HO CHI MINH CITY UNIVERSITY OF

TECHNOLOGY OFFICE FOR

INTERNATIONAL STUDY PROGRAMS

LABORATORY OF UNIT OPERATIONS

CHAPTER: GRINDING – SIEVING - MIXING

Lecturer: Phạm Hoàng Huy Phước Lợi

Class: CC01 Group: 4A

Name ID
Nguyễn Thị Thu Hồng 2053022
Trần Thị Ngọc Huyền 2053058

Ho Chi Minh City, 6 October 2023

 OUTLINE

1. PURPOSE.......................................................................................................3
2.1. Capacity and performance evaluation of the milling machine................3
2.2. Size distribution.......................................................................................4
2.3. Sieve Performance...................................................................................5
2.4. Mix Equation...........................................................................................5

3. EXPERIMENTAL METHOD....................................................................6
3.1. Mill-Turn on the machine without load to measure the no-load power. .6
3.2. Sieve.........................................................................................................6
3.3. Mix...........................................................................................................6

4. EXPERIMENTAL RESULT......................................................................6
4.1. Results......................................................................................................6
4.1.1. Grinding:...........................................................................................6
4.1.2. Sieving:..............................................................................................7
4.1.3. Mixing:..............................................................................................9
4.2. Discussion.............................................................................................13

5. REFERENCES:..........................................................................................15
1. PURPOSE
- Grinding a material, based on the sieve results, determines the size distribution of the
material after grinding, the power consumption, and the efficiency of the milling
machine.
- Sieve the material after grinding, determine the sieving efficiency, build the
cumulative distribution schema of the material after grinding, and then determine the
material size after grinding.
- Mix the two materials to determine the mixing index at various time, and build the
mixing index graph over time to determine the appropriate mixing time

2. THEORY

2.1. Capacity and performance evaluation of the milling machine

- The capacity and efficiency calculations of the milling machine through a sieve with
dimensions Dp1 (ft) and 80% of grinded materials going through a sieve with
dimensions Dpj (ft).
- P is the capacity required to reduce size of the material to Dp (Mass unit/minute), i =
∞.

P=K b
√ 1
Dp

- Wi is required energy to grind the material from a very large size to 100µm
(KWh/ton).
- Relationship between Wi and Kb (the Bond coefficient depends on the type of material
and milling machine).

60 W i=K b
√ 1
100 ×10−3
60 W i
→ K b= ≈ 19 W i
√ 10
1
→ P=19 W i
√ Dp
1 1
w h ere P 1=19 W i ∧P 2=19 W i
√ Dp1 √ Dp2

- Milling capacity of 1 ton - material over 1 minute from Dp1 to Dp2:

P=P 2−P1=19 W i
1
( −
√DP2 √DP1
1
)
- T is productivity (ton/min). The capacity (power consumption) for milling T tons/min
from Dp1 to Dp2:
 P=P 2−P1=19 W i
( 1
−
1
√DP2 √DP1 )
T ,( KW )

where Dp1, Dp2 is the size of material and product, nm.

- If dry grinding, P multiplied by 4/3
Consumption capacity of the motor of milling machine:
P '=UIcos φ

Where U: voltage (V)

I: electrical current (A)
cos φ: factor of capacity
- Milling efficiency:
P
H= 100 %
P'

2.2. Size distribution

−d ∅ b
=K D p
d Dp

Where ø: cumulative mass on Dp

Dp: particle size
K, b: coefficients to describe particle distribution
- Differential from ø = ø1 to ø = ø2 corresponding with Dp = D1 and Dp = D2, we have:
K
∅ 2−∅ 1=
b+1
( D p 1 −D p 2 )
b+1 b+1

- We consider between the nth sieve and the (n+1)th sieve, and assume use the standard
sieve which has Dpn - 1 / Dpn = r = const.
−K b +1
∆ ∅ n=∅ n −∅ n −1= ( D + Db+1
b+ 1 pn pn−1)

and replace Dpn – 1 = r.Dpn, we get:

b +1
K .(r −1) b +1
∆ ∅ 'n= D pn =K ' Db+1
pn
b+1

b+1
K .(r −1)
with K '=
b+1
or log ∆∅ n=(b+ 1)log log D pn+ log K '
- K’ and b are determined by ploting a logarithmic graph of ∆øn versus Dpn. Determine
(b+1) and K’ => K and b.
 2.3. Sieve Performance
J
E= × 100
Fa

F: Initial material mass input the sieve, (g)

J: Material mass under the sieve, (g)
a: Particle ratio can pass through the sieve, (%)
- F.a is determined in the laboratory:
+ Sift F can find out J1. Take the remaining material on the F – J1 sieve and sift again
can infer J2, continue to take the rest material on the F – (J1 + J2) and again sift it.
+ The total J1 + J2 + J3 + … will asymptote F.a
+ Sieve performance is 100% if J1 = F.a.

