Chapter 7
PSYCHROMETRY
1
PSYCHROMETRY
The study of the properties of air and vapour
pertaining to air conditioning problems is called
psychrometry.
Dry air: it is the mixture of nitrogen, oxygen and small
percentage of other gases. Air contains 79% nitrogen,
21% oxygen by volume and has a molecular weight of
29.
Moist air: it is the mixture of dry air and water vapour.
The amount of water vapour varies according to the
temperature of air and reaches saturation at one point.
Saturated air mixture: saturated air mixture is a
mixture of dry air and water vapour in which the partial
pressure of the vapour is equal to the saturation
pressure of water at the temperature of the mixture.
2
Unsaturated air mixture: it is a mixture of dry air and
super heated water vapour. The partial pressure of
vapour being less than the saturation pressure of
water at the temperature of the mixture.
Super saturated air mixture: it is a mixture of dry air
and water vapour in which the partial pressure of the
water vapour is greater than the saturation pressure
of water at the temperature of the mixture.
Dew point temperature: when the unsaturated air is
cooled at constant pressure, the mixture reaches
saturation temperature corresponding to the partial
pressure of water vapour. This temperature at which
condensation of the vapour begins resulting in
formation of liquid droplets or dew when the
mixture is cooled is called dew point temperature.
3
Absolute humidity or specific humidity or humidity
ratio:
The ratio of mass of vapour to the mass of dry air is
called absolute humidity. Denoted by ‘ω’.
4
Dalton’s law of partial pressures: the total
pressure of the mixture of gases is the sum of the
partial pressures exerted by each gas when it
occupies the same volume of the mixture at the
same temperature of the mixture.
Dry bulb temperature: it is the equilibrium
temperature of the mixture indicated by an
ordinary thermometer denoted by Tdb.
Wet bulb temperature: it is the temperature
indicated by a wet bulb thermometer which has
its temperature sensitive element (bulb) covered
with a wick soaked in water. It is denoted by Twb.
5
6
Relative humidity: it is defined as the ratio of
partial pressure of water vapour in a mixture to
the saturation pressure of water at dry bulb
temperature.
using perfect gas relation, pv.vv = pg.vg
Hence
Hence relative humidity is also defined as the
ratio of the mass of water vapour in a certain
volume of moist air at a given temperature to
the mass of water vapour in the same volume of
saturated air at the same temperature.
7
Degree of saturation or saturation ratio: it is
defined as the ratio of specific humidity of actual
air (ω) to the specific humidity of saturated air
(ωg) at the same temperature.
8
Enthalpy of moist air:
Enthalpy of moist air = enthalpy of dry air + enthalpy of
water vapour associated.
h = hair + ω hvapour = Cp tdb + ω hvapour
But,
hair = Cpa tdb = 1.005 tdb
Where Cpa = is the specific heat of air = 1.005 kJ/kg/K
hvapour = hg + Cps (tdb – tdp)
= 2500 + 1.88tdb
Cps = Specific heat of water vapour = 1.88kJ/kg/K & tdp = 0
Therefore, h = 1.005 tdb + ω(2500 + 1.88tdb) kJ/kg of dry air
9
Carrier equation: When DBT and WBT are given, for
calculating the partial pressure of water vapour in
air many equations have been proposed of which
Dr. Carriers equation is most widely used.
Where,
(pg)wb = saturation pressure at wet bulb temperature.
pv = partial pressure of water vapour
pg = partial pressure of saturated vapour
p = total pressure of moist air
tdb = dry bulb temperature, 0C
twb = wet bulb temperature, 0C
10
Psychrometric chart:
As the calculations of various properties of
moist air are tedious and time consuming, all
the essential data for complete thermodynamic
and psychrometric analysis of air conditioning
processes can be summarized and can be
presented in the form of a chart.
Such a chart which makes it possible to obtain
the necessary information readily for
engineering calculations related to moist air
known as a psychrometric chart.
A psychrometric chart is constructed for a given
mixture pressure, using dry bulb temperature
and the specific humidity as co-ordinates.
11
At a given mixture pressure, the vapour pressure pv
is a function of specific humidity only and hence
there is only one value of pv for each value of ω.
