Calculating Power/Torque transmitted between 2 pulleys

[Al_Pa_Cone]

Homework Statement

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.

Homework Equations

3. The Attempt at a Solution [/B]
(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

 

[haruspex]

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

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

 

[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?

 

[Al_Pa_Cone]

 

[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?

 

[haruspex]

View attachment 111539

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.

 

[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 don't have some of the values given, here is one example, most apear to require transposition for two unknown values in solving the equation?

 

[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?

 

[Al_Pa_Cone]

I also made an attempt at solving the actual power part. I can't figure this one out?

 

[haruspex]

View attachment 111660
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?

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?

 

[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?

 

[haruspex]

I prefer to type up my working out in Microsoft Word

Ok, but maybe there is a more commenter-friendly way of exporting the text into a post? Can Word create LaTeX?

I have highlighted my answer for the linear speed of the second pulley in red

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?

 

[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?

 

[haruspex]

but i do not understand your meaning creating LaTeX?

See if this helps: https://www.dessci.com/en/products/MathType/popup_tex_in_word.htm

The first pulley is much smaller than the second so would turn at more than twice the rate. How would I show this?

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).

 

[Al_Pa_Cone]

See if this helps: https://www.dessci.com/en/products/MathType/popup_tex_in_word.htm

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?

 

[haruspex]

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?

Good.
So how much power is transmitted?

 

[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.

 

[Al_Pa_Cone]

Also giving the same results

 

[Al_Pa_Cone]

Scrap that last attempt:

 

[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.

 

[Al_Pa_Cone]

Dammit! I thought I had it on my last attempt.

So.. the Load torque. Is the load torque a resitance to turning? e.g a lawn mower would have a turning blade hitting the grass and the grass is resisting the blade from turning would be the load torque? or is it the Load which is on the belt tensioning the two pulleys? I think this is where I remain confused.

Surely It can't be as simple as T = 200 Nm
P = Tω
So 200 * 9.8175 = 1963.5 Watts or 1.96 kW?

 

[haruspex]

So.. the Load torque. Is the load torque a resitance to turning? e.g a lawn mower would have a turning blade hitting the grass and the grass is resisting the blade from turning would be the load torque?

Yes.

Surely It can't be as simple as T = 200 Nm, P=Tω

It is.

 

[Al_Pa_Cone]

Great! Finally a correct answer! I think my initial answer for b is ok but I know I have made a mistake in c part 1. Can you advise anymore?

There are also 2 more parts to this question I haven't yet posted. Thanks

 

[haruspex]

Great! Finally a correct answer! I think my initial answer for b is ok but I know I have made a mistake in c part 1. Can you advise anymore?

There are also 2 more parts to this question I haven't yet posted. Thanks

Yes, b looks right.
For c(i), you pointed out that the tension ratio could be increased. But then consider what that means for F₁ and F₂. F₁ is still at max, right? So, physically, what would you do to the belt to make use of the increased friction?

 

[Al_Pa_Cone]

In my notes I have wrote:
If there is a higher coefficient of friction, the belt grips more to the pulley which has a negative effect on the slack side of the belt (causing the slack side to become more slack) as it carrys it over the pulley and transfers the power to the 2nd pulley. Although the ratio of tension increasing to 53.48 from 24.13 would suggest ot me, the belt has become tighter on the slack side?
While F_1 is still at the maximum tension of 4000 N, I would expect the belt to become slacker on the slack side with an increase in the coefficient of friction?

 

[haruspex]

If there is a higher coefficient of friction, the belt grips more to the pulley which has a negative effect on the slack side of the belt (causing the slack side to become more slack)

I think that is wrong.
Once there is enough static friction to prevent slipping, increasing the coefficient of friction (and changing nothing else) has no effect. Remember that the static frictional force F_s is not necessarily equal to F_Nμ_s. That is only an upper limit on the force. Increasing μ_s does not of itself change F_s.
Given a higher coefficient, what can you change and still get an adequate frictional force?

 

[Al_Pa_Cone]

I have a few options within the ratio of tension equation which I could change.

e^(μθ/sinα) =F_1/F_2
The angle of lap (θ) is fixed due to the size of the pulleys being fixed.
The groove angle (sinα) again is fixed at 20^o

The exponential term cannot be changed.
So we would need to change the belt Tension in F_1/F_2
Then using the ratio of tension equation I should be able to work out F_2.

I will spend some time tomorrow working on this and post my results.
Thanks

 

[haruspex]

So we would need to change the belt Tension

Right.

 

[Al_Pa_Cone]

Ok I have made an attempt at the full question:

I have labeled each part of my method for reference. I hope this helps if I have made any mistakes?

 

[Al_Pa_Cone]

Scrap that last attempt, I have spend some time today working on this. here are my workings with reference tags to each part.

Does this look acceptable?

 

[Al_Pa_Cone]

Can anyone offer any advice with this?

 

[haruspex]

Can anyone offer any advice with this?

Sorry for the delay, wanted to get my thoughts clear on this...

I still have a quibble with the answer to b. Increasing the friction coefficient does not in itself increase the tension ratio; it only increases the maximum ratio that can be obtained. I.e. it allows you to increase the load.
The belt is elastic. The total of the two tensions is governed by that coefficient and the ratio between the belt's relaxed length and its path length around the pulleys. If the load is increased it will increase the difference in the tensions, keeping the total constant, and thus increase the ratio.
Note that if the higher tension is already at its max to avoid breaking, but there is plenty of friction coefficient, the way to increase the load that can be handled is to slacken the belt.

Likewise, in your answer to d, the causality is not quite right. You write it as though increasing the ratio makes the slack side slacker. You also imply that the stronger frictional grip makes the slack side slacker. Neither is the case. The stronger friction increases the maximum possible ratio, not the actual ratio. That allows you to increase the load. The increased load increases the tension difference, and hence the actual ratio.

I'll take a closer look at your equations.

 

[haruspex]

I'll take a closer look at your equations

For the calculation of max torque without changing friction, and limited by breaking tension/2:
It follows from my preceding remarks that in order to achieve this maximum the geometry would have be adjusted so that the sum of the tensions is 8000+331.54N. That is, under zero load, the tension would be (8331.54/2) N throughout.

 

[haruspex]

15057.02W should be rounded to 15.06kW, not 15.05

 

[haruspex]

For ii), you write that increasing the angle reduces the power. Again, it does not necessarily reduce the actual power; it reduces the maximum power.

For the graph, you need to be clear on what is kept constant. E.g. if the geometry is fixed then the tension sum is fixed.