Gas Mixtures

MATTER

is

anything that has mass and takes upMATTER space FACT can be Water can be canfound be naturally on earth as a freezes to solid, can liquid, be condenses and a gas. to solid liquid gas

melts to evaporates to MIXTURES are can be

Two or more separated kinds of matter put together by can be sorting filtration solutions using sieves using magnets made when which floating vs. sinking one kind of cannot be evaporation material dissolves separated by distillation in another filtration chromatography

Many thermodynamic applications involve mixtures of ideal gases. That is, each of the gases in the mixture individually behaves as an ideal gas. In this section, we assume that the gases in the mixture do not react with one another to any significant degree.

We restrict ourselves to a study of only ideal-gas mixtures. An ideal gas is one in which the equation of state is given by

PV  mRT or PV  NRuT Air is an example of an ideal gas mixture and has the following approximate composition.

Component % by Volume N2 78.10 O2 20.95 Argon 0.92 CO2 + trace elements 0.03

4 Definitions

Consider a container having a volume V that is filled with a mixture of k different gases at a pressure P and a temperature T.

A mixture of two or more gases of fixed chemical composition is called a nonreacting gas mixture. Consider k gases in a rigid container as shown here. The properties of the mixture may be based on the mass of each component, called gravimetric analysis, or on the moles of each component, called molar analysis.

k gases

T = Tm V = Vm P = Pm m = mm

The total mass of the mixture mm and the total moles of mixture Nm are defined as k k mm  mi and Nm   Ni i1 i1 5 The composition of a gas mixture is described by specifying either the mass fraction mfi or the mole fraction yi of each component i.

mi Ni mfi  and yi  mm Nm Note that k k mfi = 1 and  yi  1 i1 i1 The mass and mole number for a given component are related through the molar mass (or molecular weight). mi  Ni Mi

To find the average molar mass for the mixture Mm , note k k mm  mi  Ni Mi  Nm Mm i1 i1

Solving for the average or apparent molar mass Mm k k mm Ni Mm    Mi   yi Mi (kg / kmol) Nm i1 Nm i1 6 The apparent (or average) gas constant of a mixture is expressed as

Ru Rm  (kJ / kg  K) Mm

Can you show that Rm is given as k Rm  mfi Ri i1 To change from a mole fraction analysis to a mass fraction analysis, we can show that

yi Mi mfi  k  yi Mi i1

To change from a mass fraction analysis to a mole fraction analysis, we can show that

mfi / Mi yi  k mfi / Mi i1

7 Volume fraction (Amagat model)

Divide the container into k subcontainers, such that each subcontainer has only one of the gases in the mixture at the original mixture temperature and pressure.

Amagat's law of additive volumes states that the volume of a gas mixture is equal to the sum of the volumes each gas would occupy if it existed alone at the mixture temperature and pressure. k Amagat's law: Vm  Vi (Tm , Pm ) i1

The volume fraction of the vfi of any component is V (T , P ) vf  i m m i V and m k vfi = 1 i1 8 For an ideal gas mixture Ni RuTm NmRuTm Vi  and Vm  Pm Pm Taking the ratio of these two equations gives

Vi Ni vfi    yi Vm Nm The volume fraction and the mole fraction of a component in an ideal gas mixture are the same.

Partial pressure (Dalton model)

The partial pressure of component i is defined as the product of the mole fraction and the mixture pressure according to Dalton’s law. For the component i

Pi  yi Pm k Dalton’s law: Pm   Pi (Tm ,Vm ) i1

9 Now, consider placing each of the k gases in a separate container having the volume of the mixture at the temperature of the mixture. The pressure that results is called the component pressure, Pi' .

Ni RuTm NmRuTm Pi ' and Pm  Vm Vm

Note that the ratio of Pi' to Pm is

Pi ' Vi Ni    yi Pm Vm Nm For ideal-gas mixtures, the partial pressure and the component pressure are the same and are equal to the product of the mole fraction and the mixture pressure.

