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 50 o C. Under these conditions, we treat air as an ideal gas with constant specific heats. Taking C = 1.005 kJ/kg

K, the enthalpy of the dry air is p a given by (assuming the reference state to be 0 o C 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 50 o C. 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 h v
( 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.
2

For more information and animations illustrating this topic visit the Animation
Library developed by Professor S. Bhattacharjee, San Diego State University, at this link. test.sdsu.edu/testhome/vtAnimations/index.html
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 v

P sat @ T mix
3

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.
4

Definitions
Dew Point, T dp
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 12.95
o C.
Steam
375
325
275
225
175
1.491 kPa 125
75
T
DP
25
-25
0 2 4 10
5
6 s [kJ/kg-K]
8

Relative Humidity, ϕ
 
Mass of vapor in air
Mass of in saturated air
 m m g v

P v
P g
P v and P g are shown on the following T-s diagram for the water-vapor alone.
Steam
125
75
T m
T dp
25 P g
= 3.169 kPa
P v
= 1.491 kPa o Vapor State
-25
0 2 4 6 s [kJ/kg-K]
Since
P g

P v
,
 
1 or 100%,
 
P v

P g
8 10 kPa kPa

6

Absolute humidity or specific humidity (sometimes called humidity ratio),

 


Mass of water vapor in air
Mass of dry air
 m v m a
PVM v v
/ ( R T u
)
P VM a
/ ( R T )

P M v v
P M a
P
P v a

P v
P

P v
Using the definition of the specific humidity, the relative humidity may be expressed as
 

P
 
) P g and
 

P g
P
 
P g
Volume of mixture per mass of dry air, v v

V
 m a m R T m m m
/ P m m a
After several steps, we can show (you should try this) v

V
  m a v a
R T a m
P a
7

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
Mass flow rate of dry air, m
 m a
 m v
 m a
( 1
 m v )
 m a m a m a
( 1
 
)
Based on the volume flow rate of mixture at a given state, the mass flow rate of dry air is a

V
 v
3 m / s
3 m / kg a
 kg s a
Enthalpy of mixture per mass dry air, h h

H m
 m a
H a

H v
 h a
  h v m a
 m h
 m h m a
8

Example 14-1
Atmospheric air at 30 o C, 100 kPa, has a dew point of 21.3
o C. Find the relative humidity, humidity ratio, and h of the mixture per mass of dry air.
 
P v

P g
2.548
kPa
4.247
kPa

0.6
or 60%
9

  kPa
( .
) kPa
 kg v kg a h
 h a
  h v

C T
 
( .
.
T )

 kJ kg a
 o
C
( 30 o
C )

0 01626 kg kg a v ( .

.
( o
C )) kJ kg v kJ kg a
10

Example 14-2
If the atmospheric air in the last example is conditioned to 20 o C, 40 percent relative humidity, what mass of water is added or removed per unit mass of dry air?
At 20 o C, Pg = 2.339 kPa.
P v
 
P g

0 4 2 339 kPa )

0 936 kPa w


P
P v

P v
 kg v kg a
( kPa
.
) kPa
The change in mass of water per mass of dry air is m v , 2
 m v , 1  
2
 
1 m v , 2
 m v , 1  m a m a
.
 
0 01038 kg kg a v kg
) kg a v
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 25 o C, 0.1 MPa, 50 percent relative humidity. If the mixture is cooled at constant pressure to 10 o C, 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
At 25 o C, P sat s

P v ,1
 
P
1 g ,1

0.5(3.170
kPa )

1.585
kPa
T dp ,1

T sat @ P v

13.8
o
C
12

w
1

P v , 1
P

P v , 1
 kPa
( .
) kPa
 kg v kg a
Therefore, when the mixture gets cooled to T

P v ,2
= P g,2
2
= 1.228 kPa.
= 10 o C < T dp,1
, the mixture is saturated, w
2

0.622
P v , 2
P

P v , 2

0.622

0.00773
kg v kg a
The change in mass of water per mass of dry air is m v , 2
 m v , 1  
2
 
1 m a
1.228
kPa kPa
 
.
kg
) kg a v
  kg v kg a
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 17 m 3 /min, determine the required rate of heat transfer.
Cooling fluid
In Out
Insulated flow duct
Atmospheric air
T
1
= 30 o C

