How Does Pulley Size Impact Efficiency in Large Scale Lifting Systems?

[gloo]

If one was to make a pulley to lift something massive like a ship, I am assuming the masses involved of the ship and the counterweight, will mean decreased efficiency due to :

1. Higher friction between in the sheave and axel area.
2. larger cable/robe mass on the longer end (once the pulley moves more one dir'n)
3. larger rope causing bending around the pulley.

Questions:

1. What is the loss of efficiency with a larger pulley system requirement like this?
2. What are possible advantages of a larger pulley system in terms of efficiency? For instance, i read that a larger diameter pulley can result in a large increase in efficiency for a pulley system.

Thanks.

 

[FireStorm000]

I can start to answer your question by simplifying it. Consider a single pull with a mass on one end, and the other connected to some mechanism which will measure both work and displacement.

Now, the efficiency will depend on the coefficient of friction, the force of tension on the rope (effectively the mass*gravity), and the size of the wheel thing. Work is force over distance. This means our efficiency is determined by how far the wheel moves against the axle and the coefficient of friction. How much the rope moves is some multiple of the distance the wheel moves against the axle.

Assume that the wheel has a radius of 1m, the axle one of .1m. Let's assume you want to lift a 1000kg object 1 meters(1 radian of rotation). The coefficient of friction for lubricated steel on steel is .16 ^{http://www.engineeringtoolbox.com/friction-coefficients-d_778.html"} .

Now, the 1000kg object exerts a force mg=f=10,000N on the pulley; the wheel moves against the axle a distance of .1m, meaning 1.6kJ was lost to friction. W = F*(mu)*d.
work done to raise the object is W=mgh, or 10KJ
that is an efficiency W_useful-W_F / W_Tof (10000-1600)/(10000+1600) = 72.41%

To give some useful equations:
let K = R_wheel/R_axle
u is the coefficient of friction
F is the force on the pulley, in this case mg
Efficiency is: (useful work - wasted work) / (useful work + wasted work)
Total work is: useful work + wasted work
Useful work is: (Fh)
Wasted work is: (F)(hk)(u)
Thus, efficiency for a (single) pulley is given by: (1-kmu)/(1+kmu)

You should be able to apply that to your more complex pulley system. It's just more pulleys, and the forces should be some multiple of mg. Shouldn't be too much more work. I'd imagine the efficiency when lifting a ship to be less than 10%, just based on the large axles needed to lift something so heavy, and the enormous mass involved.

 

[gloo]

I can start to answer your question by simplifying it. Consider a single pull with a mass on one end, and the other connected to some mechanism which will measure both work and displacement.

Now, the efficiency will depend on the coefficient of friction, the force of tension on the rope (effectively the mass*gravity), and the size of the wheel thing. Work is force over distance. This means our efficiency is determined by how far the wheel moves against the axle and the coefficient of friction. How much the rope moves is some multiple of the distance the wheel moves against the axle.

Assume that the wheel has a radius of 1m, the axle one of .1m. Let's assume you want to lift a 1000kg object 1 meters(1 radian of rotation). The coefficient of friction for lubricated steel on steel is .16 ^{http://www.engineeringtoolbox.com/friction-coefficients-d_778.html"} .

Now, the 1000kg object exerts a force mg=f=10,000N on the pulley; the wheel moves against the axle a distance of .1m, meaning 1.6kJ was lost to friction. W = F*(mu)*d.
work done to raise the object is W=mgh, or 10KJ
that is an efficiency W_useful-W_F / W_Tof (10000-1600)/(10000+1600) = 72.41%

To give some useful equations:
let K = R_wheel/R_axle
u is the coefficient of friction
F is the force on the pulley, in this case mg
Efficiency is: (useful work - wasted work) / (useful work + wasted work)
Total work is: useful work + wasted work
Useful work is: (Fh)
Wasted work is: (F)(hk)(u)
Thus, efficiency for a (single) pulley is given by: (1-kmu)/(1+kmu)

You should be able to apply that to your more complex pulley system. It's just more pulleys, and the forces should be some multiple of mg. Shouldn't be too much more work. I'd imagine the efficiency when lifting a ship to be less than 10%, just based on the large axles needed to lift something so heavy, and the enormous mass involved.

wow! So you are saying with something as heavy as a ship, you would lose 90 percent of the efficiency to friction?? Someone said the average elevator is pretty efficient and that the counterweight used is around 90 percent efficiency or something? They assume give or take a full capacity of around 10,000kg! How is it that their efficiency is almost 90 percent? Is it that they are adding in the regenerative power back into the equation?

But what about the argument that the larger the pulley diameter the large the efficiency factor.

