The per-unit impedance of a single-phase electric load is 0.3. The base power is 500 kVA, and the base voltage is 13.8 kV. a. Find the per-u

Question

The per-unit impedance of a single-phase electric load is 0.3. The base power is 500 kVA, and the base voltage is 13.8 kV. a. Find the per-unit impedance of the load if 1 MVA and 24 kV are selected as base values. b. Find the ohmic value of the impedance.

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Acacia 3 years 2021-08-15T20:02:11+00:00 1 Answers 13 views 0

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  1. thuthuy
    0
    2021-08-15T20:04:06+00:00
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    Answer:

    114.26

    Explanation:

    a)Formula for per unit impedance for change of base is

    Zpu2= Zpu1×(kV1/kV2)²×(kVA2/kVA1)

    Zpu2: New per unit impedance

    Zpu1: given per unit impedance

    kV1: give base voltage

    kV2: New bas votlage

    kVA1: given bas power

    kVA2: new base power

    In the question

    Zpu2=??

    Zpu1= 0.3

    kV2=24kV

    kV1= 13.8 kV

    kVA2= 1MVA ×1000= 1000 kVA

    kVA1=500kVA

    Zpu2= 0.3(13.8/24)²×(1000/500)

    Zpu2= 0.198

    b) to find ohmic impedance we will first calculate base value of impedance(Zbase). So,

    Zbase= kV²/MVA

      Zbase= 13.8²/(500/1000)

      Zbase=380.88

    Now that we have base value of impedance, Zbase, we can calculate actual ohmic value of impedance(Zactual) by using the following formula:

    Zpu=Zactual/Zbase

    0.3= Zactual/380.88

    Zactual= 114.26 ohms

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