brief teaching notes on refrigeration and Air conditioning

1. Refrigeration and Air Conditioning
Tsegaye Getachew1∗
1
Department of Mechanical Engineering, Wolaita Sodo University, tsegaye.getachew@wsu.edu.et
April 21, 2023
Contents
1 Introduction 1
1.1 Basic Refrigeration and Air Conditioning System Components . . . . . . . . . . . . . . . . . . . . . . 1
1.1.1 Terms, Definition, and Applications of RAC (Refrigeration and Air Conditioning System . . . 2
1.2 System Performance of Heat Pump and Refrigeration Cycles . . . . . . . . . . . . . . . . . . . . . . . 4
1.3 The refrigeration cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.3.1 Vapor Compression Refrigeration System Cycle (VCRS) . . . . . . . . . . . . . . . . . . . . . . 4
1.3.2 Vapor Expansion Refrigeration System Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.3.3 Air Cycle (AC) or Air Conditioning Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.4 Solved Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.5 Review Question . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2 PSYCHROMETRY 7
2.1 Solved Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.2 Review Question . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
3 Air-Conditioning 11
3.1 Solved Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3.2 Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
4 Load Calculation 15
4.1 Calculating heating and cooling loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
4.2 Methods for estimating cooling and heating loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
4.2.1 Cooling Load Calculation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
4.3 Solved Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
4.4 Multiple Choice Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
List of Figures
1 Components of Refrigeration and Air conditioning System . . . . . . . . . . . . . . . . . . . . . . . . 1
2 Sensible and Latent heat of water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
3 Isothermal wall heating, exponential Temp. Decay a. Mean surface Temp. b. Log Mean Temp. . . . . 4
4 Refrigerator versus heat pump . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
5 Basic circuit diagram of vapor absorption cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
6 Air Cooling Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
7 a. wet bulb temperature measurement b.laboratory psychrometer c. Various Air Conditioning Processes
across Mollier chart d. psychrometric chart/Mollier Diagram e. Mollier Diagram Reading . . . . . . . 8
8 Typical Air Conditioning Process Across Mollier Diagram a. Sensible heating circuit b. sensible heating
Mollier diag. c. Sensible cooling circuit d. sensible cooling Mollier diag. . . . . . . . . . . . . . . . . . 9
9 Bypass factor a. circuit diagram b. Mollier diagram c. heating coil d. cooling coil . . . . . . . . . . . 9
10 Solved Problem a. Comfort AC b. Industrial AC c. Winter AC . . . . . . . . . . . . . . . . . . . . . 13
11 Cooling load components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
12 A typical summer air conditioning system having a cooling coil with a by-pass factor of none . . . . . 19
I

2. List of Tables
1 Common Refrigerants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2 Cooling Load Factor (CLF) for glass with interior shading and located in north latitudes . . . . . . . 17
3 Total heat gain, sensible heat gain fraction from occupants . . . . . . . . . . . . . . . . . . . . . . . . 18
4 Table showing typical load of various types of appliances. . . . . . . . . . . . . . . . . . . . . . . . . . 19
II

3. 1 Introduction
Refrigeration is the process of removing heat. Refrigeration is described as a technique for lowering a system’s
temperature below that of its surroundings and holding it there by continuously removing heat from it. Refrigerator
is the name of the appliance that creates the cooling effect. Air-conditioning achieved by controlling temperature,
humidity, ventilation, and air quality within bounds established by the needs of the air conditioned enclosure, air
conditioning is a type of air treatment.
Working principle of refrigeration
1. Heat is constantly transferred from a body in a high-temperature region to a body in a low-temperature region.
2. According to the second rule of thermodynamics, heat transfer from a system with a lower temperature to an
environment with a higher temperature is achievable through some external methods.
3. The working fluid undergoes phase transitions from liquid to vapour upon heat rejection and from liquid to
vapour upon heat absorption.
4. A cooling effect is produced when the working fluid transitions from a liquid to a vapor phase.
Its foundation is the second law of thermodynamics. According to the Clausius Statement, heat cannot move from
a body that is warmer to one that is colder without some external assistance. In refrigeration, heat is continually
rejected to the environment at a higher temperature while being continuously removed from the system at a lower
temperature. This is accomplished by utilizing an outside source, such as a compressor (or) a pump. A working fluid or
medium known as refrigerant is used to remove heat from a system operating at a lower temperature. The refrigerant
radiates heat away into the hot environment. Freon, ammonia, CO2, SO2, hydrocarbon refrigerant, methylene chloride,
ethylene, ethane, air, and water are only a few of the possible refrigerants.
1.1 Basic Refrigeration and Air Conditioning System Components
Principal refrigeration components of vapor compression refrigeration cycle are given in the figure 1. Few components
can vary with different refrigeration system depending on the construction purpose of either addition or removal of
heat.
Evaporator
Compressor
Expansion Device
Condenser
Saturated
Vapor
Liquid-Vapor
M
ix.
Superheated
Vapor
Low
Pressure
Saturated
Liquid
H
igh
Pressure
Figure 1: Components of Refrigeration and Air conditioning System
The evaporator’s function is to use the liquid refrigerant to remove extra heat from the product. Low pressure
causes the liquid refrigerant inside the evaporator to boil. The rate at which heat is transferred from the product to
the liquid refrigerant in the evaporator and the rate at which the compressor removes the low-pressure vapor from
the evaporator define the amount of this pressure. To allow for heat transmission, the liquid refrigerant’s temperature
must be lower than that of the object being cooled. After being transferred, the compressor uses the suction line to
take liquid refrigerant from the evaporator. When the liquid refrigerant reaches the evaporator coil, it has transformed
into vapor.
The evaporator’s low temperature, low pressure, and overheated vapor is transferred through the suction line, com-
pressed to high pressure by the compressor, and then discharged into the condenser without changing its gaseous
state. In order to raise the pressure of the refrigerant to a point where the saturation temperature of the discharge
refrigerant is higher than the temperature of the available cooling medium, heat is added to the refrigerant during
this process. This allows the superheated refrigerant to condense at normal ambient conditions. Rotary, hermetic,
1

4. and semi-hermetic compressors are among the various types of reciprocating compressors that are used for various
applications.
The condenser’s function is to transfer heat from the refrigerant to the ambient air. It can quickly transport heat
from the air near the pipe to the gas or vapor, turning them into liquid. Because they release heat while operating,
condensers frequently operate at high temperatures. The second law of thermodynamics states that spontaneous
heat transmission in a closed system is one-way, going from high temperature to low temperature, which means that
microscopic particles carrying heat may only exacerbate disorder.
Expansion Valve:The mechanism, which is designed to pass the most liquid refrigerant feasible to achieve a good
refrigeration effect, is used to feed the high-pressure liquid from the condenser to the evaporator. To have the least
amount of pressure drop, the liquid line needs to be suitably sized. The throttling mechanism regulates the flow
of refrigerant while also reducing pressure. Additionally, it lowers the pressure from the discharge pressure to the
evaporator pressure without causing the pressure refrigerant’s condition to change. Capillary tubes and thermostatic
expansion valves are examples of throttling apparatus. For applications up to around 10 refrigeration tons, the capil-
lary tube is the most often utilized throttling device.
Vapor Compression: (VCR) system is the most affordable and straightforward form of cooling equipment currently
in use. VCR systems are applied in industrial sectors like oil refineries, petrochemical factories, food processing busi-
nesses, and natural gas plants to remove existing heat when cooling spaces is required. The heat is transferred from
the cold reservoir to the hot reservoir in these systems, which operate in the opposite direction of the Carnot cycle.
The device uses mechanical energy to go through stages of compression and expansion to provide the cooling effect.
Ammonia is the primary refrigerant used in conventional (VCR) systems when phases transition from liquid to gas
and vice versa. In the compressor, the refrigerant is first adiabatically compressed. The pressure and temperature
of the refrigerant will rise. Heat from the refrigerant is rejected to the atmosphere by the condenser, which receives
it at constant pressure. After entering an expansion device, the refrigerant will expand and its temperature and
pressure will decrease. This low-temperature refrigerant evaporates when it enters the evaporator while keeping the
pressure constant, producing cooling. The heat that the evaporator absorbs is known as the refrigeration effect. As
the low-temperature refrigerant vapor enters the compressor, the cycle is then repeated.
Refrigerant: The operating substance utilized in refrigeration systems is referred to as ”refrigerant”. The most widely
used refrigerants at the moment are a mixture of hydrocarbons and halogens sold under a variety of brand names,
including freon, genetron, arcton, isotron, frigen, etc. These are either methane- or ethane-based and contain chlorine
or fluorine atoms in place of hydrogen carbon atoms. It was discovered in the middle of the 1970s that CFCs (chloro-
fluorocarbons) increase not only the amount of ultraviolet radiation that enters the earth’s atmosphere, but also act
as a barrier to the escape of infrared radiation to space, which adds to the greenhouse effect and subsequently, global
warming. Because of this, many nations have outlawed or gradually phased out the usage of some CFCs. The (ODP)
of R-12 is most severely harmed by fully halogenated CFCs, including R-11, R12, and R115. The most common
refrigerant is chlorine-free R-134a, which is used mostly in refrigerators and air conditioners for homes.
Properties of a good refrigerant:
1. very low boiling point and very low freezing point.
2. high latent heat of evaporation associated with less work requirement of the system
3. low saturation pressure.
4. good thermal conductivity for rapid heat transfer
5. non-toxic, non-inflammable, non-corrosive to the working parts.
6. economical for both in initial cost and maintenance cost
7. chemically stable for temperature variations and neutral for lubricating oil
8. Leaks should be easily detected, located
9. Should operate under low pressure
The common refrigerants and their typical properties given in Table 1 where some hydrocarbon refrigerants naming
following the naming convention.
1.1.1 Terms, Definition, and Applications of RAC (Refrigeration and Air Conditioning System
ˆ Refrigeration EffectIt is the quantity of cooling that a refrigeration system generates. It is described as the
rate of heat removal from the area (or system) that needs to be cooled during a cycle. It is also known as
”refrigerator capacity.” In either kW or kJ/s, it is measured.
ˆ cryogenics: Cryogenics studies the causes and consequences of extremely low temperatures. The name is
derived from the Greek words ”kryos” and ”genic,” which mean ”to produce” and ”frost,” respectively. A
definition like that would allow for all temperatures below zero degrees Celsius to be included.
2

