Solution Manual to Mass Transfer Processes P. A. Ramachandran rama@wustl.edu March 25, 2018 1 Companion to Mass Transfer Processes, P. A. Ramachandran. Contents 0.1 Introduction page 6 1 Chapter 1 7 1.1 Answers to Review Questions 7 1.2 Solutions to Problems 10 2 Chapter 2 21 2.1 Answers to Review Questions 21 2.2 Solutions to Problems 23 3 Chapter 3 35 3.1 Answers to Review Questions 35 3.2 Solutions to Problems 37 4 Chapter 4 48 4.1 Answers to Review Questions 48 4.2 Solutions to Problems 50 5 Chapter 5 58 5.1 Answers to Review Questions 58 5.2 Solutions to Problems 60 6 Chapter 6 67 6.1 Answers to Review Questions 67 6.2 Solutions to Problems 69 7 Chapter 7 77 7.1 Answers to Review Questions 77 7.2 Solutions to Problems 80 8 Chapter 8 90 8.1 Answers to Review Questions 90 8.2 Solutions to Problems 92 3 Companion to Mass Transfer Processes, P. A. Ramachandran. 4 Contents 9 Chapter 9 105 9.1 Answers to Review Questions 105 9.2 Solutions to Problems 107 10 Chapter 10 117 10.1 Answers to Review Questions 117 10.2 Solutions to Problems 119 11 Chapter 11 129 11.1 Answers to Review Questions 129 11.2 Solutions to Problems 132 12 Chapter 12 137 12.1 Answers to Review Questions 137 12.2 Solutions to Problems 140 13 Chapter 13 146 13.1 Answers to Review Questions 146 13.2 Solutions to Problems 148 14 Chapter 14 155 14.1 Answers to Review Questions 155 14.2 Solutions to Problems 157 15 Chapter 15 170 15.1 Answers to Review Questions 170 15.2 Solutions to Problems 172 16 Chapter 16 179 16.1 Answers to Review Questions 179 16.2 Solutions to Problems 182 17 Chapter 17 192 17.1 Answers to Review Questions 192 17.2 Solutions to Problems 193 18 Chapter 18 200 18.1 Answers to Review Questions 200 18.2 Solutions to Problems 203 19 Chapter 19 221 19.1 Answers to Review Questions 221 19.2 Solutions to Problems 223 Companion to Mass Transfer Processes, P. A. Ramachandran. Contents 5 20 Chapter 20 228 20.1 Answers to Review Questions 228 20.2 Solutions to Problems 230 21 Chapter 21 240 21.1 Answers to Review Questions 240 21.2 Solutions to Problems 242 22 Chapter 22 251 22.1 Answers to Review Questions 251 22.2 Solutions to Problems 253 23 Chapter 23 257 23.1 Answers to Review Questions 257 23.2 Solutions to Problems 259 24 Chapter 24 268 24.1 Answers to Review Questions 268 24.2 Solutions to Problems 271 25 Chapter 25 276 25.1 Answers to Review Questions 276 25.2 Solutions to Problems 278 26 Chapter 26 287 26.1 Answers to Review Questions 287 26.2 Solutions to Problems 290 27 Chapter 27 296 27.1 Answers to Review Questions 296 27.2 Solutions to Problems 298 28 Chapter 28 304 28.1 Answers to Review Questions 304 28.2 Solutions to Problems 307 29 Chapter 29 317 29.1 Answers to Review Questions 317 29.2 Solutions to Problems 319 30 Chapter 30 325 30.1 Answers to Review Questions 325 30.2 Solutions to Problems 327 Companion to Mass Transfer Processes, P. A. Ramachandran. 6 Contents 0.1 Introduction This solution manual contains answers to review questions for all chapters and solutions to most problems in the book. If any clarification is needed, please feel free to e-mail me. The feedback from instructions is most appreciated together with pointing out any errors which will go into updating this solution book. I will be very happy to supply additional details on any of these problems if needed. Please note that I have taken care in the preparation of this manual, but I make no expressed or implied warranty of any kind and assume no responsibility for errors or omissions. No liability is assumed for incidental or consequential damages in connection with or arising out of the use of the information contained herein. Companion to Mass Transfer Processes, P. A. Ramachandran. 