motor power in kw

You need more data.

• Supply voltage and frequency?
• What is it connected to - a 'king big flywheel?
• Load torque at 100rpm?

380 volt 50 hz yes load is 100 rpm

But at what torque? Motor power is the product of torque and speed!

yes i want to clear this our load is 800 kg with the shaft that connected to reduction unit this shaft speed is 100 rpm and the reduction unit is 15:1 we want to change the motor with this shaft speed. so what is the motor power or what are the formulas

Yeah, fine. Look, if you can't obtain the torque then we're at cross-purposes and I can't help any further. I'll be at the bus stop on Nethermayne if you want me <unsubscribes>.

<change the motor>

Why not change the motor like for like? What's wrong with it - is it burnt out? What does the original equipment manufacturer's [OEM] manual say? Is there enough information on the details plate to obtain a replacement based upon the original serial number and a telephone call to the OEM?

With the available data it is impossible to select a motor as pointed out by others.Simple thing will be to take the name plate details and buy a motor of the rating mentioned in name plate if motor is damaged or old.Otherwise what are your reasons for replacing the motor and that too trying to calculate the rating from the beginning?

I'm more curious than anything, but need more information.

ok, 800kgs, but what is 800kgs? the weight of the "tool" you have sitting on the bearings? so the total mass of the device is 800kgs, but rotating on the bearings is 20kgs? or are you talking that you need 800kgs to get your device rotating?

Is the 800kg the tension on a cable that is connected to a spool? or if you put a 1 meter (3 foot) bar on the shaft, it takes 800kgs to move it? <- this is then calculated to be your torque..

Shaft speed? 100RPM, is this before or after the 15:1 gearbox? then the motor is either running at 100RPM or 1500RPM

If you get the value in HP, then to convert to kW its approx 754W per HP (been over 10 years since I was doing motor control...)

I think you don't understand exactly what are you asked for. See these pages. They will help you.

http://www.kelvin.es/pdfs/ParPotenciaIng.pdf

http://www.kelvin.es/producto.php?id=94

I already read the answers to your question.

Unfortunatly I still think the most of the participants answer just for joke, knowledge=0

To know the necessary power in CV, HP, kW or other, first we need to now the necessary torque to put the load in motion; of course all this depends if is a rotating or linear load.

I believe that what you want (or have) is a winch. If so, then the torque will be

T=d.L/2 where

T=torque (Nm)

d=wich drum diameter (m)

L=load (N (=kg.9,81)

Then P=(T.n)/(9550.eff) where

P=power (kW)

n=rpm

eff=efficiency

Since you have a gearbox of 15/1, and most probably is a worm gearbox, it's the efficiency must be around 80%

So, if data is:

- drum diameter 300mm

- load 800kg

- rotational speed 100rpm (whis indicates that the motor will be 4 pole (1500rpm syncronous nominal speed) and the gerabox ratio 15/1 => 100rpm in the output)

then

T=d.L/2 =(300/1000).(800.9,81)/2 = 1177,2N

Then

P=(T.n)/(9550.eff)= (1177,2.100)/(9550.0,8)= 15,4kW

The near standard motor is 15 ou 18,5kW so is a question of some lack in torque or an extra torque (if the gearbox's efficiency is correct and also de drum diameter)

Of course this approximate, because the torque depends on the diameter which depends on the levels winding cable (if empty the drum is 300mm - in the bottom - but full is 400mm, the matter is different) but the calculation is the same.

There's another formula for lifting but isn't considering the gearbox (pure theoretical torque, not the practical we need).

This is a mechanical answer; may be others but this is what I do everyday

Hope it helps

Please: What do you mean by "worm gearbox"?