gas-liquid system

Equation Section 4© May 22, 2008, Cerro, Higgins & WhitakerEquation Section 5

Chapter 5

Gas-Liquid Systems In the previous chapter, we began our study of macroscopic mass and mole balances for multicomponent systems. There we encountered a variety of measures of concentration and we summarize these measures as

mass of species per unit volumeA

Aρ = (5-1a)

(5-1b) 1

, total mass densityA N

AA

=

=

ρ = ρ∑ , mass fractionA Aω = ρ ρ (5-1c)

, molar concentrationA A Ac MW= ρ (5-1d)

(5-1e) 1

, total molar concentrationA N

AA

c c=

=

= ∑ or , mole fractionA A Ay x c c= (5-1f)

In the analysis of gas-phase systems it is often important to relate the concentration to the pressure and temperature. This is done by means of an equation of state, often known as a p-V-T relation. In this chapter we will make use of the ideal gas relations; however, many processes operate under conditions such that the ideal gas laws do not apply and one must make use of more general p-V-T relations. Non-ideal gas phase behavior will be studied in a subsequent course in thermodynamics. 5

.1 Ideal Gas-Phase Behavior

For an N-component ideal gas mixture we have the following relations , 1, 2, ...A Ap V n RT A N= = (5-2)

Here Ap is the partial pressure of species A and R is the gas constant. Values of the gas constant in different units are given in Table 5-1. If we sum Eq. 5-2 over all N species we obtain

pV nRT= (5-3)

where

1

A N

AA

p=

=

= p∑ (5-4)

1

A N

AA

n=

=

= n∑ (5-5)

Equations 5-2 through5-5 are sometimes referred to as Dalton's Laws.

108

Gas-Liquid Systems 109

Table 5-1. Numerical values of the gas constant, R

Numerical value Units 8.314 m3 Pa/ mol K 8.314 J/mol K 0.08314 liter bar/mol K 82.06 atm-cm3 /mol K 1.987 cal/mol K

In Figure 5-1 we have illustrated a constant pressure, isothermal mixing process. In the compartment

constantpressure

constantpressure

VA

species A

VB

species B

a) before mixing

b) after mixing

mixture of andA B

Figure 5-1. Constant pressure, isothermal mixing process containing species A illustrated in Figure 5-1a, we can use Eq. 5-2 to obtain

A ApV n RT= (5-6)

while in the compartment containing species B we have

B BpV n RT= (5-7)

Upon removal of the partition and mixing, we have the situation illustrated in Figure 5-1b. For that condition, we can use Eq. 5-3 to obtain

Chapter 5 110

( )A BpV n n RT= + (5-8)

where the volume is given by

mixA BV V V V= + + ∆ (5-9)

By definition, an ideal gas mixture obeys what is known as Amagat's Law, i.e.

(5-10) mix 0 , Amagat's LawV∆ =

Amagat’s law can also be expressed as

(5-11) 1

, Amagat's LawA N

AA

V V=

=

=∑In the gas phase the mole fraction is generally denoted by Ay while Ax is reserved for mole fractions in the liquid phase. For ideal gas mixtures it is easy to show, using Eqs. 5-1f, 5-2 and 5-3, that the mole fraction is given by A Ay p p= (5-12)

This is an extremely convenient representation of the mole fraction in terms of the partial pressure; however, one must always remember that it is strictly valid only for an ideal gas1.

EXAMPLE 5.1 Flow of an ideal gas in a pipeline A large pipeline is used to bring natural gas from Oklahoma to Nebraska as illustrated in Figure 5.1. Natural gas, consisting of methane with small amounts of ethane, propane, and other

Figure 5.1. Transport of natural gas from Oklahoma to Nebraska low molecular mass hydrocarbons, can be assumed to behave as an ideal gas at ambient temperature. At the pumping station in Glenpool, OK, the pressure in the 20-inch pipeline is 2900 psia and the temperature is T . At the receiving point in Lincoln, NE, the pressure is down to 2100 psia and the temperature is . The mass average velocity of the gas at Glenpool s 50 ft/s. Assuming ideal gas behavior and the gas being 100% methane, compute the following:

o 90 F=T1 45 F=

i

a) Mass averaged velocity at the end of the pipeline b) Mass and molar flow rates at both ends of the pipeline.

This problem has been presented in terms of a variety of units, and it is often convenient to express all variables in terms of SI units as follows:

1 Non-ideal gases are discussed briefly in Section 5.3.


( ) 6890 Pa2900 psia 19.98 MPapsia

× =

( ) 6890 Pa2100 psia 14.47 MPapsia

× =

( ) 5 C90 F 32 F 32.2 C 305.4 K9 F

− × = =

( ) 5 C45 F 32 F 7.2 C 280.4 K9 F

− × = =

( ) 0.3048 m50 ft/s 15.24 m/sft

× =

( )o 10.0254 m20 in 0.508 m

inD D= = × =

( )22

2o 1

0.508 m0.2027 m

4 4DA A

ππ= = = =

Since the natural gas is assumed to be pure methane, we can obtain the molecular mass from Table 2B in Appendix B where we find . The volume per mole of methane can be determined using the ideal gas law given by Eq. 5-2 along with the value of the gas constant found in Table 5-1. The volume per mole at the entrance is determined as

4CH 16.043 g/molMW =

( )

3

4 3oo 6

m Pa8.314 305.4 Kmol Kˆ 1.2 10 m / mol

19.98 10 PaV RTVn p

−= = = = × (1)

while the volume per mole at the exit is given by

4 311 1.53 10 m / molV RTV

n p−= = = × (2)

The gas densities at the entrance and exit of the pipeline are given by

44

CH 3 3CH o 4 3

o

16.043 g/mol( ) 133,700 g/m 133.7 kg/mˆ 1.2 10 m /mol

MW

V −ρ = = = =×

(3)

44

CH 3CH 1

1( ) 104.8 kg/mˆ

MW

Vρ = = (4)

To perform a mass balance for the pipeline, we begin with species mass balance given by

( )A A Ad dV dA r dVdt

ρ + ρ ⋅ =∫ ∫ ∫v nV A V

A

4

(5)

Since there are no chemical reactions and no accumulation, the first and last terms in this result are zero. The control volume is constructed in the obvious manner, thus there is an entrance in Glenpool, OK and an exit in Lincoln, NE. Since the mass average velocity and the diameter of the pipeline are given, we express the mass balance as

