Question: The output
voltage and power of the DC shunt motor is 230V and 7.5 kW respectively. When the field
current is 1.2 A the speed of the generator is 1450 rev/min. Total armature resistance 0.55
Ohm. This generator is used as a DC motor when output voltage is 220V and power
delivered by the motor is 6 kW. The efficiency of the motor at this condition is 82%.

Calculate the speed of the motor when field current and magnetized flux resistance
constant in each condition.

I am still not really sure I
follow each part. I think in some cases you mean generator and in others motor, also in some
places output should probably read input.

1. As a GENERATOR

V = 230 V
P = 7500 W Field current = 1.2 Amps

Armature current = 7500/230 –
1.2 = 32.60 – 1.2 = 31.4 Amps

Speed = 1450 rpm = 24.17 rps

Angular velocity (w) = 151.8 radians per second

Emf generated = 230 = Banw = BAn x 151.8 BAn = 1.515 (a constant for your
apparatus)

(A is the area of the rotating coil with n turns)

2. As a MOTOR

New back emf = Banw_{1} = 1.515 where w_{1} is the new angular velocity

Input = 220 V Output
power = 6000 W Efficiency = 82% So input power = 7317 W

Current in coil =
7217/220 – 1.2 = 33.25 – 1.2 = 32.06 Amps

Back emf = V – IR = 220 – 32.06x0.55 =
202.37 = BAnv

Angular velocity w_{1} = 202.37/1.151
= 133.6 radians per second

Speed = 1275 rpm