120/208 Single Phase Power Calculations

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carpboy

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The voltages on the two lines are 120° apart. The CTs tell the meters what the current is and what its phase relationship is to the line voltage. The current times the voltage in absolute value terms is the apparent power. The part of the current that is in phase with the voltage times the voltage is the real power.
Yes, I understand, but I and the meter engineers are trying to figure this out exactly.

BTW, I have a neutral. I saw this mentioned in a later response.

Imagine this scenario: The only load is 10A, say a dryer heating element. The apparent power is 10*208= 2080 VA. Now say the load is 2 10A 120V devices, say two hair driers that happen to be on opposite legs. here S = 2*120*10 = 2400 VA. The pair of CT's can each indicate 10A yet there are two different VA values.

My conundrum is how does a metering device calculate S when there can be different V values?
 

carpboy

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Location
PA
211009-045 EDT

In a 3 wire system consider one wire as a reference. Take two wattmeters and connective the appropriate (meaning to get correct phasing) one voltage wire of each wattmeter to any line to be assumed to be a common, and connect the other voltage wires, one of one meter to a "hot wire" and the other volt wire to the other "hot wire". Then take each wattmeter current coil and put it in the "hot" line associated with the voltage coil of that meter. Again appropriate phasing. The sum of the two wattermeter readings is the total power to the loads whatever they may be.

I believe this is discussed in M. B. Stout's book on "Electrical Measurements".

.
Would you be able to post a sketch? I don't quite follow your text description.

Thanks.
 

carpboy

Member
Location
PA
The meter must have a voltage reference to neutral in addition to the voltage reference to each 'hot' to meter correctly.

If the meter is only measuring line to line voltage, then it can't even meter power correctly, let alone apparent power.

Imagine a customer with only resistive loads. If the meter only measures line to line voltage, then a customer could draw 10A L-L (2080W) or 10A L-N on each leg (2400W) but I am pretty sure in the latter case the meter would only register 1800W.

-Jon
I do have a neutral landed on the meter. The software shows all the voltage values and they are correct. How did you arrive at 1800?
 

carpboy

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That answers OP's question, IMO. If no reference to the common (neutral) his metering is only going to be accurate for line to line loads.

He does need something that compares phasing of voltage to current to read real power, just multiplying current by amps will give him the apparent power he mentioned.
I have a neutral and the meter is fully capable of phase angle and real power display. My problem is the apparent is all over the place and I can't reconcile my V and I reading to its apparent power reading.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
Imagine this scenario: The only load is 10A, say a dryer heating element. The apparent power is 10*208= 2080 VA. Now say the load is 2 10A 120V devices, say two hair driers that happen to be on opposite legs. here S = 2*120*10 = 2400 VA. The pair of CT's can each indicate 10A yet there are two different VA values.

My conundrum is how does a metering device calculate S when there can be different V values?

The metering device can figure this out because it doesn't simply measure RMS current or RMS voltage; the meter also measures the phase angles.

In the most common setup, the meter will measure the 2 phase-neutral voltages and the 2 phase currents. The meter is almost certainly measuring _instantaneous power_ and accumulating the energy used over time, but this is better understood if you think of the system as measuring the phase angles.

In the example you just gave, when you have two 120V L-N resistive loads, the current is exactly in phase with the supplied 120V. When you have a single 208V L-L load, the current is exactly in phase with the supplied 208V, which has a _different_ phase then the 120V L-N.
So in the first case the meter will be measuring
2 * 10A * 120V * 1.0 = 2400 W (2 power meter elements * 10A * 120V * 1.0 power factor)
In the second case the meter will be measuring
2 * 10A * 120V * 0.866 = 2080W (2 power meter elements * 10A * 120V * 0.866 pf)

Hope that helps.
Jon
 

carpboy

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Location
PA
It's called single phase for historical reasons. I prefer the term "open-wye".

What it really is, is two out of three of the phase wires from a 3-phase wye system, that are 120 degrees out of phase. Most of the loads in each dwelling unit are 120V phase-to-neutral loads (lighting and receptacles), while a select few loads are dual-rated to work on either 208V or 240V. Like heating elements in ovens, stoves, and clothes dryers, and A/C. Unlike a 120/240V system where balancing all loads will eliminate neutral current, a 120/208V open-wye will have to have current on the neutral any time there are 120V loads, even with balancing.

