Question 7.1: A 25-kW 125-V separately-excited dc machine is operated at a......

a. From Eq. 7.17, with V_t = 128  V  \text{and}  E_a = 125  V, the armature current is

V_a= E_a±I_a R_a \quad \quad \quad (7.17)

I_a=\frac{V_t  –  E_a}{R_a}=\frac{ 128  –  125}{0.02}= 150  A

in the motor direction, and the power input at the motor terminal is

V_tI_a = 128 × 150 = 19.20  kW

The electromagnetic power is given by

E_aI_a = 125 × 150 = 18.75  kW

In this case, the dc machine is operating as a motor and the electromagnetic power is hence smaller than the motor input power by the power dissipated in the armature resistance.

Finally, the electromagnetic torque is given by Eq. 7.16:

T_{mech}=\frac{E_aI_a}{ω_m}=K_a Φ_d I_a \quad \quad \quad (7.16)

T_{mech}=\frac{E_aI_a}{ω_m}=\frac{18.75 × 10^3}{100π}= 59.7  N · m

b. In this case, E_a  \text{is larger than}  V_t and hence armature current will flow out of the machine, and thus the machine is operating as a generator. Hence

I_a=\frac{E_a  –  V_t}{R_a}=\frac{125  –  124}{0.02}= 50  A

and the terminal power is

V_t I_a=124 × 50 = 6.20  kW

The electromagnetic power is

E_aI_a = 125 × 50 = 6.25  kW

and the electromagnetic torque is

T_{mech} =\frac{6.25 × 10^3}{100π} = 19.9  N · m