Answer to Question #220180 in Mechanical Engineering for Brian

Question

 A belt pulley <b><i>system</i></b> uses a flat belt of cross section area 800mm squared and density <b><i>1200 kg</i></b>/m cubed. Angle of lap is 160 degrees on the smaller pulley. Coefficient of friction is 0.3. Maximum stress allowed in the belt is 3N/mm squared. Calculate; (i) Maximum power transmitted by varying speed. (ii) Speed at which it occurs. (iii The initial tension of the belt

Answer 1

Part i

$P_{max}=σ_b⋅A⋅k⋅v\\ P_{max}=3*800*0.3⋅0.2887*10^3*\frac{160}{360}\\ P_{max}=92384 W\\ P_{max}= 92.384 kW$

Part ii

$v= \sqrt{\frac{\sigma_p-\sigma_b}{3 \rho}}\\ v= \sqrt{\frac{(3-0)*10^3}{3 * 12000}}\\ v=0.2887 m/s$

$v= \frac{\pi d_1N}{60}\\ v= \frac{\pi \sqrt\frac{4A}{\pi}N}{60}\\ 0.2887*10^3= \frac{\pi \sqrt\frac{4*800}{\pi}N}{60}\\ N = \frac{0.2887*10^3}{ \frac{\pi \sqrt\frac{4*800}{\pi}}{60}}\\ N= 172.76 rpm$

Part iii

$\frac{T_1}{T_2}=e^{\mu \theta}\\ \frac{T_1}{T_2}=e^{0.3* \frac{160}{360}\pi}\\ \frac{T_1}{T_2}=1.52\\ Also, \\ 3000=\frac{T_1+T_2}{2}\\ \therefore T_1 = 1.52T_2= 1.52(6000-T_1)\\ T_1 = 9120-1.52T_1\\ T_1 = \frac{9120}{2.52}\\ T_1=3619.05 N$