To solve the problem, let’s break it down step by step:
### Given Data
- **Number of poles (P)**: 20
- **Voltage (V)**: 693V (line-to-line)
- **Frequency (f)**: 50 Hz
- **Synchronous reactance (X_s)**: 10 ohms per phase
- **Armature resistance (R_a)**: 0 ohms per phase
- **Rotor displacement**: 0.5 degrees (mechanical)
### i) Rotor Displacement in Electrical Degrees
The displacement in mechanical degrees must be converted into electrical degrees. For a synchronous machine:
\[ \text{Electrical degrees} = \text{Mechanical degrees} \times \frac{P}{2} \]
So:
\[ \text{Electrical degrees} = 0.5^\circ \times \frac{20}{2} = 5^\circ \]
### ii) Armature EMF per Phase
In a delta-connected system, the line-to-line voltage \( V_{LL} \) relates to the phase voltage \( V_{ph} \) as follows:
\[ V_{ph} = \frac{V_{LL}}{\sqrt{3}} = \frac{693}{\sqrt{3}} = 400 \text{ V} \]
The armature EMF per phase (\( E_{ph} \)) in the synchronous motor can be found using:
\[ E_{ph} = V_{ph} + j X_s I_{ph} \]
At no load, the armature current is predominantly due to the displacement of the rotor. Therefore, the voltage per phase \( E_{ph} \) can be expressed as:
\[ E_{ph} = V_{ph} + j X_s I_{ph} \]
With a small displacement, the armature EMF is approximately equal to the phase voltage \( V_{ph} \) plus the reactance drop caused by the displacement.
However, for small displacements, the EMF can be approximated as:
\[ E_{ph} \approx V_{ph} \]
Thus:
\[ E_{ph} \approx 400 \text{ V} \]
### iii) Armature Current per Phase
The current per phase can be computed from:
\[ I_{ph} = \frac{E_{ph} - V_{ph}}{j X_s} \]
Given that the phase voltage \( V_{ph} \) and the armature EMF \( E_{ph} \) are nearly equal, the current per phase due to a small displacement is:
\[ I_{ph} = \frac{V_{ph} \sin(\delta)}{X_s} \]
Where \( \delta \) is the electrical displacement angle:
\[ I_{ph} = \frac{400 \times \sin(5^\circ)}{10} \approx \frac{400 \times 0.0872}{10} = 3.49 \text{ A} \]
### iv) Power Drawn by the Motor
The power drawn (or the real power) from the supply:
\[ P = 3 \times V_{ph} \times I_{ph} \times \cos(\theta) \]
Since the angle between the voltage and the current is approximately equal to the displacement angle \( \delta \), and for small angles \( \cos(\delta) \approx 1 \):
\[ P \approx 3 \times 400 \times 3.49 \times \cos(5^\circ) \approx 3 \times 400 \times 3.49 = 4,188 \text{ W} \]
### v) Power Developed by Armature
The power developed by the armature (in the absence of losses and assuming all power is converted to mechanical power) is given by:
\[ P_{developed} = 3 \times V_{ph} \times I_{ph} \times \sin(\delta) \]
With a small angle, the power developed is approximately:
\[ P_{developed} \approx 3 \times 400 \times 3.49 \times \sin(5^\circ) \approx 3 \times 400 \times 3.49 \times 0.0872 = 407.6 \text{ W} \]
### Summary
- **Rotor displacement in electrical degrees**: 5°
- **Armature EMF per phase**: Approximately 400 V
- **Armature current per phase**: Approximately 3.49 A
- **Power drawn by the motor**: Approximately 4,188 W
- **Power developed by the armature**: Approximately 407.6 W