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How does a generator not stop itself?
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I'm having trouble reconciling my understanding of motors and generators.
Let's examine a simple DC motor at rest.
Applying a voltage ($V_applied$) to the terminals generates a large current and a corresponding large torque which rapidly accelerates the motor shaft and any mechanical load. Let's say the rotor spins counter-clockwise (CCW). As the rotor speeds up, an increasingly larger back-EMF ($E$) restricts the current in the windings until motor reaches mechanical equilibrium with the load (no more excess torque to accelerate the load) or until $E = V_applied$ (the motor's no load max speed).
What happens if we suddenly remove the applied voltage and externally spin the motor at a constant speed in the same direction (CCW)? Now the back-EMF becomes the voltage source, and current flows in the opposite direction (assuming we've shorted the circuit, $I = E/R_windings$).
Does this current generate a counter-torque that resists the externally-applied rotation of the motor? If so, does this torque and the external power source (such as windmill) reach some equilibrium state like in the motor case?
Side question: In a powerplant where steam spins a generator to produce very big EMF, does the "grid" serve as an additional resistance in the generator's circuit that limits the current?
I've read that motor manufacturers will often characterize their motors using other motors. To test a motor's torque/speed performance, they will mechanically load it with another motor and feed the power generated from the second motor back into the driving motor. Is the 2nd motor's counter-torque (as hypothesized above) what provides a mechanical load on the first motor? Otherwise, the first motor would just spin a dummy mass.
motor generator
asked 3 hours ago
techSultan
224213
add a comment |Â
up vote
1
down vote
favorite
I'm having trouble reconciling my understanding of motors and generators.
Let's examine a simple DC motor at rest.
Applying a voltage ($V_applied$) to the terminals generates a large current and a corresponding large torque which rapidly accelerates the motor shaft and any mechanical load. Let's say the rotor spins counter-clockwise (CCW). As the rotor speeds up, an increasingly larger back-EMF ($E$) restricts the current in the windings until motor reaches mechanical equilibrium with the load (no more excess torque to accelerate the load) or until $E = V_applied$ (the motor's no load max speed).
What happens if we suddenly remove the applied voltage and externally spin the motor at a constant speed in the same direction (CCW)? Now the back-EMF becomes the voltage source, and current flows in the opposite direction (assuming we've shorted the circuit, $I = E/R_windings$).
Does this current generate a counter-torque that resists the externally-applied rotation of the motor? If so, does this torque and the external power source (such as windmill) reach some equilibrium state like in the motor case?
Side question: In a powerplant where steam spins a generator to produce very big EMF, does the "grid" serve as an additional resistance in the generator's circuit that limits the current?
I've read that motor manufacturers will often characterize their motors using other motors. To test a motor's torque/speed performance, they will mechanically load it with another motor and feed the power generated from the second motor back into the driving motor. Is the 2nd motor's counter-torque (as hypothesized above) what provides a mechanical load on the first motor? Otherwise, the first motor would just spin a dummy mass.
motor generator
asked 3 hours ago
techSultan
224213
They way they test generators gives you the clue. Spinning a generator gives a voltage (EMF) as output, but it takes more mechanical energy to spin the generator than you get out of the system.
– Jack Creasey
3 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm having trouble reconciling my understanding of motors and generators.
Let's examine a simple DC motor at rest.
Applying a voltage ($V_applied$) to the terminals generates a large current and a corresponding large torque which rapidly accelerates the motor shaft and any mechanical load. Let's say the rotor spins counter-clockwise (CCW). As the rotor speeds up, an increasingly larger back-EMF ($E$) restricts the current in the windings until motor reaches mechanical equilibrium with the load (no more excess torque to accelerate the load) or until $E = V_applied$ (the motor's no load max speed).
What happens if we suddenly remove the applied voltage and externally spin the motor at a constant speed in the same direction (CCW)? Now the back-EMF becomes the voltage source, and current flows in the opposite direction (assuming we've shorted the circuit, $I = E/R_windings$).
