Mass Transfer

Abstract

This chapter on Mass Transfer starts with an introduction in Sect. 5.1, followed by Sect. 5.2 on molecular diffusion. Section 5.3 provides detailed coverage of interphase mass transfer coefficients. Section 5.4 covers gas absorption including liquid–gas ratios, and height of transfer units (HTU), number of transfer units (NTU). A comprehensive coverage of distillation including all aspects of the McCabe–Thiele method is provided in Sect. 5.5. Section 5.6 includes definition of terms used in psychrometric charts and illustrates the use of psychrometric charts in various applications while Sect. 5.7 covers cooling towers. Section 5.8 looks at drying operations with illustrations of detailed calculations related to drying. Section 5.9 covers the unit operation of liquid–liquid extraction. Both single-stage and multistage leaching operations are examined in Sect. 5.10. Section 5.11 covers the topic of adsorption of fluids using solid adsorbents. Crystallization and related calculations are examined in Sect. 5.12. Section 5.13 covers the topic of evaporation including single-effect and multiple-effect evaporators.

Appendices

Practice Problems

5.1.1 Practice Problem 5.1

In an absorption tower, sulfur dioxide is being absorbed by water. At a particular location in the tower, the partial pressure of sulfur dioxide in the air stream was measured to be 45 mm Hg, and concentration of sulfur dioxide in water was determined to be 0.0385 lbmol SO₂/ft³soln. The tower operating conditions are

1 atm and 70 °F. The individual mass transfer coefficients are:

$$ {\displaystyle \begin{array}{l}{k}_L=0.985\ \textrm{lbmol}\ {\textrm{SO}}_2/\textrm{hr}\hbox{-} {\textrm{ft}}^2\hbox{-} \left(\textrm{lbmol}\ {\textrm{SO}}_2/{\textrm{ft}}^3\textrm{soln}\right)\ \textrm{and}\\ {}{k}_G=0.195\ \textrm{lbmol}\ {\textrm{SO}}_2/\textrm{hr}\hbox{-} {\textrm{ft}}^2\hbox{-} \textrm{atm}\end{array}} $$

The equilibrium relationship is

$$ {P}_{Ai}=0.137{C}_{Ai},{P}_{Ai}\left(\textrm{atm}\right),{C}_{Ai}\left(\textrm{lbmol}\ {\textrm{SO}}_2/{\textrm{ft}}^3\textrm{soln}\right). $$

Determine:

• A. the percentage of the total resistance in the liquid film and comment on the result obtained.

• B. ${P}_{AG}-{P}_A^{\ast }$

• C. ${C}_A^{\ast }-{C}_{AL}$

• D. P_AG − P_Ai, and C_Ai − C_AL

5.1.2 Practice Problem 5.2

Carbon dioxide is absorbed from an air stream using a pure amine solution (molecular weight 32 kg/kmol) as a solvent. The amine solution is fed from the top of the tower (diameter 0.85 m) at a mass flow rate of 0.8 kg/s. The gas mixture containing 6 mole% carbon dioxide enters from the bottom of the tower at a molar flow rate of 0.0045 kmol/s. The desired mole fraction of carbon dioxide in the exit air stream is 1.25%. The equilibrium relationship is Y_A1 = 5.95X_A1 where Y_A1 is the mole ratio of carbon dioxide to air in the gas stream and X_A1 is the mole ratio of carbon dioxide to amine in the liquid stream. Determine:

• A. the mole fraction of carbon dioxide in the amine solution leaving the tower

• B. the ratio ${\left(\frac{L_S}{G_S}\right)}_{\textrm{actual}}\textrm{to}\ {\left(\frac{L_S}{G_S}\right)}_{\textrm{min}}$

5.1.3 Practice Problem 5.3

Given the McCabe–Thiele diagram for a binary distillation system, label all the lines/curves in the diagram with standard nomenclature and determine:

• A. the minimum reflux ratio.

• B. the ratio between the actual reflux ratio and the minimum reflux ratio.

• C. the value of q and the feed condition.

• D. the liquid to vapor ratio, that is, L/V, in the rectifying section.

• E. the boil-up ratio, that is, V^′/B referring to the reboiler.

5.1.4 Practice Problem 5.4

A wet solid with a mass of 5 kg and 10% moisture content is dried in a batch drier to final moisture content of 0.05 kg water/kg dry solid, a process which takes 4.8 hrs. The surface area of the wet solid is 1.8 m² and the equilibrium moisture content is 0.02 kg water/kg dry solid. Assuming a constant rate of drying, determine the drying rate in kg/m².min.

5.1.5 Practice Problem 5.5

The rate of water removal during the constant drying period of 0.73 hr of a wet solid is 8 lbm H₂O/min-ft². During this period, the moisture content is reduced from an initial value of 0.35 lbm H₂O/lbm dry solid to critical value of 0.14 lbm H₂O/lbm dry solid. The surface area of the wet solid is 2.25 ft². The equilibrium moisture content is 0.04 lbm H₂O/lbm dry solid. Determine:

• A. the mass of the dry solid.

• B. the total drying time required to achieve a final moisture content of 0.07 lbm H₂O/lbm dry solid.

5.1.6 Practice Problem 5.6

A dryer uses hot air to dry wet feed with 20% moisture content on a wet basis to a final moisture content of 10%, also on a wet basis. To accomplish this, 4225 cfm (cubic feet per minute) of air at 50 °F and 50% relative humidity is preheated to 120 °F before being fed to the dryer. The air leaves the dryer at 85 °F. Determine the mass flow rate (lbm/min) of wet solid that can be processed by the dryer.

5.1.7 Practice Problem 5.7

It is desired to extract ethylene glycol (E) from 250 lbm of a feed solution containing 60% (by mass) of ethylene glycol and 40% water (W). The solvent used is 150 lbm of pure furfural (F). The relevant equilibrium ternary phase diagram is given below.

Determine:

• A. the composition of the mixture.

• B. the mass and composition of the extract and the raffinate.

