Air-Vapor Mixtures
and Air-Conditioning
Processes
1
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Lecture Objectives:
To give introduction and
basic background material
for Air Conditioning
Applications
2
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Learning Outcomes:
Students will be able to:
• Understand important terminology such as: humidity ratio,
relative humidity, wet bulb and dry bulb temperature.
• Calculate quantities for air-vapor mixtures such as the dew
point temperature, relative humidity and humidity ratio.
• Do mass and energy balance for air conditioning processes
and apply these to important applications.
• Use the psychrometric chart to obtain properties of air-vapor
mixtures.
3
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Part 1
4
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Assumption for the Specific heat of air in A/C
applications
The Cp of air can be assumed to be constant at 1.005 kJ/(kg ·
°C) in the temperature range -10 to 50°C with an error under
0.2 percent.
This range (from - 10
to 50°C) is the one
considered in A/C
applications. So, this
assumption of
constant Cp will be
valid in what follows.
5
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Dry air, unsaturated air-vapor mixture, and
saturated air-vapor mixture
Preliminary Remarks:
It is reasonable to consider air and vapor to behave
like ideal gases;
Dry Air contains no water vapor, but atmospheric air
does.
Pair vapor mixture = Pdry air + Pvapor
OR in short notation  P = Pa + Pv
where:
Pv is the partial pressure (or simply vapor pressure) of
water vapor in atmospheric air.
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Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Dry air, unsaturated air-vapor mixture, and
saturated air-vapor mixture
This value is taken form
Table A-4
in your textbook of Cengel
For saturated air, the vapor pressure is equal to the
saturation pressure of water at the given
temperature of the air-vapor mixture; SEE Above
Figure.
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Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Assumption for the Specific enthalpy of the vapor in
the air-vapor mixture
Recall: the specific enthalpy
of an IDEAL GAS is a function
of Temperature ONLY.
The region where the vapor behaves like an ideal gas is enclosed by dotted
circle lines.
So, inside this circled region lines of constant temperature, are also lines of
constant enthalpy; (whenever the temperature remains constant, so does the
enthalpy for IDEAL GASES.
Outside this circled region, the vapor cannot be considered as ideal gas 
specific enthalpy is not only a function of Temperature, BUT also, of pressure.
8
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Assumption for the Specific enthalpy of the vapor in
the air-vapor mixture
For low the vapor in the circled
region (in which the vapor
behaves like and ideal gas) we
can take hvapor at any temperature
below 50oC to be as:
hv (at T) = hg(T)
See FIGURE; At temperatures below 50C, the h = constant lines
coincide with the T = constant lines in the superheated vapor region of
water. 
This range of temperature is the one relevant for A/C applications.
SO, for Superheated vapor at low pressure (i.e. low partial pressure of
vapor) at T below 50oC, we can treat vapor as IDEAL GAS 
This will be our assumption in the material to follow on Air Conditioning.
9
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Some important notes on Ideal Gas Mixtures
Recall for ideal gas mixtures :
all components in the mixture have the same volume as
that of the mixture 
For Air  Vapor mixture we have :
Vvapor  Vair  Vm ixture
Since we talk of Ideal Gas Mixtures, we can apply the
ideal gas equation of state for each gas component in
the mixture 
Pdry airVair  mdry air RairTair
PvaporVvapor  mvaporRvaporTvapor
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Vvapor  Vair  Vmixture  V
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Important NOTE for Flow Systems
For a flow of ideal gas mixtures :
all components in the mixture have
the same volume as that of the mixture 
For Air  Vapor mixture we have :
Vvapor  Vair  Vmixture
Pdry airVair  m dry air RairTair
PvaporVvapor  m vapor RvaporTvapor 
Vvapor  Vair  Vmixture  V
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Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Specific Humidity (or the humidity ratio: )
Specific humidity (or humidity ratio):
 = mv /ma
Using the ideal gas equation of state for both the
Pdry airV  mdry air RairTair
vapor and for dry air 
PvaporV  mvaporRvaporTvapor
= 0.622 Pv / Pa = 0.622 Pv / (P – Pv)
Remember: P in the above equation is the total
pressure of the air+vapor mixture = Pa +Pv
12
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Relative Humidity: 
Relative humidity:
Pv

