3 Liquid-liquid Extraction

Staged Liquid-Liquid Extraction and Hunter Nash Method

E_n = extract leaving stage n. This could refer to the mass of the stream or the composition of the stream.

F = solvent entering extractor stage 1. This could refer to the mass of the stream or the composition of the stream.

n = generic stage number

N = Final stage. This is where the fresh solvent S enters the system and the final raffinate R_N leaves the system.

M = Composition of the mixture representing the overall system. Points (F and S) and (E_1 and R_N) must be connected by a straight line that passes through point M. M will be located within the ternary phase diagram.

P = Operating point. P is determined by the intersection of the straight line connecting points (F, E_1) and the straight line connecting points (S, R_N). Every pair of passing streams must be connected by a straight line that passes through point P. P is expected to be located outside of the ternary phase diagram.

R_n = raffinate leaving stage n. This could refer to the mass of the stream or the composition of the stream.

S = solvent entering extractor stage N. This could refer to the mass of the stream or the composition of the stream.

S/F = mass ratio of solvent to feed

(x_i)_n = Mass fraction of species i in the raffinate leaving stage n

(y_i)_n = Mass fraction of species i in the extract leaving stage n

Schematic for multistage liquid-liquid extraction process.
Process schematic for multistage liquid-liquid extraction.

 

Determining number of stages N when (1) feed rate; (2) feed composition; (3) incoming solvent rate; (4) incoming solvent composition; and (5) outgoing raffinate composition have been specified/selected.

  1. Locate points F and S on the ternary phase diagram. Connect with a straight line.
  2. Do a material balance to find the composition of one species in the overall mixture. Use this composition to locate point M along the straight line connection points F and S. Note the position of point M.
  3. Locate point R_N on the ternary phase diagram. It will be on the equilibrium curve. Draw a straight line from R_N to M and extend to find the location of E_1 on the equilibrium curve.
  4. On a fresh copy of the graph, with plenty of blank space on each side of the diagram, note the location of points F, S, and R_N (specified/selected) and E_1 (determined in step 3).
  5. Draw a straight line between F and E_1. Extend to both sides of the diagram. Draw a second straight line between S and R_N. Note the intersection of these two lines and label as “P”.
  6. Determine the number of equilibrium stages required to achieve the desired separation with the selected solvent mass.

– Stream R_N is in equilibrium with stream E_N. Follow the tie-lines from point R_N to E_N.

– Stream E_N passes stream R_{N-1}. Connect point E_N to operating point P with a straight line, mark the location of R_{N-1}.

– Stream R_{N-1} is in equilibrium with stream E_{N-1}. Follow the tie-lines from stream R_{N-1} to E_{N-1}.

– Stream E_{N-1} passes stream R_{N-2}. Connect E_{N-1} to operating point P with a straight line, mark the location of R_{N-2}.

– Continue in this manner until the extract composition has reached or passed E_{1}. Count the number of equilibrium stages.

 

Watch this two-part series of videos from LearnChemE that shows how to use the Hunter Nash method to find the number of equilibrium stages required for a liquid-liquid extraction process.

 

Example

1000 kg/hr of a feed containing 30 wt% acetone, 70 wt% water. The solvent is pure MIBK. We intend that the raffinate contain no more than 5.0 wt% acetone.  How many stages will be required for each proposed solvent to feed ratio in the table below?

\bold{S/F} \bold{S} (kg/hr) \bold{(x_A)_M} target \bold{(y_A)_1} \bold{N}
1.0
2.0
0.2

Hunter Nash Method for Finding Smin, Tank Sizing and Power Consumption for Mixer-Settler Units

Staged LLE: Hunter-Nash Method for Finding the Minimum Solvent to Feed Ratio

E_n = extract leaving stage n. This could refer to the mass of the stream or the composition of the stream.

F = solvent entering extractor stage 1. This could refer to the mass of the stream or the composition of the stream.

n = generic stage number

N = Final stage. This is where the fresh solvent S enters the system and the final raffinate R_N leaves the system.

M = Composition of the overall mixture. Points (F and S) and (E_1 and R_N) are connected by a straight line passing through M.

P = Operating point. Every pair of passing streams must be connected by a straight line that passes through P.

R_n = raffinate leaving stage n. This could refer to the mass of the stream or the composition of the stream.

S = solvent entering extractor stage N. This could refer to the mass of the stream or the composition of the stream.

S/F = mass ratio of solvent to feed

S_{\rm min}/F = Minimum feasible mass ratio to achieve the desired separation, assuming the use of an infinite number of stages.

