Question 18.6: A sieve-plate column operating at atmospheric pressure is to......

Physical properties of methanol: Molecular weight is 32, normal boiling point is 65°C, and the density of vapor is

\rho_{V}={\frac{32\times273}{22.4\times338}}=1.15\mathrm{~kg/m}^{3}

From Perry, Chemical Engineers’ Handbook, 6th ed., p. 3-188, the density of liquid methanol is 810\,\mathrm{kg/m^{3}} at o°c and 792 kg/m³ at 20°C. At 65°C, the estimated density {{\rho}}_{L} is 750 kg/m³. Lange’s Handbook of Chemistry, 9th ed., 1956, p. 1650, gives the surface tension of methanol at 20 and 100°C. By interpolation, at 65°C, σ = 19 dyn/cm.

(a) Vapor velocity and column diameter In Fig. 18.28 the abscissa is

\frac{L}{V}\left(\frac{\rho_{V}}{\rho_{L}}\right)^{1/2}=\frac{3.5}{4.5}\left(\frac{1.15}{750}\right)^{1/2}=3.04\times10^{-2}

For 18-in. plate spacing,

K_{v}=0.29=u_{c}\biggl(\frac{\rho_{V}}{\rho_{V}-\rho_{L}}\biggr)^{1/2}\biggl(\frac{20}{\sigma}\biggr)^{0.2}

Allowable vapor velocity:

u_{c}=0.29\bigg(\frac{750-1.15}{1.15}\bigg)^{1/2}\bigg(\frac{19}{20}\bigg)^{0.2}

=7.32\,\mathrm{ft/s\;or\;2.23\;m/s}

Vapor flow rate:

V = D(R + 1) = 4.5D

={\frac{5800\times4.5}{3600\times1.15}}=6.30\ m^{3}/s

Cross-sectional area of column:

Bubbling area= 6.30/2.23 = 2.83 m²

If the bubbling area is 0.7 of the total column area,

Column area= 2.83/0.7 = 4.04 m²

Column diameter:

D_{c}=\left({\frac{4\times4.04}{\pi}}\right)^{1/2}=2.27\,\mathrm{m}

(b) Pressure drop The plate area of one unit of three holes on a triangular {\frac{3}{4}}\mathbf{i{n}} pitch is {{\frac{1}{2}}}\times{\frac{3}{4}}({\frac{3}{4}}\times{\sqrt{3}}/2)=9{\sqrt{3}}/64{}{\mathrm{~in}}^{2}. The hole area in this section (half a hole) is \textstyle{\frac{1}{2}}\times\pi/4\times(_{4}^{1})^{2}=\pi/128\ \mathrm{in}\,^{2}. Thus the hole area is \pi/128\times64/9\sqrt{3}=0.1008, or 10.08 percent of the bubbling area.
Vapor velocity through holes:

u_{0}=2.23/0.1008=22.1~\mathrm{m/s}

Use Eq. (18.58) for the pressure drop through the holes. From Fig. 18.27, C_{0}=0.73. Hence

h_{d}=\left(\frac{u_{0}^{2}}{C_{0}^{2}}\right)\left(\frac{\rho_{V}}{2g\rho_{L}}\right)=51.0\left(\frac{u_{0}^{2}}{C_{0}^{2}}\right)\left(\frac{\rho_{V}}{\rho_{L}}\right)              (18.85)

h_{d}={\frac{51.0\times22.1^{2}\times1.15}{0.73^{2}\times750}}=71.7\,\mathrm{mm\,methanol}

Head of liquid on plate:

Weir height: h_{\mathrm{w}}=2\times25.4=50.8\ \mathrm{mm}

Height of liquid above weir: Assume the downcomer area is 15 percent of the column area on each side of the column. From Perry, 6th ed., page 1-26, the chord length for such a segmental downcomer is 1.62 times the radius of the column, so

L_{w}=1.62\times2.23/2=1.81\ \mathrm{m}

Liquid flow rate:

q_{L}={\frac{5800\times4.5}{750\times60}}=0.58~{\mathrm{m}}^{3}/{\mathrm{min}}

From Eq. (18.60),

h_{o w}=43.4\left(\frac{q_{L}}{L_{w}}\right)^{2/3}               (18.60)

h_{ow}=43.4(0.58/1.81)^{2/3}=20.3\;\mathrm{mm}

From Eq. (18.59), with β = 0.6,

h_{l}=\beta(h_{w}+h_{o w})                    (18.59)

h_{l}=0.6(50.8+20.3)=42.7\,\mathrm{mm}

Total height of liquid [from Eq. (18.57)]:

h_t=h_d+h_l  (18.57)

h_{t}=71.7+42.7=114.4{{ \ m m}}

(c) Froth height in downcomer Use Eq. (18.62). Estimate h_{f,L}=10 mm methanol. Then

Z_c=2 \beta\left(h_w+h_{o w}\right)+h_d+h_{f, L}  (18.62)

Z_{c}=(2\times42.7)+71.7+10=167.1\,\mathrm{mm}

From Eq. (18.63),

Z={\frac{Z_{\circ}}{Φ_d}}                (18.63)

Z=167.1/0.5=334\,\mathrm{mm}\,(13.2\,\mathrm{in})