Relation between P2, Pc, and Pm

The rotor input P₂, 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 P₂ 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_sas,

P₂ = T x ω_s where ω_s= (2πN_s)/60 rad/sec

P₂ = 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 P₂ and P_m is rotor copper loss P_c.

P_c= P₂ - 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/P₂ = s as (N_s- N)/N_s = slip s

Rotor copper loss P_c = s x Rotor input P₂

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

Now P₂ - P_c = P_m

P₂ - sP₂ = P_m

(1 - s)P₂ = 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 E₂² R₂)/(R₂² +(s X₂)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_2r² x R₂
       but I_2r = (s E₂)/√(R₂² +(s X₂)²), hence substituting above

      Now as per    P₂: 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)