5. Table 1: Common Refrigerants
# Refrigerant Properties
1
Ammonia NH3
It has a boiling point of 33.5 o
C and It’s the most
widely used refrigerant in vapour absorption
highly inflammable, irritating and corrosive and due
to the later two reasons, it’s not suitable for food
industries
Used in large industries such as ice manufacturing
plant packaging plants for cold storage
2 CO2 Has a boiling point of -77.6 o
C and used where space
consideration (compact)is important such as ships
because of its specific volume
3
SO2
has Boiling Temperature of −10o
C and it’s Colour-
less, and suffocating chemical with irritating odour.
Historically, this refrigerant was utilized in household
refrigerators in the past
4
Freon-12
has a boiling point of −29.8o
C, non-flammable, non-
explosive, non corrosive and odourless
widely accepted refrigerant for various applications
and used in small capacity equipment such as domes-
tic refrigerators, water coolers, and air-conditioner.
5
Freon-22
has a normal boiling point of −40.8o
C and more com-
paratively high-pressure refrigerant than Freon-12
employed for air-conditioners in large capacity
plants, food freezing, freeze drying
ˆ Ton of Refrigeration or Units of Refrigeration: is the unit of refrigeration, the amount of heat absorbed in
order to produce one ton of ice in 24 hours from water, whose initial temperature 0o
C. 1Ton of refrigeration
= 210kJ/minor3.517kW
ˆ Ice Making Capacity: the ability of a refrigerating system to make ice. In other words, it is the capacity of
a refrigerating system to remove heat from water to make ice.
ˆ Co-efficient of Performance:, (COP) the performance measure of the refrigeration system, the ratio or the
amount of heat removed from a given space to the work supplied to achieve the heat removal. The design of
the system or the specific purpose configuration determines the formula of COP to be used from Equation 1, see
Figure 4 for better clarity. The computation of work from the change in heat transfer follows the second law of
thermodynamics
COPR =
Desired Output
Work supplied to the system
=
QL
W
=
QL
QH − QL
=
TL
TH − TL
for refrigerator
COPHP =
Desired Output
Work supplied to the system
=
Q
W
=
QH
QH − QL
=
TH
TH − TL
for heat pump
(1)
ˆ Relative COP: the ratio of actual COP to the theoretical COP of a refrigerator.
COPRelative =
Actual COP
TheoreticalCOP
(2)
The theoretical COP always > 1.
ˆ Application of RAC: Preservation, and conservation of food and food industries; Industries of chemical
processing; Air conditioning systems for industrial and residential use; Drying Plants, etc.
ˆ Sensible and latent heat: A mathematical and engineering abstraction to add up internal energy and flow
work make up enthalpy. Sensible heat is a change in enthalpy may be felt as a shift in temperature, change
of enthalpy with change in temperature. The unit of change in enthalpy per degree of temperature change, or
the specific heat capacity, is given in kJ/(kg K), where K is kilograms. If enthalpy doesn’t with temperature
temperature change, additional heat absorbed as Latent heat: a change in state (solid to liquid, liquid to
gas, or vice versa). This is measured in kJ/kg but depends on the boiling point see figure 2. [1]. For water,
the latent heat of freezing is 334 kJ/kg as shown in the figure 2 and the specific heat capacity averages 4.19
kJ/(kg K). The quantity of heat to be removed from 1 kg of water at 30o
C in order to turn it into ice at 0o
C is:
Q = Qsensible + Qlatent = 4.19(30–0) + 334 = 459.7kJ. Mark here also that the temperature scale not converted
to SI unit-Kelvin because the required temperature value is the difference and therefore conversion doesn’t alter
it.
3

6. Figure 2: Sensible and Latent heat of water
ˆ Heat Transfer: The average rate of heat transfer to or from a refrigerant fluid flowing in a tube can be modelled
as forced convection graphically shown in the figure 3 [2] and can be computed by using equation 3. The h inthe
equation 3 known as thermal transimitance measured in W/(mK
).
Figure 3: Isothermal wall heating, exponential Temp. Decay a. Mean surface Temp. b. Log Mean Temp.
Q̇ = hA∆Tln (3)
where the logarithmic mean temperature can be calculated by using 4
∆Tln =
∆Tmax − ∆Tmin
ln ∆Tmax
∆Tmin
=
Tout − Tin
ln(∆Tout/∆Tin)
where Q̇ = fhA∆Tln = ṁCP ∆T
(4)
1.2 System Performance of Heat Pump and Refrigeration Cycles
The COP for refrigerator and heat pump are related as seen in Eq. 1 COPR = 1 + COPHP
1.3 The refrigeration cycle
The common refrigeration cycle comprise vapor compression cycle, vapor absorption cycle, steam jet refrigeration
system(old), air cycle, and thermoelectric cooling.
1.3.1 Vapor Compression Refrigeration System Cycle (VCRS)
As shown in the figure 1, a liquid refrigerant boils and condenses within the bounds of its freezing point and critical
temperature. The transition between the liquid and gaseous states—at a temperature depends on its pressure. The
latent heat of evaporation must be obtained during boiling, and it must be released when condensing. At the lower
temperature and pressure, heat is applied to the fluid, supplying the latent heat necessary to cause it to boil and
transform into a vapour. The next step is to mechanically compress this vapour to a higher pressure and a saturation
temperature where its latent heat can be rejected, causing it to transform back into a liquid. The heat transferred to
4

7. Refrigerator
Warm Env’t
Refrigrated
Cold Space
Heat Pump
Warm
House
Cold Env’t
Wnet,in
Required Input
Q
H
Heat
Ejected
to
Q
L
Desired
Output
Wnet,in
Required Input
Q
H
Desired
Output
Q
L
Heat
Absorped
to
Figure 4: Refrigerator versus heat pump
the working fluid in the evaporator or the difference in enthalpies between the fluid entering and the vapour leaving the
evaporator, will have the overall cooling effect via an Expansion Valve. A portion of the fluid will flash off into vapour
to eliminate the energy required for this cooling when the pressure is reduced at the sudden volume enlargement valve,
which must result in a proportional drop in temperature.
1.3.2 Vapor Expansion Refrigeration System Cycle
VARS are in the same category as refrigeration systems that use vapour compression. The required input for absorption
systems is heat, as opposed to vapour compression refrigeration systems, which require temperature drop (with an aid
of expansion valve). The VARS constructed with an absorber, heat exchanger and vapor generator, see figure
5 in place of compressor compared to VCRS in the figure 1. Because of this, these systems are often known as heat-
operated or thermal energy-driven systems. Wet absorption systems are another name for typical absorption systems
Figure 5: Basic circuit diagram of vapor absorption cycle
since they absorb refrigerant using liquids. VARS are favored when low-grade energy sources like solar energy or waste
heat are available since they operate on low-grade thermal energy. They are environmentally beneficial because they
use refrigerants like water or ammonia.
VARS does not involve moving part (compressor) or electrical input work and therefore relatively quieter operating
with less wear/tear, but has less COP. Although, VARS may utilize waste heat recovery that give it edge/advantage
over VCRS.
5

