Energy dissipation when a lift stops

In summary: The brake is there to stop the car from coming into contact with the sheave at the bottom of the lift, and to provide a safety lock. When the brake is released, it can dissipate only the kinetic energy of the high-speed input (motor rotor, the brake drum itself, and the shaft). The combination of car+load+counterweight+ropes, when in motion, has a considerable momentum, but its kinetic energy can't reach back to the drum because the unidirectional characteristic of the worm gear drive prevents that. Energy is dissipated by the brake and the motor, in that order.
  • #1
Majorana
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Lifts/elevators of traditional design (i.e. not of the "gearless" type, like Kone's EcoDisc ®) are driven by a hoist that uses a worm drive gear between the electric motor and the sheave. The drum brake is always mounted on the motor shaft (high speed). As you know, worm drive gears are (except very particular applications) unidirectional, i.e. the direction of transmission is not reversible: the output (sheave) cannot drive the input (motor shaft). If the brake is released but no power is supplied to the motor, the lift would remain stationary under the sole action of the worm gear. The brake serves only to stop the hoist when the car is exactly aligned with the floor, and to provide a safety lock.
But you see, in this way, when the brake closes, it can dissipate only the kinetic energy of the high-speed input (motor rotor, the brake drum itself, and the shaft). The combination of car+load+counterweight+ropes, when in motion, has a considerable momentum, but its kinetic energy can't reach back to the drum because the unidirectional characteristic of the worm gear drive prevents that.
So my question is: where exactly is dissipated the kinetic energy of the "big masses" (car+counterweight)? We are speaking of A LOT of energy. It must be converted to heat somewhere (remember, I am still speaking of traditional designs, not electronic) and such conversion must happen by means of friction, but if not in the drum brake, where?...
 
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  • #2
Hello Majorana, :welcome:

Ambitious and interesting avatar/pseudonym !

I would search in the worm wheel area: as you deduce, the kinetic energy can't get past that.
The nice thing is that any dissipative heat is easily given off to the surrounding air.

But this is just a physicist's first guess; I'll gladly trade it for an expert opinion :smile: !
 
  • #3
@BvU Thank you very much! And of course, I wouldn't be deserving to look at Ettore Majorana's shoes, forget about tying them :redface:
But he was fascinating under many respects, not only because of his disapparance. By the way, a few weeks ago I got to know a (distant) relative of him, her grandma was Ettore's cousin...:oldsmile:
I guess that the energy dissipation thing could be easily solved by asking a manufacturer of elevator hoists, but those people are not very eager to discuss the details of their trade... :olduhh: By the way, the worm gear in elevator hoists ALWAYS works immersed in oil, to minimize friction, wear and tear... so the question about "friction" and "heat" is - IMHO - really puzzling...
 
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  • #4
I think you're missing something here

Worm gears are unidirectional because they're typically not efficient.. Worm gears with a steep worm gear angle are bidirectional, but not efficient at it.
So because of the inefficiency of the worm gear, the brake actually doesn't need to dissipate all the heat, any force it applies to the worm shaft gets multiplied.. So the two places that generate heat will be the brake and the gears, which dissipate it into oil, casing, then air... I'd guess at a 25/75 ratio one way or the other, depending on design... There's also a good chance the motor is powered by a variable frequency drive that can brake as well, which would probably be smoother than a friction brake, and puts less heat into that area (the VFD will usually have a large resistor to dump excess power into)
 
  • #5
Majorana said:
We are speaking of A LOT of energy. It must be converted to heat somewhere (remember, I am still speaking of traditional designs, not electronic) and such conversion must happen by means of friction, but if not in the drum brake, where?...
Right back at the prime mover which is the motor.

