Indicated Power
Consider a braking system with a 7 m long brake arm and a rotational speed of 25 rev/s is subjected to a 42 N force at the end of the brake arm. If the friction power is 10 kW, what is the indicated power in kW?
Expand Hint
Indicated power:
$$$\dot{W}_i=\dot{W}_b+\dot{W}_f$$$
where
$$\dot{W}_f$$
is the friction power and
$$\dot{W}_b$$
is the brake power.
Hint 2
Brake Power:
$$$\dot{W}_b=2\pi \tau \upsilon =2\pi FR\upsilon $$$
where
$$\tau$$
is torque,
$$\upsilon $$
is rotational speed,
$$F$$
is the force at the end of the brake arm, and
$$R$$
is the brake arm’s length.
First, let’s determine the brake power:
$$$\dot{W}_b=2\pi \tau \upsilon =2\pi FR\upsilon $$$
where
$$\tau$$
is torque,
$$\upsilon $$
is rotational speed,
$$F$$
is the force at the end of the brake arm, and
$$R$$
is the brake arm’s length.
$$$\dot{W}_b=2\pi (42N)(7m)(25rev/s)=46,158\:W=46.158\:kW$$$
Indicated power:
$$$\dot{W}_i=\dot{W}_b+\dot{W}_f$$$
where
$$\dot{W}_f$$
is the friction power and
$$\dot{W}_b$$
is the brake power.
Thus, solving for indicated power:
$$$\dot{W}_i=46.158\:kW+10\:kW=56.2\:kW$$$
56.2 kW
Time Analysis
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137. Brake Power
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