Common mechanical operations

1. Common Mechanical Operation
Section 3
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
∗ 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. 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 particle size distribution
follows the expression:
• 𝜑𝑐 = 0.0192𝐷𝑝
2 − 0.682𝐷𝑝 + 1.254

5. 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 particle size distribution
follows the expression:
• 𝜑𝑐 = 0.0192𝐷𝑝
2 − 0.682𝐷𝑝 +
1.254
• 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?

6. 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 particle size distribution
follows the expression:
• 𝜑𝑐 = 0.0192𝐷𝑝
2 − 0.682𝐷𝑝 +
1.254
• Estimate the dimensions of the bin
as well as its minimum bottom
opening diameter. Assume mass
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)

7. 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 particle size distribution follows
the expression:
• 𝜑𝑐 = 0.0192𝐷𝑝
2 − 0.682𝐷𝑝 + 1.254
• Estimate the dimensions of the bin
as well as its minimum bottom
opening diameter. Assume mass
flow to occur.

8. 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 particle size distribution
follows the expression:
• 𝜑𝑐 = 0.0192𝐷𝑝
2 − 0.682𝐷𝑝 +
1.254
• Estimate the dimensions of the bin
as well as its minimum bottom
opening diameter. Assume mass
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= 11.57m
• FinalDimensions:
• Dia= 5m, the conical bottom
height is 6.5m and the cylindrical
part is 13.5m

9. 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 particle size distribution
follows the expression:
• 𝜑𝑐 = 0.0192𝐷𝑝
2 − 0.682𝐷𝑝 +
1.254
• Estimate the dimensions of the bin
as well as its minimum bottom
opening diameter. Assume mass
flow to occur.
• Dmin = 2.45 (80+Dp max) tan (alpha r)
• Using the equation and substituting
𝜑𝑐 = zero
• Dp max = 1.95 mm
• Dmin= 141.5 mm

10. 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 particle size distribution
follows the expression:
• 𝜑𝑐 = 0.0192𝐷𝑝
2 − 0.682𝐷𝑝 +
1.254
• 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(ℎ𝑎𝑙𝑓 𝑎𝑝𝑒𝑥 𝑎𝑛𝑔𝑙𝑒)
sin2.5(ℎ𝑎𝑙𝑓 𝑎𝑝𝑒𝑥 𝑎𝑛𝑔𝑙𝑒)
= 0.682
Mass flow = 0.682*1600* 9.81*(0.283)2.5
= 146 kg/s

11. 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