4CH o o o CH 1 1 1( ) v ( ) vA Aρ = ρ (6)

Chapter 5 112

The only unknown in this result is the velocity of the gas at the exit of the pipeline. Solving for we obtain 1v

( )4

4

3CH oo

1 o 31 CH 1

( ) 133.7 kg/mv v 15.24 m/s 19.44 m/s( ) 104.8kg/m

AA

ρ= = =

ρ (7)

The mass flow rate is a constant given by

( ) ( ) ( )4

3 2o 1 CH o o o( ) v 133.7 kg/m 15.24 m/s 0.2027 m 413kg/sm m A= = ρ = = (8)

from which we determine the constant molar flow to be

4

4

CH o 3o 1

CH

( ) 413kg/s 25.74 10 mol/s16.043g/mol

mM M

MW= = = = × (9)

In order to use Eq. 5-3 to estimate the density of a pure gas, we multiply by the molecular mass and arrange the result in the form

n MW p MWV R

ρ = =T

(5-13)

For an ideal gas mixture, one uses the definitions of the total mass density of a mixture (Eq. 5-1b) and total pressure (Eq. 5-4) with Eq. 5-2 to obtain

1 1 1

A N A N A NA A A A

AA A A

n MW p MW p MWV R T

= = =

= = =

ρ = ρ = = =∑ ∑ ∑ R T (5-14)

Here we have used MW to represent the mean molecular mass defined by

1 1

1A N A N

A A AA A

AMW p MW yp

= =

= =

= =∑ ∑ MW

(5-15)

These results are applicable when molecule-molecule interaction is negligible and this occurs for many gases under ambient conditions. At low temperatures and high pressures, gases depart from ideal gas behavior, and under those conditions one should use more accurate equations of state, such as those

resented in a standard course on thermodynamics2. p

EXAMPLE 5.2 Molecular mass of air

The air we breathe has a composition that depends on position. Air pollution sources abound and these sources add minute amounts of chemicals to the atmosphere. Combustion of fuels in cars and power plants are a source for sulfur dioxide, oxides of nitrogen, and carbon monoxide. Chemical industries add pollutants such as ammonia, chlorine, and even hydrogen cyanide to the atmosphere. In some locations, there are minute amounts of other gases such as argon, helium, and radon. The standard dry air, for the purpose of combustion computations, is assumed to be a mixture of 79% by volume of nitrogen and 21% by volume of oxygen. In this example, we want to determine the average molecular mass of standard air. For an ideal gas, a volume percentage is also a percentage of the number of moles of a species with respect to the total number of moles. Thus, the volume percentages of nitrogen and oxygen can be simply translated to molar fractions:

2. Sandler, S.I. 2006, Chemical, Biochemical, and Engineering Thermodynamics, 4th edition, John Wiley and Sons, New York.


79% by volume of nitrogen → 2N 0.79y =

21% by volume of oxygen → 2O 0.21y =

We can use Eq. 5-15 to compute the average molecular mass of standard air according to

2 2 2 2N N O O 28.85 g/molairMW MW y MW y MW= = + =

5.2 Liquid Properties and Liquid Mixtures When performing material balances for liquid systems, one must have access to reliable liquid properties. Unlike gases, the densities of liquids are weak functions of pressure and temperature, i.e., large changes in pressure and temperature result in small changes in the density. The changes in liquid density due to changes in pressure are determined by the coefficient of isothermal compressibility which is defined by

coefficient of isothermal 1 compresibility Tp

∂ρκ = = ρ ∂

(5-16)

Changes in liquid density due to changes in temperature are determined by the coefficient of thermal expansion which is defined by

coefficient of thermal 1 expansion pT

∂ρβ = = − ρ ∂

(5-17)

The coefficient of thermal expansion, β , is defined with a negative sign since the density of most liquids decreases with increasing temperature. Using a negative sign in the definition of the thermal expansion coefficient makes β positive for most liquids. There is an interesting counterexample, however, and it is the density of liquid water at low temperatures. For liquid water, 0β < for the range of temperature between 4 C and the freezing point of water, 0 C. If it were not for this feature, water in lakes and rivers would freeze from the bottom during the winter, and this would destroy most aquatic life. The density of ideal liquid mixtures is computed using Amagat’s law. Assuming that the total volume of a mixture is equal to the sum of the volume of the components of the mixture, we obtain

o1 1

A N A NA

AAA A

mV= =

= =

= =ρ

V∑ ∑ (5-18)

Here V is the volume of the mixture, while Am and oAρ represent the masses and densities of the pure

components. Equation 5-18 can be used to compute the density of the mixture according to

o o1 1

1A N A N

A A

A AA A

m mV m

= =

= =

ρ = = =ω

ρ ρ∑ ∑ (5-19)

Non-ideal behavior of liquid mixtures is a very complex topic. At low to moderate pressures, this version of Amagat’s law for liquids is a satisfactory approximation. When some of the components of a liquid mixture are above their boiling points, or if the components are polar, the use of Eqs. 5-18 or 5-19 may give significant errors3.

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

Chapter 5 114

5.3 Vapor Pressure of Liquids If we study the p-V-T characteristics of a real gas using the experimental system shown in Figure 5-2, we find the type of results illustrated in Figure 5-3. In the system illustrated in Figure 5-2, a single component is contained in a cylinder immersed in a constant temperature bath. We can increase or decrease the pressure inside the cylinder by simply moving the piston. When the specific volume, that is the volume per mole, is sufficiently large, the distance between molecules is large enough (on the average) so that molecular interaction becomes unimportant. For example, at the temperature T , and a large value of V , we observe ideal gas behavior. This is illustrated by the fact that at a fixed temperature we have

3

/ n

constantpV n = (5-20)

However, as the pressure is increased (and the volume decreased) in the system illustrated in Figure 5-3, a point is reached where liquid appears and the pressure remains constant as the volume continues to decrease. This pressure is referred to as the vapor pressure and we will identify it as . Obviously the vapp

pressure p

volume V

moles n

Constant Temperature Reservoir

Figure 5-2. Experimental study of p-V-T behavior vapor pressure is a function of the temperature and knowledge of this temperature dependence is crucial for the solution of many engineering problems.


