This document provides rules and examples for calculating mechanical operations related to bulk solids storage and transport. It includes formulas for calculating volume and mass flow rate of cones, as well as vertical and lateral pressures at the bottom of a bin. An example calculates these pressures for a given bin geometry. It also discusses belt conveyor design, including belt and idler specifications and formulas for power calculations. The final example uses these formulas to calculate the motor power and storage bin dimensions needed for a given solid transportation rate and production period.

1. Common Mechanical Operation
Section 5
Eng. Kareem H. Mokhtar

2. Rules
• Volume of cone= pi/12 * D2 * H
• Mass flow rate= C* (bulk density) * 𝑔 ∗ 𝐷5/2
• C= (pi/6) *
(1−𝐶𝑜𝑠1.5(ℎ𝑎𝑙𝑓 𝑎𝑝𝑒𝑥 𝑎𝑛𝑔𝑙𝑒)
sin2.5(ℎ𝑎𝑙𝑓 𝑎𝑝𝑒𝑥 𝑎𝑛𝑔𝑙𝑒)
• Pv=
𝐷∗𝑔∗𝑏𝑢𝑙𝑘 𝑑𝑒𝑛𝑠𝑖𝑡𝑦
4 (tan 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 ∗ 𝑘
∗ (1 − 𝑒−4∗tan 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 ∗𝑘∗
𝐻
𝐷)
• Pl = k *Pv
• K=
1−sin(𝑎𝑙𝑝ℎ𝑎 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 )
1+sin(𝑎𝑙𝑝ℎ𝑎 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 )

3. Rules

4. Example 1
• IF a bin has diameter of 5m, the conical bottom height is 6.5m and
the cylindrical part is 13.5m, find the vertical and lateral pressures at
the bottom given that
• bulk density 1600 kg/m3
• particle density of 2500 kg/m3
• angle of repose = 35°
• angle of friction on steel = 25o
• angle of internal friction = 38o

5. Answer
• K= K=
1−sin(𝑎𝑙𝑝ℎ𝑎 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 )
1+sin(𝑎𝑙𝑝ℎ𝑎 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 )
• K=
1−sin(38 )
1+sin(38 )
= 0.238
• Pv=
𝐷∗𝑔∗𝑏𝑢𝑙𝑘 𝑑𝑒𝑛𝑠𝑖𝑡𝑦
4 (tan 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 ∗ 𝑘
∗ (1 − 𝑒−4∗tan 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 ∗𝑘∗
𝐻
𝐷)
• Pv =
5∗9.81∗1600
4 (tan 25 ∗0.238
∗ (1 − 𝑒−4∗tan 25 ∗0.238∗
𝐻
5 ) = 123 kPa
• Pl = 29300 Pa
• P liquid

6. Example 2
• If the previous question had a fluid inside it, what would be the
pressure at the bottom of the cylindrical part?
• P = density * g * h = 1600*9.81*13.5 = 211 kPa

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

8. 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

9. 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)

10. 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

11. Belt conveyor

12. Example 3
• A belt conveyor is used to transport solids to a bin, the rate of
transportation of solid is 50 tons per hour, find the driving motor
power needed then find the diameter needed for the bin to store a 9
hour production of crushed material.
• bulk density 1600 kg/m3
a particle density of 2500 kg/m3
an angle of repose = 35°
angle of friction on steel = 25o
angle of internal friction = 38o.

13. Answer
• B = 0.323 m = 350 mm
• X total = 14mm
• Mass of belt = 6.37 kg
• At spacing 1300mm  no. of rollers = 32 roller
• Mass of idler roller per length = 9.8 kg/m
• Mass of solid per unit length = 13.889 kg/ m
• Power = 7 Hp

14. Answer
• Bin diameter= 5 m
• Top section  h = 1.75 and Volume at top= 11.45 m3
• Bottom section (Mass flow)  Hb= 6.5 m and Volume = 42.5m3
• Cylinder = 11.57 m
• So the dimensions are
• Dia= 5m, the conical bottom height is 6.5m and the cylindrical part is
13.5m