A 200V , DC shunt machines has an armature resistances 0.5 ohm and field resistance of 200 ohm. The machine in running at 1000 грm as a motor drawing 31 A from the supply mains. Calculate the speed at which the machine must be driven to achieve this as generator
2 Answers
To solve this, we need to determine the speed required for the machine to achieve a given current as a generator, assuming it's already running at a specific speed as a motor.
Here’s a step-by-step process to find the speed at which the machine must be driven to achieve the same current as a generator:
### Given Data:
- Supply Voltage, \( V \) = 200 V
- Armature Resistance, \( R_a \) = 0.5 Ω
- Field Resistance, \( R_f \) = 200 Ω
- Current drawn from supply mains, \( I \) = 31 A (as a motor)
- Speed (as a motor), \( N_m \) = 1000 rpm (as given in the problem context)
### As a Motor:
1. **Calculate the armature current, \( I_a \):**
\[ I_a = I - \frac{V}{R_f} \]
Here, \( I \) is the total current drawn from the supply, and the field current is given by \( \frac{V}{R_f} \).
\[ I_f = \frac{V}{R_f} = \frac{200}{200} = 1 \text{ A} \]
\[ I_a = 31 - 1 = 30 \text{ A} \]
2. **Determine the back EMF, \( E \), when operating as a motor:**
\[ E = V - I_a \cdot R_a \]
\[ E = 200 - 30 \cdot 0.5 = 200 - 15 = 185 \text{ V} \]
### As a Generator:
To achieve the same current (31 A) as a generator, the back EMF \( E_g \) must be the same, but the terminal voltage \( V_g \) will now be the same as the supply voltage \( V \).
3. **Calculate the speed required for the generator operation:**
The relationship between the speed and the back EMF is given by:
\[ \frac{N_g}{N_m} = \frac{E_g}{E_m} \]
where \( E_g \) and \( E_m \) are the back EMFs as a generator and motor, respectively, and \( N_g \) and \( N_m \) are the speeds of the machine as a generator and motor, respectively.
Since the back EMF \( E_g \) is the same as \( E_m \):
\[ \frac{N_g}{N_m} = \frac{E_g}{E_m} = \frac{185}{185} = 1 \]
Therefore:
\[ N_g = N_m \]
So, the machine must be driven at the same speed as a generator to achieve the same current, which is \( 1000 \text{ rpm} \).
Here’s a step-by-step process to find the speed at which the machine must be driven to achieve the same current as a generator:
### Given Data:
- Supply Voltage, \( V \) = 200 V
- Armature Resistance, \( R_a \) = 0.5 Ω
- Field Resistance, \( R_f \) = 200 Ω
- Current drawn from supply mains, \( I \) = 31 A (as a motor)
- Speed (as a motor), \( N_m \) = 1000 rpm (as given in the problem context)
### As a Motor:
1. **Calculate the armature current, \( I_a \):**
\[ I_a = I - \frac{V}{R_f} \]
Here, \( I \) is the total current drawn from the supply, and the field current is given by \( \frac{V}{R_f} \).
\[ I_f = \frac{V}{R_f} = \frac{200}{200} = 1 \text{ A} \]
\[ I_a = 31 - 1 = 30 \text{ A} \]
2. **Determine the back EMF, \( E \), when operating as a motor:**
\[ E = V - I_a \cdot R_a \]
\[ E = 200 - 30 \cdot 0.5 = 200 - 15 = 185 \text{ V} \]
### As a Generator:
To achieve the same current (31 A) as a generator, the back EMF \( E_g \) must be the same, but the terminal voltage \( V_g \) will now be the same as the supply voltage \( V \).
3. **Calculate the speed required for the generator operation:**
The relationship between the speed and the back EMF is given by:
\[ \frac{N_g}{N_m} = \frac{E_g}{E_m} \]
where \( E_g \) and \( E_m \) are the back EMFs as a generator and motor, respectively, and \( N_g \) and \( N_m \) are the speeds of the machine as a generator and motor, respectively.
Since the back EMF \( E_g \) is the same as \( E_m \):
\[ \frac{N_g}{N_m} = \frac{E_g}{E_m} = \frac{185}{185} = 1 \]
Therefore:
\[ N_g = N_m \]
So, the machine must be driven at the same speed as a generator to achieve the same current, which is \( 1000 \text{ rpm} \).
To determine the speed at which a 200V DC shunt machine must be driven to achieve the same current and torque as a generator, we need to consider how the machine operates in motor and generator modes. Here’s a step-by-step approach:
### Given Data:
- Supply Voltage (\( V \)) = 200 V
- Armature Resistance (\( R_a \)) = 0.5 Ω
- Field Resistance (\( R_f \)) = 200 Ω
- Current drawn from the supply mains as a motor (\( I_{motor} \)) = 31 A
- Speed of the motor (\( N_{motor} \)) = 1000 rpm
### Motor Operation Analysis:
1. **Calculate the Field Current (\( I_f \)):**
The field current is determined by the field resistance and the supply voltage. Since the machine is shunt-connected, the field winding is directly connected across the supply voltage.
