Download REVISED Design of Cold Storage Facility for Beef Meat in Padre Garcia, Batangas (1)

DESIGN OF A 76 kW COLD STORAGE FACILITY FOR BEEF MEAT LOCATED AT
BRGY. PAYAPA, PADRE GARCIA, BATANGAS
A Design Project
Presented to the Faculty of Mechanical Engineering Department
College of Engineering
BATANGAS STATE UNIVERSITY
THE NATIONAL ENGINEERING UNIVERSITY
Alangilan Campus
Alangilan, Batangas City
In Partial Fulfillment
of the Requirements for the Course of
ME 417 - REFRIGERATION SYSTEMS
By:
ABREU, JESTER CEDRICK A.
BROSOTO, LYN MARIZ T.
DE CLARO, VOIE JERRSON M.
MALATA, MARC REY R.
GONZALES, ANGELYN R.
2023
CHAPTER I
INTRODUCTION TO PROPOSED PROJECT
This chapter presents the problem and background of the study to give the initial idea
about the nature of the design project. This also includes the objectives, parameters of the design,
conceptual paradigm of the design project and the definition of terms relevant in this project.
INTRODUCTION
Many in the Philippines can be exchanged by traders, wherein the municipality of Padre
Garcia, Batangas is said to be the home of the largest livestock cattle trade in the country with
more than 150 livestock auction markets. Livestock trading has been the main source of income
of the residents as well as the local government. Around 300 traders are conducting businesses at
the Livestock Auction Market coming from different places such as Marinduque, Romblon,
Mindoro, Bicol Region, Pangasinan, CALABARZON, and NCR. On a market day, 2,000 head of
cattle, 1,500 head of carabao, 700 head of goat, 220 horses, and 800 head of chickens are being
traded. In terms of cattle, 80% are sold which are bought for slaughter, and a small percentage of
the sold cattle (about 5%) are for fattening. The 20% remaining that are left unsold are cared for
through the paalaga system which will be traded on the next market day.
However, given the volume of cattle slaughtered each day, there is a possibility that it
will not be consumed immediately, resulting in the spoiling of the beef meat. Bacterial growth in
beef meat is facilitated by the high temperature, which ultimately results in spoilage. So it is
recommended that the remaining beef carcass be placed in the cold storage area to prevent
spoilage and be sold the following day.
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Food preservation is the process to retain the food quality, control its spoilage, avoid
oxidation of fats (rancidity), maintain nutritional value, texture, and the overall quality of the
meat for a longer time. Beef can be preserved frozen or chilled in a variety of ways. Chilling is a
short form of preservation which typically lasts around 3-5 days. It is the process of lowering the
temperature to a level slightly above freezing point. The chilling chamber’s air temperature
should be kept between 0°C and 8°C. The air speed in a cooling area is between 0.75 and 1.5m/s.
Increasing air speed will shorten the cooling period that can result in an increase of operational
cost due to fan power consumption. However, to avoid weight loss of meat during chilling, the
80%-90% humidity should be maintained.
On the other hand, freezing is a longer form of preservation which can last for weeks up
to months. It is the way of transforming water content in food into crystals in order to avoid
spoilage and growth of microbes in meat. Moreover, meat can be chilled first in the refrigerator
before freezing but it requires different conditions for different kinds of meat. To maintain a
constant temperature, relative humidity, and air circulation, adequately frozen meats are moved
from the freezer to storage chambers that have these conditions in place. Due to the low
temperature in the chamber and proper packaging of products, contamination is essentially
non-existent, therefore tainting should not be a problem. However, the storage compatibility of
frozen meats and other livestock products is typically low.
Specified conditions are necessary for the refrigeration process. The purpose of
refrigeration, which makes use of the liquid refrigerant, is to remove heat from a controlled area
and distribute it to an absorber plate. In a cold storage, which consists of a number of refrigerated
chambers, all perishable goods can be cooled, frozen, and kept in storage. In building the cold
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storage, some essential substructural factors and characteristics should be met such as, selection
for convenient transportation site and industrial water and electricity.
BACKGROUND OF THE STUDY
Padre Garcia in Batangas, known for being the capital of cattle trading, has been living
up to its reputation. According to the official website of Padre Garcia, Batangas (2022), the
municipal council established the cattle market or bakahan in the town’s public market in 1952,
located in Jose P. Rizal St, Poblacion, Padre Garcia. Despite being in direct competition with the
established livestock market in the neighboring town of Rosario, the bakahan thrived and
became the largest livestock auction market in Southern Tagalog. The town then earned the title
of "Cattle Trading Capital of the Philippines".
Previously, the local industry's income through cattle trading reached PHP8 million in
2016 and sustained the growth up to PHP11.8 million by 2017 (Geñosa, 2018). By this means,
more facilities will be required to maximize the cattle trading through expanded exportation of
beef meat. There could still be a potential increase in income if a substantial amount of beef meat
can be preserved and stored properly. Besides a slaughterhouse, one of the essential facilities in
the meat industry is the cold storage since a certain range of low temperatures are needed for
meat to avoid early spoilage. Based on the list of cold storage warehouses as of May 27, 2022,
from the National Meat Inspection Service (2022), there has been no cold storage facility
constructed specifically in the town of Padre Garcia.
Correspondingly, an idea of proposing a design of a cold storage facility for beef meat in
Padre Garcia has emerged. After visiting the target location, an adequate lot is available to
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establish and build the proposed cold storage facility. Due to the proximity to the Padre Garcia
Auction Market, where cattle are being slaughtered, an advantage can be envisioned. The
potential location of the facility is adjacent to the Padre Garcia Municipal Slaughterhouse located
in Payapa, Padre Garcia, Batangas.
Upon having an interview with the locals and architect Rick Sandoval, the researchers
learned that the said slaughterhouse can only transport meat to the local markets, nearby cities,
and provinces since the facility is classified only as double A, there has been no cold storage
included. Architect Sandoval stated that meats that are designated for long transport should be
frozen in a cold storage first to lower the pH level of the meat. Additionally, cold storage nearby
will result in higher cost and time efficiency regarding the transport of meat as there will be no
need for double transport. The transport for the live cattle is costly, so if there is already a cold
storage adjacent to the slaughterhouse, the transport cost would be lesser since it will be
transported in primal cuts. The quality of beef meat for exportation will be secured also.
The proposed design of a cold storage facility for beef meat is expected to accommodate
the necessity of meat preservation for the whole Padre Garcia, potentially capable and permitted
for local and international transport.
OBJECTIVE OF THE DESIGN PROJECT
The main objective of this proposal is to design a beef meat cold storage facility for 30
metric tons of cattle carcasses from Padre Garcia Municipal Slaughterhouse located at Brgy.
Payapa, Padre Garcia, Batangas. Specifically, this proposal aims to attain the following:
1. To design a beef cold storage facility according to the following considerations:
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1.1 Location of the proposed cold storage facility
1.2 Capacity Storage
1.3 Size of the Equipment
1.4 Dimension of the Facility
2. To optimize the refrigeration system and to determine the specifications for the
following machine parts:
2.1 Evaporator
2.2 Compressor
2.3 Condenser
2.4 Expansion Valve
2.5 Types of Refrigerant
3. To select materials/equipment in consideration of the following loads:
3.1 Transmission Load
3.2 Product Load
3.3 Internal Load
3.4 Equipment Load
3.5 Infiltration Load
4. To test the economic feasibility of a beef meat facility's proposed refrigeration
system in consideration of:
4.1 Initial Project Cost
4.2 Operational Cost
4.3 Maintenance Cost
4.4 Payback period and Rate of Return
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PARAMETERS OF THE DESIGN
A. Established Parameters
1. Thermal Conductivity
Thermal conductivity is a material's ability to transfer heat, measured in W/m•K. It is
based on Fourier's law of heat propagation and depends on the temperature difference and
material properties. Heat moves from high to low temperatures until equilibrium is reached.
Thermal conductivity is the opposite of thermal resistivity and is induced by molecule motion
and touch.
2. Thermal Resistance
Thermal resistance measures a structure's ability to withstand heat transmission. For a
cold storage facility, it's crucial to maintain a steady internal temperature to preserve the quality
of stored meat and save energy. Materials with high heat resistance, like insulated aluminum
frames, should be used for walls, ceilings, and floors. Thermal barriers and an effective cooling
system can also enhance the building's thermal resilience. Regular maintenance, such as cleaning
condenser coils and changing filters, is necessary for optimal cooling system performance.
3. Specific Heat Capacity
Specific heat capacity is the amount of energy needed to raise a substance's temperature
by one degree Celsius. In cold storage, materials with high specific heat capacity help maintain a
consistent internal temperature, which is important for livestock safety. Insulated metal walls
have lower heat capacity compared to materials like concrete, so adding insulation or thermal
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bulk can minimize temperature fluctuations. The cooling system must be designed to maintain a
consistent temperature regardless of external factors. The proper enclosure of the structure can
also help mitigate the impact of heat capacity on efficiency.
Table No. 1
Established Parameters for Materials
Thermal / Heat Properties
Material
Type of Material
Insulated Panels
Thermal
Conductivity
Thermal
Resistance
Specific Heat
Capacity
Polyurethane foam
0.018 to 0.038
W/m·K
0.026 to 0.055
m2·K/W
1200 to 1600
J/kg·K
Flooring
Epoxy Coatings
0.1 to 0.4 W/m·K
0.0025 to 0.01
m2·K/W
1000 to 1500
J/kg·K
Air Curtain
PVC Strip Curtains
0.14 to 0.17
W/m·K
0.018 to 0.021
m2·K/W
1000 to 1400
J/kg·K
Copper Type T
Thermocouples
401 W/m·K
Relatively low
due to its high
conductivity
386 J/kg·K
Constantan Type T
Thermocouples
22 W/m·K
relatively high
due to its lower
conductivity
410 J/kg·K
Type T
Thermocouples
Thermocouples
Sensor
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Evaporator
Aluminum
200-250 W/m·K
0.002-0.003
m2·K/W
900-1000
J/kg·K
Condenser
Copper
350-400 W/m·K
0.0015-0.0035
m2·K/W
385-390
J/kg·K
Expansion
Valve
Stainless Steel
15-20 W/m·K
0.02-0.04
m2·K/W
500-600
J/kg·K
Compressor
Cast Iron
30-70 W/m·K
0.02-0.05
m2·K/W
460-520
J/kg·K
Shelving
Stainless Steel
16-24 W/m·K
0.0001-0.0006
m2·K/W
460-510
J/kg·K
Lighting
LED Lighting
10-200 W/m·K
0.1-10 m²·K/W
0.7-1.2 J/kg·K
Door
Insulated Metal
0.03-0.05 W/m·K
0.025-0.035
m2·K/W
840-1200
J/kg·K
Walls
Insulated Metal
Panels (IMPs)
0.02-0.06 W/m·K
0.017-0.050
m2·K/W
840-1200
J/kg·K
Back-up Power
Supply
Battery Backup
Systems
0.1-0.5 W/m·K
0.01-0.05
m2·K/W
700-1200
J/kg·K
Trolley with
Sliding Hooks
Stainless Steel
16.3 to 21.5
W/m·K
0.046 to 0.055
m2·K /W
500 to 550
J/kg·K
Based on the data provided in the table, several materials with specific thermal and heat
properties can be used to build a beef meat cold storage.
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The insulation panels made of polyurethane foam, the insulated metal panels for walls,
and the insulated metal door, have low thermal conductivity and high thermal resistance, which
helps to prevent heat transfer and maintain the desired temperature inside the storage. The
flooring material, epoxy coatings, also has moderate thermal properties and can withstand heavy
equipment traffic. The air curtain made of PVC strip curtains, on the other hand, can help to
maintain the temperature by reducing the exchange of air between the inside and outside of the
storage. In addition, the type T thermocouples sensors made of copper and constantan can help to
measure the temperature accurately. The evaporator made of aluminum, the condenser made of
copper, the expansion valve made of stainless steel, and the compressor made of cast iron, all
have varying thermal properties and functions that are essential for the operation of the cold
storage system. Lastly, the stainless-steel shelving, Trolley with Sliding Hooks, LED lighting,
and battery backup system are also important components to consider for building a beef meat
cold storage. They have specific thermal and heat properties that contribute to the overall
efficiency of the system.
Overall, building a beef meat cold storage requires careful consideration of the thermal
and heat properties of the materials used. The materials should be chosen based on their ability to
maintain a consistent temperature, prevent heat transfer, and ensure the overall efficiency of the
system.
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Table No. 2
Established Parameters for the Product
Product
Beef Meat
Type of
Product
Carcass
Thermal / Heat Properties
Thermal
Conductivity
Thermal
Resistance
Specific
Heat
Capacity
Freezing
Point
0.43 and 0.52
W/m·K
1.9 and 2.3
m2·K /W
3.4 and 3.8
kJ/kg·K
-1.7 °C
Table 2 provides the thermal properties of beef meat carcass, which includes thermal
conductivity, thermal resistance, and specific heat capacity. The thermal conductivity of beef
meat carcass ranges between 0.43 and 0.52 W/m·K, indicating its ability to conduct heat. The
thermal resistance of beef meat carcass ranges between 1.9 and 2.3 m2·K/W, indicating its
resistance to heat flow. The specific heat capacity of beef meat carcass ranges between 3.4 and
3.8 kJ/kg·K, indicating its ability to store heat. The freezing point of beef meat is -1.7°C. These
properties are important in determining the optimal storage conditions for beef meat to maintain
its quality and safety.
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B. Overview of Proposed Design
Figure 1. Block Process Diagram of the Facility
The block process diagram of the facility that can be seen on Figure 1 shows the process
of how the beef carcass flows inside the cold storage facility. All the carcass that will undergo
reception and dispatch will be checked by a meat inspector before going to the chiller room,
cutting room, and to the cold storage.
In consideration of the design calculation of the project, this section provides a thorough
discussion of the process parameters that will affect the design.
1.
Maintaining Temperature for Different Types of Cold Storage
According to the National Meat Inspection Services, the functionality of a
cold storage facility that is treated with the highest significance is its ability to
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reach and maintain the desired temperature at all times. This is due to the fact that
temperature control is considered to be a food safety measure which is very
crucial in operating these types of facilities. Hence, this is also what is being held
with utmost importance in the design of this proposed cold storage facility.
However, the maintaining temperature of the different sections of a cold storage
facility is not the same all throughout. The temperature requirements for each of
the different areas of the facility are discussed as follows:
1.1 Ante-Room
The ante-room serves as the pre-cooling area for the beef carcasses and its
main purpose is to prepare the meat before entering the chiller room. Ante-rooms
ease out the change in temperature from room temperature to a colder temperature
and vice versa to avoid creating condensation which is not desirable in meat
preservation. This means that ante-rooms are also used when removing the meat
from the chiller room before being sent for transportation. Ante-rooms are not
meant to be used as the actual storage for meat since the temperatures in this room
are not the ones that are capable of preserving meat and are only meant to be used
as a temporary holding area. The maintaining temperatures for ante-rooms range
from 10°C up to 12°C. Additionally, this is where the receiving bay and the
loading bay are usually located in a typical cold storage facility.
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1.2 Chiller Room
This room serves as the storage area for cattle carcasses in a cold storage
facility which makes it occupy a big land area in a cold storage facility. This is
where the carcasses are held after leaving the ante-room and these rooms are the
ones that have the capacity to reach and maintain the required temperature for
effective meat preservation for meats that will be unloaded after a couple of days.
Chiller rooms can be multiple and can have different sizes depending on the
amount of storage required for carcasses. The maintaining temperature for chiller
rooms ranges from 4°C up to 0°C.
1.3 Cold Storage
Cold storage rooms are the main storage room for the finished products
which are meat chunks that are sealed and packed in boxes that are waiting to be
unloaded from the storage facility. The freezing storage room also occupies a lot
of space since it can fill up easily since this storage is the most economical in
housing products on a long-term basis. The maintaining temperature of freezing
storage ranges from -18°C or below.
For this design project, the above-mentioned areas of a cold storage
facility are all being taken into consideration together with their maintaining
temperatures.
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2.
Relative Humidity
It is ideal for every cold storage to have controlled humidity inside the
facilities. When there is higher humidity, the moisture content is also higher
which is detrimental when preserving meat. A cold storage facility is a contained
environment and has a lot of moisture inside it helps spoilage bacteria to grow
which damages the freshness of the meat and eventually causes it to spoil. On the
other hand, having too little humidity causes the air in the cold room to absorb the
moisture from the products being stored. This causes the meat to deteriorate faster
which removes its freshness and eventually leads it to be condemned. The optimal
relative humidity for meat preservation must be between 80 and 90 percent at 4°C
to have a balance between the development of bacteria and weight loss due to
moisture absorption by the air.
For this design project, the relative humidity is affected by different
factors such as the type of product to be stored, the number of products to be
stored, the temperatures that the products will be contained in, and the total heat
load will all be considered.
3.
Duration Time for the Product to Reach the Maintaining Temperature
According to the food code of the Food and Drug Administration, the food
must be cooled down from 57°C to 21°C in less than six hours. After reaching the
said temperature, it must then be cooled down from 21°C to 5°C within the span
of four hours. This is an important guideline that helps avoid the possibility of
dangerous bacterial growth that would spoil the product. Furthermore, the faster
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the product can be cooled down optimally and properly, the freshness of the meat
will be better preserved.
For the proposed design of the cold storage facility, the required time
duration for the food temperature reduction will be taken into consideration to
effectively preserve the freshness of the cattle’s meat.
4.
pH Level Measurement of the Product
For the effective elimination of microbiological and enzymatic activity
which ensures the stability of the product when preservation, it is vital to
determine the pH level of the meat when storing it in cold storage. The pH level
of meat also dictates its quality since it affects the water-holding capacity of meat
which is considered to be one of the most important factors in determining the
quality of meat. According to the Food and Agriculture Organization of the
United Nations the lowest and optimal pH level for cattle carcasses before
preservation is set to be around 5.6 to 5.8.
The pH level of the products that will be stored in the proposed cold
storage facility will also be accounted for to ensure that the facility can effectively
maintain the optimal pH level of meat while being stored in the facility.
5.
The Temperature of the Products Upon Entry
The entry temperature of the products is also an important process
parameter that should be considered to effectively determine the cooling capacity
of the cold storage. When it comes to sensitive goods such as meat, sudden
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changes in temperature can cause the unwanted degradation of the quality of
meat. It is necessary that the change in temperature happens gradually and this
will only be possible if the entry temperature of the products is known because it
would be the basis of how low the temperature should be when the cold storage
receives the product.
6.
The Air Pressure Inside the Cold Storage Facility
Another important consideration for the design of cold storage is
understanding air pressure equalization. Whenever the cold rooms are open, the
warmer air outside the room enters and this needs to be handled to avoid
unbalanced air pressure. Building up a pressure differential between the outside
and the inside of the room would cause resistance or a locking effect making it
difficult to open the doors of the rooms. Furthermore, having these pressure
differences causes a lot of fatigue to the components of the cold rooms which is
not desirable.
For this project, air pressure equalization will also be considered together
with the appropriate pressure relief ventilation system that would regulate the
pressure inside the rooms and would help in keeping the temperatures constant.
C. Design Assumptions
In this section, we will discuss the concepts of transmission load, product load, internal
load, equipment load, infiltration load and total accumulated load. These concepts are crucial in
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understanding the various types of heat loads that contribute to the overall cooling requirements
in a given system.
1. Transmission Load
It is the measurement of sensible heat gain through the floor, walls, and
roof from the temperature difference across these surfaces, into consideration of
their thickness and kind of material used in the construction of the building and
cold chambers, as well as the physical specifications of the cold storage facility. It
will also take into consideration the temperature differences between the outside
temperature and inside temperature of the building. Heat always flows from hot
to cold. The internal temperature of the cold room is obviously a lot colder than
its surroundings, so heat is always trying to enter because of temperature
differences. If the cold store is exposed to direct sunlight then the heat transfer
will be higher so an additional correction will need to be applied to allow for this.
2. Product Load
The product load refers to the heat or cooling load generated by the
products stored within a specific space, such as a cold storage facility. Different
products have varying thermal properties, which affect the amount of heat they
contribute to the overall cooling requirements. Factors such as product
composition, temperature, and mass influence the product load calculations.
Accurately estimating the product load is crucial for determining the appropriate
refrigeration capacity and energy requirements of the cold storage system.
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3. Internal Load
Internal load refers to the heat gain or cooling demand generated within a
space due to various factors such as occupants, lighting, and appliances. It
encompasses the heat produced by people, electrical devices, lighting fixtures,
and other sources of heat inside a building or room. The internal load significantly
influences the cooling or heating requirements and affects the overall energy
consumption and comfort levels within the space. Proper management and
consideration of internal loads are essential for efficient HVAC system design and
operation.
3.1 Occupant Load
Occupant load refers to the heat load generated by the presence of
people within a designated space. The number of occupants, their
activities, and the duration of occupancy contribute to the internal heat
load. Human activities and metabolic rates produce heat that affects the
cooling requirements of a space. Accurately estimating the occupant load
is essential for designing effective cooling systems, ensuring comfort, and
maintaining appropriate indoor temperatures.
3.2 Lighting Load
Lighting load refers to the heat generated by artificial lighting
sources within a space. Traditional light bulbs and fluorescent lamps emit
heat as a byproduct of producing light. This heat adds to the overall
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internal load and affects the cooling requirements of the area.
Energy-efficient lighting options, such as LEDs, generate less heat and can
help reduce the lighting load. Proper consideration of lighting load is
crucial for achieving energy-efficient designs and optimal cooling system
sizing.
4. Equipment Load
Equipment load refers to the heat generated by mechanical and electrical
equipment used in a building or facility. This includes the heat produced by
machinery, motors, pumps, fans, refrigeration systems, and other equipment.
Equipment load affects the cooling requirements and energy consumption of the
system. Proper sizing, selection, and placement of equipment are necessary to
account for the heat generated by these systems and ensure efficient cooling
system design and operation. Managing the equipment load helps optimize energy
usage, reduce operating costs, and maintain a comfortable indoor environment.
4.1 Fan Motors
Fan motors are essential components of cooling systems, such as
air conditioning and refrigeration units. These motors generate heat during
operation, contributing to the overall equipment load. The size and number
of fan motors, as well as their energy consumption, affect the cooling
requirements and energy calculations. Accurate estimation of the heat
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generated by fan motors is crucial for sizing the cooling system and
ensuring efficient operation.
4.2 Meat Trolley Load
In certain cold storage facilities, meat trolleys are used for storing
and transporting meat products. These trolleys contribute to the equipment
load due to their thermal properties and the heat they emit. The size,
material, and temperature of the meat trolleys affect the amount of heat
they add to the overall cooling requirements. Proper consideration of the
meat trolley load is vital for accurately estimating the energy requirements
and refrigeration capacity of the cold storage system.
5. Infiltration Load
Infiltration load refers to the heat load caused by the exchange of air
between the inside and outside of a controlled space. Air infiltration occurs
through gaps, cracks, or improperly sealed openings in the building envelope. The
infiltration load is influenced by factors such as temperature differences, wind
speed, and the quality of insulation. Proper insulation, sealing, and ventilation
design are essential to minimize the infiltration load and optimize the cooling
requirements.
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6. Total Accumulated Load
The total accumulated load represents the sum of all the individual heat
loads discussed above. It includes the product load, internal load (occupant load
and lighting load), equipment load (fan motors and meat trolley load), and
infiltration load. Calculating the total accumulated load is crucial for determining
the overall cooling requirements and sizing the refrigeration capacity of a cold
storage system. Accurate assessment of the total accumulated load ensures
efficient and effective cooling system design and operation.
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CONCEPTUAL PARADIGM OF THE STUDY
Figure 2. Conceptual Paradigm of the Study
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The conceptual paradigm exhibits the operational process of how the proponents will
analyze and execute the design. This section shows the flow of the process and how each part is
dependent on the other based on the IPO (Input-Process-Output).
The input stage is concerned with the Knowledge requirements, System components, and
Design Considerations. In the knowledge requirements, the proponents gather the needed
information and requirements, essential for a better understanding of how the principle of cold
storage facility works, through a variety of sources, such as books, journals, research, and
interviews. In the system components, the proponents identified the needed system elements
needed to operate the refrigeration cycle according to the operational parameters.
In the process stage, the proponents use the gathered data and information through
research and interviews to conceptualize a design for the cold storage facility. Upon finishing the
concept of the design, calculating the specific dimensions of the facility and components is
necessary, such as the area to be occupied by the whole facility including the partitions of each
building and others. After determining the dimensions needed, the design operational parameters
should be considered next with inclusion of such parameters as the temperatures, pressures, and
other parameters required and used in the cold storage facility. When the necessary parameters
are determined, materials requirements and the corresponding specification should be thoroughly
chosen and identified. To test the design efficiency of the systems to be used in the cold storage
facility, performance evaluations should be executed.
The expected output is an efficient and cost effective design of an operational cold
storage facility with well-defined dimensions and design parameters.
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DEFINITION OF TERMS
Carcass. This refers to a dead body of an animal, such as a cattle carcass from a slaughterhouse
which is the product refrigerated in the proposed cold storage.
Cold Storage Facility. This is the type of storage space intended to store the beef meat having
been designed using refrigeration systems to maintain a consistently low temperature appropriate
for the aforementioned product. This facility includes different rooms of different temperatures
to attain different storing purposes.
Compressor. This is the device that decreases the volume of the refrigerant gas or vapor causing
it to increase in pressure and temperature. This is also responsible for the circulation of the
refrigerant throughout the system.
Condenser. It is used to cool the refrigerant that the compressor sends out, which is in a gaseous
condition with high temperature and pressure. Condenser works to cool down in order to
liquidize the refrigerant.
Evaporator. This is the device that is responsible for heat absorption from the closed space and
the evaporation of the refrigerant.
Expansion Valve. Its main function is to control the refrigerant flow from the system's
high-pressure side to the low-pressure side.
Humidity. It refers to the amount of moisture present in the air and is controlled in a cold storage
in maintaining the quality or condition of the beef meat.
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Refrigerant. This is the substance used in the refrigeration systems which absorbs the heat from
the closed space being cooled and then transfers it to another location.
Refrigeration Systems. To solve the proposed problem concerning removal of heat from a
refrigerated space, the proposed cold storage, to maintain the target low temperature for the beef
meat, this concept is applied wherein evaluation and determination of the highest performing
refrigeration system design to use is attained.