• C. the fraction of the solute recovered (also known as recovery)

5.1.8 Practice Problem 5.8

Oil is extracted from a solid substrate fed to a multistage leaching system at the rate of 700 kg/hr. The solids contain 28% oil, and the rest are insoluble solids. Pure organic solvent is fed in a direction counter to the flow of the solids. The requirements specified for the leaching system are as follows:

• The solids leaving the system should contain no more than 40 kg/hr of unextracted oil.

• The underflow from each stage has 2 kg inerts/kg solution.

• The mass fraction of the solute in the product overflow solution should be no less than 0.20.

Determine:

• A. The mass flow rate and composition of the underflow solution.

• B. The mass flow rate of the overflow solution.

• C. The mass flow rate of the pure solvent required.

• D. The number of stages required assuming a straight line equilibrium relationship, y = x.

5.1.9 Practice Problem 5.9

Methane is adsorbed by activated carbon from a hydrocarbon gas mixture at atmospheric pressure and 20 °C containing 2 mole% methane. The following equilibrium data is available, where,

PCH4 (kPa)	20	40	60	80	100
mS/mA(g/g)	0.13	0.24	0.35	0.42	0.50

$$ {\displaystyle \begin{array}{l}{P}_{{\textrm{CH}}_4}=\textrm{partial}\ \textrm{pr}.\textrm{of}\ {\textrm{CH}}_4,\textrm{kPa},\textrm{and}\\ {}\left({m}_S/{m}_A\right)=\textrm{g}\ {\textrm{CH}}_4\ \textrm{adsorbed}/\textrm{g}\ \textrm{AC}\end{array}} $$

Determine the Freundlich relationship and the volume (L) of methane adsorbed by 100 g of adsorbent.

Solutions to Practice Problems

5.1.1 Practice Problem 5.1

5.1.1.1 Solution

Determine m from the given equilibrium relationship, P_Ai = 0.137C_Ai, ⇒

$$ m=0.137\ \textrm{atm}/\left(\textrm{lbmol}\ {\textrm{SO}}_2/{\textrm{ft}}^3\textrm{soln}\right) $$

• A. Determine the overall liquid mass transfer resistance using Eq. 5.5 and the known values.

$$ {\displaystyle \begin{array}{l}\frac{1}{K_L}=\frac{1}{k_L}+\frac{1}{mk_G}=\frac{1}{0.985\frac{\textrm{lbmol}}{\textrm{hr}-{\textrm{ft}}^2\left(\textrm{lbmol}\ {\textrm{SO}}_2/{\textrm{ft}}^3\textrm{soln}\right)}}\\ {}\kern2.75em +\frac{1}{\left(0.137\frac{\textrm{atm}\ }{\textrm{lbmol}\ {\textrm{SO}}_2/{\textrm{ft}}^3\textrm{soln}}\right)\left(0.195\frac{\textrm{lbmol}}{\textrm{hr}-{\textrm{ft}}^2\left(\textrm{atm}\right)}\right)}\\ {}\kern7em =38.447\ \left(\textrm{lbmol}\ {\textrm{SO}}_2/{\textrm{ft}}^3\textrm{soln}\right)/\left(\textrm{lbmol}/\textrm{hr}\hbox{-} {\textrm{ft}}^2\right)\Rightarrow \\ {}{K}_L=0.0260\ \textrm{lbmol}/\textrm{hr}\hbox{-} {\textrm{ft}}^2/\left(\textrm{lbmol}\ {\textrm{SO}}_2/{\textrm{ft}}^3\textrm{soln}\right)\end{array}} $$

Determine the resistance in the liquid film.

$$ {\displaystyle \begin{array}{l}\frac{1}{k_L}=\frac{1}{0.985\frac{\textrm{lbmol}}{\textrm{hr}-{\textrm{ft}}^2\left(\textrm{lbmol}\ {\textrm{SO}}_2/{\textrm{ft}}^3\textrm{soln}\right)}}\\ {}=1.015\ \left(\textrm{lbmol}\ {\textrm{SO}}_2/{\textrm{ft}}^3\textrm{soln}\right)/\left(\textrm{lbmol}/\textrm{hr}\hbox{-} {\textrm{ft}}^2\right)\end{array}} $$

Calculate the percentage of the total (overall) resistance in the liquid film.

$$ {\displaystyle \begin{array}{l}\%\textrm{resist}.\textrm{in}\ \textrm{liq}\ \textrm{film}=\frac{\frac{1}{k_L}}{\frac{1}{K_L}}\times 100\\ {}=\left(\frac{1.015\frac{\textrm{lbmol}\ {\textrm{SO}}_2/{\textrm{ft}}^3\textrm{soln}}{\textrm{lbmol}/\textrm{hr}\hbox{-} {\textrm{ft}}^2}}{38.447\frac{\textrm{lbmol}\ {\textrm{SO}}_2/{\textrm{ft}}^3\textrm{soln}}{\textrm{lbmol}/\textrm{hr}\hbox{-} {\textrm{ft}}^2}}\right)(100)\\ {}=2.64\%\end{array}} $$

Note: The result obtained indicates that approximately 97% of the total resistance to mass transfer is in the gas film, implying a gas-phase controlled mass-transfer process.

• B. Calculate ${P}_A^{\ast }$ using Eq. 5.8.

$$ {\displaystyle \begin{array}{l}{P}_A^{\ast }={mC}_{AL}\\ {}=\left(0.137\frac{\textrm{atm}}{\frac{\textrm{lbmol}\ {\textrm{SO}}_2}{{\textrm{ft}}^3\textrm{soln}}}\right)\left(0.0385\frac{\textrm{lbmol}\ {\textrm{SO}}_2}{{\textrm{ft}}^3\textrm{soln}}\right)\\ {}=0.0053\ \textrm{atm}\end{array}} $$

Determine ${P}_{AG}-{P}_A^{\ast }$.

$$ {\displaystyle \begin{array}{l}{P}_{AG}-{P}_A^{\ast }=\left(45\ \textrm{mm}\ \textrm{Hg}\times \frac{1\ \textrm{atm}}{760\ \textrm{mm}\ \textrm{Hg}}\right)-0.0053\ \textrm{atm}\\ {}=0.0539\ \textrm{atm}\end{array}} $$

