Question #248952

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)


1
Expert's answer
2021-10-11T07:01:35-0400

d1=2m,d2=1m,x=5m,m=2kgmd_1 =2 m , d_2 = 1m , x =5m, m =2 \frac{kg}{m}

T=4000N,μ=0.5,N1=300rpmT =4000N, \mu= 0.5, N_1= 300 rpm

velocity of the belt,

v=πd1N160=π×2×30060=31.41msv=\frac{πd1N}{160}=\frac{π×2×300}{60}= 31.41 \frac{m}{s}

centrifugal tension,

TC=mv2=2(31.41)2=1973.17N.T_C = mv^2 =2(31.41)2 =1973.17N.

T1=TTC=40001973.17=2026.83NT_1 =T- T_C =4000-1973.17=2026.83N

for an open belt drive,

sinα=(r1r2)x=(10.5)5,α=5.710sinα=\frac{(r1−r2)}{x} =\frac{(1−0.5)}{5} ,\alpha=5.71^0

Angle of lap on the smaller pulley

θ=18002α=1800(2×5.71)=168.580θ = 1800-2α =180^0-(2×5.71) =168.58^0

=168.58×(π/180)=2.942rad.=168.58 ×(π/180) =2.942 rad.

T2 = Tension in the slack side of the belt,

2.3log(T1T2)=μθ=0.3×2.942=0.88262.3log(\frac{T1}{T2})=μθ =0.3×2.942 =0.8826

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


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