How to calculate required motor HP 8

[PaulKraemer]

Hi,

I have an application in which I will be driving an 8" diameter roll that will be supported on each side by steel frames with fixed position precision bearings. I will be using a VFD to drive an AC motor with a 60:1 gearbox and an additional 2:1 belt/pulley reduction. Other than having to turn the roll itself at a constant speed, there is no additional load. From past experience, I feel pretty confident that a 1/2 HP motor will be able to do this job.

I'd like to know what calculations I can do based on the information above (and any other information I can request from the motor, gearbox, roll, and bearing manufacturers) that would allow me to determine with a higher level of confidence whether the 1/2 HP motor I have in mind will in fact work for this application.

Any advice will be greatly appreciated. Thanks in advance.

Paul

 

[itsmoked]

Since the main problem seems to be static why not coat the roller with rubber. Now there will be no sliding of the film across the roller. But now you say "there's more static"!

There are a huge number of static killers out there. You mount an ionizing bar where ever you need the static zapped and it's 100% eliminated.

[URL unfurl="true"]http://www.exair.com/index.php/products/static-eliminators/gen4-ib/p8003.html[/url]

Ultra-High-Speed Static Ionizers


Check the two movies on static issues and note that the ionization bar kicks butt in the second video. Note the plethora of bars available.

[URL unfurl="true"]https://www.ccsteven.com/cc/product-category/static-bars/[/url]

Keith Cress
kcress -

http://www.flaminsystems.com

 

[PaulKraemer]

Hi Bill, Keith, CompositePro, and Muthu,

Thank you for your continued support. The suggestions of using a clutch or other "loose" or "slippy" drive mechanism makes total sense to me if given my 60:1 gearbox reduction and my roll diameter, I was running the motor at a speed that (without slipping) resulted in the roll surface speed being faster than my web speed. This is not the case (and I apologize for not adequately explaining my application). As I believe Keith and Compositepro alluded to, if I insist on using the gearbox, I would have to implement a drive system with some type of closed-loop, master/follower, electronic gearing type speed controls. I believe this is exactly what I am doing. I will try to explain using my original illustration, which I have pasted again below...

...I actually have two separate drive systems with closed-loop speed controls, one for the Pull/Laminating Rolls and one for the Coat Roll. As it is the Pull/Laminating Rolls that determine web speed, I configure the Pull/Laminating Roll Drive System as the "Master" Drive. The Pull/Laminating Roll Drive System consists of a DC motor fitted with a 60 PPR ring-kit along with a DC regenerative drive board and a closed-loop speed controller. The speed controller allows operators to enter a speed set point in feet/minute. Internally, the speed controller does some scaling using roll diameter, gear reduction, and ring-kit PPR to relate ft/min to motor RPM, but the end result is that it controls motor RPM so that the actual roll surface speed will equal the entered set-point. As long as the gap between the two Pull/Laminating Rolls is set properly so that there is no web slippage, this ensures that my web speed will be the same as the entered speed set point.

Just like the Pull/Laminating Roll Drive System, the Coat Roll Drive System consists of a DC motor fitted with a 60 PPR ring-kit along with a DC regenerative drive board and a closed-loop speed controller. The only difference is that I configure the Coat Roll Drive System as a "Follower" Drive. In this configuration, the Coat Roll speed controller allows the operators to enter a speed set point as a ratio of the Pull/Laminating Roll speed. Typically this is always 1:1 because while the Pull/Laminating Rolls actually determine web speed, we just want to match the surface speed of the Coat Roll to the web speed so as not to create drag. (Sometimes we might adjust this to 0.999:1 or 1.001:1 just to compensate for any stretching or shrinking that might happen in the dryer). Internally, the Coat Roll speed controller monitors the feedback not only from the Coat Roll motor, but also the master feedback from the Pull/Laminating Roll motor, and (doing the necessary scaling), controls Coat Roll motor RPM to achieve the desired speed match.

Anyway, this control scheme works fine. I have absolutely no problem getting the speeds to match using this type of drive system with gearboxes. I have gotten this to work many times with different gearbox ratios, sometimes with an additional belt/pulley reduction in addition to the gearbox reduction. I choose my gearbox and belt-pulley reduction based on the particular customer's required speed range.

The reason I originally posted this question is because I strongly suspect that I routinely over-size the motors I use because I do not know how to calculate required motor horsepower based on what I am asking the motors to do.

