Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Chapter 13: Gas-Vapor Mixtures and Air-Conditioning
We will be concerned with the mixture of dry air and water vapor. This mixture is often
called atmospheric air.
The temperature of the atmoshperic 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/(kg K), 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)
ha
F
kJ I
C T G J
T in
kg
HK
L1005
kJ O kJ
M
.
T
P
kg

C
N
Q kg
pa
o
C
a
o
a
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.
The saturated vapor value of the enthalpy is a function of temperature and can be
expressed as
Chapter 13-1
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
F
kJ I
h  h (T )  25013
.  182
. T G J T in
Hkg K
v
g
o
C
v
NOTE: For the dry air-water vapor mixture,
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 occurs. Therefore, the partial pressure of the water vapor
can never be greater than the saturation pressure corresponding to the temperature of the
mixture.
Definitions
Dew Point, TDP: 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.
Chapter 13-2
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
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.
Steam
375
325
275
T [C]
225
175
1.491 kPa
125
75
TDP
25
-25
0
2
4
6
8
10
12
s [kJ/kg-K]
Relative Humidity,  :


mass of vapor in air
m
 v
mass of in saturated air mg
Pv
Pg
Pv and Pg are shown on the following T-s diagram for the water-vapor alone.
Chapter 13-3
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Steam
125
T [C]
75
Tm
25
o Vapor State
Pg = 3.169 kPa
Tdp
Pv = 1.491 kPa
-25
0
2
4
6
8
10
12
s [kJ/kg-K]
Since Pg  Pv ,   1 or 100%
Absolute Humidity or Specific Humidity (sometimes called humidity ratio),:
mass of water vapor in air mv

mass of dry air
ma
PVM / ( Ru T ) Pv M v
 v v

PVM
Pa M a
a
a / ( Ru T )
P
Pv
 0.622 v  0.622
Pa
P  Pv

Using the definition of the specific humidity, the relative humidity may be expressed as

0.622Pg
P
and  
(0.622   ) Pg
P  Pg
Volume of Mixture per Mass of Dry Air, v:
v
V
m R T /P
 m m m m
ma
ma
Chapter 13-4
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
After several steps, we can show (you should try this)
v
V
 va
ma
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 :
mv
m  ma  mv  ma (1  )  ma (1   )
ma
a :
Mass Flow Rate of Dry Air, m
Based on the volume flow rate of mixture at a given state, the mass flow rate of dry air is
V
m a 
v
m3 / s
kga

m3 / kga
s
Enthalpy of Mixture Per Mass Dry Air, h:
Hm Ha  Hv ma ha  mv hv


ma
ma
ma
 ha  hv
h
Chapter 13-5
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Example
Atmospheric air at 30oC, 100 kPa has a dew point of 21.3 oC. Find the relative humidity,
humidity ratio, and h of the mixture per mass of dry air.
T
o
Since TDP = 21.3 C, Pv = 2.548 kPa.
At T = 30 oC, Pg = 4.246 kPa.
s

Pv 2.548 kPa

 0.6 or 60%
Pg 4.246 kPa
  0.622
2.548 kPa
kg
 0.01626 v
(100  2.548) kPa
kgv
h  ha   hv
 C p , aT   (2501.3  1.82T )
 1.005
kg v
kJ
kJ
o
o
(30
C
)