Vapor Compression Cycle. This governs the design for the proposed refrigeration system
wherein it consists of a compressor, a condenser, an expansion valve, and an evaporator, wherein
a liquid refrigerant circulates. This refrigerant changes states, which causes the system to absorb
and release heat, bringing the temperature of the conditioned space to decrease.
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CHAPTER II
This chapter presents the review of the proposed application of the refrigeration system
and its processes. the involved refrigeration plant subsystems and plant equipment considered in
the overall design of the cold storage facility..
Review of the Proposed Application of Refrigeration System and Its Processes
Refrigeration systems are now as ever-present as they were before. It can be said that
they are now an integral part of modern living. These systems have proven their value in society
both at the personal level up to the industrial sectors. Nowadays, most people consider
refrigerators and air conditioners as a necessity because of the improvement that they can give to
the quality of living of an individual. Refrigerators help preserve food and air conditioners cool
down the temperature of the room to its desired temperature. These are only some examples of
how refrigeration systems are being utilized on a personal level. On the other hand, when it
comes to the industrial level, these systems have a magnitude of applications that are vital to a lot
of industries. According to Princy (2021), the application of refrigeration systems in industries
can be categorized into 5 major categories. These categories include district cooling, electricity
production, chemical and petrochemical industries, pharmaceuticals industry, and lastly, the food
and beverage industry.
For district cooling, the main concept that governs this industry is the generation of
cooling streams, which are mainly chilled water, with the use of various types of refrigeration
systems in a centralized plant. These generated cooling streams are then distributed to separate
places like residential and commercial areas, offices, venues, and many other closed areas that
require cooling. Utilizing district cooling constitutes a lot of environmental and economic
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benefits to the community. This is due to the higher efficiency that is being achieved by
centralized cooling systems as compared to individual refrigeration systems for each
infrastructure. Comparable to district cooling, the application of refrigeration systems in the
industry of electricity production also provides a favorable advantage. This industry which is
primarily based on the combustion of fuels takes advantage of refrigeration systems by using
them to regulate the temperature of multiple machines that are involved in generating electricity.
These also result in higher efficiency for electricity generation.
When it comes to chemical and petrochemical industries, refrigeration systems are
considered to be a necessity in obtaining their products. Some of the processes where these
systems are utilized include condensations, crystallizations, and distillations which are done by
removing heat. Commonly, these industries have large-scale cooling plants that are designed to
be used in their processes. In addition to these industries, the pharmaceutical industry is also one
of the industries that are heavily reliant on refrigeration systems. Every process that is being
done in the pharmaceutical industry requires strict conditions that are vital for its success. These
strict conditions include temperature regulation which is why this industry uses sophisticated
refrigeration systems that allow for precise adjustments of temperature.
The last major industry that utilizes refrigeration systems is the food and beverage
industry. In this industry, several chains of refrigeration systems are used since it is necessary
that the temperature is maintained at its optimal parameter to assure food safety. According to
Camp (2021), the main process involved in this industry is food preservation which is made
possible by cooling systems. The products of this industry are fish, poultry, meat, dairy, fruits,
and vegetables which are meant to be kept at low temperatures to preserve the optimal conditions
of the products prior to their consumption.
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This design project falls under the applications of refrigeration systems in the food and
beverage industry. Specifically, it is concerned with the design of a cold storage facility for cattle
meat in the town of Padre Garcia, Batangas. The refrigeration systems will be applied to preserve
large quantities of cattle meat while being stored in the cold storage facility. The type of
refrigeration system that will be used in this design is the mechanical-compression refrigeration
system. This system works under the first law of thermodynamics which states that energy,
which is heat for this application, cannot be created nor destroyed. It can only be transferred by
doing work which is what this refrigeration system does.
The process of refrigeration begins with the absorption of heat from the area where it is
not wanted. For this application, the unwanted heat is absorbed from inside the cold storage
rooms where the meat will be stored and preserved. Once the heat absorption is done, it will then
be pumped to another location where the heat can be radiated away. The name
mechanical-compression refrigeration comes from the mechanical work being done by the pump
and compressor to transfer the unwanted heat away. The conventional form of this type of
refrigeration can be referred to as the vapor compression cycle, also known as a reverse Rankine
cycle.
As explained by Wright (2022), the reverse Rankine cycle uses a refrigerant that absorbs
and releases heat. This refrigerant begins as a high-pressure liquid that enters an expansion valve
before entering the evaporator where it will boil and absorb the heat around the evaporator as it
turns to vapor. Then, this heated refrigerant vapor will then go through a compressor that will
pump it to the condenser where it will condense back into liquid due to its high vapor pressure
causing it to expel the absorbed heat via a radiator or a heat exchanger. After being reverted back
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to its liquid form, the refrigerant will now be sent back again to the expansion valve which starts
the refrigeration cycle again.
Refrigeration Plant Subsystems
Material Handling/Conveying
In creating an efficient and well organized food distribution system in a cold storage
facility, it requires correct working and materials handling and conveying. Cold storage is a
complex application for material handling of equipment because of the need in transportation of
goods such as beef meat for instance from cold storage to designated refrigerated areas of the
owners locally or overseas. A food storage and distribution facility needs high quality equipment
for speed and accuracy to protect the product from temperature changes. The possible equipment
that can meet cold storage facility management are Reach Truck, Four-wheel electric trucks, End
Rider, Three-wheel stand-up. multi-level order selector, and robotic lift truck which are types of
Forklift trucks.
Food handling and conveying offers unique challenges for facility workers in material
handling equipment. When material handling equipment such as forklift abruptly moves from
cold to warm environment, condensation can build within the components and cause serious
damage to its components. Therefore, forklifts should have specialized technology like
thermostatically controlled heaters to protect the equipment. Forklifts can be difficult to drive
when the workers are wearing thick gloves when entering cold storage therefore, the
management should invest in high quality gloves which have ergonomically designed controls
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that are easy to grip when wearing thermal production gear. To protect the product from
contamination, safety regulations are critical for every operation. Using lubricants that are
non-toxic and USDA inspection ready, as well as lithium-ion batteries that reduce fumes and
spills are excellent options to reduce contamination and product recalls. Therefore, Material
handling and conveying is very crucial to avoid frequent breakdowns and loss of productivity
and efficiency in a cold storage facility.
Waste Management
1. Wastewater Treatment
Wastewater is involved in almost all the cleaning processes of the products and
the facility itself. However, not all kinds of wastewater can be sent simply on the sewages
because there is a high possibility of being heavily contaminated with pathogens and
toxic chemicals, especially if the origin is the remnants of dead organisms. In such
facilities as slaughterhouses and cold storage of meat, the accumulated wastewater not
only contains toxic chemicals but also stinks which can be hostile to the olfactory system
of those working in the facility and the community nearby. With this, it may be
concerning in terms of the environment and health of the community.
According to Matt French (2018), one of the effective systems for wastewater
controls in facilities like a cold storage of meat is the Dissolved Air Flotation (DAF)
system. The purpose of this system is to clarify the wastewater by removing the
biochemical oxygen demand, greases, oils, suspended solids, and metals so that the only
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substance directed to the sewer system is liquid. In this system, the contaminants are
removed by injecting air under pressure and dissolving the air in the wastewater. After
releasing it in a flotation tank at atmospheric pressure, the air released creates tiny
bubbles that stick to suspended materials, causing it to float to the water's surface where
it may be scraped off using a skimmer.
2. Condemnation Pit
When the whole carcass is declared unfit for human consumption, all the parts are
condemned. When only certain body parts are deemed unfit for human consumption,
carcasses can be partially condemned. Since the remaining good parts of the carcass are
still suitable for human eating, it can be directed to the cold storage.
Based on the Section 13 of the Administrative Order No. 19, Series of 2010,
Guidelines on Good Hygienic Slaughtering Practices for Locally Registered Meat
Establishments, “In handling of condemned carcasses and/or any parts, a condemnation
pit shall be provided for disposal of condemned material.”
Processing Equipment
The following are several equipment used in the facility for better efficiency in the
processing system:
1. Digital Weighing Scale
This measuring device used to determine weight or mass is a digital electronic
weighing scale. A load cell with a strain gauge is how it operates. Analog scales use
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springs to indicate an object's weight, whereas digital scales translate an object's weight
force into an electric signal (Crown Scales, 2019). This is used in consideration for the
load capacities of the equipment handling the products.
2. Forklift
A forklift is an industrial vehicle with a front-mounted, power-operated forked
platform that can be lifted and lowered to be inserted beneath goods to lift or transport it
(Torcan Lift Equipment, 2017).
3. Meat Trolley
These trolleys are intended to be used in the delivery, production, and storage
rooms for transferring hanging carcasses.
4. Railing System
An overhead meat rail system is suitable for walk-in cold rooms, such as cold
storages for cattle or pork carcass, being a strengthened framework essential for meat
handling. Hooks are mounted in the rail system where the cattle carcass is hung. With the
inclusion of integrated hoists, scales, and trolleys, they are suitable for minimizing hard
lifting and are excellent for optimizing stock rotation (Angel Refrigeration, 2023).
Maintenance
Regular maintenance ensures the continuity of operation of the facility as it reduces the
possibility of equipment failures. Having a maintenance plan helps in faster identification and
prevention of problems before it worsens.
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Plant Equipment
A variety of plant equipment is necessary to guarantee the correct operation of a beef
meat cold storage facility. Mechanical, electrical, plumbing, and safety systems are among them,
with a special emphasis on mechanical systems. Mechanical systems in a beef meat cold storage
facility are crucial for maintaining the necessary temperature and humidity levels to keep the
meat fresh and safe for eating. Understanding the various types of plant equipment and their
responsibilities in a beef meat cold storage facility is critical for guaranteeing the facility's proper
operation and maintenance. The lists are as follows:
Mechanical Plant Equipment
1. Compressor
Compressors are essential components of refrigeration systems because they
compress refrigerant gasses and move them through the system. Depending on the
capacity of the refrigeration system, they come in a variety of sizes and kinds. According
to David (2021), the compressor is the "heart" of every commercial cooling system. It
compresses refrigerant gas, allowing it to flow through the remainder of the system. If the
compressor is damaged for any reason, the remaining system components will be less
efficient and more susceptible to mechanical failure. Similarly, if the compressor fails, the
entire system would shut down. Meatpacking and processing businesses have some of the
greatest performance per square foot requirements of any industrial cooling application.
With this in mind, facility leaders seeking to maximize the value of their packing floor
carefully match compressors to their requirements, therefore increasing end-to-end
efficiency.
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2.
Evaporator Coils
These are in charge of transmitting heat from the refrigerant gas to the air in the
storage facility. They are usually found within the storage area and might take the form of
coils or plates. According to the SkillCat Team (2021), evaporators are made up of a
series of coils that are loaded with refrigerant. The evaporator's major component is its
coils. Evaporator coils expand the overall surface area across which air may flow. This
accelerates the heat removal procedure. Increasing surface area always accelerates both
heating and cooling.
3. Condenser Coils
These are positioned outside of the storage room and are in charge of discharging
the heat received by the refrigerant gas into the atmosphere. They can be either air or
water cooled. Furthermore, one of two types of heat exchangers utilized in a basic
refrigeration loop is the condenser coil. This component receives high-temperature,
high-pressure vaporized refrigerant from the compressor. The condenser extracts heat
from the heated refrigerant vapor gas vapor until it condenses into a saturated liquid state,
a process known as condensation (Super Radiator Coils, 2021).
4. Fans and Blowers
Fans and blowers are utilized to move air throughout the storage room and keep
temperatures stable. They can be axial or centrifugal fans and can be installed within or
outside the storage room. However, as Raihan (2015) points out, fans and blowers are
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both mechanical devices used for air circulation. They are distinguished by the fact that a
fan circulates air over an entire room or space, whereas a blower exclusively concentrates
on a specific or specified region.
5. Pumps and Valves
Pumps and valves are used in the refrigeration system to control the flow of
refrigerant gas and other fluids. They can take the shape of centrifugal or positive
displacement pumps, as well as different valves. According to Conger (2023), the pumps
also assist in adjusting the load imposed on the evaporator, which is decided by the
cooling demands of the warehouse, while the expansion valve assists in regulating the
pressure, temperature, and amount of refrigerant discharged into the following
component, the separator.
Electrical Plant Equipment
1. Refrigeration Units
These are the most critical pieces of equipment required for beef meat cold
storage. They are in charge of keeping the storage facility's temperature within the
permitted range. Depending on the size and capacity of the storage facility, refrigeration
units come in a variety of sizes and types. Cold temperatures keep food fresher for longer.
According to How Stuff Works (n.d.), the primary principle behind refrigeration is to
slow down the action of bacteria (which all food has) so that the germs can ruin the food
for a longer period of time.
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2. Temperature Sensors
Temperature sensors are devices installed on machinery that monitor ambient
temperatures, analyze information, and determine the influence of heat conditions (On
Up Keep, 2023). These are used to keep track of the temperature in the storage facility
and guarantee that it stays within the acceptable range. They may be strategically
positioned around the storage space to produce precise readings.
3. Thermostats
According to Foster (2023), a refrigerator thermostat/cold control is essentially
the brains of the refrigerator cooling system—it controls the show. Thermostats are
commonly found inside refrigerators and contain a knob that allows users to change the
temperature. The thermostat maintains the temperature specified by the user by managing
the flow of power to the compressor.
These are utilized to maintain the proper
temperature range by controlling the temperature of the refrigeration unit.
4. Insulation
Proper insulation is critical for keeping the correct temperature range and
lowering energy expenses. Panels or spray foam insulation can be used for insulation.
According to Muchu (2019), refrigerating machinery is used to remove heat from air
conditioning or refrigerated areas, and the objective of insulation is to prevent the heat
from flowing or seeping back into the space as much as possible. The effectiveness of
this function is primarily determined by the efficiency of the insulation and the method in
which it is installed.
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5. Lighting
Sufficient lighting is required for meat inspection and for staff to traverse the
storage facility. LED lighting is a popular choice since it uses less energy and generates
less heat. According to Cold Chain Management (2021), lighting accounts for 10-12% of
overall power use in a typical cold shop. Conventional bulbs produce light by heat, and
this heat, which accounts for 100% of the rated energy consumption of the light, must be
removed by the refrigeration plant.
6. Backup Power Supply
A backup power supply is required to keep the cold storage facility operational in
the event of a power loss. As a backup power source, a generator or battery backup might
be employed. When the utility is unavailable, backup systems employ local generation at
the facility site to provide electricity. The backup power system may or may not be linked
to the utility grid (NEMA, 2013).
7. Defrosting Systems
Defrosting systems are required to remove any ice accumulation on the
evaporator coils and guarantee optimal refrigeration system operation. Electric heaters or
hot gas defrost devices are two examples. According to Danfoss (2013), electric
defrosting is the most popular and straightforward defrosting method for cold storage
rooms. The air cooler just has to be fitted with electric heaters and linked to power.
Because it consumes a lot of energy, this is a more expensive defrosting approach in
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terms of energy. Electric defrosting, on the other hand, is highly controllable and may be
the only viable defrosting alternative.
8. Electrical Control Panels
These are utilized to regulate and monitor the operation of the refrigeration
system and other storage facility equipment. They may be tailored to the facility's unique
requirements. An electrical control panel is an enclosure, usually a metal box or a plastic
molding, that houses critical electrical components that regulate and monitor a variety of
mechanical operations (Electrical Safety UK, 2023).
Plumbing Plant Equipment
1. Floor Drains
According to the Maine Department of Environmental Protection (2020), floor
drains are collecting locations that capture wash water and other liquid pollutants from a
work area and transport them away for disposal through pipes or ditches. To prevent the
collection of water or other liquids in beef meat cold storage facilities, appropriate
drainage is required. Floor drains are consequently essential for facilitating efficient
drainage and preventing moisture accumulation.
2. PVC Pipes
In cold storage facilities, PVC pipes are typically utilized to transport fluids such
as water to and from the refrigeration unit or other equipment. PVC pipes are used in
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sewage and wastewater transportation, as well as drains and vents connected with
buildings and equipment (PVC Pipe Supplies, n.d.).
3. Water Filtration Systems
Water filtering systems may be required to produce clean water for a variety of
applications, including cleaning and processing equipment. According to Aqua Care
(2016), water filter systems eliminate unpleasant tastes and odors from mains water to
offer clean, fresh-tasting water right from your faucet. This is especially crucial for beef
meat, as polluted water may lead to deterioration and bacterial development.
4. Water Supply Lines
Water supply lines are required to provide water to different portions of the
facility, including sinks and hose stations.
According to the National Center for
Environmental Health (2014), unless the hot water line is insulated, hot and cold-water
lines should be roughly 6 inches apart. This prevents the cold-water line from picking up
heat from the hot water line.
5. Backflow Preventers
A backflow preventer is a necessary component of your plumbing system.
(Mechanical, 2022). Backflow preventers prevent water from leaking back into water
supply pipes and potentially compromising the water supply. Building laws and health
standards demand this. The device is implemented to avoid drinking water contamination
from other sources owing to backflow by restricting water flow to one direction.
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Safety Plant Equipment
1. Fire suppression systems
Because of the usage of electrical equipment and the possibility of chemical
reactions, cold storage facilities may be at danger of fire. To guard against fire, fire
suppression systems such as sprinklers, extinguishers, or foam suppression systems may
be installed. Mellon (2022) recommends installing a dry or pre-action fire sprinkler
system in the cold space, as well as a water-based fire sprinkler system throughout the
remainder of the building. This combination of two types of fire sprinkler systems
provides the best possible protection for both types of areas.
2. Personal protective equipment (PPE)
Employees at a cold storage facility should be equipped with proper PPE to
protect them from freezing temperatures, slips and falls, and potential chemical exposure.
Properly cared for and cleaned PPE can help to avoid the hazards listed above. Thermal
socks, boots, trousers, jackets, and coveralls are all appropriate cold storage PPE and
should be worn at all times (Chanaud, 2020).
3. Emergency lighting and alarms
In the case of a power outage or an emergency, it is critical to have backup lights
and alarms that may assist personnel in navigating the property and securely evacuating.
Emergency lighting systems, according to Eich (2017), play a critical role in maintaining
buildings safe for public usage. Emergency lighting systems must be carefully planned
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and built to offer a highly dependable system that illuminates the evacuation path in the
event of a power loss or other malfunction.
4. Safety barriers and signage
Safety signs warn warehouse workers of potential risks so that they can avoid
them (OSHA, 2019). To prevent collisions and other incidents, safety barriers can be
utilized to divide various regions of the facility. Signage may also be used to
communicate critical safety information, such as the location of emergency exits and the
appropriate usage of personal protective equipment (PPE).
5. Ventilation systems
Cold storage facilities may employ ammonia as a refrigerant, which is dangerous
if it seeps into the air. Proper ventilation systems may aid in the management of ammonia
and other pollutants in the air, therefore protecting staff from exposure. Furthermore,
according to the Canadian Centre for Occupational Health and Safety (2023), ventilation
is employed in the workplace to reduce exposure to airborne pollutants. It is often used to
eliminate pollutants such as fumes, dusts, and vapours from working environments in
order to promote a healthy and safe working environment. Ventilation can be achieved
naturally (for example, by opening a window) or mechanically. (e.g., fans or blowers).
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Chapter III
System Diagram and Equipment Sizing
Design Trade-offs
In this part, it will discuss the various types of design options that can be utilized in
designing the Beef meat cold storage facility. It will discuss the differences among the stated
three (3) design options. The outcomes of the several design alternatives will be compared, and
the best design option will be chosen as the final recommendation for the proposed design based
on the findings of the comparison.
In developing a Beef meat cold storage facility, there are a variety of design options to consider.
The following are the design options that can be considered in the proposed design.
A. Dual Compressor, Multi Expansion Valve, Multi Evaporator Cycle
A dual compressor, multi expansion valve, and multi evaporator cycle are
components and configurations used in refrigeration and air conditioning systems. These
systems involve two compressors that work together, each with its own cooling capacity
or operating range. This arrangement allows for more efficient operation and better
control over system performance, making them suitable for larger-scale applications or
situations with variable cooling demands.
A dual compressor system refers to a refrigeration or air conditioning system that
has two compressors working in tandem. Each compressor is typically responsible for a
different cooling capacity or operating range. This setup allows for more efficient
43
operation and better control of the system's performance. Dual compressor systems are
often employed in larger-scale applications or where variable cooling demands exist.
In a multi expansion valve system, multiple expansion valves are integrated into
the refrigeration or air conditioning system. These valves regulate the flow of refrigerant
into the evaporator, enabling it to expand and absorb heat from the surroundings. With
multiple expansion points throughout the system, each valve can be controlled
independently. This setup offers precise temperature control, improved system efficiency,
and the ability to meet varying cooling requirements in different areas or sections of a
building or system.
A multi evaporator cycle refers to a refrigeration or air conditioning system that
incorporates multiple evaporators. The evaporator is responsible for absorbing heat from
the surrounding space or substance, causing the refrigerant to change from a liquid to a
vapor state. By using multiple evaporators, the system can simultaneously cool different
areas or objects. This configuration is particularly beneficial in applications like
commercial refrigeration, where various products or display cases require different
cooling temperatures.
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Figure 3. P-h Diagram of Dual Compressor, Multi Expansion Valve, Multi
Evaporator Cycle
Figure 4. Process Diagram of Dual Compressor, Multi Expansion Valve, Multi
Evaporator Cycle
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A Multiple Evaporator, Multiple Expansion Valve system with 2 compressors as
shown in figures 3 and 4, it involves a complex process flow to efficiently cool multiple
areas or applications.
The process begins with two compressors, which are responsible for compressing
the refrigerant gas. The compressed refrigerant is then directed to a condenser where it
releases heat and converts into a high-pressure liquid. From the condenser, the
high-pressure liquid refrigerant flows into a receiver, which acts as a storage vessel.
The liquid refrigerant is then distributed to multiple evaporators, each serving a
specific area or application. These evaporators are connected in parallel and operate
independently. The refrigerant enters each evaporator at a controlled pressure, typically
regulated by expansion valves.
The expansion valves play a crucial role in controlling the flow of refrigerant into
the evaporators. They regulate the refrigerant's pressure and reduce it to a level suitable
for evaporation. As the refrigerant flows through the expansion valves, it undergoes a
phase change from a high-pressure liquid to a low-pressure vapor.
Inside the evaporators, the low-pressure refrigerant absorbs heat from the
surrounding environment, causing the evaporation process. This heat absorption cools the
respective areas or applications connected to each evaporator. The resulting vapor then
exits the evaporators and returns to the compressors for the cycle to continue.
To maintain the overall system balance and efficiency, various sensors and
controls monitor parameters such as temperature, pressure, and flow rates. These
measurements are used to adjust the operation of the compressors, expansion valves, and
other components to optimize cooling performance across the multiple evaporators.
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In summary, the process flow of a Multiple Evaporator, Multiple Expansion Valve
system with 2 compressors involves compressing the refrigerant, condensing it into a
high-pressure liquid, distributing it to multiple evaporators through expansion valves,
absorbing heat in the evaporators, and returning the vapor to the compressors. The system
relies on careful control and monitoring to ensure efficient cooling for multiple areas or
applications.
B. Triple Compressor, Individual and Multiple Expansion Valve, Multiple Evaporator
Cycle
A multiple evaporator system is a setup in refrigeration or air conditioning that
incorporates more than one evaporator. Each evaporator is responsible for extracting heat
from its specific area or substance. This arrangement enables independent cooling of
different spaces or objects concurrently. Multiple evaporator systems are commonly
employed in applications like multi-zone air conditioning or refrigeration systems with
distinct cooling requirements.
An individual expansion valve system uses a separate expansion valve for each
evaporator within a refrigeration or air conditioning system. Each expansion valve
controls the flow of refrigerant into its respective evaporator, regulating the cooling
capacity and temperature of that particular area. Conversely, a multiple expansion valve
system employs multiple expansion valves within a single system, allowing independent
control of refrigerant flow and temperature for different evaporators or zones. This
configuration offers more precise temperature control and enhanced efficiency.
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A three compressor system refers to a refrigeration or air conditioning
configuration that involves the use of three compressors working together. Each
compressor typically operates within a specific cooling capacity or range. The inclusion
of three compressors ensures efficient cooling performance and improved management of
the system's operation. This setup is commonly utilized in large-scale applications with
diverse cooling requirements or in situations that necessitate redundancy and backup
capabilities.
To summarize, a multiple evaporator system utilizes multiple evaporators to cool
different areas or objects simultaneously. Individual and multiple expansion valve
systems utilize separate or multiple expansion valves, respectively, to regulate refrigerant
flow and temperature control for evaporators. A three compressor system employs three
compressors for enhanced cooling performance and system management.
Figure 5. P-h Diagram of Triple Compressor, Individual and Multiple Expansion Valve,
Multiple Evaporator Cycle
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Figure 6. Process Diagram of Triple Compressor, Individual and Multiple Expansion
Valve, Multiple Evaporator Cycle
A Multiple Evaporator system with Individual and Multiple Expansion Valves,
along with 3 compressors as shown in figures 5 and 6, entails a complex process flow
that allows for efficient cooling across multiple applications or areas. Let's examine the
steps involved in this process.
The system begins with three compressors, responsible for compressing the
refrigerant gas. Each compressor operates independently and is dedicated to a specific set
of evaporators or applications. The compressed refrigerant then flows into a condenser,
where it releases heat and transforms into a high-pressure liquid.
From the condenser, the high-pressure liquid refrigerant enters a receiver, which
acts as a storage vessel for the refrigerant. The liquid refrigerant is then directed to
individual and multiple expansion valves. The individual expansion valves control the
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flow of refrigerant to specific evaporators or applications, while the multiple expansion
valves regulate the refrigerant flow to a group of evaporators.
As the refrigerant passes through the expansion valves, its pressure is reduced,
enabling the phase change from a high-pressure liquid to a low-pressure vapor. Each
expansion valve ensures that the refrigerant enters the corresponding evaporators at the
appropriate pressure for optimal cooling performance.
The low-pressure vapor refrigerant enters the evaporators, where it absorbs heat
from the surrounding environment, causing the evaporation process to occur. The cooling
effect of this heat absorption cools the specific areas or applications connected to each
evaporator. The resulting vapor then exits the evaporators and returns to the compressors
for the cycle to continue.