8. 1.3.3 Air Cycle (AC) or Air Conditioning Cycle
Any gas, when compressed, rises in temperature. Conversely, if it is made to do work while expanding, the temperature
will drop. AC use is made of the sensible heat only (although it is, of course, the basis of the air liquefaction process).
Typical single stage schematic diagram is shown in the figure 6. The main application for this cycle is the air-
conditioning and pressurization of aircraft. The turbines used for compression and expansion turn at very high speeds
to obtain the necessary pressure ratios and, consequently, are noisy. The COP is lower than with other systems The
Figure 6: Air Cooling Cycle
other type of refrigeration cycle is Thermoelectric cooling. The Peltier cooling effect occurs when an electric current
flows through junctions of different metals and results in a drop in temperature at one junction and an increase at the
other. The development of appropriate semiconductors has recently enabled advancements in this cooling technique.
Due to the high electric currents needed, applications are constrained in size, and the most viable usage are compact
cooling systems for use in the military, aircraft, and laboratories.
1.4 Solved Example
1. Heat is to be removed at a temperature of −5o
C and rejected at a temperature of 35o
C. What is the Carnot or
Ideal COP? [1]
Solution −5o
C becomes 268 K and 35o
C becomes 308 K (to the nearest K) by adding linear 273 kelvin scale
to each. Using equation 1, yields
COP =
T2
T1 − T2
=
268
35 − (−5))
= 6.7 Note here that in the computation of∆T
Theoretical COP because COP > 1 yields the same result in both o
C K
35 − (−5)) = 308K − 268K = 40
2. A fluid evaporates at 3o
C and cools water from 11.5o
C to 6.4o
C. Use equation 3 to calculate the logarithmic
mean temperature difference and what is the heat transfer if it has a surface area of 420 m2
and the thermal
transmittance is 110 W/m2
K? Solution: Using Equation 4 and 3
LMTD = ∆Tln =
∆Tmax − ∆Tmin
ln ∆Tmax
∆Tmin
=
(11.3 − 3) − (8.5 − 3)
ln (11.3−3)
8.5−3
= 5.566K
Q = (hA) × LMTD = 420 × 110 × 5.566 = 257, 000W or 257kW
Again, as the necessary computation solely involves differences, the temperature was not converted to the SI
unit. In addition, many of these ideals will change in practice. The saturation temperature will fluctuate as a
result of a pressure drop along a pipe carrying a fluid that is either boiling or condensing. The values of heat
transfer will vary with temperature for some liquids. Due to these factors, not every heat transfer application
can be precisely predicted by the formula.
1.5 Review Question
1. Consider figure below. Using equation 1 calculate the theoretical net work required to bring forth the operation
in the case of refrigerator assuming that the heat load QL = 257kW.
6

9. R
11.3o
C
3o
C
HP
11.3o
C
3o
C
Wnet,in
Q
H
Q
L
Wnet,in
Q
H
Q
L
2. Calculate the network in the case of heat pump in the figure above considering the heat load QH = 257kW
2 PSYCHROMETRY
The study of the characteristics of air-water vapour mixtures is known as psychrometry. The air in the atmosphere is
thought to be a combination of dry air and water vapor. The air in the atmosphere, also known as atmospheric air,
typically contains some water vapor (or moisture). Dry air is defined as air without any water vapor. The atmospheric
air can then be thought of as an ideal gas mixture, whose pressure is equal to the sum of the partial pressures of dry
air and water vapor, Pa and Pv, respectively. P = Pa + Pv, where P is the total pressure
The term vapor pressure typically refers to the partial pressure of water vapor. It is the pressure that water vapor
would experience on its own, at atmospheric air’s temperature and volume. Air is relative and specific humidity: The
most sensible approach is to state explicitly how much water vapor there is in a unit amount of dry air. Absolute or
specific humidity, commonly known as humidity ratio, is shown by the following symbol:ω for given Pg = PT =TS at
ω =
mv
ma
=
PvV/RvT
PaV/RaT
=
Pv/Rv
Pa/Ra
= 0.622
Pv
Pa
=
0.622Pv
P − Pv
in kg water vapor/ kg dry air)
(5)
The humidity can also express the relative humidity ϕ. The relative humidity ranges from 0 for dry air to 1 for
saturated air. The amount of moisture air can hold depends on its temperature. Therefore, the relative humidity of
air changes with temperature even when its specific humidity remains constant.
ϕ =
ωP
(0.622 + ω)Pg
and ω =
0.622ϕPg
P − ϕPg
(6)
The total enthalpy (an extensive property) of atmospheric air is the sum of the enthalpies of dry air and the water
vapor:
H =Ha + Hv = maha + mvhv Dividing by ma gives
h =
H
ma
= ha +
mv
ma
hv = ha + ωhv
=ha + ωhg in kJ/kg of dry air
(7)
Dry Bulb Temperature, or DBT(tdb): the true temperature of a gas or gas combination as measured using an
error-free thermometer. Wet of the wet bulb: WBT(twb): the temperature measured by an accurate thermometer
with a distilled water-moistened wick. The dew point TemperatureDPT(tdp): the temperature at which liquid
droplets begin to form as soon as the humid air is constantly cooled. Psychrometers: pieces of equipment that
concurrently measure the wet and dry bulb temperatures. A psychrometric chart shows the many thermodynamic
properties of air graphically. It helps to reduce the number of calculations needed to determine the qualities of air
that are needed in the field of air conditioning. The charts created by various air conditioning manufacturers differ
slightly, but overall, they are all the same. Normal atmospheric pressure for drawing the psychrometric chart is 760
mm Hg (or 1.01325 bar). See figure Air conditioning technique: Air conditioning systems are made to meet human
body needs. Air-conditioning operations are necessary to keep a residential area or an industrial facility at the proper
temperature and humidity. The process of air conditioning include: raising the temperature, simply decreasing it,
humidifying (adding moisture), and dehumidifying (removing moisture)as shown in the figure 7 (c). Air conditioning
involves getting the air to the right temperature and humidity, often two or more of these steps are required.
Sensible Heating: As depicted in Fig. 8 (a), suppose that air at temperature td1 travels over a heating coil at
temperature td3. It should be noticed that the temperature of the air (td2) when it exits the heating coil will be lower
than td3.On the psychrometric chart, the sensible heating process is represented by a horizontal line 1-2 running from
7

10. (a) (b) (c)
(d) (e)
Figure 7: a. wet bulb temperature measurement b.laboratory psychrometer c. Various Air Conditioning Processes across Mollier chart d.
psychrometric chart/Mollier Diagram e. Mollier Diagram Reading
left to right, as shown in Fig. 8 (b). The surface temperature of the heating coil is represented by point 3 in the
graph. The enthalpy difference (h2 − h1) as illustrated in Fig. 8 (b) can be used to calculate the heat absorbed by the
air during sensible heating from the psychrometric chart. It should be remembered that the specific humidity remains
unchanged throughout sensible heating (i.e., W1 = W2). Heat added by sensible heating obtained from eq. 8,
q = h2 − h1
= cpa (td2 − td1) + Wcps (td2 − td1)
= (cpa + Wcps) (td2 − td1) = cpm td2 − ṫd1
subst. the term (cpa + Wcps)
the humid specific heat with (cpm)
taking (cpm) = 1.022kJ/kgK
∴ q = 1.022 (td2 − td1) kJ/kg
(8)
The value of humid specific heat (cpm) The value is taken as 1.022 kJ/kgK. Sensible Cooling: Sensible cooling is
the process of reducing air temperature without affecting the specific humidity of the air. As indicated in Fig. 8 (d),
let air at temperature td1 to pass over a cooling coil at temperature td3. It should be noticed that the temperature of
the air (td2) when it exits the cooling coil will be higher than td3. A horizontal line 1 − 2 running from right to left on
the psychrometric chart represents the sensible cooling process. The heat rejected by sensible cooling calculated from
8

11. Figure 8: Typical Air Conditioning Process Across Mollier Diagram a. Sensible heating circuit b. sensible heating Mollier diag. c.
Sensible cooling circuit d. sensible cooling Mollier diag.
9: Heat rejected,
q = h11 − h22
= Cpa (td1 − td2) + WCps (td1 − td2)
= (Cpa + WCps) (td1 − td2) = Cpm (td1 − td2)
The humid spec. heat term (Cpa + WCps)
(Cpm) with value 1.022 kJkgK
∴ Heat rejected, q = 1.022 (td1 − td2) kJ/kg
(9)
When air passes over a coil, some of it (say x kg also known as by-pass factor: the ratio of loss in cooling or heating
to ideal cooling or heating) just by-passes unaffected while the remaining (1 - x) kg comes in direct contact with the
coil. The by-pass factor of a cooling coil decreases with decrease in fin spacing and increase in number of rows. The
temperature of the air corning out of the apparatus (td2) will be less than td3 in case the coil is a heating coil and
more than td3 or (tds) in case the coil is a cooling coil. x is obtained by balancing the enthalpies as seen in eq. 10
and Fig. 10. [shanmukha2020bypass].
(a) (b) (c) (d)
Figure 9: Bypass factor a. circuit diagram b. Mollier diagram c. heating coil d. cooling coil
xcpmtd1 + (1 − x)cpmtd3
= 1 × cpmtdd
x (td3 − td1) = td3 − td2
x =
td3 − td2
td3 − td1
BPFheating coil =
td3 − td2
td3 − td1
BPFcooling coil =
td2 − td3
td1 − td3
ηheating/cooling coil = 1 − x = 1 − BPFheating/cooling coil
(10)
9

