Evaluating Motor Specifications for a Conveyor

[oho11]

TL;DR Summary

I have a gear and sprocket machine with the details of the motor and want to see if it will be enough or I need something more

Hi everyone!

I am in a bit of a halt right now as I don't really know where to begin with this problem,

There is a chain and sprocket machine that I have the details of the motor of and majority of the dimensions of but I don't know how to start, I tried to look at some of the posts on here but I still feel a bit confused, I have attached some photos of the information I have and what I have done so far which I know is not much but hence why I wanted to come on here and ask for help,

KEEP IN MIND I am not asking for it to be solved or anything, I simply want to learn about mechanics and would appreciate some foundation on where to start from and work my way up, I have taken some courses but I don't fully know everything that needs to be taken into consideration of hence why I am glad to find this website since a lot of people seem to be very helpful

If there is any additional information required please let me know and I will get them for you since I am not sure what other dimensions might be needed,

Thanks in advance!

 

[A.T.]

TL;DR Summary: I have a gear and sprocket machine with the details of the motor and want to see if it will be enough or I need something more

Enough for what? What is the machine supposed to do?

 

[oho11]

Enough for what? What is the machine supposed to do?

I apologize for not specifying, I want the machine to be able to move at approximately 7 meters per minute, so it will be conveying a panel, which I am not sure of the weight but I can try to get it for you across but for now lets say we neglect it, at 7 meters per minute.

Hope that helps!

 

[A.T.]

I apologize for not specifying, I want the machine to be able to move at approximately 7 meters per minute, so it will be conveying a panel, which I am not sure of the weight but I can try to get it for you across, at 7 meters per minute.

You will need to know the resistance to the motion, which likely depends on the weight of the panel. The force of resistance times the speed, gives you the output power. If the gearing is given, you can use it to compute the speed the motor must operate at, and check its specs if it can supply enough power (plus losses) at that speed.

 

[Devin-M]

I noticed in you notes it said motor output 1000Nm with a 137:1 gear reduction, so you want the final output torque to be 137000Nm?

 

[oho11]

You will need to know the resistance to the motion, which likely depends on the weight of the panel. The force of resistance times the speed, gives you the output power. If the gearing is given, you can use it to compute the speed the motor must operate at, and check its specs if it can supply enough power (plus losses) at that speed.

I know this might sound dumb but if I know the weight of the chain would I just need to calculate F=ma to find the force of resistance or is there another way to do that?

 

[oho11]

I noticed in you notes it said motor output 1000Nm with a 137:1 gear reduction, so you want the final output torque to be 137000Nm?

View attachment 344733

That is what was written on the specifications of the motor, I just provided it in case it would mean some help

 

[Devin-M]

I’d personally start by determining the required KM or motor size constant which is defined as the torque per square root of resistive power loss:

https://en.m.wikipedia.org/wiki/Motor_constants

This will give you an idea of how large a motor you’ll need. In general larger motors can produce more output power for the same conversion efficiency.

 

[oho11]

I’d personally start by determining the required KM or motor size constant which is defined as the torque per square root of resistive power loss:

https://en.m.wikipedia.org/wiki/Motor_constants

This will give you an idea of how large a motor you’ll need. In general larger motors can produce more power for the same efficiency.

View attachment 344734

Thank you so much I will definitely look into this as I didn't think about this part!

But how would I be able to apply this to my problem after I find the motor constant?

 

[Baluncore]

There is a chain and sprocket machine ...

It seems you have a chain conveyor transporting panels.

Energy will be needed to overcome friction and to initially accelerate the panels as they are placed on the conveyor.

1. You need to estimate, or measure, the power required to keep the conveyor running.
2. What does a panel weigh, and what is the maximum rate they might be placed on the conveyor?
3. Is the conveyor level, or does it rise or fall along its path?

 

[oho11]

It seems you have a chain conveyor transporting panels.

Energy will be needed to overcome friction and to initially accelerate the panels as they are placed on the conveyor.

1. You need to estimate, or measure, the power required to keep the conveyor running.
2. What does a panel weigh, and what is the maximum rate they might be placed on the conveyor?
3. Is the conveyor level, or does it rise or fall along its path?

Thank you for your reply!

To answer your last two question;

1) I am not sure about the panel weight but I will do my best to get that, but for now I would like to say there is no panel as I want to see if the motor I have specified is enough to move the chain first

2) the conveyor is level yes, it will follow a horizontal path and the only time it will turn is at the end for it to come back around

And by power required do you mean the P=Fv formula?