12
Saturation curve: the saturation line represents the
states of saturated air at different temperatures.
Wet bulb and dew point temperatures fall on this
curve. Relative humidity on the curve is 100%.
Relative humidity lines: these are the curves with
humidity ranging from maximum 100% to
minimum 0%.
Constant specific volume lines: these are inclined
lines and are equally spaced. The WBT lines are
much flatter than constant enthalpy lines.
DBT lines: these are vertical lines and are equally
spaced. DBT increases as we move from left to
right.
Specific humidity lines: these are horizontal lines
and value of ω increases from bottom towards top
of chart.
13
Sensible heating:
Heating of air without
addition or subtraction of
moisture is called sensible
heating.
This can be achieved by
passing the air over a heating
coil.
The heat added increases the
DBT of air.
This is useful in winter air
conditioning.
The heat added is given by,
Qs = ma (h2 – h1)
14
Sensible cooling:
Cooling of air without
addition or subtraction of
moisture is called sensible
cooling.
This can be achieved by
passing the air over a cooling
coil.
This is useful in summer air
conditioning.
The heat removed is given by,
Qr = ma (h1 – h2)
15
Cooling and dehumidification:
Water vapour may be removed from air by cooling it
below its dew point temperature.
As a result of the cooling process a portion of the vapour
in the air is condensed. Dehumidification will take place
along with cooling when the saturation temperature of
the cooling coil is below the dew point temperature of
the cooling air.
16
In the above process warm air t1 enters the cooling coil
maintained at temperature t2.
The surface temperature of the cooling coil t2 is lesser
than the dew point temperature of the incoming air, t4.
Under ideal conditions air leaves at t2, but due to
inefficient cooling it leaves at a higher temperature t3.
17
The temperature t2 corresponding to point 2 on the
saturation curve is known as apparatus dew point,
ADP. The ratio of actual heating/cooling to the ideal
heating/cooling is known as by-pass factor.
In this case it is given by
18
Adiabatic humidification: if humidification is
carried out adiabatically, the energy required for
the evaporation of the added moisture must
come from the entering air.
19
As the dry bulb temperature of air decreases during
adiabatic humidification process, it is also known as
evaporative cooling process or cooling with
adiabatic humidification of air.
When warm air is passed through a spray chamber,
part of water is vaporized and carried away with air.
This results in humidification of air.
20
Mixing process: In this process two or more streams
are mixed to produce a stream with desirable state of
temperature and relative humidity.
21
Making energy balance,
ma1(ha1 + ω1hv1) + ma2(ha2 + ωhv2) = ma3(ha3 + ω3hv3)
Making mass balance
ma1 + ma2 = ma3
Making moisture balance
ma1 ω1 + ma2 ω2 = ma3 ω3
Combining we get,
22
Requirement of Comfort air conditioning:
The following 5 factors determine the comfort
feeling of the people in an air conditioned space.
1. Supply of O2 and removal of CO2.
2. Removal of body moisture dissipated by the
occupants.
3. To provide sufficient air movement and air
distribution in the occupied space.
4. To maintain the purity of air by removing odour
and dust.
23
1. Oxygen supply: human body takes in oxygen
through the system and gives out CO2. Normally
each person requires nearly 0.65m3 of O2 per hour
and produces 0.2m3 of CO2.
The percentage of CO2 in atmosphere is about
0.6% and it is necessary to maintain this to ensure
comfort for easy breathing. Thus the quantity of air
supply to an air conditioned space should be
regulated properly to see that percentage of CO2
should not be exceeded.
2. Heat removal: it is a well known fact that human
beings dissipate good amount of heat to the
atmosphere during breathing etc.
24
The atmosphere should be capable of absorbing
the heat dissipated by persons otherwise
discomfort exist.
Thus sufficient circulation of air should be provided
through proper ventilation system to avoid rise in
temperature of air in the air conditioned space.
3. Moisture removal: A moisture loss of up to 50%
from human body is commonly observed.
This should be properly taken into account while
