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Forum Index : Electronics : twin 36W fluro power useage
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| rgormley Senior Member  Joined: 22/02/2006 Location: AustraliaPosts: 245 |
i just measured the power (amps) that my twin 36 watt fluro tube draws.. 0.86amps thats around 206.4 watts fo the 2 tubes plus ballast is this about correct??? i thought fluros were better with engery, or is it a case of the light output versus power usage is better?? the lighting in my room draws more power than my computer!! |
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Gizmo Admin Group  Joined: 05/06/2004 Location: AustraliaPosts: 5030 |
I just tested a single fluro with one of those Jaycar power meters. It said 0.4 amps, for a single 36 watt bulb. But is also said it was using 34 watts, and the power factor was 33. Fluro's are capacitive, and I think its the power factor thats giving the higher amp rating. I dont know enough about AC loads to go any further, but I do know anything inductive ( motor ) or cacitive will give a false amp reading. Glenn The best time to plant a tree was twenty years ago, the second best time is right now. JAQ |
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| RossW Guru  Joined: 25/02/2006 Location: AustraliaPosts: 495 |
It's not necessarily a "false" reading. You're probably too used to DC where amps *DO* something useful. A reactive load (capactive or inductive) draws current out of phase to the voltage. The further out of phase, the lower the "power factor". (The cosine of the phase difference). Taking an entirely reactive load, you will measure current (lets say 1 amp) and voltage (lets say 240V) and you would be inclined to say "that's 240 watts". DC you'd be right, but in AC it's volts * amps * powerfactor. Purely reactive (inductive is 90 degrees lagging, capacative is 90 degress leading) will draw current, but is "wattless" power. It draws amps that means you need larger wire, but it doesn't actually do any work. From 90 degrees to 0 degrees (in phase, Unity power factor, as you'd find with a resistive load) is a gradual transition where the closer you get to 0 degrees, the more of your amps are doing useful work. Does that make any sense?! |
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oztules Guru Joined: 26/07/2007 Location: AustraliaPosts: 1686 |
Rossw "but it doesn't actually do any work. " I hear this, but it makes me wonder.... If I have an inductive load of .8pf, and I turn it on, and it requires 8 real amps, that means I must supply 10 apparent amps across the supply lines. 2 of these amps are not used because of the vector thinggy, and 8 are used in phase with the voltage and we get real watts out...... Now if we had superconductors, I assume we would have no loss in the line due to the 2 unused out of phase amps, but in the real world, we have resistance. As you say, we need thicker wire. Does this imply that we may do some work heating up the real world supply wires?... ands does that mean we have to generate more real power than what the real load is using? Putting it another way, if instead of needing to get the VARs from the supply, if we put caps at the load and "generated them locally", does this actually mean we lessen the load on the generator (less line loss which must surely cost real power... maybe?) This stuff makes my head swim a bit. Just when I think I got it..... I aint got it.... I guess what I'm trying to find out is.... do we expend more power generating real power for a poor pf load (including real world supply lines) than for a equivalent kw load of unity pf. I know in theory vars are wattless, but in say the grid, do they cost real watts...... I'm going round in circles aren't I 
.......oztules Village idiot...or... just another hack out of his depth |
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| rgormley Senior Member Joined: 22/02/2006 Location: AustraliaPosts: 245 |
yes my brain dribbled from my ears after i read Rosss`s post... so could i just be happy that the twin fluro is using 72 watts (plus whatever the ballast uses) or is it really using 0.86A x 240Cac =206.4 watts !! |
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| RossW Guru Joined: 25/02/2006 Location: AustraliaPosts: 495 |
Current draw will be 10 REAL amps. Load will be 2400 VA but work would only be 1920 Watts Absolutely. Because the resistive component is in phase. Yes. No such thing as a free lunch. Yes indeed. And that is why (industry particularly) spends so much time and money on powerfactor correction. No, you're completely correct. What people really should do is measure the amps. They are the important thing. Your inverter will pull much more current from your batteries to supply a 500W load at 0.5PF than a 500W load at unity PF. To the original poster: if your fluro is NOT a "HPF" fitting (they will usually be branded such, and have a largish powerfactor correction capacitor inside), see if you can (by experiment and your jaycar powermeter) find a suitable PFC to get your PF "near unity" (0.95 or better). Monitor the DC power into your inverter with the fluro running barefoot vs with the PFC and report back! |