P
1
=100 kPa
1
= 80%
V 1
= 17m 3 /min
Condensate at 20 o C
T
2
= 20 o C

P
2
= 98 kPa
2
= 95%
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?
P v ,1
 
P
1 g ,1

0.8(4.247
kPa )

3.398
kPa
T dp ,1

T sat @ P v

26.01
o
C
So for T
2
= 20 o C < T dp
, 1, some water-vapor will condense. Let's assume that the condensed water leaves the air conditioner at 20 o C. Some say the water leaves at the average of 26 and 20 o C; however, 20 o C is adequate for our use here.
net
  inlets
 i
(

V
2
2
 gz ) i
  net
  exits
 e
(

V 2

2 gz ) e
Neglecting the kinetic and potential energies and noting that the work is zero, we get

     net
 a 1 a 1
 v 1 v 1
 a 2 a 2
 v 2 v 2
 l 2 l 2
Conservation of mass for the steady-flow control volume is
 inlets m i
  exits e
15

For the dry air:
   a 1
  a 2 a
For the water vapor:
 v 1
  v 2
  l 2
The mass of water that is condensed and leaves the control volume is
 l 2
  v 1
 
 v 2
 ( a
 
1

2
)
Divide the conservation of energy equation by , then a
Q net
 h a 1
  h
1 v 1
 h a 2
  h
2 v 2

(
 
1

2
) h l 2 a net
 h a 2
 h a 1
  h
2 v 2
  h
1 v 1

(
 
1

2
) h l 2 a net

C pa
( T
2

T
1
)
  h
2 v 2
  h
1 v 1

(
 
1

) h
2 l 2 a
16

Now to find the

's and h 's.

1

0.622
P v 1
P
1

P v 1

0.622(3.398)
P v 2

0.02188
kg v kg a
 
P
2 g 2
 kPa )

2 222 kPa

2


P v 2
P
2

P v 2

0 622 2 222 )
.
kg v kg a
17

Using the steam tables, the h 's for the water are h v 1
 kJ
2555.6
kg v h v 2
 kJ
2537.4
kg v h l 2
 kJ
83.91
kg v
The required heat transfer per unit mass of dry air becomes
Q net m a

C pa
( T
2

T
1
)
  h
2 v 2
  h
1 v 1

(
 
1 2
) h l 2

1.005
kJ
 o kg C a
 o
C

0.01443
kg kg v a
(2537.4
kJ kg v
)
 kg kJ
  kg kJ
0.02188
v (2555.6
) (0.02188 0.01443) v (83.91
) kg kg kg kg a v a v
 
28.73
kJ kg a
18

The heat transfer from the atmospheric air is q out


Q net m a
 
28.73
kJ kg a
The mass flow rate of dry air is given by v
1

R T a 1
P a 1

0.287
kJ kg a

K
 m 3
0.90
kg a kPa
K kJ a
V


1 v
1 m a
 m
3
17
0.90
min m
3 kg a

18.89
kg a min
Q out
 m q a out

18.89
kg a min
(28.73
kJ kg a
)
1min 1 kWs
60 s kJ

9.04
kW
0.2843
Tons

2.57
Tons of refrigeration kW
19

The Adiabatic Saturation Process
Air having a relative humidity less than 100 percent flows over water contained in a
 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
 net
  inlets
 i
(

V
2
2
 gz ) i
  net
  exits
 e
(

V

2
2
 gz ) e
Neglecting the kinetic and potential energies and noting that the heat transfer and work are zero, we get
 a 1 a 1
  v 1 v 1
  l 2 l 2
  a 2 a 2
  v 2 v 2
Conservation of mass for the steady-flow control volume is
 inlets m i
  exits e
21

For the dry air:
   a 1
  a 2 a
For the water vapor:
 v 1
  l 2
  v 2
The mass flow rate water that must be supplied to maintain steady-flow is,
 l 2
  v 2
 
 v 1
 ( a
 
2

1
)
Divide the conservation of energy equation by , then a h a 1
  h
1 v 1

(

2
 
) h
1 l 2
 h a 2
  h
2 v 2
What are the knowns and unknowns in this equation?
22

Solving for

1
Since

1 is also defined by

1

 h a 2
 h a 1

(

2 h h v 1
 l
( h h v 2
 l
2
)
2
C pa
( T
2

T
1
)
  h
2 fg 2
)
( h h g 1
 f 2
)