To maintain system balance and efficiency, various sensors and controls
continuously monitor parameters such as temperature, pressure, and flow rates. These
measurements are used to regulate the operation of the compressors, expansion valves,
and other components, ensuring optimal cooling performance across the multiple
evaporators and applications.
In summary, the process flow of a Multiple Evaporator system with Individual
and Multiple Expansion Valves, along with 3 compressors, involves compressing the
refrigerant, condensing it into a high-pressure liquid, directing it to individual and
multiple expansion valves, absorbing heat in the evaporators, and returning the vapor to
the compressors. The system relies on precise control and monitoring to achieve efficient
cooling across multiple areas or applications.
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C. Triple Compressor, Multiple Expansion Valve, Multiple Evaporator Cycle
The triple compressor cycle is a refrigeration or air conditioning process that
involves the use of three compressors. This cycle typically includes multiple expansion
valves and multiple evaporators as well. Working together, the three compressors
compress the refrigerant and circulate it throughout the system. The refrigerant undergoes
phase changes, absorbing heat from the surroundings in the evaporator and releasing it in
the condenser. This cycle enables efficient cooling and precise temperature control in
various areas or zones.
A triple compressor system refers to a refrigeration or air conditioning setup that
incorporates three compressors operating in conjunction. Each compressor is designed to
function within a specific cooling capacity or operating range. The inclusion of three
compressors enables effective cooling performance and improved system management.
Such configurations are commonly found in larger-scale applications with diverse
cooling requirements or where redundancy and backup capabilities are important.
A multiple expansion valve system involves the use of multiple expansion valves
within a refrigeration or air conditioning system. These expansion valves regulate the
flow of refrigerant into the evaporator, controlling the cooling capacity and temperature
in different sections of the system. By utilizing multiple expansion valves, the system
achieves precise temperature control, enhanced energy efficiency, and the ability to meet
varying cooling demands in different zones or evaporators.
A multiple evaporator system refers to a refrigeration or air conditioning setup
that incorporates more than one evaporator. Each evaporator is responsible for extracting
heat from its specific area or substance. This arrangement allows for independent cooling
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of different spaces or objects simultaneously. Multiple evaporator systems are commonly
utilized in applications such as multi-zone air conditioning or refrigeration systems with
specific cooling requirements.
In summary, a triple compressor system employs three compressors for efficient
cooling performance. A multiple expansion valve system utilizes multiple expansion
valves to achieve precise temperature control and improved energy efficiency. A multiple
evaporator system enables independent cooling of different spaces or objects. The triple
compressor cycle incorporates three compressors, multiple expansion valves, and
multiple evaporators in its refrigeration or air conditioning operation.
Figure 7. P-h Diagram of Triple Compressor, Multiple Expansion Valve, Multiple
Evaporator Cycle
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Figure 8. Process Diagram of Triple Compressor, Multiple Expansion Valve,
Multiple Evaporator Cycle
The process flow of a system with multiple evaporators, multiple expansion
valves, and three compressors as shown in figures 7 and 8 typically involves the
following steps.
Initially, the refrigerant, which is in a low-pressure state, enters the first
compressor. The first compressor raises the pressure and temperature of the refrigerant
and pumps it into the condenser. In the condenser, the refrigerant releases heat to the
surroundings and undergoes a phase change from a high-pressure vapor to a
high-pressure liquid.
The high-pressure liquid refrigerant then flows through multiple expansion
valves, each serving a specific evaporator. These expansion valves control the flow of
refrigerant into each evaporator, regulating the cooling capacity of the system. The
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expansion valves reduce the pressure and temperature of the refrigerant, allowing it to
evaporate inside each evaporator.
As the refrigerant evaporates in the evaporators, it absorbs heat from the
surroundings, providing cooling effects. The evaporated refrigerant, now in a
low-pressure vapor state, is then sucked into the three compressors.
Each compressor further raises the pressure and temperature of the vapor,
preparing it for the next cycle. The compressed vapor is then directed back to the
condenser, where the heat release process occurs again, completing the cycle.
Overall, the multiple evaporators, multiple expansion valves, and three
compressors work in synchronization to provide cooling to different areas or zones
simultaneously, optimizing the efficiency and performance of the refrigeration system.
Heat Load and Thermodynamic Calculations
a. Design Data
Considering the layout, it primarily consists of an Ante Room, Chiller Room and
Cold Storage with additional complementary rooms depending on the requirements such
as admin office, meat inspection office, cleaning room, toilets, and equipment room;
specification of the equipment is essential to calculate for the cooling load of the system.
In general, the ideal construction type is a single-story as it is less costly, lighter
construction, and it can be easily designed in accordance with the required refrigerator
specification and easier handling process. It will serve as the guide and basis to solve the
54
important parameters in the cold storage facility. The material with its specifications
presented and it will be used for the design of the beef meat cold storage facility.
Wall
The wall required on a cold facility needs for a more strategic construction to
maintain the inside condition. For this proposed design, we will be using XPS insulation
boards that are lightweight and could offer greater thermal insulation than others.
Specified below are all of the important details and specifications.
Table No. 3
Specification of Wall Panel
Wall Element
Thickness,m
Thermal Conductivity
(k), W/m-k
Thermal
Resistance
20 Gauge 304
Stainless Steel
0.00095
16.2
0.00005864197531
Extruded
Polystyrene
Board
0.14
0.038
3.684210526
20 Gauge 304
Stainless Steel
0.0095
16.2
0.00005864197531
Concrete Slab
0.2
0.87
0.2298850575
0.3019
33.308
2.861581289
Total
Legend
55
Figure 8. Heat Transfer Model of the Walls
Roof
A roof’s purpose is to maintain a building's interior dry and protected. For a cold
storage requirement, it must be vapor-tight, water-tight, energy-efficient, andis being
highly used in different food processing facilities where temperature control is essential.
Table No. 4
Specification of Roof Panel
Wall Element
Legend
Thickness,m
Thermal Conductivity
(k), W/m-k
Thermal
Resistance
20 Gauge 304
Stainless Steel
0.00095
16.2
0.00005864197531
Extruded
Polystyrene Board
0.14
0.038
3.684210526
56
20 Gauge 304
Stainless Steel
Total
0.0095
16.2
0.00005864197531
0.1419
32.438
3.68432781
Figure 9. Heat Transfer Model of the Roof
Floor
Due to the continuous cyclic movement and heavy loadings, the floor must be
strong enough and can withstand any varying temperature which is predominant in any
manufacturing and processing industries. Also, floors must be well-insulated to reduce
heat transmission from the ground below and to block the transmission of water vapor
through any external surfaces.
57
Table No. 5
Specification of Floor
Wall Element
Thickness,m
Thermal Conductivity
(k), W/m-k
Thermal
Resistance
Reinforced
Concrete
0.15
3.2
0.046875
Extruded
Polystyrene
Board
0.14
0.038
3.684210526
Concrete Slab
0.2
0.87
0.2298850575
0.49
4.108
3.960970584
Total
Legend
Figure 9. Heat Transfer Model of the Floor
58
b. Heat Load
The following types of heat loads are examined, which are essential for calculating the
overall cooling loads and determining the refrigerating capacity.Also, The formula for
determining the quantity of energy that would be required or removed to maintain the
temperature in cold storage will be presented.
1. Transmission Load
Transmission load refers to the heat gain or loss through the building envelope,
including walls, roofs, windows, doors, and floors. It accounts for the thermal transfer
that occurs between the indoor and outdoor environments. The transmission load is
influenced by factors such as the insulation properties of the building materials, the
surface area of the envelope, temperature differentials, and weather conditions. To
calculate the transmission load, the following formula can be used:
𝑄 = 𝑈·
𝐴 (𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 − 𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒) 24
1000
Where:
Q - heat load (kWh/day)
U - heat transfer coefficient (𝑊/𝑚2 · 𝐾)
A - walls, roof, and floor surface are m2
External Temp. - ambient air temperature (°𝐶)
Internal Temp. - inside the chamber temperature (°𝐶)
24 - no. of hours in a day
59
1000 - conversion from Watt (W) to Kilowatt (kW)
2. Product Load
The product load is the heat gain or cooling demand generated by the stored
products within a specific space, such as a cold storage facility. Estimating the product
load is crucial for determining the refrigeration capacity and energy requirements to
maintain the desired temperature. To calculate the product load, the following formula
can be used:
𝑄 =
𝑚𝐶𝑝 (𝑇𝑒𝑚𝑝. 𝐸𝑛𝑡𝑒𝑟 − 𝑇𝑒𝑚𝑝. 𝑆𝑡𝑜𝑟𝑎𝑔𝑒)
3600
Where:
Q - heat load (kWh/day)
m - the mass of added products (kg)
𝑐𝑝 - specific heat capacity of the products (kJ/kg °C)
Temp. Enter - temperature of the entering products (°C)
Temp. Storage - the temperature inside the storage (°C)
3. Internal Load
The internal load is the heat, within the facility, that is given off by the people
working in the cold room and lighting.
60
3.1 Occupant Load
The occupant load is the heat given off by the people entering the
cold chamber facility. In order to determine the load, the number of
people, time, and heat will be considered
𝑄 =
(𝑂𝑐𝑐𝑢𝑝𝑎𝑛𝑡𝑠)(𝑇𝑖𝑚𝑒)(𝐻𝑒𝑎𝑡)
1000
Where:
Q - heat load (kWh/day)
Occupants - Number of People Inside
Time - Working hours (hr)
Heat - heat loss per person per hour (W)
3.2 Lighting Load
The lighting load is the heat generated by the lights inside the cold
chamber facility. To determine the lighting load we can use the formula:
𝑄 =
(𝐿𝑎𝑚𝑝𝑠)(𝑇𝑖𝑚𝑒)(𝑊𝑎𝑡𝑡𝑎𝑔𝑒)
1000
Where:
Q - heat load (kWh/day)
Lamps- number of lamps installed
Time- operating hours per day
Wattage- power rating of the lamp
61
4. Equipment Load
In this part, equipment used in cold chamber facilities such as fan motors in the
evaporator and crates must be considered. The heat removed from it must be computed.
This can be computed as:
4.1 Fan Motors
Heat generation of the fan motors in the evaporator. For this, we can use
the formula:
𝑄 =
(𝐹𝑎𝑛𝑠)(𝑇𝑖𝑚𝑒)(𝑊𝑎𝑡𝑡𝑎𝑔𝑒)
1000
Where:
Q - heat load (kWh/day)
Fans- number of fans installed
Time- fan daily run hours (hrs)
Wattage- power rating of the fan motors (Watt)
1000 = convert from watts to kW
4.2 Meat Trolley Load
𝑄 = 𝑚𝐶𝑝𝑛∆𝑇
Where:
Q- heat load (kWh/day)
m- the mass of the trolley with sliding hooks(kg)
𝐶p - specific heat capacity of the trolley with sliding hooks (kJ/kg °C)
n- number of trolley with sliding hooks
62
∆𝑇 - change in temperature (°C)
5. Infiltration Load
The infiltration heat load is the amount of heat absorbed through the windows,
doors, and walls with higher enthalpy that entered the refrigerated area. To calculate the
infiltration load, it will consider the number of volume changes each day, the volume of
the cold storage, energy, and the inlet and outdoor temperature.
𝑄 =
𝑣𝑜𝑙𝑢𝑚𝑒 𝑥 𝑒𝑛𝑒𝑟𝑔𝑦 𝑥 𝑐ℎ𝑎𝑛𝑔𝑒 𝑥 ∆𝑇
3600
Where:
Q - heat load (kWh/day)
Volume - the volume of the cold storage (𝑚3)
Energy = Energy per cubic meter (kJ/m³ ℃)
Change - number of volume changes per day
ΔT - the difference between the air temperature outside and air temperature inside (°C)
3600 - conversion of kJ to kWh
6. Total Accumulated Load
To obtain the value for the total accumulated load, all cooling load parameters
will be added. The formula to be used is:
𝑇𝐴𝐿 = 𝑇𝐿 + 𝑃𝐿 + 𝐼𝑛𝑡𝐿 + 𝐸𝐿 + 𝐼𝑛𝑓𝐿
Where:
TAL = Total Accumulated Load (kWh/day)
63
TL - Transmission Load (kWh/day)
PL - Product Load (kWh/day)
IntL - Internal Load (kWh/day)
EL - Equipment Load (kWh/day)
InfL - Infiltration Load (kWh/day)
c. Calculation Results
Summary of Heat Load Calculations
i.
Total Transmission Load
Table No. 6
Transmission Load Calculations for Ante Room
Parameters
Value
Unit
Area of Front Side Wall
132.4
m2
Area of Back Side Wall
99.2
m2
Area of Right SIde Wall
99.2
m2
Area of Left Side Wall
69.6
m2
Area of Roof
168.53
m2
Area of Floor
168.53
m2
Area of Door
0.03
m2
64
Internal Temperature
283.15
K
Ambient Air Temperature
298.15
K
Chiller Room Temperature
273.15
K
Cold Storage Temperature
255.15
K
Heat Transfer Coefficient on Wall (U)
0.2554792071
W/m2K
Heat Transfer Coefficient on Roof (U)
0.271419931
W/m2K
Heat Transfer Coefficient on Floor (U)
0.2524633745
W/m2K
Heat Transfer Coefficient on Door (U)
0.05666666667
W/m2K
Heat Load Results
Wall
-4.04188544
kWh/day
Roof
16.46726435
kWh/day
Floor
15.3171549
kWh/day
Door
0.00204
kWh/day
Total Transmission Load
27.74457381
kWh/day
The table above presents the different transmission load on the wall, roof, floor, and door.
The results differ mainly because of the materials being used in the construction and how large
the area is. For the total transmission load of Ante room, the calculated Transmission load is
27.74457381 kWh/day.
65
Table No. 7
Transmission Load Calculations for Chiller Room
Parameters
Value
Unit
Area of Front Side Wall
24
m2
Area of Back Side Wall
24
m2
Area of Right Side Wall
40
m2
Area of Left Side Wall
40
m2
Area of Roof
80
m2
Area of Floor
80
m2
Area of Door
10.15
m2
Internal Temperature
273.15
K
Ambient Air Temperature
298.15
K
Ante Room Temperature
283.15
K
Cold Storage Temperature
255.15
K
Heat Transfer Coefficient on Wall (U)
0.2554792071
W/m2K
Heat Transfer Coefficient on Roof (U)
0.271419931
W/m2K
Heat Transfer Coefficient on Floor (U)
0.2524633745
W/m2K
Heat Transfer Coefficient on Door (U)
0.3625
W/m2K
Heat Load Results
66
Wall
11.52722183
kWh/day
Roof
6.25351521
kWh/day
Floor
4.847296791
kWh/day
Door
1.7661
kWh/day
Total Transmission Load
24.39413383
kWh/day
The table above presents the different transmission load on the wall, roof, floor, and door.
The results differ mainly because of the materials being used in the construction and how large
the area is. For the total transmission load of the Chiller room, the calculated Transmission load
is 24.39413383 kWh/day.
Table No. 8
Transmission Load Calculations for Cold Storage
Parameters
Value
Unit
Area of Front Side Wall
18
m2
Area of Back Side Wall
18
m2
Area of Right Side Wall
40
m2
Area of Left Side Wall
40
m2
Area of Roof
60
m2
Area of Floor
60
m2
Area of Door
10.15
m2
Internal Temperature
255.15
K
67
Ambient Air Temperature
298.15
K
Ante Room Temperature
283.15
K
Chiller Room Temperature
273.15
K
Heat Transfer Coefficient on Wall (U)
0.2554792071
W/m2K
Heat Transfer Coefficient on Roof (U)
0.271419931
W/m2K
Heat Transfer Coefficient on Floor (U)
0.2524633745
W/m2K
Heat Transfer Coefficient on Door (U)
0.39
W/m2K
Heat Load Results
Wall
23.59401574
kWh/day
Roof
10.94365162
kWh/day
Floor
10.17932326
kWh/day
Door
5.320224
kWh/day
Total Transmission Load
39.85789135
kWh/day
The table above presents the different transmission load on the wall, roof, floor, and door.
The results differ mainly because of the materials being used in the construction and how large
the area is. For the total transmission load of Cold Storage, the calculated Transmission load is
39.85789135 kWh/day.
68
ii.
Total Infiltration Load
Table No. 9
Infiltration Load Calculations for Ante Room
Parameters
Value
Unit
Volume Changes per Day
4
Cold Storage Volume
775.238
m3
Energy per cubic meter
2
kJ/m3kg
Ambient Temperature
25
℃
Internal Temperature
10
℃
Infiltration Load
25.84126667
kWh/day
The table above presents the parameters in computing for the infiltration load of the cold
storage which includes the volume changes per day, volume of the cold storage, energy per cubic
meter of air, and the ambient air and internal air temperature. Thus, the computed value of
infiltration load for AnteRoom is 25.84126667 kWh/day
Table No. 10
Infiltration Load Calculations for Chiller Room
Parameters
Value
Volume Changes per Day
4
Unit
69
Cold Storage Volume
368
m3
Energy per cubic meter
2
kJ/m3℃
Ambient Temperature
10
℃
Internal Temperature
0
℃
Infiltration Load
8.177777778
kWh/day
The table above presents the parameters in computing for the infiltration load of the cold
storage which includes the volume changes per day, volume of the cold storage, energy per cubic
meter of air, and the ambient air and internal air temperature. Thus, the computed value of
infiltration load for Chiller Room is 8.177777778 kWh/day
Table No. 11
Infiltration Load Calculations for Cold Storage
Parameters
Value
Unit
Volume Changes per Day
4
Cold Storage Volume
285
m3
Energy per cubic meter
2
kJ/m3℃
Ambient Temperature
0
℃
Internal Temperature
-18
℃
Infiltration Load
-5.077333333
kWh/day
70
The table above presents the parameters in computing for the infiltration load of the cold
storage which includes the volume changes per day, volume of the cold storage, energy per cubic
meter of air, and the ambient air and internal air temperature. Thus, the computed value of
infiltration load for Cold Storage is -5.077333333 kWh/day.
iii.
Total Internal Load
Table No. 12
Internal Load Calculations for Ante Room
Parameters
Value
Unit
Lighting
Number of Lights
10
Wattage of Lights
50
W
Time of Operation
24
hrs
Lighting Load
12
kWh/day
Occupant
Number of Person Working Inside
4
Time of Operation
3
hrs
Average Heat Loss per person
565
Watt/hr
Occupant Load
6.78
kWh/day
71
Total Internal Load
18.78
kWh/day
The table above presents the parameters in computing the internal load where in it is the
summation of the lighting load and occupant load inside the Anteroom. Thus, the internal load at
the anteroom is 18.78 kWh/day.
Table No. 13
Internal Load Calculations for Chiller Room
Parameters
Value
Unit
Lighting
Number of Lights
5
Wattage of Lights
50
W
Time of Operation
24
hrs
Lighting Load
6
kWh/day
Occupant
Number of Person Working Inside
10
Time of Operation
2
hrs
Average Heat Loss per person
565
Watt/hr
Occupant Load
11.3
kWh/day
Total Internal Load
17.3
kWh/day
72
The table above presents the parameters in computing the internal load where in it is the
summation of the lighting load and occupant load inside the chiller room. Thus, the internal load
at chiller room is 17.3 kWh/day
Table No. 14
Internal Load Calculations for Cold Storage
Parameters
Value
Unit
Lighting
Number of Lights
5
Wattage of Lights
50
W
Time of Operation
24
hrs
Lighting Load
6
kWh/day
Occupant
Number of Person Working Inside
10
Time of Operation
2
hrs
Average Heat Loss per person
565
Watt/hr
Occupant Load
11.3
kWh/day
Total Internal Load
17.3
kWh/day
The table above presents the parameters in computing the internal load where in it is the
summation of the lighting load and occupant load inside the cold storage. Thus, the internal load
at cold storage is 17.3 kWh/day.
73
iv.
Total Product Load
Table No. 15
Product Load Calculations for Ante Room
Parameters
Value
Unit
Mass
340
kg
Specific Heat
3.4
kJ/kg-K
Temperature of Entering Products
298.15
K
Internal Temperature
283.15
K
Product Load
4.816666667
kWh/day
The table above presents the different parameters in computing for the product load in the
cold storage which includes the mass of the beef carcass entering the ante room, specific heat
capacity of the beef, temperature of entering product, and the internal temperature of the Ante
room. Thus, the product load is 4.816666667 kWh/day.
Table No. 16
Product Load Calculations for Chiller Room
Parameters
Value
Unit
Mass
17000
kg
Specific Heat
3.4
kJ/kg-K
74
Temperature of Entering Products
283.15
K
Internal Temperature
273.15
K
Product Load
160.5555556
kWh/day
The table above presents the different parameters in computing for the product load in the
cold storage which includes the mass of the beef carcass entering the chiller room, specific heat
capacity of the beef, temperature of entering product, and the internal temperature of the Chiller
room. Thus, the product load is 160.5555556 kWh/day.
Table No. 17
Product Load Calculations for Cold Storage
Parameters
Value
Unit
Mass
17000
kg
Specific Heat above freezing
3.4
kJ/kg-K
Latent Heat Fusion
233
kJ/kg
Specific Heat Below Freezing
1.67
kJ/kg-K
Freezing Point of Beef
271.45
K
Temperature of Entering Products
275.15
K
Internal Temperature
257.15
K
Product Load
1256.115833
kWh/day
75
The table above presents the different parameters in computing for the product load in the
cold storage which includes the mass of the beef carcass entering the cold storage, specific heat
capacity above freezing of the beef, latent heat fusion, specific heat capacity below freezing,
freezing point of beef, temperature of entering product, and the internal temperature of the Cold
Storage. Thus, the product load is 1256.115833 kWh/day.
v.
Equipment Load Calculation
Table No. 18
Equipment Load Calculations
Parameters
Value
Unit
Ante Room
7.2
kWh/day
Chiller Room
21.78333333
kWh/day
Cold Storage
33.45
kWh/day
Equipment Load
62.43333333
kWh/day
The table above presents the parameters in computing the equipment load in cold storage.
It is the summation of the load accumulated from the fans, plastic crates, and racks in three
rooms. Thus, the equipment load is 62.43333333 kWh/day.
76
Summary of Heat Loads
Table No. 19
Summary of Heat Loads in Ante Room
Type of Load
Value
Unit
Transmission Load
27.74457381
kWh/day
Product Load
4.816666667
kWh/day
Internal Load
18.78
kWh/day
Equipment Load
7.2
kWh/day
Infiltration Load
25.84126667
kWh/day
Total Cooling Load
84.38250715
kWh/day
Refrigeration Cooling Capacity
3.867531578
kW
Tons of Refrigeration
1.09998054
TOR
The table above presents the result of the calculation for the total cooling load of the ante
room of cold storage based on the different parameters. Thus, the total cooling load is
84.38250715 kWh/day. After dividing the number of hours per day, the total cooling capacity is
now 3.867531578 kW. Thus, the proposed design of Ante Room has 1.09998054 TOR.
77
Table No. 20
Summary of Heat Loads in Chiller Room
Type of Load
Value
Unit
Transmission Load
24.39413383
kWh/day
Product Load
160.5555556
kWh/day
Internal Load
17.3
kWh/day
Equipment Load
21.78333333
kWh/day
Infiltration Load
8.177777778
kWh/day
Total Cooling Load
232.2108005
kWh/day
Refrigeration Cooling Capacity
10.64299502
kW
Tons of Refrigeration
3.027017925
TOR
The table above presents the result of the calculation for the total cooling load of the cold
storage based on the different parameters. Thus, the total cooling load is 232.2108005 kWh/day.
After dividing the number of hours per day, the total cooling capacity is now 10.64299502 kW.
Thus, the proposed design of Ante Room has 3.027017925 TOR.
Table No. 21
Summary of Heat Loads in Cold Storage
Type of Load
Value
Unit
Transmission Load
39.85789135
kWh/day
78
Product Load
1256.115833
kWh/day
Internal Load
17.3
kWh/day
Equipment Load
33.45
kWh/day
Infiltration Load
-5.077333333
kWh/day
Total Cooling Load
1341.646391
kWh/day
Refrigeration Cooling Capacity
61.49212627
kW
Tons of Refrigeration
17.48922818
TOR
The table above presents the result of the calculation for the total cooling load of the cold
storage based on the different parameters. Thus, the total cooling load is 1341.646391 kWh/day.
After dividing the number of hours per day, the total cooling capacity is now 61.49212627 kW.
Thus, the proposed design of Ante Room has 17.48922818 TOR.
Individual Equipment Design Sizing
A. Compressors
Table No. 22
Compressor Sizing
Parameters
Design Option 1
Design Option 2
Design Option 3
Bore at Compressor 1
62.26 mm
36.62 mm
36.62 mm
Stroke at Compressor
62.26 mm
36.62 mm
36.62 mm
79
1
Bore at Compressor 2
124 mm
58.55 mm
55.42 mm
Stroke at Compressor
124 mm
58.55 mm
55.42 mm
Work Compressor 1
1.727 kW
0.3151 kW
0.3151 kW
Work Compressor 2
11.93 kW
1.437 kW
1.218 kW
Work Compressor 3
-
11.93 kW
11.93 kW
2
B. Condenser/Evaporator
Table No. 23
Condenser/Evaporator Sizing
Parameters
Design Option 1
Design Option 2
Design Option 3
Length
1.401 m
1.434 m
1.414 m
Heat Rejected
75.53 kW
75.16 kW
73.34 kW
Inside Diameter of Tube
77.927 mm
77.927 mm
77.927 mm
Outside Diameter of Tube
88.9 mm
88.9 mm
88.9 mm
Inside Diameter of Annulus
102.26 mm
102.26 mm
102.26 mm
Outside Diameter of
114.30 mm
114.30 mm
114.30 mm
3 in
3 in
3 in
Annulus
Nominal Diameter of Tube
80
Nominal Diameter of
4 in
4 in
4 in
Annulus
C. Expansion Valve
Table No. 24
Expansion Valve Sizing
Parameters
Design Option 1
Design Option 2
Design Option 3
Pressure Drop 1
355.7 kPa
355.7 kPa
355.7 kPa
Pressure Drop 2
121.9 kPa
121.9 kPa
121.9 kPa
Pressure Drop 3
148.3
148.3
148.3
Pressure Drop 4
121.9 kPa
-
-
Individual Equipment Design Catalog
A. Compressors
Table No. 25
Design Option 1 Compressor Catalog
Tag No.
CMP-011
Parameter
s
B1
Computed Model No.
