Relationship between motor amps and force through the cutting blade 2

[Vipul19]

I have a cutting operation where I had a 14" blade with 80 teeth, spinning a 1800 rpm with 10HP motor. The material being cut was pushed through at 8.3ft/min. We measured the amp load on the motor.
The force required to push the material through was approx 800lbs.

Then we changed the blade to 36 teeth and pushed the material through at 20 ft/min and measured the amp load it was approx 50% of before. I can't measure the force anymore, but would like to know if there is a way to calculate?

 

[bill318]

Great physics question but I think more information would be needed to answer it.

Assumptions:
material being cut was identical in both tests.

What is known from the test results:
1) In the second test, motor current was cut in half indicating force (motor torque + blade friction) were also cut in half.

2) More material was processed in one minute with half the motor energy required.

What is missing to figure out the feed force on the material in second test:
1) How much force is required just to move the material through the machine due to frictional drag alone? For example does it take 700lbs of force just to slide the material through even if no blade is installed in the machine?

I'm not great at physics but it looks like you almost have enough information to figure it out!

 

[electricpete]

Yes, good points. In addition, motor current is not proportional to load in the low-load ranges where the magnetizing current dominates. Decrease by half between scenario 1 and scenario 2 tells you the that scenario 2 load was 50% or less of scenario 1 (could be close to 0 load under scenario 2 if scenario 2 is close to no-load current). If you tell us nameplate data and current measurements, someone here might be able to give you a better guess of whether it is 50% or someting lower (assuming no change in voltage).

By the way I'm assuming induction motor... you might want to confirm that.

=====================================
(2B)+(2B)' ?

 

[waross]

Do you want to know badly enough to be willing to spend some money on a Watt-meter? It will give a much more accurate indication of the load. (Still not perfect but much better than an ammeter.)

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[Vipul19]

Ok, here is what I have recorded:

Initial Setup
Force through motor was 700lbs.
Force to overcome frictional drag is 100lbs
Original blade was 15" diameter and had 80 teeth
Motor amp reading was 13.2amps
Feed rate was 8.3 ft/min

Motor plate information is:

256T
20 HP
TYPE SCE
3 ph, 575V, 60 HZ
3510 RPM
SF 1.15
90.2 eff
PF 93.0
Nema B
TEFC

Modified setup:
New blade is 16" diameter and has 36 teeth.
Feed rate is 20ft/min
amp reading is 8.9amps
Nothing else is changed.

Any magical formulas are appreciated....

 

[zlatkodo]

It would be good to know the value of no-load currents, and with what instrument you measure the currents? Maybe a clamp-meter?
Zlatkodo

 

[Vipul19]

Sorry - Idle is at 3.7 amps and it was with a clamp meter.

Thanks again.

 

[zekeman]

You can't be serious.

How can you expect to get any reliable answer when you change blades of different geometry and sharpness.

In the 1st experiment, I see you pushed thru the material to the 20HP power limit and got 1/3 of the of the thruput vs the second blade at 1/2 the power.

The only unequivocal conclusion is the second blade is more efficient by factor of 6. Or perhaps the first blade was dull or the configuration of the tooth geometry was vastly different.Or perhaps the tooth size and spacing (doubled); I'd be surprised if tooth spacing alone made this difference.

You can only get a formula if you take on one blade at a time and for various thruputs get corresponding HP at various stages of blade wear (which may not be easy to quantize)


Keep in mind that Two parameters, namely blade wear condition and cut roughness would have to be in the equation and they ain't easy to quantize, not to mention the thickness of the material being cut

BTW, what is the purpose of getting this magical formula.

 

[electricpete]

Attached is results of fitting motor equivalent circuit to your data, and using that equivalent circuit to estimate performance. Slide 1 is the equivalent circuit parameters. Slide 2 shows how the model/parameters match the targets. Slide 4 shows the model doesn’t work good for high-slip as expected since no locked-rotor conditions were specified and deep bar effect was not modeled.

Slide 3 could be used to estimate your loading:
8.9A => ~48% load
13.2A => ~73% load

Don’t let the apparent complexity of the model fool you… it’s just a swag and only as good as the input. Adding no-load data as zlatkodo suggested definitely helped improve the estimate. Half-load efficiency/p.f. likewise would help but probably not needed for this purpose.