designing a air conditioning unit.
The ventilation system must be capable of
maintaining an RH of below 70%.
25
4. Air movement:
In addition to providing air motion, proper air
distribution is very important.
Air distribution is defined as a uniform supply of
air to an air conditioned system.
Air movement without proper air-distribution is
permissible for local cooling sensation known as
draft.
A velocity of about 8m/min associated with
temperature differential of 10C do not result in
noticeable draft.
Velocities
greater
than
this
produces
uncomfortable drafting conditions.
26
5. Purity of air:
It is important to maintain quality of air in any airconditioned space.
Odour, dust, toxic gases and bacteria are
considered for defining the purity of air.
The various factors which makes air impure are:
i. The evaporation on the surface of the body adds
odour to the air.
ii. The smoke from the surroundings which has a bad
effect on nose, eyes and heart.
iii. The toxic gases are objectionable as they cause
irritation.
27
Summer air conditioning systems
(for Hot and dry outdoor conditions): this system is
used when outdoor conditions are hot and dry.
That means atmospheric temperature is higher than
the comfort temperature with less moisture content
(lesser RH).
28
Atmospheric air at condition ‘1’ (higher DBT and lower
RH) enters the air dampers and passes over the cooling
coil via an air filter.
Temperature of the air is reduced to condition ‘2’ in the
cooling coil.
Air now enters an adiabatic humidifier and is passed
over water eliminators.
29
Air enters the conditioned space at condition ‘3’. Point
‘4’ represents the condition of air after passing over the
cooling coil if efficiency of the coil were 100%.
Condition line 1-2-4 represents the changes in DBT of
air when passes over the cooling coil and the condition
line 2-3-5 represents the changes in DBT of cooled air in
the adiabatic humidifier.
30
Summer air conditioning systems:
Hot and humid outdoor conditions: this system is used
when the out door condition is hot and humid, like in
coastal areas. That means the atmospheric temperature is
higher but at the same time contains large moisture
content. This condition eliminates the need for an
adiabatic humidifier as shown in fig.
31
Atmospheric air at condition ‘1’ enter the air filter via air
dampers. Filtered air is cooled when it passes over the
cooling coil at condition ‘3’. Condition line 1-2-3
represents the changes in DBT of air.
While passing over the cooling coil with an efficient coil
the air would have been cooled to 2.
32
As the air temperature is still below the required comfort
condition, it is now passed over a heating coil.
The air coming out of the coil is at condition ‘5’ which is
delivered to the space.
Line 3-5-4 represents changes in DBT of air while it passes
over the heating coil.
Point ‘4’ represents the maximum temperature that could
have been attained by using an efficient heating coil.
33
Winter air conditioning system:
During winter air in the atmosphere is at lower
temperature than the required conditions.
Also the relative humidity may be more or lesser than the
actual relative humidity required for human comfort. Fig
shows an arrangement for such a system where dry
conditions prevail in the atmosphere.
34
Cold air from the atmosphere is filtered in an air filter
before it passes over the heating coil.
Air at condition ‘1’ is heated to ‘3’ in the heating coil.
However due to losses, condition of air leaving the heat
coil is ‘2’. Condition line 1-2-3 represents the changes in
the DBT of air while it passes over the heating coil.
Hot air at ‘2’ now enters the humidifier and it is cooled to
‘4’.
35
Any water particles suspended in air is removed by the
water eliminator.
As the humidity and temperature of air have still not
reached comfort conditions air is heated again up to point
‘7’ by another heating coil.
Due to heat losses the final temperature of air entering
the conditioned space is at point ‘6’.
36
Problem 1. Moist air at 30°C,1.01325 bar has a relative
humidity of 80%. Determine without using the
psychrometry chart Partial pressures of water vapour
and air , Specific humidity , Specific Volume and Dew
point temperature
(V.T.U. July2004)
Solution: At 30C from t able ps  4.2461kPa
p

