QUESTION 20 A belt drive system with a tension in the tight side equal to 987 N, the tension in the slack side is equal to 529 N of the belt. The larger pulley rotates at 312 rev/min with a diameter equal to 0.59 m. Compute the Power transmitted by the Belt drive system in (W)?
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To compute the power transmitted by the belt drive system, we need to know the torque and rotational speed of the system. Show more…
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The power transmitted by a belt drive is proportional to $T v-\frac{w v}{g}$, where $v=$ speed of the belt, $T=$ tension on the driving side and $\omega=$ weight per unit length of belt. Find the speed at which the transmitted power is a maximum.
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Further problems F.8
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The power transmitted between two shafts 3.5 meters apart by a cross belt drive around the two pulleys 600 mm and 300 mm in diameter is 6 kW. The speed of the larger pulley (driver) is 220 rpm. The permissible load on the belt is 25 N/mm width of the belt, which is 5 mm thick. The coefficient of friction between the smaller pulley surface and the belt is 0.35. Determine: 1. the necessary length of the belt; 2. the width of the belt, and 3. the necessary initial tension in the belt. [Ans. 8.472 m; 53 mm; 888 N]
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