Chief Delphi - CIM Quiz 2

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- - CIM Quiz 2 (http://www.chiefdelphi.com/forums/showthread.php?t=87726)

Quote:

Originally Posted by Ether (Post 983084)

Well done !

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

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.

Quote:

Originally Posted by AdamHeard (Post 983079)

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).

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.

Quote:

Originally Posted by Ether (Post 983080)

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.

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.

Quote:

Originally Posted by Tom Line (Post 983120)

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.

Quote:

Originally Posted by Tom Line (Post 983120)

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

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

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?

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

Quote:

Originally Posted by joek (Post 983185)

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

How did you come up with these numbers?

Quote:

Originally Posted by Chris is me (Post 983186)

How did you come up with these numbers?

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

Quote:

Originally Posted by Ether (Post 983082)

Pretty close. Show your calculations so we can discuss.

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

Quote:

Originally Posted by joek (Post 983190)

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

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....

Quote:

Originally Posted by Chris is me (Post 983197)

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

15600/5310 = 2.94

Quote:

Originally Posted by rahilm (Post 983193)

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

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.

Quote:

Originally Posted by EricVanWyk (Post 983078)

CIM puts out 337W @ 2655 RPM
FP puts out 184W @ 7500 RPM
674W Total is ~7411 RPM fan speed.

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

Quote:

Originally Posted by Ether (Post 983203)

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

... 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!