Calculating Total Electric Power Input kW 3 phase Motor 1

[Iomcube]

I am attaching motor nameplate (its an Cl2 compressor)


The running ampere on the meter (it varies with throughput) is now 320A

According to this post (https://www.eng-tips.com/threads/actual-work-380v-vs-460v.91809/post-344509) , total electric kW input will be

=1.732 x 320 Amp x 380 Volt x 1 kW / 1000 W = 210 kW

& if I multiple it by 24 hr I can get kWh

The problem is that nameplate shows a rated power of 185 kW however mine calculate answer is higher than this?

From another source I am quoting this

If the machine is 380 V rated it means that it had a rated phase to neutral voltage of 220 V.

So this means the correct answer will be
=320 Amp x 380 Volt x 1 kW / 1000 W = 121.6 kW

 

[waross]

380V x 1.73 x .0.87 x 0.959 x 320A / 1000 = 148 kW Actual Mechanical Power Out
380V x 1.73 x .0.87 x 0.959 x 336.9A / 1000 = 185 kW Maximum Rated Mechanical Power Out
380V x 1.73 x .0.87 x 336.9A / 1000 = 193 kW Maximum Electrical Power In.

 

[MintJulep]

Power factor (cos(theta)) is given as 0.87.

210 * 0.87 = 182

 

[Iomcube]

@waross

380V x 1.73 x 320A / 1000 = 210.4 kW Actual Electrical Power In ??

 

[waross]

Iomcube​

380V x 1.73 x 320A / 1000 = 210.4 kW Actual Electrical Power In ??​

Probably not.
On the other hand, given as both PF and eff vary with variations in loading, any use of those values at less than 100% load is probably inacurate;
Some of my calculations may be more illustrative than definitive.
When I recognize that my results are in error, I am hesitant to challenge your erroneous results.

 

[Iomcube]

@waross
My electric foreman simply stated this (running Electric consumption):

380V x 320A / 1000 = 121.6 kW​

 

[davidbeach]

Your foreman is missing the root(3) factor.

 

[waross]

My electric foreman simply stated this (running Electric consumption):

The boss is not always right but he is always the boss.

Your foreman is missing the root(3) factor.

And he is missing the PF.
Your foreman is using the formula for a DC motor.
It doesn't work for an AC motor.

 

[edison123]

Output KW = √3 x Input voltage x Input current x power factor x efficiency = √3 x 0.38 x 336.9 x 0.87 x 0.959 = 185 KW
Input KW = Output KW / Efficiency = 185/0.959 = 192.9 KW