Reversing direction on a spinning cylinder of water

[billinr]

I have an application (clothes washer) which requires a drive design such that:

The drum contains 170 lbs of water and clothes, spinning at 95 rpm. The drum spins at this speed for 1 second, then decelerates for 0.25 seconds and reverses direction.

This process repeats for 30 minutes.

My problem is that I haven't a clue as to how to construct the torque curve to show the deceleration and direction change, and my task is to design the drive belt.

The drive pulley for this drum is aluminum, and is 234mm in diameter.

Could anyone assisst in providing the torque values seen at the pulley?

Thank you

 

[Bob S]

This is a very interesting problem. Because the force on the water includes both gravity and the centrifugal force, the moment of inertia depends on the rpm of the drum. Another interesting parameter is that the drum with water is not a rigid body; the water does not accelerate and decelerate as fast as the drum itself. Maybe this is negligible when the drum is full of laundry. We are missing one critical parameter: the diameter of the drum.
Bob S

 

[billinr]

Apologies for leaving out that piece of information.

The drum diameter is 563mm.
There are baffels within the drum, so I would probably increase the torque requirement (maybe 10%?).
The drum and pulley share the same axis.

Thank you

 

[Bob S]

I forgot to ask: Is the drum rotating about a vertical or horizontal axis? And about what % of drum volume is filled with water?
Bob S

 

[billinr]

The drum axis is vertical.
At max load (stated), drum is approximately 80% full.

 

[billinr]

Perhaps I did not read you question correctly.
The drum holds 90.6L of water. The remainder of the weight is clothing.

 

[Bob S]

Hi billinr
The peak angular velocity is about ω = 9.95 radians per second, and the peak centrifugal force at the outside radius is about 27.7 m/sec² (~2.82 g’s). The moment of inertia of a full load (using 80 Kg load), including an allowance for the washer drum, is about I = 3.9 Kg-m². The total stored energy at peak RPM is about ½ I ω² = 193 joules, so the average power required during the 0.25-second acceleration phase is 772 watts. The average torque during the 0.25 sec acceleration cycle is L = Idω/dt = I ω/0.25 sec = 155 Newton-meters. If the torque is constant during the acceleration phase, the peak power is roughly P= ωL = 1540 watts.

This required repetitive torque over the complete acceleration cycle cannot be handled by an induction motor. Furthermore, the deceleration phase will require extracting 193 joules, prior to acceleration in the opposite direction. This extracted energy probably should not be absorbed by the motor itself, but is best handled by a brushless dc (BLDC) motor (with a permanent magnet rotor and a Hall-Effect-sensor-switched stator coil), with a ~20:1 RPM reduction ratio. The BLDC motor can operate as part of a kinetic energy recovery system and transfer the extracted energy somewhere else as electrical energy until the acceleration phase.

There are two difficult problems. Kinetic or electrical energy has to be repetitively extracted and stored somewhere for ~ ½ second. And the energy recovery motor probably should have a ~20:1 speed reduction ratio (with friction losses).

I hope this helps.

Bob S

 

[billinr]

Bob S.

Thank you for all of that information! That will be a big help in finishing this design.
While I really cannot speak to the motor design, I can see where all of that kinetic energy has to go somewhere... conversations I have had tell tales of self-destructing mounts, broken belts, etc.
That's what I have to fix next! At least now I know why...

Thanks again. I really appreciate all of your help.

 

[Bob S]

If I were designing this, I would probably use a brushless dc permanent magnet motor in the ~ 1 to 2 HP range, together with IGBTs (insulated gate bipolar transistors), a battery capable of storing ~ 300 joules, and a microprocessor to control the torque, going up and going down. I would talk to Bodine Electric about brushless dc gearmotors. See
http://www.bodine-electric.com/

ps. The engine in my car can produce only about 200 Newton meters (Nm) of torque. I normally use about 30 Nm to drive at 30 mph. 30 Nm x 2000 RPM = 6300 watts = 8.4 HP.
Bob S