Servo KW??

I want to calculate the KW of this servo motor, the serial number is MSK061C-0600 BOSCH REXROTH, the catalogue doesn't supply the KW information, but it does supply the rated current and the maximum current. The rated current is 7.7 amps and the max current is 34.7 amps.

I've used KW = 1.732 x 380V x 7.7amps.

I don't know the PF or efficiency, I just need an approx KW value as i'm using this value to work out the heat loading from the motor, the KW value i came up with is 5KW, but this seems large for a servo motor.

Thanks for any help

Sparky

The first thing you will need to do is get your hands on a torque/speed curve for the motor. You will propbably see that the rated torque falls off rather quickly as speed increases. This is due to internal eddy current losses in the motor as the rotor speed increases. The 7.7A current value you listed is probably stall current, which is the current required to bring the motor to it's thermal limit with a locked rotor. This is the highest current level you will be able to apply to the motor without exceeding it's thermal limit. In fact, if you follow the motor curve far enough to the right you will see a point where the maximum continuous current goes to zero.

However, even this information won't give you what you want if you are looking for heat loading. Your equation assumes that all the energy going to the motor is converted to heat. If this were the case the motor shaft wouldn't turn. You are looking for the motor losses. As I stated above, this can be a bit difficult to find due to internal motor losses that have nothing to do with current being applied to the motor.

Your best bet may be to contact Indramat about this. They should be able to tell you the thermal capacity of the motor. You can then use this as your worst case heat generation.

Keith

Using 1.732 assumes the servo is using 3 phase but it could be 2,3,4, or 5 phase (maybe more, these I have dealt with). I am not qualified enough to explain it but as kamenges mentions it is more involved. The best idea may be to contact Bosch/Indramat and see if they do not already have the information you need.

Here is a Motion Control Handbook offered by Danaher: http://www.danahermotion.com/education/learn_about_mc/mchandbook/

What do you mean by heat loading? What specifically are you trying to size?
The current rating is related to the DC bus where the PWM is generated, not the incoming 3 phase.
You can calculate mechanical power from the torque and the speed. You can find the related calculations in the Indradrive System manual.

I require the heat loading of all the motors on a particular production line, because we are looking at installing localised air conditioning and i have to build up a heat map of the production line to give to the A/C engineers.

I've been intouch with Bosch Rexroth and the formula they gave me was (speed x torque) / 9550 to give me a KW value, but as kamenges said this assumes that all the energy is converted to heat, the information that i give the A/C guys will be worse case scenario and then some.

thanks for the help.

I talked to my Rexroth rep in this area and he ran this back through the guys at Rexroth US. They came back with this:

From Indramat Rep:

Here is the calculation provided by the factory:

For zero speed, the only applicable losses will be I^2R.

The basic formula will be:

I^2R * (rt(3) / 2)

(I-squared x R x root(3) / 2)

The current is rms. So, you'll have to convert for the MHD, MKE or MKD specs (peak of waveform given x .707 Vs RMS (shown in the specifications on new series MSK motors)).

The resistance we specify is phase-to-phase. That's where the divide by 2 factor comes from.

Root(3) to get 3-phase.

I checked the formula against one of our motors, MSK060C-0300 from the DVD manual and got 69.2 watts. This seems correct for this size motor.

Click to expand...

I should have figured this intuitively but I'm getting sloppy. Stall current will bring the motor to rated max temperature at zero speed. All the energy going into the motor is being turned into heat. Stall is just the easiest place to calculate this value since no other losses are present.

Keith