12. Humidification is the process of introducing moisture into the air without modifying its dry bulb temperature.
Dehumidification is the process of removing moisture from the air without changing the dry bulb temperature. It
should be noticed that during humidification, the specific humidity likewise rises from W1 to W2, as shown in Fig. 9
(b) , and the relative humidity rises from 2 to 1. In the reverse process, Relative humidity falls from 1 to 2 during
dehumidification, and specific humidity falls from W1 to W2, as seen in Fig.9 b. It should be noticed that the
intercept (h2 − h1) on the psychrometric curve indicates the change in enthalpy during humidification. Because the
dry bulb temperature of the air during humidification stays constant, so does its sensible heat. It follows logically that
the rise in moisture content equal to ∆W,(where ∆W = W2 − W1) or (W1 − W2 for dehumidification) kg per kg of
dry air is what causes the change in enthalpy per kg of dry air, which is what is referred to as a latent heat transfer
(LH) given in equation 11 for the given of dry air flowing in m3/min v .
LH = h2 − h1 = hfg(W2 − W1)
= v × 1.2 × 2500 × ∆W = 3000v × ∆WkJ/min
(11)
The air is humidified by adding or spraying steam, hot water, or cold water into the atmosphere. The following two
techniques can be used to obtain the humidification:
1. Direct approach: With this method, water is sprayed into the room to be air-conditioned in a highly atomized
state. It is not very effective to obtain humidification using this method.
2. Indirect approach: In this method, an air-washer, is used to inject water into the air in the air-conditioning
plant. The room that needs air conditioning is subsequently given this conditioned air.
The other way around air conditioning can either be combination of (heating or cooling) and (humidification and
dehumidification), respectively, and Adiabatic Mixing. Sensible Heat Factor is the ratio of sensible heat to the sum
of sensible and latent heats. In the adiabatic mixing streams, the heat transfer with the surroundings is usually
small, and thus the mixing processes can be assumed to be adiabatic. Mixing processes normally involve no work
interactions, and the changes in kinetic and potential energies, if any, are negligible.
(a) (b)
mass of dry air ṁa1 + ṁa2 = ṁa3
mass of water vapor ω1ṁa1 + ω2ṁa2 = ω3ṁa3
energy ṁa1 h1 + ṁa2 h2 = ṁa3 h3
ṁa1
ṁa2
=
ω2 − ω3
ω3 − ω1
=
h2 − h3
h3 − h1
(12)
2.1 Solved Examples
1. Atmospheric air at a dry bulb temperature of 16o
C and 25% relative humidity passes through a furnace and
then through a humidifier, in such a way that the final dry bulb temperature is 30o
C and 50% relative humidity.
Find the heat and moisture added to the air. Also determine the sensible heat factor of the-process.
solution Heat added to the air: First of all, mark the initial condition of air i.e. at dry bulb temperature 16
degree Celsius and 25 percent relative humidity on the psychrometric chart at point 1, as shown in Fig. Then
mark the final condition of air at 30 degree Celsius dry bulb temperature and 50 percent relative humidity on
the psychrometric chart at point 2. Now locate the point A by drawing horizontal line through point 1 and
vertical line through point
10

13. h1 = 23kJ/kg of dry air hA = 38kJ/kgof dry air
h2 = 64kJ/kgof dry air Heat added = h2 − h1 = 64 − 23 = 41kJ/kgof dry air Ans.
2. Determine From the psychrometric chart, we find that enthalpy of air at point 1 in the question 1, Solution:
W1 = 0.0026kg/kg of dry air and specific humidity in the air at point 2 , W2 = 0.0132kg/kg of dry air ∴ Moisture
added to the air
= W2 − W1 = 0.0132 − 0.0026 = 0.0106kg/kg of dry air Ans.
sensible heat factor of the process,
SHF =
hA − h1
h2 − h1
=
38 − 23
64 − 23
= 0.366 Ans.
2.2 Review Question
1. Saturated air leaving the cooling section of an air-conditioning system at 14o
C at a rate of 50m3
/minis mixed
adiabatically with the outside air at 32o
C and 60 percent relative humidity at a rate of 20m3
/min. Assuming
that the mixing process occurs at a pressure of 1 atm, determine the specific humidity, the relative humidity,
the dry-bulb temperature, and the volume flow rate of the mixture.
2. Distinguish between sensible heat and latent heat with an appropriate sketch
3. Explain heating with humidification and cooling with dehumidification
4. describe A/C processes across Mollier diagram
3 Air-Conditioning
The study of air conditioning, i.e., supplying and sustaining desirable internal atmospheric conditions for human
comfort regardless of exterior conditions, is an area of engineering science. Air conditioning for industrial use, food
processing, food storage, and other topics are also covered.
A/C variables Conditioned air: The following four elements are crucial for comfortable air conditioning: air’s temper-
ature, air humidity, air quality, and the flow of air.
Air Conditioning System Equipment
1. Circulation fan:-The primary function of this fan is to flow air into and out of the room.
2. A/C unit: It is a unit that consists of cooling and dehumidifying operations for summer air conditioning or
heating and humidification activities for winter air conditioning.
3. supply ducts: The supply duct transfers the conditioned air from the circulating fan to the appropriate place in
the space to be air conditioned.
4. Supply outlets: These are grills that distribute conditioned air uniformly throughout the room.
5. Return outlets are the apertures in a room’s surface that allow room air to enter the return duct.
6. Filters: The primary function of filters is to remove dust, grime, and other potentially hazardous germs from
the air.
Air Conditioning System Classification : The following are some broad categories of air conditioning systems:
1. According to the purpose, there are two types of air conditioning systems: comfort air conditioning systems and
industrial air conditioning systems.
11

14. 2. Depending on the time of year There are three types of air conditioning systems: winter air conditioning, summer
air conditioning, and year-round air conditioning.
3. Depending on the equipment configuration, a unitary air conditioning system and a central air conditioning
system
1.System of Comfort Air Conditioning: The air is brought to the desired dry bulb temperature and relative
humidity for human health, comfort, and efficiency in comfort air conditioning. If there is insufficient data for the
desired condition, it is assumed to be 21o
C dry bulb temperature and 50 percent relative humidity. In general,
the sensible heat factor is kept as follows: = 0.9 for home or private office For a busy restaurant or workplace
= 0.8 Auditorium or cinema hall = 0.7 Ballroom, dance hall, and so on = 0.6 Comfort air conditioning can be
used in homes, workplaces, stores, restaurants, theaters, hospitals, and schools, among other places. 2. Industrial
Air Conditioning System It is an important system of air conditioning these days in which the inside dry bulb
temperature and relative humidity of the air is kept constant for proper working of the machines and for the proper
research and manufacturing processes. Some of the sophisticated electronic and other machines need a particular dry
bulb temperature and relative humidity. Sometimes, these machines also require a particular method of psychrometric
processes. This type of air conditioning system is used in textile mills, paper mills, machine-parts manufacturing
plants, tool rooms, photo-processing plants etc.
3. System of Winter Air Conditioning: The air is heated in winter air conditioning, which is usually accompanied
by humidification. The system’s schematic configuration. The outside air passes via a damper and mixes with the
recirculated outside air (which is obtained by a fan from the conditioned space). The Filter mixed air is filtered to
eliminate dirt, dust, and other contaminants. The air is now sent via a preheat coil to prevent water from freezing and
to control the evaporation of water in the humidifier. Following that, the air is forced to flow through a reheat coil
in order to reach the desired dry bulb temperature. A fan now supplies the conditioned air to the conditioned room.
The exhaust fans or ventilators expel a portion of the used air from the conditioned space to the atmosphere. The
exhaust fans or ventilators expel a portion of the used air from the conditioned space to the atmosphere. The residual
utilized air (known as re-circulated air) is conditioned once again. To compensate for the loss of conditioned (or used)
air through exhaust fans or ventilation from the conditioned area, outside air is drawn and mixed with re-circulated
air.
4.Air Conditioning System for Summer: It is the most common type of air conditioning, and it cools and
dehumidifies the air. The outside air passes through the damper and mixes with re-circulated air (from the conditioned
area). The mingled air is filtered to eliminate dirt, dust, and other pollutants. The air is now forced to flow through
a cooling coil. The coil is substantially cooler than the acceptable dry bulb temperature of the air in the conditioned
space. Cooled air flows through a perforated membrane, losing moisture in the condensed form, which is collected
in a sump. The air is then forced to travel through a heating coil, which gently warms it. A fan now supplies the
conditioned air to the conditioned room. The exhaust fans or ventilators expel a portion of the used air from the
conditioned space to the atmosphere. The residual utilized air (known as re-circulated air) is conditioned once again.
To compensate for the loss of conditioned (or utilized) air due to exhaust fans or ventilation from the conditioned
room, outside air is drawn and mixed with re-circulated air.
5.All-Year Air Conditioning System: The year-round air conditioning system should include both summer and
winter air conditioning equipment. A diagram of a modern summer year-round air conditioning system. The outside
air enters the damper and mixes with the Damper re-circulated air (which comes from the conditioned room). The
mingled air is filtered to eliminate dirt, dust, and other pollutants. The cooling coil in summer air conditioning
works to chill the air to the appropriate temperature. Dehumidification is achieved by running the cooling coil at a
temperature lower than the dew point (apparatus dew point). In the winter, the cooling coil is turned off and the
heating coil is activated to heat the air. During the dry season, spray humidifiers are also used to humidify the air.
System of Integrated Air Conditioning The constructed air conditioner in a unitary air conditioning system is
installed in or near the room to be conditioned. The most popular type of one-room conditioner is a unitary system,
which sits in a window or wall opening and has internal controls. A fan blows interior air over the evaporator, which
cools it. As a second fan blows over the conditioner, the outside air is heated. Heat is provided from the room and
discharged to the environment throughout this procedure. A big house or building may contain numerous such units,
allowing each area to be cooled independently. There are two types of unitary air conditioning systems.
1. windows unit: These air conditioners have a tiny capacity of 1TR to 3TR and are referred to through a window
or wall. They are solely used to condition the air in one room. When the room is larger, two or more units are
employed.
2. Vertical Packed Units/PTAC systems or vertical packed units: These air conditioners have a larger capacity of
5 to 20TR and are located near the space to be conditioned. They can be used in a restaurant, a bank, or a
small office. PTACS/wall split air conditioning systems/ductless systems used in hotels feature two independent
units, the evaporative unit on the inside and the condensing unit on the outside, connected by tubing that runs
through the wall. This permits each room to be neighboring in its own right. In cold weather, a PTAC system
can be modified to offer heating, either directly with an electric strip, gas, or other heaters, or indirectly by
reversing the refrigerant flow to heat the inside and draw heat from the outer air, transforming the air into a
12