2.4. Mix Equation

- When mixing the amount an of A with the amount b of B, will create a homogeneous
mixture. Component A and B in ideal mixture:
a
● With A: C A=
a+ b
b
● With B: C B=
a+ b
- These components will be the same in any volume of the mixture. However, this ideal
mixture just can be reached when mixing time increases until infinite and doesn't have
any factor against the mix process.
- In reality, time can’t advance infinitely, so components A and B will be different in
part of different volumes.
- To evaluate the homogeneity of the mixture, we consider the mean squared
difference.
- If in V1, the volume composition of A and B is C1A, C1B respectively, then the mean
squared difference of the ideal mixture is:

√
N

∑ ( C A −C iA )2
i=1
sA =
N−1

√
N

∑ ( C B−C iB )2
i=1
s B=
N −1

With CA, CB is the composition of A, B in the mixture. We can see the more ideal the
mixture is, the smaller the sA, sB is. sA, sB depend on many factors but the most
 important is the time. The relation between s and time is described by the graph
(assume other factors do not change).
- In reality, depending on the requirement of s, we can determine a suitable time. To
evaluate the homogeneous of the mixture, we can see another value is the mix index:
σe
I s=
s

with σe: standard deviation

σ e=
√ C A CB
n

√
C A C B (N−1)
→ I s= N
n . ∑ ( C A −CiA ) 2
i=1

n: the number of particles in case of mixed loose material.

3. EXPERIMENTAL METHOD

3.1. Mill-Turn on the machine without load to measure the no-load power
- Give 200g of rice to the mill
- Turn on the input screw, press the meter, measure the maximum load current intensity
- When the current is back to no-load value, press the meter to determine the grinding
time

3.2. Sieve
- Determine sieve efficiency at 0.2 mm size: sieve 80g of ground rice 5 times, each
time 5 minutes, and weigh the rice through the sieve each time.
- Determine the size distribution of the material after grinding: sieve 80g grinded rice
through various sieves of different sizes for 20 minutes and weigh accumulated rice in
each sieve.

3.3. Mix
- Put 1.5kg of green beans and 3kg of soybeans into the machine
- Turn on the machine
- Stop the machine 6 times (5s, 15s, 30s, 60s, 120s, 300s) and take 8 samples each time
according to the chart. Count each particle in the sample.
4. EXPERIMENTAL RESULT

4.1. Results

4.1.1. Grinding:
Table 1.
Electrical current (A)
Mass Time
No load Load

200g 38.5s 3.0 3.6

-Power consumption of milling machine:

( )
−6
4 1 1 4 1 1 200 x 10
P=P 2−P1= x 19 W i − T = x 19 x 13 x ( − )x =0.1559( KW )
3 √ DP 2 √ D P 1 3 √0.25 √ 1.5 30 /60

- Power Consumption P’:

P’ = U x I x cosφ = 220 x 3.6 x 0.8 = 633.6 (W) = 0.6336 (kW)

- Efficiency of the milling machine H:

P 0.1559
H= x 100 % = x 100 % = 24.61 %
P' 0.6336

4.1.2. Sieving:
a) Determination of sieve efficiency 0.2mm: Mass F=80g
Table 2: Experimental results to determine the sieving efficiency

No. of times Time (min) Mass distribution (g) ∑ 𝐽i

1 5 27.07 27.07

2 10 0.50 0.50

3 15 0.20 0.20

4 20 0.12 0.12

5 25 0.09 0.09

Chart 1. Chart of ƩJi following a number of time

 Total material through sieve after 5 times:

n
Fa=∑ J i =J 1+ J 2+ J 3+ J 4 +J 5=¿ ¿ 27.98 (g)
i=1

Determination of sieve efficiency 0.25mm

J1 27.07
E= x 100= x 100 = 96.75%
Fa 27.98

b) Sieve analysis results: Mass F=80g

Table 3: The size distribution of the material after grinding
Mass Log ( log Dn
Sieve sizes Acummulated
Time Mass (g) distribution ∆∅n¿
Dpn (mm) mass ϕ (g)
(g)

