A 20 pole, 693V, 50 Hz, 3 phase, delta connected synchronous motor is operating at no load with normal excitation. It has armature resistance per phase of zero and synchronous reactance of 10 ohm. If the rotor is restarted by 0.5 degree (mechanical) from its synchronous position, compute i) rotor displacement in electrical degrees ii) armature emf/phase iii) armature current /phase, iv) power drawn by the motor and v) power developed by armature.
2 Answers
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
### 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
To solve the problem, we need to use the following given data:
- **Number of Poles (P):** 20
- **Voltage (V):** 693 V (line-to-line)
- **Frequency (f):** 50 Hz
- **Delta Connected Synchronous Motor**
- **Synchronous Reactance (Xs):** 10 Ω
- **Armature Resistance (Ra):** 0 Ω
- **Rotor Displacement:** 0.5 degrees (mechanical)
Let’s calculate the requested quantities step by step:
### I. Rotor Displacement in Electrical Degrees
For a synchronous motor:
1. **Mechanical Degree per Pole Pair** is given by:
\[
\text{Electrical Degrees per Mechanical Degree} = \frac{P}{2}
\]
2. **Electrical Displacement** is:
\[
\text{Rotor Displacement (in electrical degrees)} = \text{Mechanical Displacement} \times \frac{P}{2}
\]
Substituting the values:
\[
\text{Rotor Displacement (in electrical degrees)} = 0.5 \times \frac{20}{2} = 0.5 \times 10 = 5^\circ
\]
### II. Armature EMF/Phase
In a synchronous motor, the armature EMF (E) can be found from the equation considering the synchronous reactance (Xs), which we use to express the relationship between the applied voltage (V) and the induced EMF:
1. **Line-to-Line Voltage (V_LL)** for a delta-connected motor is related to phase voltage (V_Ph) by:
\[
V_{LL} = \sqrt{3} \times V_{Ph}
\]
Therefore:
\[
V_{Ph} = \frac{V_{LL}}{\sqrt{3}} = \frac{693}{\sqrt{3}} \approx 400 \text{ V}
\]
2. **Armature EMF per Phase** in a synchronous motor is related to the phase voltage and the reactance, considering the displacement of the rotor and that the motor is operating at no load with normal excitation:
Since the motor is at no load, the EMF per phase \( E_{\text{ph}} \) approximately equals the phase voltage of the system, which is 400 V.
\[
E_{\text{ph}} \approx V_{\text{ph}} = 400 \text{ V}
\]
### III. Armature Current/Phase
To find the armature current, we need to use the rotor displacement which affects the back EMF induced. For a synchronous motor:
1. **Phase Current (I_ph)** is given by the following formula where \( \delta \) is the rotor displacement in electrical degrees:
\[
I_{\text{ph}} = \frac{E_{\text{ph}} \sin \delta}{X_s}
\]
Substituting the values:
\[
I_{\text{ph}} = \frac{400 \sin 5^\circ}{10}
\]
\[
\sin 5^\circ \approx 0.087
\]
\[
I_{\text{ph}} = \frac{400 \times 0.087}{10} \approx 3.48 \text{ A}
\]
### IV. Power Drawn by the Motor
The power drawn by the motor is given by the formula:
1. **Power per Phase**:
\[
P_{\text{ph}} = V_{\text{ph}} \times I_{\text{ph}} \times \cos \phi
\]
For a synchronous motor operating at no load, the power factor is close to unity.
