Ohms law, power factor & reality.

Discussion in 'General Electronics Chat' started by Sherlockohms, Apr 8, 2012.

  1. Sherlockohms

    Thread Starter New Member

    Apr 8, 2012
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    :eek:Hi there,
    I am a 30+ yr HVAC?R service technician who has ben considered one of the best wherever I worked, I always used Ohms law A X V = W etc for troubleshooting.
    I knew transformers didnt apply to resistance and ohms laws but never had it throw me on the job, I have a co worker who adamantly insists power factor in a residential home is making a accurate reading of power used impossible, he cant explain how that afffects me using ohms law to see if a circuits open or shorted etc. and i read its imaginary power, and we are billed for real power and PF is insignifigant in residential and isnt even a factor in residential.
    yet i see CFL bulbs have a 50% PF yet get rebates?
    Is the ohms law i always used no onger accurate to my line of work?/ I never had a circuit use more than it read or had breakers blow etc, How do i wxplain this to myself or him to stop his insisting yet unable to elaborate, i see what he keeps insisting about the cosign to the vector blah blah but that changes nothing in my little world< or does it?
     
  2. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    A X V = W is not even ohm's law. Ohm's law is V=I.R, and I.V=P is the law for power, but anyway power factor should not matter most of the time. If you have a household with lots of large switch mode supplies or inductive motors then the average current will be a little larger than expected for the power rating, but that's about it.
     
  3. #12

    Expert

    Nov 30, 2010
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    Don't worry about it and don't argue with him. I've been doing HVAC for 40 years and Ohms Law works well enough to get them repaired.

    One thing to know is that your amp-clamp reading times your voltmeter reading is going to show a little high. The motors in the fans and compressor are more efficient than they appear to be on simple instruments. Still, I've never measured and calculated A X V that violated the SEER rating on the nameplate. If the specs ever get so stringent that you need to figure cos theta, you can be sure the local wholesaler will be able to sell you a true watt meter that will work in the range you need.

    In another 10 or 20 years you'll be like me...too old to do the work...and it won't matter any more to either of us. I still do consulting because I can do the math. The people that hire me can do the math too, but they are not confident in themselves. All I sell them is confidence. They know I never lie and I never mess up the math, and they can take it to the bank or a court of law. I hope life gets that easy for you, too.
     
  4. shortbus

    AAC Fanatic!

    Sep 30, 2009
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    @sherlockohms - Why are you YELLING at us?
     
  5. Adjuster

    Well-Known Member

    Dec 26, 2010
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    This may be a hard fight for you to win, especially if your colleague has genuine knowledge of the subject, and more still if he is more highly qualified (I would concede that being qualified does not necessarily make him right, but managers may be inclined to think so).

    The problem of an experienced worker using approximate methods coming agaist someone else, typically younger or with a more theoretical training who wants to do things "by the book" is an old one: both workers may have a point, but unfortunately their different backgrounds may make it hard to appreciate each other's arguments.

    Having been around long enough to see this kind of thing from both sides of the fence, I also would counsel you to play it down, or at least to think about the possible consequences before continuing to make an issue of it

    Finally note that though I agree that CFLs have many disadvantages, even a PF of 0.5 is nowhere near low enough to make their VA consumption as high as that of equivalent tungsten bulbs.
     
  6. jimkeith

    Active Member

    Oct 26, 2011
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    @Sherlockohms--just study a little AC theory--it will make you an even better technician--it is obvious that you already understand some of this by knowing that DC resistance is different than AC resistance in transformers and that there are parameters such as power factor, imaginary power, true power, etc.

    Otherwise, jut put up with him--or learn from him...
     
  7. strantor

    AAC Fanatic!

    Oct 3, 2010
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    +1. if you hit the books and come back to him and explain it to him in the layman's terms that he couldn't seem to muster, you can both be right, both be on the same page, but you'll be the one who really knows what they're talking about. After that "in your face" discussion, you can then school him in your 30 years worth of "oh, and by the way, it doesn't matter". Book closed, sherlockohms is the undisputed victor of the battle of knowledge & egos.
     
  8. crutschow

    Expert

    Mar 14, 2008
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    And remember that the PF doesn't affect the power consumed, only the value of the current. And the only affect of that is to slightly increase the IR loss in the lines between the generator and the device.
     
    Last edited: Apr 9, 2012
  9. GetDeviceInfo

    Senior Member

    Jun 7, 2009
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    the math is real and is always at work in the scheme of things. However, as successful practioner, you have likely learnt and apply certain 'rules of thumb' that tolerate the subtleties at play. In field service, the ability to get to the bottom of, and resolve problems is of greater importance than trying to explain the details. Looking at your partners prespective, he's trying to justify his knowledge, which is good. Together, you may make a good team, as long as you both can respect each others contributions.
     
