can someone recommend a really TRUE RMS meter?

[The Phantom]

On Fri, 20 Jun 2008 06:23:28 -0700 (PDT), panfilero <panfilero@gmail.com> wrote:

On Jun 18, 3:08 pm, Phil Hobbs <pcdhSpamMeSensel...@pergamos.net
wrote:
panfilero wrote:
"You can't just multiply an RMS current and RMS voltage measurement to
get power." - RL

So.... I've read that Vrms*Irms = Pavg..... but..... you're saying I
need to include the phase shift?! Isn't the phase shift implied in the
waveform? I have two meters hooked up, and I'm reading a Vrms and an
Irms, if I multiply these together don't I get the Apparent Power?
Aren't RMS values scalars? No phaeses involved.... and isn't
Vrms*Irms = P apparent?

You really do have to worry about the phase shift. When the voltage
and current are in phase, the power is going in the same direction as
the current. When they're out of phase, the power is going in the
opposite direction. From the wall socket's perspective, that's the
difference between a motor and a generator.

If you imagine a load that drew some enormous current right when the
voltage was near 0, and much less current near the peaks, you could be
off by a factor of 100 or more if you just multiplied RMS voltage times
RMS current. The same is true of a purely reactive load, in which the
voltage is 1/4 cycle out of phase with the current. For instance, this
is what it would look like with a capacitor across the line.

I___ ___
/ X \ V
/ / \ \
/---/---\---\---/----------------------
/ \ \ /
/ \ X
-- --- --
| | | | |
| | | | |
1 2 3 4

In quarter-cycle 1, I >0 and V < 0, so the VI < 0 : power is going from
the capacitor into the wall socket.
In QC 2, I>0 and V>0, so VI > 0: the power is going from the wall socket
into the capacitor.
In QC 3, I<0 and V>0, so VI < 0 again, and
In QC 4, I<0 and V<0, so VI > 0.

Since the waveforms are symmetric, the net power transferred over a
cycle is zero.

Cheers,

Phil Hobbs

Hmmmmm..... ok, well my voltages will definitely be out of phase since
there's some reactance involved.... and I agree with everything you
are saying here, but this is looking at just one cycle of each.... I'm
assuming a meter is constantly sampling the signal and so the RMS
value is some sort of average that is constantly being updated, and
the average is maybe taken over a hundred cycles or somethign like
that not just one cycle, so do you think that since the meter is
taking an rms average of many cycles that somehow the phase shift
would then be included in there when I multiply the values together?
I'm fine with the fact that I'm getting the reacitve power (the
negative power going back to the source) in my readings, I'm fine with
the fact that I'm getting apparent power instead of true/active
power... but I only have meters and am trying to avoid the phase
issue.... I'm using the two-wattmeter method to do this (thanks
Phantom)

What you're doing isn't the two-wattmeter method. You are using two separate
meters and two separate meters don't make one wattmeter:

http://en.wikipedia.org/wiki/Wattmeter

The essence of an electrodynamometer wattmeter is that the movement has two
coils, a current coil and a voltage coil. The torque that deflects the needle
is proportional to the instantaneous product of the current and voltage. The
mechanical inertia of the movement smooths out the pulsations that might occur
due to the fact that the instantaneous
power varies from 0 to some magnitude rapidly.

The reactive power is dealt with by the fact that when the power is returning to
the source, the needle would ordinarily be deflected backward, so the average of
forward and reverse power is what needle indicates.

When you use two separate RMS responding meters, you are in fact measureing
apparent power. The real power will always be less than or equal to the
apparent power. If the load is purely resistive, they will be equal. If the
load is more or less reactive, the real power will be less than the apparent
power, and sometimes the real power may be MUCH less than the apparent power.

If you apply, for example, a 24 VAC stimulus (using a transformer for safety) to
a 10 microfarad capacitor (not an electrolytic), the apparent power would be
2.17 volt-amperes. I have such a capacitor lying around in my junk box, and it
has an ESR of 50 milliohms. The real power drawn in this case would only be 4.5
milliwatts. This is almost a one thousand to one ratio of apparent power to
real power.

Furthermore, you have what amounts to a three-phase load. If you let one lead
of the motor circuit be reference, and measure the voltage and current in the
other two with respect to that reference with separate RMS responding meters,
you will not be carrying out the two-wattmeter method. You have to use actual
wattmeters, or their equivalent, a two channel scope with trace math.

The three phase nature of your load additionally complicates things. I think
that since your load is probably inductance + resistance, the two separate
measurements probably add up to the apparent power, but I'd have to think about
a little to be sure.

J.