# 4.10: Problems

- Page ID
- 44484

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

## Section 4.1

1. Use Equation \((4.1.20)\) to obtain a macroscopic *mole balance* for species \(A\) in terms of a moving control volume. Indicate how your result can be used to obtain Equation \((4.1.17)\).

## Section 4.2

2. Determine the mass density, \(\rho\), for the mixing process illustrated in Figure \(4.2.1\).

3. A liquid hydrocarbon mixture was made by adding 295 kg of benzene, 289 kg of toluene and 287 kg of p-xylene. Assume there is no change of volume upon mixing, i.e., \(\Delta V_{mix} = 0\), in order to determine:

- The species density of each species in the mixture.
- The total mass density.
- The mass fraction of each species.

4. A gas mixture contains the following quantities (per cubic meter) of carbon monoxide, carbon dioxide and hydrogen: carbon monoxide, 0.5 kmol/m\(^{3}\), carbon dioxide, 0.5 kmol/m\(^{3}\), and hydrogen, 0.6 kmol/m\(^{3}\). Determine the species mass density and mass fraction of each of the components in the mixture.

5. The species mass densities of a three-component (\(A\), \(B\), and \(C\)) liquid mixture are: acetone, \(\rho_{A} = 326.4 \ kg/m^{ 3}\), acetic acid, \(\rho_{B} = 326.4 \ kg/m^{ 3}\), and ethanol, \(\rho_{C} = 217.6 \ kg/m^{ 3}\). Determine the following for this mixture:

- The mass fraction of each species in the mixture.
- The mole fraction of each species in the mixture.
- The mass of each component required to make one cubic meter of mixture.

6. A mixture of gases contains one kilogram of each of the following species: methane \((A)\), ethane \((B)\), propane \((C)\), carbon dioxide \((D)\), and nitrogen \((E)\). Calculate the following:

- The mole fraction of each species in the mixture
- The average molecular mass of the mixture

7. Two gas streams, having the flow rates and properties indicated in Table \(\PageIndex{1}\), are mixed in a pipeline. Assume *perfect mixing*, i.e. no change of volume upon mixing, and determine the composition of the mixed stream in moles/m\(^{3}\).

Stream #1 | Stream #2 | |
---|---|---|

Mass flow rate | 0.226 kg/s | 0.296 kg/s |

methane |
0.48 kg/m\(^{3}\) | 0.16 kg/m\(^{3}\) |

ethane |
0.90 kg/m\(^{3}\) | 0.60 kg/m\(^{3}\) |

propane |
0.88kg/m3 | 0.220 kg/m\(^{3}\) |

8.Develop a representation for the *mole fraction* of species \(A\) in an \(N\)-component system in terms of the mass fractions and molecular masses of the species. Use the result to prove that the mass fractions and mole fractions in a binary system are equal when the two molecular masses are equal.

9. Derive the total mass balance for an arbitrary moving control volume beginning with the species mass balance given by Equation \((4.1.20)\).

## Section 4.3

10. The species velocities, in a *binary system*, can be decomposed according to

\[ \mathbf{v}_{A} = \mathbf{v} + \mathbf{u}_{A} , \quad \mathbf{v}_{B} = \mathbf{v} + \mathbf{u}_{B}\]

in which \(\mathbf{v}\) represents the mass average velocity defined by Equation \((4.2.17)\). One can use this result, along with the definition of the mass average velocity, *to prove* that

\[ \mathbf{v}_{A} = \mathbf{v}_{B} + \left(\frac{1}{1-\omega_{A} } \right)\mathbf{u}_{A} \label{2}\]

This means that the *approximation*, \(\mathbf{v}_{A} \approx \mathbf{v}_{B}\) requires the restriction

\[ \left|\mathbf{u}_{A} \right|<<(1-\omega_{A} )\left|\mathbf{v}_{A} \right|\]

Since \(1-\omega_{A}\) is always less than one, we can always satisfy this inequality whenever the mass diffusion velocity is small compared to the species velocity, i.e.,

\[ \left|\mathbf{u}_{A} \right|<<\left|\mathbf{v}_{A} \right|\]

For the sulfur dioxide mass transfer process illustrated in Figure \(4.3.2\), this means that the approximation

\[ \mathbf{v}_{ \ce{SO2} } \cdot \mathbf{k}\approx \mathbf{v}_{air} \cdot \mathbf{k}\]

is valid whenever the mass diffusion velocity is restricted by

\[ \left|\mathbf{u}_{\ce{SO2} } \cdot \mathbf{k}\right|<<\left|\mathbf{v}_{\ce{SO2} } \cdot \mathbf{k}\right|\]

In many practical cases, this restriction is satisfied and all species velocities can be approximated by the mass average velocity.

