Differences

This shows you the differences between two versions of the page.

--- ref:fault_current [2025/04/14 12:05] – ↷ Page moved from reference:fault_current to ref:fault_current Alan Shea
+++ ref:fault_current [Unknown date] (current) – removed - external edit (Unknown date) 127.0.0.1
@@ Line 1: / Line 1: @@
-======Estimating Fault Current Availability=====
-Short-circuit current available in bolted-fault conditions. The system fault current is the sum of all connected source fault currents.
-=====Utility Transformers====
-For utility transformers, a worst-case estimate of available short-circuit current in a three-phase fault can be determined by dividing the transformer ampere rating by the impedance of the transformer.((https://www.jcalc.net/transformer-short-circuit-fault-current-calculator)) This assumes an infinite utility bus with unlimited current; as this is rarely if ever the case, actual fault current will be limited by available utility bus current.
-\\
-===Three-Phase Fault==
-<WRAP col2>
-<m 14>I ~ = ~ {kVA * 1000}/{sqrt{3} * V}</m>
-<m 14>I3p ~ = ~ I / {Z1}</m>
-<m 14>I3p ~=~ {kVA * 1000}/{sqrt{3} * V * Z1}</m>
-</WRAP>
-\\
-For example, a 630kVA/400VAC, 5.76% impedance transformer would be capable of delivering approximately 15,800 amperes of fault current into a three-phase fault circuit. \\
-<WRAP col2>
-<m 14>I ~ = ~ {630kVA * 1000}/{sqrt{3} * 400} ~ = ~ 910 ~ Amps</m>
-<m 14>I3p ~ = ~ 910 / 0.0576 ~ = ~ 15,833 ~ Amps</m>
-</WRAP>
-\\
-----
-===Phase to Phase Fault==
-<WRAP col2>
-<m 14>Ipp ~=~ I / {(Z1 + Z2)}</m>
-\\ \\
-<m 14>Ipp ~=~ {kVA * 1000} / {V * (Z1 + Z2)}</m>
-\\ \\
-<m 14>Ipp ~=~ {630 * 1000} / {400 * (0.0576 + 0.0576)} ~=~ 13,671 ~Amps</m>
-\\
-</WRAP>
-\\
-----
-===Phase to Neutral Fault==
-<WRAP col2>
-<m 14>Ipn ~=~ {3 * I}/ {(Z0 + Z1 + Z2)}</m>
-\\ \\
-<m 14>Ipn ~=~ {3*VA} / {sqrt{3} * V * (Z0 + Z1 + Z2)}</m>
-\\ \\
-<m 14>Ipn ~=~ {3 * (630 * 1000)}/{sqrt{3} * 400 * (0.0576 + 0.0576 + 0.04896)} ~=~ 16,618 ~ Amps</m>
-\\
-</WRAP>
-\\
-----
-\\
-<WRAP center 650px>
-| **Voltage:** |V = V<sub>LL</sub>  |
-|  <m 14>sqrt{3}</m>  | <m 14>1.732</m> |
-| **Positive Sequence Impedance:** | <m 14>Z1 ~=~ {Transformer Impedance %}/100</m>|
-| **Negative Sequence Impedance:** | <m 14>Z2 ~=~ Z1</m>|
-| **Zero Sequence Impedance:**  | <m 14>Z0 ~=~ 0.85 * Z1</m>((For [[https://en.wikipedia.org/wiki/Transformer#Cores|core-type transformers]] with delta-star-earth connections.))|
-</WRAP>
-=====Generators====
-The three-phase bolted fault current from a generator is roughly the rated ampacity of the generator, divided by the subtransient reactance of the machine at that rating. Subtransient reactance is noted on the manufacturer's performance data sheet for that generator, using the "submittal data" on the data sheet, not the implemented or de-rated ampacity of the installation. The Cummins Power Application Manual on Gensets contains calculations for converting that submittal data to another rating in the section on calculating fault currents.((//Application Manual -- Liquid Cooled Generator Sets T030f//, Rev May 2010, pg 5-36 / p113))
-===Three-Phase Bolted Fault===
-For example, with a 200kW/480vAC machine with 12% subtransient reactance (//X"d//), the available fault current at the generator terminals is approximately 2,500 Amps. If this generator is paralleled with two other identical machines, the total available from all generator sets is the sum of the available current from each machine, or 7,500 Amps.
-\\
-<WRAP col2>
-<m 14>I ~ = ~ {kW * 1000}/{1.732 * V * PF}</m>
-\\  \\
-<m 14>I ~ = ~ {200kW * 1000}/{1.732 * 480V *0.8PF} ~ = ~ 300 ~ Amps</m>
-\\
-</WRAP>
-\\
-<WRAP col2>
-<m 14>I3p ~ = ~ I / {X prime prime d}</m> \\
-\\
-<m 14>I3p ~ = ~ 300 / 0.12 ~ = ~ 2,500 ~ Amps</m>
-</WRAP>
-\\
-===Single Phase - Neutral Bolted Fault===
-A single phase-neutral bolted fault may have higher short-circuit current than a Line-Line fault on wye-connected generators with 2/3 pitch windings. The product of the three phase currents divided by the sum of the Subtransient Reactance (//X"d//), Negative Sequence Reactance (//X2//) and the Zero Sequence Reactance (//X<sub>0</sub>//) gives the available phase-neutral fault current.((Caterpillar EDS 70.4, //Generator Winding Pitch and Harmonics//, pg. 6))
-For a generator with full load current at 200 kW, 480 volts, 60 Hz of 300 amps:
-<WRAP col2>
-  * Subtransient Reactance (//X"d//) = 0.110 pu
-  * Negative Sequence Reactance (//X2//)  = 0.110 pu
-  * Zero Sequence Reactance (//X<sub>0</sub>//) = 0.019 pu
-    * //Note: "pu" = per unit, otherwise considered as percent.//
-\\
-<m 14>Ipn ~=~ {3 * I} / {X prime prime d + X2 + X0}</m>
-\\ \\ \\ \\
-<m 14>Ipn ~=~ {3 * 300Amps}/(0.110 + 0.110 + 0.019) ~ = ~ 3,766 Amps</m> \\
-</WRAP>
-\\
-----
-===References===
-  * [[https://web.archive.org/web/20131130123633/http://cumminspower.com/en/technical/application/|Cummins Power, Application Engineering Manuals]]
-    * [[https://web.archive.org/web/20131130123633/http://cumminspower.com/www/literature/applicationmanuals/t030.pdf|Liquid Cooled Generator Application Manual [PDF - 10Mb]]]
-    * [[https://web.archive.org/web/20131130123633/http://cumminspower.com/www/en/technical/application/t-016.pdf|Paralleling Application Manual [PDF - 10Mb]]]

MetaTek

User Tools

Site Tools

Differences

Page Tools