Does this current generate a counter-torque that resists the externally-applied rotation of the motor? If so, does this torque and the external power source (such as windmill) reach some equilibrium state like in the motor case?
Side question: In a powerplant where steam spins a generator to produce very big EMF, does the "grid" serve as an additional resistance in the generator's circuit that limits the current?
I've read that motor manufacturers will often characterize their motors using other motors. To test a motor's torque/speed performance, they will mechanically load it with another motor and feed the power generated from the second motor back into the driving motor. Is the 2nd motor's counter-torque (as hypothesized above) what provides a mechanical load on the first motor? Otherwise, the first motor would just spin a dummy mass.
motor generator
asked 3 hours ago
techSultan
224213
I'm having trouble reconciling my understanding of motors and generators.
Let's examine a simple DC motor at rest.
Applying a voltage ($V_applied$) to the terminals generates a large current and a corresponding large torque which rapidly accelerates the motor shaft and any mechanical load. Let's say the rotor spins counter-clockwise (CCW). As the rotor speeds up, an increasingly larger back-EMF ($E$) restricts the current in the windings until motor reaches mechanical equilibrium with the load (no more excess torque to accelerate the load) or until $E = V_applied$ (the motor's no load max speed).
What happens if we suddenly remove the applied voltage and externally spin the motor at a constant speed in the same direction (CCW)? Now the back-EMF becomes the voltage source, and current flows in the opposite direction (assuming we've shorted the circuit, $I = E/R_windings$).
Does this current generate a counter-torque that resists the externally-applied rotation of the motor? If so, does this torque and the external power source (such as windmill) reach some equilibrium state like in the motor case?
Side question: In a powerplant where steam spins a generator to produce very big EMF, does the "grid" serve as an additional resistance in the generator's circuit that limits the current?
I've read that motor manufacturers will often characterize their motors using other motors. To test a motor's torque/speed performance, they will mechanically load it with another motor and feed the power generated from the second motor back into the driving motor. Is the 2nd motor's counter-torque (as hypothesized above) what provides a mechanical load on the first motor? Otherwise, the first motor would just spin a dummy mass.
motor generator
motor generator
asked 3 hours ago
techSultan
224213
asked 3 hours ago
techSultan
224213
asked 3 hours ago
techSultan
224213
asked 3 hours ago
techSultan
224213
asked 3 hours ago
techSultan
224213
224213
They way they test generators gives you the clue. Spinning a generator gives a voltage (EMF) as output, but it takes more mechanical energy to spin the generator than you get out of the system.
– Jack Creasey
3 hours ago
add a comment |Â
They way they test generators gives you the clue. Spinning a generator gives a voltage (EMF) as output, but it takes more mechanical energy to spin the generator than you get out of the system.
– Jack Creasey
3 hours ago
They way they test generators gives you the clue. Spinning a generator gives a voltage (EMF) as output, but it takes more mechanical energy to spin the generator than you get out of the system.
– Jack Creasey
3 hours ago
They way they test generators gives you the clue. Spinning a generator gives a voltage (EMF) as output, but it takes more mechanical energy to spin the generator than you get out of the system.
– Jack Creasey
3 hours ago
add a comment |Â
2 Answers
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3
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How does a generator not stop itself?
It does, under a later assumption that you made:
What happens if we suddenly remove the applied voltage and externally spin the motor at a constant speed in the same direction (CCW) … assuming we've shorted the circuit, $I = E/R_windings$ …
Shorting the terminals of a generator/motor results in maximum opposing torque. The mechanical energy is converted into heat in the motor windings. (This is the maximally violent way of braking a machine driven by an electric motor.)
If, instead, you leave the circuit open, then there is no current, hence approximately no torque, and the back-EMF maintains a voltage across the open terminals. (This is a normal and non-destructive situation, just like an outlet with nothing plugged in.) An ideal model with no losses would keep spinning at the same speed without any mechanical or electrical power input.
