cooling load calculation

DESIGN OF AIR CONDITIONING SYSTEM FOR CAD LAB

B.N. COLLEGE OF ENGINEERING, PUSAD Page 1

Chapter 1

Introduction

Refrigeration may be defined as the process of achieving and maintaining a temperature

below that of the surroundings, the aim being to cool some product or space to the

required temperature. One of the most important applications of refrigeration has been

the preservation of perishable food products by storing them at low temperatures.

Refrigeration systems are also used extensively for providing thermal comfort to human

beings by means of air conditioning. Air Conditioning refers to the treatment of air so as

to simultaneously control its temperature, moisture content, cleanliness, odour and

circulation, as required by occupants, a process, or products in the space. The subject of

refrigeration and air conditioning has evolved out of human need for food and comfort,

and its history dates back to centuries. The development of refrigeration and air

conditioning industry depended to a large extent on the development of refrigerants to

suit various applications and the development of various system components. At present

the industry is dominated by the vapour compression refrigeration systems, even though

the vapour absorption systems have also been developed commercially. Refrigeration and

air conditioning involves various processes such as compression, expansion, cooling,

heating, humidification, de-humidification, air purification, air distribution etc. In all

these processes, there is an exchange of mass, momentum and energy.

The primary function of an air conditioning system is to maintain the conditioned

space at required temperature, moisture content with due attention towards the air motion,

air quality and noise. The required conditions are decided by the end use of the

conditioned space, e.g. for providing thermal comfort to the occupants as in comfort air

conditioning applications, for providing suitable conditions for a process or for

manufacturing a product as in industrial air conditioning applications etc. The reason

behind carrying out cooling and heating load calculations is to ensure that the cooling and

heating equipment designed or selected serves the intended purpose of maintaining the

required conditions in the conditioned space. Design and/or selection of cooling and

heating systems involve decisions regarding the required capacity of the equipment


B.N.COLLEGE OF ENGINEERING, PUSAD Page 2

selected, type of the equipment etc. By carrying out cooling and heating load calculations

one can estimate the capacity that will be required for various air conditioning equipment.

For carrying out load calculations it is essential to have knowledge of various energy

transfers that take place across the conditioned space, which will influence the required

capacity of the air conditioning equipment. Cooling and heating load calculations involve

a systematic step-wise procedure by following which one can estimate the various

individual energy flows and finally the total energy flow across an air conditioned

building.



Chapter 2

Thermodynamic cycles

2.1 Gas Cycles:

In a typical gas cycle, the working fluid (a gas) does not undergo phase change;

consequently the operating cycle will be away from the vapour dome. In gas cycles, heat

rejection and refrigeration take place as the gas undergoes sensible cooling and heating.

2.2 Vapour Cycles:

In a vapour cycle the working fluid undergoes phase change and refrigeration effect is

due to the vaporization of refrigerant liquid. If the refrigerant is a pure substance then its

temperature remains constant during the phase change processes. However, if a zeotropic

mixture is used as a refrigerant, then there will be a temperature glide during vaporization

and condensation. Since the refrigeration effect is produced during phase change, large

amount of heat (latent heat) can be transferred per kilogram of refrigerant at a near

constant temperature. Hence, the required mass flow rates for a given refrigeration

capacity will be much smaller compared to a gas cycle. Vapour cycles can be subdivided

into vapour compression systems, vapour absorption systems, vapour jet systems etc.

Among these the vapour compression refrigeration systems are predominant.

1. Vapour Compression System:

As mentioned, vapour compression refrigeration systems are the most commonly used

among all refrigeration systems. As the name implies, these systems belong to the general

class of vapour cycles, wherein the working fluid (refrigerant) undergoes phase change at

least during one process.



Fig. 2.1 Vapour Compression Refrigeration System

In a vapour compression refrigeration system, refrigeration is obtained as the

refrigerant evaporates at low temperatures. The input to the system is in the form of

mechanical energy required to run the compressor. Hence these systems are also called as

mechanical refrigeration systems. Vapour compression refrigeration systems are

available to suit almost all applications with the refrigeration capacities ranging from few

Watts to few megawatts. A wide variety of refrigerants can be used in these systems to

suit different applications, capacities etc. The actual vapour compression cycle is based

on Evans-Perkins cycle, which is also called as reverse Rankine cycle.

Figure 2.1 shows the basic components of a vapour compression refrigeration

system which consists of an evaporator, compressor, condenser and an expansion valve.

The refrigeration effect is obtained in the cold region as heat is extracted by the

vaporization of refrigerant in the evaporator. The refrigerant vapour from the evaporator

is compressed in the compressor to a high pressure at which its saturation temperature is

greater than the ambient or any other heat sink. Hence when the high pressure, high

temperature refrigerant flows through the condenser, condensation of the vapour into

liquid takes place by heat rejection to the heat sink. To complete the cycle, the high



pressure liquid is made to flow through an expansion valve. In the expansion valve the

pressure and temperature of the refrigerant decrease. This low pressure and low

temperature refrigerant vapour evaporates in the evaporator taking heat from the cold

region. It should be observed that the system operates on a closed cycle. The system

requires input in the form of mechanical work. It extracts heat from a cold space and

rejects heat to a high temperature heat sink.

A refrigeration system can also be used as a heat pump, in which the useful output

is the high temperature heat rejected at the condenser. Alternatively, a refrigeration

system can be used for providing cooling in summer and heating in winter. Such systems

have been built and are available now.



Chapter 3

Main Components of Air-conditioning system:

1. Compressor

2. Condenser

3. Expansion device

4. Evaporator

1. Compressor:

A compressor is the most important and often the costliest component (typically 30 to 40

percent of total cost) of any vapour compression refrigeration system (VCRS). The

function of a compressor in a VCRS is to continuously draw the refrigerant vapour from

the evaporator, so that a low pressure and low temperature can be maintained in the

evaporator at which the refrigerant can boil extracting heat from the refrigerated space.

The compressor then has to raise the pressure of the refrigerant to a level at which it can

condense by rejecting heat to the cooling medium in the condenser.

2. Condensers:

In condensers the refrigerant vapour condenses by rejecting heat to an external fluid,

which acts as a heat sink. Normally, the external fluid does not undergo any phase

change, except in some special cases such as in cascade condensers, where the external

fluid (another refrigerant) evaporates. Next to compressors, proper design and selection

of condensers is very important for satisfactory performance of any refrigeration system.

3. Expansion device:

An expansion device serves normally two purposes:

One is the thermodynamic function of expanding the liquid refrigerant from the

condenser pressure to the evaporator pressure. The other is the control function which

may involve the supply of the liquid to the evaporator at the rate at which it is evaporated.



4. Evaporator:

An evaporator, like condenser is also a heat exchanger. In an evaporator, the refrigerant

boils or evaporates and in doing so absorbs heat from the substance being refrigerated.

The name evaporator refers to the evaporation process occurring in the heat exchanger.



Chapter 4

Human Comfort

According to the American Society of heating, Refrigeration and Air Conditioning

Engineers, Air conditioning is the process of treating air so as to control simultaneously

its temperature, humidity, cleanliness and distribution to meet the requirements of

conditioned space. Air Conditioning is often used to improve an industrial process or to

maintain human comfort. In an industrial system, the conditions to be maintained are

determined by the nature of the process or material being handled. In comfort system,

however, conditions are determined by the requirements of the human body.

4.1 Body Comfort

What is it that makes a person feel hot or cold? The human body burns food to provide

heat and energy in a process called metabolism, in much the same way an automobile

engine burns gasoline, providing heat and energy. The excess heat we generate must be

given off from our body at a rate necessary to maintain our normal temperature of 98.60F

or 370C, this means when the surrounding temperature is higher than 37 deg C body

receives heat and when temperature of surrounding is less than 37 deg c body is rejecting

heat. Thermodynamically speaking, the ideal human comfort exists when rate of heat

production becomes equal to heat loss.

