air conditioning - Elements of Heat Engines

air conditioning - Elements of Heat Engines
3
AIR CONDITIONING
3.1 Introduction
Air conditioning may be defined as the simultaneous control of tem perature, humidity
(moisture content), motion of air, (movement and circulation), and purity (filtering and
purification) of air within an enclosed space for the purpose of :
. . Producing comfortable and healthy conditions for the occupants (persons) in
residences, theatres, office buildings, hospitals, railway coaches, etc.
. . Increasing efficiency of the employees working in factories, offices and stores.
Conditioned air has a comfort effect, a health effect, and a psychological or
emotional effect on human beings, and
. . Making possible more complete environmental condition for manufacturing processes
in'industries - such industries include printing, textile manufacturing, chem ical,
pharm aceutical, etc.
3.2
Facto rs in A ir Conditioning
Complete air conditioning provides automatic control of the following factors for all
outdoor (outside) weather conditions for both summer and winter :
- Remove the dust, dirt, soot and germs from the outside air entering the air
conditioning plant.
- Cool the air in summer and heat the air in winter.
- Decrease the humidity (moisture content in the air) in summer and increase the
humidity in winter.
- Circulate and distribute the conditioned air evenly and pleasantly throughout the
building at all tim es.
- Ozonate or ionize the air for the elimination (removal) of odours (unpleasant sm ell)
after it has been conditioned.
3.2.1 Clean A ir : Clean air is desirable for breathing, for the reason that the dust
in the air covers the air passages of the nose, throat and lungs, which seriously affects
their healthy functioning. Dirt is objectionable also because it serves as a carrier of germs.
In the manufacture of food products, drugs and-other chem icals, the dirt may be dangerous
to health. In the finishing of automobiles and furnitures and in many other manufacturing
processes, the elimination (removal) of dirt is quite necessary. In the air cooling of electric
generators in power plants, dirt must be removed as it may choke the narrow air passages
in winding and thus, cause serious overheating.
3.2.2
Tem perature Control : Control of temperature, which means the maintenance
of any desired temperature (temperature of 25°C is recommended for maximum comfort)
within an enclosed space even when the outside temperature is either above or below
the desired room temperature. This means, either addition or removal of heat from and
within the enclosed space as the occasion might demand (i.e. to increase the temperature
in winter and to decrease the temperature in summer).
3.2.3 Hum idity Control : Control of humidity, which means adding humidity or
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ELEM EN TS OF H EAT EN G IN ES Vol. Ill
moisture to the air, a s is necessary during the winter, and removing moisture from air
during summer, in order to produce comfortable and healthy conditions within an enclosed
space. The water vapour in air, usually called “moisture", is an important but quite variable
constituent of the atm osphere; its amount in outside air depends on the weather conditions.
When air contains the maximum amount of water vapour that it can hold, it is said to
be saturated. At a particular temperature, only a fixed amount o f water vapour can be
held by one kg o f dry air. If the temperature of saturated water vapour drops down
sufficiently, i.e., below its saturation temperature, some of the vapour will get condensed
and if it is raised, the capacity of the air to hold water vapour will be increased.
Humidity in air is necessary for comfort. During cold weather a person in a room
often feels uncomfortable even though the thermometer indicates what is considered a
normally comfortable tem perature. This condition is caused by air currents and insufficient
water vapour in the air. Moisture is therefore, an important factor in any of the applications
of air conditioning. Proper conditioning of air during hot weather increases not only the
business but also the efficiency of employees. In some manufacturing processes, removal
of dirt from air is of secondary importance, so that air conditioning is then mainly for the
control of the humidity of the air used for ventilation or manufacturing purposes. Air
conditioning has been found to improve the quality and to reduce the cost of manufacture
of automobiles, drugs and pharm aceuticals, electric goods, textiles etc.
3.2.4 A ir Motion Control : Control of air motion which includes the distribution of
air in an air conditioned space. This is necessary to keep even temperature through out
the conditioned space. This factor also concerns the question of draught (air current)
which affect^ the circulation of the air.
3 .2.5 Rem oval of Odour : In air conditioning definite amount of Ozone should be
introduced in the air conditioned room for de-odourizing and refreshing the air. For this
reason, air is passed through Ozonator after it has been conditioned. Ozonator is a sm all
electric apparatus which is scientifically designed to produce and control the exact amount
of Ozone required in the system . Ozone is used in air conditioning as an odour eliminator.
3.3
Term s used in A ir Conditioning
In the study of air conditioning, the following terms should be well understood :
Dry air is a mixture of the constituent gases which comprise atmospheric air excluding
water vapour. On an average, composition of dry air by volume is 20-95% oxygen, 78-09%
nitrogen, 0-93% argon, 0-03% carbon dioxide and traces of other gases. Normally, argon
is grouped with nitrogen and carbon dioxide is ignored and is generally stated that
volumetric composition of air is 21% oxygen and 79% nitrogen. Molecular weight of dry
air is taken as 28-966 or 29.
Moist air is a mixture of dry air and water vapour. The amount of water vapour in
moist air depends on degree of saturation. The weight of water vapour in moist air is
expressed in grammes or kg per kg of dry air.
The water vapour in air is usually called “moisture” and is an important factor in any
of the applications of air conditioning. Its amount in the outside air depends on the
weather conditions. When a mixture of air and water vapour contains the maximum amount
of water vapour that it can hold, it is said to be saturated. If the temperature of the
mixture of air and water vapour is higher than the saturation temperature, the vapour is
superheated. If the temperature of air drops below its saturation temperature, some of
the water vapour will be condensed.
According to Dalton’s law there is, in ’m echanical’ mixture of gases, a pressure for
each gas, which is called its partial pressure. The sum of all these partial pressures is
the total pressure of the mixture. The partial pressure of the individual constituent (water
AIR CONDITIONING
119
vapour and air) is the pressure that it displays, and it has the sam e temperature as that
of the mixture.
The water vapour that is contained in or mixed with atmospheric air, is the humidity.
The weight of water vapour that the air in a given place can hold, depends on its
temperature and is independent of the pressure of the air. It varies with the partial
pressure of water vapour.
v
Specific humidity or humidity ratio is the weight of water vapour in moist air per kg
of dry air.
The actual weight of water vapour contained in a unit volume of air is called the
absolute humidity. Weight of water vapour is expressed in grains or grammes per cubic
metre. It may be noted that one kg is equal to 15,430 grains.
Relative humidity (R .H .) is the ratio of weight of the water vapour in the air in a
given space at a given tem perature, to the weight of water vapour that the air in the
same space and at the sam e temperature would contain when fully saturated. To illustrate
if 50% is the relative humidity (R .H .) of air at a given temperature, it will then contain
half as much moisture as it would contain when fully saturated. One kg of dry air at
25°C would contain 20-2 gm of water vapour (from tables of properties of air) when fully
saturated (100% R .H .). Hence, air at 25° C and R .H . 50% would contain 10.1 gm of
water vapour.
Relative humidity of air varies with the temperature. With the increase of temperature,
R.H . will decrease and vice versa. To illustrate, air at 25°C and R .H . 50% is heated to
30°C. Now one kg of dry air at 30°C would contain 27 gm of water vapour (from tables)
when fully saturated. Hence, R .H . of air at 30°C when heated from 25°C and 50% R .H .
would be 10-1/27 = 0-375, i.e., 37-5%. Again suppose that air at 25°C and 50% R .H . is
cooled to 15°C. Now, one kg of dry air at 15°C would contain 10-8 gm of water vapour
(from tables) when fully saturated. Hence, R.H . of air at 25°C and R .H . 50% When cooled
to 15°C would be 10-1/10-8 = 0-935, i.e., 93-5%.
Relative humidity of 50% and dry-bulb temperature of 25°C is recommended for
comfort.
Dry-bulb (D .B.) temperature is the temperature indicated by an ordinary mercury
themometer. It is the m easure of sensible enthalpy of air. It has no relation to the
condition of the air so far as the humidity or the water vapour content is concerned.
Wet-bulb (W .B.) temperature is the measure of enthalpy of air. It is the temperature
indicated by a thermometer whose bulb is covered by a piece of wet (water moistened)
wick.
The wet-bulb thermometer, has its bulb covered with a piece of clean soft cloth or
wick, which is dipped in a sm all basin (cup) of water; this keeps the bulb moistened
(wet). If there is much water vapour present in the air outside (i.e . atmospheric air has
a high R .H .), there will be little evaporation from the moisture (contained in the wick)
near the bulb of the thermometer, and therefore, only a sm all cooling effect will be
produced. If the air outside is dry (less water vapour in air), more rapid evaporation will
take place and lower will be the wet-bulb temperature (i.e. there will be more difference
between the dry-bulb temperature and wet-bulb temperature). Thus, the difference in
reading between the dry-bulb temperature and wet-bulb temperature is the m easure of
the amount of water vapour in air. Both the thermometers are read when steady conditions
have been attained and the relative humidity is found by using Table 3 -1 or Psychrom etric
chart (fig. 3 -1 ).
.T h e difference between wet-bulb temperature and dry-bulb temperature is known as
wet-bulb depression.
For example, the relative humidity of air having 30°C dry-bulb temperature and 22°C
wet-bulb temperature, is 50% from table 3-1. The wet-bulb depression is 30-22 = 8°C.
ELEM EN TS OF H EAT EN G IN ES Vol.
120
Tab le 3-1
P sychro m etric table of percentage relative hum idity of air
Reading of
dry-bulb
thermometer
°c
-10
- 8
- 6
- 4
- 2
0
2
4
6
B
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
42
44
46
48
50
52
54
56
58
60
62
64
66
68
70
\
Difference between reading of dry-bulb and wet-bulb thermometers, °C
(wet-bulb depression)
1.0
67
70
73
77
80
82
84
85
86
87
88
89
90
91
91
91
91
92
92
93
93
93
93
93
94
94
94
95
95
95
95
95
95
95
95
95
95
95
95
95
95
2.0
3.0
4.0
38
45
51
57
62
65
68
71
73
75
76
78
80
81
81
82
83
84
84
85
86
86
87
87
88
88
89
90
90
90
90
90
90
90
90
90
90
91
91
91
91
10
22
32
38
44
48
53
58
60
63
65
68
70
71
72
73
74
75
76
77
78
79
80
81
82
82
83
84
84
84
84
84
84
85
85
85
85
86
86
86
86
15
22
29
33
39
43
47
51
54
57
60
62
64
66
68
69
70
71
72
74
75
75
76
77
78
79
79
79
79
80
80
81
81
81
82
82
82
82
82
5.0
9
17
20
25
31
36
40
44
48
51
54
56
58
60
62
64
65
66
68
69
69
70
71
72
73
74
74
74
75
76
77
77
77
78
78
78
78
78
6.0
•
12
18
25
30
34
38
42
46
48
51
54
56
58
59
61
62
63
64
65
66
68
68
70
70
70
71
72
73
73
73
74
74
75
75
75
7.0
7
14
20
24
29
34
38
41
44
46
49
51
53
55
57
58
59
60
61
62
63
64
65
66
67
68
68
69
69
70
70
71
71
71
8.0
9.0
10
11
12
4
10
15
20
25
30
34
36
40
43
45
47
50
52
53
54
55
56
58
59
60
60
62
63
64
64
65
65
66
67
67
68
68
11
18
23
27
30
34
37
40
42
44
46
48
49
50
51
53
54
55
56
58
59
60
60
61
61
62
63
63
64
64
10
16
20
24
28
31
34
37
40
42
44
45
46
47
49
50
51
52
54
55
56
57
58
58
59
60
61
61
61
8
13
17
22
26
29
32
35
37
39
40
42
43
45
46
47
48
49
51
52
53
54
55
56
57
57
58
58
9
11
16
20
24
27
30
32
34
36
38
40
42
43
45
46
47
48
49
50
51
52
53
54
54
55
55
AIR CONDITIONING
121
As the a ir is cooled the relative humidity increases. If the temperature is lowered
sufficiently there after a point will be reached at which the relative humidity would be
100% or the air would be fully saturated. This is termed the dew point (D .P.) or saturation
temperature. A further lowering of the temperature of the air below this point, causes
water vapour to condense in form of particles of water.
