Torque and power requirement for roller

In summary: Force = mass ....Acceleration= velocity/timeVelocity= radius. Angular velocityAngular velocity = (rpm.2.pi )/ 60The roller needs to run @ 10 rpm. Using the above formulas I've calculated this.Hence angular velocity= 1.05 rad/secsVelocity = 0.045 x 1.05= 0.05 m/secsAcceleration= 0.05/3 = 0.02 m/sq.secs (0 to 10 rpm in 3 secs)Force = (20+6).0.02x 9.8= 4.0 N ( 20 kg fabric roll, 6 kg roller self weight )
  • #1
saruvanan
10
0
We have roller setup which is to be driven by a motor. The roller dia is 90 mm. A shaft which is connected to its end is 25 mm of dia. The roller weight is around 10 kg. On the periphery of the roller there are some 40 kgs of mass is acting. The roller rpm is 10. What is the torque required to run this roller and power required by motor. Thanks in advance
 
Engineering news on Phys.org
  • #2
Once the system you describe is operating it will take little energy to keep it going.
What is the purpose of the roller ? What does it do ?
Where will the motor energy go when the system is operating ?
Is there a belt ?
 
  • #3
The roller is used to drive another roller by surface contact. The driven roller is placed directly over the driver and it sits by means of gravity. The driven roller weight is 40 kg. The driving roller weight is 10 kg. The motor is to be directly connected at the driver,s roller end. No belt. Thanks for replying baluncore. If posible i need calculation steps
 
  • #4
Where does the energy go?
You have still not identified why this is being done.
 
  • #5
Baluncore, its difficult application, hard to explain words. Can you help me with calculations?
 
Last edited by a moderator:
  • #6
Power is energy flow per second. That requires work to be done on something.
You have not specified where the energy will go and so there can be no power calculation.
What is between the rollers?
 
  • #7
saruvanan said:
Baluncore, its difficult application, hard to explain words. Can you help me with calculations?

Why are you refusing to show any work on your part? Is this for schoolwork? We do not do your schoolwork for you here...
 
  • #8
saruvanan said:
The roller is used to drive another roller by surface contact. The driven roller is placed directly over the driver and it sits by means of gravity. The driven roller weight is 40 kg. The driving roller weight is 10 kg. The motor is to be directly connected at the driver,s roller end. No belt. Thanks for replying baluncore. If posible i need calculation steps

P=Tω
power = torque . angular velocity

To calc power required (for a constant speed) you need to figure out/measure the opposing torque due to whatever loads are acting and know what speed the roller needs to run at.
 
  • #9
Let me rephrase my question

Dimensions
We’ve a roller of length 400mm & Dia of 90mm. At both ends of the roller a shaft is fitted. Shaft dia is 25 mm & length is 100mm each side. On the 25mm Dia shaft, two flanged bearings are fitted. The total roller & shaft weight is 6 kgs.

Working Principle:

Over the roller, a fabric in a roll form is loaded. The fabric roll sits over this roller by means of gravity. The roller has to drive the fabric roll on its circumference & feed the fabric end to another machine.

Problem:

1. What is the torque required to drive the roller at the shaft end.

2. What HP of motor, we should select to drive the shaft, if the motor is directly connected (without any gear box) to the 25mm dia shaft
 
  • #10
saruvanan said:
Let me rephrase my question

Dimensions
We’ve a roller of length 400mm & Dia of 90mm. At both ends of the roller a shaft is fitted. Shaft dia is 25 mm & length is 100mm each side. On the 25mm Dia shaft, two flanged bearings are fitted. The total roller & shaft weight is 6 kgs.

Working Principle:

Over the roller, a fabric in a roll form is loaded. The fabric roll sits over this roller by means of gravity. The roller has to drive the fabric roll on its circumference & feed the fabric end to another machine.

Problem:

1. What is the torque required to drive the roller at the shaft end.

2. What HP of motor, we should select to drive the shaft, if the motor is directly connected (without any gear box) to the 25mm dia shaft

Thread is locked for Moderation. Please PM me with the answers to my questions.
 
Last edited:
  • #11
Thread reopened for now.

