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A belt drive consists of a V-belt working on a grooved pulley, with an angle of lap of 150°. The cross sectional area of the belt is 750 mm^2 the groove angle is 40°, and the coefficient of mu =0.15 . the density of the belt material is 1200 kg/m^3 and its maxium safe stress is 7MN/m^2 Calculate the maximum power that can be transmitted at a belt speed of 30 m/s.

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Based on the available information, the maximum power that can be transmitted by the belt drive at a belt speed of 30 m/s is 157,500 Watts or 157.5 kW.

To calculate the maximum power that can be transmitted by the belt drive, consider the maximum tension in the belt and the belt speed.

The maximum tension in the belt can be determined using the maximum safe stress and the cross-sectional area of the belt.

Given information:

Cross-sectional area of the belt (A) = 750 mm^2 = 750 * 10^-6 m^2

Coefficient of friction (μ) = 0.15

Density of the belt material (ρ) = 1200 kg/m^3

Maximum safe stress (σ) = 7 MN/m^2

Belt speed (v) = 30 m/s

Step 1: Calculate the maximum tension in the belt (T):

T = σ * A

T = 7 MN/m^2 * 750 * 10^-6 m^2

T = 5250 N

Step 2: Calculate the maximum power (P) that can be transmitted:

P = T * v

P = 5250 N * 30 m/s

P = 157,500 W

Therefore, the maximum power that can be transmitted by the belt drive at a belt speed of 30 m/s is 157,500 Watts or 157.5 kW.

User Vahid Pazirandeh
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