Next we direct our attention to the mass transfer process at the gas-liquid interface illustrated in Figure \(4.3.3\). If we assume that there is no mass transfer of air into or out of the liquid phase, we can *prove* that

\[ \mathbf{u}_{\ce{SO2} } \cdot \mathbf{n} = \left(1-\omega_{\ce{ SO2} } \right)\mathbf{v}_{\ce{SO2} } \cdot \mathbf{n} , \quad \text{ at the gas-liquid interface} \label{7}\]

Under these circumstances, the mass diffusion velocity is *never small* compared to the species velocity for practical conditions. Thus the type of approximation indicated by Equation \((4.3.7)\) is *never valid* for the component of the velocity *normal* to the gas-liquid interface. As a simplification, we can treat the sulfur dioxide-air system as a binary system with species \(A\) representing the sulfur dioxide and species \(B\) representing the air.

(a) Use the definition of the mass average velocity given by Equation \((4.2.17)\) to prove Equation \ref{2}.

(b) Use \(\mathbf{v}_{ air} \cdot \mathbf{n} = \mathbf{v}_{B} \cdot \mathbf{n} = 0\) in order to prove Equation \ref{7}.

## Section 4.4

11. A three component liquid mixture flows in a pipe with a mass averaged velocity of \( \mathbf{v} = 0.9 \ m/s\). The density of the mixture is \(\rho\) = 850 kg/m\(^{3}\). The components of the mixture and their mole fractions are: n-pentane, \(x_{P} = 0.2\), benzene, \(x_{B} = 0.3\), and naphthalene, \(x_{N} = 0.5\). The diffusion fluxes of each component in the streamwise direction are: pentane, \(\rho_{P} u_{P} = 1{ .564}\times { 10}^{-6} \ kg/m^{ 2} { s}\), benzene, \(\rho_{B} u_{B} = 1{ .563}\times 1{ 0}^{-6} \ kg/m^{ 2} { s}\), and naphthalene, \(\rho_{N} u_{N} = -3.127\times { 10}^{-6} \ kg/m^{ 2} { s}\). Determine the diffusion velocities and the species velocities of the three components. Use this result to determine the molar averaged velocity, \(\mathbf{v}^{*}\). Note that you must use eight significant figures in your computation.

## Section 4.5

12. Sometimes heterogeneous chemical reactions take place at the walls of tubes in which reactive mixtures are flowing. If species \(A\) is being consumed at a tube wall because of a chemical reaction, the concentration profile may be of the form

\[ c_{A} (r) = c_{A}^{ o} \left[1-\phi \left({r / r_{ o} } \right)^{2} \right]\]

Here \(r\) is the radial position and \(r_{ o}\) is the tube radius. The parameter \(\phi\) depends on the net rate of production of chemical species at the wall and the molecular diffusivity, and it is bounded by \(0\leq \phi \leq 1\). If \(\phi\) is zero, the concentration across the tube is uniform at the value \(c_{A}^{ o}\). If the flow in the tube is laminar, the velocity profile is given by

\[ \mathbf{v}_{z} (r) = 2\langle \mathbf{v}_{z} \rangle \left[1-\left({r / r_{ o} } \right)^{2} \right]\]

and the volumetric flow rate is

\[ Q = \langle \mathbf{v}_{z} \rangle \pi r_{ o}^{ 2}\]

For this process, determine the molar flow rate of species \(A\) in terms \(c_{A}^{ o}\), \(\phi\), and \(\langle \mathbf{v}_{z} \rangle\). When \(\phi = 0.5\), determine the bulk concentration, \(\langle c_{A} \rangle_{b}\),and the area-averaged concentration, \(\langle c_{A} \rangle\). Use these results to determine the difference between \(\dot{M}_{A}\) and \(\langle c_{A} \rangle Q\).

13. A flash unit is used to separate vapor and liquid streams from a liquid stream by lowering its pressure before it enters the flash unit. The feed stream is pure liquid water and its mass flow rate is 1000 kg/hr. Twenty percent (by weight) of the feed stream leaves the flash unit with a density \(\rho\) = 10 kg/m\(^{3}\). The remainder of the feed stream leaves the flash unit as liquid water with a density \(\rho\) = 1000 kg/m\(^{3}\) Determine the following:

- The mass flow rates of the exit streams in kg/s.
- volumetric flow rates of exit streams in m\(^{3}\)/s.

