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[Slayinasian]

A local utilty vehicle production company have consldered to use In one of Its deslgn a multl-disc clutch consisting of 9 steel discs and 8 bronze dlscs, with effective outside and inslde diameter of 220 mm and 160 mm, respectively, If the shaft turns at 1500 rpm and an axial force of 600 Newtons Is applied, fInd the horsepower capaclty, considering a coefficient of friction of 0.25.

A.48,41 hp
B.47.59 hp
C. 52.48 hp
D.50.40 hp

 

[🤖]

To calculate the horsepower capacity of the mult-disc clutch, we can use the formula:

Horsepower (HP) = (Torque × RPM) / 5252

First, we need to calculate the torque produced by the axial force applied on the clutch. The torque can be calculated using the formula:

Torque (T) = Force (F) × Radius (r)

The radius can be calculated as the difference between the outside and inside diameters divided by 2:

Radius (r) = (Outside Diameter - Inside Diameter) / 2

Given:
Outside Diameter = 220 mm
Inside Diameter = 160 mm
Axial Force (F) = 600 N
Coefficient of Friction (µ) = 0.25

Radius (r) = (220 mm - 160 mm) / 2
= 30 mm

Now, we can calculate the torque:

Torque (T) = 600 N × 0.03 m
= 18 Nm

Next, we need to convert the RPM to radians per second:

RPS (Radians per Second) = RPM × (2π / 60)

RPS = 1500 rpm × (2π / 60)
= 157.08 rad/s

Finally, we can calculate the horsepower capacity:

Horsepower (HP) = (Torque × RPS) / 5252
= (18 Nm × 157.08 rad/s) / 5252
≈ 0.5402 HP

Thus, the horsepower capacity of the mult-disc clutch is approximately 0.54 HP.

Since none of the given answer choices match this result, it seems there might be an error in the problem statement or calculations.