This heat transfer takes place constantly, every second, every day of the year, in

three ways by convection, by radiation and by evaporation.

1. Convection

When heat is given off by convection, the air close to our body becomes warmer

than the air away from the body. Since warm air is lighter than cool air, it floats upward.

As it does, it is replaced by the cooler air. As this cooler air absorbs body heat, it too

floats upward.



2. Radiation

The second way the body loses heat is by radiation. Heat radiates directly from the

body to any cooler object, just as the rays of the sun travel through space to warm the

surface of the earth. Heat may be lost from the bodys skin to wall, ceiling, or any object

which is cooler than the body.

3. Evaporation

Evaporation is the third way the body gives off heat. Moisture or perspiration is

discharged through the pores of the skin. As this moisture evaporates it absorbs heat from

the body. In other words, it cools the body by transferring body heat to the surrounding

air. One can readily feel the effect of evaporation by rubbing alcohol on the skin. Because

alcohol vaporizes at a lower temperature than perspiration, the body feels cooler.

Evaporation from the body goes on constantly whether or not we sense it. When drops of

perspiration can be seen, the body is producing more heat than it can reject by

evaporation. This may occur when the moisture content of the air, in other words, the

relative humidity becomes too high for the air to accept water vapor at the rate needed.

All three methods of giving off heat, convection, radiation and evaporation are

normally used at the same time. However, depending on surrounding conditions, one

method may be called upon to do a major share of the job.

Temperature, relative humidity, and air motion are three conditions that affect the

bodys ability to reject heat. Changes in each of these surrounding conditions will speed

up or slow down convection, radiation, or evaporation.

1. Temperature

Heat always flows from a place of higher temperature to one of lower temperature.

The greater the temperature difference, the faster is the flow of heat. If the difference is

too great, the body may loose heat more rapidly than it should; discomfort is the result,

we feel cold. Conversely, higher the air temperature, slower is the rate of heat transfer.

As the air temperature approaches body temperature, the body loses heat less rapidly



through convection. If the heat cant be dissipated, we start to feel hot. Thus air

temperature has an important effect on comfort.

Fig4.1 ASHRAE Comfort Zone



2. Relative Humidity

As mentioned earlier there are conditions other than surrounding temperatures that

affect the bodys ability to reject heat. Relative humidity is a measure of how much

moisture is in the air. It is an indication of the airs ability to absorb more moisture. For

example, when we say that the air has a relative humidity of 50%, we are saying that the

air at this specific temperature contains half the amount of moisture it can actually

absorb. Relative humidity of 100% indicates that the air contains all the moisture it can

hold at its present temperature. Air at 100% relative humidity is commonly called

saturated. Because relative humidity has a direct correlation to temperature, any time

the temperature is raised or lowered the relative humidity or moisture content of air will

also change. Cool air has less capacity to hold moisture than warm air. When surrounding

air has a low relative humidity, the body is able to give off more heat through

evaporation. Conversely, when the relative humidity is high, the body is less able to give

off heat.

Experience has shown that while the acceptable conditions of comfort vary from

person to person, temperatures somewhere between 72 and 780F and 50% relative

humidity are satisfactory to most.

3. Air Motion

Air motion is the third condition that affects the heat rejection from the body. One

result of air motion is an increase in the rate of evaporation. As we have seen,

evaporation depends on ability of the air to absorb moisture. Air moves across the body

forces away the saturated air, allowing more moisture to evaporate from the skin, cooling

it. If there were no air motion, the layer of air closest to the body would soon approach

saturation. Its relative humidity will increase to the point where it could no longer absorb

water vapor. At this point evaporation from body would almost stop and discomfort

would result.

Air motion also speeds up the convection process by removing the warn air close to

the body and carrying away the heat from walls, ceilings, and other surfaces surrounding

the body.



When outdoors, we must depend on changes of clothing and the whims of nature fro

comfort. If the combination of temperature, relative humidity, and air motion happens to

be just right, and if these conditions allow our bodies to reject excess heat and no more,

we feel comfortable. If the combination isnt right, we feel uncomfortable.

The definition given in ASHRAE standard 55 for human comfort is Thermal

Comfort is that condition of mind that expresses satisfaction with thermal environment.

Although this definition leaves open what is meant by condition of mind or satisfaction,

but it correctly emphasizes that the judgment of comfort is a cognitive process involving

many inputs influenced by physical, physiological, psychological, other processes.The

conscious mind appears to reach conclusions about thermal comfort and discomfort from

direct temperature and moisture sensations from the skin, deep body temperatures, and

the efforts necessary to regulate body temperatures.

In general, comfort occurs when body temperatures are held within narrow ranges,

skin moisture is low, and the physiological effort needed for regulation is minimized.

Some of the possible behavioral actions to reduce discomfort are altering clothing,

altering activity, changing posture or location, changing the thermostat setting, opening a

window, complaining or leaving the space.

Surprisingly, although regional climate conditions, living conditions, and cultures

differ widely throughout the world, the temperature that people choose for comfort under

like conditions of clothing, activity, humidity, and air movement has been found to be

very similar.



Chapter 5

Cooling Load Estimating

While designing air conditioning systems, the main objective is to maintain designed

conditions in the specified space. If the air conditioning calls for reducing temperature

and humidity, we need cooling. In order to do so, we have to pump out heat from the

space, to a place where we are able to reject this heat.

In order to pump out heat we need external energy to operate an air conditioning

system. In an air conditioning system design, the air, which is circulating in the space to

be cooled, picks up this heat and in turn gets heated. The warm air is then mixed with

certain quantity of fresh outside air and the mixture is then cooled in an air handler

housing cooling coil, where it gets cooled, dried, filtered and is then supplied to the space

for picking up heat again. The cycle thus continues. The purpose of load estimating is

therefore establishing correct quantity of air at a particular temperature which will offset

the heat load of a space to be cooled & maintain required space conditions is the primary

objective. In order to arrive at this quantity of air, we need to first estimate the cooling

load imposed on the space that needs to be neutralized. There are many

independent/dependant variables which influence cooling load & it is necessary to

understand these, if cooling load is to be estimated accurately.

5.1 Sources of Heat generation

We shall now look at the various sources that contribute to the cooling load.

These can be grouped in three categories

1. External loads

2. Internal loads

3. Other loads

4. Process loads



1. EXTERNAL LOADS- Originate from heat sources outside or external to

conditioned spaces

The main contributing sources are

a. Heat gain by conduction through walls, roofs, windows and through internal

partitions, ceilings and floors.

b. Solar heat gain through glass radiation and conduction

c. Outside air load through ventilation and infiltration

2. INTERNAL LOADS- These include

a. People

b. Lighting

c. Appliances and equipment

3. OTHER LOADS

These are system generated heat sources and may comprise of

a. Supply duct leakage and heat gain

b. Heat gain from fan motor in the air handler

There is another way to look at the distribution of the heat contributing sources

The cooling load components can thus be divided in to

a. Sensible heat load components- These components would always tend to cause

increase in dry bulb temperatures of the space. They are from walls, windows, roofs,

lights, solar heat gain as well as from people, appliances, equipment and

ventilation/infiltration air.

b. Latent heat load components- Latent heat results when moisture is entering the

space and causes humidity to increase. The factors contributing to latent heat load are

people, appliances as well as infiltration/ventilation air.

A load component may be totally sensible or latent or a combination of the two such as

people, appliances, and air which contribute to both sensible and latent heat load.



5.4 Estimation of Loads:

1. Estimation of external loads:

a) Heat transfer through opaque surfaces: This is a sensible heat transfer process. The

heat transfer rate through opaque surfaces such as walls, roof, floor, doors etc. is given by:

Qopaque = U. A. CLTD

Where

U is the overall heat transfer coefficient and

A is the heat transfer area of the surface on the side of the conditioned space

For interior air conditioned rooms surrounded by non-air conditioned spaces, the CLTD

of the interior walls 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, with all of them at the same temperature, the

CLTD values of the walls of the interior room will be zero.

