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Chapter Two
Psychrometrics of Air Condition Processes
2.1- Psychrometric Chart
Fig.(2.1) Psychrometric chart
It is a graphical representation of various thermodynamic properties of moist air. The
psychrometric chart is very useful for finding out the properties of air (which are required in the
field of air condition) and eliminate lot of calculations. There is a slight variation in the charts
prepared by different air-conditioning manufactures but basically they are all alike. The
psychrometric chart is normally drawn for standard atmospheric pressure of 760 mm of Hg
(or 1.01325 bar).
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In a psychrometric chart, dry bulb temperature is taken as abscissa and specific heat i.e.
moisture content as ordinate, as shown in Fig.(2.1). Now the saturation cure is drawn by plotting the
various saturation points at corresponding dry bulb temperatures. The saturation curve represent
100% relative humidity at various dry bulb temperatures. It also represents the wet bulb and dew
point temperatures.
Though the psychrometric chart has a number of details, yet the following lines are important
from subject point of view:
1. Dry bulb temperature lines: The dry bulb temperature lines are vertical i.e. parallel to the
ordinate and uniformly spaced as shown in Fig.(2.2). Generally, the temperature range of these lines
on psychrometric chart is from -6ºC to 45ºC. The dry bulb temperature lines are drawn with
difference of every 5ºC and up to the saturation curve as shown in the figure. The values of dry bulb
temperatures are shown on the saturation curve.
2. Specific humidity or moisture content lines: The specific humidity (moisture content) lines are
horizontal i.e. parallel to the abscissa and are also uniformly spaced as shown in Fig.(2.3).
Generally, moisture content range of these lines on psychrometric chart is from 0 to 30 g / kg of dry
air (or from 0 to 0.030 kg / kg of dry air). The moisture content lines are drawn with a difference of
every 1 g (0.001 kg) and up to the saturation curve as shown in the figure.
3. Dew point temperature lines: The dew point temperature lines are horizontal i.e. parallel to the
abscissa and non-uniformly spaced as shown in Fig.(2.4). At any point on the saturation curve, the
dry bulb and dew point temperatures are equal.
The values of dew point temperatures are generally given along the saturation curve of the chart
as shown in the figure.
Fig.(2.2) Dry bulb temperature lines
Fig.(2.3) Specific humidity lines
Fig.(2.4) Dew point temperature lines
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4. Wet bulb temperature lines: The wet bulb temperature lines are inclined straight lines and nonuniformly spaced as shown in Fig.(2.5). At any point on the saturation curve, the dry bulb and wet
bulb temperatures are equal.
The values of wet bulb temperatures are generally given along the saturation curve of the chart
as shown in the figure.
5. Enthalpy (total heat) lines: The enthalpy (or total heat) lines are inclined straight lines and
uniformly spaced as shown in Fig.(2.6). These lines are parallel to the wet bulb temperature lines,
and are drawn up to the saturation curve. Some of these lines coincide with the wet bulb
temperature lines also.
The values of total enthalpy are given on a scale above the saturation curve as shown in the
figure.
6. Specific volume lines: The specific volume lines are obliquely inclined straight lines and
uniformly spaced as shown in Fig.(2.7). These lines are drawn up to the saturation curve.
The values of volume lines are generally given at the base of the chart.
7. Relative humidity lines: The relative humidity lines are curved and follow the saturation curve.
Generally, these lines are drawn with values of relative humidity 10%, 20%, 30% etc. and up to
100%. The saturation curve presents 100% relative humidity. The values of relative humidity lines
are generally given along the lines themselves as shown in Fig.(2.8).
Fig.(2.5) Wet bulb temperature lines
Fig.(2.7) Specific volume lines
Fig.(2.6) Enthalpy lines
Fig.(2.8) Relative humidity lines
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Example 2.1. For a sample of air having 22ºC DBT, relative humidity 30 percent at barometric
pressure of 760 mm of Hg, calculate: 1.Vapour pressure, 2. Humidity ratio, 3. Vapour density, and
4.Enthalpy.
Verify your result by psychrometric chart.
Solution: Given :td =22ºC ; φ =30%=0.3 ; pb=760 mm of Hg=760×133.4= 101384N/m2=1.01384
bar
1.Vapour pressure
Let
pv=Vapour pressure
From steam tables (table 1.4), we find that the saturation pressure of vapour corresponding to
dry bulb temperature of 22ºC is
ps=0.026448 bar
We know that relative humidity ( φ ),
p
pv
0 .3 = v =
ps 0.026448
∴
pv=0.3×0.02642=0.007934 bar
2.Humidity ratio
We know that humidity ratio,
pv
0.622 × 0.007934
W = 0.622 ×
=
pb − pv 1.01384 − 0.007934
= 0.0049 kg / kg of dry air
3.Vapour density
We know that vapour density,
W ( pb − pv ) 0.0049(1.01384 − 0.007934 )10 5
ρv =
=
Ra Td
287(273 + 22)
3
= 0.00582 kg / m of dry air
4.Enthalpy
We know that enthalpy,
h = (1.007t d − 0.026 ) + W (2501 + 1.84t d )
= (1.007×22-0.026) + 0.0049× (2501+1.84×22)
=34.58 kJ/kg dry air
Verification from psychrometric chart
The initial condition of air i.e. 22ºC dry bulb temperature and 30% relative humidity is marked
on the psychrometric chart at point A as shown in Fig.(2.9)
From point A, draw a horizontal line meeting the humidity ratio line at C. From the
psychrometric chart, we find that the humidity ratio at point C,
W≈5 g/kg of dry air ≈ 0.005 kg/kg of dry air
Fig.(2.9)
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We also find from psychrometric chart that the specific volume at point A is 0.843 m3/kg of dry
air.
∴ Vapour density, ρv=W/ρa=0.005/0.843=0.0059 kg/m3 of dry air
Now from point A, draw a line parallel to the wet bulb temperature line meeting the enthalpy
line at point E. Now the enthalpy of air as read from the chart is 34.8 kJ/kg of dry air.
2.2 Psychrometric Processes
The various psychrometric processes involved in air conditioning to vary psychrometric
properties of air according to the requirement are as follows:
1. Adiabatic mixing of air streams
2. Sensible heating
3. Sensible cooling
4. Humidification and dehumidification
5. Cooling and adiabatic humidification
6. Cooling and humidification by water injection
7. Heating and humidification
8. Humidification by steam injection
9. Adiabatic chemical dehumidification
2.2.1 Adiabatic Mixing of Two Air Streams
When two quantities of air having different enthalpies and different specific humidities are
mixed, the final condition of the air mixture depends upon the masses involved, and on the enthalpy
and specific humidity of each the constituent masses which enter the mixture.
Now consider two air streams 1 and 2 mixing adiabatically as shown in Fig.(2.10)(a)
Let
m1= Mass of air entering at 1,
h1= Enthalpy of air entering at 1,
W1= Specific humidity of air entering at 1,
m2, h2, W2=Corresponding values of air entering at 2, and
m3, h3, W3=Corresponding values of air mixture at 3.
Fig(2.10)Adiabatic mixing of two air streams
Assuming no loss of enthalpy and specific humidity during the air mixture process, we have for
the mass balance of dry air,
m1 + m2 = m3 ………………… (2.1)
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For the energy balance,
m1 h1 + m2 h2 = m3 h3 ………………… (2.2)
and for the mass balance of water vapour,
m1 W1 + m2 W2 = m3 W3 ………………… (2.3)
Substituting the value of m3 from equation (2.1) in equation (2.2),
m1 h1 + m2 h2 = (m1 + m2 ) h3 = m1 h3 + m2 h3
or m1 h1 − m1 h3 = m2 h3 − m2 h2
m1 (h1 − h3 ) = m2 (h3 − h2 )
∴
m2 h1 − h3
=
………………… (2.4)
m1 h3 − h2
Or
h3 =
m1h1 + m2 h2
………………… (2.5)
m1 + m2
Similarly, substituting the value of m3 from equation (2.1) in equation (2.3), we have
m2 W1 −W3
=
………………… (2.6)
m1 W3 −W2
m W + m2W2
Or
W3 = 1 1
……………………… (2.7)
m1 + m2
Now from equations (2.4) and (2.6)
m2 h1 − h3 W1 − W3
∴
=
=
………………… (2.8)
m1 h3 − h2 W3 − W2
The adiabatic mixing process is represented on the psychrometric chart as shown in Fig.(2.10)
(b). The final condition of the mixture (point 3) lies on the straight line 1-2. The point 3 divides the
line 1-2 in the inverse ratio of the mixing masses.
