# 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.

Let T = Gross torque developed by motor in N-m.

We know that the torque and power are related by the relation,

P = T x ?

where P = Power

and ? = angular speed

= (2?N)/60 , N = speed in r.p.m.

Now input to the rotor P_{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,

P_{2} = T x ?_{s } where ?_{s }= (2?N_{s})/60 rad/sec

P_{2} = T x (2?N_{s})/60 where N_{s} is in r.p.m. ………..(1)

The rotor tries to deliver this torque to the load. So rotor output is gross mechanical power developed P_{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}.

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

P_{m }= T x (2?N)/60 ………….(2)

The difference between P_{2} and P_{m }is rotor copper loss P_{c}.

P_{c }= P_{2} – P_{m }= T x (2?N_{s}/60) – T x (2?N/60)

P_{c }= T x (2?/60)(N_{s }– N) = rotor copper loss ………..(3)

Dividing (3) by (1),

P_{c}/P_{2} = s as (N_{s }– N)/N_{s }= slip s

Rotor copper loss P_{c} = s x Rotor input P_{2}

Thus total rotor copper loss is slip times the rotor input.

Now P_{2} – P_{c} = P_{m}

P_{2} – sP_{2} = P_{m }

(1 – s)P_{2} = P_{m}

Thus gross mechanical power developed is (1 – s) times the rotor input

The relationship can be expressed in the ratio from as,

The ratio of any two quantities on left hand side is same as the ratio of corresponding two sides on the right hand side.

This relationship is very important and very frequently required to solve the problems on the power flow diagram.

**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,

and T_{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,