Figure 5-3. p-V-T behavior of methane In a course on thermodynamics you will learn that the Clausius-Clapeyron equation gives a reasonably accurate representation of the vapor pressure as a function of temperature. The Clausius-Clapeyron equation can be expressed as

o

o, ,

1 1exp vapA vap A vap

Hp p

R T T ∆

= − −

(5-21)

in which ,A vapp represents the vapor pressure at the temperature T. We have used o,A vapp to represent the

vapor pressure at the reference temperature T , while represents the molar heat of vaporization. A more accurate, but empirical, expression for the vapor pressure is given by Antoine’s equation

o vapH∆

( )10 ,log ( )A vap

Bp AT

= −θ +

(5-22)

in which ,A vapp is determined in mm Hg and T is specified in C. The coefficients A, B, and θ are given in Table 2B of the Appendix B for a variety of compounds. Note that Eq. 5-22 is dimensionally incorrect and must be used with great care as we indicated in our discussion of units in Sec. 2.3

EXAMPLE 5.3: Vapor pressure of a single component In this example we wish to estimate the vapor pressure of methanol at using the Clausius-Clapeyron equation. The heat of vaporization of methanol is ∆ = at the normal boiling point of methanol, 33 . The heat of vaporization is a function of temperature and pressure. The data given for the heat of vaporization is for the temperature

. At this temperature, the vapor pressure of methanol is equal to atmospheric pressure. In order to estimate the vapor pressure at , we use the normal boiling temperature as the reference temperature. Normally we would want to compute the value of the

25 C8426 cal/molevapH

7.8 K

337.8 K = 64.6 CT =25 C

Chapter 5 116

heat of vaporization at 25 using a thermodynamic relationship and then use an average value for in Eq.5-21. For the purpose of this example, we will estimate the vapor pressure at

using the heat of vaporization at . All variables can be converted into SI units as follows:

C

erature

vapH∆C25 64.6 C

273.16337.8 K

oM,vapp

(8426

( )

101= =

4.186 J/c

M,p 371 J/mo

m Pa/ma expvap

8.07 157= =

M,vap146

238= −

Ap x=

A

Temp : 25 C + = 298.16 K oT =

1 atm ,300 Pa

cal/mol) ( al) 35,271 J/molvapH∆ = =

Substitution of these results into Eq. 5-21 gives

352 l 1 1101,300 P 19,112 Pa298.2 K 337.8 K8.314 ol K

= − − =

Vapor pressures estimated using the Clausius-Clapeyron equation can exhibit substantial errors with respect to experimental values of vapor pressure. This is caused by the fact that the assumptions made in the development of this equation are not always valid. The semi-empirical equation known as Antoine’s equation has the advantage that it is based on the correlation of experimental values of the vapor pressure.

EXAMPLE 5.4. Vapor pressure of single components using Antoine’s equation

In this example, we determine the vapor pressure of methanol at 25 C using Antoine’s equation, Eq. 5-22, and compare the result with the vapor pressure computed in Example 5.3. The numerical values of the coefficients in Antoine’s equation are obtained from Table II in the appendix, and they are given by

246 , 4.99 , 238.86A B θ =

Substitution of these values into Eq. 5-22 gives

,574.99log 8.072 2.10342.86 25

p =+

and the vapor pressure of methanol at 25 C is = 126.9 mmHg = 16,912 Pa. The result

computed using the Clausius-Clapeyron equation was, = 19,112 Pa, thus the two results differ by 11%. The results of this example clearly indicate that it is misleading to represent calculated values of the vapor pressure to five significant figures.

M,vapp

M,vapp

5.3.1 Mixtures The behavior of vapor-liquid systems having more than one component can be quite complex; however, some mixtures can be treated as ideal. In an ideal, vapor-liquid multi-component system, the partial pressure of species A in the gas phase is given by

(5-23) ,vapA Ap

Here p is the partial pressure of species A in the gas phase, Ax is the mole fraction of species A in the liquid phase, and ,A vapp is the vapor pressure of species A at the temperature under consideration. It is important to remember that Eq. 5-23 is an equilibrium relation; however, when the condition of local thermodynamic equilibrium is valid Eq. 5-23 can be used to calculate values of Ap for dynamic systems.


For an ideal gas, the gas-phase mole fraction can be expressed as

A Ay p p= (5-24)

and this result can be used with Eq. 5-23 to obtain a relation between the gas and liquid-phase mole fractions that is given by

( ),vapA A Ay x p= p (5-25)

This result is sometimes referred to as Raoult’s law. For a two-component system we can use Eq. 5-25 along with the constraint on the mole fractions

(5-26) 1 , 1A B A Bx x y y+ = + =

to obtain the following expression for the mole fraction of species A in the gas phase:

1 (

AB AA

A AB

xyxα

=1)+ α −

(5-27)

Here ABα is the relative volatility defined by

,vap

,vap

AAB

B

pp

α = (5-28)

For a dilute binary solution of species A, one can express Eq. 5-27 as

(5-29) , for ( 1)A AB A A ABy x x= α α − <<1

1<<

and this special form of Raoult’s law is often referred to as Henry’s law. For an N-component system, one can express Henry’s law as (5-30) , forA A A Ay K x x=

Here AK is referred to as the Henry’s law constant even though it is not a constant since it depends on the composition of the liquid, i.e. ( ), , ....A B C NK x x x= F (5-31)

This treatment of gas-liquid systems is extremely brief and devoid of the rigor that will be encountered in a comprehensive discussion of phase equilibrium. However, we now have sufficient information to solve a few simple mass balance problems that involve gas and liquid phases. 5

.4 Equilibrium Stages

Mass transfer of a chemical species from one phase to another is an essential feature of the mixing and purification processes that are omnipresent in the chemical and biological process industries. A comprehensive analysis of mass transfer requires an understanding of the prerequisite subjects of fluid mechanics, thermodynamics and heat transfer; however, there are some mass transfer processes that can be approximated as equilibrium stages and these processes can be analyzed using the techniques presented in this text. Most students are familiar with an equilibrium stage when it is carried out in a batch-wise manner, since this is a common purification technique used in organic chemistry laboratories.