Just consider everything being frictionless so no part of the machinery dissipates energy by friction.
( In any case one cannot rely upon parasitic dissipative effects as being reliably constant for elevator performance over the life of the elevator )
How would you change the speed of the elevator?
Look at the worm gear/big gear and how the teeth mesh,
To accelerate in one direction, faces "a" of the two gears have to be in contact.
To accelerate in the opposite direction, faces "b" of the teeth have to be in contact, where face "b" is the other side of the tooth from side "a".
Is it any different whether the elevator is starting from motionless and moving in either direction.
Than from when the elevator is moving, and the elevator is required to accelerate gaining speed or loosing speed.

The motor will provide the required force on the gear faces by attempting to increase its rotational velocity, or decrease.
In either case there is a heat build up within the motor.
 
  • #6
@256bits The worm gear drive is almost frictionless, nevertheless it is actually unidirectional (except in very particular applications where it's designed so as to be bidirectional, like - if memory doesn't betray me - in the Torsen differential). Proof is that if you manually release the brake, the elevator would not move at all. So, again, I can't see how the kinetic energy of car+load+counterweight can possibly reach back past the worm drive. Your reasoning about the "A" or "B" side of the teeth being in contact depending on whether the system is accelerating or decelerating seems wrong. I think that the side (A or B) of the teeth that comes in contact, only depends on whether the weight of car+load is more or less of the weight of the counterweight [by design, the weight of the counterweight is equal to the weight of the empty car + 40% to 50% of the rated lifting capacity] : the hoist will only "pull" or "loosen", but since it can't "push" on a rope, the side of the teeth being in contact does not change, IMHO.
 
  • #7
@Rx7man My reasoning was all about traditional elevators, where the motor is the asynchronous squirrel-cage type without any electronics that can possibly dissipate energy. As I just replied to 256bits, a worm gear drive is unidirectional not because it's particularly inefficient, but rather by its very characteristics. A drive in oil bath (like the ones used in elevator hoists) can run almost frictionless if you drive it the "right" way, but can hold as steady as if it were welded if you try to drive it from output to input, even if the high speed shaft is left completely free.
Nevertheless, the energy of the car+load+counterweight must go somewhere: most probably into the oil, case, and then air, as you supposed.
 
  • #8
I agree that it'll always be the same face of the teeth in contact with each other, for this application.

OK, so if the motor just freewheels and doesn't do any braking, the energy will have to go into the oil and brake.

But about them being nearly frictionless, I don't think that's quite right.. I know that in excavator final drives, they have planetary 2 or 3 stage reduction gears, and there's no chance of driving them from the 'wrong' end.. just like a worm gear, they won't freewheel.
If you look at a worm gear, and for this example and to make life easy, let's say the capstan/drum is the same diameter as the driven gear of the worm drive, so that the cable pull is the same as the force against the worm gear.
If you look at the way the gear is cut, you'll notice it's cut on an angle, and this angle is the key, and it determines the diameter of the worm... Let's say we have a 1/2" pitch on the screw, and it's 2" in diameter.. That will make the angle tan(.5/(2pi)) = 4.5degrees.. and the bigger the diameter, the less angle.
Looking at it from the driven side, the component of the force acting perpendicular to the teeth will be very nearly the weight on the rope (that's why I chose the same size capstan/drum as worm gear)... however the component of the force acting to rotate the worm is very small because of the steep angle...
So with a 4.5* cut gear, the the amount of force contributing to rotating the worm is tan(4.5) or 0.078.. 7.8%
I don't know what a reasonable coefficient of friction is between the gears, but as long as it's greater than .078 it will be impossible to turn.. and .078 isn't much of a coefficient of friction.
 