Example 2.2. 300 m3/min of fresh air at 30 ºC (DBT) dry bulb temperature and 50% RH is to be
mixed with 800 m3/min of recirculated air at 22 ºC (DBT) dry bulb temperature and 10 ºC dew point
temperature. Determine the enthalpy, specific volume, humidity ratio, and dew point temperature of
the mixture.
Solution. Given: v1=300 m3/min; td1=30 ºC; φ1 =50%; v2=800 m3/min; td2=22 ºC; tdp2=10 ºC
Enthalpy of the mixture
Let
h3= Enthalpy of the mixture
The condition of recirculated air at 22 ºC DBT and 10 ºC dew point temperatures is marked on
the psychrometric chart at point 2 as shown in Fig.(2.11). Now mark the condition of fresh air at 30
ºC dry bulb temperature and 50% relative humidity at point 1 as shown in the figure. Join 1 and 2.
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P sychrom etrics of A ir C ondition Processes
h1
50%
h3
1
h2
W1
3
tdp3
W3
10 ºC
W2
2
0.876
m3/kg
vs3
0.846
m3/kg
22
30
Fig(2.11)
From the psychrometric chart, we find that enthalpy of air at point 1,
h1=64.6 kJ/kg of dry air
Enthalpy of air at point 2,
h2=41.8 kJ/kg of dry air
Specific humidity of air at point 1,
W1= 0.0134 kg/kg of dry air
Specific humidity of air at point 2,
W2= 0.0076 kg/kg of dry air
Specific volume at point 1,
vs1= 0.877 m3/kg of dry air
and Specific volume at point 2,
vs2= 0.846 m3/kg of dry air
We know that mass of fresh air at point 1,
v
300
m1 = 1 =
= 342.07 kg / min
v s1 0.877
and mass of recirculated air at point 2,
v
800
= 945.6 kg / min
m2 = 2 =
v s 2 0.846
We know that
m1 h3 − h2
342.07 h3 − 41.8
=
or
=
m2 h1 − h3
945.6 64.6 − h3
h3=47.86 kJ/kg of dry air
Ans.
∴
Specific volume, humidity ratio, and dew point temperature of the mixture
Plot point 3 on line joining the points 1 and 2 corresponding to enthalpy h3=47.86 kJ/kg of dry
air, as shown in Fig.(2.11).
From point 3 on psychrometric chart, we find that specific volume of the mixture at point 3,
vs3= 0.855 m3/kg of dry air
Ans.
Humidity ratio of the mixture at point 3,
W3= 0.0092 kg/kg of dry air
Ans.
And dew point temperature of the mixture at point 3,
tdp3≈12.7 ºC
Ans.
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2.2.2 Sensible Heat Factor
Actually, the heat added during a psychrometric process may be split up into sensible heat and
latent heat. The ratio of the sensible heat to the total heat is known as sensible heat factor (briefly
written as SHF) or sensible heat ratio (briefly written as SHR). Mathematically,
Qs
Sensible heat
SH
SHF =
=
=
………………(2.9)
Total heat
SH + LH Qs + Ql
where
SH=Sensible heat, and
LH=Latent heat
:
. 1
. !" ##$ % & ' ( &) $ *+ , -.2
.&) /&0 / 12 3 &) /&0 #$ , 43 &) 5 *+ ,$3 3
SHR
∆ht
∆hl
1
∆hs
Parallel
State point
2
Sensible heat ratio line
Fig(2.12) Sensible heat ratio line
2.2.3 Sensible Heating
The heating of air, without any change in its specific humidity, is known as sensible heating.
Let air at temperature td1 passes over a heating coil of temperature td3, as shown in Fig.(2.13)a. It
may be noted that the temperature of air leaving the heating coil (td2) will be less than td3. The
process of sensible heating, on the psychrometric chart, is shown by a horizontal line 1-2 extending
from left to right as shown in Fig.(2.13) b. The point 3 represents the surface temperature of the
heating coil.
The heat absorbed by the air during sensible heating may be obtained from the psychrometric
chart by the enthalpy difference (h2-h1) as shown in Fig.(2.13) b. It may be noted that the specific
humidity during the sensible heating remains constant (i.e.W1=W2). The dry bulb temperature
increases from td1 to td2 and relative humidity reduces from φ1 to φ2 as shown in Fig.(2.13) b. The
amount of heat added during sensible heating may also be obtained from the relation:
Heat added,
q = h2 − h1
= c pa (t d 2 − t d 1 ) + W c ps (t d 2 − t d 1 )
= (c pa + W c ps )(t d 2 − t d 1 ) = c pm (t d 2 − t d 1 )
Chapter Tw o
P sychrom etrics of A ir C ondition Processes
The term (c pa + W c ps ) is called humid specific heat (cpm) and its value is taken as 1.022 kJ/kg K.
∴ Heat added
q = 1.022 (t d 2 − t d 1 ) kJ / kg ………….(2.10)
Fig(2.13) Sensible Heating
Notes: 1- For sensible heating, steam or hot water is passed through the heating coil. The heating
coil may be electric resistance coil.
2- The sensible heating of moist air can be done to any desired temperature.
2.2.4 Sensible Cooling
The cooling of air, without any change in its specific humidity, is known as sensible cooling.
Let air at temperature td1 passes over a cooling coil of temperature td3, as shown in Fig.(2.14)a. It
may be noted that the temperature of air leaving the heating coil (td2) will be more than td3. The
process of sensible cooling, on the psychrometric chart, is shown by a horizontal line 1-2 extending
from right to left as shown in Fig.(2.14) b. The point 3 represents the surface temperature of the
cooling coil.
Fig(2.14) Sensible Cooling
The heat rejected by the air during sensible cooling may be obtained from the psychrometric
chart by the enthalpy difference (h1-h2) as shown in Fig.(2.14) b. It may be noted that the specific
humidity during the sensible cooling remains constant (i.e. W1=W2). The dry bulb temperature
reduces from td1 to td2 and relative humidity increases from φ1 to φ2 as shown in Fig.(2.14) b. The
amount of heat rejected during sensible cooling may also be obtained from the relation:
Heat rejected,
q = h1 − h2
= c pa (t d 1 − t d 2 ) + W c ps (t d 1 − t d 2 )
= (c pa + W c ps )(t d 1 − t d 2 ) = c pm (t d 1 − t d 2 )
Chapter Tw o
P sychrom etrics of A ir C ondition Processes
The term (c pa + W c ps ) is called humid specific heat (cpm) and its value is taken as 1.022 kJ/kg K.
q = 1.022 (t d 1 − t d 2 ) kJ / kg
∴ Heat rejected,
Generally, Heat rejected,
q = m × (h1 − h2 ) ………………(2.11)
Notes: 1- For sensible cooling, the cooling coil may have refrigerant, cooling water or cool gas
flowing through it.
2- The sensible cooling can be done only upto the dew point temperature (tdp) as shown in
Fig.(2.14) b. The cooling below this temperature will result in the condensation of moisture.
2.2.5 By-pass Factor of Heating and Cooling Coil
We already discussed that the temperature of the air coming out of the apparatus (td2) will be
less than *td3 in case the coil is a heating coil and more than td3 in case the coil is a cooling coil.
