Power requirement to drive a tumble dryer

Well, we need a ton more information. Where are you, in the US most electric motors run 3600 rpm, (60Hz) and most anywhere else it's 3,000rpm, (50 Hz.) What size is your pully? What size in your drum?

My suggestion is to figure out what you need to acheive your 30 rpm, 120 to 1 reduction or 100 to 1? Then attack the problem of horsepower required for the 60 kg.

Good luck..

Thank you for the suggestion. The different parameters such as sizes of pulley, reduction ratio in the gear box, shaft diameter on drum, weight of drum can all be known and calculated. What other factors, besides the weight of the garments, we need to consider to rotate the drum in order to provide a torque on trhe drum.

I'd say that there's not too much except the weight of the load, the size of the drum, the resistance (atrict) of the drum, and the efficiency of your gearbox and transmission untill your motor shaft.

In the point of maximum load, you'd be dealing with your drum radius x 60 kg x 9.81 = "X" Nm of torque, at 30 rpm = Y power required at the device.

Multiply by the system losses coeficients, check some torque losses over time (bearing wear, ...) translate into a safety margin, and you got it.

1. Initially wet>- total load may be 90Kg
2. Initially sticky>- might go uphill as a big packet and Drop and jerk and up and drop>>> time when you need maximum power. Design for this worst phase.
3. Towards full dry, hardly any up>down>jerk>fall torque. No concern here -when #2 is already taken care of.

Worst Phase Power Requirement:

• 30 rpm-- so, half turn=1second-- through which 90Kg rises 1 diameter high
• We assume diameter 0.9m
• Power=90x0.9Kg-m/second. Which will be almost 720 watt.=nearly 1HP
• Most Induction Motors can take 2Xoverload torque.
• You can get away with 1/2 HP.(Notify"Spin well before tumbling")