  10. Sherlockohms

    Thread Starter New Member

    Apr 8, 2012
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    Watts = Volts x Amps
    Volts = Watts ÷ Amps
    Amps = Watts ÷ Volts
    We can use this equation to do lots of common
    electrical computations. For example, if a motor is
    drawing 20 amps at 12 volts, we know that it’s using
    energy at the rate of 240 watts (watts = volts x amps). If
    a lightbulb is drawing 100 watts at 4 amps, we can
    conclude that the voltage is 25 volts (volts = watts ÷
    amps). And if we have a 150 watt lightbulb running on
    120 volts, we know that it is drawing 1.25 amps (amps
    = watts ÷ volts).
    Well all i know is this is one of several formulas on the ohms wheel I see with numerous formulas for resistance etc, Pragmatics,semantics, are a sign to me they dont know, i have yet to have anyone show ne how this imaginary power, which i knew of theoretically, however never saw it affect my meters reading, sizing a potential relay checking mfd of a cap in the system,
    sizing and troubleshooting 24v transformers, low voltage controls, line voltage controls or ddc. I have no time for engineer magazine mechanics telling me a meter isnt accurate or useful in my systematic analysis,I know LRA is a millisecond or high amp draw, but that amount of surge is a fractionof a penny, I saw PF factor correction devices that lowered anamp draw as their meter was set up to read reactive power, watts stayed the same!
    I wanted to see how specidfically this makes my meter and basic ohms law theory innacturate or not useful! I know a transformers a short by ohms law and am not expecting it to be ohmic, i am talking troubleshooting like amp draws on a heat strip or a breaker popping or a gfi tripping etc, nobody has yet show me how the old ohms law equations are innacurate, maybe if my meter measured into the 1/10th of a watt? it would calibrate it, but its irrelevant fromeverything I ever saw, or every troubleshooting systematic analysis never said look at the heat strip guesss what its resistance is and guess if its drawing too much etc etc, I have found even more bizzare argumentative things this guy vehemently insists upon like checking to ground for power? its ok for safey but to grounds not reliable tecnique , as you know, and he says a circuit without a load is open, not a direct short
    and as the atmosphere has impedence 337 ohms ? etc? he says no circuits are ever eally opened or closed or shorted! lol! i say you control the loads by opening a switch! he says loads come on even with the switch open by arcing?? I never saw this, he says knob and tube did it? I never see knob and tube since 1978, and never saw a circuit disobey laws of physics.
    I know ohmic, and non ohmic etc, but power factor making a accurate reading is bull so far I know a engineer needs to know about it to size a feed etc but me as a hvac tecjh in the real world see RLA and LRA and size potential relays find shorts and troubleshoot as always with my amprobe!He says open switches are not open and circuits without loads are not a short/but open! I asked him what if I lay a wrench across my car battery? he says it will get hot so its a load lol!! he cant see a overloaded circuit is different than a directly shorted one, one has to many loads , the other none, both pop breakers so he says they are al open and shorted for 2 seconds lol ridiculous, i ohm a motors leads to find common run start and speeds still without any new issues, what is the real issue with my meter and sytematic analysis, its not relevant or if so, so negligible as to be insignifigant.
     
  11. Sherlockohms

    Thread Starter New Member

    Apr 8, 2012
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    i PUT ME IN CAPS SO YOU SEE HIS MORONTIA, AND THIS GUYS SERIOUS!

    SAY I LAY A WRENCH ACROSS THE BATTERY IN A CAR? WHAT THEN?


    I can tell you the wrench will get hot
    which tells us that the wrench draws current and therefore the wrench is a load. Kenzz

    NO KENZZ, A SHORT IS A OVERLOAD AND TRIPS THE BREAKER!
    A LOAD WONT POP THE BREAKER EVERY TIME YOU TRY TO TURN IT ON !


    With the switch open there is no load An open curcuit is a curcuit with no load
    Geez, how simple is this? put a crowbar across a 12v battery The crowbar gets warm doesn't it? The crowbar is a big-ass load. It draws current. I can tell you the wrench will get hot Kenzz

    Hey ,you say a circuit that
    has no load it will short when energized, Wrong 100% it will be open when the breaker pops again in 2 seconds so for 2 seconds its a
    short not every day though! Geez you are a hack! . many circuits have no loads a light is not a load its a light !

