Answer to Question #248952 in Mechanical Engineering for mazon47

Question

An open belt drive consists of two pulleys which are 5 m apart. The diameters of the larger pulley and smaller pulley are 2 m and 1 m respectively. The allowable maximum tension of the belt is 4 kN. The coefficient of friction between the belt and the pulley is 0.4 and the mass of the belt is 2kg/m length. Calculate:

3.1. The torque on the shafts when the larger pulley rotates at 300 RPM. (7)

3.2. The power transmitted. (3)

3.3. The speed in RPM at which maximum power can be transmitted. (5) 3.4. The maximum power that can be transmitted. (3)

Answer 1

$d_1 =2 m , d_2 = 1m , x =5m, m =2 \frac{kg}{m}$

$T =4000N, \mu= 0.5, N_1= 300 rpm$

velocity of the belt,

$v=\frac{πd1N}{160}=\frac{π×2×300}{60}= 31.41 \frac{m}{s}$

centrifugal tension,

$T_C = mv^2 =2(31.41)2 =1973.17N.$

$T_1 =T- T_C =4000-1973.17=2026.83N$

for an open belt drive,

$sinα=\frac{(r1−r2)}{x} =\frac{(1−0.5)}{5} ,\alpha=5.71^0$

Angle of lap on the smaller pulley

$θ = 1800-2α =180^0-(2×5.71) =168.58^0$

$=168.58 ×(π/180) =2.942 rad.$

T2 = Tension in the slack side of the belt,

$2.3log(\frac{T1}{T2})=μθ =0.3×2.942 =0.8826$

$T2 = T1/2.438 =(2026.83/2.438) =831.34N.$