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x:f Reynolds Electric
Available Fault Current Calculation
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Utility Fault Current 23,132 amperes WA = 150
E = 208
I = kVA x 1000 = trans. FLA trans. FLA = 4;16j
E x 1.732
Inca = trans. FLA x 100 _ Li. - ...4161
transformer Z 1.8 %
l+• l co=ampere short-circuit current RMS symmetrical. Lori = ,23;1321 amperes
Three Phase 208/120
Point to Point Method t.._._-...._..----_—_._..—_
Length (distance) L = 20 'Aluminum in Nonmetallic F
f'factor= 1.732 x L x I . (ASC) Isca = I 23,1321
N X C X E L-N #conductors per phase N = 1
Phase conductor constant C = 1. 23,492 Phase Conductor I i
Volt Line to Line E L-L = 208 Volt
f = 1 0..1641
Neutral conductor constant C = . 23,492 Neutral Conductor 1
Multiplier Volt Line to Neutral E L-N = 120 Volt
f = [M0328&
M = 1
1 +f Line to Line M = D:859
Line to Neutral M = L. 0'7531
.r' Fault Current at Service Equipment
Isca x M = fault current at terminals of main disconnect L- L= _x,19;88 amperes
Isce x M = fault current at terminals of main disconnect L- N = °X117,4,1,6. amperes
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Fault Current from Service disc. to panel " " 1 Copper in Nonmetallic Rac
`f I Thre
Three Phase Feeder 'Length (distance) L = 22
1.732 x L x I (ASC) , 198731 Phase 11,416
!+y if'factor=
,; .
N x C X E L-N #conductors per phase N = 1
Phase conductor constant C = 13;923 Phase Conductor I'
, Yr` Volt Line to Line E L-L
"r2081 volt
f 0.2611
r
Neutral conductor constant C 13;923 Neutral Conductor
Volt Line to Neutral E L-N = to �1201 Volt
(6e Multiplier f 074591
M = 1 Line to Line
1 +f Line to Neutral M = 0;686.