I'm intest of emonTxV3 ,... but , in the documentation , it is limited on "240 Volt" ; and I don't unsterdand WHY? If I know the Voltage is stable => I can set a mutiplication factor ? ( 380V / 220V = 1.7 )
br
etienne
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emonTxV3Submitted by etienne on Wed, 23/12/2015 - 14:04
I'm intest of emonTxV3 ,... but , in the documentation , it is limited on "240 Volt" ; and I don't unsterdand WHY? If I know the Voltage is stable => I can set a mutiplication factor ? ( 380V / 220V = 1.7 )
br etienne » |
Re: emonTxV3
Can you explain your question a little more clearly?
There are too many things I have to guess to answer you. Please read the FAQ and then give more details.
Re: emonTxV3
I want to use a emonTxV3 => to take energy measurement on electrical box
3 phases + NEUTRAL
380 volts
100 ampère
in the emonTxV3 documentation => limitation of the use is "220 V"
question : I can set a mutiplication factor ? ( 380V / 220V = 1.7 ) to calculate the energy ?
tks
etienne
Re: emonTxV3
It is a well-established principle of electrical engineering that to measure the power in a multi-phase installation, you need one less wattmeter than the number of wires. Therefore, as you have a 4-wire supply, you need 3 wattmeters. That means 3 pairs of voltage and current measurements. The easiest way to do that is to treat your neutral wire as the reference and measure (or know) the phase-neutral voltage and measure the currents in each phase conductor.
But from your email address, I think you live in Belgium, and I believe there that your phase-phase voltage is 380 V, and your phase-neutral voltage is 220 V.
If you are measuring the voltage, you connect your ac adapter between line and neutral. If you are not measuring it but using the nominal voltage, you use 220 V. You install a CT on each phase conductor. The sum of the apparent powers, which using your voltages is 220 × I1 + 220 × I2 + 220 × I3, will give you the total power.
If you use the phase-phase voltage of 380 V, you will get the wrong answer.
It is not essential that you use the neutral wire as the reference, you could use for example phase 1 as the reference, then you would need a CT on each of phase 2, phase 3 & neutral, and the total apparent power would be (using your voltages again) 380 × I2 + 380 × I3 + 220 × IN.
You can multiply by 1.7 if you wish, but I think your mathematics will be wrong if you do. The voltage to use if the ac adapter is not available is set at 240 V because that is the UK standard voltage.
Re: emonTxV3
Hi,
thank you for these clear explanation.
Br
etienne