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 ω

where P = Power

and ω = angular speed

= (2πN)/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. ω

_{s }as, P

_{2}= T x ω_{s }where ω_{s }= (2πN_{s})/60 rad/sec P

_{2}= T x (2πN_{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 ω and not ω_{s}. P

_{m}= T x ω where ω = (2πN)/60 P

_{m }= T x (2πN)/60 .............(2) The difference between P

_{2}and P_{m }is rotor copper loss P_{c}. P

_{c }= P_{2}- P_{m }= T x (2πN_{s}/60) - T x (2πN/60) P

_{c }= T x (2π/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

We have seen earlier that

T = (k s E

and it mentioned that k = 3/(2π n

The rotor copper losses can be expressed as,

P

but I

Now as per P

P

Now P

= T x (2πN/60)

Now N = N

but N

So substituting in the above equation,

Comparing the two torque equations we can write,

_{sh}< T due to mechanical losses.__1.1 Derivation of k in Torque Equation__

T = (k s E

_{2}^{2}R_{2})/(R_{2}^{2}+(s X_{2})^{2})and it mentioned that k = 3/(2π 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 aboveNow 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 ω= T x (2πN/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,

Comparing the two torque equations we can write,

excellent explanation by the author...

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