• C. Determine ${C}_A^{\ast }$ using Eq. 5.9.

$$ {C}_A^{\ast }=\frac{P_{AG}}{m}=\frac{45\ \textrm{mm}\ \textrm{Hg}\times \frac{1\ \textrm{atm}}{760\ \textrm{mm}\ \textrm{Hg}}}{0.137\frac{\textrm{atm}}{\frac{\textrm{lbmol}\ {\textrm{SO}}_2}{{\textrm{ft}}^3\textrm{soln}}}}=0.4322\ \textrm{lbmol}\ {\textrm{SO}}_2/{\textrm{ft}}^3\textrm{soln} $$

Calculate ${C}_A^{\ast }-{C}_{AL}$.

$$ {\displaystyle \begin{array}{l}{C}_A^{\ast }-{C}_{AL}=0.4322\frac{\textrm{lbmol}\ {\textrm{SO}}_2}{{\textrm{ft}}^3\textrm{soln}}-0.0385\frac{\textrm{lbmol}\ {\textrm{SO}}_2}{{\textrm{ft}}^3\textrm{soln}}\\ {}=0.3937\ \textrm{lbmol}\ {\textrm{SO}}_2/{\textrm{ft}}^3\textrm{soln}\end{array}} $$

Alternately, ${C}_A^{\ast }-{C}_{AL}$ can be calculated using Eq. 5.7, that is, ${N}_A={K}_G\left({P}_{AG}-{P}_A^{\ast}\right)={K}_L\left({C}_A^{\ast }-{C}_{AL}\right)$. However, this approach requires the calculation of K_G first using Eq. 5.4, that is, $\frac{1}{K_G}=\frac{1}{k_G}+\frac{m}{k_L}$. Obviously, this would be a more laborious process.

• D. To calculate P_AG − P_Ai, and C_Ai − C_AL, first calculate the molar flux using Eq. 5.7 and the result for ${C}_A^{\ast }-{C}_{AL}$ obtained in the solution to part “C.”

$$ {\displaystyle \begin{array}{l}{N}_A={K}_L\left({C}_A^{\ast }-{C}_{AL}\right)\\ {}=\left(0.0260\frac{\frac{\textrm{lbmol}}{\textrm{hr}\hbox{-} {\textrm{ft}}^2}}{\frac{\textrm{lbmol}\ {\textrm{SO}}_2}{{\textrm{ft}}^3\textrm{soln}}}\right)\left(0.3937\frac{\textrm{lbmol}\ {\textrm{SO}}_2}{{\textrm{ft}}^3\textrm{soln}}\right)\\ {}=0.0102\ \textrm{lbmol}/\textrm{hr}\hbox{-} {\textrm{ft}}^2\end{array}} $$

Determine P_AG − P_Ai, and C_Ai − C_AL using Eq. 5.3.

$$ {\displaystyle \begin{array}{l}{N}_A={k}_G\left({P}_{AG}-{P}_{Ai}\right)={k}_L\left({C}_{Ai}-{C}_{AL}\right)\Rightarrow \\ {}{P}_{AG}-{P}_{Ai}=\frac{N_A}{k_G}=\frac{0.0102\frac{\textrm{lbmol}}{\textrm{hr}-{\textrm{ft}}^2}}{0.195\frac{\frac{\textrm{lbmol}}{\textrm{hr}-{\textrm{ft}}^2}}{\textrm{atm}}}=0.0523\ \textrm{atm}\\ {}{C}_{Ai}-{C}_{AL}=\frac{N_A}{k_L}=\frac{0.0102\frac{\textrm{lbmol}}{\textrm{hr}-{\textrm{ft}}^2}}{0.985\frac{\frac{\textrm{lbmol}}{\textrm{hr}-{\textrm{ft}}^2}}{\frac{\textrm{lbmol}\ {\textrm{SO}}_2}{{\textrm{ft}}^3\textrm{soln}}}}=0.0104\ \textrm{lbmol}\ {\textrm{SO}}_2/{\textrm{ft}}^3\textrm{soln}\end{array}} $$

5.1.2 Practice Problem 5.2

5.1.2.1 Solution

Draw the schematic diagram for the process with nomenclature consistent with that used in Fig. 5.3.

• A. Calculate the cross-section area of the tower.

$$ {A}_{CS}=\left(\frac{\pi }{4}\right){\left(0.85\ \textrm{m}\right)}^2=0.5674\ {\textrm{m}}^2 $$

Calculate the molar flux of the gas entering the tower on a solute-free basis.

$$ {\displaystyle \begin{array}{l}{G}_S={G}_1\left(1-{y}_{a1}\right)=\left(\frac{0.0045\frac{\textrm{kmol}\ \textrm{mix}}{\textrm{s}}}{0.5674\ {\textrm{m}}^2}\right)\left(\left(1-0.06\right)\frac{\textrm{kmol}\ \textrm{air}}{\textrm{kmol}\ \textrm{mix}}\right)\\ {}=0.0074\ \textrm{kmol}\ \textrm{air}/\textrm{s}\hbox{-} {\textrm{m}}^2\end{array}} $$

Calculate the molar flux of amine solvent entering the absorption tower using its molecular weight and cross section area of the tower. Since pure solvent is entering the tower, L_S = L₁ and X_A2 = 0.

$$ {\displaystyle \begin{array}{l}{L}_S={L}_1=\frac{{\dot{m}}_{a\min e}}{M_{a\min e}\times {A}_{cs}}=\frac{0.8\frac{\textrm{kg}}{\textrm{s}}}{\left(32\frac{\textrm{kg}}{\textrm{kmol}}\right)\left(0.5674\ {\textrm{m}}^2\right)}\\ {}=0.0441\ \textrm{kmol}/\textrm{s}\hbox{-} {\textrm{m}}^2\end{array}} $$

Calculate Y_A1 and Y_A2 using Eq. 5.11.

