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5.8: Problems

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    44492
  • Problems marked with the symbol \(\ddagger\) will be difficult to solve without the use of computer software.

    Section 5.2

    1. Show that the mole fraction in an ideal gas mixture can be expressed as \(y_{A} = {p_{A} / p}\).

    2. Demonstrate that the volume percent of a mixture is the same as the mole percent for an ideal gas mixture.

    3. Assuming ideal gas behavior, determine the average molecular mass of a mixture made of equal amounts of mass of chlorine, argon, and ammonia.

    Section 5.3

    4. A liquid mixture of hydrocarbons has 40% by weight of cyclohexane, 40% of benzene, and 20% toluene. Assuming that volumes are additive compute the following:

    1. species densities of the components in the mixture.
    2. overall density of the mixture
    3. concentrations of the components in moles/m\(^{3}\)
    4. mole fractions of the components in the mixture.

    Section 5.4

    5. Determine the vapor pressure, in Pascal, of ethyl ether at 25 C and at 30 C. Estimate the heat of vaporization of ethyl ether using these two vapor pressures and the Clausius-Clapeyron equation.

    6. Determine the vapor pressure of methanol at 25 C and compare it to that of ethanol at the same temperature. Consider the ethanol-methanol system to be an ideal solution in the liquid phase and an ideal gas mixture in the vapor phase. Determine the mole fraction of methanol in the vapor phase when the liquid phase mole fraction is 0.50. If the liquid phase is allowed to slowly evaporate, will it become richer in methanol or ethanol? Here you are asked to provide an intuitive answer concerning the composition of the liquid phase during the process of distillation. In Chapter 8 a precise analysis of the process will be presented.

    7. Determine the vapor-liquid equilibrium curve of a binary mixture of glycerol and acetone. Plot the mole fraction of acetone in the vapor phase versus the mole fraction of acetone in the liquid phase at one atmosphere (760 mm Hg).

    8. Use Eqs. \((5.4.10)\) and \((5.4.11)\) order to derive \((5.4.12)\).

    9. Demonstrate that Equation \((5.4.15)\) is valid for an ideal system containing three components when an appropriate constraint is imposed.

    10. Make use of Equation \((5.4.6)\) to develop the general form of Henry’s law given by Eqs. \((5.4.15)\) and \((5.4.16)\). Begin with the general representations for the chemical potentials

    \[ \mu_{A, gas} = \mathscr{G}(T,p,y_{A} , y_{B} , \text{ etc.}) , \quad \mu_{A, liquid} = \mathscr{F}(T,p,x_{A} , x_{B} , \text{ etc.})\]

    and use a Taylor series expansion for \(\mu_{A, gas}\)(see Problem 31) to obtain

    \[ \mu_{A, gas} = \left. \mathscr{G}\right|_{y_{A} = 0} + \left. \frac{\partial \mathscr{G}}{\partial y_{A} } \right|_{y_{A} = 0} (y_{A} ) + \left. \frac{\partial^{2} \mathscr{G}}{\partial y_{A}^{2} } \right|_{y_{A} = 0} (y_{A} )^{2} + .....\]

    Since the chemical potential of species \(A\) is zero when there is no species \(A\) present, this simplifies to

    \[ \mu_{A, gas} = \left. \frac{\partial \mathscr{G}}{\partial y_{A} } \right|_{y_{A} = 0} (y_{A} ) + \left. \frac{\partial^{2} \mathscr{G}}{\partial y_{A}^{2} } \right|_{y_{A} = 0} (y_{A} )^{2} + .....\]

    Restricting this development to dilute solutions of species \(A\), we can impose \(y_{A} <<1\) in order to express the chemical potential in the gas phase as

    \[ \mu_{A, gas} = \left. \frac{\partial \mathscr{G}}{\partial y_{A} } \right|_{y_{A} = 0} (y_{A} )\]

    Develop a similar representation for the liquid phase and show how these special representations for the chemical potential lead to the generalized form of Henry’s law given by Equation \((5.4.15)\).