If an organic reaction produces a desired product that is soluble in an organic phase and an undesirable product that is soluble in an aqueous phase, the product can be purified by liquid-liquid extraction as illustrated in Figure 5-4. In the first step of step of this process, the mixture from a reactor is placed in a separatory funnel. Water is added, the system is agitated, and the phases are allowed to equilibrate and separate. The amount of the undesirable product in the organic phase is reduced by an amount related to the volumes of the organic and aqueous phases and the distribution coefficient defined by

Chapter 5 118

solubility in the organic phase (moles of /vol)distribution coefficientsolubility in the aqueous phase (moles of /vol)A

AKA

= = (5-32)

The analysis of the process illustrated in Figure 5-4 is relatively simple provided that the following conditions are valid: (1) There are negligible changes in the volumes of the organic and aqueous phases

Figure 5-4. Batch-wise liquid-liquid extraction because of the mass transfer process, and (2) the linear equilibrium relation given by Eq. 5-32 is valid. If the batch process illustrated in Figure 5-4 is repeated n times, the concentration of species A in the organic phase is given by

( ) ( ) 1

1

A Afinal initialaq

A org

nc cV

n K V

=

+

(5-33)

Here we have used V to represent the volume of the aqueous phase and V to represent the volume of the organic phase. Equation 5-33 indicates that repeated batch-wise extractions can be used to reduce the concentration of species A in the organic phase to arbitrarily small values.

aq org

EXAMPLE 5.5 Mass transfer in a gas-liquid system Sulfur dioxide is “scrubbed” from a stream of air in the absorption column illustrated in Figure 5.5a. The column is designed to treat 1200 ft3/min of the contaminated air stream, and the mole fraction of sulfur dioxide in the incoming air stream is 0.02. The air stream leaving the scrubber is assumed to contain a negligible amount of SO . By negligible, we mean that ( ) can be set equal to zero provided that small causes produce small effects

2 2SO 4y4. This approximation is

consistent with the idea that the air stream leaving the scrubber is in equilibrium with the water entering the scrubber. Under these circumstances, the scrubber is considered to be an “equilibrium stage” and we will study equilibrium stages in more detail in Sec. 5.6.

4 See page 4 in Birkhoff, G. 1960, Hydrodynamics: A Study in Logic, Fact, and Similitude, Princeton University Press, Princeton, New Jersey.


Figure 5.5a. Sulfur dioxide scrubber The volumetric flow rate of the pure water entering the scrubber is 20 ft3/min. Given these conditions, we want to know the mole fraction of SO in the water stream leaving the system and the molar fraction of water in the air leaving the system. The scrubber operates at atmospheric pressure and the temperature is constant at . In determining the mole fraction of SO in the water, we will ignore the reaction with water to form sulfurous acid, and it will be left as an exercise for the student to explore the impact of this assumption. In addition to this simplification, we will assume that the transfer of air from the gas stream to the liquid stream is negligible. This last assumption is justified by the small solubility of both nitrogen and oxygen in water at room temperature.

2

20 C 2

To construct a control volume for this system, we follow the rules described previously and make cuts where information is given and required. One such cut is illustrated in Figure 5.5b where we have shown a cut where information is given, i.e., at the incoming stream of water. Information is also given at the incoming contaminated air stream, and we are required to determine the both amount of water in the outgoing air stream and the amount of sulfur dioxide in the outgoing water stream. In conclusion, cuts must be made at all four streams that are entering and leaving the system. When these cuts are joined where 0Ac ⋅ =v n , we obtain the control volume illustrated in Figure 5.5c where we have constructed the surface of the control volume in a manner that makes it visible. In reality, the surface of the control volume, in regions other than at

Chapter 5 120

Figure 5.5b. Cut for sulfur dioxide scrubber the entrances and exits, is located at the solid-air interface where the molar flux of all species is zero. For convenience, we transform given flow rates into SI units:

( )( )3 3 3 3 3

1 3

1200ft / min 0.0283m /ft m m0.566 , = 0.00943 60s/min s s

Q Q= =

For steady-state conditions and no chemical reaction, we make use of the molecular species balance in the form

(1) 0Ac dA⋅ =∫ v nA

Here we have replaced the species velocity Av with the mass average velocity v on the basis of the ideas discussed in Secs. 4.3 through 4.5. Evaluation of the four terms associated with Eq. 1 for the sulfur dioxide leads to

Sulfur dioxide: (2) 2 2SO SO1 1 2 2( ) ( )c Q c Q=

It will be left as an exercise (see Problem 5-9) for the student to develop arguments supporting the idea that air can be treated as a single species for this particular process. Under these circumstances, use of Eq. 1 for air leads to

Air: air air1 1 4 4( ) ( )c Q c Q= (3)

This result requires the assumption that the absorption of nitrogen, oxygen, etc., in the water is


Figure 5.5c. Control volume for sulfur dioxide scrubber negligible. The molar concentration of SO can be expressed in terms of the mole fraction according to

2

(4a) 2 2SO SO , in the gas phasec y c=

(4b) 2 2SO SO , in the liquid phasec x c=

where c is the total molar concentration. Use of Eqs. 4 in Eq. 2 leads to

Sulfur dioxide: 2 2SO SO1 1 1 2 2 2( ) ( )y c Q x c Q= (5)

and a similar representation for Eq. 3 provides

Air: air air1 1 1 4 4( ) ( )c y Q c y Q4= (6)

A little thought will indicate that this latter result can be expressed as

(7) 2SO1 1 1 4 H O 41 ( ) 1 ( )c y Q c y − = − 2 4Q

Finally, the balance of water leads to

Water: ( ) ( ) ( )2 2 23 H O 3 2 H O 2 4 H O3 2c c c 44

x Q x Q y= + Q (8)

Since the temperature and pressure are assumed to be constant, the total molar concentration in the gas (air) streams can be treated as a constant

31 4 3

101300Pa 0.0416kmol/m8.314 Pa m /mol K 293.16 K

c c p RT= = = = (9)

Chapter 5 122

Finally, we must determine the vapor pressure of water at using Antoine’s equation. From Appendix II, Antoine’s constants for water are:

20 C

7.94915, 1657.46, 227.03A B C= = =

Substitution of these results into Antoine’s equation5 gives the vapor pressure of water at , and we can use Raoult’s law to compute the mole fraction of water in stream #4 as

20 C

2

2 2

oH Oo

H O H O 417.36mm Hg , ( ) = 0.0228p

p yp

= = (10)

Here we have assumed that the water vapor in stream #4 is in equilibrium with the pure liquid water entering the scrubber at . This type of simplification will be explored carefully in subsequent courses on heat and mass transfer.

