 # Commercial Practicability - the Sandwich Belt System and the Pocket Belt System

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[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

1. Level ground
2. 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

1. Power required to convey the material from lower drum to upper drum or vice versa= Pg=== 0KW[pic 8][pic 6][pic 7]
2. 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

1. Power required to convey the material from lower drum to upper drum or vice versa= Pg=== 15.69KW[pic 22][pic 20][pic 21]
2. 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