Emporia Energy Community › Share Your Emporia Experience › Difference between Emporia 50A CT and Fluke Clamp meter
- This topic has 4 replies, 3 voices, and was last updated 1 year, 8 months ago by keikeoki.
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Pete6Member
I just did a quick test with my Fluke Clamp Meter and the newly installed CT on the same circuit. I chose the furnace breaker since this has just one item connected to it and it is easy to turn on and off. Also it is a medium size load. The Vue 2 showed 6.79A on when displaying Amps/Sec whilst the Fluke showed 8.6A. Bothe showed zero Amps when the furnace was off. Gotta say I trust the Fluke here.
Can this be verified by others and, can the value be adjusted?
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djwakeleeMember
The Vue2 doesn’t show you actual amps. It takes the real power, and divides by 120V. Emporia added Amps to the software by customer request (originally the app didn’t have it), but fundamentally Amps is not communicated from the device -> cloud -> app. As designed, the Vue 2 is an energy and power meter – not a volt or ampmeter (even though it does those low-level readings to get the power).
The measurement you make with your fluke is only showing amps. It does not show you the phase relationship with voltage. Unless you measure the current on a purely resistive circuit (electric hot water tank, dryer, range, heater), power factor comes into play, and you will have different power readings. Your furnace is not a pure resistive load (it has inductance and is a reactive load), so VA and W will be different. Emporia is lying with the Amps figure, but the power (in W) is correct.
If you want to understand more, do some reading on apparent (VA) and real power (W). If it doesn’t hurt your brain, it may shed some light on things for you. But the fundamental issue is that Amps display on the Vue 2 assumes a power factor of 1.0 and voltage of 120.0V. This is not generally the case so the Amps readings won’t match. But the actual power displayed (W), energy (kWh), and $ is correct for the loads.
- This reply was modified 2 years, 1 month ago by djwakelee.
- This reply was modified 2 years, 1 month ago by djwakelee.
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Pete6Member
Yes, I understand VA is just that. Volt x Amps = Watts. Yes again, Amps are much easier to determine with a DC voltage and a pure resistive load. Coming from the Vue 2 CT is a small AC voltage. Looking at the circuit board, it looks as if this is half wave rectified and smoothed (not sure because I have not traced out a full schematic). This is always going to be approximate given varying load types.
My Fluke is a fairly old model that I have had for decades and I do not think it delivers a true RMS amperage reading. However over the years it has been pretty consistent measuring everything from incandescent light bulbs and heaters to 5HP motors. I trust it.
Given what you say about how the Vue 2 derives its AMPs reading I am not at all surprised at the difference. I need the Amps reading on a per second basis to calculate phase (well the 180 degree difference between the ends of a transformer winding) so as to know what I can switch on or off when running on my 7,000 Watt inverter. This was the reason I bought the Vue 2.
I shall try this test again with a much newer true RMS clamp meter this weekend when a friend brings his one to my house. More then.
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djwakeleeMember
Just to clarify, Volts x Amps does not equal Watts. For AC, Volts x Amps x Power Factor = Watts (real power). Volts x Amps = VA (apparent power). The Vue 2 will give you real power (W), whereas using a DVM for current and voltage you will only see apparent power (VA).
A true RMS vs. non-true RMS DVM will definitely yield different results also. You’d indeed want to use that, and then just measure a purely resistive load like a heater circuit. Assuming your AC voltage is close to 120/240V, the Vue2’s ‘calculated’ Amps should match your true RMS DVM Amps reading within a few percent.
- This reply was modified 2 years, 1 month ago by djwakelee.
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keikeokiMember
The power factor of a typical hvac system is .8 to .9. So 6.79 amps converted from a true power reading is correct. to turn that into actual; amps you need a power factor meter to give you the pc to adjust to your measurements. It sounds to me like your fluke and Emporia are in agreement, you just don’t know the power factor to use to compare the two numbers.
The fact is Emporia is guessing the amps based on a power factor of 1. Your actual power factor will be lower because the inductive loads do not consume power for the full sine wave cycle. Your fluke is showing the actual measured amps, which cannot be used by itself to derive accurate watts. V * A =VA not watts. If your furnace was electrric only, that .8 power factor would be a lot closer to 1 because the main load would be resistive. The resistive heating elements have a pf of 1, motors are closer to .8, and the computer circuit boards are closer to .65. Doing a direct conversion from VA to watts is not possible without a power factor measurement, and your fluke doesn’t do that… Even my new DVM’s won’t calculate pf, it takes a power/energy meter to do that. Nothing wrong with your fluke, the problem is it is the wrong kind of meter for this. Funny thing, as often as engineers with vom’s are doing math in watts, you would think the VOM’s would include either direct readings in watts, or a power factor measurement so you could do it yourself.
I used to manage data centers, and it is a larger problem with computers as they are in the .65 range so going off of VA and you are going to be really inaccurate. You can just plug in a guess as to the power factor, and even though it is a guess it is more accurate than not making any pc adjustment.
Do this, clamp onto the resistive element, VA*1=watts. Clamp onto each motor, VA*.8=watts. Clamp onto the transformer powering the electronics, VA*.65=watts. Add those values together and see if emporia agrees. Ignore Emporia’s amps, it isn’t a real measurement, only look at watts.
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