1 Chapter 1 1.1 Answers to Review Questions 1. What is meant by concentration jump? The concentration becomes discontinuous as we cross from one phase to another and this known as the concentration jump. 2. What is meant by the continuum assumption and what are its implications in model development? Continuum assumption assumes that the matter is continuously distributed in space and ignores the atomic/molecular nature of matter. This permits us to assign a point value to the variables such as concentration and so on. The information on the interaction of molecules is however lost and has to be supplemented by suitable constitutive model. 3. Indicate some situations where the continuum models are unlikely to apply. Continuum models are unlikely to apply if there are not sufficient number of molecules in the volume of interest. The number should be large so that the statistical average can be assigned to the ensemble of molecules. For example in high vacuum system or a plasma reactor the number of molecules are small and continuum description may not be adequate. 4. Give an example of a system where the total concentration is (nearly) constant. Gas mixture at constant total pressure and temperature. 5. Give an example of a system where the mixture density is nearly constant. Liquid mixtures of compounds with similar chemical nature. 6. Can average molecular weight be a function of position? Yes. Since in a diffusing binary system, the mole fraction of A varies along the position and correspondingly the molecular weight will vary as a function of position. 7. What information is missing in the differential models based on the continuum assumption? Information on the transport rate caused by molecular motion is missing in the context of continuum models. 8. Why are constitutive models needed in the context of differential models? Molecular level transport occurs due molecular motion and is not modeled in the continuum level of modeling. In order to incorporate these effects, a constitutive model is needed to quantify the diffusion flux. 7 Companion to Mass Transfer Processes, P. A. Ramachandran. 8 Chapter 1. Chapter 1 9. Does the Fick’s law apply universally to all systems? No. It is specific since it is the result of the of molecular level interactions. These are system specific. It is an accurate model for binary gas mixture at low or moderate pressures. However it is commonly used as a first level model for a large class of problems. 10. Write a form of Fick’s law using partial pressure gradient as the driving force. Using C A = p A /R g T we can write Fick’s law using partial pressure gradient as: J Ax = − D A R g T dp A dx 11. What is meant by the invariant property of the flux vector? A vector remains same if the coordinates are rotated or if different coordinate system (e.g., cylindrical) is used and this is known as the invariance property of the flux vector. 12. The combined flux is partitioned into the sum of the convection and diffusion flux. Is this partitioning unique? No; It is not unique. It will depends on how the mixture velocity is defined and thereby what part of the combined flux is allocated to the convection flux. 13. State the units of the gas constant if the pressure is expressed in bar instead of Pa, (ii) if expressed in atm. Gas constant will have a value of 8 . 205 × 10 − 5 m 3 atm / mol K if pressure is expressed atm. It will have a value of 8 . 314 × 10 − 5 if pressure is in bars. 14. Express the dissolved oxygen concentration in Example in p.p.m. The concentration of dissolved oxygen was calculated as 0.2739 mol/m 3 . This can be converted to gm/gm using the molecular weight (32 g/g) and also using the liquid density of water. The result is 8 . 467 × 10 − 6 . This corresponds to 8.467 p.p.m. 15. What is a macroscopic level model? What information needs to be added here in addition to the conservation principle? A larger control volume or the whole reactor or separator is taken as the control volume in a macroscoipc model. The control volume does not tend to zero. Hence the local information is lost and needs to be added as additional closure information in addition to the conservation law. 16. How is mass transfer coefficient defined? Why is it needed? The flux across a surface is not available in the meso- or macro-models since it depends on the local concentration gradient in accordance to Fick’s law. Hence the flux is represented as a product of a mass transfer coefficient and a suitably defined driving force. It it therefore needed in the meso- and macro- scale model. 