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#### Sunking

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The units conversion factor is certainly 746 W/hp. What Sunking calculated and presented as "w/hp efficiency" was a combination of the inverse of efficiency and the units conversion factor, which I agree (as I suggested earlier) is bizarre.
Bizarre to most people I suppose, or perhaps genius to others. It depends on how you approach and solve a problem. I do quite a bit of Renewable Energy work using Off-Grid Solar which is extremely inefficient. At the very best on paper is 66% and realistically less than 50% when put into application. So if you want 1 Kwh usable, means the panels have to generate 1.5 to 2 Kwh. So with a 4 Sun Hour would require say 1.5 Kwh / 4 Sun Hours = 375 watts to 500 watts of panels to give you 1 Kwh usable. The priority is different where you would not be concerned too much about efficiency.

So tell me which is easier and takes less fingers and toes to figure out. Teacher ask:

We have a motor that develops 10 HP, is 74.6% efficient, and uses 1000 watts per HP. How much electrical power is required to run the motor?

A Nerd would pull out a slide rule like he was taught, go to work pushing buttons using traditional percentage method to solve the problem, and finally figured it out about 5 to 10 seconds after I raised my hand and said "10 Kw!" Guess who got an A+ from the teacher.

#### MattsAwesomeStuff

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Guess who got an A+ from the teacher.
Again, I find that bizarre. The "genius" of this solution exists only when the question is perfectly framed for that exact answer, as it's 74.6% efficient and there are 746 watts in a horsepower. Any other numbers and you're doing just as much calculation, only now you have unit conversion on top of it.

A sensible real world problem would say: "We have a 7460 watt motor that is 74.6% efficient, how many watts does it take to run?" With the answer being "Gee, 10kw."

But at that point all you're calculating is efficiency, which has almost nothing to do with motors and such specifically. It could be relevant to anything.

...

I've heard other esoteric things, like, the way you know that a contactor is rated for motor loads is if it shows a Horsepower rating in addition to the wattage rating. I don't know if that's true, but it seems silly. If it's rated for motor use, it should just say that, or say Inductive, or, something. Not just change the units and imply that that means motors.

#### Sunking

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A sensible real world problem would say: "We have a 7460 watt motor that is 74.6% efficient, how many watts does it take to run?" With the answer being "Gee, 10kw."
My answer certainly would not be 10 Kw as you answered incorrectly. Not sure what your beef is, I have not responded to anything you have said up until now in this thread. Motors are rated either in Output Mechanical Horse Power, or Input Electrical Power. Correct answer is 7.46 Kw @ 7.46 HP. Wattage is the thermal limits you want to know and that comes from Watts. I clearly understand the units of measure and know how to re-arrange them to suit my taste.

The point of the question was it gave you data to solve the problem 3 different ways. One was humorously the Academic method dividing out 10 x 746 watts / .746 = 10 Kw. Or just simple Applied Methodology using 10 x 1000 watts = 10 Kw.

One universal characteristic of a DC Motors we know is efficiency is on the order of 70 to 80 percent. Method to my Bizarre Madness is assume Worse Case (70%) efficiency and you are in the ball park every time with 1000 watts/HP and can never be disappointment with the details of the final results. It will work and exceed expectations. It is a KISS method of design.

As you were saying......

#### brian_

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Motors are rated either in Output Mechanical Horse Power, or Input Electrical Power.
They are rated in output (mechanical) power, and there is of course a corresponding input (electrical) power for the rated conditions. Input power will be expressed in watts or a multiple (kW); you are accustomed to seeing output power expressed in horsepower (which is traditional and still used, especially in North America), but it would be in watts or a multiple (kW) for anyone working in consistent S.I. units.

I did a quick search for euro motor specs, and the first thing I found was a catalog for the Marathon Electric Motors IEC IE3 Motor Range, which shows (in the Technical Data charts starting on page 12) output power in kW (that's kW only, there is no mention of horsepower).

You might also be interested in the EU's ecodesign requirements for electric motors, which consistently uses the proper power units (kW, not hp) for motor output.

A lot of people are hung up on "horsepower", reading a lot of meaning into the use of this unit which is often not justified. It is not limited to mechanical power, and is not the only power unit which is appropriate for mechanical power.

#### MattsAwesomeStuff

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Not sure what your beef is,
Huh? I have none.

It's odd to me that you think "someone has a different opinion than me, why don't they like me?" and presume some motive of aggression, rather than simply, "I said somethign and someone said something in reply". It's a discussion forum. Just commenting. I find that way of doing things bizarre. It's not an attack, I'm not telling you that you can't or that you're wrong. It's just something that came to mind based on what you said.

I suppose it's a neat trick that with DC motors you can kind of ballpark that 1hp of output requires 1kw of input, but, beyond a chuckle, it's a bit funny to be using them in tandem, one to refer to output the other input, when they're both the same thing.

Probably a generational thing too.

#### Sunking

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I did a quick search for euro motor specs, and the first thing I found was a catalog for the Marathon Electric Motors IEC IE3 Motor Range, which shows (in the Technical Data charts starting on page 12) output power in kW (that's kW only, there is no mention of horsepower).

Thanks for the links. After 3 pages here is where I have arrived at. From an electrical point of view there is no difference between supplying power for a AC motor on fixed line voltage, or a DC Series motor speed controller. Determine maximum continuous load current at a fixed voltage say 100 amps which is the same method you would use for an AC motor using FLA value. Likewise in an AC circuit we want to account LRA to properly size conductors and coordinate current limiting because sometimes LRA can be as much as 6 x FLA. All though Major and I danced around we come to the same methodology. Size Controller 4 - 6 x FLA aka DC Max Continuous Load Current. Rest takes care of itself assuming the mechanical train driver got his part right.

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