# 8.6: Problems

- Page ID
- 44511

## Section 8.1

1. A tank containing 200 gallons of saturated salt solution (3 lb\(_m\) of salt per gallon) is to be diluted by the addition of brine containing 1 lb\(_m\) of salt per gallon. If this solution enters the tank at a rate of 4 gallons per minute and the mixture leaves the tank at the same rate, how long will it take for the concentration in the tank to be reduced to a concentration of 1.01 lb\(_m\) per gallon?

2. A salt solution in a perfectly stirred tank is washed out with fresh water at a rate such that the average residence time, \(\mathscr{V}/Q\), is 10 minutes. Calculate the following:

a) The time in minutes required to remove 99% of the salt originally present

b) The percentage of the original salt removed after the addition of one full tank of fresh water.

3. Two reactants are added to a stirred tank reactor as illustrated in Figure \(\PageIndex{1}\). In the inlet stream containing reactant #2 there is also a miscible liquid catalyst which is added at a level that yields a concentration *in the reactor* of 0.002 moles per liter. The volumetric flow rates of the two reactant streams are equal and the total volumetric rate entering and leaving the stirred tank is 15 gallons per minute.

It has been decided to change the type of catalyst and the change will be made by a substitution of the new catalyst for the old as the reactant and catalyst are continuously pumped into the 10,000 gallon tank. The inlet concentration of the new catalyst is adjusted to provide a final concentration of 0.0030 moles per liter when the mixing process is operating at steady state. Determine the time in minutes required for the concentration of the new catalyst to reach 0.0029 moles per liter.

4. Three perfectly stirred tanks, each of 10,000 gallon capacity, are arranged so that the effluent of the first is the feed to the second and the effluent of the second is the feed to the third. Initially the concentration in each tank is \(c_{ o}\). Pure water is then fed to the first tank at the rate of 50 gallons per minute. You are asked to determine:

a) The time required to reduce the concentration in the first tank to 0.10 \(c_{ o}\)

b) The concentrations in the second and third tanks at this time

c) A general equation for the concentration in the \(n^{th}\) tank in the cascade system illustrated in Figure \(\PageIndex{2}\).

In order to solve an ordinary differential equation of the form

\[ \frac{d\langle c_{A} \rangle }{dt} + g(t)\langle c_{A} \rangle =f(t) \label{4.1}\]

explore the possibility that this *complex equation* can be transformed to the *simple equation* given by

\[ \frac{d}{dt} \left[a(t)\langle c_{A} \rangle \right]=b(t) \label{4.2}\]

We refer to this as a simple equation because it can be integrated directly to obtain

\[ a(t)\langle c_{A} \rangle =\left[a(t) \langle c_{A} \rangle \right]_{t=0} + \int_{\xi = 0}^{\xi = t}b(\xi )d\xi \label{4.3}\]

The new functions, \(a(t)\) and \(b(t)\) can be determined by noting that

\[ \frac{1}{a(t)} \frac{da}{dt} =g(t) , \quad b(t)=a(t)f(t) \label{4.4}\]

Any initial condition for \(a(t)\) will suffice since the solution for \(\langle c_{A} \rangle\) does not depend on the initial condition for \(a(t)\). Use of dummy variables of integration is essential in order to avoid confusion.

5. Develop a solution for Eqs. \((8.1.16)\) and \((8.1.17)\) when \(f(t)\) is given by

\[ f(t)=\begin{cases} c_{A}^{ o} + \left(c_{A}^{1} -c_{A}^{ o} \right) t / \Delta t , & 0<t<\Delta t \\ c_{A}^{1} , & t\geq \Delta t \end{cases} \label{5.1}\]

This represents a process in which the original steady-state concentration is \(c_{A}^{ o}\) and the final steady-state concentration is \(c_{A}^{1}\). The *response time* for the mechanism that creates the change in the concentration of the incoming stream is \(\Delta t\) while the response time of the tank can be thought of as the *mean residence time*, \({\it \tau }\). The time required to approach within 1% of the new steady state is designated as \(t_{1}\), and we express this idea as

\[ \langle c_{A} \rangle =c_{A}^{1} + 0.01 \left(c_{A}^{ o} -c_{A}^{1} \right) , \quad t=t_{1} \label{5.2}\]

For \(\Delta t=0\) we know that \(t_{1} /{\it \tau }=4.6\) on the basis of Equation \((8.1.15)\). In this problem, we want to know what value of \(t_{1} /{\it \tau }\) is required to approach within 1% of the new steady state when \(\Delta t/{\it \tau }=0.2\).

