Relationship between total Rotor Power and Losses

Click For Summary
SUMMARY

The relationship between total rotor power (Pr), mechanical power output (Pm), and power loss in the rotor (Pc) is defined by the equations Pr : Pm : Pc = 1 : 1-S : S. The total power into the rotor is calculated using the formula Pr=I'22R'2/S, while mechanical power output is given by Pm=I'22R'2(1-S)/S. The power loss in the rotor is represented as Pc=I'22R'2S. The discussion confirms that the sum of useful output power and wasted power equals the input power, establishing a clear understanding of rotor power dynamics.

PREREQUISITES
  • Understanding of rotor dynamics and power equations
  • Familiarity with electrical engineering concepts, specifically rotor resistance (R'2)
  • Knowledge of the significance of slip (S) in rotor performance
  • Basic algebraic manipulation skills for equation simplification
NEXT STEPS
  • Research the derivation of rotor power equations in electrical machines
  • Learn about the impact of slip (S) on rotor efficiency and performance
  • Explore advanced topics in rotor resistance and its effect on power loss
  • Investigate methods for optimizing rotor design to minimize losses
USEFUL FOR

Electrical engineers, students studying rotor dynamics, and professionals involved in optimizing rotor performance in electric machines will benefit from this discussion.

Joe85
Messages
29
Reaction score
3
Homework Statement
Demonstrate that the relationship between total rotor power, rotor losses
and the mechanical power generated, in terms of rotor current and
resistance, by the rotor of an induction motor is of the form 1 : (1 – s) : s
Relevant Equations
Below
Equations:

Pr : Pm : Pc = 1 : 1-S : S

Total Power into rotor: Pr=I'22R'2/S

Mechanical Power Output: Pm=I'22R'2(1-S)/S

Power loss in rotor: Pc=I'22R'2So I'm a little skeptical of my answer here. Seems a little too easy, which normally means I've missed to mark by about 100 miles.

Attempt:

Pr : Pm : Pc = I'22R'2/S : I'22R'2(1-S)/S : I'22R'2

Dividing each term through: I'22R'2/S

Pr :

(I'22R'2/S)/(I'22R'2/S)

∴ (I'22R'2/S) × (S/I'22R'2)

∴ (I'22R'2S)/(I'22R'2S)

= 1

Pm :


[I'22R'2(1-S)/S]/ [I'22R'2/S]

∴ [I'22R'2(1-S)/S] × [S/I'22R'2]

∴ [I'22R'2(1-S)S]/[SI'22R'2]

Cancelling S: [I'22R'2(1-S)]/[I'22R'2]

Cancelling I'22R'2:

= 1-SPc :

I'22R'2/ (I'22R'2/S)

∴ I'22R'2 × (S/I'22R'2)

∴ (I'22R'2S)/I'22R'2

Cancelling I'22R'2:

= S

Pr : Pm : Pc = 1 : 1-S : SSeems very straight forward and perhaps they are expecting a little more explanation on where the original equations are derived from?

Once again, any guidance would be greatly appreciated.
 
Physics news on Phys.org
Looks good.
It is basically trivial. The sum of useful output power plus wasted power must be the input power, so adding the last two terms must produce the first term. s is defined as fraction, so you know the first to last term ratio. And that's it.
 
  • Like
Likes   Reactions: Joe85
Thank you, Sir.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
8
Views
3K
  • · Replies 15 ·
Replies
15
Views
3K