15. heat pump. While room air conditioning offers the most flexibility when it comes to cooling rooms, it is typically
more expensive than a central air conditioning system.
7.Central Air Conditioning System: the most common type of air conditioning system, used when cooling capacity
of 25TR or more is necessary. It is used when the air flow is greater than 300 m3/min or when multiple zones in a
structure require A/C. Air-conditioning is used in food preparation and processing sectors. Used in hospitals to offer
patients with comfortable surroundings. And many more industries, such as textiles, printing, photography, and so
on. A human being uses an air-conditioning system for commercial purposes. For example, at theaters, departmental
storerooms, and so on. Many modes of transportation, including cars, trains, aircraft, and ships, use air-conditioning
systems. This ensures that the passengers are comfortable. For a specific purpose, the air-conditioning system utilized
in television centers, computer centers, and museums.
3.1 Solved Examples
1. An air conditioning plant is required to supply 60m3
of air per minute at a DBT of 21o
C and 55% RH. The
outside air is at DBT of 28o
C and 60% RH. Determine the mass of water drained and capacity of the cooling
coil. Assume the air conditioning plant first to dehumidify and then to cool the air.
Solution Given v2 = 60m3
/min, td2 = 21o
C, ϕ2 = 55%, tdt = 28o
C, ϕ1 = 60%
Mass of water drained First of all, mark the initial condition of air at 28o
C dry bulb temperature and 60%
(a) (b) (c)
Figure 10: Solved Problem a. Comfort AC b. Industrial AC c. Winter AC
relative humidity on the 1h psychrometric chart as point 1 , as shown in Fig. 10a.. Now mark the final condition
of air at 21o
C dry bulb temperature % and 55% relative humidity as point 2 . From the psychrometric chart,
we find that Specific humidity of air at point 1 , W1 = 0.0141kg/kg of dry air Specific humidity of air at point
2 , W2 = 0.0084kg/kg of dry air and specific volume of air at point 2 , vs2 = 0.845m3/kg of dry air We know
that mass of air circulated, ma = v2
vn
= 60
0.845 = 71kg/min ∴ Mass of water drained
= ma (W1−W2) = 71(0.0142 − 0.0084) = 0.412kg/min
= 0.412 × 60 = 24.72kg/h Ans.
Capacity of the cooling coil From the psychrometric chart, we find that Enthalpy of air at point 1 ,
h1 = 64.8kJ/kg of dry air
and enthalpy of air at point 2 ,
h2 = 42.4kJ/kg of dry air
∴ Capacity of the cooling coil
= ma (h1 − h2) = 71(64.8 − 42.4) = 1590.4kJ/min
= 1590.4/210 = 7.57 TR Ans.
2. Following data refers to an air conditioning system to be designed for an industrial process for hot and wet
climate: Outside conditions = 30o
C DBT and 75% RH Required inside conditions = 20o
C DBT and 60% RH
The required condition is to be achieved first by cooling and dehumidifying and then by heating. If 20 nil of air
is absorbed-by the plant every minute, find : 1. capacity of the cooling coil in tones of refrigeration; 2. capacity
of the heating coil in kW; 3. amount of water removed per hour; and 4. By-pass factor of the heating coil, if its
surface temperature is 35C
.
Solution: Given: tdt = 30o
C, ϕ1 = 75%, td3 = 20o
C, ϕ3 = 60%; vl = 20m3
/min, td4 = 35o
C 1.Capacity of the
cooling coil in tones of refrigeration First of all, mark the initial condition of air at 30o
C dry bulb temperature
and 75% relative humidity on the psychrometric chart as point 1, as shown in Fig. 10 (b) . Then mark the final
13

16. condition of air at 20o
C dry bulb temperature and 60% relative humidity oil the chart as point 3.
Now locate the points 2’ and 2 on the saturation curve by drawing horizontal lines through points 1 and 3 as
shown in Fig. 2. On the chart, the process 1-2’ represents the sensible cooling. 2’-2represents dehumidifying
process and 2-3 represents the sensible heating process. From the psychrometric chart, we find that the specific
volume of air at point 1. vx1 = 0.886m3
/ kg of dry air ; Enthalpy of air at point 1, h1 = 81.8kJ/kg of dry air
and enthalpy of air at point 2, h2 = 34.2kJ/kg of dry air
ma =
v1
vs1
=
20
0.866
= 22.6kg/min
∴ Capacity of the cooling coil
= ma (h1 − h2) = 22.6(81.8 − 34.2) = 1075.76kJ/min
= 1075.76/210 = 5.1 TR Ans.
2. Capacity of the heating coll in kW From the psychrometric ehart, we find that enthalpy of air at point 3 ,
h3 = 42.6kJ/kg of dry air
∴ Capacity of the heating coil
− ma (h1 − h2) − 22.6(42.6 − 34.2) − 189.84Id/min
− 189.84/60 − 3.16kW Ans.
3. Amount of water removed per hour From the psychrometric chart, we find that specific humidity of air at
point 1 ,
W1 = 0.0202kg/kg of dry air
and specific humidity of air at point 2.
W2 = 0.0088kg/kg of dry air
∴ Amount of water removed per hour
− ma (W1 − W2) − 22.6(0.0202 − 0.0088) − 0.258kg/min
= 0.258 × 60 = 15.48kg/h Ans.
3. The amount of air supplied to an air conditioned hall is 300m3
/min. The atmospheric conditions are 35o
C DBT
and 55% RH. The required conditions are 20o
C DBT and 60%RH. Find out the sensible heat and latent heat
removed from the air per minute. Also find sensible heat factor for the system.
Solution:Given v1 = 300m3
/min; tdt, = 35o
C, ϕ1 = 55%, td2 = 20o
C; ϕ2 = 60% First of all, mark the initial
condition of air at 35o
C dry bulb temperature and 55% relative humidity on the psychrometric chart at point
1, as shown in Fig. 10 (c). Now mark the final condition of air at 20o
C dry bulb temperature and 60% relative
humidity on the chart as point 2. Locate point 3 on the chart by drawing horizontal line through point 2 and
vertical line through point 1.
From the psychrometric chart, we find that specific volume of air at point 1, vs1 = 0.9m3
/kg of dry air ∴ Mass
of air supplied,Sensible heat removed from the air From the psychrometric chart, we find that enthalpy of air
at point 1, h1 = 85.8kJ/kgofdryair Enthalpy of air at point 2, h2 = 42.2kJ/kg of dry air and enthalpy of air
at point 3, h3 = 57.4kJ/kg of dry air We know that sensible heat removed from the air, SH = ma(h3 − h2) =
333.3(57.4 − 42.2) = 5066.2kJ/min Ans. Latent heat removed from the air We know that latent heat removed
from the air, LH = ma(h1 − h3) = 333.3(85.8 − 57.4) = 9465.7kJ/min Ans,
We know that sensible heal factor for the system SHF = SH
T otalHeat = SH
SH+LH , SHF = 5066.2
5066.2+9465.7 = 0.348
Ans
3.2 Review Questions
1. An air conditioning plant is to be designed for a small office for winter conditions with the following data: Outdoor
conditions = 10o
C DBT and 8o
C WBT, Required indoor conditions = 20o
CDBTand60% RH , Amount of air
circulation = 0.3m3
/min/person , Seating capacity of the office = 50 persons , The required condition is achieved
first by heating and then by adiabatic humidifying. Find: I. Heating capacity of the coil in kW and the surface
temperature, if the by-pass factor of the coil is 0.32 and 2. capacity of the humidifier.
2. Explain Various types and application areas of systems of Air conditioning
4 Load Calculation
Heating and cooling load calculations are performed to estimate the required capacity of heating and cooling systems
to keep the conditioned space at the appropriate temperature.
14

