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Mohammed Kamel

Jr Member
  • Content Count

    3
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About Mohammed Kamel

  • Rank
    Intern

Previous Fields

  • Engineering Field
    Power systems analysis and distribution
  • License
    PE
  • Calculator
    Casio
  • Discipline
    Electrical

Profile Information

  • Gender
    Male
  • Location
    Makkah, KSA
  1. IbB is equal to IaA but lagged by angle 120 degree, hence, IbB = 70 * (Cos(-140) + j Sin(-140)) = -53.623-j45.
  2. This problem can be solved easily by applying KVL for loop (a-A-B-b) as it is without converting it into single phase values where, IaA = 65.78-j23.94, IbB = -53.62-j45 ( IbBI = 70, angle = -140), Vab = (65.78-j23.94)*(5+j10) + 12500 - (-53.62-j45)*(5+j10) Vab = 12886.47 + j1299.3 IVabI = 12952 volt
  3. I think this question may issued considering the permissions under article 440.41 (A) :"In case the motor controller is rated in horsepower but is without one or both of the foregoing current ratings, equivalent currents shall be determined from the ratings as follows. Table 430.248, Table 430.249, and Table 430.250 shall be used to determine the equivalent full-load current rating. Table 430.251(A) and Table 430.251(B) shall be used to determine the equivalent locked-rotor current ratings.", and article 430.83(D).
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