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There is a plethora of information on motor sizing on the internet discussing the rated voltage, current, torque and power of the motor, and how these parameters should be selected based on the load the motor's driving. Some articles went deeper discussing the moment and the torque. Then, HP = (torque x speed)/5252.

But how often is all this information available? If you're sizing a motor for a conveyor in an industrial plant, what other data other than the voltage (likely 480V in north america) and the mass of the load do you have readily available. Companies that manufacture conveyor systems probably have software tools that calculate the motor size for them.

But for someone who doesn't have this software, how does one go about determining what size motor they need to drive a 6000lbs load over a conveyor of length 20 feet. Let's assume 1800RPM since that's what most conveyor motors at our plant are, and 480V 3phase across the line start.

I guess I could start with calculating the torque required: torque = force x radius. Is the radius in question that of the conveyor roll? And force = mass x acceleration. Mass = 6000lbs. Acceleration = (1800 - 0)/t. Assuming starting from rest and going to full rated speed. Not sure what "t" is since the motor is direct on line start (and not controlled by a VFD so there is no ramp up time)

I want to get a good grasp on the subject so a detailed answer would be much appreciated.

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    \$\begingroup\$ No matter how you look at it there will be a ramp up time of some sort. You cannot go from 0 RPM to 1800 RPM in zero time. Also remember that for optimal behavior you want the rotational inertial of the motor to be equal to the rotational inertia of the load where the motor shaft is joined to the load. \$\endgroup\$ – Michael Karas Feb 14 at 3:40
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    \$\begingroup\$ If you're sizing a motor for a conveyor in an industrial plant, you call motor vendors, and they will rush all necessary information to you, maybe including a trip of their application engineer to your location, who will guide and teach you how to use their motors to their best performance. \$\endgroup\$ – Ale..chenski Feb 14 at 5:07
  • \$\begingroup\$ You first understand your load - you have the equipment, so work out starting load / torque, running load etc etc speed required etc Only then can you start looking for the solution ie which motor... \$\endgroup\$ – Solar Mike Feb 14 at 8:22
  • \$\begingroup\$ I'm voting to close this question as off-topic because it is a problem of mechanical rather than electrical engineering \$\endgroup\$ – Chris Stratton Feb 14 at 15:33
  • \$\begingroup\$ I'd say it's both. You need electrical knowledge for knowing the FLA and voltage of the motor, and mechanical knowledge for knowing the torque requirements, etc. \$\endgroup\$ – vasiqshair Feb 16 at 15:32
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The torque doesn't have anything to do with accelerating the motor, it's much more about accelerating the load. How fast is your belt moving?

You say you have 6000 lb — presumably that's the total load on the 20 feet of belt at any given time. Suppose your belt is moving at 2 feet/second — that means that any given bit of load spends 10 seconds on the belt, and that you're adding 600 lb/second to what the belt is carrying. THAT is the mass that needs to be accelerated from 0 to 2 fps, requiring torque from the motor to do so.

Multiply those two numbers together, and you get 1200 ft-lb/sec2, which is a direct measurement of the force on the belt. Multiply by the radius of the drive roller in order to get the torque, and from there you can figure out the gearing to match it to motor torque and speed (and therefore power).

There are other power losses associated with moving the belt, so incorporate a design margin into your final selection. If your conveyor is not level, then you also need to add the force required to lift the load against gravity to your calculation.

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It's a mechanical function, not an electrical one. There are mechanical factors you need to know that have nothing to do with the electrical requirements. Once you KNOW the HP required for the task, then the problem shifts to being electrical in nature.

For simple inclined Conveyors: HP=((P x B)+(P+M)x F x S)/33,000 Where: B: Sine of angle of incline F: Coefficient of friction HP: Horsepower M: Overall Belt Weight P: Product weight (lbs.) S: Conveyor Speed-Feet per minute

None of that is electrical...

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