Re: Power Measuring!! Need Assistance

On Tue, 13 Dec 2005 03:59:28 GMT, Ignoramus12137
<ignoramus12137@xxxxxxxxxxxxxxxxxxxx> wrote:

>On 12 Dec 2005 19:31:51 -0800, @donis <adonis@xxxxxxxxxxx> wrote:
>> Hi Folks, Im working on a power meter design using the ADE7756 and
>> right now im working in the interfacing stage(taking the current and
>> voltage values) . My question is quite simple but right now im a little
>> bit confused. Im my country we use the North American 3-wire
>> configuration we have a Neutral , L1 and L2. From Neutral to ground
>> the voltage is 120VAC
>
>Scary stuff!!!
>
>(hint, that is wrong)
>
>> , L1 to L2 is 240VAC. I know that apparent power
>> equals S=VxI. So far so good with my concepts. My confusion is
>> regardings the power readings.
>
>Be careful applying this rule when you have alternating current. It is
>incorrect to multiply average voltage (RMS) by average current (RMS),
>in case of any but resistive loads.
>
>That is because current may be out of phase with voltage, as it
>happens with motors.
>
>> IF I have a 120V load the current will
>> return thru the Neutral Wire. and if I have a 240 Load the current will
>> return thru one of the lines. My question is if I have both loads
>> connected(120V load and 240V) and S=VxI , what would be my total power
>> value? what voltage should I use for sensing this (120 or 240), what
>> would be the value of my current? would the current value be a
>> vectorial sum? . Please include any information that might help me in
>> my design. Thanks in Advance guys..
>
>I am not sure if you should really be designing a power meter, if you
>are weak on concepts of power etc.
>
>If you have 3 currents, I1, I2, and In (In stands for neutral)
>
>And resistive load
>
>The current In = I1 - I2
>
>110V power is 110*In = 110*(I1-I2)
>220V power is 220*(I1-In) = 220*I2
>
>The sum is Power = 110*(I1-I2) + 220*I2 = 110*(I1+I2)
>

That's only true if the line voltages are symmetric to neutral, which
they never really are. Some domestic electric meters assume this: they
wind both current-sensing coils around a common laminated core, and
use a single potential coil from line-to-line.

It's better to compute power properly,

P = Van * Ia + Vbn * Ib

where the multiplies are instantaneous products, of course.

John


.

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