You commonly see this on apartment buildings, where the site as a whole gets the full 3-phase supply, and each dwelling unit gets two out of three phases from it. The common loads panel likely gets a full three phase service. Each dwelling unit gets a "single phase" (open wye) service, with staggered pair of phases across the system. Such that remainder-1 unit numbers (1, 4, 7, 10 etc) will get phases AB, remainder-2 unit numbers (2, 5, 8, 11 etc) get phases CA, and the divisible-by-3 unit numbers (3, 6, 9, 12 etc) get phases BC.
This is my scenario exactly. In regards to neutral current, the only way I see to do this is to measure neutral current, call that all 120V load, then subtract that from the CT value and call that 208V load.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
I do have a neutral landed on the meter. The software shows all the voltage values and they are correct. How did you arrive at 1800?

As I described in my analysis of your two examples with the neutral included, the meter knows both the magnitude and the phase angle of both voltage and current.

In the (broken) case of a meter that only knows the L-L voltage (no neutral reference) then if you had L-N loads their current would be out of phase with the L-L voltage. If you have 10A L-N on each of two phases, but only know the L-L voltage, then the meter would think you had 2080 VA apparent power (simply the 10A * 208V). I am not sure how such a meter would respond for the real power measurement; my guess is 1800W because of the 30 degree phase difference between the actual current flow and the L-L voltage.

-Jon
 

petersonra

Senior Member
Location
Northern illinois
Occupation
engineer
This is my scenario exactly. In regards to neutral current, the only way I see to do this is to measure neutral current, call that all 120V load, then subtract that from the CT value and call that 208V load.
The power meter measures voltage of both lines including the phase angle. It also measures the current in both lines including the phase angle. What more does it need to calculate the true and apparent power?
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
This is my scenario exactly. In regards to neutral current, the only way I see to do this is to measure neutral current, call that all 120V load, then subtract that from the CT value and call that 208V load.

You are on the right track in terms of questioning how the meter differentiates between L-L loads and L-N loads. But you don't need to actually measure the neutral current.

The rule (which gar already posted) is that 3 wires feeding a load (any three wires, could be the 'open-wye' system, could be a 3 phase delta, could be 3 wire two phase) then to measure power you need only two 'watt elements' meaning you need to measure two of the three currents and two of the 3 voltages, and then multiply the _instantaneous_ voltage and current for each 'watt element'.

If you think about it, you can see that with any two current measurements you can calculate the third (by the Kirchhoff Current Law) and with any two voltage measurements you can calculate the third (by the Kirchhoff Voltage Law), so you can see that 2 of 3 measurements is sufficient.

-Jon
 

carpboy

Member
Location
PA
The metering device can figure this out because it doesn't simply measure RMS current or RMS voltage; the meter also measures the phase angles.

In the most common setup, the meter will measure the 2 phase-neutral voltages and the 2 phase currents. The meter is almost certainly measuring _instantaneous power_ and accumulating the energy used over time, but this is better understood if you think of the system as measuring the phase angles.

In the example you just gave, when you have two 120V L-N resistive loads, the current is exactly in phase with the supplied 120V. When you have a single 208V L-L load, the current is exactly in phase with the supplied 208V, which has a _different_ phase then the 120V L-N.
So in the first case the meter will be measuring
2 * 10A * 120V * 1.0 = 2400 W (2 power meter elements * 10A * 120V * 1.0 power factor)
In the second case the meter will be measuring
2 * 10A * 120V * 0.866 = 2080W (2 power meter elements * 10A * 120V * 0.866 pf)

Hope that helps.
Jon
Jon, that helps a ton, I think I can see a solution here.

You are saying in essence the meter is looking at Vln and then applying the phase shift (30*) if it sees the I out of phase with the V? That seems to be a valid way to differentiate the source voltage for a current. How about when there is a mixed voltage load, resistive? Still just the same, apply the phase angle, whatever it is? I guess that then also applies to a mixed reactive load, say a dryer motor? Again, just measure V and I and apply the pf?

