V.A calcs
- Thread starter normbac
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Buck Parrish
Senior Member
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- NC & IN
chris kennedy
Senior Member
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- Miami Fla.
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- 60 yr old tool twisting electrician
That would be a PF around .75. Not a very efficient fixture IMO.
LarryFine
Master Electrician Electric Contractor Richmond VA
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- Henrico County, VA
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- Electrical Contractor
Yes, the ballast does consume a bit of power itself. Ballasted fixtures should always be figured by the ballast current rating, and not the bulb/tube wattage. Some fluorescents actually have a lower line consumption than the tube wattage would suggest.
LLSolutions
Senior Member
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- Long Island, NY
It may be a combination of the ballast and igniting it.
chris kennedy
Senior Member
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- Miami Fla.
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- 60 yr old tool twisting electrician
OK.
This is IMO the least efficient luminaire I have seen recently. PF of .9 is reasonable for a modern fixture.
LarryFine
Master Electrician Electric Contractor Richmond VA
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- Henrico County, VA
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- Electrical Contractor
Efficiency and PF are not synonymous.
chris kennedy
Senior Member
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- Miami Fla.
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- 60 yr old tool twisting electrician
Care to share?
Power consumption of the bulb is 50W with a PF of 1. The power into the ballast is 68 VA at some unknown power factor. So this 68 VA goes two places - to the bulb and out of the ballast as heat. Since one figure is in VA and the other is watts we don't know how much of the difference is power loss in the ballast vs. power factor.
chris kennedy
Senior Member
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- Miami Fla.
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- 60 yr old tool twisting electrician
50W ?.75pf=6.666.... That's pretty close.
gar
Senior Member
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- Ann Arbor, Michigan
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- EE
081214-2030 EST
chris:
If I take a very good capacitor, these can be very low in loss, and select its value of capacitance to have a capacitive reactance of 100 ohms, then connect it in series with a 100 ohm resistor. To this series combination I apply a voltage to cause 1 A of current flow. The input electrical power is 100 W and the output power is 100 W in heat. There is no other heat loss in this series combination. Thus, the efficiency is 100%. The phase angle is 45 deg, and the input voltage to the series combination is 141.4 V. The input VA calculation is 141.4 VA, and the PF is 0.707.
I chose to use a capacitor in this example so I did not have to process the series resistance in an inductor. With an inductor we have substantial power loss in its internal resistance, and in a practical illustration we would have less than 100 % efficiency if the intended load was the 100 ohm resistor.
Efficiency and power factor are not the same. However, bad power factor does influence the efficiency of the power distribution system even if it did not effect the efficiency of my capacitor resistor example.
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chris:
If I take a very good capacitor, these can be very low in loss, and select its value of capacitance to have a capacitive reactance of 100 ohms, then connect it in series with a 100 ohm resistor. To this series combination I apply a voltage to cause 1 A of current flow. The input electrical power is 100 W and the output power is 100 W in heat. There is no other heat loss in this series combination. Thus, the efficiency is 100%. The phase angle is 45 deg, and the input voltage to the series combination is 141.4 V. The input VA calculation is 141.4 VA, and the PF is 0.707.
I chose to use a capacitor in this example so I did not have to process the series resistance in an inductor. With an inductor we have substantial power loss in its internal resistance, and in a practical illustration we would have less than 100 % efficiency if the intended load was the 100 ohm resistor.
Efficiency and power factor are not the same. However, bad power factor does influence the efficiency of the power distribution system even if it did not effect the efficiency of my capacitor resistor example.
.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
081214-2056 EST
normbac:
The ballast is an inductive component of an equivalent circuit of an inductor and resistor. It is also not linear.
This load causes a phase shift of the load current relative to the applied voltage. When you draw the resulting vector diagrams for the voltages and currents you find that as soon as you deviate from a purely resistive load that the VA value will be larger than the power dissipated in the resistance. Thus, a power factor less than 1.0 .
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normbac:
The ballast is an inductive component of an equivalent circuit of an inductor and resistor. It is also not linear.
This load causes a phase shift of the load current relative to the applied voltage. When you draw the resulting vector diagrams for the voltages and currents you find that as soon as you deviate from a purely resistive load that the VA value will be larger than the power dissipated in the resistance. Thus, a power factor less than 1.0 .
.
There's more than 50 watts of power going into the ballast. All of the ballast losses that turn into heat are watts. For example if the PW was 0.9, 68 VA * .9 PF = 61.2 watts going into the ballast, of which 50 W go to the lamp and 11.2 W are losses in the ballast.
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