1

P v 1
P P
1
 v 1
We can solve for P v 1
.
P v 1


 
1
Then, the relative humidity at state 1 is

1

P v 1
P g 1
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: h f 2

67.2
kJ kg v h v 1

2544.7
kJ kg v h fg 2

2463.0
kJ kg v
From the above analysis

1

C pa
( T
2

T
1
)
 
2 h fg 2
( h g 1
 h f 2
)
 kJ
1.005
o kg C a
  o
C
 kg kJ
0.0115
v (2463.0
) kg kg a v kJ kg v
 kg
0.00822
kg a v
25

We can solve for P v 1
.
P v 1


P
1 1
0.622
 
1

0.00822(100 kPa )

1.3
kPa
Then the relative humidity at state 1 is

1

P v 1

P P g 1 sat
P v 1
@ 24 o
C

1.3
kPa
3.004
kPa

0.433
or 43.3%
The enthalpy of the mixture at state 1 is h
1
 h a 1
  h
1 v 1

C T
1
 
1 h v 1

1.005
kJ kg a
 o
C
(24 o
C
 kg kg v a
2544.7
kJ kg v

45.04
kJ kg a
26

Wet-Bulb and Dry-Bulb Temperatures
In normal practice, the state of atmospheric air is specified by determining the wetbulb 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 wetbulb 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 24 o C, the wet-bulb temperature is 16 o C, and atmospheric pressure is 100 kPa.
From the psychrometric chart read
 
44%
  g v kg a
 kg v kg a h
 kJ
46 kg a m
3 v
 kg a
NOTE: THE ENTHALPY READ FROM THE PSYCHROMETRIC CHART
IS THE TOTAL ENTHALPY OF THE AIR-VAPOR MIXTURE PER UNIT
MASS OF DRY AIR. h= H/m a
= h a
+ ωh v
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
Psychrometric Diagram
0.050
0.045
Pressure = 101.3 [kPa]
0.040
0.035
0.8
0.030
30 C
0.025
0.6
0.020
0.015
0.010
0.005
h
1
0 C h
3
=48 kJ/kga h
2
=37 kJ/kga
10 C
20 C
3
0.4
0.2
2
0.000
-10 -5 0 5 10 15 20 25 30 35 40 v
1
=0.793 m^3/kga
T [C]

3
=0.0091kgv/kga

1
=

2
=0.0049 kgv/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
 inlets m i
  exits e
32

For the dry air
   a 1
  a 2 a
For the water vapor (note: no water is added or condensed during simple heating)
   v 1 v 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 a
  h v we obtain
  in
 out
 in
  a 1
  a 2
 in
  a
(
2
 h
1
)
33

Now to find the m a and h 's using the psychrometric chart.
At T
1
= 5oC,

1
= 90%, and T
2
= 24 o C:
The mass flow rate of dry air is given by a
V


1 v
1
34

m a
 m
3
60 min m
3 kg a
 kg a min
1 min

60 s
The required heat transfer rate for the heating section is kg a s
Q in
 kg
.
( 37 17 s a

) kJ kg a
1 kWs kJ
 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, T
3
= 25 o C 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 i
  inlets exits e
For the dry air
 a 2
 m a 3
 m a
For the water vapor (note: water is added during the humidification process)
 v 2
  s
  v 3
 s
  v 3
  v 2
 s
  a
(
 
3

2
)
 kg a s
( .

.
) kg kg a v
 kg v s
36

Neglecting the kinetic and potential energies and noting that the heat transfer and work are zero, the conservation of energy yields
E

E
 in
 out
 a 2
  s s
  a 3
 s s
  a
(
3
 h
2
)
Solving for the enthalpy of the steam,
 ( a

3
 
2
) h s
  a
(
3
 h
2
) h s
 h
3
 h
2

3
 
2
37

h s
 kJ kg a kg v kg a
 kJ
2750 kg v
At P s
= 1 MPa and h s
= 2750 kJ/kg v
, T s
= 179.88
o C and the quality x s
= 0.985.
See the text for applications involving cooling with dehumidification, evaporative cooling, adiabatic mixing of airstreams, and wet cooling towers.
38