0.01626
(2501.3

1.82(30
C
))
kg a o C
kg a
kg v
 71.71
kJ
kg a
Example
If the atmospheric air in the last example is conditioned to 20oC, 40% relative humidity,
what mass of water is added or removed per unit mass of dry air?
At 20oC, Pg = 2.339 kPa.
Chapter 13-6
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Pv  Pg  0.4(2.339 kPa )  0.936 kPa
Pv
0.936 kPa
 0.622
P  Pv
(100  0.936) kPa
kg
 0.00588 v
kga
w  0.622
The change in mass of water per mass of dry air is
mv , 2  mv ,1
ma
mv , 2  mv ,1
ma
  2  1
 (0.00588  0.01626)
 0.01038
kgv
kga
kgv
kga
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
Atmospheric air is at 25oC, 0.1 MPa, 50% 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
Chapter 13-7
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
At 25oC, Psat = 3.169 kPa, and with  1 =50%
Pv ,1   1 Pg ,1  0.5(3169
.
kPa )  15845
.
kPa
Tdp ,1  Tsat @ Pv  138
. oC
w1  0.622
Pv ,1
P  Pv ,1
 0.01001
 0.622
15845
.
kPa
(100  15845
.
) kPa
kgv
kga
Therefore, when the mixture get cooled to 10oC , the mixture is saturated, and  2 =
100%. Pv,2 = Pg,2 = 1.228 kPa.
w2  0.622
Pv , 2
P  Pv , 2
 0.00773
 0.622
1228
.
kPa
(100  1228
. ) kPa
kgv
kga
The change in mass of water per mass of dry air is
mv , 2  mv ,1
ma
  2 1
 (0.00773  0.01001)
 0.00228
kgv
kga
kgv
kga
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.
Chapter 13-8
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
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:
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.
Cooling fluid
In
Out
Insulated Flow
Duct
Atmospheric
Air
T2=20oC
T1=30oC
P1=100kPa
 1=80%
P2=98kPa
Condensate
At 20oC
 2=95%
V 1=200m3/s
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?
Pv ,1   1 Pg ,1  0.8(4.246 kPa )  3.396 kPa
Tdp ,1  Tsat @ Pv  26.01o 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, 20oC is adequate for our use here.
Apply the conservation of energy to the steady-flow control volume
Chapter 13-9
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
2
2
V
V
 i (h 
 e (h 
Q net   m
 gz)i  Wnet   m
 gz) e
2
2
inlets
exits
Neglecting the kinetic and potential energies and noting that the heat transfer and work
are zero, we get
 a1ha1  m
 v1hv1  m
 a 2 ha 2  m
 v 2 hv 2  m
 l 2 hl 2
Q net  m
Conservation of mass for the steady-flow control volume is
 m   m
i
inlets
e
exits
For the dry air:
 a1  m
 a2  m
a
m
For the water vapor:
 v1  m
 v2  m
 l2
m
Thus
 l2  m
 v1  m
 v2
m
 a ( 1   2 )
m
Divide the conservation of energy equation by
m a , then
Chapter 13-10
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Q net
 ha1   1hv1  ha 2   2 hv 2  ( 1   2 )hl 2
m a
Q net
 ha 2  ha1   2 hv 2   1hv1  ( 1   2 )hl 2
m a
Q net
 C pa (T2  T1 )   2 hv 2   1hv1  ( 1   2 )hl 2
m a
Now to find the 's and h's.
1 
0.622 Pv1 0.622(3.396)

P1  Pv1
100  3.396
kgv
 0.02187
kga
Pv 2   2 P2
 (0.95)(2.339 kPa )  2.222 kPa
2 
0.622 Pv 2 0.622(2.222)

P2  Pv 2
98  2.222
kgv
 0.01443
kga
Chapter 13-11
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Using the steam tables the h's for the water are:
kJ
kgv
kJ
hv 2  25381
.
kgv
kJ
hl 2  83.96
kgv
hv1  2556.3
The required heat transfer per unit mass of dry air becomes
Q net
 C pa (T2  T1 )   2 hv 2   1hv1  ( 1   2 )hl 2
m a
 1005
.
kJ
kgv
kJ
o
(
20

30
)
C

0
.
01443
(
25381
.
)
o
kga  C
kga
kgv
kgv
kJ
kg
kJ
(2556.3
)  (0.2187  0.01443) v (83.96
)
kga
kgv
kga
kgv
kJ
 8.622
kga
 0.02187
The heat transfer from the atmospheric air is
qout
Q net
kJ

