# Calculating Power/Torque transmitted between 2 pulleys

1. Jan 13, 2017

### Al_Pa_Cone

1. The problem statement, all variables and given/known data

Q1. A pulley 150 mm diameter is driven directly by an electric motor at 250 revs min–1. A V-belt is used to transmit power from this pulley to a second pulley 400 mm diameter against a load of 200 Nm.

The distance between the centre of the pulleys is 600 mm, the included angle of the pulley groove = 40°, the coefficient of friction between the belt and pulley is 0.4 and the ultimate strength of the belt is 8 kN

(a) Calculate the actual power transmitted to the second pulley.

(b) Calculate the power which can be transmitted if the maximum tension in the belt is limited to half of the ultimate strength of the belt.

(c) What would be the effect of the following factors on the maximum power which can be transmitted (give reasons for your answer):

(i) increasing the coefficient of friction
(ii) increasing the included angle of the pulley groove.

2. Relevant equations

3. The attempt at a solution

(a) Calculate the actual power transmitted to the second pulley.

(b) Calculate the power which can be transmitted if the maximum tension in the belt is limited to half of the ultimate strength of the belt.

(c) What would be the effect of the following factors on the maximum power which can be transmitted (give reasons for your answer):

(i) increasing the coefficient of friction

(ii) increasing the included angle of the pulley groove.

Does any of this look correct? and can anyone advise, If I have missed something or miscalculated?

Thanks

2. Jan 13, 2017

### haruspex

You seem to have used the rotation rate of the first pulley and the torque load of the second pulley.

3. Jan 13, 2017

### Al_Pa_Cone

I was unsure how to take the given value 200 MPa and I was missing a torque, so i suspected this was the Torque I needed to carry out my calculations. How would I find the Torque given the results I have for everything else?

4. Jan 13, 2017

### Al_Pa_Cone

5. Jan 13, 2017

### Al_Pa_Cone

This is as far as I can get with a simultaneous equation. I think i need to factor out f1 but not sure how?

6. Jan 13, 2017

### haruspex

I don't think you have understood the structure of the first part. You can calculate the rotation rate of the second pulley, assuming no slipping. That allows you to find the actual power delivered, but still with that predicate. Separately, you can find the maximum power that can be delivered, based on the coefficient of friction. Only by considering both can you arrive at the actual power.

7. Jan 16, 2017

### Al_Pa_Cone

I try to follow the examples and in lessons. I have had a read through all the lesson content and example questions but I cannot find any information on how to find the rotation speed of the second pulley? I have already obtained the speed of the driving pulley of diameter 150 mm at 24.13 rad sec^-1 and I have the diameter of the second pulley at 400mm, but I do not have an equation to solve the difference in speed transmitted? Also once I obtain this value I do not have an equation to use it when I am looking for the actual power? I have collected a fair amount of information already, can it be done with what I have in my notes?

I have some similar example lessons where I can swop out the values to suit my question but I dont have some of the values given, here is one example, most apear to require transposition for two unknown values in solving the equation?

8. Jan 16, 2017

### Al_Pa_Cone

Does this look good for the rotation rate of the second pulley, assuming there is no slip? and I believe at somepoint I should deduct the 200 N m Load on the second pulley?

9. Jan 16, 2017

### Al_Pa_Cone

I also made an attempt at solving the actual power part. I cant figure this one out?

10. Jan 16, 2017

### haruspex

I'd prefer you did not post your working as images, but if you wish to do so, please number your equations so that I can refer to them in replies.

You seem to have assumed the second pulley is rotating at the same rate as the first (even though this is what you needed to calculate) and deduced a different linear speed (5.236 ms-1).
Since the belt is not slipping, what can you say about the two linear speeds?

11. Jan 16, 2017

### Al_Pa_Cone

Apologies for the screen shots, I prefer to type up my working out in Microsoft Word because its quicker to type up equations. I will add a reference to each one in future.

I have highlighted my answer for the linear speed of the second pulley in red? The speed for the first pulley was 1.9635 m s^-1 therefore I have attempted to show the difference in rotating speeds of the two pulleys. However I had worked out on paper a different method.

I took the diameter of each pulley 0.150 m and 0.4 m and multiplied them by 2pi/60 to get 0.0157 rad sec^-1 and 0.0419 rad sec^-1 but I thought this was totally wrong so I scrapped it. I do not have an equation to work this rotational difference out. Can you suggest one?

12. Jan 16, 2017

### haruspex

Ok, but maybe there is a more commenter-friendly way of exporting the text into a post? Can Word create LaTeX?
But to do that you assumed it had the same rotational speed as the first pulley, which is clearly false.
Think about the linear speed of the belt. How does that relate to the rotational speed of a pulley of radius r?

13. Jan 16, 2017

### Al_Pa_Cone

I have tried to copy and paste the equations from Microsoft word but it looses all format when transfered. I am sorry but i do not understand your meaning creating LaTeX? I presume its a format I can copy into?

I can see now what you mean with regards to the rotational speed. The first pulley is much smaller than the second so would turn at more than twice the rate. How would I show this? Could it be the 250 revs min / diameter of each pulley?

14. Jan 16, 2017

### haruspex

See if this helps: https://www.dessci.com/en/products/MathType/popup_tex_in_word.htm
As I wrote, think about the linear speed of the belt. If a pulley radius r rotates at rate w, what is the linear speed of the belt around it (assuming no slipping).

15. Jan 17, 2017

### Al_Pa_Cone

Ahh I see, it is a formula understood by microsoft to write up equations. I have usually open a box to write in my equation and then type them in using common keystrokes to produce the equation or use the symbol section for greek representation symbols. I will stick to numbering my equations for reference when posting my answers.

So back to the question:
As the driving pulley is 150 mm diameter or 75 mm radius and the measurement of linear speed is in meters per second, I would use the radius in meters of the driving pulley, or 0.075 m. The equation I would use to work out Linear speed is given as v = wr.
w was provided in the question as 250 rev min but I needed the answer to be converted to seconds. So 250 * 2pi/60 = 26.18 radians per second.

Therefore the linear speed of the belt would be 26.18 * 0.075 = 1.9635 meters per second( Assuming no slip). This would obviously be the same for both pulleys, however the driven pulley is 166% (83/50) larger than the driving pulley therefore a full rotation takes longer. So using the previous equation v = wr

1.9635(v) = w * 0.2(r) transpose for w gives 1.9635/0.2 = w.
As a pulley radius r rotates at rate w then the driven pulley rotates at 9.8175 radians per second.

Does This look ok?

16. Jan 17, 2017

### haruspex

Good.
So how much power is transmitted?

17. Jan 17, 2017

### Al_Pa_Cone

Continuing my attempt at solving this, following my answer to the rotational speed of the driven pulley. I have screen shot my work but referenced each method part 1 to 4.

18. Jan 17, 2017

### Al_Pa_Cone

Also giving the same results

19. Jan 17, 2017

### Al_Pa_Cone

Scrap that last attempt:

Last edited: Jan 17, 2017
20. Jan 17, 2017

### haruspex

No, in all those attempts you have assumed maximum tension. Part a) does not say maximum tension.
You know the rotation rate of the second pulley, and you know load torque.