Calculating Power/Torque transmitted between 2 pulleys

  • Thread starter Al_Pa_Cone
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In summary: 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?In summary, the question involves calculating the power transmitted between two pulleys, with given values for diameter, distance between
  • #36
haruspex said:
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.
For this I hav tried to make my answer clear by adding a statement to the bottom of the graph, The statement is as follows:

Here is a plotted Graph showing the results of the changes effected by the coefficient of friction and the change in pulley groove angle on the ratio of tension. The middle results 24.13 and 15.06 represent the original measurements taken prior to changes. Then to the left is a positive change to the coefficient of friction which reflects a rise in maximum power transmitted. To the right represents the change in groove angle as increasing this decreases the maximum power transmitted. They both run parallel to the final power transmitted proving therefore, a negative change to the ratio of tension reflects a negative change to the power transmitted to the second pulley and a positive change will also reflect a positive power output.

haruspex said:
15057.02W should be rounded to 15.06kW, not 15.05

Good spot, I should have noticed that one!
haruspex said:
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.

From the question:

(b) 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.

I have tried to follow an example question in the lesson and work through my results by substitution of the known figures:
upload_2017-1-20_12-3-8.png

I have Highlighted in the red boxes, The figure which is given to me as the ultimate strength of the belt? So 8000N in my case. Following the method provided for working this out:
upload_2017-1-20_12-5-21.png

In my method I substituted the coefficient of friction, and increased the pulley groove angle to obtain the results. Would I have been better off reworking the soloution without using the ultimate strength as my starting value?

and finally:
haruspex said:
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.

I have changed my statement, included some of your text and I think It may now be ok?Solution:
The coefficient of friction is governed to a mainly by the fact that a pulley


surface has to be mostly smooth or the belt would wear too quickly.

There are semi-liquid compounds available which can help preserve belts and to make the belt surface 'sticky' so that the coefficient of friction will remain at a reasonable value.



If the coefficient of friction was improved by this change, as it increases resistance in the contacting surfaces of the belt and the pulley, it increases the maximum ratio of tension that can be obtained. I.e. it allows you to increase the load. This would reflect a positive increase in the effective tangential force

The total of the two tensions is governed by that coefficient and the ratio between the belt's ‘slack’ side and its ‘tight’ side 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.

It is understood that more power is transmitted if the 'effective tension', i.e. the difference in tension between the tight and slack sides of the belt (F1 - F2) can be increased
 
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  • #37
Al_Pa_Cone said:
The total of the two tensions is governed by that coefficient
"That" coefficient referred to the elasticity, which you have not mentioned.
Al_Pa_Cone said:
governed to a mainly
That got past the proofreading.
Al_Pa_Cone said:
If the coefficient of friction was improved
There used to be a mood called the subjunctive...
 
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  • #38
Al_Pa_Cone said:
Here is a plotted Graph showing the results of the changes effected by the coeff
How about mentioning that in each case it is assumed that the initial (no load) tension is optimised? Maybe show that value (tension sum/2) in the array?
 
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  • #39
I apologise for being sloppy with my statements. I am trying to work through this assignment while I am in work. Also I have an unavoidable distraction, with a Bose speaker playing the local radio station right above my head. I have no control over the volume and this frequently effects my thinking!

I will work through this as soon as I get a chance.

By the way, my last mechanical principals assignment has been marked. Thanks for your help in the Macaulays method question as I obtained a distinction mark!
 
  • #40
Al_Pa_Cone said:
I obtained a distinction mark!
Well done us.
 
  • #41
I have reattempted my statement for:

(b) 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

haruspex said:
"That" coefficient referred to the elasticity, which you have not mentioned.

That got past the proofreading.

There used to be a mood called the subjunctive...
upload_2017-1-23_11-9-26.png
 
  • #42
haruspex said:
How about mentioning that in each case it is assumed that the initial (no load) tension is optimised? Maybe show that value (tension sum/2) in the array?

If I reworked the answer to this question including the 3 examples I have provided (from the change in coefficient of friction and the change in pulley groove angle), how could I give my answer for maximum power transmitted using the equations I have available?

I would take a guess at maximum power (P) = Sum of Optimised Tensions (T) x The Angular velocity of the belt at pulley 2 (ω)

So P= Tω

Where before T = (F_1-F_2)r ... now I would use (F_1+F_2)/2 for optimised tension, then substitute into the 1st equation as T = [(F_1+F_2)/2]r

and the results for T would then work in P=Tω for maximum power?

If this looks correct, I will plot graph of my results and rework the answer.

Thanks
 
  • #43
Can anyone advice if this method is the correct method to pursue?
 
  • #44
Al_Pa_Cone said:
Where before T = (F_1-F_2)r ... now I would use (F_1+F_2)/2 for optimised tension
No, you misunderstand my comments about optimised tensions. You correctly calculated the maximum achievable power earlier.

My concern is that in order to achieve it the system has to operate with certain ideal values, F1, F2, for the two tensions. How is that to be achieved in practice? Once the system is set up physically, their sum won't change during operation.
The way, therefore, is to tension the whole belt to (F1+F2)/2 before starting up the motor. When peak load torque occurs, the two tensions will be at their ideal values.
 
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  • #45
So I can figure out (F_1+F_2)/2 but at what point in my equations do I begin to use the value? I think all the equations I have are relevant to the pulleys being in motion
 
  • #46
Al_Pa_Cone said:
So I can figure out (F_1+F_2)/2 but at what point in my equations do I begin to use the value? I think all the equations I have are relevant to the pulleys being in motion
You don't need to use that value in your equations. You can deduce the value from your equations. I.e. you can show that in order to achieve the maximum power you calculated, the tension sum must be this value.
The use of that is in the practicalities of achieving this maximum: the mechanic would have to set the belt up to have that tension sum in the first place. If she makes the belt slacker it will slip before reaching the power you calculated; if tauter, it will break before reaching that power.
 

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