## Input Power

Consider a brushless AC motor spins at 1,000 rpm to deliver 40 N∙m of mechanical torque. How much power (W) is required to use the AC motor if its efficiency is 0.8?

Expand Hint
Efficiency of a machine:
$$\eta =\frac{P_{out}}{P_{in}}$$$where $$P_{out}$$ is the machine’s output power and $$P_{in}$$ is the machine’s input power. Hint 2 Mechanical power in a rotating machine: $$P=T \omega_m$$$
where $$T$$ is the mechanical torque, and $$w_m$$ is the angular velocity.
Since angular velocity is in rad/s, let’s convert the rotational speed from the problem statement:
$$\omega_m=\frac{2\pi}{60}n$$$where $$n$$ is the motor’s speed in rpm. $$\omega_m=\frac{2\pi}{60}\cdot 1,000=104.7\:rad/s$$$
Mechanical power in a rotating machine:
$$P=T \omega_m$$$where $$T$$ is the mechanical torque, and $$w_m$$ is the angular velocity. $$P= (40N\cdot m)(104.7\:rad/s)=4,188.8\:W$$$
Efficiency of a machine:
$$\eta =\frac{P_{out}}{P_{in}}$$$where $$P_{out}$$ is the machine’s output power and $$P_{in}$$ is the machine’s input power. For a motor, $$P_{in}$$ is the active component of the electrical power input, and $$P_{out}$$ is the mechanical power output. For a generator, it’s vice versa. $$0.8 =\frac{4,188.8W}{P_{in}}$$$
$$P_{in} =\frac{4,188.8W}{0.8}=5,236\:W$$\$
5,236 W
Similar Problems from FE Sub Section: Rotating Machines (General)

Similar Problems from FE Section: AC Power