20 C

On the basis of Eqs. 2 to 10 we can complete the degree-of-freedom table, given by Table 5.5, in order to conclude that we have a solvable problem.

Table 5.5. Degrees-of-Freedom Stream Variables compositions N x M = 12 flow rates M = 4 Generic Degrees of Freedom (A) 16 Number of Independent Balance Equations mass/mole balance equations N = 3 Eqs. 5, 7, and 8 Number of Constraints for Compositions 4

Generic Specifications and Constraints (B) 7 Specified Stream Variables compositions 0 flow rates 2 Constraints for Compositions 4 Auxiliary Constraints 3 Eqs. 9 and 10 Particular Specifications and Constraints (C) 9

Degrees of Freedom (A - B - C) 0

To develop a solution to this problem, we note that Eq. 5 takes the form

(11) 2 2

41 SO 1 1 2 SO 2 24.7110 kmol/s( ) ( )c y Q c x Q−= =

while the mole balance of air, Eq. 7, reduces to.

5. See Sec. 2.4 for a discussion of the convenience unit mm Hg as a measure of pressure.


(12) 2 2SO1 1 1 4 H O 41 ( ) 0.0231 kmol/s 1 ( )c y Q c y − = = − 4Q

Since we know the total molar concentration and the mole fraction of water in the exit air stream, we can compute the flow rate as

[ ]3

4 43kmol m0.0231 Kmol/s 0.0416 1 0.0228 ; 0.5682

smQ Q= − = (13)

The mole balance for water given by Eq. 8 reduces to

( ) ( ) ( )2 23 H O 3 4 H O 4 2 H O3 40.5225kmol/sc x Q c y Q c x Q− = =

2 22 (14)

We add now Eqs. 11 and 14, using the composition constraint, to get

( ) ( )2 22 2 2 22 20.523 kmol/s H O SOc x Q c x Q c Q= + = 2 2

=

(15)

It is easy now to compute the mole fractions of water and sulfur dioxide in stream #2 using Eqs. 11, 14, and 15. This leads to

(16) 2 2H O SO2 2( ) 0.9991 , ( ) 0.0009x x=

In the previous example, we studied a process for which the use of molar quantities was appropriate. In other situations, mass fractions and mass flow rates are more appropriate and in some cases it is convenient to use both. The following example is an illustration of the latter situation.

EXAMPLE 5.6. Use of air to dry wet solids

In Figure 5.6 we have illustrated a counter-current air drier. The solids entering the drier contain 20% water on a mass basis and the mass flow rate of the wet solids entering the drier is 1000 lbm/hr. The dried solids contain 5% water on a mass basis. The air leaving the drier is assumed to be in equilibrium6 with the entering wet solids and this leads to a partial pressure of water vapor that is equivalent to 200 mm Hg. If the partial pressure of water vapor in the fresh air entering the drier is equivalent to 10 mm Hg, what is the total molar flow rate of fresh air entering the system? Assume that the drier operates at a pressure of one atmosphere.

wet solids,20% water

dry solids,5% water

dry air wet air

drier

Figure 5.6a. Air drier To construct a control volume for the analysis of this system, we need only make cuts where information is given and required. These cuts are illustrated in Figure 5.6b and they lead to the control volume shown in Figure 5.6c.

6 . This type of assumption is generally associated with what is known as an “equilibrium stage”.

Chapter 5 124

drier

Figure 5.6b. Cuts for the construction of a control volume We begin this problem with the species macroscopic mole balance for a steady-state process in the absence of chemical reaction and note that the species velocity, Av , can be replaced with the mass average velocity, v, at entrances and exits to obtain

(1) 0Ac dA⋅ =∫ v nA

It is important to remember that the molar concentration can be expressed as

, gas streamsA Ac y c= (2)

where Ay is the mole fraction of species A and c is the total molar concentration. The form given

wet solids,20% water

dry solids,5% water

dry air wet air

drier

controlvolume

12

3 4

Figure 5.6c. Control volume for the analysis of the drier by Eq. 2 is especially useful in the analysis of the air stream; however, for the wet solids stream it is convenient to work in terms of mass rather than moles and make use of

, wet solid streamsAA

Ac

MWρ

= (3)

If we let species A be water and apply Eq. 1 to the control volume shown in Figure 5.6c, we obtain


Water:

( ) ( )2 2 2 2

1 2

2

3

H O H O H O H O

H O

molar flow rate of water molar flow rate of water entering with the solid leaving with the solid

molar flow rate of water enterin

A A

A

MW dA MW dA

y c dA

ρ ⋅ + ρ ⋅

⋅

∫ ∫

∫

v n v n

v n2

4

H O

molar flow rate of water leaving with the airg with the air

0A

y c dA+ ⋅ =∫ v n

+

(4)

Here we have used Eqs. 2 and 3 in order to arrange the fluxes in forms that are convenient, but not necessary, for this particular problem, and we can express those fluxes in terms of averaged quantities to obtain

Water: 2 2

2 22 2

H O 1 1 H O 2 2H O 3 3 H O 4 4

H O H O( ) ( )

Q Qy M y M

MW MW⟨ρ ⟩ ⟨ρ ⟩

− + − + 0= (5)

In this representation of the macroscopic mole balance for water, we have drawn upon the analysis presented in Sec. 4.5. Specifically, we have imposed the following assumptions:

Gas streams: constantc =⋅v n (6a)

Wet solid streams: constant=⋅v n (6b)

in which “constant” means constant across the area of the entrances and exits. The three phases contained in the wet solid streams are illustrated in Figure 5.6d for stream #1. The total density in these streams consists of the density of the solid, the water, and the air, and this density can be written explicitly as

2H Osolid air⟨ρ⟩ = ⟨ρ ⟩ + ⟨ρ ⟩ + ⟨ρ ⟩ (7)

The mass fraction of water in the wet solids is defined by

2

2

H OH O

⟨ρ ⟩⟨ω ⟩ =

⟨ρ⟩ (8)

and use of this representation in Eq. 5 leads to

Water: 2 2

2 22 2

H O 1 1 H O 2 2H O 3 3 H O 4 4

H O H O( ) ( )

m my M y M

MW MW⟨ω ⟩ ⟨ω ⟩

− + − + 0=

2 2Q

(9)