## Section 4.6

14. Show that Equation \((4.6.2)\) results from Equation \((4.6.1)\) when either \(c \mathbf{v} \cdot \mathbf{n}\) or \(x_{A}\) is constant over the area of the exit.

15. Use Equation \((4.6.2)\) to prove Equation \((4.6.3)\).

16. Derive Equation \((4.6.5)\) given that either \(\rho \mathbf{v} \cdot \mathbf{n}\) or \(\omega_{A}\) is constant over the area of the exit.

17. Prove Equation \((4.6.9)\).

## Section 4.7

18. Determine \(\dot{M}_{3}\) and the unknown mole fractions for the distillation process described in Sec. 4.7 subject to the following conditions:

Table \(\PageIndex{2}\)

Stream #1 | Stream #2 | Stream #3 |
---|---|---|

\(\dot{M}_{1} = 1200\) mol/hr | \(\dot{M}_{2} = 250\) mol/hr | \(\dot{M}_{3} = ?\) |

\(x_{A} = 0.3\) | \(x_{A} = 0.8\) | \(x_{A} = ?\) |

\(x_{B} = 0.2\) | \(x_{B} = ?\) | \(x_{B} = 0.25\) |

\(x_{C} = ?\) | \(x_{C} = ?\) | \(x_{C} = ?\) |

19. A continuous filter is used to separate a clear filtrate from alumina particles in a slurry. The slurry has 30% by weight of alumina (specific gravity of alumina = 4.5). The cake retains 5% by weight of water. For a feed stream of 1000 kg/hr, determine the following:

- The mass flow rate of particles and water in the input stream
- The volumetric flow rate of the inlet stream in m\(^{3}\)/s.
- The mass flow rate of filtrate and cake in kg/s.

20. A BTX unit, shown in Figure \(\PageIndex{1}\), is associated with a refinery that produces benzene, toluene, and xylenes. Stream #1 leaving the reactor-reforming unit has a volumetric flow rate of 10 m\(^{3}\)/hr and is a

mixture of benzene \((A)\), toluene \((B)\), and xylenes \((C)\) with the following composition:

\[\langle c_{A} \rangle_{1} = { 6,000} \ mol/m^{ 3} , \quad \langle c_{B} \rangle_{1} = 2,000 \ mol/m^{ 3} , \quad \langle c_{C} \rangle_{1} = 2,000 \ mol/m^{ 3}\nonumber\]

Stream (1) is the feed to a distillation unit where the separation takes place according to the following specifications:

- 98% of the benzene leaves with the distillate stream (stream #2).
- 99% of the toluene in the feed leaves with the bottoms stream (stream #3)
- 100% of the xylenes in the feed leaves with the bottoms stream (stream #3).

Assuming that the volumes of components are additive, and using the densities of pure components from Table I in the Appendix, compute the concentration and volumetric flow rate of the distillate (stream #2) and bottoms (stream #3) stream leaving the distillation unit.

21. A standard practice in refineries is to use a holding tank in order to mix the light naphtha output of the refinery for quality control. During the first six hours of operation of the refinery, the stream feeding the holding tank at 200 kg/min had 30% by weight of n-pentane, 40% by weight of n-hexane, 30% by weight of n-heptane. During the next 12 hours of operation the mass flow rate of the feed stream was 210 kg/min and the composition changed to 40% by weight of n-pentane, 40% by weight of n-hexane, and 20% by weight of n-heptane. Determine the following:

The average density of the feed streams

The concentration of the feed streams in moles/m\(^{3}\).

After 12 hours of operation, and assuming the tank was empty at the beginning, determine:

The volume of liquid in the tank in m\(^{3}\).

The concentration of the liquid in the tank, in mol/m\(^{3}\).

The partial density of the species in the tank.

22. A distillation column is used to separate a mixture of methanol, ethanol, and isopropyl alcohol. The feed stream, with a mass flow rate of 300 kg/hr, has the following composition:

Table \(\PageIndex{3}\)

Component | Species mass density |
---|---|

methanol |
395.5 kg/m\(^{3}\) |

ethanol |
197.3 kg/m\(^{3}\) |

isopropyl alcohol |
196.5 kg/m\(^{3}\) |

Separation of this mixture of alcohol takes place according to the following specifications:

(a) 90% of the methanol in the feed leaves with the distillate stream

(b) 5% of the ethanol in the feed leaves with the distillate stream

(c) 3% of the isopropyl alcohol in the feed leaves with the distillate stream.

Assuming that the volumes of the components are additive, compute the concentration and volumetric flow rates of the distillate and bottom streams.

