chapter 14: gas-vapor mixtures and...
TRANSCRIPT
Chapter 14
Gas-Vapor Mixtures and Air-Conditioning
Study Guide in PowerPoint
to accompany
Thermodynamics: An Engineering Approach, 5th edition
by Yunus A. Çengel and Michael A. Boles
2
We will be concerned with the mixture of dry air and water vapor. This mixture is
often called atmospheric air.
The temperature of the atmospheric air in air-conditioning applications ranges from
about –10 to about 50oC. Under these conditions, we treat air as an ideal gas with
constant specific heats. Taking Cpa = 1.005 kJ/kgK, the enthalpy of the dry air is
given by (assuming the reference state to be 0oC where the reference enthalpy is
taken to be 0 kJ/kga)
The assumption that the water vapor is an ideal gas is valid when the mixture
temperature is below 50oC. This means that the saturation pressure of the water
vapor in the air-vapor mixture is below 12.3 kPa. For these conditions, the enthalpy
of the water vapor is approximated by hv(T) = hg at mixture temperature T. The
following T-s diagram for water illustrates the ideal-gas behavior at low vapor
pressures. See Figure A-9 for the actual T-s diagram.
3
The saturated vapor value of the enthalpy is a function of temperature and can be
expressed as
Note: For the dry air-water vapor mixture, the partial pressure of the water vapor in
the mixture is less that its saturation pressure at the temperature.
P Pv sat Tmix @
4
Consider increasing the total pressure of an air-water vapor mixture while the
temperature of the mixture is held constant. See if you can sketch the process on the
P-v diagram relative to the saturation lines for the water alone given below. Assume
that the water vapor is initially superheated.
P
v
When the mixture pressure is increased while keeping the mixture temperature
constant, the vapor partial pressure increases up to the vapor saturation pressure at
the mixture temperature and condensation begins. Therefore, the partial pressure of
the water vapor can never be greater than its saturation pressure corresponding to
the temperature of the mixture.
5
Definitions
Dew Point, Tdp
The dew point is the temperature at which vapor condenses or solidifies when cooled
at constant pressure.
Consider cooling an air-water vapor mixture while the mixture total pressure is held
constant. When the mixture is cooled to a temperature equal to the saturation
temperature for the water-vapor partial pressure, condensation begins.
When an atmospheric air-vapor mixture is cooled at constant pressure such that the
partial pressure of the water vapor is 1.491 kPa, then the dew point temperature of
that mixture is 12.95oC.
0 2 4 6 8 10 1212
-25
25
75
125
175
225
275
325
375
s [kJ/kg-K]
T [
C]
1.491 kPa
Steam
TDP
6
Relative Humidity, ϕ
Mass of vapor in air
Mass of in saturated air
m
m
P
P
v
g
v
g
Pv and Pg are shown on the following T-s diagram for the water-vapor alone.
0 2 4 6 8 10 1212
-25
25
75
125
s [kJ/kg-K]
T [
C]
Pv = 1.491 kPa
Pg = 3.169 kPa
Steam
o
Tdp
Tm Vapor State
Since P P orP
P
kPa
kPag v
v
g
,.
.. 1 100%,
1491
3169047
7
Absolute humidity or specific humidity (sometimes called humidity ratio),
Mass of water vapor in air
Mass of dry air
m
m
PVM R T
PVM R T
P M
P M
P
P
P
P P
v
a
v v u
a a u
v v
a a
v
a
v
v
/ ( )
/ ( )
. .0 622 0 622
Using the definition of the specific humidity, the relative humidity may be expressed
as
P
Pand
P
P Pg
g
g( . )
.
0 622
0 622
Volume of mixture per mass of dry air, v
vV
m
m R T P
ma
m m m m
a
/
After several steps, we can show (you should try this)
vV
mv
R T
Pa
aa m
a
8
So the volume of the mixture per unit mass of dry air is the specific volume of the dry
air calculated at the mixture temperature and the partial pressure of the dry air.
Mass of mixture
m m m mm
mma v a
v
a
a ( ) ( )1 1
Mass flow rate of dry air, ma
Based on the volume flow rate of mixture at a given state, the mass flow rate of dry
air is
/
/m
V
v
m s
m kg
kg
sa
a
a 3
3
Enthalpy of mixture per mass dry air, h
hH
m
H H
m
m h m h
m
h h
m
a
a v
a
a a v v
a
a v
9
Example 14-1
Atmospheric air at 30oC, 100 kPa, has a dew point of 21.3oC. Find the relative
humidity, humidity ratio, and h of the mixture per mass of dry air.
2.5480.6 60%
4.247
v
g
P kPaor
P kPa
0 6222 548
100 2 548001626.