17. What is meant by a cross-sectional average concentration? What is a cup mixing or the bulk concentration? Companion to Mass Transfer Processes, P. A. Ramachandran. Chapter 1 9 In the cross-section average the local concentration is weighted by the local area and integrated over the cross-section. One then divides this by the total area to get the cross-sectional average. In the cup mixing average, the local concentration is weighted by the local volumetric flow rate and integrated over the cross-section. One the divides this by the total volumetric flow rate to get the cup mixing concentration. 18. What assumptions are involved in plug flow model? The cup mixing concentration is assumed to be the same as the cross-sectional average concentration for a plug flow idealization. 19. What assumptions are involved in a completely backmixed model? The average concentration in the reactor and the exit concentration are assumed to be the same in a completely backmixed reactor. 20. What additional closure is needed for mass transport in turbulent flow? Why? The contribution of the turbulent diffusivity (eddy diffusivity) should be added in addition to molecular diffusivity to calculate the flux across a con- trol surface. This extra term arises due to the fluctuations in velocity causing additional transport. 21. What is an ideal stage contactor? How do you correct if the stage is not ideal? The exit streams leaving a two phase contactor are assumed to be in equilibrium in an ideal stage contactor. If the stage is not ideal, it is corrected by using a stage efficiency factor. 22. What is the dispersion coefficient and where is it needed? Dispersion coefficient connects the cup mixing and cross-sectional averages by using a Fick’s type of relation. It is needed to close the mesoscopic models in systems where a chemical reaction is taking place, e. g., tubular flow reactors. Companion to Mass Transfer Processes, P. A. Ramachandran. 10 Chapter 1. Chapter 1 1.2 Solutions to Problems 1. Mass fraction to mole fractions: Show that mass fractions can be con- verted to mole fractions by the use of the following equation: y i = ω i M i ¯ M (1.1) Derive an expression for dy i as a function of dω i values. Do this for a binary mixture. Expression for multicomponent mixture becomes rather unwieldy! Solution: The mass fraction has the units of kg i / kg total. The molecular weight of gas i has the unit of kg i / mol i Dividing these we get ω i M i which has the units of mol i/ kg total. Then dividing by the average molecular weight ¯ M which has the unit of kg total/ mol total, we get mol i/ mol total which is the mole fraction. Hence the relation is verified. To get dy i one should note that ¯ M is a function of ω i . ¯ M = ( 1 ω A /M A + ω B /M B ) Hence y A = ω A M A ( 1 ω A /M A + ω B /M B ) and ω B = 1 − ω A . Differentiating and after some algebraic manipulations we get the following relation for dy i as a function of dω i values. dy A = ¯ M 2 M A M B dω A 2. Mole fraction to mass fractions: Show that mole fractions can be con- verted to mass fractions by the use of the following equation: ω i = y i M i ¯ M (1.2) Derive an expression for dω i as a function of dy i values for a binary mixture. Solution: The mole fraction has the units of mol i / mol total. The molecular weight gas i has the unit of kg i / mol i Multiplying these we get kg i/ mol total and then dividing by the average molecular weight which has the unit of kg total/ mol total, we get kg i/ kg total which is the mass fraction. Hence the relation is verified. To get dω A one should note that ¯ M is a function of y i . We have for a binary : ¯ M = y A M A + (1 − y A ) M B Companion to Mass Transfer Processes, P. A. Ramachandran. Chapter 1 11 Using this in the expression for ω A and differentiating it and simplifying the algebra, we get M A M B ¯ M 2 dy A which is the required relation connecting the mass fraction gradient and the mole fraction gradient. 