Let 1 kg of air at temperature td1 is passed over the coil having its temperature (i.e. coil
temperature surface) td3 as shown in Fig. (2.15).
A little consideration will show that when air passes over a coil, some of it (say x kg) just bypasses unaffected while the remaining (1-x) kg comes in direct contact with the coil. This by-pass
process of air is measured in terms of a by-pass factor. The amount of air that by-passes or by-pass
factor depends upon the following factors:
1. The number of fins provided in a unit length i.e. the pitch of the cooling coil fins;
2. The number of rows in a coil in the direction of flow ; and
3. The velocity of flow air.
It may be noted that by-pass factor of a cooling coil decreases with decrease in fin spacing and
increase in number of rows.
Fig(2.15) By-pass factor
Balancing the enthalpies, we get
x c pm t d 1 + (1 − x )c pm t d 3 = 1× c pm t d 2
x (t d 3 − t d 1 ) = t d 3 − t d 2
...(where c pm = Specific humid heat )
td 3 − td 2
…………………..(2.12)
t d 3 − t d1
where x is called by-pass factor of the coil and is generally written as BPF. Therefore, by-pass
factor for heating coil,
∴
x=
BPF =
td 3 − td 2
td 3 − td1
…………………..(2.13)
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P sychrom etrics of A ir C ondition Processes
* Under ideal condition, the dry bulb temperature of the air leaving the apparatus ( t d 2 ) should be equal to that of the
coil ( t d 3 ). But it is not so, because of the inefficiency of the coil. This phenomenon is known as by-pass factor.
Similarly, by-pass factor for cooling coil,
t −t
BPF = d 2 d 3 …………………..(2.14)
td1 − td 3
The value of x (BPF) also is obtained by balancing the enthalpies as follows:
h2 = x h1 + (1 − x )h3
for heating coil
h − h2
x= 3
……………………….. (2.15)
h3 − h1
for cooling coil
h −h
x= 2 3
………………….. (2.16)
h1 − h3
Note: The performance of a heating or cooling coil is measured in terms of a by-pass factor. A coil
with low by-pass factor has better performance.
2.2.6 Efficiency of Heating and Cooling Coils
The term (1-BPF) is known as efficiency of coil or contact factor.
∴
Efficiency of the heating coil,
t −t
t −t
η H = 1 − BPF = 1 − d 3 d 2 = d 2 d 1
td 3 − td1
td 3 − td1
Similarly, efficiency of the cooling coil,
ηC = 1 −
t d 2 − t d 3 t d1 − t d 2
=
…………………..(2.17)
t d1 − t d 3 td1 − t d 3
Example 2.3. In a heating application, moist air enters a steam heating coil at 10º C, 50% RH and
leaves at 30º C. Determine the sensible heat transfer, if mass flow rate of air is 100kg of dry air per
second. Also determine the steam mass flow rate if steam enters saturated at 100º C and condensate
leaves at 80º C.
Solution. Given: td1=10º C ; φ1 =50% ; td2=30º C; ma=100kg/s ; ts= 100º C; tC= 80º C
Sensible heat transfer
First, mark the initial condition of air, i.e. 10 ºC dry bulb temperature and 50% relative humidity on
the psychrometric chart at point 1, as shown in Fig.(2.16). Draw a constant specific humidity line
from point 1 to intersect the vertical line drawn through 30º C dry bulb temperature at point 2. the
line 1-2 represents sensible heating of air.
From the psychrometric chart, we find that
enthalpy at point 1,
h1=19.5 kJ/kg of dry air
Fig(2.16)
and enthalpy at point 2,
h2=40 kJ/kg of dry air
We know that sensible heat transfer,
Q=ma(h2-h1)=100(40-19.5)=2050 kJ/s
Ans.
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Steam mass flow rate
From steam tables (table (1.4)), corresponding to temperature of 100º C, we find that enthalpy of
saturated steam,
hg=2675.44 kJ/kg
and enthalpy of condensate, corresponding to 80 ºC,
hf=335 kJ/kg
∴ Steam mass flow rate
Q
2050
=
=
= 0.876 kg / s
hg − h f 2675.44 − 335
= 0.876 × 3600 = 3153 kg / h
Ans.
Example 2.3. The air enters a duct at 10 ºC and 80% RH at the rate of 150 m3/min and is heated to
30 ºC without adding or removing any moisture. The pressure remains constant at 1 atmosphere.
Determine the relative humidity of air at exit from the duct and the rate of heat transfer.
Solution. Given: td1=10º C; φ1 =80%; v1=150 m3/min; td2=30º C; p=pb= 1 atm=101.325 kPa
Relative humidity of air at exit
First, mark the initial condition of air, i.e. 10 ºC dry bulb temperature and 80% relative
humidity on the psychrometric chart at point 1, as shown in Fig.(2.17). Since air is heated to 30 ºC
without adding or removing any moisture, therefore it is a case of sensible heating. Draw a constant
specific humidity line from point 1 to intersect the vertical line drawn through 30º C dry bulb
temperature at point 2. The line 1-2 represents sensible heating of air.
Fig(2.17)
From the psychrometric chart, we find that the relative humidity of air exit i.e. at point 2,
φ2 ≈23.5%
Rate of heat transfer
From the psychrometric chart, we also find that the specific volume of air at point 1,
υ1=0.81 m3/kg of dry air
enthalpy of air at point 1,
h1=26 kJ/kg of dry air
and enthalpy of air at point 2,
h2=45.5 kJ/kg of dry air
We know that amount of air supplied,
Chapter Tw o
ma =
v1
υ1
=
P sychrom etrics of A ir C ondition Processes
150
= 185.2 m 3 / min
0.81
The amount of air supplied may also be obtained as discussed below:
From steam tables (table (1.4)), pressure of vapour corresponding to dry bulb temperature of 10 ºC,
we find that saturation pressure of vapour,
ps1=0.001228 bar
We know that partial pressure of vapour,
pv1= φ1 × ps1=0.8×0.001228=0.00982 bar
…. (Q φ1 = pv1 / ps1 )
(
pb − pv1 )v1 (1.013 − 0.00982)10 5 × 150
ma =
=
= 185.2 m 3 / min
Ra Td 1
287 (273 + 10)
∴
Rate of heat transfer
=ma(h2-h1)=185.2(45.5-26)=3611.4 kJ/min
Ans.
2.2.7 Humidification and Dehumidification
The addition of moisture to the air, without change in its dry bulb temperature, is known as
humidification. Similarly, removal of moisture from the air, without change in its dry bulb
temperature is known as dehumidification. The heat added during humidification process and heat
removed during dehumidification process is shown on the psychrometric chart in Fig.(2.18) a and b
respectively.
It may be noted that in humidification, the relative humidity increases from φ1 to φ2 and
specific humidity also increase from W1 to W2 as shown in Fig.(2.18) a. Similarly, in
dehumidification, the relative humidity decrease from φ1 to φ2 and specific humidity also increase
from W1 to W2 as shown in Fig.(2.18) b.
Fig.(2.18) Humidification and Dehumidification
It may be noted that in humidification, change in enthalpy is shown by the intercept (h2-h1) on
the psychrometric chart. Since the dry bulb temperature of air during the humidification remains
constant, therefore its sensible heat also remains constant. It is thus obvious, that the change in
enthalpy per kg of dry air due to the increased moisture content equal to (W2 -W1) per kg of dry air
is considered to cause a latent heat transfer (LH). Mathematically,
LH = (h2 − h1 ) = h fg (W2 − W1 ) …………………..(2.18)
where hfg is the latent heat transfer of vaporization at dry bulb temperature( t d 1 ).
Notes:
1. For dehumidification, the above equation may be written as:
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P sychrom etrics of A ir C ondition Processes
LH = (h1 − h2 ) = h fg (W1 − W2 ) …………………..(2.19)
2. Absolute humidification and dehumidification processes are rarely found in practice. These
are always accompanied by heating and cooling processes.