Here we have identified the mass flow rates of the wet solid streams according to

1 1 1 2,m Q m= ⟨ρ⟩ = ⟨ρ⟩ (10)

A little thought (see Problem 5-12) will indicate that a mass balance for the solid material leads to

Solid: (11) 2H O 1 1 H O 2 21 1m − ⟨ω ⟩ = − ⟨ω ⟩ 2

m

and this result can be used in Eq. 7 to obtain

2 2 2

2 2 2 2

H O 4 H O 2 H O 13 4

H O 3 H O 3 H O 2 H O

( )( ) ( ) 1

y1M M

y y MWm

⟨ω ⟩ − ⟨ω ⟩ = + − ⟨ω ⟩

(12)

Chapter 5 126

A molar balance for the air will allow us to eliminate 4M from this result and the calculation of

3M easily follows. This will be left as an exercise for the student (see Problem 5-13).

air

solid

water

drier

Figure 5.6d. Wet solids entering the drier

5.5 Saturation, Dew Point and Bubble Point of Liquid Mixtures When a pure component A in the liquid phase is in equilibrium with a gas, the vapor pressure of the component in the gas phase is equal to the vapor pressure of the pure component.

,A A vapp p= (5-34)

This result is obtained by letting in Eq. 5-23 so that the liquid is pure component A. In general, when air is the gas phase we will assume that the vapor pressure in the gas phase adjacent to the liquid is saturated with the liquid component and that the concentration of the air in the liquid is negligible. When the liquid phase is a mixture, Raoult’s law (Eq. 5-23) must be used to compute the composition of the gas phase. If a liquid mixture is in equilibrium with its own vapors, then the overall pressure is equal to the sum of the vapor pressures of the individual components.

1Ax =

,1

A N

A vap AA

p p=

=

= ∑ x (5-35)

When a liquid mixture is heated, the vapor pressure of the components in the mixture increases and the sum of the partial pressures, given by Eq. 5-35, increases accordingly. If the liquid mixture is in contact with air at atmospheric pressure, the partial pressure of the components of the mixture is also given by Eq. 5-35. When the sum of the partial pressures of the components of the mixture is equal to the atmospheric pressure, the liquid mixture boils. The difference between a liquid mixture and a pure liquid is that the boiling temperature of a mixture is not constant. For a mixture in equilibrium with its own vapors, the bubble point of a mixture is the pressure at which the liquid starts to vaporize. Similarly, for a vapor mixture the dew point of the mixture is the pressure at which the vapors start to condense. The use of the terms has been extended when the liquid is in contact with air, i.e. it is customary to call bubble point at the temperature at which the liquid mixture starts boiling and dew point the temperature at which the first condensed liquid appears.


EXAMPLE 5.7. Bubble point of a water-alcohol mixture.

A mixture of ethanol, , and water is slowly heated under well-stirred conditions in an open beaker. Using Antoine’s equation to determine the vapor pressure of the components and Raoult’s Law to estimate the partial pressures, we can estimate the bubble point of this mixture, i.e. the temperature at which the first bubbles will start forming at the bottom of the beaker as well as the composition of the first bubbles.

Et 0.5x =

The vapor pressure of pure ethanol and water can be computed using Antoine’s equation. The partial pressure of the components in the gas phase in equilibrium with the liquid mixture are computed using Raoult’s Law given by Eq. 5-25. The bubble point will be determined as the temperature at which the sum of the partial pressures of ethanol and water is equal to atmospheric pressure.

(1) 2 2 2H O Et H O H O, Et Et, 760mmHg 0vap vapp p p x p x p+ − = + − =

This problem, in principle, can be solved by substitution of Antoine’s equation for the vapor pressures of the pure components in Eq. 1, and then solving for the temperature. A much simpler route consists of guessing values of the temperature until we satisfy Eq. 1. This procedure is easily done using a spreadsheet where one of the columns are values of the temperature and the columns to the right are values of the partial pressure of the components and the residue of Eq. 1. The values of Antoine’s coefficient for the components, from Appendix II are:

Water: 7.94915, 1657.46, 227.03A B C= = =

Ethanol: 8.1629, 1623.22, 227.03A B C= = =

and computed values of the vapor pressure are given in the following table:

Computation of dew point of Ethanol and water mixture

Temp 2H Op Ethp Residue Degrees C mmHg mmHg mmHg

60 74.7483588 175.7127 −509.539 70 116.9527 270.8179 −372.229 80 177.72766 405.8736 −176.399 86 225.542371 511.0501 −23.4076

86.8 232.662087 526.6464 −0.6915 86.9 233.565109 528.6235 2.188576 87 234.471058 530.6067 5.077746 90 263.047594 593.0441 96.09172

We could continue the computation by inserting additional rows between the temperatures

and T . However, for the purpose of this example we will accept the boiling point of the mixture as T .

86.8 CT = 86.9 C=86.85 0.05 C= ±

5.5.1 Humidity In air-water mixtures the humidity is often used as the measure of concentration. The humidity is defined by

mass of waterhumidity =mass of dry air

(5-36)

It will be left as an exercise for the student to show that

Chapter 5 128

( )

2 2

2

H O H O

H Oair

humidity =MW p

MW p p− (5-37)

where p is the total pressure and is the partial pressure of the water vapor. This result can be derived

from the definition given by Eq. 5-36 only if the air-water mixture is treated as an ideal gas. The percent relative humidity is often used as a measure of concentration since our personal comfort may be closely connected to this quantity. It is defined by

2H Op

2

2

H OoH O

% relative humidity = 100p

p× (5-38)

where is the vapor pressure of water. When the percent relative humidity is 100% the air is

completely saturated and the addition of further water will result in condensation. Values of the vapor pressure of water are listed in Table 5-2.

