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

 

[waross]

At 320 Amps, the current is at about 95.2%.
That does not imply 95.2% loading.
The efficiency typically increases as the load decreases and peaks at around 66% to 75% of full load.
If this alone is considered, the load may be greater than 95.2% at 320 Amps.
However that is not the only factor.
The Amps must be multiplied by the cosine of the angle of displacement between the voltage and the current.
This is given as cos(theta) = 0.87
Unfortunately the PF value is valid only at full load and decreases as the load decreases.

Are these errors serious?
It depends.
Three things to consider,
1. When trouble shooting and/or analyzing motor problems and issues, knowing and understanding the relationships between the various factors is much more important then knowing the precise values.
2. The errors decrease as the motor load approaches 100%
3. The voltage is seldom exactly the nominal voltage and the voltage error may be greater than the other errors.

There is not enough information given on a typical nameplate to accurately determine the Cos(theta) or efficiency at less than 100% load.
Nevertheless, we knowingly use the known inaccurate values as the best available.

It is common practice to use the nameplate values for loading between about 75% and 100% because;
"Good enough is good enough."

In some sites it is possible to extract an accurate kW value from the kWhr meter.
It is possible with some never meters to also extract accurate information as to VAR consumption.
With accurate Current, kW and VAR values it is possible to calculate the cos(theta), removing one uncertainty.

 

[Iomcube]

@edison123

Calculations at RUNNING conditions

Output KW = √3 x Input voltage x Input current x power factor x efficiency = √3 x 0.38 x 320 x 0.87 x 0.959 = 175.7 KW
Input KW = Output KW / Efficiency = 185/0.959 = 183.2 KW

as @waross indicated efficiency is load dependant but real value I cannot get so I assumed it to be constant. Also thanks for this information

At 320 Amps, the current is at about 95.2%.
That does not imply 95.2% loading.

 

[waross]

Be careful sharing that information with your boss.
It may be a matter of "Being right" versus "Your career".