The difference between the dry bulb temperature and dew-point tem perature is known
as dew point depression.
Effective tem perature is an index of sensation of warmth produced by the combined
effect of air tem perature, humidity, and air motion. Sensation of warmth depends not only
on the temperature of the surrounding air, as shown by ordinary thermometer, but also
upon the wet-bulb temperature (humidity), air movementand radiation effects. Effective
temperature cannot be measured directly, but can be determined from dry-bulb and
wet-bulb tem peratures and air motion observations, by referring to effective temperature
charts or tables. These charts or tables are prepared as a result of a series of studies
at research laboratories. Effective temperature can also be determined em pirically.
Effective temperature is an index of degree o f warmth experienced by a person on
exposure to different combinations of air temperature, humidity (moisture content o f air),
and air movement. To illustrate, the effective temperature of an air conditioned plant is
15°C, when the conditioned air produces a sensation of warmth like that experienced in
slow moving air (at 6 m per min.) at 15°C.
Psychrom eter is an instrument to m easure the psychrometric properties of air. The
most commonly used type employs two thermometers, one dry-bulb and other wet-bulb.
The wet-bulb temperature reading in still air is erroneous. This reading is 10 to 20 of
the bulb depression round the temperature of adiabatic saturation. The sling psychrom eter
is a common type most generally used for checking conditions on job. The instrument is
rotated by hand to get air movement across the bulb of the thermometer. In the aspiration
psychrometer, air across the stationary bulb is drawn with the help of a operated aspirator
or a sm all blower like those used in portable hair driers.
Psychrometry deals with relation of air and water vapour or humidity. This relationship
is shown by a series of curves that make up the psychrometric chart.
Psychrometric chart attached (fig. 3-1) shows the relationship between the following
fundamental properties of air, nam ely, (a) dry-bulb temperature, (d) dew point temperature,
(c) wet-bulb tem perature, and (d) relative humidity. When any two of these four properties
are known, the other two can be determined directly from the psychrometric chart without
any mathematical calculation.
Lines of dry-bulb temperature are vertical. The values of dry-bulb temperature are
given on tne base of the chart.
Lines of dew point temperature are horizontal. The saturation line (curve) is drawn
by plotting the dry-bulb temperature (abscissa) versus moisture content in kg per kg of
dry air necessary to saturate (ordinate). Under saturation conditions, the dry-bulb temperature
becomes dew point temperature. The weight of moisture in kg to saturate 1 kg of dry
air at a given temperature is determined by following the horizontal line from the given
(desired) dew point to the right. The values of dew point temperature (saturation temperature)
are given along the saturation curve of the chart.
The diagonal lines running downward frpm the saturation curve across the chart are
lines of wet-bulb temperature. The values of wet-bulb temperature are given along the
saturation curve of the chart.
The curved lines that follow the sam e direction as the saturation curve are the lines
122
ELEM EN TS OF H EAT EN G IN ES Vol. Ill
of relative humidity. The values of relative humidity are given along the lines of relative
•humidity.
Lines of constant volume are obliquely running downward to the right of the saturation
curve. The values of specific volume (m3/kg) are given on the base of the chart.
To determine the relative humidity of air for a given dry-bulb temperature and wet-bulb
tem perature, draw a vertical line from the given dry-bulb temperature to intersect the
given wet-bulb temperature diagonal line. The point of intersection gives the R .H .
To determine the relative humidity of air for a given dry-bulb temperature and dew
point tem perature, follow the horizontal line from the given dew point temperature to
intersect the given vertical dry-bulb temperature line. The point of intersection gives the
R .H .
To determine the wet bulb-temperature for a given dry-bulb temperature and dew
point tem perature, from the point of intersection of vertical dry-bulb temperature line and
horizontal dew point temperature line, follow the diagonal wet-bulb temperature line to
saturation curve and read the wet-bulb temperature.
To determine the dew point temperature of air for given dry-bulb temperature and
relative humidity, follow the vertical line from the given dry-bulb temperature to intersect
the given relative liumidity curve. From the point of intersection, follow the horizontal dew
point temperature lina to the saturation curve and read the temperature.
To determine the enthalpy content or enthalpy of air for given dry-bulb temperature
and relative humidity, from the point of intersection of vertical dry-bulb temperature line
and relative humidity curve, draw a line parallel to diagonal wet-bulb temperature line
upto the enthalpy scale and determine the enthalpy of air.
To determine the volume of air for a given dry-bulb temperature and relative humidity,
from the point of intersection of dry-bulb temperature and relative humidity curve, draw
a line parallel to constant volume line and determine the specific volume on volume scale
given on the base of the chart.
FigCire 3-2 shows the psychrometric chart (on reduced scale) on which the different
psychrometric processes are shown. Line ab represents sensible cooling at constant
moisture content or at constant humidity ratio
of air. Thus, in this process, dew point
temperature rem ains contant, while relative
humidity increases. The final temperature
cannot t>e below the initial dew point tempera­
ture but the relative humidity will be raised.
If we consider line b-a or e - g which
represents heating without changing the
moisture content (i.e . humidity ratio) of air
wherein the relative humidity is lowered.
Line a -c or d -e which is parallel to the
wet bulb lines from the entering-air condition
to the saturation curve is the adiabatic satura­
tion process line. "Adiabatic" means no heat
added or removed externally and ‘saturation’
implies adding of moisture. This process is carried out in practice by sparying water in
air stream . Evaporative cooling closely approaches adiabatic saturation process. Evaporative
cooling is more effective in hot dry climate when the cooler humid condition is more
desirable than the hot dry condition. Generally, the leaving wet-bulb temperature cannot
AIR CONDITIONING
123
be lower than the spray water tem perature. Thus, relative humidity is raised during this
process.
Line a - f and a - d respectively represents humidification process. During these
processes, dry-bulb temperature rem ains constant and humidity increases in humidification
and decreases in dehumidication process.
Line h - j represents the dehumidification process accom plished by cooling and
dehumidification, or summer air conditioning process wherein temperature of air leaving
is reduced and consequently relative humidity is increased.
Problem-1 : Using the psychrom etric chart, determine the relative humidity of air having
2 $ C dry-bulb temperature and 18°C wet-bulb temperature.
On the psychrometric chart mark the dry-bulb temperature of 25°C on the horizontal
scale and draw a vertical line through it. Then find the wet-bulb temperature 18°C on
the saturation curve and a diagonal line through it to intersect the vertical dry-bulb
temperature line. The point of intersection is between 50% and 55% relative humidity
curves. Hence, the relative humidity of the given air is 52%.
Problem -2 : If the air has a dry-bulb temperature of 30°C and the relative humidity
50%, find the dew point temperature of the air.
On the psychrom etric chart mark the dry-bulb temperature of 30°C on the horizontal
scale and draw a vertical line through it to intersect 50% relative humidity curve. The
point of intersection represents the quantity of moisture contained in each kg of air which
can be read on the right-hand side vertical scale by drawing the horizontal line through
the above intersection point. Now, if the air is cooled without changing the moisture
content, it will be represented by the horizontal line passing through the above intersection
point. It will be found that the horizontal line will intersect the saturation curve at a point
where temperature is 18.3°C. This is the dew point temperature of the given air.
3.4 Processes in A ir C onditioning
In order to achieve true or complete air conditioning, the following processes (operations)
have to be carried out, and hence, in the equipment of the air conditioning plant, the
units performing these duties are used :
3.4.1.
A ir Purifications The first operation in air conditioning is cleaning of outside
air so as to make it free from dust, dirt, soot, germs, and other im purities. This is
achieved by passing outside air through air
filters, which are made out of expanded mesh
(wire net), glass wool, hemp, fibres etc. Before
th6 air filters are fitted they are coated with
viscous substances. When the outside air is
passed through the wet air filters or viscous air
filters (fig. 3 -3 ), impurities are caught and clean
air (free from dust etc.) is allowed to enter the
conditioning plant. Wet filters can be removed,
cleaned with cleaning compound, allowed to dry,
then dipped in oil and refitted. Sometimes, dry
air filters are used which can be removed,
cleaned and renewed whenever necessary, If
the proportion of dust in the outside air is very
high, water filters (air washers) are employed
for cleaning the air.
Water filters or Air washers : Water can
Fig. 3 - 3 Viscous (wet type) air filter
124
ELEM EN TS OF H EAT EN G IN ES Vol. Ill
easily carry away the impurities which can be wetted by it, from the air. But the impurities
like carbon particles which cannot be wetted by water, are not cleaned from the air by
it. Thus, water can work as a filtering agent.
The spray type of filtering method provides some humidity control. The water is
sprayed in the passage of air to be cleaned in the form of finely divided particles. These
water particles will come in contact with dirt and will remove most of the dirtfrom the
air. This method is mostly used in large units.
The second method is to use loosely woven cotton cloth. This cloth is wetted with
water and then placed in the passage of air
over the wire fram es so that the air may be
drawn through when it is placed in a plane
perpendicular to the plane of motion of the air
(fig. 3 -4 ), or it may pass over the cloth when
it is placed in a plane, parallel to the plane of
motion of air. The cloth is kept wet by means
of drip of water slowly falling from the perforated
pipe at the top or by wick action if one end of
the cloth is dipped into the vessel containing
water.
In the third type of water filter, condensing
coil is used. If this coil is operating below
dew-point temperature of the air to be condi­
tioned, most of vapour of the air will condense
u
on the outer surface of the coil in the form of
Fig. 3-4 Watet filter or Air washer.
water particles. The air passing over these water
particles will lose much of its dirt. When sufficient moisture is formed, it drips down with
dirt and is drained out through the drain pipe.
