The driveshaft of a building’s air-handling fan is turned at 300 RPM by a belt running on a 0.3-m-diameter pulley. The net force applied by the belt on the pulley is 2000 N. determine the torque applied by the belt on the pulley, in N.m, and the power transmitted, in kW.
torque=F*D/2=2000n(0.15m)=300 N*m
Power = Torque*angular speed
=(300 N*m)(300 RPM) (2*PI/60)
=9424 W = 9.42 kW
These are the answers in the back of the text.
torque= force*radius
power= torque*angular speed where
angular speed= 300*2PI/60 rad/sec
this will give you power in watts.
To determine the torque applied by the belt on the pulley, we can use the following formula:
Torque (τ) = Force (F) × Radius (r)
First, we need to calculate the radius of the pulley. The diameter of the pulley is given as 0.3 m. The radius is half of the diameter, so the radius (r) = 0.3 m / 2 = 0.15 m.
Now we can calculate the torque:
τ = 2000 N × 0.15 m
τ = 300 N·m
Therefore, the torque applied by the belt on the pulley is 300 N·m.
To determine the power transmitted, we can use the formula:
Power (P) = Torque (τ) × Angular Velocity (ω)
The angular velocity (ω) is given in RPM (revolutions per minute), so we need to convert it to radians per second (rad/s).
ω = 300 RPM × 2π rad/rev × (1 min/60 s)
ω ≈ 31.42 rad/s
Now we can calculate the power:
P = 300 N·m × 31.42 rad/s
P ≈ 9426 W
To convert the power to kilowatts (kW):
P ≈ 9.43 kW
Therefore, the power transmitted by the belt is approximately 9.43 kW.
torque=300Nm
power=31.41 Watts