62.26 mm
Specifications
GPT18RG Capacity: 1731 W
Manufacturer
Cost
Cubigel
$179.19
81
S1
62.26 mm
Displacement: 18
cm3
Power: 0.5 hp
Application: HBP
CPR Cooling: F
Wc1
Voltage
1.727 kW
Frequency:200-22
0/230V 50/60Hz
~1
Motor: CSR
Weight: 12.84 kg
B2
124 mm
Cooling Capacity:
S2
124 mm
13.52 kW
Displacement: 28
m3/h
CMP-012
2DB-50X
Wc2
Capacity Control:
Copeland
$993.15
None, Step,
11.93 kW
Inverter
Power Supply:
400V/3Ph/50/Hz
Table No. 26
Design Option 2 Compressor Catalog
Tag No.
Parameters Computed Model No.
B1
CMP-021
S1
36.62 mm
36.62 mm
B35G5
Specifications
Capacity: 366 W
Displacement: 3.5
Manufacturer
Cost
Cubigel
$108.84
82
cm3
Power: 0.1 hp
Application: HBP
CPR Cooling: S/F
Wc1
0.3151 kW
Voltage
Frequency:110-11
5V 60Hz - 1
Motor: CSIR
Weight: 4.9 kg
B2
58.55 mm
Capacity: 1706 W
S2
58.55 mm
Displacement:
14.32 cm3
Power: 0.5 hp
Application: HBP
CMP-022
GPY14RDa CPR Cooling: F
Wc2
Cubigel
$159.23
Copeland
$993.15
Voltage
1.437 kW
Frequency:115V
60Hz - 1
Motor: CSIR
Weight: 12.03 kg
B3
124 mm
Cooling Capacity:
S3
124 mm
13.52 kW
Displacement: 28
m3/h
CMP-023
2DB-50X
Wc3
11.93 kW
Capacity Control:
None, Step,
Inverter
Power Supply:
400V/3Ph/50/Hz
83
Table No. 27
Design Option 3 Compressor Catalog
Tag No.
Parameters Computed Model No.
Specifications
B1
36.62 mm
Capacity: 366 W
S1
36.62 mm
Displacement: 3.5
Manufacturer
Cost
Cubigel
$108.84
Cubigel
$137.23
Copeland
$993.15
cm3
Power: 0.1 hp
Application: HBP
B35G5
CMP-031
CPR Cooling: S/F
Voltage:110-115V
Wc1
0.3151 kPa
Frequency: 60Hz 1
Motor: CSIR
Weight: 4.9 kg
B2
55.42 mm
Capacity: 1372 W
S2
55.42 mm
Displacement:
12.10 cm3
Power: 3/8 hp
Application:
CMP-032
GPY12RDa
Wc2
1.218 kPa
HMBP
CPR Cooling: F
Voltage: 115V
Frequency:60Hz 1
Motor: CSIR
Weight: 12.03 kg
CMP-033
B3
124 mm
S3
124 mm
Capacity: 13.52
2DB-50X
kW
Displacement: 28
84
m3/h
Capacity Control:
Wc3
None, Step,
11.93 kPa
Inverter
Power Supply:
400V/3Ph/50/Hz
B. Condenser
Table No. 28
Design Option 1 Condenser Catalog
Tag No.
Parameters Computed Model No.
L
1.401 m
Qr
75.53 kW
Inside
Diameter (T)
Outside
Diameter (T)
CND-011
Cost
Leg Mount
88.9 mm
TCVM
Type
102.26 mm
(A)
050.1-13-E
-N
Vertical
Length: 2644 mm
Depth: 1050 mm
Height: 938 mm
Thermocoil
LTD.
$2372.5
Net Weight: 168
kg
Outside
Diameter
Manufacturer
77.927 mm
Inside
Diameter
Specifications
114.30 mm
(A)
ND for Tube
3 in
85
ND for
Annulus
4 in
Table No. 29
Design Option 2 Condenser Catalog
Tag No.
Parameters
Computed
L
1.434 m
Qr
75.16 kW
Inside
Diameter (T)
Outside
Diameter (T)
Diameter
Manufactu
rer
Cost
Leg Mount
88.9 mm
Vertical
TCVM Type
102.26 mm 050.1-13-EN
(A)
Outside
Diameter
Specifications
77.927 mm
Inside
CND-021
Model No.
Length: 2644
mm
Thermocoi
Depth: 1050 mm
l LTD.
$2372.5
Height: 938 mm
Net Weight: 168
114.30 mm
kg
(A)
ND for Tube
ND for
Annulus
3 in
4 in
86
Table No. 30
Design Option 3 Condenser Catalog
Tag No. Parameters Computed
L
1.414 m
Qr
73.34 kW
Model No. Specifications Manufacturer
Cost
Inside
Diameter
77.927 mm
(T)
Base Mount:
Outside
Diameter
CND-031
Horizontal
88.9 mm
Length: 2644
(T)
TCHM
mm
Inside
Type
Depth: 1050
Thermocoil
mm
LTD.
Diameter
102.26 mm 050.1-13-CN
(A)
114.30 mm
(A)
ND for
Tube
ND for
Annulus
Height: 938
mm
Outside
Diameter
$2304
Net Weight:
177 kg
3 in
4 in
87
C. Evaporator
Table No. 31
Design Option 1, 2 and 3 Evaporator Catalog
Tag No. Parameters Computed Model No.
Specifications
Manufacturer
Cost
Stefani Spa
$1246
No. of Fans: 3
Capacity: 3.8 kW
Power: 219 W
BOREA H
EVP-001
RE1
3.8 kW
25-3 E 7,5
A AC 04S
Fan Speed: 1300
rpm
Weight: 43 kg
Length: 1275 mm
Height: 396 mm
Width: 360 mm
No. of Fans: 3
Capacity: 10.8
kW
BOREA H
EVP-002
RE2
10.8 kW 35-3 D 6,5
A AC 04S
Power: 495 W
Fan Speed: 1340
rpm
Stefani Spa
$1424
Stefani Spa
$2136
Weight: 84.8 kg
Length: 1846 mm
Height: 575 mm
Width: 491 mm
BOREA H
EVP-003
RE3
61.6 kW 50-4 G 4 A
AC 04S
No. of Fans: 4
Capacity: 61.6
kW
Power: 2720 W
88
Fan Speed: 1300
rpm
Weight: 354 kg
Length: 4000 mm
Height: 844 mm
Width: 633 mm
D. Expansion Valve
Table No. 32
Design Option 1 Expansion Valve Catalog
Tag No.
Parameters
EXP-011
Pressure
Computed Model No.
Specifications
Cost
Type: CGX
355.7 kPa
Inlet Connection: 5/16”
Drop 1
Solder
2315D
Outlet Connection: ⅜”
Solder
$192
Net Weight: 0.15 kg
Factory Pressure Setting:
0.206+- 0.02 MPa
EXP-012
Pressure
Type: CGX
121.9 kPa
Inlet Connection: ½” Flare
Drop 2
2348B
Outlet Connection: ⅝” Flare
Net Weight: 0.85 kg
$265
Factory Pressure Setting:
0.142 MPa
EXP-013
Pressure
148.3
2348B
Type: CGX
$265
89
Inlet Connection: ½” Flare
Drop 3
Outlet Connection: ⅝” Flare
Net Weight: 0.85 kg
Factory Pressure Setting:
0.142 MPa
EXP-014
Pressure
Type: CGX
121.9 kPa
Inlet Connection: ½” Flare
Drop 4
2348B
Outlet Connection: ⅝” Flare
Net Weight: 0.85 kg
$265
Factory Pressure Setting:
0.142 MPa
Table No. 33
Design Option 2 Expansion Valve Catalog
Tag No.
Parameters
EXP-021
Pressure
Computed Model No.
Specifications
Cost
Type: CGX
355.7 kPa
Inlet Connection: 5/16”
Drop 1
Solder
2315D
Outlet Connection: ⅜”
Solder
$192
Net Weight: 0.15 kg
Factory Pressure Setting:
0.206+- 0.02 MPa
EXP-022
Pressure
Drop 2
Type: CGX
121.9 kPa
2348B
Inlet Connection: ½” Flare
Outlet Connection: ⅝”
$265
Flare
90
Net Weight: 0.85 kg
Factory Pressure Setting:
0.142 MPa
EXP-023
Pressure
Type: CGX
148.3 kPa
Inlet Connection: ½” Flare
Drop 3
Outlet Connection: ⅝”
2348B
Flare
$265
Net Weight: 0.85 kg
Factory Pressure Setting:
0.142 MPa
Table No. 34
Design Option 3 Expansion Valve Catalog
Tag No.
Parameters
EXP-031
Pressure
Computed Model No.
Specifications
Cost
Type: CGX
355.7 kPa
Inlet Connection: 5/16”
Drop 1
Solder
2315D
Outlet Connection: ⅜”
Solder
$192
Net Weight: 0.15 kg
Factory Pressure Setting:
0.206+- 0.02 MPa
EXP-032
Pressure
Type: CGX
121.9 kPa
Inlet Connection: ½” Flare
Drop 2
2348B
Outlet Connection: ⅝”
$265
Flare
Net Weight: 0.85 kg
91
Factory Pressure Setting:
0.142 MPa
EXP-033
Pressure
Type: CGX
148.3 kPa
Inlet Connection: ½” Flare
Drop 3
Outlet Connection: ⅝”
2348B
Flare
$265
Net Weight: 0.85 kg
Factory Pressure Setting:
0.142 MPa
92
Chapter IV
Engineering Design Collection and Cost Analysis
1. Cost Benefit Analysis
Initial Project Cost
In this section, the initial project cost of the project is presented for each of the
design trade-offs. These include the cost of materials, components, equipment, and other
significant items that are needed to complete the 76 kW cold storage facility for beef in
the municipality of Padre Garcia, Batangas.
A. Initial Project Cost of Materials for Design Options 1,2, and 3
Table No. 35
Initial Project Cost of Materials for Project Construction
Materials
Quantity
Price per Quantity
Total
$889.60
$889.60
500 sqm
$195.71/sqm
$97,855
45 tons(45,000 kg)
$4.22/40 kg
$4,747.50
67.5 tons(67,500 kg)
$0.55/30 kg
$1,237.50
135 tons(135,000)
$0.85/30 kg
$3,825.00
Permits
Land
Ten Wheeler Truck of
Cement
Dump Truck of Sand
Dump Truck of
Gravel
93
Concrete Hollow
10,750 pcs
$0.22/block
$2,365.00
50 (6 m)
$96.37/pc
$4,818.18
20
$15.93/pc
$318.55
Deformed Bar
2500
$0.57/pc
$1,434.55
Concrete Nail #3
25 kg
$3/kg
$75
Concrete Nail #2
25 kg
$2.75/kg
$68.75
PVC Strip Curtain
26 strips
$3/pc
$78
4
$363.31/pc
$1,453.24
500 sqm
$16.93/sqm
$8,465
500 sqm
$16.93/sqm
$8,465
Door
3
$120.90/pc
$362.73
Toilet Bowl
2
$65.23/pc
$130.55
Basin Sink
1
$89.09/pc
$89.09
LED Lights (50W)
10
$4.23/pc
$42.3
LED Bulb (20W)
10
$3.25/pc
$32.5
Block #4
Wide Flange Beam
Stainless Steel Angle
Bar
PUF Insulated Sliding
Door
PUF Insulation Roof
Panel
(0.35 mm thickness)
PUF Insulation Wall
Panel
(0.35 mm thickness)
94
Miscellaneous
$5,000
Total Cost
$141,752.86
B. Initial Project Cost of Equipment for Design Options 1,2, and 3
Table No. 36
Initial Project Cost of Equipment
Equipment
Quantity
Price per Quantity
Total
2
$29500/pc
$59,000
Plastic Crates
500
$5/pc
$2,500
Office Chair
14
$26.69/pc
$373.66
Table
4
$35.58/pc
$142.32
Trolley
6
$80/pc
$480
Forklift 4-5 Ton - H
Series
Total Cost
$62,495.98
95
C. Initial Project Cost of Instrumentation and Control Equipment for Design
Options 1, 2, and 3
Table No. 37
Initial Project Cost of Instrumentation and Control Equipment
Equipment
Quantity
Price per Quantity
Total
Electrical Boxes and Conduit
10
$13.50/pc
$135
Fire Extinguisher
6
$26.50/pc
$159
Thermocouple Sensor
4
$12.50
$50
Uninterrupted Power Supply
1
$2,000
$2,000
Generator
1
$650
$650
Total Cost
$2,994
D. Initial Project Cost of Components for Design Options 1
Table No. 38
Initial Project Cost of Components for Design Option 1
Tag No.
Component
Model No
Quantity
Price per
Total
Quantity
EVP-001
BOREA H 25-3
E 7,5 A AC 04S
EVP-002
1
$1246
$1246
1
$1246
$1246
BOREA H 35-3
D 6,5 A AC 04S
96
EVP-003
Evaporator
BOREA H 50-4
G 4 A AC 04S
CMP-011
CMP-012
CND-011
Compressor
Condenser
1
$2136
$2136
GPT18RG
1
$179.19
$179.19
2DB-50X
1
$993.15
$993.15
TCVM Type
1
$2372.5
$2372.5
050.1-13-E-N
EXP-011
2315D
1
$192
$192
EXP-012
2348B
1
$265
$265
2348B
1
$265
$265
2348B
1
$265
$265
-
1
$20,838
$20,838
EXP-013
EXP-014
PSA-011
Expansion
Valve
Piping System
Assembly
Total Cost
$29,997.84
E. Initial Project Cost of Components for Design Options 2
Table No. 39
Initial Project Cost of Components for Design Option 2
Tag No.
Component
Model No.
Quantity
Price per
Total
Quantity
BOREA H 25-3 E
EVP-001
7,5 A AC 04S
1
$1246
$1246
97
BOREA H 35-3 D
EVP-002
6,5 A AC 04S
Evaporator
EVP-003
Compressor
CMP-023
CND-021
$1424
$1424
1
$2136
$2136
B35G5
1
$108.84
$108.84
GPY14RDa
1
$159.23
$159.23
2DB-50X
1
$993.15
$993.15
TCVM Type
1
$2372.5
$2372.5
2315D
1
$192
$192
2348B
1
$265
$265
2348B
1
$265
$265
-
1
BOREA H 50-4 G
4 A AC 04S
CMP-021
CMP-022
1
Condenser
050.1-13-E-N
EXP-021
EXP-022
Expansion
Valve
EXP-023
PSA-021
Piping System
$22061
Assembly
Total Cost
$31,222.72
98
F. Initial Project Cost of Components for Design Options 3
Table No. 40
Initial Project Cost of Components for Design Option 3
Tag No.
Component
Model No.
Quantity
Price per
Total
Quantity
EVP-001
BOREA H 25-3 E
7,5 A AC 04S
EVP-002
1
$1246
$1246
1
$1424
$1424
1
$2136
$2136
B35G5
1
$108.84
$108.84
GPY12RDa
1
$137.23
$137.23
2DB-50X
1
$993.15
$993.15
TCHM Type
1
$2304
$2304
2315D
1
$192
$192
2348B
1
$265
$265
$265
$265
BOREA H 35-3 D
Evaporator
EVP-003
6,5 A AC 04S
BOREA H 50-4 G
4 A AC 04S
CMP-031
CMP-032
Compressor
CMP-033
CND-031
Condenser
050.1-13-C-N
EXP-031
EXP-032
Expansion
EXP-033
Valve
2348B
1
PSA-031
Piping System
-
1
$21785
Assembly
Total Cost
$30,856.22
99
G. Project Management Chart for Cold Storage Facility Construction
Table No. 41
Project Management Chart
Engineering
PIC/
No. of
Activities
Equipment
PIC
Legal Advisor
Securing of
Permit
Site Survey
Preliminary
Design
Timeline
Rate/Day
Total
1
1 month
$300
$9,000
Surveyor
2
2 weeks
$250
$7,000
Architect
1
1 month
$400
$36,000
Civil Engineer
1
1 month
$400
$36,000
1
1 month
$400
$36,000
1
2 weeks
$500
$7,000
1
2 weeks
$500
$7,000
1
1 month
$300
$9,000
Foreman
3
6 months
$150
$27,000
Carpenter
6
6 months
$120
$21,600
Mechanical
Engineer
Design
Professional
Checking and
Mechanical
Consultation
Engineer
Design
Checking and
Consultation
Professional
Civil Engineer
Material
Procurement
Procurement
Officer
Project
Construction
100
Mason
12
6 months
$80
$14,400
15
6 months
$60
$10,800
1
6 months
$600
$108,000
Electrician
2
3 months
$250
$45,000
Plumber
2
2 months
$250
$30,000
2
3 months
$300
$54,000
1
2 months
$500
$30,000
1
6 months
$400
$72,000
1
12 months
$600
$216,000
General
Laborers
Construction
Project
Supervision
Manager
Electrical
Installation
Plumbing
Installation
HVAC
HVAC
Installation
Technician
Cold Storage
Equipment
Equipment
Supplier
Utilities
(Electricity,
Water)
Utility Service
Provider
Project
Project
Management
Manager
Total Cost
$775,800
101
H. Man Power, Materials and Equipment, Utilities Cost for Construction
Table No. 42
Man Power, Materials and Equipment, Utilities Cost for Construction
Timeline
1 month
1 month
2 months
1 month
3 months
3 months
2 months
3 months
2 weeks
Engineering
PIC/
No. of PIC
Rate/Day
Total Cost
Activities
Equipment
2
$200
$12,000
3
$150
$13,500
2
$180
$21,600
5
$180
$27,000
Carpenters
4
$200
$72,000
Painting
Painters
2
$150
$9,000
Flooring
Flooring
installation
crew
3
$180
$32,400
Equipment
Installation
installation
crew
4
$200
$72,000
2
$250
$7,000
Site
preparation
Excavator
Foundation
Concrete
work
mixer
Structural
Welding
framing
machine
Roofing
Interior
finishing
Final
inspections
Roofing
crew
Inspectors
Total
$266,500
102
I. Total Initial Project Cost for Each Design Tradeoff
Table No. 43
Total Initial Project Cost for Design Option 1
Expenditure
Cost
Initial Project Cost of Materials
$141,752.86
Initial Project Cost of Equipment
$62,495.98
Initial Project Cost of Components
$29,997.84
Initial Project Cost of Instrumentation and
$2,994
Control Equipment
Project Management Cost
$775,800
Cost of Facility Construction
$266,500
Total Cost
$1,279,540.68
J. Total Initial Project Cost for Each Design Tradeoff
Table No. 44
Total Initial Project Cost for Design Option 2
Expenditure
Cost
Initial Project Cost of Materials
$141,752.86
Initial Project Cost of Equipment
$62,495.98
103
Initial Project Cost of Components
$31,222.72
Initial Project Cost of Instrumentation and
$2,994
Control Equipment
Project Management Cost
$775,800
Cost of Facility Construction
$266,500
Total Cost
$1,280,765.56
K. Total Initial Project Cost for Each Design Tradeoff
Table No. 45
Total Initial Project Cost for Design Option 3
Expenditure
Cost
Initial Project Cost of Materials
$141,752.86
Initial Project Cost of Equipment
$62,495.98
Initial Project Cost of Components
$30,856.22
Initial Project Cost of Instrumentation and
$2,994
Control Equipment
Project Management Cost
$775,800
Cost of Facility Construction
$266,500
Total Cost
$1,280,399.06
104
Operation and Maintenance Cost
In this section, the operation and maintenance cost of the project is presented for each of
the design tradeoffs. These include the operation and maintenance cost of the components and
equipment of the 76 kW cold storage facility for beef in the municipality of Padre Garcia,
Batangas.
A. Operation and Maintenance Cost of Design Option 1
Table No. 46
Operational Cost of Design Option 1
Schedule
Tag No.
Component
Annual
Total Annual
Operation Cost
Cost
$252.71
$252.71
$571.20
$571.20
14 hrs/day
EVP-001
14 hrs/day
EVP-002
14 hrs/day
EVP-003
$3138.71
$3138.71
14 hrs/day
CMP-011
$430.42
$430.42
14 hrs/day
CMP-012
$7235.19
$7235.19
14 hrs/day
CND-011
$1042.57
1042.57
14 hrs/day
EXP-011
-
-
14 hrs/day
EXP-012
-
-
14 hrs/day
EXP-013
-
-
Evaporator
Compressor
Condenser
Expansion Valve
105
24 hrs/day
PSA-011
Piping System
-
-
Assembly
Total Cost
$12,670.80
Table No. 47
Maintenance Cost for Design Option 1
Maintenance
Tag No.
Component
PIC
Schedule
No. of
Rate/Da
Total
PIC
y
Annual
Cost
EVP-001
Semi-Annually
EVP-002
Evaporator
Evaporator
Technician
1
$90
$180
Compressor
Compressor
1
$90
$360
1
$90
$360
1
$90
$180
1
$90
$1080
EVP-003
CMP-011
Quarterly
Quarterly
CMP-012
CND-011
Technician
Condenser
Condenser
Technician
EXP-011
Semi-Annually
EXP-012
Expansion
Expansion
Valve
Valve
Technician
EXP-013
Monthly
PSA-011
Piping System
Plumbing
Assembly
Technician
106
Total Cost
$2,160
B. Operation and Maintenance Cost of Design Option 2
Table No. 48
Operational Cost of Design Option 2
Schedule
Tag No.
Component
12 hrs/day
Annual
Total Annual
Operation Cost
Cost
$216.61
$216.61
$489.60
$489.60
$2690.33
$2690.33
$73.79
$73.79
$368.93
$368.93
$6201.59
$6201.59
$1042.57
$1042.57
-
-
-
-
-
-
EVP-001
12 hrs/day
EVP-002
Evaporator
12 hrs/day
EVP-003
12 hrs/day
CMP-021
12 hrs/day
CMP-022
12 hrs/day
CMP-023
12 hrs/day
CND-021
12 hrs/day
EXP-021
12 hrs/day
EXP-022
12 hrs/day
EXP-023
Compressor
Condenser
Expansion Valve
107
24 hrs/day
PSA-021
Piping System
-
-
Assembly
Total Cost
$11,083.42
Table No. 49
Maintenance Cost for Design Option 2
Maintenance
Tag No.
Component
PIC
Schedule
No. of
Rate/Da
Total
PIC
y
Cost
1
$90
$180
EVP-001
Semi-Annually
EVP-002
Evaporator
Evaporator
Technician
EVP-003
CMP-021
Semi-Annually
CMP-022
Compressor
Compressor
Technician
1
$90
$180
Condenser
Condenser
1
$90
$360
1
$90
$180
CMP-023
Quarterly
CND-021
Technician
EXP-021
Semi-Annually
EXP-022
EXP-023
Expansion
Expansion
Valve
Valve
Technician
108
Monthly
PSA-021
Piping
Plumbing
System
Technician
1
$90
$1080
Assembly
Total Cost
$1,980
C. Operation and Maintenance Cost of Design Option 3
Table No. 50
Operational Cost of Design Option 3
Schedule
Tag No.
Component
Annual
Total Annual
Operation Cost
Cost
$216.61
$216.61
$489.60
$489.60
12 hrs/day
EVP-001
12 hrs/day
EVP-002
12 hrs/day
EVP-003
$2690.33
$2690.33
12 hrs/day
CMP-031
$73.79
$73.79
12 hrs/day
CMP-032
$276.70
$276.70
12 hrs/day
CMP-033
$6201.59
$6201.59
12 hrs/day
CND-031
$1057.05
$1057.05
12 hrs/day
EXP-031
-
-
12 hrs/day
EXP-032
-
-
12 hrs/day
EXP-033
-
-
Evaporator
Compressor
Condenser
Expansion Valve
109
24 hrs/day
PSA-031
Piping System
-
-
Assembly
Total Cost
$11,005.67
Table No. 51
Maintenance Cost for Design Option 3
Maintenance
Tag No.
Component
PIC
Schedule
No. of
Rate/Day
Total Cost
1
$90
$180
PIC
EVP-001
EVP-002
Evaporator
Evaporator
Technician
EVP-003
Semi-Annually
CMP-031
Semi-Annually
CMP-032
Compressor
Compressor
Technician
1
$90
$180
Condenser
Condenser
1
$90
$360
1
$90
$180
CMP-033
Quarterly
CND-031
Technician
EXP-031
Semi-Annually
EXP-032
EXP-033
Expansion
Expansion
Valve
Valve
Technician
110
Monthly
PSA-031
Piping
Plumbing
System
Technician
1
$90
$1080
Assembly
Total Cost
$1,980
Overall Project Cost
In this section, the overall project cost of the project is presented for each of the design
tradeoffs. These include the initial project cost and operation and maintenance cost of the
components and equipment of the 76 kW cold storage facility for beef in the municipality of
Padre Garcia, Batangas.
