Sheet1rpm motor1450ratio sprocket1ratio gearbox50dia drum273mmrpm roll29rpmkeliling drum857.22mmspeed belt414.32mm/s0.41m/s52Capasity coal silo(Cap)=40m3=44000kg/m3Conveyor capacity(Cc)=20t/h=5.56kg/secBelt speed(V)=0.41m/secConveyor height(H)=11mConveyor length(L)=24.5mMass of a set of idlers(m'i)=20KgIdler spacing(l')=0.8mLoad due to belt(mb)=25Kg/mInclination angle of the conveyor()=19degCoefficient of friction(f)=0.02Start-up factor(Ks)=1.5Drive efficiency(Kd)=0.8Friction factor(Cr)=15Breaking strength loss factor(Cv)=0.75Acceleration due to gravity(g)=9.81m/sec2Load due to the conveyed materials(mm)=13.41Kg/m.Calculation:First, we will use theeqn.1.2for finding out the load due to idlers:Load due to idlers (mi): This can be calculated as below:mi=(mass of a set of idlers) / (idlers spacing)=m'I /l'=25kg/mfinding out the belt tension in steady state:Tb=1.37*f*L*g*[2*mi+ (2*mb + mm)*cos ()] + (H*g*mm)=1463.94192316NThe belt tension while starting the system can be calculated by usingTbs=Tb*Ks=2195.9NFor calculating the power at drive pulley,Pp=(Tb*V)/1000=0.60654KWWe will use theeqn.1.5estimate the size of the motor:Pm=Pp/KdPm=0.7582KWWe will use theeqn.1.6to find out the acceleration of the motor:A=(Tbs Tb)/ [L*(2*mi + 2*mb+mm)]=0.2634396772m/sec2Lastly, we will use theeqn.1.7to find out the belt breaking strengthBs=(Cr*Pp)/ (Cv*V)=29.28N/mm=2.986E-05kg/mmThisBsvalue is used to select the conveyor belt from the manufacturers catalogue

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