$$ {\displaystyle \begin{array}{l}{Y}_{A1}=\frac{y_{A1}}{1-{y}_{A1}}=\frac{0.06}{1-0.06}=0.064\\ {}{Y}_{A2}=\frac{y_{A2}}{1-{y}_{A2}}=\frac{0.0125}{1-0.0125}=0.0127\end{array}} $$

Calculate the mole fraction of carbon dioxide in the amine leaving the tower using Eq. 5.10 after first calculating the mole ratio, moles of carbon dioxide to moles of amine in the liquid using the material balance equation (Eq. 5.16).

$$ {X}_{A1}=\frac{x_A}{1-{x}_A}\Rightarrow 0.0086=\frac{x_{A1}}{1-{x}_{A1}}\Rightarrow {x}_{A1}=0.0085 $$

B. Determine the exit mole ratio of ammonia solute to water in equilibrium $\left({X}_{A1}^{\ast}\right)$ with the entrance mole ratio of ammonia to air (Y_A1) using the given equilibrium relationship.

$$ {\displaystyle \begin{array}{l}{Y}_{A1}=5.95{X}_{A1}^{\ast}\Rightarrow \\ {}{X}_{A1}^{\ast }=\frac{Y_{A1}}{5.95}=\frac{0.064}{5.95}=0.0108\end{array}} $$

Calculate (L_S/G_S)_min using Eq. 5.18.

$$ {\left(\frac{L_S}{G_S}\right)}_{\textrm{min}}=\frac{\left({Y}_{A1}-{Y}_{A2}\right)}{\left({X}_{A1}^{\ast }-{X}_{A2}\right)}=\frac{0.064-0.0127}{0.0108-0}=4.75 $$

Calculate (L_S/G_S)_actual by substituting the known values.

$$ {\left(\frac{L_S}{G_S}\right)}_{\textrm{actual}}=\frac{0.0441\frac{\textrm{kmol}\ \textrm{amine}}{\textrm{s}\hbox{-} {\textrm{m}}^2}}{0.0074\frac{\textrm{kmol}\ \textrm{air}}{\textrm{s}\hbox{-} {\textrm{m}}^2}}=5.96 $$

Calculate the ratio ${\left(\frac{L_S}{G_S}\right)}_{\textrm{actual}}/\kern0.5em {\left(\frac{L_S}{G_S}\right)}_{\textrm{min}}.$

$$ \textrm{Required}\ \textrm{Ratio}=\frac{{\left(\frac{L_S}{G_S}\right)}_{\textrm{actual}}}{{\left(\frac{L_S}{G_S}\right)}_{\textrm{min}}}=\frac{5.96}{4.75}=1.25 $$

5.1.3 Practice Problem 5.3

5.1.3.1 Solution

All the relevant lines/curves have been labeled as shown in the figure.

• A. From the diagram, the compositions of feed, distillate, and the bottom product are 0.42, 0.93, and 0.06, respectively, that is,

$$ {x}_F=0.42,{x}_D=0.93,\textrm{and}\kern0.5em {x}_B=0.06, $$

The y-intercept of the R_min-line is 0.47. Therefore, from Eq. 5.34,

$$ \frac{x_D}{R_{\textrm{min}}+1}=0.47\Rightarrow \frac{0.93}{R_{\textrm{min}}+1}=0.47\Rightarrow {R}_{\textrm{min}}=0.98 $$

• B. Extend the rectification line to the y-axis as shown in the figure. This line is the R_actual-line. From the diagram, the y-intercept of the R_actual-line is 0.32.

Therefore, from Eq. 5.34,

$$ \frac{x_D}{R_{\textrm{actual}}+1}=0.32\Rightarrow \frac{0.93}{R_{\textrm{actual}}+1}=0.32\Rightarrow {R}_{\textrm{actual}}=1.91 $$

Therefore, the required ratio of actual reflux to minimum reflux is

$$ \textrm{Ratio}=\frac{R_{\textrm{actual}}}{R_{\textrm{min}}}=\frac{1.91}{0.98}=1.95 $$

• C. From the figure, determine the slope of the q-line.

$$ {\textrm{Slope}}_{q-\textrm{line}}=\frac{\Delta y}{\Delta x}=\frac{28\ \textrm{units}}{-35\ \textrm{units}}=-0.80 $$

From Eq. 5.43,

$$ \textrm{slope}\ \textrm{of}\ q\hbox{-} \textrm{line}=\frac{q}{q-1}=-0.80\Rightarrow q=0.44 $$

From Fig. 5.7, since 0 < q < 1, the feed condition is liquid–vapor mixture.

• D. From material balance for the column, Eq. 5.33 is

$$ {y}_{n+1}=\left(\frac{L_n}{L_n+D}\right){x}_n+\left(\frac{D}{L_n+D}\right){x}_D\kern0.36em \textrm{and}\kern0.36em {L}_n+D={V}_{n+1} $$

Equation 5.34 for the operating line for the rectification section is

$$ {y}_{n+1}=\left(\frac{R}{R+1}\right){x}_n+\frac{x_D}{R+1} $$

(5.34)

Assuming constant molal overflow, the suffixes can be dropped, the preceding equations can be combined to obtain the following result.

$$ {\displaystyle \begin{array}{l}y=\left(\frac{L}{V}\right)x+\left(\frac{D}{L_n+D}\right){x}_D=\left(\frac{R}{R+1}\right)x+\frac{x_D}{R+1}\Rightarrow \\ {}\frac{L}{V}=\frac{R_{\textrm{actual}}}{R_{\textrm{actual}}+1}=\frac{1.91}{1.91+1}=0.66\end{array}} $$

E. The slope of the operating line for the strip** section can be obtained from the diagram and then the material balance relationships (Eqs. 5.36, 5.37, and 5.38) for the strip** section can be used to obtain the following results and eventually the ratio, V′/B.