In this particular case, 2-pole motor with low no-load current, the current ended up pretty darned close to linear. As you can see by comparison to the blue line in slide 3.

Maybe others can provide simpler ways to provide estimates. I used the spreadsheet because it was already available.


=====================================
(2B)+(2B)' ?

 

[Vipul19]

I know I am asking for too much. We have been really getting into the blade design and found a few interesting things:

1. The original blade was not dull at all. Apparently, because of the large number of teeth, the "chip load" was very low. This is a parameter that quantifies the amount each tooth is cutting. When it is too low, the blade is not actually cutting, but "burning" through the material. With a more optimal/higher chip load, the teeth will cut better, require less force, but wear out much faster (since they now are actually cutting).

- So sharpness was not really different.
- Teeth geometry and quantity were different
- blade diameter was different.

I was hoping that there was a way to say that if 700lbs of force requires the motor to work at 13.2amps, then 8.9 amps means X amount of force through the blades. It doesn't need to be exact but is it linear? how are they related.

The reason for wanting to know this is that we are in the process of designing a automated carriage system. When we base it on the 700lbs of force at a slow speed of 8.3ft/min, we end up with 10HP motor and 49:1 gear ratio to get that range. With the changes in blade and apparent reduction in force through the cutters, it looks like I could redesign for something like slow speed of 15 ft/min and somewhere around 500lbs of force... I would then end up with a lower gear ratio, maybe a lower HP motor.

 

[Vipul19]

Thanks electricalpete!

I was hoping that it was linear, and it looks good enough for me I think.

Thanks again.

 

[electricpete]

A simple relationship assuming constant efficiency and neglecting effects of leakage reactance is:
Current = sqrt(NLA^2 + X^2 *(FLA^2-NLA^2))
Where NLA = no-load amps, FLA = full-load amps, X = load as fraction of full load.

Let me think about whether we can add a simple correction for leakage reactance.

=====================================
(2B)+(2B)' ?

 

[Vipul19]

Based on what you did, is the following correct:

From your ppt:
13.2A = 73% load = 700lbs of pusshing
8.9A = 48%load

= 700x48/73 = 460lbs of force?

 

[electricpete]

All I can say is that at 8.9A, the torque devivered by the motor is approximately a fraction (48/73) times what the torque was at 13.2A.

I don't understand the mechanical problem well enough to comment on relationship between torque and force.

=====================================
(2B)+(2B)' ?

 

[zekeman]

"= 700x48/73 = 460lbs of force?"


I don't think so.

From my perspective the correlation that I think exists is a linear relationship of feed rate and power for a given cutter. My reasoning is that it takes a fixed amount of work to cut a certain length ,so if you double the feed rate you double the power to a point. The force is only incidental to moving the feed and has no role in the cutting process. And I think if you try to move too fast you end up like case your one where the chips cannot clear fast enough for normal cutting.
So, for each cutter there is a limit feed rate that allows normal cutting. I would be surprised if you got that lucky that there is also a linear relationship with force.

BTW, you are making too much of the current. What you care about is power and you can get that with a power meter.

By this model, it doesn't look like there is a strong relationship between force and power, certainly not when you're pushing the feed to the limit.

 

[waross]

In the sawmill industry there is a close relationship between the volume of wood being removed in a cut and the cutting power required.
An old 11/32" circular blade used three times as much power for the same cut as a new 1/8" blade. Sorry, I can't remember the force required to push through a cut.
Tooth loading will give a maximum feed rate for a given thickness of material.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[lukin1977]

I will do a new test in scenario 2 this time measuring the force to push the wood through the blade. I think it is the easiest and most accurate way to do the correct desing

 

[lukin1977]

I meant if I was you

 

[MikeHalloran]

Since woodcutting blades normally have a lot of positive front rake, I don't understand why so much force was necessary to feed the wood into the blade. It would be necessary if you were using a metalcutting blade with substantial negative front rake.
You didn't report a rake angle. Did you measure it?

For research/development purposes, I'd remove table friction from the equation by using a ball bearing carriage to carry the wood toward the blade.

If you're going to be comparing different cutting tools, you should be measuring the material removal rate, which normalizes at least some of the variables you appear to have neither measured nor controlled.




Mike Halloran
Pembroke Pines, FL, USA

 

[Vipul19]

Thanks lukin - I thought you were volunteering :->

Also thanks to Mike. I assume by removal rate, you mean #teeth x tip speed?