Answer to Question #220180 in Mechanical Engineering for Brian

Question #220180

 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


1
Expert's answer
2021-07-27T02:38:01-0400

Part i

"P_{max}=\u03c3_b\u22c5A\u22c5k\u22c5v\\\\\nP_{max}=3*800*0.3\u22c50.2887*10^3*\\frac{160}{360}\\\\\nP_{max}=92384 W\\\\\nP_{max}= 92.384 kW"


Part ii

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

"v= \\frac{\\pi d_1N}{60}\\\\\nv= \\frac{\\pi \\sqrt\\frac{4A}{\\pi}N}{60}\\\\\n0.2887*10^3= \\frac{\\pi \\sqrt\\frac{4*800}{\\pi}N}{60}\\\\\nN = \\frac{0.2887*10^3}{ \\frac{\\pi \\sqrt\\frac{4*800}{\\pi}}{60}}\\\\\nN= 172.76 rpm"


Part iii

"\\frac{T_1}{T_2}=e^{\\mu \\theta}\\\\\n\\frac{T_1}{T_2}=e^{0.3* \\frac{160}{360}\\pi}\\\\\n\\frac{T_1}{T_2}=1.52\\\\\nAlso, \\\\\n3000=\\frac{T_1+T_2}{2}\\\\\n\\therefore T_1 = 1.52T_2= 1.52(6000-T_1)\\\\\nT_1 = 9120-1.52T_1\\\\\nT_1 = \\frac{9120}{2.52}\\\\\nT_1=3619.05 N"


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