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#### Sunking

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OK this is just a drill to make sure I got my noodle wrapped around Series Wound motors and choosing the right Controller. Thought I had it figured out but after another thread has me double checking. My experience with motors is all Industrial AC Induction motors on a fixed 60 hz operating frequency and some VFD’s thrown in once in a while.

So here goes, someone throws a motor on the bench and ask what can it do? All we have is nameplate data that reads:

Operating Voltage = 48 vdc
Peak HP = 10
Continuous HP = 4.14

First thing it tells me I would be using 48 volts x 88 amps = 4224 watts. From that if I divide 4224 watts / 4.14 HP = 1020 w/hp efficiency. Here is where I get stuck and might be making an errors. This is a real motor with real numbers and it is recommended with a 400 amp controller. That does not jive with a Peak 10 HP motor. If I assume a 1020 w/hp efficiency, a 10 HP would be roughly 10,200 watts. Realistically I know efficiency will not be that high, so say 70% efficiency 10 HP is 10.700 watts. My protein computer says 10.700 watts / 48 volts = 222 amps. At 400 amps input would be 19,200 watts. Efficiency??

Having gone through all that and from a little research both methods are all wet. At least for selecting controller amperage. Everything I have been taught is you want as much as connecting the battery directly to the Terminals will produce called Stall Current at 0 RPM. If Stall Current is not listed in the specs, one can use a DRLO making several measurements while rotating shaft, and then averaging resistance to determine Stall Current = Battery Voltage / Motor Resistance.

So what is right if any?

Engineer inside me says, just a little less than Stall Current, and let the motor take care of the rest when it spins up. Allow for voltage sag of 25% with current limiting to protect the motor.

#### Sunking

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It looks like you're calculating an "electrical power to produce mechanical power" factor, in watts/horsepower. As for actual efficiency:

Yes essentially Input Power so I could calculate how much input power input required for 10 HP

input: 48 volts x 88 amps = 4224 watts (as above)
output: 4.14 hp = 4.14 x 746 = 3088 watts
efficiency = output / input = 3088 W / 4224 W = 73%

That's not good, but I don't know if any better can be expected.

The "Load Torque" and "Load Speed" do correspond to the listed "Continuous HP".

Yes I had to use torque and rpms to extrapolate Continuous HP and it worked out to 4.14 HP. I also extrapolated 10 HP power by using the constant 1020 watts /HP x 10 = 10,200 watts, then factored out current of 222 amps peak.

So where have I gone wrong? Pretty sure I have made an incorrect assumption. I calculated current required based Input Power of 222 amps, so a 200 to 250 amp controller would work. However the manufacture recommends a 400 Amp controller with the motor. OK it occurred to me manufacture might be specifying Stall Current rather than based on 10 HP wattage.

I know peak power values of a motor is a Thermal Limit so max power might be debatable if more cooling factors in. Manufacture did not publish Peak Power watts for motor @ 10 hp.

You are throwing me off converting on mechanical power. In my minds eye Input Power is what a designer needs to know right?

#### Sunking

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Hi Sunking, brian_,

Sun... You know units, right?
The same is true for power. Both watts and horsepower are units of measure for power. It does not matter in what form the power is, like electrical, or mechanical, or radiant, or average, or peak. Both hp & watts measure the same thing: Power. One horsepower = 746 watts. Here's a good review of power and its units of measure.
https://en.m.wikipedia.org/wiki/Power_(physics)
Now that is funny stuff Major, glad you chimed in. I am an EE so think I have units of measure down after 40 years. Also, beware of ratings. Rarely will rated power translate to useful efficiency values. Manufacturers often use worst case tolerance limits on nameplates, or nominal values. And for a motor, peak power (output power) may not be a thermal limit at all.

And the controller current limit is just that. It does not relate to motor stall current or motor rated power. It is the same "5hp rated" motor if I use a controller having a 200, 400, or 1000 ampere current limit.
This is where I am at Major, I fully understand if you use a 400 amp controller, is the Maximum Current the Controller cill Sink. Does not mean the motor will ever use that high of a current, just means it can if the Resistance is low enough for the applied voltage.

Riddle me this. Everything I know about Series Wound Motors Draw maximum current at 0 RPM's aka Stall Current. The Stall Current is simple Ohm's Law of Battery Voltage / Motors Resistance = Stall Current. Do you agree with that? There is no current higher than Stall Current.

