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View Full Version : Hablo usted POWER FACTOR!!?


piercho
03/06/2002, 09:14 PM
This is a major irritation for me, so I just HAVE to vent.

In AC, V X A = VA (volt-amps), not watts. V X A X PF = W. AC amps and volts are vector quantities and their magnitudes only multiply directly if they are in perfect phase. The value used to get from VA to W is PF, or power factor.

A common example: the HX-NPF 250W H37 ballast recommended in the link "MH for less than $100", and sold by PFO in their 250WEYE ballast. Described on this Advance data sheet: http://www.advancetransformer.com/ecom_PDFS/out/9071568311.pdf

The ballast draws 285W, so it's amp draw is (by your formula) 2.4A, right? Not correct. Check out the power factor on the data sheet. 0.5. The worst case (running) amp draw is 285/(.5 X 120V) =4.75A. Typical running amps is 4.20, according to the data sheet. For this particular ballast circuit, starting amps can be even worse at 6.85A.

Lets say I was budgeting a 15A house electrical circuit. If I took the 2.4A value, I'd believe I could fit 6 of these ballasts on a single circuit. I reality, if I was starting the ballasts off of the same timer, I could only fit 2. For amp budgeting purposes, in this case, A=W/V is off by 185%.

Cap and coil (CWA) magnetic ballasts have a PF of .9, and some electronic ballasts have a PF approaching unity. So, in those cases, using VA for W does not produce such large errors. But there is still an error. Most ballasts have a max amp draw printed on them, and that is the value that should be used to calculate amp draw.

There are devices that draw LOTS more current than the watts that they consume reveal. IMO, these HX-NPF ballasts are probably the worst of them for the devices that are in typical use for the reef aquarium. Also, you only pay for (as a residential customer) W, not VA, so the V X A = W is wrong there, too.

You've given me a lot of help over the last few months so I am very reluctant to criticize.:) But I've been fighting this issue since I came to the board and putting it in print as fact just makes this erroneous notion harder to correct.

piercho
03/06/2002, 09:22 PM
BTW, the effort you put in to this board is amazing, so don't let one anal EE with his knickers in a twist put you off. Your web site and these columns, as well as you personal attention to the DIY board ahve been a big help to me. Thanks.

Snailman
03/07/2002, 05:50 AM
I know that reactive loads may not have a power factor of unity. I thought long and hard about trying to explain power factor and could not figure out a good way without also haveing to explain electronics 101 and AC ohms law.

MiNdErAsR
03/07/2002, 09:33 AM
Piercho,
If one was to measure the load of a component in a real time situation you would get the reading in amperage. Since this component is running in a specific application, would Power Factor (PF) still need to be used in the V x A = W equation? For example, if a water pump (such as a GenX Mak4) was measured to be drawing 0.8a wouldn't the PF be a non-issue, as opposed to calculating using the given max draw on the pump placard?

BTW, excellent post. :)

piercho
03/09/2002, 07:59 PM
IMO, the amperage you are reading is the typical current draw for your application. If you wanted to use this measurement to determine the typical watts consumed by the pump, you could say 0.8A*120V=96VA. The pump is consuming 96 watts or less, depending on how close it's power factor is to unity.

Going the other way, saying that a pump rated at 96W is drawing no more than 0.8A could be underestimating the current the pump draws.

It's been close to a decade since I studied motors in the lab so I have no recollection of what their PFs typically are.

If you want to use the measurement to budget a circuit, keep in mind "in-rush currents" when the pump first turns on. I don't know anything about aquarium pump wiring. I assume that they have no current regulation so when the motor is static and power is initially applied the current will be much greater than the normal running current.

Snailman
03/09/2002, 08:40 PM
Originally posted by piercho
Lets say I was budgeting a 15A house electrical circuit. If I took the 2.4A value, I'd believe I could fit 6 of these ballasts on a single circuit. I reality, if I was starting the ballasts off of the same timer, I could only fit 2. For amp budgeting purposes, in this case, A=W/V is off by 185%.

If you were an electrican rather than an EE you would also know that the most constant draw you should have on a 15A circuit is 12A. Also many electric things draw more current when they start so loading a circuit to 96% of it's capacity at run load and starting the entire load at once there is only 4% head room for starting current even if the PF is unity.

piercho
03/31/2002, 01:14 PM
Snailman: 12A sounds like a practical value for maximal loading of a 15A breaker to avoid tripping it when several devices power up at once. I see where it could be construed I was saying that a circuit can be loaded at it's breaker value in the origonal post, and I agree doing that in our application would be foolish. As far as in-rush currents in coils (motors, magnetic ballasts, whatever), I alluded to that in my previous post.

I'm sticking with my main point: people who budget circuits based on W=V X A can get into trouble. I don't think you or anyone else should publish that formula as fact when it's not. I'm not trying to argue with you. I'm just trying to give you feedback that I've seen people make assumptions based on that formula that lead to mistakes, like buying ballasts that overload their tank circuit when they could have purchased ballasts to perform the same function that don't.

I apologize that my thread lead was hostile and the origonal post came off as a lecture. PEACE.