CIM Quiz 2
 

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

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]

Re: CIM Quiz 2
 

Re: CIM Quiz 2
 

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.

Re: CIM Quiz 2
 

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.

Re: CIM Quiz 2
 

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.

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.

Re: CIM Quiz 2
 

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.

Re: CIM Quiz 2
 

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

Re: CIM Quiz 2
 

Definitely misread the question.

Re: CIM Quiz 2
 

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.

Re: CIM Quiz 2
 

Re: CIM Quiz 2
 

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).

Re: CIM Quiz 2
 

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

... 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.

Re: CIM Quiz 2
 

and two gear ratios makes four.

Re: CIM Quiz 2
 

Pretty close. Show your calculations so we can discuss.

Re: CIM Quiz 2
 

Well done !

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

Re: CIM Quiz 2
 

I pulled it from http://www.usfirst.org/uploadedFiles...or%20Curve.pdf

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

Re: CIM Quiz 2
 

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.

Re: CIM Quiz 2
 

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.

Re: CIM Quiz 2
 

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).

Re: CIM Quiz 2
 

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.

Re: CIM Quiz 2
 

Does not compute.

For the FP @ 10.5 volts and 35 amps it says
0.222 Nm, 623.7 rad/sec... = 138 watts, not 180 watts.

The spreadsheet is pretty busy. Am I reading it wrong?

Re: CIM Quiz 2
 

cim w/ 1:12 gearbox, fp w/ 1:6

Re: CIM Quiz 2
 

How did you come up with these numbers?

Re: CIM Quiz 2
 

my head, since the RPm of an Fp motor is roughly double that of a cim, the cim will have to travel twice as fast

Re: CIM Quiz 2
 

CIM max power is 337 watts @ 2655 RPM
FP max power is 184 watts @ 7500 RPM

Total max power is 521 watts
sqrt(521*81500) = 6516.25

6516.25/2655 = 2.45
6516.25/7500 = 0.869 = 1/1.15

Re: CIM Quiz 2
 

But why increase either motor's speed by a factor of 6? What led to that decision?

(though it's worth noting that it's closer to 3.5 times....

Re: CIM Quiz 2
 

15600/5310 = 2.94

Re: CIM Quiz 2
 

Looks good.

The data I have here shows the FP max power 184W at 7800 rpm. The FP would draw slightly less current and run cooler if geared for 7800 instead of 7500.

Re: CIM Quiz 2
 

I just noticed: how did you add 337+184 and get 674 ?

Re: CIM Quiz 2
 

... Ill advisedly.

Wow. Not sure what happened there. I'll have to check the excel sheet I made, tomorrow.

Edit to add:
Apparently I mis-clicked in excel, and selected the CIM motor's output power twice. Oops!