6. Volume 2 of the Guidelines for Incorporating Safety and Health into Engineering Curricula published by the Joint Council for Health, Safety, and Environmental Education of Professionals (JCHSEEP) introduces, without derivation, the following equation for the determination of concentration of contaminants inside a room:

\[\frac{G - Q^{*} C}{G} =e^{- Q^{*}t/V} \nonumber\]

where:

\(C\) = concentration of contaminant at time t | \(G\) = rate of generation of contaminant |

\(Q\) = effective rate of ventilation. | \(Q^* = Q/K\) |

\(V\) = volume of room enclosure | \(K\) = design distribution ventilation constant |

\(t\) = length of time to reach concentration \(C\). |

Making the appropriate assumptions, derive the above equation from the material balance of contaminants. What would be a proper set of units for the variables contained in this equation?

7. Often it is convenient to express the transient concentration in a perfectly mixed stirred tank in terms of the dimensionless concentration defined by

\[C_{A} =\frac{\langle c_{A} \rangle - c_{A}^{ o} }{c_{A}^{1} - c_{A}^{ o} } \nonumber\]

Represent the solution given by Equation \((8.1.15)\) in terms of this dimensionless concentration.

8. A perfectly mixed stirred tank reactor is illustrated in Figure \(\PageIndex{3}\). A feed stream of reactants enters at a volumetric flow rate of \(Q_{ o}\) and the volumetric flow rate leaving the reactor is also \(Q_{ o}\). Under steady state conditions the tank is half full and the volume of the reacting mixture is \(V_{ o}\). At \(t=0\) an inert species is added to the system at a concentration \(c_{A}^{ o}\) and a volumetric flow rate \(Q_{1}\). Unfortunately, someone forgot to change a downstream valve setting and the volumetric flow rate leaving the tank remains constant at the value \(Q_{ o}\). This means that the tank will overflow at \(t=V_{ o} /Q_{1}\). While species \(A\) is *inert* in terms of the reaction taking place in the tank, it is *mildly toxic* and you need to predict the concentration of species \(A\) in the fluid when the tank overflows. Derive a general expression for this concentration taking into account that the concentration of species \(A\) in the tank is zero at \(t=0\).

## Section 8.2

9. Show how Equation \((8.2.6)\) can be used to derive Equation \((8.2.7)\).

10. A perfectly mixed batch reactor is used to carry out the reversible reaction described by

Chemical kinetic schema: \[A+E \quad \overset{\xrightarrow{k_1}}{\overleftarrow{k_2}} \quad B \label{10.1}\]

and use of mass action kinetics provides the chemical kinetic rate equation given by

\[ R_{A} =- \underbrace{ k_{1} c_{A} c_{E} }_{\text{second order reaction} }+\underbrace{ k_{2} c_{B} }_{\text{first order decomposition}} \label{10.2}\]

If species \(E\) is present in great excess, the concentration \(c_{E}\) will undergo negligible changes during the course of the process and we can define a pseudo first order rate coefficient by

\[ k'_{1} =k_{1} c_{E} \label{10.3}\]

Given initial conditions of the form

I.C. \[c_{A} = c_{A}^{ o} , \quad c_{B} = c_{B}^{ o} , \quad t=0 \label{10.4}\]

use the pseudo first order rate coefficient to determine the concentration of species \(A\) and \(B\) as a function of time.

