How would superconducting motors/generators work?

Click For Summary
SUMMARY

Superconducting motors and generators enhance performance by utilizing superconductors to expel magnetic fields, thereby generating Lorentz forces in motors and induced EMF in generators. The discussion highlights the potential of superconducting windings to create stronger magnetic fields, which, when combined with superconducting rotors, can lead to increased torque and speed due to higher currents. Numerous resources, including articles and studies, confirm the viability and benefits of superconducting electric machines.

PREREQUISITES
  • Understanding of superconductivity principles
  • Familiarity with Lorentz force and electromagnetic induction
  • Knowledge of electric motor and generator design
  • Basic concepts of magnetic fields and their interactions
NEXT STEPS
  • Research superconducting electric machines and their applications
  • Explore the design and engineering of superconducting rotors
  • Learn about the impact of superconducting materials on torque and speed
  • Investigate existing superconducting motor projects and their performance metrics
USEFUL FOR

Engineers, researchers, and students interested in advanced motor and generator technologies, particularly those focusing on superconductivity and its applications in electrical engineering.

PhiowPhi
Messages
203
Reaction score
8
How can superconductors improve motor/generator's performance?

A superconductor will expel a magnetic field, how can it generate a Lorentz force in the case of a motor, and a induced EMF in the case of a generator?

It's makes sense that we can only improve the windings of the electromagnet, to make it superconducting generating higher fields. But what about the rotor of a motor if it's a superconducting... having higher currents would lead to greater torques & speeds,due to the combination of a stronger field & higher current?
 
Engineering news on Phys.org
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 37 ·
2
Replies
37
Views
7K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
0
Views
2K