## 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**

**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**

**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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**Rotating Machines (General)**499. Torque Units

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Similar Problems from FE Section:

**AC Power**110. Motor Shaft RPM

118. Power Factor

302. Synchronous Speed

397. Induction Motor

401. Motor Slip

405. Motor Poles

499. Torque Units

500. Mechanical Power

501. Sync Speed

505. Input Power

508. Generator Efficiency