In practice, if you are using a generator to do some work then there is some load on it that is neither an open circuit nor a short circuit, and so the behavior will be intermediate between these two cases.
Dynamic braking (which includes regenerative braking) is the practice of using an electric motor to stop a vehicle or machine. The simplest way to do it is to connect a suitable resistor between the terminals; then, unlike the short-circuit case, most of the energy is dissipated in the resistor. Regenerative braking uses a rechargeable battery instead (with intermediate power converters managing the voltage/current); just as the motor can be a generator, the battery can absorb instead of supply power, as long as the intermediate systems are designed to allow this.
answered 3 hours ago
Kevin Reid
4,97511633
add a comment |Â
up vote
1
down vote
Think of an ideal permanent magnet brushed DC machine (either a motor or generator, they are equivalent) as a sort of electromechanical transformer, that converts between the mechanical domain and the electrical domain. (Brushless DC and wound field are fundamentally the same, but have extra details that make reasoning about them more difficult, as are AC machines which are even more difficult)
Neglecting losses, torque and current are proportional, and voltage and speed are also proportional.
Power, measured in watts, is the product of voltage and current, and is also the product of angular speed (radians/second) and torque (Newton metres). Power is conserved when it's converted between one side and the other side of the electrical machine.
Once you have those principles embedded in your brain, it's possible to reason 'what happens if' this or that is done to a motor.
If you spin the motor, you generate a proportional voltage on the terminals. If there's no current flowing, then no torque is felt on the shaft. If you connect a load, a current flows, so a torque has to be driven. The motor will slow down if the engine is not powerful enough to supply that torque at that speed.
answered 1 hour ago
Neil_UK
70.3k273155
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
How does a generator not stop itself?
It does, under a later assumption that you made:
What happens if we suddenly remove the applied voltage and externally spin the motor at a constant speed in the same direction (CCW) … assuming we've shorted the circuit, $I = E/R_windings$ …
Shorting the terminals of a generator/motor results in maximum opposing torque. The mechanical energy is converted into heat in the motor windings. (This is the maximally violent way of braking a machine driven by an electric motor.)
If, instead, you leave the circuit open, then there is no current, hence approximately no torque, and the back-EMF maintains a voltage across the open terminals. (This is a normal and non-destructive situation, just like an outlet with nothing plugged in.) An ideal model with no losses would keep spinning at the same speed without any mechanical or electrical power input.
In practice, if you are using a generator to do some work then there is some load on it that is neither an open circuit nor a short circuit, and so the behavior will be intermediate between these two cases.
Dynamic braking (which includes regenerative braking) is the practice of using an electric motor to stop a vehicle or machine. The simplest way to do it is to connect a suitable resistor between the terminals; then, unlike the short-circuit case, most of the energy is dissipated in the resistor. Regenerative braking uses a rechargeable battery instead (with intermediate power converters managing the voltage/current); just as the motor can be a generator, the battery can absorb instead of supply power, as long as the intermediate systems are designed to allow this.
answered 3 hours ago
Kevin Reid
4,97511633
add a comment |Â
up vote
3
down vote
How does a generator not stop itself?
It does, under a later assumption that you made:
What happens if we suddenly remove the applied voltage and externally spin the motor at a constant speed in the same direction (CCW) … assuming we've shorted the circuit, $I = E/R_windings$ …
Shorting the terminals of a generator/motor results in maximum opposing torque. The mechanical energy is converted into heat in the motor windings. (This is the maximally violent way of braking a machine driven by an electric motor.)
If, instead, you leave the circuit open, then there is no current, hence approximately no torque, and the back-EMF maintains a voltage across the open terminals. (This is a normal and non-destructive situation, just like an outlet with nothing plugged in.) An ideal model with no losses would keep spinning at the same speed without any mechanical or electrical power input.
In practice, if you are using a generator to do some work then there is some load on it that is neither an open circuit nor a short circuit, and so the behavior will be intermediate between these two cases.