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  • #9
Ok, I agree with you! So the energy actually gets dissipated in the worm gear, or more precisely, in the thin oil layer separating the worm groove from the tooth of the gear wheel. Thank you! :bow:
 
  • #10
Majorana said:
@256bits The worm gear drive is almost frictionless, nevertheless it is actually unidirectional (except in very particular applications where it's designed so as to be bidirectional, like - if memory doesn't betray me - in the Torsen differential). Proof is that if you manually release the brake, the elevator would not move at all. So, again, I can't see how the kinetic energy of car+load+counterweight can possibly reach back past the worm drive. Your reasoning about the "A" or "B" side of the teeth being in contact depending on whether the system is accelerating or decelerating seems wrong. I think that the side (A or B) of the teeth that comes in contact, only depends on whether the weight of car+load is more or less of the weight of the counterweight [by design, the weight of the counterweight is equal to the weight of the empty car + 40% to 50% of the rated lifting capacity] : the hoist will only "pull" or "loosen", but since it can't "push" on a rope, the side of the teeth being in contact does not change, IMHO.
Certainly the counter weight system is designed so that the rope is always in tension.

I wrote as if the system was in balance which I had thought would have been easier to comprehend how the motor must take up the work load in either direction of movement, such that different faces up the gears would be in contact. In the designed unbalanced system so that there is always a tension in the rope, yes, the same face should always be in contact.

These systems are not capable of energy recovery, just by their nature, as you say the worm gear cannot be driven by your term "unidirectional". Which by the way does not mean that the larger gear can not be driven in either direction.

Perhaps, to see it more clearly, just have a mass attached to a rope hanging down from a drum, around of which the rope can be wound. The drum is connected to larger gear of the worm gear drive. Since the weight of the mass provides a tension to the rope and a subsequent torque to the drum, the same face of the gears will always be in contact.

To lift the mass, the motor provides a torque to accelerate the mass to a certain speed. The contact force between the teeth is greater than when the mass is at rest, or moving at constant velocity ( it would be the same if the mating gear face contact was frictionless ). Since the motor is not a perfect prime mover, it will heat up during the acceleration phase. ( during the constant velocity phase of travel, not as much heat will be produced within the motor ).

To decelerate the mass from upward movement, the gear faces will have to have a lessor force between them than that at constant speed. The motor will have to work less hard depending upon the required rate of deceleration, but it will always have to keep turning. ( If it stops turning, the upward movement of the mass will decrease the tension in the rope to 0, which we do not want, and its movement could then simply be calculated as a free body in a gravitational field ). Gravitation acts to slow the upward movement of the mass.

To lower the mass, the motor again has to turn, but in the opposite direction. During the downwards acceleration phase, the gears faces have a lessor force upon them, and gravitation can be thought of as providing the acceleration force upon the mass. ( It should be mentioned, that it is the difference weight of the mass and the force from the tension of the rope which determines whether the direction of acceleration of the mass for this case and that above, just so we can be clear about that ). To decrease the downwards velocity of the mass to zero, the motor has to provide more torque, so that the gear face force increases, so that the tension is greater than the weight of the object. Again, as before, the motor, not being a perfect prime mover, will heat up.
 
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  • #11
First of all, I wanted to thank every contributor for helping me to understand that tricky aspect of elevator hoists! :cool:
I failed to write something in the proper way:

"the hoist will only "pull" or "loosen", but since it can't "push" on a rope, the side of the teeth being in contact does not change"

I should have specified "...does not change with the direction of motion, only with load, the tipping point being whether the weight of (car+load) exceeds or not the weight of the counterweight", but I guess it was clear enough anyway.

Now cooking my next thread... :oldbiggrin:
 
  • #12
If the counterweight is on the same rope, but the other end of it, it would be possible for the teeth to be meshing on opposite faces, and there'd still always be pull on the elevator rope, though it really makes no difference as to where the energy ends up
 
  • #13
Thanks for a very interesting thread. Personally, I'm still trying to work out exactly which energy goes where and when.

What I wanted to add, is a comment about irreversible worm drives. I thought this was the case, until a few years ago I met a worm drive that seemed to reverse fairly easily. It was a small unit with nylon gears and, I think, silicone grease lubrication. Unfortunately I don't have it to hand to measure. But before posting a comment, I thought I ought to do a bit of research and came across this brief article, Self-locking Worm Gears, which may interest you and makes any further comment by me redundant.
 
