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# Power kW needed on conveyor system

## Power kW needed on conveyor system

(OP)
Hi there all,

I'm trying to find the formula to calculate what electric motor kW is needed for the system.

People have told me it's around 1.5kW (2HP) but they can't give me a formula to calculate it for myself or for future use.

The total weight of the product on the conveyer will be 2 tons of steel parts, 5m in length belt it needs to travel on, the driving drum diameter is 300mm and it needs to go around 10 meters/minute.

That is all the information that I have and need a more or less kW selection on this.
Is there anyone that could possibly help me with the formula ??
It doesn't have to be very accurate/precise as we will go one size up anyway on the size of the kW, so a formula that could give me the more or less kW output would be greatly appreciated.

### RE: Power kW needed on conveyor system

power = force x velocity

assuming belt slides on a slider bed conveyor and belt weighs 100 kg and ignoring return belt roller friction and bearing friction

power ~ 0,4 friction factor for sliding belt on conveyor bed x 2.100 kg x 9,81 m/s2 x 10 m/min x 1 min/60 sec = 1.373 N-m/sec = 1,4 kW

### RE: Power kW needed on conveyor system

What about the 2 tons of product on the conveyor? 280 kW?

je suis charlie

### RE: Power kW needed on conveyor system

Not a decimal point? Oh well.

je suis charlie

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