(x_i)_n = Mass fraction of species i in the raffinate leaving stage n

(y_i)_n = Mass fraction of species i in the extract leaving stage n

 

P_{\rm min} = Point associated with the minimum feasible S/F for this feed, solvent and (raffinate or extract) composition. P_{\rm min} is the intersection of the line connecting points (R_N, S) and the line that is an extension of the upper-most equilibrium tie-line.

 

Determining minimum feasible solvent mass ratio (S_{\rm min}/F) when (1) feed composition; (2) incoming solvent composition; and (3) outgoing raffinate composition have been specified/selected.

  1. Locate points S and R_N on the phase diagram. Connect with a straight line.
  2. Extend the upper-most tie-line in a line that connects with the line connecting points (S and R_N). Label the intersection P_{\rm min}.
  3. Find point F on the diagram. Draw a line from P_{\rm min} to F and extend to the other side of the equilibrium curve. Label E_1@S_{\rm min}.
  4. On a fresh copy of the phase diagram, label points F, S, R_N and E_1@S_{\rm min}. Draw one line connecting points S and F and another line connecting points E_1@S_{\rm min}
  5.  and R_N. The intersection of these two lines is mixing point M. Note the composition of species i at this location.
  6. Calculate

(5.1)   \begin{displaymath} \frac{S_{\rm min}}{F}=\frac{(x_i)_F-(x_i)_M}{(x_i)_M-(x_i)_S} \end{displaymath}

Example

We have a 1000 kg/hr feed that contains 30 wt% acetone and 70 wt% water. We want our raffinate to contain no more than 5.0 wt% acetone. What is the minimum mass of pure MIBK required?

Liquid-Liquid Extraction: Sizing Mixer-settler Units

\Phi_C = volume fraction occupied by the continuous phase

\Phi_D = volume fraction occupied by the dispersed phase

\mu_C = viscosity of the continuous phase (mass time-1 length-1)

\mu_D = viscosity of the dispersed phase (mass time-1 length-1)

\mu_M = viscosity of the mixture (mass time-1 length-1)

\rho_C = density of the continuous phase (mass volume-1)

\rho_D = density of the dispersed phase (mass volume-1)

\rho_M = average density of the mixture (mass volume-1)

 

D_i = impeller diameter (length)

D_T = vessel diameter (length)

H = total height of mixer unit (length)

N = rate of impeller rotation (time-1)

N_{\rm Po} = impeller power number, read from Fig 8-36 or Perry’s 15-54 (below) based on value of N_{Re} (unitless)

(N_{\rm Re})_C = Reynold’s number in the continuous phase = inertial force/viscous force (unitless)

P = agitator power (energy time-1)

Q_C = volumetric flowrate, continuous phase (volume time-1)

Q_D = volumetric flowrate, dispersed phase (volume time-1)

V = vessel volume (volume)

 

Tank and impeller sizing

(5.2)   \begin{displaymath} {\rm residence\; time} = \frac{V}{Q_C+Q_D} \end{displaymath}

Geometry of a cylinder

(5.3)   \begin{displaymath} V = \frac{{\pi}D_T^2H}{4} \end{displaymath}

General guidelines

(5.4)   \begin{displaymath} \frac{H}{D_T}=1 \end{displaymath}

(5.5)   \begin{displaymath} \frac{D_i}{D_T}=\frac{1}{3} \end{displaymath}

Impeller power consumption:

(5.6)   \begin{equation*} P=N_{Po}N^3D_i^5{\rho}_m \end{equation*}

(5.7)   \begin{equation*} N_{Re}=\frac{D_i^2N{\rho}_M}{{\mu}_M} \end{equation*}

(5.8)   \begin{equation*} {\rho}_M={\rho}_C{\Phi}_C+{\rho}_D{\Phi}_D \end{equation*}

(5.9)   \begin{equation*} {\mu}_M=\frac{{\mu}_C}{{\Phi}_C}\left[1+\frac{1.5{\mu}_D{\Phi}_D}{{\mu}_C+{\mu}_D}\right] \end{equation*}

Modeling Mass Transfer in Mixer-Settler Units

\Delta\rho = density difference (absolute value) between the continuous and dispersed phases (mass volume-1)

\phi_C = volume fraction occupied by the continuous phase

\phi_D = volume fraction occupied by the dispersed phase

\mu_C = viscosity of the continuous phase (mass time-1 length-1)

\mu_D = viscosity of the dispersed phase (mass time-1 length-1)

\mu_M = viscosity of the mixture (mass time-1 length-1)

\rho_C = density of the continuous phase (mass volume-1)

\rho_D = density of the dispersed phase (mass volume-1)

\rho_M = average density of the mixture (mass volume-1)

