First we determine the motor's efficiency - a ratio of the mechanical output power and the electrical input power:
[tex]E = \frac{P_out}{P_in}=\frac{14920W}{U\cdot I}=\frac{14920W}{400V\cdot 40A}=0.93[/tex]
The output mechanical power is directly proportional to torque and angular speed. From this relationship we can isolate the torque:
[tex]P_{out}=\tau \cdot\omega=E\cdot P_{in}=E\cdot I \cdot U\\\implies\tau = \frac{I \cdot U \cdot E}{\omega}=\frac{I\cdot U\cdot E\cdot 60}{2\pi\cdot \omega_{rpm}}[/tex]
whereby the last expression uses the angular speed in rpm units.
(a) we calculate the torque, given that the motor draws 1.2 times the rated current, i.e., 48 Amps:
[tex]\tau = \frac{I\cdot U\cdot E\cdot 60}{2\pi\cdot \omega_{rpm}}=\frac{48A\cdot 400V\cdot0.93\cdot 60}{2\pi \cdot 400 rpm} = 426.3 Nm[/tex]
(b) use formula for the acceleration time:
[tex]t_{start} = \frac{j\cdot \omega_{rpm}}{\frac{60}{2\pi}\cdot \tau}=\frac{7.5\frac{kg}{m^2}\cdot 400rpm}{9.55\cdot 426.3kg\frac{m^2}{s^2}}\approx 0.74s[/tex]
The startup time will be 0.74s
