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Your generator constant is 2 V per Webber-Hz, or 2 V/(Wb Hz) = 2 (V s)/Wb. You somehow got a value of 5. The rest of your calculations are probably off due to that. Added I didn't have much tim...
Answer
#5: Post edited
by
Olin Lathrop
·
2023-05-15T17:31:58Z (over 1 year ago)
- Your generator constant is 2 V per Webber-Hz, or 2 V/(Wb Hz) = 2 (V s)/Wb. You somehow got a value of 5. The rest of your calculations are probably off due to that.
- <h2>Added</h2>
- I didn't have much time when I wrote the above, but can delve into this more now.
- From your first measurement, we can see that the pure generator inside this unit produces 2 V/(Wb Hz). Since you are using the fixed exitation of 1 Wb throughout this question, we can simplify that to 2 V/Hz for this case.
- Now the question is how fast the unit will spin without load when 10 V are applied. You know that the armature resistance is 1 Ω and that the unit draws 60 W when the 10 V is applied.
- We'll use the simple model of an ideal motor/generator with the armature resistance in series. To get an idea of speed, we need to know the voltage applied to the internal ideal motor. To get that we find the voltage drop across the resistance, then subtract that from the applied 10 V.
- The current from a 10 V source delivering 60 W is (60 W)/(10 V) = 6 A. The voltage drop from that current across the internal resistance is then (6 A)(1 Ω) = 6 V. That means the voltage across the internal ideal motor/generator is (10 V) - (6 V) = 4 V.
If the motor were ideal, then its speed would be (4 V)/(2 V/Hz) = 2 Hz = 120 RPM. The actual speed of a real motor would be less due to mechanical friction.One way to get a handle on the friction losses would be to know the mechanical power put into the motor when tested as a generator to produce the 10 V of the first measurement. Since that wasn't given, the best answer is <i>"A bit under 120 RPM."</i>.
- Your generator constant is 2 V per Webber-Hz, or 2 V/(Wb Hz) = 2 (V s)/Wb. You somehow got a value of 5. The rest of your calculations are probably off due to that.
- <h2>Added</h2>
- I didn't have much time when I wrote the above, but can delve into this more now.
- From your first measurement, we can see that the pure generator inside this unit produces 2 V/(Wb Hz). Since you are using the fixed exitation of 1 Wb throughout this question, we can simplify that to 2 V/Hz for this case.
- Now the question is how fast the unit will spin without load when 10 V are applied. You know that the armature resistance is 1 Ω and that the unit draws 60 W when the 10 V is applied.
- We'll use the simple model of an ideal motor/generator with the armature resistance in series. To get an idea of speed, we need to know the voltage applied to the internal ideal motor. To get that we find the voltage drop across the resistance, then subtract that from the applied 10 V.
- The current from a 10 V source delivering 60 W is (60 W)/(10 V) = 6 A. The voltage drop from that current across the internal resistance is then (6 A)(1 Ω) = 6 V. That means the voltage across the internal ideal motor/generator is (10 V) - (6 V) = 4 V.
- Since the ideal generator inside this motor is producing 4 V, we know that the motor speed is (4 V)/(2 V/Hz) = 2 Hz = 120 RPM.
#4: Post edited
by
Olin Lathrop
·
2023-05-12T16:18:53Z (over 1 year ago)
- Your generator constant is 2 V per Webber-Hz, or 2 V/(Wb Hz) = 2 (V s)/Wb. You somehow got a value of 5. The rest of your calculations are probably off due to that.
- <h2>Added</h2>
- I didn't have much time when I wrote the above, but can delve into this more now.
From your first measurement, we can see that the pure generator inside this unit produces 2 V/(Wb Hz). Since you are using the fixed exitation of 1 Wb throughout this question, we can simplify this to 2 V/Hz for this case.- Now the question is how fast the unit will spin without load when 10 V are applied. You know that the armature resistance is 1 Ω and that the unit draws 60 W when the 10 V is applied.
- We'll use the simple model of an ideal motor/generator with the armature resistance in series. To get an idea of speed, we need to know the voltage applied to the internal ideal motor. To get that we find the voltage drop across the resistance, then subtract that from the applied 10 V.
- The current from a 10 V source delivering 60 W is (60 W)/(10 V) = 6 A. The voltage drop from that current across the internal resistance is then (6 A)(1 Ω) = 6 V. That means the voltage across the internal ideal motor/generator is (10 V) - (6 V) = 4 V.
- If the motor were ideal, then its speed would be (4 V)/(2 V/Hz) = 2 Hz = 120 RPM. The actual speed of a real motor would be less due to mechanical friction.
