Question 27.6: In the crystallizer of Example 27.5, a growth rate G of 0.00......

The screen opening of a 20-mesh standard screen is, from Appendix 20, 0.0328 in., or 0.00273 ft. This dimension can be used for L, and the shape factor a [Eq. (27.16)] is assumed to be unity. From Example 27.5, the volume flow rate of mother liquor in the product magma is

$v_p=a L^3$    (27.16)

$Q=\frac{44,520}{82.5}=540~ ft ^3 / h$

Since, when z = 3, $L_{pr}$ = 0.00273, Eq. (27.28) gives for the drawdown time and the volume of liquid in the crystallizer

$z \equiv \frac{L}{G \tau}$   (27.28)

$\tau=\frac{L_{p r}}{3 G}=\frac{0.00273}{3 \times 0.0018}=0.506~ h \quad V_{ c }=0.506 \times 540=273 ~ft ^3$

The total magma volume is 273/0.85 = 321 ft³, or 2400 gal.
The nucleation rate is, from Eq. (27.44), since C = 10,000 lb/h,

$B^{\circ}=\frac{C n_c}{m_c V_c}=\frac{C}{6 a \rho_c(G \tau)^3 V_c}=\frac{9 C}{2 a \rho_c V_c L_{p r}^3}$    (27.44)

By Eq. (27.40), the zero-size particle density is n° = 7.72 × 10$^{7}$/0.0018 = 4.289 × 10$^{10}$ nuclei/ft$^4$ The value of L/Gτ is L/(0.0018)(0.506) = 1.1 × 10³ L. The equation for the number-density distribution is, from Eq. (27.27),

$B^{\circ}=G n^{\circ}$    (27.40)

$\begin{aligned}& \int_{n^{\circ}}^n \frac{d n}{n}=-\frac{1}{G \tau} \int_0^L d L \\ & \ln \frac{n^{\circ}}{n}=\frac{L}{G \tau}\end{aligned}$    (27.27)

$\log n=\log n^{\circ}-\frac{1.1 \times 10^3 L}{2.3026}=10.632-4.777 \times 10^2 L$

This equation is plotted in Fig. 27.16.

The screen analysis is found by reading ordinates from Fig. 27.15c for values of z corresponding to mesh openings. For example, for the 20-mesh point, where z = 3, $x_m$ is 0.35. In general,

$z=\frac{L}{\tau G}=\frac{L}{0.506 \times 0.0018}=1098 L$

Details are given in Table 27.2.