Demister Design

Abstract

In this chapter, demister is covered.

The allowable flow velocity in vertical and horizontal demisters is determined.

The special features for demisters at high pressures are also covered.

Very small droplets are separated by a droplet enlargement.

The falling velocity of droplets in gas is determined.

Appendix: Calculation of the Falling Velocity of Droplets in Air or Gas

First, the Archimedes number Ar, according to Formulas 11.11 and 11.11a, is calculated and therefore by using Eqs. (11.12a) and (11.12c) the Reynolds number Re is determined.

The falling velocity for the different Reynolds number regions is calculated with Formula 11.13a to 11.13c. Alternatively, the falling velocity can be calculated using the Eq. 11.13d for Re >1.

• 1. Archimedes number Ar

  $$ Ar = \frac{{d^{3} \cdot g \cdot \rho_{\text{G}} (\rho_{\text{Fl}} - \rho_{\text{G}} )}}{{\eta_{\text{G}}^{2} }} $$

  (11.11)
  $$ Ar = \frac{{Re^{2} \cdot (\rho_{\text{Fl}} - \rho_{\text{G}} )}}{{Fr \cdot \rho_{\text{G}} }} $$

  (11.11a)

  Fr = Froude number

• 2. Reynolds number Re

  $$ Re = \frac{{w_{\text{Fall}} \cdot d{ \cdot }\rho_{\text{G}} }}{{\eta_{\text{G}} }} $$

  (11.12)

  Because the falling velocity is unknown the Reynolds number has to be determined using the Archimedes number. The following relationships are valid for different regions:

  $$ Ar < 3.6\quad Re = \frac{Ar}{18} $$

  (11.12a)
  $$ 3.6 < Ar < 83000\quad Re = \left( {\frac{Ar}{13.9}} \right)^{1/1.4} $$

  (11.12b)
  $$ Ar > 83000\quad Re = 1.73 * \sqrt {Ar} $$

  (11.12c)

• 3. Falling velocity w _Fall

  $$ w_{\text{Fall}} = \frac{{Re \cdot \eta_{\text{G}} }}{{d \cdot \rho_{\text{G}} }}\quad ({\text{m}}/{\text{s}}) $$

  (11.13)

• 3.1 For the region Re < 0.2 Stoke’s Law is valid

  $$ w_{\text{Fall}} = \frac{{d^{2} (\rho_{Fl} - \rho_{G} ) \cdot g}}{{18 \cdot \eta_{G} }}\quad ({\text{m}}/{\text{s}}) $$

  (11.13a)

• 3.2. In the region Re = 500 to Re = 150,000 the relationships of Newton are valid

  $$ w_{\text{Fall}} = 5.48 \cdot \sqrt {\frac{{d \cdot (\rho_{\text{Fl}} - \rho_{\text{G}} )}}{{\rho_{\text{G}} }}} \quad ({\text{m}}/{\text{s}}) $$

  (11.13b)

• 3.3. In the intermediate region 0.2 < Re < 500 the following equation is valid

  $$ \begin{aligned} w_{\text{Fall}} & = \sqrt {\frac{4}{3}.\frac{d}{c}.\frac{{g.(\rho_{Fl} - \rho_{G} )}}{{\rho_{\text{G}} }}} \quad ({\text{m}}/{\text{s}}) \\ C & = 18.5/Re^{0.6} \\ \end{aligned} $$

  (11.13c)

• 3.4. Alternatively, for Re > 1

  $$ w_{\text{Fall}} = \frac{{0.153 \cdot g^{0.71} \cdot d^{1.143} \cdot (\rho_{\text{fl}} - \rho_{\text{G}} )^{0.714} }}{{\rho_{\text{G}}^{0.286} \cdot \eta^{0.429} }}\quad ({\text{m}}/{\text{s}}) $$

  (11.13d)

• d = droplet diameter (m)

• g = acceleration of gravity (9.81 m²/s)

• η _G = gas viscosity (mPa)

• ϱ _G = gas density (kg/m³)

• ϱ _G = liquid density (kg/m³)

Example 1

$\varvec{d} = {\mathbf{0}}.{\mathbf{1}}\,{\mathbf{mm}}\quad\varvec{\rho}_{{\mathbf{G}}} = {\mathbf{1}}.{\mathbf{2}}\,{\mathbf{kg}}/{\mathbf{m}}^{{\mathbf{3}}} \quad\varvec{\eta}_{{\mathbf{G}}} = {\mathbf{0}}.{\mathbf{015}}\,{\mathbf{mPa}}\quad\varvec{\rho}_{{{\mathbf{Fl}}}} = {\mathbf{1}}.{\mathbf{000}}\,{\mathbf{kg}}/{\mathbf{m}}^{{\mathbf{3}}}$

