air conditioning duct design

1. Chapter Three
Psychrometric chart use and
requirement calculations

2. DRY AND ATMOSPHERIC AIR
Air in the atmosphere normally contains some water vapor
(or moisture) and is referred to as atmospheric air. By
contrast, air that contains no water vapor is called
dry
air.
The temperature of air in air-conditioning applications
ranges from about -10 to about 50°C. In this range, dry air
can be treated as an ideal gas with a constant c value of
p
1.005 kJ/kg · K
The enthalpy and enthalpy change of dry air can be
determined from
where T is the air temperature in °C and ΔT is the change
in temperature.

3.  At 50°C, the saturation pressure of water is 12.3 kPa.
 At pressures below this value, water vapor can be
treated as an ideal gas. even when it is a saturated
vapor.
 Therefore, water vapor in air behaves as if it existed
alone and obeys the ideal-gas relation Pv=RT.
 Then the atmospheric air can be treated as an ideal-gas
mixture whose pressure is the sum of the
partial
pressure of dry air P and that of water vapor P
a v
 Since water vapor is an ideal gas, the enthalpy of water
vapor is a function of temperature only.
 This can be observed from the T-s diagram of water
where the constant enthalpy lines coincide with
constant-temperature lines at temperatures below 50°C.

4. At temperatures below
50°C, the h=constant
lines coincide with the
T=constant lines in the
superheated
region of water.
vapor
 Therefore, the enthalpy The enthalpy of water vapor at
0°C is 2500.9 kJ/kg. The
average c value of water vapor
in the temperature range -10 to
50°C can be taken to be 1.82
kJ/kg · °C. Then the enthalpy of
water vapor can be determined
approximately from
of water vapor in air
can be taken to be equal to the
enthalpy of saturated vapor at
the same temperature. That is,
p

5. SPECIFIC AND RELATIVE HUMIDITY OF AIR
 The amount of water vapor in the air can be specified in various
ways. the specific humidity (ω) and relative humidity (Ø)
 As the mass of water vapor present in a unit mass of dry
air called absolute or specific humidity (also called humidity
ratio)
denoted by ω:
 Ra=0.2870& Rv= 0.4615 kJ/kg · K
The amount of water vapor in
saturated air at a specified
temperature and pressure can
be determined from equations
above by replacing P by P , the
 the vapor pressure may be
given by the following
equation :P =Ps-PA(T – T )
v g
saturation pressure of water at V db wb
that temperature
example
Where
A-is constant =6.66X10-4 ºC-1
T &T -dry and wet bulb Temp
db wb
Ps-saturation pressure at T
P-total pressure
wb

6.  The amount of moisture in the air has a definite effect
on how comfortable we feel in an environment.
 And this comfort level depends more on the amount of
moisture the air holds (m ) relative to the maximum
v
amount of moisture the air can hold at the same
temperature (m ). The ratio of these two quantities is
g
called the relative humidity (Ø)
 Where
 Combiningω andØ

7. enthalpy of air
 the enthalpy of air is expressed in terms of the
enthalpies of the dry air and the water vapor.
 Since the amount of dry air in the air–water-vapor
mixture remains constant, and the amount of water
vapor changes enthalpy of atmospheric air is expressed
per unit mass of dry air instead of per unit mass of the
air–water vapor mixture.

8. Example-1
A 5mX5mX3m room contains air at 25°C and 100kPa at a relative
humidity of 75 percent.
Determine (a) the partial pressure of dry air, (b) the specific humidity,
(c) the enthalpy per unit mass of the dry air, and (d ) the masses of
the dry air and water vapor in the room.
Solution
(a) P =P –P ,
a v
P =ØP =ØP (0.75)(3.1698 kPa)=2.38 kPa
v g sat @ 25°C
Therefore Pa=100KPa-2.38KPa=97.62KPa
(b) The specific humidity ω=0.622Pv/(p-Pv)=0.0152 kg H2O/kg dry air
(c) The enthalpy of air per unit mass of dry air is h=h + ωh =C T + ωh
g
a v p
=(1.005kJ/Kg. 0C)(250C)+(0.0152)(2546.5KJ/Kg)
=63.8kJ/Kg dry air The enthalpy of water vapor (2546.5 kJ/kg) could
also be determined from the approximation given by
Both the dry air and the water vapor fill the entire room completely.
Therefore, the volume of each gas is equal to the volume of the room:
Va=Vv=V=5X5X3=75m3

9. The masses of the dry air and the water vapor are determined from
the ideal gas relation applied to each gas separately:
ma=PaVa/RaT=(97.62KPa)(75m3)/[(0.287KPa.m3/Kg.K)(298K)]=85.6Kg
mv=PvVv/RvT=2.38KPa)(75m3)/[(0.4615KPa.m3/Kg.K)(298K)]=1.3Kg
Mass of the water can also be determined from mv=ωma
=(0.0152)(85.6Kg)=1.3Kg

10. DEW POINT TEMPERATURE
 The dew-point temperature T is defined as the
dp
temperature at which condensation begins when the air
is cooled at constant pressure.
 T is the saturation temperature of water
dp
corresponding to the vapor pressure:
The ordinary temperature and
the dew-point temperature of
saturated air are identical.

