Frank: It is not clear why your requirement is for such constant rpm (must be between 59 and 61 rpm to satisfy some standard??)
Crispin is correct that a synchronous motor can do that with simple gearing - but your 1/3 hp motor at 1725 rpm is almost certainly an induction motor. Except for clocks, almost all industrial motors are induction - as Andrew was describing (they are much cheaper). With a speed sensor, voltage control, and the right set of gears, you can probably get the constant 60 rpm you want with your existing induction motor. I presume such speed controllers are on the market at reasonable cost. It may be cheaper to buy the synchronous type of motor if you have to maintain a specified speed. If you get the synchronous motor, your deviation from 60 rpm will be measured probably in thousandths or millionths of a cycle - not +/- one cycle. This is courtesy of your local electrical supplier, who gets fined a lot if the grid frequency is ever off by very much Ron (taught a motor course once) ----- Original Message ----- From: "Frank Shields" <[email protected]> To: "Discussion of biomass cooking stoves" <[email protected]> Sent: Tuesday, February 12, 2013 2:48:56 PM Subject: Re: [Stoves] Calculating the RPM Dear Crispin and Kevin, I wonder what they consider ‘full load’ when rating the motor. Must be some torque percentage of the HP value or something? Way more complicated than I thought. Electronic slippage (?) Interesting along with calculating the speed based on 60 cycles per sec ….. Thanks For the info. Frank From: Stoves [mailto:[email protected]] On Behalf Of Crispin Pemberton-Pigott Sent: Tuesday, February 12, 2013 11:15 AM To: 'Discussion of biomass cooking stoves' Subject: Re: [Stoves] Calculating the RPM Dear Frank The motor will not slow down under no load if you drop the voltage because it is a synchronous motor. You would have to change the frequency of the AC to change the speed. That said, if it was lowered enough the motor couldn’t ‘keep up’ to the 60 cycle synchronous rotation and would slip, slowing down as a result, however that slippage is strongly dependent on the load which would be really inconvenient. Being a square box the speed would actually change 4 times per rotation as the load alternately came and went. I suspect that the motor is rated at 1725 RPM under full load not actually ‘all the time’. Under no load it might be very close to 1800 (60 cycles per second * 60 seconds per minute / 2 sets of windings gives 1800). As it is geared down so much I suspect the motor will not be working hard at all. Given that you want about a 30:1 reduction and there is a general rule that you should not step down more than 7:1 at a time, I suggest you get 2 sets of 1-1/2” and 8” pulleys in series which will give you 28.4:1. If your speed is 1725, the shaft will turn at 60.64 under load. That is pretty close. It might go a tiny bit faster but the pulleys are not all that accurate anyway so you would have to measure it to know. It is slow enough you can count while watching. With such an installation you should use Goodyear belts (the ones that are fan belts but look like they have big teeth). If you drop the voltage it will only have to slip a little (magnetically speaking) in order to drop one or two RPM but it will fluctuate a little during the turn because the box is square. I have a book here for calculating how much power can be transmitted through the belt contact angle on the small pulley based on their centre distances, if it matters. I don’t think it will. Pick a centre distance that matches a common belt. Regards Crispin Dear Frank 1: Box shape and diameter are VERY important. At "critical speed", for a given combination of rotational speed and container diameter, centrifugal force will hold the pellets stationary against the container wall, and there will be zero tumbling and pellet wear. A machine running at say 95% critical speed will create very little abrasion, but a machine running at say 60% of critical Speed will create a great deal of abrasion effect, because of the cascading action. 2: Do you want to turn the drum/container/box for a fixed time, OR a fixed number of revolutions? 3: One could calculate the required HP, but the simplest and least costly thing is to try the 1/3 HP motor, and if it does not overheat after say 4 consecutive tests, it is good for the job. If it does overheat, then you have proven that you need a bigger motor. Best wishes, Kevin ----- Original Message ----- From: Frank Shields To: 'Discussion of biomass cooking stoves' Sent: Tuesday, February 12, 2013 2:30 PM Subject: [Stoves] Calculating the RPM Greetings Stovers, Building equipment for determining the hardness of pellets and biochar I need something to turn a 9 kg box at a speed of 60 RPM +/- 1. I have a 1/3 rd HP motor that is rated at 1725 RPM Fitted with a 2” pulley attached to the motor and going to an 8” pulley gives me a shaft turning 431 RPM Then another 2” pulley going to an 8” pulley on another shaft gives me 108 RPM I plan to adjust the 108 down to 60 using a voltage regulator. Is this a good plan? And is there another factor I need to include for the weight of the box I am turning? Thanks Frank Frank Shields Control Laboratories, Inc. 42 Hangar Way Watsonville, CA 95076 (831) 724-5422 tel (831) 724-3188 fax www.controllabs.com _______________________________________________ Stoves mailing list to Send a Message to the list, use the email address [email protected] to UNSUBSCRIBE or Change your List Settings use the web page http://lists.bioenergylists.org/mailman/listinfo/stoves_lists.bioenergylists.org for more Biomass Cooking Stoves, News and Information see our web site: http://www.bioenergylists.org/ _______________________________________________ Stoves mailing list to Send a Message to the list, use the email address [email protected] to UNSUBSCRIBE or Change your List Settings use the web page http://lists.bioenergylists.org/mailman/listinfo/stoves_lists.bioenergylists.org for more Biomass Cooking Stoves, News and Information see our web site: http://www.bioenergylists.org/
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