Engineering Relationship between total Rotor Power and Losses

[Joe85]

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:

P_r : P_m : P_c = 1 : 1-S : S

Total Power into rotor: P_r=I'₂²R'₂/S

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

Power loss in rotor: P_c=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:

P_r : P_m : P_c = I'₂²R'₂/S : I'₂²R'₂(1-S)/S : I'₂²R'₂

Dividing each term through: I'₂²R'₂/S

P_r :

(I'₂²R'₂/S)/(I'₂²R'₂/S)

∴ (I'₂²R'₂/S) × (S/I'₂²R'₂)

∴ (I'₂²R'₂S)/(I'₂²R'₂S)

= 1

P_m :

[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-SP_c :

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

P_r : P_m : P_c = 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.

 

[mfb]

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.

 

[Joe85]

Thank you, Sir.