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Thread: Load estimation

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    Load estimation

    Please correct me if I'm wrong. During short circuit fault, the fault current that flows in a cable become twice as much as the load current.

    Now using Method 2 Installation if I'm estimating cable size for plugs with estimated power of 3000W, I get 13.04A which I multiply by 2 to get 26.08A Ic current. The table 6.2(a) on SANS 10142 leads me to 4mm.
    Does this mean that if I use a 2.5mm cable, it will get burnt immediately in the event of short circuit with an overcurrent/overload protection (circuit breaker) failure?

    Why do we need to multiply load current by 2 during Load Estimation?

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    Quote Originally Posted by Reginald View Post
    Please correct me if I'm wrong. During short circuit fault, the fault current that flows in a cable become twice as much as the load current.
    Then please allow me to correct you.

    During a short circuit fault, the potential short-circuit current (PSC) is limited by the line resistance, and is bound to be waaay more than the overcurrent protection provided by the circuit breaker for the circuit, let alone the actual load current.

    PSC thus determines your maximum over-current rating of the circuit breaker (the kA rating marked on the circuit beaker), not the overcurrent protection rating.

    Quote Originally Posted by Reginald View Post
    The table 6.2(a) on SANS 10142 leads me to 4mm.
    Does this mean that if I use a 2.5mm cable, it will get burnt immediately in the event of short circuit with an overcurrent/overload protection (circuit breaker) failure?
    If the overcurrent/overload (circuit breaker) protection fails and the current runs much higher than the cable is rated for, it's just a matter of time...

    Normally in the event of a short circuit fault though, the current will run so high the fault will trip the next layer of protection really quickly and no significant harm to the wiring should be done.

    Quote Originally Posted by Reginald View Post
    Why do we need to multiply load current by 2 during Load Estimation?
    There's a general principle in engineering to overspec tolerances by a factor of at least 2 under expected standard conditions. It's a safety factor to allow for non-standard conditions and variances...

    For some issues, that over-specification safety margin will be much higher.
    The trouble with opportunity is it normally comes dressed up as work.

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    Diamond Member AndyD's Avatar
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    Quote Originally Posted by Reginald View Post
    Please correct me if I'm wrong. During short circuit fault, the fault current that flows in a cable become twice as much as the load current.
    The short circuit current has no relationship to the load connected to the circuit. By definition a short circuit is when there is a circuit with no useful load connected in it. The only thing that limits the current flow during a short circuit condition is the resistance of the actual wiring, which is very low, hence short circuit currents are very high.

    Quote Originally Posted by Reginald View Post
    Now using Method 2 Installation if I'm estimating cable size for plugs with estimated power of 3000W, I get 13.04A which I multiply by 2 to get 26.08A Ic current. The table 6.2(a) on SANS 10142 leads me to 4mm.
    Does this mean that if I use a 2.5mm cable, it will get burnt immediately in the event of short circuit with an overcurrent/overload protection (circuit breaker) failure?
    I'm not entirely understanding your scenario. Are you trying to design a single circuit that can run two 3kW heaters?
    You always size your circuit breakers according to your cable size, cable length and installation method. The circuit breaker's job is to protect the installation cabling from overload damage.
    During a short circuit condition we want a very large current to flow because we want the circuit to be disconnected very quickly by the fuse or the circuit breakers. If the short circuit current is too low then there's a danger that the circuit breaker could take several minutes to disconnect and damage may be inflicted upon the wiring or cabling.

    Quote Originally Posted by Reginald View Post
    Why do we need to multiply load current by 2 during Load Estimation?
    Can you quote the reg that states this?
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    When I did my apprenticeship, City & Guilds, (back when electricians were the only tradesmen allowed to ware a bowler hat) we used fuse wire, 20 amp for 1.5mm, 32 for 2.4mm and 45 for 4.0mm wire, I never had a wire burn out under direct short fault condition.
    Now under the SA regulations the main faults I find are loose/burned connections, this is usually caused because the wire is too big for the connector it is inserted in and can not be tightened down properly, any one else come across this problem?
    Apparently there is nothing that cannot happen today.

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