Activated Carbon filters : In this type of filter, activated carbon in the charcoal form
is used. A number of sheets of carbon are placed horizontally with a certain gap between
two adjacent sheets in the two vertical supports. This structure is again placed into a
framework which will fit exactly over the structure. This type of filter will remove solid
particles as well as bacteria. Thus, the use of carbon filter is extensively increasing in
the field of air conditioning and refrigeration with a great success. The charcoal will absorb
as much as 50 per cent of its weight of foreign gases. This type of filter when used for
a long time becomes inactive, and to reactivate it, it is baked at 500°C to drive out the
absorbed gases. They are also available in the form of canisters.
3.4.2 Tem perature Control : The second important process
in air conditioning is to
control or to regulate the dry-bulb temperature by various psychrometric processes. This
is attained by sim ple heating or cooling which may be associated with humidification or
dehumidification processes.
Heating of air means raising the dry-bulb temperature. It can be attained by passing
air over the heating surfaces. The air heating surfaces are located outside the room to
be conditioned. The air heated to the desired temperature flows to the room by the fan
action. In any heating system , air is the ultimate carrier of the heat, while steam or hot
water may act as intermediate carrier, between the furnace (source of heat) and the air.
Electric heaters are also installed in the path of air leading to the conditioned space
for heating the air. If the relative humidity of the outside air is of such a range that the
desired R .H . as well as the inside dry-bulb temperature can be obtained by merely
heating the air, then the process becomes simplified. Pre-calculated quantity of heat is
AIR CONDITIONING
125
added directly to air to raise its temperature to the desired value. However, in most cases
the desired condition of air is difficult to attain in this manner.
In the indirect method, the air is cooled down to the dew point temperature of the
air in its final conditioned form and then the mixture is heated by a heater to obtain the
necessary higher dry-bulb temperature and the proper relative humidity. Th is method
needs greater amount of heat for heating.
Cooling of air m eans lowering of the dry-bulb temperature. It can be attained by
passing the air over the evaporator coils of the refrigerating system in a sim ple case, as
is done in the room air conditioner. Here, the relative humidity aspect i& neglected or is
of such order that it gets adjusted by itself. However, in large installations, cooling is
obtained by injecting chilled water in the air passage to lower its temperature to dew
point temperature of the air in its final condition and later on it is heated by a heater
to the desired temperature. Th is shows that even for controlling the temperature (i.e .
heating and cooling), the processes of humidification and dehumidification autom atically
get involved.
We have seen that it is possible to bring about adiabatic saturation of air. Here, the
sensible heat of air is reduced (i.e . dry-bulb temperature is lowered) but this heat recovered
as latent enthalpy is utilized for evaporation of water into vapour (m oisture content increases
and therefore, the dew point is raised). It may be noted that the wet-bulb temperature
does not change since the enthalpy of air is unaltered.
In winter it may be necessary to heat the air as well as to humidify it .' For this
purpose, spray water temperature must be above the dry-bulb temperature of the entering
air. We must also supply heat continuously to spray water so that the temperature of
water may not fall to the wet-bulb temperature of air (spray water re-circulated without
heating will soon fall to wet-bulb tem perature). This implies that the circulating water must
be heated.
There are two types of cooling coils in general use for air conditioning purposes.
They are direct expansion type and secondary refrigeration type. In case of direct expansion
type, the refrigerant is expanded directly through the expansion valve, i.e . throttle valve,
and then it is passed into the coils for the cooling effect to be produced by it. In this
type, the transfer of heat from air to refrigerant in the air conditioning plant takes place,
and it may be parallel flow, counter flow or cross-flow . In most of the cases, they are
provided with fans so as to increase the effective heat transfer surface. For larger air
conditioning plants generally secondary refrigeration system is used. In this system , chilled
water is used for producing cooling effect. This water is cooled to the required temperature
in the evaporator of a refrigeration system employed in a centralized place. This chilled
water from the evapourator is distributed to the required place for producing the cooling
effect for air-conditioning purposes. Sometimes, brine is also used instead of chilled water,
because water freezes at 0°C while brine does not, and hence it can work satisfactory
below 0°C also.
3.4.3
Hum idity Control : The third important process in air conditioning is to control
humidity. This is achieved by the processes of himidification (increasing humidity) or
dehumidification (decreasing humidity).
Humidification is accomplished by addition of steam to air but common practice is to
supply warm water. Because humidification is desired in winter and at the sam e time
heat must be added to make up heat losses from the conditioned space, common practice
is to heat and humidity the air at the sam e time.
Humidification by steam may be divided into two catagories:
126
ELEM EN TS O F H EAT EN G IN ES Vol. Ill
. .
A direct type in which water is sprayed in highly atomized state into the room
itself which is to be air conditioned.
. .
An indirect type which employs special humidifier as described hereafter. These
devices consist of :
— a chamber of casing which is provided with a system of spray nozzles,
— a tank at the bottom for collecting the spray water, and
— an eleminator section at the delivery end of the chamber, for removal of
drops of entrapped moisture from the air leaving the chamber.
Humidification may be accom plished in three different w ays with an air washer :
— Using recirculated spray water without prior heating of air,
— Pre-heating the air and washing it with recirculted water, and
— Using heated spray water.
Dehumidification is the reduction of water vapour content of air. Dehumidification can
be accom plished by the use of an air washer or by use of absorbents. It is normally
done by cooling the outside air below the dew point temperature so that it loses moisture
by condensation. Moisture removal is obtained when the spray water temperature is lower
than the dew point of the entering air.
Dehumidification can also be achieved by using absorbants. Absorbants are the
m aterials having capacity for absorbing moisture. The m aterials used as absorbants are:
activated alum ina, silica gel and calcium chloride. These are useful only for sm all room
air conditioners (window units).
Dehumidification system can also consist of an air washer in which very much chilled
(cold) water, whose temperature is kept equal to the dew point temperature of the room
Fig. 3 - 5
Dehumidification system with air washer*
^
air, is sprayed on fresh air coming out from the filter (1 -2). By this, the outside air is
cooled below its dew point temperature (2-3) and hence its surplus moisture is separated
out and any extra moisture is removed in the eliminator (3 -4 ). However, the dry-bulb
temperature of air leaving the eliminator is much below the dry-bulb temperature desired
in the conditioned space. Hence, a heating element (4—5) is necessary in the dehumidification
system . Common practice is to use a heater, either steam or gas operated and is placed
after the air washer chamber (fig. 3-4). The air is then supplied to the conditioned space
127
AIR CONDITIONING
by fans or blowers.
Thus, it will be seen that spFay chamber (air washer) through which air passes on
its way to the conditioned space can act as air cooler, dehumidifier or air cleaner (w asher),
depending upon the purpose for which it is required.
Fig. 3 - 6
#J,„
Open tray humidifier.
Fig* 3 - 7
Spray nozzle type humidifier.
>
—*
Spray humidifying is one of the most common method of increasing the humidity of
air passing through water sprays in the air washer chamber. If sufficient time is available,
air will leave the air w asher fully saturated. This is only an ideal case and in practice
the saturation achieved is about 95% .'
The various typers of humidifiers used for humidification of air are :
In Open tray humidifier a large surface of the water is exposed to air to be humidified.
This is shown in fig. 3 -6 . The warm air flowing through the duct is passed over the
larger surface of water which is contained in the tray. The level of the water in the tray
is controlled by the float valve. The water pipe connects the tank to the water tray. A
metal float, floating on the top of the water surface and projecting in the passage of
warm air gets warmed and it transfers the heat to the water in the tray. The water
Revolving
drum
Woolen wicks
Float
Oullet pipe
Heater coil
Inlet pipe
Cold w ater
outlet
Fig. 3 - 9
Fig. 3 - 8
Shallow pan and hot water pr
steam radiator type humidifier.
Revolving wicks type
humidifier.
128
ELEM EN TS O F H EAT EN G IN ES Vol. Ill
temperature is raised and its evaporation is accelerated to humidify the air.
Shallow tray
gases
Fig. 3 -1 0
Float valve Humidified
air
In spray nozzle type humidifier,
humidification of the air can also be
done by using spray nozzle in the warm
air duct as shown in fig. 3 -7 . The water
is led to the spray nozzle under pressure
from the tank through the pipe. The
water is atomized and mixed with warm
air flowing in the duct. The water flow
is controlled by the valve.
In shallow pan and hot water or
steam radiator type humidifier, cold air
passing in the duct when passes over
the heater co ils, is warmed. The heater
coils contain hot water led through the
pipe and the cold water is taken out
through the pipe. There is a pan placed
on the top of the heater coils, so that
the water contained in the pan gets
heated and steam formed due to heating
gets mixed with the flow of the air as
Shallow pan and furnace type humidifier. shown in fig. 3-8.
Revolving wicks types humidifier, is shown in fig. 3-9. In the duct of warm air, a
revolving, drum is placed. The circum ference of this drum is covered with woollen wicks
placed along the axis of rotation. The tray contains water. The water is led through the
pipe and the level of the water is controlled by the float. The surplus water is drained
out through the pipe, when the drum rotates. The lower portion of the drum is dipped
into the water in the tray. The wetted w icks after coming out of the water come in contact
with warm air passing over them. The
w ater va p o rise s and the a ir gets
humidified. This process goes on con­
Inspection
tinuously so long as the drum rotates.
cover \
The rotation of the drum also helps to
a little extent in producing the draught
of the warm air. A certain amount of
water is splashed out by the wicks into
To space
the space of warm air and the air gets
to be
humidified.
humidified
In shallow pan and furnace type
humidifier method of humidification, a
shallow tray is provided on the top of
the coal furnace. This tray contains water
and the level of the water is controlled
by the float valve. The water is supplied
from the tank through the pipe. The cold
air which is warmed by the flue gases
passes over the tray. This warm air will
Electric heater
evaporate water to some extent. The
tray being on the top the furnance, most
of the heat of evaporation is obtained
AIR CONDITIONING
129
from the flue gases. Thus, the evaporated water vapour will humidity the air.
Electrically operated space humidifier is shown in fig. 3-11. It has a cabinet in which
a water tray and an electric resistance heater are installed. When the switch is operated,
the heater generates heat which is utilised to heat the water contained in the tray. The
steam generated is mixed with air and forced out of the humidifier by the draught produced
by the fan. Th is fan is controlled by the control switch. The water level in the tray is
controlled autom atically by the float valve. The water is fed into the tray from the tank
through the pipe. The air which flows from space 4o be humidified, is directed to enter
into the cabinet by the louvers. Louvers discharge the air in a particular direction into
space. Both louvers are gravity controlled. There is a cover on the top of the cabinet
for inspection.
3.4.4
C ontrol o f A ir M otion : A fan is used to circulate the air and maintain a
constant pressure at all outlets to the building or room. Ducts (pipes) are provided along
the length of room to carry the conditioned air evenly over the room and to return it to
the cooling element of air conditioned plant. The ducts are fitted with outlets called grilles
through which air com es out. The grilles are provided with shutters (which have variable
openings) for controlling the quantity of conditioned air.