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| oztules Guru Joined: 26/07/2007 Location: AustraliaPosts: 1686 |
Thanks Ross, "No, you're completely correct".... more by good luck than good management I think. Just to make sure... In the example above where VA equals 2400 and the load equals 1920watts.... How much power does it mean we actually have to generate... in watts or petrol for the genny. Is it 2400 + line loss + genny coils @ 10A or 1920 + line loss and genny coils @ 10A I'm thinking the latter but not completely sure. I'm trying to quantify this: "What people really should do is measure the amps. They are the important thing. Your inverter will pull much more current from your batteries to supply a 500W load at 0.5PF than a 500W load at unity PF. " You get up too early for me. ...........oztules Village idiot...or... just another hack out of his depth |
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sPuDd Senior Member Joined: 10/07/2007 Location: AustraliaPosts: 251 |
G'day all, I was helping a friend install some new 36W round tube fluros in his house. They had a 0.1uF X2 type 275VAC rated MKT capacitor across the supply terminals. The Jaycar power meter showed about 0.8PF with the cap in, and about 0.3PF with the cap out. The Jaycar meter shows the Watts used as a result of Volts x Amps x PowerFactor. So both times the 'diplayed' power was the same, but the actual power expected to be measured by the energy suppliers meter will be lower for a lower power factor. Going back to what rgormley was asking, assuming his fluro has about a 0.33PF then: 0.33PF x 240V X 0.86A =~ 68W which is about 2 x 36W tubes, give or take a little. sPuDd.. It should work ...in theory |
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| oztules Guru Joined: 26/07/2007 Location: AustraliaPosts: 1686 |
Gee Spudd, "The Jaycar meter shows the Watts used as a result of Volts x Amps x PowerFactor. So both times the 'diplayed' power was the same, but the actual power expected to be measured by the energy suppliers meter will be lower for a lower power factor. " This is the sort of thing I don't quite get. As I understand, my old fashioned dial type grid/utility meter uses a series coil to generate a magnetic field proportional to current..and a parallel coil to get a magnetic field in proportion to the voltage (240). It directs these two fields onto a rotatable plate, which rotates due to the eddy currents induced into it. (makes it spin). It has a permanent magnet to give some calibtrated drag.... so we measure Inphase power that causes the disk to spin against a load... we measure power. So a poor power factor should not produce any discernible variation to charged power consumption.....??? as the current in the VARs is out of phase with the voltage, and so does not interact with this magnetic system.... perhaps... maybe... If the VARs component is wattless, how would this measure that? Well.... I'm still not getting it 
......oztules Village idiot...or... just another hack out of his depth |
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GWatPE Senior Member Joined: 01/09/2006 Location: AustraliaPosts: 2127 |
Something to ponder on by the engineers. A fluoro fitting that measures 32W [by jaycar powermeter] connected to the mains. Round tube type, not long tube, with no caps, only a ballast. This same fitting when connected to my inverter consumes only 1.22ADC with the battery sitting on 26.8V. Gordon. become more energy aware |
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| sPuDd Senior Member Joined: 10/07/2007 Location: AustraliaPosts: 251 |
Long story short: Poor power factor (inductive or capacitive) is not really a problem for a grid consumer with an old type utility meter. It is detrimental for the grid supplier who has to provide more power but cannot charge for it. (Also bad for the environment yada yada...) Poor power factor may be a problem for some inverters, I do remember Gizmo having to remove the PF caps in his fluros to help his little inverter run properly. On a side note, I recall my old mans Arc welder was quite a vintage transformer type unit that had a power factor so bad that while welding the utility meter disk would vibrate like it was going to fall off! But it would never advance forward :-) sPuDd.. It should work ...in theory |
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| RossW Guru Joined: 25/02/2006 Location: AustraliaPosts: 495 |