3. In air conditioning, the latent heat load per minute is given as:
LH = ma ∆ h = ma h fg ∆W = ν ρ h fg …………………..(2.20)
where
ν = Rate of dry air flowing in m3/s.
ρ = Density of moist air =1.2 kg/m3 of dry air.
hfg= Latent heat of vaporization = 2450 kJ/kg, and
∆W = Difference of specific humidity between the entering and leaving conditions of
air=( W2 − W1 ) for humidification and (W1-W2) for dehumidification.
Substituting these values in the above expression, we get
LH = ν × 1.2 × 2500 × ∆W = 2940ν × ∆W kJ / s or kW …………………..(2.21)
2.2.8 Cooling and Dehumidification
This process is generally used in summer air conditioning to cool and dehumidify the air. The
air passed over a cooling coil or through a cold water spray. In this process, the dry bulb
temperature as well as the specific humidity of air decreases. The final relative humidity of the air is
generally higher than of the entering air. The dehumidification of air is only possible when the
effective surface temperature of the cooling coil (i.e. td4) is less than the dew point temperature of
the air entering the coil (i.e. tdp1). The effective surface temperature of the coil is known as
apparatus dew point (briefly written as ADP). The cooling and dehumidification process is shown
in Fig.(2.19).
Fig.(2.19) Cooling and dehumidification
Let
td1= Dry bulb temperature of air entering the coil,
tdp1= Dew point temperature of air entering = td3, and
td4 = Effective surface temperature or ADP of the coil.
Under ideal conditions, the dry bulb temperature of the air leaving the cooling coil (i.e. td4)
should be equal to the surface temperature of the cooling coil (i.e. ADP), but it is never possible due
to inefficiency of the cooling coil. Therefore, the resulting condition of air coming out of the coil is
Chapter Tw o
P sychrom etrics of A ir C ondition Processes
shown by a point 2 on the straight line joining the points 1 and 4. The by-pass factor in this case is
given by
BPF =
td 2 − td 4
t − ADP
= d2
…………………..(2.22)
td1 − td 4
t d 1 − ADP
W2 − W4
h − h2
= 4
…………………..(2.23)
W1 − W4
h1 − h2
Actually, the cooling and dehumidification process follows the path as shown by a dotted curve
in Fig.(2.19) a, but for calculation of psychrometric properties, only end points are important. Thus
the cooling and dehumidification process shown by a line 1-2 may be assumed to have followed a
path 1-A (i.e. dehumidification) and (i.e. cooling) as shown in Fig.(2.19)a. We see that the total heat
removed from the air during the cooling and dehumidification process is
Also
BPF =
q = h1 − h2 = (h1 − hA ) + (hA − h2 ) = LH + SH
where
LH = h1 − hA = Latent heat removed due to condensation of vapour of
reduced moisture content (W1-W2), and
SH = hA − h2 =Sensible heat removed.
We know that sensible heat factor,
SHF =
Sensible heat
SH
h − h2
=
= A
…………………..(2.24)
Total heat
SH + LH h1 − h2
Note: The line 1-4 (i.e. the line joining the point of entering air and the apparatus dew point) in Fig.
(2.19) b is known as sensible heat factor line.
Example 2.4. The atmospheric air at 30 ºC dry bulb temperature and 75% relative humidity enters a
cooling coil at the rate 200 m3/min. the coil dew point temperature is 14 ºC and the by-pass factor of
the coil is 0.1. Determine: 1. the temperature of air leaving the cooling coil; 2. the capacity of the
cooling coil in tones of refrigeration and in kilowatt; 3. the amount of water vapour removed per
minute; and 4. the sensible heat factor for the process.
Solution. Given: td1=30 ºC ; φ1 =75% ; v1=200 m3/min ; ADP= td4=14 ºC; BPF=0.1
1. Temperature of air leaving the cooling coil
Let
td2= Temperature of air leaving the cooling coil
First, mark the initial condition of the air, i.e. 30 ºC dry bulb temperature and 75% relative
humidity on the psychrometric chart at point 1, as shown in Fig.(2.20). From the psychrometric
chart, the dew point temperature of the entering air at point 1.
tdp1= 25.2 ºC
Since the coil dew point temperature (or ADP) is less than the dew point temperature of entering
air, therefore it a process of cooling and dehumidification.
We know that the by-pass factor,
Chapter Tw o
P sychrom etrics of A ir C ondition Processes
W1
t − td 4
t − ADP
BPF = d 2
= d2
td1 − td 4
t d 1 − ADP
0 .1 =
W2
t d 2 − 14
30 − 14
td2= 15.6
Ans.
∴
Fig(2.20)
2. Capacity of the cooling coil
The resulting condition of air coming out of the coil is shown by point 2, on the line joining
the points 1 and 4, as shown in Fig.(2.20). The line 1-2 represent the cooling and dehumidification
process which may be assumed to have followed the path 1-A (i.e. dehumidification) and A-2 (i.e.
cooling). Now from psychrometric chart, we find that
Water vapour in the entering air or the specific humidity of entering air at point 1,
W1=0.0202 kg/kg of dry air
Water vapour in the leaving air or the specific humidity of leaving air at point 2,
W2=0.011 kg/kg of dry air
Specific volume of entering air at point 1,
υ1=0.886 m3/kg of dry air
Enthalpy of entering air at point 1,
h1=82 kJ/kg of dry air
Enthalpy air at point A,
hA=58 kJ/kg of dry air
and enthalpy of leaving air at point 2
h2=43.5 kJ/kg of dry air
We know that mass of air flowing through the cooling coil,
v
200
ma = 1 =
= 225.7 kg / min
υ1 0.886
∴ Capacity of cooling coil in tones of refrigeration
=ma(h1-h2)=225.7(82-43.5)=8690.7 kJ/min
=8690.7/210=41.38 TR
Ans.
… (Q 1 TR=210 kJ/min)
and capacity of the cooling coil in kilowatt
=8690.7/60=144.845 kW
Ans.
3. Amount of water vapour removed per minute
We know that amount of water vapour removed
=ma(W1-W2)=225.7(0.0202-0.011)=2.076 kg/min
Ans.
4. Sensible heat factor for the process
We know that sensible heat factor,
hA − h 2 58 − 43.5
SHF =
=
= 0.377
h1 − h 2 82 − 43.5
Ans.
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2.2.9 Methods of Obtaining Humidification and Dehumidification
The humidification is achieved either by supplying or spraying steam or hot water or cold
water into the air. The humidification may be obtained by the following two methods:
1. Direct method: In this method, the water is sprayed in a highly atomized state into the room
to be air-conditioned. This method of obtaining humidification is not very effective.
2. Indirect method: In this method, the water is introduced into the air in the air-conditioning
plant, with the help of an air washer, as shown in Fig.(2.21). This conditioned air is then
supplied to the room to be air-conditioned. The air washer humidification may be
accomplished in the following three ways:
a. By using re-circulated spray water without prior heating of air.
b. By pre-heating the air and then washing it with re-circulated water, and
c. By using heated spray water.
Fig(2.21) Air washer
The dehumidification may be accomplished with the help of an air-washer or by using
chemicals. In the air washer system, the outside or entering air is cooled below its dew point
temperature so that it looses moisture by condensation. The moisture removal is also accomplished
when the spry water is chilled water and its temperature is low than the dew point temperature of
the entering air. Since the air leaving the air washer has its dry bulb temperature much below the
desired temperature in the room, therefore a heating coil is placed after the air-washer. The
dehumidification may also be achieved by using chemicals which have the capacity to absorb
moisture in them. Two types of chemicals known as absorbents (such as calcium chloride) and
adsorbents (such as silica gel and activated alumina) are commonly used for this purpose.