Does not mean mean or imply you should size the controller to handle Stall Current, but is the only way to get maximum torque at very low RPM's. My thinking is select a Controller to some value less than Stall Current for protection. How much I do not know and the reason for this journey. Having said that I do not think any motor speed controllers would allow to apply Full Battery (100% PWM modulation) instantaneously to drive Stall Current. At least the Alltrax Controllers I am familiar with will not do that. They walk the voltage up over a period of time of say a few seconds, so by the time the Controller hits 100% modulation, the motor RPM's will have already spooled up to a few hundred RPM's to where Stall Current is not possible due to back EMF.

I know how to find Rm and have the equipment. Lock the Rotor down, apply 10 amps, and measure the voltage on the motor terminals. I even have a 4-point DRLO that does the same thing. But me thinks I will get a resistance so low on a 48 volt golf cart motor, would exceed any controller current limit out there. Just trying to figure out how to size correctly without a lot of over kill. Say 300 amp is more than sufficient and no reason to pay up for 400 amps. Part of me suspects Stall Current maybe higher than some of the controller current limit. That is what I am trying to nail down.

I work with AC motors all the time, and that is easy. Voltage specified is the most efficient operating voltage, and the other two numbers I want to know if LRA (which is the exact same thing as Stall Current on a DC motor), and FLA (full load current). Example for a 480 volt motor maybe specified at 470 volts, FLA = 40 amps. That tells me I need to size my wire so I incur a 10 volt drop with 40 amps, and coordinate breakers for LRA start up current if needed. LRA can be as low as 2 x FLA for soft start, and as high as 6 x FLA.

#### Sunking

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The resistance of a DC series motor is very low - as in 1/100th of an ohm (WAG) - so it only takes 10v to get 1000 amps
I think you might be off by a magnitude 10. in my application. Size is going to have a lot to do with it. Most motors I am speaking of are 2 to 4 hp motors. Larger motors will have lower resistance. But I agree it will be low, and with the size of motor I am working with resistance will be 1/10th. I could be wrong.

Funny thing on another forum a member posted some specs from D&D Motors on a Controller / Motor Combo, a 48 volt golf cart motor and 400 amp Controller. It came with some dynamo test.

Rm was listed at 0.0891 Ohms.
Maximum Continuous Load current = 92 Amps
4.1 Kw Continuous, 12 Kw Peak.

Ignoring voltage sag Stall Current would be 48 / .09 = 533 Amps. Interesting enough Stall Current is specified @ 412 amps which is likely taking voltage sag in a 48 volt battery system into consideration. All the resistances in a golf cart is fairly uniform and known. So you could use fixed resistance constant into an equation that is realistic.

The "stall current" is simply dependent on the voltage

Voltage and Resistance.

#### Sunking

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You clearly do not understand power units (hp & watt) and efficiency calculation.

Don't be silly. The specs state both continuous mechanical hp (Torque and RPM) output = 4.14 hp and specifies electrical input power of 88 amps @ 48 volts. Efficiency = 3088 / 4224 x 100 = 73.1%. Now if I assume an efficiency of 73.1%, to get 1 mechanical hp out will cost me 746 watts / .731 = 1020 watts Electrical Input Power.

From the Specs Max Peak HP is 10 Hp. I assume that is mechanical HP and if I assume 73.1 Efficiency Input Power required is 10 x 1020 watts = 10,200 watts. My assumptions may be wrong, but my math is spot on.

At 73.1% efficiency requires 1020 watts for every HP mechanically out. So how is that wrong?

Disagree. The DC stall current using the equivalent series resistance of the motor is almost always, maybe always, higher, usually much higher, than the current limit of the controller. On any full throttle start, current limit is active. After maybe the first few pulse cycles (like 10 ms), there is no ramp function.

Funny stuff because you just agreed with me. I said Stall Current = Battery Voltage / Motor Resistance. That is what I asked,and you cannot disagree with that, it is in all university subject matter on motors and boils down to simple Ohm's Law.

Now I agree with you DC Stall Current might, maybe, could, and/or will be higher than the Controller. Allow me to remind you the whole point of the thread was to pair a Controller/Motor. Question boils down to this.

Do you push a motor to DC Stall Current to obtain maximum start-up torque, or do you want to limit it and how much? Should not affect speed, just very low speed torque. I would think limiting Stall Current would be worth giving up some low end torque?