 

[Devin-M]

Thank you so much I will definitely look into this as I didn't think about this part!

But how would I be able to apply this to my problem after I find the motor constant?

Once I had determined the necessary Km or motor size constant, I’d look at the gear ratio and input voltage the motor to determine how fast that motor needs to turn to make the machine move at the proper speed. To confirm the motor is up to the task, you’d want to look at the Kv (max unloaded rpm per volt) and Kt (torque per amp). In SI units these are the inverse of each other (higher torque per amp gives lower max unloaded rpm per volt).

 

[Devin-M]

Once I had determined the necessary Km or motor size constant, I’d look at the gear ratio and input voltage the motor to determine how fast that motor needs to turn to make the machine move at the proper speed. To confirm the motor is up to the task, you’d want to look at the Kv (max unloaded rpm per volt) and Kt (torque per amp). In SI units these are the inverse of each other (higher torque per amp gives lower max unloaded rpm per volt).

 

[Devin-M]

Once you verify the motor is big enough for efficiency and to not overheat, and that it can turn fast enough considering your max input voltage, then its a matter of how many amps you put through the motor to get the torque you need.

 

[Devin-M]

A final consideration might be whether or not you can alter the gear ratio to get the motor’s operating speed close to the speed that it’s most efficient. The motor puts out the same resistive loss as it turns faster & faster at the same torque giving higher output power for the same resistive loss at high speeds, but “iron losses” start to increase rapidly as the motor turns quicker, so there is going to be a certain sweet spot rpm for motor efficiency if choosing the gear ratio is an option. Unfortunately iron losses are much more difficult to calculate than resistive losses, so determining the most efficient rpm usually requires experimentation on a dynamometer with each individual motor.

 

[jrmichler]

It seems you have a chain conveyor transporting panels.

Energy will be needed to overcome friction and to initially accelerate the panels as they are placed on the conveyor.

1. You need to estimate, or measure, the power required to keep the conveyor running.
2. What does a panel weigh, and what is the maximum rate they might be placed on the conveyor?
3. Is the conveyor level, or does it rise or fall along its path?

This.

This is a machine design problem. Expanding on what @Baluncore said, the general procedure is as follows:
1) Decide what speed the conveyor needs to run -- how fast to move the panels in m/sec.
2) Calculate the RPM of the conveyor input shaft.
3) Figure out the necessary torque of the conveyor input shaft. This can be difficult to calculate because you would need to calculate the friction from every single moving part PLUS gravity effects if moving the load uphill PLUS acceleration if the panels are accelerated by the conveyor. Generally, the easiest and most accurate method is to look at the motor sizes on similar conveyors.
4) Now that you know the RPM and torque required at the motor/reducer output shaft, buy a motor/reducer rated for at least that torque and RPM at the output shaft. You want the RPM correct, and the rated torque larger than what you estimated in order to have a safety factor.
5) Hook it up and go.

This assumes that you want the conveyor to run at a particular speed. If you want to control the speed, design for the maximum speed and use speed controls to slow it down. Note that none of this requires you to know about motor constants -- just the RPM and torque.

 

[russ_watters]

It's common for shorter/smaller conveyors to assume a high friction coefficient of .75-1 on which to base the motor size. That's a lot but when dealing with a small motor it doesn't hurt you/cost much to oversize the motor by a factor of 2+.

 

[oho11]

This.

This is a machine design problem. Expanding on what @Baluncore said, the general procedure is as follows:
1) Decide what speed the conveyor needs to run -- how fast to move the panels in m/sec.
2) Calculate the RPM of the conveyor input shaft.
3) Figure out the necessary torque of the conveyor input shaft. This can be difficult to calculate because you would need to calculate the friction from every single moving part PLUS gravity effects if moving the load uphill PLUS acceleration if the panels are accelerated by the conveyor. Generally, the easiest and most accurate method is to look at the motor sizes on similar conveyors.
4) Now that you know the RPM and torque required at the motor/reducer output shaft, buy a motor/reducer rated for at least that torque and RPM at the output shaft. You want the RPM correct, and the rated torque larger than what you estimated in order to have a safety factor.
5) Hook it up and go.