Pg
where:
Pg is the saturation pressure of water at
temperature Tmixture.
Also this means:
Pg is the partial pressure of water vapor when
the Air+Vapor mixture is fully saturated with
water vapor @ Tmixture.
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Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Relative Humidity: 
Also we can express relative humidity in terms of mass ratio as :
mv

mg
mv: is the actual mass of vapor in the air-vapor mixture.
mg is maximum vapor mass that the air can hold at a given
temperature 
This happens when the air is fully saturated with vapor.
At that condition of maximum vapor mass content in the air,
the partial pressure of vapor id the saturation pressure of
water at the given temperature of the mixture.
14
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Relative Humidity: 
REMEMBER that we can apply the IDEAL GAS Equation of STATE
for the Vapor both at the actual condition as it exists in the air/vapor
mixture, AND also at the condition of full saturation (i.e. when the
vapor mass content in the air is maximum; mg,max).
This means:
PvV  mv RvaporTairmixture
Pg ,sat @TairmixtureV  mg ,max RvaporTairmixture
By dividing the above two equations we can show that:
mv Pv


m g Pg
15
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Relative Humidity: 
Important note:

mv Pv

m g Pg
Note: in the above equation mg represents the maximum
vapor mass the air can hold at a certain temperature (i.e. at
the saturation condition).
So, if you try to inject water vapor into the saturated
air+vapor mixture, the vapor will not stay as gas, but will
fall down as liquid (i.e. Cannot be held in the mixture as
gas), BECAUSE THE MIXTURE IS SATURATED ALREADY
WITH VAPOR.
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Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Important Notes
Relative humidity () ranges from 0 to 1 (0% to
100%)
For saturated air  = 100%
For dry air  = 0%
Any water vapor added to a saturated air at a given
temperature will condense.
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Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Example
Consider a room filled with air vapor mixture with the following
data:
P = 101 kPa; T=23oC; Dimensions: 10m x 6m x 3.5m
Relative humidity: 60 percent.
1.Is the air/vapor mixture saturated? Justify your answer!
Determine:
2.The partial pressure of water vapor in the room.
3.The mass of dry air in the room.
4.The specific humidity.
5.The humidity ratio.
6.The Partial pressure of dry air.
7.The mass of vapor in the room.
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Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Important NOTE for Flow Systems
If you have a flow system for air+vapor mixture,
you can ALSO write the humidity ratio equation in
terms of flow rate quantities, as:
m v

 m v   m a
m a
Similarly, you also can write the ideal gas equation of
state in terms of flow rate quantities:
19
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Example
Consider a steady flow of ideal gas mixture with the following
data:
P = 101 kPa; T=23oC; Relative humidity: 60 percent. Flow rate is 10
m3 /min
Determine:
1. The partial pressure of dry air and that of water vapor.
2. The humidity ratio.
3. The mass flow rate of the water vapor.
20
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Recall:
Putting the Ideal Gas Equation of State in
terms of Time-rate Quantities:
Below relations are valid as long as we have justified
reason to assume IDEAL GAS Mixture.
Pdry airVair  m dry air RairTair
PvaporVvapor  m vapor RvaporTvapor
Vvapor  Vair  Vmixture  V
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Dr. Salah Addin B. Al-Omari, MEDept., UAEU
End of Part 1
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Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Enthalpy of Air-Vapor Mixture
H = Ha + Hv = ma ha + mv hv
mv
recall :  
ma
Divide the above equation by the mass of dry air (ma)

h = H / ma = ha +mv hv / ma = ha +  hv
But (as mentioned earlier) at low pressures for
temperatures below 50 oC,
hv is approximately = hg at the
given temperature 
h  ha   hg
Unit of h is : J/kg dry air
23
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Summary of the moist Air Relations
Recall: The enthalpy of moist
(atmospheric) air is expressed (by
convention; (or by definition) per unit
mass of dry air, NOT as the usual
practice we adopted so far by
considering the total mass of the
system (in this case the total mass of
the moist air mixture).
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Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Dew-Point Temperature of Air-vapor
mixture
Dew-Point Temperature: the temperature at which
condensation begins when the air is cooled at constant
pressure.
Pv
   Pv   Pg
Pg
where : Pg  Psat @Tair Vapor mixture ;
from Table A - 4
Then : Tdew  Tsaturation @ Pv ;
From Table A - 5
25
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Example on Dew-Point Calculations
EXAMPLE:
Moist air with relative
humidity of 60% at a
temperature of 30oC.
What is its dew-point
temperature?
26
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Solution of the Dew-Point Temperature
Example
Pv