Estimation of CLTD values of floor and roof with false ceiling could be tricky. For floors

standing on ground, one has to use the temperature of the ground for estimating CLTD.

However, the ground temperature depends on the location and varies with time. ASHRAE

suggests suitable temperature difference values for estimating heat transfer through ground. If

the floor stands on a basement or on the roof of another room, then the CLTD values for the

floor are the temperature difference across the floor (i.e., difference between the temperature

of the basement or room below and the conditioned space). This discussion also holds good

for roofs which have non-air conditioned rooms above them. For sunlit roofs with false

ceiling, the U value may be obtained by assuming the false ceiling to be an air space.

However, the CLTD values obtained from the tables may not exactly fit the specific roof.

Then one has to use his judgment and select suitable CLTD values.

b) Heat transfer through fenestration: Heat transfer through transparent surface such as a

window, includes heat transfer by conduction due to temperature difference across the

window and heat transfer due to solar radiation through the window. The heat transfer

through the window by conduction is calculated by using the same equation as for heat



transmitted through opaque surface except CLTD being equal to the temperature difference

across the window and Area is equal to the total area of the window.

Qopaque = (U). (A). (CLTD)

Where,

U= Overall heat transfer coefficient

A=Area of window

The heat transfer due to solar radiation through the window is given by:

Qtrans= (Aunshaded). (SHGFmax). (SC). (CLF)

Where,

Aunshaded

is the area exposed to solar radiation,

SHGFmax

the maximum Solar Heat Gain Factor

SC is the Shading Coefficient,

And CLF is the Cooling Load Factor.

The Cooling Load Factor (CLF) accounts for the fact that all the radiant energy that

enters the conditioned space at a particular time does not become a part of the cooling load1

instantly. As solar radiation enters the conditioned space, only a negligible portion of it is

absorbed by the air particles in the conditioned space instantaneously leading to a minute

change in its temperature. Most of the radiation is first absorbed by the internal surfaces,

which include ceiling, floor, internal walls, furniture etc. Due to the large but finite thermal

capacity of the roof, floor, walls etc., their temperature increases slowly due to absorption of

solar radiation. As the surface temperature increases, heat transfer takes place between these

surfaces and the air in the conditioned space. Depending upon the thermal capacity of the

wall and the outside temperature, some of the absorbed energy due to solar radiation may be

conducted to the outer surface and may be lost to the outdoors. Only that fraction of the solar

radiation that is transferred to the air in the conditioned space becomes a load on the building,



the heat transferred to the outside is not a part of the cooling load. Thus it can be seen that the

radiation heat transfer introduces a time lag and also a decrement factor depending upon the

dynamic characteristics of the surfaces. Due to the time lag, the effect of radiation will be felt

even when the source of radiation, in this case the sun is removed. The CLF values for

various surfaces have been calculated as functions of solar time and orientation and are

available in the form of tables in ASHRAE Handbooks.

c) Ventilation for Indoor Air Quality (IAQ):

The quality of air inside the conditioned space should be such that it provides a

healthy and comfortable indoor environment. Air inside the conditioned space is polluted

by both internal as well as external sources. The pollutants consist of odours, various

gases, volatile organic compounds (VOCs) and particulate matter. The internal sources of

pollution include the occupants (who consume oxygen and release carbon dioxide and

also emit odors), furniture, appliances etc, while the external sources are due to impure

outdoor air. Indoor Air Quality (IAQ) can be controlled by the removal of the

contaminants in the air or by diluting the air. The purpose of ventilation is to dilute the air

inside the conditioned space. Ventilation may be defined as the supply of fresh air to the

conditioned space either by natural or by mechanical means for the purpose of

maintaining acceptable indoor air quality. Generally ventilation air consists of fresh

outdoor air plus any re-circulated air that has been treated. If the outdoor air itself is not

pure, then it also has to be treated before supplying it to the conditioned space.

d) Infiltration:

Infiltration may be defined as the uncontrolled entry of untreated, outdoor air directly

into the conditioned space. Infiltration of outdoor air into the indoors takes place due to wind

and stack effects. The wind effect refers to the entry of outdoor air due to the pressure

difference developed across the building due to winds blowing outside the building. The

stack effect refers to the entry of outdoor air due to buoyancy effects caused by temperature

difference between the indoor and outdoors. Though infiltration brings in outdoor air into the

building similar to ventilation, in many commercial buildings efforts are made to minimize it,

as it is uncontrolled and uncertain. Some of the means employed to control infiltration

include use of vestibules or revolving doors, use of air curtains, building pressurization and



sealing of windows and doors. It is very difficult to estimate the exact amount of infiltration

as it depends on several factors such as the type and age of the building, indoor and outdoor

conditions (wind velocity and direction, outdoor temperature and humidity etc.). However,

several methods have been proposed to estimate the amount of infiltration air. Sometimes,

based on type of construction, buildings are classified into loose, average or tight, and

infiltration is specified in terms of number of air changes per hour (ACH). One ACH is equal

to the airflow rate equal to the internal volume of the occupied space per hour. The ACH

values are related to the outside wind velocity and the temperature difference between the

indoor and outdoors. Infiltration rates are also obtained for different types of doors and

windows and are available in the form of tables in air conditioning handbooks.

The sensible heat transfer rate due to ventilation and infiltration, Qs,vi

is given by:

The latent heat transfer rate due to ventilation and infiltration, Ql,vi

is given by:

In the above equations:

mo and Vo are the mass flow rate and volumetric flow rates of outdoor air due to

ventilation and infiltration, cpm is the average specific heat of moist air, h

fg is the latent heat

of vaporization of water, To and T

i are the outdoor and indoor dry bulb temperatures and W

o

and Wi are the outdoor and indoor humidity ratios. Thus from known indoor and outdoor

conditions and computed or selected values of ventilation and infiltration rates, one can

calculate the cooling and heating loads on the building. The sensible and latent heat transfer

rates as given by the equations above will be positive during summer (heat gains) and

negative during winter (heat losses).

Though the expressions for heat transfer rates are same for both ventilation and

infiltration, there is a difference as far as the location of these loads are considered. While

heat loss or gain due to infiltration adds directly to the building cooling or heating load, heat

loss or gain due to ventilation adds to the equipment load.



Estimation of Internal Load

I. People:

a) Qsensible = N.(Sensible heat gain).CLF

N= no. of people

Sensible heat gain = 70 w

(For SHG refer table 3, 1989 ASHRAE HANDBOOK, For people sited in office doing very

light work)

CLF=0.61

(For CLF refer table 40, 1989 FUNDAMENTAL HANDBOOK)

(Operation of CAD LAB from 10am to 6pm will be 8 hours and after a practical of 2 hour

new bath students will enter in the LAB.)

b) Qlatent =N.(Latent heat gain).CLF

N= no. of Persons

Latent heat gain=45 w

(For LHG refer table 3,1989 ASHRAE HANDBOOK.)

CLF=1 (for CLF refer table 27, 1989 ASHRAE HANDBOOK.)

II. LIGHT:

Q= Input CLF

= (40 26) 0.82

(For CLF Refer table 48, 1989 ASHRAE HANDBOOK.)

III. APPLIANCES:

A) COMPUTERS, FAN, PRINTER

Qsensible= Heat gainCLF

(For CLF refer table48,1989 FUNDAMENTAL ASHRAE HANDBOOK.)