$$ {\displaystyle \begin{array}{l}\textrm{Slope}=\frac{\Delta y}{\Delta x}=\frac{16\ \textrm{units}}{9\ \textrm{units}}=1.78=\frac{L^{\prime }}{L^{\prime }-B}=\frac{V^{\prime }+B}{V^{\prime }}=\frac{\frac{V^{\prime }}{B}+1}{\frac{V^{\prime }}{B}}\Rightarrow \\ {}\frac{V^{\prime }}{B}+1=(1.78)\left(\frac{V^{\prime }}{B}\right)\Rightarrow 0.78\left(\frac{V^{\prime }}{B}\right)=1\Rightarrow \\ {}\frac{V^{\prime }}{B}=1.28\end{array}} $$

5.1.4 Practice Problem 5.4

5.1.4.1 Solution

First, calculate the mass of completely dry solid, designated as “dry solid.”

$$ {\displaystyle \begin{array}{l}{m}_S=\left(5\ \textrm{kg}\ \textrm{wet}\ \textrm{solid}\right)\left(0.90\frac{\textrm{kg}\ \textrm{dry}\ \textrm{solid}}{\textrm{kg}\ \textrm{wet}\ \textrm{solid}}\right)\\ {}=4.5\ \textrm{kg}\ \textrm{dry}\ \textrm{solid}\end{array}} $$

Express the initial moisture content in terms of kg H₂O/kg dry solid. Therefore,

$$ {\displaystyle \begin{array}{l}{X}_i=\left(\frac{5\ \textrm{kg}\ \textrm{wet}\ \textrm{solid}}{4.5\ \textrm{kg}\ \textrm{dry}\ \textrm{solid}}\right)\left(\frac{0.10\ \textrm{kg}\ {\textrm{H}}_2\textrm{O}}{\textrm{kg}\ \textrm{wet}\ \textrm{solid}}\right)\\ {}=0.1111\ \textrm{kg}\ {\textrm{H}}_2\textrm{O}/\textrm{kg}\ \textrm{dry}\ \textrm{solid}\end{array}} $$

Calculate the constant drying rate using Eq. 5.50a. Substitute X_f in place of X_c

$$ {\displaystyle \begin{array}{l}{R}_c=\frac{m_s\left({X}_i-{X}_f\right)}{At_c}\\ {}=\frac{\left(4.5\ \textrm{kg}\ \textrm{dry}\ \textrm{solid}\right)\left(0.1111\frac{\textrm{kg}\ {\textrm{H}}_2\textrm{O}}{\textrm{kg}\ \textrm{dry}\ \textrm{solid}}-0.05\frac{\textrm{kg}\ {\textrm{H}}_2\textrm{O}}{\textrm{kg}\ \textrm{dry}\ \textrm{solid}}\right)}{\left(1.8\ {\textrm{m}}^2\right)\left(4.8\ \textrm{hrs}\right)}\\ {}=0.0318\ \textrm{kg}\ {\textrm{H}}_2\textrm{O}/{\textrm{m}}^2\cdot \textrm{hr}\end{array}} $$

5.1.5 Practice Problem 5.5

5.1.5.1 Solution

• A. Using the given information and Eq. 5.50a for constant drying rate, determine the mass of the dry solid as shown.

$$ {\displaystyle \begin{array}{l}{R}_c=\frac{m_s\left({X}_i-{X}_c\right)}{At_c}\Rightarrow \\ {}{m}_s=\frac{R_c{At}_c}{\left({X}_i-{X}_c\right)}\\ {}=\frac{\left(8.5\ \frac{\textrm{lbm}\ {\textrm{H}}_2\textrm{O}}{\min \hbox{-} {\textrm{ft}}^2}\right)\left(2.25\ {\textrm{ft}}^2\right)\left(0.73\ \textrm{hrs}\times \frac{60\ \min }{\textrm{hr}}\right)}{\left(0.35\frac{\textrm{lbm}\ {\textrm{H}}_2\textrm{O}}{\textrm{lbm}\ \textrm{dry}\ \textrm{solid}}-0.14\frac{\textrm{lbm}\ {\textrm{H}}_2\textrm{O}}{\textrm{lbm}\ \textrm{dry}\ \textrm{solid}}\right)}\\ {}=3989\ \textrm{lbm}\ \textrm{dry}\ \textrm{solid}\end{array}} $$

• B. Calculate the total drying time required using Eq. 5.54.

$$ {\displaystyle \begin{array}{l}{t}_{\textrm{total}}=\kern-0.9em \left(\frac{m_s}{AR_c}\right)\left[\left({X}_i-{X}_c\right)+\left({X}_c-{X}_e\right)\ln \left(\frac{X_c-{X}_e}{X_f-{X}_e}\right)\right]\Rightarrow \\ {}=\kern-0.9em \left(\frac{3989\ \textrm{lbm}\ \textrm{dry}\ \textrm{solid}}{\left(2.25\ {\textrm{ft}}^2\right)\left(8.5\ \frac{\textrm{lbm}\ {\textrm{H}}_2\textrm{O}}{\min \hbox{-} {\textrm{ft}}^2}\right)}\right)\left[\begin{array}{l}\left(0.35-0.14\right)+\left(0.14-0.04\right)\times \\ {}\kern4.75em \ln \left(\frac{0.14-0.04}{0.07-0.04}\right)\end{array}\right]\\ {}=\kern-0.8em 68.91\ \min \left(1.1485\ \textrm{hrs}\right)\end{array}} $$

• Note: The units for moisture content are lbm H₂O/lbm dry solid, and they are not shown for the sake of the brevity of the presentation of the equation.

5.1.6 Practice Problem 5.6

5.1.6.1 Solution

A schematic diagram for the process is drawn as shown.

