Common mechanical operation section 6

1. Common Mechanical Operation
Section 6
Eng. Kareem H. Mokhtar

2. Question 5
• In the manufacture of water based paints, the ingredients (pigment +
adhesive) are suspended in water to form a slurry. The average
density of solids = 2300 kg/m3, and the solid content by weight is
25%. This slurry is to be ground in an available batch ball mill of
length = diameter = 1.25 m. Porcelain balls of 40 mm diameter are
used (Specific gravity = 2.6). They fill 50% of the mill and the void
ratio between these balls = 45%.
• Calculate the mass of slurry in the mill and the number of balls
required.

3. • To determine the mass of slurry
• The density of slurry:
• Assume 1 kg of slurry it will contain 0.25kg of solid and 0.75kg water
• Volume of solid= 0.25/2300 = 1.09 *10-4 m3
• Volume of water = 0.75/1000 = 7.5 *10-4 m3
• Density of slurry = mass/volume = 1/ (1.09 *10-4 + 7.5 *10-4 ) =
1164.65 kg/m3

4. • Mass of feed = (pi/4)*(1.25)2*1.25 * 0.5*1164.6*0.45 = 402 kg
• Mballs = 1096.8 kg
• To get the number of balls
• Total volume of balls = 1096.8/2600 = 0.42 m3
• Volume of one ball = (4/3)* pi* (0.02)3= 3.35 * 10-5 m3
• No. of balls = 0.42/(3.35 * 10-5 ) = 12537 balls

5. Question 6
• The increase in specific surface of a feed of rock is proportional to the
square root of the total number of revolutions of a batch ball mill. 15 tons
of crushed rock of specific gravity = 2.6 and shape factor = 1.5, are fed to
the mill. They correspond to 2” spheres. The mill is 1.5 m in diameter and
2” grinding steel balls are used. When the mill is operated at a speed of 15
rpm for 2 hours, the product possesses the screen analysis shown below.
• If it is required to increase the specific surface by 25% operating at 65%
of the critical speed, what would be the time required for grinding?
• In the former case, estimate the theoretical power consumption if the
work index = 10.

6. Answer
• Aw =
6 𝑠ℎ𝑎𝑝𝑒 𝑓𝑎𝑐𝑡𝑜𝑟
𝑑𝑒𝑛𝑠𝑖𝑡𝑦
*
1
𝐷𝑣𝑠
• Feed  Aw1 = (6*1.5)/2600 * (1/0.0508) = 0.068 m2/kg
• Aw2= 11.97 m2/kg (from screen analysis)
• P = 4/3 * 36 * 10* (15000/2*3600) * (1/ sqrt (0.00048) – 1/ sqrt(0.0508))
• = 82 Kwatt

7. Answer
• Aw2 – Aw1 = k sqrt( N * t )
• From 1 – 2
• 11.97 – 0.068 = k sqrt( 15*2*60)
• So k = 0.28
• From 2 – 3
• 11.97*1.25 – 11.97 = 0.28 sqrt(N * t)
• N = 42/ sqrt( 1.5-0.0508) = 35 rpm
• Actual N = 0.65 * 35
• Time = 5 min

8. Question 7
• A batch rod mill is operated by a 10 hp motor 80% of which are
transmitted to the mill. The power required to drive the empty mill =
3 hp, and the grinding efficiency = 50%. The feed is equivalent to 20
mm spheres, and the product is assumed to follow the approximate
relation:
• φ = 1 – (D/Dmax)2
• Where Dmax is the maximum particle size in product = 2 mm.
• Calculate the time required to grind one ton of this batch (Work
index = 12).

9. Answer
• 10 hp motor 80% of which are transmitted to the mill
• The power required to drive the empty mill = 3 hp, and the grinding
efficiency = 50%.
• The feed is equivalent to 20 mm spheres
• φ = 1 – (D/Dmax)2
• Where Dmax is the maximum particle size in product = 2 mm.
• Work index = 12).
• Calculate the time required to grind one ton

10. Answer
• Power = (10*0.8 -3)*0.5 = 2.5 Hp *735 = 1837.5 watt
•
• 1837.5 = (4/3)* 36* 12 * (1000/t) (1/Db2 – 1/Db1 )
• Db1 = 0.02
• Db2 --> 0.2 = 1- (Db2/0.002)2  Db2 = 0.00179 m
• T = 5194 seconds = 1.44 hours