    Section 5.5

    11. An equi-molar mixture of ethanol and ethyl ether is kept in a closed container at 103 KPa and 95 C. The temperature of the container is slowly reduced to the dew point of the mixture. Determine:

    1. What is the dew point temperature of the mixture?
    2. What is the pressure of the container at the dew point temperature of the mixture?
    3. What is the composition of the first drop of liquid at the dew point?

    12. A liquid mixture of n-hexane (mole fraction equal to 0.32) and n-heptane is heated until it begins boiling. Find the bubble point at \(p = 760\) mm Hg. What are the mole fractions in the vapor when the mixture begins to boil?

    13. A vapor mixture of benzene and toluene is slowly cooled inside a constant volume vessel. Initially the pressure inside the vessel is 300 mm Hg and the temperature is 70 C. As the vessel is cooled, the pressure inside the vessel decreases. Assume the vapor behaves like an ideal gas and take the dew point of the mixture to be 60 C. What is the mole fraction of benzene in the initial vapor mixture?

    14. Given Equation \((5.5.4)\) as the definition of the point humidity, explore the possible definitions for the area averaged humidity and the volume averaged humidity. Refer to Example \(5.6.2\) for guidance.

    15. Derive Equation \((5.5.5)\) from Equation \((5.5.4)\).

    16. Consider a day when the percent relative humidity is 70%, the temperature is 80 F and the barometric pressure is 1 atm. What is the humidity, mole fraction of water in the air, and dew point of the air?

    17. A mole of air is sampled from the atmosphere when the atmospheric pressure is 765 mm Hg, the temperature is 25 C, and relative humidity is 75%. The sample of air is placed inside a closed container and heated to 135 C and then compressed to 2 atm. What are the relative humidity, the humidity, and the mole fraction of water in the compressed air?

    18. A humidifier is used to introduce moisture into air supplied to an office building during winter days. Outside air at atmospheric pressure and 5 C is introduced into the heating system at a rate of 100 m3/min, on a dry air basis. The relative humidity of the outside air is 95%, and the heating system delivers warm air into the building at 20 C. How much water must be introduced into the warm air, in kg/min, the keep the relative humidity inside the building at 75%?

    19. The modified mass fraction, \(\Omega_{A}\), is defined by

    \[\Omega_{A} = {\rho_{A} / \sum_{B = 1}^{B = N-1}\rho_{B} } \nonumber\]

    Use this definition to develop relations analogous to Eqs. \((5.5.8)\) and \((5.5.10)\).

    Section 5.6

    20. Demonstrate that if the batch process illustrated in Figure \(5.6.1\) is repeated \(N\) times, the concentration of species \(A\) in the organic phase is given by

    \[ (c_{A\gamma } )_{n} = {(c_{A\gamma } )_{o} / \left(1 + A\right)^{n} } \label{1a}\]

    in which A is the absorption factor defined by

    \[ A = {V_{\beta } / \kappa_{eq,A} V_{\gamma } }\]

    Here \(V_{\gamma }\) represents the original volume of the organic phase, and \(V_{\beta }\) represents the volume of the aqueous phase used in each of the \(N\) steps in the process.

    21. The result indicated by Equation \ref{1a} in Problem 20 is based on the condition that the concentration of species \(A\) is zero in the aqueous phase used in the extraction process. Repeat the analysis of the batch extraction process assuming that the concentration of species \(A\) in the original aqueous phase is \((c_{A\beta } )_{o}\).

    22. In the analysis of the batch liquid-liquid extraction process illustrated in Figure \(5.6.1\) the equilibrium relation was given as

    Equilibrium relation: \[c_{A\gamma } = \kappa_{eq,A} (c_{A\beta } )\]

    Illustrate how \(\kappa_{eq,A}\) is related to the Henry’s law equilibrium coefficient given by \(K_{eq,A}\) in Equation \((5.4.15)\). Use \(y_{A\gamma }\) to represent the mole fraction of species \(A\) in the organic phase and \(x_{A\beta }\) to represent the mole fraction of species \(A\) in the aqueous phase.