@saruvanan -- do you understand what billy_joule wrote in post #8? Can you post a picture of your setup or a similar one?
 
  • #12
@Billy_joule
Power = torque. Angular velocity
Torque = radius . Force
Force = mass. Acceleration
Acceleration= velocity/time
Velocity= radius. Angular velocity
Angular velocity = (rpm.2.pi )/ 60
The roller needs to run @ 10 rpm.
Using the above formulas I've calculated this.
Hence angular velocity= 1.05 rad/secs
Velocity = 0.045 x 1.05= 0.05 m/secs
Acceleration= 0.05/3 = 0.02 m/sq.secs (0 to 10 rpm in 3 secs)
Force = (20+6).0.02x 9.8= 4.0 N ( 20 kg fabric roll, 6 kg roller self weight )
Now torque= 4.0 x 0.045 = 0.18 N.m

This is the calculation I've done. Not sure its right. I cannot attach a picture of the setup. Perhaps a drawing i can attach
 
  • #13
attached a conceptual drawing

This is a cinceptual design
 

Attachments

  • IMG_20140911_095532.JPG
    IMG_20140911_095532.JPG
    17.7 KB · Views: 1,869
  • #14
saruvanan said:
@Billy_joule
Power = torque. Angular velocity
Torque = radius . Force
Force = mass. Acceleration
Acceleration= velocity/time
Velocity= radius. Angular velocity
Angular velocity = (rpm.2.pi )/ 60
The roller needs to run @ 10 rpm.
Using the above formulas I've calculated this.
Hence angular velocity= 1.05 rad/secs
Tangential Velocity = 0.045 x 1.05= 0.05 m/secs
Acceleration= 0.05/3 = 0.02 m/sq.secs (0 to 10 rpm in 3 secs)
Bolded word mine.

These calculations are all correct.

It's a little strange that you specified the rpm and not the tangential velocity. I would've expected you to need the fabric to be fed at a certain rate as that is the entire purpose of the machine. 5 cm of fabric per second seems mighty slow to me, but then again you haven't told us the purpose of the machine... Why did you select 10 rpm? Why did you select 3 seconds? why not 6? That would half the motor cost/size/energy use..


Force = (20+6).0.02x 9.8= 4.0 N ( 20 kg fabric roll, 6 kg roller self weight )
Now torque= 4.0 x 0.045 = 0.18 N.m

These are wrong.
The weight of rolls acts through the center of rotation so the 'r' in the T=Fr equation is zero. That is to say they make no contribution to torque about the shaft.

You need to calculate the moment of inertia ('I') of the rotating masses.
http://en.wikipedia.org/wiki/Moment_of_inertia

Then you can calculate the torque required for the acceleration:
http://en.wikipedia.org/wiki/Angular_acceleration


But what you are really need to know is the steady state operation- how much power you need to keep the motor turning at your target rpm. for that you need to know where energy is going (eg the opposing torques from friction etc); not a trivial task.

EDIT: I now see your design. How does it compare with other similar machines?

Many people have invested millions of dollars and man hours in designing machines just like that...Do you think you'll compete with those designs? Why do you want to reinvent the wheel? Why not get (or copy...) an off the shelf design with years of R&D already done for you?
 
  • #15
billy_joule said:
... Why do you want to reinvent the wheel? Why not get (or copy...) an off the shelf design with years of R&D already done for you?


Ummm - do they give these designs away for free ?
 
  • #16
Re; post #13.
As the fabric unrolls at a set speed determined by the bottom roller, it rises up between two other pinch rollers.
Fabric cannot be pushed uphill, so it must be pulled up by the two pinch rollers. All the work of moving the fabric must be done by the upper pinch rollers. The bottom "driving" roller is not driving and will therefore not require power.
 
  • #17
Dnomyar said:
Ummm - do they give these designs away for free ?


google image search "*machine you want to know about*" and look at the images.
I bet every single image returned will have a better design than the OP's.

You don't need the original engineering drawings to copy a simple design. a single photo or a 20 second explanation will do. This isn't cutting edge confidential technology. They have probably been made the same for 50 years!
 