Chapter 6

Duct Design

6.1 The chief requirements of an air conditioning duct system are:

1. It should convey specified rates of air flow to prescribed locations

2. It should be economical in combined initial cost, fan operating cost and cost of

building space

3. It should not transmit or generate objectionable noise

Generally at the time of designing an air conditioning duct system, the required airflow

rates are known from load calculations. The location of fans and air outlets are fixed

initially. The duct layout is then made taking into account the space available and ease of

construction. In principle, required amount of air can be conveyed through the air

conditioning ducts by a number of combinations. However, for a given system, only one

set results in the optimum design. Hence, it is essential to identify the relevant design

parameters and then optimize the design.

6.2 General rules for duct design:

1. Air should be conveyed as directly as possible to save space, power and material

2. Sudden changes in directions should be avoided. When not possible to avoid sudden

changes, turning vanes should be used to reduce pressure loss

3. Diverging sections should be gradual. Angle of divergence 20o

4. Aspect ratio should be as close to 1.0 as possible. Normally, it should not exceed 4

5. Air velocities should be within permissible limits to reduce noise and vibration

6. Duct material should be as smooth as possible to reduce frictional losses

6.3 Classification of duct systems:

Ducts are classified based on the load on duct due to air pressure and turbulence. The

classification varies from application to application, such as for residences, commercial

systems, industrial systems etc. For example, one such classification is given below:

Low pressure systems: Velocity 10 m/s, static pressure 5 cm H2O (g)



Medium pressure systems: Velocity 10 m/s, static pressure 15 cm H2

O (g)

High pressure systems: Velocity > 10 m/s, static pressure 15



iv. From the duct layout, dimensions and airflow rates, find the dynamic pressure losses

for all the bends and fittings

v. Select a fan that can provide sufficient FTP for the index run

vi. Balancing dampers have to be installed in each run. The damper in the index run is

left completely open, while the other dampers are throttled to reduce the flow rate to the

required design values.

The velocity method is one of the simplest ways of designing the duct system for both

supply and return air. However, the application of this method requires selection of

suitable velocities in different duct runs, which requires experience. Wrong selection of

velocities can lead to very large ducts, which, occupy large building space and increases

the cost, or very small ducts which lead to large pressure drop and hence necessitates the

selection of a large fan leading to higher fan cost and running cost. In addition, the

method is not very efficient as it requires partial closing of all the dampers except the one

in the index run, so that the total pressure drop in each run will be same.

6.5 Estimation of pressure loss in ducts:

As air flows through a duct its total pressure drops in the direction of flow. The pressure

drop is due to:

1. Fluid friction

2. Momentum change due to change of direction and/or velocity

1. Fluid Friction

The pressure drop due to friction is known as frictional pressure drop or friction loss,

pf. The pressure drop due to momentum change is known as momentum pressure

drop or dynamic loss, pd. Thus the total pressure drop p

t is given by:



In general in air conditioning ducts, the fluid flow is turbulent. It is seen from the above

equation that when the flow is turbulent, the friction factor is a function of Reynolds

number, hydraulic diameter and inner surface roughness of the duct material. Of the

different materials, the GI sheet material is very widely used for air conditioning ducts.

Taking GI as the reference material and properties of air at 20o

C and 1 atm. pressure,

the frictional pressure drop in a circular duct is given by:

Where Qair is the volumetric flow rate of air in m3

/s, L is the length and D is the inner

diameter of the duct in meters, respectively.

2. Dynamic losses in ducts:

Dynamic pressure loss takes place whenever there is a change in either the velocity or

direction of airflow due to the use of a variety of bends and fittings in air conditioning ducts.

Some of the commonly used fittings are: enlargements, contractions, elbows, branches,

dampers etc. Since in general these fittings and bends are rather short in length (< 1 m), the

major pressure drop as air flows through these fittings is not because of viscous drag

(friction) but due to momentum change. Pressure drop in bends and fittings could be

considerable, and hence should be evaluated properly. However, exact analytical evaluation

of dynamic pressure drop through actual bends and fittings is quite complex. Hence for

almost all the cases, the dynamic losses are determined from experimental data. In turbulent

flows, the dynamic loss is proportional to square of velocity. Hence these are expressed as:

Where,

K is the dynamic loss coefficient, which is normally obtained from experiments.



Chapter 7

Calculations

1. BASIC ANGLES:

Declination Angle:

For calculation of Declination Angle

d= [23.47sin360 (284+N)]/365

N=no of days starting from Jan1 say, for Jan1=1

For Jan20=20

For March=65

d= [23.47sin360 (284+65)]/365

d= - 6.38

Hour Angle:

At

12:00 pm=0 1:00 am =195

1:00 pm =15 2:00 am =210

2:00 pm =30 3:00 am =225

3:00 pm =45 4:00 am =240

4:00pm =60 5:00 am =255



5:00 pm =75 6:00 am =270

6:00 pm =90 7:00 am =285

7:00 pm =105 8:00 am =300

8:00 pm =120 9:00 am =315

9:00 pm =135 10:00 am =330

10:00 pm =150 11:00 am =345

12:00 pm =360=0 12:00 am =180

For Pusad,

1. Hour Angle at 12:00 pm=0

2. Declination angle for 21 June.

d=23.47 sin [360(284+N)]/365

=23.47 sin [360(284+172)]/365

d=23.46

(N=no. of days)

3. Latitude angle =19 from Net for Pusad.

4. Altitude angle, =1[cosl.cosh.cosd+sinl.sind]

= 1[cos (19).cos (0).cos(23.46)+sin(19).sin(23.46)]

= 1[(0.19) (1) (0.917) + (0.32) (0.99)]



=1 [0.861+0.1248]

= 80.33 .. [Cross check it by page 1-24]

5. Zenith angle () =(/2-)

=9.67

6. Solar azimuth angle ():

= 1[cosl.sind- cosd.cosh.sinl]/cos

But l



For west side;

= [ ( 62.79+90)]

=27.21

For North side;

=-62.79

For East side;

= [-(62.79-90)]

=207.21

9. Direct Radiation from sun

=A exp (-B/SIN)..w/m2

A= Apparent solar radiation

=1080 for summer

B=Atmospheric extinction coeff.

=0.21 for summer

IDN=1080 exp [(-0.21)/sin]

10. Diffuser Radiation From sky.

Id=C.IDN.FWSw/m2

a. C=const for cloudless sky.

C=0.135 for mid suffer.



b. Fws=view factor or configuration factor

Fws= (1+cos)/2.. (=Tilt angle)

[For horizontal surface =o, For vertical surface=90]

For horizontal surface;

Fws= (1+coso)/2

=2/2

=1

For vertical surface;

Fws= (1+cos90)/2

=1/2

=0.5

Id =0.135.IDN .FWS

11. Reflected short wave (solar) radiation (Ir)

Ir = (IDN+ID)g.FWG

g=reflectivity of the ground

FWS=View factor.

FWS= (1-COS)/2

For horizontal =o

For vertical =90



For horizontal,

FWG= (1-COS90)/2

=0

For vertical,

FWG= (1-COS90)/2

=0.5

2. AMOUNT OF HEAT AVAILABLE (PRODUCE) FOR PUSAD:

Latitude=19

Reflectivity=0.6

Hour angle=0

Declination=+23.5

Considering Cloudless condition:

Wall azimuth angle ()=0south facing

=180north facing

=-90east facing

North side:

Altitude=80.33.. (Calculated)

Solar azimuth angle () = 0. (As < )

Now,

Wall solar azimuth angle,

=180-( + )

=180-(0+180)

=0



Incidence angle,

= 1(cos.cos)

= 1(cos80.33.cos0)

=80.33

1. Direct Radiation:

=IDN.cos

IDN=A.exp (-B/sin)

=1080 exp (-0.21/sin80.33)

IDN=872.78w/m2

Direct Radiation =IDN.cos

=872.78cos (80.33)

Direct Radiation =146.6w/m2

2. Diffuse Radiation(Id):

View factor, FWS= (1+cos)/2

Vertical surface, =90

FWS= (1/2)