In most air conditioning plants, whether for summer cooling or winter heating, a large
part of total conditioned air supplied to the conditioned space is recirculated. The air that
has been removed from the conditioned space is reconditioned in the air conditioning
plant and then returned to the conditioned space. Usual practice is to allow 5 to 10 per
cent of fresh (outside) air to be drawn in by the fan and mix it with the room air to be
recirculated.
3.5
Cooling Load Calculation
Before an air conditioning system can be designed for either summer or winter
operation, the cooling and heating demands on the system must be determined. In winter,
buildings constantly lose heat to the outside air and in summer heat from outdoors leaks
into the building. Moisture in air, being in vapour form, also moves easily in and out of
the occupied space through cracks in doors and windows. Heat in considerable quantity
may be produced in space which is to be air conditioned and thus, becomes a part of
the load. The air conditioning loads may conveniently be considered under different heads,
the summer cooling load and the winter heating load. The summer cooling load is the
amount of heat which must be removed per hour to produce and maintain inside design
conditions within the space. Conversely, the winter heating load is the amount of heat
which must be supplied per hour to maintain inside design conditions within the space.
Cooling load calculations deal with heat loads of two types :
. . Sensible heat load, which as it flows into or is produced in a space, will tend to
cause a temperature rise in a space, and
. . Latent heat load in the form of moisture, which although does not cause the
temperature rise, but does change the condition of air in the space, resulting in
a higher relative humidity.
Depending on sources of heat, load may be classified as external, if it comes from
outside the conditioned space which may be either latent or sensible, and internal, if it
is produced within the space to be conditioned. Th is, too, may be either in the form of
sensible or latent load.
Sensible heat load includes :
. . Heat transm ission through the building structure as a result of conduction, convection
and also solar radiation.
HE 3-9
130
ELEM EN TS OF H EAT EN G IN ES Vol. Ill
. . Heat brought inside with the outside air which is introduced for ventilation.
. . Heat produced by occupants (persons).
. . Heat produced in the space by lights, electrical appliances and motors.
. . Heat to be extracted from the material or products brought into the space in case
of industrial air conditioning jobs.
. . M iscellaneous heat gains (load) due to supply and return duct (pipe) heat gains,
supply duct leakages, and fan motor heat gains.
,
Latent heat (m oisture) includes :
— Heat from outside air which includes air that is introduced for ventilation, and air
that infiltrates into the space.
— Heat from occupants (persons).
— Heat from cooking, hot baths, or other vapourisation processes in the space.
— Heat from products, or material brought into the space.
Sensible heat loads includes the amount of heat required to offset heat losses by
building structure, and heat to warm the outside air, i.e . ventilation and infiltration (leakage)
air which comes into the buildings. Ordinarily, no allowance is made for sensible heat
produced in the space by occupants, lights or appliances. The latent heat load is that
required to evaporate enough moisture to make up inside design conditions. No allowance
is ordinarily made for moisture produced within the air conditioned space. If a humidifier
is not provided as a part of the air conditioning equipment, there will be no latent heat
load for winter air conditioning.
Now, let us consider the contribution of each factor described above. In order to
evaluate the values, reference may be made to the standard cooling load calculation
tables giving values under different conditions.
3.5.1 Heat Tran sm issio n through Building Structure : We know by law of ther­
modynamics that if the temperature on one side of wall differs from that on the other
side, heat flows from higher temperature side to the lower temperature side. The heat
transferred under steady heat flow through a given part of structure is equal to the overall
heat transfer coefficient multiplied by the total surface area and the difference between
the inside and outside air tem peratures. The area of windows and doors is also to be
calculated separately. Thus, it depends on material of wall construction and consequently
on thermal conductivity, specific heat, density, character of m aterial, room air temperature,
convection heat transfer between inside surface and room air, and radiant heat transfer
between inside surface and the other sources of heat in the room.
3 .5.2 S o lar R adiation : Radiation is almost always present on roofs and ceiling. It
may account for as much as 50% of the total summer load. Thus, it is important to
estim ate this load very accurately.
3.5.3 Heat G ain from O utside air : Outside air in the conditioned space is due to
ventilation air, infiltration (air leakage) through w alls and cracks around doors and windows,
and also leakage through door openings. During summer air is hot and dry with respect
to the inside design conditions, and it adds to both the sensible heat and latent heat
loads. In well constructed houses and buildings, window and door crack infiltration in
summer is sm all unless exhaust fans are operating within the space. This low infiltration
is the result of negligible wind velocity in summer. Ventilation code allows certain minimum
number of air changes to meet with this demand. Latent heat gains from outside air in
kJ/hr can be calculated from specific humidity of outside air and room air, and air quantity.
Since wind velocity pressures are ordinarily much greater in winter than in summer, cold
AIR CONDITIONING
131
air infiltrates building in ' large amounts in winter, and the infiltration heat load is of
considerable magnitude.
3.5.4 Load due to O ccupants : In many air conditioning installations, occupant load
is of little or no consequence. It is standard practice to omit this factor entirely for
residential installations. However, in commercial buildings such as offices and retail stores,
occpant load is appreciable. W hile in case of theatres, auditoriums and dance halls, the
occupant load may be a well dominating factor. The amount of heat which people give
off to the surrounding air depends on their number and the degree of m uscular activity
at the time/ The heat dissipated (given off) w ill be partly sensible and partly latent.
3.5.5 Electric Appliances : To the space load must be added the heat produced
and dissipated by lights, motors and other appliances. Heat from lights and motors is
sensible, but certain appliances such as cloth driers, restaurant steam tables, etc., produce
both sensible heat and latent heat loads. Th is load is calculated by the conversion of
electric units into heat units.
3.5.6 Product Load : In almost all industrial air conditioning applications, products or
material brought into the space for processing, fabrication or storage make up a considerable
portion of sensible load as well as latent load. G enerally, this consists of several
sub-divisions, depending upon the product and temperature involved. Each of these
sub-divisions, which must be calculated separately are : chilling above freezing, cooling
below freezing, and product reaction or respiration heat. In order to evaluate above, one
must know the entering-product temperature, the final product tem perature, the weight of
product to be chilled, the time required for chilling, the individual size of product handled,
and the specific heat of the product.
3.5.7 M iscellaneous Heat Gains : Certain heat gains are inherent within the air
conditioning system itself and though more often they can be neglected, a careful analysis
of these should be made before deciding to do so. These include : (1 ) supply duct and
return duct heat gains when both are located outside the conditioned space which can
be prevented by insulating the duct, (ii) supply duct air leakage which is function of duct
design workmanship exercised in fabricating and installing the duct system , and (iii) fan
and pump heat gains wherein the electrical energy is converted directly into heat energy,
adding heat to the air stream .
3.6 A ir-C onditioning System
Air conditioning system s may be classified as central system and unitary system . A
central system is one in which the components of the system are all grouped together
in one central room and the conditioned air is distributed from there to the spaces to be
conditioned through extensive duct (pipe) work. Unitary system makes use of factory
assembled units which are usually installed in or nearby to the space to be conditioned.
These systems may be further classified as : (i) Cooling (summer air conditioning) system s,
(ii) Heating (winter air conditioning) system s, and (iii) Combination of above two system scalled year round air conditioning system , as is done in central plant where for cooling
either vapour compression or vapour absorption refrigeration plant is used, and for heating
either steam or electric heater is used. Sometimes, heat pumps are used wherein both
the functions, viz. cooling and heating are performed one at a time and are economical
only when they are used as coolers in summer and heaters in winter. Heat pumps are
best suited for locations where climate in winter and summer is ^ trem e, i.e . winters are
very cold and summers are extremely hot.
The advantages of the unitary system over the central system are :
— Saving in installation and field assem bly labour.
132
ELEM EN TS O F H EAT EN G IN ES Vol. Ill
— Exact requirements of each separate room is met, whereas a central system can
not meet the individual needs of separate room.
— Zoning or distribution system of duct work is eliminated.
—
Only those rooms
which need cooling will have their
unit running,
wh
central system , plant will have to run all the time for the sake of only few rooms
to be conditioned.
— Failure of one unit puts off air conditioning only in one room. This disadvantage
of central system is many a time prevented by installing two plants instead of
one of full capacity, so that when one plant is out of order, atleast a few room s,
can be served by the other plant.
— Starting from only a few rooms, the air conditioning may be progressively extended
to other rooms without spending large amount of money to start with.
The advantages of the central system over the unitary system are :
. . Lower initial cost.
.
. It can be housed
(installed) quite awayfrom rooms in a place which is not so
useful otherwise.
. . Noise and vibration problems are not so acute.
. . Better accessibility for maintenance.
3.7
A ir C onditioning Units and Plants
Outdoor air should enter the system through some type of rain proof opening equipped
with an arrangement to keep out rain and snow as the case may be. Air which is
surprisingly ladden with dust must be passed through some type of air cleaner or filter
as described earlier. Care must be exercised in its selection.
Before cleaning, it may be necessary to heat the air for better humidification. A heater
of proper capacity and size must be used. It may be an electric, steam or gas heater.
The heater must have proper automatic control. The outdoor air intake duct (pipe) may
be joined to the duct bringing air for recirculation. The two (fresh air and recirculating
air) may join before the heating stage or afterwards. Designer must select this point
carefully.
Conditioner is a very important unit and is available in various types and designs.
The interior of the conditioner may resemble a huge automobile radiator with coils inside
having fin s. Chilled water is circulated through these coils. Another type of conditioner is
known as air washer and consists of a number of spray nozzles or otherwater atomising
devices, such that the chamber is filled with a fog of tinywater particles.Conditioner
must be selected carefully so that it may add or remove proper amount of moisture
and heat. It must also be of proper size to handle the quantity of air involved. Refrigerating
unit is required to produce chilled water which is to be used in the process.
The air is moved through the ducts by centrifugal fan or blower. All fans produce
certain amount of noise and great care should be taken in selecting the fan so as to
keep the noise within perm issible lim its. Since the fan does the work of moving the air,
it must be of sufficient capacity to overcome the resistance to circulation of air. Ducts
are normally prepared from sheet metal. The ducts supplying conditioned air are called
supply ducts. Conditioned air from ducts is supplied through outlets known as grilles.
G rilles are available in large varieties of shapes and designs and when carefully selected,
helps considerably in the proper distribution of the air.
Here, the operating features of a few, most used, air conditioning units are described.
I
t
AIR CONDITIONING
133
3.7.1
W indow type Room A ir C onditioner : It is used to condition the air of a
' particular space, such as office room, bed room of a house, drawing office etc. It cools
| the air and som etim es dehumidifies it. It operates autom atically once it is put into operation.
Window room air conditioner consists of a casing which is divided into two parts by
a partition : out door part and indoor part (fig. 3-12). The partition has a sm all opening
at the top. The outdoor part (part of unit extending outside of the room) consists of
hermetically sealed motor driven com pressor, condenser, motor driven fan, and a tray.
