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Thread: Question on Fault Current

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    Question on Fault Current

    I can do the calculations to my main panels but there is a 75kva transformer where I am stepping down from 480 to 208. I'm not sure if I start over at this point the same as the power company to figure the 120/208 panels downstream.

    Thanks

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    Quote Originally Posted by KWH View Post
    I can do the calculations to my main panels but there is a 75kva transformer where I am stepping down from 480 to 208. I'm not sure if I start over at this point the same as the power company to figure the 120/208 panels downstream.

    Thanks
    Here's how to calculate fault current across a transformer:
    https://forums.mikeholt.com/showthread.php?t=171709

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    You could stop where you are, and call for all downstream panels to have an SCCR of 10,000 amps. The short circuit current available at the secondary of the transformer will be under 4,000 amps. This can be calculated by taking the rated secondary side current (75,000 / 360 = 208 amps), and then dividing by the percent impedance (assume 5.75%), and you get 3623 amps.
    Charles E. Beck, P.E., Seattle
    Comments based on 2017 NEC unless otherwise noted.

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    Quote Originally Posted by charlie b View Post
    You could stop where you are, and call for all downstream panels to have an SCCR of 10,000 amps. The short circuit current available at the secondary of the transformer will be under 4,000 amps. This can be calculated by taking the rated secondary side current (75,000 / 360 = 208 amps), and then dividing by the percent impedance (assume 5.75%), and you get 3623 amps.
    Transformers smaller than 500KVA are usually supplied with impedances less than 5.75%. I would have used 3.5% for a maximum fault current of 6kA, which is still less than the standard breaker AIC of 10kA.
    Just because you can, doesn't mean you should.

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    Quote Originally Posted by charlie b View Post
    You could stop where you are, and call for all downstream panels to have an SCCR of 10,000 amps. The short circuit current available at the secondary of the transformer will be under 4,000 amps. This can be calculated by taking the rated secondary side current (75,000 / 360 = 208 amps), and then dividing by the percent impedance (assume 5.75%), and you get 3623 amps.
    The reason that works out that way, is as follows. When you combine the formula after the f-factor and M-multiplier intermediate steps of the fault current across transformers calculation, you get the following:
    Is = Ip*Vp/(Vs*(1 + √3*Ip*Vp*Z/K))

    Where:
    Ip and Is are available fault currents on the primary side and secondary side respectively
    Vp and Vs are nominal line-to-line voltages on the primary side and secondary side respectively
    Z is the impedance expressed as a decimal (rather than percent)
    K is the Volt-Ampere rating, i.e. 1000*the KVA rating.

    You notice that Ip*Vp appear as a package deal in this formula. Call this X, and the formula becomes:
    Is = X/(Vs*(1 + √3*X*Z/K))

    Simplify:
    Is = X*K/(Vs*(K + √3*X*Z))

    If you select values for K, Z, and Vs, and graph this curve, you notice that it takes off to a cruising altitude as X gets large. There is a limiting maximum value it can approach, which is calculated by dropping K from (K + √3*X*Z), due to its insignificance as X gets large, and cancelling the X on top and bottom.
    Is = X*K/(Vs*(K + √3*X*Z))
    Is = X*K/(Vs*(0 + √3*X*Z))
    Is = K/(√3*Z*Vs))

    Conclusion:
    Is = K/(√3*Z*Vs))

    Notice that K/(√3*Vs) is how you calculate secondary current at full KVA. Then divide by Z as a decimal, and get the limiting value of fault current. This limiting behavior is useful for calculating fault current at the service point, because you may not know the primary side voltage or fault current, so it gives you a worst case scenario. Primary-side behavior is also a moving target as the utility upgrades substations, introduces new customers, adjusts transformer primary-side taps, or whatever else they may do.
    Is = K/(√3*Z*Vs))

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    The POCO transformer supplying 480 V it could be a 2500 kVA HV/480 V and 6% short-circuit impedance. So the short-circuit current at 480 V transformer terminals will be 50 kA.
    At the 208 V side you are 2 impedances in series: of 2500 kVA 6% transformer and of 75 kVA 3% transformer.
    Z1=0.208^2/2.5*6%=0.001038 Ω and Z2=0.208^2/0.075*3%=0.017306 Ω.
    Ztot=0.018344 Ω
    Isc208V=208/SQRT(3)/0.018344=6546.5 A

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    Help me understand your Formula please

    Quote Originally Posted by Julius Right View Post
    The POCO transformer supplying 480 V it could be a 2500 kVA HV/480 V and 6% short-circuit impedance. So the short-circuit current at 480 V transformer terminals will be 50 kA.
    At the 208 V side you are 2 impedances in series: of 2500 kVA 6% transformer and of 75 kVA 3% transformer.
    Z1=0.208^2/2.5*6%=0.001038 Ω and Z2=0.208^2/0.075*3%=0.017306 Ω.
    Ztot=0.018344 Ω
    Isc208V=208/SQRT(3)/0.018344=6546.5 A
    Julius
    Question 1
    Z = ( V^2/S) * %Z
    Origin of this formula ?

    Question 2
    Straight addition of Z1 & Z2...
    Does not Z1 have to be reflected through 2nd transformer by turns ratio BEFORE you add ?

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    1) Simply algebra. By definition the p.u. short-circuit impedance it is: uk%=Zk/Zrat
    Zk=transformer short-circuit impedance [ohm] Zrat =transformer rated impedance[ohm]
    Then Zk=uk%*Zrat
    Zrat=Vrat/(Irat.√3)
    Srat=√3.Irat.Vrat
    √3.Irat=Srat/Vrat
    Zrat=Vrat/(Srat/Vrat)
    Zrat=Vrat^2/Srat
    Zk=uk%.Vrat^2/Srat
    2) Z1'[referred to V2]=Z1*(V2^2/V1^2)
    Then Zk'1=uk%*V1^2/S*(V2/V1)^2=uk%*V2^2/S
    If you take for Zk calculation the voltage V2 instead of V1 you'll get the referred to 2
    impedance.Z1 and Z2 are both calculated for 208 V.

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