10 Other properties of ideal-gas mixtures

The extensive properties of a gas mixture, in general, can be determined by summing the contributions of each component of the mixture. The evaluation of intensive properties of a gas mixture, however, involves averaging in terms of mass or mole fractions: k k k Um  Ui  miui   Niui (kJ) i1 i1 i1 k k k Hm   Hi  mihi   Nihi (kJ) i1 i1 i1 k k k Sm   Si  mi si   Ni si (kJ / K) i1 i1 i1 and k k um  mfiui and um   yiui (kJ / kg or kJ / kmol) i1 i1 k k hm  mfihi and hm   yihi (kJ / kg or kJ / kmol) i1 i1 k k sm  mfi si and sm   yi si (kJ / kg K or kJ / kmol K) i1 i1 k k Cv, m  mfiCv, i and Cv, m   yiCv, i i1 i1 k k 11 Cp, m  mfiCp, i and Cp, m   yiCp, i i1 i1 These relations are applicable to both ideal- and real-gas mixtures. The properties or property changes of individual components can be determined by using ideal-gas or real-gas relations developed in earlier chapters.

Ratio of specific heats k is given as

Cp,m Cp,m km   Cv,m Cv,m The entropy of a mixture of ideal gases is equal to the sum of the entropies of the component gases as they exist in the mixture. We employ the Gibbs-Dalton law that says each gas behaves as if it alone occupies the volume of the system at the mixture temperature. That is, the pressure of each component is the partial pressure.

For constant specific heats, the entropy change of any component is

12 The entropy change of the mixture per mass of mixture is

The entropy change of the mixture per mole of mixture is

13 In these last two equations, recall that

Pi, 1  yi, 1 Pm, 1

Pi, 2  yi, 2 Pm, 2 Example 13-1

An ideal-gas mixture has the following volumetric analysis

Component % by Volume

N2 60 CO2 40

(a)Find the analysis on a mass basis.

For ideal-gas mixtures, the percent by volume is the volume fraction. Recall

yi  vfi

14 Comp. yi Mi yiMi mfi = yiMi /Mm kg/kmol kg/kmol kgi/kgm

N2 0.60 28 16.8 0.488 CO2 0.40 44 17.6 0.512 Mm = yiMi = 34.4

(b) What is the mass of 1 m3 of this gas when P = 1.5 MPa and T = 30oC?

Ru Rm  (kJ / kg  K) Mm kJ 8.314 kJ  kmol  K  0.242 kg 34.4 kg  K kmol

PmVm mm  RmTm 1.5MPa(1m3 ) 103 kJ  (0.242kJ / (kg  K))(30  273)K m3 MPa  20.45 kg

15 (c) Find the specific heats at 300 K.

Using Table A-2, Cp N2 = 1.039 kJ/kgK and Cp CO2 = 0.846 kJ/kgK 2 Cp,m  mfiCp,i  (0.488)(1.039)  (0.512)(0.846) 1 kJ  0.940 kgm  K kJ Cv,m  Cp,m  Rm  (0.940  0.242) kgm  K kJ  0.698 kgm  K

16 (d) This gas is heated in a steady-flow process such that the temperature is increased by 120oC. Find the required heat transfer. The conservation of mass and energy for steady-flow are

m1  m 2  m  m1h1  Qin  m 2h2  Qin  m (h2  h1)

 m Cp,m (T2  T1) The heat transfer per unit mass flow is Q q  in  C (T  T ) in m p,m 2 1 kJ  0.940 (120 K) kgm  K kJ  112.8 17 kgm (e) This mixture undergoes an isentropic process from 0.1 MPa, 30oC, to 0.2 MPa.

Find T2.

The ratio of specific heats for the mixture is C 0.940 k  p,m   1.347 Cv,m 0.698 Assuming constant properties for the isentropic process

(f) Find Sm per kg of mixture when the mixture is compressed isothermally from 0.1 MPa to 0.2 MPa.

18 But, the compression process is isothermal, T2 = T1. The partial pressures are given by Pi  yi Pm The entropy change becomes

For this problem the components are already mixed before the compression process. So, yi,2  yi,1 Then,

19 2 sm  mfi si i1 kg kJ kg kJ  (0.488 N2 )(0.206 )  (0.512 CO2 )(0.131 ) kg kg  K kg kg  K m N2 m CO2 kJ  0.167 kgm  K

Why is sm negative for this problem? Find the entropy change using the average specific heats of the mixture. Is your result the same as that above? Should it be?

(g) Both the N2 and CO2 are supplied in separate lines at 0.2 MPa and 300 K to a mixing chamber and are mixed adiabatically. The resulting mixture has the composition as given in part (a). Determine the entropy change due to the mixing process per unit mass of mixture.