A. Overall Project Cost for Design Option 1
Table No. 52
Overall Project Cost for DO1
Expenditure
Cost
Initial Project Cost of Materials
$141,752.86
Initial Project Cost of Equipment
$62,495.98
Initial Project Cost of Components
$29,997.84
Initial Project Cost of Instrumentation and
$2,994
Control Equipment
111
Project Management Cost
$775,800
Cost of Facility Construction
$266,500
Operational Cost
$12,670.80
Maintenance Cost
$2,160
Total Cost
$1,294,371.48
B. Overall Project Cost for Design Option 2
Table No. 53
Overall Project Cost for DO2
Expenses
Cost
Initial Project Cost of Materials
$141,752.86
Initial Project Cost of Equipment
$62,495.98
Initial Project Cost of Components
$31,222.72
Initial Project Cost of Instrumentation and
$2,994
Control Equipment
Project Management Cost
$775,800
Cost of Facility Construction
$266,500
Operational Cost
$11,083.42
Maintenance Cost
$1,980
112
Total Cost
$1,293,828.98
C. Overall Project Cost for Design Option 3
Table No. 54
Overall Project Cost for DO3
Expenses
Cost
Initial Project Cost of Materials
$141,752.86
Initial Project Cost of Equipment
$62,495.98
Initial Project Cost of Components
$30,856.22
Initial Project Cost of Instrumentation and
$2,994
Control Equipment
Project Management Cost
$775,800
Cost of Facility Construction
$266,500
Operational Cost
$11,005.67
Maintenance Cost
$1,980
Total Cost
$1,293,384.73
113
Assessment of Techno-Economic Viability of the Project
Figure 10. Cash Flow Diagram of Design 1
Figure 11. Cash Flow Diagram of Design 2
114
Figure 12. Cash Flow Diagram of Design 2
Rate of Return (ROR)
𝑅𝑂𝑅 =
𝑁𝑒𝑡 𝐴𝑛𝑛𝑢𝑎𝑙 𝑃𝑟𝑜𝑓𝑖𝑡
𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐼𝑛𝑣𝑒𝑠𝑡𝑒𝑑
𝑊ℎ𝑒𝑟𝑒:
𝑁𝑒𝑡 𝐴𝑛𝑛𝑢𝑎𝑙 𝑃𝑟𝑜𝑓𝑖𝑡 = 𝑇𝑜𝑡𝑎𝑙 𝐴𝑛𝑛𝑢𝑎𝑙 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 − 𝑇𝑜𝑡𝑎𝑙 𝐴𝑛𝑛𝑢𝑎𝑙 𝐶𝑜𝑠𝑡
𝑇𝑜𝑡𝑎𝑙 𝐴𝑛𝑛𝑢𝑎𝑙 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 (𝑇𝐴𝑅) = 𝑅𝑒𝑛𝑡/𝐷𝑎𝑦 𝑥 365 𝑑𝑎𝑦𝑠
𝑇𝑜𝑡𝑎𝑙 𝐴𝑛𝑛𝑢𝑎𝑙 𝐶𝑜𝑠𝑡 (𝑇𝐴𝐶) = 𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 + 𝑂&𝑀
115
𝑊ℎ𝑒𝑟𝑒:
● 𝑂&𝑀 − 𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝑀𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒
𝑖
⎤
𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 = (𝐶𝐼 − 𝑆𝑉)⎡⎢
𝑛
⎥
(1+𝑖)
−1
⎣
⎦
𝑊ℎ𝑒𝑟𝑒:
● 𝐶𝐼 − 𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐼𝑛𝑣𝑒𝑠𝑡𝑒𝑑
● 𝑆𝑉 − 𝑆𝑎𝑙𝑣𝑎𝑔𝑒 𝑉𝑎𝑙𝑢𝑒
𝑆𝑎𝑙𝑣𝑎𝑔𝑒 𝑉𝑎𝑙𝑢𝑒 = 0. 10(𝐸𝑞𝑢𝑖𝑝𝑚𝑒𝑛𝑡 + 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 + 𝑂𝐸 + 𝐶𝐹 + 𝐼𝐶𝐸)
𝑊ℎ𝑒𝑟𝑒:
● 𝐸𝑞𝑢𝑖𝑝𝑚𝑒𝑛𝑡 = 𝐶𝑜𝑠𝑡 𝑜𝑓 𝐸𝑞𝑢𝑖𝑝𝑚𝑒𝑛𝑡
● 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 = 𝐶𝑜𝑠𝑡 𝑜𝑓 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠
● 𝐶𝐹 − 𝐶𝑜𝑠𝑡 𝑜𝑓 𝐹𝑎𝑐𝑖𝑙𝑖𝑡𝑦 𝐶𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛
● 𝐼𝐶𝐸 − 𝐼𝑛𝑠𝑡𝑟𝑢𝑚𝑒𝑛𝑡𝑎𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝐶𝑜𝑛𝑡𝑟𝑜𝑙 𝐸𝑞𝑢𝑖𝑝𝑚𝑒𝑛𝑡
A. Rate of Return for Design Option 1
For the Depreciation of DO1:
0.17
⎤
𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 = ($1, 294, 371. 48 − $36, 198. 78)⎡⎢
20
⎥
⎣ (1+0.17) −1 ⎦
𝑖 = 17%
116
𝑛 = 20 𝑦𝑒𝑎𝑟𝑠
𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐼𝑛𝑣𝑒𝑠𝑡𝑒𝑑 (𝐶𝐼) = $1,279,540.68
𝑆𝑎𝑙𝑣𝑎𝑔𝑒 𝑉𝑎𝑙𝑢𝑒 = 0. 10($62, 495. 98 + $29, 997. 84 + $266, 500 + $2, 994)
𝑆𝑎𝑙𝑣𝑎𝑔𝑒 𝑉𝑎𝑙𝑢𝑒 (𝑆𝑉) = $36,198.78
𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 = $9,675.80
For the Total Annual Cost (TAC) of DO1:
𝑂&𝑀 = $12,670.80 + $2,160
𝑂&𝑀 = $14,830.80
𝑇𝑜𝑡𝑎𝑙 𝐴𝑛𝑛𝑢𝑎𝑙 𝐶𝑜𝑠𝑡 (𝑇𝐴𝐶) = $9, 675. 80 + $14, 830. 80
𝑇𝑜𝑡𝑎𝑙 𝐴𝑛𝑛𝑢𝑎𝑙 𝐶𝑜𝑠𝑡 (𝑇𝐴𝐶) = $24,506.60
For the Total Annual Revenue (TAR) of DO1:
𝑇𝑜𝑡𝑎𝑙 𝐴𝑛𝑛𝑢𝑎𝑙 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 (𝑇𝐴𝑅) = $800/𝑑𝑎𝑦 𝑥 365 𝑑𝑎𝑦𝑠
𝑇𝑜𝑡𝑎𝑙 𝐴𝑛𝑛𝑢𝑎𝑙 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 (𝑇𝐴𝑅) = $292,000
For the Net Annual Profit (NAP) of DO1:
𝑁𝑒𝑡 𝐴𝑛𝑛𝑢𝑎𝑙 𝑃𝑟𝑜𝑓𝑖𝑡 = $292, 000 − $24, 506. 60
𝑁𝑒𝑡 𝐴𝑛𝑛𝑢𝑎𝑙 𝑃𝑟𝑜𝑓𝑖𝑡 (𝑁𝐴𝑃) = $267,493.40
117
For the Rate of Return (ROR) of DO1:
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑅𝑒𝑡𝑢𝑟𝑛 (𝑅𝑂𝑅) =
$267,493.4
$1,294,371.48
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑅𝑒𝑡𝑢𝑟𝑛 (𝑅𝑂𝑅) = 20.66589106%
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑅𝑒𝑡𝑢𝑟𝑛 (𝑅𝑂𝑅) = 20.67%
B. Rate of Return for Design Option 2
For the Depreciation of DO2:
0.17
⎤
𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 = ($1, 293, 828. 98 − $36, 321. 27)⎡⎢
20
⎥
⎣ (1+0.17) −1 ⎦
𝑖 = 17%
𝑛 = 20 𝑦𝑒𝑎𝑟𝑠
𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐼𝑛𝑣𝑒𝑠𝑡𝑒𝑑 (𝐶𝐼) = $1,293,828.98
𝑆𝑎𝑙𝑣𝑎𝑔𝑒 𝑉𝑎𝑙𝑢𝑒 = 0. 10($62, 495. 98 + $31, 222. 72 + $266, 500 + $2, 994)
𝑆𝑎𝑙𝑣𝑎𝑔𝑒 𝑉𝑎𝑙𝑢𝑒 (𝑆𝑉) = $36,321.27
𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 = $9,670.69
For the Total Annual Cost (TAC) of DO2:
𝑂&𝑀 = $11,083.42 + $1,980
118
𝑂&𝑀 = $13,063.42
𝑇𝑜𝑡𝑎𝑙 𝐴𝑛𝑛𝑢𝑎𝑙 𝐶𝑜𝑠𝑡 (𝑇𝐴𝐶) = $9, 670. 69 + $13, 063. 42
𝑇𝑜𝑡𝑎𝑙 𝐴𝑛𝑛𝑢𝑎𝑙 𝐶𝑜𝑠𝑡 (𝑇𝐴𝐶) = $22,734.11
For the Total Annual Revenue (TAR) of DO2:
𝑇𝑜𝑡𝑎𝑙 𝐴𝑛𝑛𝑢𝑎𝑙 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 (𝑇𝐴𝑅) = $800/𝑑𝑎𝑦 𝑥 365 𝑑𝑎𝑦𝑠
𝑇𝑜𝑡𝑎𝑙 𝐴𝑛𝑛𝑢𝑎𝑙 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 (𝑇𝐴𝑅) = $292,000
For the Net Annual Profit (NAP) of DO2:
𝑁𝑒𝑡 𝐴𝑛𝑛𝑢𝑎𝑙 𝑃𝑟𝑜𝑓𝑖𝑡 = $292, 000 − $22, 734. 11
𝑁𝑒𝑡 𝐴𝑛𝑛𝑢𝑎𝑙 𝑃𝑟𝑜𝑓𝑖𝑡 (𝑁𝐴𝑃) = $269,265.89
For the Rate of Return (ROR) of DO2:
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑅𝑒𝑡𝑢𝑟𝑛 (𝑅𝑂𝑅) =
$269,265.89
$1,293,828.98
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑅𝑒𝑡𝑢𝑟𝑛 (𝑅𝑂𝑅) = 20.81155193 %
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑅𝑒𝑡𝑢𝑟𝑛 (𝑅𝑂𝑅) = 20.81 %
119
C. Rate of Return for Design Option 3
For the Depreciation of DO3:
0.17
⎤
𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 = ($1, 293, 384. 73 − $36284. 62)⎡⎢
20
⎥
⎣ (1+0.17) −1 ⎦
𝑖 = 17%
𝑛 = 20 𝑦𝑒𝑎𝑟𝑠
𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐼𝑛𝑣𝑒𝑠𝑡𝑒𝑑 (𝐶𝐼) = $1,293,384.73
𝑆𝑎𝑙𝑣𝑎𝑔𝑒 𝑉𝑎𝑙𝑢𝑒 = 0. 10($62, 495. 98 + $30, 856. 22 + $266, 500 + $2, 994)
𝑆𝑎𝑙𝑣𝑎𝑔𝑒 𝑉𝑎𝑙𝑢𝑒 (𝑆𝑉) = $36,284.62
𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑡𝑖𝑜𝑛 = $9,667.55
For the Total Annual Cost (TAC) of DO3:
𝑂&𝑀 = $11,005.67 + $1,980
𝑂&𝑀 = $12,985.67
𝑇𝑜𝑡𝑎𝑙 𝐴𝑛𝑛𝑢𝑎𝑙 𝐶𝑜𝑠𝑡 (𝑇𝐴𝐶) = $9, 667. 55 + $12, 985. 67
𝑇𝑜𝑡𝑎𝑙 𝐴𝑛𝑛𝑢𝑎𝑙 𝐶𝑜𝑠𝑡 (𝑇𝐴𝐶) = $22,653.22
For the Total Annual Revenue (TAR) of DO3:
𝑇𝑜𝑡𝑎𝑙 𝐴𝑛𝑛𝑢𝑎𝑙 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 (𝑇𝐴𝑅) = $800/𝑑𝑎𝑦 𝑥 365 𝑑𝑎𝑦𝑠
120
𝑇𝑜𝑡𝑎𝑙 𝐴𝑛𝑛𝑢𝑎𝑙 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 (𝑇𝐴𝑅) = $292,000
For the Net Annual Profit (NAP) of DO3:
𝑁𝑒𝑡 𝐴𝑛𝑛𝑢𝑎𝑙 𝑃𝑟𝑜𝑓𝑖𝑡 = $292, 000 − $22, 653. 22
𝑁𝑒𝑡 𝐴𝑛𝑛𝑢𝑎𝑙 𝑃𝑟𝑜𝑓𝑖𝑡 (𝑁𝐴𝑃) = $269,346.78
For the Rate of Return (ROR) of DO3:
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑅𝑒𝑡𝑢𝑟𝑛 (𝑅𝑂𝑅) =
$269,346.78
$1,293,384.73
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑅𝑒𝑡𝑢𝑟𝑛 (𝑅𝑂𝑅) = 20.82495438 %
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑅𝑒𝑡𝑢𝑟𝑛 (𝑅𝑂𝑅) = 20.82 %
Payback Period (PBP)
𝑃𝐵𝑃 =
𝐼𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡 − 𝑆𝑎𝑙𝑣𝑎𝑔𝑒 𝑉𝑎𝑙𝑢𝑒
𝑁𝑒𝑡 𝐴𝑛𝑛𝑢𝑎𝑙 𝐶𝑎𝑠ℎ 𝐹𝑙𝑜𝑤
𝑊ℎ𝑒𝑟𝑒:
𝑁𝑒𝑡 𝐴𝑛𝑛𝑢𝑎𝑙 𝐶𝑎𝑠ℎ 𝐹𝑙𝑜𝑤 = 𝑇𝑜𝑡𝑎𝑙 𝐴𝑛𝑛𝑢𝑎𝑙 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 − 𝑇𝑜𝑡𝑎𝑙 𝐴𝑛𝑛𝑢𝑎𝑙 𝐶𝑜𝑠𝑡
𝑆𝑎𝑙𝑣𝑎𝑔𝑒 𝑉𝑎𝑙𝑢𝑒 = 0. 10(𝐸𝑞𝑢𝑖𝑝𝑚𝑒𝑛𝑡 + 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 + 𝑂𝐸 + 𝐶𝐹 + 𝐼𝐶𝐸)
𝑊ℎ𝑒𝑟𝑒:
● 𝐸𝑞𝑢𝑖𝑝𝑚𝑒𝑛𝑡 = 𝐶𝑜𝑠𝑡 𝑜𝑓 𝐸𝑞𝑢𝑖𝑝𝑚𝑒𝑛𝑡
121
● 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 = 𝐶𝑜𝑠𝑡 𝑜𝑓 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠
● 𝐶𝐹 − 𝐶𝑜𝑠𝑡 𝑜𝑓 𝐹𝑎𝑐𝑖𝑙𝑖𝑡𝑦 𝐶𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛
● 𝐼𝐶𝐸 − 𝐼𝑛𝑠𝑡𝑟𝑢𝑚𝑒𝑛𝑡𝑎𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝐶𝑜𝑛𝑡𝑟𝑜𝑙 𝐸𝑞𝑢𝑖𝑝𝑚𝑒𝑛𝑡
A. Payback Period (PBP) of Design Option 1
For the Net Annual Cash Flow of DO1:
𝑁𝑒𝑡 𝐴𝑛𝑛𝑢𝑎𝑙 𝐶𝑎𝑠ℎ 𝐹𝑙𝑜𝑤 = $292, 000 − $24, 506. 60
𝑁𝑒𝑡 𝐴𝑛𝑛𝑢𝑎𝑙 𝐶𝑎𝑠ℎ 𝐹𝑙𝑜𝑤 = $267,493.40
For the Salvage Value (SV) of DO1:
𝑆𝑎𝑙𝑣𝑎𝑔𝑒 𝑉𝑎𝑙𝑢𝑒 (𝑆𝑉) = $36,198.78
For the Payback Period (PBP) of D01:
𝑃𝐵𝑃 =
$1,294,371.48 − $36,198.78
$267,493.40
𝑃𝐵𝑃 = 4.70 years
𝑃𝐵𝑃 = 4 years and 9 months
B. Payback Period (PBP) of Design Option 2
For the Net Annual Cash Flow of DO2:
𝑁𝑒𝑡 𝐴𝑛𝑛𝑢𝑎𝑙 𝐶𝑎𝑠ℎ 𝐹𝑙𝑜𝑤 = $292, 000 − $22, 734. 11
122
𝑁𝑒𝑡 𝐴𝑛𝑛𝑢𝑎𝑙 𝐶𝑎𝑠ℎ 𝐹𝑙𝑜𝑤 = $269,265.89
For the Salvage Value (SV) of DO2:
𝑆𝑎𝑙𝑣𝑎𝑔𝑒 𝑉𝑎𝑙𝑢𝑒 (𝑆𝑉) = $36,321.27
For the Payback Period (PBP) of D02:
𝑃𝐵𝑃 =
$1,293,828.98 − $36,321.27
$269,265.89
𝑃𝐵𝑃 = 4.67 years
𝑃𝐵𝑃 = 4 years and 8 months
C. Payback Period (PBP) of Design Option 3
For the Net Annual Cash Flow of DO3:
𝑁𝑒𝑡 𝐴𝑛𝑛𝑢𝑎𝑙 𝐶𝑎𝑠ℎ 𝐹𝑙𝑜𝑤 = $292, 000 − $22, 653. 22
𝑁𝑒𝑡 𝐴𝑛𝑛𝑢𝑎𝑙 𝐶𝑎𝑠ℎ 𝐹𝑙𝑜𝑤 = $269,346.78
For the Salvage Value (SV) of DO3:
𝑆𝑎𝑙𝑣𝑎𝑔𝑒 𝑉𝑎𝑙𝑢𝑒 (𝑆𝑉) = $36,284.62
For the Payback Period (PBP) of D03:
𝑃𝐵𝑃 =
$1,293,384.73 − $36,284.62
$269,346.78
123
𝑃𝐵𝑃 = 4.67 years
𝑃𝐵𝑃 = 4 years and 8 months
Design Selection Based on Pareto Analysis
Table No. 55
Pareto Analysis for Design Tradeoff Selection
Criteria
DO1
DO2
DO3
Preferred
Design
Option
Coefficient of
583.6%
582.6%
592%
DO3
Rate of Return
20.67%
20.81 %
20.82 %
DO3
Operational
$12,670.80
$11,083.42
$11,005.67
DO3
$2,160
$1,980
$1,980
DO2 or
Performance
Cost
Maintenance
Cost
DO3
Overall Project $1,294,371.48 $1,293,828.98
$1,293,384.73
DO3
Cost
Payback
4 years and 9
4 years and 8
4 years and 8
DO2 or
Period
months
months
months
DO3
Best Design Option:
DO3
124
The table presents a Pareto analysis for design tradeoff selection, evaluating three
different design options (DO1, DO2, and DO3) based on several criteria. The preferred design
option for each criterion and the corresponding values are also provided.
The first criterion is the coefficient of performance, which measures the efficiency of the
design options. DO1 has a coefficient of performance of 583.6%, DO2 has 582.6%, and DO3 has
592%. Among the three options, DO3 has the highest coefficient of performance, making it the
preferred design option for this criterion.
The second criterion is the rate of return, which indicates the profitability of each design
option. DO1 has a rate of return of 20.67%, DO2 has 20.81%, and DO3 has 20.82%. Once again,
DO3 has the highest rate of return, making it the preferred design option for this criterion.
The third criterion is the operational cost, representing the expenses associated with
running the design options. DO1 has an operational cost of $12,670.80, DO2 has $11,083.42, and
DO3 has $11,005.67. In this case, DO3 has the lowest operational cost, making it the preferred
design option.
The fourth criterion is the maintenance cost, which refers to the expenses required for
maintaining the design options. DO1 has a maintenance cost of $2,160, DO2 has $1,980, and
DO3 also has $1,980. Here, both DO2 and DO3 have the same maintenance cost, making either
of them a valid choice.
The fifth criterion is the overall project cost, which encompasses all expenses associated
with the design options. DO1 has an overall project cost of $1,294,371.48, DO2 has
125
$1,293,828.98, and DO3 has $1,293,384.73. DO3 has the lowest overall project cost, making it
the preferred design option for this criterion.
The last criterion is the payback period, indicating the time it takes for the project to
recoup its initial investment. DO1 has a payback period of 4 years and 9 months, DO2 has 4
years and 8 months, and DO3 also has 4 years and 8 months. Both DO2 and DO3 have the same
payback period, making either of them suitable options.
Thus, considering all the criteria and their corresponding values, the best design option
overall is DO3 which is the Triple Compressor, Multiple Expansion Valve, Multiple Evaporator
Cycle. It achieves the highest coefficient of performance, rate of return, and the lowest
operational cost and overall project cost. Additionally, it shares the lowest maintenance cost with
DO2 and has a competitive payback period. Therefore, DO3 is the preferred design option based
on the analysis of these criteria.
126
Chapter V
Engineering Design Collection and Cost Analysis
Conclusion
Refrigeration technology plays a vital role in the preservation and storage of perishable
goods, such as beef meat. The efficient design of cold storage facilities is essential to maintain
the quality and safety of the stored products. In this project, the design options were evaluated,
and after careful consideration, Design Option 3, which incorporates the Triple Compressor,
Multiple Expansion Valve, Multiple Evaporator Cycle, has been selected as the most suitable
choice for the 76 kW cold storage facility for beef meat located at Brgy. Payapa, Padre Garcia,
Batangas. This design offers several advantages, including enhanced temperature control,
improved energy efficiency, and increased operational flexibility, making it an optimal solution
for the specific requirements of the beef meat cold storage facility.
The Triple Compressor, Multiple Expansion Valve, Multiple Evaporator Cycle design
option provides significant benefits in terms of temperature control. The utilization of multiple
evaporators allows for independent cooling zones within the cold storage facility, ensuring
precise temperature management for different beef meat products. This feature is particularly
advantageous as it enables the facility to accommodate varying temperature requirements, such
as chilling and freezing, for different cuts of beef. By maintaining optimal temperatures
throughout the storage area, this design option minimizes the risk of spoilage, bacterial growth,
and product degradation, thereby ensuring the quality and safety of the beef meat.
127
Moreover, the incorporation of three compressors in the system offers improved energy
efficiency. With this setup, the refrigeration load can be distributed among the compressors based
on the cooling requirements of each evaporator. By optimizing the workload, the system can
operate at higher efficiency levels, resulting in reduced energy consumption and operating costs.
The ability to modulate the compressors based on the actual cooling demand also enhances the
system's responsiveness and adaptability, making it well-suited for the dynamic nature of a cold
storage facility that experiences varying loads over time.
Additionally, the integration of multiple expansion valves allows for better control over
the refrigerant flow and pressure levels within the system. This feature enables precise
adjustment of the cooling capacity for each evaporator, ensuring that the required cooling is
delivered accurately to different zones of the facility. By efficiently distributing the refrigerant
flow, this design option optimizes the cooling process, prevents imbalances, and reduces the risk
of excessive frost formation or inadequate cooling. As a result, the Triple Compressor, Multiple
Expansion Valve, Multiple Evaporator Cycle design enhances the overall performance and
reliability of the cold storage facility.
Furthermore, to conclude as well, the chosen best design option which is the has an
evident result that it’s the best among the two other design options which are the Dual
Compressor, Multi Expansion Valve, Multi Evaporator Cycle and Triple Compressor, Individual
and Multiple Expansion Valve, Multiple Evaporator Cycle based on the Pareto analysis. Pareto
Analysis for the design tradeoff selection, it is evident that DO3 emerges as the preferred design
option for the DO3 criterion. In terms of the coefficient of performance, DO3 exhibits the highest
value among the options, boasting a remarkable 592%. Moving on to the rate of return, DO3
once again demonstrates its superiority by achieving a rate of return of 20.82%. On the other
128
hand, when considering operational costs, DO3 stands out as the most economical choice, with
costs amounting to $11,005.67. In terms of maintenance costs, both DO2 and DO3 share the
same value, with maintenance costs of $1,980. DO1 incurs slightly higher maintenance costs at
$2,160. Nevertheless, the cost differential between DO2 and DO3 is not significant in this
criterion. Considering the overall project cost, DO3 once again proves to be the more favorable
option, with a cost of $1,293,384.73. Lastly, the payback period analysis reveals that both DO2
and DO3 share the same payback period of 4 years and 8 months, outperforming DO1's payback
period of 4 years and 9 months.
In conclusion, the selection of Design Option 3, which incorporates the Triple
Compressor, Multiple Expansion Valve, Multiple Evaporator Cycle, for the beef meat cold
storage facility at Brgy. Payapa, Padre Garcia, Batangas, offers numerous advantages. It provides
enhanced temperature control by utilizing multiple evaporators to accommodate varying
temperature requirements for different cuts of beef, minimizing spoilage and ensuring product
quality and safety. The inclusion of three compressors improves energy efficiency by optimizing
the workload and modulating the compressors based on cooling demand. The integration of
multiple expansion valves enhances refrigerant flow control, preventing imbalances and ensuring
accurate cooling distribution. Overall, this design option delivers a reliable, efficient, and flexible
solution for the specific needs of the beef meat cold storage facility, maximizing the preservation
and storage of the perishable goods.
129
Recommendation
1. Evaluate all the design options by thoroughly analyzing each design option, considering
factors such as efficiency, cost-effectiveness, operational requirements, and feasibility.
Assess the advantages and disadvantages of each option to make an informed decision.
2. Prioritize performance and cost-efficiency wherein focus on design options that offer the
best balance between performance and cost. Consider criteria such as energy efficiency,
temperature control capabilities, maintenance costs, and overall project costs to identify
the design option that provides optimal results.
3. Consult and seek for the advice of experts who are experienced engineers, refrigeration
specialists, and industry professionals who can provide valuable insights and
recommendations. Their expertise can help optimize the design and ensure that it aligns
with industry standards and best practices.
4. Consider future scalability and flexibility, anticipate potential changes in the facility's
requirements or expansion plans. Opt for a design option that offers flexibility and
adaptability to accommodate future growth or modifications, minimizing the need for
costly retrofitting or redesign in the future.
5. Review local regulations and standards that ensure the chosen design option complies
with local regulations, safety standards, and environmental guidelines. Compliance is
crucial to avoid penalties and ensure the long-term sustainability of the facility.
6. Regardless of the chosen design option, proper maintenance and monitoring are essential
to ensure optimal performance and longevity. Implement a proactive maintenance plan
and monitor the facility regularly to address any issues promptly and prevent costly
breakdowns or inefficiencies.
130
Appendices
131
Appendix A
(References)
132
Camp, K. (2021, November 29). Major Applications of Industrial Refrigeration Systems and Its
Importance to Different Industries. Hillphoenix. Retrieved March 19, 2023, from
https://www.hillphoenix.com/major-applications-of-industrial-refrigeration-systems-and-i
ts-importance-to-different-industries/
Evans, P. (2017, December 26). Cooling Load Calculation - Cold Room. The Engineering
Mindset. Retrieved March 19, 2023, from
https://theengineeringmindset.com/cooling-load-calculation-cold-room/
French, M. (2022, August 30). Wastewater Controls for Cold Storage Facility. AMPS Industrial
Controls. Retrieved March 14, 2023, from
https://ampsic.com/wastewater-controls-cold-storage/
Geñosa, M. (2018, July 26). Batangas town sustains country's 'cattle trading capital' tag.