Also, what if we used several pulleys (single fixed for instance) to reduce the mass needed for one big pulley? Would that increase the total efficiency by decreasing the friction from the mass?

Btw, thank you very much for such a thorough answer. I am not a physics or engineering student so it take a bit for me to follow all the math details. I did complete grade 13 physics but very very rusty at this point.

 

[FireStorm000]

First things first, I botched the eq for efficency:
not:(useful work - wasted work) / (useful work + wasted work)
but rather:(useful work) / (useful work + wasted work)
Edit: And the one for k
k = rsmall/rbig

I'd imagine so. Let's imagine a ship with mass 50000metric tons. That's 500000000kg. Let's assume a 5m radius pulley with a .2 radius axle, lifting 10m:
k=25
m=500,000,000kg
u=.16
(1)/(1+mu/k)=3.125e-7
I imagine due to needing more pulleys that efficiency would be worse, as one of the pulleys still has to support the full weight of the ship, thus the force on one of the pulleys is the same, and you have to add in more inefficiencies.

Edit: also, using your numbers for an elevator
1/(1+rsml*u*m/rbig)
1/(1+.01*.16*(5000(load)+5000)(counter)/1)=.04%
HOWEVER, that equation *does not* take into account recycled energy
You'd have to go back and use the basic equations to derive a new equation for that one. In short, distance matters in the new one. Because of the counterwieght, energy is only lost due to friction, because it takes no energy to fight gravity (if load = counter)

 

[gloo]

First things first, I botched the eq for efficency:
not:(useful work - wasted work) / (useful work + wasted work)
but rather:(useful work) / (useful work + wasted work)
Edit: And the one for k
k = rsmall/rbig

I'd imagine so. Let's imagine a ship with mass 50000metric tons. That's 500000000kg. Let's assume a 5m radius pulley with a .2 radius axle, lifting 10m:
k=25
m=500,000,000kg
u=.16
(1)/(1+mu/k)=3.125e-7
I imagine due to needing more pulleys that efficiency would be worse, as one of the pulleys still has to support the full weight of the ship, thus the force on one of the pulleys is the same, and you have to add in more inefficiencies.

Edit: also, using your numbers for an elevator
1/(1+rsml*u*m/rbig)
1/(1+.01*.16*(5000(load)+5000)(counter)/1)=.04%
HOWEVER, that equation *does not* take into account recycled energy
You'd have to go back and use the basic equations to derive a new equation for that one. In short, distance matters in the new one. Because of the counterwieght, energy is only lost due to friction, because it takes no energy to fight gravity (if load = counter)

1. When you say one pulley still has to support the ship, I assume you misunderstood my suggestion of using several fixed pulleys. So imagine 3 or 4 fixed to a ceiling with it's own counterweight but the 3 or 4 is either 1/3 or 1/4 of using one big counterweight; this is to reduce the burden one a large counterweight and distribute the lifting among several pulleys. It would make sense if the additional mass adds an exponential factor to the friction which I am kind of assuming no? Maybe it's wrong if you use a constant for the co-efficient of friction?!

2. so the calculation should have been 86% efficiency? [10,000/(10,000 + 1600)] and not 72.41 % for the original 1000kg example you used?

 

[gloo]

First things first, I botched the eq for efficency:
not:(useful work - wasted work) / (useful work + wasted work)
but rather:(useful work) / (useful work + wasted work)
Edit: And the one for k
k = rsmall/rbig

I'd imagine so. Let's imagine a ship with mass 50000metric tons. That's 500000000kg. Let's assume a 5m radius pulley with a .2 radius axle, lifting 10m:
k=25
m=500,000,000kg
u=.16
(1)/(1+mu/k)=3.125e-7
I imagine due to needing more pulleys that efficiency would be worse, as one of the pulleys still has to support the full weight of the ship, thus the force on one of the pulleys is the same, and you have to add in more inefficiencies.

Edit: also, using your numbers for an elevator
1/(1+rsml*u*m/rbig)
1/(1+.01*.16*(5000(load)+5000)(counter)/1)=.04%
HOWEVER, that equation *does not* take into account recycled energy
You'd have to go back and use the basic equations to derive a new equation for that one. In short, distance matters in the new one. Because of the counterwieght, energy is only lost due to friction, because it takes no energy to fight gravity (if load = counter)

--ok, you know what? I am totally lost by your equations. As mentioned before, I am not a physicist and not great with equations. But your corrections and the answers you derive has me stumped?

questions: 1. what does this 1/(1+rsml*u*m/rbig), stand for again?
2. which one is correct? in the first you said Rwheel/Raxel... and in the second you corrected it to rsmall/rbig? I assume since you got 25 in the second example (radius of pulley is 5, radius of axel is .2) , then the correct equation is k=rbig/rsmall?!?