11. Example-2
In cold weather, condensation frequently occurs on the
inner surfaces of the windows due to the lower air
temperatures near the window surface. Consider a house
that contains air at 20°C and 75 percent relative
humidity. At what window temperature will the moisture
in the air start condensing on the inner surfaces of the
windows?
Solution
 The saturation pressure of water at 20°C is
Psat=2.3392kPa
 The temperature distribution in a house, in general, is
not uniform. When the outdoor temperature drops in
winter, so does the indoor temperature near the walls
and the windows. Therefore, the air near the walls and

12.  the windows remains at a lower temperature than at the
inner parts of a house even though the total pressure
and the vapor pressure remain constant throughout the
house. As a result, the air near the walls and the
windows undergoes a P constant cooling process until
v=
the moisture in the air starts condensing. This happens
when the air reaches its dew-point temperature which is
determined from
Where Pv=ØPg@200C=(0.75)(2.3392KPa)=1.754KPa
thus Tdp=Tsat@1.754KPa=15.40C
Note that the inner surface of the window should be
maintained above 15.4°C if condensation on the window
surfaces is to be avoided.

13. Summary

14. PSYCHROMETRIC CHART
 In air conditioning interpolations make use of prepared
tables of values time consuming, hence information that
is sufficiently precise for the solution of most air
conditioning problems can usually be obtained-from
graphs or charts.
 The commonly air conditioning chart applicable and
relates temperature, humidity, enthalpy, and certain
other properties of moist air i.e specific volume is called
psychrometric chart.
PSYCHROMETRIC CHART Is a readable
chart used in sizing of typical air
conditioning systems which involve numerous
calculations.

15. The basic features of the
psychrometric chart includes
dry-bulb temperatures shown
on the horizontal axis, and
the specific humidity shown
on the vertical axis.
As can be seen from the above diagram
 On the left end of the chart, there is a curve (called the saturation
line) it is the curve of 100 percent relative humidity.
 Lines of constant wet-bulb temperature have a downhill appearance to
the right.
 Lines of constant specific volume (in m3
/kg dry air) look similar,
except they are steeper.
 Lines of constant enthalpy (in kJ/kg dry air) lie very nearly parallel to
the lines of constant wet-bulb temperature.
 In some charts the constant wet-bulb-temperature lines are used as
constant-enthalpy lines.
The dew-point temperature is determined by drawing a horizontal line from the
specified state to the left until it intersects the saturation line

16. ... Psychrometric Chart: Temperature
 Dry Bulb Temperature (T
;
DB
°C or F)
Wet Bulb Temperature(TWB; °C or F)
 The temperature of air
measured by a thermometer
freely exposed to the air but
shielded from radiation and
moisture.
 It is the minimum temperature that the
moist air could achieve if enough water
was added to achieve saturation (RH =
100%).
 It is the true air
temperature we “feel.”  Also called the adiabatic saturation
temperature.

17. ...Psychrometric Chart: Temperature
 Dew Point Temperature (T
;
DP
°C or F)
• The temperature at which
saturation is reached (RH =
100%) when the moisture content
of the air (W) stays constant.
• In other words, T is the
DP
temperature at which water will
begin to condense out of moist
air.
• Condensation occurs when:
 T < T
air DP
 E.g. When the temperature of
cold drink is below the dew-point
temperature of the surrounding
air, it “sweats.”

18. ...Psychrometric Chart; Water content: Specific
humidity/ absolute humidity
 Humidity Ratio/Specific Humidity (kg H O/kg DA
or
2
lb H O/lb
DA)
2
•Expressed as a ratio of water vapour content to total
amount of dry air.

19. ...Psychrometric Chart; Water content:
Relative humidity
 Relative Humidity (φ/RH; %)
•The ratio of actual vapour pressure to saturation vapour
pressure at the same temperature.
•Simply, it is a measure of how much water is in the air
versus how much water the air can hold at the same
temperature.

20. ...Psychrometric Chart; Energy content:
Enthalpy
 Enthalpy (h; kJ/kg dry air or BTU/lb air)
 Enthalpy represents the amount of sensible and latent energy
contained in the moist air. Lines of constant enthalpy and constant
wet-bulb are the same on this chart but values are read off
separate scales.
 Example: For air condition point (P) the enthalpy is read at point A.
The sensible heat component can be read at point B, corresponding
to the enthalpy of dry air at the same temperature. The remainder,
i.e. A - B, is the latent heat content.

21. ...Psychrometric Chart: Volume
 Specific Volume (ν;
m
3/kg dry
air or ft /lb dry
air)
3
•This is the volume of the moist
air mixture (volume occupied by
both dry air and water vapour)
versus the unit mass of dry air.
•At higher temperatures, the air
molecules are more energetic
causing the volume of the moist
air mixture to expand and the
density to decrease.
•The specific volume of air is the
inverse of density (ν = 1/ρ)

22. ...Psychrometric Chart: Saturation line
 Saturated air line or dew point temperature line
•Saturation line is a temperature condition at which water
will begin to condense out of moist air.
•Given air at a certain dry bulb temperature and relative
humidity, if the temperature is allowed to decrease, the
air is no longer able to hold as much moisture.