\[
I_f = \frac{V}{R_f} = \frac{200\text{ V}}{200\text{ Ω}} = 1\text{ A}
\]
2. **Calculate the Armature Current (\( I_a \)):**
The armature current is the total current drawn minus the field current.
\[
I_a = I_{motor} - I_f = 31\text{ A} - 1\text{ A} = 30\text{ A}
\]
3. **Calculate the Armature Voltage (\( V_a \)):**
The armature voltage is the voltage across the armature, which is the supply voltage minus the voltage drop across the armature resistance.
\[
V_a = V - I_a \cdot R_a = 200\text{ V} - 30\text{ A} \cdot 0.5\text{ Ω} = 200\text{ V} - 15\text{ V} = 185\text{ V}
\]
The voltage \( V_a \) is the effective voltage applied to the armature when the machine is running as a motor.
### Generator Operation Analysis:
In generator mode, the same armature current (\( I_a \)) will be present, but the armature voltage will be different. For the machine to deliver the same current as a generator, the induced voltage in the armature must balance the supply voltage when including the armature resistance drop.
1. **Induced Voltage (\( E_g \)) Calculation:**
The induced voltage in the generator mode will be given by:
\[
E_g = V_a + I_a \cdot R_a = 185\text{ V} + 30\text{ A} \cdot 0.5\text{ Ω} = 185\text{ V} + 15\text{ V} = 200\text{ V}
\]
So, to draw 31 A from the mains as a generator, the induced voltage \( E_g \) must also be 200 V.
2. **Calculate the Speed of the Generator (\( N_g \)):**
The speed of a DC machine is directly proportional to the induced voltage and inversely proportional to the flux. Given that the field current \( I_f \) remains constant, the flux will be constant as well.
The speed is related to the voltage in a linear relationship:
\[
\frac{N_{generator}}{N_{motor}} = \frac{E_g}{V_a}
\]
Given that \( E_g \) and \( V_a \) are equal to 200 V, and considering that \( N_{motor} = 1000\) rpm:
\[
N_{generator} = N_{motor} = 1000\text{ rpm}
\]
Therefore, the speed of the machine must be the same as in the motor mode.
### Conclusion:
To achieve the same current and torque as a generator, the machine must be driven at the same speed as it was when operating as a motor, which is 1000 rpm.
### Given Data:
- Supply Voltage (\( V \)) = 200 V
- Armature Resistance (\( R_a \)) = 0.5 Ω
- Field Resistance (\( R_f \)) = 200 Ω
- Current drawn from the supply mains as a motor (\( I_{motor} \)) = 31 A
- Speed of the motor (\( N_{motor} \)) = 1000 rpm
### Motor Operation Analysis:
1. **Calculate the Field Current (\( I_f \)):**
The field current is determined by the field resistance and the supply voltage. Since the machine is shunt-connected, the field winding is directly connected across the supply voltage.
\[
I_f = \frac{V}{R_f} = \frac{200\text{ V}}{200\text{ Ω}} = 1\text{ A}
\]
2. **Calculate the Armature Current (\( I_a \)):**
The armature current is the total current drawn minus the field current.
\[
I_a = I_{motor} - I_f = 31\text{ A} - 1\text{ A} = 30\text{ A}
\]
3. **Calculate the Armature Voltage (\( V_a \)):**
The armature voltage is the voltage across the armature, which is the supply voltage minus the voltage drop across the armature resistance.
\[
V_a = V - I_a \cdot R_a = 200\text{ V} - 30\text{ A} \cdot 0.5\text{ Ω} = 200\text{ V} - 15\text{ V} = 185\text{ V}
\]
The voltage \( V_a \) is the effective voltage applied to the armature when the machine is running as a motor.
### Generator Operation Analysis:
In generator mode, the same armature current (\( I_a \)) will be present, but the armature voltage will be different. For the machine to deliver the same current as a generator, the induced voltage in the armature must balance the supply voltage when including the armature resistance drop.
1. **Induced Voltage (\( E_g \)) Calculation:**
The induced voltage in the generator mode will be given by:
\[
E_g = V_a + I_a \cdot R_a = 185\text{ V} + 30\text{ A} \cdot 0.5\text{ Ω} = 185\text{ V} + 15\text{ V} = 200\text{ V}
\]
So, to draw 31 A from the mains as a generator, the induced voltage \( E_g \) must also be 200 V.
2. **Calculate the Speed of the Generator (\( N_g \)):**
The speed of a DC machine is directly proportional to the induced voltage and inversely proportional to the flux. Given that the field current \( I_f \) remains constant, the flux will be constant as well.
The speed is related to the voltage in a linear relationship:
\[
\frac{N_{generator}}{N_{motor}} = \frac{E_g}{V_a}
\]
Given that \( E_g \) and \( V_a \) are equal to 200 V, and considering that \( N_{motor} = 1000\) rpm:
\[
N_{generator} = N_{motor} = 1000\text{ rpm}
\]
Therefore, the speed of the machine must be the same as in the motor mode.
### Conclusion:
To achieve the same current and torque as a generator, the machine must be driven at the same speed as it was when operating as a motor, which is 1000 rpm.