 8.622
m a
kga
The mass flow rate of dry air is given by
Chapter 13-12
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
V1
m a 
v1
v1 
Ra T1

Pa1
kJ
(30  273) K 3
m kPa
kga  K
(100  3.396) kPa
kJ
0.287
m3
 0.886
kga
m3
200
kga
min
a 
m
 2251
.
3
m
min
0.886
kga
kga
kJ 1 min 1kWs
Q out  m a qout  2251
.
(8.622
)
min
kga 60s kJ
 32.35 kW  9.196 Tons
The Adiabatic Saturation Process
Air having a relative humidity less than 100% flows over water contained in a wellinsulated duct. Since the air has  < 100%, some of the water will evaporate and the
temperature of the air-vapor mixture will decrease.
Chapter 13-13
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
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, make-up 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
2
2
V
V
 i (h 
 e (h 
Q net   m
 gz)i  Wnet   m
 gz) e
2
2
inlets
exits
Neglecting the kinetic and potential energies and noting that the heat transfer and work
are zero, we get
m a1ha1  m v1hv1  m l 2 hl 2  m a 2 ha 2  m v 2 hv 2
Conservation of mass for the steady-flow control volume is
 m   m
i
inlets
e
exits
Chapter 13-14
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
For the dry air:
 a1  m
 a2  m
a
m
For the water vapor:
 v1  m
 l2  m
 v2
m
Thus
 l2  m
 v2  m
 v1
m
 a ( 2   1 )
m
Divide the conservation of energy equation by
m a , then
ha1   1hv1  ( 2   1 )hl 2  ha 2   2 hv 2
What are the knowns and unknowns in this equation?
Solving for 1
ha 2  ha1   2 (hv 2  hl 2 )
1 
 2 (hv1  hl 2 )
C pa (T2  T1 )   2 h fg 2

 2 (hg1  h f 2 )
Since 1 is also defined by
 1  0.622
Pv1
P1  Pv1
We can solve for Pv1.
Chapter 13-15
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Pv1 
 1 P1
0.622   1
Then the relative humidity at state 1 is
1 
Pv1
Pg1
Example:
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.
Insulated Flow
Duct
Atmospheric
Air
T2=20oC
T1=24oC
P1=100kPa
 1= ?
P2=100kPa
Make-up Water
At 16oC
1 = ?
h1 = ?
Chapter 13-16
 2=100%
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
R
||P  P  1818
. kPa
  100% U
|
0.622 P
0.622(1818
. )
 

V
S
.
T  16 C W
|| P  P 100  1818
kg

0
.
0115
|T
kg
v2
g2
2
v2
o
2
2
2
v2
v
a
Using the steam tables:
kJ
kgv
kJ
hv1  2545.4
kgv
kJ
h fg 2  2463.6
kgv
h f 2  76.2
From the above analysis
1 
C pa (T2  T1 )   2 h fg 2
 2 (hg1  h f 2 )
b
g
kJ
kgv
kJ
o
16

24
C

0
.
0115
(
2463
.
4
)
o
kga
kgv
kga C

kJ
(2545.4  67.2)
kgv
kg
 0.00820 v
kga
1005
.
Chapter 13-17
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
We can solve for Pv1.
Pv1 
 1 P1
0.622   1
0.00820(100kPa )

0.622  0.00820
 13
. kPa
Then the relative humidity at state 1 is
Pv1
Pv1
1 

Pg1 Psat @ 24 o C
13
. kPa

 0.436 or 43.6%
2.985 kPa
The enthalpy of the mixture at state 1 is
h1  ha1   1hv1
 C pa T1   1hv1
 1005
.
kJ
kgv
kJ
o
(
24
C
)