24. Reading the Psychrometric chart
If any two of the dry bulb temperature ( T ), wet bulb temperature
db
(T ), dew point temperature ( T ), specific humidity (ω ), relative
wb dp
humidity (Ø), specific volume ( v ), and enthalpy or total heat content
(H ), seven properties of an air water vapour mixture are known, the
others can readily be found from the chart.

25. Example 1
Consider a room that contains air at 1 atm, 35°C, and 40
percent relative humidity. Using the psychrometric chart,
determine
(a) the specific humidity, (b) the enthalpy, (c) the wet-bulb
temperature, (d ) the dew-point temperature, and (e) the
specific volume of the air.
Solution
given total pressure, the state of atmospheric air is
completely specified by two independent properties such as
the dry-bulb temperature and the relative humidity. Other
properties are determined by directly reading their values
at the specified state.
(a) The specific humidity is determined by drawing a
horizontal line from the
specified state to the right until it intersects with the ω
axis, ω= 0.0142 kg H2O/kg dry air

26. (b) The enthalpy of air per unit mass of dry air is determined by drawing a
line parallel to the h =constant lines from the specific state until it intersects
the enthalpy scale, giving
h= 71.5 kJ/kg dry air
(c) The wet-bulb temperature is determined by drawing a line parallel to the
Twb constant lines from the specified state until it intersects the saturation
line, giving
T =24° C
wb
(d ) The dew-point temperature is determined by drawing a horizontal line
from the specified state to the left until it intersects the saturation line,
giving T = 19. 4° C
dp
(e) The specific volume per unit mass of dry air is determined by noting the
distances between the specified state and the v constant lines on both sides
of the point. The specific volume is determined by visual interpolation to be v
=0. 893 m3/kg dry air

27. Example 2
 A sample of moist air has a T of 43 ċ and T of 29 ċ ,
db WB
using the psychrometric chart find :
a- Specific humidity
b- Relative humidity
c- Dew point temperature
d- Specific enthalpy
e- Specific volume

28. Examples 3

29. Changing the Condition of Air
 The heating, cooling, humidifying, and dehumidifying
processes which take place in air conditioning all change
the air from a condition represented by an initial state
point on the chart to some condition represented by
another state point on the chart.
There are different processes possible.
1. Constant latent heat process ( indicated by constant moisture
content and constant dew point temperature ).
2. Constant sensible heat process ( indicated by constant dry bulb
temperature ).
3. combination of sensible heat and latent heat change And Constant
relative humidity process ( all other factors change ).
4. Constant enthalpy (total heat ) process or adiabatic process (
indicated by constant wet bulb temperature ).
5. Mixture of air quantities at different conditions.

30. 1. Constant latent heat process (Sensible heating, sensible cooling)

31. 2. Constant sensible heat process
(Humidification, dehumidification)

32. 3. Combination of
sensible
latent
and
heat
change processes

33. 4. Constant enthalpy (total heat ) process
[Adiabatic cooling]
The process of adding latent heat and removing sensible
heat at constant enthalpy.

34. Examples :
Air at a state of T = 14 ċ , RH= 50% is passed through a
DB
heating coil . The T is increased up to 42 ċ . The
DB
moisture content remain constant in this process.
Find : a) T of the exit air. b) The dew point temperature.
WB
c) The sensible heat added by the heating coil for 1.0
kg/s of air .
a) 19.5ċ , b) 3.9ċ , c) 28.6 kW}

35. 5. Mixture of air quantities at different
conditions

36.  Adiabatic mixing of different quantities of air in two
different states at constant pressure. The conditions of
the mixing state may be found by the following relations
and as shown in the figure above
T = (m T + m T ) / (m + m ) or ;
3 1 1 2 2 1 2
h = (m h + m h ) / (m + m ) or;
3 1 1 2 2 1 2
ω = (m ω + m ω ) / (m + m ) ;
3 1 1 2 2 1 2
It is acceptable practice in air conditioning to use volume
ratio rather than mass ratio:
T = ( v T + v T ) / (v + v ) ; similarly
3 1 1 2 2 1 2
h = ( v h + v h ) / (v + v ) ; and similarly for ω
3 1 1 2 2 1 2
ω = ( v ω + v ω ) / (v + v ) ;
3 1 1 2 2 1 2
where m is mass flow rate in kg/s
v is volume flow rate in m3/s
T is dry bulb temperature in 0C
ω is specific humidity and h is enthalpy

37. Example :
 Two air streams are mixed the first at T =21ċ , T =
DB WB
14ċ and the second at T = 28ċ , T = 20 ċ with mass
DB WB
flow rates of 1 kg/s and 3 kg/s for the first and second
respectively .
 Find the moisture content ,enthalpy ,and the T for the
DB
mixture and plot the process on the psychrometric
chart .
(answers : 0.01 kgwv/kgda ,52.15 kJ/kg , 26.25 ċ )