Edit: apply pf, I mean cos(30) regardless of reactance as this is apparent power.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
Jon, that helps a ton, I think I can see a solution here.

You are saying in essence the meter is looking at Vln and then applying the phase shift (30*) if it sees the I out of phase with the V? That seems to be a valid way to differentiate the source voltage for a current. How about when there is a mixed voltage load, resistive? Still just the same, apply the phase angle, whatever it is? I guess that then also applies to a mixed reactive load, say a dryer motor? Again, just measure V and I and apply the pf?

Edit: apply pf, I mean cos(30) regardless of reactance as this is apparent power.

Essentially yes. Please keep in mind that the meter is not actually measuring the phase angles and calculating the power factor but imagining that the meter is doing so is a useful narrative.

If you have a mixed load (some L-L, some L-N) or mixed load power factors (some resistive, some inductive) the same thing applies. Both of the sensors will see some current with some phase angle and will calculate the total power from that.

Under the hood, what the metering element is measuring the product of voltage * current moment by moment. Not RMS voltage, RMS current and phase angle, but the _instantaneous_ values of voltage and current and taking their product. But the integral of this product is directly related to the phase angles.

-Jon
 

GoldDigger

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Location
Placerville, CA, USA
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Retired PV System Designer
Yes, I understand, but I and the meter engineers are trying to figure this out exactly.

BTW, I have a neutral. I saw this mentioned in a later response.

Imagine this scenario: The only load is 10A, say a dryer heating element. The apparent power is 10*208= 2080 VA. Now say the load is 2 10A 120V devices, say two hair driers that happen to be on opposite legs. here S = 2*120*10 = 2400 VA. The pair of CT's can each indicate 10A yet there are two different VA values.

My conundrum is how does a metering device calculate S when there can be different V values?
The simple answer is that the meter is sensitive to phase angle between the line current and the reference voltage.

A 208V line to line resistive load has the current in phase with the voltage.

If you consider just the current on each of the hot leads with respect to the 120V line to neutral voltage you will see that the current is not in phase with the line to neutral voltage. That reduction in real power lets the sum of the two 120V power calculations to equal the single 208V power calculation.

When you have two separate resistive line to neutral resistive loads each line's current will be in phase with the voltage, so multiplying the current times the voltage gives the real power.
 

Hv&Lv

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-
Occupation
Engineer/Technician
Under the hood, what the metering element is measuring the product of voltage * current moment by moment. Not RMS voltage, RMS current and phase angle, but the _instantaneous_ values of voltage and current and taking their product. But the integral of this product is directly related to the phase angles.

-Jon
👍

Excellent way of putting it.
Just to add from from Bill Hardy:

“Today all modern digital meters measure voltage and current waveforms by digitizing them using many samples per cycle. This has changed the approaches that we can use for measuring power quantities. There are two fundamentally different approaches that can be used: 1) computation in the time domain and 2) computation in the frequency (Fourier) domain. Provided that the waveforms are repetitive over the number of cycles in the FFT, both approaches give identical answers for active power (P) and apparent power (S). There is no direct computation for reactive power in the time domain.
Measurement of active power in the time domain is very straight forward. We need simply sum up the product of instantaneous voltage and current over the period of time we desire (provided that the time period is an integral number of cycles).”
 

Besoeker3

Senior Member
Location
UK
Occupation
Electrical Engineer
If you consider just the current on each of the hot leads with respect to the 120V line to neutral voltage you will see that the current is not in phase with the line to neutral voltage. That reduction in real power lets the sum of the two 120V power calculations to equal the single 208V power calculation.
That seems to be a sticking point. If they are not in phase then they can't be single phase.
 

carpboy

Member
Location
PA
If you have a mixed load (some L-L, some L-N) or mixed load power factors (some resistive, some inductive) the same thing applies. Both of the sensors will see some current with some phase angle and will calculate the total power from that.
You have guessed my next step in the process, the mixed voltages and the addition of reactance.

Let me know if the following is in error:

The meter is referencing both LN and LL voltages. The phase angle between these two waveforms is a constant 30*.
The VLN waveform is used as a 'reference'.

