120/208 Single Phase Power Calculations

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carpboy

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I have a tenant metering system where the service is 120/208. There is a meter for each tenant that has a CT on each leg and voltage reference to the meter for each leg.

How does the meter calculate the apparent power in a situation like this? The load could be 120 V or 208 V. I've read one instance where it was stated that real power could be measured and then apparent back calculated. But either method needs to know the voltage components.

Thanks.
 
carpboy said:
I have a tenant metering system where the service is 120/208. There is a meter for each tenant that has a CT on each leg and voltage reference to the meter for each leg.

How does the meter calculate the apparent power in a situation like this? The load could be 120 V or 208 V. I've read one instance where it was stated that real power could be measured and then apparent back calculated. But either method needs to know the voltage components.

Thanks.
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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.
 
petersonra said:
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.
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I think maybe he didn't ask his question with correct terminology?

If you have a 120 volt load current isn't in phase with any of the line to line voltages is it? Don't you need a reference to neutral to have anything in phase to that voltage measurement? Isn't this why they typically use 5 and 7 jaw meters on wye systems.
 
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".

.
 
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
 
gar said:
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".

.
Click to expand...
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.
 
5 jaw meters (12s) have the fifth clip for neutral reference.
The metering equation is

P=Va*Ia+Vb*Ia (vectors) ( iPhone, can’t get the arrows over the variables)

9S and 16 s is basically the same equation, just add C phase V&I
 
211009-1055 EDT

The general rule is:

If there are N wires in a multiphase circuit, then it only requires N-1 wattmeters to measure total power.

.
 
gar said:
211009-1055 EDT

The general rule is:

If there are N wires in a multiphase circuit, then it only requires N-1 wattmeters to measure total power.

.
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What’s odd is the fact the most popular and abundant meter doesn’t follow that rule.
 
carpboy said:
How does the meter calculate the apparent power in a situation like this? The load could be 120 V or 208 V.

Thanks.
Click to expand...
I had a similar question just last week and the answer I came up with is that at peak voltage on a 120V circuit, the current will be at a peak (discounting for power factor) 120V to neutral, and a 208V circuit will have peak current when the sine waves of each circuit are 104V to neutral, so AP=IR is a self adjusting measurement.

I hope what I said makes sense, remember I'm an uneducated old sparky. :)
 
oldsparky52 said:
I had a similar question just last week and the answer I came up with is that at peak voltage on a 120V circuit, the current will be at a peak (discounting for power factor) 120V to neutral, and a 208V circuit will have peak current when the sine waves of each circuit are 104V to neutral, so AP=IR is a self adjusting measurement.

I hope what I said makes sense, remember I'm an uneducated old sparky. :)
Click to expand...

The nominal voltages we use, aren't really the peak voltages. They are the RMS voltages, that are a special kind of time averaging. The same goes with current values. The motive for defining RMS voltage, is to calculate an equivalent DC voltage and DC current that would produce the same average power as the actual AC voltage waveform.

120V to neutral is really +170V to neutral at is high peak, and -170V to neutral at its low peak. The peak voltage is sqrt(2)*the RMS voltage.
 
Besoeker3 said:
So it isn't single phase.
Others may not agree.
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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.
 
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kwired said:
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.
Click to expand...

Most meters do measure real power by default, unless it is specifically set up to keep track of reactive power, which is typically only something you see on larger services where reactive power intensive loads are expected. Your meter will accumulate the instantaneous product of voltage and current over time.

For a typical residential customer, where the meter is simple and only measures real power, adding power factor correction won't reduce your meter's totals.
 
Carultch said:
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.
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Thank you for that Sir. The two of the three phases are not in phase.
That's how I'd call it too. UK and many other countries use single phase. Commercial and industrial use, as a rule, 400V/230V three-phase.
 
Carultch said:
The nominal voltages we use, aren't really the peak voltages. They are the RMS voltages, that are a special kind of time averaging. The same goes with current values. The motive for defining RMS voltage, is to calculate an equivalent DC voltage and DC current that would produce the same average power as the actual AC voltage waveform.

120V to neutral is really +170V to neutral at is high peak, and -170V to neutral at its low peak. The peak voltage is sqrt(2)*the RMS voltage.
Click to expand...
Other than the numbers, does that change what I said? If so, in what way? Thanks.
 
211009-1503 EDT

Relative to post #9 for single phase center tapped supplies I believe the power company meters, when electro mechanical meters were used, had one voltage coil (total line to line voltage), two current coils, and one multiplier disk. A multiplier disk takes the instantaneous product of current and voltage (instantaneous product ( power ) ) converts this to speed of rotation, and instantaneous rotation is integrated mechanically to give a kWH accumulation. The assumption is made that the two line to neutral voltages are sufficiently equal that 1/2 that voltage gives adequate accuracy.

.
 
gar said:
211009-1503 EDT

Relative to post #9 for single phase center tapped supplies I believe the power company meters, when electro mechanical meters were used, had one voltage coil (total line to line voltage), two current coils, and one multiplier disk. A multiplier disk takes the instantaneous product of current and voltage (instantaneous product ( power ) ) converts this to speed of rotation, and instantaneous rotation is integrated mechanically to give a kWH accumulation. The assumption is made that the two line to neutral voltages are sufficiently equal that 1/2 that voltage gives adequate accuracy.

.
Click to expand...
imbalance in voltage and/or current, or unequal power factors in each leg can cause errors up to (realistically) >5%, even in the new electronic meters.
the older mechanical ones could have errors higher. Years ago we tested some and could get all the way up to ~8-10% error
 
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