=0.5

Id=C.IDN.FWS

=0.135 827.78 0.5

Id=55.87 w/m2



3. Reflected Radiation(Ir):

= reflectivity = 0.6

View factor= (1-cos)/2


FWG= (1-cos90)/2

=0.5

Ir= (IDN+ID) g.FWG

= (872.78+55.87) 0.6 0.5

Ir=278.595 w/m2

Total incident radiation (IT):

IT =IDN.cos+Id+ Ir

=146.6+55.87+278.595

IT=481.065w/m2

Area of North side wall=520sq.ft

=520/10.76sq.m

IT= (481.065520)/10.76

ITN=23.2kw



East Side:

Altitude ==80.33

Solar azimuth angle==0

Wall azimuth angle=-90=

Wall solar azimuth angle ()

=180-( + )

=180-(0+90)

=270

Incidence angle

() = 1(cos.cos)

() = 1(cos80.33.cos270)

()=90

1. Direct radiation =IDN.cos

Direct radiation=0 [as cos90=0]

2. Diffuse radiation (Id):

View factor, FWS= (1+COS)/2


FWS=1/2=0.5



Id =C.IDN.FWS

IDN =A. exp (-B/sin)

=1080 exp (-0.21/sin80.33)

IDN=872.78 .w/m2

Id=0.135 872.78 0.5

Id=58.91w/m2

3. Reflected Radiation (Ir):

=0.6

=Reflectivity

View factor, FWG = (1-cos/2)

For vertical, =90

FWG= (1-cos90/2)

=0.5

Ir= (IDN+Id).g.FWG

= (872.78+58.91)0.60.5

= 279.5 w/m2

Total incident radiation:

IT=IDN.cos90+Id+Ir

IT=0+58.91+279.5

IT=338.41 w/m2

Area of east side wall=250sq.ft



IT=338.41 23.23

IT=7.86 w/m2

3. External Load Calculations:

Amount Of heat transmitted in Lab by following way,

1. Through wall.

2. Through Glass.

3. Through Roof.

4. Through Floor.

1. Through wall:

CAD wall =200 mm common brick, so take a wall of B type.

(ASHRAE Fundamentals Handbook1989 PG 26.39)

A. Heat transfer through North side wall:

CLTDadj=CLTDtable+ (Tav-29) (27 p.no,ch34,ASHRAE handbook)

=8+ (35-29) (ASHRAE hand book 1989,p no26.37)

=14

A=36.91m2

Qnorth=Uwall.Awall.CLTDadj



=1.7136.9114

=883.625w

B. Heat transfer through East side wall.

A=22.4m2

CLTDadj=CLTDtable+ (Tav-29)

=15+ (35-29)

=21

Qeast =UwallAwallCLTDadj

=1.722.421

=804.384w

C. Heat transfer through south side wall:

A=52.65m2

Q=UAT

=1.7152.65(35-20)

=1350.47w

This is shaded wall hence CLTD is not used.

D. Heat transfer through west side wall:

A=21.96m2

Q=UAT



=1.7121.96(35-20)

=563.27w

This is shaded wall hence CLTD is not used.

2. Through Window

A. North Side:

qconduction=UA(CLTD)

(U value taken from table5,pg.no611,ASHRAE2001 fundamental).

(CLTD value taken from 1989ASHRAE handbook,Table29,pg.no26.39,at

12 noon).

SINGLE GLASS WITH INTERNAL SHADING 6mm.

(ASHARAE-2001 Fundamentals.)

=54.28w

For one windows heat transfer through glass by conduction= 54.28w

For 4 window=454.28

=217.12w

Heat transfer through fenestration

Qsolar=ASCSHGFCLF

=2.360.5 186 0.89



=195.34w

(For SC refer table5, Pg.no611, ASHRAE2001 fundamental)

(For SHGF refer chapter 26, table34)

(For CLF refer chapter26, table39)

There are 4 windows on North side

Q=4 195.34

=781.36w

B. East Side:

Heat transfer through glass due to conduction

qconduction =U.A.CLTD

=4.6 2.36 5

=54.28w

(For U refer table5, pg. no.611, ASHRAE2001 fundamental)

(For CLTD refer chapter26 1989 ASHRAE handbook)

There are 2 windows on east side

qconduction= 2 54.28

=108.56w

Heat transfer due to fenestration:



Qsolar =A.SC.SHGFCLF

=2.36 0.5 663 0.27

(Depth of inset is zero hence total area of window is considered)

(For SC refer table5, pg.no. 611, ASHRAE2001 fundamental)

(For SHGF Refer table34, chapter24, 1989ASHRAE handbook)

(For CLF refer chapter26, table39)

=211.232w

There are 2 windows on east si de.

Qsolar=2 211.232

=422.464w

C. Through Roof:

A=155 m2

Q=U.A. (CLTD)

=0.761 155 38 20

(For U Refer Table29, pg.no26.35, 1989ASHRAE handbook, at solar

time12pm)

Q=1887.28w



D. Through Floor:

Q=UAT

=4.101 155 35 20

=9534.825w

(Floor is the slab of room which is unconditioned hence T is considered.)

4. Internal Load:

A. People:

a) Qsensible = N.(Sensible heat gain).CLF

N= 40 no. of people

Sensible heat gain = 70 w

(For SHG refer table 3, 1989 ASHRAE HANDBOOK,

people sited in office doing very light work)

CLF=0.61

(For CLF refer table 40,1989 FUNDAMENTAL

HANDBOOK)

(Operation of CAD LAB from 10am to 6pm will be 8 hours

and after a practical of 2 hour new bath students will enter in

the LAB.)

Qsensible =40 70 0. 61

= 1708 w

b) Qlatent =N.(Latent heat gain).CLF



N= 40 Persons

Latent heat gain=45 w

(For LHG refer table 3,1989 ASHRAE HANDBOOK.)

CLF=1 (for CLF refer table 27,1989 ASHRAE

HANDBOOK.)

Qlatent =40 45 1

=1800 W

B. Appliances:

a) COMPUTERS:

[CRT: 1 Wipro, 2 Acer][LCDs: 33 HP]

Qsensible= Heat gainCLF

Heat Gain= [ (CRT 15 Watt)+(LCD 14 Watt)+(LCD

19Watt) ]

= [ (5 255) + 19 153 + 15 255 ]

= 1275+2907+3825

=8007 w

Qsensible= 8007 0.78

(For CLF refer table48,1989 FUNDAMENTAL ASHRAE HANDBOOK.)

=6245.46 W



b) FAN:

Qsensible= Heat Gain CLF

= (12 55) 0.78

(For CLF refer table 48, 1989 FUNDAMENTAL ASHRAE HANDBOOK)

=514.8 W

c) PRINTER:

[Company: Epson FX-1170]

Qsensible = Heat Gain CLF

= (1 92) 0.78

(For CLF refer table48,1989 FUNDAMENTAL ASHRAE HANDBOOK)

=71.76 W

d) LIGHT:

Q= Input CLF

= (40 26) 0.82

(For CLF Refer table 48, 1989 ASHRAE HANDBOOK.)

= 852.8 W

5. Ventilation Load :

For meeting place ventilation required=3.5 l/sec/person.

Total outdoor air required:

VO,V=3.5No.of person.



=3.5 40

=140 m3/s

Mass flow rate of ventilation air:

MO,V=Volumetric flow rate /specific volume

For outside condition:

DBT=44C,RH=20%

VO =Specific volume

=0.915 m3/kg

WO=Moisture content

=0.015 kg/kg dry air.

Ho=specific enthalpy

=74 kJ/kg da

Mo,v=0.14/0.915

=0.153 kg/s

Sensible Heat Transfer Due To Ventilation Is Given By:

QS,V=MO,V.CPM(TO-TI)

=0.153 1.0216(44-20)



QS,V=3.75 kw

Latent heat transfer due to ventilation is given by;

QL,V=MO,V.hfg(wo-wi)

For inside conditions:

DBT=20C

RH=50%

WI=0.0075

WO=0.015

QL,V=0.153 2501 (0.015-0.0075)

QL,V=2.86 kw

Hence heat transfer through ventilation.