11. When molecular species \(A\) and \(B\) combine to form a product, one often adopts the chemical kinetic schema given by

\[ A + B \xrightarrow{k} \text{ products} \label{11.1}\]

Use of mass action kinetics then leads to a chemical kinetic rate equation of the form

\[ R_{A} =- k c_{A} c_{B} \label{11.2}\]

One must always look upon such rate expressions as hypotheses to be tested by experimental studies. For a homogeneous, liquid-phase reaction, this test can be carried out in a batch reactor which is subject to the initial conditions

I.C.1 \[c_{A} =c_{A}^{ o} , \quad t=0 \label{11.3}\]

I.C.2 \[c_{B} =c_{B}^{ o} , \quad t=0 \label{11.4}\]

Use the macroscopic mole balances for both species \(A\) and \(B\) along with the stoichiometric constraint

\[ R_{A} =R_{B} \label{11.5}\]

in order to derive an expression for \(c_{A}\) as a function of time. In this case one is forced to assume perfect mixing so that \(\langle c_{A} c_{B} \rangle\) can be replaced by \(\langle c_{A} \rangle \langle c_{B} \rangle\).

12. A batch reactor illustrated in Figure \(\PageIndex{4}\) is used to study the irreversible, decomposition reaction

\[ A \xrightarrow{k} \text{ products} \label{12.1}\]

The proposed chemical kinetic rate equation is

\[ R_{A} =- k c_{A} \label{12.2}\]

and this decomposition reaction is catalyzed by sulfuric acid. To initiate the batch process, a small volume of catalyst is placed in the reactor as illustrated in Figure \(\PageIndex{4}\). At the time, \(t=0\), the solution of species \(A\) is added at a volumetric flow rate \(Q_{ o}\) and a concentration \(c_{A}^{ o}\). When the reactor is full, the stream of species \(A\) is shut off and the system proceeds in the normal manner for a batch reactor. During the start-up time, the volume of fluid in the reactor can be expressed as

\[ \mathscr{V}(t)=V_{ o} + Q_{ o} t \label{12.3}\]

and the final volume of the fluid is given by

\[ \mathscr{V}_{ 1} =V_{ o} + Q_{ o} t_{1} \label{12.4}\]

Here \(t_{1}\) is the start-up time. In this problem you are asked to determine the concentration of species \(A\) during the start-up time and all subsequent times. The analysis for the start-up time can be simplified by means of the transformation

\[ y(t)=\langle c_{A} \rangle \mathscr{V}(t) \label{12.5}\]

and use of the initial condition

I.C. \[y=0 , \quad t=0 \label{12.6}\]

After you have determined \(y(t)\) you can easily determine \(\langle c_{A} \rangle\) during the start-up period. The concentration at \(t=t_{1}\) then becomes the initial condition for the analysis of the system for all subsequent times. Often one simplifies the analysis of a batch reactor by assuming that the time required to fill the reactor is *negligible*. Use your solution to this problem to identify what is meant by “negligible” for this particular problem.

13. In the perfectly mixed continuous stirred tank reactor illustrated in Figure \(\PageIndex{5a}\), species \(A\) undergoes an irreversible reaction to form products according to

\[ A \xrightarrow{k} \text{ products, } \quad R_{A} =-k c_{A} \label{13.1}\]

The original volume of fluid in the reactor is \(V_{ o}\) and the original volumetric flow rate *into and out of* the reactor is \(Q_{ o}\). The concentration of species \(A\) *entering* the reactor is fixed at \(c_{A}^{ o}\) and under steady state operating conditions the concentration at the exit (and therefore the concentration in the reactor) is \(\langle c_{A} \rangle\). Determine the concentration \(\langle c_{A} \rangle\) under steady state operating conditions.

Because of changes in the downstream processing, it is necessary to reduce the concentration of species \(A\) leaving the reactor. This is to be accomplished by increasing the volume of the reactor by adding pure liquid to the reactor, as illustrated in Figure \(\PageIndex{5b}\), at a volumetric flow rate \(Q_{1}\) until the desired reactor volume is achieved. During the transient period when the volume is given by \(\mathscr{V}\left(t\right)=V_{ o} +Q_{1} t\), the exit flow rate is constant at \(Q_{ o}\). In this problem you are asked to determine the exit concentration during this transient period.

