Common mechanical operations

1. Common Mechanical Operation
Section 4
Eng. Kareem H. Mokhtar

2. Belt conveyor - Bridth
(4 mm)
(2 mm)

3. Belt conveyor- no of rollers
• Idlers spacing
• If 2000 kg/m3  1100 mm to 1500 mm
• If heavier  1000, 1300 mm
• No of idler rollers on one side= [length of belt/spacing]-1

4. Belt conveyor- power calculations
F= 0.02 – 0.04
T3 = 1.1 T2
So we need to calculate mb (mass of belt), midlers (mass of idlers), and m(mass of solid) Per unit length
(If not given= pi, if given do not forget to transform it)

5. Belt conveyor – masses per unit length
Midlers (mass of one idler) =
7800 kg/m3
Mass of idler per unit length =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑖𝑑𝑙𝑒𝑟 ∗ 𝑁𝑜.𝑜𝑓 𝑖𝑑𝑙𝑒𝑟
𝐿𝑒𝑛𝑔𝑡ℎ
Mb (mass of belt per unit length) = B * x * density of belt
1000 - 1500 kg/m3
Mass of solids per unit length = mass flow rate / velocity
Velocity  assume 1 m/s if not stated

6. Belt conveyor

7. Let us solve

8. Sheet 3