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
Time Analysis
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