\sigma = interfacial tension between the continuous and dispersed phases
(mass time-2)

a = interfacial area between the two phases per unit volume (area volume-1)

c_{D,\rm in}, c_{D,\rm out} = concentration of solute in the incoming or outgoing dispersed streams (mass volume-1)

c^*_D = concentration of solute in the dispersed phase if in equilibrium with the outgoing continuous phase (mass volume-1)

D_C = diffusivity of the solute in the continuous phase (area time-1)

D_D = diffusivity of the solute in the dispersed phase (area time-1)

D_i = impeller diameter (length)

D_T = vessel diameter (length)

d_{vs} = Sauter mean droplet diameter; actual drop size expected to range from 0.3d_{vs}-3.0d_{vs} (length)

E_{MD} = Murphree dispersed-phase efficiency for extraction

g = gravitational constant (length time-2)

H = total height of mixer unit (length)

k_c = mass transfer coefficient of the solute in the continuous phase (length time-1)

k_D = mass transfer coefficient of the solute in the dispersed phase (length time-1)

K_{OD} = overall mass transfer coefficient, given on the basis of the dispersed phase (length time-1)

m = distribution coefficient of the solute, \Delta c_C/\Delta c_D (unitless)

N = rate of impeller rotation (time-1)

(N_{\rm Eo})_C = Eotvos number = gravitational force/surface tension force (unitless)

(N_{\rm Fr})_C = Froude number in the continuous phase = inertial force/gravitational force (unitless)

N_{\rm min} = minimum impeller rotation rate required for complete dispersion of one liquid into another

(N_{\rm Re})_C = Reynold’s number in the continuous phase = inertial force/viscous force (unitless)

(N_{\rm Sh})_C = Sherwood number in the continuous phase = mass transfer rate/diffusion rate (unitless)

(N_{\rm Sc})_C = Schmidt number in the continuous phase = momentum/mass diffusivity (unitless)

(N_{\rm We})_C = Weber number = inertial force/surface tension (unitless)

Q_D = volumetric flowrate of the dispersed phase (volume time-1)

V = vessel volume (volume)

 

Calculating N_{\rm min}

(6.1)   \begin{equation*} \frac{N_{\rm min}^2{\rho}_MD_i}{g{\Delta}{\rho}}=1.03\left(\frac{D_T}{D_i}\right)^{2.76}({\phi}_D)^{0.106}\left(\frac{{\mu}_M^2{\sigma}}{D_i^5{\rho}_Mg^2({\Delta}{\rho})^2}\right)^{0.084} \end{equation*}

(6.2)   \begin{equation*} {\rho}_M={\rho}_C{\phi}_C+{\rho}_D{\phi}_D \end{equation*}

(6.3)   \begin{equation*} {\mu}_M=\frac{{\mu}_C}{{\phi}_C}\left(1+\frac{1.5{\mu}_D{\phi}_D}{{\mu}_C+{\mu}_D}\right) \end{equation*}

 

Estimating Murphree efficiency for a proposed design

Sauter mean diameter

(6.4)   \begin{equation*} {\rm if}\;\; N_{\rm We} < 10,000,\; d_{vs}=0.052D_i(N_{\rm We})^{-0.6}\exp({4{\phi}_D}) \end{equation*}

(6.5)   \begin{equation*} {\rm if}\;\; N_{\rm We} >10,000,\; d_{vs}=0.39D_i(N_{\rm We})^{-0.6} \end{equation*}

(6.6)   \begin{equation*} N_{\rm We}=\frac{D_i^3N^2{\rho}_C}{\sigma} \end{equation*}

mass transfer coefficient of the solute in each phase

(6.7)   \begin{equation*} k_D=\frac{6.6D_D}{d_{vs}} \end{equation*}

(6.8)   \begin{equation*} k_C=\frac{(N_{\rm Sh})_CD_c}{d_{vs}} \end{equation*}

(6.9)   \begin{align*} & (N_{\rm Sh})_C =  1.237\times 10^{-5}(N_{\rm Sc})_C^{1/3}(N_{\rm Re})_C^{2/3}(\phi_D)^{-1/2}\\ & (N_{\rm Fr})_C^{5/12}\left(\frac{D_i}{d_{vs}}\right)^2 \left(\frac{d_{vs}}{D_T}\right)^{1/2}(N_{\rm Eo})_C^{5/4} \end{align*}

(6.10)   \begin{equation*} (N_{\rm Sc})_C=\frac{{\mu}_C}{{\rho}_CD_C} \end{equation*}

(6.11)   \begin{equation*} (N_{\rm Re})_C=\frac{D_i^2N{\rho}_C}{{\mu}_C} \end{equation*}

(6.12)   \begin{displaymath} (N_{\rm Fr})_C=\frac{D_iN^2}{g} \end{displaymath}

(6.13)   \begin{equation*} (N_{\rm Eo})_C=\frac{{\rho}_Dd_{vs}^2g}{{\sigma}} \end{equation*}

