Friction Power
Consider a braking system with a 5 m long brake arm and a rotational speed of 30 rev/s is subjected to a 40 N force at the end of the brake arm. If the indicated power is 45 kW, what is the friction 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.
$$$\dot{W}_b=2\pi (40N)(5m)(30rev/s)=37,699.11\:W=37.699\:kW$$$
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 friction power:
$$$\dot{W}_f=\dot{W}_i-\dot{W}_b=45\:kW-37.699\:kW=7.3\:kW$$$
7.3 kW
Time Analysis
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Similar Problems from FE Sub Section: Cycles and Processes
137. Brake Power
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475. Brake Force
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137. Brake Power
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