In a drop extractor, liquid droplets of approximately uniform size and spherical shape are formed at a series of nozzles and rise countercurrently through the continuous phase which is flowing downwards at a velocity equal to one half of the terminal rising velocity of the droplets. The flowrates

Question

In a drop extractor, liquid droplets of approximately uniform size and spherical shape are formed at a series of nozzles and rise countercurrently through the continuous phase which is flowing downwards at a velocity equal to one half of the terminal rising velocity of the droplets. The flowrates of both phases are then increased by 25 per cent. Because of the greater shear rate at the nozzles, the mean diameter of the droplets is, however, only 90 per cent of the original value. By what factor will the overall mass transfer rate change? It may be assumed that the penetration model may be used to represent the mass transfer process. The depth of penetration is small compared with the radius of the droplets and the effects of surface curvature may be neglected. From the penetration theory, the concentration $C_{A}$ at a depth y below the surface at time t is given by:

$\frac{C_{A}}{C_{AS}} =\text{ erfc }\left[\frac{y}{2\sqrt{(Dt)} } \right]$ where erfc $X =\frac{2}{\sqrt{\pi } } \int_{X}^{\infty }e^{-x^{2}}dx$

where $C_{AS}$ is the surface concentration for the drops (assumed constant) and D is the diffusivity in the dispersed (droplet) phase. The droplets may be assumed to rise at their terminal velocities and the drag force F on the droplet may be calculated from Stokes’ Law, F = 3πμ du.

[latex]\frac{C_{A}}{C_{AS}} = ext{ erfc }\left[\frac{y}{2\sqrt{(Dt)} } ight][/latex] where erfc [latex] X =\frac{2}{\sqrt{\pi } } \int_{X}^{\infty }e^{-x^{2}}dx[/latex]

where [latex]C_{AS}[/latex] is the surface concentration for the drops (assumed constant) and D is the diffusivity in the dispersed (droplet) phase. The droplets may be assumed to rise at their terminal velocities and the drag force F on the droplet may be calculated from Stokes’ Law, F = 3πμ du.

Accepted Answer

Case 1: For a volumetric flowrate Q₁, the numbers of drops per unit time is:

[latex]\frac{Q_{1}}{(\frac{1}{6} )\pi d^{3}} =\frac{6Q_{1}}{\pi d^{3}_{1}}[/latex]

The rising velocity is given by a force balance:

[latex]3\pi \mu d_{1u}=\frac{1}{6} \pi d^{3}_{1}( ho _{1}- ho _{2})g[/latex]

or: [latex]u_{1}=\frac{d_{1}^{2}g}{18\mu } ( ho _{1}- ho _{2})=Kd_{1}^{2}[/latex] relative to continuous phase

The downward liquid velocity is[latex]\frac{1}{2} Kd^{2}_{1}[/latex]

The upward droplet velocity relative to container is:

[latex]Kd^{2}_{1}-\frac{1}{2} Kd^{2}_{1}=\frac{1}{2} Kd^{2}_{1}[/latex]

and the time of contact during rise through height H is:

[latex]t_{c}=\frac{H}{\frac{1}{2}Kd_{1}^{2} }[/latex] (i)

The mass transfer rate is: -D(∂[latex]C_{A}[/latex]/∂y).

Thus: [latex]\frac{1}{C_{AS}} \frac{∂C_{A}}{∂y} =\frac{∂}{∂y} \left\{erfc\frac{y}{2\sqrt{Dt} } ight\} =\frac{∂}{∂y} \left\{\frac{2}{\sqrt{\pi } } \int_{y/2\sqrt{Dt} }^{\infty }e^{-y^{2}/4Dt}d\left(\frac{y}{2\sqrt{Dt} } ight) ight\} [/latex]

[latex]=\frac{∂}{∂y} .\frac{1}{2\sqrt{Dt} } -\frac{2}{\sqrt{\pi } } \int_{y}^{\infty }e^{-y^{2}/4Dt}dt[/latex].

[latex]\frac{∂C_{A}}{∂y} =-\frac{C_{AS}}{\sqrt{\pi Dt} } e^{-y^{2}/4Dt}\left(\frac{∂C_{A}}{∂y} ight) _{y=0}=-\frac{C_{AS}}{\sqrt{\pi Dt} } [/latex]

The mass transfer rate at the surface is: (moles/area × time).

[latex]-D\left(-\frac{C_{AS}}{\sqrt{\pi Dt} } ight) =\sqrt{\frac{D}{\pi t} } C_{AS}[/latex]

The mass transfer in time [latex]t_{e1}[/latex] is:

[latex]\sqrt{\frac{D}{\pi } } C_{AS}\int_{0}^{t_{e1}}{} t^{-1/2}dt=2\sqrt{\frac{D}{\pi } } t_{e1}^{1/2}C_{AS}=Kt_{e1}^{1/2}[/latex]

Substituting from equation (i):
Mass transfer in moles per unit area of drop [latex]\propto \frac{\sqrt{2H} }{\sqrt{K}d_{1} } [/latex]

The mass transfer per drop is proportional to:

[latex]\sqrt{2} \sqrt{\frac{H}{K} } d_{1}^{-1}d^{2}_{1}\propto \sqrt{2} \sqrt{\frac{H}{K} } d_{1}[/latex]

The mass transfer per unit time = Mass transfer per drop × drops/time

or: proportional to: [latex]\sqrt{2} \sqrt{\frac{H}{K} } d_{1} imes \frac{6Q_{1}}{\pi d_{1}^{3}} \propto 8.48\sqrt{\frac{H}{K} } \frac{Q_{1}}{\pi d^{2}_{1}} [/latex]

Case 2: diameter = 0.9d = d₂, Q₂ = 1.25Q₁
The number of drops per unit time is:

[latex]\frac{6Q_{2}}{\pi d^{3}_{2}} =\frac{7.5Q_{1}}{\pi d^{3}_{1}}[/latex]

Rising velocity = K [latex]d^{2}_{1} = K(0.9d^{2}_{1} = 0.81Kd^{2}_{1}[/latex]

Downward liquid velocity = [latex]\frac{5}{4} imes (\frac{1}{2} Kd^{2}_{1})=0.625Kd^{2}_{1}[/latex]relative to continuous phase
Rising velocity relative to container = 0.81 K[latex]d^{2}_{1} - 0.625 Kd^{2}_{1}=0.185Kd^{2}_{1}[/latex]

Contact time, [latex]t_{e2} = \frac{H}{0.185Kd_{1}^{2}} [/latex]

Mass transfer in time [latex]t_{e2}[/latex] per unit area[latex]\propto kt_{e_{2}}^{1/2}\propto \sqrt{\frac{H}{0.185K} } \frac{1}{d_{1}} \propto 2.325\sqrt{\frac{H}{K} } d_{1}^{-1}[/latex]

Mass transfer per drop [latex]\propto 2.325\sqrt{\frac{H}{K} } d_{1}^{-1} imes (0.9d_{1})^{2}\propto 1.883\sqrt{\frac{H}{K} } d_{1}[/latex]

Mass transfer per unit time [latex]\propto 1.883\sqrt{\frac{H}{K} } d_{1} imes \frac{10.29Q_{1}}{\pi d_{1}^{3}} [/latex]

[latex]\propto 19.37\sqrt{\frac{H}{K} } \frac{Q_{1}}{\pi d_{1}^{2}}[/latex]

Thus, the factor by which mass transfer rate is increased is: (19.37/8.48) = 2.28

Question 10.43: In a drop extractor, liquid droplets of approximately unifor...

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