IEC Electrical Load Study

You need to stop referring to motors in kw. A motor is rated in HP. A 10 HP motor is not a 10 kw motor.

"IEC Motor gives the Shaft Power in KW at the nameplate Efficiency." Yes but that is not the HP or the input power. You must input more power that you get out. That's efficiency.

"e.g. Nameplate : 3 phase, 10 KW motor, fla = ? ( KW / ( KV x 1.732 x pf ) ?" No. If the 10 kw is the output power then FLA = kw/(volts x 1.73 x pf x eff)

"PF = 0.70, Efficiency = 0.85 or fla = ( KW x I.732 x pf x efficiency ) ?" Yes if the 10 kw is the shaft power. It would be easier to go to a motor mfg and look up the FLA of the motor.

"Load Factor given as 0.75."

"What is the Power by the Motor ?" LF Has nothing to do with the power the motor consumes.

"10 KW x 0.75 LF = 7.5 KW ". No. Look up the definition of load factor. If you have 12 motors connected and only 9 motors are running at any one time, then the load factor is 0.75. 9/12 = 0.75 or 12 motors x 0.75 = 9 motors.

I do not see Efficiency required to calculate the Consumed Power ? It is if you only have the output power.

"Input Power is rated for 10 KW ? and not 10 KW / Efficiency ?" Incorrect.
Input power = output power / efficiency. or 10 kw / 0.70 = 14.3

"Efficiency to me is the Electrical Power Output to the Mechanical Load.

and that is ( 10 KW x Efficiency )" No. You have been using 10 kw as the output power in your post. Make up your mind.

What is the IEC definition of " Absorbed KW " as it would relate to Consumed Power ? I don't know. Look it up.

I assume that the Motor Full Load Amps is derived from the 10 KW ? In a way.

FLA = KVA / (volts x 1.73 x PF)

Jeff

PLease respond.

Thank you very much for your response.

Everything is now crystal clear. Appreciate your help and explaination.

Just confused by the different terms, not much exposure to IEC motors.

in defination of terms between the IEC and NEMA standards.

For example your reference to " Load Factor " is really the defination of

" Diversity Factor "

For the Load Study the goal is to calculate the " Consumed KW " and the term

in this regard for " Load Factor " is defined as the " the ration between the motor

shaft available full load output and the actual load drawn by the mechanical equipment load "

What this all apparently means is that " Consumed KW " =

( Motor Rated Load KW / Motor Nameplate Efficiency ) x ( Load Factor )

The IEC uses KW for motor sizes, NEMA uses HP

Thanks

Jeff

For example your reference to " Load Factor " is really the definition of

" Diversity Factor "

Sometimes these terms are interchanged but I think you are corrrect.