A shunt motor has the following ratings: 75 kW, 240 V, 1750 rpm, 300 A. The moto

Question

A shunt motor has the following ratings: 75 kW, 240 V, 1750 rpm, 300 A. The motor is supplied with 240 V source (constant) at its armature terminals. Its armature resistance is 0.0144 [O]. At rated current Ia = 300 A, the motor runs at rated speed of 1750 rpm.
(a) Compute the output power Pout, torque T, and speed ?m of the motor at armature current of Ia = 0 [A].
(b) Repeat the solution for armature current Ia = 150 [A].
Units: Find the speed ?m in [rad/s], the ouput power Pout in [kW], and the torque T in [N m].
Conversion factors: 1 rpm = 1 rev/min. 1 rev = 2p rad. 60 s = 1 min.

Explanation / Answer

Given Data: V=240v, Rated current Ia=300A, Rated Speed = 1750rpm, Armature resistance(Ra) = 0.0144.

For Shunt Motor, the following equations are valied
V=E+IaRa ---(Eq1),

E ---(Eq2),

Torque Ia ---(Eq3)

--> At Rated speed and Rated current

     V=E+IaRa ==> 240 = E +300*0.0144 ==> E= 235.68

(a) Ia = 0A means no load. At no load output power and output torque are zero(neglecting fincticion & rotational losses )

    From (Eq2) No load speed = 240*1750/235.68 = 1782.08 rpm

(b) For Ia = 150A, V=E+IaRa ==> 240 = E +150*0.0144 ==> E= 237.84

Pout =E*Ia = 237.84*150 = 35.67kW

From (Eq2), Speed at 150A = 237.84*1750/235.68 = 1766.04 rpm

Speed in rad/sec = 2*3.14*1766.04/60 = 184.94 rad/sec

Torque = Pout/speed(in rad/sec) = 35676/184.94 = 192.91 Nm

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