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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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A three-phase, 50Hz salient-pole synchronous motor has a per-phase dir...
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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.
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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