2

oH Op

5.5.2 Modified mole fraction In general, the most useful measures of concentration are the molar concentration Ac and the species density Aρ . Associated with these concentrations are the mole fraction defined by Eq. 5-1f and the mass fraction defined by Eq. 5-1c. Sometimes it is convenient to use a modified mole fraction or mole ratio

Table 5-2. Vapor Pressure of Water as a Function of Temperature

T, C Vapor Pressure, mm Hg T, F Vapor Pressure, in. Hg 0 4.579 32 0.180 5 6.543 40 0.248 10 9.209 50 0.363 15 12.788 60 0.522 20 17.535 70 0.739 25 23.756 80 1.032 30 31.824 90 1.422 35 42.175 100 1.932 40 55.324 110 2.596 45 71.88 120 3.446 50 92.51 130 4.525 55 118.04 140 5.881 60 149.38 150 7.569 65 187.54 160 9.652 70 233.7 170 12.199 75 289.1 180 15.291 80 355.1 190 19.014 85 433.6 200 23.467 90 525.76 212 29.922 95 633.90 220 34.992 100 760.00 230 42.308 105 906.07 240 50.837 110 1074.56 250 60.725 115 1267.98 260 72.134 120 1489.14 270 85.225 125 1740.93 280 100.18 130 2026.16 290 117.19 135 2347.26 300 136.44


which is based on all the species except one. If we identify that one species as species N, we express the modified mole fraction as

1

1

B N

A A BB

X c c= −

=

= ∑ (5-39)

When it is convenient to work in terms of this modified mole fraction, one usually needs to be able to convert from Ax to AX and it will be left as an exercise for the student to show that this relation is given by

( )1 , 1, 2,...A A NX x x A= − = N (5-40)

In some types of analysis it is convenient to choose species N to be species A. Under those circumstances we need to express Eq. 5-39 as

1

B N

D DBB A

BX c=

=≠

= c∑ (5-41)

and Eq. 5-39 takes the form

( )1 , 1, 2,...D D AX x x D= − = N (5-42)

Similar relations can be developed for the modified mass fraction AΩ and they will be left as exercises for the student. It is important to note that the sum over all species of the modified mass or mole fraction is not one.

EXAMPLE 5.8 Condensation of water in humid air On a warm spring day in Baton Rouge, LA, the atmospheric pressure is 755 mm Hg, the temperature is 80 F, and the humidity is 80%. A large industrial air conditioner treats 1000 kg/hr of air (dry air basis) and lowers the air temperature to 15 C. How much liquid water, in kg/h, is removed from the air by the air conditioning system? It is convenient to transform all the units to a consistent set of units, such as the SI system of units. In this case the transformation is given by

755 mm Hg 101,300 Pa755 mm Hg 100,634 Pa760 mm Hg/atm atm

p = = =

( ) ( )80 F 32 F 5C 9 F 26.7C 299.8KT = − × = =

From Table 5-2 we find the saturation vapor pressure of water at 80 F and at 15 C to be given by Saturation pressure at 80 F: (1.032 in Hg) (25.4 mm/in) (101,300 Pa/760 mm Hg) = 3,494 Pa Saturation Pressure at 15 C: (12.788 mm Hg) (101,300 Pa/760 mm Hg) = 1705 Pa The vapor pressure of water in air at 15 C is assumed to be at saturation, i.e. at the bubble point of a mixture of air and water. This is a standard assumption during condensation and it is based on the assumption of local thermodynamic equilibrium. The vapor pressure of water at 80 F is only 80% of the value at saturation and this leads to Vapor pressure of water at 80 F: ( )0.80 3494 Pa 2795 Pa× = We use these values to compute the humidity at the input and output conditions:

Chapter 5 130

( )

( )

2 2

2

H O H O

air H O

2 2

humidity (80 F)

18.015 g H O / mol 2,795Pa g H O0.017928.84gair/mol 100,634 2,795 Pa g air

M p

M p p= =

−

×= =

× −

( )

2 218.015 g H O 1,705 Pa g H Ohumidity (15C) 0.010828.84g air 100,634 1,705 Pa gair

×= =

× −

The difference between input humidity and output humidity gives the amount of grams of water condensed per gram of dry air.

2 2g H O g H O g water condensed0.0179 0.0108 0.0071g air gair g dry air

− =

Multiplying the grams of water condensed per gram of dry air by the load of the air conditioning unit in kilograms of dry air per unit time, we obtain the amount of water condensed per unit time:

g water condensed kg dry air kg water condensed0.0071 1000 7.1g dry air hour hour

× =

E

XAMPLE 5.9. Humid air flow

Humid air exits a dryer at atmospheric pressure, 75 C, 25% relative humidity, and at a olumetric flow rate of 100 m3/min. In this example we wish to determine: v

a) Absolute humidity of the air in kg water/kg air. b) Molar flow rates of water and dry air.

The vapor pressure of water at 75 C is found in Table 5-2 to be . The density of mercury is found in Table I of the Appendix. We convert all parameters into SI units according to

o 289.1 mm Hgwp =

( ) ( ) ( )2 3w

289.1mmHg0.25 9.81m/s 13546kg/m 9605Pa1000 mm/m

p = =

and we use Eq. 5-26 to compute the molar fraction of water in the air as

9605 0.095101,300

ww

p Payp Pa

= = =

In order to determine the absolute humidity, we use Eq. 5-36

mass of water mass of water/volumehumidity =mass of dry air mass of dry air/volume

=

and this leads to an expression for the humidity given by

( )( )

( )

humidity(1 )

18.05 kg water/kmol 0.0950.066kg water / kg air

28.84 kg air/kmol (1 0.095)

w w w w

a a a w

MW y MW yMW y MW y

= =−

= =−


Assuming that water vapor and air behave as ideal gases at atmospheric pressure, we use the ideal gas law given by Eq. 5-3 to compute the total concentration of the mixture. The concentration of the gas mixture is the total number of moles of air and water per unit volume of the mixture. This can be expressed as

33

101,300Pa 35 mol/mm Pa8.314 348.16Kmol K

n pcV RT

= = = =×

(1)

This result gives the total number of moles of gas per unit volume of mixture. If one considers a material volume of a flowing gas mixture, and the gases satisfy the ideal gas law, the concentration of the flowing mixture in the material volume should be equal to the concentration computed using Eq. 1. In order to determine the molar flow rates of water and dry air, we note that

(2a) 3 30.095 35mol/m 100m /min

332.5 mol water/min

w w wM c Q y cQ= = = × ×

=

(2b) (1 ) 3167.5molair /mina a a wM c Q y cQ y cQ= = = − =

5.6 Staged Processes In Sec. 5.3 we examined several systems that consisted of a single contacting process in which equilibrium conditions were assumed to exist at the exit streams. Knowing when the condition of equilibrium is a reasonable approximation requires a detailed study of the heat and mass transfer processes taking place. These details will be studied in subsequent courses where it will be shown that the condition of equilibrium is a reasonable approximation for many mass transfer processes.

