UNIVERSITY OF THE EAST

MANILA CAMPUS
COLLEGE OF ENGINEERING
MECHANICAL ENGINEERING DEPARTMENT

Centralized Air Conditioning System of Semi-Conductor Manufacturing Facility

In Partial Fulfillment of the course subject NME 4102: Air-Conditioning and Ventilation
Systems

Submitted by:
Hadap, John Israel F.
Ocampo, John Rey T.

Submitted to:
Engr. Diosdado Doctor

August 25, 2021

 TABLE OF CONTENTS

INTRODUCTION

CHAPTER 1: Building Description

1.1 Building Location

1.2 Floorplan Layout
1.3 Zoning
1.3.1 Zone 1
1.3.2 Zone 2
1.3.3 Zone 3
1.3.4 Zone 4
1.3.5 Zone 5

CHAPTER 2: Cooling Load Calculation

3.1 Design Conditions

3.2 Heat Transmission from structure

3.2.1 Design Parameters

3.2.2 Heat Transmission from structure calculation

3.3 Solar Radiation

3.3.1 Solar load through transparent surfaces

3.3.2 Design Parameters

3.3.3 Solar Load through transparent surfaces calculation

3.4 Solar Load through opaque surfaces

3.4.1 Design Parameters

3.4.2 Solar load through opaque surfaces calculation

3.5 People Heat Load

3.5.1 Design Parameters

 3.5.2 People load calculations

3.6 Heat gains through infiltration of air leakage

3.6.1 Design parameters

3.6.2 Heat gains through infiltration calculation

3.7 Lighting Load

3.7.1 Design Parameters

3.7.2 Lighting load calculation

3.8 Heat Gain from appliances

3.8.1

3.8.2

3.9

CHAPTER 4: Air Distribution System

4.1 Mixing System

4.2 Duct Sizing

4.2.1 Design Parameters

4.2.2 Branch ducting calculation

4.2.3 Main ducting calculation

4.3 Pressure losses

4.3.1 Design parameters

4.3.2 Pressure losses calculation

CHAPTER 5: Air Conditioning System

5.1 Chiller

5.1.1 Liquid receiver model selection

 5.2 Air handler

5.4.1 Air handler model specification

GLOSSARY OF TERMS

REFERENCES
 Introduction

Heating, Ventilation, and Air Conditioning, also known as HVAC is an evolving

trend that once started from a simple solution for prehistoric problems. For instance,
cavemen used to live under the rocks. They did not know the idea of constructing a
house. They also discovered that air is cooler underground. They are the inventors of
the first geothermal cooled homes. Similarly for the Egyptians, considering that they
used to hang wet reeds in windows to cool the air naturally as the air passes through
the plant, which becomes the basis for water-cooled air conditioning. Moving on until the
mid-1800s, the idea of cooling cities during high-temperature month is formally
introduced by an inventor named Dr. John Gorrie. Dr. Gorrie secured a patent for his
device capable of automatic control for air temperature and humidity using cooling coils.
Nowadays, modern practices enable the HVAC systems to stay hidden inside the house
or establishment frame. This technique allows the system to function at the maximum
possible comfortability to the people under less noise.

The HVAC's primary purpose is to provide the people their needs towards
thermal comfort and good air quality. The heating and air conditioning function by
moving the warm air, whether inside or outside the establishment to attain the desired
heat intensity while ventilation is responsible for the exchange of air within a space to
provide better air quality indoors. The air quality treatment requires the removal of
elements such as smoke, dust, odor, carbon dioxide, and airborne bacteria.

This HVAC system development plan is particularly focused on a 2021.72 square

meter testing area of a semi-conductor manufacturing facility, that aims to provide
efficient yet beneficial HVAC strategies. This requires several computations starting
from the summary of area allocation in which sectors of the manufacturing facility are
divided. Followed by the cooling load calculation in which different factors for HVAC
design parameters are to be considered, such as the geographical location of the
establishment, site orientation based on cardinal direction, heat transmission from the
structure, heat loads from the people, lighting, and appliances. This also includes solar
radiation and others that may affect the design dramatically.
 Afterward, an effective and mathematically calculated air distribution system
sufficient for the semi-conductor manufacturing facility will then be provided, this
includes the mixing system, duct sizing, and pressure losses appropriate for the proper
ventilation of the establishment. Moreover, followed by the design for the air distribution
system, the air conditioning system will then focus on chiller model selection and air
handler model specification. Lastly, the developed plan strategy for the HVAC system
will then be compared to its previous model to determine the improvements in its
design.
 Chapter I

Building Description

The first chapter provides the semi-conductor manufacturing facility descriptions,

beneficial for designing a good air-conditioning and ventilation system. Such
descriptions are the building location, floor plan layout, and the zone division.

1.1 Building Location

The design of a centralized air-conditioning system is made for a semi-conductor

manufacturing facility. It can be found in one of the most dynamic regions in the country,
the province of Laguna. Located in Laguna Technopark, on the western section. Figure
1.1 presents the vicinity around the facility.

Source: Google Maps 2021 Imagery

 Figure 1.1 Vicinity Map

Most of the buildings around the facility also work with electronic components,
Shown on Figure 1.1. A world-class economic processing zone with big factories, one of
those factories is where we will apply our centralized air-conditioning system. Sitting on
a 10.8-hectare lot, lies the semi-conductor manufacturing facility. An approximation of
the lot is shown in Figure 1.2 to better interpret the location of the facility.

Source: Google Maps 2021 Imagery

Figure 1.2 Lot Area of Amkor Technology

 These buildings located in Laguna, comprises 2 out of 4 of their assembly and
test factories, namely Building P3 and Building P4. Whereas, the centralized air-
conditioning system will only be applied to section P3 which focuses on testing of “ready
for market” semiconductors. The section provided has a total area of 2021.72 square
meter (m2) and is highlighted on Figure 1.3.