.
( . ).
kPa
kPa
kg
kg
v
a
h h h
C T T
kJ
kg CC
kg
kgC
kJ
kg
kJ
kg
a v
p a
a
o
o v
a
o
v
a
, ( . . )
. ( ) . ( . . ( ))
.
25013 182
1005 30 0 01626 25013 182 30
7171
10
Example 14-2
If the atmospheric air in the last example is conditioned to 20oC, 40 percent relative
humidity, what mass of water is added or removed per unit mass of dry air?
At 20oC, Pg = 2.339 kPa.
P P kPa kPa
wP
P P
kPa
kPa
kg
kg
v g
v
v
v
a
0 4 2 339 0 936
0 622 0 6220 936
100 0 936
0 00588
. ( . ) .
. ..
( . )
.
The change in mass of water per mass of dry air is
m m
m
v v
a
, ,2 1
2 1
m m
m
kg
kg
kg
kg
v v
a
v
a
v
a
, ,( . . )
.
2 10 00588 0 01626
0 01038
11
Or, as the mixture changes from state 1 to state 2, 0.01038 kg of water vapor is
condensed for each kg of dry air.
Example 14-3
Atmospheric air is at 25oC, 0.1 MPa, 50 percent relative humidity. If the mixture is
cooled at constant pressure to 10oC, find the amount of water removed per mass of
dry air.
Sketch the water-vapor states relative to the saturation lines on the following T-s
diagram. T
s
At 25oC, Psat = 3.170 kPa, and with = 50% 1
,1 1 ,1
,1 @
0.5(3.170 ) 1.585
13.8v
v g
o
dp sat P
P P kPa kPa
T T C
12
wP
P P
kPa
kPa
kg
kg
v
v
v
a
1
1
1
0 622 0 62215845
100 15845
0 01001
. ..
( . )
.
,
,
Therefore, when the mixture gets cooled to T2 = 10oC < Tdp,1, the mixture is saturated,
and = 100%. Then Pv,2 = Pg,2 = 1.228 kPa. 2
,2
2
,2
1.2280.622 0.622
(100 1.228)
0.00773
v
v
v
a
P kPaw
P P kPa
kg
kg
The change in mass of water per mass of dry air is
m m
m
kg
kg
kg
kg
v v
a
v
a
v
a
, ,
( . . )
.
2 1
2 1
0 00773 0 01001
0 00228
13
Or as the mixture changes from state 1 to state 2, 0.00228 kg of water vapor is
condensed for each kg of dry air.
Steady-Flow Analysis Applied to Gas-Vapor Mixtures
We will review the conservation of mass and conservation of energy principles as
they apply to gas-vapor mixtures in the following example.
Example 14-3
Given the inlet and exit conditions to an air conditioner shown below. What is the
heat transfer to be removed per kg dry air flowing through the device? If the volume
flow rate of the inlet atmospheric air is 200 m3/min, determine the required rate of
heat transfer.
14
Before we apply the steady-flow conservation of mass and energy, we need to decide
if any water is condensed in the process. Is the mixture cooled below the dew point
for state 1?
,1 1 ,1
,1 @
0.8(4.247 ) 3.398
26.01v
v g
o
dp sat P
P P kPa kPa
T T C
So for T2 = 20oC < Tdp, 1, some water-vapor will condense. Let's assume that the
condensed water leaves the air conditioner at 20oC. Some say the water leaves at
the average of 26 and 20oC; however, 20oC is adequate for our use here.
Apply the conservation of energy to the steady-flow control volume
( ) ( )Q m hV
gz W m hV
gznet i
inlets
i net e
exits
e
2 2
2 2
Neglecting the kinetic and potential energies and noting that the work is zero, we get
Q m h m h m h m h m hnet a a v v a a v v l l 1 1 1 1 2 2 2 2 2 2
Conservation of mass for the steady-flow control volume is
m mi
inlets
e
exits
15
For the dry air:
m m ma a a1 2
For the water vapor:
m m mv v l1 2 2
The mass of water that is condensed and leaves the control volume is
( )
m m m
m
l v v
a
2 1 2
1 2
Divide the conservation of energy equation by , then ma
( )
( )
Q
mh h h h h
Q
mh h h h h
net
a
a v a v l
net
a
a a v v l
1 1 1 2 2 2 1 2 2
2 1 2 2 1 1 1 2 2
( ) ( )
Q
mC T T h h hnet
a
pa v v l 2 1 2 2 1 1 1 2 2
16
Now to find the 's and h's.