3. Average molecular weight: At a point in a methane reforming furnace we have a gas of the composition: CH 4 = 10%; H 2 = 15%; CO = 15% and H 2 O = 10% by moles. Find the mass fractions and the average molecular weight of the mixture. Find the density of the gas. Solution: Assume the rest is nitrogen which was not specified as part of the problem. Calculations of this type are best done in EXCEL spreadsheet or using a simple MATLAB snippet. % mass fraction calculations y = [0.1 0.15 0.15 0.1 0.5] M = [ 18 2 28 18 28 ] % note g/mol unit Mbar = sum (y .*M) omega = y.*M/Mbar %% The results are Mbar = 22.100 g/mol omega = [ 0.081448 0.013575 0.190045 0.081448 0.633484 ] 4. Average molecular weight variations: Two bulbs are separated by a long capillary tube which is 20cm long. On one bulb we have pure hydrogen while at the other bulb we have nitrogen. The mole fraction profile varies in a linear manner along the length of the capillary. Calculate the mass fraction profile and show that the variation is not linear. Also calculate the average molecular weight as a function of the length along the capillary. Solution: Mole fraction profile is given as linear. Hence at any location the mole fraction can be calculated as a linear interpolation between the two end point values. For example at distance X = 4cm, the mole fraction of hydrogen is (1 − 4 / 20) = 0.8. Correspondingly the average molecular weight at this point is 0 . 8 × 2 × 10 − 3 + 0 . 2 × 28 × 10 − 3 = 7.2 g/mol. The mass fraction at this point is 2 × 0 . 8 / ¯ M = 0 . 22. Similar calculations can be done at other points. For example at distance x = 16 cm , we get the mass fraction as 0.0175. The mass fraction profile is seen to be not linear while the mole fraction is. Similarly the average molecular weight is a function of position. Companion to Mass Transfer Processes, P. A. Ramachandran. 12 Chapter 1. Chapter 1 5. Mass fraction gradient: For a diffusion process across a stagnant film, the mole fraction gradient of the diffusing species was found to be constant. What is the mass fraction gradient? Is this also linear? The mixture is benzene-air. Solution: The mole fraction gradient is constant. Hence the mole fraction profile is linear. The average molecular weight at any location is given as the weighted average of the two species: ¯ M = 78 X + 29(1 − X ) where X is a scaled distance at the two ends. (Here we assume X = 0 is air while X = 1 is benzene.) The mass fraction profile is related to the mole fraction profile as: ω A = M A x A M A x A + (1 − x A ) M B This is found to be nonlinear (due to the terms in the denominator). Corre- spondingly the mass fraction gradient is also nonlinear. 6. Total concentration in a liquid mixture: Find the total molar concen- tration and species concentrations of 10% ethyl alcohol by mass in water at room temperature. Solution: The total molar concentration is equal to density of the mixture divided by the average molecular weight of the mixture. Density of alcohol is 0.7935 g/cm 3 . The average density at a mass fraction of 0.1 is calculated by interpolation using the following relation: 1 ρ = ω A ρ 0 A + ω B ρ 0 B The superscripts indicate pure component values. The calculated value is 0 . 9746 g / cm 3 . It may be noted that the solution is not ideal and hence the density is to be found from partial molar volume considerations for a more accurate result. The density from internet data base is 0 . 98187 g/cm 3 which is the more accu- rate value. The average molecular weight is calculated from the following equation: 1 ¯ M = ∑ ω A M A The value is found as 19.16 g/mol. Hence the total molar concentration, C of the mixture is ρ/ ¯ M = (0 . 