Question A2.15: A flat belt is required to transmit 35 kW from a pulley of 1...
A flat belt is required to transmit 35 kW from a pulley of 1.5 m effective diameter running at 300 rpm. The angle of contact is spread over 11/24 of th e circumference and the coefficient of friction between belt and pulley surface is 0.3. Taking centrifugal tension into account, determine the width of the belt. Take belt thickness as 9.5 mm, density 1100 kg/m³ and permissible stress as 2.5 MN/m². [KU, June 2010]
Given : P = 35 kW, d = 1.5 m, n = 300 rpm, \theta=\frac{11}{24} \times 360=165^{\circ}, \mu=0.3, t=9.5 \mathrm{~mm}\rho=1100 \mathrm{~kg} / \mathrm{m}^3, \sigma_w=2.5 \mathrm{\ MN} / \mathrm{m}^2
Belt speed, ν=\frac{\pi d n}{60}=\frac{\pi \times 1.5 \times 300}{60}=23.56 \mathrm{~m} / \mathrm{s}
Let b = width of belt, mm
Mass of belt per metre length, m=b t \times 1 \times \rho=b \times 9.5 \times 10^{-6} \times 1 \times 1100=0.0104 .5 \times b \mathrm{~kg} / \mathrm{m}
Centrifugal tension, T_c=m ν^2=0.01045 \times 6 \times(23.56)^2=5.8 \times b \mathrm{~N}
Maximum tension in belt, \mathrm{T}=\sigma_w \times b t=2.5 \times b \times 9.5=23.75 \times b \mathrm{~N}
Tension on right side of belt, T_1=T-T_C=23.75 \times b-5.8 \times b=17.95 \times b \mathrm{~N}
Now,
\begin{gathered}\frac{T_1}{T_2}=\exp (\mu \theta)=\exp \left\lgroup 0.3 \times \frac{\pi}{180} \times 165\right\rgroup=\exp (0.86394)=2.3725 \\T_2=\frac{T_1}{2.3725}=\frac{17.95 \times b}{2.3725}=7.566 \times b\end{gathered}
Power transmitted, P=\frac{\left(T_1-T_2\right) ν}{10^3}
35=\frac{(17.95-7.566) \times b \times 23.56}{10^3} b=143 \mathrm{~mm}