I have four 150 watt hps lights installed on a building they pull 3.2 amps each with they came with no capacitors. the specifications on the transformer states if i use a 55uf cap the lights will pull 1.5 amps each. can i put one cap on the line to these lights to do the same thing if so what size cap for all four lights?
150 watt hps
- Thread starter domnic
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mbrooke
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Yup, caps across the incoming line will work. 55uF at something like 370 volts will work. Just make sure they are rated for HID fixtures.
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Are you really saying that if each luminaire wants a 55uf cap you can put one 55uf cap on the line feeding four luminaires and get the same effect? Your answer appears that way.
mbrooke
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Incoming line to the fixture, which would be in the fixture itself. Not the feeder. Also notice the mention of HID rated capacitor.
mbrooke
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Every fixture would need a 55uF capacitor inside of it. Is there a spot for a capacitor?
mbrooke
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One cap on the feeder would not produce enough leading current to cancel the reactive lagging current. Each cap is sized to cancel out most of the lagging current from the ballast.
If you did want to use on cap you can do that, but it would have to be much larger, and current would only go down at the point of attachment.
FWIW don't sweat the caps. Ive done the same
jumper
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Lots I do not know, but every HID fixture I have installed had its own cap.
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Try 220. Typo?
mbrooke
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How much uF would really be needed in this case if the cap was at one point?
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4x55=220uF.
mopowr steve
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No
Residential electric meter registers Watts=volts x amps x power factor. The same reason power saving units sold don't really save money. They just like to show you how amperage goes down but don't show you power factor going up thus nullifying end result.
Residential electric meter registers Watts=volts x amps x power factor. The same reason power saving units sold don't really save money. They just like to show you how amperage goes down but don't show you power factor going up thus nullifying end result.
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It does not work that way. Voltage and current are what you measure and voltage does not change. The phase angle between voltage and current changes, and the projection of the current vector onto the voltage axis gives the real power.
kwired
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Volts applied is still same no matter what the PF is.
If your input amps is 3.2 amps and voltage is 120 then VA is 384. If PF is .50 then Watts is 192, VAR is 332.
To correct to .95 PF you need a capacitor with a 269 VAR (don't know right off how to convert that to microfarads). This would leave you still with 120 volts input and 192 watts, but VA will drop to 202, with remaining VAR of 63.
As far as where to place in the circuit, where do you want PF correction? This is light enough load that it really wouldn't have much impact on the branch circuit unless it was a really long run, then it would help with voltage drop because actual current is lower, and would lower line losses. put all four lights on one long run and your current goes up 4 times, but is reduced if correction is on load end.
Becomes a question of what are your concerns and what you may be trying to accomplish. Outside of line losses watts remains the same either way and that is what you are billed for unless you do have power factor penalties assessed to you.
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