Does tension in a belt drive system affect torque requirements?

[hihiip201]

HiI would like to know, does the torque required to turn a pulley increases if you increase the tension?
I have a system like this:pulley motor pulley
car and a belt goes around this system, the motor is included in the system via a gear that's attached to its shaft, so as the motor turns, so does the gear, and the belt would drive the pulleys and move the car to overcome friction.my question is, if I have a very strong tension in my belt that goes around this system, will it increase the torque required for the motor to turn the gear?

I don't think it could as that would imply the higher the tension the more energy consumption, but I really don't see where else the energy would be converted to if I run the motor at the same speed. I am trying to do a force analysis on the gear but it is proving to be troublesome as I don't really understand how tension in a belt/rope works as it is moving around a pulley system, this is beyond what I've learned in my physics, and even engineering courses, and I would appreciate it if someone can enlighten me.

 

[hihiip201]

My own attempt of explaining this is: for a perfectly inelastic rope/belt, as soon as the motor starts moving, it will tighten one side of the belt and loosen the other, hence creating a differences in torque that will drive the pulleys and the cars. Hence no matter how tight the belt was loaded, that tension is "gone" and replaced by the tension necessary to overcome the friction in the system, or accelerate the components in the system. That is, for inelastic rope only.
though experimentally, (this is an actual set up that I have for my project).
it seems that tightening the tension harder does increases the torque required for the motor as the motor actually stopped working, or perhaps this is only due to the increase in friction in pulleys due to higher tension?

 

[jryer]

If the belt system was tighten to the point where the pulley shafts were actually bent and the bearings started to bind, then yes additional torque would be needed to turn the pulleys. This assumes the belt wouldn't break nor slip with all that additional tension placed on it, and the bearings were only partially binding.

 

[hihiip201]

If the belt system was tighten to the point where the pulley shafts were actually bent and the bearings started to bind, then yes additional torque would be needed to turn the pulleys. This assumes the belt wouldn't break nor slip with all that additional tension placed on it, and the bearings were only partially binding.

in that case the energy is dissipated in the friction between binded bearing tracks?

so in the case where bearing is not binded, no matter how tight i tighten the belt, torque required of the motor should still be dependent of the friction in the system? (in an ideal case)

thank you for answering!

 

[jryer]

Yup, the energy is being dumped primarily into the ball bearings and case.

Just be aware any increase of friction in the pulley system will place a greater load on the engine, thereby either reducing engine output (torque) or require the engine to generate more output (torque) to compensate. In other words, friction may not only come from binded bearings. A flaky belt, a loose bolt allowing pulley mounts to shake, mis-alignments, etc all can contribute to friction increase.

in that case the energy is dissipated in the friction between binded bearing tracks?

so in the case where bearing is not binded, no matter how tight i tighten the belt, torque required of the motor should still be dependent of the friction in the system? (in an ideal case)

thank you for answering!

 

[hihiip201]

Yup, the energy is being dumped primarily into the ball bearings and case.

Just be aware any increase of friction in the pulley system will place a greater load on the engine, thereby either reducing engine output (torque) or require the engine to generate more output (torque) to compensate. In other words, friction may not only come from binded bearings. A flaky belt, a loose bolt allowing pulley mounts to shake, mis-alignments, etc all can contribute to friction increase.

I think we do have some misalignment problems, but how does misallignment usually translate into increase in friction?

Also, would increase tension increase friction in any way?thank you

 

[jryer]

I think we do have some misalignment problems, but how does misallignment usually translate into increase in friction?

Also, would increase tension increase friction in any way?


thank you

If the pulleys are not aligned to each other then the belt is having to make a slight turn to track onto the next pulley. Anytime a turn occurs in nature, things tend to bunch up. As they bunch up they start banging into each other and this collision creates heat. Kind of like rubbing your hands together real fast, eventually you get some heat from the friction.

 

[hihiip201]

If the pulleys are not aligned to each other then the belt is having to make a slight turn to track onto the next pulley. Anytime a turn occurs in nature, things tend to bunch up. As they bunch up they start banging into each other and this collision creates heat. Kind of like rubbing your hands together real fast, eventually you get some heat from the friction.

so is it like: if the a belt is supposed to be in x = 0 and all pulleys should align in that straight line.

but if pulley A is in x = 0 and B is in x = 3 mm, then the belt would collide with the parts that force the belt into alignment with the pulley, so the belt would be rubbing through these parts as it travels. ( I can see how if the two pulleys are not aligned you will need to force the belt to stay in place when it is at a pulley, we actually had quite a problem with this before)

I hope I am not saying stupid things.

 

[jryer]

If we assume the belt starts at 0 on the Y axis, and is moving along the X axis of a x/y graph, then yes Y + 3 would create friction because the belt is being diverted along the Y axis. Verbally I can express this a lot better than my cheesy math skills can. I believe at this point we are referring to kinetic friction, but I could be wrong.

Check out

http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html (Friction and Surface Roughness -- kinetic friction)

 

[hihiip201]

If we assume the belt starts at 0 on the Y axis, and is moving along the X axis of a x/y graph, then yes Y + 3 would create friction because the belt is being diverted along the Y axis. Verbally I can express this a lot better than my cheesy math skills can. I believe at this point we are referring to kinetic friction, but I could be wrong.

Check out

http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html (Friction and Surface Roughness -- kinetic friction)

Yea I think kinetic friction is what I'm referring to when i say parts are "forcing" the belt as the belt continue to move.