17. 4.1 Calculating heating and cooling loads
Heating load calculations are performed to predict the heat loss from the building during the winter in order to
determine the required heating capabilities. During the winter, the peak heating load usually comes before daybreak,
and the external conditions do not fluctuate greatly during the season. The heat load estimates are performed under
steady-state settings (no solar radiation and constant external conditions) while ignoring internal heat sources.
This cautious approach results in a minor overestimation of heating capacity. For more precision, the thermal capacity
of the walls and interior heat sources are included, which complicates the situation. The unsteady state processes are
used to estimate cooling loads since the peak cooling demand occurs throughout the day and the outside conditions
vary greatly during the day owing to solar radiation. Furthermore, all internal sources contribute to cooling loads,
and ignoring them would result in an overestimate of the needed cooling capacity and the likelihood of being unable
to sustain the required indoor conditions. As a result, cooling load estimates are fundamentally more difficult since
they entail solving unsteady equations with unsteady boundary conditions and interior heat sources.
There is a balance point for any building where the solar radiation (Qsolar) and internal heat generation rate (Qint)
exactly balance the heat losses from the building. As a result of the sensible heat balance equation, at the balanced
condition for the product of overall heat transfer coefficient and A is heat transfer area of the building,UA, given in
equation 13:
Qsolar + Qint = UA(Tin − Tout solving for outdoor temp. yields
Tout,bal = Tin −
(Qsolar + Qint)sensible
UA
(13)
When Tout Tout,bal, cooling of the building is required.
when Tout Tout,bal, then heating the building is required. when Tout = Tout,bal, no cooling or heating is required.
The balanced external temperature for residential buildings (with fewer inside heat sources) can range from 10 to
18o
C.This indicates that if the balanced outdoor temperature is 18o
C, then when the outdoor temperature reaches
18o
C, a cooling system is required. This means that buildings require cooling not only in the summer, but also in
the spring and fall. If the building is highly insulated (low UA) and/or internal loads are high, the balanced outdoor
temperature will decrease according to the energy balancing equation 13, resulting in an extended cooling season and
a shortened heating season.
Thus a smaller balanced outdoor temperature implies higher cooling requirements and smaller heatingrequirements,
and vice versa. For commercial buildings with large internal loads and relatively smaller heat transfer areas, the
balanced outdoor temperature can be as low as 2o
C, implying a lengthy cooling season and a small heatingseason.
If there are no internal heat sources and if the solar radiation is negligible, then from the heat balance equation,
Tout,bal = Tin, this implies that if the outside temperature exceeds the required inside temperature (say, 25o
C for
comfort) then there is a need for cooling otherwise there is a need for heating.
4.2 Methods for estimating cooling and heating loads
The heating/cooling load calculations entail a systematic, sequential approach for determining the needed system
capacity by accounting for all building energy flows. In practice, building loads are calculated using a range of
approaches ranging from simple rules of thumb to complicated Transfer Function approaches. Typical rules-of-thumb
procedures for cooling loads, for example, indicate the required cooling capacity based on floor space or occupancy.
Such guidelines are important for estimating the size and cost of equipment. The major conceptual flaw of rule-of-
thumb techniques is the assumption that building design makes no difference. As a result, the rules for a poorly
planned building are often the same. More accurate load estimation methods combine analytical methods with
empirical results derived from actual data, such as the use of Cooling Load Temperature Difference (CLTD) for
estimating fabric heat gain and Solar Heat Gain Factor (SHGF) for estimating heat transfer through fenestration.
These approaches are extensively utilized by air conditioning engineers because they produce reasonably accurate
findings and can be performed manually in a very short period of time. TFM (Transfer Function Method):Computers
are required for accuracy. They are, however, costly and time demanding, and are used to estimate cooling loads for
large commercial or institutional buildings.
4.2.1 Cooling Load Calculation
The amount of a building’s cooling demand ranges from zero to a maximum value. The design cooling load is a load
that is close to but not always the maximum magnitude. The design cooling load takes into account all of the loads in
a building under a certain set of assumptions. The following are the assumptions: A long-term statistical database is
used to select design outside conditions. The circumstances will not necessarily represent any specific year, but will be
representative of the building’s location. The solar radiation load on the building is assessed for clear sky conditions.
The occupancy of the facility is considered to be at full design capacity. All building equipment and appliances are
assumed to be in generally good working order. The cooling load of a building can be separated into two categories:
Heat carried through the building envelope (walls, roof, floor, windows, doors, etc.) is an example of an external load.
Heat generated by inhabitants, equipment, and lighting constitutes internal load. Any building’s overall cooling load
includes both obvious and latent load components. The sensible load influences dry bulb temperature, whereas the
15

18. latent load influences conditioned space moisture content. Buildings can also be characterized as either outwardly
or internally loaded. The cooling load on an externally loaded building is mostly due to heat transfer between the
surroundings and the internal conditioned area.
Because the daily environmental conditions vary, so does the external cooling demand. Internal heat generating
sources such as humans, appliances, or processes account for the majority of the cooling load in a building. In general,
internal heat sources may remain relatively constant, and because heat transfer from changeable surroundings is
substantially smaller than internal heat sources, the cooling load of an internally loaded building may remain quite
constant. Knowing whether the building is externally or internally loaded is critical for good system design. Heat
transfer by conduction through the building walls, roof, floor, doors, and so on, as well as heat transfer by radiation
through fenestration such as windows and skylights, are examples of external loads. These are all examples of sensible
heat transfers. In addition to these, the external load includes heat transfer owing to infiltration, which includes both
sensible and latent components. The heat transfer caused by ventilation is a load on the system, not the structure.
Internal loads include sensible and latent heat transmission from inhabitants, products, processes, and appliances, as
well as sensible heat transfer from lighting and other equipment. Figure 11 below shows various components that
constitute the cooling load on a building. [3]
Figure 11: Cooling load components
Table 2 below shows typical data on required cooling capacities based on the floor area or application [4]. The
Table 2: Cooling Load Factor (CLF) for glass with interior shading and located in north latitudes
Solar Time h
Direction of Sunlight window facing
N NE E SE S SW W NW Horiz.
6 0.73 0.56 0.47 0.30 0.09 0.07 0.06 0.07 0.12
7 0.66 0.76 0.72 0.57 0.16 0.11 0.09 0.11 0.27
8 0.65 0.74 0.80 0.74 0.23 0.14 0.11 0.14 0.44
9 0.73 0.58 0.76 0.81 0.38 0.16 0.13 0.17 0.59
10 0.80 0.37 0.62 0.79 0.58 0.19 0.15 0.19 0.72
11 0.86 0.29 0.41 0.68 0.75 0.22 0.16 0.20 0.81
12 0.89 0.27 0.27 0.49 0.83 0.38 0.17 0.21 0.85
13 0.89 0.26 0.26 0.33 0.80 0.59 0.31 0.22 0.85
14 0.86 0.24 0.24 0.28 0.68 0.75 0.53 0.30 0.81
15 0.82 0.22 0.22 0.25 0.50 0.83 0.72 0.52 0.71
16 0.75 0.20 0.20 0.22 0.35 0.81 0.82 0.73 0.58
17 0.78 0.16 0.16 0.18 0.27 0.69 0.81 0.82 0.42
18 0.91 0.12 0.12 0.13 0.19 0.45 0.61 0.69 0.25
cooling load calculations are carried out based on the CLTD/CLF method suggested by ASHRAE. Estimation of
external loads: Heat transfer through opaque surfaces: This is a sensible heat transfer prrocess. The heat transfer
rate through opaque surfaces such as walls, roof, floor, doors etc. and it is given by eq. 14. [5]
QOpaque = UA × CLTD (14)
where U is the overall heat transfer coefficient and A is the heat transfer area of the conditioned space’s surface. The
cooling load temperature differential is denoted by CLTD. CLTD must be taken from the CLTD tables for sunny
surfaces. If the actual conditions differ from those on which the CLTD tables are generated, the values received from
16

