Relationship between total Rotor Power and Losses

In summary: I appreciate the clarification. In summary, the equations state that the ratio of power into the rotor, mechanical power output, and power loss in the rotor is 1:1-S:S, where S is the defined fraction. This is due to the fact that the sum of useful output power and wasted power must equal the input power.
  • #1
Joe85
29
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.
 
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  • #2
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 Joe85
  • #3
Thank you, Sir.
 

1. What is the relationship between total rotor power and losses?

The total rotor power is directly related to the losses in a system. As the losses increase, the total rotor power required to maintain the system also increases.

2. What are the types of losses that affect the total rotor power?

There are various types of losses that can affect the total rotor power, such as friction losses, electrical losses, core losses, and stray losses.

3. How do losses affect the efficiency of a system?

Losses decrease the efficiency of a system because they require additional power to compensate for the energy lost. This results in a decrease in the overall efficiency of the system.

4. Can losses be reduced to increase the total rotor power?

Yes, losses can be reduced by improving the design of the system and using more efficient materials. This can increase the total rotor power required to maintain the system.

5. Is there a specific equation that relates the total rotor power and losses?

Yes, the equation is P = Ploss + Poutput, where P is the total rotor power, Ploss is the sum of all losses, and Poutput is the output power of the system.

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