1.000 3.08 3.08 0.0385 0.0385 -1.4145 0.0000

0.315 48.83 51.91 0.6489 0.6104 -0.2144 -0.5017

20
mins
0.250 4.50 56.41 0.7051 0.0562 -1.2503 -0.6021

0.150 10.95 67.36 0.8420 0.1369 -0.8636 -0.8239

0.100 3.93 71.29 0.8911 0.0491 -1.3089 -1.0000

Chart 2. Chart of size distribution of the material on sieve

 4.1.3. Mixing:

5s 15s 30s 60s 120s 300s CA CB

Sample
S G S G S G S G S G S G

1 107 50 71 39 29 30 46 28 8 29 18 11

2 40 49 72 41 35 17 61 30 13 9 19 12

3 34 85 47 74 43 40 40 30 21 19 15 19

4 51 14 58 24 15 22 38 41 34 15 16 9

5 42 23 45 30 25 40 44 64 32 13 19 12

6 25 35 40 37 53 52 63 56 26 21 25 6

7 42 23 68 27 24 6 46 24 19 21 14 9

8 25 86 36 80 25 13 44 85 16 17 11 9

TIME 5" CA = 0.67

Sample S G CiA CiA-CA (CiA-CA)2 Ʃ(CiA-CA)2 n σe Is

1 107 50 0.6815 0.0115 0.0013 0.5139 731 0.0174 0.0643

 2 40 49 0.4494 -0.2206 0.0487

3 34 85 0.2857 -0.3843 0.1477

4 51 14 0.8770 0.207 0.0428

5 42 23 0.6462 -0.0238 0.0057

6 25 35 0.4167 -0.2533 0.0642

7 42 23 0.6462 -0.0238 0.0057

8 25 86 0.2252 -0.4448 0.1978

TIME 15" CA = 0.67

Sample S G CiA CiA-CA (CiA-CA)2 Ʃ(CiA-CA)2 n σe Is

1 71 39 0.6455 -0.1483 0.0220

2 72 41 0.6372 0.0558 0.0031

3 47 74 0.3884 -0.2816 0.0793

4 58 24 0.7073 0.0373 0.0014

0.1355 673 0.0182 0.1306
5 45 30 0.6000 -0.07 0.0049

6 40 37 0.5195 -0.1505 0.0227

7 68 27 0.7158 0.0458 0.0021

8 36 80 0.3103 -0.3597 0.1294

TIME 30" CA = 0.67

Sample S G CiA CiA-CA (CiA-CA)2 Ʃ(CiA-CA)2 n σe Is

1 29 30 0.4915 -0.1785 0.0319 0.2508 469 0.0218 0.1150

 2 35 17 0.6731 0.0031 0.0000096

3 43 40 0.5181 -0.1519 0.0231

4 15 22 0.4054 -0.2646 0.0700

5 25 40 0.3846 -0.2854 0.0815

6 53 52 0.5048 -0.1652 0.0273

7 24 6 0.8000 0.1300 0.0169

8 25 13 0.6579 -0.0121 0.0001

TIME 60" CA = 0.67

Sample S G CiA CiA-CA (CiA-CA)2 Ʃ(CiA-CA)2 n σe Is

1 46 28 0.6216 -0.0484 0.0023

2 61 30 0.6703 0.0003 0.0000001

3 40 30 0.5714 -0.0986 0.0097

4 38 41 0.4810 -0.1890 0.0357

0.1365 740 0.0173 0.1241
5 44 64 0.4074 -0.2626 0.0690

6 63 56 0.5294 -0.1406 0.0198

7 46 24 0.6571 -0.0129 0.0002

8 44 85 0.3411 -0.3289 0.1082

TIME 120" CA = 0.67

Sample S G CiA CiA-CA (CiA-CA)2 Ʃ(CiA-CA)2 n σe Is
 1 8 29 0.2162 -0.4538 0.2059

2 13 9 0.5909 -0.0790 0.0063

3 21 19 0.5250 -0.1450 0.0210

4 34 15 0.6939 0.0239 0.0006

0.3214 313 0.0266 0.1244
5 32 13 0.7111 0.0411 0.0017

6 26 21 0.5532 -0.1168 0.0136

7 19 21 0.4750 -0.195 0.0380

8 16 17 0.4848 -0.1852 0.0343

TIME 300" CA = 0.67

Sample N X CiA CiA-CA (CiA-CA)2 Ʃ(CiA-CA)2 n σe Is

1 18 11 0.6207 -0.0493 0.0024

2 19 12 0.6129 -0.0571 0.0033

3 15 19 0.4412 -0.2288 0.0524

4 16 9 0.6400 -0.03 0.0009

0.0991 224 0.0315 0.2647
5 19 12 0.6129 -0.0571 0.0033

6 25 6 0.8065 0.1365 0.0186

7 14 9 0.6087 -0.0613 0.0038

8 11 9 0.5500 -0.1200 0.0144

 Chart 4. Chart of mixing index following time

4.2. Discussion
1) Discussing the adaptation of Bond’s Law in predicting the milling efficiency,
especially focus on the other theories:

Based on the milling theories mentioned in 2.1.2, we can see that:

