Question

A-V belt is to transmit 20 kW from a 250 mm pitch diameter sheave to a 900 mm diameter pulley. The center distance between the two shafts is 1000 mm. The groove angle is 40' and the coefficient of friction for the belt and sheave is 0.2 and the coefficient of friction between the belt and flat pulley is 0.2. The cross-section of the belt is 40 mm wide at the top, 20 mm wide at the bottom and 25 mm deep. The density of the belt is 1000 kg/m and the allowable tension per belt is 1000 N. Find the number of belts required.

Answer 1

According to the data given in the question following calculations are made,

Given, Tranomit Power = 20kW ď Sheare diameter = 300 250 mm 'D' Pulley diameter = 900 mm Centre distance C - 1m = 1000 mm 'O' Godore Angle = 40° f' coefficient of friction = 0.2 po Density - 1000 kg/m3 Allowable tension 'P' = 1000N Assuming N = 1440 rpm Lg = 180 – 2 Sti" (-) = 180 -2sir"(20250) 2 = 142° = 2.48 radians. fas/ sin(0/2) (0-2)(248) sin(402) So, é = e = 4.263 - 40mm 125 mm K-20mm Avea of cross-section = 2 (20+4(25) = 750 mm2

Mass = density x volume = Sal M = (1000)(750)x16*X 1 kg/m .m= 0.75 kg/m Velocity v = ADN = 1(250) (1440). – 18.85mg 1000 60x1000 So, mye? = 0.75 (18.85) = 266. 48 Pi- miez - efas/sin(9) P,-266.48 – 4.263 .= eo P2 - mu P2-266.48 Pi = 1000 N Solving, P2 = 438.55 N Power ber belt = (P-2) ve – (1000-438.55)(18.85) (kW) 1000 = 10.58 KW Number of belts required = 20KW - 1.89 1000 10.58 kW | 1.89 2 2 Belts