Source:
Google Maps 2021
Imagery

Figure 1.3 Area of

the design
application

Area to be
conditioned

1.2 Floor Plan

On the first section, a floor plan from the highlighted area from Figure 1.3 is
shown. It is a single-storey building used to manufacture semi-conductors. Having a
total area of 2021.72 m2, the provided floor plan will be the basis of the design of a
centralized air-conditioning system – Shown in Figure 1.4.
 Figure 1.4 Semi-Conductor Manufacturing Facility’s Floor Plan

1.3 Zoning

The zoning section provides the zone description together with its area allocation
and orientation. The division of zone is beneficial for the design of air-conditioning and
ventilation system due to the facility’s lack of walls separating each testing procedures,
in which makes the facility wide open. The zoning is also used to produce a more
precise outcome.
 Figure 1.5 represents the division of zones of the testing area of a semi-
conductor manufacturing site. It enables separate strategies of air-conditioning and
ventilation system to be applied depending of the zone inclusions.
Zone 5 Zone 3

Zone 4

Zone 2
Zone 1

Figure 1.5 Zoning Layout

Zone 1

The first zone is divided into five (5) different segments, all chosen area are not
directly involved on the testing procedure of a semi-conductor. Such areas are the
following; Service lobby, machine room, smock room, maintenance room, and electrical
room. The zone one covers a total floor area of 414.94 m2.
 Figure 1.6 Zoning Layout of Zone 1

Zone 2

Zone two consist of three (3) segments, chosen segments are within the range of
loading and unloading location. The area also serves as the main passage for the
workers and materials needed for the testing and packaging of semi-conductor. The
zone covers a total floor area of 618.3 m2.

Figure 1.7 Zoning Layout of Zone 2

Zone 3
 The zone is particularly focused on the semi-conductor testing, in which involves
the usage of different sets of machineries and manpower. The zone is divided into two
(2) segments, loading and unloading area, and the burn-in testing area. The testing
area can detect early failures in the product and enables the semi-conductor to reduce
the potential defects and conflicts during the usage of the product. The burn-in
procedure typically ranges for a time period of 48 to 168 hours. The zone three covers a
total floor area of 381.84 m2.

Figure 1.8 Zoning Layout of Zone 3

Zone 4

Zone four consist of wafer probing equipment used for the process of testing
each die performance and status of a semi-conductor. The portion contains one (1)
handling equipment and four (4) Eagle Test System (ETS-1) which is designed to
perform a high-volume production testing of integrated circuits. The zone four covers a
total floor area of 308.84 m2.

Figure 1.9 Zoning Layout of Zone 4

Zone 5

The zone five includes different sets of handling machineries used to produce an
excellent packaging for semi-conductor that passes the testing procedures on zones
four and five. Zone five covers a total floor area of 298.29 m2.
 Figure 1.10 Zoning Layout of Zone 5
Table 1.1 Zoning Description

Zone No. Orientation Segment Area

West Service Lobby 177.08 m2
Machine Room 20.40 m2
1 Smock Room 73.87 m2
South
Maintenance Room 8.52 m2
Electrical Room 135.07 m2
BIB Cart Parking Area 33.34 m2
2 South Handler Parking Area 50.41 m2
Loading & Unloading Area 534.55 m2
Burn-In Loading & Unloading Area 99.09 m2
3 East
Burn-In Testing Area 282.26 m2
4 North Water Probing Area 308.84 m2
5 North Semi-Conductor Handling Area 298.29 m2
Total 2021.72

Table 1.1 presents the summary of five (5) different zones with its corresponding
area allocation; it also shows the area dimension together with its orientation for a
specific portion, in a semi-conductor manufacturing facility having a total floor area of
2021.72 m2.
 Chapter II

Cooling Load Calculations

The second chapter involves calculations of the design parameters beneficial for
the application of air-conditioning system strategies that is used for the testing portion of
a semi-conductor manufacturing facility. Such computations are the following;
conduction through external surfaces, transmission load, and others that may affect the
cooling load value. The chapter will also include the tabulated data from the
computation that is used for the evaluation of the air-conditioning design outcome.

2.1 Design Temperature

Upon the application of air-conditioning and ventilation system for the testing site
of semi-conductor manufacturing facility, factors such as the inside and outside dry bulb
temperatures and humidity ratio should first be considered. It will help on further
computations to evaluate the necessary amount of cooling loads to be applied in the
structure.

The semi-conductor manufacturing facility which is located at Binan Laguna had

an outside temperature value of 39 oC as highest temperature recorded during the
month of May 2013 and 23 oC for the lowest temperature recorded during the month of
January 2010. The data covers the year frame of 2010 up to 2020.
Source: Metroblue Meteorological Service

Figure 2.1 Climate Average

Table 2.1 considers the highest temperature recorded during year 2010 to 2020
from the Figure 2.1 as the main basis of the facility’ air-conditioning design.

Table 2.1 Meteorological Data

Temperature
Dry Bulb 39 oC
Relative Humidity 72 %
kg moisture
Humidity Ratio 0.027
kg dry air
Source: Stoecker 1979, Time and Date
 With the use of dry bulb and relative humidity provided for year 2010 to 2020 in
the province of Laguna, humidity ratio can now be identified with the use of
psychometric chart at a standard barometric pressure of 101.325 kPa. The gathered
readings are critical for the calculation of design temperature and relative humidity
needed in the semi-conductor manufacturing facility.

kg mo .
≈ 0.032
kg d . air

Figure 2.2 Psychrometric Chart

By using the data from the Table 2.1, the outside relative humidity can now be
calculated with the use of equation pertaining to humidity ratio (W).

ps
W =0.622 ( Stoecker , 1979)
pt − ps

Where:

kg mo .
W =Humidity Ratio; W =0.032
kg d .air
 pt =atmospheric pressure ; pt =101.325 kPa

ps =∂ pressure of the moisture

By substitution of values to humidity ratio equation,

0.032
kg moisture
kg dry air
=0.622
( ps
101.325 kpa−p s )
ps =4.96 kPa

Now, for the equation of relative humidity.

existing ∂ pressure of the moisture ps

RH = = (Stoecker, 1979).
saturation pressure of the pure water at the same temp . p sat

The saturation pressure ( p sat ) of the pure water at the same temperature can be
found in the book of Stoecker 2nd edition, Table A-1 using interpolation.

psat at dry bulb temp . of 39 ℃

psat =6.9995 kPa∨7 kPA

By using the relative humidity equation:

ps
RH = (Stoecker , 1979)
p sat

4.22 kPa
RH = × 100 %
7 kPa

RH =60.29 %

The 2007 ASHRAE handbook – HVAC application was used in determining the
standard temperature range for a semi-conductor manufacturing facility. The handbook
provided the dry bulb and relative humidity value necessary for a good air-conditioning
design.
 Source: ASHRAE, 2007

Figure 2.3 Temperature and Humidity for Industrial Air Conditioning

Overall, the summary of design temperature and relative humidity are presented
at the Table 2.2 for evaluation. The mean of relative humidity range provided by 2007
ASHRAE handbook was used for the inside desired value.