Philippine News Agency. Retrieved March 10, 2023, from
https://www.pna.gov.ph/articles/1042712
Guide: Material Handling Equipment for Food Distribution and Cold Storage – LiftOne. (2022,
March 11). LiftOne. Retrieved March 19, 2023, from
https://www.liftone.net/blog/food-cold-storage-warehouse-guide/
Home. (n.d.). YouTube. Retrieved March 19, 2023, from
https://www.google.com/search?q=inurl%3Ahttps%3A%2F%2Fwww.techtarget.com%2
Fsearchdatacenter%2Fdefinition%2FMechanical-refrigeration&rlz=1C1ONGR_enPH101
1PH1011&oq=inurl%3Ahttps%3A%2F%2Fwww.techtarget.com%2Fsearchdatacenter%2
Fdefinition%2FMechanical-refriger
Industrial Refrigeration System: Applications in Diverse Industrial Spheres. (2021, August 6).
Research Dive. Retrieved March 19, 2023, from
133
https://www.researchdive.com/blog/different-applications-of-industrial-refrigeration-syst
ems-in-varied-industries
Lingayat, A. (2015, December 22). Material Handling Operations in a Cold Store.
ColdChainManagement.org. Retrieved March 19, 2023, from
https://coldchainmanagement.org/2015/12/22/material-handling-operations-in-a-cold-stor
e/
Manual on meat cold store operation and management. (n.d.). Manual on meat cold store
operation and management. Retrieved March 19, 2023, from
https://www.fao.org/3/T0098E/T0098E02.htm
Manual on simple methods of meat preservation. (n.d.). Manual on simple methods of meat
preservation. Retrieved March 19, 2023, from
https://www.fao.org/3/x6932e/X6932E05.htm
Meat Rail, Hanging & Storage Systems |. (n.d.). Angel Refrigeration. Retrieved March 19, 2023,
from https://www.angelrefrigeration.co.uk/category/meat-rail-hanging-storage-systems/
National Meat Inspection. (2022, May 30). List of Licensed Cold Storage Warehouse.
NATIONAL MEAT INSPECTION SERVICE. Retrieved March 10, 2023, from
https://www.nmis.gov.ph/images/pdf/accredited_list/2022/may/csw.pdf
Official Site of Padre Garcia. (2022, June 13). History of Padre Garcia. Padre Garcia, Batangas.
Retrieved March 10, 2023, from https://www.padregarcia.gov.ph/about/history
Pressure Relief Vents in Cool Rooms & Freezers. (n.d.). Cold-Rite. Retrieved March 19, 2023,
from https://www.cold-rite.com.au/post/pressure-relief-vents-in-cool-rooms-freezers
Two-Stage Cooling Process Poster. (n.d.). State Food Safety. Retrieved March 19, 2023, from
https://www.statefoodsafety.com/Resources/Resources/two-stage-cooling-process
134
What is Digital Electronic Weighing Scale and How it works? (2019, May 14). Crown Scales.
Retrieved March 19, 2023, from
https://www.crownscales.co.in/blog/what-is-digital-electronic-weighing-scale-and-how-it
-works
What is Forklift? Working Mechanism & Where it is used? (n.d.). Torcan Lift Equipment.
Retrieved March 19, 2023, from
https://torcanlift.com/what-is-forklift-working-mechanism-where-it-is-used/
135
Catalogs
For Compressor:
● GPT18RG
Source:https://clivere.com/images/documents/pdf/cubigel-katalog.pdf
136
● 2DB-50X
Source:https://www.elektronika-sa.com.pl/en/products/refrigeration/compressors/reciprocating/di
scus/2DB-50X
137
● B35GB
● GPY14RDa
Source:https://clivere.com/images/documents/pdf/cubigel-katalog.pdf
138
● GPY12RDa
Source:https://clivere.com/images/documents/pdf/cubigel-katalog.pdf
139
For Evaporator:
● BOREA H 25-3 E 7,5 A AC 04S
Source: https://www.stefaniexchangers.com/download/stefani-spa-catalogue-borea-EN.pdf
140
● BOREA H 35-3 D 6,5 A AC 04S
Source: https://www.stefaniexchangers.com/download/stefani-spa-catalogue-borea-EN.pdf
141
● BOREA H 50-4 G 4 A AC 04S
Source: https://www.stefaniexchangers.com/download/stefani-spa-catalogue-borea-EN.pdf
142
For Expansion Valves:
● 2348B
● 2315D
Source:https://www.saginomiya.co.jp/en/auto/dl/expv_en.pdf
143
For Condenser
Source: https://thermocoil.co.za/Thermocoil%20Catalogue%202015%20-Condensers.pdf
144
Appendix B
(Design Library)
145
Final Plant Layout (Design Option 3)
146
Piping Arrangement and Connection Assembly
147
Plant Layout Options (Design Option 1)
148
Plant Layout Options (Design Option 2)
)
149
Plant Layout Options (Design Option 3)
150
Appendix C
(Related Computation)
151
Related Computation
Determining Design Trade Offs:
Determining design trade offs with EES Software to determine the cycle with the highest
coefficient of performance:
1. Solving Two-Stage Evaporator Cycle
Where: Multiple Compressor and Multiple Expansion Valve Cycle yields the highest
COP. This cycle was then used as the basis for determining design trade offs at Three-Stage
Evaporator Cycle.
152
● Single Compressor and Individual Expansion Valve
f$='R134a'
RE_1= 25[kJ/s]
RE_2= 40[kJ/s]
T_evap1= 0 [C]
T_evap2= -18[C]
T_cond= 10[C]
x_3=0
x_6=1
x_7=1
"State Point 3"
h_3=Enthalpy(f$,T=T_cond,x=x_3)
"State Point 4"
h_4=h_3
"State Point 5"
h_5=h_4
"State Point 6"
h_6=Enthalpy(f$,T=T_evap1,x=x_6)
"State Point 7"
153
h_7=Enthalpy(f$,T=T_evap2,x=x_7)
"State Point 8"
h_8=h_6
"For mass flowrate"
m_dot_r=m_dot_7+m_dot_8
RE_2=m_dot_7*(h_7-h_5)
RE_1=m_dot_8*(h_6-h_4)
"State Point 1"
(m_dot_7*h_7)+(m_dot_8*h_8)=m_dot_r*h_1
P_1=P_sat(f$,T=T_evap2)
s_1=Entropy(f$,P=P_1,h=h_1)
"State Point 2"
P_2=P_sat(f$,T=T_cond)
s_2=s_1
h_2=Enthalpy(f$,P=P_2,s=s_2)
"For Work of Compression"
W_dot_c=m_dot_r*(h_2-h_1)
154
"For COP"
COP=((RE_1+RE_2)/W_dot_c)*100[%]
● Multiple Compressor and Individual Expansion Valve
"Given"
f$ = 'R134a'
Tevap_1 = 0 [C]
x_1=1 []
T_6 = 10 [C]
P_6=P_sat(f$,T=T_6)
RE_1 = 25 [kW]
RE_1 = m_dot_1*(h_1-h_7)
Tevap_2 = -18 [C]
Tevap_2 = T_3
RE_2 = 40 [kW]
RE_2 = m_dot_3*(h_3-h_8)
x_3 = 1 []
x_6 = 0 []
"State Point 1"
h_1 = Enthalpy(f$,T=Tevap_1,x=x_1)
s_1 = Entropy(f$,T=Tevap_1,x=x_1)
155
"State Point 2"
h_2 = Enthalpy(f$,P=P_6,s=s_2)
s_2 = s_1
"State Point 3"
h_3 = Enthalpy(f$,T=T_3,x=x_3)
s_3 = Entropy(f$,T=T_3,x=x_3)
"State Point 4"
h_4 = Enthalpy(f$,P=P_6,s=s_4)
s_4 = s_3
"State Point 5"
m_dot_2*h_2 + m_dot_4*h_4 = (m_dot_2 + m_dot_4)*h_5
m_dot_2 = m_dot_1
m_dot_4 = m_dot_3
"State Point 6"
h_6 = Enthalpy(f$,P=P_6,x=x_6)
"State Point 7"
h_7 = h_6
"State Point 8"
h_8 = h_6
156
"Solving for the Power Required of the Compressor 1"
W_dot_c1 = m_dot_1*(h_2-h_1)
"Solving for the Power Required of the Compressor 2"
W_dot_c2= m_dot_3*(h_4-h_3)
"Solving for the Total Power Required of the Compressor "
W_cTotal = W_dot_c1 + W_dot_c2
"Solving for the COP of the System"
COP = ((RE_1+RE_2)/(W_dot_c1 + W_dot_c2))*100[%]
● Single Compressor and Multiple Expansion Valve
f$='R134a'
RE_1= 25[kJ/s]
RE_2= 40[kJ/s]
T_evap1= 0 [C]
T_evap2= -18[C]
T_cond= 10[C]
x_3=0
x_5=0
x_8=1
x_7=1
"State Point 3"
h_3=Enthalpy(f$,T=T_cond,x=x_3)
157
"State Point 4"
h_4=h_3
"State Point 5"
h_5=Enthalpy(f$,T=T_evap1,x=x_5)
"State Point 6"
h_6=h_5
"State Point 7"
h_7=Enthalpy(f$,T=T_evap1,x=x_7)
"State Point 8"
h_8=Enthalpy(f$,T=T_evap2,x=x_8)
"State Point 9"
h_9=h_7
"For mass flowrate"
m_dot_r=m_dot_7+m_dot_8
RE_2=m_dot_7*(h_7-h_4)
RE_1=m_dot_8*(h_8-h_6)
158
"State Point 1"
(m_dot_7*h_7)+(m_dot_8*h_8)=m_dot_r*h_1
P_1=P_sat(f$,T=T_evap2)
s_1=Entropy(f$,P=P_1,h=h_1)
"State Point 2"
P_2=P_sat(f$,T=T_cond)
s_2=s_1
h_2=Enthalpy(f$,P=P_2,s=s_2)
"For Work of Compression"
W_dot_c=m_dot_r*(h_2-h_1)
"For COP"
COP=((RE_1+RE_2)/W_dot_c)*100[%]
● Multiple Compressor and Multiple Expansion Valve
"Multi E.V and Multiple Compressor"
"Known"
f$='R134a'
RE_1=25[kJ/s]
RE_2=40[kJ/s]
T_evap1=0[C]
T_evap2=-18[C]
T_cond=10[C]
P_1=P_sat(f$,T=T_evap1)
P_7=P_1
P_8=P_1
159
P_2=P_sat(f$,T=T_cond)
P_4=P_2
P_5=P_2
P_6=P_2
P_3=P_sat(f$,T=T_evap2)
P_9=P_3
m_dot_r=m_dot_1+m_dot_2
x_1=1[]
x_3=1[]
x_6=0[]
x_8=0[]
"State Point 1"
h_1=Enthalpy(f$,P=P_1,x=x_1)
s_1=Entropy(f$,P=P_1,x=x_1)
"State Point 2"
s_2=s_1
h_2=Enthalpy(f$,P=P_2,s=s_2)
"State Point 3"
h_3=Enthalpy(f$,P=P_3,x=x_3)
s_3=Entropy(f$,P=P_3,x=x_3)
"State Point 4"
s_4=s_3
h_4=Enthalpy(f$,P=P_4,s=s_4)
"State Point 5"
h_5*m_dot_r=h_2*m_dot_1+h_4*m_dot_2
"State Point 6"
h_6=Enthalpy(f$,P=P_6,x=x_6)
"State Point 7"
h_7=h_6
160
"State Point 8"
h_8=Enthalpy(f$,P=P_8,x=x_8)
"State Point 9"
h_9=h_8
"Solution"
RE_2=m_dot_2*(h_3-h_9)
RE_1=m_dot_1*(h_1-h_7)
W_dot_c1=m_dot_1*(h_2-h_1)
W_dot_c2=m_dot_2*(h_4-h_3)
COP=((RE_1+RE_2)/(W_dot_c1+W_dot_c2))*100[%]
2. Solving Three-Stage Evaporator Cycle
Figure above is the different configuration based on the previously determined optimum
two-stage cycle. The top three configurations with the highest yield of COP are the following:
161
A. Dual Compressor, Multi Expansion Valve, Multi Evaporator Cycle
B. Triple Compressor, Individual and Multiple Expansion Valve, Multiple
Evaporator Cycle
C. Triple Compressor, Multiple Expansion Valve, Multiple Evaporator Cycle
EES CODES:
"Multi E.V and Multiple Compressor 3 Stage"
"Known"
f$='R134a'
RE_1=383[kJ/s]
RE_2=232[kJ/s]
RE_3 = 86[kJ/s]
T_evap1=-18[C]
T_evap2=10[C]
T_evap3=0[C]
T_cond=25[C]
T_1=T_evap1
T_3=T_evap2
T_5=T_evap3
T_8=T_cond
T_10=T_1
x_1 = 1 []
x_3 = 1 []
x_5 = 1 []
x_8 = 0 []
x_10 = 0 []
"State Point 1"
h_1=Enthalpy(f$,T=T_1,x=x_1)
s_1=Entropy(f$,T=T_1,x=x_1)
"State Point 2"
P_2=P_sat(f$,T=T_cond)
162
s_2=s_1
h_2=Enthalpy(f$,P=P_2,s=s_2)
"State Point 3"
h_3=Enthalpy(f$,T=T_3,x=x_3)
s_3=Entropy(f$,T=T_3,x=x_3)
"State Point 4"
P_4=P_sat(f$,T=T_cond)
s_4=s_3
h_4=Enthalpy(f$,P=P_4,s=s_4)
"State Point 5"
h_5=Enthalpy(f$,T=T_5,x=x_5)
s_5=Entropy(f$,T=T_5,x=x_5)
"State Point 6"
P_6=P_sat(f$,T=T_cond)
s_6=s_5
h_6=Enthalpy(f$,P=P_6,s=s_6)
"State Point 7"
m_dot_r*h_7=m_dot_1*h_2+m_dot_2*h_4+m_dot_3*h_6
"State Point 8"
h_8=Enthalpy(f$,T=T_8, x=x_8)
"State Point 9"
h_9=h_8
"State Point 10"
h_10=Enthalpy(f$,T=T_10,x=x_10)
"State Point 11"
h_11=h_10
"State Point 12"
h_12=h_11
163
"For Mass Flow Rates"
RE_3=m_dot_3*(h_5-h_12)
RE_2=m_dot_2*(h_3-h_11)
RE_1=m_dot_1*(h_1-h_9)
m_dot_r=m_dot_1+m_dot_2+m_dot_3
Qr = m_dot_r*(h_7-h_8)
W_dot_c3=m_dot_3*(h_6-h_5)
W_dot_c2=m_dot_2*(h_4-h_3)
W_dot_c1=m_dot_1*(h_2-h_1)
COP=((RE_1+RE_2+RE_3)/(W_dot_c1+W_dot_c2+W_dot_c3))*100[%]
"Multi E.V and Multiple Multiple Evaporator Compressor 3 Stage"
"Known"
f$='R134a'
RE_1=383[kJ/s]
RE_2=232[kJ/s]
RE_3 = 86[kJ/s]
T_evap1=-18[C]
T_evap2=10[C]
T_evap3=0[C]
T_cond=25[C]
T_1 = T_evap1
T_3 = T_evap2
T_5 = T_evap3
T_8 = T_cond
P_8 = P_sat(f$,T=T_8)
x_1 = 1 []
x_3 = 1 []
x_5 = 1 []
x_8 = 0 []
x_10 = 0 []
x_12 = 0 []
164
"State Point 13"
h_12=h_13
"For RE"
RE_1 = m_dot_1*(h_1-h_13)
RE_2 = m_dot_2*(h_3-h_11)
RE_3 = m_dot_3*(h_5-h_8)
"For Wc"
W_dot_c1=m_dot_1*(h_2-h_1)
W_dot_c2=m_dot_2*(h_4-h_3)
W_dot_c3=m_dot_2*(h_6-h_5)
"For Mass Total"
m_total = m_dot_1 + m_dot_2 +m_dot_3
"For COP"
COP=((RE_1+RE_2+RE_3)/(W_dot_c1+W_dot_c2+W_dot_c3))*100[%]
"For QR"
QR=m_total*(h_7-h_8)
"Multi Evaporator, Multi Expansion Valve and Muti Compressor"
"Given"
f$ = 'R134a'
T_1 = 10 [C]
T_2 = 25 [C]
T_3= 0 [C]
T_5=-18[C]
RE_1=86 [kJ/s]
RE_2=232[kJ/s]
RE_3=383[kJ/s]
x_1=1[]
x_3=1[]
x_5=1[]
x_9=0[]
x_12=0[]
P_2=P_sat(f$,T=T_2)
P_2=P_4
165
P_2=P_6
P_2=P_9
"State Point 1"
h_1=Enthalpy(f$,T=T_1,x=x_1)
s_1=Entropy(f$,T=T_1,x=x_1)
"State Point 2"
s_1=s_2
h_2=Enthalpy(f$,P=P_2,s=s_2)
"State Point 3"
h_3=Enthalpy(f$,T=T_3,x=x_3)
s_3=Entropy(f$,T=T_3,x=x_3)
"State Point 4"
s_3=s_4
h_4=Enthalpy(f$,P=P_2,s=s_4)
"State Point 5"
h_5=Enthalpy(f$,T=T_5,x=x_5)
s_5=Entropy(f$,T=T_5,x=x_5)
"State Point 6"
s_5=s_6
h_6=Enthalpy(f$,P=P_2,s=s_6)
"State Point 7"
h_7*m_dot_t=h_1*m_dot_1+h_3*m_dot_3+h_5*m_dot_5
m_dot_t=m_dot_1+m_dot_3+m_dot_5
"State Point 9"
h_9=Enthalpy(f$,T=T_2,x=x_9)
"State Point 10"
h_10=h_9
"State Point 11"
h_11=h_10
166
"State Point 12"
h_12=Enthalpy(f$,T=T_3,x=x_12)
"State Point 13"
h_13=h_12
"Solution"
RE_3=m_dot_1*(h_5-h_13)
RE_2=m_dot_3*(h_3-h_11)
RE_1=m_dot_5*(h_1-h_10)
W_dot_c1=m_dot_1*(h_2-h_1)
W_dot_c2=m_dot_3*(h_4-h_3)
W_dot_c3=m_dot_5*(h_6-h_5)
COP=((RE_1+RE_2+RE_3)/(W_dot_c1+W_dot_c2+W_dot_c3))*100[%]
"Multi Evaporator, Multi Expansion Valve and Dual Compressor 1n2 combi"
"Given"
f$ = 'R134a'
T_cond = 25 [C]
T_e1= 10 [C]
T_e2=0[C]
T_e3= -18 [C]
RE_1=86 [kJ/s]
RE_2=232[kJ/s]
RE_3=383[kJ/s]
x_3=0[]
x_5=0[]
x_8=0[]
x_9=1[]
x_7=1[]
x_12=1[]
P_1=P_sat(f$,T=T_e2)
P_8=P_1
P_6=P_1
P_9=P_1
P_10=P_1
167
P_12=P_sat(f$,T=T_e3)
P_12=P_11
P_7=P_sat(f$,T=T_e1)
P_4=P_7
P_5=P_7
P_2=P_sat(f$,T=T_cond)
P_3=P_2
P_13=P_2
P_14=P_2
h_3=Enthalpy(f$,P=P_3,x=x_3)
h_4=h_3
h_5=Enthalpy(f$,P=P_5,x=x_5)
h_6=h_5
h_8=Enthalpy(f$,P=P_8,x=x_8)
h_11=h_8
h_7=Enthalpy(f$,P=P_7,x=x_7)
h_10=h_7
h_9=Enthalpy(f$,P=P_9,x=x_9)
h_12=Enthalpy(f$,P=P_12,x=x_12)
s_12=Entropy(f$,P=P_12,h=h_12)
s_13=s_12
h_13=Enthalpy(f$,P=P_13,s=s_13)
h_1*m_dot_12=h_9*m_dot_2+h_10*m_dot_1
s_1=Entropy(f$,P=P_1,h=h_1)
s_2=s_1
h_2=Enthalpy(f$,P=P_2,s=s_2)
h_14*m_dot_t=h_2*m_dot_12+h_13*m_dot_3
m_dot_t=m_dot_1+m_dot_2+m_dot_3
m_dot_12=m_dot_1+m_dot_2
"For Mass Flowrates"
RE_3=m_dot_3*(h_12-h_11)
RE_2=m_dot_2*(h_9-h_6)
168
RE_1=m_dot_1*(h_7-h_4)
"Solution"
W_dot_c1=(m_dot_12)*(h_2-h_1)
W_dot_c2=m_dot_3*(h_13-h_12)
COP=((RE_1+RE_2+RE_3)/(W_dot_c1+W_dot_c2))*100[%]
"2 Compressor, Lower Indiv E.V., Multi Evaporator"
f$='R134a'
RE_1=86[kJ/s]
RE_2=232[kJ/s]
RE_3=383[kJ/s]
T_evap_1= 10 [C]
T_evap_2 = 0 [C]
T_evap_3 = -18 [C]
T_cond = 25 [C]
T_1=T_evap_1
T_3=T_evap_2
T_6=T_evap_3
T_9=T_cond
P_9 = P_sat(f$,T=T_9)
P_3=P_sat(f$,T=T_3)
x_1=1[]
x_3=1[]
x_6=1[]
x_9=0[]
x_11=0[]
m_total=m_dot_1+m_dot_2+m_dot_3
m_dot_comb = m_dot_1+m_dot_2
"State Point 1"
h_1=Enthalpy(f$,T=T_1,x=x_1)
169
"State Point 2"
h_2=h_1
"State Point 3"
h_3=Enthalpy(f$,T=T_3,x=x_3)
"State Point 4"
(m_dot_comb)*h_4=m_dot_2*h_3 +m_dot_1*h_2
s_4=Entropy(f$,P=P_3,h=h_4)
"State Point 5"
s_5=s_4
h_5=Enthalpy(f$,P=P_9,s=s_5)
"State Point 6"
h_6=Enthalpy(f$,T=T_6,x=x_6)
s_6=Entropy(f$,T=T_6,x=x_6)
"State Point 7"
s_7=s_6
h_7=Enthalpy(f$,P=P_9,s=s_7)
"State Point 8"
m_total*h_8=m_dot_3*h_7 + (m_dot_comb)*h_5
"State Point 9"
h_9=Enthalpy(f$,P=P_9,x=x_9)
"State Point 10"
h_10=h_9
"State Point 11"
h_11=Enthalpy(f$,T=T_1,x=x_11)
"State Point 12"
h_11=h_12
"State Point 13"
h_12=h_13
170
RE_1=m_dot_1*(h_1-h_10)
RE_2=m_dot_2*(h_3-h_12)
RE_3=m_dot_3*(h_6-h_13)
"Wc"
W_dot_c1 = m_dot_3*(h_7-h_6)
W_dot_c2 = (m_dot_1+m_dot_2)*(h_5-h_4)
"COP"
COP = ((RE_1+RE_2+RE_3)/(W_dot_c1 + W_dot_c2)) * 100
"QR"
QR=m_total*(h_8-h_9)
"Given"
f$ = 'R134a'
T_cond = 25 [C]
T_evap1=10[C]
T_evap2 = 0 [C]
T_evap3= -18 [C]
RE_1=86 [kJ/s]
RE_2=232[kJ/s]
RE_3=383[kJ/s]
P_2=P_sat(f$,T=T_cond)
P_1=P_sat(f$,T=T_evap2)
x_6=1[]
x_7=1[]
x_11=1[]
x_3=0[]
x_9=0[]
"State Point 3"
h_3=Enthalpy(f$,T=T_cond,x=x_3)