11
1 1
0.622 0.622(3.398)
100 3.398
0.02188
v
v
v
a
P
P P
kg
kg
P P
kPa kPa
P
P P
kg
kg
v g
v
v
v
a
2 2 2
22
2 2
0 95 2 339 2 222
0 622 0 622 2 222
98 2 222
0 01443
( . )( . ) .
. . ( . )
.
.
17
Using the steam tables, the h's for the water are
1
2
2
2555.6
2537.4
83.91
v
v
v
v
l
v
kJh
kg
kJh
kg
kJh
kg
The required heat transfer per unit mass of dry air becomes
2 1 2 2 1 1 1 2 2( ) ( )
1.005 (20 30) 0.01443 (2537.4 )
0.02188 (2555.6 ) (0.02188 0.01443) (83.91 )
8.627
netpa v v l
a
o v
o
a a v
v v
a v a v
a
QC T T h h h
m
kgkJ kJC
kg C kg kg
kg kgkJ kJ
kg kg kg kg
kJ
kg
18
The heat transfer from the atmospheric air is
8.627netout
a a
Q kJq
m kg
The mass flow rate of dry air is given by
mV
va
1
1
3
11
1
3
0.287 (30 273)
(100 3.398)
0.90
a a
a
a
kJK
R T kg K m kPav
P kPa kJ
m
kg
3
3
200min 222.2
min0.90
aa
a
mkg
mm
kg
19
1min 1222.2 (8.627 )
min 60
31.95 9.08
aout a out
a
kg kJ kWsQ m q
kg s kJ
kW Tons
The Adiabatic Saturation Process
Air having a relative humidity less than 100 percent flows over water contained in a
well-insulated duct. Since the air has < 100 percent, some of the water will
evaporate and the temperature of the air-vapor mixture will decrease.
20
If the mixture leaving the duct is saturated and if the process is adiabatic, the
temperature of the mixture on leaving the device is known as the adiabatic
saturation temperature.
For this to be a steady-flow process, makeup water at the adiabatic saturation
temperature is added at the same rate at which water is evaporated.
We assume that the total pressure is constant during the process.
Apply the conservation of energy to the steady-flow control volume
( ) ( )Q m hV
gz W m hV
gznet i
inlets
i net e
exits
e
2 2
2 2
Neglecting the kinetic and potential energies and noting that the heat transfer and
work are zero, we get
m h m h m h m h m ha a v v l l a a v v1 1 1 1 2 2 2 2 2 2
Conservation of mass for the steady-flow control volume is
m mi
inlets
e
exits
21
For the dry air:
m m ma a a1 2
For the water vapor:
m m mv l v1 2 2
The mass flow rate water that must be supplied to maintain steady-flow is,
( )
m m m
m
l v v
a
2 2 1
2 1
Divide the conservation of energy equation by , then ma
h h h h ha v l a v1 1 1 2 1 2 2 2 2 ( )
What are the knowns and unknowns in this equation?
Solving for 1
12 1 2 2 2
1 2
2 1 2 2
1 2
h h h h
h h
C T T h
h h
a a v l
v l
pa fg
g f
( )
( )
( )
( )
22
Since 1 is also defined by
11
1 1
0 622
.P
P P
v
v
We can solve for Pv1.
PP
v11 1
10 622
.
Then the relative humidity at state 1 is
11
1
P
P
v
g
23
Example 14-4
For the adiabatic saturation process shown below, determine the relative humidity,
humidity ratio (specific humidity), and enthalpy of the atmospheric air per mass of dry
air at state 1.
24
Using the steam tables:
2
1
2
67.2
2544.7
2463.0
f
v
v
v
fg
v
kJh
kg
kJh
kg
kJh
kg
From the above analysis
2 1 2 2
1
1 2
( )
( )
1.005 16 24 0.0115 (2463.0 )
(2544.7 67.2)
0.00822
pa fg
g f
o v
o
a va
v
v
a
C T T h
h h
kgkJ kJC
kg kgkg C
kJ
kg
kg
kg
25
We can solve for Pv1.
1 11
10.622
0.00822(100 )
0.622 0.00822
1.3
v
PP
kPa
kPa
Then the relative humidity at state 1 is
1 11
1 @24
1.30.433 43.3%
3.004
o
v v
g sat C
P P
P P
kPaor
kPa
The enthalpy of the mixture at state 1 is
1 1 1 1
1 1 1
1.005 (24 ) 0.00822 2544.7
45.04
a v
pa v
o v
o
a a v
a
h h h
C T h
kgkJ kJC
kg C kg kg
kJ
kg
26
Wet-Bulb and Dry-Bulb Temperatures
In normal practice, the state of atmospheric air is specified by determining the wet-
bulb and dry-bulb temperatures. These temperatures are measured by using a device
called a psychrometer. The psychrometer is composed of two thermometers
mounted on a sling. One thermometer is fitted with a wet gauze and reads the wet-
bulb temperature. The other thermometer reads the dry-bulb, or ordinary,
temperature. As the psychrometer is slung through the air, water vaporizes from the
wet gauze, resulting in a lower temperature to be registered by the thermometer. The
dryer the atmospheric air, the lower the wet-bulb temperature will be. When the
relative humidity of the air is near 100 percent, there will be little difference between
the wet-bulb and dry-bulb temperatures. The wet-bulb temperature is approximately
equal to the adiabatic saturation temperature. The wet-bulb and dry-bulb
temperatures and the atmospheric pressure uniquely determine the state of the
atmospheric air.