Checking of the falling velocity using Eq. (11.13c):

$$ \begin{aligned} w_{\text{Fall}} & = \sqrt {\frac{4}{3} \cdot \frac{{0.1 \cdot 10^{ - 3} }}{10.48} \cdot \frac{9.81 \cdot (1000 - 1.2)}{1.2}} = 0.32\,{\text{m}}/{\text{s}} \\ C & = 18.5/2.575^{0.6} = 10.48 \\ \end{aligned} $$

$$ \begin{aligned} Ar & = \frac{{(0.6 \cdot 10^{ - 3} )^{3} \cdot 9.81 \cdot 1.2(1000 - 1.2)}}{{(15 \cdot 10^{ - 6} )^{2} }} = 52.26 \\ Re & = \left( {\frac{52.26}{13.9}} \right)^{1/14} = 2.575 \\ w_{\text{Fall}} & = \frac{{2.575 \cdot 15 \cdot 10^{ - 6} }}{{0.1 \cdot 10^{ - 3} \cdot 1.2}} = 0.32\,{\text{m}}/{\text{s}} \\ \end{aligned} $$

Example 2: Data as in Example 1, but ρ _G = 0.1 kg/m³

$$ Ar = 4. 3 5 9\quad Re = 0. 4 3 6 $$

$$ w_{\text{Fall}} = \frac{{0.436 \cdot 15 \cdot 10^{ - 6} }}{{0.1 \cdot 10^{ - 3} \cdot 0.1}} = 0.65\,{\text{m}}/{\text{s}} $$

Example 3: Data as in Example 1, but ρ _G = 10 kg/m³

$$ Ar = 4 3 1. 6\quad Re = 1 1. 6 4 $$

$$ w_{\text{Fall}} = \frac{{11.64 \cdot 15 \cdot 10^{ - 6} }}{{0.1 \cdot 10^{ - 3} \cdot 10}} = 0.17\,{\text{m}}/{\text{s}} $$

Conclusion

Falling velocity increases with decreasing gas density (in a vacuum)!

Example 4: Check of Example 1 using Eq. (11.13d)

$$ w_{\text{Fall}} = \frac{{0.153 * 9.81^{0.71} * 0.0001^{1.143} * (1000 - 1.2)^{0.714} }}{{1.2^{0.286} * (15 * 10^{ - 6} )^{0.429} }} $$

$$ w_{\text{Fall}} = 0.32\,{\text{m/s}} $$

Example 5: Check of Example 2 using Eq. (11.13c)

$$ \begin{aligned} w_{\text{Fall}} & = \sqrt {\frac{4}{3} \cdot \frac{0.1 \cdot 10^{-3} }{30.44} \cdot \frac{9.81 \cdot (1000 - 0.1)}{0.1}} = 0.65\,{\text{m}}/{\text{s}} \\ c & = 18.5/0.436^{0.6} = 30.44 \\ \end{aligned} $$

Example 6: Data as in Example 1, but d = 2 mm

$$ \begin{aligned} Ar & = \frac{{(2 \cdot 10^{ - 3} )^{3} \cdot 9.81 \cdot 1.2(1000 - 1.2)}}{{(15 \cdot 10^{ - 6} )^{2} }} = 418,058 \\ Re & = 1.73 \cdot \sqrt {418058} = 1118.6 \\ w_{\text{Fall}} & = 1118.6 \cdot \frac{{15 \cdot 10^{ - 6} }}{{2 \cdot 10^{ - 3} \cdot 1.2}} = 7\,{\text{m}}/{\text{s}} \\ \end{aligned} $$

Check of the falling velocity using Eq. (11.13b):

$$ w_{\text{Fall}} = 5.48 \cdot \sqrt {\frac{0.002 \cdot (1000 - 1.2)}{1.2}} = 7\,{\text{m}}/{\text{s}} $$

Check of the Archimedes number using Eq. (11.11a):

Fr = Froude number

$$ \begin{aligned} Fr & = \frac{{w^{2} }}{d \cdot g} = \frac{{7^{2} }}{0.002 \cdot 9.81} = 2497.4 \\ Ar & = \frac{{1118.6^{2} \cdot (1000 - 1.2)}}{2497.4 \cdot 1.2} = 417013 \\ \end{aligned} $$

• Results tables

See Tables 11.2 and 11.3; Fig. 11.11.

Table 11.2 ρ _G = 1.2 kg/m³ ρ _Fl = 995 kg/m³ η _G = 0.018 mPa

Table 11.3 ρ _G = 0.1 kg/m³ ρ _Fl = 995 kg/m³ η _G = 0.018 mPa

Fig. 11.11

Falling velocities of water droplets in air rat different gas densities