"State Point 4"
h_4=h_3
171
"State Point 5"
h_5=h_4
"State Point 6"
h_6=Enthalpy(f$,T=T_evap1,x=x_6)
"State Point 7"
h_7=Enthalpy(f$,T=T_evap2,x=x_7)
"State Point 8"
h_8=h_6
"State Point 9"
h_9=Enthalpy(f$,T=T_evap2,x=x_9)
"State Point 10"
h_10=h_9
"State Point 11"
h_11=Enthalpy(f$,T=T_evap3,x=x_11)
s_11=Entropy(f$,T=T_evap3,x=x_11)
"State Point 12"
s_12=s_11
h_12=Enthalpy(f$,P=P_2,s=s_12)
"For mass flowrates"
RE_1=m_dot_1*(h_6-h_4)
RE_2=m_dot_2*(h_7-h_5)
RE_3=m_dot_3*(h_11-h_10)
m_dot_r=m_dot_1+m_dot_2+m_dot_3
"State Point 1"
h_1*(m_dot_1+m_dot_2)=h_7*m_dot_2+h_8*m_dot_1
s_1=Entropy(f$,P=P_1,h=h_1)
"State Point 13"
h_13*m_dot_r=h_2*(m_dot_1+m_dot_2)+h_12*(m_dot_3)
172
"State Point 2"
s_2=s_1
h_2=Enthalpy(f$,P=P_2,s=s_2)
"For the Work of Compression"
W_dot_c1=(m_dot_1+m_dot_2)*(h_2-h_1)
W_dot_c2=(m_dot_3)*(h_12-h_11)
"For the Heat Rejected"
Q_dot_r=m_dot_r*(h_13-h_3)
"For the COP"
COP=((RE_1+RE_2+RE_3)/(W_dot_c1+W_dot_c2))*100[%]
"“Lower Indi EV 2N3 COMBI”"
f$='R134a'
RE_1=86 [kJ/s]
RE_2=232 [kJ/s]
RE_3=383[kJ/s]
T_cond=25[C]
T_e1=10[C]
T_e2=0[C]
T_e3=-18[C]
x_3=1[]
x_11=1[]
x_12=1[]
x_6=0[]
x_8=0[]
P_1=P_sat(f$,T=T_e3)
P_10=P_1
P_12=P_1
P_13=P_1
P_11=P_sat(f$,T=T_e2)
P_8=P_11
P_9=P_11
P_3=P_sat(f$,T=T_e1)
P_7=P_3
173
P_5=P_sat(f$,T=T_cond)
P_2=P_5
P_4=P_5
P_6=P_5
m_dot_t=m_dot_1+m_dot_23
m_dot_23=m_dot_2+m_dot_3
h_3=Enthalpy(f$,P=P_3,x=x_3)
s_3=Entropy(f$,P=P_3,x=x_3)
s_4=s_3
h_4=Enthalpy(f$,P=P_4,s=s_4)
h_6= Enthalpy(f$,P=P_3,x=x_6)
h_7=h_6
h_8=Enthalpy(f$,P=P_8,x=x_8)
h_9= h_8
h_10=h_8
h_11= Enthalpy(f$,P=P_11,x=x_11)
h_12=Enthalpy(f$,P=P_12,x=x_12)
h_13=h_11
h_1*m_dot_23=h_12*m_dot_3+h_13*m_dot_2
s_1=Entropy(f$,P=P_1,h=h_1)
s_2=s_1
h_2=Enthalpy(f$,P=P_2,s=s_2)
h_5*(m_dot_t)=h_2*m_dot_23+h_4*m_dot_1
RE_1=m_dot_1*(h_3-h_7)
RE_2=m_dot_2*(h_11-h_9)
RE_3=m_dot_3*(h_12-h_10)
W_c1=m_dot_1*(h_4-h_3)
W_c2=m_dot_23*(h_2-h_1)
COP=((RE_1+RE_2+RE_3)/(W_c1+W_c2))*100[%]
"MEV, 2N3 COMBI"
174
f$='R134a'
RE_1=86 [kJ/s]
RE_2=232 [kJ/s]
RE_3=383[kJ/s]
T_cond=25[C]
T_e1=10[C]
T_e2=0[C]
T_e3=-18[C]
x_3=1[]
x_12=1[]
x_13=1[]
x_6=0[]
x_8=0[]
x_10=0[]
m_dot_t=m_dot_1+m_dot_23
m_dot_23=m_dot_2+m_dot_3
P_13=P_sat(f$,T=T_e3)
P_1=P_13
P_11=P_13
P_14=P_13
P_12=P_sat(f$,T=T_e2)
P_9=P_12
P_10=P_12
P_3=P_sat(f$,T=T_e1)
P_7=P_3
P_8=P_3
P_5=P_sat(f$,T=T_cond)
P_2=P_5
P_4=P_5
P_6=P_5
h_3=Enthalpy(f$,P=P_3,x=x_3)
s_3=Entropy(f$,P=P_3,x=x_3)
175
s_4=s_3
h_4= Enthalpy(f$,P=P_4,s=s_4)
h_6=Enthalpy(f$,P=P_6,x=x_6)
h_7=h_6
h_8=Enthalpy(f$,P=P_8,x=x_8)
h_9=h_8
h_10=Enthalpy(f$,P=P_10,x=x_10)
h_11=h_10
h_12=Enthalpy(f$,P=P_12,x=x_12)
h_13=Enthalpy(f$,P=P_13,x=x_13)
h_14=h_12
h_1*m_dot_23=h_13*m_dot_3+h_14*m_dot_2
s_1=Entropy(f$,P=P_1,h=h_1)
s_2=s_1
h_2=Enthalpy(f$,P=P_2,s=s_2)
h_5*m_dot_t=h_2*m_dot_23+h_4*m_dot_1
"RE"
RE_1=m_dot_1*(h_3-h_7)
RE_2=m_dot_2*(h_12-h_9)
RE_3=m_dot_3*(h_13-h_11)
"W_c"
W_c1=m_dot_1*(h_4-h_3)
W_c2=m_dot_23*(h_2-h_1)
Q_r=m_dot_t*(h_5-h_6)
COP=((RE_1+RE_2)/(W_c1+W_c2))*100[%]
"“Upper Indi EV 2N3 COMBI”"
f$='R134a'
RE_1=86 [kJ/s]
RE_2=232 [kJ/s]
RE_3=383[kJ/s]
176
T_cond=25[C]
T_e1=10[C]
T_e2=0[C]
T_e3=-18[C]
x_3=1[]
x_11=1[]
x_12=1[]
x_6=0[]
x_9=0[]
P_1=P_sat(f$,T=T_e3)
P_10=P_1
P_12=P_1
P_13=P_1
P_11=P_sat(f$,T=T_e2)
P_9=P_11
P_3=P_sat(f$,T=T_e1)
P_7=P_3
P_8=P_3
P_5=P_sat(f$,T=T_cond)
P_2=P_5
P_4=P_5
P_6=P_5
m_dot_t=m_dot_1+m_dot_23
m_dot_23=m_dot_2+m_dot_3
h_3=Enthalpy(f$,P=P_3,x=x_3)
s_3=Entropy(f$,P=P_3,x=x_3)
s_4=s_3
h_4=Enthalpy(f$,P=P_4,s=s_4)
h_6= Enthalpy(f$,P=P_3,x=x_6)
h_7=h_6
h_8=h_6
h_9=Enthalpy(f$,P=P_9,x=x_9)
h_10=h_9
h_11= Enthalpy(f$,P=P_11,x=x_11)
h_12=Enthalpy(f$,P=P_12,x=x_12)
177
h_13=h_11
h_1*m_dot_23=h_12*m_dot_3+h_13*m_dot_2
s_1=Entropy(f$,P=P_1,h=h_1)
s_2=s_1
h_2=Enthalpy(f$,P=P_2,s=s_2)
h_5*(m_dot_t)=h_2*m_dot_23+h_4*m_dot_1
RE_1=m_dot_1*(h_3-h_7)
RE_2=m_dot_2*(h_11-h_8)
RE_3=m_dot_3*(h_12-h_10)
W_c1=m_dot_1*(h_4-h_3)
W_c2=m_dot_23*(h_2-h_1)
COP=((RE_1+RE_2+RE_3)/(W_c1+W_c2))*100[%]
Computation for Equipment Sizing of Design Option 1 (Dual Compressor, Multi Expansion
Valve, Multi Evaporator Cycle)
A. Compressor
"COMPRESSOR - 1N2 COMBI MEV"
"Given"
f$ = 'R134a'
T_cond = 30 [C]
T_e1= 10 [C]
T_e2=0[C]
T_e3= -18 [C]
RE_1= 3.766898295 [kJ/s]
RE_2=10.59522511 [kJ/s]
RE_3=61.43888144[kJ/s]
x_3=0[]
x_5=0[]
x_8=0[]
x_9=1[]
x_7=1[]
x_12=1[]
178
P_1=P_sat(f$,T=T_e2)
P_8=P_1
P_6=P_1
P_9=P_1
P_10=P_1
P_12=P_sat(f$,T=T_e3)
P_12=P_11
P_7=P_sat(f$,T=T_e1)
P_4=P_7
P_5=P_7
P_2=P_sat(f$,T=T_cond)
P_3=P_2
P_13=P_2
P_14=P_2
h_3=Enthalpy(f$,P=P_3,x=x_3)
h_4=h_3
h_5=Enthalpy(f$,P=P_5,x=x_5)
h_6=h_5
h_8=Enthalpy(f$,P=P_8,x=x_8)
h_11=h_8
h_7=Enthalpy(f$,P=P_7,x=x_7)
h_10=h_7
h_9=Enthalpy(f$,P=P_9,x=x_9)
h_12=Enthalpy(f$,P=P_12,x=x_12)
s_12=Entropy(f$,P=P_12,h=h_12)
s_13=s_12
h_13=Enthalpy(f$,P=P_13,s=s_13)
h_1*m_dot_12=h_9*m_dot_2+h_10*m_dot_1
s_1=Entropy(f$,P=P_1,h=h_1)
s_2=s_1
h_2=Enthalpy(f$,P=P_2,s=s_2)
h_14*m_dot_t=h_2*m_dot_12+h_13*m_dot_3
179
m_dot_t=m_dot_1+m_dot_2+m_dot_3
m_dot_12=m_dot_1+m_dot_2
"For Mass Flowrates"
RE_3=m_dot_3*(h_12-h_11)
RE_2=m_dot_2*(h_9-h_6)
RE_1=m_dot_1*(h_7-h_4)
"Solution"
W_dot_c1=(m_dot_12)*(h_2-h_1)
W_dot_c2=m_dot_3*(h_13-h_12)
COP=((RE_1+RE_2+RE_3)/(W_dot_c1+W_dot_c2))*100[%]
"Solution"
k=1.07[]
c=0.05[]
N=1780[1/min]*Convert(1/min,1/s)
"COMPRESSOR 1"
sv_1=Volume(f$,P=P_1,h=h_1)
v_d1=m_dot_12*(sv_1)
v_d1=(pi/4)*((B1)^2)*(S1)*(N)
B1=S1
v_1=v_dot_c1+v_d1
v_dot_c1=v_d1*(c)
v_c1=v_d1*(c)*(1/N)
w_c1=((k)/(k-1))*P_1*v_1*( (((P_2)/(P_1))^( (k-1)/(k) )) -1)
180
"COMPRESSOR 2"
sv_12=Volume(f$,P=P_12,h=h_12)
v_d2=m_dot_3*sv_12
v_d2=(pi/4)*((B2)^2)*(S2)*(N)
B2=S2
v_2=v_dot_c2+v_d2
v_dot_c2=v_d2*(c)
v_c2=v_d2*(c)*(1/N)
w_c2=((k)/(k-1))*P_12*v_2*( (((P_13)/(P_12))^( (k-1)/(k) )) -1)
B. Condenser
"1N2 COMBI MEV"
"Given"
f$ = 'R134a'
T_cond = 30 [C]
T_e1= 10 [C]
T_e2=0[C]
T_e3= -18 [C]
RE_1=3.766898295 [kJ/s]
RE_2=10.59522511[kJ/s]
RE_3=61.43888144 [kJ/s]
x_3=0[]
x_5=0[]
x_8=0[]
x_9=1[]
x_7=1[]
x_12=1[]
P_1=P_sat(f$,T=T_e2)
P_8=P_1
P_6=P_1
P_9=P_1
181
P_10=P_1
P_12=P_sat(f$,T=T_e3)
P_12=P_11
P_7=P_sat(f$,T=T_e1)
P_4=P_7
P_5=P_7
P_2=P_sat(f$,T=T_cond)
P_3=P_2
P_13=P_2
P_14=P_2
h_3=Enthalpy(f$,P=P_3,x=x_3)
h_4=h_3
h_5=Enthalpy(f$,P=P_5,x=x_5)
h_6=h_5
h_8=Enthalpy(f$,P=P_8,x=x_8)
h_11=h_8
h_7=Enthalpy(f$,P=P_7,x=x_7)
h_10=h_7
h_9=Enthalpy(f$,P=P_9,x=x_9)
h_12=Enthalpy(f$,P=P_12,x=x_12)
s_12=Entropy(f$,P=P_12,h=h_12)
s_13=s_12
h_13=Enthalpy(f$,P=P_13,s=s_13)
h_1*m_dot_12=h_9*m_dot_2+h_10*m_dot_1
s_1=Entropy(f$,P=P_1,h=h_1)
s_2=s_1
h_2=Enthalpy(f$,P=P_2,s=s_2)
h_14*m_dot_t=h_2*m_dot_12+h_13*m_dot_3
m_dot_t=m_dot_1+m_dot_2+m_dot_3
m_dot_12=m_dot_1+m_dot_2
182
"For Mass Flowrates"
RE_3=m_dot_3*(h_12-h_11)
RE_2=m_dot_2*(h_9-h_6)
RE_1=m_dot_1*(h_7-h_4)
"Solution"
W_dot_c1=(m_dot_12)*(h_2-h_1)
W_dot_c2=m_dot_3*(h_13-h_12)
COP=((RE_1+RE_2+RE_3)/(W_dot_c1+W_dot_c2))*100[%]
"Solution for Heat Exchanger"
Q_r=U*A_o*LMTD
T_14=Temperature(f$,P=P_14,h=h_14)
Q_r=m_dot_t*(h_14-h_3)
Rf_i=0.000352[]
Rf_o=0.000088[]
k=13.46
"For R134a"
h_i=200[W/m^2-K]
"For Water"
h_o=100[W/m^2-K]
T_r_in=T_14
T_r_out=25[C]
T_w_in=25[C]
T_w_out=30[C]
DELTAT_out=T_w_out-T_r_out
DELTAT_in=T_r_in-T_w_in
"LMTD"
LMTD=(DELTAT_out-DELTAT_in)/(ln(DELTAT_out/DELTAT_in))
183
"U"
U=1/(R_t)
R_t=R_1+R_2+R_3+R_4+R_5
R_1=1/(h_i*A_i)
R_2=Rf_i/A_i
R_3=(ln(OD_t/ID_t))/(2*pi*k*L)
R_4=Rf_o/A_o
R_5=1/(h_o*A_o)
A_o=pi*(OD_t)*L
A_i=pi*(ID_t)*L
"DIAMETERS FOR TUBE, ID_t and OD_t
v_14=Volume(f$,P=P_14,h=h_14)
v_dot_r=m_dot_t*v_14
V_econ_r=7[ft/s]*Convert(ft/s,m/s)
V_econ_r=(v_dot_r)/(((ID_t)^2)*(pi/4))"
ID_t=0.077927[m]
OD_t=0.0889[m]
ND_t=3[in]
"DIAMETERS FOR ANNULUS, ID_a and OD_a
Q_r=m_dot_w*4.187 [kJ/kg-K]*5[K]
v_dot_w=m_dot_w*(1/997)[m^3/kg]
V_econ_w=7.25[ft/s]*Convert(ft/s,m/s)
V_econ_w=(v_dot_w)/((((ID_a)^2)-(OD_t)^2)*(pi/4))"
IID_a=0.10226[m]
OD_a=.11430[m]
ND_a=4[in]
184
C. Piping
"PIPING - 1N2 COMBI"
DELTAT=5[K]
eta_p=0.87[]
e=0.066[]
i=0.17[]
N=20[]
L=85.6487[m]*Convert(m,ft)
//D=0.05[m]
DELTAZ=0[]
"Given"
f$ = 'R134a'
T_cond = 30 [C]
T_e1= 10 [C]
T_e2=0[C]
T_e3= -18 [C]
Q_a_1= 3.766898295 [kJ/s]
Q_a_2=10.59522511 [kJ/s]
Q_a_3=61.43888144[kJ/s]
x_3=0[]
x_5=0[]
x_8=0[]
x_9=1[]
x_7=1[]
x_12=1[]
P_1=P_sat(f$,T=T_e2)
P_8=P_1
P_6=P_1
P_9=P_1
P_10=P_1
P_12=P_sat(f$,T=T_e3)
P_12=P_11
P_7=P_sat(f$,T=T_e1)
P_4=P_7
P_5=P_7
185
P_2=P_sat(f$,T=T_cond)
P_3=P_2
P_13=P_2
P_14=P_2
h_3=Enthalpy(f$,P=P_3,x=x_3)
h_4=h_3
h_5=Enthalpy(f$,P=P_5,x=x_5)
h_6=h_5
h_8=Enthalpy(f$,P=P_8,x=x_8)
h_11=h_8
h_7=Enthalpy(f$,P=P_7,x=x_7)
h_10=h_7
h_9=Enthalpy(f$,P=P_9,x=x_9)
h_12=Enthalpy(f$,P=P_12,x=x_12)
s_12=Entropy(f$,P=P_12,h=h_12)
s_13=s_12
h_13=Enthalpy(f$,P=P_13,s=s_13)
h_1*m_dot_12=h_9*m_dot_2+h_10*m_dot_1
s_1=Entropy(f$,P=P_1,h=h_1)
s_2=s_1
h_2=Enthalpy(f$,P=P_2,s=s_2)
h_14*m_dot_t=h_2*m_dot_12+h_13*m_dot_3
m_dot_t=m_dot_1+m_dot_2+m_dot_3
m_dot_12=m_dot_1+m_dot_2
"For Mass Flowrates"
Q_a_3=m_dot_3*(h_12-h_11)
Q_a_2=m_dot_2*(h_9-h_6)
Q_a_1=m_dot_1*(h_7-h_4)
"Q_r"
Q_r=m_dot_t*(h_14-h_3)
186
"m_dot_w"
Cp=4.187[kJ/kg-K]
Q_r=m_dot_w*(Cp)*DELTAT
"v_dot_w"
rho=997[kg/m^3]
v_dot_w=(m_dot_w)/rho
"AC_p"
D_ft=D*Convert(m,ft)
C_1=22.5[]
s=1.14[]
IC_p=(((1+e)^(2023-1980))*C_1*D_ft^(s))*l
IC_p=AC_p*((1-(1+i)^(-N))/(i))
"AC_h"
F=7[]
IC_h=IC_p*F
IC_h=AC_h*((1-(1+i)^(-N))/(i))
"AC_m"
b=0.02[]
AC_m=b*(IC_p+IC_h)
"AC_e"
AC_e=P*t*c
P=WP/(eta_p)
"WP"
WP=(gamma*v_dot_w*TDH)*Convert(J/s,kJ/s)
gamma=rho*9.81[m/s^2]
TDH=(DELTAZ)+HL_p+HL_f
L_m=L*Convert(ft,m)
"HL_p"
HL_p=(f_f*L_m*V^2)/(2*D*9.81[m/s^2])
187
V=(v_dot_w)/((pi/4)*(D)^2)
f_f=MoodyChart(Re,epsilon/D)
epsilon=0.0015[m]
mu=3.2044[kg/m-h]
Re=V*D/kv
kv=(mu/rho)*Convert(m^2/h,m^2/s)
"HL_f"
HL_f=((28*0.9+4*5.6+4*0.4)*(V^2))/(2*9.81[m/s^2])
t=365*24[hr]
c=0.23[$/kW-hr]
"Total Annual Cost, AC_t"
AC_t=AC_p+AC_h+AC_m+AC_e
Computation for Equipment Sizing of Design Option 2 (Triple Compressor, Individual and
Multiple Expansion Valve, Multiple Evaporator Cycle)
A. Compressor
"COMPRESSOR - 3C UPPER"
"Given"
f$ = 'R134a'
188
T_cond = 30 [C]
T_e1 = 10 [C]
T_e2= 0 [C]
T_e3=-18[C]
RE_1= 3.766898295 [kJ/s]
RE_2=10.59522511 [kJ/s]
RE_3=61.43888144[kJ/s]
x_1=1[]
x_3=1[]
x_5=1[]
x_8=0[]
x_11=0[]
P_2=P_sat(f$,T=T_cond)
P_2=P_4
P_2=P_6
P_2=P_8
P_2=P_7
P_1=P_sat(f$, T=T_e1)
P_1=P_9
P_3=P_sat(f$, T=T_e2)
P_3=P_10
P_3=P_11
P_5=P_sat(f$, T=T_e3)
P_5=P_12
"State Point 1"
h_1=Enthalpy(f$,P=P_1,x=x_1)
s_1=Entropy(f$,P=P_1,x=x_1)
"State Point 2"
s_2=s_1
h_2=Enthalpy(f$,P=P_2,s=s_2)
189
"State Point 3"
h_3=Enthalpy(f$,P=P_3,x=x_3)
s_3=Entropy(f$,P=P_3,x=x_3)
"State Point 4"
s_4=s_3
h_4=Enthalpy(f$,P=P_4,s=s_4)
"State Point 5"
h_5=Enthalpy(f$,P=P_5,x=x_5)
s_5=Entropy(f$,P=P_5,x=x_5)
"State Point 6"
s_5=s_6
h_6=Enthalpy(f$,P=P_6,s=s_6)
"State Point 7"
h_7*m_dot_t=h_2*m_dot_1+h_4*m_dot_2+h_6*m_dot_3
m_dot_t=m_dot_1+m_dot_2+m_dot_3
"State Point 8"
h_8=Enthalpy(f$,P=P_8,x=x_8)
"State Point 9"
h_9=h_8
"State Point 10"
h_10=h_9
"State Point 11"
h_11=Enthalpy(f$,P=P_11,x=x_11)
"State Point 12"
h_12=h_11
"Solution"
RE_1=m_dot_1*(h_1-h_9)
RE_2=m_dot_2*(h_3-h_10)
RE_3=m_dot_3*(h_5-h_12)
190
W_dot_c1=m_dot_1*(h_2-h_1)
W_dot_c2=m_dot_2*(h_4-h_3)
W_dot_c3=m_dot_3*(h_6-h_5)
COP=((RE_1+RE_2+RE_3)/(W_dot_c1+W_dot_c2+W_dot_c3))*100[%]
"Solution for Compressor"
k=1.07[]
c=0.05[]
N=1780[1/min]*Convert(1/min,1/s)
"COMPRESSOR 1"
sv_1=Volume(f$,P=P_1,h=h_1)
v_d1=m_dot_1*(sv_1)
v_d1=(pi/4)*((B1)^2)*(S1)*(N)
B1=S1
v_1=v_dot_c1+v_d1
v_dot_c1=v_d1*(c)
v_c1=v_d1*(c)*(1/N)
w_c1=((k)/(k-1))*P_1*v_1*( (((P_2)/(P_1))^( (k-1)/(k) )) -1)
"COMPRESSOR 2"
sv_3=Volume(f$,P=P_3,h=h_3)
v_d2=m_dot_2*sv_3
v_d2=(pi/4)*((B2)^2)*(S2)*(N)
B2=S2
v_2=v_dot_c2+v_d2
v_dot_c2=v_d2*(c)
v_c2=v_d2*(c)*(1/N)
w_c2=((k)/(k-1))*P_3*v_2*( (((P_4)/(P_3))^( (k-1)/(k) )) -1)
191
"COMPRESSOR 3"
sv_5=Volume(f$,P=P_5,h=h_5)
v_d3=m_dot_3*sv_5
v_d3=(pi/4)*((B3)^2)*(S3)*(N)
B3=S3
v_3=v_dot_c3+v_d3
v_dot_c3=v_d3*(c)
v_c3=v_d3*(c)*(1/N)
w_c3=((k)/(k-1))*P_5*v_3*( (((P_6)/(P_5))^( (k-1)/(k) )) -1)
B. Condenser
"CONDENSER - 3C UPPER"
"Given"
f$ = 'R134a'
T_cond = 30 [C]
T_e1 = 10 [C]
T_e2= 0 [C]
T_e3=-18[C]
RE_1= 3.766898295 [kJ/s]
RE_2=10.59522511 [kJ/s]
RE_3=61.43888144[kJ/s]
x_1=1[]
x_3=1[]
x_5=1[]
x_8=0[]
x_11=0[]
P_2=P_sat(f$,T=T_cond)
P_2=P_4
P_2=P_6
P_2=P_8
P_2=P_7
192
P_1=P_sat(f$, T=T_e1)
P_1=P_9
P_3=P_sat(f$, T=T_e2)
P_3=P_10
P_3=P_11
P_5=P_sat(f$, T=T_e3)
P_5=P_12
"State Point 1"
h_1=Enthalpy(f$,P=P_1,x=x_1)
s_1=Entropy(f$,P=P_1,x=x_1)
"State Point 2"
s_2=s_1
h_2=Enthalpy(f$,P=P_2,s=s_2)
"State Point 3"
h_3=Enthalpy(f$,P=P_3,x=x_3)
s_3=Entropy(f$,P=P_3,x=x_3)
"State Point 4"
s_4=s_3
h_4=Enthalpy(f$,P=P_4,s=s_4)
"State Point 5"
h_5=Enthalpy(f$,P=P_5,x=x_5)
s_5=Entropy(f$,P=P_5,x=x_5)
"State Point 6"
s_5=s_6
h_6=Enthalpy(f$,P=P_6,s=s_6)
"State Point 7"
h_7*m_dot_t=h_2*m_dot_1+h_4*m_dot_2+h_6*m_dot_3
m_dot_t=m_dot_1+m_dot_2+m_dot_3
"State Point 8"
h_8=Enthalpy(f$,P=P_8,x=x_8)
193
"State Point 9"
h_9=h_8
"State Point 10"
h_10=h_9
"State Point 11"
h_11=Enthalpy(f$,P=P_11,x=x_11)
"State Point 12"
h_12=h_11
"Solution"
RE_1=m_dot_1*(h_1-h_9)
RE_2=m_dot_2*(h_3-h_10)
RE_3=m_dot_3*(h_5-h_12)
W_dot_c1=m_dot_1*(h_2-h_1)
W_dot_c2=m_dot_2*(h_4-h_3)
W_dot_c3=m_dot_3*(h_6-h_5)
COP=((RE_1+RE_2+RE_3)/(W_dot_c1+W_dot_c2+W_dot_c3))*100[%]
"Solution for Condenser"
Q_r=U*A_o*LMTD
T_7=Temperature(f$,P=P_7,h=h_7)
Q_r=m_dot_t*(h_7-h_8)
Rf_i=0.000352[m^2-K/W]
Rf_o=0.000088[m^2-K/W]
k=13.46[]
"For R134A"
h_i=200[W/m^2-K]
194
"For Water"
h_o=100[W/m^2-K]
T_r_in=T_7
T_r_out=25[C]
T_w_in=25[C]
T_w_out=30[C]
DELTAT_out=T_w_out-T_r_out
DELTAT_in=T_r_in-T_w_in
"LMTD"
LMTD=(DELTAT_out-DELTAT_in)/(ln(DELTAT_out/DELTAT_in))
"U"
U=1/R_t
R_t=R_1+R_2+R_3+R_4+R_5
R_1=1/(h_i*A_i)
R_2=Rf_i/A_i
R_3=(ln(OD_t/ID_t))/(2*pi*k*L)
R_4=Rf_o/A_o
R_5=1/(h_o*A_o)
A_o=pi*(OD_t)*L
A_i=pi*(ID_t)*L
"DIAMETERS FOR TUBE, ID_t and OD_t
v_7=Volume(f$,P=P_7,h=h_7)
v_dot_r=m_dot_t*v_7
V_econ_r=7[ft/s]*Convert(ft/s,m/s)
V_econ_r=(v_dot_r)/(((ID_t)^2)*(pi/4)) "
ID_t=0.077927[m]
OD_t=0.0889[m]
ND_t=3[in]
"DIAMETERS FOR ANNULUS, ID_a and OD_a
Q_r=(m_dot_w)*(4.187 [kJ/kg-K])*(5[K])
v_dot_w=m_dot_w*(1/997)[m^3/kg]
V_econ_w=7.25[ft/s]*Convert(ft/s,m/s)
195
V_econ_w=(v_dot_w)/((((ID_a)^2)-(OD_t)^2)*(pi/4)) "
ID_a=0.10226[m]
OD_a=.11430[m]
ND_a=4[in]
c. Piping
"PIPING - 3C UPPER"
DELTAT=5[K]
eta_p=0.87[]
e=0.066[]
i=0.17[]
N=20[]
L=90.5599[m]*Convert(m,ft)
//D=0.05[m]
DELTAZ=0[]
"Given"
f$ = 'R134a'
T_cond = 30 [C]
T_e1 = 10 [C]
T_e2= 0 [C]
T_e3=-18[C]
Q_a_1= 3.766898295 [kJ/s]
Q_a_2=10.59522511 [kJ/s]
Q_a_3=61.43888144[kJ/s]
x_1=1[]
x_3=1[]
