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Ks = the start-up factorSizing of the motor: The minimum motor power can be calculated as:Pm = Pp/Kdeqn.1.5Where,Pm is in Kw.Pp = the power at drive pulley in KwKd = Drive efficiency.Acceleration : The acceleration of the conveyor belt can be calculated as:A= (Tbs Tb)/ [L*(2*mi + 2*mb+mm)]eqn.1.6Where,A is in m/sec2Tbs = the belt tension while starting in N.Tb = the belt tension in steady state in N.L = the length of the conveyor in meters.mi = Load due to the idlers in Kg/m.mb = Load due to belt in Kg/m.mm = Load due to the conveyed materials in Kg/m.Belt breaking strength: This parameter decides the selection of the conveyor belt. The belt breaking strength can be calculated as:Bs= (Cr*Pp)/ (Cv*V)..eqn.1.7Where,Bs is in Newton.Cr = friction factorCv = Breaking strength loss factorPp = Power at drive pulley in Newton.V = belt speed in m/sec.An Example of Conveyor Belt CalculationsInput data:Conveyor capacity (Cc) = 1500 t/h = 416.67 Kg/secBelt speed (V) = 1.5 m/secConveyor height (H) = 20 mConveyor length (L) = 250 mMass of a set of idlers (mi) = 20 KgIdler spacing (l) = 1.2 mLoad due to belt (mb) = 25 Kg/mInclination angle of the conveyor () = 5 0Coefficient of friction (f) = 0.02Start-up factor (Ks) = 1.5Drive efficiency (Kd) = 0.9Friction factor (Cr) = 15Breaking strength loss factor (Cv) = 0.75Calculation:First, we will use the eqn.1.2 for finding out the load due to idlers:mi = (20/1.2) = 16.67 Kg/mWe will use the eqn.1.1 for finding out the belt tension in steady state:Tb = 1.37*0.02*250*9.81*[16.67+ {2*25+ (416.67/1.5)}*cos (5)] + (20*9.81* (416.67/1.5)) = 77556.88 N.The belt tension while starting the system can be calculated by using the eqn.1.4:Tbs = 1.5 * 77556.88 = 116335.32 NFor calculating the power at drive pulley, we will use the eqn.1.3:Pp = (77556.88*1.5)/ 1000 = 116.335 KwWe will use the eqn.1.5 estimate the size of the motor:Pm = 116.35/0.9 = 129.261 KwWe will use the eqn.1.6 to find out the acceleration of the motor:A = (116335.32 - 77556.88)/ [250*{(2*16.67) + (2*25) + (416.67/1.5)}]= 0.429 m/sec2Lastly, we will use the eqn.1.7 to find out the belt breaking strength:Bs = (15*116.35) / (0.75*1.5) = 1551.33 N/mmThis Bs value is used to select the conveyor belt from the manufacturers catalogue.ConclusionThe conveyor belt calculations methodology discussed in the article is to be used only for the guidance on calculating the initial conveyor design parameters; the final design must be validated by using the FEA or other similar tools before building the prototype.ReferenceDunlop Conveyor belt technique design and calculationBridgestone Conveyor Belt Design Manual

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Ks = the start-up factorSizing of the motor: The minimum motor power can be calculated as:Pm = Pp/Kdeqn.1.5Where,Pm is in Kw.Pp = the power at drive pulley in KwKd = Drive efficiency.Acceleration : The acceleration of the conveyor belt can be calculated as:A= (Tbs Tb)/ [L*(2*mi + 2*mb+mm)]eqn.1.6Where,A is in m/sec2Tbs = the belt tension while starting in N.Tb = the belt tension in steady state in N.L = the length of the conveyor in meters.mi = Load due to the idlers in Kg/m.mb = Load due to belt in Kg/m.mm = Load due to the conveyed materials in Kg/m.Belt breaking strength: This parameter decides the selection of the conveyor belt. The belt breaking strength can be calculated as:Bs= (Cr*Pp)/ (Cv*V)..eqn.1.7Where,Bs is in Newton.Cr = friction factorCv = Breaking strength loss factorPp = Power at drive pulley in Newton.V = belt speed in m/sec.An Example of Conveyor Belt CalculationsInput data:Conveyor capacity (Cc) = 1500 t/h = 416.67 Kg/secBelt speed (V) = 1.5 m/secConveyor height (H) = 20 mConveyor length (L) = 250 mMass of a set of idlers (mi) = 20 KgIdler spacing (l) = 1.2 mLoad due to belt (mb) = 25 Kg/mInclination angle of the conveyor () = 5 0Coefficient of friction (f) = 0.02Start-up factor (Ks) = 1.5Drive efficiency (Kd) = 0.9Friction factor (Cr) = 15Breaking strength loss factor (Cv) = 0.75Calculation:First, we will use the eqn.1.2 for finding out the load due to idlers:mi = (20/1.2) = 16.67 Kg/mWe will use the eqn.1.1 for finding out the belt tension in steady state:Tb = 1.37*0.02*250*9.81*[16.67+ {2*25+ (416.67/1.5)}*cos (5)] + (20*9.81* (416.67/1.5)) = 77556.88 N.The belt tension while starting the system can be calculated by using the eqn.1.4:Tbs = 1.5 * 77556.88 = 116335.32 NFor calculating the power at drive pulley, we will use the eqn.1.3:Pp = (77556.88*1.5)/ 1000 = 116.335 KwWe will use the eqn.1.5 estimate the size of the motor:Pm = 116.35/0.9 = 129.261 KwWe will use the eqn.1.6 to find out the acceleration of the motor:A = (116335.32 - 77556.88)/ [250*{(2*16.67) + (2*25) + (416.67/1.5)}]= 0.429 m/sec2Lastly, we will use the eqn.1.7 to find out the belt breaking strength:Bs = (15*116.35) / (0.75*1.5) = 1551.33 N/mmThis Bs value is used to select the conveyor belt from the manufacturers catalogue.ConclusionThe conveyor belt calculations methodology discussed in the article is to be used only for the guidance on calculating the initial conveyor design parameters; the final design must be validated by using the FEA or other similar tools before building the prototype.ReferenceDunlop Conveyor belt technique design and calculationBridgestone Conveyor Belt Design Manual

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