# 4.27: Untitled Page 82

## Chapter 4

 1 . 1616

1 . 16484 1 20005

.

1

A

0 . 1152

0 0677

.

1 2326

.

(4‐152)

0 . 0464

1 0971

.

0 0325

.



The solution procedure that led from Eq. 4‐142 to the final answer can be expressed in compact form according to,

1

1

1

A u = b ,  A Au = A b ,  Iu = A b ,

1

u = A b

(4‐153)

and in terms of the details of the inverse matrix the last of these four equations can be expressed as

(4‐154)

Given that computer routines are available to carry out the Gauss‐Jordan algorithm, it should be clear that the solution of a set of linear mass balance equations is a routine matter.

4.10 Problems

Section 4.1

4‐1. Use Eq. 4‐20 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 Eq. 4‐17.

Section 4.2

4‐2. Determine the mass density,  , for the mixing process illustrated in Figure 4‐2.

4‐3. 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., V

 0 , in order to determine:

mix

1. The species density of each species in the mixture.

2. The total mass density.

3. The mass fraction of each species.

4‐4. A gas mixture contains the following quantities (per cubic meter) of carbon monoxide, carbon dioxide and hydrogen: carbon monoxide, 0.5 kmol/m3, carbon

Multicomponent systems

146

dioxide, 0.5 kmol/m3, and hydrogen, 0.6 kmol/m3. Determine the species mass density and mass fraction of each of the components in the mixture.

4‐5. The species mass densities of a three‐component ( A, B, and C) liquid mixture are: acetone,

3

  326 4 kg/m , acetic acid,

3

  326 4 kg/m , and ethanol,

A

.

B

.

3

  217 6 kg/m . Determine the following for this mixture:

C

.

1. The mass fraction of each species in the mixture.

2 The mole fraction of each species in the mixture.

3. The mass of each component required to make one cubic meter of mixture.

4‐6. 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:

1. The mole fraction of each species in the mixture

2. The average molecular mass of the mixture

4‐7. Two gas streams, having the flow rates and properties indicated in Table 4.7, 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 mol/m3.

Table 4.7. Composition of gas streams

Stream #1

Stream #2

Mass flow rate

0.226 kg/s

0.296 kg/s

methane

0.48 kg/m3

0.16 kg/m3

ethane

0.90 kg/m3

0.60 kg/m3

propane

0.88kg/m3

0.220 kg/m3

4‐8.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.

4‐9. Derive the total mass balance for an arbitrary moving control volume beginning with the species mass balance given by Eq. 4‐20.

Section 4.3

4‐10. The species velocities, in a binary system, can be decomposed according to

147