    23. In order to justify the simplification indicated by Equation \((5.6.13)\) we need estimates of \({\dot{M}_{1}} -{\dot{M}_{4}}\) and \({\dot{M}_{2}} -{\dot{M}_{3}}\). If we let species \(B\) represent the organic phase (the \(\gamma\)-phase), and we assume that none of this species is transferred to the aqueous phase (the \(\beta\)-phase), a mole balance for species \(B\) takes the form

    Species \(B\): \[- (y_{B} )_{2} {\dot{M}_{2}} + (y_{B} )_{3} {\dot{M}_{3}} = 0 \]

    Use this species mole balance along with the definition

    \[ (y_{B} )_{3} = (y_{B} )_{2} + \Delta (y_{B} )\]

    to obtain an estimate of \({\dot{M}_{2}} -{\dot{M}_{3}}\). Use this estimate to identify the conditions that are required in order that the molar flow rates of the \(\gamma\)-phase are constrained by \({\dot{M}_{2}} -{\dot{M}_{3}} <<{\dot{M}_{3}}\)

    24. Small amounts of an inorganic salt contained in an organic fluid stream can be removed by contacting the stream with pure water as illustrated in Figure \(\PageIndex{1}\). The process requires that the organic and aqueous streams be contacted in a mixer that provides a large surface area for mass transfer, and then separated in a settler. If the mixer is efficient, the two phases will be in equilibrium as they leave the settler and you are to assume that this is the case for this problem. You are given the following information:

    1. Organic stream flow rate: 1000 \(lb_m\)/min
    2. Specific gravity of the organic fluid: \(\rho_{org} /\rho_{\ce{H2O}} = 0.87\)
    3. Salt concentration in the organic stream entering the mixer: \((c_{A} )_{org}\) = 0.0005 moles/liter
    4. Equilibrium relation for the inorganic salt: \((c_{A} )_{org} = \kappa_{eq,A} (c_{A} )_{aq}\) where \(\kappa_{eq,A} = 1/60\)

    Here \((c_{A} )_{aq}\) represents the salt concentration in the aqueous phase that is in equilibrium with the salt concentration in the organic phase, \((c_{A} )_{org}\). In this problem you are asked to determine the mass flow rates of the water stream that will reduce the salt concentration in the organic stream to 0.1, 0.01 and 0.001 times the original salt concentration. The aqueous and organic phases are to be considered completely immiscible, i.e., only salt is transferred between the two phases. In addition, the amount of material transferred is so small that the volumetric flow rates of the two streams can be considered constant.

    image
    Figure \(\PageIndex{1}\): Liquid-liquid extraction

    25. In this problem, we examine the process of recovering fission materials from spent nuclear fuel rods. This is usually referred to as reprocessing of the fuel to recover plutonium (Pu) and the active isotope of uranium (\(\ce{U235}\)). Reprocessing can be done by separation of the soluble isotope nitrates from a solution in nitric acid by a solvent such as a 30% solution of tributyl phosphate (TBP) in dodecane (\(\ce{C12H26}\)) in which the nitrates are preferentially soluble. Industrial reprocessing of nuclear fuels is done by countercurrent operation of many liquid-liquid separation stages. These separation stages consist of well-mixed contacting tanks, where \(\ce{UO2} (\ce{NO3} )_{2} \)is exchanged between two immiscible liquid phases, and separation tanks, where the organic and aqueous phases are separated. A schematic of a separation stage is shown in Figure \(\PageIndex{2a}\).

    image
    Figure \(\PageIndex{2a}\): Liquid-liquid separation stage for reprocessing