2.2.10 Air Washer
Figure (2.21) shows the schematic representation of an air washer. It involves the flow of air
through a spray of water. During the course of flow, the air may be cooled or heated, humidified or
dehumidified, or simply adiabatically saturated, depending on the mean surface temperature of
water. The water is, accordingly, externally cooled or heated or simply recirculated by pump. Make
up water is added for any loss in the case of humidification of air. Eliminator plates are provided to
minimize the loss of water droplets.
Figure (2.22) shows the thermodynamic changes of state of air along paths 1-2 in air washer,
depending on the mean surface temperature of water droplets tsurf. which is equal to the actual
temperature of water tw.
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P sychrom etrics of A ir C ondition Processes
Thus, the droplets of water act as wetted surface, and both sensible and latent heat transfer take
place. Their directions depend on the temperature and vapour pressure potentials. The following
processes are possible:
Process 1-2A: Heating and humidification (tsurf. > td)
The mean surface temperature of water is greater than the dry bulb temperature of air. The water
is externally heated.
Process 1-2B: Humidification (tsurf. =td)
The mean surface temperature of water is equal to the dry bulb temperature of air. The enthalpy of
air increases. Hence the water is required to be externally heated.
Process 1-2C: Cooling and humidification (tw<tsurf. <td)
The mean surface temperature of water is less than the dry bulb temperature of air but greater
than the wet bulb temperature of air. Though the air is cooled, its enthalpy increases as a result of
humidification. The water is, therefore, required to be externally heated.
Process 1-2D: Adiabatic saturation (tsurf. =tw)
This is the case of pumped recirculation of water without any external heating or cooling as
discussed in Sec. 1.5. The recirculated water reaches the equilibrium temperature which is equal to
the thermodynamic wet bulb temperature of air.
Process 1-2E: Cooling and humidification (tdp<tsurf. <tw)
The process is similar to 1-2C with the difference that the enthalpy of air decreases in this case.
Accordingly, water is required to be externally cooled.
Process 1-2F: Cooling (tsurf. = tdp)
The temperature of water is equal to the dew point temperature of air. water is required to be
externally cooled.
Process 1-2G: Cooling and dehumidification (tsurf. < tdp)
The mean surface temperature of water is lower than the dew point temperature of air. Ai is
simultaneously cooled and dehumidified. The process is exactly similar to that of a cooling and
dehumidifying coil. Again, the limiting process is along the condition line tangent to the saturation
line drawn from initial state 1.
It is thus seen that the air washer affords means for a year-round air-conditioning system.
W1
tdp1
tw1
td1
Fig.(2.22) Range of psychrometric processes with air washer
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The effectiveness or humidifying efficiency of the air washer or spray chamber is given by
Actual drop inDBT Actual drop in sp. humidity
=
Ideal drop inDBT
Ideal drop in sp. humidity
t −t
W − W1
= d 2 d1 = 2
…………………..(2.25)
t s − t d 1 Ws − W1
ηH =
2.2.11 Humidification by injection of water or steam
Let water, or steam, at certain temperature be injected and sprayed into a flowing air stream with
the help of nozzles as shown in Fig.(2.23).
When use steam or water and by assume that all water is evaporating or drift the vapour with air
and not condensate steam on the walls of the channel and assume adiabatic process, we get on the
following equation of energy balance:
m2
t2
W2
h2
m1
t1
W1
h1
m1h1 + mw h3 = m2 h2 …………………..(2.26)
For mass balance for dry air
m1=m2=m …………………..(2.27)
for water vapour
m1W1 + mw = m2W2 …………………..(2.28)
Or
mw = m(W2 − W1 )
……………………(2.29)
also
mw h3 = m(h2 − h1 )
……………………(2.30)
By dividing Equ.(2.30)on Equ.(2.29), we get:
h −h
∆h
h3 = 2 1 =
…………………..(2.31)
W2 − W1 ∆W
as shown in Fig.(2.24)
Humidification by water injection
(1-a) constant enthalpy humidification
(1-b) adiabatic humidification
(1-c) humidification with water at 100 ºC
Humidification by steam injection
(1-d) isothermal humidification with steam at 100 ºC
(1-e) humidification with steam at 234 ºC
and 30 bar and hs=2803 kJ/kg
as shown in Fig.(2.25)
h3
Fig.(2.23)Steam or water injection
Fig.(2.24) Gradation of enthalpy of water or
steam injection on the protractor
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2.2.12 Cooling and Humidification by Water Injection (Evaporative Cooling)
Let water at a temperature tl is injected into the flowing stream of dry air as shown in Fig.(2.26).
The final condition of air depends upon the amount of water evaporation. When the water is
injected at the temperature equal to the wet bulb temperature of the entering air (tw1), then the
process follows the path of constant wet bulb temperature line, as shown by the line 1-2 in
Fig.(2.26 )b.
Fig.(2.26) Cooling and Humidification by water injection
Example 2.5. 200 m3 of air per minute is passed through the adiabatic humidifier. The condition of
air at inlet is 40 ºC dry bulb temperature and 15% relative humidity and the outlet condition is 25 ºC
dry bulb temperature and 20 ºC wet bulb temperature. Find the dew point temperature and the
amount of water vapour added to the air per minute.
Solution. Given: v1=200 m3/min; td1=40 ºC ; φ1 =15% ; td2=25 ºC; tw2=20 ºC
First, mark the inlet condition of air at 40 ºC dry bulb temperature and 15% relative humidity on
the psychrometric chart at point 1, as shown in Fig.(2.27). Now mark the outlet condition of air at
25ºC dry bulb temperature and 20 ºC wet bulb temperature, as point 2. The line 1-2 represents the
adiabatic humidification.
Dew point temperature
On the psychrometric chart, draw a horizontal line through point 2 up to the saturation curve. From
the chart, we find that dew point temperature,
tdp2=17.6 ºC
Ans.
Amount of water vapour added to the air per minute
From the psychrometric chart,
we find that specific volume of air at point 1,
υ1=0.896 m3/kg of dry air
Specific humidity at point 1,
W1=0.007 kg/kg of dry air
and specific humidity at point 2,
W2=0.1026 kg/kg of dry air
We know that mass of air supplied,
v
200
ma = 1 =
= 223.2 kg / min
υ1 0.896
∴ Amount of water vapour added to the air
=ma(W2- W1)
Fig.(2.27)
=223.2(0.0126-0.007) =1.25 kg/min
Ans.
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P sychrom etrics of A ir C ondition Processes
Example 2.6. Determine the final dry bulb temperature and relative humidity of air washer with
recirculated spray water if the air is initially at dry bulb temperature 35 ºC and 50% relative
humidity as it enters an air washer which has humidifying efficiency of 85 percent.
Solution. Given: td1=35 ºC ; φ1 =50% ; ηH=85%=0.85
First, mark the initial condition of air at 35 ºC dry bulb temperature and 50% relative humidity
on the psychrometric chart at point 1, as shown in Fig.(2.28). The wet bulb temperature of the
entering air as read from the psychrometric chart is
tw1=26.1 ºC
Final dry bulb temperature
Let td2 = final dry bulb temperature of the air
leaving the air washer
We know that humidifying efficiency of an air washer (ηH),
t −t
35 − t d 2
0.85 = d 1 d 2 =
t d 1 − t d 3 35 − 26.1
td2 = 35-0.85×8.9=27.435 ºC
Ans.
∴
Fig.(2.28)
Final relative humidity
On the constant wet bulb temperature line 1-3, mark point 2 such that td2 = 27.435 ºC. Now the
relative humidity of the air leaving the air washer (corresponding to point 2) as read from the
psychrometric chart is
φ2 =90%
Ans.
2.2.13 Heating and Humidification by Steam Injection
The steam is normally injected in to the air in order to increase its specific humidity as shown in
Fig.(2.29) a). This process is used for the air conditioning of textile mills where high humidity is to
be maintained. The dry bulb temperature of air changes very little during this process, as shown on
the psychrometric chart in Fig.(2.29) b).