The entering solid has 20% moisture content on a wet basis. Using this information and assuming ${\dot{m}}_{\textrm{ws}}$ as the mass flow rate of the wet solid fed to the dryer, express the mass flow rate of bone-dry solid in terms of the mass flow rate of the wet solid fed to the dryer.

$$ {\displaystyle \begin{array}{l}{\dot{m}}_{\textrm{bds}}=\left(\frac{{\dot{m}}_{\textrm{ws}}\textrm{lbm}\ \textrm{wet}\ \textrm{solid}\ }{\textrm{hr}}\right)\left(\frac{0.8\ \textrm{lbm}\ \textrm{bone}\ \textrm{dry}\ \textrm{solid}}{\textrm{lbm}\ \textrm{wet}\ \textrm{solid}}\right)\\ {}=\left(0.8{\dot{m}}_{\textrm{ws}}\right)\ \textrm{lbm}\ \textrm{bone}\ \textrm{dry}\ \textrm{solid}/\textrm{hr}\end{array}} $$

The partially dried solid coming out of the dryer has a moisture content of 10%. Based on this information, express the mass flow rate of the partially dry solid in terms of the mass flow rate of the wet solid fed to the dryer.

$$ {\displaystyle \begin{array}{l}{\dot{m}}_{\textrm{pds}}=\left(\frac{\left(0.8{\dot{m}}_{\textrm{ws}}\right)\textrm{lbm}\ \textrm{bone}\ \textrm{dry}\ \textrm{solid}\ }{\textrm{hr}}\right)\left(\frac{1\ \textrm{lbm}\ \textrm{partially}\ \textrm{dry}\ \textrm{solid}}{0.90\ \textrm{lbm}\ \textrm{bone}\ \textrm{dry}\ \textrm{solid}}\right)\\ {}=\left(0.89{\dot{m}}_{\textrm{ws}}\right)\ \textrm{lbm}\ \textrm{partially}\ \textrm{dry}\ \textrm{solid}/\textrm{hr}\end{array}} $$

Obtain the properties of moist air by locating the state points of air on the psychrometric chart.

State point 1 (50 °F, 50% rh):

$$ {\omega}_1=0.004\ \textrm{lbm}\ {\textrm{H}}_2\textrm{O}/\textrm{lbm}\ \textrm{d}\textrm{ry}\ \textrm{a}\textrm{ir},{v}_1=12.9\ {\textrm{ft}}^3/\textrm{lbm}\ \textrm{d}\ \textrm{a} $$

State points 2 and 3 (120 °F, ω₁ = ω₂ = ω₃, sensible heating at const. moist.):

$$ {\omega}_1={\omega}_2={\omega}_3=0.004\ \textrm{lbm}\ {\textrm{H}}_2\textrm{O}/\textrm{lbm}\ \textrm{dry}\ \textrm{air} $$

To locate state point 4, move along the constant wet bulb temperature line (because of adiabatic saturation of air) from state point 3 to the given final dry bulb temperature of 85 °F. At state point 4,

$$ {\omega}_4=0.012\ \textrm{lbm}\ {\textrm{H}}_2\textrm{O}/\textrm{lbm}\ \textrm{dry}\ \textrm{air}. $$

Using the specific volume of the air entering the pre-heater, calculate the mass flow rate of dry air, ${\dot{m}}_a$, which remains constant throughout the processes in the pre-heater and dryer.

$$ {\dot{m}}_a=\frac{{\dot{V}}_1}{v_1}=\frac{4225\frac{{\textrm{ft}}^3}{\min }}{12.9\frac{{\textrm{ft}}^3}{\textrm{lbm}\ \textrm{d}\ \textrm{a}}}=327.52\ \textrm{lbm}\ \textrm{d}\ \textrm{a}/\min $$

The increase in moisture content of air due to the drying process is ${\dot{m}}_a\left({\omega}_4-{\omega}_3\right)$, where ω₃ and ω₄ are moisture contents in air before and after the drying process. Calculate the mass of water evaporated and absorbed by air by mass balance for water in the solids and hence calculate the mass flow rate of the wet solid fed to the dryer, ${\dot{m}}_{\textrm{ws}}$, by substituting all remaining known values.

$$ {\displaystyle \begin{array}{l}\textrm{mass}\ \textrm{of}\ \textrm{water}\ \textrm{vapor}\ \textrm{absorbed}\ \textrm{by}\ \textrm{air}\\ {}\kern7em =\textrm{mass}\ \textrm{of}\ \textrm{water}\ \textrm{evaporated}\ \textrm{from}\ \textrm{wet}\ \textrm{solid}\\ {}\kern7em =\textrm{mass}\ \textrm{of}\ \textrm{wet}\ \textrm{solid}\ \textrm{in}\hbox{--} \textrm{mass}\ \textrm{of}\ \textrm{partially}\ \textrm{dry}\ \textrm{solid}\ \textrm{out}\end{array}} $$

$$ {\displaystyle \begin{array}{l}{\dot{m}}_a\left({\omega}_4-{\omega}_3\right)={\dot{m}}_{\textrm{ws}}-{\dot{m}}_{\textrm{pds}}={\dot{m}}_{\textrm{ws}}-0.89{\dot{m}}_{\textrm{ws}}=0.11{\dot{m}}_{\textrm{ws}}\\ {}\Rightarrow {\dot{m}}_{\textrm{ws}}=\frac{{\dot{m}}_a\left({\omega}_4-{\omega}_3\right)}{0.11\frac{\textrm{lbm}\ {\textrm{H}}_2\textrm{O}\ \textrm{removed}}{\textrm{lbm}\ \textrm{wet}\ \textrm{solid}}}\\ {}=\frac{\left(327.52\frac{\textrm{lbm}\ \textrm{d}\ \textrm{a}}{\min}\right)\left(0.012\frac{\textrm{lbm}\ {\textrm{H}}_2\textrm{O}}{\textrm{lbm}\ \textrm{d}\ \textrm{a}}-0.004\frac{\textrm{lbm}\ {\textrm{H}}_2\textrm{O}}{\textrm{lbm}\ \textrm{d}\ \textrm{a}}\right)}{0.11\frac{\textrm{lbm}\ {\textrm{H}}_2\textrm{O}\ \textrm{removed}}{\textrm{lbm}\ \textrm{wet}\ \textrm{solid}}}\\ {}=23.82\ \textrm{lbm}\ \textrm{wet}\ \textrm{solid}/\min \end{array}} $$

5.1.7 Practice Problem 5.7

5.1.7.1 Solution

• A. Determine the composition of the mixture by using overall and component mass balances. FD represents the feed solution, F represents the furfural solvent, and M represents the mixture. Subscript F represents furfural, subscript W represents water, and subscript EG represents ethylene glycol.