Dynamic braking (which includes regenerative braking) is the practice of using an electric motor to stop a vehicle or machine. The simplest way to do it is to connect a suitable resistor between the terminals; then, unlike the short-circuit case, most of the energy is dissipated in the resistor. Regenerative braking uses a rechargeable battery instead (with intermediate power converters managing the voltage/current); just as the motor can be a generator, the battery can absorb instead of supply power, as long as the intermediate systems are designed to allow this.
answered 3 hours ago
Kevin Reid
4,97511633
add a comment |Â
up vote
3
down vote
up vote
3
down vote
How does a generator not stop itself?
It does, under a later assumption that you made:
What happens if we suddenly remove the applied voltage and externally spin the motor at a constant speed in the same direction (CCW) … assuming we've shorted the circuit, $I = E/R_windings$ …
Shorting the terminals of a generator/motor results in maximum opposing torque. The mechanical energy is converted into heat in the motor windings. (This is the maximally violent way of braking a machine driven by an electric motor.)
If, instead, you leave the circuit open, then there is no current, hence approximately no torque, and the back-EMF maintains a voltage across the open terminals. (This is a normal and non-destructive situation, just like an outlet with nothing plugged in.) An ideal model with no losses would keep spinning at the same speed without any mechanical or electrical power input.
In practice, if you are using a generator to do some work then there is some load on it that is neither an open circuit nor a short circuit, and so the behavior will be intermediate between these two cases.
Dynamic braking (which includes regenerative braking) is the practice of using an electric motor to stop a vehicle or machine. The simplest way to do it is to connect a suitable resistor between the terminals; then, unlike the short-circuit case, most of the energy is dissipated in the resistor. Regenerative braking uses a rechargeable battery instead (with intermediate power converters managing the voltage/current); just as the motor can be a generator, the battery can absorb instead of supply power, as long as the intermediate systems are designed to allow this.
answered 3 hours ago
Kevin Reid
4,97511633
How does a generator not stop itself?
It does, under a later assumption that you made:
What happens if we suddenly remove the applied voltage and externally spin the motor at a constant speed in the same direction (CCW) … assuming we've shorted the circuit, $I = E/R_windings$ …
Shorting the terminals of a generator/motor results in maximum opposing torque. The mechanical energy is converted into heat in the motor windings. (This is the maximally violent way of braking a machine driven by an electric motor.)
If, instead, you leave the circuit open, then there is no current, hence approximately no torque, and the back-EMF maintains a voltage across the open terminals. (This is a normal and non-destructive situation, just like an outlet with nothing plugged in.) An ideal model with no losses would keep spinning at the same speed without any mechanical or electrical power input.
In practice, if you are using a generator to do some work then there is some load on it that is neither an open circuit nor a short circuit, and so the behavior will be intermediate between these two cases.
Dynamic braking (which includes regenerative braking) is the practice of using an electric motor to stop a vehicle or machine. The simplest way to do it is to connect a suitable resistor between the terminals; then, unlike the short-circuit case, most of the energy is dissipated in the resistor. Regenerative braking uses a rechargeable battery instead (with intermediate power converters managing the voltage/current); just as the motor can be a generator, the battery can absorb instead of supply power, as long as the intermediate systems are designed to allow this.
answered 3 hours ago
Kevin Reid
4,97511633
answered 3 hours ago
Kevin Reid
4,97511633
answered 3 hours ago
Kevin Reid
4,97511633
answered 3 hours ago
Kevin Reid
4,97511633
4,97511633
add a comment |Â
add a comment |Â
up vote
1
down vote
Think of an ideal permanent magnet brushed DC machine (either a motor or generator, they are equivalent) as a sort of electromechanical transformer, that converts between the mechanical domain and the electrical domain. (Brushless DC and wound field are fundamentally the same, but have extra details that make reasoning about them more difficult, as are AC machines which are even more difficult)
Neglecting losses, torque and current are proportional, and voltage and speed are also proportional.
