Forums » Electrical Engineering

Power dissipation of a circuit breaker

    • 27 posts
    March 4, 2019 12:33 PM PST

    How can I calculate or find out the power dissipation  of a circuit breaker?

  • March 4, 2019 2:45 PM PST

    measure the voltage across each pole (probably in the 10s of millivolts range)

    measure the current through each pole

    multiply each pole V X I

    Add the three

     

    Just curious - what is the context?

     

    • 200 posts
    March 6, 2019 6:09 PM PST

    Thanks for the suggestion of this topic, Terry.

    The procedure advised by Carl Coulter is correct for 3-phase AC Circuit Breaker in order to find the instantaneous power dissipation (at the time of measurement).

    If both of you could allow me to add few notes for the sake of knowledge of all Members:

    - For a DC Circuit Breaker, probably in the Positive pole, only one (pole) measurement is necessary - however it is necessary to open the circuit (in this same pole) to measure the current;

    - For a AC Circuit Breaker, all three (pole) phases measurements are necessary - however there is no need to open the circuit to measure the current - use clamp meter (or clamp ammeter);

    - The voltage is measured from input to output connectors (bolts) of the Circuit Breaker for better precision, even better after to check if the connectors are all correctly tightened. Carl Coulter advised tens of millivolt for a GOOD Circuit Breaker, but if you are concerned about the power dissipation of a CB we could believe that you are suspecting something is wrong, probably the CB is heating, and as often we can find so many times you can also find voltage drops across a CB from 1 Vac to 10 Vac(!!!) even for a 110Vac phase;

    - As the current (load) varies excessively and consequently voltage drop across CB poles, a good advice is to calculate the resistivity of CB poles, a more stable variable, and it is possible to do periodically monitor the quality of each pole of CB:

    Rp1 = Vd1 / Ii1 ;
    Rp2 = Vd2 / Ii2 ;
    Rp3 = Vd3 / Ii3 ;

     

    Where:

    - Rp is the CB pole resistivity which may rise due to arcing and oxidation and consequent carbonization or deterioration;
    - Vd is the voltage drop across the CB pole;
    - Ii is the instantaneous current flowing to the system across the CB pole;

    - For frequent monitoring it needs to measure just one (easier) of variables for each pole, next near time: the voltage drop (Vd) or the instantaneous current (Ii). We need to measure both again when we suspect the resistivity of a pole is increasing faster, let's say, in 3 or 6 months. So for power dissipation the calculations would be:


    If voltage drop measurements were provided:

    Pd1 = Vd1^2 / Rp1 ;
    Pd2 = Vd2^2 / Rp2 ;
    Pd3 = Vd3^2 / Rp3 ;

    Or if instantaneous current measurements were provided:

    Pd1 = Ii1^2 x Rp1 ;
    Pd2 = Ii2^2 x Rp2 ;
    Pd3 = Ii3^2 x Rp3 ;

    - The total power dissipation of a CB is the sum of all its poles power dissipation:

    Pcb = Pd1 + Pd2 + Pd3 ;

    Hoping this could be of any help, Regards.


    This post was edited by Alex de Moura at March 6, 2019 6:11 PM PST
    • 1 posts
    May 12, 2019 2:43 AM PDT

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    • 73 posts
    May 17, 2019 7:30 AM PDT

    This data is usually published for medium voltage breakers.