Dear Shri Patil,

Your query is not clear. You have not mentioned the name of the equipment. You are just saying to drive 5000 kgs at 500 rpm. Technically these do not convey any factual information.

Regards,

Ishwar G Mulani.

Author of Book : Engineering Science and Application Design for Belt Conveyors.

Author of Book : Belt Feeder Design and Hopper Bin Silo

Advisor / Consultant for Bulk Material Handling System & Issues.

Email : parimul@pn2.vsnl.net.in

Tel.: 0091 (0)20 25882916 ■

Dear Shri Patil,

You should be able to obtain calculation methods and charts from the websites of the major motor suppliers. Check the sites for Reliance, SEW, Marathon, et cetera.

These will help you to determine your answers and to ensure that all considerations are taken into account.

Regards, ■

Your probelm is ill defined.

Work is defined by force times distance. Power is work per unit of time. kW = (Pull x Velocity) / (drive efficiency); where Pull is in kN (force) and Velocity is in m/s (distance per unit of time)

So, we can assume you are defining 5000 kg x k = Pull x conversion factor "k" of kg to kN.

Drive efficieny according to your design < 100% or >1.00

Since you are posting this in the belt conveyor forum, your motor speed must be referenced to the belt surface speed. Therefore, you need to provide this extra piece to the puzzle. One such method to set the pulley diameter and reducer ratio for you 500 rpm as reference to the belt speed. ■