9784 g/cm 3 ) / (19 . 16 g/mol ) = 0 . 0509 mol/cm 3 = 50866 mol/m 3 . 7. Effect of coordinate rotation on flux components: In Figure 1.1, the flux vector is 2 e x + e y . Now consider a coordinate system which is rotated by an angle θ . Find this angle such that the flux component N Ay is zero. What is the value of N Ax in this coordinate system? Companion to Mass Transfer Processes, P. A. Ramachandran. Chapter 1 13 Solution: The original unit vectors in the two coordinates are related to the unit vectors in the new coordinates as follows: e x = e x ; new cos θ − e y ; new sin θ e y = e x ; new sin θ + e y ; new cos θ The original vector is 2 e x + e y and this gets is transformed to: e x ; new [2 cos θ sin θ ] + e y ; new [ − 2 sin θ + cos θ ] New components are therefore [2 cos θ sin θ ] and [ − 2 sin θ + cos θ ]. If the flux component N Ay in the new coordinates has to be zero, then − 2 sin θ + cos θ = 0 The angle of rotation must then be such that tan θ = 1 / 2. Hence θ = π/ 4. The value of N Ax is then 2 cos θ sin θ . 8. Flux vector in cylindrical coordinates: Define flux vector in terms of its components in cylindrical coordinates. Sketch the planes over which the components act. Show the relations between these components and the com- ponents in Cartesian coordinates. Solution: Flux vector is the same in cylindrical coordinates but the components are different. The flux vector is represented as: N A = e r N Ar + e θ N Aθ + e z N Az The component N Ar can be viewed as the mass crossing a unit area in a plane normal to the r-direction in cylindrical coordinates. The other components can be viewed in a similar manner. Thus, for example, N Aθ is the moles crossing a plane perpendicular to the θ direction. The components are related to those in Cartesian by the following relations from vector transformation rules. e r = (cos θ ) e x + (sin θ ) e y e θ = ( − sin θ ) e x + (cos θ ) e y e z = e z 9. Flux vector in spherical coordinates: Define flux vector in terms of its components in spherical coordinates. Sketch the planes over which the compo- nents act. Show the relations between these components and the components in Cartesian coordinates. Show the relations between these components and the components in Carte- sian coordinates. Solution: Companion to Mass Transfer Processes, P. A. Ramachandran. 14 Chapter 1. Chapter 1 . The flux vector in spherical coordinates is represented as: N A = e r N Ar + e θ N Aθ + φ z N Aφ The components are interpreted as mass crossing a unit area in a plane normal to the direction indicated in the subscript. For example, N Aθ is the moles crossing a plane perpendicular to the θ direc- tion. The components are related to those in Cartesian by the following relations from vector transformation rules. e r = (sin θ cos φ ) e x + (sin θ sin φ ) e y + (cos θ ) e z e θ = (cos θ cos φ ) e x + (cos θ sin φ ) e y + ( − sin θ ) e z e φ = ( − sin φ ) e x + ( − cos φ ) e y + (0) e z 10. Different forms of the Henry’s law constant: Express the Henry’s constants reported in Table 1.4 as H i,pc and H i,cp . Solution: The relation for mole fraction in the liquid is x A equals C A /C tot Using this in the Herny law p A = H A x A = ( H A /C tot ) x A Hence the Henry coefficient in pressure-concentration unit is given as H pc = H A /C tot Henry’s law in pressure-concentration form for hydrogen is therefore: H H,pc = (7 . 099 × 10 4 ; atm] / (55000; mol / m 3 ) = 1 . 2907atm m 3 / mol In the concentration-pressure unit, it is the reciprocal of this quantity, 0 . 7748mol / atm m 3 Values for other gases are: Oxygen = 0.7744; CO 2 = 0 . 0296; ammonia = 5 . 