 p  0.8 x 4.2461 3.397kPa
ps
0.622p
3.397

 0.622x
p  p
101.325 3.397
 0.213kg/kg of dry air.
Corresponding to Pv =3.397 kPa from tables,
we get dew point temperature = 28.9°C
37
Problem 2: Atmospheric air at 101.325 kPa ha 30°C DBT and
15°C DPT. Without using the psychometric chart, using the
property values from the table, Calculate Partial pressure
of air and water vapour, Specific humidity , Relative
humidity,Vapour density and Enthalpy of moist air
Solution:
p  101.325kpa  1.01325bar
DBT  30C , DP T  15C
From table
Corresponding t o DBT  30C, we haveps  0.042461bar
Corresponding t o DP T  15C, we havep  0.017051bar
Partial pressure of air  p - p  1.01325 0.017051
 0.984274bar
38
p
0.622x0.017051
Specifichum idity 0.622

pa
0.984274
 0.01077kJ/
kg of dry air
p
0.017051
Re lative hum idity

 0.4015
ps 0.042461
Ent halphy 1.005t db
 40.15%
  ( 2500 1088t db )
 1.005x30 0.010775(2
500 1.88x30)
 57.69kJ/kgof dry air
RT
Specific volumeof dry air,a 
P
0.2872x303

 0.874m3 / kg
0.98425x100
 0.010775
Vapourdensity  w 

 0.12kg / m3
a
0.847
39
Problem 3:
Air at 30°C DBT and 25°C WBT is heated to 40°C. if
the air is 300 m3/min, find the amount of heat
added/min and RH and WBT of air. Take air pressure
to be 1 bar
Solution:
At 25°C WBT from tables page no 14
Pvs(wbt)=0.03166 bar
40
 p  ( PVS ) wbt
( p  pswbt )(tdb  t wb )

1547 1.44t wb
(1 - 0.03166)(3
0 - 25)
 0.031661547 1.44x 25
 0.0284bar
p
1  0.622
p  p
 0.0284 
 0.622

 1  0.0284
 0.0179kJ / kg of dry air
41
At 40C DBT
PVS  0.07375bar
During sensible heating and p rem ainconst ant
p  0.0284bar
p
0.0284
RH   

ps
0.07375
 0.385 38.5%
H 2  1.005x 40  0.0179( 2500 1.88x 40)
 86.29kJ/kgof dry air
( p  p )V
3
Weight of 300m / min ofair 
RT
(1- 0.0284)x30
0x102

 335.18kg / min
0.287x303
 Heat added/min  335.18(86.
29 - 76)  3449kJ/min
From chart WBT 27.2C
42
4) One stream of air at 5.5m3/min at 15°C and 60% RH
flows into another stream of air at 35m3/min at 25°C and
70%RH, calculate for the mixture 1) Dry bulb temperature,
2) Wet bulb temperature 3) Specific Humidity and
4) Enthalpy
Solution: For air at 15°C and 60%RH, V=5.5m3/min
 ps  0.017051bar
p
RH   
 p  0.6 x0.017051 0.01023bar
ps
(p - p )V (1.01325 0.01023) x102 x5.5
Mass of air 

RT
0.287x 288
m1  6.672kg / min
0.622p
0.622x0.01023
1 

(p - p )
(1.01325 0.01023)
 0.006343kg / kg of dry air
43
H1  1.005t db  1 (2500 1.88t db )
 1.008x18 0.006343(2
500 1.88x15)
 34.12J/kgof dry air
For air at 25C and 70% RH, V  35m3 / min,Ps  0.03169bar
p
  RH 
 p  0.03169x0.7  0.02218bar
ps
2
(1.01325- 0.02218x10
x35)
Mass of air 
 m 2  40.55kg. min
0.287x 298
0.622x0.02218
2 
 0.01392kg / kg of dry air
(1.01325 0.02218)
H 2  (1.005x 25)  0.01392(2500 1.88x 25)
H 2  60.59kJ / kg of dry air
Mass of dry air/Unit mass of moist air
m a1 
m1
6.672

 6.6299
1  1 1  0.006343
44
m2
40.55
Since m a2 

 39.993
1  2 1  0.01392
Then enthalpyof themixedair,
ma1 ( H1 )  ma 2 ( H 2 )
6.6299(34.
12)  39.993(60.
56)
H mix 

m1  m2
6.672 40.55)
 55.96kJ/kg of dry air
Specific Humidityof themixedair,
ma1 (1 )  ma 2 (2 ) (6.6299x0.
006343) (39.993x0.
01932)
mix 