Table 2.2 Summary of Design Temperature and Relative Humidity

Temperature Relative Humidity

Outside 39 oC 60.29 %
Inside 20 oC 45 %
 Hallway 25 oC 45 %

2.2 Air Film/ Air Resistance

The thermal resistance of air or air film is small relative to the thermal resistance
of most wall assemblies. But is still considered on solving the over-all resistance of the
design. Air Film, is an imaginary film on the surface on a building, either exterior or
interior. On the interior wall surface, the thermal resistance of air is shown in Table 2.3
as still air. It is dependent of the position of the surface and the direction of heat
transfer. On the exterior wall surface, air is moved by wind, hence the thermal
resistance is also dependent with windspeed. Table 2.3 shows the standardized value
of surface resistance on the interior and exterior wall. (ASHRAE, 2009)

Thermal
Heat Transfer
Resistance, R
Direction
( hr ∙ f t 2 ∙ ℉ /BTU )
Vertical Still Air Horizontal 0.68
7.5 mph Wind Any 0.25
Source: ASHRAE, 2009 – Surface Resistance

Table 2.3 Air Resistance Value

2.2 Solar Heat Gain through Exterior Structure

Heat gain by a direct contact from sunrays affects the inside temperature of the
establishment the most. Whereas, all surfaces of the manufacturing facility having an
open contact to the sun will be considered throughout the computation to determine the
required amount of cooling load in order to attain the desired inside temperature. Such
surfaces involved are the external walls, roof and windows. The equation provided
below will be used as the main basis to compute the thermal conduction through
exterior structures.

Q=U × A ×CLT D c (PITA, 2002)

 Where:

Q=cooling load for roof , wall ,∨glass

U =Overall heat transfer coefficient for roof , wall ,∨glass

A=Area of roof , wall ,∨glass

CLT D c =Corrected cooling load temperature difference

Conduction Through Transparent Surfaces

Transparent surfaces typically glass windows are considered as an important

matter in designing a good air-conditioning system. It is because of the thin and
translucent characteristic of the glass that allows the solar heat to simply pass through
the establishment. The specification of glass materials used for the whole testing facility
of semi-conductor manufacturing site is shown in the Table 2.5 Table 2.4. It includes the
zone where the glass is located, type of glass, number of windows to be used, the
thickness, and its thermal conductivity.

Table 2.4 Glass Specifications

Zon Thermal
Type Description Quantity
e Conductivity
6mm Acrylic/ Polycarb (Swinging
3 Single Glazing – Insulated Fiberglass/ Vinyl 1 4.18 W/(m K)
Frame)
3.2 mm glass (Sliding –
5 Single Glazing 2 5.20 W/(m K)
Wood/Vinyl Frame)
Source: 2017 ASHRAE Handbook – Fundamentals and Stoecker 1979

Figure 2.2 shows the type of window being considered for each zone. However,
the portion of the semiconductor manufacturing facility whereas the design will be
applied only consist a total of three (3) windows based on the floorplan layout presented
in chapter one.

Source: 2017 ASHRAE Handbook – Fundamentals and Stoecker 1979

Figure 2.4 Window Specification.

Figure 2.5 presents the glass specification used for zones 3 and 5 for the
establishment. It also demonstrates a decrease in temperature as the heat passes
through the glass. However, there will be no considered solar heat gain through
transparent surfaces in the facility due to the lack of external windows on the floorplan
layout that has a direct contact to the sun, since the windows are only located in
between zones as shown in the zoning description on chapter one. Yet, the glass and
other transparent surfaces specification that can be found within the site are still
gathered for the computation of the transmission load.
 T outside

T inside

Figure 2.5 Glass Material

Solar Heat Gain Through Exterior Walls

The external wall design specifications are shown in the Figure 2.6 to better
visualize the wall composition. The type of material, thermal resistance and wall
thickness are also included to be used in cooling load computation. However, the
cooling load required for the conduction through facility’s ceiling will be disregarded on
the computation, since the facility is a two-storey building and the main facility to be
airconditioned is located on the ground level.
Figure 2.6 Exterior Wall Composition

CEMENT PLASTER

LIGHTWEIGHT AGGREGATE

CEMENT PLASTER FINISH

Figure 2.6 Exterior Wall Composition

 The exterior wall specification of semiconductor manufacturing facility is
composed of three (3) different internal layers excluding the outside and inside air films.
Figure 2.7 shows the wall material description based on their respective order.

Figure 2.7 Exterior Wall Thickness

OUTSIDE & INSIDE AIRFILM

 Figure 2.7 Exterior Wall Thickness

The external wall composition is used as the main basis in the computation for
the conduction through external wall surfaces that passes through the establishment.
Table 2.5 presents the material resistance data of the wall whereas all necessary data
such as the thermal conductivity, thickness and thermal resistance of the wall material
are indicated. Those value will set the limit of the heat that passes through the walls
from the outside environment into a lower temperature.

Table 2.5 Material Resistance Data

Thermal Thermal
Thickness (x)
Surface Type Conductivity (k) Resistance (R)
(mm) 2
W /(m∙ K ) (m ∙ k )/W
Outside Air film - - - 0.029
Gypsum
Exterior Material 10 0.16 -
wallboard
Lightweight
Concrete 127 0.20 -
aggregate
Insulating Mineral wool
25 0.04 1.940
Material fiber
Cement
Interior Material 15 0.72 -
plaster
Inside Air film - - - 0.120
Source: 2017 ASHRAE Handbook – Fundamentals and Stoecker 1979

Table 2.5 Material Resistance Data

Thickness Thermal Thermal

Surface Type (x) Conductivity (k) Resistance (R)
2
(mm) W /(m∙ K ) (m ∙ k )/W
Outside Air film - - - 0.029
Exterior Material Cement plaster 10 0.72 -
Concrete Lightweight 127 0.20 -
 aggregate
Cement plaster
Interior Material 15 0.72 -
finish
Inside Air film - - - 0.120
Source: 2017 ASHRAE Handbook – Fundamentals and Stoecker 1979

In determining thermal resistance for the overall heat transfer coefficient of

exterior wall:

x
R=
k

For Exterior Material (Cement Plaster):

R=
10 mm ( 10001 mmm ) =0.014 m ∙ k
2

W W
0.72
m∙ k

By following the same solving procedure for the remaining wall thermal
resistance value:

2
m ∙k
Rtotal= ( 0.029+0.014+ 0.635+0.02+0.120 )
W
2
m ∙k
Rtotal=0.818
W

While, for the computation of the wall overall heat transfer coefficient that will be
used in the computation of thermal transmission load.