Fig.(2.29)Heating and Humidification by steam injection
Chapter Tw o
P sychrom etrics of A ir C ondition Processes
Let
ms= Mass of steam supplied,
ma= Mass of dry air entering,
W1= Specific humidity of entering air,
W2= Specific humidity of leaving air,
h1= Enthalpy of entering air,
h2= Enthalpy of leaving air, and
hs= Enthalpy of steam injected into the air,
Now for the mass balance,
m
…………………(2.23)
W2 = W1 + s
ma
and for heat balance,
m
h2 = h1 + s × hs = h1 + (W2 − W1 ) × hs
ma
…. [From equation (2.23)]
Example 2.7. The atmospheric air at 25 ºC dry bulb temperature and 12 ºC wet bulb temperatures is
flowing at the rate of 100 m3/min through the duct. The dry saturated steam at 100 ºC is injected
into the air steam at the rate of 72 kg per hour. Calculate the specific humidity and enthalpy of the
leaving air. Also determine the dry bulb temperature, wet bulb temperature and relative humidity of
the leaving air.
Solution. Given: td1=25 ºC ; tw1=12 ºC v1=100 m3/min; ts=100 ºC; ms=72 kg/h = 1.2 kg/min
Specific humidity of the leaving air
Let
W2= Specific humidity of the leaving air.
First, mark the initial condition of air at 25 ºC dry bulb temperature and 12 ºC wet bulb
temperature on the psychrometric chart at point 1, as shown in Fig.(2.30). Now from the
psychrometric chart, we find that the specific volume of air at point 1,
υ 1= 0.849 m3/kg of dry air
Specific humidity of air at point 1,
W1= 0.0034 kg/kg of dry air
Enthalpy of air at point 1,
h1= 34.2 kJ/kg of dry air
We know that mass of air flowing,
v
100
ma = 1 =
= 117.78 kg / min
υ1 0.849
Fig.(2.30)
We know that
m
1 .2
W2 = W1 + s = 0.0034 +
= 0.0135 kg / kg of dry air
Ans.
ma
117.78
Enthalpy of the leaving air
Let
h2= enthalpy of the leaving air.
From steam table (1.4), we find that enthalpy of dry-saturated steam corresponding to 100 ºC
hs= 2675.44 kJ/kg
m
1 .2
We know that h2 = h1 + s × hs = 34.2 +
× 2675.44 = 61.45 kJ / kg of dry air
Ans.
ma
117.78
Dry bulb temperature, wet bulb temperature and relative humidity of the leaving air.
Chapter Tw o
P sychrom etrics of A ir C ondition Processes
Mark the condition of leaving air on the psychrometric chart as point 2 corresponding to
W2= 0.0135 kg/kg of dry air and h2= 61.45 kJ/kg of dry air. Now from the psychrometric chart
corresponding to point 2,
Dry bulb temperature of the leaving air,
td2=26.1 ºC
Wet bulb temperature of the leaving air,
tw2=21.1 ºC
and relative humidity of the leaving air,
φ2 =64%
Ans.
Note: Solve this example by using the protractor
2.2.14 Heating and Dehumidification - Adiabatic Chemical Dehumidification
This process is mainly used in industrial air conditioning and can also be used for some comfort
air conditioning installations requiring either a low relative humidity or low dew point temperature
in the room.
In this process, the air is passed over chemicals which have an affinity for moisture. As the air
comes in contact with these chemicals, the moisture gets condensed out of the air and gives up its
latent heat. Due to the condensation, the specific humidity decreases and the heat condensation
supplies sensible heat for heating air and thus increasing its dry bulb temperature. The processes,
which is the reverse of adiabatic saturation process, is shown by the line 1-2 on the psychrometric
chart as shown in Fig.(2.31). The path followed during the process is along the constant wet bulb
temperature line or constant enthalpy line.
Actual
adsorption
dehumidification
W1
W2
W3
Fig.(2.31)Heating and Dehumidification
The effectiveness or efficiency of the dehumidifier is given as
ηH =
Actual increase in dry bulb temperature t d 3 − t d 1
=
Ideal increase in dry bulb temperature t d 2 − t d 1
Notes:
1. In actual practice, the process is accompanied with a release of heat called heat of adsorption,
which is very large. Thus the sensible heat gain of air exceeds the loss of latent heat and the process
is shown above the constant wet bulb temperature line in Fig.(2.31).
2. Two types of chemicals used for dehumidification are absorbents and absorbents. The
absorbents are substances which can take up moisture from air and during this process change it
chemically, physically or in both respects. This includes water solutions or brines of calcium
Chapter Tw o
P sychrom etrics of A ir C ondition Processes
chloride, lithium chloride, lithium bromide and ethylene glycol. These are used as air dehydrators
by spraying or otherwise exposing a large surface of the solution in the air stream.
The adsorbents are substances in the solid state which can take up moisture from the air and
during this process do not change it chemically or physically. These include silica gel (which is a
form of silicon dioxide prepared by mixing fused sodium silicate and sulphuric acid) and activated
alumina (which is a porous amorphous form of aluminium oxide).
Example 2.8. Saturated air at 21 ºC is passed through a drier so that its final relative humidity is
20%. The drier uses silica gel adsorbent. The air is then passed through a cooler until its final
temperature is 21 ºC without a change in specific humidity. Determine: 1. the temperature of air at
the end of the drying process; 2. The heat rejected during the cooling process; 3. the relative
humidity at the end of the cooling process; 4. the dew point temperature at the end of the drying
process; and 5.the moisture removed during the drying process.
Solution. Given: td1= td3=21 ºC; φ2 =20%
1. Temperature of air at the end of the drying process
First, mark the initial condition of air at 21 ºC dry bulb temperature upto the saturation curve
(because the air is saturated) on the psychrometric chart at point 1, as shown in Fig.(2.32). Since the
drying process is a chemical dehumidification process, therefore, it follows a path along the
constant wet bulb temperature or the constant enthalpy line as shown by the line 1-2 in Fig.(2.32).
Now mark the point 2 at relative humidity of 20%.
From the psychrometric chart, the temperature at the end of drying process at point 2,
td2=38.5 ºC
2. Heat rejected during the cooling process
The cooling process is shown by the line 2-3 on the psychrometric chart as shown in Fig.(2.32).
From the psychrometric chart, we find that enthalpy of air at point 2,
h2=61 kJ/kg of dry air
and enthalpy of air at point 3,
h3=43 kJ/kg of dry air
∴ Heat rejection during the cooling process
= h2- h3=61-43 =18 kJ/kg of dry air
Ans.
Fig.(2.32)
3. Relative humidity at the end of the cooling process
From the psychrometric chart, we find that relative humidity at the end of the cooling process
(i.e. at point 3),
φ3 =55%
Ans.
4. Dew point temperature at the end of the drying process
From the psychrometric chart, we find that the dew point temperature at the end of the drying
process,
tdp2=11.6 ºC
Ans.
5. Moisture removed during the drying process.
Chapter Tw o
P sychrom etrics of A ir C ondition Processes
From the psychrometric chart, we find that the moisture in air before the drying process at
point 1,
W1=0.0157 kg/kg of dry air
And moisture in air after the drying process at point 2,
W2=0.0084 kg/kg of dry air
∴ Moisture removed during the drying process
= W1- W2=0.0157-0.0084=0.0073 kg/kg of dry air
Ans.
2.3 Cycles of air conditioning
To design the air conditioning system for any space and for any season of the year, the internal
and external condition of the place must be taken in the account. All the air conditioning cycles may
be used to obtain the internal condition of the space. The most important air conditioning cycles are
the summer air conditioning cycle and the winter air conditioning cycle.