m1  m2
6.672 40.55
 0.01268kg/kg of dry air
But H mix  1.005t db  mix (2500 1.88t db )
55.96  1.005xtdb  0.01234(2500 1.88t db )  t db  24.42C
DBT of t hemixt ure 24.42C
From chart WBT 19C , RH  67%
45
Problem 5:
An air conditioning system is designed under the
following conditions
Outdoor conditions: 30°C DBT, 75% RH
Required indoor conditions: 22°C DBT,70% RH
Amount of Free air circulated 3.33 m3/s
Coil dew point temperature DPT=14°
The required condition is achieved first by cooling
and dehumidification and then by heating.
Estimate
The capacity of the cooling coil in tons of refrigeration
Capacity of the heating coil in kW
The amount of water vapour removed in kg/hr
46
Solution:
Locate point' a' 30C DBT ,75%RH out door condit ion
Locate point' d' 22C DBT ,70%RH required condit ion
Locate point' b' 14C DP T ,coil surface t emperat re
u
Join ab at d, draw a horizont alline t o cut t he
line ab at pointc.
ac  coolingand dehumidificat ion
cd  heat ing
47
From chart
H a  83kJ / kg of air , H b  40kJ / kg of air
H d  53kJ / kg of air , H c  48kJ / kg of air ,
Wa  0.0202kg / kg of dry air
Wc  Wd  0.0118kg / kg of dry air , Vsa  0.88m / kg
3
V
3.33
Mass of air 

 3.78kg / s
Va
0.88
ma (H a  H c )
Capacit yof coolingcoil 
3. 5
3.78(83  48)

 37.84tons of refrigeration
3. 5
Capacit yof heat ingcoil  m a ( H d  H c )
 3.78(53- 48)  18.92kW
48
Am ountof water vapour removed ma (a  d )3600
 3.78(0.020
2 - 0.0118)360
0
 114.3kg/hr
49
Problem 6:
A summer air conditioning system for hot and humid
weather (DBT=32°Cand 70% RH)
Consists in passing the atmosphere air over a cooling
coil where the air is cooled and dehumidified. The air
leaving the cooling coil is saturated at the coil
temperature. It is then sensibly heated to the required
comfort condition of 24°C and 50%RH by passing it over
an electric heater then delivered to the room.
Sketch the flow diagram of the arrangement and
represent the process undergone by the air on a
skeleton psychometric chart and determine
1. The temperature of the cooling coil
2. The amount of moisture removed per kg of dry air in
the cooling coil.
3. The heat removed per kg of dry air in the cooling coil
and
4. The heat added per kg of dry air in the heating coil 50
From chart
H a  86kJ / kg of air
H b  38kJ / kg of air
H c  48.5kJ / kg of air
a  0.021kg / kg of dry air
b  0.0092kg / kg of dry air
51
The t emperat re
u of t hecoolingcoil  Tb  13C
Am ountof moist ureremoved a  b
 0.021- 0.0092 0.0108kg/kg of dry air
Heat removed H a  H b  86  38  48 kJ/kg of dry air
Heat added  H c  H b  48.5  38  10.5kJ/kg of dry air
Locatepoint' a' 32C , 70%RH out door condit ion
Locatepoint' c' 24C DBT ,50%RH required condit ion
At c draw a horizont alline t o cut t hesat urat ion
line at point' b'
Join ab
ab  coolingand dehumidificat ion
bc  heat ing
52
Problem 7
It is required to design an air conditioning plant for an
office room with the following conditions.
Outdoor conditions: 14°CDBT, 10°CWBT
Required conditions: 20°CDBT,60% RH
Amount of air circulated 0.3m3/min/person
Starting capacity of the office= 60
The required condition is achieved first by heating and
then by adiabatic humidifying. Determine the
following.
Heating capacity of the coil in kW and the surface
temperature required, if the by pass factor of the
coil is 0.4 Capacity of the humidifier.
53
Locate point' a'14C , and10CWBT (out door condition)
Locate point' c' 20C DBT ,60%RH required condition
At a draw a horizontalline
At ' c' draw a constantenthalpyline to cut thehorozontalline
at point' b'
Join ab
ab  heating
bc  adiabatichumidification
54
From chart
H a  30kJ / kg of air , H b  H c  43kJ / kg of air
 a  b  0.006kg / kg of dry air
c  0.00875kg/kg of dry air
Specific volomeVsa  0.8175m 3 / kg
0.3x60
Volum eof air supplied  V 
 0.3m 3 / sec
60
V
0.3
Weight of air supplied m a 