    In this process an aqueous solution of uranil nitrate [\(\ce{UO2} (\ce{NO3} )_{2} \)] is one of the feed streams to the separation stage, and the mass flow rate of the aqueous feed phase is, \(\dot{m}_{1} = 400\) kg/hr. The second feed stream is an organic solution of tributyl phosphate (TBP) in dodecane (\(\ce{C12H26}\)) which we assume to be a single component. The organic and inorganic phases are assumed to be immiscible, thus only the uranil nitrate is transferred from one stream to the other. The process specifications are indicated in Figure \(\PageIndex{2b}\), and for this problem it is the mass flow rates that you are asked to determine.

    image
    Figure \(\PageIndex{2b}\): Specified stream variables

    26. A gas stream consisting of air with a small amount of acetone is purified by contacting with a water stream in the contacting device illustrated in Figure \(5.6.4\). The inlet gas stream (Stream #2) contains one percent acetone and has a molar flow rate of 30 kmol/h. The molar flow rate of the pure water stream (Stream #1) is 90 kmol/h. The process operates at constant temperature and pressure, and the process equilibrium relation is given by

    Process Equilibrium Relation: \[(y_{A} )_{3} = K_{eq,A} (x_{A} )_{4}\]

    where \(K_{eq,A} = 2.53\). Assume that water and air are immiscible and that the molar flow rates entering and leaving the equilibrium stage are constant. In this problem you are asked to

    1. Determine the absorption factor for this equilibrium stage.
    2. Determine the mole fractions of acetone in the two exit streams.

    27. The concept of an equilibrium stage is a very useful tool for the design of multi-component separations, and a typical equilibrium stage for a distillation column is shown in Figure \(\PageIndex{3}\). A liquid stream, S1, flowing downward encounters a vapor stream, S2, flowing upward. We assume that the vapor

    image
    Figure \(\PageIndex{3}\): Sketch of an equilibrium stage process

    and liquid streams exchange mass inside the equilibrium stage until they are in equilibrium with each other. Equilibrium is determined by a ratio of the molar fractions of each component in the liquid and vapor streams according to

    Equilibrium relation: \[K_{A} = \frac{(y_{A} )_{4} }{(x_{A} )_{3} } \quad A = 1, 2, .., N \nonumber\]

    The streams leaving the stage, S3 (liquid), and S4 (vapor) are in equilibrium with each other and therefore satisfy the above relation. The ratio of the molar flow rates of the output streams is a function of the energy balance within the stage. In this problem we assume that the ratio of the liquid output molar flow rate to the vapor output molar flow rate, \({\dot{M}_{3}} / {\dot{M}_{4}}\), is given. Assuming that the compositions, i.e. the mole fractions of the components and the molar flow rates of the input streams, S1 and S2, are known, and the equilibrium constant for one of the components is given, develop the mass balances for a two component vapor-liquid equilibrium stage.

    28. A single stage, binary distillation process is illustrated in Figure \(\PageIndex{4}\). The total molar flow rate entering the unit is \(\dot{M}_{ 1}\) and the mole fraction of species \(A\) in this liquid stream is \((x_{A} )_{1}\). Heat is supplied in order to generate a vapor stream, and the ratio, \({\dot{M}_{2}} /{\dot{M}_{3}} = \beta\), depends on the rate at which heat is supplied. At the vapor-liquid interface, we can assume local thermodynamic equilibrium (see Equation \((5.4.6)\)) in order to express the vapor-phase mole fraction in terms of the liquid-phase mole fraction according to

    Equilibrium relation: \[y_{A} = \frac{\alpha_{AB} x_{A} }{1+x_{A} (\alpha_{AB} -1)} , \quad \text{ at the vapor-liquid interface} \nonumber\]

    Here \(\alpha_{AB}\) represents the relative volatility. If the distillation process is slow enough, one can assume that the vapor and the liquid leaving the distillation unit are in equilibrium; however, at this point in our studies we do not know what is meant by slow enough. In order to proceed with an approximate solution to this problem, we replace the equilibrium relation with a process equilibrium relation given by