5.7 Problems* Section 5.1 5-1. Show that the mole fraction in an ideal gas mixture can be expressed as A Ay p p= . 5-2. Assuming ideal gas behavior, determine the average molecular mass of a mixture made of equal amounts of mass of chlorine, argon, and ammonia. Section 5.2 5-3. 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:

(a) partial densities of components in the mixture. (b) overall density of the mixture (c) concentration of components in moles/m3 (d) molar fractions of components in the mixture.

* Problems marked with the symbol will be difficult to solve without the use of computer software.

Chapter 5 132

Section 5.3 5-4. 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. 5-5. 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. 5-6. Determine the vapor-liquid equilibrium curve of a binary mixture of acetone and benzene. 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). 5-7. Use Eqs. 5-25 and 5-26 order to derive 5-27. 5-8. Demonstrate that Eq. 5-30 is valid for an ideal system containing three components, and think about replacing the constraint with something more appropriate. 1Ax << Section 5.4 5-9. Consider air to consist of nitrogen and oxygen and indicate under what circumstances the mole balances for these two components can be added to obtain the special form

0air dAc =⋅∫ v nA

that was used in Example 5.5. Here the molar concentration of air is defined by c c . 2 2N Oair c= +

5-10. In Example 5.5 we used a control volume located outside of the scrubber. This control volume is illustrated in Figure 5.5c, and in this problem you are asked to construct a suitable control volume that lies insider the scrubber. 5-11. In Example 5.5 we used a macroscopic mole balance to determine the mole fraction of SO2 in the exit water stream. In actual fact the exit water stream contains two molecular species that involve sulfur dioxide, i.e. and H S . This occurs because of the reaction 2SO 2 O3

3 2 2 2SO H O H SO+

A more precise analysis of this problem would make use of a total macroscopic mole balance for and . Repeat the analysis given in Example 5.5 for the molar concentration of species C where

2SO

2H SO3

C A Bc c c= +

Here cA represents the concentration of and c represents the concentration of . Be careful to consider the reaction rate term in the species mole balance and comment on your treatment of the species velocities,

2SO B 2H SO3

Av and . Bv 5-12. Derive the form of the solid phase mass balance given by Eq. 8 in Example 5.6. 5-13. Complete the analysis in Example 5.6 in order to determine the total molar flow rate of fresh air entering the drier.


5-14. A gas mixture leaves a solvent recovery unit as illustrated in Figure 5.14. 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% , 4% 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

, O and

2CO 2O

2CO 2 2N are insoluble in benzene, thus the liquid phase is pure benzene.

Figure 5.14. Recovery-condenser system 5-15. 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 5.15. 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 o assume that this is the case for this problem. You are given the following information: t

a) Organic stream flow rate: 1000 lbm/min b) Specific gravity of the organic fluid: ρ ρ

2H O/ 0org = .87

c) Salt concentration in the organic stream entering the mixer: ( )A orgc = 0.0005 moles/liter

d) Equilibrium relation for the inorganic salt: ( ) ( )A aq eq A orgc K c= where 60eqK = Here ( )A aqc represents the salt concentration in the aqueous phase that is in equilibrium with the salt

concentration in the organic phase, ( )A orgc . 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.

Chapter 5 134

Figure 5.15. Liquid-liquid extraction 5-16. 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, 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 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 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 5.16a.

2 3UO (NO )2

Figure 5.16a. Liquid-liquid separation stage for reprocessing In this process an aqueous solution of uranil nitrate, is one of the feed streams to the separation stage, and the mass flow rate of the aqueous feed phase is, . The second feed stream is an organic solution of TBP in dodecane, which we assume to be a single component. The organic and inorganic phases are assumed to be completely immiscible, thus only the uranil nitrate is transferred from one stream to the other. The process specifications are indicated in Figure 5.16b, and for this problem it is the mass flow rates that we wish to determine.

2 3UO (NO )2

1 400kg/hrm =


Figure 5.16b. Specified stream variables 5-17. 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 5.17. A liquid stream, S1, flowing downward encounters a vapor stream, S2, flowing upward. We assume that the vapor

Figure 5.17. 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

,4

,31, 2, ..,A

AA

yK A N

x= =

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, 3 / 4M M , 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. 5-18. A single stage, binary distillation process is illustrated in Figure 5.18. The total molar flow rate entering the unit is 1M and the mole fraction of species A in this liquid stream is 1( )Ax . Heat is supplied in

order to generate a vapor stream, and the ratio, 2 3/M M = β , depends on the rate at which heat is supplied. At the vapor-liquid interface, we can assume local thermodynamic equilibrium in order to express the vapor-phase mole fraction in terms of the liquid-phase mole fraction according to

Chapter 5 136

, at the vapor-liquid interface1 ( 1)

AB AA

A AB

xyxα

=+ α −

Here ABα 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

32

3

( )( ) , process equilibrium relation1 ( ) ( 1)

AB AA

A AB

xyxα

=+ α −

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 ( in terms of (2)Ay 1)Ax and examine three special cases: . 0 , 0 , 1AB ABα → β→ α =

Figure 5.18. Single stage binary distillation 5-19. A saturated solution of calcium hydroxide enters a boiler as shown in Figure 5.19 and a fraction, ϕ, of the water entering the boiler is vaporized. Under these circumstances a portion of the calcium hydroxide precipitates and you would like to know the mass fraction of this suspended solid calcium hydroxide in the liquid stream leaving the boiler. The solubility can be expressed as

2 2solubility = g of Ca(OH) g of H OS =

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. Develop a general solution for the mass fraction of the suspended solid in the liquid stream leaving the boiler in terms of ϕ and S. For ϕ = 0.50, 0.21, and 0.075, determine the mass fraction of suspended solid when . 32.5 10S −= ×

Figure 5.19. Precipitation of calcium hydroxide in a boiler


Section 5.5 5-20. 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:

(a) What is the dew point temperature of the mixture? (b) What is the pressure of the container at the dew point temperature of the mixture? (c) What is the composition of the first drop of liquid at the dew point?

5-21. 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 molar fractions of the vapor when the mixture starts boiling? 5-22. 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? 5-23. 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? 5-24. 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 air? 5-25. 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%? Section 5.6 5-26. 5-27.

gas-liquid system

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