Just to complete the description of my application, Muthu touched on the fact that this application sounds similar to a cassette recorder where the tape has to travel at a constant speed, but the Unwind and Rewind spools have to vary in speed as diameter changes. I think this is a good analogy. While the Pull/Laminating Roll Drive (assisted slightly by the Coat Roll Drive) handles my speed control function, I implement tension control (rather than speed control) for my Unwind and Rewind Shafts as described below.

For unwind tension, I mount an electro-magnetic brake on the Unwind Shaft and I have a tension sensing roll (not shown in my illustration) in the Unwind Web path. I use a closed-loop tension controller to monitor the feedback from the tension sensing roll and to regulate the brake voltage (allowing it to slip) as necessary to keep tension at set point while web speed remains constant.

Similarly, for rewind tension, I mount an electro-magnetic clutch on the Rewind Shaft and I have a tension sensing roll (not shown in my illustration) in the Rewind Web path. I drive the input to the clutch slightly faster than what would be necessary to keep up with web speed at minimum spool diameter). Then I use a closed loop tension controller to monitor the feedback from the tension sensing roll and to regulate the clutch voltage (allowing it to slip) as necessary to keep tension at setpoint while web speed remains constant.

Again, this all has worked fine for me for quite some time. I would just like to become better able to estimate my motor HP requirements to (hopefully) avoid over-sizing as I am sure I have done in the past.

As I mentioned, for the project that sparked my question, I have already ordered the 1/2 HP motor and 60:1 gearbox for the Coat Roll, and it seems like I could have gotten away with something smaller. The project that requires this Coat Roll only includes an add-on coating station for an existing machine (built by others). The information I have provided about the Pull/Laminating Rolls and Rewind are based on what I consider typical for the machines we build. In this case, my Pull Roll motor (with a 60:1) gearbox would be directly driving the (chrome-plated) Pull/Laminating roll on the left, and indirectly driving (by surface contact) the (silicone covered) Pull/Laminating Roll on the right. It would also be pulling the two films (one from the Primary Unwind and one from the Backing Film Unwind). In some cases, I use the same motor to also drive the Rewind Shaft (with electro-magnetic clutch). Based on past experience, I would be totally confident that a 1-1/2 HP motor would be more than capable of this task, and I would be reasonably confident that 1 HP would also be fine. While I expect I might be able to go even smaller, not knowing how to do the necessary calculations would make me hesitant to try.

It seems to me that there must be a way that I can calculate a rough estimate of my required motor HP in an application like this. I have a feeling that using the weight of my rolls, my tension ranges, my diameter ranges, and possibly other factors, I should be able to calculate my torque requirement for the driven roll. What is unclear to me is considering that I will be using a gearbox, how do I relate my torque requirement for the driven roll to a HP requirement for my motor.

I am going to do some googling/reading on gearbox sizing (with a focus on learning how application torque requirements relate to gearbox specs and motor specs). If anyone here can point towards some useful material or provide any additional advice, I'd greatly appreciate it.

You all have been amazingly helpful.

Best regards,
Paul

 

[edison123]

Paul

The torque is proportional to the speed reduction in your gear box. Half the speed, twice the torque; one third the speed, three times the torque and so on.

You have meticulously designed and successfully time tested a special purpose machine for a specific application. Why chance it by chasing the 'power saving' squirrel down that rabbit hole? At such fractional HP's, the power saving would be minuscule since it saves only the no-load loss (aka iron loss) and since the load loss is fixed more by the load than by the motor.

As a matter of fact, an oversized motor in your case provides the cushion for those occasional overloads without burning out the motor. Also, an oversized motor runs cooler thereby extending its useful life considerably.

To coin a cliché, let sleeping dogs lie.

Muthu

http://www.edison.co.in

 

[waross]

The idea with the torque drive is to supply enough torque to almost start the roller turning. Then the film may be able to turn the roll.
More torque than that and you may over-drive the roll and cause the film to droop.
On the other hand, if the film is able to turn the roll, you may be able to fit a pinch roll and let the film turn the roll without a motor.

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

 

[PaulKraemer]

Thank you Bill and Muthu for your follow-ups. The idea of the torque drive is very interesting, and I think this and/or a pinch roll is worth considering in situations where the budget is not there for the motor, gearbox, and drive/control system capable of a master/follower electronic gearing type implementation like I am accustomed to doing.

Muthu - your point about letting a sleeping dog lie is well taken. I wouldn't change a proven design that has worked in the past just to save a little money on motor cost or energy savings.

The reason I am still interested in learning how to do these calculations is because I am rarely fortunate enough to get to repeat a proven design over and over again. All of the equipment we design/build performs basically the same function (roll-to-roll coating), but it is all custom and the specific requirements for every project is different.