2.3.1. The summer typical air conditioning cycle
Figure (2.33) represent a typical air conditioning cycle in summer for comfort purpose. Point (r)
represents internal room condition while point (o) represents the external design condition. The
mixing of these two streams occur due to ventilation purpose at point (m), after that the mixture
enters a cooling coil to cooling it and reduce the moisture content to be at point (a). The point (a)
depends on the by-pass-factor and the apparatus dew point (c). If the air is supplied directly to the
room then the process is (ra) (the sensible heat ratio curve), or the air is heated sensibly to a point
(s) the supplying point. The air after that supplied to the room. The sensible heating from (a) to (s)
is caused by the heat from the fan or distributing ducts.
Fig.(2.33) Typical summer air conditioning cycle
2.3.2. The winter typical air conditioning cycle
Figure (2.34) represent a typical air conditioning cycle in winter for comfort purpose. Point (r)
represents internal room condition while point (o) represents the external design condition, and
point (m) represents the mixture condition, from point (m) the mixture is heated sensibly to the
supplied point (s), then to the room by the path (rs), or from point (s) to point (a) by humidification
process and then sensible cooling from (a) to (r) due to the heat losses through the ducts and other
sources.
Chapter Tw o
P sychrom etrics of A ir C ondition Processes
Wr=Wa
Wm=Ws
Fig.(2.34) Typical winter air conditioning cycle
Example 2.9.An air conditioned auditorium is to be maintained at 27°C dry bulb temperature and
60% relative humidity. The ambient condition is 40°C dry bulb temperature and 30°C wet bulb
temperature. The total sensible heat load is 100 000 kJ/h and the total latent heat load is 40 000
kJ/h. 60% of the return air is recirculated and mixed with 40% of make-up air after the cooling coil.
The condition of air leaving the cooling coil is at 18°C. Determine 1.Room sensible heat factor;
2.The condition of air entering the auditorium; 3.The amount of make-up air; 4.Apparatus dew
point; and 5.By-pass factor of the cooling coil. Show the processes on the psychometric chart.
Solution. Given: td4=27 ºC; φ4 =60% ; td1=40 °C; tw1=30 °C; RSH=100000 kJ/h; RLH=40000kJ/h;
td2=18 ºC
1.Room sensible heat factor
We know that room sensible heat factor,
RSH
100000
RSHF =
=
= 0.714
RSH + RLH 100000 + 40000
2. Condition of air entering the auditorium
The line diagram for processes involved in the air conditioning of an auditorium is shown in Fig.(
). These processes are shown on the psychometric chart as discussed below:
Fig.(2.35)
First, mark the ambient condition of air (outside air) at 40 ºC dry bulb temperature and 30 °C wet
bulb temperature on the psychrometric chart at point 1, as shown in Fig.(2.36). Now mark the
condition of air in the auditorium at 27°C dry bulb temperature and 60% relative humidity, as point
4.
Mark the calculated value of SHR=0.714 on the sensible heat factor scale on the protractor and
draw line from the center of the protractor to value of SHR as shown in Fig.(36). Now from point 4,
Chapter Tw o
P sychrom etrics of A ir C ondition Processes
draw a line 4-5(known as RSHF line) parallel to the line drawn in the protractor. Since the condition
of air leaving the cooling coil is at 18°C, therefore, mark point 2 such that td2=18 ºC. Join points 1
and 2 and produce upto point 6 on the saturation curve. The line 1-2-6 is given that 60% of the air
from the auditorium is recirculated and mixed with 40% of the make-up air after the cooling.
RSHF
Fig.(2.36)
The condition of air entering the auditorium is given by point 3. From the psychrometric chart, we
find that at point 3,
ma , 4
m
Q
t3 = a , 2 t 2 +
t 4 = 0.4 × 18 + 0.6 × 27 = 23.4 oC
ma ,3
ma ,3
Wet bulb temperature
tw3=19.7 °C
and relative humidity
φ3 =71%
3. Amount of make-up air
From the psychrometric chart, we find that enthalpy of air at point 4,
h4=61.5 kJ/kg of dry air
and enthalpy of air at point 3,
h3=56.5 kJ/kg of dry air
Room total heat RSH + RLH
ms =
=
h4 − h3
h4 − h3
100000 + 40000
=
= 28000 kg / h
61.5 − 56.5
Since the make-up air is 40% of supply air, therefore mass of make-up air
=0.4×28000=11200 kg/h
Ans.
4.Apparatus dew point
From the psychrometric chart, we find that the apparatus dew point of the cooling coil at
point 6 is
ADP= td6=13.3 °C
Ans.
5. By pass factor of the cooling coil
We know that by-pass factor of the cooling coil,
t − ADP 18 − 13.3
BPF = d 2
=
= 0.176
Ans.
t d 1 − ADP 40 − 13.3
Chapter Tw o
P sychrom etrics of A ir C ondition Processes
Example 2.10.Air at 10°C dry bulb temperature and 90% relative humidity is to be heated and
humidified to 35°C dry bulb temperature and 22.5°C wet bulb temperature. The air is pre-heated
sensibly before passing to the air washer in whish water is recalculated. The relative humidity of the
air coming out of the air washer is 90%. This air is again reheated sensibly to obtain the final
desired condition. Find : 1.The temperature to which the air should be preheated. 2. The total
heating required; 3.The make up water required in he air washer; and 4.the humidifying efficiency
of the air washer.
Solution. Given: td1=10 ºC; φ1 =90% ; td2=35°C; tw2=22.5°C
First, mark the initial condition of air at 10 ºC dry bulb temperature and 90% relative humidity,
on the psychrometric chart at point 1, as shown in Fig.(2.37). Now mark the final condition of air at
35°C dry bulb temperature and 22.5°C wet bulb temperature at point 2.
From point 1, draw a horizontal line to represent sensible heating and from point 2 draw
horizontal line to intersect 90% relative humidity curve at point B. Now from point B, draw a
constant wet bulb temperature line, which intersects the horizontal line drawn through point 1 at
point A. The line 1-A represents preheating of air, line AB represents humidification and line B-2
represents reheating to final condition.
1.Temperature to which the air should be preheated
From the psychrometric chart, the temperature to which the air should be preheated
(corresponding to point A) is
tdA=32.6 ºC
Ans.
2.Total heating required
From the psychrometric chart,
we find that enthalpy of air at point 1,
h1=27.2 kJ/kg of dry air
Enthalpy of air at point A,
hA=51 kJ/kg of dry air
and enthalpy of air at point 2,
h2=68 kJ/kg of dry air
we know that heat required for preheating of air
= hA- h1=51-27.2=23.8 kJ/kg of dry air
and heat required for reheating of air
Fig.(2.37)
= h2- hB=68-51=17 kJ/kg of dry air
… (Q hB = hA)
∴ Total heat required
=23.8+17=40.8 kJ/kg of dry air
Ans.
3. Make up water required in the air washer
From the psychrometric chart, we find that specific humidity of entering air,
W1=0.00068 kg/kg of dry air
and specific humidity of leaving air,
W2=0.0122 kg/kg of dry air
∴ Make up water required in the are washer
= WB- WA= W2- W1
=0.0122-0.00068=0.0054 kg/kg of dry air
4.Humidifying efficiency of the air washer
From the psychrometric chart, we find that
tdB=19.1 ºC
and tdB' =18 ºC
We know that humidifying efficiency of the air washer
Actual drop in DBT t dA − t dB
ηH =
=
Ideal drop in DBT
t dA − t dB '
32.6 − 19.1
= 0.924 or 92.4%
32.6 − 18
Chapter Tw o
P sychrom etrics of A ir C ondition Processes
Example 2.11. In order to compare the space conditions produced and the relative energy
requirements of bypass and reheat systems, consider a space that has a sensible load of ∑Qs=205kW
and a latent load of ∑QL=88kW when the space is maintained at a dry bulb temperature of 25°C and
the outdoor condition are 35°C dry bulb temperature and 40% relative humidity. The space dry bulb
condition is to be met by using either of the systems shown in Figs.(2.38) or (2.40). Supply air is to
be introduced to the space at a flow rate of 30kga/s.The flow rate of exhaust air is 4.5 kga/s. The exit
condition of the systems cooling coil are a dry bulb temperature of 10°C and relative humidity of
95%. For by-pass system shown in fig. (2.38)determine:(a) the relative humidity in the space and
(b) the required system cooling capacity. For the reheat system shown in fig. (2.40)determine (c) the
relative humidity in the space , (d) the rate of heat required for the reheat coil, and (e) the required
system cooling capacity.