    Process equilibrium relation: \[(y_{A} )_{2} = \frac{\alpha_{AB} (x_{A} )_{ 3} }{1+(x_{A} )_{3} (\alpha_{AB} -1)} , \quad \text{ between the exit streams} \nonumber\]

    Use of this relation means that we are treating the system shown in Figure \(\PageIndex{4}\) as an equilibrium stage. Given a detailed study of mass transfer in a subsequent course, one can make a judgment concerning the conditions that are required in order that this process equilibrium relation be satisfactory. For the present, you are asked to use the above relation to derive an implicit expression for \((y_{A} )_{2}\)in terms of \((x_{A} )_{1}\) and examine three special cases: \(\alpha_{AB} \to 0\), \(\beta \to 0\), \(\alpha_{AB} = 1\).

    image
    Figure \(\PageIndex{4}\): Single stage binary distillation

    29. A saturated solution of calcium hydroxide enters a boiler as shown in Figure \(\PageIndex{5}\) and a fraction, \(\varphi\), of the water entering the boiler is vaporized. Under these circumstances a portion of the calcium hydroxide precipitates and you would like to determine the mass fraction of this suspended solid calcium hydroxide in the liquid stream leaving the boiler. Assume that no calcium hydroxide leaves in the vapor stream, that none accumulates in the boiler, and that the temperature of the liquid entering and leaving the boiler is a constant. Assume that the solid carbon hydroxide leaving the boiler is in equilibrium with the dissolved carbon dioxide, i.e., the boiler is an equilibrium stage. The solubility is often expressed as;

    Equilibrium relation: \[\text{ solubility} = S = \text{ g of Ca(OH)}_{2}/\text{g of } \ce{H2O} \nonumber\]

    however, a more precise description can be constructed.

    In this problem you are asked to develop a general solution for the mass fraction of the suspended solid in the liquid stream leaving the boiler in terms of \(\varphi\) and \(S\). For \(\varphi\) = 0.50, 0.21, and 0.075, determine the mass fraction of suspended solid when \(S = 2.5\times 10^{-3}\).

    image
    Figure \(\PageIndex{5}\): Precipitation of calcium hydroxide in a boiler

    30. In problem 29 an equilibrium relation between solid calcium hydroxide and dissolved calcium hydroxide was given by

    Equilibrium relation: \[\text{ solubility} = S = \text{ g of Ca(OH)}_{2} / \text{g of } \ce{H2O} \label{1s}\]

    In Sec. 5.4.1 we expressed the general gas/liquid equilibrium relation for species \(A\) in terms of the chemical potential and for a solid/liquid system we would express Equation \((5.4.6)\) as

    Equilibrium relation: \[(\mu_{A} )_{solid} = (\mu_{A} )_{liquid} \label{2s}\]

    For the process considered in Problem 29, we assume that the solid phase is pure calcium hydroxide so that Equation \ref{2s} takes the form

    Equilibrium relation: \[\mu_{A, solid}^{o} = \mu_{A, liquid} \label{3s}\]

    The description of phase equilibrium phenomena in terms of the chemical potential will be the subject of a subsequent course in thermodynamics; however, at this point one can illustrate how Eqs. \ref{1s} and \ref{3s} are related.