## Section 8.3

14. Consider the process studied in Example \(8.2.1\), subject to an initial condition of the form,

I.C. \[c_{A} =c_{A}^{ o} , \quad c_{B} =c_{B}^{ o} , \quad t=0 \nonumber\]

and determine the concentration of species \(A\) as a function of time.

15. For the process studied in Example \(8.2.1\), assume that the equilibrium coefficient and the first order rate coefficient have the values

\[ K_{eq} =10^{-1} , \quad k_{1} =10 \ { min}^{-1} \label{15.1}\]

and determine the time, \(t_{1}\) required for the concentration of species \(A\) to be given by

\[ \langle c_{A} \rangle =c_{A}^{ o} + 0.99\left(\langle c_{A} \rangle_{eq} - c_{A}^{ o} \right) , \quad t=t_{1} \label{15.2}\]

Here \(\langle c_{A} \rangle_{eq}\) represents the equilibrium concentration.

16. In this problem we consider the heated, semi-batch reactor shown in Figure \(\PageIndex{6}\) where we have identified the vapor phase as the \(\gamma\)-phase and the liquid phase as the \(\beta\)-phase. This reactor has been designed to determine the chemical kinetics of the dehydration of t-butyl alcohol (species \(A\)) to produce isobutylene (species \(B\)) and water (species \(C\)). The system is initially charged with t-butyl alcohol; a catalyst is then added which causes the dehydration of the alcohol to form isobutylene and water. The isobutylene escapes through the top of the reactor while the water and t-butyl alcohol are condensed and remain in the reactor. If one measures the concentration of the t-butyl alcohol in the liquid phase, the rate of reaction can be determined and that is the objective of this particular experiment.

The analysis begins with the fixed control volume shown in Figure \(\PageIndex{6}\) and the general macroscopic balance given by

\[ \frac{d}{dt} \int_{\mathscr{V}}c_{D} dV + \int_{\mathscr{A}}c_{D} \mathbf{v}_{D} \cdot \mathbf{n} dA=\int_{\mathscr{V}}R_{D } dV , \quad D=A, B, C \label{16.1}\]

The moles of species \(A\) (t-butyl alcohol) in the \(\gamma\)-phase can be neglected, thus the macroscopic balance for this species takes the form

t-butyl alcohol: \[\frac{d}{dt} \int_{V_{\beta } (t)}c_{A\beta } dV =\int_{V_{\beta } (t)}R_{A\beta } dV \label{16.2}\]

and in terms of average values for the concentration and the reaction rate we have

t-butyl alcohol: \[\frac{d}{dt} \left[\langle c_{A\beta } \rangle V_{\beta } (t)\right]=\langle R_{A\beta } \rangle V_{\beta } (t) \label{16.3}\]

If we also assume that the moles of species \(B\) (isobutylene) and species \(C\) (water) are negligible in the \(\gamma\)-phase, the macroscopic balances for these species take the form

isobutylene: \[\frac{d}{dt} \left[\langle c_{B\beta } \rangle V_{\beta } (t)\right] + \dot{M}_{B\gamma } =\langle R_{B\beta } \rangle V_{\beta } (t) \label{16.4}\]

water: \[\frac{d}{dt} \left[\langle c_{C\beta } \rangle V_{\beta } (t)\right]=\langle R_{C\beta } \rangle V_{\beta } (t) \label{16.5}\]

This indicates that alcohol and water are retained in the system by the condenser while the isobutylene

leaves the system at a molar rate given by \(\dot{M}_{B\gamma }\). The initial conditions for the three molecular species are given by

IC. \[\langle c_{A} \rangle =c_{A}^{ o} , \quad t=0 \label{16.6a}\]

IC. \[\langle c_{B} \rangle =0 , \quad t=0 \label{16.6b}\]

IC. \[\langle c_{C} \rangle =0 , \quad t=0 \label{16.6c}\]

Since the molar rates of reaction are related by

\[ R_{C\beta } =-R_{A\beta } , \quad \text{ and } \quad R_{B\beta } =-R_{A\beta } \label{16.7}\]

we need only be concerned with the rate of reaction of the t-butyl alcohol. If we treat the reactor as *perfectly mixed*, the mole balance for t-butyl alcohol can be expressed as

\[ R_{A\beta } =\frac{dc_{A\beta } }{dt} + \frac{c_{A\beta } }{V_{\beta } (t)} \frac{dV_{\beta } (t)}{dt} \label{16.8}\]