This assumes that you want the conveyor to run at a particular speed. If you want to control the speed, design for the maximum speed and use speed controls to slow it down. Note that none of this requires you to know about motor constants -- just the RPM and torque.

Thank you so much for this!

I have already decided that the panel must be 0.12 m/sec (converted from 7 m/min),

When you say the rpm of the shaft do you mean the one produced from the motor? Since if that is the case I have already shown that the motor outputs a speed of 10.5 to 11 rpm,

The third part where you mentioned that I need to find necessary torque is where I am having an issue, I found some information that's mentioned for my system it requires a 5 kW motor, but there is no other information such as gear reduction ratio, output speed, etc.

And yes I will be running it at 7 or 8 m/min max and have it slowed down,

If there is any formulas that you can suggest I start off from that would be great!

 

[Devin-M]

I have already shown that the motor outputs a speed of 10.5 to 11 rpm

If the gear ratio is 137:1 you’re saying the final output is 11/137= 0.08rpm?

Are you sure it isn’t 11*137 = 1507rpm (motor)?

 

[oho11]

If the gear ratio is 137:1 you’re saying the final output is 11/137= 0.08rpm?

Are you sure it isn’t 11*137 = 1507rpm (motor)?

I have attached the specifications of the motor below, that is what I have been provided with

 

[jrmichler]

When you say the rpm of the shaft do you mean the one produced from the motor?

No, I mean this shaft:

2) Calculate the RPM of the conveyor input shaft.

Given the conveyor speed and the diameter of the roll that drives the conveyor belt, calculate the RPM of that roll.

 

[oho11]

No, I mean this shaft:

Given the conveyor speed and the diameter of the roll that drives the conveyor belt, calculate the RPM of that roll.

Alright so I found the rpm to be 45.5 by doing the following

(D/19108)*rpm=conveyor speed

With D being 70 mm and the conveyor speed i'm looking for is a max of 0.166667 (or 10 meters per minute)

I just have a question, does that mean that the rpm of the motor (where I attached the specifications for above) is before the gearbox reduces it?

 

[Devin-M]

Alright so I found the rpm to be 45.5

So if the final output rpm is 45.5rpm, and you have a 137:1 gear reduction between the motor and final output, then the motor spins 45.5 * 137 = 6233rpm

 

[Baluncore]

An AC induction motor, without a speed controller, should have a maximum possible speed of; (supply_Hz * 60) = 3600 RPM.
I suspect there is a factor of π or 2 missing somewhere.

 

[Devin-M]

If the motor has magnets and a DC power supply (BLDC motor) you would look at the motor’s Kv constant (max unloaded rpm per volt) If your DC power supply is 100v, then a 65-70Kv motor should work. If the DC supply is 48v, then you’d need a 6500rpm / 48v = 135kv or higher motor.

The Kv of the motor depends on how many turns of copper are wrapped around each stator pole in the motor. It could be wound with less turns of thicker wire or more turns of thinner wire. Less turns of thicker wire gives higher KV. More turns of thinner wire gives lower KV, but since KT (torque per amp) is the inverse of KV in SI units, more turns gives more torque per motor amp. I say “motor amp” because the speed controller does a voltage change between the input dc voltage and the voltage sent to the motor via altering the duty cycle, and so you will generally have more amps running through the motor than being drawn from the DC supply, because the speed controller reduces the “effective” voltage sent to the motor to control the speed.

 

[Devin-M]

A word of caution about top motor speed, some modern BLDC speed controllers use a different more modern control algorithm called “FOC” short for field oriented control that makes the motor run more quietly and efficiently by outputting a sinusoidal waveform, but beware this might reduce the peak speed of the motor by very roughly, if my memory serves about 15%-20%, from the top unloaded speed implied by the dc input voltage multiplied by the motor kv. This is because the sinusiodal waveform reduces the maximum effective voltage output by controller below the numerical value of the DC input supply voltage. The workaround for using FOC without a reduction in top speed (to get the rpm you need) is to choose a slightly higher kv motor.

 

[jrmichler]

Alright so I found the rpm to be 45.5 by doing the following

That's the same that I get assuming 10 meters per minute and a 70 mm diameter conveyor drive roll. The next step is:

3) Figure out the necessary torque of the conveyor input shaft. This can be difficult to calculate because you would need to calculate the friction from every single moving part PLUS gravity effects if moving the load uphill PLUS acceleration if the panels are accelerated by the conveyor. Generally, the easiest and most accurate method is to look at the motor sizes on similar conveyors.

