Motor Sizing Made Easy

There are different ways of selecting electric motors for specific applications. Perhaps the simplest way of approaching motor selection, though, is to ascertain the mechanical or physical requirements of the job and make the electrical requirements match them. For example, if constrained by space or weight requirements, initially select a motor within those parameters. Then, try to use mechanical means (pulleys, gears, gear heads, speed reducers, and so forth) to meet the mechanical requirements.

Designers typically first settle on either an ac or dc motor or gearmotor. Gearmotors are ac or dc motors typically used for higher torque and lower rotational speed. Knowing the torque and speed requirements will help in determining if an ac or dc motor is required.

One of the mechanical limiting factors of electric motors is the bearings. Motors that use bearings will typically last longer than those using bushings. They also typically handle more perpendicular loading to the shaft (radial load), whether horizontally or vertically.

No matter how much torque the motor can generate, it will eventually hit a crossing point where either torque falls off as speed rises or the motor can only maintain a given torque by rotating more slowly. Once these torque versus speed qualities have been established, then you can play with the numbers using the aforementioned accessories.

Let us take a concrete example of a Baldor Electric Co. dc motor that generates 11,500 rpm with a 1-in.-pitch-diameter pulley. This configuration will produce a linear speed of 36,128 ipm, or 3,011 fpm, or 602 ips. The pulley size, of course, could be changed to vary the speed or torque. However, some applications might need slower motors with a gearbox. It’s a number game; as the speed requirement rises, the load capability drops, and vice versa.

Consider the example of applying this motor in a conveyor or tangential drive system. Further assume the need to spray 1 fl oz of material over an 18 × 14-in. area using a spray tip which produces 0.050 gallons/min or 0.1067 fl-oz/sec at 40 psi.

Motor selection starts by first finding the needed speed (or velocity), and torque. Next comes acceleration, determined by establishing the amount of time required for the move and then solving for shaft speed in rpm.

One determines time in this case by dividing the amount of material to be dispersed by the rate of dispersion, or 1 fl-oz/0.1067 fl-oz/sec = 9.372 sec. To determine linear velocity, divide the length of material by elapsed time, or 18 in./9.372 sec = 1.9206 ips.

In many cases, velocity is the operational requirement which dictates the motor size and/or type. Examples include the speed with which you can transfer a part from one location to another, the rate at which you can fill a container or remove material, or a dispersal rate for sprayed fluid.

To find the rotational velocity in rpm corresponding to this linear velocity, we first convert inches/minute to inches/second and then convert to revolutions. In this example the pulley diameter is 1.003 in. That gives 1.9203 ips × 60 sec/min × 1 rev/(1.003 in. × π) = 36.57 rev/min or 0.6 rev/sec.