    I cut a lampcord as you say to do and connected the bared ends with solder and taped them too, I then
    plugged in the cord and of course its open so as I saids nothing happens! what happens is exactly how lightning travels up not down!
    when the wires are seperated they will then arc as a set of points do and this is why they electric tarnish! Kenzz


    MY POSITION IS BELOW

    Every once in a while theres a SURGE in your electrical lines caused by things like your refrigerator kicking in.

    But this isnt meaningful amount of electricity, because the savings happens for only a fraction of a second., because the surge is so small it doesn't even amount to a fraction of a penny, but if you could be charged for it, you would have been.) it happens for only a fraction of a second, so small you can't really measure it.
    Let's assume you had a 100% spike every three hours and that it lasts a millisecond, which is pretty long for a surge.
    ( being generous.) Three hours is 10,800 seconds. So:

    Without Spikes
    With Spikes
    * 10,800 seconds at 120V

    * 10,799.999 seconds at 120V, plus
    * 0.001 seconds at 240V

    So that means instead of using 10,800 seconds worth of 120V electricity, you're using 10,800.001 seconds of electricity. Put another way, every three hours you use an extra 0.001 of electricity. Now, if you think that's a lot, do you think you would save a lot of energy by turning off your lights for 0.001 seconds? Do you think you will pay a lot more for electricity by keeping them on for an extra 0.001 seconds?
     
  12. Sherlockohms

    Thread Starter New Member

    Apr 8, 2012
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    While we are on the subject, there isn't any
    true open and short circuits in our world. An open switch for instance has a small electrical capacitance and can conduct some current at high frequencies. Also even the shortest of conductors one uses trying to create a "short circuit" has inductance and therfore impedance. For low level service techs they can speak of opens and shorts but these are very LOOSE terms as what they observe as a short circuit often has some resistance. It is silly to argue about these terms... just look at who is using the term and their expertise or lack thereof to evaluate the TRUE state of the problem. Consumer walks in with busted appliance and says it has a short in it that means nothing more than the item is not working. An electrician says there is a short, that means excessive current is flowing and causing a problem... nothing is to be assumed about how "good" the "short" is or how much current is flowing... you can assume it is above what is expected... That is all...Kenzz
     
  13. Sherlockohms

    Thread Starter New Member

    Apr 8, 2012
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    My coworker now asserts 1,500-2,600 % efficient heat pumps!LOL!
    alleging a COP 30!!!
    --------------------------
    According to the laws of thermodynamics, nothing can be more than 100% efficient. A heat pump does not generate heat or cold, but moves heat from one place to another. In this case, it is able to supply 2.55 times as much heat as the energy input (by extracting it from outside), achieving a COP of 2.55, which is effectively 255% efficiency. So it is right as to COP 30 = 3000% efficiency! HOWEVER a COP of 30 is as preposterous as a COP of 60 which was the previous delusional figure tossed about to allegedly corect me then, now its 1/2 off but still imbecilic to state if you are not a moron. remember 100% gives 3.45 btu heat per watt power used. Typical EER for residential central cooling units = 0.875 × SEER. SEER is a higher value than EER for the same equipment.[1] A more detailed method for converting SEER to EER uses this formula: EER = -0.02 × SEER² + 1.12 × SEER[2] Note that this method is used for benchmark modeling only and is not appropriate for all climate conditions.[2] A SEER of 13 is approximately equivalent to an EER of 11, and a COP of 3.43, which means that 3.43 units of heat energy are removed from indoors per unit of work energy used to run the air conditioner or heat pump. SO SEE WHAT HIS COP OF 30 DO YOU KNOW WHAT IT EQUATES TO IN THE RERAL WORLD?? A COP OF 3.43 = EER OF 11, SO A COP OF 30 - EER 100 + !!!! THIS IS FAR TOO MUCH FOR A PERSON WHO CANT DO MATH PERCENTAGES AS you kenzz who defends such morontia
     
  14. strantor

    AAC Fanatic!

    Oct 3, 2010
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    Haha I guess you found your outlet. Got that out of your system? I think factoring in the heat energy to the heat pump efficiency formula is wrong. Just because the pump is transferring energy does not mean that the energy transferred has anything to do with the pump. If that were the case, then every gas station pump in the country is a marvelous wonder of overunity!

    With regards to everything else you typed, I skimmed through it but wasn't able to tell who was who so I don't know which one of you I won't let near anything electrical in my house.
     
  15. GetDeviceInfo

    Senior Member

    Jun 7, 2009
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    Frank, is that you? Please call into the shop, the boys have been worried ever since you wandered off the other day mumbling something about the ohmic revolution...
     
    panic mode likes this.
  16. shortbus

    AAC Fanatic!

    Sep 30, 2009
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    You guys, play nice with Loosie's big brother. :D:D:D
     
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