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#1 |
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Member
![]() Join Date: Feb 2009
Location: Memphis
Posts: 13
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Motor Hz Calculations
I need some help. I am programming a spice auger system. I want to base how fast the auger should run on the required spice (gram/min, value known). I have my auger motor, I have all the name plate data on the motor.
So I know: My motor runs 60Hz at 2500 rpm, it has a 10:1 reducer I know what my gram/revolution needs to be for my auger. For this case 8.16 g/rev I also know how much spice I need to run. The value is in gram/min. What I need is to know the calculation to get to how many Hz to send to my motor to run. |
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#2 |
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Member
![]() Join Date: Apr 2002
Location: NC
Posts: 1,430
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I'm curious, is this a universal motor with brushes? That is the only motor type that I've seen run 2500 RPM at 60 Hz. I'm more familiar with regular AC motors at 60 Hz. that run either approx. 3550 RPM (two pole construction) or approx. 1725 RPM (4 pole construction).
In any event, at max gearbox output RPM (250) multiply by the grams per Rev. (8.16) to get the max grams/min. (2040). That would also = your motor max RPM (2500). Divide the 2500 rev. by 60 Hz. Each Hz. would = 34 gram/min. As you can see, this is not a formula but an attempt to explain the math. I hope this helps a little. Last edited by jrwb4gbm; September 13th, 2011 at 05:14 PM. |
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#3 |
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Member
![]() Join Date: Jan 2007
Location: Michigan
Posts: 1,055
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So you have:
8.16 Grams/Revolution of your ball screw. Motor speed, 2500 Revolutions/Minute (60HZ) 10:1 reduction You want: Grams/Minute of spice output At 60HZ, your ball screw is spinning at 2500/10 or 250 Revolutions/Minute which gives you 2040 Grams/Minute of spice output. At 30HZ, Your ball screw is spinning at 1250/10 or 125 Revolutions/Minute which gives you 1020 Grams/Minute of spice output. (Drive Command in Hz) / 60 * 2500 * (1/10) * 8.16 = Output in Grams/Minute |
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#4 |
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Member
![]() Join Date: May 2002
Location: Kalamazoo MI
Posts: 1,915
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As jrwb4... points out, there is something fundementally wrong with the statement " at 60hz the speed is 2500rpm. Not if it is an induction motor! Better check that before you go any further.
60hz is going to give you either 3600rpm, 1800rpm, 1200rpm, or some other even interger divided into 7200rpm. It could be a four pole motor running at 2500rpm but the frequency will be around 78hz, not 60. |
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#5 |
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Lifetime Supporting Member
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could be a gear reduction motor ?
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" You can teach a dog to sit and lie down but then he justs gets lazy"
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#6 |
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Lifetime Supporting Member + Moderator
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That is an odd base speed indeed.
So you have.... 60Hz/2500mRPM * 10mRPM/1aRPM * 1aREV/8.16grams so Hz = 0.0294 * X (Grams/Min) In my experience, I have found augers to be horribly non-linear with speed. If your motor is open loop you will also get variation just from the loading changes. |
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#7 |
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Lifetime Supporting Member
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maybe feedback? encoder wouldnt want my secret sauce to change flavors
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" You can teach a dog to sit and lie down but then he justs gets lazy"
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#8 |
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Lifetime Supporting Member + Moderator
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what are you using to program this with?
place one or two tags on the Auger into a counter control the revs from there
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![]() Give all the answers - People seldom learn. Guide People to the posibilities - Their Answers will come easily. - Their Knowledge will grow. They will then teach others Regards Ian Trost |
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#9 |
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Member
![]() Join Date: Feb 2009
Location: Memphis
Posts: 13
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here is the motor name plate data.
SPEC-33-2328Z148G1 HP-.5 RPM-2500 Frame-42CZ Type-3327P Volts-90 Amps-5 Serial-WO608280049 Gear ratio-10-1 iant, I am using RSlogix! |
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#10 |
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Member
![]() Join Date: May 2002
Location: Kalamazoo MI
Posts: 1,915
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That's a permanent magnet DC motor. Note the 90V. There must be a 120VAC powered DC drive ahead of the motor for speed control.
That complicates your auger feed rate calculations because the frequency is not controlling your motor shaft speed. The DC output voltage from the DC drive is. And, unless the DC drive has encoder or tach feedback from the motor shaft, the speed regulation under motor load variations will be as much as 5%. At any rate, if your auger can be trusted to output 8.16g of spice per revolution over the whole speed range, then it follows from the gearbox ratio that it will output .816g per motor revolution. Since 90VDC equals 2500rpm on the motor, the feed rate could also be expressed as: .816g/1(mtr rev) x 2500(mtr rev)/90VDC = 22.67g/VDC This is, as mentioned above, only about 95% accurate (5% error) but that might be close enough for your purposes especially if the load on the auger is constant. The error will be worse under about 100rpm on the motor too so it would be better to operate above that speed. If this works out to be unacceptably sloppy, you might want to consider using a stepper motor to run your auger. They are not expensive and can be controlled simply with very little error. Good luck! |
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#11 |
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Member
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Looking at the name plate data, I would guess you have a DC motor, or possibly a universal motor if it is actually running on AC. Definately single phase.
If this is the case, then you cannot effectively adjust speed by changing the hz. You would have to vary the voltage. Also, you would also not get an accurate adjustment unless you have some sort of feedback, probably an encoder would be best. Have you actually got this motor running now and just want to adjust the speed? Or are you setting up the motor for the first time, and have not run it yet? Also, what make of motor do you have? |
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#12 |
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Member
![]() Join Date: Feb 2009
Location: Memphis
Posts: 13
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There is a DC drive controlling the motor. Sorry I went from what the mechanics told me and spent time looking at the system today. So there is a drive board from MINARIK, SHANKLIN and here are the specs on the drive DRIVE, VARIABLE SPEED, ELEC, AC INVERTER, 1 HP, 115 VAC, 50/60 HZ INPUT, 90/180 VDC OUTPUT.
I am looking through the drive manual right now! |
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