19. the table must be adjusted.
The CLTD value is simply equal to the temperature differential across the wall or roof for surfaces that are not
illuminated or have little thermal mass (such as doors). For example, the CLTD value for external doors is just the
difference between the design outside and indoor dry bulb temperatures. Tout − Tin
The CLTD of the interior walls of interior air conditioned rooms surrounded by non-air conditioned spaces is equal to
the temperature difference between the surrounding non-air conditioned space and the conditioned space. Obviously,
if an air conditioned room is surrounded by other air conditioned rooms, all of which are set to the same temperature,
the CLTD values of the interior room’s walls will be zero.
CLTD for floor and roof values with a false ceiling could be challenging. When determining CLTD for floors that
stand on the ground, use the temperature of the ground. The ground temperature, on the other hand, is affected
by location and time. For evaluating heat transfer via ground, ASHRAE recommends appropriate temperature
difference values. If the floor is supported by a basement or another room’s roof, the CLTD for the floor is the
temperature differential across the floor (that is, the temperature difference between the basement or room below and
the conditioned space). b) Heat transfer through fenestration: Heat transfer through a transparent surface, such as a
window, involves conduction heat transfer due to temperature differences across the window and solar radiation heat
transmission through the window. The heat transfer due to solar radiation through the window is given by:
Qtrans = AUnshaded × SHGFmax × SC × CLF (15)
where Aunshaded is the area exposed to solar radiation, SHGFmax and SC are the maximum Solar Heat Gain Factor
and Shading Coefficient, respectively, and CLF is the Cooling Load Factor. The un-shaded area has to be obtained
from the dimensions of the external shade and solar geometry. SHGFmax and SC are obtained from ASHRAE tables
based on the orientation of the window, location , month of the year and the type of glass and internal shading device.
Cooling Load Factor/CLF: The Cooling Load Factor (CLF) compensates for the fact that all radiant energy
entering the conditioned area at a given time does not immediately become a part of the cooling load1. When solar
radiation enters the conditioned space, only a small amount of it is instantly absorbed by the air particles, resulting
in a minor temperature change. The majority of the radiation is absorbed by the internal surfaces, which include
the ceiling, floor, internal walls, furniture, and so on. Because of the high but finite thermal capacity of the roof,
floor, walls, and so on, their temperature rises slowly as a result of solar radiation absorption. Only the fraction of
solar radiation transported to the air in the conditioned room creates a strain on the building; heat transferred to the
outside does not contribute to the cooling load. As can be observed, radiation heat transfer introduces a time lag as
well as a decrement factor based on the dynamic properties of the surfaces. c) Heat transfer due to infiltration: Heat
transfer due to infiltration consists of both sensible as well as latent components. The sensible heat transfer rate due
to infiltration is given by equation 16
Qs,inf = moCp,m(To − Ti) = vρocp,m(To − Ti) (16)
where V̇o is the infiltration rate in m3
/s , ρo and cp,m are the density and specific heat of the moist, infiltrated air,
respectively. T0 and Ti are the outdoor and indoor dry bulb temperatures. The latent heat transfer rate due to
infiltration is given by Eq. 17
Ql,inf = mohi,g(Wo − Wi) = vρohi,g(Wo − Wi) (17)
where hfg is the latent heat of vaporization of water, Wo and Wi are the outdoor and indoor humidity ratio, respectively.
The infiltration rate depends upon several factors such as the tightness of the building that includes the walls, windows,
doors etc and the prevailing wind speed and direction. The infiltration rate is obtained by using either the air change
method or the crack method (i) The infiltration rate by air change method is given by:
Vo = ACH × V/3600inm3
/s (18)
where ACH is the number of air changes per hour and V is the gross volume of the conditioned space in m3
. Normally
the ACH value varies from 0.5 ACH for tight and well-sealed buildings to about 2.0 for loose and poorly sealed
buildings. For modern buildings the ACH value may be as low as 0.2 ACH.
(ii) The infiltration rate by the crack method is given by:
Vo = A.C.H∆Pn
inm3
/s (19)
where A is the effective leakage area of the cracks, C is a flow coefficient which depends on the type of the crack
and the nature of the flow in the crack, ∆P is the difference between outside and inside pressure (Po-Pi) and n is an
exponent whose value depends on the nature of the flow in the crack. The value of n varies between 0.4 to 1.0, i.e.,
0.4 ≤ n ≤ 1.0.
The pressure difference ∆P arises due to pressure difference due to the wind (∆Pwind), pressure difference due to the
stack effect (∆Pstack) and pressure difference due to building pressurization (∆Pbld).
∆P = ∆Pwind + ∆Pstack + ∆Pbld (20)
Representative infiltration rate figures for various types of windows, doors, walls, and so on have been measured
and are published in tabular form in air conditioning design handbooks. d) Variable external loads:-In addition to
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20. the aforementioned loads, if the cooling coil has a positive by-pass factor (BPF ¿ 0), some ventilation air enters the
conditioned space directly, becoming part of the building cooling load. Because of the bypassed ventilation air, the
sensible and latent heat transfer rates, and can be calculated using equtations 16 and 17 substituting the ventilation
rate and BPF is the by-pass factor of the cooling coil. Furthermore, sensible and latent heat transmission to the
building happens as a result of heat transfer and air leakage in the supply ducts. Depending on the specifics of the
supply air ducts, a safety factor is frequently specified to accommodate for this.
If the supply duct has a supply air fan with a motor, the power input to the fan forms part of the building’s external
sensible load. If the duct contains an electric motor that drives the fan, the efficiency of the fan motor must also be
considered when estimating the cooling load. Most of the time, the power input to the fan is unknown at this stage
since the volume of supplied air required is unknown. To account for this aspect, it is first assumed that the supply
fan contributes around 5% of the room’s sensible cooling load, and cooling loads are then assessed. This value is then
corrected at the end when the actual fan selection done
Estimating the internal load:Internal loads include: loads caused by inhabitants, loads caused by illumination,
loads caused by equipment and appliances, and loads caused by products stored or processes done in the conditioned
space. Internal cooling load owing to occupants: The internal cooling load due to inhabitants includes both sensible
and latent heat components. The rate of sensible and latent heat transmission is mostly determined by the population
and activity level of the inhabitants as seen in eq. 21
Qs,occ = no.people × senible heat gain per person × CLF
Ql,occ = no.people × latent heat gain per person
(21)
The table 3 below depicts typical values of total heat gain from occupants as well as sensible heat gain fraction as a
function of activity in an air-conditioned environment. When the conditioned space temperature rises, the perceptible
percentage of total heat gain falls while the latent heat gain rises, and vice versa.
Table 3: Total heat gain, sensible heat gain fraction from occupants
Activity Total heat gain, W Sensible heat gain fraction
Sleeping 70 0.75
Seated, quiet 100 0.60
Standing 150 0.50
Walking @ 3.5 kmph 305 0.35
Office work 150 0.55
Teaching 175 0.50
Industrial work 300 to 600 0.35
Qs,light = Installed wattage × Usage Factor × CLF
Ql,Appliences = Installed Wattage × Usage Factor × CLF general
Ql,Appliences = Installed Wattage × Latent fraction for appliances in the Table 4
(22)
A typical ballast factor value of 1.25 for lighting is taken for fluorescent lights, while it is equal to 1.0 for incandescent
lamps. Other equipment, such as computers and printers, have a sensible heat transfer load that is assessed based
Table 4: Table showing typical load of various types of appliances.
Appliance Sensible load, W Latent load, W Total load, W
Coffee brewer, 0.5 gallons 265 65 330
Coffee warmer, 0.5 gallons 71 27 98
Toaster, 360 slices/h 1500 382 1882
Food warmer/m² plate area 1150 1150 2300
on the rated power consumption. Because the radiative heat transfer from these devices is often insignificant due to
lower operating temperatures, the CLF value for these equipments can be considered as 1.0. When the equipment is
powered by electric motors that are also kept inside the conditioned environment, the efficiency of the electric motor
must be considered.
Though the equations appear to make estimating cooling load due to appliance and equipment simple, a large amount
of uncertainty is introduced due to the usage factor and the difference between rated (nameplate) power consumption
at full loads and actual power consumption at part loads. If the conditioned area is used to store products (for example,
cold storage) or to carry out specific procedures, the sensible and latent heat emitted by these specific products and/or
processes must be included to the internal cooling loads.
Air conditioning and refrigeration handbooks contain information on the sensible and latent heat release rates of a
wide range of live and dead items typically stored in cold storage. These tables can be used to estimate the needed
cooling capacity of cold storages.
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21. Using the aforementioned equations, one can calculate the sensible (Qs,r), latent (Ql,r), and total cooling load (Qt,r)
of the buildings. Because the load due to illuminated surfaces varies with solar time, it is advisable to compute the
cooling loads at different solar periods and use the maximum load to estimate system capacity. The building’s Room
Sensible Heat Factor (RSHF) can be calculated using the sensible and total cooling loads. Estimation of the system’s
cooling capacity: To determine the system’s needed cooling capacity, take into consideration the sensible and latent
loads owing to ventilation, leakage losses in the return air ducts, and heat added due to the return air fan (if any).
Load on the system as a result of vented air: The by-pass factor of the cooling coil is X. The cooling load on the coil
as a result of sensible heat transfer from the vented air is then provided by:
Qs, vent = ṁvent (1 − X) · cp,m (T0 − Ti) = V̇vent ρ0(1 − X) · cp,m (T0 − Ti) (23)
where Where mvent and vvent are the mass and volumetric flow rates of the ventilated air and X is the by-pass factor
of the coil. The amount of sensible and latent heat transfer rates caused by air leakage from or into the system is
determined by the efficacy of the sealing supplied, as well as the condition of the external and return air. Because
the load due to the return air duct, including the return air fan (Qreturn duct), is unknown a priori, an initial value is
assumed and calculations are undertaken. At the end, this figure 12 is adjusted to account for actual leakage losses
and return fan power usage.
Load on coil due to leakage: The following figure 12 shows a typical summer air conditioning system with a cooling
coil of none-zero by-pass factor. Now, the total sensible load on the coil (Qs,c) is calculated by adding the total sensible
Figure 12: A typical summer air conditioning system having a cooling coil with a by-pass factor of none
load on the building (Qs,r), ventilation sensible load (Qs,vent), and return air duct and fan sensible load (Qs,retrunduct)
as seen in Equation. Similarly, the total latent load on the coil (Ql,c) is calculated by adding the total latent load on
the building (Ql.r), ventilation latent load (Ql,vent), and return air duct and fan latent load (Ql,retrunduct), that is:
Qs,c = Qs,r + Qs, vent + Qs, return duct Total Sensible Cooling Load
Q1,C = QI,J + Q1, vent + Q1 Ireturn duct Total Latent Cooling Load
(24)
Finally the required cooling capacity of the system which is equal to the total load on the coil by adding up the
two. From the given data, one may calculate the sensible heat factor for the coil (CSHF), draw the process line on
the psychrometric chart, and obtain the needed coil Apparatus Dew Point Temperature (coil ADP). In the end, a
reasonable safety factor is usually utilized to account for uncertainties in occupants, equipment, external infiltration,
external circumstances, and so on.
Heating Load calculation: For evaluating building heating demands, traditional steady-state circumstances are
assumed, and interior heat sources are ignored. The process for calculating heating load thus becomes pretty straight-
forward. Only sensible and latent heat losses from building walls, roof, ground, windows, and doors due to infiltration
and ventilation must be estimated. Equations identical to those used for cooling load estimates are utilized, but the
CLTD values are simply replaced by the design temperature differential between the conditioned interior and the
outdoors.
Because a constant state is assumed, the system’s necessary heating capacity is equal to the entire heat loss from
the building. As a result, depending on the unique circumstance, a suitable and economically justified strategy for
estimating heating loads must be chosen.
4.3 Solved Example
1. A building has a U-value of 0.5W/m2
K and a total exposed surface area of 384m2
. The building is subjected
to an external load (only sensible) of 2 kW and an internal load of 1.2 kW (sensible). If the required internal
temperature is 25o
C, state whether a cooling system is required or a heating system is required when the external
19