 Pv   Pg
Pg
where : Pg  Psat @T
o
air Vapor mixture  30 C
from Table A - 4 : Psat @T
o
air Vapor mixture  30 C
 4.2469 kPa;
 Pv  0.6  4.2469  2.548 kPa
 Table A  5 @ Pv  2.548 kPa 
Tsat @ Pv  21.2o C  Tdew point tem perature
Note: later we will use Chart to solve the above example,
instead of equations.
27
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Example
Consider a room filled with air vapor mixture with the following data:
P = 101 kPa; T=23oC; Dimensions: 10m x 6m x 3.5m
Relative humidity: 60 percent.
1.Is the air/vapor mixture saturated? Justify your answer!
Determine:
2.The partial pressure of water vapor in the room.
3.The mass of dry air in the room.
4.The specific humidity.
5.The humidity ratio.
6.The Specific enthalpy of the mixture per unit mass of dry air.
7.The dew point temperature.
8.The Partial pressure of dry air.
9.The mass of vapor in the room.
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Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Wet-Bulb Temperature
FIGURE
A simple arrangement to measure the
wet-bulb temperature.
Note 1:
Using a simple thermometer without
having its bulb covered with a cotton
wick soaking water we measure what
is called the “dry bulb temperature”.
Note 2:
When we measure the wet bulb
temperature using the setup shown in
the Figure, the air surrounding the
cotton wick is fully saturated with water
vapor; i.e. it has relative humidity ϕ=
100%
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Dr. Salah Addin B. Al-Omari, MEDept., UAEU
The Psychrometric chart.
This is the chart that gives the thermo
physical properties of air + vapor
mixtures, such as:
The specific enthalpy (per unit mass of
dry air), the dry and wet bulb
temperatures, the relative humidity, the
humidity ratio, the specific volume, and
the dew point.
FIGURE
Schematic for a psychrometric chart.
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Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Psychrometric Chart for Air+Vapor mixture at a pressure of 101.325 kPa
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Dr. Salah Addin B. Al-Omari, MEDept., UAEU
More detailed Psychrometric chart (Available in
your book)
FIGURE
Actual psychrometric chart for air+vapor mixture at total mixture pressure of 101 kPa.
32
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Air-Conditioning Processes
FIGURE
Various airconditioning
processes.
33
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Example
Repeat the previous example BUT use the psychrometric Chart, instead of
the equations.
Compare between your results based on the chart and those based on the
equations.
Consider a room filled with air vapor mixture with the following data:
P = 101 kPa; T=23oC; Dimensions: 10m x 6m x 3.5m; Relative humidity: 60 percent.
Determine:
2.The partial pressure of water vapor in the room.
3.The mass of dry air in the room.
4.The specific humidity.
5.The humidity ratio.
6.The Specific enthalpy of the mixture per unit mass of dry air.
7.The dew point temperature.
8.The Partial pressure of dry air.
9.The mass of vapor in the room.
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Dr. Salah Addin B. Al-Omari, MEDept., UAEU
EXAMPLE on Use of the psychrometric chart
Consider a room that contains air at 1 atm, 22oC
and 40% relative humidity. Using the
psychrometric chart, determine:
1. The humidity ratio (specific humidity).
2. The enthalpy (per unit mass of dry air).
3. The wet-bulb temperature.
4. The dew-point temperature.
5. The specific volume of the air.
HW: repeat the solution of the above example, BUT use Equations;
NOT the psychrometric chart. Compare the results of both approaches!!
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Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Air-Conditioning Processes: Mass Balance
1
2
FIGURE
During simple heating, specific humidity
 remains constant, but relative
 humidity  decreases.
Dry air mass balance : Water mass balance:
 m a ,in   m a ,exit
 m w,in   m w,exit
Water mass balance can also be written as :
 m a ,in in   m a ,exit exit
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Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Keep in mind:
During simple heating
in = out
Air-Conditioning Processes: Energy Balance
Steady Flow Energy Balance gives (1st law) :
Q in  W in   m in hin  Q out  W out   m exit hexit
For simple heating 
Q  m h  h 
in
a
exit
in
For simple cooling 
Q  m h  h 
out
37
a
in
exit
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
EXAMPLE on Simple Heating Processes; part of example 14-6
An air-conditioning system is to take outdoor air at 10oC and
30% relative humidity at steady rate of 45 m3/min and condition
it to 25oC without adding any moisture to it .
Determine:
1. What do we call this kind of air conditioning process?
2. The relative humidity at the exit.
3. The humidity ratio at the exit.
4. The rate at which heat is added to the air.
5. Show the process on a psychrometric chart.
6. Solve this problem in two ways: using the
psychrometric chart and using the basic
relevant equations.
38
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Air-Conditioning Processes: Simple Cooling
FIGURE
During simple cooling,
specific humidity