23. A mixture of ethanol \((A)\) and water \((B)\) is separated in a distillation column. The volumetric flow rate of the feed stream is 5 m\(^{3}\)/hr. The concentration of ethanol in the feed is \(c_{A} = 2,800 \ mol/m^{ 3}\). The distillate leaves the column with a concentration of ethanol \(c_{A} = 13,000 \ mol/m^{ 3}\). The volumetric flow rate of distillate is one cubic meter per hour. How much ethanol is lost through the bottoms of the column, in kilograms of ethanol per hour?

24. A ternary mixture of benzene, ethylbenzene, and toluene is fed to a distillation column at a rate of \(10^{5}\)mol/hr. The composition of the mixture in % moles is: 74% benzene, 20% toluene, and 6% ethylbenzene. The distillate flows at a rate of \(75\times 10^{3}\)mol/hr. The composition of the distillate in % moles is 97.33 % benzene, 2% toluene, and the rest is ethylbenzene. Find the molar flow rate of the bottoms stream and the mass fractions of the three components in the distillate and bottoms stream.

25. A complex mixture of aromatic compounds leaves a chemical reactor and is fed to a distillation column. The mass fractions and flow rates of distillate and bottoms streams are given in Table \(\PageIndex{4}\). Compute the molar flow rate and composition, in molar fractions, of the feed stream.

(kg/hr) | \(\omegaup\)\(_{Benzene}\) | \(\omegaup\)\(_{Toluene}\) | \(\omegaup\)\(_{Benzaldehide}\) | \(\omegaup\)\(_{BenzoicAcid}\) | \(\omegaup\)\(_{MethylBenzoate}\) | |
---|---|---|---|---|---|---|

Distillate | 125 | 0.1 | 0.85 | 0.03 | 0.0 | 0.02 |

Bottoms | 76 | 0.0 | 0.05 | 0.12 | 0.8 | 0.03 |

26. A hydrocarbon feedstock is available at a rate of \(10^{6}\)mol/hr, and consists of propane (\(x_{A} = 0.2\)), n-butane (\(x_{B} = 0.3\)), n-pentane (\(x_{C} = 0.2\)) and n-hexane (\(x_{D} = 0.3\)). The distillate contains all of the propane in the feed to the unit and 80% of the pentane fed to the unit. The mole fraction of butane in the distillate is \(y_{B} = 0.4\). The bottom stream contains all of the hexane fed to the unit. Calculate the distillate and bottoms streams flow rate and composition in terms of mole fractions.

## Section 4.8

27. It is possible that the process illustrated in Figure \(4.8.3\) could be analyzed beginning with Control Volume I rather than beginning with Control Volume II. Begin the problem with Control Volume I and carry out a degree-of-freedom analysis to see what difficulties might be encountered.

28.\(\ddagger\) In a glycerol plant, a 10% (mass basis) aqueous glycerin solution containing 3% NaCl is treated with butyl alcohol as illustrated in Figure \(\PageIndex{2}\). The alcohol fed to the tower contains 2% water on a mass basis. The raffinate leaving the tower contains all the original salt, 1.0% glycerin and 1.0% alcohol. The extract from the tower is sent to a distillation column. The distillate from this column is the alcohol containing 5% water.

The bottoms from the distillation column are 25% glycerin and 75% water. The two feed streams to the extraction tower have equal mass flow rates of 1000 \(lb_m\) per hour. Determine the output of glycerin in pounds per hour from the distillation column.

## Section 4.9

29.\(\ddagger\) In Sec. 4.8 the solution to the distillation problem was shown to reduce to solving the matrix equation, \(\mathbf{ Au} = \mathbf{ b}\), in which

\[\mathbf{ A} = \begin{bmatrix} {1} & {1} & {1} \\ {0.045} & {0.069} & {0.955} \\ {0.091} & {0.901} & {0.041} \end{bmatrix} , \quad \mathbf{ u} = \begin{bmatrix} {\dot{m}_{2} } \\ {\dot{m}_{3} } \\ {\dot{m}_{4} } \end{bmatrix} , \quad \mathbf{ b} = \begin{bmatrix} {1000} \\ {500} \\ {300} \end{bmatrix}\nonumber\]

