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A three-phase, 50Hz salient-pole synchronous motor has a per-phase direct-axis reactance (Xd) of 0.8 pu and a per phase quadrature-axis reactance (Xq) of 0.6pu. Resistance of the machine is negligible. It is drawing full-load current at 0.8 pf (leading). When the terminal voltage is 1pu, perphase induced voltage, in pu, is __________.
    Correct answer is '1.6086'. Can you explain this answer?
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    Solution:

    Given data:

    Xd = 0.8 pu
    Xq = 0.6 pu
    Power factor (pf) = 0.8 (leading)
    Resistance (R) = 0

    To find:

    Per-phase induced voltage (E), in pu

    Formula:

    The per-phase induced voltage of a synchronous motor is given by:

    E = V + IaZ

    Where,
    V = Terminal voltage per phase
    Ia = Armature current per phase
    Z = Impedance of the motor per phase

    The impedance of the motor per phase can be represented as:

    Z = R + jX

    Where,
    R = Resistance per phase
    X = Reactance per phase

    The reactance per phase can be represented as:

    X = Xd + Xqcos2θ

    Where,
    θ = Power factor angle

    Calculation:

    Given data:

    Xd = 0.8 pu
    Xq = 0.6 pu
    pf = 0.8 (leading)
    R = 0
    V = 1 pu

    From the given data, we can calculate the value of θ as:

    cosθ = pf
    cosθ = 0.8 (leading)
    θ = cos⁻¹(0.8)
    θ = 36.8699°

    The reactance per phase can be calculated as:

    X = Xd + Xqcos2θ
    X = 0.8 + 0.6cos(2×36.8699°)
    X = 1.0137 pu

    The impedance per phase can be represented as:

    Z = R + jX
    Z = 0 + j1.0137

    The armature current per phase can be calculated as:

    pf = cosθ = Re(IaZ) / |IaZ|
    0.8 = Re(IaZ) / |IaZ|
    |IaZ| = Re(IaZ) / 0.8

    |IaZ| = V / √(R² + X²)

    |IaZ| = 1 / √(0² + 1.0137²)

    |IaZ| = 0.9867 pu

    The armature current per phase is leading the terminal voltage, so it can be represented as:

    Ia = |IaZ| ∠-θ
    Ia = 0.9867 ∠-36.8699°

    Now, we can calculate the per-phase induced voltage as:

    E = V + IaZ
    E = 1 + (0.9867 ∠-36.8699°) × (0 + j1.0137)
    E = 1.6086 pu ∠39.8054°

    Therefore, the per-phase induced voltage, in pu, is 1.6086.
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    A three-phase, 50Hz salient-pole synchronous motor has a per-phase direct-axis reactance (Xd) of 0.8 pu and a per phase quadrature-axis reactance (Xq) of 0.6pu. Resistance of the machine is negligible. It is drawing full-load current at 0.8 pf (leading). When the terminal voltage is 1pu, perphase induced voltage, in pu, is __________.Correct answer is '1.6086'. Can you explain this answer?
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