1
U= 2
m ∙k
0.818
W

W
U =1.22 2
m K
 The surface area that will undergo thermal transmission process is provided in
Table 2.8. One example computation for wall area is also given below for clarification.

For Hallway :W A =4 m× 48 m

2
W A =192 m

By following the same solving procedure for the remaining wall area:

Table 2.6 Dimensions of Walls Exposed to Sunlight

Wall Wall
Zone Wall Area
Segment Orientation Height Lengt
No. (m2)
(m) h (m)
Hallway West 4 48 192
Service Lobby South 4 7.5 30
Machine Room South 4 4.3 17
1
Smock Room South 4 9 36
South 12 48
Electrical Room 4
East 10.2 40
BIB Cart Parking Area East 4 2.85 11
2
Handler Parking Area South 4 6.3 25
Burn-In Loading &
East 4 7 28
Unloading Area
3
North 11.5 46
Burn-In Testing Area 4
East 27.6 110
4 Water Probing Area North 4 18 72
Semi-Conductor
5 North 4 14.8 59
Handling Area
Source: Pita, 2002

By applying the thermal conduction through external surfaces equation to

determine the required cooling load value for each zone. Having an overall heat transfer
W
coefficient (U) of 1.22 2 for every wall segment due to similar wall configuration.
m K
 For the sample computation of conduction though external walls for zone 1
hallway:

Q=U × A ×CLT D c

W
Q=1.22 2
( 240 m2 ) ( 14 ) K
m K

Q=3279.14 W

Similar process for the remaining values will be used for further computation.
Hence, the computed data is listed on the Table 2.7 excluding the overall heat transfer
coefficient (U) due to the similar wall configuration value.

Table 2.7 Conduction Through External Walls

Thermal
Zone Wall Area CLTDc
Segment Orientation Conduction
No. (m2) (K )
(W)
Hallway West 192 14 2268
Service Lobby South 30 19 480.94
Machine Room South 17 19 275.74
1 Smock Room South 36 19 577.13
Maintenance Room - - - -
South 48 19 769.5
Electrical Room
East 40 19 769.5
BIB Cart Parking
East 11 19 654.075
Area
Handler Parking
2 South 25 19 182.76
Area
Loading &
- - - -
Unloading Area
3 Burn-In Loading &
East 28 19 -
Unloading Area
Burn-In Testing North 46 19 448.88
 North 46 19 1769.85
Area
East 110 19
4 Water Probing Area North 72 19 737.44
Semi-Conductor
5 North 59 19 1154.25
Handling Area
Total 10088
Source: Pita, 2002

Table 2.7 Solar Heat Gain Through External Walls

Thermal
Zone Wall Area CLTDc
Segment Orientation Conduction
No. (m2) (K )
(W)
Hallway West 192 14 3279.36
Service Lobby South 30 19 695.4
Machine Room South 17 19 394.06
1 Smock Room South 36 19 834.48
Maintenance Room - - - -
South 48 19 1112.64
Electrical Room
East 40 19 927.2
BIB Cart Parking
East 11 19 254.98
Area
Handler Parking
2 South 25 19 579.5
Area
Loading &
- - - -
Unloading Area
Burn-In Loading &
East 28 19 649.04
Unloading Area
3
Burn-In Testing North 46 19 1066.28
Area East 110 19 2549.8
4 Water Probing Area North 72 19 1668.96
 Semi-Conductor
5 North 59 19 1367.62
Handling Area
Total 15379.32
Source: Pita, 2002

2.3 Thermal Transmission Load

Thermal transmission load section discusses the sensible heat load that affects
the cooling load in different ways. That includes the gain or loss of heat passing through
each facilities section. It includes the doors, walls, transparent surfaces, and others
where temperature difference occurs. However, the transmission load on the ceiling of
semiconductor manufacturing facility is not considered due to the 2 nd floor level of the
establishment where room temperature is also based from 20 oC.

The primary equations that will be involved in the thermal transmission load:

QTT =UA ( t o−t i ) (Stoecker, 1979)

1 x
U= ¿ ; R= (Stoecker, 1979)
R total
k

Where:
QTT = Amount of thermal transmission load , ( W )

U =Overall heat−transfer coefficient ,

( mW∙ K )
2

A=Surface area,(m¿ ¿2) ¿

t o−t i=outside−inside temperature difference ,( K)

( )
2
¿ m ∙k
Rtotal=Total thermal resistance ,
W

x=thickness of wall(m)

k =thermal conductivity ( m∙WK )

 The semiconductor manufacturing facility consist only two (2) zones which has a
wall that separates each manufacturing rooms/ segments. These zones are zones 1
and 3 whereas, zone 1 consist rooms that are not directly involved in the semiconductor
testing such as hallway, service lobby, machine room, smock room, maintenance room
and electrical room. While zone 3, whereas burn-in testing area can be located. The
remaining zones such as 2, 4 and 5 are not divided by walls based on the facilities
floorplan layout which will not produce transmission load due its temperature difference
amounting to zero.

Table 2.8 Transmission Load Through Walls

Overall
Heat Surface
Wall Transmission
Transfer Temperature
Zone Segment Orientation Area Load
Coefficient Difference -
(m2) (W)
W ¿−t i (K )
( 2
¿
m ∙K
Hallway West 0.693 192 25-20 665.28
Service
South 0.693 30 25-20 103.95
Lobby
Machine
South 0.693 17.2 25-20 59.59
1 Room
Smock
South 0.693 36 25-20 124.74
Room
Electrical South 0.693 48 25-20 166.32
Room East 0.693 40.8 25-20 141.37
Loading &
2 Unloading East 0.693 61.6 213.44
Area
Burn-In North 0.693 28 25-20 97.02
3
Testing Area East 0.693 110 25-20 381.15
5 Semi- North 0.693 76 25-20 263.34
Conductor
 Handling
Area
Total 2216.21
Source: 2017 ASHRAE Handbook – Fundamentals and Stoecker 1979

Glass Thermal Transmission Load

The thermal transmission load on the transparent surfaces only involves two (2)
sets of windows that can be located in between zones. These two (2) windows will be
used for the computation of the amount of heat transfer from a zone to another. The
windows are located in zones 3 and 5 as shown in the floorplan layout (Chapter 1).
Table 2.9 shows the transmission load through the establishment windows.

