A measurment error?

I didn't see where they claim to measure reactive power. Or how accurately they measure real power for that matter.

There are a couple of possibilities:

This thing measures amps only and makes an assumption about power factor and voltage to calculate watts.

Its possible to measure the phase angle of the conductor's voltage using the capacitive coupling between the conductor and CT secondaty winding. The voltage itself can't be measured accurately, but the phase angle between V and I can, so power factor can be calculated. An asumed value for voltage magnitude is used.

The web page installation instructions are too simplistic and there are a couple of voltage inputs into the metering.

--
Paul Hovnanian     mailto:Paul@Hovnanian.com
------------------------------------------------------------------
If life was fair, Elvis would be alive and all the impersonators
would be dead. -- Johnny Carson

Yes. Power will be the reading multiplied by some constant in the range (+1, -1). Probably, for typical household appliances (water heater, refrigerator, incandescent lamps) the constant will be very near +1.

By definition, a CURRENT transformer cannot measure or indicate power.

"Robert Baer"

** Under circumstances where the AC supply voltage is known and PF is likely to have only a minor effect on the total energy consumption - it most certainly can give useful indications.

.... Phil

amdx schrieb:

Hello,

they write no batteries are to be changed, therefore the transmitter with the connected clamps should have a power supply. If there is a power supply, the voltage can be measured too. The display has its own power supply, it is possible to measure the voltage within the display and to do the math to calculate power there.

Bye

It says on the page I posted above, "The transmitter's D size batteries should last up to 7 years before replacement." There is a video showing installation, it doesn't show any physical connection to the wiring. So, my thinking at this point, they measure current, assume voltage and power factor* and calculate power with those numbers.

*I send the an email and see if they assume anything other than 100% PF. Mike

amdx schrieb:

Hello,

but there is the power supply for the display unit, they can measure the voltage there and they can do the necessary math, assume the power factor and assume the second voltage is the same as the measured one.

Bye

Cheap "power meters" usually measure current and assume both line voltage and some typical power factor, like 120 volts and 0.8 for example. Their current sensors tend to be bad as well.

John

I got a response back already, "The units assume 100% powerfactor and a level 120 volts."

could make it."

John

Yes John, it is the simple way to do an inaccurate measurement that is easy for an inexperienced homeowner to hookup. Although they already have the homeowner inside the circuit breaker panel putting the clamps around the feed lines, it would not be much more difficult to stick a wire in a circuit breaker and another in the neutral bar. BUT now they're getting into rules and regulations. It's always something! MikeK

Pulling the panel off gets you into "rules and regulations".

I know some people on this site that should never go inside a panel..

tm

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"Indication" is *not* "measure".

amdx schrieb:

Hello,

it would have been easier for the homeowner if they put the voltage measurement within the display unit. But for a correct measurement three wires are necessary to measure the two voltages on both phases.

Bye

It would be easier, but if it was plugged into a heavily loaded line it might drop 2, 3 or 4 volts and there goes the accuracy. Getting current phase at the circuit breaker box and voltage phase at the display and making a comparision of phase shift thru an RF link ? I think that's asking for more error.

I figured someone on this group could do it with just two wires, I think the 180* phase shift between L1 and L2 is pretty constant. MikeK

amdx schrieb:

Hello,

a pretty constant phase shift of 180 degrees does not guarantee a constant voltage ratio of 1.000 between L1 and L2. The load is not very well balanced between L1 and L2 for all time of the day.

Bye

Noted, MikeK