    The development begins with a general representation for the chemical potential at some fixed temperature and pressure. This is given by

    \[ \mu_{A, liquid} = \mathscr{F}(T,p,x_{A} , x_{B} , \text{ etc.})\]

    where \(x_{A}\) is the mole fraction of species \(A\) (calcium hydroxide) in the liquid phase. A Taylor series expansion about \(x_{A} = 0\) leads to (see Problem 31)

    \[ \mu_{A, liquid} = \left. \mathscr{F}\right|_{x_{A} = 0} + \left. \frac{\partial \mathscr{F}}{\partial x_{A} } \right|_{x_{A} = 0} (x_{A} ) + \left. \frac{\partial^{2} \mathscr{F}}{\partial x_{A}^{2} } \right|_{x_{A} = 0} (x_{A} )^{2} + ..... \label{4s}\]

    The first term in this expansion is zero and when the mole fraction of species \(A\) is small compared to one, \(x_{A} <<1\), we can make use of a linear form of Equation \ref{4s} given by

    \[ \mu_{A, liquid} = \left. \frac{\partial \mathscr{F}}{\partial x_{A} } \right|_{x_{A} = 0} (x_{A} ) \label{5s} \]

    In this problem you are asked to use Equation \ref{3s} along with Equation \ref{5s} and the approximation

    \[ x_{A} = \frac{c_{A} }{c_{B} } , \quad \text{ when }c_{A} <<c_{B}\]

    to derive Equation \ref{1s}. Here \(c_{A}\) represents the molar concentration of calcium hydroxide and \(c_{B}\) represents the molar concentration of water. In terms of species \(A\) and species \(B\), it will be convenient to express the solubility in the form

    Equilibrium relation: \[\text{ solubility} = S = {\rho_{A} / \rho_{B} }\]

    and note that this can be related to the molar form by use of \(c_{A} MW_{A} = \rho_{A}\) and \(c_{B} MW_{B} = \rho_{B}\).

    31. In the previous problem we made use of a Taylor series expansion to obtain a simplified expression for the chemical potential as a function of the mole fraction. A Taylor series expansion is a powerful tool for predicting the value of a function at some position \(x = b\) when information about the function is only available at \(x = a\). If we think about the definite integral given by

    \[ f(b) = f(a) + \int_{x = a}^{x = b}\left({df / dx} \right) dx \label{1d}\]

    we see that we can determine \(f(b)\), given \(f(a)\), only if we know the derivative, \(df/dx\) everywhere between \(a\) and \(b\). To see how we can use Equation \ref{1d} to determine \(f(b)\) using only information at \(x = a\), we first make use of the change of variable, or transformation, defined by

    \[ \eta = x - b\]

    This leads to the relations

    \[ dx = d\eta , \quad {df / dx} = {df / d\eta }\]

    which can be used to express Equation \ref{1d} in the form

    \[ f(b) = f(a) + \int_{\eta = a-b}^{\eta = 0} \left( df / d\eta \right) d\eta \label{4d}\]

    Remember that the rule for differentiating a product

    \[ \frac{d}{d\eta } \left(UV\right) = U\frac{dV}{d\eta } + V\frac{dU}{d\eta }\]

    is the basis for the technique known as integration by parts where it is employed in the form

    \[ U\frac{dV}{d\eta } = \frac{d}{d\eta } \left(UV\right) - V\frac{dU}{d\eta } \label{6d}\]

    If we make use of the two representations

    \[ U = \frac{df}{d\eta } , \quad V = \eta\]

    Equation \ref{6d} provides the following identity for the derivative of \(f\) with respect to \(\eta\):

    \[ \frac{df}{d\eta } = \frac{d}{d\eta } \left[\frac{df}{d\eta } \eta \right] - \eta \frac{d}{d\eta } \left[\frac{df}{d\eta } \right]\]

    Use this result in Equation \ref{4d} to produce the second term in the Taylor series expansion for \(f(b)\) about \(f(a)\). Repeat this entire procedure to extend the representation for \(f(b)\) to obtain the third term in the Taylor series expansion and thus illustrate how one obtains a representation for \(f(b)\) that involves only the function and its derivatives evaluated at \(x = a\). Keep in mind that the only mathematical tools used in this derivation are the definition of the definite integral and the rule for differentiating a product10.

