Question 8.5: Estimate the symmetrical short-circuit current at the second......

Estimate the symmetrical short-circuit current at the secondary side of a
13.8 kV208Y/120V13.8 \ kV-208Y/120-V transformer rated at 1000 kVA1000 \ kVA with Z = 5% and supplied by a utility having 15 kVA15 \ kVA fault current capabilities.

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First, the MVASCMVA_{SC} values for both the utility and the transformer are estimated. For the utility (using the utility distribution voltage and the utility fault current),

MVASC,U=313.8 kV15 kVA=358.52 MVAMVA_{SC,U}=\sqrt{3}*13.8 \ kV*15 \ kVA=358.52 \ MVA

For the transformer using Equations 8.13 and 8.14,

Y=100Z(%)Y=\frac{100}{Z(\%)}                                 (8.13)

MVASC=MVAYMVA_{SC}=\frac{MVA}{Y}                     (8.14)

MVASC,T=1000 kVA0.05=20 MVAMVA_{SC,T}=\frac{1000 \ kVA}{0.05}=20 \ MVA

Since the utility source and the transformer are in series, the total MVASCtotMVA_{SC-tot} value can be estimated using Equation 8.16:

1/MVASCtot=1/MVASC,1+1/MVASC,2++1/MVASC,n1/MVA_{SC-tot}=1/MVA_{SC,1}+1/MVA_{SC,2}+ …+1/MVA_{SC,n}                  (8.16)

1/MVASCtot=1/MVASC,U+1/MVASC,T=1358.52+120=0.052791/MVA_{SC-tot}=1/MVA_{SC,U}+1/MVA_{SC,T}=\frac{1}{358.52}+\frac{1}{20}=0.05279

Thus,

1/MVASCtot=10.05279=18.94 MVA1/MVA_{SC-tot}=\frac{1}{0.05279}=18.94 \ MVA

Therefore, the symmetrical short-circuit current at the secondary side of the transformer can be calculated using Equation 8.17 with a line voltage of EL=208 VE_L=208 \ V:

ISC=MVASCtot3ELI_{SC}=\frac{MVA_{SC-tot}}{\sqrt{3}E_L }                (8.17)

ISC=18.94 MVA3(208 V)=52,583 AI_{SC}=\frac{18.94 \ MVA}{\sqrt{3}(208 \ V) } =52,583 \ A

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