For 120V resistive loads only: At the maximum of the LN voltage waveform, S=V*I*cos(theta), theta being the phase difference of the current and VLN waveforms. Voltage and current waveforms here are coincident.

For 208V resistive loads only: S=V*I*cos(30)=V*I*0.866. Since VLN is 'reference' this 30* is the phase difference of the two voltage waveforms.

Slightly unsure territory:

Mixed 208V and 120V loads, all resistive: Simply add the two together. The meter looks at V max and then calculates as above, but there are 2 Voltage/current waveforms per cycle (120V and 208V), the two S values get added together in a single cycle.

Very unsure territory:

120V load with inductance: Say 10* lag. S should remain unchanged, it is apparent power. As this lagging phase angle is introduced, how does the meter logic work? S should be simply V*I, but since the V and I waveforms aren't coincident how does the meter know which I peak to use for this calculation? Adding in the phase angle component will reduce the apparent power calculation. If the calculation is made at VLNmax, the I value will be < Imax. Since there is only one current waveform in this case, does the meter just grab the Imax value irrespective to phase angle, then do the math?

208V load with inductance: Same as 120V, single waveform, grab Imax and do the usual math?

Mixed voltages with inductance: This one has me totally baffled. Since there are two I waveforms I don't know how the meter would associate the two Imax's with the appropriate voltage. Related to that, how does it calculate phase angle between V and I when there are two waveforms of each present?

Thanks for holding my hand through this.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
That seems to be a sticking point. If they are not in phase then they can't be single phase.

Please, we've been over this many times. Everyone here knows that in the system being discussed the L-L voltage is not in phase with the L-N voltages. Everyone here knows that it is _not_ truly a single phase system.

The only 'sticking point' is that it is common practice in North America to use the name 'single phase' for this particular system that is two legs of a three phase wye system. It is virtually never used to supply loads that depend upon the phase angle difference. It is essentially only used to supply residential services.

We might as well call both 120/240 systems and these 120/208 systems 'split residential supplies' and never say anything about 'phase'.

-Jon
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
120V load with inductance: Say 10* lag. S should remain unchanged, it is apparent power. As this lagging phase angle is introduced, how does the meter logic work? S should be simply V*I, but since the V and I waveforms aren't coincident how does the meter know which I peak to use for this calculation? Adding in the phase angle component will reduce the apparent power calculation. If the calculation is made at VLNmax, the I value will be < Imax. Since there is only one current waveform in this case, does the meter just grab the Imax value irrespective to phase angle, then do the math?

208V load with inductance: Same as 120V, single waveform, grab Imax and do the usual math?

Mixed voltages with inductance: This one has me totally baffled. Since there are two I waveforms I don't know how the meter would associate the two Imax's with the appropriate voltage. Related to that, how does it calculate phase angle between V and I when there are two waveforms of each present?

The above question goes to the heart of why I said 'a useful way to think about the problem, but not what is truly happening under the hood' about considering phase angle.

If you mix both L-L loads and L-N loads, there are _not_ two current waveforms with different peaks that the meter can somehow measure. There is only a single current flow which is the combination of all the connected loads. It is a rule for adding sine waves of the same frequency but different amplitude and phase that the resulting sum is also a sine wave of the same frequency. So you throw whatever sinusoidal loads you have at the system (L-N, L-L, whatever phase angle) and what you get is a sine wave current of some strange phase angle, not necessarily in line with the L-L voltage or the L-N voltage. The phase angle that you would use for the calculation is whatever the composite current flow has.

But again remember that what is happening 'under the hood' is _not_ taking the phase angle of the voltage and current waveforms and doing power factor math. Thinking of it in terms of power factor math is useful but not what the meter is actually doing. At no time does the meter try to figure out the time offset of the waveform peaks or zero crossings or any of those other timing tricks. It is true that the rules of sinewaves mean that what the meter is actually doing (multiplying voltage and current on a continuous basis) gives the same results as the power factor rules. Thus the power factor rules are useful for understanding but shouldn't be stretched too far.

To go further needs video :)

-Jon


Of course the meter could be detecting the phase angle of this composite current flow,
 
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