Solution. The nomenclature of Fig.(2.38) will be used for parts (a) and (b). Standard atmospheric
pressure is assumed and the solution carried out using psychrometric chart. States 4 and 6 can be
located from given information. States 6,7, and 2b fall on the mixing line connecting 6 and 2b. The
slope of this line is determined by observing that it is also the space-condition line from state 1 to
state 2. The sensible-heat ratio for the space-condition line is given by:
Qs
205
SHR =
=
= 0.7
Qs + Ql 205 + 88
(a) State 2 is located at the intersection of the line with an SHR=0.7 drawn through state 6 and
td2=25 ºC. The resulting point falls at φ2 =50%. Thus using the system shown in Fig.(2.35), the
space conditions will be td2=25 ºC and φ2 =50%.
(b) State 5 is on the mixing line connecting state 4 and 2a.(Note that states 2, 2a, 2b, and 3 are the
same) In order to determine the division of the return air between mºa,2a and mºa,2b, locate state 7,1.
Qs = m c p (t r − t s )
205=30×1.02×(25- td1)
⇒ td1=18.3 ºC
Where tr=td2 , ts=td1, and cp=1.02 kJ/kg .K
m2b h1 − h6 t1 − t 6
=
=
m6 h2 − h1 t 2 − t1
From Fig.(2.38) we observing that mºa,2b= mºa,1- mºa,6 and that mºa,5 = mºa,6
Q
Q
m5 h2 − h1 t 2 − t1 25 − 18.3
=
=
=
= 0.45
m1 h2 − h6 t 2 − t 6
25 − 10
Thus
mºa,5 =0.45×30=13.5 kg/s
mºa,2b=30-13.5=16.5 kg/s
Also, from Fig.(2.38) it is seen that
mºa,2a= mºa,1- mºa,3- mºa,2b=30-4.5-16.5=9 kg/s
ma , 4t 4 − ma , 2 a t 2 a
4.5 × 35 − 9 × 25
= 28.3 oC
ma ,5
13.5
Thus, state 5 is located at the intersection of the line connecting 2a and 4 and a dry bulb temperature
28.3 ºC.
from chart ⇒ h5 = 57.8 kJ / kg dry air
The required system cooling capacity
Q
t5 =
=
Chapter Tw o
5
P sychrom etrics of A ir C ondition Processes
Q6 = mo 5 × (h5 − h6 )
=13.5×(57.8-28.3)=398.25 kW
SHR
4
5
2,2a,2b,3
1,7
6
Fig.(2.38)
SHR
Fig.(2.39)
The nomenclature of Fig.(2.40) will be used to analyze the reheat system.
(c) The dry bulb temperature of the supply air is found in the same manner as it was in part (b) of
this example. Therefore, td1=18.3 ºC. The state 1,7 is located by following a sensible-heating
process line (constant W) beginning at state 6 and ending at td1= td7=18.3 ºC. State 2 is located by
drawing the space condition line beginning at state 1 and ending at the prescribed space dry bulb
temperature of td2=25 ºC. In part (a) of this example the SHR for the space-condition line was
calculated to be SHR =0.70. Locating state 2 on the psychrometric chart, we read φ2 =45%. Thus it
is seen that use of the reheat system results in a lower relative humidity in the space than does the
use of the bypass system (45% vs. 50%).
From psychrometric chart h1,7=37.1 kJ/kg of dry air
(d) The reheat energy rate is
o
6 Q7 = m 5 × (h7 − h6 )
=30×(37.1-28.3)=264 kW
(e) State 5 is the intersection of td5 with the adiabatic mixing process line connecting states 2 and 4.
Notice that in the reheat system the entire supply-air flow rate passes through the cooling coil.
Therefore, mºa,5 = mºa,1=30 kg/s and
mºa,2= mºa,1- mºa,3=30-4.5=25.5 kg/s
from chart h2=46.5 kJ/kg of dry air
ma , 4t 4 − ma , 2 t 2 4.5 × 35 − 25.5 × 25
Q
t5 =
=
= 26.5 oC
ma ,5
30
After locating state 5, we read h5=50.2 kJ/kg of dry air
The system cooling capacity required is
Chapter Tw o
5
P sychrom etrics of A ir C ondition Processes
Q6 = m o 5 × (h5 − h6 )
=30×(50.2-28.3)=657 kW
SHR
4
5
2,3
6
1,7
SHR
Fig.(2.41)
Fig.(2.40)
The result of example are summarized in table
Result or Set point
Bypass System
Reheat System
Space dry-bulb temperature
Space relative humidity
Required cooling capacity
Required reheat
Total energy rate required
25 ºC
50%
398.25kW
---398.25kW
25 ºC
45%
657 kW
264 kW
921 kW
Chapter Tw o
P sychrom etrics of A ir C ondition Processes
Example 2.13. The below figure show layout of winter system. Sensible load for the space 30 kW
and the space lose humidity at rate 8 kg/hr. Indoor design condition DBT=22 ºC and 30% Rh and
outdoor design condition DBT=5 ºC and WBT=2 ºC. The air supplied to the conditioned space at
rate 10200 m3/hr and the air humidified after supplied to the conditioned space by injection of
saturated steam at temperature 100 ºC. Rate of the outdoor ventilation air (fresh air) 2000 m3/hr.
Determine: 1. Condition of supply air to the conditioned space (DBT & RH); 2.Consumption rate of
steam in humidifier; 3. Load of heating load; 4. Draw the cycle on the psychrometric chart.
Exhaust
air
Supply
air
Humidifier
Conditioned
space
Return
air
Steam
at 100 ºC
Heating
coil
Outdoor
air
Fig.(2.42)
Solution:
QL = m o w h fg , hfg=2450 kJ/kg
QL=8/3600×2450=5.444 kW
Qs
30
=
= 0.846 ≈ 0.85
Qs + QL 30 + 5.44
Qs = m c p (t s − t r )
Where tr=td1 , ts=td5, and cp=1.02 kJ/kg .K
SHR =
Qs = 1.22 v o (t s − t r )
30 =1.22×(10200/3600)×( td5-22)
td5 =30.67 ºC
from the psychrometric chart, υ1=0.843 m3/kg dry air
υ2=0.792 m3/kg dry air
m1=m5-m2
v1=v5-v2
ma1 =
v1
=
v2
=
υ1
10200 − 2000
= 9727.16 kg / h
0.843
2000
= 2525.25 kg / h
υ 2 0.792
m5= m1+m2=9727.16+2525.25= 12252.41kg/h=3.4 kg/s
m1t1 − m2 t 2 9727.16 × 22 − 2525.25 × 5
Q
t3 =
=
= 18.494 oC ≈ 18.5o C
m1 + m2
9727.16 + 2525.25
From psychrometric chart
td4 =30.4 ºC, h4=43.5 kJ/kg d.a. , h5=47.1 kJ/kg d.a.
hs from table(1.4)(at t=100 ºC) = 2675.44 kJ/kg
ma 2 =
Chapter Tw o
P sychrom etrics of A ir C ondition Processes
By using value of hs at the protractor, from psychrometric chart, as shown in Fig.(2.43)
RH5=19% , h3=32 kJ/kg d.a.
By using energy balance on the humidifier
m4 h4+ms hs=m5 h5
,
3.4×43.5+ms×2675.44=3.4×47.1
ms=4.575×10-3=16.47 kg/h
Qheating coil = m o 3 × (h4 − h3 )
=3.4×(43.5-32)=39.1 kW
SHR
SHR
hs
1
5
3
4
2
Humidifier
Fig.(2.43)