27
The Psychrometric Chart
For a given, fixed, total air-vapor pressure, the properties of the mixture are given in
graphical form on a psychrometric chart.
The air-conditioning processes:
28
29
Example 14-5
Determine the relative humidity, humidity ratio (specific humidity), enthalpy of the
atmospheric air per mass of dry air, and the specific volume of the mixture per mass
of dry air at a state where the dry-bulb temperature is 24oC, the wet-bulb temperature
is 16oC, and atmospheric pressure is 100 kPa.
From the psychrometric chart read
44%
8 0 0 008
46
08533
. .
.
g
kg
kg
kg
hkJ
kg
vm
kg
v
a
v
a
a
a
30
Example 14-6
For the air-conditioning system shown below in which atmospheric air is first heated
and then humidified with a steam spray, determine the required heat transfer rate in
the heating section and the required steam temperature in the humidification section
when the steam pressure is 1 MPa.
31
The psychrometric diagram is
-10 -5 0 5 10 15 20 25 30 35 40
0.000
0.005
0.010
0.015
0.020
0.025
0.030
0.035
0.040
0.045
0.050
T [C]
Hu
mid
ity R
ati
oPressure = 101.3 [kPa]
0.2
0.4
0.6
0.8
0 C
10 C
20 C
30 C
Psychrometric Diagram
1
2
3
1= 2=0.0049 kgv/kga
3=0.0091kgv/kga
h2=37 kJ/kga
h1=17 kJ/kga
h3=48 kJ/kga
v1=0.793 m^3/kga
Apply conservation of mass and conservation of energy for steady-flow to process
1-2.
Conservation of mass for the steady-flow control volume is
m mi
inlets
e
exits
32
For the dry air
m m ma a a1 2
For the water vapor (note: no water is added or condensed during simple heating)
m mv v1 2
Thus,
2 1
Neglecting the kinetic and potential energies and noting that the work is zero, and
letting the enthalpy of the mixture per unit mass of air h be defined as
h h ha v
we obtain
( )
E E
Q m h m h
Q m h h
in out
in a a
in a
1 2
2 1
33
Now to find the and h's using the psychrometric chart. ma
At T1 = 5oC, 1 = 90%, and T2 = 24oC:
The mass flow rate of dry air is given by
mV
va
1
1
34
min
.
.min
min.m
m
m
kg
kg
s
kg
sa
a
a a
60
0 793
75661
601261
3
3
The required heat transfer rate for the heating section is
. ( )
.
Qkg
s
kJ
kg
kWs
kJ
kW
ina
a
1261 37 171
2522
This is the required heat transfer to the atmospheric air. List some ways in which this
amount of heat can be supplied.
At the exit, state 3, T3 = 25oC and 3 = 45%. The psychrometric chart gives
35
Apply conservation of mass and conservation of energy to process 2-3.
Conservation of mass for the steady-flow control volume is
m mi
inlets
e
exits
For the dry air
m m ma a a2 3
For the water vapor (note: water is added during the humidification process)
( )
. ( . . )
.
m m m
m m m
m m
kg
s
kg
kg
kg
s
v s v
s v v
s a
a v
a
v
2 3
3 2
3 2
1261 0 0089 0 0049
0 00504
36
Neglecting the kinetic and potential energies and noting that the heat transfer and
work are zero, the conservation of energy yields
( )
E E
m h m h m h
m h m h h
in out
a s s a
s s a
2 3
3 2
Solving for the enthalpy of the steam,
( ) ( )m h m h h
hh h
a s a
s
3 2 3 2
3 2
3 2
(48 37)
(0.0089 0.0049)
2750
as
v
a
v
kJ
kgh
kg
kg
kJ
kg
37
At Ps = 1 MPa and hs = 2750 kJ/kgv, Ts = 179.88oC and the quality xs = 0.985.
See the text for applications involving cooling with dehumidification, evaporative
cooling, adiabatic mixing of airstreams, and wet cooling towers.