This document discusses common mechanical operations used to transport materials, including belt conveyors, bucket elevators, screw conveyors, and pneumatic transportation. It provides equations and parameters to calculate the mass, power, and other specifications for each type of conveyor. The document includes examples of calculations for belt conveyors, bucket elevators, and screw conveyors based on given parameters. It assumes typical values like belt and material densities, idler spacing, bucket spacing, screw pitch, and filling ratios that can be used in calculations.

1. Common Mechanical Operation
Section 3
Eng. Kareem H. Mokhtar

2. Belt conveyor
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

3. Belt conveyor
Mb (mass of belt per unit length) = B * x * density of belt
1000 - 1500 kg/m3
(4 mm)
(2 mm)

4. Belt conveyor
Midlers (mass of idlers)
7800 kg/m3
Mass of idler per unit length =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑖𝑑𝑙𝑒𝑟 ∗ 𝑁𝑜.𝑜𝑓 𝑖𝑑𝑙𝑒𝑟
𝐿𝑒𝑛𝑔𝑡ℎ
Idlers spacing if 2000 kg/m3  1100 mm to 1500 mm
if heavier  1000, 1300 mm
Length / spacing

5. Belt conveyor
Mass of solids per unit length = mass flow rate / velocity
Velocity  assume 1 m/s if not stated

6. Bucket Elevator
Assume
Chain speed --> 0.2 – 1 m/s (usually 0.5m/s)
Spacing  300 mm to 450 mm

7. Let us solve
• A) Lecture questions
• B) The sheet

8. Sheet 3

9. Sheet 3 Q1
• B= 319 mm  400 mm
• X total = 14mm
• Mass of belt = 6.16 kg
• At spacing 1300mm  no. of
rollers = 33 roller
• Mass of idler roller per length =
9.23 kg
• Mass of solid per unit length =
11.11 kg/ m
T1 = 5500N
T4 = 10125.17N
Power = 8 Hp

10. Sheet 3 Q2
• A bucket elevator is used to lift 20 ton / hr of crushed alumina (of
average size = 10 mm) to a height of 10 m. to be dropped into a bin.
The bulk density is 2000 kg / m³. Select suitable buckets and calculate
the power required to drive the belt.

11. Sheet 3 Q2 – Final Answer
• At 300 mm spacing
• C = 2.08 L
• Td = 2463.4 N
• P = 3 hp

12. F free flowing = 2.5
otherwise = 4
Screw Conveyors

13. Screw Conveyors
• The pitch of the screw is taken as equal to its diameter (D) for free flowing
non-abrasive solids and to (0.8 D) otherwise
• The shaft rotates on bearings that are usually fixed by hangers to a plate
closing the trough. These hangers are spaced by about 3 meters
• filling ratio (F), which represents the ratio between the actual cross
section filled by solids to the total cross sectional area.
• F = 0.4 for free flowing solids.
F = 0.3 for free flowing mildly abrasive material
F = 0.25 for slow flowing mildly abrasive material
F = 0.13 for slow flowing highly abrasive material.

14. Sheet 3 – Q3
• Speed = 1.33 RPS
• F= 0.4
• P= D
• D= 230 mm
• Trough = 235 mm
• P= 7 hp

15. Pneumatic Transportation
We then assume a duct diameter D (inch) and get the air flow rate in m3.min-1

16. Pneumatic Transportation

17. Pneumatic Transportation

18. Sheet 3- Q4
• Q= 79.38 m3/min
• Delta p=4.38
• You solve