Induction motors: magnetizing currents, output torque...etc

[SSP1987]

Hi ,
Few basic questions from a non electrical guy like me.
I am trying to select motors for dryers in Pharma industry. ( viscous paste like / powder mixing applications )
Now the no load current drawn by most motors is around 50 percent of full load current. Does it mean 50% of motor torque is consumed just to keep only the motor running? Does it depend on the load or is it independent of load on motor?
And does using a VFD influence this value ?
For eg: If an application uses a 5hp 1440 rpm motor, ( full load motor torque = 24Nm ) then if the torque needed to run only the motor is 50% = 12 Nm, then is only 12 Nm available to run the actual machine? Conversely, if 24 Nm is needed on the driving shaft , then should a 10 hp motor full load torque= (49 Nm) or bigger size be selected after considering this 50 percent torque loss?
I am assuming the values given in standard motor catalogs is inclusive of all losses. Or is it otherwise ?
Any help is appreciated.

 

[jpts]

No-load current of induction machine is mostly needed to produce the magnetic field, and small portion of it to cover the losses inside the motor (stator/rotor winding, stator core losses and windage and friction losses). You don't need to take these into account when matching the load. If the machine nameplate has full-load torque of 24 Nm, this is what you get from the shaft when applying rated current

 

[SSP1987]

Ok.
Meaning , the no load current even if to the tune of 50 percent of Full load current is to be disregarded when selecting and sizing a motor ?
Should it be safely assumed that entire torque as per motor catalog is available for use at motor output shaft ?
This torque times gear ratio and gear efficiency will be available at gearbox outlet ? Does a significant torque go into running only the motor itself ?

 

[SSP1987]

I will attach a sketch shortly to easily view the problem at hand.

 

[LionelHutz]

Motor datasheet torque is always the torque available at the output shaft.

 

[waross]

You are overthinking this.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[cranky108]

Most motors can be turned by hand, meaning the friction losses are not that much. Likely just the bearings. Other losses are windage, or how much air it moves to keep it cool. And core losses which generally should not be that great.
I recall numbers of above 65% efficiency in total when I was in school. Newer motors are likely more efficient.

Magnetizing current, is not real power losses, and will not show up on your electric bill as a power consumed. It will show up as a capacity charge, but can be compensated to fix that. Or a VFD can do that.

 

[BrianPetersen]

To clarify something, most of the magnetising current will be out of phase, indicated by a power factor measurement showing a low power factor. Ideally all the magnetising current is 90-degrees out of phase and the power factor will be zero if no mechanical power is being drawn (which means no "real" electrical power will be drawn), but in reality there are some losses due to electrical resistance, windage, etc.