Thus, power per phase:
\[
P_{\text{ph}} = 400 \times 3.48 \times \cos 5^\circ
\]
\[
\cos 5^\circ \approx 0.996
\]
\[
P_{\text{ph}} \approx 400 \times 3.48 \times 0.996 \approx 1380 \text{ W}
\]
2. **Total Power Drawn**:
Since the motor is delta-connected, the total power drawn is:
\[
P_{\text{total}} = 3 \times P_{\text{ph}} \approx 3 \times 1380 \approx 4140 \text{ W}
\]
### V. Power Developed by the Armature
The power developed by the armature is given by:
1. **Power Developed**:
\[
P_{\text{developed}} = \frac{E_{\text{ph}} \times I_{\text{ph}} \times \sin \delta}{X_s}
\]
The power developed by the armature equals the power drawn by the motor because there is no loss (armature resistance is zero):
\[
P_{\text{developed}} = P_{\text{total}} \approx 4140 \text{ W}
\]
### Summary
1. **Rotor Displacement in Electrical Degrees:** 5°
2. **Armature EMF/Phase:** 400 V
3. **Armature Current/Phase:** 3.48 A
4. **Power Drawn by the Motor:** 4140 W
5. **Power Developed by Armature:** 4140 W
- **Number of Poles (P):** 20
- **Voltage (V):** 693 V (line-to-line)
- **Frequency (f):** 50 Hz
- **Delta Connected Synchronous Motor**
- **Synchronous Reactance (Xs):** 10 Ω
- **Armature Resistance (Ra):** 0 Ω
- **Rotor Displacement:** 0.5 degrees (mechanical)
Let’s calculate the requested quantities step by step:
### I. Rotor Displacement in Electrical Degrees
For a synchronous motor:
1. **Mechanical Degree per Pole Pair** is given by:
\[
\text{Electrical Degrees per Mechanical Degree} = \frac{P}{2}
\]
2. **Electrical Displacement** is:
\[
\text{Rotor Displacement (in electrical degrees)} = \text{Mechanical Displacement} \times \frac{P}{2}
\]
Substituting the values:
\[
\text{Rotor Displacement (in electrical degrees)} = 0.5 \times \frac{20}{2} = 0.5 \times 10 = 5^\circ
\]
### II. Armature EMF/Phase
In a synchronous motor, the armature EMF (E) can be found from the equation considering the synchronous reactance (Xs), which we use to express the relationship between the applied voltage (V) and the induced EMF:
1. **Line-to-Line Voltage (V_LL)** for a delta-connected motor is related to phase voltage (V_Ph) by:
\[
V_{LL} = \sqrt{3} \times V_{Ph}
\]
Therefore:
\[
V_{Ph} = \frac{V_{LL}}{\sqrt{3}} = \frac{693}{\sqrt{3}} \approx 400 \text{ V}
\]
2. **Armature EMF per Phase** in a synchronous motor is related to the phase voltage and the reactance, considering the displacement of the rotor and that the motor is operating at no load with normal excitation:
Since the motor is at no load, the EMF per phase \( E_{\text{ph}} \) approximately equals the phase voltage of the system, which is 400 V.
\[
E_{\text{ph}} \approx V_{\text{ph}} = 400 \text{ V}
\]
### III. Armature Current/Phase
To find the armature current, we need to use the rotor displacement which affects the back EMF induced. For a synchronous motor:
1. **Phase Current (I_ph)** is given by the following formula where \( \delta \) is the rotor displacement in electrical degrees:
\[
I_{\text{ph}} = \frac{E_{\text{ph}} \sin \delta}{X_s}
\]
Substituting the values:
\[
I_{\text{ph}} = \frac{400 \sin 5^\circ}{10}
\]
\[
\sin 5^\circ \approx 0.087
\]
\[
I_{\text{ph}} = \frac{400 \times 0.087}{10} \approx 3.48 \text{ A}
\]
### IV. Power Drawn by the Motor
The power drawn by the motor is given by the formula:
1. **Power per Phase**:
\[
P_{\text{ph}} = V_{\text{ph}} \times I_{\text{ph}} \times \cos \phi
\]
For a synchronous motor operating at no load, the power factor is close to unity.
Thus, power per phase:
\[
P_{\text{ph}} = 400 \times 3.48 \times \cos 5^\circ
\]
\[
\cos 5^\circ \approx 0.996
\]
\[
P_{\text{ph}} \approx 400 \times 3.48 \times 0.996 \approx 1380 \text{ W}
\]
2. **Total Power Drawn**:
Since the motor is delta-connected, the total power drawn is:
\[
P_{\text{total}} = 3 \times P_{\text{ph}} \approx 3 \times 1380 \approx 4140 \text{ W}
\]
### V. Power Developed by the Armature
The power developed by the armature is given by:
1. **Power Developed**:
\[
P_{\text{developed}} = \frac{E_{\text{ph}} \times I_{\text{ph}} \times \sin \delta}{X_s}
\]
The power developed by the armature equals the power drawn by the motor because there is no loss (armature resistance is zero):
\[
P_{\text{developed}} = P_{\text{total}} \approx 4140 \text{ W}
\]
### Summary
1. **Rotor Displacement in Electrical Degrees:** 5°
2. **Armature EMF/Phase:** 400 V
3. **Armature Current/Phase:** 3.48 A
4. **Power Drawn by the Motor:** 4140 W
5. **Power Developed by Armature:** 4140 W