Table 2.9 Transmission Load Through Windows

Heat Surface
Transfer Area temperature Transmission
Zone Type Description 2
Coefficient (m ) difference - Load (W)
2
(W /m ∙ K ) ¿−t i (K )
6mm Acrylic/
Polycarb
Single (Swinging –
3 4.18 3 20-20 0
Glazing Insulated
Fiberglass/
Vinyl Frame)
3.2 mm
glass
Single
5 (Sliding – 5.20 3 25-20 78 x 2
Glazing
Wood/Vinyl
Frame)
Total 156
Source: 2017 ASHRAE Handbook – Fundamentals and Stoecker 1979
Door Thermal Transmission Load

The testing portion of the semiconductor manufacturing facility has a total of

fifteen (15) doors which covers the overall zones. The door thermal transmission load
will change depending on its location and the door specification. Doors with high
thermal transmission load are usually located on the outer most part of the
establishment whereas exposed to a 39 oC temperature. There are two main door
specifications used in the establishment in which shown in the Figure 2.4. The door
dimensions provided below are in the unit of millimeter (mm). The doors specification
used in the establishment are identical, which is made of foam insulated steel stab with
metal edge in steel frame, having 6% single glazing (22 x 8 in. lite) (2009 ASHRAE
Handbook).

Figure 2.8 Single & Double Door Dimensions

The thermal load transmission through the establishment’s door is provided in

the Table 2.10 whereas the necessary parameters such as the type of door to be used,
its quantity, and the heat transfer coefficient of the door are included.
 Table 2.10 Transmission Load Through Doors

Heat Transfer
Transmission
Zone Segment Type Quantity Coefficient
2
Load (W)
(W /m ∙ K )
Hallway - - - -
Single
Service Lobby 1 0.44 13.31
Door
Single
Machine Room 1 0.44 13.31
Door
1
Double
Smock Room 4 0.44 106.44
Door
Single
Maintenance Room 2 0.44 53.22
Door
Electrical Room - - - -
Double
BIB Cart Parking Area 1 0.44 13.31
Door
2 Handler Parking Area - - - -
Loading & Unloading
- - - -
Area
Burn-In Loading & Double
1 0.44 13.31
Unloading Area Door
3
Double
Burn-In Testing Area 2 0.44 53.22
Door
4 Water Probing Area - - - -
Semi-Conductor Handling
5 - - - -
Area
Total 266.12
Note: The dash (-) in the Table 2.10 means that the segment is an open area
Source: 2017 ASHRAE Handbook – Fundamentals and Stoecker 1979
and does not require any use of doors.
 Table 2.11 Summary for Transmission Load

T. L. T. L.
T. L.
through through
Zone Segment Orientation through Total (W)
windows doors
walls (W)
(W) (W)
Hallway West 2430 - - 665.28
Service Lobby South 379.69 - 13.31 117.26
Machine Room South 217.69 - 13.31 72.9
Smock Room South 455.63 - 106.44 231.18
1
Maintenance
- - - 53.22 53.22
Room
South 607.5 - - 166.32
Electrical Room
East 607.5 - - 141.37
BIB Cart Parking
East - - 13.31 13.31
Area
Handler Parking
2 South - - - 0
Area
Loading &
- - - - 213.44
Unloading Area
Burn-In Loading
& Unloading East - - 13.31 13.31
3 Area
Burn-In Testing North 354.375 - 53.22 150.24
Area East 1397.25 - - 381.15
Water Probing
4 North - - - 0
Area
Semi-Conductor
5 North - 156 - 419.34
Handling Area
Total 2638.32
 Table 2.11 present the total transmission load passing through establishment’s
internal surfaces. The table also indicated each load for heat that passes through
internal walls, windows, and doors having a total of 2638.32 watts.

2.4 Lighting Load

Another factor affecting the amount of cooling load required is the lighting load,
whereas it emits heat and produces light simultaneously. The lighting load computation
will include the calculation of the required number of lumens per square area in order to
determine the power rating necessary for the computation of lighting load. It will also
help to determine the number of required lightbulbs involving the calculation of required
number of lumens per area. The design for the number of lumens per square area will
be set as 350 as the standard value. Figure 2.9 presents the type of light bulb to be
used on the whole testing portion of semiconductor manufacturing facility.

Source: Philips (https://www.lighting.philips.com.ph/consumer/p/led-bulb/8718699619046/specifications)

Figure 2.9 LED Light Bulb Specification

Table 2.12 presents the calculated value for the required lumens per light bulb
using the standard lumens per square area of 200. The equation used on the table is
presented below:

No . of lumens
No . of Light Bulb=
Lumens per bulb

For the sample computation of the minimum number of light bulbs required for
hallway:
 34650 cd
No . of Light Bulb=
cd
2150
bulb

No . of Light Bulb=16.11≈ 16 pieces of 24 W LED Light

By following the sample computation for the remaining segments.

Table 2.12 Number of Light Bulb Required

Standard
Wall Floor Lumens Required No. of
Zone
Segment Area Area per Area Lumens Light
No.

( )
(m2) (m2) cd (cd ) Bulb
2
m
Hallway 240 68 180 12240 6
Service Lobby 37.5 28 180 5040 2
Machine Room 21.5 15 180 2700 1
1 Smock Room 45 27 180 4860 2
Maintenance
- 7.2 180 1296 1
Room
Electrical Room 60 75 180 13500 6
BIB Cart Parking
51 34.16 180 6148.8 3
Area
Handler Parking
2 14.25 42 180 7560 4
Area
Loading &
237.5 480 180 86400 40
Unloading Area
Burn-In Loading
3 & Unloading 195.5 330 180 59400 28
Area
Water Probing
4 57.5 255 180 45900 21
Area
 Semi-Conductor
5 90 275.5 180 49590 23
Handling Area
Source: Philips

The equation for determining cooling load due to heat gain from lighting is:

Q=3.4 ×W × BF ×CLF ; (PITA , 2002)

Where:

Q=Cooling Load ¿ lighting

W =Lighting Capacity

B=Ballast Factor

CLF =Cooling Load factor for Lighting

Source: Stoecker 1979

Figure 2.10 Cooling Load Factor for Lighting

Figure 2.10 shows the table used for identifying the amount of CLF needed for
semiconductor manufacturing operation which it involves 16 full hours of operation.