Our "most standard" machine has an eight inch diameter (16 inch wide) Coat Roll and 8 inch diameter (16 inch wide) Pull/Laminating Rolls, with two 3 inch diameter rewind Air Shafts capable of winding spools up to 20 inch diameter. As we have built this a few times, I know that a 1 HP motor (w/ 60:1 gearbox) for the Coat Roll, a 1-1/2 HP motor (w/ 60:1 gearbox) for the Pull/Laminating Rolls, and a 1-1/2 HP motor to drive both rewinds (using electro-magnetic clutches) works fine. Even though I feel like my motors might have more horsepower than I truly need, I would not change this if I am fortunate enough to have an opportunity to build this machine again.

Quite often, however, we take on projects that are quite different than our standard machine. Currently, we are beginning our design phase for a very small machine which would have just a single drive. My Pull Rolls (no laminating function on this one) will be 3 inch diameter and 12 inches wide. My single Rewind will only have to handle spools from 6 - 10" diameter. Where as with the "standard" machine I mentioned above I used three separate motors/drives, for this small machine, I hope to use the same motor/drive for the Rewind that I will use for the Pull Rolls. On this machine, there will not be a driven Coat Roll. This will be a 2020 version of a machine that was built by others in 1990 (just two inches wider). The 1990 machine drives uses a 1/8 HP motor, so I am leaning towards 1/4 HP just to be on the safe side considering the extra 2 inches of width. While I feel comfortable in making this decision, I still wish I knew how to back it up with some calculations.

Another project I am working on at the moment is to add a dedicated motor/drive to handle two Rewind Shafts on an existing machine. These rewind shafts currently share the motor/drive that handles the Pull/Laminating Rolls, but a new process requirement will make it necessary for the Rewinds to have an independent drive. In this case, space on the existing machine is very limited. For this reason, it will be to my advantage to use as (physically) small a motor/gearbox that I can comfortably get away with.

For reasons like this, I really want to make an effort to get an understanding of how to (mathematically rather than gut feeling) estimate my required HP in various circumstances. I will always try to err on the side of caution (over-sizing), but I feel making this effort will be well worth my while, even if for no reason other than trying to get a better understanding of the systems/equipment I work with.

You guys have all been so helpful. If you don't mind, I may try to get started on some calculations for the project on which I will be adding a rewind drive and post them here to see if you think I am on the right track.

I really appreciate your help.

Thanks again,
Paul

 

[Compositepro]

I guess none of us who have replied have answered your your basic question of how you calculate required horsepower. This is because the answer is so basic to engineers that we assumed you were interested in more subtle details. Horsepower is a unit of power. Power is the rate at which energy is consumed. Energy is force times distance, ft-lb (ft. traveled x pounds force pushing). Do not get confused by the fact that torque has apparently the same units of lb-ft (pounds force twisting x ft. of lever arm).

For your application hp is torque (required to turn your roll) x angular speed (rpm). Get a basic physics book and the first chapter on mechanics will explain this. You can also find tutorial on the internet like those at Khan Academy.

 

[itsmoked]

To further Comp's statement we're not really answering your fundamental question because it's not a mathematically straight forward calculation. Answering the question of, "how much horsepower does it take to turn this roller" has many possibilities of answers based on more specific details.

1) Are you seeking steady state hp?
The bearings are all that matters in your case.

2) The hp to accelerate the roller up to what speed?
When you accelerate the roller you are storing energy in it that once speed is achieved no longer demands further energy input. But your app likely requires no real acceleration.

3) How fast to that speed? This increases the torque demanded.
The faster you want to get the roller to speed the more torque demanded. More torque can be achieved most simply by a bigger motor. However it can also be achieved with a gear reduction.

4) Drive train efficiency is never 100%. One set of gears is at least a 3 percent loss. Belts have losses too. So this muddies the water further.

5) Then you have the load contribution of the material across the roller. Is the roller contributing to driving the material? If it is, that number for the drag is critical to the answer. No math is going to provide that number for you! It has to be empirically measured.

Your particular situation is not amenable to "calculation" because the variability of the needed hp/torque calculation is completely swamped by the motion system itself as compared to the actual work needed to be done.

As an example, a sand conveyor is relatively easy to calculate. You can pile W amount of weight on a foot of conveyor belt. The conveyor belt is d long. The belt weight can also be calculated though it could be a wash due to the returning belt paying back with gravity. You then have the angle of the belt which provides the height the sand is being hoisted. These few facts directly gives the fundamental hp required to do the job.

