CT question

[mopowr steve]

I originally posted about this in campfire chat but think maybe it will get more attention here.

My question is, on a sub meter running on 120v with only one element (CT) and I would like to measure current from a load that incidently is on an opposing phase (180*) could I run that wire through the CT in opposite direction therefore placing its current draw in phase with the originating power supply line?

 

[ggunn]

I originally posted about this in campfire chat but think maybe it will get more attention here.

My question is, on a sub meter running on 120v with only one element (CT) and I would like to measure current from a load that incidently is on an opposing phase (180*) could I run that wire through the CT in opposite direction therefore placing its current draw in phase with the originating power supply line?

It would flip the phase of the signal, sure, but I can't say what it would do to your data.

 

[gar]

151024-2335 EDT

mopowr steve"

Restate your original question and and be complete in exactly what you want to measure.

.

 

[mopowr steve]

Looking at Levitons Mini-Meter as a sub meter for campsites. 120v model with 1 CT

Each campsite has 2 receptacles ( 1-30a and 1-20a @ 120v) on opposite legs of a single phase source.

My thought is to run the feed wires to receptacles through the CT. But because one receptacle is fed by opposite phase by running that feed wire through the CT in the opposite direction make current phase in polarity with the 1st receptacles phase therefore being an additive sum of currents instead of a difference in currents. Right?

So the question is, will a watt-hour meter that utilizes 120v accurately read wattage if done this way?

Do any of you have experience with these?

 

[big john]

Only experience I have is where I've seen someone do that accidentally. Yes, the current was additive, I assue the power increased linearly, too, but I can't swear.

 

[gar]

151025-0846 EDT

The answer is yes that you wiring scheme will measure the total real power of your two phases provided that the instrument does a good job of calculating the instantaeous product of current and voltage and averaging this over time. In turn kWh is measured if the instrument can integrate the instantaneous power over time. This works only if the two phases are 180 degs out of phase, and the voltage magnitude on each side of neutral is about the same. Thus, if sourced from a three phase wye system there will be a considerable error in the data from the phase that the voltage is not derived from.

The primary error problem with a current transformer system is from uncorrected phase shift in the current transformer on non-resistive loads. A TED system (The Energy Detective) has this phase shift problem. The "Kill-A-Watt" uses a resistive shunt and does not have this phase shift problem. But you can not use a "Kill-A-Watt" for your application.

.

 

[mopowr steve]

Thanks gar,
I did realize that 3phase would be a problem but you brought up a good point about voltage on both legs to neutral need to be relatively the same or could lead to an erroneous calculation. Definiatly gives me something to consider, where it may be necessary to use the 240v model with 2-CT's. Space within enclosures is a premium as long as CT's don't take up to much room.

Love to hear anyone else's take on this, or suggestions from any-one who has done campground sub-metering

 

[GoldDigger]

Just the comment that a standard 120/240 meter has no connection to the neutral and has the exact same issue about assuming the two L-N voltages to be equal.
The error is small enough that neither POCO nor the regulators worry about it. (The fact that any error caused by voltage drop will always be in POCO's favor may have something to do with it too.)

 

[gar]

151025-2240 EDT

mopowr steve:

What power company meters do for single phase power monitoring from a center tapped transformer is to measure voltage line-to-line and measure the current in each line and sum those currents to instantaneously add. You are doing the correct summation by reversing the wire direction of one of the two wires you are running thru the current transformer.

In the TED System of power measurement they use two current transformers and properly add their outputs. Two CTs are used because it is not usually possible to run both hot lines thru one CT.

Whether you use 240 or 120 as the input voltage is only a calibration difference. The assumption is that the two 120 V values that add to 240 are sufficiently well balanced that measuring one is a good predictor of the 240 value. The TED System can be setup either way.

To do a more accurate measurement requires two separate power measurement circuits. One for each side of the neutral, transformer center tap.

,

 

[ggunn]

151025-2240 EDT

mopowr steve:

What power company meters do for single phase power monitoring from a center tapped transformer is to measure voltage line-to-line and measure the current in each line and sum those currents to instantaneously add. You are doing the correct summation by reversing the wire direction of one of the two wires you are running thru the current transformer.

For L-N loads maybe but with a balanced L-L load on a 240V single phase service the current in L1, the current in L2, and the current through the load are all the same. If you sum the currents in L1 and L2 the result is twice the current through the load. It seems to me that it is the power, not the current, that should be summed.

 

[gar]

151026-0951 EDT

ggunn:

You argument is not correct, and I don't have a good source for a reference on the subject.

Unfortunately M. B. Stout in his instrumentation book does not discuss the induction kWh meter. I looked on the Internet via Google and did not find a good reference.

The instantaneous power in a load is
p = i*v
The energy consumed in a load over time is
the integral of p with respect to time over the desired time period.

Consider a center tapped single phase power source and two wattmeters, one for each half of the secondary. One meter's voltage coil is fed from phase A of the source and its current coil is in series phase A's hot line. This means the voltage coil is connected between neutral and the appropriate phase hot conductor. Similar connection for phase B and its power meter. The two loads consist of identical 100 W resistors. One resistor on phase A and the other on B.

First the mid point of the two resistors is connected to the transformer neutral. Each power meter reads 100 W. Next disconnect the resistors midpoint from the transformer neutral. Is there any change in the wattmeter readings? No.

Can you extend this to the kWh meter, and your argument? Your argument only relates to calibration of the meter.

If you look at the wiring diagrams for single phase kWh meters you will see there is no connection to the neutral. It is a pass thru conductor, if there is one in the meter itself. Otherwise it is just a pass thru in the meter base.

.

 

[GoldDigger]

For L-N loads maybe but with a balanced L-L load on a 240V single phase service the current in L1, the current in L2, and the current through the load are all the same. If you sum the currents in L1 and L2 the result is twice the current through the load. It seems to me that it is the power, not the current, that should be summed.

As long as the voltage is constant, summing the current then multiplying by voltage is the same as multiplying by voltage and then summing.
If you multiply by 240 you do then have to divide by two, which the meter does.

 

[K8MHZ]

Why would there be any difference between using an 'apartment' meter with a single receptacle on each leg for a camper vs. multiple loads on each leg like in an apartment?

There would be a CT on each leg going to a single meter in each instance, so I don't see where there would be a problem.

 

[ggunn]

As long as the voltage is constant, summing the current then multiplying by voltage is the same as multiplying by voltage and then summing.
If you multiply by 240 you do then have to divide by two, which the meter does.

...i.e., summing the power. I agree.

 

[gar]

151026-1210 EDT

ggunn:

A true power meter measures the instantaneous product of current and voltage. Thus, it only extracts the in-phase component of the current.

Summing the currents before multiplication does not create any problem with measuring real power. Phase A and B voltages are sufficiently alike that it does not matter whether you use one 120 V source as the voltage reference or you use the two 120 V sources together, 240 V. It is just a calibration difference by a factor of 2.

ptotal = i1*v1 + i2*v2
since v1 = v2
ptotal = v1 ( i1+i2 )
If you want v1+v2 to be the source, then
ptotal = ( i1+i2 ) * ( v1+v2 ) / 2
or
ptotal = ( i1+i2 ) * ( 2*v1 ) / 2 = ( i1+i2 ) * ( v1*2 / 2 ) = ( i1+i2 ) * v1

.