Outdoor portion is also divided into two parts by a partition which has an opening on
the left hand side. Indoor part (part of unit extending into the room) consists of evaporator,
motor driven fan, remote bulb refrigerant control and control panel, an 3ir—filter, power
connection, and a tray. Th is indoor portion is also divided into two parts by a partition
which has an opening on the right hand side. Two trays are connected by a pipe line.
Condenser is connected to the evaporator by a capillary tube control through the refrigerant
filter. Evaporator is connected to the compressor by a suction pipe line. The air conditioner
is fitted in the window such that the outdoor portion rem ains outside the window sill. The
indoor portion is covered in front with partitions fitted with shutters, which can be set at
different inclinations. Sim ilarly, back of the outdoor part is fitted with fixed partitions.
When the unit is in operation, low pressure vapour is drawn from the envaporator
through the suction pipe line to the hermetic compressor where it is com pressed, from
low pressure to high pressure, and delivered to the condenser. When heat is removed
from the refrigerant vapour by condensation, liquid refrigerant is collected in the lower
coils. This liquid refrigerant passes through the filter into the capillary tube control and
then flows to the evaporator coils at low pressure.
Fig. 3 -1 2
Window room air conditioner.
The liquid refrigerant at low pressure inside the evaporator coils rapidly boils and
picks up evaporation enthalpy from the evaporator surface. A motor driven fan draws air
from the room, through air filter from the lower portion of the unit and forces it over the
evaporator coils where it (air) is cooled and circulated back into the conditioned room.
Moisture from the circulating air is removed to some extent. The moisture flows from the
134
ELEM EN TS O F H EA T EN G IN ES Vol. Ill
surface of the evaporator coils in the downward direction and drips into the pan at the
bottom of the evaporator. The moisture of this pan is directed to flow through the pipe
shown in the fig. 3-12 into the pan which is placed in the outdoor portion. This water
of the moisture in the pan evaporates to some extent and helps in cooling the compressor
and condenser.
The com pressor and condenser are mounted in such a way that the fan in the
compressor and condenser compartment draws outside air from the lower portion of the
unit, circulates it over the condenser coil and discharges it outside from the upper portion
of the unit. The air during its passage over the condenser takes away the enthalpy of
the vapour refrigerant and condenses it into liquid form. When the required temperature
is reached inside the room, the unit automatically stops. This is accomplished by the
control panel.
3.7.2
Room D ehum idifier : It is the device which reduces the moisture content o
air inside the room. A compressor type room dehumidifier is shown in fig. 3-13. This
dehumidifier works on the principle of refrigeration cycle. A s the moisture content is to
be removed from air, it (air) is cooled so that moisture may condense and drip out
through drain pipe. Moreover, the air which has been cooled below the required room
temperature must be heated to the desired room temperature. So if this cooled air is
passed over the condenser coils, it will be heated and the vapour inside the coils will
be condensed to liquid. Hence, it is clear that the air drawn from the conditioned room
must first be passed over the evaporator coils and then over the condenser coils.
Fig. 3 -1 3
Room dehumidifier.
Com pressor type room dehumidifier consists of a wooden cabinet with perforated
covers in front and back as shown in the fig. 3-13. Outside the cabinet there is hum idity
AIR CONDITIONING
135
control and electric sw itch. Inside the cabinet there is -a hermetic com pressor which is
connected to the evaporator and condenser. Condenser and evaporator are mutually
connected'by a capillary tube which acts as the refrigerant control. It also has a fan,
driven by the motor which is controlled by the motor control. The suction pipe line from
the evaporator to compressor passes through the frost control. There is trough having a
drain pipe.
On switching on the switch, motor-compressor runs and draws the low pressure
vapour through the suction line from the evaporator. This vapour is com pressed and
delivered at high pressure into the condenser where vapour is condensed and liquid
refrigerant collects in the lower coils of the condenser and flows through the filter into
the capillary tube. The refrigerant after passing through the capillary tube enters the
evaporator at low pressure. In the evaporator, the liquid refrigerant boils rapidly and
picks up latent enthalpy from the evaporator surface of the coils. Hence, the enthalpy
from the air passes to the refrigerant and moisture is condensed on the outside surface
of the evaporator coils. This condensed moisture slowly flqyvs in the downward direction
to the bottom of the evaporator coils and drips into the condensate trough (tray) from
where it drains out through the pipe connected at its bottom. Thus, air flowing through
the evaporator coils is cooled as well as dehumidified. A motor driven fan shown in the
figure, produces a draught of air and a considerable amount of air passes through and
again picks up heat, so that the air leaving the dehumidifier is at about the sam e
temperature as when it enters the humidifier.
The hum idity control permits the dehumidifier to operate until the desired relative
humidity is reached, then it autom atically shuts off the m achine.
The frost control placed in the suction line between the evaporator and the compresor,
shuts-off the com pressor motor at high enough temperature so that the evaporator coils
will not freeze over and stop the flow of air through it.
The arrows marked from left to right show the direction of flow of air through the
dehumidifier.
3.7.3
Packaged A ir C onditioner : This type of air conditioning unit may be console
type or of remote unit type. The console type is placed in the room below the window
sill, and having suitable fresh air inlet openings provided in the w alls. The refrigeration
unit possesses sufficient capacity to cool the air to the dew point temperature of the
conditioned air. Th is air is passed through the washer to saturate it and then heated
over the tempering coils which are electrically heated. The air discharging over the
tempering coils m ixes with the recirculated air to deliver the required conditioned air. The
unit may be adjusted for 100 % recirculation of air or for any fractional refrigeration as
desired. The mixing of fresh conditioned air and recirculating air takes place in a plenum
chamber, the dampers of which are servo-controlled from the operating panel.
Packaged type units are generally vertically placed and may be equipped with discharge
registers or discharge ducts.
Most of the units have air cooled condensers’ and therm etically sealed units with an
independent stream of cooling air obtained from a separate fan. This air also cools the
motor and compressor. During cooling of the air over the evaporator, some of its moisture
condenses in a drip tray from which it is led to the condenser fan (scopes) which throws
it over the condenser coils to evaporate it and throw it outdoors. Some of the large units
have water cooled condenser which makes it less portable due to water connections. The
evaporator may be fin and tube type. This type of unit is available in 5, 7.5, 10 and 15
tonnes capacity.
3.7.4. Central A ir C onditioning Plant : All the air conditioning plants described so •
136
ELEM EN TS O F H EAT EN G IN ES Vol. Ill
far are factory assem bled, but the central plant has to be assem bled at the site. A central
and year round air conditioning system is one which usually serves a single large building
such as an auditorium, theatre, or a m ulti-storied office or residential building. A central
system , depending on its type, may contain several and usually most of the following
equipment, all assem bled on the site:
-C o ils for cooling or dehumidifying, either for chilled water or for direct expansion of
the refrigerant.
-Sp rays for cooling and dehumidifying air or for washing the air.
-C o ils for heating, supplied with steam or hot water or electric heaters.
-A ir cleaning equipment, i.e . filters, such as dry, viscous, odour removing and masking
equipment and germ icidal lam ps.
-Blow er/s (fan/s) and its driving motor/s.
-A wide variety of control devices for controlling various air conditioning parameters
on the control panel.
«
A central system serves different rooms through extensive duct work with individual
control in each room.
Layout : Im purities, in the form of either solids or gases, must be removed from the
outside air during the air conditioning process. Solids may be separated from the air by
using m echanical filters or water filters (water w ashers). The majority of impure gases,
including carbon dioxide, are soluble in water and may be removed by scrubbing (water
washing).
Before the air is purified in filters and scrubbers, it may be necessary to heat the
air for better humidification. For this purpose, a heating element (preheater) of proper
size and capacity, is arranged before the filter. The heating element is equipped with
proper automatic controls to have the automatic regulation of the system . This preheater
may be necessary only for winter humidification. This preheater may be removed for
summer cooling (dehum idification.)
From this point onward, air passes through the cooling element (dehum idifier). The
cooling element may be air washer in which several rows of nozzles are installed to
develop a water spray through the air washer chamber. The cold water (whose temperature
is below dew point) entering the dehumidifier passes through an arrangement of pipes
that acts as header for the nozzles. When the air is passed through the dehumidifier, it
(air) picks up the moisture (if required) and at the sam e time gets cooled down to dew
point temperature. In cooling down to dew point temperature, surplus moisture in air is
separated out. However, the'dry-bulb temperature of the air leaving the air washer is
much below the dry-bulb temperature desired in the conditioned space. Hence, heating
of air to the desired dry-bulb temperature is necessary. This is done by passing the cold
and fully saturated (100% R .H .) air through a heating element (reheater) which is fitted
after the cooling element. The heating element may consist of steam radiator, hot water
radiator, or electric heater. Before the air is passed through the heater, it is passed
through an eliminator (fig. 3-14), which removes entrapped water from the air. This heating
of fully saturated air to the desired dry-bulb temperature brings the relative humidity to
the required value.
Som etim es, the aditorium air is not completely used up in the sense that it has still
got higher level of oxygen content which can suitably be used by the occupants (persons
occupying the room) but its temperature has increased to the extent which is not tolerable.
In such a situation, a large part of the room air (1’) is mixed with the outside fresh air
(1 ) and mixture is treated in the air conditioning plant and finally supplied to the room
137
AIR CONDITIONING
after heating to required temperature. Such a process is termed as by-pass method.
This method reduces the load on the filtering element and sometimes on the dehumidifier
Motor.
Refrigerating
plant
«
*
Dampers
Fig. 3-14 Layout and equipment of the year round central air conditioning plant.
and heating co ils, if the two varieties are mixed after the heating unit, depending on the
condition of the air mixed. The two varieties are mixed after the heating coils by providing
a branch pipe connection between points 1’ and 3 of fig. 3-15. Automatic dampers are
installed at points 1' and 3 in order to control the amount of auditorium air to be
by-passed.
Air from
auditorium
Air
leakage
Auditorium
-4 = r
Conditioned
air
Freshair
A.C.plant
1
f
Heater
3
Fan
f
Condensate
Fig. 3-15 By-pass system.
The conditioned air leaving the heater is moved through ducts (pipes) by a fan to
the conditioned space (auditorium). All fans produce certain amount of noise and hence
great care has to be exercised in making the selection of the fan so as to keep the
noise within perm issible lim its. Since, the fan does the job of circulation of air, it must
be of sufficient size to overcome the resistance to circulation of air in the ducts. The air
then flows along the supply ducts (pipes) provided along the length of the room
138
ELEM EN TS O F H EAT EN G IN ES Vol. Ill
(fig. 3-1 4 ). They are usually made of sheet metal and have round, square, or rectangular
cross-section. If air is sim ply let free in the room, its distribution will not be uniform over
the whole room. Hence, the ducts help to carry the air evenly to all places in the room.
The ducts are fitted with outlets, called grilles, through which air comes out. They are
provided with shutters which have variable openings, and hence the quantity of conditioned
air can be controlled. They also help to direct the air in the required direction.
