Power requirement

[brettball]

Please refer to the attached .pdf sketch.
I have a component of mass 7.5 (74730N) Tonne mounted in a frame.
The frame is connected to a shaft mounted on bearings.
The shaft carries a large sprocket which is driven by a transmission chain via a small sprocket connected to the motor.
The component/shaft is required to rotate at 0.6 rpm (0.01rps).
My question is, what is the power requirement of the motor?
Although this is not my forte, I have made an attempt and have come up with 4.123Kw.

 

[desertfox]

Hi brettball

Any chance you can give the maths that led you to the 4kw motor, from my calculations you need nothing anywhere near that size unless there is something I don't know.
I calculated the moment of inertia of the 7 Tonne disk to be 567kgm^2 however I don't know what the inertia of your supporting frame, friction in your drive etc might be.
Just looking at your sketch from a mechanical point of view, it looks like a lot of mass to be hanging off what looks like a relatively thin shaft.I assume you have done or will do all the calculations to support your design.

regards

desertfox

 

[tygerdawg]

You're missing some vital information. Make sure your dynamic analysis contains everything:
(1) acceleration & deceleration times....this helps determine peak torque requirements. Just turning at constant speed requires lower torque.
(2) Is your mass moment of inertia modelling a cylinder? It may not be.
(3) Total Torque will be (acceleration torque) + (deceleration torque) + (torque to turn an inertia in steady state) + (torque to move an inertia) + (torque to overcome friction) + (this torque) + (that torque) + (the other torque). Identify them and add them all up.
(4) Rotational and translational inertia of ALL masses: sprockets, chains, bearings....every mass you must move. This feeds into the torque required from the motor output shaft.

So then you've finally accounted for all the torque required of the motor. POWER will be the amount of torque required over a period of time.

From my little Inch-Standard unit cheat sheet: Horsepower = (Torque X RPM)/(63025 X efficiency)

KW = HP X 0.75

Then oversize it for factor of safety, robustness, number of starts/stops per hour, heating effects, etc. A little extra torque (and the bigger windings, frames, and bearings to support it) don't really cost that much extra. And your motor will only use as much juice as is necessary.

TygerDawg
Blue Technik LLC
Virtuoso Robotics Engineering

http://www.bluetechnik.com

 

[brettball]

Thanks desertfox.
This is not really my forte and my mechanics knowledge is limited. It is a project I have been asked to get involved with. I am not too proud to show my inadequacies so my theory and calculations are as follows.I treated it as a simple lever where to turn the mass 74730N(7.5t) on radius 885(1770/2) requires at large sprocket radius 314(628/2) a force of 210624N(835/314* 74730). If the large sprocket rotates at 0.01rps(0.6rpm) then the small sprocket rotates at 0.0419rps(628/150*.01). Therefore in 1s the small sprocket moves 0.0197M (.15*pi*.0419). I reckon force on the small sprocket is same as on the large one ie.210624N. So power at small sprocket=(210624*0.0197)Nm/s = 4149W.
The drawing is only schematic and some of my colleagues are
working on the frame design.
Please don't hesitate to embarrass me.
Cheers,
Brett.

 

[imcjoek]

You may be working this a bit backwards.

Your motor will likely want to run ~1750rpm. If the maximum speed of your drum is to be 0.6rpm, then you can get away with 2900:1 (!) reduction. Your pulley/chain setup only gives you 628/150 ~= 4.2.

So you probably want a gear reducer somewhere around 690:1 in combination with your chain drive (be careful of the order when sizing!).

Looking at a 1HP Marathon BlackMax vector motor as an example, it has a F.L. Torque of 3 ft-lb (and tons more at "stall"). So 3ft-lb * 2900 = enough to turn your house upside down.

You need an idea of:
1) inertia and desired acceleration rates
2) bearing drag
3) any other significant drag

Then you can see if this torque is enough.

 

[desertfox]

hi brettball

I assumed the centre of gravity of the disc your rotating to be on the centre line of rotation and not offset.
Best way to describe it would be your 7.5 Tonne mass spinning like a coin on a table and I worked out the inertia from that assuming of course it is a disc

I=MR^2/4 = 7.5*10^3*1.25^2/4 = 2929.68kgm^2

Now to calculate the torque on the shaft required to start from zero rpm to 0.6 rpm I used this relationship:-

T=I(W2-W1)/t where I is 2929.68

t = time to reach 0.6rpm from
zero

W2 = final angular velocity
W1 = intial angular velocity

If I assume that we allow the motor 10 seconds to get the disc turning at 0.6 rpm you need a torque of 110Nm and a power of 7W.
I made a mistake in my first post with the moment of inertia but this one given here should be correct.
Now of course I have ignored the sprocket ratio's,supporting frame and their inertia's, friction,efficiency etc..
All I have done is calculated the torque required at the shaft to turn the disc and bring it upto speed in a given time of 10 seconds from which I have obtained the power.
Someone needs to decide how quickly the disc reaches its final speed as that will have a large affect on motor power.

regards

desertfox

 

[brettball]

Thanks,
I will re-visit taking your replies on board.
I now have a much clearer view of this.

Regards,
Brett.

 

[drawoh]

brettball,

You have a 7.5[ ]tonne disc rotating in a cantilevered frame. You do realize that by the time you have a frame that is structurally rigid and strong enough, your rotated mass will be way over 7.5[ ]tonnes. Bearings strong enough to handle that load could affect your power requirements.

If something breaks, who or what will the 7.5+[ ]tonnes land on? Even a broken drive chain could be exciting. This is not a good project to learn mechanical design on. Is there not someone at your site who knows this stuff?

JHG

 

[TheTick]

Main points:

[ol][li]The only power you need to keep the load at speed is what is necessary to overcome friction.[/li]
[li]Additional power is required to start (and stop?) the load. More power means more acceleration. Power requirement determined by how fast you need load to start or stop.[/li][/ol]

 

[chicopee]

Drawoh has an excellent point about the bearings. The cantilevered load imposes a high load on the bearings and the shaft.

 

[brettball]

We have spent a lot of time on the structural design ie.
frame and bearing capacity which is more in line with our expertise and have had it verified by our local university
facility via FE analysis. Just trying to get our heads around the motor and drive requirements.
Really appreciate your comments.

Regards,
Brett.