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    And Reactance of Is Connected To The Generating Station Bus-Bars Through 5 MVA Step-Up Transformer Having A

    Uploaded by

    James Santos
    1. The document calculates the short-circuit current (SCC) at two points on an electrical system with a transmission line, transformer, and generator. The SCC is 2847 A at the load end and 2886 A at the transformer terminals. 2. It then analyzes a second system with multiple transformers and generators to find the SCC at point F. The calculations determine the SCC would be 2320 A for a 3-phase fault at location F.

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    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd

    And Reactance of Is Connected To The Generating Station Bus-Bars Through 5 MVA Step-Up Transformer Having A

    Uploaded by

    James Santos
    275 views7 pages
    1. The document calculates the short-circuit current (SCC) at two points on an electrical system with a transmission line, transformer, and generator. The SCC is 2847 A at the load end and 2886 A at the transformer terminals. 2. It then analyzes a second system with multiple transformers and generators to find the SCC at point F. The calculations determine the SCC would be 2320 A for a 3-phase fault at location F.

    Original Title

    NEE-515-3phasefault

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    1. The document calculates the short-circuit current (SCC) at two points on an electrical system with a transmission line, transformer, and generator. The SCC is 2847 A at the load end and 2886 A at the transformer terminals. 2. It then analyzes a second system with multiple transformers and generators to find the SCC at point F. The calculations determine the SCC would be 2320 A for a 3-phase fault at location F.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    275 views7 pages

    And Reactance of Is Connected To The Generating Station Bus-Bars Through 5 MVA Step-Up Transformer Having A

    Uploaded by

    James Santos
    1. The document calculates the short-circuit current (SCC) at two points on an electrical system with a transmission line, transformer, and generator. The SCC is 2847 A at the load end and 2886 A at the transformer terminals. 2. It then analyzes a second system with multiple transformers and generators to find the SCC at point F. The calculations determine the SCC would be 2320 A for a 3-phase fault at location F.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    You are on page 1of 7

    1.

    Problem:

    A 3-phase transmission line operating at 10 kV and having a resistance of 1 Ω and reactance of


    4 Ω is connected to the generating station bus-bars through 5 MVA step-up transformer having a
    reactance of 5%. The bus-bars are supplied by a 10 MVA alternator having 10% reactance. Calculate
    the short-circuit kVA fed to symmetrical fault between phases if it occurs reactance. Calculate the
    short-circuit kVA fed to symmetrical fault between phases if it occurs

    (a) at the load end of transmission line


    (b) at the high voltage terminals of the transformer

    Using the Sb = 10 MVA, Eb = 10 kV at the transmission line, we have

    ZL Eb 2 (10kV ) 2
    Z Lpu = Z base = = = 10 Ω
    Z base Sb (10 MVA)

    1+ j4
    Z Lpu = = 0.1 + j 0.4 Ω
    10

    To solve for the base voltages

    Eb 10kV 2Eb
    = ; Eb1 = but Eb = 10 kV
    Eb1 2kV 10

    2(10)
    E b1 = = 2 kV
    10

    For the transformer per unit reactance;

    2
    E  S 
    X Tpunew = X Tpuold  bold   bnew 
     Ebnew   Sbold 
    2
     10   10 
    X Tpu = j 0.05     = j 0.1 pu
     10   5 
    For the generator subtransient reactance per unit;

    2
    E  S 
    X gpunew = X gpuold  bold   bnew 
     Ebnew   S bold 
    2
     2   10 
    X gpunew = j 0.1    = j 0.1 pu
     2   10 

    if we let VLpu = 1.0 pu, then Espu = VLpu = 1.0 pu at no-load.

    So for a fault at point A.


    then by KVL, we have

    Espu − I 3φ FApu ( j 0.1 + j 0.1 + 0.1 + j 0.4) = 0

    Espu 1.0∠0°
    I 3φ FApu = =
    ( j 0.1 + 0.1 + j 0.4) ( j 0.1 + j 0.1 + 0.1 + j 0.4)

    I 3φ FApu = 1.644∠ − 80.54° pu

    3S b 3(10 x106 )
    I base = = = 1732.05 Amps
    Eb 10 x103

    I 3φ FA = 1.644(1732.05)

    I 3φ FA = 2847.47 Amps

    So for a fault at point B.

    then by KVL, we have

    Espu − I 3φ FBpu ( j 0.1 + j 0.1) = 0

    Espu 1.0∠0°
    I 3φ FBpu = =
    ( j 0.1 + j 0.1) j 0.2

    I 3φ FApu = 5.0∠ − 90° pu


    Sb (10 x106 )
    I base = = = 577.35 Amps
    3Eb 3(10 x103 )

    I 3φ FB = 5.0(577.35)

    I 3φ F = 2886.75 Amps

    2. Problem:

    Figure shows the single line diagram of a 3-phase system. The percentage reactance of each
    alternator is based on its own capacity. Find the short-circuit that will flow into a complete 3-phase
    short-circuit at F.

    Using the Sb = 60 MVA, Eb = 13.2 kV at the point F, and solving for the base voltages we have

    Eb1 66 kV 66Eb
    = ; E b1 = but Eb = 13.2 kV
    Eb 13.2 kV 13.2

    66(13.2)
    E b1 = = 66 kV
    13.2
    Eb 2 12 kV 12 (66)
    = ; Eb2 = but Eb2 = 12 kV
    Eb1 66 kV 66

    For reactances of transformers

    2
     13.2   60 
    X T1pu = j 0.06     = j 0.06 pu
     13.2   60 
    2
     66   60 
    X T2pu = j 0.05     = j 0.06 pu
     66   50 

    For reactances of generators

    2
     12   60 
    X g1pu = j 0.30     = j1.20 pu
     12   15 
    2
     12   60 
    X g2pu = j 0.50     = j1.50 pu
     12   20 

    For reactances of transmission lines

    Eb 2 (66kV ) 2
    Z base = = = 72.6 Ω
    Sb (60 MVA)

    j 25
    Z LINEpu = = j 0.3444 pu
    72.6
    if we let VLpu = 1.0 pu, then Espu = VLpu = 1.0 pu at no-load.

    So for a fault at point F.

    then by KVL, we have

    Espu − I 3φ Fpu ( j 0.6667 + j 0.06 + j 0.3444 + j 0.06) = 0

    Espu 1.0∠0°
    I 3φ FBpu = =
    ( j 0.6667 + j 0.06 + j 0.06 + j 0.3444) j1.1311
    I 3φ Fpu = 0.8841∠ − 90° pu

    Sb (60 x106 )
    I base = = = 2624.32 Amps
    3Eb 13.2 x103

    I 3φ F = 0.8841(2624.32)

    I 3φ F = 2320.16 Amps

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