Mechanical engineer's kVA question

[bill nye scienceguy!]

Hi all.

I'm a mechanical engineer by training and a manufacturing engineer by trade. I currently find myself in the position of being asked to fill out a spreadsheet containing, amongst other things, machine tools' power consumption in kVA.

Occasionally an OEM will be helpful and provide this directly in the datasheet but more often I'm given volts, frequency and phase. How can I reliably approximate a value for kVA given these variables or what else do I need to know in order to calculate?

Many thanks!

 

[Jony130]

You need to know the current that is drawn by the machine to find power.

 

[Mech_Engineer]

If I'm understanding the units correctly kVA is just kW, so as long as the supplier specifies the machine's power consumption you can convert to kW.

 

[bill nye scienceguy!]

Sorry, it's been a long day. Of course I need to know current.

I'm not entirely sure of the relationship between kVA and kW beyond that they're both measures of power. For example, in the following calculator 480kW is seen to be equivalent to 600kVA. What equation sits behind the buttons on the screen? Also, in 'power requirement calculator' it mentions something called a power factor - can someone give me a quick explanation on what this is? What is the significance of working in 1- or 3-phase wrt a difference in power?

http://dieselserviceandsupply.com/power_calculator.aspx

If it's possible to explain this without using complex numbers I'd be much obliged...

 

[russ_watters]

The wiki on this is good, but in short, since AC electricity is a sine wave, if the waves for volts and amps are not aligned, you aren't multiplying peak (rms actually) volts by peak amps.

But many electrical devices are sized based on their amps and supply voltage, so you still need to know the APPARENT power and that's what kVA is.

And apparent power and actual power are related by power factor.

 

[Jony130]

Read this and use "Power triangle" instead of complex numbers.
http://www.allaboutcircuits.com/vol_2/chpt_11/2.html

 

[bill nye scienceguy!]

Ok, this is starting to make sense. To check my understanding: if something has a power rating of 600kVA and 480kW that means there's a power factor of 0.8? And in words that means that means that there is 600kVA total power in the system and 480kW is dissipated by the load.

Assuming I have that right, how do I calculate the power factor from first principles ie not from kW/kVA? Let's say I know that the current is 3-phase or single phase but I don't know the impedance angle.

 

[russ_watters]

It can't easily be calculated from first principles. It has to be estimated or looked up in a table.

 

[bill nye scienceguy!]

Ok, so is it safe to assume a power factor of 0.8 in estimating kVA? I say 0.8 since this seems to be quite common.

 

[russ_watters]

Different devices have different power factors in different situations. .8 works for a reasonably well loaded motor: inductive loads are subject to power factor, resistive loads are not.

 

[bill nye scienceguy!]

I just asked google about inductive loads and I got this.

http://www.markallen.com/teaching/ucsd/147a/lectures/lecture3/3.php

It explains that inductive loads involve magnetic fields eg in a motor. It goes on to say that resistive loads convert electrical energy into another form of energy eg in a lightbulb. So I can see that whenever I have a motor I need to take into account power factor.

This is maybe an adjunct at this point but - does this mean that in the case of a lightbulb the apparent power is equal to the dissipated power? Or is it not appropriate to view a resistive load in those terms?

 

[russ_watters]

You are correct: when power factor is 1, actual and apparent power are equal.