Overall mass balance:

$$ {\displaystyle \begin{array}{l}\textrm{M}=\textrm{FD}+\textrm{F}=250\ \textrm{lbm}+150\ \textrm{lbm}=400\ \textrm{lbm}\\ {}\Rightarrow {x}_{\textrm{F},\textrm{M}}=\frac{\textrm{F}}{\textrm{M}}=\frac{150\ \textrm{lbm}}{400\ \textrm{lbm}}=0.375\end{array}} $$

Mass balance for water:

$$ {\displaystyle \begin{array}{l}\left(\textrm{FD}\right){x}_{\textrm{W},\textrm{F}\textrm{D}}+\textrm{F}{x}_{\textrm{W},\textrm{F}}=\textrm{M}{x}_{\textrm{W},\textrm{M}}\Rightarrow \\ {}{x}_{\textrm{W},\textrm{M}}=\frac{\left(\textrm{FD}\right){x}_{\textrm{W},\textrm{F}\textrm{D}}+\textrm{F}{x}_{\textrm{W},\textrm{F}}}{\textrm{M}}=\frac{\left(250\ \textrm{lbm}\right)(0.40)+\left(150\ \textrm{lbm}\right)(0)}{400\ \textrm{lbm}}=0.25\end{array}} $$

$$ {x}_{\textrm{EG},\textrm{M}}=1-{x}_{\textrm{F},\textrm{M}}-{x}_{\textrm{W},\textrm{M}}=1-0.375-0.25=0.375 $$

• B. Locate the mixture point, M, on the given ternary phase diagram using the preceding compositions. Then, draw the tie line through the mixture point, M, and determine the locations and compositions of the extract, E, and the raffinate, R, on the equilibrium isotherm as shown. The extract, being rich in the solvent, is located closer to the pure solvent on the equilibrium curve and the raffinate is located on the opposite side. On the diagram, EG represents ethylene glycol to make it distinct from E, which represents the extract, and FD represents the feed to make it distinct from F, which represents furfural.

From the phase diagram, the compositions of the extract and raffinate are

$$ {y}_{\textrm{EG},\textrm{E}}=0.19,{y}_{\textrm{W},\textrm{E}}=0.07,{y}_{\textrm{F},\textrm{E}}=0.74 $$

$$ {x}_{\textrm{EG},\textrm{R}}=0.49,{x}_{\textrm{W},\textrm{R}}=0.35,{x}_{\textrm{F},\textrm{R}}=0.16 $$

The amounts of the extract and raffinate can be determined from mass and component balances as shown.

Overall mass balance:

$$ \textrm{R}=\textrm{M}-\textrm{E}=400\ \textrm{lbm}-\textrm{E} $$

Mass balance for water:

$$ {\displaystyle \begin{array}{l}\textrm{E}{y}_{\textrm{W},\textrm{E}}+\textrm{R}{x}_{\textrm{W},\textrm{R}}=\textrm{M}{x}_{\textrm{W},\textrm{M}}\Rightarrow \textrm{E}{y}_{\textrm{W},\textrm{E}}+\left(400-\textrm{E}\right){x}_{\textrm{W},\textrm{R}}=400{x}_{\textrm{W},\textrm{M}}\Rightarrow \\ {}\textrm{E}=\frac{400\left({x}_{\textrm{W},\textrm{M}}-{x}_{\textrm{W},\textrm{R}}\right)}{y_{\textrm{W},\textrm{E}}-{x}_{\textrm{W},\textrm{R}}}=\frac{\left(400\ \textrm{lbm}\right)\left(0.25-0.35\right)}{0.07-0.35}=143\ \textrm{lbm}\end{array}} $$

$$ \textrm{R}=400\ \textrm{lbm}-\textrm{E}=400\ \textrm{lbm}-143\ \textrm{lbm}=257\ \textrm{lbm} $$

• C. The fraction of the solute recovered can be calculated as follows:

$$ \textrm{Recovery}=1-\frac{\left({x}_{\textrm{EG},\textrm{R}}\right)\left(\textrm{R}\right)}{\left({x}_{\textrm{EG},\textrm{F}}\right)\left(\textrm{FD}\right)}=1-\frac{(0.49)\left(257\ \textrm{kg}\right)}{(0.60)\left(250\ \textrm{kg}\right)}=0.1605 $$

5.1.8 Practice Problem 5.8

5.1.8.1 Solution

Draw the schematic diagram for the system as shown.

• A. Calculate the mass flow rate and composition of the underflow solution, L_N as shown.

$$ {\displaystyle \begin{array}{l}{N}_{\textrm{N}}=2\frac{\textrm{kg}\ \textrm{B}}{\textrm{kg}\ \textrm{soln}.}=\frac{B_{\textrm{N}}}{L_{\textrm{N}}}\overset{B_{\textrm{N}}=504\ \textrm{kg}/\textrm{hr}}{\to}\\ {}{L}_{\textrm{N}}=\frac{504\frac{\textrm{kg}\ \textrm{B}}{\textrm{hr}}}{2\frac{\textrm{kg}\ \textrm{B}}{\textrm{kg}\ \textrm{soln}.}}=252\ \textrm{kg}.\textrm{soln}\ \left(\textrm{A}+\textrm{C}\right)/\textrm{hr}\end{array}} $$

$$ {y}_{\textrm{N}}=\frac{{\textrm{C}}_{\textrm{N}}}{L_{\textrm{N}}}=\frac{40\frac{\textrm{kg}}{\textrm{hr}}}{252\frac{\textrm{kg}}{\textrm{hr}}}=0.1587 $$

• B. Calculate the mass flow rate of the solution, V₁.

