Section 2

Common Mechanical Operation
Section 2
Eng. Kareem H. Mokhtar

Rules
• M= bulk density * volume = bulk density * pi/12 * D2 * H
• Tan ( alpha r) =
2𝐻
𝐷
• D =
3 24 𝑚𝑎𝑠𝑠 ∗cot(𝑎𝑙𝑝ℎ𝑎 𝑟 )
𝑝𝑖 ∗𝑏𝑢𝑙𝑘 𝑑𝑒𝑛𝑠𝑖𝑡𝑦
• Dmin = 2.45 (80+Dp max) tan (alpha r)
• Mass flow rate= C* (bulk density) * 𝑔 ∗ 𝐷5/2
• C= (pi/6) *
(1−𝐶𝑜𝑠1.5(ℎ𝑙𝑎𝑓 𝑎𝑝𝑒𝑥 𝑎𝑛𝑔𝑙𝑒)
sin2.5(ℎ𝑙𝑎𝑓 𝑎𝑝𝑒𝑥 𝑎𝑛𝑔𝑙𝑒)

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
2𝐻
𝐷
D =
Volume of cone =
𝑝𝑖
12
∗ 𝐷2
∗ 𝐻

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 = ?
bulk density * pi/4 * D2 * H =
2𝐻
𝐷
Mass= 1700* pi/4 * (6)2 * (6/2) * tan(45) =
48 Tons

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

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
bulk density * pi/12 * D2 * H
2𝐻
𝐷
D =
Volume of cone =
𝑝𝑖
12
∗ 𝐷2
∗ 𝐻

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)
2𝐻
𝐷
• Tan (alpha r) = 2 * (H-h)/5
• 180= 25h+(pi/12)*52* (5/2) tan (35)
• h= 6.74m
• H= 8.5 m

Sheet 2
• It is required to design a cylindrical
bin to store 450 tons of soda ash of
bulk density 1600 kg/m3, a particle
density of 2500 kg/m3 and an angle of
repose = 35°, angle of friction on
steel = 25o and angle of internal
friction = 38o.
• Estimate the dimensions of the bin
as well as its minimum bottom
opening diameter. Assume mass flow
to occur.
• What would be the solid flow rate
when the bottom opening
diameter is twice the minimum?
• The differential screen analysis is
shown in the table.
Mesh size 4 / 6 6 / 10 10 / 20 20 / 40
xi 0.00 0.35 0.45 0.20

Sheet 2
• 450 tons of soda ash
• 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
• The differential screen analysis
is shown in the table.
opening diameter. Assume mass
flow to occur.
• What would be the solid flow rate
when the bottom opening
diameter is twice the minimum?
Mesh size 4 / 6 6 / 10 10 / 20 20 / 40
xi 0.00 0.35 0.45 0.20

Sheet 2
flow to occur.
V total = 450000/1600=281.25 m3
Assume cylindrical volume:
281.25 = (pi/4) * D2 * H
Let H=3D
Then D= 4.9 m around 5 m
Top Section:
Tan ( alpha r) = 2 h /D  h = 1.75
Volume at top= 11.45 m3
Bottom section (Mass flow)
Mesh size 4 / 6 6 / 10 10 / 20 20 / 40
xi 0.00 0.35 0.45 0.20

Sheet 2
flow to occur.
Mesh size 4 / 6 6 / 10 10 / 20 20 / 40
xi 0.00 0.35 0.45 0.20

Sheet 2
flow to occur.
• Hlaf apex angle = 20
• Tan(half apex angle) = D/2hb
• Hb= 6.5 m
• Volume at bottom = pi/12 * 52 * 6.5
= 42.5 m3
• Cylindrical section
• Vol= 281.25-11.45-42.5=227.3 m3
• 227.3= (pi/4) * 52 * H
• H= 11.2
• So the total volume is 19.75m
• Final: Dia= 5m, the concical
bottom height is 6.5m and the
cylindrical part is 13.5m
Mesh size 4 / 6 6 / 10 10 / 20 20 / 40
xi 0.00 0.35 0.45 0.20

Sheet 2
flow to occur.
• Dmin = 2.45 (80+Dp max) tan (alpha r)
• Dp max = 6/10 mesh = 2.489mm
• Dmin= 141.5 mm
Mesh size 4 / 6 6 / 10 10 / 20 20 / 40
xi 0.00 0.35 0.45 0.20

Sheet 2
• What would be the solid flow rate when
the bottom opening diameter is twice
the minimum
• D = 2 Dmin = 283 mm
• Mass flow rate=
C* (bulk density) * 𝑔 ∗ 𝐷5/2
• C= (pi/6) *
(1−𝐶𝑜𝑠1.5(ℎ𝑙𝑎𝑓 𝑎𝑝𝑒𝑥 𝑎𝑛𝑔𝑙𝑒)
sin2.5(ℎ𝑙𝑎𝑓 𝑎𝑝𝑒𝑥 𝑎𝑛𝑔𝑙𝑒)
= 0.682
Mass flow = 0.682*1600* 9.81*(0.283)2.5
= 146 kg/sMesh size 4 / 6 6 / 10 10 / 20 20 / 40
xi 0.00 0.35 0.45 0.20

Sheet 2
• In the above problem, calculate
the vertical and the lateral
pressure on the bottom of the
cylindrical part of the bin. Also
calculate the equivalent vertical
pressure exerted by a liquid
having the same (bulk) density
of the solid.
• Pv=
𝐷∗𝑔∗𝑏𝑢𝑙𝑘 𝑑𝑒𝑛𝑠𝑖𝑡𝑦
4 (tan 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 ∗ 𝑘
∗
𝑒−4∗tan(𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛)∗𝑘∗𝐻/𝐷
• Pl = k *Pv
• K=
1−sin(𝑎𝑙𝑝ℎ𝑎 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 )
1+sin(𝑎𝑙𝑝ℎ𝑎 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 )
K= 0.238
Pv= kPa
Pl= 29.3 kPa
Pliq = 211 kPa

Section 2

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Section 2