The designers then find the drag of the rollers (probably from a book, rule of thumb, or a belt provider) which they add into the hp requirement. When done they multiply in a safety factor of perhaps 1.3 to cover any unexpected drag.

In your very different case you're turning the roller so slowly that acceleration, inertia, and bearing drag are all essentially nothing or are so fleeting as to be completely inconsequential. By far the largest consumer of hp in your case is the gearbox/belts aspect followed by the fan on the motor. The material across the roller is not even a load.

So, asking how to calculate the hp requirements in your application is.. incalculable. Instead it would normally be done one of three ways.

1) You'd assemble the roller and load up the machine and pull the web with a spring scale to get the work needed to be done. Then you'd double that and work back to the motor size.

Or

2) You'd work from experience having a feel for the size needed.

or

3) You'd go with the smallest size normally found in the industry.

Note that none of these embraces "calculations".

Hence, you see why we aren't showing you a way do it. We ourselves wouldn't be bothering.

Now, if you aren't price sensitive and want an excellent source for everything related to motion control talk to Mike at [URL unfurl="true"]http://kilroywashere.com/[/url]. He is a great guy with vast motion control knowledge. He helped me thru a huge switch from active pneumatics to servo drives. He helped me -in detail- with how to use their Kollmorgen servo systems. Those systems have outstanding user interfaces with amazing built-in metrics and setup wizards that help with any conceivable use of a servo drive system including crazy variable electronic geared applications. Contact Mike with your application info and you'll get a concise solution quickly. Warn Mike that I sent you.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[waross]

A Watt-meter may be a good investment. Watts is a much better indication of motor load than Amps.
Measure the Watts draw on your successful machines.
This will help you to develop some judgement as too approximately how much power you motors are using.
In the end, judgement and experience may outweigh calculations.

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

 

[edison123]

Paul

The manufacturers of commonly used driven equipments like fans, pumps and compressors do lot of theoretical calculations, model studies and actual load tests before they publish their performance curves.

In the absence of such a professional set up to do these complex studies, measuring the power from your already supplied equipments and present ones in your factory is a quick guide for your fixing the power of your motor. As Bill said, a good power quality analyzer would give you plenty of info about your equipment power needs.

As I said before, at such a fractional HP range, neither the cost nor the size is going to be the deal breaker in your overall system cost and space. Over-sizing the motors is quite normal in industries since they are the critical parts and are most likely to break down early and often.

Muthu

http://www.edison.co.in

 

[PaulKraemer]

Thank you Muthu, Bill, and Keith for your continued support. It is greatly appreciated.

I now understand the complexity of what I have been hoping to "calculate". A watt-meter or power quality analyzer sounds like a great idea.

For my current project in which I am adding a motor/gearbox to drive two rewind Air Shafts, I am leaning towards erring on the side of caution (over-sizing) and going with a 1 HP motor and a 30:1 gearbox. The rating data (based on 1750 RPM input) listed for the (Dodge/Baldor Tiger-2 Size 20) gearbox I have in mind is as follows:

30:1 gearbox:
Mechanical input HP = 0.96
Thermal input HP = 1.49
Output torque (lb.in) = 802
Mechanical output HP = 0.74
Output overhung load (lbs) = 1560

I am curious about the meaning of the "Mechanical input HP" specification. This Tigear-2 Size 20 gearbox is available in different reductions starting at 5:1 and going all the way up to 60:1 (whereas I am leaning towards 30:1).

Anyway, closest to my preferred 30:1 option, there is a 25:1 option and a 40:1 option. The rating data for these is listed as follows:

25:1 gearbox:
Mechanical input HP = 1.11
Thermal input HP = 1.59
Output torque (lb.in) = 788
Mechanical output HP = 0.88
Output overhung load (lbs) = 1560

40:1 gearbox:
Mechanical input HP = 0.76
Thermal input HP = 1.29
Output torque (lb.in) = 801
Mechanical output HP = 0.56
Output overhung load (lbs) = 1560

Does this "Mechanical input HP" spec mean that I should avoid using a motor with more HP than the "Mechanical input HP" spec? (i.e. make sure that motor HP <= Mechanical Input HP)

Or does it mean that I should use a motor with at least as much HP the "Mechanical input HP"? (i.e. make sure that motor HP >= Mechanical Input HP)

Thanks again and best regards,
Paul

 

[DM61850]

You should be able to apply 1/2 HP or 323 W by hand. You'll get tired pretty fast but you should be able to get it moving at least. Riding a bicycle at 20 mph is about 180 watts. It will be hard but if you can't move it by hand I would be worried that 1/2 hp is not enough. I don't know if that helps you in design, though.