QVENTILATION=QS,V+QL,V

=3.75+2.86

=6.61 kw

6. Infiltration Load:

Infiltration rate,

MInf=Density of air (ACHVolume of the room)/3600

=1.095(1 155 3)/3600



MInf=O.14 kg/sec

QS=MInfCpm(To-TI)

=0.14 1.0216 44 20

=3.43 kw

QL=MInf.hfg(WO-WI)

=0.14 2501 0.015 0.0075

=2.62 kw

QInf=QS+QL

=3.43+2.62

=6.05 kw

SENSIBLE HEAT GAIN IN CAD LAB:

1. Heat gain through North side wall = 883.625 w

2. Heat gain through East side wall = 804.384w

3. Heat gain through South side wall=1350.47w

4. Heat gain through West side wall=563.27 w

5. Heat gain through North side window=998.48w

6. Heat gain through East side window =531.02 w

7. Heat gain from roof =1887.28 w

8. Heat gain through floor =9534.825 w

9. Heat due to Occupants =1708 w

10. Heat due to Light =852.8 w



11. Heat due to Computers =6245.46 w

12. Heat due to Fan =514.8 w

13. Heat due to Printer =71.76 w

14. Heat due to ventilation =3750 w

15. Heat due to infiltration =3430 w

16. Taking 3% duct gain.

17. Taking 5% leakage loss.

TOTAL SENSIBLE HEAT GAIN =35770 W

= 35.77

LATENT HEAT GAIN :

1.Person =1800 watt

=1.8kw

2.ventilation =2.86kw

3.Infiltration =2.62kw

4. Taking 3% duct gain=

Total latent heat=7.66 kw

Total heat load =35.77+ 7.66

=43.43 kw

Total TR= 43.43/3.517



=12.34

=13 TR of Refrigeration

As,

RSH=35.77kw

RLH=7.49kw

RSHF= RSH/(RSH+RLH)

=35.77/(35.77+7.49)

=0.82

Now calculate ERSH

ERSH=RSH+BF(OASH)

OASH=0.0204cmmda(to-tr)

=0.0204 0.14 60 (44 20)

OASH=4.11kw

Similarly

OALH=50cmmda(wo-wr)

=50 0.14 60 0.0117 0.0076

OALH=1.722kw

So, Grand sensible Heat (GSH);

RSH+OASH=35.77+4.11



GSH=39.88kw

GLH=RLH+OALH

=7.49+1.722

GLH=9.212kw

GSHF=GSH/(GSH+GLH)

=(37.236)/(37.236+9.002)

GSHF=0.805 kw

ERSH=35.77+0.1(4.11)

ERSH=36.181 kw

ERLH=7.49+0.1(1.722)

ERLH=7.66 kw

ESHF=ERSH/(ERSH+ERLH)

=36.181/(36.181+7.66)

ESHF=0.82

Coil ADP=7 (From Psychometric chart)

cmmda =ERSH/(0.0204(1-BF)(t2-tadp))

=33.537/(0.0204(1-0.1)(20-7))

cmmda =140.51m3/min



cmmda=cmmoa+cmmra

140.51=(0.14 60) +cmmra

Cmmra=140.51-8.4

=132.11m3/min

t3 =(cmm1t1+cmm2t2)/(cmm1+cmm2)

=(8.4 44 + 132.11 20)/(8.4 + 132.110)

=21.43C

BF=(t4-tadp)/(t3-tadp)

0.1=t4-7/21.43-7

t4=8.443C

Condition of air entering the cooling coil state3;

=DBT=21.43C

WBT=8C

Room ADP=8C

Condition of air entering the room:

DBT=8.44C

WBT=8

Quantity of air supplied to the room=8430.6 m3/hr



Duct Calculations

1. Area of main duct (position A):

A=Quantity of air supplied/velocity.

A=Q/V

=2.34/8

=0.292 m2

A=0.292

A=/4D2

D=0.61m

2. Area of main duct ()

A=Q/V

= [2.34-(0.334 2)]/8

A=0.209m2

=/2 D2

D=0.51m

3. Area of main Duct (k):

A=Q/A

=0.0.125

=/4 D2



D=0.4m

4. Area of main Duct (k):

A=Q/V

=0.336/8

A=0.042

A=/4.D2

D=0.231m

5. Area of supply Ducts:

[C, E, G, B, D, F]

A=Q/V

=0.334/8

A=0.042 m2

=/4 D2

D=0.231m

DYNAMIC COEFFICIENT:

1. AS/AC=0.209/0.292 =0.715

Ab1/Ac=0.042/o.292 =0.143

Qb1/Qc=0.334/2.34 =0.14

Cb=1.2 (ASHRAE HANDBOOK FUNDAMENTAL 2001,P.N.839)



2. As/Ac=Aj/Ai=0.125/0.209=0.59

Ab1/Ac =As/Ai=0.042/0.209=0.2

Qb1/Qc=0.334/1.672=0.19

Cb=1.56 (ASHRAE HANDBOOK FUNDAMENTAL

2001,P.N.839)

3. As/Ac=Ak/Aj=0.042/0.125=0.336

Ab1/Ac =As/Aj=0.042/0.125=0.336

Qb1/Qc=0.334/1.004=0.332

Cb=2.44 (ASHRAE HANDBOOK FUNDAMENTAL

2001,P.N.839)

ROUND FITTING:

D=230 mm

r/D=1.5

Co=0.11 (ASHRAE HANDBOOK FUNDAMENTAL

2001,P.N.816)

Calculation of Pressure Drop:

1) Dynamic loss due to fire damper

Co=Dynamic loss coefficient=0.12

(ASHRAE HANDBOOK FUNDAMENTAL 2001,P.N.819)

Dynamic loss (PUFD) =Co .(v2/2)



=0.12(1.2 82/2)

=4.608

2) Dynamic loss due to Screen

A0/A1=1 (Same duct dia.)

Consider, n=0.9

Where,

n= free area ratio of screen

A0= Cross sectional Area of duct

A1= Cross sectional Area of duct where screen is located.

(ASHRAE HANDBOOK FUNDAMENTAL 2001,P.N.819)

Dynamic loss coefficient Co=0.14

Dynamic loss = Co (v2/2)

=0.14(1.2 82/2)

= 5.376 Pa

Frictional loss:

1. Section A-B

PA-B=PA1F+PB1F

PA1F= (0.022243Q1.852 L)/D4.973

= (0.022243 2.3411.8523.55)/(0.61)4.973

PA1F=4.45 Pa

PB1F= (0.022243 0.3341.852 2.26)/(0.231)4.973

=9.63 Pa



Dynamic pressure drop from upstream to branch is given by:

Pu-b=Cu-b (2/2)

=1.2(1.2 82/2)

Pu-b=46.08 Pa

Dynamic pressure drop at Branch to Diffuser:

P=C (v2/2)

=0.11[(1.282)/2

=4.22 Pa

Dynamic pressure drop at Diffuser:

P=15 Pa (from IIT Notes)

Total pressure drop at section (A-B)

PAB=4.608+5.376+4.45+9.63+46.08+4.22+15

= 89.36 Pa

2. SECTION A-C

Same as section A-B

PA-B=PA-C

=79.38 Pa



3. Section A-I-D

PA-I-D=PAF+PIF+PDF+PU-B+PU-C+PU-D+PU E+PUd

PAF=4.45 Pa

PIF= (0.022243QAIR1.852L)/(D)4.973

= (0.0222431.6721.8523.55)/(0.51)4.973

= 5.82 Pa

PDF= (0.022243QAIR1.8522.26)/(0.231)4.973

=9.64Pa

PUB=46.08Pa

PUC=46.08Pa

PUD=CU-D[(v2)/2]

