[pic 1]
NATIONAL INSTITUTE OF TECHNOLOGY, ROURKELA
B.Tech. 5th Sem Mining Engineering Autumn- 2018
Subject: Material Handling in Mines (MN-334)
Assignment-2
Submitted by:- Dundra Vikranth
Roll.No:- 116MN0583
Instructor: Prof. D.S.Nimaje
Q.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/m
Material carrying capacity of belt conveyor(T) = 80 Kg/sec
Span of transportation(L) = 2.7 Km
Friction coefficient of belt, bearing of idlers(µi) = 0.3
Friction coefficient of material and belt(µm) = 0.32
Speed of belt conveyor(v) = 2.95 m/s
Efficiency 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 m
Height upto which material is to be raised or lowered(h)=20 m
From figure 1.1
Tanδ==7.4074×10-3[pic 3]
δ =0.4244º[pic 4]
cosδ ~ 1.0
sinδ ~ 0.0
Thus,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 =Pm
iii) Power required to counteract friction between belt and idlers(Top and Bottom)=Pe
CASE-A:- When material will be conveyed on level ground.
Height upto which material is to be raised or lowered(h)= 0 m
- Power 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 = pm
Pm===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 + pm
Po= 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 m
- Power 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 = pm
Pm=== 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