For sample computation for hallway :

Q=3.4 ×W × BF ×CLF =3.4 × ( 24 Watts ×6 Bulb ) ×1 × 0.89

 Q=435.744 Watts

Therefore, by following the same solving procedure, the value for total occupants’
load during the working hours in Semi-Conductor Manufacturing Site within the testing
facility is shown in Table 2.13 below

Table 2.13 Lighting Load

Lighting
Ballast
Zone Segment Type Quantity CLF Load (W)
Factor

Hallway 24 W LED 6 - 0.89 435.744

Service Lobby 24 W LED 2 - 0.89 145.248

Machine Room 24 W LED 1 - 0.89 72.624

1
Smock Room 24 W LED 2 - 0.89 145.248

Maintenance Room 24 W LED 1 - 0.89 72.624

Electrical Room 24 W LED 6 - 0.89 435.744

BIB Cart Parking Area 24 W LED 3 - 0.89 217.872

2 Handler Parking Area 24 W LED 4 - 0.89 290.496

Loading & Unloading Area 24 W LED 40 - 0.89 2904.96

Burn-In Loading & Unloading

24 W LED 28 - 0.89 2033.472
3 Area
Burn-In Testing Area 24 W LED 21 - 0.89 1525.104
4 Water Probing Area 24 W LED 23 - 0.89 1670.352
Semi-Conductor Handling
5 24 W LED 6 - 0.89 435.744
Area
Total 10385.232
 Table 2.13 displays the cooling load necessary for the lighting load amounting to
a total number of 10385.232 watts.

2.5 Occupants Load

The heat gain from people is composed of two parts, sensible heat and the latent
heat resulting from perspiration. Some of the sensible heat may be absorbed by the
heat storage effect, but not the latent heat. The equations for cooling loads from
sensible and latent heat gains from people are:

For sensible heat gain:

Qs =q s × n ×CLF ;(PITA ,2002)

For latent heat gain:

Ql=ql ×n ;(PITA ,2002)

Where:

Qs =Sensible heat gain, (W )

Ql=Latent heat gain ,(W )

q s=Sensible heat gain per person , ( person

W
)
q l=Latent heat gain per person , ( person
W
)
n=Number of occcupants

CLF =Cooling Load Factor

The number of people displayed in the site floorplan is used for the calculation of
the number of occupants. However, 50% of the total occupants displayed in the floor
plan will be added for allowance, this will help the cooling load calculation to be
consistent on the actual load of people during the working hours.

Figure 2.11 presents the sensible and latent heat gain within the manufacturing
facility during working hours. The values provided will be used as a variable in
calculating total occupant’s load.

Source: 2017 ASHRAE Handbook – Fundamentals

Figure 2.11 Sensible and Latent Heat Gain from Occupants

By converting the units provided by the 2017 ASHRAE handbook - fundamental

from English to SI.

Table 2.14 Sensible and Latent Heat Gain from Occupants

Degree of Activity Sensible Heat (W) Latent Heat (W)

Light bench work 80.59 139.20
Walking (Approx. 3mph) 109.90 183.17
Heavy Work 169.98 254.97
Heavy Machine Work (Lifting) 186.10 282.81

On the other hand, the CLF for people is shown in Figure 2.12 in which a total of
10 hours in space and time after entry will be used, having an overall cooling load factor
of 0.89 for people within the facility. Meanwhile, below also shows the sample
computation on hallway for the values presented in the Table 2.14.

Source: Pita - 2002

Figure 2.12 Cooling Load Factor for People

For sensible heat gain :

Qs =q s × n ×CLF =109.90W × 10 ×0.89

Qs =978.11Watts

For latent heat gain:

Ql=ql ×n=183.17 W ×10

Ql=2549.7 Watts

For total heat gain :

Qt =Qs +Ql =( 978.11+2549.7 ) Watts

Qt =3527.81Watts

Therefore, by following the same solving procedure, the value for total occupants’
load during the working hours in Semi-Conductor Manufacturing Site within the testing
facility is shown in Table 2.13 below.

Table 2.13 Occupant’s Load

Total
No. of Sensible Latent
Zone Segment CLF Occupants
Occupants Heat (W) Heat (W)
Load (W)
 Hallway 10 0.89 978.11 1831.70 2809.81

Service Lobby 7 0.89 502.08 502.08 1004.15

Machine Room 2 0.89 143.45 278.40 421.85

1
Smock Room 5 0.89 489.06 915.85 1404.91

Maintenance Room 3 0.89 453.85 764.91 1218.76

Electrical Room 4 0.89 662.52 1131.24 1793.76

BIB Cart Parking Area 2 0.89 195.62 219.80 415.42

2 Handler Parking Area 2 0.89 302.56 339.96 642.52

Loading & Unloading Area 20 0.89 3312.58 3722.00 7034.58

Burn-In Loading & Unloading

9 0.89 1490.66 1674.90 3165.56
3 Area
Burn-In Testing Area 7 0.89 1058.98 1784.79 2843.77
4 Water Probing Area 10 0.89 717.25 805.90 1523.15
Semi-Conductor Handling
5 15 0.89 2484.44 4242.15 6726.59
Area
Total 31004.82

2.6 Equipment Load

The heat output from motors and the equipment driven by them results from the
conversion of the electrical energy to heat. The proportion of heat generated that is
gained by the air-conditioned space depends on whether the motor and driven load are
both in the space or only one of them is (PITA, 2002). Below is the equipment load
equation provided by the handbook.

Q=P × N ; (PITA , 2002)

 Where:

P=Equipment power rating

N=No . of equipment

Table 2.14 Equipment Load

Equipment
Equipment
Equipment Description Quantity Power
Load
Rating
Zone 1
PC System 2 60 120
Zone 2
RASCO JAGUAR Handler 4 1500 6000
SYNAX SX1101 Handler 3 1500 4500
MCT 5100 DUAL Probing Machine 9 2250 20250
MCT 5100 ePRO Handler 3 1500 4500
PC System 8 60 480
EPSON
Tester 2 1500 3000
NS800 Series
EPSON
Photo Printer 1 500 500
ASL3000
SRM MAV-068 Handler 1 1150 1150
EPSON
Handler 11 1200 13200
NS6040
Zone 3
DCW-336F-MP550 Burn-In Oven 7 1500 10500
DCA-336F-MP Burn-In Oven 7 1500 10500
MP-1406F Burn-In Oven 5 800 4000
MP-206F-WAS Burn-In Oven 9 1500 13500
MP-336F Burn-In Oven 2 1000 2000
DC 136 Burn-In Oven 1 3500 3500
DC-1506-WAS Burn-In Oven 1 3220 3220
AEHR MAX2B Burn-In Oven 3 500 1500
AEHR ATS-12100 Burn-in Oven 13 900 11700
PC SYSTEM 2 60 120
Zone 4
MCT 5100 DUAL Probing Machine 1 800 800
Wafer Inspection
MTT ETS-1 4 1200 4800
Equipment
PC System 4 60 240
Zone 5
 MCT 3608 E Handler 2 500 1000
RASCO MAV-0 Handler 1 1500 1500
RASCO MAV-044 Handler 1 1500 1500
RASCO ePRO Handler 3 580 1740
AETRIUM 5050
Handler 1 2000 2000
MAV -040
SMR MAV-02. Handler 3 950 2850
MCT AETRIUM
Handler 9 800 7200
EPRO
MCT 3608 ePRO Handler 17 500 8500
ASECO (S2-006
Handler 2 1500 3000
SENTRY 20)
MCT 5100 ePRO Handler 4 400 1600
PC System 10 60 600
Total 151570

