## Relation between P2, Pc, and Pm

The rotor input P_{2}, rotor copper loss P_{c} and gross mechanical power developed P_{m} are related through the slip s. Let us derive this relationship.

_{2}is from stator side through rotating magnetic field which is rotating at synchronous speed N

_{s}.

So torque developed by the rotor can be expressed interms of power

input and angular speed at which power is inputted i.e. ?_{s }as,

_{2}= T x ?

_{s }where ?

_{s }= (2?N

_{s})/60 rad/sec

_{2}= T x (2?N

_{s})/60 where N

_{s}is in r.p.m. ………..(1)

_{m}and torque T. But rotor gives output at speed N and not N

_{s}. So from output side P

_{m}and T can be related through angular speed ? and not ?

_{s}.

_{m}= T x ? where ? = (2?N)/60

_{m }= T x (2?N)/60 ………….(2)

_{2}and P

_{m }is rotor copper loss P

_{c}.

_{c }= P

_{2}– P

_{m }= T x (2?N

_{s}/60) – T x (2?N/60)

_{c }= T x (2?/60)(N

_{s }– N) = rotor copper loss ………..(3)

_{c}/P

_{2}= s as (N

_{s }– N)/N

_{s }= slip s

_{c}= s x Rotor input P

_{2}

_{2}– P

_{c}= P

_{m }

_{2}– sP

_{2}= P

_{m }

_{2}= P

_{m }

**Key Point**:

The torque produced by rotor is gross mechanical torque and due to

mechanical losses entire torque can not be available to drive load. The

load torque is net output torque called shaft torque or useful torque

and is denoted as T

_{sh}. It is related to P

_{out}as,

_{sh}< T due to mechanical losses.

__1.1 Derivation of k in Torque Equation__

We have seen earlier that

T = (k s E_{2}^{2} R_{2})/(R_{2}^{2} +(s X_{2})^{2})

and it mentioned that k = 3/(2? n_{s}) . Let us see its proof.

The rotor copper losses can be expressed as,

P_{c }= 3 x I_{2r}^{2} x R_{2}

but I_{2r} = (s E_{2})/?**(**R_{2}^{2} +(s X_{2})^{2}**)**, hence substituting above

Now as per P_{2 }: P_{c } : P_{m } is 1 : s : 1-s ,

P_{c}/P_{m }= s/(1-s)

Now P_{m }= T x ?

= T x (2?N/60)

Now N = N_{s }(1-s) from definition of slip, substituting in above,

but N_{s}/60 = n_{s } in r.p.m.

So substituting in the above equation,

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