Here it is understood that the mass flow rates have been made dimensionless by dividing by \({ lb}_{ m} /{ hr}\). In addition to the matrix \(\mathbf{A}\), one can form what is known as an *augmented matrix*. This is designated by \(\mathbf{ A}\dot{\dot{\cdot }}\mathbf{ b}\) and it is constructed by adding the column of numbers in \(\mathbf{b}\) to the matrix \(\mathbf{A}\) in order to obtain

\[\mathbf{ A}\dot{\dot{\cdot }}\mathbf{ b} = \begin{bmatrix} {1} & {1} & {1} & {.} & {1000} \\ {0.045} & {0.069} & {0.955} & {.} & {500} \\ {0.091} & {0.901} & {0.041} & {.} & {300} \end{bmatrix}\nonumber\]

Define the following *lists* in *Mathematica* corresponding to the rows of the augmented matrix, \(\mathbf{ A}\dot{\dot{\cdot }}\mathbf{ b}\).

\[\begin{align*} & R1 && = && \left\{ {1,} {1,} {1,} {1000} \right\} \\ & R2 && = && \left\{ {0.045,} {0.069,} {0.955,} {500} \right\} \\ & R3 && = && \left\{ {0.091,} {0.901,} {0.041,} {300} \right\} \end{align*}\nonumber\]

Write a sequence of *Mathematica* expressions that correspond to the elementary row operations for solving this system. The first elementary row operation that given Equation \((4.9.29)\) is

\[R2 = (-0.045) R2 + R1\nonumber\]

Show that you obtain an augmented matrix that defines Equation \((4.9.33)\).

30.\(\ddagger\) In this problem you are asked to continue exploring the use of Mathematica in the analysis of the set of linear equations studied in Sec. 4.9.2, i.e., \(\mathbf{ Au} = \mathbf{ b}\) where the matrices are defined by

\[\mathbf{ A} = \begin{bmatrix} {1} & {1} & {1} \\ {0.045} & {0.069} & {0.955} \\ {0.091} & {0.901} & {0.041} \end{bmatrix} , \quad \mathbf{ u} = \begin{bmatrix} {\dot{m}_{2} } \\ {\dot{m}_{3} } \\ {\dot{m}_{4} } \end{bmatrix} , \quad \mathbf{ b} = \begin{bmatrix} {1000} \\ {500} \\ {300} \end{bmatrix}\nonumber\]

(1) Construct the augmented matrix \(\mathbf{ A}\dot{\dot{\cdot }}\mathbf{ I}\) according to

\[\mathbf{ A}\dot{\dot{\cdot }}\mathbf{ I} = \begin{bmatrix} {1} & {1} & {1} & {.} & {1} & {0} & {0} \\ {0.045} & {0.069} & {0.955} & {.} & {0} & {1} & {0} \\ {0.091} & {0.901} & {0.041} & {.} & {0} & {0} & {1} \end{bmatrix}\nonumber\]

and use elementary row operations to transform this augmented matrix to the form

\[\mathbf{ A}\dot{\dot{\cdot }}\mathbf{ I} = \begin{bmatrix} {1} & {0} & {0} & {.} & & & \\ {0} & {1} & {0} & {.} & & {\it B} & \\ {0} & {0} & {1} & {.} & & & \end{bmatrix}\nonumber\]

Show that the elements represented by \(B\) make up the matrix \(\mathbf{B}\) having the property that \(\mathbf{ B} = \mathbf{ A}^{-1}\). Use your result to calculate \(\mathbf{ u} = \mathbf{ A}^{-1} \mathbf{ b}\).

(2) Show that the inverse found in Part 1 satisfies \(\mathbf{ A} \mathbf{ A}^{-1} = \mathbf{ I}\).

(3) Use *Mathematica*’s built-in function **Inverse** to find the inverse of \(\mathbf{A}\).

(4) Use *Mathematica*’s **RowReduce** function on the augmented matrix \(\mathbf{ A}\dot{\dot{\cdot }}\mathbf{ b}\) and show that from the row echelon form you can obtain the same results as in (1).

(5) Use *Mathematica*’s **Solve** function to solve \(\mathbf{ Au} = \mathbf{ b}\).

- Lavoisier, A. L. 1777,. Memoir on Combustion in General,
*Mémoires de L’Academie Roayal des Sciences*592-600.↩ - Toulmin, S.E. 1957, Crucial Experiments: Priestley and Lavoisier, Journal of the History of Ideas,
**18**, 205-220.↩ - Here it should be clear that \(R_A\) represents both the
*creation*of species \(A\) (when \(R_A\) is positive) and the*consumption*of species \(A\) (when \(R_A\) is negative).↩ - See Section 2.1.1.↩
- Whitaker, S. 2012, Mechanics and Thermodynamics of Diffusion, Chem. Engng. Sci.
**68**, 362-375.↩