26.8 Volts * 1.22 Amps = 32.69 watts. You are surprised because??? |
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| RossW Guru Joined: 25/02/2006 Location: AustraliaPosts: 495 |
The power required to be made will depend on many factors. Offhand its been too long ago for me to recall exactly what it is. I slept in this morning too! (I generally start by 6am and finish between 11pm and midnight - except when things break like yesterday, was 4am before I went to bed, 5am I got to sleep, 5:30 I had to get back up!) |
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| GWatPE Senior Member Joined: 01/09/2006 Location: AustraliaPosts: 2127 |
Hi Ross, I was trying to point out that the battery only has to provide the same amount of power. I should point out that the AC current was 0.36A. The VA would be 86W, and clearly the inverter did not have to supply this much power. In the grid system, the transmission is affected by the power factor. I have been told with good reliability that the power utility decides how and what to charge for power. For domestic use with power factor usually above 0.8, they charge for kWhr. For large consumers with lower power factor, they charge for kVA. These consumers will benefit from power factor correction to bring the kVA closer to the kW. Gordon. become more energy aware |
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| BjBlaster Regular Member  Joined: 04/04/2008 Location: AustraliaPosts: 55 |
That is a very efficient inverter.... Check out my projects here in: Bj's Shed |
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| RossW Guru Joined: 25/02/2006 Location: AustraliaPosts: 495 |
Lets say for a moment his meters are dead accurate. 26.8V could be as high as 26.899999V, and 1.22 amps could be as high as 1.2299999 amps. That makes 33.087 watts. If the inverter is 95% efficient at that, the output would be 31.43 watts. If the voltmeter and the ammeter were out by as little as 2%, the figures would allow for 32 watts to be delivered to the load with an inverter efficiency of only 93%. There is also nothing that says the load is exactly 32 watts, nor that the jaycar powermeter is anything even aproaching "dead accurate"! In fact, reading from its spec sheet, "Measure power accuracy: +/- 4% or +/- 10W" So lets not be silly here huh? |
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| GWatPE Senior Member Joined: 01/09/2006 Location: AustraliaPosts: 2127 |
Thanks guys, If I had intended to quote numbers, I would have used and quoted readings from the 5 1/2 digit calibrated HP reference meter and measured and calculated all the numbers for each test. I was merely pointing out the AC current alone will give different calculated power when power factor is not 1, even though the power supplied by the source close to equaled the power consumed by the load. In this case power transmission had no influence. The measured 240VAC current was 0.36A The VA was 86W, but the W was only 32W. Gordon. become more energy aware |
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| RossW Guru Joined: 25/02/2006 Location: AustraliaPosts: 495 |
Not wishing to be argumentative Gordon, but the mistake seems to be over the mis-interpretation of "volts * amps = watts". Also, "VA was 86W" is absolute nonsense. "VA was 86" is fine, but NOT 86W. Once you include powerfactor, the 86VA becomes 32W, which is fine. *ding*. I think I see a glimmer of what you may have been getting at? Were you saying that "86VA * pf = 32W" on the AC side, but it still only drew 32 *watts* from the DC side, not the *CURRENT* of 86VA? |
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| GWatPE Senior Member Joined: 01/09/2006 Location: AustraliaPosts: 2127 |
Most readers will have access to digital volt meters. They will be able to measure independently, AC volts and AC amps, and DC volts, and DC amps, but not necessarily power factor. Most will calculate the power by multiplication in each instance. The example here highlights the problems with this approach in AC measurements when the pf is not 1. ACVolts x ACamps without a knowledge of pf does not equal W, unlike the familiar DCvolts x DCamps. Gordon. become more energy aware |
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| BjBlaster Regular Member Joined: 04/04/2008 Location: AustraliaPosts: 55 |
Ok, This is slightly off topic, but still within the realms of this discussion.... How can we measure Power Factor? How does that little Jaycar meter do it? I want to measure AC current to work out my house usage in Watts to compare it to my grid fed solar contribution in watts, but can't work out how to measure PF on the fly?? Bj Check out my projects here in: Bj's Shed |
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