Answer:
The magnitude of the starting line current (|I_start|) is approximately
316.41 A.
Step-by-step explanation:
To calculate the starting line current in the case of part-winding starting for the given 3-phase squirrel-cage induction motor, we need to consider the part-winding connection and use the equivalent circuit model.
In a part-winding starting scenario, the stator winding is divided into two sections, and during the starting period, only one section is energized while the other remains open. This reduces the impedance seen by the motor during starting, which helps in reducing the starting current.
First, let's calculate the equivalent impedance of the motor referred to as the stator for the part-winding connection.
Total stator impedance per phase (Zs) = Rr + jXr + (Xs / 2) [Considering the star-connected winding with two sections in parallel]
Zs = 1.5 + j1.7 + (2 / 2) = 1.5 + j1.7 + 1 = 2.5 + j1.7 Ohms
Now, we can calculate the starting line current (I_start) using the following formula:
I_start = V_phase / Zs
where V_phase is the phase voltage, which can be calculated as:
V_phase = V_line / √3
Given that V_line is 1600V and the motor is a 3-phase motor (balanced system), the line voltage (V_line) is equal to the phase voltage (V_phase).
V_phase = 1600V / √3 ≈ 923.88V (rounded to two decimal places)
Now, we can calculate the starting line current:
I_start = 923.88V / (2.5 + j1.7) Ohms
To compute the magnitude of the starting line current, we use the formula for the modulus of a complex number:
|I_start| = |V_phase / Zs| = |923.88V / (2.5 + j1.7)| ≈ |923.88V / 2.92 < 34.99°| ≈ 316.41V (rounded to two decimal places)
So, the magnitude of the starting line current (|I_start|) is approximately 316.41 A.