Motor Power Ratings and Motor Upgrades

First you need to calculate you power requirement as accurately as possible. If the load is indeed 230 kW then your 220 kW installed power would be adequate with a very small overload and well within the motor capability.

Apart from that I am not sure what formula you are looking for as the 110 kW rated motor will give you 110 kW. Safety factor and drive losses are up to you with a minimum of 10% for a conveyor system I would guess.

DEAR:-

ALL YOU HAVE TO DO IS JUST CHECK THE AMPARES ON THE CONVEYOR DURING RUNNING WITH LOAD IF U WANT TO GET THE POWER OF THE MOTOR THAT U WILL HAVE TO BUY RESPECTIVELY TO YOUR REQUIRMENT SO P=V*I. AS U R SAYING THAT THE LOAD OF UR CONVEYOR IS ABOUT 230KW THEN GET THE DRIVE N MOTOR OF 250 KW COZ IN CASE OF OVER LOAD IT WILL NOT DAMAge any thing okz so thats so simple

Just keep it simple and replace it with a 3to 5 HP 220V motor.

If I am not mistaken a 3-5 HP motor might just be a tad on the small side. Let us assume that everything is 100% efficient and if I remember correctly 1 HP is 746 watts. If your load requires 250 KW allowing for a margin for safety.

Doing the math

HP = 250 KW / 746 Watts

= 250,000/746

= 335.12

If I am wrong please forgive me, if I am correct apart from assuming 100% efficiency please let me know.

Thanx - Paul

WOW...must be one LARGE conveyor. (lignite coal transfer?) From my "shade tree" for a 220 KW motor, 2 x 200 hp? (I'd expect to see 458 amps on a 480 3 phase line. A whole lot of ponies. If the system is operational, I'd look at the incoming voltage for a good power factor & voltage while under load and note the fully loaded current. Compare these values with the motor nameplate for full load amperage. Ideally (for me) the amperage to the motor should run (while loaded) slightly below the nameplate amperage. If the motor draws excess current (above nameplate amperage) note the ratio. Up-sizing the rated horsepower by the same ratio (+ a tad) should solve the motor longevity.

(only 999 between a watt and a kilowatt)

Thank you all for the information. Maybe I was not entire clear with my question. What I'm after is 220 KW is delievered by 2 x 110KW motor. Would it be correct to assume that a 2 x 120KW motor will deliever 240KW?? And yes, I've not taken the losses and efficiency into consideration but I'm sure I can easily up the power with bigger motor should the need be required. Thanks

yes you are absolutely right. 2 X120kW will give you a total of 240kW output power, but bare in mind that do not underload the motor because motor efficiency will drop if the motor is operating underload

Motor "ratings" are the MECHANICAL power output ratings. So yes, if you have determined that the load needs 240kW of MECHANICAL power, then 2 x 120kW motor is series will run it, minimally. There are mechanical power losses in whatever your motors are using to connect to the conveyor belts, i.e. V belts or clutches, cou8plings etc. that must be considered. I would not cut it that close. the comment about lowering efficiency if the motor is not fully loaded is valid, but a bit miscontrued. Mdern motor peak efficiency tends to not drop significantly until you run at below 60-70% of full load. Most motor mfrs will provide you with an efficiency curve that will show you that.

Here is an example from Weg, notice that the efficiency remains stable down to 60% load, and the added losses don't really become too severe until 40% load:

But if you undersize again and the motor stalls, the efficiency is ZERO!