    32. A gas mixture leaves a solvent recovery unit as illustrated in Figure \(\PageIndex{6}\). The partial pressure of benzene in this stream is 80 mm Hg and the total pressure is 750 mm Hg. The volumetric analysis of the gas, on a benzene-free basis, is 15% \(\ce{CO2}\), 4% \(\ce{O2}\) and the remainder is nitrogen. This gas is compressed to 5 atm and cooled to 100 F. Calculate the percentage of benzene condensed in the process. Assume that \(\ce{CO2}\), \(\ce{O2}\) and \(\ce{N2}\) are insoluble in benzene, thus the liquid phase is pure benzene.

    image
    Figure \(\PageIndex{6}\): Recovery-condenser system

    33. Derive the form of the solid phase mass balance given by Equation \((11)\) in Example \(5.6.8\).

    34. In this problem we wish to complete the analysis given in Example \(5.6.8\) in order to determine the total molar flow rate of fresh air entering the dryer. To accomplish this we must first derive the macroscopic mole balance for air which can then be applied to the dryer illustrated in Example \(5.6.8\).

    Part I.

    Consider air to consist of nitrogen and oxygen and determine under what conditions the mole balances for these two components can be added to obtain the special form

    \[ \frac{d}{dt} \int_{\mathscr{V}}c_{air} dV + \int_{\mathscr{A}}c_{air} \mathbf{v} \cdot \mathbf{n} dA = 0 \label{1g}\]

    in which \(c_{air} = c_{\ce{N2} } +c_{\ce{O2}}\). Begin the analysis with the macroscopic mole balances given by

    \[ \frac{d}{dt} \int_{\mathscr{V}}c_{A} dV + \int_{\mathscr{A}}c_{A} \mathbf{v}_{A} \cdot \mathbf{n} dA = \int_{\mathscr{V}}R_{A} dV\]

    \[ \frac{d}{dt} \int_{\mathscr{V}}c_{B} dV + \int_{\mathscr{A}}c_{B} \mathbf{v}_{A} \cdot \mathbf{n} dA = \int_{\mathscr{V}}R_{B} dV\]

    and identify the simplifications that are necessary in order to obtain Equation \ref{1g}.

    Part II.

    Use Equation \ref{1g} to determine the molar flow rate of the incoming air indicated by \({\dot{M}_{2}}\). Clearly identify the process equilibrium relation that must be imposed in order to solve this problem.

    Section 5.7

    35. A sequential analysis of the multi-stage extraction process illustrated in Figure \(5.7.2\) led to the relation between \((y_{A} )_{1}\) and \((y_{A} )_{N+1}\) given by Equation \((5.7.25)\). That result was based on the special condition given by \((x_{A} )_{o} = 0\). In this problem you are asked to develop a general form of Equation \((5.7.25)\) that is applicable for any value of \((x_{A} )_{o}\).

    36. In Example \(5.7.1\) what value of \({ \dot{M}_{\gamma } }\) can be used instead of \({\dot{M}_{\gamma }} = { 30 } \) kmol/h so that the exit condition of \((y_{A} )_{1} = 0.001\) is satisfied exactly?

    37. In Example \(5.7.1\) an implicit equation for the absorption factor was given by

    \[1+A+A^{2} +A^{3} +A^{4} +A^{5} +A^{6} = 20.0 \nonumber\]

    The solution for \(A\) can be obtained by the methods described in Appendix B. Use at least one of the following methods to determine the value of the absorption coefficient:

    1. The bi-section method
    2. The false position method
    3. Newton’s method
    4. Picard’s method
    5. Wegstein’s method

    38. Point #2 in Figure \(5.7.8\) is represented by an equation given in Sec. 5.7.1. Identify the equation.

    39. Point #3 in Figure \(5.7.8\) is represented by an equation given in Sec. 5.7.1. Identify the equation.

    40. Point #4 in Figure \(5.7.8\) is represented by an equation given in Sec. 5.7.1. Identify the equation.

    41. Point #5 in Figure \(5.7.8\) is represented by an equation given in Sec. 5.7.1. Identify the equation.

    42. In Example \(5.7.2\) the \(\beta\)-phase mole fraction entering the \(N^{th}\)stage was listed as \((x_{A} )_{N} = 0.00386\). Indicate how this mole fraction was determined and verify that the molar flow rate of acetone leaving in the liquid stream remains unchanged because of the change in \({\dot{M}_{\beta }}\).

