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Relationship between total Rotor Power and Losses

  • Engineering
  • Thread starter Joe85
  • Start date
22
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

Homework 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'2


So 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-S


Pc :


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 : S


Seems 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.
 

Answers and Replies

33,564
9,295
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.
 
22
3
Thank you, Sir.
 

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