4545 × 10 4 . in pressure-concentration unit. 11. Henry’s law constants: Unit conversions: Henry’s law constant for O 2 and CO 2 are reported as 760.2 L. atm/mol and 29.41 L. atm /mol. What is the form of the Henry’s law used? Convert to values for the other forms shown in the text. Solution: From the units we deduce that the Henry’s law is reported in pressure- concentration form. To get the value in pressure-mole fraction form we use C A = x A C where C is the total concentration in the liquid. Value of 55 mol/L is used for the total concentration in water. Hence the Henry’s law in pressure-mole fraction form Companion to Mass Transfer Processes, P. A. Ramachandran. Chapter 1 15 is: H = (760 . 2L . atm / mol)(55mol / L) = 4 . 18 × 10 4 atm Similar calculation for CO 2 shows the value of 1 . 617 × 10 3 atm 12. Solubility of CO 2 : Henry’s constant values for CO 2 is shown below as a function of temperature. Temperature, K 280 300 320 H, bar 960 1730 2650 Fit an equation of the type: ln H = A + B/T What is the physical significance of the parameter B ? Find the solubility of pure CO 2 in water at these temperature. Solution: The plot ln H as 1 /T should be linear and the constants A and B can be found from this plot. However, it is best to use a linear regression model in MATLAB. The following MATLAB code is useful for the linear regression and can be used in general for other problems. % Linear regression: problem 1.12 CO2 solubility data. % P = POLYFIT(X,Y,N) finds the coefficients of a polynomial P(X) of % degree N that fits the data Y best in a least-squares sense. P is a % row vector of length N+1 containing the polynomial coefficients in % descending powers, P(1)*X^N + P(2)*X^(N-1) +...+ P(N)*X + P(N+1). x = [ 280 300 320 ] %% temperature values y = [960 1730 2650 ] %% henry values in bars Y = log (y) X = 1./x P = polyfit (X, Y, 1 ) % linear fit here. Y1 = P(2) + P(1) * X %% fitted constants %%% comparison plots plot ( X,Y, ’*’ ); hold on plot (X, Y1) delh = 8.314 * P(1) % heat of solution Answer= -19 kJ/mol % solutions: B = -2.2792e+003; % A= 15.0215 The fitted constants are found to be B = − 2 . 2792 e + 003 and A = 15 . 0215 The parameter B is a measure of heat of solution in accordance with van’t Hoff equation: B = ∆ H s R Companion to Mass Transfer Processes, P. A. Ramachandran. 16 Chapter 1. Chapter 1 From the fitted data we find the heat of solution of CO 2 as -19 kJ/mol which is close to experimental value of − 19 . 4kJ / mol reported in the literature. 13. Vapor pressure calculations: the Antoine equation: The Antoine con- stants for water are: A = 8 . 07131; B=1730.63; C = 233 . 426 in the units of mm Hg for pressure and deg C for temperature. Convert this to a form where pressure is in Pa and temperature is in deg K; Also rearrange the Antoine equation to a form where temperature can be calculated explicitly. This represents the boiling point at that pressure. What is the boiling point of water at Denver, CO (mile high city)? Solution: p (Pa) = p (mm) 1 . 0135 × 10 5 P a/atm 760 mm/ 1 atm Taking log log 10 p (mm Hg) = log 10 p (Pa) − log(0 . 0075) = log 10 p (Pa) − 2 . 1249 Using this in Antoine equation and substituting for log 10 p (mm Hg) we get log 10 p (Pa) = 2 . 1229 − B C + T ( K ) − 273 or log 10 p (Pa) = 10 . 1962 − 1730 . 63 T ( K ) − 39 . 57 which is the required relation with pressure in Pa and temperature in deg K. The equation can be rearranged to be explicit in temperature and the following relation is obtained: T = B log 10 p − A − C The pressure at Denver at normal condition is given as: p/p 0 = exp( − M air gh R g T ) from the equation of hydrostatics. Here p O is the pressure at sea level. M w = 20 × 10 − 3 kg/mol and h = elevation = 1 mile = 1600 m. Hence p = 367 mm Hg. Substituting in the Antoine equation we find the boiling point of water as 95 deg Celsuis. 14. van’t Hoff relation: Given the Antoine constants for a species, can you calculate the heat of vaporization of that species? Find this value for water from the data given in Problem 1.12. Solution: One should use the van’t Hoff relation as a first approximation. The relation is: d ln P ; vap dT = ∆ H ; vap R g T 2 Companion to Mass Transfer Processes, P. A. Ramachandran.