- The Surfacial Theory of P. R. Rittinger: It can only be precisely applicable under the
condition that the energy provided per weight unit of solid is not too considerable, and
can be used to measure the real milling process with Kr determined practically on the
milling machine, which is the same as the real one. Because of the constrained energy
condition and the complex Kr determination due to the necessity of determining the
factor based on each identified material and machine, this theory does not have high
practicality in milling process prediction.
- The Volumetric Theory of Kick: On the basis of stress analysis theory of the elasticity
in the elastic limit. This theory does not have high practical value due to the
complicated Kr determination.
- Bond’s Law: This is the highest practical value theory in the milling efficiency
prediction. Because work factor Wi has included the inner friction inside the milling
machine. At the same time, it has small errors when implementing the calculation of
the efficiency on the different milling machines of the same type and with the dry and
wet milling processes. Therefore, this theory is very convenient for calculation.
 2) Give the comments on the efficiency of sifting and milling. Compare with the result
in the book. Explain.
❖ Milling efficiency is calculated based on: E = 24.61 %
❖ The weight of material used for milling M (by scale).
- Milling time (by stopwatch).
- The current with maximum load measured by Ammeter. The efficiency can be
concluded to be pretty low.
- Objective reasons:
● The efficiency of the machine is not high.
- Subjective reasons:
● The error when weighing the materials.
● Time adjustment is not precise.
However, these reasons cause very small errors which do not affect the result.
● Due to the measurement after sifting: Because the material is very light-
weighted and has small particles, it is easy to disperse into the surrounding
environment (by wind). Besides, an amount of material is still stuck on the
surface of the sieve due to its small size.
This has the most effect on the error of the result.
❖ Sieving: E = 96.75%
Efficiency is calculated based on:
- The weight of material that went through the sieve after the first sifting J1.
- The amount of material that might go through sieve J.a, is measured in chart 1.
Efficiency is pretty high, thanks to
● The humidity in the material is low.
● The suitable thickness of the material on the sieve can make the material go
through much faster and more easily.
● The flat surface of the sieves.

3) Discus about the reliability of the result and the most effective factors:
❖ Grinding:
The reliability of milling results is low.
The most effective factors (are as said above in the question 2)
❖ Sieving:
The reliability of sifting is high.
The most effective factors affect the result:
- The low humidity of the material.
- The suitable thickness of the material on the sieves.
- The flat surface of sieves.
- The leftovers of the material on the sieve from the previous sifting.
- The simple calculation also helps to reduce the error in the calculation.
❖ Mixing:
- The beans size distribution: Because of the different sizes between the soya beans and
green beans affect badly on the process.
- Time mixing: The time measurement implemented is imprecise. But it is not
considerable.
- Specific weight of material: Green beans and soybeans have approximate specific
weights, which make the process convenient.
- Breakability: These beans are not easily breakable so the process operates more
easily.
- The samples are taken from different positions (such as the diaphragm), which
ensures the unique property of the sample.
- The sample calculation helps to avoid error.
4) Comment on the way of taking the samples in the mixing experiment:
The samples in the experiment are taken at 6 different times: 5", 15", 30", 60",
120" vaø 300". At each time, we take 8 samples as the diagram:
1 2 3
4 5
6 7 8
We have to take the samples at those above positions to ensure the analysis of all
beans, which can make the samples unique so that the results will give higher
accuracy. Because during the mixing process, not all the positions have the same
bean distribution. Therefore, we have to take a lot of positions to calculate the
average. The beans only have the most uniform distribution at some time in the
mixing process.
Besides, we have to take the samples at 6 different times to analyze the change of
the mixing index according to time. Since then, we can find the time that the
beans reach the highest mixing index. That is the time we should implement the
bean mixture to meet the highest uniform level.
5) Comment of the reliability of the result and the elements that affect the most to the
mixing experiment:
The reliability of the mixing result is pretty high.
The elements affect significantly the mixing result:
○ The beans distribution: Due to the different sizes between green beans and soya
beans, the mixing process is affected badly.
○ Mixing time: is determined by the stopwatch (manually), so there will be an error. But
it is not considerable.
○ The specific weight of the material: Because green beans and soybeans have
approximate weight so they can support the mixing process.
○ Breakability: Green beans and soya beans do not have breakability, which makes the
process operate more easily.
○ Samples are taken at positions (as in the diagram) so that the unique property of the
sample is ensured, which makes the result have higher accuracy.
○ The result calculation is simple so the error is too small to mention.
5. REFERENCES:
[1] Vũ Bá Minh – Hoàng Minh Nam, “ Quá trình và thiết bị trong công nghệ hóa học –
Tập 2:
Cơ học vật liệu rời”, Nhà xuất bản Khoa học và Kỹ thuật, Hà Nội, 1998
[2] Bộ môn Quá trình – Thiết bị, “ Thí nghiệm Quá trinh – Thiết bị”, Đại học Bách Khoa
Tp.HCM, 2003