  • #14
Rx7man said:
though it really makes no difference as to where the energy ends up

We are not at odds here at all.
Both places, gear face due to the changing friction and at the motor due to the changing load.
 
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  • #15
@Merlin3189 That article basically says exactly what I said... and the numbers I pulled out of my arse are even darned closed to the example they use... I wasn't going to complicate things with static and dynamic friction coefficients, but it certainly does apply.

From that article...
Consider a worm gear speed reducer with a lead angle of 5 deg and a static coefficient of friction of 0.13. The arctan of the coefficient of friction is 7.4 deg, and is the friction angle. Because the friction angle is larger than the lead angle, a worm gear with a 5-deg lead angle is considered statically self-locking. If, however, this self-locking reducer is subjected to shock and vibration, the friction coefficient between worm and gear may suddenly drop. If it drops to 0.08, then the friction angle (arctan of 0.08) drops to 4.6 deg, which is now less than the lead angle of 5 deg. During the time that the friction coefficient is 0.08, the gears are no longer self-locking and back-driving can occur. Once started, backdriving usually continues because the friction coefficient decreases with increasing speed.

Also in that article, it says to never rely on the self locking of a worm gear for critical applications and to always add a brake.. hence why the elevator has a brake.. Also, the brake will positively stop the elevator at a given point rather than waiting for the motor to spin down.
 
  • #16
@Merlin3189 Thank you for the very interesting article!
As I said in the opening of this thread, worm gear drives are normally unidirectional, but as that property derives just from the "exacerbation" of friction effects due to the gear ratio, that can make backdrive virtually, or practically, impossible. But it also means that a worm gear drive can be designed so as backdrive is actually possible. It's just a matter of numbers. I fully agree with the statement quoted in the article:
It is usually impractical to design irreversible worm gearing with any security. If irreversibility is desired, it is recommended that some form of brake be employed
And, without a mechanical brake, stopping and keeping the car exactly aligned to the floor would be really difficult (even with electronic drives).
Once started, backdriving usually continues because the friction coefficient decreases with increasing speed
That phenomenon is known (in English literature) as "friction of rest". In my language, we use a term that I find more self-explicative: translated to English, it would be "friction of first separation". It renders the idea of the friction suddenly dropping as the two surfaces start to move one respect to the other. The first decrease in friction is (usually) much larger than further decreases as speed increases.

One good example of worm gear designed to work in both ways can be found in the Torsen differential (image: www.arstechnica.de)

torsena_ani.gif
 

1. How is energy dissipated when a lift stops?

When a lift stops, the kinetic energy of the lift and its passengers is converted into heat energy through a process called friction. This occurs as the lift's brakes are applied and the wheels or cables of the lift rub against each other, creating resistance and dissipating the energy.

2. Why is energy dissipation important when a lift stops?

Energy dissipation is important when a lift stops because it prevents sudden and potentially dangerous impacts. Without proper energy dissipation, the lift and its passengers could experience a sudden jolt or even a crash landing, resulting in injuries.

3. How does the design of a lift impact energy dissipation when it stops?

The design of a lift can impact energy dissipation in several ways. A lift with more efficient braking systems or shock-absorbing components will be able to dissipate energy more effectively. The weight and speed of the lift can also affect energy dissipation, as heavier and faster lifts will require more energy to be dissipated when stopping.

4. Are there any safety measures in place to ensure energy dissipation when a lift stops?

Yes, there are several safety measures in place to ensure proper energy dissipation when a lift stops. These include regular maintenance and inspections of the lift's braking systems, as well as emergency brakes and sensors that can detect abnormalities and stop the lift if necessary.

5. Is energy dissipation when a lift stops a concern for energy efficiency?

Yes, energy dissipation when a lift stops is a concern for energy efficiency. The more energy that is dissipated when a lift stops, the more energy will need to be used to power the lift and bring it back up to speed. Therefore, it is important for lifts to have efficient energy dissipation systems in order to save energy and reduce operating costs.

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