## Machine Efficiency

Consider a brushless AC motor requires 10 kW of power to deliver 10 N∙m of mechanical torque. What is the machine’s efficiency if its rotational speed is 2,000 rpm?

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. $$\omega_m=\frac{2\pi}{60}\cdot 2,000=209.3\:rad/s$$$
Mechanical power in a rotating machine:
$$P=T \omega_m$$$where $$T$$ is the mechanical torque, and $$w_m$$ is the angular velocity. $$P= (10N\cdot m)(209.3\:rad/s)=2,093\:W$$$
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. $$\eta =\frac{2,093W}{10,000W}=0.21$$$
0.21
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