Va 0.8175
 0.3669kg/sec
Capacityof theheatingcoil  m a ( H b  H a )
 0.3669(43- 30)  4.77kW
From chart Tb  26.5C
Let coil surface temperatu
re be Td
55
Td  Tb
By passing factor
 0.4Td  5.6  Td  26kJ
Td  Ta
Td  26.5
0 .4 
 Td  34.8C
Td  1.4
Capacityof thehumidifier  m a (c  b ) x3600
 0.3669(0.00875- 0.006)3600
 3.63kg/hour
56
Problem 8
An air conditioned system is to be designed for a hall of
200 seating capacity when the following conditions are
given:
Atmospheric condition = 300C DBT and 50% RH
Indoor condition = 220C DBT and 60% RH
Volume of air required = 0.4m3/min/person
The required condition is achieved first by chemical
dehumidification and after that by sensible cooling.
Find the following .
1. DBT of the air leaving the dehumidifier.
2. The quantity of water vapour removed in the
dehumidifier per hour.
3. The capacity of cooling coil in tons of refrigeration.
4. Surface temperature of the coil if the by pass factor of
the coil is 0.25.
57
Solution:
Locate point ‘a’, 300C DBT, 50% RH, the atmospheric condition.
Locate point ‘c’, 220C DBT, 60% RH, the required indoor
condition.
“Since chemical dehumidification process follows constant
enthalpy line”
at a draw a line parallel to constant enthalpy line.
At ‘c’ draw a constant  line to cut the previous line at point b.
DBT of air leaving the dehumidifier Tb = 40.50C
From chart
Hb = Ha = 65kJ/kg, a = 0.013 kg/kg of dry air
Hc = 45 kJ/kg, b = 0.009 kg/kg of dry air, Vsa = 0.875 m3/min
Volume of air = 200 X 0.4 = 80 m3/min
Wa = Weight of air = V/Vsa = 80/0.875 = 91.42 kg/min
58
Quantity of water vapour removed/hour =
Wa(a-b)60
= 91.42(0.13-0.009)60 = 21.94 kg/hr
Capacity of cooling coil = Wa(Ha-Hb)/ (60 X 3.5)
= 91.42(65-45)/(60 X 3.5)
= 8.7 tons
By pass factor = (Tc-Td)/( Tb-Td) = 0.25
Td = Temperature of cooling coil = 15.830C
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Problem 9
An air conditioned system is to be designed for a
cinema hall of 1000 seating capacity when the
following conditions are given:
Outdoor condition = 110C DBT and 70% RH
Required indoor condition = 200C DBT and 60% RH
Amount of air required = 0.3m3/min/person
The required condition is achieved first by heating,
then by humidifying and finally by heating. The
condition of air coming out of the humidifier is 75%
RH.
Find the following .
Heating capacity of the first heater in kW and condition
of the air coming out of the first heater in kW and
condition of the air
60
Locate point ‘a’, 300C DBT, 50% RH, the atmospheric condition.
Locate point ‘c’, 220C DBT, 60% RH, the required indoor
condition.
“Since chemical dehumidification process follows constant
enthalpy line”
at a draw a line parallel to constant enthalpy line.
At ‘c’ draw a constant  line to cut the previous line at point b.
DBT of air leaving the dehumidifier Tb = 40.50C
From chart
Hb = Ha = 65kJ/kg, a = 0.013 kg/kg of dry air
Hc = 45 kJ/kg, b = 0.009 kg/kg of dry air
Vsa = 0.875 m3/min
Volume of air = 200 X 0.4 = 80 m3/min
Wa = Weight of air = V/Vsa = 80/0.875 = 91.42 kg/min
61
Quantity of water vapour removed/hour
= Wa(a-b)60
= 91.42(0.13-0.009)60 = 21.94 kg/hr
Capacity of cooling coil
= Wa(Ha-Hb)/ (60 X 3.5) = 91.42(65-45)/(60 X 3.5)
= 8.7 tons
By pass factor = (Tc-Td)/( Tb-Td) = 0.25
Td = Temperature of cooling coil = 15.830C
62