Induction motor LRA is not "the exact same thing as Stall Current on a DC motor." Yes, both values are the respective maximum current draw for the motor across the mains. That's where the similarity ends, IMO.
Well let others decide for themselves. I can define to textbook definitions easily looked up

LRA = Locked Rotor Current aka Start-Up current is the amount of current the Rotor will draw when the Rotor is at 0 RPM or locked. As the RPM's spools up, back EMF will press back on the voltage applied to the Rotar and current will follow Ohm's Law for AC circuits. LRA = Applied Voltage / Rotor Impedance. Sound familar?

DC Stall Current occurs at 0 RPM when full battery is applied. As RPM's spools up Back EMF presses back against the battery, lowering voltage and current of Ohm's Law. DC Stall Current = Battery Voltage / Motor Resistance.

Sounds the same to me. I am just stupid. But you are right, they are different. One use Resistance to determine Current, and the other uses Impedance which will be a little higher than it would be for DC. What do I know?

#### Sunking

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w/hp efficiency. O.K. I get it now. My bad. It is as brian_ says; an inverse efficiency and units conversion mixed into one convenient mess.

No problem, I just took a different road to get where I was going. I do understand the efficiency is going to be different at Peak HP and Continuous HP. Most likely Peak HP Efficiency will be lower than Continuous HP. Perhaps you can comment on that?

Whole point of the exercise is to determine how to size a Controller to a Motor. Example I know for fact the motor in my example would work with 100 amp controller. I would still have 4.22Kw Input Power @ 73.1% efficiency (4.14 HP). However Peak Input and Output would be severely limited with a 4.8 KW peak Input Power. Great for motor life, not so good for acceleration.

So this has me going to different directions. First direction I assume if a manufacture list maximum Peak Input Power is say 12 Kwh, (thermal limit) rated nominal Voltage = 48 volts, then one would need at least a 12,000 watts / 48 volts = 250 Amp Controller to get to 12 Kw.

Then when I see the motor manufacture says to use a 400 amp controller is telling me something else is at play. My educated guess from knowing quite a bit about batteries and wiring there is going to be significant voltage sag incurred on the batteries and wiring. Example say you are using Trojan T-105 batteries have an Internal Resistance of 3 milli-ohms x 8 in series is 24 milli-ohms, and 10 feet of 4 AWG resistance is 2.5 milli-ohms giving a total of 27 milli-ohms. With 400 amps flowing is 10.8 volts, just call it 11 volt sag dropping 48 to 37 volts. Just so happens 37 volts x 400 amps 14.8 Kw which is in the ball park of 12 Kw the thermal limit of the motor.

My last thought is select a controller capable of delivering Stall Current based on pure Ohms Law of the motor resistance.

Boils down to how big of a controller does it take.

It all depends on your design objectives. The drag racer may well put more value in the launch torque than the golfer.

I agree 100%. A drag racer has no problem smoking a motor, that is what they are suppose to do, push the limit to the breaking point. That is why there are sponsors, someone to fill the trailer up with spare motors, controllers, and batteries. For me I want some quickness, balanced with motor/controller longevity.

So Major if you can help with that, I would appreciate it.

#### Sunking

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Major and Duncan, thx.

Duncan you got me puzzled a bit. Add what switches where to control current limit? I understand how a current limiter works electrically, and I know many of the newer higher end controllers allow you to select what current limit you want with either software or dip switches in the controller. So do you mean move the dip switches outside the controller to the Dash? I am missing something.

Major thank you for taking the time to explain your design logic. I have no problem with it. It sounds reasonable.

What got me started thinking and grinding on this as I look at upgraded replacement 48 volt golf cart motors, I noticed the specs really do not say anything and the motor nameplate even gives less information. However a few motors like the one I used in the example had some numbers in the specs that could be crunched. Anyway all the after market motors I see suggest a controller size, and being the train driver I am wanted to know how to size the Controller.

Coming from the AC Motor side of the house we do not use speed controllers, just raw line voltage. (VFD not included in statement). Knowing AC motors and DC motors are virtually identically the same except for commutation, your 4x or more continuous current makes sense too me. Going back to LRA and FLA example in an AC Motor you will find LRA can be as high as 6 x FLA if no Soft Start is present. So if you use the motor example I used with a max continuous current rating = 88 amps, the 400-Amp Controller now makes sense, and is in the ballpark of 4 to 6 times normal load current. That I can accept and live with.

Again, thank you Major and Duncan.

#### 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.

#### 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......

#### 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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