This is a spin off from another thread.

I have a limited understanding of electricity and power and am hoping others can further my understanding.

I have a Watt Node meter that says my heat pump uses about 4800 watts when running.

I put my clamp meter on the lines running to my heat pump and measured 19.5 amps.

I measured the voltage and read  248 v

Since watts = volts x amps,   19.5 x 248 = 4836 watts.

So now I am wondering,  when measuring amps with a clamp meter on a 240 volt motor do you need to measure each leg and add them together?

If so,  my above calculation is wrong as I only measured one leg.

No - 1 leg should suffice - go ahead and measure the other - they should be the same.

Watts = Volts x Amps only in case of DC power and in case of AC with unity power factor - purely resistive loads such as water heater, incandescent light bulb, coffee maker, clothes iron, etc.

Where an AC load includes motors or transformers, Watts = Volts x Amps x PF. Power factor varies from 0.0 to 1.0 depending on the nature of the load.

Neglecting power factor and calculating Watts from volts and current will lead to inaccurately high calculated wattage.

I wonder if your Wattnodemeter calculates PF and adjusts - based on your post it may not.

Hope this helps.

Which is more likely?

The watt node doesn't adjust for PF or my PF =1 ?

Good question.

My understanding up until now has been that compressor motors exhibit lagging power factor - PF<1

I surfed Bill's link to Wattnode (I assume you have similar unit) and it quite emphatically stated that it adjusts for PF

So, for now, I don't know - maybe motors have gotten better about power factor lately

Engineer,

Thanks for your input.

My Wattnode is the same brands as Bill's.

The specs on my Heat Pump say that it should draw 4 kW. So if I add the fan and loop pumps, I guess that this all adds up.

FWIW, I put my Kill-a-Watt on the house fridge - PF = 0.97 with compressor running. That supports the idea that compressor motors have become better regarding power factor

igshpa--    dmd=  .85 X ( vx I ) for single phase

3 phase is 1.732 x .85

 

Electrical demand is then used to find eff and capacity

what is dmd?

So, should I infer from that that a rule-of-thumb PF for compressors is 0.85?

That would be handy to know, and reconcile the amps I read for my unit against WF's specs for Wattage at my waterside conditions.

Whether it is a rule of thumb or not,,,, im not sure. It is a set value in the IGHSPA Handbook under water to air unit startup sheet to find capacity ect.

I use it to calculate capacity, cop, ect