Table 2.14 presents the total equipment load for the testing and packaging sector
of a semiconductor manufacturing facility. Some of the equivalent value shown for each
equipment rating are derived from the machine’s manual and manufacturer. However,
some equipment that shows no power rating information from the said sources are
obtained from alternative machines that has similar equipment configuration. For
instance, due to lack of information regarding to the power rating of Rasco MAV-0 on its
equipment manual and different HVAC handbooks, the power rating being considered is
from MCT SO 2800 that has the same function and technical configuration but different
brand and model.

2.7 Infiltration

Infiltration is defined as the flow of outside air into the building in either through
cracks around windows, other unintentional openings, or through the normal use of
exterior doors. (ASHRAE, 2009). It has a significant influence on the cooling load for it
produces both sensible and latent heat gain to the rooms (Pita, 2002).

The amount of sensible heat gain can be calculated using the sensible heat
equation:

Qs =M C p ΔT (Pita , 2002)
 Where:

Qs =The heat required ¿ cool outdoor ( warm ) air ¿ room temperature , BTU /hr

M =Weight of flow rate of outdoor air infiltration ,lb/hr

BTU
C p=Specific heat of air , ∙F
lb

Δ T =Temperature difference between outdoor∧indoor air

The equation above from Pita, gives us a value of (m) measured in lb/hr.
However, flow rates in HVAC are usually measured in ft 3/min (CFM). Hence, converting
the lb/hr value of (m) to ft3/min gives us the equation:

Qs =1.1 x CFM x ΔT (Pita ,2002)

Where:

Qs =Sensible heat gain¿ infiltration , BTU /hr

3
CFM =Air infiltration flow rate , f t /min

Δ T =Temperature difference between outdoor∧indoor air , ℉

Outdoor air, infiltrating the building, is often less humid than the inside air. At
some point, the air humidity ratio in the building may fall to an unacceptable level for
comfort. To maintain comfortable air humidity ratio, water vapor must be added. The
addition of this moisture requires latent heat vaporization of water (Pita, 2002). The
equation for the latent heat is expressed in the equation:

Ql=0.68 x CFM x ( W ' i−W ' o ) (Pita, 2002)

Where:

Ql=Latent heat required for infiltration , BTU /hr

3
CFM =Air infiltration Rate , f t /min
 ' '
W i∧W o=Indoor∧outdoor humidity ratio , grains water /lb dry air

In solving the CFM, doors and windows are both considered and calculated by
using “crack method”. Pita describes the crack method as the accurate estimate of the
rate of infiltration per foot of crack opening established by using data from various
energy standards. Figure 2.13 presents the typical allowable design air infiltration rates
through exterior doors and windows.

Source: Pita – 2002

Figure 2.13 Infiltration rates

The equation to calculate CFM from cracks on windows is shown below:

CFM =I ∙ L

Where:

3
CFM =Air infiltration flow rate , f t /min

CFM
I =Infiltration rate through cracks ,
ft

L=Length of crack on windows , ft

There are a total of 3 windows in the semi-conductor facility. Table 2.15 presents
a data table for these windows in which the CFM from the crack openings are
calculated. In addition, the design of the windows is a standard double-hung type 3 ft W
x 4 ft H window. Where the length of crack is:

L=3 ( 3 ) +2 ( 4 )=17 ft
Zone Type Infiltration Rate Quantity Length of crack, ft CFM, ft3/min

3 Double-hung 0.37 CFM/ft of crack 1 17 6.29

5 Double-hung 0.37 CFM/ft of crack 2 17 12.58

Table 2.15 CFM of Windows

Source: Pita – 2002

The calculation of CFM from doors also came from the Infiltration rates of Pita,
shown on Figure 2.14.

Source: Pita - 2002

Figure 2.14 Infiltration rates for door usage

For a sample computation, the calculation for the infiltration load on zone 3 is
presented below whereas,

QT =Qs +Ql

For the sensible heat:

Qs =1.1 x CFM x Δ T

( )
3
ft
Qs =1.1 6.29 ( 77−68 ) ℉
min

( )
BTU 1 Watts
Qs =62.271 =18.26 Watts
hr BTU
3.41
hr
 For the latent heat:

Ql=0.68 x CFM x ( W ' i−W ' o )

( )
3
ft grains water
Ql=0.68 6.29 ( 0.032−0.027 )
min lb dry air

( )
BTU 1 Watts
Q l=0.0214 =0.0063 Watts
hr BTU
3.41
hr

For total infiltration load:

QT =18.26+0.0063

QT =18.27 Watts

By following the same solving procedure for zone 5:

Table 2.16 Infiltration Load

Sensible Heat Latent Heat Total Infiltration Total Infiltration

Zone
(BTU/hr) (BTU/hr) Load (BTU/hr) Load (Watts)
3 62.271 0.0214 62.29 18.27
5 124.54 0.0428 124.58 36.53
Total 54.80

2.8 Ventilation

The ventilation helps to improve indoor air quality by removing moisture, smoke,
smells, heat, dust, airborne bacteria, carbon dioxide, and other pollutants, as well as
controlling temperature and replenishing oxygen. The sensible and latent heat of
outside air is usually greater that that of the room air, so it becomes part of the cooling
load (Pita 2002). By the using the provided equation for both latent and sensible heat to
determine the amount of ventilation load necessary to attain the design indoor
temperature of the building.