This indicates that we need to know both \(c_{A\beta }\) and \(V_{\beta }\) as functions of time in order to obtain *experimental values* of \(R_{A\beta }\). The volume of fluid in the reactor can be expressed as

\[ V_{\beta } (t)=n_{A\beta } \bar{ v}_{A} + n_{B\beta } \bar{ v}_{B} + n_{C\beta } \bar{ v}_{C} \label{16.9}\]

in which \(n_{A\beta }\), \(n_{B\beta }\) and \(n_{C\beta }\) represent the moles of species \(A\), \(B\) and \(C\) in the \(\beta\)-phase and \(\bar{ v}_{A}\), \(\bar{ v}_{B}\) and \(\bar{ v}_{C}\) represent the partial molar volumes.

To develop a useful expression for \(R_{A\beta }\), *assume* that the liquid mixture is ideal so that the partial molar volumes are constant. In addition, *assume* that the moles of species \(B\) in the liquid phase are negligible. On the basis of these assumptions, show that Equation \ref{16.8} can be expressed as

\[ R_{A\beta } =\frac{d c_{A\beta } }{dt} \left\{1 + \frac{c_{A\beta } \left(\bar{ v}_{A} -\bar{ v}_{C} \right)}{\left[1 - c_{A\beta } \left(\bar{ v}_{A} -\bar{ v}_{C} \right)\right]} \right\} \label{16.10}\]

This form is especially useful for the interpretation of *initial rate data*, i.e., experimental data can be used to determine both \(c_{A\beta }\) and \(dc_{A\beta } /dt\) at \(t=0\) and this provides an experimental determination of \(R_{A\beta }\) for the initial conditions associated with the experiment.

An alternate approach^{10} to the determination of \(R_{A\beta }\) is to measure the molar flow rate of species \(B\) that leaves the reactor in the \(\gamma\)-phase and relate that quantity to the rate of reaction.

## Section 8.4

17. When Eqs. \((8.4.9)\) and \((8.4.10)\) are valid, Equation \((8.4.11)\) represents a valid result for the chemostat shown in Figure \(\PageIndex{10}\). One can divide this equation by a constant, \(m_{cell}\), to obtain Equation \((8.4.15)\); however, the average mass of a cell, \(m_{cell}\), in the chemostat may not be the average mass of a cell in the incoming stream. If \(m_{cell}\) is different than \((m_{cell} )_{1}\), Equation \((8.4.15)\) is still correct, but our *interpretation* of \(\langle n\rangle_{1}\) is not correct. Consider the case, \((m_{cell} )_{1} \neq m_{cell}\), and develop a new version of Equation \((8.4.15)\) in which the incoming flux of cells is interpreted properly in terms of the number of cells per unit volume in the incoming stream.

18. Develop a general solution for Eqs. \((8.4.22)-(8.4.23)\) and consider the behavior of the system for \(D<{\it \mu }\), \(D={\it \mu }\) and \(D>{\it \mu }\).

## Section 8.5

19. Repeat the analysis of binary batch distillation when Raoult’s law is applicable, i.e., the *process equilibrium relation* is given by

\[\langle y_{A} \rangle_{exit} =\frac{\alpha_{eff} \langle x_{A} \rangle }{1 + \langle x_{A} \rangle \left(\alpha_{eff} -1\right)} \nonumber\]

Here \(\alpha_{eff}\) is the effective relative volatility and is temperature dependent; however, in this problem you may assume that \(\alpha_{eff}\) is constant. Use your result to construct figures comparable to Figure \(\PageIndex{12}\) for \(x_{A}^{ o} =0.1\) and \(x_{A}^{ o} =0.5\) with values of \(\alpha_{eff}\) given by \(1/4\), \(1/2\), 1, 2, and 4.

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