Question 28.1: The screen analysis shown in Table 28.2 applies to a sample ......
The screen analysis shown in Table 28.2 applies to a sample of crushed quartz. The density of the particles is 2650 kg/m³ (0.00265 g/mm³), and the shape factors are a = 2 and \Phi_s= 0.571. For the material between 4-mesh and 200-mesh in particle size, calculate (a) A_w in square millimeters per gram and N_w in particles per gram, \text { (b) } \bar{D}_V, \text { (c) } \bar{D}_s,(d) \bar{D}_w \text {, and }(e) N_i for the 150/200-mesh increment. (f) What fraction of the total number of particles is in the 150/200-mesh increment?
| TABLE 28.2 Screen analysis |
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| Mesh | Screen opening D_{pi}, mm |
Mass fraction retained, x_i, |
Average particle diameter in increment,\bar{D}_{pi}, mm |
Cumulative fraction smaller than D_{pi} |
| 4 | 4.699 | 0.0000 | – | 1.0000 |
| 6 | 3.327 | 0.0251 | 4.013 | 0.9749 |
| 8 | 2.362 | 0.1250 | 2.845 | 0.8499 |
| 10 | 1.651 | 0.3207 | 2.007 | 0.5292 |
| 14 | 1.168 | 0.2570 | 1.409 | 0.2722 |
| 20 | 0.833 | 0.1590 | 1.001 | 0.1132 |
| 28 | 0.589 | 0.0538 | 0.711 | 0.0594 |
| 35 | 0.417 | 0.0210 | 0.503 | 0.0384 |
| 48 | 0.295 | 0.0102 | 0.356 | 0.0282 |
| 65 | 0.208 | 0.0077 | 0.252 | 0.0205 |
| 100 | 0.147 | 0.0058 | 0.178 | 0.0147 |
| 150 | 0.104 | 0.0041 | 0.126 | 0.0106 |
| 200 | 0.074 | 0.0031 | 0.089 | 0.0075 |
| Pan | – | 0.0075 | 0.037 | 0.0000 |
Density of the particles: 2650 kg/m³ (0.00265 g/mm³)
Shape factors: a = 2 and \Phi_s = 0.571
Particle size range: between 4-mesh and 200-mesh
Increment: 150/200-mesh increment
To find A_w \text { and } N_{w}, Eq. (28.4) can be written
A_w=\frac{6 x_1}{\Phi_s \rho_d \bar{D}_{p 1}}+\frac{6 x_2} {\Phi_s \rho_p \bar{D}_{p 2}}+\cdots+\frac{6 x_n}{\Phi_s \rho_p\bar{D}_{p n}}
=\frac{6}{\Phi_s \rho_p} \sum\limits_{i=1}^n \frac{x_i}{\bar{D}_{p i}} (28.4)
A_w=\frac{6}{0.5 / 1 \times 0.00265} \sum \frac{x_i}{\bar{D}_{pi}}=3965 \sum \frac{x_i}{\bar{D}_{p i}}and Eq. (28.11) as
N_w=\frac{1}{a \rho_p} \sum\limits_{i=1}^n \frac{x_i}{\bar{D}_{p i}^3}=\frac{1}{a \rho_p \bar{D}_V^3} (28.11)
N_w=\frac{1}{2 \times 0.00265} \sum \frac{x_i}{\bar{D}_{pi}^3}=188.7 \sum \frac{x_i}{\bar{D}_{p i}^3}(a) For the 4/6-mesh increment \bar{D}_{p i} is the arithmetic mean of the mesh openings of the defining screens, or, from Table 28.2, (4.699 + 3.327)/2 = 4.013 mm. For this increment x_i = 0.0251; hence x_i \bar{D}_{p i}= 0.0251/4.013 = 0.0063 and x_i \bar{D}_{p i}^3 = 0.0004.
Corresponding quantities are calculated for the other 11 increments and summed to give \sum x_i / \bar{D}_{p i} = 0.8284 and \sum x_i / \bar{D}_{p i}^3 = 8.8296. Since the pan fraction is excluded, the specific surface and number of particles per unit mass of particles 200-mesh or larger are found by dividing the results from Eqs. (28.4) and (28.11) by 1 – x_1 (since i = 1 for the pan), or 1 – 0.0075 = 0.9925. Then
\begin{aligned}& A_w=\frac{3965 \times 0.8284}{0.9925}=3309 \ mm ^2 / g \\& N_w=\frac{188.7 \times 8.8296}{0.9925}=1679 \ \text { particles } / g\end{aligned}
(b) From Eq. (28.9),
\bar{D}_V=\left[\frac{1}{\sum\limits_{i=1}^n\left(x_i / \bar{D}_{pi}^3\right)}\right]^{1 / 3} (28.9)
\bar{D}_V=\frac{1}{8.8296^{1 / 3}}=0.4238 \ mm
(c) The volume-surface mean diameter is found from Eq. (28.6):
\bar{D}_s=\frac{1}{\sum\limits_{i=1}^n\left(x_i / \bar{D}_{p i}\right)} (28.6)
\bar{D}_s=\frac{1}{0.8284}=1.207 \ mm
(d) Mass mean diameter \bar{D}_w is obtained from Eq. (28.8). For this, from the data in Table 28.2,
\bar{D}_w=\sum\limits_{i=1}^n x_i \bar{D}_{p i} (28.8)
\sum x_i \bar{D}_{p i}=\bar{D}_w=1.677 \ mm
(e) The number of particles in the 150/200-mesh increment is found from Eq.(28.11):
\begin{aligned}N_2 & =\frac{x_2}{a \rho_p \bar{D}_{p 2}^3}=\frac{0.0031}{2\times 0.00265 \times 0.089^3} \\& =836 \text { particles } / g\end{aligned}
This is 836/1679 = 0.498, or 49.8 percent of the particles in the top 12 increments.
For the material in the pan fraction the number of particles and specific surface area are enormously greater than for the coarser material, but they cannot be accurately estimated from the data in Table 28.2.