remains constant, but
relative humidity 
Increases.
39
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
EXAMPLE on Simple Cooling Processes
An air-conditioning system is to take in outdoor air at 30oC and
50% relative humidity at steady rate of 45 m3/min and condition
it to the lowest temperature possible while maintaining a simple
cooling process.
Determine:
1. The dry bulb and wet bulb temperatures at the exit.
2. The humidity ratio at the exit.
3. The relative humidity at the exit.
4. The rate at which heat is added to the air.
5. Show the process on a psychrometric chart.
6. Solve this problem in two ways: using the psychrometric
chart and using the basic relevant equations.
40
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Air-Conditioning Processes: Simple Heating
with Humidification
FIGURE
Simple Heating with
Humidification
It consists of two stages:
•
Simple Heating from 1 to 2
•
Heating while humidifying
from 2 to 3
41
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
EXAMPLE on Simple Heating with humidification
An air-conditioning system is to take in outdoor air at 10oC and
30% relative humidity at steady rate of 45 m3/min and to
condition it to 25oC and 60% relative humidity. The outdoor air
is first heated to 22oC in the heating section. Assuming the
entire process takes place at a pressure of 100 kPa, determine:
1. The rate of heat supply in the heating section and
2. The mass flow rate of the steam required in the humidifying
section.
3. At which temperature should the water be
injected in the humidification section?
42
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Air-Conditioning Processes: Simple Cooling with
Dehumidification
FIGURE
Simple Cooling with
Dehumidification
It consists of two stages:
•
Simple cooling from 1 to
the saturation line (see
Fig.)
•
Cooling while
DEhumidifying to point 2
(see Fig.)
43
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Air-Conditioning Processes: Simple Cooling with
Dehumidification
IMPORTANT NOTE
You can NOT jump directly from
point 1 to point 2 along a straight
line; in this kind of
dehumidification process you
MUST reach first the saturation
point then dehumidification (i.e.
condensation) starts while the
temperature drops until you reach
the desired point 2.
44
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Air-Conditioning Processes: Simple Cooling with
Dehumidification
Another IMPORTANT NOTE
The condensation (i.e.
dehumidification process) does not
take place at a fixed temperature;
but at a range of temperatures from
the saturation (or dew point
temperature) to the final point
(point 2).
An average temperature between
these two temperatures can be
used in calculations as an
approximation.
45
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
EXAMPLE on Simple Cooling with Dehumidification
Air enters a window air conditioner at 1 atm, 30oC, and 80%
relative humidity at a rate of 10 m3/min, and it leaves as
saturated air at 14oC.
As a matter of simplification of the analysis,
consider that the moisture in the air that
condenses during the process is also
removed at 14oC.
Determine:
1. The rate of heat removal from the air.
2. The rate of mass removal from the air.
46
Dr. Salah Addin B. Al-Omari, MEDept., UAEU
Continued
Psychrometric chart
FIGURE
For saturated air,
the dry-bulb, wetbulb, and dewpoint temperatures
are identical.
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Dr. Salah Addin B. Al-Omari, MEDept., UAEU