For sensible heat:

 Qs =1.1 ( CFM ) TC ( Pita , 2002 )

Where:

BTU
Q s =Sensible cooling loads ¿ ventilation air ,
hr

3
CFM =air ventilation rate , f t /min

TC=Temperature hange between outdoor∧inside air , F

For latent heat:

Ql=0.68 (CFM ) ( W o' −W i ' ) ( Pita , 2002 )

Where:

BTU
Ql=Latent cooling loads ¿ ventilation air ,
hr

' '
W o −W i =Outdoor ∧inside humidity ratio , grains water /lb dry air

For the CFM:

CFM =ACH ( 60V )(Pita , 2002)

Where:

ACH=Segment air change per hour

The value used for the ACH required for the CFM computation are collected from
“Americraft Manufacturing Co. INC.” Table 2.17 presents the data for ACH whereas the
all values are extracted. ACH depends on the purpose or usage of the room, whereas
some rooms on the semiconductor manufacturing facility will require higher air change
per hour than the others. For instance, the zone 3 burn in testing air requires higher air
change compared on the zone 2 which does not govern any operating equipment.

Table 2.17 Number of Air Change per Hour

Source: https://americraftmfg.com/general-ventilation-requirements/

Hence for the sample computation of CFM on zone 2 using the ACH value
presented above:

CFM =ACH ( 60V )

( ( ) )
3
3.28 ft
556.16 m3
m f t3
CFM =10 = 3270.92
60 min

Table 2.18 Air Ventilation Rate

 Zone ACH Volume (m3) CFM (ft3/min)

2 10 556.16 3270.92
3 12 330 2328.98
4 12 255 1799.65
5 10 180 1058.63
Note: Zone 1 is not included in the table since ventilation is not necessary.

Now, for the sample computation of sensible and latent cooling loads for zone 2:

QT =Qs +Ql

For the sensible heat:

Qs =1.1 ( CFM ) TC

( )
3
ft
Qs =1.1 3270.92 ( 77−68 ) ℉
min

( )
BTU 1 Watts
Q s =32382.11 =9496.22 Watts
hr BTU
3.41
hr

For the latent heat:

Ql=0.68 x CFM x ( W ' i−W ' o ) (Pita, 2002)

( )
3
ft grains water
Ql=0.68 3270.92 ( 0.032−0.027 )
min lb dry air

( )
BTU 1 Watts
Ql=11.12 =3.26 Watts
hr BTU
3.41
hr

For total infiltration load:

QT =9496.22+3.26

QT =9499.48 Watts

Table 2.18 Ventilation Load

 Sensible Heat Latent Heat Total Ventilation Total Ventilation
Zone
(BTU/hr) (BTU/hr) Load (BTU/hr) Load (W)

2 32382.11 11.12 32393.23 9499.48

3 23056.90 7.92 23064.82 6763.88
4 17816.54 6.12 17822.66 5226.59
5 10480.44 3.60 10484.04 3074.50
Total 24564.45
 Table 2.17 Overall Cooling Load Equivalent

Conduction TOTAL
Transmissio Lighting Occupancy Equipment Infiltration Ventilation
Description through External (10% FS) TOR
n Load (W) Load (W) Load (W) Load (W) Load (W) Load (W)
Surfaces (W) WATTS

Zone 1
Hallway 3279.36 665.28 435.744 2809.81 0 0 0 7909.21 2.25

Service Lobby 695.4 117.26 145.248 1004.15 0 0 0 2158.26 0.61

Machine Room 394.06 72.9 72.624 421.85 0 0 0 1057.58 0.30

Smock Room 834.48 231.18 145.248 1404.91 0 0 0 2877.40 0.82

Maintenance
- 53.22 72.624 1218.76 120 0 0 1611.06 0.46
Room
Electrical Room 1112.64 307.69 435.744 1793.76 0 0 0 4014.82 1.14
Total Cooling Load in Zone 1 19628.3 5.58
Zone 2
BIB Cart Parking
927.2 13.31 217.872 415.42 0 0 0 1731.18 0.49
Area
Handler Parking
254.98 0 290.496 642.52 0 0 0 1306.80 0.37
Area
Loading &
579.5 213.44 2904.96 7034.58 53580 0 9499.48 81193.16 23.07
Unloading Area
Total Cooling Load in Zone 2 84231.1 23.93
 Zone 3
Burn-In Loading
& Unloading - 13.31 2033.472 3165.56 60540 0 0 72327.58 20.55
Area

Burn-In Testing
649.04 531.39 1525.104 2843.77 - 18.27 6763.88 13564.60 3.85
Area

Total Cooling Load in Zone 3 16859 4.79

Zone 4

Water Probing 5840

1066.28 0 1670.352 1523.15 0 5226.59 16859.01 4.79
Area

Total Cooling Load in Zone 4 16859.01 4.79

Zone 5

Semi-Conductor
2549.8 419.34 435.744 6726.59 31490 36.53 3074.50 49205.75 13.98
Handling Area

Total Cooling Load in Zone 5 49205.75 13.98

TOTAL COOLING LOAD 255816.41 72.68

 Finally, presented on the table above, the total cooling load required for the testing portion of a semiconductor
manufacturing facility equivalent to a total of 255,816.41 wattage for the whole zones and segments having 10% factor of
safety for other minimal heat gain not included in the computation. Overall, the total cooling load in terms of Ton of
Refrigeration amounts to 72.68 TOR.

11% 7%
4%

13%
SOLAR AND TRANSMISSION
LOAD
LIGHTING LOAD
OCCUPANTS LOAD
EQUIPMENT LOAD
INFILTRATION AND VENTI-
LATION LOAD

65%

Figure 2.15 Cooling Load Distribution

For observation purposes on which factor affects the cooling load of the semiconductor manufacturing facility the
most is provided above through a pie chart in Figure 2.15. The graph displays the equivalent required cooling load in
terms of wattage for factors within the facility adding heat to the system. Therefore, as a conclusion, in order to reach the
design inside temperature of 20 oC having 39 oC on the outside of the facility, a total of 72.68 ton of refrigeration must be
supplied on the testing portion of the semiconductor manufacturing facility to attain the desired temperature.
 0%

21%

35%

4%
Zone 1
Zone 2
Zone 3
Zone 4
Zone 5

40%

Figure 2.16 Equipment Cooling Load Breakdown

Figure 2.16 presents the equipment cooling load for each zone to determine which area requires higher cooling
load due to higher quantity of equipment. Based on the figure presented above, it can be seen that zone 3 has the highest
required cooling load in terms of equipment due to the use of burn in ovens for the testing of semiconductor wafer under
extreme temperature level. On the other hand, it can also be observed that zones 1 and 4 has the lowest required cooling
load due to the small quantity of equipment required.