    43. Verify that \((x_{A} )_{N} = 0.00386\) for the conditions associated with Figure \(5.7.12b\) in Example \(5.7.2\).

    44. Verify that \((x_{A} )_{N} = 0.0403\) for the conditions associated with Figure \(5.7.12c\) in Example \(5.7.2\).

    45. Consider the process described in Example \(5.7.2\) for \({\dot{M}_{\beta }} = { 90 } \) kmol/h and \({\dot{M}_{\gamma }} = { 30 }\) kmol/h. Assume that the mole fraction of acetone in the air (\(\gamma\)-phase) entering the system is specified as \((y_{A} )_{6} = 0.010\) and assume that the mole fraction of acetone in the water (\(\beta\)-phase) entering the system is specified as \((x_{A} )_{o} = 0\). Given that there are 5 equilibrium stages, reconstruct Figure \(5.7.12a\), and use the reconstructed figure to determine \((y_{A} )_{1}\).

    46. Consider the process described in Example \(5.7.2\) for \({\dot{M}_{\beta }} = { 70 } \) kmol/h and \({\dot{M}_{\gamma }} = { 30 }\) kmol/h. Assume that the mole fraction of acetone in the air (\(\gamma\)-phase) entering the system is specified as \((y_{A} )_{6} = 0.010\) and assume that the mole fraction of acetone in the water (\(\beta\)-phase) entering the system is specified as \((x_{A} )_{o} = 0\). Given that there are 5 equilibrium stages, reconstruct Figure \(5.7.12b\) and use the reconstructed figure to determine \((y_{A} )_{1}\).

    47. Consider the process described in Example \(5.7.2\) for \({\dot{M}_{\beta }} = { 67 } \) kmol/h and \({\dot{M}_{\gamma }} = { 30 }\) kmol/h. Assume that the mole fraction of acetone in the air (\(\gamma\)-phase) entering the system is specified as \((y_{A} )_{6} = 0.010\) and assume that the mole fraction of acetone in the water (\(\beta\)-phase) entering the system is specified as \((x_{A} )_{o} = 0\). Given that there are 5 equilibrium stages, reconstruct Figure \(5.7.12c\) and use the reconstructed figure to determine \((y_{A} )_{1}\).


    1. Sandler, S.I. 2006, Chemical, Biochemical, and Engineering Thermodynamics, \(4^{th}\) edition, John Wiley and Sons, New York.↩

    2. Reid, R. C., Prausnitz, J. M., and Sherwood, T. K., 1977, The Properties of Gases and Liquids, Sixth Edition, New York, McGraw-Hill Books.↩

    3. Wisniak, J. 2001, Historical development of the vapor pressure equation from Dalton to Antoine, J. Phase Equilibrium 22, no. 6, 622-630.↩

    4. Gibbs, J.W. 1928, The Collected Works of J. Willard Gibbs, Longmans, Green and Company, London.↩

    5. Prigogine, I. and Defay, R. 1954, Chemical Thermodynamics, Longmans Green and Company, London.↩

    6. This relation is based on the general concept of thermodynamic equilibrium illustrated by Equation \((5.4.5)\).↩

    7. See Figure \(1.3.3\) for a similar description of length scales.↩

    8. The solution of implicit equations is discussed in Appendix B.↩

    9. Stein, S.K. and Barcellos, A. 1992, Calculus and Analytic Geometry, Chapters 5 and 3, McGraw-Hill, Inc., New York.↩

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