Mass balance for the solute, C:

$$ {\displaystyle \begin{array}{l}{x}_1=0.20=\frac{{\textrm{C}}_{{\textrm{V}}_1}}{{\textrm{V}}_1}\Rightarrow \\ {}{\textrm{V}}_1=\frac{{\textrm{C}}_{{\textrm{V}}_1}}{0.20}=\frac{156\frac{\textrm{kg}}{\textrm{hr}}}{0.20}=780\ \textrm{kg}/\textrm{hr}\end{array}} $$

• C. Overall mass balance for the solution (A + C) will result in the answer for the mass flow rate of the pure solvent, V_N+1.

$$ {\displaystyle \begin{array}{l}{L}_0+{V}_{N+1}={L}_N+{V}_1\Rightarrow \\ {}{V}_{N+1}={L}_N+{V}_1-{L}_0\\ {}\kern2.2em =252\frac{\textrm{kg}}{\textrm{hr}}+780\frac{\textrm{kg}}{\textrm{hr}}-196\frac{\textrm{kg}}{\textrm{hr}}\\ {}\kern2.3em =836\ \textrm{kg}/\textrm{hr}\end{array}} $$

• D. The following values can be obtained from a combination of the schematic diagram, from the preceding calculations, and from the equilibrium relationship, y = x.

$$ {\displaystyle \begin{array}{l}{x}_{N+1}=0,\kern3em {x}_{N+1}^{\ast }={y}_N=0.1587\\ {}{x}_1=0.20,\kern2.75em {x}_1^{\ast }={y}_0=1.0\end{array}} $$

Substitute all the known values into Eq. 5.66 to calculate the number of stages required.

$$ N=1+\frac{\log \left[\frac{x_{N+1}-{x}_{N+1}^{\ast }}{x_1-{x}_1^{\ast }}\right]}{\log \left[\frac{x_{N+1}-{x}_1}{x_{N+1}^{\ast }-{x}_1^{\ast }}\right]}=1+\frac{\log \left[\frac{0-0.16}{0.20-1.0}\right]}{\log \left[\frac{0-0.20}{0.16-1.0}\right]}=2.12 $$

5.1.9 Practice Problem 5.9

5.1.9.1 Solution

Obtain the log of the given data and include it in the data table.

PCH4 (kPa)	20	40	60	80	100
mS/mA (g/g)	0.13	0.24	0.35	0.42	0.50
log PCH4	1.30	1.60	1.78	1.90	2.00
log (mS/mA)	−0.89	−0.62	−0.46	−0.38	−0.30

Plot $\log \left({m}_S/{m}_A\right)\ \textrm{vs}.\log {P}_{CH_4}$ and obtain the slope of the resulting straight line.

$$ \textrm{Slope}=\frac{\Delta y}{\Delta x}=\frac{-0.30-\left(-0.80\right)}{2-1.40}=0.833=\frac{1}{n}\Rightarrow n=1.2 $$

Obtain the value of K by taking a point on the graph and substituting the known values into Eq. 5.73.

$$ {\displaystyle \begin{array}{l}\log \left(\frac{m_S}{m_A}\right)=\log K+\left(\frac{1}{n}\right)\log {P}_{{\textrm{CH}}_4}\Rightarrow \\ {}\log K=\log \left(\frac{m_S}{m_A}\right)-\left(\frac{1}{n}\right)\log {P}_{{\textrm{CH}}_4}=-0.46-\left(\frac{1}{1.20}\right)(1.80)=-1.96\\ {}\Rightarrow K={10}^{-1.96}=0.0110\end{array}} $$

Using Eq. 5.72 as a template for Freundlich equation and the parameters just obtained, the Freundlich equation applicable in this situation is

$$ \frac{m_S}{m_A}={KC}^{\frac{1}{n}}\overset{n=1.2,K=0.011}{\to}\frac{m_S}{m_A}=(0.011){\left({P}_{{\textrm{CH}}_4}\right)}^{\frac{1}{1.2}} $$

• B. For a component in an ideal gas mixture, mole fraction = pressure fraction. Apply the preceding concept to methane in the hydrocarbon mixture. Note that the total pressure is 1 atm = 101 kPa. Therefore,

$$ {y}_{{\textrm{CH}}_4}=\frac{P_{{\textrm{CH}}_4}}{P}\Rightarrow {P}_{{\textrm{CH}}_4}={y}_{{\textrm{CH}}_4}P=(0.02)\left(101\ \textrm{kPa}\right)=2.02\ \textrm{kPa} $$

From the Freundlich equation just derived, determine the mass of the solute (CH₄) adsorbed by 100 g of activated carbon (adsorbent/solvent).

$$ {\displaystyle \begin{array}{l}\frac{m_S}{m_A}=(0.011){\left({P}_{{\textrm{CH}}_4}\right)}^{\frac{1}{1.2}}\Rightarrow {m}_S=\left({m}_A\right)(0.011){\left({P}_{{\textrm{CH}}_4}\right)}^{\frac{1}{1.2}}\\ {}=\left(100\ \textrm{g}\right)(0.011){\left(2.02\ \textrm{kPa}\right)}^{\frac{1}{1.2}}\\ {}=1.9762\ \textrm{g}\ {\textrm{CH}}_4\ \textrm{adsorbed}\end{array}} $$

Calculate the moles of CH₄ adsorbed by using its molecular weight.

$$ {N}_{{\textrm{CH}}_4}=\frac{m_{{\textrm{CH}}_4}}{M_{{\textrm{CH}}_4}}=\frac{1.9762\ \textrm{g}}{16\frac{\textrm{g}}{\textrm{mol}}}=0.1235\ \textrm{mol} $$

Convert the temperature of the hydrocarbon gas mixture to its absolute value.

$$ T={20}^{{}^{\circ}}\textrm{C}+273=293\ \textrm{K} $$

Calculate the volume of methane adsorbed by using the ideal gas law.

$$ {\displaystyle \begin{array}{l}{PV}_{{\textrm{CH}}_4}={N}_{{\textrm{CH}}_4}\overline{R}T\Rightarrow \\ {}{V}_{{\textrm{CH}}_4}=\frac{N_{{\textrm{CH}}_4}\overline{R}T}{P}=\frac{\left(0.1235\ \textrm{mol}\right)\left(0.0821\frac{\textrm{L}\hbox{-} \textrm{atm}}{\textrm{mol}\cdot \textrm{K}}\right)\left(293\ \textrm{K}\right)}{1\ \textrm{atm}}=2.9708\ \textrm{L}\end{array}} $$