- One way to get a handle on the friction losses would be to know the mechanical power put into the motor when tested as a generator to produce the 10 V of the first measurement. Since that wasn't given, the best answer is <i>"A bit under 120 RPM."</i>.
- Your generator constant is 2 V per Webber-Hz, or 2 V/(Wb Hz) = 2 (V s)/Wb. You somehow got a value of 5. The rest of your calculations are probably off due to that.
- <h2>Added</h2>
- I didn't have much time when I wrote the above, but can delve into this more now.
- From your first measurement, we can see that the pure generator inside this unit produces 2 V/(Wb Hz). Since you are using the fixed exitation of 1 Wb throughout this question, we can simplify that to 2 V/Hz for this case.
- Now the question is how fast the unit will spin without load when 10 V are applied. You know that the armature resistance is 1 Ω and that the unit draws 60 W when the 10 V is applied.
- We'll use the simple model of an ideal motor/generator with the armature resistance in series. To get an idea of speed, we need to know the voltage applied to the internal ideal motor. To get that we find the voltage drop across the resistance, then subtract that from the applied 10 V.
- The current from a 10 V source delivering 60 W is (60 W)/(10 V) = 6 A. The voltage drop from that current across the internal resistance is then (6 A)(1 Ω) = 6 V. That means the voltage across the internal ideal motor/generator is (10 V) - (6 V) = 4 V.
- If the motor were ideal, then its speed would be (4 V)/(2 V/Hz) = 2 Hz = 120 RPM. The actual speed of a real motor would be less due to mechanical friction.
- One way to get a handle on the friction losses would be to know the mechanical power put into the motor when tested as a generator to produce the 10 V of the first measurement. Since that wasn't given, the best answer is <i>"A bit under 120 RPM."</i>.
#3: Post edited
by
Olin Lathrop
·
2023-05-12T16:17:07Z (over 1 year ago)
- Your generator constant is 2 V per Webber-Hz, or 2 V/(Wb Hz) = 2 (V s)/Wb. You somehow got a value of 5. The rest of your calculations are probably off due to that.
- Your generator constant is 2 V per Webber-Hz, or 2 V/(Wb Hz) = 2 (V s)/Wb. You somehow got a value of 5. The rest of your calculations are probably off due to that.
- <h2>Added</h2>
- I didn't have much time when I wrote the above, but can delve into this more now.
- From your first measurement, we can see that the pure generator inside this unit produces 2 V/(Wb Hz). Since you are using the fixed exitation of 1 Wb throughout this question, we can simplify this to 2 V/Hz for this case.
- Now the question is how fast the unit will spin without load when 10 V are applied. You know that the armature resistance is 1 Ω and that the unit draws 60 W when the 10 V is applied.
- We'll use the simple model of an ideal motor/generator with the armature resistance in series. To get an idea of speed, we need to know the voltage applied to the internal ideal motor. To get that we find the voltage drop across the resistance, then subtract that from the applied 10 V.
- The current from a 10 V source delivering 60 W is (60 W)/(10 V) = 6 A. The voltage drop from that current across the internal resistance is then (6 A)(1 Ω) = 6 V. That means the voltage across the internal ideal motor/generator is (10 V) - (6 V) = 4 V.
- If the motor were ideal, then its speed would be (4 V)/(2 V/Hz) = 2 Hz = 120 RPM. The actual speed of a real motor would be less due to mechanical friction.
- One way to get a handle on the friction losses would be to know the mechanical power put into the motor when tested as a generator to produce the 10 V of the first measurement. Since that wasn't given, the best answer is <i>"A bit under 120 RPM."</i>.
#2: Post edited
by
Olin Lathrop
·
2023-05-11T12:24:08Z (over 1 year ago)
Your units in the first equation don't make sense. Open-circuit, the output voltage of the generator should be proportional to shaft speed and to the static magnetic field. The latter is proportional to field current, not power into the field coil. Power actually goes with the square of the field current.The generator constant should be in units of EMF / (speed * current), like V/(Hz A), which is also (V s)/A. Note that you have (V s)/W.
- Your generator constant is 2 V per Webber-Hz, or 2 V/(Wb Hz) = 2 (V s)/Wb. You somehow got a value of 5. The rest of your calculations are probably off due to that.
#1: Initial revision
by
Olin Lathrop
·
2023-05-10T17:36:45Z (over 1 year ago)
Your units in the first equation don't make sense. Open-circuit, the output voltage of the generator should be proportional to shaft speed and to the static magnetic field. The latter is proportional to field current, not power into the field coil. Power actually goes with the square of the field current. The generator constant should be in units of EMF / (speed * current), like V/(Hz A), which is also (V s)/A. Note that you have (V s)/W.