=1.56[(1.282)/2]

=59.9Pa

PUe=4.22 Pa (Elbow loss)

Diffuser loss=15 pa

Total loss at A-I-D:

PA-I-D=4.608+4.45+5.82+9.64+46.08+46.08+59.9+4.22+15

=201.168 Pa



4. Section A-I-E=A-I-D

PA-I-E= 201.168 pa

5. Section A-I-J-F

PA-I-J-

F=PA1F+PI1F+PJ1F+PF1F+PU1B+PU1C+UE+PUD+PUF+FD+FSD

+Fel+Fd

FD=4.608 Pa

FSD=5.37 Pa

PA1F= 4.45 Pa

PI1F=5.82 Pa

PJ1F= (0.022243.QAIR1.852L)/(D)4.973

= (0.0222431.0041.8523.55)/(0.4)4.973

= 7.57 Pa

PF1F= (0.02243.QAIR1.852L)/D4.973

= (0.0222430.3341.852 2.26)/(0.231)4.973

PF1F=9.63 Pa

PUB=46.08 Pa

PUC=46.08 Pa



PUD= 59.9 Pa

PUE=59.9 Pa

PU-F=Cu-F (2/2)

=2.44(1.2 82/2)

PUF=93.69 Pa

PEl= 4.22 Pa

PFD=15 Pa

Total loss in section A-I-J-F = 4.608+5.37+4.45+5.82+7.57+9.63

+46.08+46.08+59.9+59.9+93.69+4.22+15

=362.31 Pa

6. Section A-I-J-G:-

Section A-I-J-F = Section A-I-J-G

=362.31 Pa

7. Section A-I-J-K-H:-

Total pressure loss at A-I-J-K-H

= PFD+PSL+PA1F+P1F+PJF+PK1F+PH1F



+PU C+PUB+PUD+PUE+PUF+PUG+PU H+ Pelbow +Pdiffuser

PFD=4.608 Pa

PSL=5.37 Pa

PA1F=4.45 Pa

P1F=5.82 Pa

PJF=7.57 Pa

PK1F= (0.022243.QAIR1,852L)/D4.973

= (0.022243 0.3341.852 3.55)/(0.231.973)

PK1F=15.1 pa

PH1F= (0.022243.QAIR1.852L)/D4.973

= (0.0222430.3341.8522.26)/(0.231)4.973

PH1F=9.64 pa

PU C=46.08 Pa

PUB=46.08 Pa

PUD=59.9 Pa

PUE=59.9 Pa

PUF=93.69 Pa

PUG=93.69 Pa



PU H=C (.v2/2)

=0.11[(1.282]/2

=4.22 Pa

ELEBOW loss, Pelbow =4.22 Pa

DIFFUSER loss, Pdiffuser =15 Pa

Total pressure loss at A-I-J-K-H :

=4.608+5.37+4.45+5.82+7.57+15.1+9.64+46.08+59.9+59.9+93.69+93.69+4

.22+4.22+15

=475.33 Pa

Thus run with maximum pressure drop in A-I-J-K-H is index run , hence

FTP required is;

FTP=PA-I-J-K-H

= 475.33 Pa

=475.33 N/m2

Calculations of Amount of Dampering required:

AMOUNT OF DAMPERING REQUIRED AT B & C.

= 475.33-89.36

=385.97 Pa



AMOUNT OF DAMPERING REQUIRED AT E & D.

=475.33-201.168

=274.164 Pa

AMOUNT OF DAMPERING REQUIRED AT G & F.

=475.33-362.31

=113.02 Pa

Fan Power:

WFAN=FTPaair/fan

=475.33 2.34/0.9

=1235.85 watt

WFAN =1.235 kW

RETURN DUCT CALCULATION:

Amount of recirculated air =132.11 m3/min

Q =2.20 m3/sec

Area of duct = Q/V

=2.2/6

=0.36 m2

Area =(/4)D2

D= 0.36/(

4)



D=0.68 m

FRICTIONAL LOSS:

PRF=(0.022243QAIR1.852L)/D4.973

=(0.0222432.21.85210)/0.684.973

=6.52 pa

DYNAMIC LOSS:

ELBOW C=0.11

P= C(.V2/2)

= 0.11(1.262/2)

P=2.376 pa

Dynamic pressure drop at diffuser, Pdiffuser =15 pa

Total loss in return duct:

= PRF+P+Pdiffuser

=6.52+2.376+15

=23.89 pa



Chapter 8

Components Selection

1. VOLTAS Specifications:



CARRIER Technical Specification:



DAIKIN INVERTER BASED SYSTEM Specification:

FDYQ100K

Overview

Features

Specifications

Controllers

Unit Indoor Unit FDYQ100KAV1A Outdoor Unit RZQ100HY4A

Rated Capacity Cool (KW) 10.0 Heat (KW) 12.1

Capacity Range Cool (KW) 5.0-11.2 Heat (KW) 5.1-12.5

C.O.P

3.24/3.50

Rated Power Input Cool (kW) 3.09 Heat (kW) 3.46

Air Flow Rate (Rated) l/s 815

Indoor Sound Level (@ 1.5M) dbA 46

ESP Settings

STD/HI

Indoor Fan Speeds

HI/LO

Dimensions (H x W x D) Outdoor (mm) 1346 x 900 x 320 Indoor (mm) 360 x 1478 x 899

Weight Outdoor (kg) 108 Indoor (kg) 59

Power Supply

3 phase, 415V, 50Hz

Compressor Type Type Hermetically sealed scroll type

Refrigerant Type R410A

Refrigerant Control

Electronic

Refrigerant Pipe Size Liquid (mm) 9.5 (Flared) Gas (mm) 15.9 (Flared)

Drain Pipe Size (ID/OD) mm ID 25mm, OD 32mm

Supply Air Conn. mm 1152 x 243 (Flange)

Return Air Conn. mm 2 x 400 (Oval)

Max Actual Pipe Length m 75

Maximum Level Difference m 30

Pre Charged Length m 30

Operating Range (Outdoor Temp)

Cooling CDB -5 to 46 Heating CWB -15 to 15.5

EPA Sound Power Level Outdoor (dBA) 65

Outdoor Sound Level Pressure dbA (C/H) 49/51



As Calculated, for Mechanical CAD Laboratory 13 TR of

Refrigeration System is required and we have three options-

1. Voltas

2. Carrier

3. Daikin

Daikin is a inverter based system so the part load performance will be

better as the refrigerant load is calculated considering outdoor temperature

44oc. But this system will be very costly. It also uses zero ODP and low

GWP refrigerant R-410a.we can go for a hybrid system of 11 TR carrier

and10 kw daikin inverter based system

Voltas and Carrier are conventional systems. For Voltas we can go for

combination of 5.5TR and 8.75TR systems. For carrier we can for

combination of 5.5TR and 8.5TR systems.

Split Air conditioning systems:

To avoid the constraint of having all parts in one package, the evaporator set may

be split from the condenser, the compressor going with either. The unit will be designed

as a complete system but the two parts are located separately and connected on site. On

some small units, flexible refrigerant piping may be provided. If the system is of a range

up to about 5 kW, coils of precharged soft copper tube, with self-sealing couplings, may

be supplied for connection within a limited distance of 515 m. This facility enables full

factory processing to be carried out to the standards of a one-piece unit. It is limited to the

availability of suitable tubing, usually 58 inch outside diameter. In such systems, the total

charge is suitable for the final assembly, and pipes should not be extended beyond the

factory-supplied length without prior consultation with the supplier.

Larger split packages must be piped on site by normal methods, and then processed and

charged as an open plant. Split unit evaporators should not be located more than 5 m

higher than their condensers. Those light sleepers who need quite while sleeping, the



ideal choice is a split air conditioning system as it hardly makes any noise. This is

because the compressor is kept outside the house.