Power, measured in watts, is the product of voltage and current, and is also the product of angular speed (radians/second) and torque (Newton metres). Power is conserved when it's converted between one side and the other side of the electrical machine.
Once you have those principles embedded in your brain, it's possible to reason 'what happens if' this or that is done to a motor.
If you spin the motor, you generate a proportional voltage on the terminals. If there's no current flowing, then no torque is felt on the shaft. If you connect a load, a current flows, so a torque has to be driven. The motor will slow down if the engine is not powerful enough to supply that torque at that speed.
answered 1 hour ago
Neil_UK
70.3k273155
add a comment |Â
up vote
1
down vote
Think of an ideal permanent magnet brushed DC machine (either a motor or generator, they are equivalent) as a sort of electromechanical transformer, that converts between the mechanical domain and the electrical domain. (Brushless DC and wound field are fundamentally the same, but have extra details that make reasoning about them more difficult, as are AC machines which are even more difficult)
Neglecting losses, torque and current are proportional, and voltage and speed are also proportional.
Power, measured in watts, is the product of voltage and current, and is also the product of angular speed (radians/second) and torque (Newton metres). Power is conserved when it's converted between one side and the other side of the electrical machine.
Once you have those principles embedded in your brain, it's possible to reason 'what happens if' this or that is done to a motor.
If you spin the motor, you generate a proportional voltage on the terminals. If there's no current flowing, then no torque is felt on the shaft. If you connect a load, a current flows, so a torque has to be driven. The motor will slow down if the engine is not powerful enough to supply that torque at that speed.
answered 1 hour ago
Neil_UK
70.3k273155
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Think of an ideal permanent magnet brushed DC machine (either a motor or generator, they are equivalent) as a sort of electromechanical transformer, that converts between the mechanical domain and the electrical domain. (Brushless DC and wound field are fundamentally the same, but have extra details that make reasoning about them more difficult, as are AC machines which are even more difficult)
Neglecting losses, torque and current are proportional, and voltage and speed are also proportional.
Power, measured in watts, is the product of voltage and current, and is also the product of angular speed (radians/second) and torque (Newton metres). Power is conserved when it's converted between one side and the other side of the electrical machine.
Once you have those principles embedded in your brain, it's possible to reason 'what happens if' this or that is done to a motor.
If you spin the motor, you generate a proportional voltage on the terminals. If there's no current flowing, then no torque is felt on the shaft. If you connect a load, a current flows, so a torque has to be driven. The motor will slow down if the engine is not powerful enough to supply that torque at that speed.
answered 1 hour ago
Neil_UK
70.3k273155
Think of an ideal permanent magnet brushed DC machine (either a motor or generator, they are equivalent) as a sort of electromechanical transformer, that converts between the mechanical domain and the electrical domain. (Brushless DC and wound field are fundamentally the same, but have extra details that make reasoning about them more difficult, as are AC machines which are even more difficult)
Neglecting losses, torque and current are proportional, and voltage and speed are also proportional.
Power, measured in watts, is the product of voltage and current, and is also the product of angular speed (radians/second) and torque (Newton metres). Power is conserved when it's converted between one side and the other side of the electrical machine.
Once you have those principles embedded in your brain, it's possible to reason 'what happens if' this or that is done to a motor.
If you spin the motor, you generate a proportional voltage on the terminals. If there's no current flowing, then no torque is felt on the shaft. If you connect a load, a current flows, so a torque has to be driven. The motor will slow down if the engine is not powerful enough to supply that torque at that speed.
answered 1 hour ago
Neil_UK
70.3k273155
answered 1 hour ago
Neil_UK
70.3k273155
answered 1 hour ago
Neil_UK
70.3k273155
answered 1 hour ago
Neil_UK
70.3k273155
70.3k273155
add a comment |Â
add a comment |Â
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They way they test generators gives you the clue. Spinning a generator gives a voltage (EMF) as output, but it takes more mechanical energy to spin the generator than you get out of the system.
– Jack Creasey
3 hours ago