 

[edison123]

Googling thermal rating of gears.

https://www.mardustrial.com/blog/un...power-your-falk-gearbox-selection-06-29-2015/

http://www.alltorquetransmissions.com/gearbox-thermal-capacity/

Muthu

http://www.edison.co.in

 

[PaulKraemer]

Thank you DM61850 for trying to put HP requirements in easy to understand terms and Muthu for the thermal rating articles.

Both of these articles make it sound like it is more typical for a gearbox to have a thermal input rating that is lower than its mechanical input rating. In my case, the Dodge Tigear-2 size 20, 30:1 reducer I am considering lists "Thermal Input HP" (1.49) as being higher than the "Mechanical Input HP" (0.96).

We are considering using this reducer with a 1 HP motor. It seems to me I am ok with the Thermal Input rating, but I am concerned about what would be the potential drawback (if any) of using a motor HP that is slightly higher than the Mechanical Input HP rating of the reducer. I called Dodge technical support, and the person I spoke to informed me that if we apply more than 0.96 HP to the reducer input, we will over-load the reducer.

The datasheet on the DC motor we have in mind lists HP = 1750 and torque = 36 lb-in (3 lb-ft). This is exactly what I would expect based on the formula HP = torque x RPM / 5252. What I am not sure of is for a "typical" DC motor, whether I would expect the same torque (3 lb-ft) at lower speeds or if I would expect greater torque at lower speeds. (I guess the question could be rephrased as whether I would expect the motor to produce constant torque at all speeds or constant HP at all speeds). I reached out to the motor manufacturer (Leeson) to get an answer to this.

It seems to me that if the motor produces no more torque than 3 lb-ft at lower speeds, then my HP at the reducer input would be less than 1 HP at lower speeds. As my application will never require motor speeds greater than 1000-1100 rpm, it seems like I might be safe to assume that I would not be in jeopardy of over-loading the reducer.

If anyone here can tell me if my logic sounds reasonable or if I am off-track, I would greatly appreciate it.

Thanks and best regards,
Paul

 

[waross]

The load demands torque and HP.
The motor produces whatever the load demands, within reason.
The motor torque rating is the limit of safe operation without overheating.
Many motors are capable of about 200% of rated torque.
At that loading, they will overheat and burn out if the protection does not take them off-line.
The gear-box must be rated to handle the load.
If he motor and gearbox are fully loaded it may be wise to select a gearbox from the next column.
If the assembly will be running lightly loaded the gearbox may be roughly matched to the motor HP.
Consider a gear-head motor.
It will be quicker to mount and the gearbox will be properly sized by the manufacturer.
It looks better and there is one less guard to to build.
Have you acquired a Watt=meter yet?

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

 

[edison123]

Take Bill's advice. Go for geared motor for your HP and final speed. Cost and space saver. No need to worry about thermal input HP and all that crap.

Muthu

http://www.edison.co.in

 

[PaulKraemer]

Hi Bill and Muthu,

Thank your for your responses. When you say gearhead-motor or geared motor, does this simply mean a motor with a built-in gearbox? (As opposed to buying a motor and a gearbox separately?)

Bill - You say the motor produces whatever torque the load demands. Considering that the torque rating of my DC motor is 36 lb-in, if it turns out that my load (the rolls I am turning, coupled to the motor using a gearbox with belts and pulleys, with whatever load is added by the film I am pulling through the machine) demands less than 36 lb-in at whatever speed I am running (considering that I am using a closed loop speed controller to regulate the DC voltage to my motor to ensure a constant motor speed that is less than 1750 RPM), that the motor will produce only the torque I need (something less than 36 lb-in) to run at this speed?

Intuitively, this seems fortunate because if the motor produced more torque than required to run at this speed, I imagine that there would be some undesired effect (possibly unwanted acceleration or unnecessary strain on my drive system mechanical components?).

I have not purchased watt-meter yet, but I want to. I have googled watt-meters and it seems as though there is a bewildering variety of choices. Is there a particular brand/type that you would recommend?

Thanks again for all your help.

Best regards,
Paul

 

[waross]

Google is your friend: "gearmotors"
US Motors
Baldor Motors
Acklands-Grainger
Bison Gear

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

 

[waross]

"Wattmeter"
If this will be used only to measure motor loads occasionally, you may use a single phase, 240 Volt Watt-Meter for $15 or $25.
For three phase measure two phases and multiply by 1.73.
OR
You may wish to spend more for a good quality power analyzer that may be used for more than checking motor loads.


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