Thus, the equipment of air conditioning system consists primarily of an air purification
element (filter) for cleaning air, a cooling element (air washer) for cooling and adjusting
the humidity, a heater to heat thte air to the required dry-bulb temperature, a fan driven
by an electric motor for circulating air, ducts for distributing air to the various parts of
the space to be conditioned. For humidification in winter additional removable heater is
placed before the filter and air washer (fig. 3-15). Thus, it will be seen that the same
air conditioning plant can be used for summer cooling (dehumidification) as well as for
winter heating (humidification).
Additional equipment of the plant consist of a refrigeration plant which produces chilled
water for the w asher, a source of heat for the heating element or heater, and a pump,
(fig. 3-15) for maintaining the circulation of chilled water through air washer.
Refrigeration plant used for air conditioning may be either working on vapour absorption
system or vapour compression system . The refrigerant used should be non-toxic, (nonpoisonous) and odourless. The freon group of refrigerants are most suited for vapour
compression system . Com pressors may be reciprocating, rotary or centrifugal type, depending
upon the capacity and the refrigerant to be handled and it may be air cooled or water
cooled. G enerally, the centrifugal compressors are used wih low pressure refrigerants,
such as F-11 and F-113, and for capacity above 100 tonnes. The condenser may be
air cooled upto 20 tonnes capacity and water cooled for capacity upto 300 tonnes. '
In practice, in some rooms such as conference cham bers, the entire volume of air
of conditioned space is generally replaced by fresh air from outside once every hour. On
the other hand, if there are few occupants and no smoking in the rooms, a part of the
air is taken from outside and a part of the air is recirculated.
3 .7.5
A ir Conditioning Plant fo r Cinem a H all : The air conditioning of a cinema
hall, public hall is not exactly an industrial problem, because it Is designed for greater
comfort of the audience. The air conditioning plant for a cinem a hall has to (i) remove
heat infiltrating (entering) through roof and w alls, heat produced by electric lights and
motors, and heat given up by the audience (human bodies give out considerable hteat),
(ii) remove dirt, sm oke, unpleasant odour, and (iii) maintain a temperature of 25° C and
R .H . of 50%.
In practice, the entire volume of air of cinema conditioned space is generally replaced
three tim es per hour by outside fresh air. For the working of the plant and its layout,
see the preceeding article and figs. 3-15 and 3-16.
Problem -3 : Atmospheric air at 4& C and 75% R.H . is to be conditioned in the air
conditioning plant to a temperature of 2!>C andm50% R.H . The method employed is to
lower the temperature of atmospheric air to dew point and then to raise it to the required
temperature by passing it through a heater and then delivering it to the air conditioned
room. Using the tables of poperties of air, find: (i) the m ass of water drained out per
kg o f dry air, and (ii) the dew point temperature.
Data from tables o f psychrometric properties of air is as under :
139
A IR CONDITIONING
Temperature of air,’C
45
43
40
25
15
14
Mass of water vapour when fully
saturated (100% R.H.) in gm/kg
of dry air
65-19
58-17
4892
20-09
1065
9-971
(i) One kg of atmospheric air at 43°C would contain 58.17 gm of water vapour (from
tables) when fully saturated (100% R .H .). Hence, air at 43°C and 75% R .H . contains 0.75
x 58.17 = 43.6 gm of water vapour per kg of dry air, i.e ., initial moisture content is 43.6
gm per kg of dry air.
One kg of air at 25°C would contain 20.09 gm of water vapour (from tables) when
fully saturated (100% R .H .). Hence, air at 25°C and 50% R .H . (i.e . conditioned air) contains
0.5 x 20.09 = 10.045 gm of water vapour per kg of dry air, i.e . final moisture content
is 10.045 gm per kg of dry air.
.*. M ass of water to be removed from atmospheric air
= 43.6 - 10.045 = 33.555 gm/kg of dry a ir. .
(ii) For dew point tem perature, refer the tables and find out temperature at which air
when fully saturated (100% R .H .) would contain 10.045 gm of water vapour per kg of
dry air (corresponding to condition of 25°C and 50% R .H .). From tables, temperature
corresponding to moisture content of 10.045 gm/kg of dry air is 14.1°C. Therefore, the
dew point temperature is 14.1#C , i.e . the temperature of atmospheric air, is first lowered
to 14.1 #C so that surplus moisture of 33.555 gm/kg of dry air is separated out and
removed.
Problem -4 : Atmospheric air at 21 °C and 40% R.H . is to be conditioned to a temperature
of 25°C. and 50% R.H . The method employed is to lower the temperature o f atmospheric
air to dew point temperature by passing it through a spray of chilled water whose
temperature is lower than dew point temperature and then to raise it to the required
temperature by passing it through a heater. Using the tables of psychrometric properties
of air, find: (i) the m ass of water added per kg of dry air, and (ii) the dew point
temperature.
Data from tables o f psychrometric properties of air is a s under :
Temperature of air.’C
30
25
21
20
15
14
Mass of water vapour when fully
saturated (100% R.H.) in gm/kg
of dry air
2723
20 09
15-66
14-7
1065
9971
(i) Amount of water vapour in atmospheric air at 21 °C and 40% R .H . is
0-4 x 15-66 = 6-264 gm/kg of dry air.
Amount of water vapour in conditioned air at 25°C and 50% R .H . is
0-5 x 20-09 = 10-045 gm/kg of dry air.
.-. M ass of water to be added = 10-045-6.264 = 3.781 gm/kg of dry a ir.
(ii) Dew point temperature is 141°C- corresponding to condition of 25°C and 50%
R .H ., i.e . corresponding to water vapour of 10-045 gm per kg of dry air.
Problem -5 : Air at 37°C and 60% R.H. is to be conditioned to a temperature of 25°C
and 50% R.H . The method employed is to lower the temperature to dew point and then
raise it to the required temperature. Using the psychrometric chart, find: (i) the dew point
temperature, and (ii) the mass of water drained out per minute when the air flow is 60
kg/min.
(i) On the psychrometric chart (fig. 3-1) mark a point A where 37°C dry-bulb
140
ELEM EN TS O F H EA T EN G IN ES Vol. Ill
temperature vertical line and 60% relative humidity curve intersect one another. Sim ilarly,
mark a point B where 25°C dry-bulb temperature vertical line and 50% relative humidity
curve intersect one another. From point B draw horizontal line upto the saturation curve
(100% R .H .) and read the dew point temperature. The dew point temperature read from ’
the chart is 14.1#C .
(ii) From points A and B draw horizontal lines upto the moisture content verticalscale
on the right hand side and read the moisture content in 1
kg of dry air. The moisture
content for point A is 0.024 kg/kg of dry air and for point B is 0.01 kg/kg of dry air.
A The amount of water drained out from 60 kg of air per minute
= 60 x (0 024 - 0 01) = 0.84 kg /m in.
Problem - 6 : Outside air at 0°C and 80% R.H . is warmed to a temperature o f 20°C.
Find: (i) the relative humidity of air, (ii) the change in volume per kg of dry air, and (iii)
increase o f enthalpy p er kg o f dry air with the help o f psychrometric chart.
(i) On the psychrom etric chart mark the dry-bulb temperature of 0°C and draw a
vertical line through it to intersect 80% R .H . line. A s the moisture is not added, the
heating will be along a constant vapour pressure or constant moisture line. This is a
typical heating path for air. Hence, through the above intersection point draw a horizontal
line to intersect the vertical line through 20°C dry-bulb temperature on the horizontal
scale. Now, read the relative humidity corresponding to the second point. This is 20.2%
R .H .
(ii) The first intersection point gives 0-778 m3 per kg of dry air while the second
intersection point gives 0.834m 3 per kg of dry air. Hence, the volume of air increases
from 0-778 m3 to 0-834 m3 per kg of dry air. Thus, the change in volume of air
= 0-834—0-778 = 0-056 m3 per kg of dry a ir.
(iii) From the first point of intersection, draw an oblique line parallel to wet-bulb
temperature line to intersect the enthalpy scale. Read the value of enthalpy which is 7
kJ per kg of dry air. Sim ilarly, from the second point of intersection draw an oblique line
to intersect the enthalpy scale. Read the value of enthalpy which is 27-5 kJ per kg of
dry air.
Thus, the increase of enthalpy = 27-5-7 = 20-5 k J per kg of dry a ir.
Problem -7 : In an air conditioning plant, 200 kg/minute of outside air at 35°C D.B.
temperature and 60% relative humidity is mixed with 800 kg/min o f recirculated air at
28°C D.B. temperature and 30% relative humidity. Find the condition of air after mixing.
On the psychrometric chart draw a straight line parallel to the constant temperature
line from 35°C D .B, temperature, to intersect 60% R .H . line at a point, and then from
this intersection point, draw a vertical line to intersect the saturation curve in a point.
Read the value of the moisture at this point. It is 21-9 gm/kg of dry air. Now, find a
point of intersection of 28°C D.B. temperature line with 30% R .H . line and draw a vertical
line to intersect the saturation curve and read the value of moisture. It is 7.5 gm/kg of
dry air.
M ass of vapour (moisture) in the mixture is given by
W!, x1 + W2 x2
Xfn =
W, + w2
where, W1 = 200 kg, W2 = 800 kg, x, = 21-9 gm/kg of dry air, and
x2 = 7-5 gm/kg of dry air.
AIR CONDITIONING
141
. (200 « a a ♦ (800 « . 7 ;g) _ 10-38 gm/kg of dry air.
xm 200 + 800
Now, mean tem perature of mixture can be obtained as m ass mean value,
m (200 x 35) ->- (800 x 28) = 2g ^
f/n =
200+ 800
Now, find a point on the saturation curve where the moisture content is 10-38 gm/kg
of dry air. From this point draw a vertical to intersect 29-4°C constant temperature line.
At this point of intersection, read the value of relative humidity, which is 40-2%
Thus, the air after mixing is at 29-4°C. D .B . temperature and with 40-2% R .H .
Problem - 8 : An auditorium room of 200 seating capacity is to be air conditioned. The
outdoor condition is 25°C dry-bulb temperature and 15°C wet-bulb temperature. The
comfort condition required' is 23°C dry-bulb temperature and 55% R.H . The quantity of
outside air supplied is 0-6 m3 per minute per person. This condition o f comfort is
achieved by humidifying the air in an air washer with recirculated water, followed by
heating with a heating coil.
Find: (i) The dry-bulb temperature and R.H . of the air leaving the humidifier, (ii) the
capacity- of the humidifier in kg/hr, and (iii) the capacity o f the heating coil in KW.
(i) Locate point ‘1’ on the psychrom etric chart
(fig. 3-16) where 25°C D .B . temperature line
and 15°C W .B. temperature line intersect. Locate
point *3’ on the chart where 55% R .H . curve
intersects 23°C D .B . temperature line. Draw a
constant specific humidity line through point ‘3’
to intersect 15°C W .B. temperature line at point
‘2’ as shown in fig. 3-16. From the psychrometric
chart (fig. 3-16) we get,
Hi = H2 = 42 kJ/kg,
W1 = 6-6 gm/kg,
Vs\ = 0-854 m3/kg, H3 = 46-5 kJ/kg,
18 °C
23’C 25'C
W2 = 9-69 gm/kg, fe =18°C.
