Common mechanical operations

1. Common Mechanical Operation
Section 2
Eng. Kareem H. Mokhtar

2. Rules
• M= bulk density * volume =
bulk density * pi/12 * D2 * H
• Tan (αr) =
2𝐻
𝐷
• D =
3 24 𝑚𝑎𝑠𝑠 ∗cot(αr)
𝑝𝑖 ∗𝑏𝑢𝑙𝑘 𝑑𝑒𝑛𝑠𝑖𝑡𝑦

3. Sheet 2
• It is required to store gravel in a
stock pile. The available floor
area is a square of dimensions
6 x 6 m2. The angle of repose is
45° and the bulk density is 1700
kg/m3. Estimate the mass of
gravel that can be stored into
that area.
• M= bulk density * volume =
bulk density * pi/12 * D2 * H
• Tan (αr) =
2𝐻
𝐷
D =
3 24 𝑚𝑎𝑠𝑠 ∗cot(αr)
𝑝𝑖 ∗𝑏𝑢𝑙𝑘 𝑑𝑒𝑛𝑠𝑖𝑡𝑦
Volume of cone =
𝑝𝑖
12
∗ 𝐷2
∗ 𝐻

4. Sheet 2
• area is a square of dimensions
6 x 6 m2
• The angle of repose is 45°
• the bulk density is 1700 kg/m3.
mass of gravel = ?
• Mass= 1700* pi/12 * (6)2 * (6/2)
* tan(45) =
• 48 Tons
• M= bulk density * volume =
bulk density * pi/12 * D2 * H =
• Tan (αr) =
2𝐻
𝐷

5. Sheet 2
• The figure shows a
compartment used for storing
bauxite ore in a factory. It
consists of a cuboid having a
square base 5 x 5 m2. It should
accommodate a three days
reserve of raw material. This is
fed to a furnace at the rate of 5
ton/hr
Bulk density = 2000 kg/m3 and angle
of repose = 35° .
• Calculate the minimum value of
dimension H in figure

6. Sheet 2
• Bulk density = 2000 kg/m3 and
angle of repose = 35° .
• square base 5 x 5 m2
• three days reserve of raw
material. This is fed to a furnace
at the rate of 5 ton/hr
• Total mass reserved = 5 *24* 3=
360 ton
• M= bulk density * volume =
bulk density * pi/12 * D2 * H
• Tan (αr) =
2𝐻
𝐷
D =
3 24 𝑚𝑎𝑠𝑠 ∗cot(αr)
𝑝𝑖 ∗𝑏𝑢𝑙𝑘 𝑑𝑒𝑛𝑠𝑖𝑡𝑦
Volume of cone =
𝑝𝑖
12
∗ 𝐷2
∗ 𝐻

7. Sheet 2
• Bulk density = 2000 kg/m3 and
angle of repose = 35° .
• square base 5 x 5 m2
• three days reserve of raw
material. This is fed to a furnace
at the rate of 5 ton/hr
• Total mass reserved = 5 *24* 3=
360 ton
• M= bulk density * volume
• So
• Volume = 360*103/2000 =
• 180 m3
• Volume = 5 * 5* h+ (pi/12) D2 *(H-h)
• Tan (αr) =
2𝐻
𝐷
• Tan (αr) = 2 * (H-h)/5
• 180= 25h+(pi/12)*52* (5/2) tan (35)
• h= 6.74m
• H= 8.5 m

8. Sheet 2
• The opposing figure shows a storage
facility consisting of a vertical wall
sided by two vertical partition walls
perpendicular to the back wall. Find
an equation relating the mass of
material to be stored to the
measurable dimensions x and L. Such
a facility is to be used to store a 2
month reserve of crushed gypsum
stone (Bulk density = 1500 kg.m-3 and
angle of repose = 38
• Estimate the required dimensions for
a daily consumption of 400 ton.
(Make any necessary assumption)

9. Solution
• Mass to be stored = 400,000 * 30 *2 =24,000,000 kg
• Mass = ρb Vb = 1500 A L
• A = 1/2* x* (x* tan(38))
• 24,000,000= 1500 *1/2* x* (x tan (38)) L = 586 x2 .L
• Assume x = L
• X = L = 35

10. Let us try it in laboratory