  • #18
billy_joule said:
google image search "*machine you want to know about*" and look at the images.
I bet every single image returned will have a better design than the OP's...!

Nope, not a one looking like the concept put forward - and Zero computations.
 
  • #19
Dnomyar said:
Nope, not a one looking like the concept put forward - and Zero computations.

Of course none look like the OP's, his idea won't work at all!

Like Baluncore explained, fabric cannot be pushed off a roll.
And generally you won't find calculations/computations with an image search.
 
  • #20
"And generally you won't find calculations/computations with an image search."
Duh - Really ? You didn't comment on the lack of suitable designs either, so you suggestion of Googling the solution wasn't so helpful.
So, since there were no computations, perhaps you don't mind the OP trying to establish the calculations then ? Perhaps he is allowed to expand his understanding a bit ?"Like Baluncore explained, fabric cannot be pushed off a roll."
Fabric 'can' be pushed off a roll. The easiest way would be to put two rollers under the fabric roll, and power them, making them rotate the fabric roll - the material will be pushed off the roll of fabric.
 
  • #21
Dnomyar said:
"And generally you won't find calculations/computations with an image search."
Duh - Really ? You didn't comment on the lack of suitable designs either, so you suggestion of Googling the solution wasn't so helpful

What lack of designs?
Mechanisms to pull sheet or tape off rolls can be found easily. Powered pinch rollers are used. a single roller can be used if the wrap angle and friction are sufficient.

here are some example I found with a "fabric roller machine" search term, better terms (like the name of the machine he wants to build) will net better results.

Hand cranked pinch rollers
http://gannetaus.com.au/UserFiles/hw11.png

powered pinch rollers:
http://i00.i.aliimg.com/photo/106970453/Fabric_Rolling_Machine.jpg

The bottom line is that you need to pull the sheet off the reel.

So, since there were no computations, perhaps you don't mind the OP trying to establish the calculations then ? Perhaps he is allowed to expand his understanding a bit ?

The calculations required appear earlier in this thread, namely P = Tω. see post #14 for further advice. Without further info from the OP little more can be said.

"Like Baluncore explained, fabric cannot be pushed off a roll."
Fabric 'can' be pushed off a roll. The easiest way would be to put two rollers under the fabric roll, and power them, making them rotate the fabric roll - the material will be pushed off the roll of fabric.

Fabric cannot be pushed up through unpowered pinch rollers. See post #16
 

1. What is torque and power requirement for a roller?

Torque is a measure of the rotational force required to move a roller, while power is the rate at which work is done. The torque and power requirement for a roller depends on factors such as the size and weight of the roller, the type and surface of the material being rolled, and the speed at which it is being rolled.

2. How is torque and power requirement calculated for a roller?

The torque requirement can be calculated by multiplying the roller's radius by the force it needs to overcome, while the power requirement can be calculated by multiplying the torque by the angular velocity of the roller.

3. Why is it important to consider torque and power requirement for a roller?

Knowing the torque and power requirement for a roller is important in order to select the appropriate motor and drive system, as well as to ensure efficient and safe operation of the roller. It also helps in determining the maximum load capacity and speed of the roller.

4. How can torque and power requirement be reduced for a roller?

The torque and power requirement for a roller can be reduced by using a smaller roller diameter, reducing the weight of the roller, and increasing the efficiency of the drive system. Additionally, using materials with lower rolling resistance can also help in reducing the torque and power requirement.

5. What are some common applications of torque and power requirement for rollers?

Torque and power requirement for rollers are important factors in industries such as manufacturing, construction, and transportation. They are used in various applications such as conveyor systems, road rollers, printing presses, and agricultural equipment.

Similar threads

  • Mechanical Engineering
Replies
12
Views
2K
  • Mechanical Engineering
Replies
15
Views
3K
  • Mechanical Engineering
Replies
5
Views
3K
Replies
4
Views
2K
Replies
3
Views
2K
  • Mechanical Engineering
Replies
4
Views
8K
  • Mechanical Engineering
Replies
7
Views
2K
Replies
5
Views
1K
  • Mechanical Engineering
Replies
19
Views
856
Replies
1
Views
2K
Back
Top