicated when the motor is running at normal load
and speed (back EMF and all that), but my concern would be with what 
happens
when you get to the end of travel, or have a very heavy load that keeps RPM
way down.  Under stall or near-stall conditions, double the voltage means
about double the current.

It's the current that makes the heat that burns out your armature.  That's
part of what the fuse is supposed to protect against.  But since it's still
current that concerns you, not power, I don't see the need to halve the
fuse's rating.  I'd think that would just cause nuisance opens.  What am I
missing here?


David Roden - Akron, Ohio, USA

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