Dry bulb temperature R .H . at point 2 is 74% and D .B . temperature
Fig. 3-16
is 18°C.
00 Mass of outside air supplied per hour
200 x 0-6 x 60
= 8,431 kg/hr.
0-854
8,431 (w2 - Wj)
Capacity of the humidifier =
,1,000
8,431 (9-69 - 6-6)
= 26-05 kg/hr.
1,000
8,431 (H3 - Hz)
(iii) Capacity of the heating coil =
3.600
_ 8,431 (46-5 - 42) _
= 10-54 kJ/sec., or 10-54 kW
3.600
Problem - 9 : The circulation of free air through the air conditioning plant is 300
m3/minute. The rate for cooling is 55 paise per tonne of refrigeration per hour while that
ELEM EN TS OF H EAT EN G IN ES Vol. Ill
142
for heating is 12 paise p er k.W .hr. Dew point
temperature of the cooling coil is 13°C.
The outdoor condition o f air is 35°C dry-bulb temperature and 26°C wet-bulb
temperature. Find which o f the following indoor conditions are economical for running the
plant, if they give the sam e comfort: (i) 25°C and 60% R .H ., and (ii). 30°C and 40%
R.H .
(i)
A ir at 25°C and 60% R .H .
The condition is achieved firstly by cooling and humidifying the air and then by heating.
First locate point p which is at 35°C D .B . temp, and 26°C W.B. tem p., then locate point
q corresponding to 25°C D.B. temperature and 60% R .H ., then locate point s of 138C
dew point tem perature, join p s and draw a constant humidity line through q to intersect
p s at r. The point r gives the condition of air entering the heating coil.
From the psychrom etric chart (fig. 3-17a), we get,
Hp = 80 kJ/kg, Hq = 55-5 kJ/kg, H,= 48 kJ/kg, Hs = 38 kJ/kg,
w1 = 17-5 gm/kg; w2 = 11-8 gm/kg and Vsp = 0-896 m3/kg.
Dry bulb temperature
,
Dry, bulb temperature
<°>
(W
Fig. 3-17
The m ass of air circulated per minute
=
= 335 kg/min. = 335 x 60 kg/hr.
The heat carried away by the cooling coil per hour
= 335 x 60 (Hp - Hr) = 335 x 6 0 (8 0 - 48) = 6 ,46,872 kJ/hr.
Now, one tonne of refrigeration = 210 x 60 = 12,600 kJ/hr.
Capacity of cooling coil in tonnes of refrigeration
-
^ ir -
5 1 3 3 9 ,onnes-
Cost of running the cooling plant/day
= 51-339 x 24 x
= 677-70 Rupees
AIR CONDITIONING
Capacity of heating coil = 335 (5i^ — ^
143
kW
Cost of running the heating plant/day
« 335
Total
48) X 24 X |
= 120-50 Rupees
cost of running the plant/day =
677-70 + 120-50 = R s. 798-20
(ii) A ir at 30°C and 40% R .H .
In this case, as shown in chart, locate point p \ q ' and s ' respectively representing
30*C dry-bulb temperature and 26°C wet-bulb temperature, 30°C dry-bulb. temperature
and 40% R .H . and 13°C dew point temperature. Join p ' to s ' and from q 'draw a line
of constant humidity intersecting the former at point r \
From psychrom etric chart (fig. 3 -17b),
Hp ' = 80 kJ/kg, Hq ' = 58 kJ/kg,
H r' = 43 kJ/kg, and Hs ' = 38 kJ/kg.
Air circulation is sam e as in the first case = 335 x 60 kg/hr.
Heat carried away by cooling coil/hr
= 335 x 6 0 (Hp ' - H r') = 335 x 60(80 - 43) = 7,43,700 kJ/hr.
Capacity of cooling coil in tonnes of refrigeration =
Thus, the cost of running the cooling plant/day
- 7'43'™ ° » 2-4 X ^
= R s.779-10
12,600
100
. . . . . .
..
335(58 - 43) 14A<
Capacity of heating coil = --------kW
Cost of running the heating plant/day
335(58 - 43) « 24
12 =
60
100
Total running cost of running the plant per day
= 779-10 + 241-20 = R s. 1,020 30
Thus, the air at 25°C D.B. temperature and 60% R .H . is economical than the air at
30°C D.B. temperature and 40% R .H .
Problem -10 : A spray cooling coil is chosen to circulate 4,500 kg/min. of chilled
water for an air conditioning plant. The inlet temperature of chilled water is 8PC. Air flows
through the coil at a rate o f 2,600 m /min. The air entering the cooling coil has condition
of 30°C dry-bulb temperature and 50% R.H. Air outlet conditions are 12°C dry-bulb
temperature and 45% R.H. Calculate: (i) the cooling load on the coil in tonnes of
refrigeration, (ii) the temperature o f the chilled water leaving the cooling coil chamber,
and (iii) the temperature of the air leaving the cooling coil chamber when its R.H . is 60%
and the outlet temperature o f the chilled water is 13°C.
(i) From psychrometric chart (fig.. 3-18),
Hf = 64 kJ/kg of dry air, /^ = 21 kJ/kg of dry air, and
Vs1 = 0.857 m3/kg of dry air.
2 600
M ass of air passing through the cooling coil per min. ■
— = 3033-84 kg.
0*857
144
ELEM EN TS O F H EA T EN G IN ES Vol.
= (jif§ p * 60 ■ 1.82,030 kg/hr.
One tonne of refrigeration =
210 x 60 = 12,600 kJ/hr.
Dry bulb temperot u re
Fig. 3-18
Total cooling load in tonnes of refrigeration =
1,62,030 (H] - Hz)
12,600
Total cooling load = 1.82,030 (64 - 21) _ 5 2 1-2 1 tonnes o f refrigeration
12,OU0
(ii) Heat lost by air per min = Heat gained by chilled water per min!
Wa (H i - Hz) = Ww (t2 - U) 4187
( where Wa = m ass of air/min )
3033-84 (64 - 21) = 4,500 (t2 - 8) x
_
« « «„
3033-84 x 43
A 2 = 4,500 x 4-187
~
+
4*187
„
14^24 C
The temperature of chilled water leaving the cooling chamber is 14-924*C.
(iii) When the temperature of chilled water at outlet is 13°C, we have by equating
enthalpies,
ma (Hi - H^)
/.
3033-84 ( 64—
64 ' *
= mw(t2 - ft)
)
= 4,500 (13-8) x 4-187
x 5 x 4-187
4,500
3 0 ^ 8 4 ------
H i - 64 - 4,5003^ .^ 4 4—
- 32 948 W *0 of dry air
Since, the condition of the air leaving the cooling coil chamber is 60% R .H ., the
intersection of the enthalpy line 32-948 kJ/kg of dry air with 60% R .H . curve on the
psychrometric chart, gives the temperature of air at outlet of the cooling coil, which is
16°C.
Problem-11 : A cooling tower is to be designed to cool 800 kg of water per minute
from temperature of 38°C to 32°C. The atmospheric conditions are 38°C dry-bulb temperature
and 26°C wet-bulb temperature. The condition o f the air leaving the cooling tower is
AIR CONDITIONING
145
3 2'C dry-bulb temperature and 80% R.H . Calculate the following:
(a) Volume o f the air to be handled by the cooling tower in m3 per minute, and
(b) Quantity o f m ake-up water in kg/hr, required by the cooling tower.
W0
Ai
(b )
la)
Fig. 3-19
(a) Referring to fig. 3-19 (b),
Let /)w1 = Enthalpy per kg of water at inlet,
ft *2 = Enthalpy per kg of water after cooling,
Hy =
Enthalpy per kg of dry air at condition (1 ) at atmosphericconditions,
Hz =
Enthalpy per kg of dry air at condition (2) leaving the cooling tower,
Wf m moisture content of air in gm/kg at atmospheric conditions,
Mfe =
moisture content of
air
leaving the cooling tower in gm/kg
W\ =
m ass of hot water supplied in kg/min.,
W2 = m ass of cold water in kg/min., and
V = volume of the air circulated or handled by cooling tower in m3/min.
The hot water gives out heat to the air and moisture contentof the air thus Increases.
Heat lost by the water = Heat gained by the air. *
Wyhwi - W2h ^ =
- (H z - H ,)
Vs1
]
The m ass of water evaported from hot water and which increases the moisture content
of air =
s1
^
1,000
kg/min.
V 'Wz - Wf'
kg/min.
1,000
Vs*
Hence; m ass of cooled water, W2 = W,
V 'Wz - wi '
1,000
Vs1
HE3-10
hv/2.
"
1
1
vs\
(H z -
H y)
146
ELEM EN TS OF H EA T EN G IN ES Vol. Ill
From the psychrom etric chart (fig. 3-19b), we have
Hy = 80 kJ/kg, Hz = 98 kJ/kg, iwy = 16-1 gm/kg, w^ = 25-3 gm/kg,
Va1 = 0-905 m3/kg,
= 159 kJ/kg, h ^ = 134 kJ/kg,
Wi = 800 kg/min (given).
1425-3 - 16-1)1
800(159) - 800 x 134
(98 - 80)
0-905 x 1,000
0-905
V x 9-2 x 134
V x 18
0-905 x 1,000
0-905
18
1,232 81
20,000 - V
0-905
905
Volume of air handled by cooling tower,
20,000
V
= 1,0795 m /m in.
18-527
(b) Amount of m ake-up water required by cooling tower in kg per hr.
V
Wz - Wy
x 60 kg/hr.
1,000
800(159 - 134) +
1,079-5
(25-3 - 16-1)
X 60 = 658-44 kg/hr.
0-905
1,000
Problem -12 An air conditioning plant is to be installed for an office having 60 occupants,
with the following specifications :
Condition of outside atmosphere
12 C dry-bulb temperature and 10 C wet-bulb
temperature.
Condition required in office after heating
and then by adiabatic humidifying .. 20°C dry-bulb temperature and 53% R.H.
Amount o f air circulated per occupant (person) .. 16 m per hour.
Calculate: (i) the capacity of heating coils in kW, (ii) the capacity of the humidifier in
kg/hr, and (iii) the relative humidity of air at the end of heating.
(i) First locate points ‘I’ and ‘3’ on psychrometric chart as shown in fig. 3-20. Point
1 is the point of intersection of a vertical lint through 12°C D .B . temperature and inclined
line through 10°C W .B. temperature and
Point 3’ is the point of intersection of vertical
H?-£ 0 2 k ;/k g
line through 20°C D .B . temperature and 53%
XJ
R .H . curve from the given data.
e
i
Now, locate point ‘2’ which is point of
intersection of constant enthalpy line through
point ‘3’ and the constant specific humidity
3 *
Constant
a
enthalpy
line through point I. From the psychrometric
line
chart (fig. 3-20), we get,
- - - - W|=WEnthalpy at point ‘I’, H* = 24.7 kJ/kg of
dry air,
Enthalpy at point ‘2’, H2 - H3 = 40.3
kJ/kg of dry air,
10*C12*C
' 20"C
27'C
fe = 27°C, specific volume of air at point I,
Dry bulb temperature
VS1 = 0.814 m3/kg, w\ (moisture content
Fig. 3-20
A IR CONDITIONING
of at point T ) =
147
5 gm/kg of dry air, and
w*3 (moisture content of air at point ‘3’) = 7.4 gm per kg of dry air.
i b * fin
(i) M ass of air circulated per hr, Wa - — — - - 1,179-4 kg/hr.
