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Capacitor under load testing

What is 2652 while we are testing capacitor under load

Comments

  • PC7060
    PC7060 Member Posts: 1,155
    What procedure are you using to test the capacitor?
  • mattmia2
    mattmia2 Member Posts: 9,575
    what? You can calculate the capacitance by measuring the voltage across the cap and current through it.
  • Jamie Hall
    Jamie Hall Member Posts: 23,168
    You know, I'm rarely really baffled by a question, but that one's got me. It's utterly meaningless as posed. Try again.
    Br. Jamie, osb
    Building superintendent/caretaker, 7200 sq. ft. historic house museum with dependencies in New England
  • SuperTech
    SuperTech Member Posts: 2,139
    PC7060 said:
    What procedure are you using to test the capacitor?
    He's measuring amperage of the start windings multiplied by 2652 then divided by the voltage across the capacitor to get the actual capacitance while the system is running.  Its a more accurate way to test a capacitor than using multimeter, because you are measuring using the voltage that the system is operating with. I've seen capacitors that check out ok with a meter fail the test outlined above.  Using this method on maintenance calls pretty much eliminates failed dual capacitor service calls for our maintenance customers.  

    I'm not sure where the 2652 comes from. I'm sure it relates to Ohms law in some way. 
    STEVEusaPAPC7060motoguy128
  • EdTheHeaterMan
    EdTheHeaterMan Member Posts: 7,720
    edited March 2021
    The actual number is not 2652 but it is close enough. It is easy to remember because 26 + 26 = 52

    this is from an article in ACHR News.

    "Since I started training with this method, I have often been asked where this calculation comes from. It is actually a simplified version of the equation to find capacitative reactance combined with a version of Ohm’s Law: Xc = 1/(2 x π x ƒ x C) and E = I x Xc.

    I’m taking the 1/(2πƒ) part, multiplying by the measure we’re looking for (microfarads or millionths of a farad). This gives us the capacitative reactance, which, multiplied by the amperage and then divided by the voltage, gives us the capacitance. So you get: Xc = 2,652 and E = 4.4 x Xc = 11,668.8, so that 11,668.8 ÷ 335 V = 34.83."

    Perfectly understandable. Clear as Mud!

    "To simplify, we can use approximate numbers and just call it amps times 2,652. Then, divide by the voltage."

    Read the full article here https://www.achrnews.com/articles/135163-testing-the-run-capacitor-while-the-system-is-running

    This goes under the "He forgot more about HVAC than you will ever know" department and I just go with it.


    Respectfully Submitted,
    Mr.Ed

    PS. sounds like @AtifMahmood is the type of person who needs to know why it works. Not just that it works. I believe we need more analytical minds in this industry. Seems that the Trades get the people that are considered Sub Standard by those from Yale and Harvard, or who work on Wall Street. Little do they know that without the Trades, they would not be able to take that elevator to their corner office.

    I digress, but I applaude AtifMahmood. Keep asking those questions. Learn as much as you can!

    Edward F Young. Retired HVAC ContractorSpecialized in Residential Oil Burner and Hydronics
    Alan (California Radiant) ForbesHomerJSmithSuperTechPC7060
  • mattmia2
    mattmia2 Member Posts: 9,575
    it comes form the definition of a farad and some algebra...
    SuperTech
  • EdTheHeaterMan
    EdTheHeaterMan Member Posts: 7,720
    edited March 2021
    mattmia2 said:

    it comes form the definition of a farad and some algebra...

    The farad was named in honor of Michael Faraday. I believe there is also a restaurant chain named after him. TGIF


    I could be wrong on that restaurant thing



    Edward F Young. Retired HVAC ContractorSpecialized in Residential Oil Burner and Hydronics
    HomerJSmithmattmia2SuperTech
  • Jamie Hall
    Jamie Hall Member Posts: 23,168
    I HATE formulae like that. It's a compilation of a number of constants -- with a couple of assumptions. Yes, it works. But -- it assumes the frequency is 60 hz, so it only works in North America. And it assumes that you want the answer in microfarads, which you usually do... but not always.

    And I might point out that it would work for either the start or run capacitors (assuming there are two) -- but only when they are in the circuit when you measure the current.
    Br. Jamie, osb
    Building superintendent/caretaker, 7200 sq. ft. historic house museum with dependencies in New England
    EdTheHeaterManBrent H.
  • mattmia2
    mattmia2 Member Posts: 9,575

    I HATE formulae like that. It's a compilation of a number of constants -- with a couple of assumptions. Yes, it works. But -- it assumes the frequency is 60 hz, so it only works in North America. And it assumes that you want the answer in microfarads, which you usually do... but not always.

    The constant really should always be written with the units and the user should cancel the units out. It doesn't solve the problem with the impedance being dependent on frequency, but it helps with most other errors.

  • GW
    GW Member Posts: 4,691
    Thanks Ed for the explanation
    I’ll make a note and start testing on the fly 
    Gary Wilson
    Wilson Services, Inc
    Northampton, MA
    gary@wilsonph.com