Another way to get the drive torque is to place the maximum load on the conveyor, wrap a long string around the drive roll, and pull the string using a spring scale. From the force to move the conveyor belt, and the radius of the drive roll (35 mm), you calculate the drive torque. Multiply the measured drive torque by a safety factor of 2 to 3 to get the design drive torque.

Now calculate the power to run the conveyor from the design drive torque and RPM (45.5 RPM). After the torque and power are well understood, then we can start to size the motor and reducer.

 

[Devin-M]

One more point about choosing the right KV BLDC motor (if you go that route). You want a high enough KV that the motor will get to the rpm you need (based on the DC supply voltage)… but why not go too high? The reason you don’t want the KV too high is because higher KV means less torque per motor amp. With a higher KV motor you’ll need more amps through the motor to get the same torque. More motor amps means your speed controller will get hot faster for the same torque with a higher KV motor. You want the KV low enough that your controller wont over heat with too many amps, but high enough that the motor can reach the proper RPM given the supply voltage. Interestingly if you have two motors of different KV’s but they have the same Km size constant, they will both produce the same copper loss waste heat inside the motor for the same torque (because more amps through the lower resistance wires produces the same heating), but the controller on the higher kv motor will be hotter.

 

[Devin-M]

It says your power requirement is 1.1kW

This motor seems to fit the bill (would require 48v DC power supply and Electronic speed controller:

https://flipsky.net/collections/e-s...ened-6384-190kv-4000w-for-electric-skateboard

Motor $119USD

48v x 140kv gives up to 6700rpm
Rated Up to 4000w (4kW)

Electronic Speed Controller $121USD:
https://flipsky.net/collections/ele...e-on-vesc6-6-with-aluminum-anodized-heat-sink

2000W 110-240AC to 48v DC Power Supply $259:

https://www.amazon.com/JINGMAIDA-Sw...uter-Project/dp/B08SQ5Z8VN/?tag=pfamazon01-20

 

[Devin-M]

With D being 70 mm

Are you sure this part’s diameter is 70mm (2.76in) and not 70cm (27.5in or 2.2ft)?

 

[oho11]

Are you sure this part’s diameter is 70mm (2.76in) and not 70cm (27.5in or 2.2ft)?
View attachment 344907

Sorry I forgot to mention that I got the 70 mm from the shaft coming out the motor, the one you have shown is the sprocket that moves the conveyor belt, I did this because @jrmichler mentioned to measure the rpm coming OUT the motor

 

[oho11]

That's the same that I get assuming 10 meters per minute and a 70 mm diameter conveyor drive roll. The next step is:

Another way to get the drive torque is to place the maximum load on the conveyor, wrap a long string around the drive roll, and pull the string using a spring scale. From the force to move the conveyor belt, and the radius of the drive roll (35 mm), you calculate the drive torque. Multiply the measured drive torque by a safety factor of 2 to 3 to get the design drive torque.

Now calculate the power to run the conveyor from the design drive torque and RPM (45.5 RPM). After the torque and power are well understood, then we can start to size the motor and reducer.

Thank you so much!

I will look into this as soon as possible and try to keep you guys updated with any progress,

Thank you to everyone who has contributed and helped out it really means a lot and I hope to come back with more questions or even participate in discussion myself

 

[Devin-M]

Sorry I forgot to mention that I got the 70 mm from the shaft coming out the motor, the one you have shown is the sprocket that moves the conveyor belt, I did this because @jrmichler mentioned to measure the rpm coming OUT the motor

If the sprocket that moves the conveyor is 702mm this changes everything.

 

[oho11]

If the sprocket that moves the conveyor is 702mm this changes everything.

I have shows how the system goes from the motor, to a sprocket moving another sprocket with the same amount of teeth, which moves that bigger sprocket. I just want to know if the motor I have specified is enough for my use, if not I will need a bigger motor obviously.

The reason for my post is that I don't really understand how torques and forces are transferred in a sprocket system, when you say it changes everything what do you mean exactly?

 

[Devin-M]

The reason for my post is that I don't really understand how torques and forces are transferred in a sprocket system, when you say it changes everything what do you mean exactly?

If the sprocket that moves the conveyor is 702mm instead of 70mm, the motor will need to turn 10x slower and output 10x as much torque.