Power transmitted to the second pulley

[Big Jock]

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.

Calculate the actual power transmitted to the second pulley.

So far I have
Angle of lap of small pulley= 2.722 radians
Angle of lap of the large pulley= 3.561 radians
Angular speed= 8 1/3pi
Driver speed = 250 revs min-1 = 25pi/3 rad s-1

 

[Simon Bridge]

Nice start - can you describe the operation of the setup in terms of energy?

 

[Big Jock]

I believe I have worked it out now and the answer should be 1963kW. Is that correct?

 

[Big Jock]

If that answer is correct Simon and I know the tensions is the tight and the slack side of the belt, how would I 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? I am a bit confused with that part and can't find any relevant info or examples to compare against...

 

[Simon Bridge]

You are doing well - have a look at:
http://www.technologystudent.com/gears1/pulley1.htm
http://en.wikipedia.org/wiki/Belt_(mechanical)#Power_transmission
... for comparison.

Remember that you have friction in your system - so some energy gets lost that way.

 

[Big Jock]

So was my answer I quoted wrong Simon?
Also still struggling with question in #4 I read the links you sent but didn't seem to help me or I have missed something along the line.

 

[Simon Bridge]

I didn't do the math - the idea was to provide you with something to compare with, so you can check for yourself.
In general, I won't do the calculation - but if you show me your reasoning I can see if what you did makes sense.

 

[Big Jock]

I calculated the speed of the second pulley and multiplied it against the torque to get the power, which was the answer I quoted. Now from what I I have read I believe that to be true.
If that is the case how do I set about tackling --the power which can be transmitted if the maximum
tension in the belt is limited to half of the ultimate strength of the
belt?
Find this very confusing but I may be staring at the correct route just need nudged in the right direction.

 

[Simon Bridge]

How did you handle the friction?

If the belt tension is not allowed to go above a certain value - what does that tell you about how to run the motor?

 

[Big Jock]

to get the power I used power = torque x angular velocity. I used 200Nm x the speed of the large pulley. Would that be correct or have I just calculated the power in the whole system??
Now tension in the tight side must be 8kN and using the 2.722 angle of lap I got a figure for the tension in the slackside.
Would this be correct for the original tight and slack side tensions. Think I can take it from there if it is...

 

[Simon Bridge]

Well - is the calculated belt tension too high or too low as it is?

 

[Big Jock]

How do I find this out?? As I am completely lost now...

 

[Simon Bridge]

Did you not just calculate belt tension above?
Are you not provided with a maximum tension figure?

 

[Big Jock]

Worked it all out Simon I think many thanks for you help

 

[Simon Bridge]

Great stuff :)

 

[PizzaWizza]

Hey, I'm currently working on the same paper.

If I calculate the Torque in the slack part of the belt (T2), using the angle of lap for the small pulley, would this be correct to calculate the power when using half of the ultimate strength of belt?