Inverter power calculations

[Colin Carpenter]

Not strictly Mitsubishi based, but I thought I'd ask here as there are some good people who view this forum.

It's a general question regarding calculation of electrical power. All my career, I've worked on a couple of simple formulae to calculate electrical power, namely

I x V for DC circuits and single phase AC circuits and 1.732 x I x V x cos phi for 3 phase AC circuits.

However, as inverter control becomes much more common and energy becomes much more expensive, we have now started looking at the energy information available from modern inverters ..... and I'm now very puzzled.

The motor in question is a 75 kW three phase motor running a dairy homogeniser, effectively a piston pump that can have a lot of restrictive back pressure enabled to smash the fat globules to smithereens.

Effectively, the machine has two modes. It can run on water with no back pressure while cleaning and it can run on dairy product with high back pressure when homogenising. In other words, the pistons can be freewheeling or working hard.

The ABB inverter works fine and it's output frequency is controlled by a 4-20 signal from the q plc and it all works fine.

However, the output power and amps that the inverter is showing don't make sense to me.

In the freewheeling state, the inverter shows 368 volts output voltage, 44 amps and 3.08 kw of power.

In the full production, high back pressure state, the inverter shows the same voltage, but 52 amps and 16.36 kw.

Which makes me think that my career long use of those equations now needs a complete rethink if those inverter amps values are correct......

Can anyone enlighten me please.

 

 

[pturmel]

A free-wheeling AC motor is moving kVARs, not kW.  Voltage has to scale (largely) with speed to correspond to self-generation (old-school Counter-EMF).  So power factor is very low when free-wheeling. Keep in mind that the drive is typically going to show you the values at the motor terminals, not the values at the drive's line terminals.

[Colin Carpenter]

So, in the freewheeling state where it's showing 44 amps at 3 kw, assuming that the "root 3 IV cos phi" equation still holds, this would equate to a power factor (cos phi) of around 0.1 to produce a stated power of around 3 kw?

I suppose what I'm really try to do is determine how much it actually costs to run the plant, in other words, how many kWh will the electricity supply company charge to run that freewheeling motor for one hour? Presumably 3 kWh?

[pturmel]

Utilities charge industrial users separately for kVARs and kW, with rate change triggers at various raw amperage levels.  Can be quite complicated.  One of the reasons drives can pay for themselves--they expose the plant just to their rectifier section's power factor, not the motor's power factor.  You'll have to do separate analyses for motors with drives versus motors that run across the line (just a starter or soft-starter).

You'll need each drive's power factor as seen by the supply bus.