0
.
00820
2545
.
4
kga o C
kga
kgv
 44.99
kJ
kga
Chapter 13-18
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Wet-Bulb and Dry-Bulb Temperature
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 call 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%, 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.
Chapter 13-19
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
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.
Chapter 13-20
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Chapter 13-21
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Example:
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%
gv
kg
 0.008 v
kga
kga
kJ
h  46
kga
  8.0
m3
v  0.853
kga
Example:
For the air-conditioning system shown below in which atmospheric air is first heated and
then humified 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.
Heating fluid
Out
In
s
Steam, P = 1 MPa
Throttle
Atmospheric
Air
T3=25oC
T1=5oC
P1=100kPa
 1=90%
T2=24oC
V 1=60m3/min
Chapter 13-22
P3=100kPa
 3=45%
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
The psychrometric diagram is
Psychrometric Diagram
0.050
0.045
Pressure = 101.3 [kPa]
0.040
Humidity Ratio
0.035
0.8
0.030
30 C
0.025
0.6
0.020
h3 =48 kJ/kga
0.015
0.010
 3 =0.0091kgv/kga
0.2
0C
-5
0.4
3
10 C
h1 =17 kJ/kga1
0.005
0.000
-10
20 C
h2 =37 kJ/kga
 1 = 2 =0.0049 kgv/kga
2
0
5
10
v1 =0.793 m^3/kga
15
20
25
30
35
40
T [C]
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   m
i
inlets
e
exits
For the dry air:
 a1  m
 a2  m
a
m
For the water vapor (Note: no water is added or condensed during simple heating):
m v1  m v 2
Thus
 2  1
Neglecting the kinetic and potential energies and noting that the work is zero, and let the
enthalpy of the mixture per unit mass of air, h, be defined as
Chapter 13-23
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
h  ha  hv
we obtain
E in  E out
Q in  m a h1  m a h2
Q in  m a (h2  h1 )
Now to find the m a and h's using the psychrometric chart.
At T1 = 5oC, 1 = 90% and T2 = 24oC:
R
|| h  17 kgkJ
|  0.0049 kg
  90% U
VS
kg
T 5 C W
||
m
||Tv  0.793 kg
kg U
    0.0049
R
kJ
kg |
h

37
V
S
kg
T
|
T  24 C
W
1
a
1
v
1
o
a
1
3
1
a
v
2
1
a
2
a
o
2
The mass flow rate of dry air is given by
Chapter 13-24
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
V1
m a 
v1
m3
60
kga
min  75.66 kga 1 min  1261
a 
m
.
m3
min 60s
s
0.793
kga
The required heat transfer rate for the heating section is
kga
kJ 1kWs

Qin  1261
.
(37  17)
s
kga kJ
 25.22 kW
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
R
|| h  48 kgkJ
|  0.0089 kg
  45% U
VS
kg
T  25 C W
||
||Tv  0.858 kgm
3
a
3
o
v
3
a
3
3
3
a
Chapter 13-25
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Apply conservation of mass and conservation of energy to process 2-3.
Conservation of mass for the steady-flow control volume is
 m   m
i
inlets
e
exits
For the dry air:
m a 2  m a 3  m a
For the water vapor (Note: water is added during the humidification process):
m v1  m s  m v 2
m s  m v 2  m v1
m s  m a ( 2   1 )
kga
kg
(0.0089  0.0049) v
s
kga
kg
 0.00504 v
s
 1261
.
Neglecting the kinetic and potential energies and noting that the heat transfer and work
are zero, the conservation of energy yields
E in  E out
m a h2  m s hs  m a h2
m s hs  m a (h2  h1 )
Solving for the enthalpy of the steam
Chapter 13-26
Chapter 13 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
m a ( 3   2 )hs  m a (h2  h1 )
h h
hs  2 1
3 2
kJ
(48  37)
kgv
hs 
kgv
(0.0089  0.0049)
kga
kJ
 2750
kgv
At Ps = 1 MPa and hs = 2750 kJ/kgv, Ts = 180oC and the quality, xs = 0.9861.
See the text for applications involving cooling with dehumidification, evaporative
cooling, adiabatic mixing of airstreams, and wet cooling towers.
Chapter 13-27