Question

An open belt, 100 mm wide, connects two pulleys mounted on parallel shafts with their centers 2.4 m apart. The diameter of the larger pulley is 450 mm and that of the smaller pulley 300 mm. The coefficient of friction between the belt and the pulley is 0.3 and the maximum stress in the belt is limited to 14 N/mm width. If the larger pulley rotates at 120 r.p.m., find the maximum power that can be transmitted

Answer 1

Explanation:

An open belt 100 mm wide connects two pulleys mounted on parallel shafts with their centres 2.4 m apart. The diameter of the larger pulley is 450 mm and that of the smaller pulley 300 mm. The coeffi- cient of friction between the belt and the pulley is 0.3 and the maximum stress in the belt is limited to 14 N/mm width. If the larger pulley rotates at 120 r.p.m., find the maximum power that can be transmitted (Ans. 2.387 kW) 5. A rough rule for leather belt is that effective tension in it, shall not exceed 15 N/mm of width for a belt of 10 mm thickness. This rule is applied to determine width of belt required to transmit 37 kW, under the following conditions: Angle of lap = 165º; Coefficient of friction = 0.3; Velocity of belt = 1500 m/min; Density of leather = 950 kg/m? Find the width of belt required. Assuming limiting friction between belt and pulley rim, find the stress in the belt. [Ans. 140 mm; 1.48 MPa]

4. An open belt 100 mm wide connects two pulleys mounted on parallel shafts with their centres 2.4 m apart. The diameter of the larger pulley is 450 mm and that of the smaller pulley 300 mm. The coeffi- cient of friction between the belt and the pulley is 0.3 and the maximum stress in the belt is limited to 14 N/mm width. If the larger pulley rotates at 120 r.p.m., find the maximum power that can be transmitted [Ans. 2.387 kW]

HOPEFULLY SAYING IT'S USEFULL

Answer 2

Maximum power that can be transmitted is 2.39 kW.

Explanation:

Max stress in the belt =

 $X= 14N/mm = 14N*100 mm/mm = 1400N$​                                            

  $v=\frac{\pi D_1N_1}{60}=\frac{\pi*0.45*120}{60}=2.827 m/sv$

 $\alpha =sin^{-1}(\frac{0.45-0.3}{2*2.4})=1.79^0$

  $\therefore 2 \alpha =3.58 ^0$

$\theta = (180-2 \alpha)*\frac{\pi}{180}=3.079 rad$

$\frac{X}{Y}=e^{\mu \theta}=e^{0.3*3.079}=2.518 \implies Y=\frac{1400}{2.518}=555.996 N$

Maximum power transmitted,

$P=(X-Y)*V=(1400-555.996)*2.827=2386W=2.39kW$