1. Scroll compressors:

Scroll compressors are orbital motion, positive displacement type compressors, in

which suction and compression is obtained by using two mating, spiral shaped, scroll

members, one fixed and the other orbiting. Figure shows the working principle of scroll

compressors. Figures8.2 and 8.3 shows the constructional details of scroll compressors.

As shown in Fig.8.1, the compression process involves three orbits of the orbiting scroll.

In the first orbit, the scrolls ingest and trap two pockets of suction gas. During the second

orbit, the two pockets of gas are compressed to an intermediate pressure. In the final

orbit, the two pockets reach discharge pressure and are simultaneously opened to the

discharge port. This simultaneous process of suction, intermediate compression and

discharge leads to the smooth continuous compression process of the scroll compressor.

One part that is not shown in this diagram but is essential to the operation of the scroll is

the anti-rotation coupling. This device maintains a fixed angular relation of 180 degrees

between the fixed and orbiting scrolls. This fixed angular relation, coupled with the

movement of the orbiting scroll, is the basis for the formation of gas compression

pockets.

As shown in Figs.8.2 and 8.3 each scroll member is open at one end and bound by a base

plate at the other end. They are fitted to form pockets of refrigerant between their

respective base plates and various lines of contacts between the scroll walls. Compressor



capacity is normally controlled by variable speed inverter drives.

Fig 8.1 Working principle of a scroll

Fig 8.2 Main parts of scroll compressor



This graphic shows the two spiral shaped intermeshing scrolls

This graphic shown above show a side view of the interior

components of the compressor

Fig 8.3 Different views of scroll compressor



Fig. 8.4 Cut view of Scroll Compressor

Currently, the scroll compressors are used in small capacity (3 to 50 kW)

refrigeration, air conditioning and heat pump applications. They are normally of hermetic

type. Scroll compressors offer several advantages such as:

1. Large suction and discharge ports reduce pressure losses during suction and

discharge

2. Physical separation of suction and compression reduce heat transfer to suction gas,

leading to high volumetric efficiency

3. Volumetric efficiency is also high due to very low re-expansion losses and

continuous flow over a wide range of operating conditions

4. Flatter capacity versus outdoor temperature curves



5. High compression efficiency, low noise and vibration compared to reciprocating

compressors

6. Compact with minimum number of moving parts

2. Air-cooled condensers:

As the name implies, in air-cooled condensers air is the external fluid, i.e., the

refrigerant rejects heat to air flowing over the condenser. Air-cooled condensers can be

further classified into natural convection type or forced convection type.

Forced convection type:

In forced convection type condensers, the circulation of air over the condenser

surface is maintained by using a fan or a blower. These condensers normally use fins on air-

side for good heat transfer. The fins can be either plate type or annular type. Forced

convection type condensers are commonly used in window air conditioners, water coolers

and packaged air conditioning plants. These are either chassis mounted or remote mounted.

In chassis mounted type, the compressor, induction motor, condenser with condenser fan,

accumulator, HP/LP cut- out switch and pressure gauges are mounted on a single chassis. It

is called condensing unit of rated capacity.

3. Thermostatic Expansion Valve (TEV)

Thermostatic expansion valve is the most versatile expansion valve and is most

commonly used in refrigeration systems. A thermostatic expansion valve maintains a

constant degree of superheat at the exit of evaporator; hence it is most effective for dry

evaporators in preventing the slugging of the compressors since it does not allow the

liquid refrigerant to enter the compressor. The schematic diagram of the valve is given in

Figure. This consists of a feeler bulb that is attached to the evaporator exit tube so that it

senses the temperature at the exit of evaporator. The feeler bulb is connected to the top of

the bellows by a capillary tube. The feeler bulb and the narrow tube contain some fluid



that is called power fluid. The power fluid may be the same as the refrigerant in the

refrigeration system, or it may be different. In case it is different from the refrigerant,

then the TEV is called TEV with cross charge. The pressure of the power fluid Pp

is the

saturation pressure corresponding to the temperature at the evaporator exit. If the

evaporator temperature is Te

and the corresponding saturation evaporator pressure is Pe,

then the purpose of TEV is to maintain a temperature Te+T

sat the evaporator exit,

where Ts is the degree of superheat required from the TEV. The power fluid senses this

temperature Te+T

s by the feeler bulb and its pressure P

p is the saturation pressure at this

temperature. The force F pexerted on top of bellows of area A

b due to this pressure is

given by:

Fp

= Ab

P p

The evaporator pressure is exerted below the bellows. In case the evaporator is large and

has a significant pressure drop, the pressure from evaporator exit is fed directly to the

bottom of the bellows by a narrow tube. This is called pressure-equalizing connection.

Such a TEV is called TEV with external equalizer, otherwise it is known as TEV with

internal equalizer. The force Fe

exerted due to this pressure P e

on the bottom of the

bellows is given by

Fe= A

b P

e

The difference of the two forces F pand F

e is exerted on top of the needle stand. There is

an adjustment spring below the needle stand that exerts an upward spring force Fs on the

needle stand. In steady state there will be a force balance on the needle stand, that is,

F s= F

p - F

e

During off-cycle, the evaporator temperature is same as room temperature

throughout, that is, degree of superheat Ts is zero. If the power fluid is the same as the

refrigerant, then P p= P

eand F

p= F

e. Therefore any arbitrarily small spring force F

s

acting upwards will push the needle stand against the orifice and keep the TEV closed. If

it is TEV with cross charge or if there is a little degree of superheat during off-cycle then

for TEV to remain closed during off-cycle, Fs should be slightly greater than (F - F).



Fig.8.5 Schematic of Thermostatic expansion valve

As the compressor is started, the evaporator pressure decreases at a very fast rate

hence the force Fe decreases at a very fast rate. This happens since TEV is closed and no

refrigerant is fed to evaporator while a compressor draws out refrigerant at a very fast

rate and tries to evacuate the evaporator. The force Fp

does not change during this period

since the evaporator temperature does not change. Hence, the difference Fp-F

e, increases

as the compressor runs for some time after starting. At one point this difference becomes

greater than the spring force Fs

and pushes the needle stand downwards opening the

orifice. The valve is said to open up. Since a finite downward force is required to open

the valve, a minimum degree of superheat is required for a finite mass flow rate.

As the refrigerant enters the evaporator it arrests the fast rate of decrease of evaporator

pressure. The movement of needle stand also slows down. The spring however gets



compressed as the needle stand moves downward to open the orifice. If Fs0

is the spring

force in the rest position, that is, off-cycle, then during open valve position

F s= F

s0+ F

s

Eventually, the needle stand reaches a position such that,

F s= F

p - F

e= A

b (Pp

P

e )

That is, F pis greater than F

eor P

pis greater than P

e. The pressure P

pand P

e are saturation

pressures at temperature (T e+ T

s) and T

e respectively. Hence, for a given setting force

Fs

of the spring, TEV maintains the difference between F pand F

eor the degree of

superheat Ts constant.

T s (F

p - F

e )

Fs

This is irrespective of the level of Pe, that is, evaporator pressure or temperature, although

degree of superheat may be slightly different at different evaporator temperatures for

same spring force, Fs. It will be an ideal case if the degree of superheat is same at all

evaporator temperatures for a given spring force.

Advantages, disadvantages and applications of TEV

The advantages of TEV compared to other types of expansion devices are:

1. It provides excellent control of refrigeration capacity as the supply of refrigerant to the

evaporator matches the demand

2. It ensures that the evaporator operates efficiently by preventing starving under high

load conditions

3. It protects the compressor from slugging by ensuring a minimum degree of superheat

under all conditions of load, if properly selected.

However, compared to capillary tubes and AEVs, a TEV is more expensive and

proper precautions should be taken at the installation. For example, the feeler bulb must

always be in good thermal contac

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