Commercial Practicability – the Sandwich Belt System and the Pocket Belt System – Term Paper – ARJUNmuskan
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Commercial Practicability – the Sandwich Belt System and the Pocket Belt System
[pic 1]NATIONAL INSTITUTE OF TECHNOLOGY, ROURKELAB.Tech. 5th Sem Mining Engineering Autumn- 2018Subject: Material Handling in Mines (MN-334)Assignment-2Submitted by:- Dundra VikranthRoll.No:- 116MN0583Instructor: Prof. D.S.NimajeQ.1. m- Mass of the belt = 20.5 Kg/m T- Material carrying capacity of belt conveyor = 80 Kg/sec L- Span of transportation = 2.7 Km μi- Friction coefficient of belt, bearing of idlers = 0.3 μm- Friction coefficient of material and belt = 0.32 v- Speed of belt conveyor = 2.95 m/s ɳ- Efficiency of motor =70% Calculate the total power required for running the belt conveyor using above data in case of Level ground Raising or lowering the materials if the height between two drums = 20 m Also calculate the mass of the material per unit length of the belt conveyor.Solution:-Given information:Mass of the belt(m) = 20.5 Kg/mMaterial carrying capacity of belt conveyor(T) = 80 Kg/secSpan of transportation(L) = 2.7 KmFriction coefficient of belt, bearing of idlers(µi) = 0.3Friction coefficient of material and belt(µm) = 0.32Speed of belt conveyor(v) = 2.95 m/sEfficiency of motor(ɳ) =70%Height upto which material is to be raised or lowered=20 m[pic 2]Blue dum is return drum and green is Drive drum.Distance between the center of these two drum is the span of transportation(L)= 2700 mHeight upto which material is to be raised or lowered(h)=20 mFrom figure 1.1Tanδ==7.4074×10-3[pic 3]δ =0.4244º[pic 4]cosδ ~ 1.0sinδ ~ 0.0Thus,we can take the normal load value as weight of material directly[pic 5]For calculation of power:-i)Power required to convey the material from lower drum to upper drum or vice versa= Pg
ii) Power required to counteract friction between material and belt =Pmiii) Power required to counteract friction between belt and idlers(Top and Bottom)=PeCASE-A:- When material will be conveyed on level ground.Height upto which material is to be raised or lowered(h)= 0 mPower required to convey the material from lower drum to upper drum or vice versa= Pg=== 0KW[pic 8][pic 6][pic 7]Power required to counteract friction between material and belt = pmPm===678.06KW[pic 11][pic 9][pic 10]3. Power required to counteract friction between belt and idlers(Top and Bottom)=Pe Pe=KW=480.54 KW[pic 14][pic 12][pic 13]4. Total power required to convey the material or output power (Po)= Pg+ Pe + pmPo= Pg+ Pe + pm=0KW+678.06KW+480.54 KW=1158.6 KW[pic 15]Efficiency of motor(ɳ) = 0.7= =[pic 16][pic 17]Input power == 1655.143 KW[pic 19][pic 18]CASE-B:- When material will be conveyed from higher level to lower level.Height upto which material is to be lowered(h)= 20 mPower required to convey the material from lower drum to upper drum or vice versa= Pg=== 15.69KW[pic 22][pic 20][pic 21]Power required to counteract friction between material and belt = pmPm=== 678.06KW[pic 25][pic 23][pic 24]3. Power required to counteract friction between belt and idlers(Top and Bottom)=Pe Pe=KW=480.54 KW[pic 28][pic 26][pic 27]4. Total power required to convey the material or output power (Po)= -Pg+ Pe + pm
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By: ARJUNmuskan
Submitted: January 7, 2019
Essay Length: 4,569 Words / 19 Pages
Paper type: Term Paper Views: 240
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