Calculations - some fundamentals

[kp516]

Hi,
that's my first post here.First of all – I do not want to ask someone to calculate something for me.

I did my best (I hope correctly) – can you briefly check it and answer my additional questions? I am especially interested in these questions – I always have some doubts during calculations.

Ok, so I need to roughly calculate carousel storage motor parameters.

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I am assuming that only 5 of boxes on one side will be loaded with 200 kg. This is the worst case.

Imagine I want to lift the lowest one up to 6 meters in let’s say 30 seconds.

W= mgh= (5*200 kg) * 10 m/s^2 * 6 m = 60000J = 60 kJ

P= W/t = 60000 J/30 s = 2000 W = 2 kW <- required motor power1) There is a gear in my sketch – does this gear has any influence to formulas for power used above?

2) By the way – if all boxes will be loaded to max can it be calculated like a lift with counterweight?

Can anyone provide more information about “counterweight effect” which “helps” motor?

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Now let’s calculate a torque_motor and speed – it seems that tutorial example is the same:

torque_motor / radius_1 = torque_2 / radius_2

torque_2 / radius_2 = weight -> torque_2 = weight * radius_2

so:

torque_motor / radius_1 = (weight * radius_2) / radius_2 -> torque_motor = weight * radius_1

so torque_motor = 10000N * 0,5m (more or less) = 5000 Nm

3) This is a minimum of course – how much additional percentage do you suggest to add considering belt efficiency etc.?

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speed (0,2m/sec)

rps_motor * radius_1 = rps_gear_2 * radius_2

rps_gear2 * 2 * pi * radius_2 = velocity_weight -> rps_gear2 = velocity_weight / (2 * pi * radius_2)

so

rps_motor = ((velocity_weight / (2 * pi * radius_2)) * radius_2) / radius_1

rps_motor = velocity_weight / (2 * pi * radius_1)4) BUT NOW PLEASE NOTICE that there is no radius_2 in both of these equations (torque and speed)! So what’s the reason of this gear (bigger pulley)?... maybe equations are in the fact are not ok?
5) By the way – if torque and speed are given now, isn’t it natural that there must be a motor power which is combined with this parameters – on the other hand – was it required to determine power at the beginning of this post?

6) Any other tips for this application?Thanks for anyone who will support me a little bit.

 

[Simon Bridge]

Hi,
that's my first post here.

Welcome to PF.

1) There is a gear in my sketch – does this gear has any influence to formulas for power used above?

Yes and no - an ideal gear won't affect the energy requirements... it does have a moment of inertia though. Anything that turns will store energy as a flywheel.

2) By the way – if all boxes will be loaded to max can it be calculated like a lift with counterweight?

Yes.

Can anyone provide more information about “counterweight effect” which “helps” motor?

In terms of your energy calculation - an equal counterweight means that when one weight goes up h meters another one goes down h meters and the total energy change is zero. i.e. the counterweight makes the energy requirement zero.

It still takes energy to get the thing turning, but, absent friction, it takes no energy to keep it moving.
Similarly - once the unbalanced load is lifted past the top, it falls down by itself: giving you energy back maybe.

3) This is a minimum of course – how much additional percentage do you suggest to add considering belt efficiency etc.?

No way to tell - this is why engineers get paid the big bucks ;)
These machines are always over-engineered. What you do is select from the available commercial motors whichever has at least the needed power, and is within your budget.

4) BUT NOW PLEASE NOTICE that there is no radius_2 in both of these equations (torque and speed)! So what’s the reason of this gear (bigger pulley)?... maybe equations are in the fact are not ok?

Seems fine - the combination of two equations that do have "radius2" in them allows that variable to be eliminated by algebra.

5) By the way – if torque and speed are given now, isn’t it natural that there must be a motor power which is combined with this parameters – on the other hand – was it required to determine power at the beginning of this post?

Yes, to use the equation to find speed you need torque ... similarly you can use the equation to find the torque needed.

 

[Merlin3189]

I am assuming that only 5 of boxes on one side will be loaded with 200 kg. This is the worst case.

Imagine I want to lift the lowest one up to 6 meters in let’s say 30 seconds.

W= mgh= (5*200 kg) * 10 m/s^2 * 6 m = 60000J = 60 kJ

P= W/t = 60000 J/30 s = 2000 W = 2 kW <- required motor power

I think you are overestimating the average power here.
Although 5 boxes on one side is the max load, it does not apply for the whole 6m lift. Once the boxes have moved up about 1.2m, the top one is at the top, then starts to fall. For the next 1.2m or so, you are lifting 3 boxes (4 up and 1 down), then for 1.2m lifting one box (3 up 2 down) and after that you get a free ride for the next 5.4m. (that's very rough because there is a not a sudden change, but a gradual one as boxes move across at the top and bottom. I didn't try to be more precise without the centre to centre distance and diameter of the wheels.)
You still need your peak torque and peak power, but the average power is much lower. I'd say the average load is under 300 kg, so your average power under 600 W. (Actually I think load is nearer 200 than 300kg, so that could be your friction allowance!)

4) BUT NOW PLEASE NOTICE that there is no radius_2 in both of these equations (torque and speed)! So what’s the reason of this gear (bigger pulley)?.

Simon is quite right about the algebra of course, but it worried me also: I would not expect that to happen. The reason is that you have a top pulley for the drive exactly the same size as the circle round which the boxes are moving. Because of this, the speed of the drive belt is exactly the same as the belt carrying the boxes and your calculation simply gives the speed of the drive belt - obviously dependent only on the drive pulley and motor rpm.

This gives a very low rotational speed for pulley 1, hence your gearbox on the motor. You may find you can specify the motor & gearbox as a unit - output speed and torque, average and peak power.

Looking at the RH diagram pulley 2 looks smaller than in the LH diagram.
Either way (unless there is some further gearing at the top pulley) the tension in the drive belt is at least the tension in the box carrier, say peak 1000kg (x g Newtons.) I'm no expert on drive belts: it just sounded a lot to me. Maybe you need to consider a chain? Apart from the strength, any slippage of a rubber belt at that tension could generate a lot of heat.

 

[kp516]

thanks, that was one of the best answers I get when asking about that question!

 

[kp516]

hmmm

just one more concern...

if there is a line around a wheel at the top, with two weights = 10 N, then what is the tension in this line?... 10 or 20 N?...

in my opinion it shall be 20N - am I right?

 

[jack action]

The torque required to turn this carousel is the friction torque only at constant speed. You have to add the inertia effect when accelerating.

Friction torque is the losses from the bearings + the losses from the chain/belt under tension (ask the manufacturer to learn how to estimate it. It should be similar to tension times some coefficient of friction times the pulley diameter).

The inertia effect (which also affect chain/belt tension) will be (I + mr²)α, where I is the mass moment of inertia of the pulleys (kg.m²), m is the combined mass of the chain/belt & the boxes (all of them) (kg), r is the radius of the pulley (m) and α is the angular acceleration of the pulley (rad/s²).

The power (W) needed will be the torque required (N.m) times the rpm (rad/s).

if there is a line around a wheel at the top, with two weights = 10 N, then what is the tension in this line?... 10 or 20 N?...

in my opinion it shall be 20N - am I right?

To better see it, unroll the line from the wheel, such that it forms a straight line. You should see that the answer is 10 N.