22. temperature is 3o
C. How the results will change, if the U-value of the building is reduced to 0.36W/m.K
Solution Ans.: From energy balance,
Tout,bal = Tin −
(Qsolar + Qint )sensible
UA
= 25 −
(2 + 1.2) × 1000
0.5 × 384
= 8.33o
C
Since the outdoor temperature at balance point is greater than the external temperature (Text Tout bal ); the
building requires heating (Ans.) When the U-value of the building is reduced to 0.36W/m.K, the new balanced
outdoor temperature is given by:
Tout,bal = Tin −
(Qsolar + Qint )sensible
UA
= 25 −
(2 + 1.2) × 1000
0.36 × 384
= 1.85o
C
Since now the outdoor temperature at balance point is smaller than the external temperature (Text Tout bal );
the building now requires cooling (Ans.) The above example shows that adding more insulation to a building
extends the cooling season and reduces the heating season.
2. An air conditioned room that stands on a well ventilated basement measures 3 m wide, 3 m high and 6 m deep.
One of the two 3 m walls faces west and contains a double glazed glass window of size 1.5 m by 1.5 m, mounted
flush with the wall with no external shading. There are no heat gains through the walls other than the one
facing west. Calculate the sensible, latent and total heat gains on the room, room sensible heat factor from the
following information. What is the required cooling capacity
Inside conditions : 25o
C dry bulb, 50 percent RH, Outside conditions : 43o
C dry bulb, 24o
C wet bulb, U-
value for wall : 1.78W/m2
.K, U-value for roof : 1.316W/m2
.K, U-value for floor : 1.2W/m2
.K, Effective
Temp. Difference (ETD) for wall: 25o
C, Effective Temp. Difference (ETD) for roof: 30o
C, U-value for glass
; 3.12W/m2
.K, Solar Heat Gain (SHG) of glass: 300W/m2
, Internal Shading Coefficient (SC) of glass: 0.86,
occumpancy : 4 (90 W sensible heat/person) (40 W latent heat/person), Lighting load : 33W/m2
of floor area,
Appliance load : 600WSensible + 300Wlatent, Infiltration : 0.5 Air Changes per Hour, Barometric pressure : 101
kPa
solution Ans.: From psychrometric chart, For the inside conditions of 25o
C dry bulb, 50 percent RH: Wi =
9, 9167 × 10 − 3kgw/kgda For the outside conditions of 43o
C dry bulb, 24o
C wet bulb: Wo = 0.0107kgw/kg,
density of dry air = 1.095kg/m3
External loads:
(a) Heat transfer rate through the walls: Since only west wall measuring 3m x 3m with a glass windows of
1.5m × 1.5m is exposed; the heat transfer rate through this wall is given by:
Qwall = Uwall × AwallETDwall = 1.78 × (9 − 2.25) × 25 = 300.38WSensible
(b) Heat transfer rate through roof:
Qroof = Uroof × Aroof ETDroof = 1.316 × 18 × 30 = 710.6WSensible
(c) Heat transfer rate through floor: Since the room stands on a well-ventilated basement, we can assume the
conditions in the basement to be same as that of the outside (i.e., 43o
C dry bulb and 24o
C wet bulb), since
the floor is not exposed to solar radiation, the driving temperature difference for the roof is the temperature
difference between the outdoor and indoor, hence:
Qfloor = Ufloor × AfloorETDfloor = 1.2 × 18 × 18 = 388.8WSensible
(d) Heat transfer rate through glass: This consists of the radiative as well as conductive components. Since no
information is available on the value of CLF, it is taken as 1.0. Hence the total heat transfer rate through the
glass window is given by:
Qglass = Aglass × [Uglass(To − Ti) + SGHmax × SC] = 2.25[3.12 × 18 + 300 × 0.86] = 706.9WSensible
(e) Heat transfer due to infiltration: The infiltration rate is 0.5 ACH, converting this into mass flow rate, the
infiltration rate in kg/s is given by:
minf = ρ × (ACH × volumeoftheroom)/3600 = 1.095 × (0.5 × 3 × 3 × 6)/3600
minf = 8.2125x10 − 3kg/s
Sensible heat transfer rate due to infiltration, Qs,inf;
Qs,inf = minf cp,m(To − Ti) = 8.2125 × 10 − 3 × 1021.6 × (43 − 25) = 151WSensible
Latent heat transfer rate due to infiltration, Ql,inf :
Ql,inf = minf hfg(Wo − Wi)
= 8.8125 × 10 − 3 × 2501 × 103(0.0107 − 0.0099)
= 16.4Wsensible
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23. Internal Loads
Load due to occupants: The sensible and latent load due to occupants are:
Qs,occ = no.ofoccupants × SHG = 4x90 = 360WQl,occ = no.ofoccupants × LHG = 4x40 = 160W
b) Load due to lighting: Assuming a CLF value of 1.0, the load due to lighting is:
Qlights = 33xfloorarea = 33x18 = 594WSensible
c) Load due to appliance:
Qs,app = 600WSensibleQl,app = 300WLatent
Total sensible and latent loads are obtained by summing-up all the sensible and latent load components (both
external as well as internal) as:
Qs,total = 300.38+710.6+388.8+706.9+151+360+594+600 = 3811.68WAns.Ql,total = 16.4+160+300 = 476.4WAns.
Total load on the building is:
Qtotal = Qs,total + Ql,total = 3811.68 + 476.4 = 4288.08WAns.
Room Sensible Heat Factor (RSHF) is given by:
RSHF = Qs,total/Qtotal = 3811.68/4288.08 = 0.889Ans.
To calculate the required cooling capacity, one has to know the losses in return air ducts. Ventilation may be
neglected as the infiltration can take care of the small ventilation requirement. Hence using a safety factor of
1.25, the required cooling capacity is: Required cooling capacity
= 4288.08x1.25 = 5360.1W ≡ 1.5TR
4.4 Multiple Choice Review Questions
For Similar Questions with answers, please visit the website [6]
1. Which of the following statements are TRUE?
(a) Steady state methods are justified in case of heatingload calculations as the outside temperatures during
winter are normally very low
(b) Steady state methods are justified in case of heatingload calculations as the peak load normally ocurs before
sunrise
(c) Steady state methods are justified in case of heatingload calculations as the outside temperature variation
is normally low during winter months
(d) Neglecting internal heat sources while calculating heatingloads underestimates the required capacity
2. Which of the following statements are TRUE?
(a) Methods based on rules-of-thumb are not always useful as they are not based on practical systems
(b) Methods based on rules-of-thumb are not always useful as they do not distinguish between a good building
design and a bad building design
(c) Methods based on Transfer Function Method are not always useful as they do not yield accurate results
(d) Methods based on Transfer Function Method are not always useful as they are complex and time consuming
3. Which of the following statements are TRUE?
(a) An internally loaded building requires a system with variable cooling capacity
(b) An externally loaded building requires a system with variable cooling capacity
(c) An auditorium is a good example of an internally loaded building
(d) A residence is a good example of an internally loaded building
4. Which of the following statements are TRUE?
(a) External loads consist of only sensible components, where as internal loads consist of both sensible and
latent components
(b) Both external and internal loads consist of sensible as well as latent components
21

24. (c) Fabric heat gain consists of both sensible and latent components
(d) Heat transfer due to occumpancy consists of both sensible and latent components
5. Which of the following statements are TRUE?
(a) Infiltration load is a part of the building load
(b) Infiltration load is not a part of the building load
(c) Infiltration rate increases as the pressure difference across the building decreases
(d) Infiltration rate is uncontrollable
References
[1] Guy F Hundy, Albert Runcorn Trott, and TC Welch. Refrigeration and air-conditioning. Butterworth-Heinemann,
2008.
[2] Y Cengel and Transfer Mass Heat. “A practical approach on Heat and Mass Transfer”. In: Heat and Mass Transfer
(2003).
[3] RS Khurmi and JK Gupta. Textbook of refrigeration and air conditioning. S. Chand Publishing, 2008.
[4] William Rudoy and Joseph F Cuba. Cooling and heating load calculation manual. Vol. 158. ASHRAE, 1979.
[5] Bereket Asgedom Nigusse. Improvements to the radiant time series method cooling load calculation procedure.
Oklahoma State University, 2007.
[6] Sanfoundry. Refrigeration and Air Conditioning MCQ (Multiple Choice Questions). 2022. url: https://www.
sanfoundry.com/1000-refrigeration-air-conditioning-questions-answers/.
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