Overall mass transfer coefficient for the solute

(6.14)   \begin{equation*} \frac{1}{K_{OD}}=\frac{1}{k_D}+\frac{1}{mk_C} \end{equation*}

Murphree efficiency

(6.15)   \begin{equation*} E_{MD}=\frac{K_{OD}aV}{Q_D}\left(1+{\frac{K_{OD}aV}{Q_D}}\right)^{-1} \end{equation*}

(6.16)   \begin{equation*} a=\frac{6\phi_D}{d_{vs}} \end{equation*}

Experimental assessment of efficiency

(6.17)   \begin{equation*} E_{MD}=\frac{c_{D,\rm in}-c_{D,\rm out}}{c_{D,\rm in}-c^*_D} \end{equation*}

 

 

Example

1000 kg/hr of 30 wt% acetone and 70 wt% water is to be extracted with 1000 kg/hr of pure MIBK. Assume that the extract is the continuous phase, a residence time of 5 minutes in the mixing vessel, standard sizing of the mixing vessel and impeller. Find the power consumption and Murphree efficiency if the system operates at N_{\rm min}, controlled at the level of 1 rev/s. Ignore the contribution of the solute and the co-solvent to the physical properties of each phase.

  • MIBK
    • density = 802 kg m-3
    • viscosity = 0.58 cP
    • diffusivity with acetone at 25°C = 2.90×10-9 m2 s-1
  • Water
    • density = 1000 kg m-3
    • viscosity = 0.895 cP
    • diffusivity with acetone at 25°C = 1.16×10-9 m2 s-1
  • The interfacial tension of water and MIBK at 25°C = 0.0157 kg s-2. Use the ternary phase diagram to find m.

Liquid-Liquid Extraction Columns

\Delta \rho = density difference (absolute value) between the continuous and dispersed phases (mass volume-1)

\mu_C = viscosity of the continuous phase (mass time-1 length-1)

\rho_C = density of the continuous phase (mass volume-1)

\rho_D = density of the dispersed phase (mass volume-1)

\sigma = interfacial tension between the continuous and dispersed phases
(mass time-2)

 

D_T = column diameter (length)

H = total height of column (length)

{\rm HETS} = height of equilibrium transfer stage (length)

m^*_C = mass flowrate of the entering continuous phase (mass time-1)

m^*_D = mass flowrate of the entering dispersed phase (mass time-1)

N = required number of equilibrium stages

u_0 = characteristic rise velocity of a droplet of the dispersed phase (length time-1)

U_i = superficial velocity of phase i (C = continuous, downward; D = dispersed, upward)  (length time-1)

V^*_i = volumetric flowrate of phase i (volume time-1)

 

(7.1)   \begin{displaymath} U_i=\frac{4V^*_i}{{\pi}D_T^2} \end{displaymath}

definition of superficial velocity

 

(7.2)   \begin{displaymath} \frac{U_D}{U_C}=\frac{m^*_D}{m^*_C}\left(\frac{{\rho}_C}{\rho_D}\right) \end{displaymath}

(7.3)   \begin{displaymath} (U_D+U_C)_{\rm actual}=0.50(U_D+U_C)_f \end{displaymath}

for operation at 50% of flooding

(7.4)   \begin{displaymath} u_0=\frac{0.01{\sigma}{\Delta}{\rho}}{{\mu}_C{\rho}_C} \end{displaymath}

for rotating-disk columns, D_T = 8 to 42 inches, with one aqueous phase

 

(7.5)   \begin{displaymath} D_T=\left(\frac{4m^*_D}{{\rho}_DU_D{\pi}}\right)^{0.5}=\left(\frac{4m^*_C}{{\rho}_CU_C{\pi}}\right)^{0.5} \end{displaymath}

(7.6)   \begin{displaymath} H = {\rm HETS}*N \end{displaymath}

 

 

 

Example

1000 kg/hr of 30 wt% acetone and 70 wt% water is to be extracted with 1000 kg/hr of pure MIBK in a 2-stage column process. Assume that the extract is the dispersed phase. Ignoring the contribution of the solute and the co-solvent to the physical properties of each phase, find the required column diameter and height.

  • MIBK
    • density = 802 kg m-3
    • viscosity = 0.58 cP
  • Water
    • density = 1000 kg m-3
    • viscosity = 0.895 cP
  • The interfacial tension of water and MIBK at 25°C = 0.0157 kg s-2.

 

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Chemical Engineering Separations: A Handbook for Students Copyright © 2021 by Monica H. Lamm and Laura R. Jarboe is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License, except where otherwise noted.