Transcript
MATTER
is
solid liquid gas
melts to
freezes to
evaporates to
condenses to
anything that has mass and takes
up space
can be
can be
can be
MATTER FACT
Water can be found naturally on earth as a solid, liquid, and a gas.
MIXTURES
Two or more kinds of matter
put together
are
solutions
separated
can be
can be
by
filtration
using sieves
using magnets
floating vs. sinking
evaporation
sorting
distillation
chromatography
one kind of material dissolves
in another
made when
cannot be separated by
filtration
which
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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 NR Tu
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
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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
m m N Nm i
i
k
m i
i
k
1 1
and
6
mfm
my
N
Ni
i
m
ii
m
and
mf yi
i
k
i
i
k
1 1
1 = 1 and Note that
The composition of a gas mixture is described by specifying either the mass fraction
mfi or the mole fraction yi of each component i.
The mass and mole number for a given component are related through the molar
mass (or molecular weight).
m N Mi i i
To find the average molar mass for the mixture Mm , note
m m N M N Mm i
i
k
i i m m
i
k
1 1
Solving for the average or apparent molar mass Mm
Mm
N
N
NM y M kg kmolm
m
m
i
m
i
i
k
i i
i
k
1 1
( / )
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The apparent (or average) gas constant of a mixture is expressed as
RR
MkJ kg Km
u
m
( / )
Can you show that Rm is given as
R mf Rm i i
i
k
1
To change from a mole fraction analysis to a mass fraction analysis, we can show
that
mfy M
y Mi
i i
i i
i
k
1
To change from a mass fraction analysis to a mole fraction analysis, we can show that
ymf M
mf Mi
i i
i i
i
k
/
/1
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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.
Amagat's law: V V T Pm i m m
i
k
( , )1
The volume fraction of the vfi of any component is
vfV T P
Vi
i m m
m
( , )
and
vf i
i
k
1
= 1
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For an ideal gas mixture
VN R T
Pand V
N R T
Pi
i u m
m
mm u m
m
Taking the ratio of these two equations gives
vfV
V
N
Nyi
i
m
i
m
i
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
P y Pi i m
Dalton’s law: P P T Vm i m m
i
k
( , )1
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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' .
PN R T
Vand P
N R T
Vi
i u m
m
mm u m
m
'
Note that the ratio of Pi' to Pm is
P
P
V
V
N
Nyi
m
i
m
i
m
i
'
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.
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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:
U U m u N u
H H m h N h
S S m s N s
m i
i
k
i i
i
k
i i
i
k
m i
i
k
i i
i
k
i i
i
k
m i
i
k
i i
i
k
i i
i
k
1 1 1
1 1 1
1 1 1
(kJ)
(kJ)
(kJ / K)
and u mf u u y u
h mf h h y h
s mf s s y s
m i i
i
k
m i i
i
k
m i i
i
k
m i i
i
k
m i i
i
k
m i i
i
k
1 1
1 1
1 1
and (kJ / kg or kJ / kmol)
and (kJ / kg or kJ / kmol)
and (kJ / kg K or kJ / kmol K)
C mf C C y C
C mf C C y C
v m i v i
i
k
v m i v i
i
k
p m i p i
i
k
p m i p i
i
k
, , , ,
, , , ,
1 1
1 1
and
and
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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
kC
C
C
Cm
p m
v m
p m
v 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
13
The entropy change of the mixture per mass of mixture is
The entropy change of the mixture per mole of mixture is
14
In these last two equations, recall that
P y P
P y P
i i m
i i m
, , ,
, , ,
1 1 1
2 2 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
y vfi i
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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?
RR
MkJ kg K
kJ
kmol Kkg
kmol
kJ
kg K
mu
m
( / )
.
.
.
8 314
34 4
0 242
mP V
R T
MPa m
kJ kg K K
kJ
m MPa
kg
mm m
m m
15 1
0 242 30 273
10
20 45
3 3
3
. ( )
( . / ( ))( )
.
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(c) Find the specific heats at 300 K.
Using Table A-2, Cp N2 = 1.039 kJ/kgK and Cp CO2 = 0.846 kJ/kgK
C mf C
kJ
kg K
p m i p i
m
, , ( . )( . ) ( . )( . )
.
1
2
0 488 1039 0512 0846
0 940
C C RkJ
kg K
kJ
kg K
v m p m m
m
m
, , ( . . )
.
0 940 0 242
0 698
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(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 m m
m h Q m h
Q m h h
mC T T
in
in
p m
1 2
1 1 2 2
2 1
2 1
The heat transfer per unit mass flow is
mC T T
kJ
kg KK
kJ
kg
inin
p m
m
m
( )
. ( )
.
, 2 1
0 940 120
112 8
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(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
kC
C
p m
v m
,
,
.
..
0 940
0 6981347
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.
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But, the compression process is isothermal, T2 = T1. The partial pressures are given
by
P y Pi i m
The entropy change becomes
For this problem the components are already mixed before the compression process.
So, y yi i, ,2 1
Then,
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s mf s
kg
kg
kJ
kg K
kg
kg
kJ
kg K
kJ
kg K
m i i
i
N
m N
CO
m CO
m
1
2
0 488 0 206 0512 0131
0167
2
2
2
2
( . )( . ) ( . )( . )
.
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.
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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
P y Pi i m
The entropy change becomes
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But here the components are not mixed initially. So,
y
y
N
CO
2
2
1
2
1
1
,
,
and in the mixture state 3,
y
y
N
CO
2
2
3
3
0 6
0 4
,
,
.
.
Then,
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Then,
s mf s
kg
kg
kJ
kg K
kg
kg
kJ
kg K
kJ
kg K
m i i
i
N
m N
CO
m CO
m
1
2
0 488 0152 0512 0173
0163
2
2
2
2
( . )( . ) ( . )( . )
.
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 kRTMixture
N2 and CO2
C = 500 m/s
T = 500oC
NOZZLE
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
va ppp
The total pressure of atmospheric air
The partial pressure of dry air
The partial pressure of vapor, increases with the amount of vapor in air
Dew-point Temperature
T
p1
T1
1
As to vapor state 1, decrease the temperature , then pressure of vapor won’t change
Dew- point
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
a
v
m
m
aa
aa
vv
vv
TRVp
TRVp
Since Vv=Va and Tv=Ta
a
v
v
a
P
P
R
R
a
v
P
P622.0
v
v
PP
P
622.0
Ra= 287.1 J/kg.K
Rv= 461.4 J/kg.K
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
g
v
m
m
TR
Vp
TRVp
v
g
v
v
g
v
p
p
The saturated pressure under the temperature of atmospheric air
In the field of air-conditioning, the reference temperature is 0℃. Then:
ha= 1.005t kJ/kg.℃
hv= 2501 + 1.863t kJ/kg.℃
h= 1.005t + ω( 2501 + 1.863t) kJ/kg.℃
2501kJ/kg means
the enthalpy at 0 ℃ 1.863kJ/kg. ℃ is the
average specific heat
of vapor
1.863kJ/kg. ℃ is the
average specific heat
of dry air
Adiabatic Saturation and Wet-Bulb Temperature
Adiabatic saturation
t1, ω1, 1 t2, ω2, 2100%
Unsaturated air Saturated air
hf
Conservation of mass
aaa mmm 21
21 vfv mmm
Conservation of energy
22221111 vvaaffvvaa hmhmhmhmhm
12 vvf mmm
2222121111 )( vvaafvvvvaa hmhmhmmhmhm
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