[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