Question
(f) Two pulleys are one 450mm diameter and the other 200mm diameter on the parallel shafts 1 .95m apart are connected by a crossed belt Find the length of the belt required and the angle of contact between the belt and each pulley. What power can be transmitted by the belt when the larger pulley rotates at 200rev/min, if the maximum possible tension in the belt is 1KN, and the coefficient of function between the belt and the pulley is 0.25.
Solution
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(268 Votos)
Rodrigo
Veterano · Tutor por 10 anos
Resposta
To find the length of the belt required and the angle of contact between the belt and each pulley, we can use the following formulas:1. Length of the belt (L):L = (D1 + D2) / 2 * sqrt((D1 + D2)^2 - 4 * D1 * D2)Where D1 and D2 are the diameters of the two pulleys.2. Angle of contact (θ):θ = arccos((D1^2 + D2^2 - L^2) / (2 * D1 * D2))Now, let's calculate the length of the belt required:D1 = 450 mm = 0.45 mD2 = 200 mm = 0.2 mDistance between the shafts = 1.95 mL = (0.45 + 0.2) / 2 * sqrt((0.45 + 0.2)^2 - 4 * 0.45 * 0.2)L ≈ 1.68 mNext, let's calculate the angle of contact between the belt and each pulley:θ = arccos((0.45^2 + 0.2^2 - 1.68^2) / (2 * 0.45 * 0.2))θ ≈ 88.8 degreesNow, to find the power transmitted by the belt, we can use the following formula:Power = (Tension * Belt Speed * Coefficient of Friction) / Belt LengthWhere Tension is the maximum possible tension in the belt, Belt Speed is the speed at which the larger pulley rotates, and Coefficient of Friction is the coefficient of friction between the belt and the pulley.Belt Speed = (2 * π * Revolutions per minute * Pulley Diameter) / 60Belt Speed = (2 * π * 200 * 0.45) / 60Belt Speed ≈ 47.1 m/sPower = (1000 * 47.1 * 0.25) / 1.68Power ≈ 7035.71 Watts or 7.04 kWTherefore, the power that can be transmitted by the belt is approximately 7.04 kW.