Now, 1 kW = 1 kJ/sec.
Capacity of heating coil in kW.
Wa (H2 - HQ
3,600
1 179*4
=
(40-2 - 24*7) = 5*8 kJ/sec. or 5 8 kW
3,600
Wa (W3 - Wi)
(ii) Capacity of the humidifier in kg/hr =
■——----I ,uuu
1,179*4(7*4 - 5)
5oo
“ 2 83 k9/hr
(iii) From psychrom etric chart, relative humidity corresponding to point ‘2’ is 23.6%
R.H.
Problem -13 : The circulation of free air through the air conditioning plant is 300 m3/minute.
The outdoor condition o f air is 35°C D.B. temperature and 26°C W.B. temperature. The
indoor condition o f air required is 25°C D.B. temperature and 60% R.H . The rate of
tooling is 55 paise per tonne o f refrigeration per hour, while that o f heating is 12 paise
per kW.hr. Dew pointtemperature o f cooling coil is 13°C. Calculate: (i) the capacity of
the cooling coil intonnes o f refrigeration, (ii) the cost o f the running the cooling plant
per day, (iii) The capacity o f the heating coil in kW., (iv) the cost of running the heating
plant per day, and (v) the total cost o f running
the air conditioning plant per day.
The indoor condition of air is achieved firstly
by cooling and humidifying the air in the cooling
coil and then by heating this air in the heating
co il. First locate points ‘p ’ and ‘q ’ on the
psychromeric chart as shown in fig. 3-21. Point
‘p 'is the point of intersection of 35°C D .B . tempera­
ture vertical line and 26°C W .B. temperature in­
clined line, and point ‘q ’ is the point of intersection
of 25°C D .B . temperature vertical line and 60%
R .H . curve. Then locate point 's' of 13°C dew
1TC
25’C 35T
point temperature as shown in fig. 3-21. Join 'ps'
and draw constant humidity line through 'q’ to
Dry bulb tem perature- >
intersect ’p s ’ at ’r ’. Point ‘r ’ gives the condition
of air entering the heating coil.
Fig. 3-21
From psychrometric chart (fig. 3-21), we get, Hp = 80*4 kJ/kg of dry air, Hr = 48*2
kJ/kg of dry air, Hq = 55*7 kJ/kg of dry air, Hs = 37*7 kJ/kg of dry air and Vsp = 0*896
m /kg of dry air.
300
(i) The m ass of air circulated per min. = ■■ „ - 335 kq
•
0*896
a
Heat carried away by cooling coil/hr.
= 335 x 60 x (Hp - Hr)
148
ELEM EN TS OF H EAT EN G IN ES Vol. Ill
= 335 x 60 x (80-4 - 48-2) =
One tonne of refirgeration = 3-517
and one tonne of refrigeration per
= 3-517 x 60 x 60 = 12,661-2
4,67,220 kJ/hr.
kJ/sec.
hour
kJ/hr.
Capacity of cooling coil in tonnes of refrigeration
4,67,220
_______
’ 1 2 ^ 6 1 * = 5 1 1 2 ,onne9
(ii) Cost of running the cooling plant per day
= 51-12
x
24
X
^
= 674-82 Rupees
(iii) Capacity of the heating coil in kW
335
=
x (55-7 - 48-2) = 41-87 kJ/sec. or 41-87 kW .
(iv) Cost of running the heating plant per day
= 41-87 x 24 x
= 120-6 Rupees.
1 uu
(v) Total cost of running the air conditioning plant per day
= 674-82 + 120-6 = 795-42 Rupees
T utorial-3
1.
Fill in the gaps with appropriate words to complete the following statem ents:
(a) Air conditioning may be defined as the simultaneous control o f
,
motion,
and
of air within air enclosed space.
(b)
the air for elimination of unpleasant odours.
(c) Actual weight of water vapour contained in a unit volume of air is ca lle d _________
humidity.
(d) W et-bulb temperature is the m easure__________ of air.
(e) Air conditioning system may be classifed a s _________ system an d ___________ system .
(f) _________ may be necessary only for winter humidification.
(g) Refrigerant used for the air conditioning system should b e____________ and____________ .
[(a) tem perature, humidity, purity, (b) Ozonate or Ionize, (c) absolute, (d) enthalpy,
(e) central, unitary, (f) preheater, (g) non-toxic, odourless.)]
2.
Choose proper phrase/s to complete the following statements:
(a) Air conditioning involves proper control of
(i) pressure; (ii) temperature; (iii) cleanliness; (iv) humidity; (v) (i), (ii) and (iii) of
the above, (vi) (ii) (iii) and (iv) of the above.
(i) pressure; (ii) temperature; (iii) cleanliness; (iv) humidity; (v) (i), (ii) and (iii) of
the above; (vi) (ii), (iii) and (iv) of the above.
(b) The method commonly adopted for dehumidifying air is
(i) cooling, (ii) heating and cooling, (iii) heating, (iv) passing steam and spray
cooling.
(c) In air-conditioning, the normal required comfort conditions are
AIR CONDITIONING
149
(i)
15°C D BT and 75% R .H ., (ii) 20°C D BT and 80% R .H ., (iii) 15°C D BT and 35%
R .H ., (iv) 22°C D BT and 60% R .H .
(d) For proper control of the percentage of°C0 2 in air supply during air-conditioning,
so that its level does not exceed 9.6% . the quantity of free air required per person
should be (i) 1-2 m3/m in., (ii) 1-2 m /hr, (iii) 0-2 m3/m in., (v) 5-3 m3/min.
(e) Dew point temperature is alw ays an indicator of
(i) dryness of air, (ii) latent heat, (iii) coolness of air, (iv) none of the above.
(f) The parameter which rem ains constant during evaporative cooling process through
an air w asher, with the sam e water recirculated again and again is (i) wet-bulb
tem perature, (ii) dry-bulb temperature, (iii) dew point tem perature, (iv) absolute
humidity, (v) relative humidity.
(g) The specific humidity during evaporative cooling process with recirculated water
spray
(i) increases, (ii) decreases, (iii) rem ains sam e, (iv) unpredictable, (v) none of the
above.
[(a) iv, (b) i, (c) iv, (d) i, (e) iii, (f) iv, (g) i]
3. (a) Explain the term air conditioning. What are its fields of application?
(b) Draw a line of difference between industrial air conditioning and air conditioning
for human comfort.
4.
Explain the following term s as applied to air conditioning:
(i) humidity, (ii) relative humidity, (iii) dry-bulb temperature, (iv) wet-bulb tempera­
ture, (v) dew point temperature, (vi) effective temperature, (vii) psychrom etry, (viii)
psychrom etric chart, and (ix) viscous air filter.
5.
Explain the significance of effective temperature with reference to air conditioning.
6. (a) What are the important factors which play decisive role in air conditioning?
(b) What are the different processes adopted to carry out air conditioning fully for
human comfort round the year?
7 .'
What are the different types of filters used for air conditioning plants ? Trace the
necessity of it.
8.
Differentiate between "partial air conditioning" and “complete air conditioning”
9.
Sketch a typical layout of a modern air conditioning plant. Explain its working.
Explain the advantages of by-pass system .
10.
Describe, giving a sketch of layout, an air conditioning plant for a public hall or
a cinem a hall. What should be the frequency of air replacement?
11. (a) Explain, with the help of a neat diagram, the working of window air conditioner,
(b) Explain, giving a neat line diagram, the working of a packaged air conditioner.
12.
Describe the different air conditioning system s. Give the advantages of unitary
system over central system .
13.
What are the different air conditioning system s? Trace the importance of central
system , stating different components and their role in conditioning the air.
14. (a) D iscuss the relative advantages and disadvantages of the central system with self
contained room coolers for residential air conditioning.
(b) Using psychrometric chart, determine the relative humidity of air having 24°C
dry-bulb temperature and 19°C wet-bulb temperature.
[62% R .H .]
150
15.
16.
17.
18.
19.
20.
ELEM EN TS O F H EA T EN G IN ES Vol. Ill
If the air has a dry-bulb temperature of 25°C and the relative humidity 50% , find
the dew point temperature of the air with the help of psychrometric chart.
[14.1°C]
Air at 40°C dry-bulb temperature and 70% R .H . is to be conditioned to a dry-bulb
temperature of 25°C and 50% R .H . The method employed is to lower the
temperature of air to dew point and thento raise it to required temperature by
passing it through a heater and then delivering it to the conditioned space. Using
the psychrom etric chart find: (i) the dew point temperature, and (ii) the m ass of
water drained out per minute when the air flow is 60 kg/min.
[(i) 14.1°C, (ii) 1.452 kg.min.]
If the dry-bulb temperature of air is 40°C and the wet-bulb temperature is 25°C,
find, using the psychrometric chart, (i) the relative humidity, (ii) the dew point
temperature of air, (iii) the enthalpy per kg of dry air, and (iv) the moisture content
per kg. of dry air.
[(i) 29% , (ii) 19.4% , (iii) 77 kJ per kg of
dry air, (iv) 14.4 gm per kg of dry air]
A Cinem a hall for 1,500 persons is to be air conditioned. Comfort conditions of
20°C dry-bulb temperature and 60% R .H . are to be achieved by cooling and
dehumidification followed by sim ple heating. The outdoor conditions are 30°C
dry-bulb temperature and 70% R .H . The quantity of outside air supplied is 20
m per hour per person. Using the psychrometric chart, find: (i) the m ass of water
eliminated per hour in dehumidifying the air, (ii) the capacity of the cooling coil
in tonnes of refrigeration, and (iii) the capacity of the heating coil in kW.
[(i) 353-8 kg/hr., (ii) 121 -9 tonnes of refrigeration, (iii) 74-8 K.W .]
A cooling tower cools 800 kg of water per minute from 35.5°C to 31.5°C. The
condition of air entering the cooling tower is 36°C dry-bulb temperature and 26°C
wet-bulb tem perature. The air leaving the cooling tower has relative-hum idity of
90% . The m ass of make-up water required is 8.55 gm per cubic meter of air.
Calculate the following:
(0 The volume of air to be handled by the cooling tower in m3/min.,
(ii) The quantity of make-up water required in kg/hr, and
(iii) The temperature of the air leaving the cooling tower.
[(i) 1,126 m3/min, (ii) 578 kg/hr. (iii) 30°C dry-bulb temperature]
D iscuss the factors which are normally involved in the selection of an air conditioning
plant.
I
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