Kilowatt hour reading for 3 phase lighting circuits

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highpowered

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I have a Levington meter series 2,000 hooked up on 6 circuits in a 208y/120volt 3 phase 4 wire system. total connected load for the circuits is 25.8amps @116volts. In 21 days the meter read 998kwhrs. By my calculations it should have read 1,452 kwhrs. if I cube it 1.73 I get what the meter reads but I dont think that that I applies to this case? If it does then my other meter is off. Any suggestions or is the meter malfunctioning or installed incorrectly. Thankyou in advance!
 
How did you come up with a single current on six circuits?

Did you add all six together?

Are any of them line to line loads?

Are they all on the same phase?
 
the phaseing is circuits #2,8a phase #4,10b phase #6,12 c phase under 3 doghnuts. I took amp readings on each circuit and added them together
25.8 then averaged the voltage reading on the 3 phase to 116volts(garage flourescent lights are the loads)

25.8*116=2,992.8watts/1,000=2.9kw

2.9kw*504 hours=1,508 kw hrs
 
highpowered said:
the phaseing is circuits #2,8a phase #4,10b phase #6,12 c phase under 3 doghnuts. I took amp readings on each circuit and added them together
25.8 then averaged the voltage reading on the 3 phase to 116volts(garage flourescent lights are the loads)

25.8*116=2,992.8watts/1,000=2.9kw

2.9kw*504 hours=1,508 kw hrs
Click to expand...

There are a couple of issues to consider with your methodology.

The first is your averaging of the voltage readings. You should use the voltage reading for each phase with its associated load current to be more precise. If the voltages were 112, 116 and 120 (averaging 116) with the associated load currents, you would get a different result than if the voltages were 114, 116 and 118 (also averaging 116.)

The second issue is that 25.8A * 116V = 2.99kVA, not kW. To calculate the kWH accurately, you would need to consider what the power factor is for each load.

The third issue is whether or not the load currents are constant/unchanging for the total 504 hours of measurement.
 
david luchini said:
There are a couple of issues to consider with your methodology.

The first is your averaging of the voltage readings. You should use the voltage reading for each phase with its associated load current to be more precise. If the voltages were 112, 116 and 120 (averaging 116) with the associated load currents, you would get a different result than if the voltages were 114, 116 and 118 (also averaging 116.)

The second issue is that 25.8A * 116V = 2.99kVA, not kW. To calculate the kWH accurately, you would need to consider what the power factor is for each load.

The third issue is whether or not the load currents are constant/unchanging for the total 504 hours of measurement.
Click to expand...
Thank you for taking time to respond I appreciate it
The parking garage lights are always on 24hrs a day. I'm not sure of how to obtain the power factor for each load please explain? I agree about the voltage my goal is to get my calculations close to the actual reading on the meter. I dont need to be dead on. Is'nt watts I*E? thanks again
 
Last edited:
120726-1602 EDT

highpowered:

Your method is reasonable for a ball park figure. But you need an estimate of the power factor. From the values you provided it looks to be about 0.69 .

We can assume the lights have a lagging power factor, and 0.69 would not be unlikely. Not enough information on the lamps.

In an AC circuit where the load is not a pure resistance, then you can expect the energy consumed to be less than predicted by V*I * time.

If you take a good quality capacitor and measure current, voltage, and power you might find the real power to be 1% of the volt-amperes measured.

.
 
gar said:
120726-1602 EDT

highpowered:

Your method is reasonable for a ball park figure. But you need an estimate of the power factor. From the values you provided it looks to be about 0.69 .

We can assume the lights have a lagging power factor, and 0.69 would not be unlikely. Not enough information on the lamps.

In an AC circuit where the load is not a pure resistance, then you can expect the energy consumed to be less than predicted by V*I * time.

If you take a good quality capacitor and measure current, voltage, and power you might find the real power to be 1% of the volt-amperes measured.

.
Click to expand...

Ok what would the equation be using the power factor estimate?
 
120726-2208 EDT

highpowered:

The definition for power factor is:

PF = real power/volt-amperes where volts and amperes are the RMS values.

If your have a linear RLC load with sine wave voltage excitation (also means the current has a sine wave shape, but maybe not in phase with the voltage), then PF is the cosine of the phase difference between the current and voltage.

For you to get an approximate value for PF you need to measure it, or get the information from someone or document with the information.

Consider a resistor and a good quality capacitor in series connected across an AC sine wave voltage source. The capacitor dissipates very little power and we shall ignore it. So all of the dissipated power is in the series resistor, and none in the capacitor. Thus, power dissipation is I2*R. The voltage across the resistor will never be greater than the source voltage. Power dissipated can not be greater than source volts times amperes. For a capacitor with a reactance equal to the resistance the voltage across the resistor will be 0.707 times the source voltage. Power dissipated is 1/2 the value of source volts times amperes. PF in this case is 0.50 .

Ask more questions.

.
 
gar said:
120726-2208 EDT

highpowered:

The definition for power factor is:

PF = real power/volt-amperes where volts and amperes are the RMS values.

If your have a linear RLC load with sine wave voltage excitation (also means the current has a sine wave shape, but maybe not in phase with the voltage), then PF is the cosine of the phase difference between the current and voltage.

For you to get an approximate value for PF you need to measure it, or get the information from someone or document with the information.

Consider a resistor and a good quality capacitor in series connected across an AC sine wave voltage source. The capacitor dissipates very little power and we shall ignore it. So all of the dissipated power is in the series resistor, and none in the capacitor. Thus, power dissipation is I2*R. The voltage across the resistor will never be greater than the source voltage. Power dissipated can not be greater than source volts times amperes. For a capacitor with a reactance equal to the resistance the voltage across the resistor will be 0.707 times the source voltage. Power dissipated is 1/2 the value of source volts times amperes. PF in this case is 0.50 .

Ask more questions.

.
Click to expand...
So there is not a meter of some type that can read it like a volt meter? the Ballast for the t12 2 bulb 26 watt flourescents would have the info on it? My knowledge with capacitors and resistors is over 10 years old. I learned in school about them but never used one.
 
highpowered said:
So there is not a meter of some type that can read it like a volt meter? the Ballast for the t12 2 bulb 26 watt flourescents would have the info on it? My knowledge with capacitors and resistors is over 10 years old. I learned in school about them but never used one.
Click to expand...

Yes, you can get a meter that can read Power Factor. Either a permanent switchboard type or a portable type such as Fluke.
 
120727-1142 EDT

highpowered:

You can buy a Kill-A-Watt EZ from Home Depot for about $30. This unit is good for 120 V loads up to 15 A. It measures RMS voltage, RMS current, watts, volt-amperes, frequency (not much use except on small generators), power factor (except my two most recent units read incorrectly on very low power factor, a good capacitor), kWh, and elapsed time. The EZ version does not loose accumulated kWh and time with loss of AC power.

You can also get my book on energy measurement.

The Kill-A-Watt will allow you to play with various loads and see how PF changes. An interesting experiment is to measure a single phase induction motor that does not use a run capacitor, 1/3 HP is a convenient size. This will have an unloaded PF of maybe 0.25 up to 0.90 as you change a power factor correction capacitor from 0 ufd to about 102 ufd. Need to use high quality polypropylene capacitors.

.
 
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