20 Take the time to apply the steady-flow conservation of energy and mass to show that the temperature of the mixture at state 3 is 300 K.

But the mixing process is isothermal, T3 = T2 = T1. The partial pressures are given by

Pi  yi Pm The entropy change becomes

21 But here the components are not mixed initially. So, y  1 N2 ,1 y  1 CO2 ,2 and in the mixture state 3, y  0.6 N2 ,3 y  0.4 CO2 ,3 Then,

22 Then, 2 sm  mfi si i1 kg kJ kg kJ  (0.488 N2 )(0.152 )  (0.512 CO2 )(0.173 ) kg kg  K kg kg  K m N2 m CO2 kJ  0.163 kgm  K If the process is adiabatic, why did the entropy increase?

Extra Assignment

Nitrogen and carbon dioxide are to be mixed and allowed to flow through a convergent nozzle. The exit velocity to the nozzle is to be the speed of sound for the mixture and have a value of 500 m/s when the nozzle exit temperature of the mixture is 500°C. Determine the required mole fractions of the nitrogen and carbon dioxide to produce this mixture. From Chapter 17, the speed of sound is given by

C  kRT NOZZLE C = 500 m/s Mixture T = 500oC

N2 and CO2

23 Answer: yN2 = 0.589, yCO2 = 0.411

Gas-Vapor Mixtures

Atmospheric air The air in the atmosphere normally contains some water vapor and is referred to atmospheric air. Air that contains no water is called dry air. Although the amount of water vapor in the air is small, it plays a major role in human comfort. Therefore it is an important consideration in air conditioning applications The dry air and vapor of atmospheric air in air conditioning application range (temperature changes from -10 ℃ to 50 ℃) can be treated as ideal gas The pressure of atmospheric air

p  pa  pv

The total pressure of atmospheric The partial air pressure of vapor, increases with the The partial amount of vapor pressure of in air dry air T p2 pmax p1

T1

s Dew-point Temperature

T

p1 1 T1 As to vapor state 1, decrease the temperature , Dew- point then pressure of vapor won’t change The dew-point Tdp is defined as the temperature at which condensation begins if the air is cooled at constant pressure The properties of air

Specific humidity of air The ratio of the mass of water vapor present in a unit mass of dry air, denoted by ω pvVv R = 287.1 J/kg.K m R T a   v  v v p V Rv= 461.4 J/kg.K ma a a RaTa

Since Vv=Va and Tv=Ta R P P P   a v  0.622 v  0.622 v Rv Pa Pa P  Pv Relative humidity of air The comfort level depends more on the amount

of moisture the air holds(mv) relative to the maximum amount of moisture the air can hold at

the same temperature(mg). The ratio of these two quantities is called the relative humidity The saturated pressure under the pvV temperature of m R T p atmospheric air   v  v  v mg pgV pg RvT P P   0.622 v  0.622 g P  P v P Pg

Enthalpy of air

H  Ha  Hv  maha  mvhv

Specific enthalpy:

mv h  ha  hv  ha hv ma 1.863kJ/kg. ℃ is the In the field of air-conditioning, the referenceaverage specific heat temperature is 0℃. Then: of dry air

ha= 1.005t kJ/kg.℃

hv= 2501 + 1.863t kJ/kg.℃

2501kJ/kg means 1.863kJ/kg. ℃ is the the enthalpy at 0 ℃ average specific heat of vapor h= 1.005t + ω( 2501 + 1.863t) kJ/kg.℃ Adiabatic Saturation and Wet-Bulb Temperature

Adiabatic saturation

Unsaturated air Saturated air

t1, ω1, 1 t2, ω2, 2100%

hf Conservation of mass

m a1  m a2  m a

m v1  m f  m v2

m f  m v2  m v1 Conservation of energy

m a1ha1  m v1hv1  m f hf  m a2ha2  m v2hv2

m a1ha1  m v1hv1  (m v2  m v1)hf  m a2ha2  m v2hv2 m a1ha1  m v1hv1  (m v2  m v1)hf  m a2ha2  m v2hv2

Dividing by gives: m a  m a1  m a2

ha1 1hv1  (2 1)hf  ha2 2hv2