# CIM Quiz 2

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The following was inspired by a recent discussion I had about analyzing a vacuum ball magnet:

The power required to drive a certain vacuum impeller is given by the equation

watts = (rpm^2)/81500

The impeller is being driven by two motors: a CIM and an FP, each through a separate gearbox.

The motors are each being driven at 12 volts.

Find the theoretical gear ratios of the two gearboxes which produce the fastest impeller speed.

[edit 11:56am] Note: for purposes of this analysis, ignore gearbox losses. [/edit]

[edit 4:54pm] Note: For this hypothetical problem, ignore the 40amp breaker and allow each motor to draw whatever current it needs (with 12V applied) in order to maximize the impeller speed. Pretend the impeller is only operated momentarily. [/edit]

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My Attempt
[spoiler]CIM : ~1:2 (1:1.97)
FP : ~5:1 (1:0.21)
[/spoiler]

At those gear ratios, the FP is not helping the CIM. It is actually acting as a generator, putting additional load on the CIM.

Post or PM your calculations if you want to discuss.

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Is the “Power” required to drive the impeller electrical watts or mechanical watts? Also, what are the units on the 81500 constant? I only ask because my first pass at this has no bounds.

The impeller is driven mechanically by the two motors (through their respective gearboxes). The formula watts = (rpm^2)/81500 gives the mechanical power input to the impeller required to spin it at the specified rpm.

Also, what are the units on the 81500 constant? I only ask because my first pass at this has no bounds.

The units on the 81500 would have to be (rpm^2)/watt.

For example, to spin the impeller at 5000 rpm would require (5000^2)/81500 = 307 watts.

Hint: this problem has a closed-form solution, although you could just set up the equations and iterate rather than solve algebraically.

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Ok. My first question is where did you get the specs for your impeller (is that provided by the manufacturer?).

Anyway: (apologies in advance - this is for 10.5 volts as that is all I had on my scratch pad from when I worked a similar problem before)

Cim @ 35 amps: .70 Nm torque, 230 W, 395 rad/s = 3770 rpm
Fish @ 35 amps: .260 Nm torque, 180 W, 686 rad/s = 6550 rpm

Total power, 230 W + 180 W = 410 W (assuming zero loss - real life I’d throw a 20% in here).

410 W= rpm^2 / 81500, rpm = 5780 rpm

3770 (cim) / 5780 = .65:1 or 1:1.53 for the cim.
6550 (fish) / 5780 = 1.13:1 for the fish.

Your request didn’t specify the maximum amperage, so I assumed 35 to be safe for our 40 amp breakers. If taken at 12 volts these will be slight different. This is based on the free speed of the motors at 10.5 volts, but truthfully it shouldn’t be /that/ far off from 12 volts.

I believe you are making the mistake of assuming the motors are providing their maximal mechanical power while spinning at their free speed.

Definitely misread the question.

Revised attempt
[spoiler]CIM - ~1:2.5 (1:2.45)
FP - ~1.2:1 (1.15:1)
[/spoiler]

Sorry for not being clearer: For this hypothetical problem, ignore the 40amp breaker and allow each motor to draw whatever current it needs (with 12V applied) in order to maximize the impeller speed. Pretend the impeller is only operated momentarily.

Tom: could you please PM me or post your 10.5 CIM and FP motor curves. Thx.

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Wherein the EE turned SE tries to be an ME
[spoiler]
CIM puts out 337W @ 2655 RPM
FP puts out 184W @ 7500 RPM
674W Total is ~7411 RPM fan speed.
CIM gearing is ~2.79
FP gearing is ~.988
[/spoiler]

A hint to all.

You have two unknowns, power and speed.

Two equations relating them to each other (one from the motor, one from the impeller).

What he did, I believe, is to pull the following data:

Cim @ 35 amps: .70 Nm torque, 230 W, 395 rad/s = 3770 rpm
Fish @ 35 amps: .260 Nm torque, 180 W, 686 rad/s = 6550 rpm

… from some 10.5 Volt motor curves that he has.

He then reasoned as follows:

• if 35 amps are supplied to each motor, they will produce the output power indicated above at the rpm’s specified above (if his 10.5 motor curve data is valid, this is valid reasoning)

• the sum of the output power from each motor is then used to calculate how fast the impeller will spin, given that power

• that impeller speed is then used to calculate what the gear ratio must be for each motor in order for each motor to be spinning at its respective specified rpm when the impeller is spinning at the speed just calculated.

It’s an interesting solution, but it’s not the solution to the problem I had in mind because it requires that the current to each motor be controlled at 35 amps. The problem I had in mind did not involve controlling the current. As stated, each motor is being driven at a constant voltage (12V), not a constant current.

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and two gear ratios makes four.

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Pretty close. Show your calculations so we can discuss.

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Well done !

(but where did you get your FP motor data from? It differs slightly from what I have)

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Looking back, I think my eyes wandered by a box or two when I read the speed.

I was making the assumption that one had already combined the two motors into one equation by gearing the FP to match the CIM (with the CIM at 1:1). I should have been more clear.

You are correct. I use the motor curve data from here:

http://www.chiefdelphi.com/media/papers/download/2593

Indeed, I try to do my math to confirm that I will get what I need without popping breakers. In addition, I’ve seen even a fresh battery drop voltage when you are pulling power to the drive train: doing your calculations at 12 volts may result in a performance problem in the field. Just our experience though, perhaps others differ.

For systems where the voltage variation would have an appreciable effect on performance, we run all math scaled to 10V (just linearly scale down the free speed and stall torque).

Does anyone know the continuous current rating for the FP’s? I’m pretty sure the CIM is about 27A, and so that impeller would make sense to work with it, but for the FP? I’d imagine replacing it every match.