I am just always under the impression that as long as all forces interact via contact force or static friction, no energy should be lost mechanically.

thanks

 

[hihiip201]

My groupmates were saying that, if you have a very tight tension belt, then as you are trying to drive it you are "going against the tension", which I found to be very hard to accept because:

if you are driving a belt with a gear that is in the middle of the belt. Say the belt is tighten as the gear turns, the gear is going against tension on one side , but another side is aiding your motion.

if one side is loosen and another is tighten as the gear is turning(say loosen to a point tension on that side is negligible), then by loosening one side the other side should only be as tight as it needs to be to overcome friction or to accelerate, it should not have any tension to overcome from the other side since it is now loosen.

 

[jryer]

if one side is loosen and another is tighten as the gear is turning(say loosen to a point tension on that side is negligible), then by loosening one side the other side should only be as tight as it needs to be to overcome friction or to accelerate, it should not have any tension to overcome from the other side since it is now loosen.

As far as negligibility, you may want to familiarize yourself with how Power is Transmitted via belts:
First figure out the belt tension for both sides:

F = Moving force in the pulley driver
T1 = tension on the tight side
T2 = tension on the slack side
e = base of the Neperian or natural logarithms (2.7182)
μ = coefficient of friction
α = angle subtended by contact surface at the center of the pulley

T1 = F * (1 + (1/(e^ua - 1)))
T2 = F * (1/(e^ua - 1))

Then you can solve for how much Power is actually being Transmitted

Power transmitted between a belt and a pulley is expressed as the product of difference of tension and belt velocity. Work out the math, it'll help answer a few things.

P= power
T1 = tension on the tight side
T2 = tension on the slack side
v = velocity (of the belt in this case)

P = (T1-T2) * v

T1 & T2 are related as:

e = base of the Neperian or natural logarithms (2.7182)
μ = coefficient of friction,
α = angle subtended by contact surface at the center of the pulley

(T1/T2) = e^ua

 

[hihiip201]

As far as negligibility, you may want to familiarize yourself with how Power is Transmitted via belts:
First figure out the belt tension for both sides:

F = Moving force in the pulley driver
T1 = tension on the tight side
T2 = tension on the slack side
e = base of the Neperian or natural logarithms (2.7182)
μ = coefficient of friction
α = angle subtended by contact surface at the center of the pulley

T1 = F * (1 + (1/(e^ua - 1)))
T2 = F * (1/(e^ua - 1))

Then you can solve for how much Power is actually being Transmitted

Power transmitted between a belt and a pulley is expressed as the product of difference of tension and belt velocity. Work out the math, it'll help answer a few things.

P= power
T1 = tension on the tight side
T2 = tension on the slack side
v = velocity (of the belt in this case)

P = (T1-T2) * v

T1 & T2 are related as:

e = base of the Neperian or natural logarithms (2.7182)
μ = coefficient of friction,
α = angle subtended by contact surface at the center of the pulley

(T1/T2) = e^ua

T1, T2, Tensions after the gear is engaged (turning), the differences between them (T1-T2) is independent of their preloaded magnitude right?

 

[jryer]

force = power/velocity

(force has other equations, we're using the one suited for this application)

Assuming you have an motor of 2 hp (1492 N-m/s) and the belt is moving at 12 mph (5.3645 m/s). Metric units are better here so ...

F = (1492 / 5.3645) = 278 Newton (62 pound force - lbf)

So now

T1 = 278 * (1 + (1/(2.7182 * 0.25 * 0.5934))) = 967.4072 Newton (217 lbf)
T2 = 278 * (1/(2.7182 * 0.25 * 0.5934)) = 689.4072 Newton (155 lbf)
P = (967.4072-689.4072) * 5.3645 = 1491.331 N-m/s

It appears the proof works because 1491.331 N-m/s = 1.999 hp (close enough)

u = coefficient of friction is a tough one, but 0.25 for a dry running belt is a good base
a = find out your pulley groove angle, convert to radians (i'm using 34 deg = 0.5934 radians)

Assuming I didn't screw things up, it appears the slack side, in this case, is 71% of the tight side. So it's fairly tight even on the slack side.

1. Just figure what horsepower your motor has (convert hp to N-m/s)
2. How fast the belt is moving (mph to m/s)

Then just replace my values above with yours, and you should get your T1 and T2 tension values

 

[jryer]

"I would like to know, does the torque required to turn a pulley increases if you increase the tension?" ... YES

Here's why. If we tighten the belt, we increase the 'coefficient of friction' in the T1 & T2 equation.
Coefficient of friction comes from contact between the belt and pulley.
An increase in Coefficient of Friction increases the Pound Force on the belt, which in turn increases the Power demand. In a nut shell, tightening the belt puts more force on the belt, requiring more Power to turn it.

T1 = F * (1 + (1/(eua - 1)))
T2 = F * (1/(eua - 1))

μ = coefficient of friction

 

[hihiip201]

"I would like to know, does the torque required to turn a pulley increases if you increase the tension?" ... YES

Here's why. If we tighten the belt, we increase the 'coefficient of friction' in the T1 & T2 equation.
Coefficient of friction comes from contact between the belt and pulley.
An increase in Coefficient of Friction increases the Pound Force on the belt, which in turn increases the Power demand. In a nut shell, tightening the belt puts more force on the belt, requiring more Power to turn it.

T1 = F * (1 + (1/(eua - 1)))
T2 = F * (1/(eua - 1))

μ = coefficient of friction

so it is absolutely NOT the case that the difference increased due to having to "go against" the preloaded tension as my group partner suggested. But rather the friction increased due to increased preloaded tensionsorry if it sounds like I am reiterating myself, it's not that I don't understand what friction does (and that kinetic friction increase as normal force increases), is just that I want to ensure that my understanding of tension and mechanics have not been defied, I'm a Junior Mechanical engineering student but sometimes I just ask question that sounds very amateur to ensure that what I think I understand is in fact correct. thanks