Power Needed to Rotate a Mass to a Speed in a Specified Time 5

[edison123]

What will be the power needed to rotate a uniform cylindrical rotor
weighing 1600 kgs
dia 600 mm
speed 3300 RPM
5 minutes acceleration time from zero speed to 3300 RPM

Any particular equation to calculate this? Will acceleration time play a role in power needed?

The cylindrical rotor will be rotating horizontally in free air on two bearings.

 

[Denial]

Your questions are fairly basic. I will reply with a suggestion rather than an answer. Rather than thinking about a problem involving rotational motion, think about a mathematically identical problem involving linear motion. Linear speed instead of rotational speed. Mass instead of rotational inertia. Etc.

 

[IRstuff]

You can calculate your final rotational kinetic energy https://en.wikipedia.org/wiki/Rotational_energy and then back out what the transferred power was.

 

[JStephen]

As stated, this sounds like a homework problem, which isn't really the point of the forums.
Based on the information given, it's entirely an energy problem: Calculate kinetic energy stored in the rotating mass and divide by time to get average power.
In the real world, you'd need to be looking at electric motor characteristics in great details. Watch for overheating, etc.
Or maybe diesel engines, with transmissions, etc.
Also ponder how you balance it and what happens if a mount breaks. And watch out for the flywheel-disintegrating issue.

 

[MintJulep]

Everything you need is here: https://phys.libretexts.org/Bookshe...n/10.09:_Work_and_Power_for_Rotational_Motion

 

[GregLocock]

Um, Muthu's been here for quite a while.

IRstuff's approach is the right one. w=3300*2*pi/60 rad/s=345 rad/s
I=1/2*m*r^2

r=.3
m=1600
so I=72 kg m^2
KE=1/2*I*w^2
KE=4284900 J
average power to accelerate the flywheel =KE/300=14.3 kW

You'd better check that it seems high. Also, getting constant power in to accelerate something is a complex issue- in a car we'd use a CVT. You'd get a much better estimate if you defined the motor characteristic and the transmission.

 

[edison123]

Thanks, Greg.

I did do some math (had to recall some college days equations) and got exactly double of that i.e. 28.68 KW.

Wonder where I am going wrong.

Got moment inertia I = 72 kg-sq m right

Torque = Moment of inertia x Angular acceleration = 82.97 N-m

Power = torque x angular speed = 28.68 KW.

Not a student, btw.

 

[edison123]

Driving motor has enough power and speed range up to 3300 RPM and the whole set up is via a big VFD.

 

[GregLocock]

Yes, you've got twice as much because your power at t=0 is 0, and increases linearly with speed. Hence peak power (which you worked out) = 2 * average power (which was my method).

In reality it depends on the torque vs speed characteristic of your motor and transmission.

 

[IRstuff]

The power you calculated applies only to the instantaneous angular rate, i.e., only at the end of the 300 seconds. Since the power must start at zero, the average power is one half of what you calculated, corresponding to the area under the curve of a linear ramp in power.

 

[edison123]

Thanks guys. But when the load is running at 3300 RPM, the motor has to deliver that peak power of 28.68 KW, right?

 

[3DDave]

At constant speed and neglecting any other load, there is no power required. Realistically, internal power losses in the motor and windage and bearing losses, but those cannot be determined from the given assumptions.

 

[Flying Whale]

You don't technically need any power once you're at full speed since you've finished the acceleration phase. But aside from the losses mentioned by Dave (which will need constant power), there's whatever loads you have on your cylinder. The rotor can't possibly just be rotating in free air, not doing anything. If it's a propeller, you'll need power to overcome pressure and drag. If it's grinding seeds, it needs to overcome that as well. I don't know what else you may use it for, but whatever tries to slow it down, the motor has to provide power against it. I'd go for setting up a smaller scale test for estimating those forces.

 

[LittleInch]

Muthu,

you're accelerating something from zero rpm to 3300, therefore for power time is important.

Also you might be asking the wrong question and actually thinking about torque?

A lot depends on how you want this mass to accelerate? a smooth RPM flat line over time ( constant acceleration until you get to 3300) or more of a parabolic curve with greater acceleration at the beginning and less at the end. If the power was constant then the torque needs to vary over the acceleration time and would probably give you you lower power, but higher torque at the beginning and less at the end.

 

[edison123]

Littleinch

Yes, if the acceleration time is increased from 5 to 10 minutes for 0 to 3300 RPM, the power required drops to half i.e. 14.34 KW.

I am looking at a linear acceleration which the VFD will be programmed to do.

Both the drive cage induction motor and VFD are rated for 132 KW.

I will be testing this set up in 4 to 5 weeks and will let you know how it goes.

Thanks for all your helpful tips.

 

[MintJulep]

I am looking at a linear acceleration

I hope you mean "linear speed profile", not linearly increasing acceleration.

But that requires constant acceleration, which you probably don't want to do.

Basically, you want the left half of this (from: https://www.digikey.jp/en/blog/motion-control-profiles-good-better-and-best)

 

[edison123]

I am looking at 3300/600 = 5.5 rpm per sec. That's linear acceleration, right?

 

[GregLocock]

I thought you said 5 minutes, 300s?

 

[MintJulep]

3300/600 = 5.5 rpm per sec. That's linear acceleration, right?

That's a constant acceleration.

It results in a linear speed v. time profile.