Post By robvds
6 pole Vs 8 pole motor, does more poles make it stronger ??
I need and would appreciate some advice.
I want to put a three phase motor and a VFD on a cold-saw that currently has a single phase motor at 940 RPM.
I also want to have a little more power so I'll go from .93kW to 1.1kW or even possiblly 1.5Kw.
The final RPM I am looking to get is between 600 & 850.
I can get an 8-pole/710 RPM motor or 6-pole/940 RPM motor.
Is it better to get an 8 pole motor and speed it up a little or a 6 pole motor and slow it down.
I just need to be able to control the speed without loosing power as the saw already stalls a bit on stainless and is just too fast.
I need your expertise.
Thanking you in advance,
Horsepower is Torque multiplied by RPM. So a motor with more poles has more torque and a lower RPM. If the saw is too fast now a lower RPM motor with more poles seems like a better choice, and keep in mind that a limiting factor is the strength of the drive system of the saw.
I agree with GeoD. Is this a direct drive cold saw? I am used to 40 to 100 rpm (14") for steel and 200 rpm for aluminum. If you slow down the motor with a vfd you will not lose torque but HP will be lower so I think going bigger is warranted.
The more poles the greater the torque so 8 pole motor will have more torque than a 6 pole. Lots of reasons why that I will not go into. However you also have to realize that the more poles the larger the physical size of the motor, more copper in the winding and therefore greater cost. Also note that speed rating of the motor is at the KW rating. The speed you list for the motor is when it's under full KW rating load. No load speed for a 50Hz 6pole motor will be 1000RPM and 8 pole is 750RPM (synchronous speed) and will "slip" down as torque requirements increase. The input current rating is also at the KW rating and will be less at no load.
dan u did not say how you would control the speed; sorta important to the answer... as stated already, vfd changes speed but does not give u more torque as you go slower like gear change does.... if your using gears to change speed, then both motors will give SAME results! if using vfd it may be another story.....
as froneck also said, more torque/lower speed is typically a larger motor physically for same hp; said another way, any motor's size will be governed by torque rating, so that slower/higher torque motor will be larger and so probably more expensive. so why do it if you are using gears and will get same exact output from either motor? for that matter, use a 4 pole (cheaper again) and gear it more - same output.
so give more info on how u plan to control speed. finally, YOU can calculate EXACTLY what you will get from either motor, we can show you how to do this fairly easily & understandably, but tell us the rest of the story first so we know which way to say it...
Thanks for the info to date.
I will be using a VFD to suit as the motor will be directly mounted to a gearbox that I can't change the gearing in.
I know little about the workings of these things and I'm not sure if I should get an 8-pole/710 RPM motor and occasionally speed it up a little using the VFD or 6-pole/940 RPM motor and always lower the speed.
I have 2 motor choices, 1.1kW or 1.5kW. I'm thinking going from .9 kW to 1.1 kW will be enough increase (about 20%) in power as I don't want to go overboard.
It's the speed up occasionally or always speed down that gets me. I guess that's what will decide the motor I get.
Thanks again for the help.
so u will be using a vfd, means you will NOT get more torque with lower speeds.... so lets use ur 0.9kw motor choice as example. u can change the math to work it for the 1.1kw too later if u want....
0.9kw/(0.746kw/hp)=1.2hp motor. So this will be either a 710 or 940rpm motor..... so.... since hp=N*T/5252....
T_of_710rpm=8.9#-ft -- from say 100-710rpm, then constant 1.2hp from 710-2000rpm
T_of_940rpm=6.7#-ft -- from say 100-940rpm, then constant 1.2hp from 940-2000rpm
So you see you will get LESS torque at lower speeds from the 940rpm motor if same HP...
So you see you will get SAME torque at higher speeds above 940rpm motor if same HP...
Since you want more torque at lower speeds, then the higher pole count motor would give this to you.... 8.9/6.7= 33% more torque to be exact....
So now you can compare the two specs in real world TORQUE terms, which is what I think you were asking to begin with. Keep in mind the 8 pole same hp motor very well should and may be in a larger frame size since nothing is free in this world; motor size is determined by TORQUE capability; I always say "speed is free," TORQUE is what you pay for....
When using VFDs with standard induction motors you need to consider the effect of Inductive Reactance. Inductive Reactance is like resistance in that it limits current, but differes in that it changes with the frequency of the applied power. What this means is that as the frequency goes up the Inductive Reactance goes up reducing the current and the magnetic field therefore the torque goes down. this is relatively harmeless except for the loss of power. Conversely if the frequency goes down the Inductive Reactance goes down increasing the current and the magnetic field and the torque goes up. this can be OK up to a point but as the current goes up so does the heating of the motor and this can destroy a motor. Another factor limiting the torque increase with lowered frequency is core saturation. Iron cores increase and focus the magnetic field of the windings but they can only allow so much field strength before they saturate and increasing field produces no more field but still increases heat loss and temperature build up. I'm not sure just how much increase/decrease os frequency is OK for the average motor but think it may be in the order of 10%-20%. Good luck with your project
cadman remember inductive reactance is a constant when the voltage increases and decreases in proportion to the frequency..... said another way, all vfd's for the 30 yrs or more have set the output voltage/freq to the constant the motor likes: said another way, the magnetic or reactive current is a pure constant from base speed down to 0 speed, making heating due to it a non issue. Above base speed of course you run out of supplying more voltage, so v/hz goes down - means the magnetic field reduces, reducing torque along a very linear HP curve - so again self limiting.
cadman - as MK has stated - virtually all drives available on the market today automatically maintain a proper V/Hz relationship. Flux Density is linear from zero Hz to motor base speed (rated terminal voltage) beyond which frequency increases while voltage is saturated at the line voltage (yielding a lower flux density that drops linearly with an increase in rotor speed [frequency]).
NEMA rated motors are designed to be started across the line with 600% inrush current - rotor / stator / iron saturation can occur with a drive, but rarely does anyone ever size a drive with enough current capacity to generate more than 150% - 200% of motor nameplate current and as such, saturation is almost never a problem (unless you purposefully increase voltage in the V/Hz relationship to exploit an increased magnetizing current component to achieve more torque at less than base speed).
Also - a clarification on # of poles - it has nothing to do with rotor length. As pole count goes up - typically the rotor diameter increases.
Thank you Mike & Motion Guru, that would be true if the voltage varied with frequency the problem would not arise. It would make sense that a comercially available VFD would take that into account. I believe I was thinking back to some similar threads in which people were devising their own equipment without consideration of all the variables.
I thank you for your time and information.
It is much appreciated. I'm sure I'll be back with further questions.