3. Are you saying that an elevator pulley is .04% efficient?

I am not good at math and physics was not my best subject, but I passed. But I can't follow what the example is here.

 

[FireStorm000]

--ok, you know what? I am totally lost by your equations. As mentioned before, I am not a physicist and not great with equations. But your corrections and the answers you derive has me stumped?

Sorry if I'd made you more confused than you were. I probably should have checked my work more thoroughly, and I changed variables on you. I never edited the first post, so only the second one on are fixed.
r_sml=r_axle
r_big=r_wheel
Always be aware the possibility that I'm wrong. I was hoping someone would come along and check my work(*cough* *cough* like soon)

1. what does this 1/(1+rsml*u*m/rbig), stand for again?

That is an equation giving efficiency. rsml and rbig are explained above. u is coefficient of friction, m is mass.

2. which one is correct? in the first you said Rwheel/Raxel... and in the second you corrected it to rsmall/rbig? I assume since you got 25 in the second example (radius of pulley is 5, radius of axle is .2) , then the correct equation is k=rbig/rsmall?!?

that is correct if we divide by k. If you want to multiply by k as one normally does, then we need to take the reciprocal. The important thing to realize is that it implies k the ratio between the wheel and axle gets bigger, efficiency increases.

Are you saying that an elevator pulley is .04% efficient?

I think it really depends on how you define efficiency. I'm going to check all my work and fix all my examples at the end of this post.

1. When you say one pulley still has to support the ship, I assume you misunderstood my suggestion of using several fixed pulleys. So imagine 3 or 4 fixed to a ceiling with it's own counterweight but the 3 or 4 is either 1/3 or 1/4 of using one big counterweight; this is to reduce the burden one a large counterweight and distribute the lifting among several pulleys.

I did indeed misunderstand your idea. Using multiple pulleys would significantly increase efficiency, comparable to increasing the ration between the wheel and axle by the same factor as the number of pulleys.

Maybe it's wrong if you use a constant for the co-efficient of friction?!

No, the coefficient of friction is a constant between any two given materials. The force of friction varies directly with the force applied. http://en.wikipedia.org/wiki/Coefficient_of_friction#Dry_friction"

2. so the calculation should have been 86% efficiency? [10,000/(10,000 + 1600)] and not 72.41 % for the original 1000kg example you used?

You're correct. You're getting this more than you give yourself credit for.

Going back to the derivation, and fixing errors as I go:

I can start to answer your question by simplifying it. Consider a single pull with a mass on one end, and the other connected to some mechanism which will measure both work and displacement.

Now, the efficiency will depend on the coefficient of friction, the force of tension on the rope (effectively the mass*gravity), and the size of the wheel thing. Work is force over distance. This means our efficiency is determined by how far the wheel moves against the axle and the coefficient of friction. How much the rope moves is some multiple of the distance the wheel moves against the axle.

Assume that the wheel has a radius of 1m, the axle one of .1m. Let's assume you want to lift a 1000kg object 1 meters(1 radian of rotation). The coefficient of friction for lubricated steel on steel is .16 ^{http://www.engineeringtoolbox.com/friction-coefficients-d_778.html"} .
Now, the 1000kg object exerts a force mg=f=10,000N on the pulley. In order to counter that, 10000N must be applied to the other end for a grand total 20000N acting on the pulley. The wheel moves against the axle a distance of .1m, meaning 1.6kJ was lost to friction. W = F_Tension*(u)*d.
work done to raise the object is W=mg$\Delta$h, or 10KJ
that is an efficiency W_useful/ W_Tof (10000)/(10000+1600) = 86%

To give some useful equations:
let K = R_axle/R_wheel. Smaller 'K's are more efficient
u is the coefficient of friction. It is a constant for any two given materials.
F is the force on the pulley, in this case 2mg
Efficiency is: (total work - work of friction) / (total work)
Total work is defined as all of the energy put into or taken out of a system
Wasted work is: (F_T)($\Delta$h*k)(u)
Total work is the sum of wasted and useful work
Thus, efficiency for a (single) pulley is given by: (mg$\Delta$h + 2mg$\Delta$hku - 2mg$\Delta$hku)/(mg$\Delta$h + 2mg$\Delta$hku)
(mg$\Delta$h)/(mg$\Delta$h + 2mg$\Delta$hku)
Efficiency=1/(1+2ku)

So in conclusion, I've now failed at this twice, but I'm fairly certain that the bold-ed equation works for a change, and should give you your 90% elevator efficiency.