x_5=1[]
x_8=0[]
x_11=0[]
196
P_2=P_sat(f$,T=T_cond)
P_2=P_4
P_2=P_6
P_2=P_8
P_2=P_7
P_1=P_sat(f$, T=T_e1)
P_1=P_9
P_3=P_sat(f$, T=T_e2)
P_3=P_10
P_3=P_11
P_5=P_sat(f$, T=T_e3)
P_5=P_12
"State Point 1"
h_1=Enthalpy(f$,P=P_1,x=x_1)
s_1=Entropy(f$,P=P_1,x=x_1)
"State Point 2"
s_2=s_1
h_2=Enthalpy(f$,P=P_2,s=s_2)
"State Point 3"
h_3=Enthalpy(f$,P=P_3,x=x_3)
s_3=Entropy(f$,P=P_3,x=x_3)
"State Point 4"
s_4=s_3
h_4=Enthalpy(f$,P=P_4,s=s_4)
"State Point 5"
h_5=Enthalpy(f$,P=P_5,x=x_5)
s_5=Entropy(f$,P=P_5,x=x_5)
"State Point 6"
s_5=s_6
h_6=Enthalpy(f$,P=P_6,s=s_6)
197
"State Point 7"
h_7*m_dot_t=h_2*m_dot_1+h_4*m_dot_2+h_6*m_dot_3
m_dot_t=m_dot_1+m_dot_2+m_dot_3
"State Point 8"
h_8=Enthalpy(f$,P=P_8,x=x_8)
"State Point 9"
h_9=h_8
"State Point 10"
h_10=h_9
"State Point 11"
h_11=Enthalpy(f$,P=P_11,x=x_11)
"State Point 12"
h_12=h_11
"For Mass Flowrates"
Q_a_1=m_dot_1*(h_1-h_9)
Q_a_2=m_dot_2*(h_3-h_10)
Q_a_3=m_dot_3*(h_5-h_12)
"Q_r"
Q_r=m_dot_t*(h_7-h_8)
"m_dot_w"
Cp=4.187[kJ/kg-K]
Q_r=m_dot_w*(Cp)*DELTAT
"v_dot_w"
rho=997[kg/m^3]
v_dot_w=(m_dot_w)/rho
"AC_p"
D_ft=D*Convert(m,ft)
198
C_1=22.5[]
s=1.14[]
IC_p=(((1+e)^(2023-1980))*C_1*D_ft^(s))*l
IC_p=AC_p*((1-(1+i)^(-N))/(i))
"AC_h"
F=7[]
IC_h=IC_p*F
IC_h=AC_h*((1-(1+i)^(-N))/(i))
"AC_m"
b=0.02[]
AC_m=b*(IC_p+IC_h)
"AC_e"
AC_e=P*t*c
P=WP/(eta_p)
"WP"
WP=(gamma*v_dot_w*TDH)*Convert(J/s,kJ/s)
gamma=rho*9.81[m/s^2]
TDH=(DELTAZ)+HL_p+HL_f
L_m=L*Convert(ft,m)
"HL_p"
HL_p=(f_f*L_m*V^2)/(2*D*9.81[m/s^2])
V=(v_dot_w)/((pi/4)*(D)^2)
f_f=MoodyChart(Re,epsilon/D)
epsilon=0.0015[m]
mu=3.2044[kg/m-h]
Re=V*D/kv
kv=(mu/rho)*Convert(m^2/h,m^2/s)
"HL_f"
HL_f=((26*0.9+3*5.6+4*0.4)*(V^2))/(2*9.81[m/s^2])
t=365*24[hr]
199
c=0.23[$/kW-hr]
"Total Annual Cost, AC_t"
AC_t=AC_p+AC_h+AC_m+AC_e
Computation for Equipment Sizing of Design Option 3 (Triple Compressor, Multiple
Expansion Valve, Multiple Evaporator Cycle)
A. Compressor
"COMPRESSOR - 3C MEV"
"Known"
f$='R134a'
RE_1= 3.766898295 [kJ/s]
RE_2=10.59522511 [kJ/s]
RE_3=61.43888144[kJ/s]
T_evap1=-18[C]
T_evap2=0[C]
T_evap3=10[C]
T_cond=30[C]
T_1 = T_evap1
T_3 = T_evap2
T_5 = T_evap3
T_8 = T_cond
200
P_8 = P_sat(f$,T=T_8)
x_1 = 1 []
x_3 = 1 []
x_5 = 1 []
x_8 = 0 []
x_10 = 0 []
x_12 = 0 []
"State Point 1"
h_1=Enthalpy(f$,T=T_1,x=x_1)
s_1=Entropy(f$,T=T_1,x=x_1)
"State Point 2"
s_2=s_1
h_2=Enthalpy(f$,P=P_8,s=s_2)
"State Point 3"
h_3=Enthalpy(f$,T=T_3,x=x_3)
s_3=Entropy(f$,T=T_3,x=x_3)
"State Point 4"
s_4=s_3
h_4=Enthalpy(f$,P=P_8,s=s_4)
"State Point 5"
h_5=Enthalpy(f$,T=T_5,x=x_5)
s_5=Entropy(f$,T=T_5,x=x_5)
"State Point 6"
s_6=s_5
h_6=Enthalpy(f$,P=P_8,s=s_6)
"State Point 7"
h_7*m_total = m_dot_1*h_6+m_dot_2*h_4+m_dot_3*h_2
"State Point 8"
h_8=Enthalpy(f$,P=P_8,x=x_8)
201
"State Point 9"
h_9=h_8
"State Point 10"
h_10=Enthalpy(f$,T=T_5,x=x_10)
"State Point 11"
h_11=h_10
"State Point 12"
h_12=Enthalpy(f$,T=T_3,x=x_12)
"State Point 13"
h_12=h_13
"For RE"
RE_1 = m_dot_1*(h_5-h_9)
RE_2 = m_dot_2*(h_3-h_11)
RE_3 = m_dot_3*(h_1-h_13)
"For Wc"
W_dot_c1=m_dot_1*(h_6-h_5)
W_dot_c2=m_dot_2*(h_4-h_3)
W_dot_c3=m_dot_3*(h_2-h_1)
"For Mass Total"
m_total = m_dot_1 + m_dot_2 +m_dot_3
"For COP"
COP=((RE_1+RE_2+RE_3)/(W_dot_c1+W_dot_c2+W_dot_c3))*100[%]
"Solution"
k=1.07[]
c=0.05[]
N=1780[1/min]*Convert(1/min,1/s)
"COMPRESSOR 1"
202
sv_5=Volume(f$,T=T_5,h=h_5)
P_5=P_sat(f$,T=T_5)
P_6=P_sat(f$,T=T_8)
v_d1=m_dot_1*(sv_5)
v_d1=(pi/4)*((B1)^2)*(S1)*(N)
B1=S1
v_1=v_dot_c1+v_d1
v_dot_c1=v_d1*(c)
v_c1=v_d1*(c)*(1/N)
w_c1=((k)/(k-1))*P_5*v_1*( (((P_6)/(P_5))^( (k-1)/(k) )) -1)
"COMPRESSOR 2"
sv_3=Volume(f$,T=T_3,h=h_3)
P_3=P_sat(f$,T=T_3)
P_4=P_sat(f$,T=T_8)
v_d2=m_dot_2*sv_3
v_d2=(pi/4)*((B2)^2)*(S2)*(N)
B2=S2
v_2=v_dot_c2+v_d2
v_dot_c2=v_d2*(c)
v_c2=v_d2*(c)*(1/N)
w_c2=((k)/(k-1))*P_3*v_2*( (((P_4)/(P_3))^( (k-1)/(k) )) -1)
"COMPRESSOR 3"
sv_1=Volume(f$,T=T_1,h=h_1)
P_1=P_sat(f$,T=T_1)
P_2=P_sat(f$,T=T_8)
v_d3=m_dot_3*sv_1
v_d3=(pi/4)*((B3)^2)*(S3)*(N)
B3=S3
v_3=v_dot_c3+v_d3
v_dot_c3=v_d3*(c)
v_c3=v_d3*(c)*(1/N)
203
w_c3=((k)/(k-1))*P_1*v_3*( (((P_2)/(P_1))^( (k-1)/(k) )) -1)
B. Condenser
"CONDENSER - 3C MEV"
"Known"
f$='R134a'
RE_1= 3.766898295 [kJ/s]
RE_2=10.59522511 [kJ/s]
RE_3=61.43888144[kJ/s]
T_evap1=-18[C]
T_evap2=0[C]
T_evap3=10[C]
T_cond=30[C]
T_1 = T_evap1
T_3 = T_evap2
T_5 = T_evap3
T_8 = T_cond
P_8 = P_sat(f$,T=T_8)
x_1 = 1 []
x_3 = 1 []
x_5 = 1 []
x_8 = 0 []
x_10 = 0 []
x_12 = 0 []
"State Point 1"
h_1=Enthalpy(f$,T=T_1,x=x_1)
s_1=Entropy(f$,T=T_1,x=x_1)
"State Point 2"
s_2=s_1
h_2=Enthalpy(f$,P=P_8,s=s_2)
204
"State Point 3"
h_3=Enthalpy(f$,T=T_3,x=x_3)
s_3=Entropy(f$,T=T_3,x=x_3)
"State Point 4"
s_4=s_3
h_4=Enthalpy(f$,P=P_8,s=s_4)
"State Point 5"
h_5=Enthalpy(f$,T=T_5,x=x_5)
s_5=Entropy(f$,T=T_5,x=x_5)
"State Point 6"
s_6=s_5
h_6=Enthalpy(f$,P=P_8,s=s_6)
"State Point 7"
h_7*m_dot_total = m_dot_1*h_6+m_dot_2*h_4+m_dot_3*h_2
"State Point 8"
h_8=Enthalpy(f$,P=P_8,x=x_8)
"State Point 9"
h_9=h_8
"State Point 10"
h_10=Enthalpy(f$,T=T_5,x=x_10)
"State Point 11"
h_11=h_10
"State Point 12"
h_12=Enthalpy(f$,T=T_3,x=x_12)
"State Point 13"
h_12=h_13
"For RE"
RE_1 = m_dot_1*(h_5-h_9)
205
RE_2 = m_dot_2*(h_3-h_11)
RE_3 = m_dot_3*(h_1-h_13)
"For Wc"
W_dot_c1=m_dot_1*(h_6-h_5)
W_dot_c2=m_dot_2*(h_4-h_3)
W_dot_c3=m_dot_3*(h_2-h_1)
"For Mass Total"
m_dot_total = m_dot_1 + m_dot_2 +m_dot_3
"For COP"
COP=((RE_1+RE_2+RE_3)/(W_dot_c1+W_dot_c2+W_dot_c3))*100[%]
"Solution for Condenser"
Q_r=U*A_o*LMTD
P_7=P_8
T_7=Temperature(f$,P=P_7,h=h_7)
Q_r=m_dot_total*(h_7-h_8)
Rf_i=0.000352[m^2-K/W]
Rf_o=0.000088[m^2-K/W]
k=13.46[]
"For R134A"
h_i=200[W/m^2-K]
"For Water"
h_o=100[W/m^2-K]
T_r_in=T_7
T_r_out=25[C]
T_w_in=25[C]
T_w_out=30[C]
DELTAT_out=T_w_out-T_r_out
DELTAT_in=T_r_in-T_w_in
206
"LMTD"
LMTD=(DELTAT_out-DELTAT_in)/(ln(DELTAT_out/DELTAT_in))
"U"
U=1/R_t
R_t=R_1+R_2+R_3+R_4+R_5
R_1=1/(h_i*A_i)
R_2=Rf_i/A_i
R_3=(ln(OD_t/ID_t))/(2*pi*k*L)
R_4=Rf_o/A_o
R_5=1/(h_o*A_o)
A_o=pi*(OD_t)*L
A_i=pi*(ID_t)*L
"DIAMETERS FOR TUBE, ID_t and OD_t
v_7=Volume(f$,P=P_7,h=h_7)
v_dot_r=m_dot_total*v_7
V_econ_r=7[ft/s]*Convert(ft/s,m/s)
V_econ_r=(v_dot_r)/(((ID_t)^2)*(pi/4)) "
ID_t=0.077927[m]
OD_t=0.0889[m]
ND_t=3[in]
"DIAMETERS FOR ANNULUS, ID_a and OD_a
Q_r=(m_dot_w)*(4.187 [kJ/kg-K])*(5[K])
v_dot_w=m_dot_w*(1/997)[m^3/kg]
V_econ_w=7.25[ft/s]*Convert(ft/s,m/s)
V_econ_w=(v_dot_w)/((((ID_a)^2)-(OD_t)^2)*(pi/4)) "
ID_a=0.10226[m]
OD_a=0.11430[m]
ND_a=4[in]
207
C. Piping
"PIPING - 3C MEV"
DELTAT=5[K]
eta_p=0.87[]
e=0.066[]
i=0.17[]
N=20[]
L=90.2781[m]*Convert(m,ft)
//D=0.05[m]
DELTAZ=0[]
"Known"
f$='R134a'
RE_1= 3.766898295 [kJ/s]
RE_2=10.59522511 [kJ/s]
RE_3=61.43888144[kJ/s]
T_evap1=-18[C]
T_evap2=0[C]
T_evap3=10[C]
T_cond=30[C]
T_1 = T_evap1
T_3 = T_evap2
T_5 = T_evap3
T_8 = T_cond
P_8 = P_sat(f$,T=T_8)
x_1 = 1 []
x_3 = 1 []
x_5 = 1 []
x_8 = 0 []
x_10 = 0 []
x_12 = 0 []
208
"State Point 1"
h_1=Enthalpy(f$,T=T_1,x=x_1)
s_1=Entropy(f$,T=T_1,x=x_1)
"State Point 2"
s_2=s_1
h_2=Enthalpy(f$,P=P_8,s=s_2)
"State Point 3"
h_3=Enthalpy(f$,T=T_3,x=x_3)
s_3=Entropy(f$,T=T_3,x=x_3)
"State Point 4"
s_4=s_3
h_4=Enthalpy(f$,P=P_8,s=s_4)
"State Point 5"
h_5=Enthalpy(f$,T=T_5,x=x_5)
s_5=Entropy(f$,T=T_5,x=x_5)
"State Point 6"
s_6=s_5
h_6=Enthalpy(f$,P=P_8,s=s_6)
"State Point 7"
h_7*m_dot_total = m_dot_1*h_6+m_dot_2*h_4+m_dot_3*h_2
"State Point 8"
h_8=Enthalpy(f$,P=P_8,x=x_8)
"State Point 9"
h_9=h_8
"State Point 10"
h_10=Enthalpy(f$,T=T_5,x=x_10)
"State Point 11"
h_11=h_10
209
"State Point 12"
h_12=Enthalpy(f$,T=T_3,x=x_12)
"State Point 13"
h_12=h_13
"For RE"
RE_1 = m_dot_1*(h_5-h_9)
RE_2 = m_dot_2*(h_3-h_11)
RE_3 = m_dot_3*(h_1-h_13)
"For Wc"
W_dot_c1=m_dot_1*(h_6-h_5)
W_dot_c2=m_dot_2*(h_4-h_3)
W_dot_c3=m_dot_3*(h_2-h_1)
"For Mass Total"
m_dot_total = m_dot_1 + m_dot_2 +m_dot_3
"Q_r"
Q_r=m_dot_total*(h_7-h_8)
"m_dot_w"
Cp=4.187[kJ/kg-K]
Q_r=m_dot_w*(Cp)*DELTAT
"v_dot_w"
rho=997[kg/m^3]
v_dot_w=(m_dot_w)/rho
"AC_p"
D_ft=D*Convert(m,ft)
C_1=22.5[]
s=1.14[]
IC_p=(((1+e)^(2023-1980))*C_1*D_ft^(s))*l
IC_p=AC_p*((1-(1+i)^(-N))/(i))
"AC_h"
210
F=7[]
IC_h=IC_p*F
IC_h=AC_h*((1-(1+i)^(-N))/(i))
"AC_m"
b=0.02[]
AC_m=b*(IC_p+IC_h)
"AC_e"
AC_e=P*t*c
P=WP/(eta_p)
"WP"
WP=(gamma*v_dot_w*TDH)*Convert(J/s,kJ/s)
gamma=rho*9.81[m/s^2]
TDH=(DELTAZ)+HL_p+HL_f
L_m=L*Convert(ft,m)
"HL_p"
HL_p=(f_f*L_m*V^2)/(2*D*9.81[m/s^2])
V=(v_dot_w)/((pi/4)*(D)^2)
f_f=MoodyChart(Re,epsilon/D)
epsilon=0.0015[m]
mu=3.2044[kg/m-h]
Re=V*D/kv
kv=(mu/rho)*Convert(m^2/h,m^2/s)
"HL_f"
HL_f=((30*0.9+3*5.6+4*0.4)*(V^2))/(2*9.81[m/s^2])
t=365*24[hr]
c=0.23[$/kW-hr]
"Total Annual Cost, AC_t"
AC_t=AC_p+AC_h+AC_m+AC_e
211
212
Heat Load Calculations
I.
Transmission Load
𝑄 = 𝑈·
𝐴 (𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 − 𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒) 24
1000
𝑈 =
1
𝑅
𝑊ℎ𝑒𝑟𝑒:
𝑄 = ℎ𝑒𝑎𝑡 𝑙𝑜𝑎𝑑 (𝑘𝑊ℎ/𝑑𝑎𝑦)
2
𝑈 − ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 (𝑊/𝑚 · 𝐾)
2
𝐴 − 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑤𝑎𝑙𝑙𝑠, 𝑟𝑜𝑜𝑓, 𝑎𝑛𝑑 𝑓𝑙𝑜𝑜𝑟 (𝑚 )
𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑤𝑎𝑙𝑙 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 (𝐾)
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 𝑖𝑛𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝑐ℎ𝑎𝑚𝑏𝑒𝑟 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 (𝐾)
24 − 𝑛𝑜. 𝑜𝑓 ℎ𝑜𝑢𝑟𝑠 𝑖𝑛 𝑎 𝑑𝑎𝑦
1000 − 𝑐𝑜𝑛𝑣𝑒𝑟𝑠𝑖𝑜𝑛 𝑓𝑟𝑜𝑚 𝑊𝑎𝑡𝑡 (𝑊) 𝑡𝑜 𝐾𝑖𝑙𝑜𝑤𝑎𝑡𝑡 (𝑘𝑊)
𝑅 = 𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 (𝐾/𝑊)
A. Walls of Cold Storage:
Wall 1
𝑅 = 3. 914212868 𝐾/𝑊
𝐴 = 40 𝑚
2
𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 298. 15 𝐾
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 255. 15 𝐾
𝑄 =
1
3.914212868 𝐾/𝑊
2
·
40 𝑚 (298.15 𝐾− 255.15 𝐾) 24
1000
= 10. 54618167 𝑘𝑊ℎ/𝑑𝑎𝑦
213
Wall 2
𝑅 = 3. 914212868 𝐾/𝑊
𝐴 = 40 𝑚
2
𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 283. 15 𝐾
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 255. 15 𝐾
𝑄 =
2
1
3.914212868 𝐾/𝑊
40 𝑚 (283.15 𝐾− 255.15 𝐾) 24
1000
·
= 6. 867281088 𝑘𝑊ℎ/𝑑𝑎𝑦
Wall 3 and 4
𝑅 = 3. 914212868 𝐾/𝑊
𝐴 = 5𝑚
2
𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 283. 15 𝐾
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 255. 15 𝐾
𝑄 =
1
3.914212868 𝐾/𝑊
2
·
5 𝑚 (283.15 𝐾− 255.15 𝐾) 24
1000
(2) = 1. 716820272 𝑘𝑊ℎ/𝑑𝑎𝑦
Wall 5 and 6
𝑅 = 3. 914212868 𝐾/𝑊
𝐴 = 13 𝑚
2
𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 283. 15 𝐾
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 255. 15 𝐾
𝑄 =
1
3.914212868 𝐾/𝑊
2
·
13 𝑚 (283.15 𝐾− 255.15 𝐾) 24
1000
(2) = 4. 463732707 𝑘𝑊ℎ/𝑑𝑎𝑦
214
Total Heat Load in Cold Storage = 23.59401574 kWh/day
B. Walls of Chiller Room:
Wall 1
𝑅 = 3. 914212868 𝐾/𝑊
𝐴 = 40 𝑚
2
𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 283. 15 𝐾
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 273. 15 𝐾
𝑄 =
1
3.914212868 𝐾/𝑊
2
40 𝑚 (283.15 𝐾− 273.15 𝐾) 24
1000
·
= 2. 452600388 𝑘𝑊ℎ/𝑑𝑎𝑦
Wall 2
𝑅 = 3. 914212868 𝐾/𝑊
𝐴 = 40 𝑚
2
𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 298. 15 𝐾
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 273. 15 𝐾
𝑄 =
1
3.914212868 𝐾/𝑊
2
40 𝑚 (298.15 𝐾− 273.15 𝐾) 24
1000
·
= 6. 867281088 𝑘𝑊ℎ/𝑑𝑎𝑦
Wall 3 and 4
𝑅 = 3. 914212868 𝐾/𝑊
𝐴 = 16 𝑚
2
𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 283. 15 𝐾
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 273. 15 𝐾
215
𝑄 =
1
3.914212868 𝐾/𝑊
2
·
16 𝑚 (283.15 𝐾−273.15 𝐾) 24
1000
(2) = 1. 962080311 𝑘𝑊ℎ/𝑑𝑎𝑦
Wall 5 and 6
𝑅 = 3. 914212868 𝐾/𝑊
𝐴 = 8𝑚
2
𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 283. 15 𝐾
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 273. 15 𝐾
1
3.914212868 𝐾/𝑊
𝑄 =
2
·
8 𝑚 (283.15 𝐾− 273.15 𝐾) 24
1000
(2) = 0. 9810401554 𝑘𝑊ℎ/𝑑𝑎𝑦
Total Heat Load in Chiller Room = 11.52722183 kWh/day
C. Walls of Ante Room:
Wall 1
𝑅 = 3. 914212868 𝐾/𝑊
𝐴 = 40 𝑚
2
𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 273. 15 𝐾
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 283. 15 𝐾
𝑄 =
1
3.914212868 𝐾/𝑊
2
·
40 𝑚 (273.15 𝐾− 283.15 𝐾) 24
1000
=
− 2. 452600388 𝑘𝑊ℎ/𝑑𝑎𝑦
Wall 2
𝑅 = 3. 914212868 𝐾/𝑊
𝐴 = 40 𝑚
2
𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 255. 15 𝐾
216
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 283. 15 𝐾
𝑄 =
1
3.914212868 𝐾/𝑊
2
·
40 𝑚 (255.15 𝐾− 283.15 𝐾) 24
1000
=
− 6. 867281088 𝑘𝑊ℎ/𝑑𝑎𝑦
Wall 3 and 4
𝑅 = 3. 914212868 𝐾/𝑊
𝐴 = 33. 2 𝑚
2
𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 273. 15 𝐾
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 283. 15 𝐾
𝑄 =
1
3.914212868 𝐾/𝑊
2
·
33.2 𝑚 (273.15 𝐾−283.15 𝐾) 24
1000
(2) =
− 4. 071316645 𝑘𝑊ℎ/𝑑𝑎𝑦
(2) =
− 9. 339502279 𝑘𝑊ℎ/𝑑𝑎𝑦
Wall 5 and 6
𝑅 = 3. 914212868 𝐾/𝑊
𝐴 = 27. 2 𝑚
2
𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 255. 15 𝐾
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 283. 15 𝐾
𝑄 =
1
3.914212868 𝐾/𝑊
2
·
27.2 𝑚 (255.15 𝐾− 283.15 𝐾) 24
1000
Wall 7 and 8
𝑅 = 3. 914212868 𝐾/𝑊
2
𝐴 = 18. 8 𝑚
𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 298. 15 𝐾
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 283. 15 𝐾
217
1
3.914212868 𝐾/𝑊
𝑄 =
2
18.8 𝑚 (298.15 𝐾− 283.15 𝐾) 24
1000
·
(2) = 3. 458166548 𝑘𝑊ℎ/𝑑𝑎𝑦
Wall 9 and 10
𝑅 = 3. 914212868 𝐾/𝑊
𝐴 = 72 𝑚
2
𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 298. 15 𝐾
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 283. 15 𝐾
𝑄 =
2
1
3.914212868 𝐾/𝑊
·
72 𝑚 (298.15 𝐾− 283.15 𝐾) 24
1000
(2) = 13. 2440421 𝑘𝑊ℎ/𝑑𝑎𝑦
Wall 11 and 12
𝑅 = 3. 914212868 𝐾/𝑊
𝐴 = 10. 8 𝑚
2
𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 298. 15 𝐾
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 283. 15 𝐾
𝑄 =
1
3.914212868 𝐾/𝑊
2
·
10.8 𝑚 (298.15 𝐾− 283.15 𝐾) 24
1000
(2) = 1. 986606315 𝑘𝑊ℎ/𝑑𝑎𝑦
Total Heat Load in Ante Room = -4.04188544 kWh/day
D. Roof of Cold Storage:
𝑅 = 3. 68432781 𝐾/𝑊
𝐴 = 60 𝑚
2
𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 283. 15 𝐾
218
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 255. 15 𝐾
𝑄 =
2
1
3.68432781 𝐾/𝑊
·
60 𝑚 (283.15 𝐾− 257.15 𝐾) 24
1000
= 10. 94365162 𝑘𝑊ℎ/𝑑𝑎𝑦
Total Heat Load in Cold Storage = 10.94365162 kWh/day
E. Roof of Chiller Room:
𝑅 = 3. 68432781 𝐾/𝑊
𝐴 = 80 𝑚
2
𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 285. 15 𝐾
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 273. 15 𝐾
𝑄 =
1
3.68432781 𝐾/𝑊
2
·
80 𝑚 (285.15 𝐾− 275.15 𝐾) 24
1000
= 6. 25351521 𝑘𝑊ℎ/𝑑𝑎𝑦
Total Heat Load in Chiller Room = 6.25351521 kWh/day
F. Roof of Ante Room:
𝑅 = 3. 68432781 𝐾/𝑊
𝐴 = 168. 53 𝑚
2
𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 298. 15 𝐾
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 283. 15 𝐾
𝑄 =
1
3.68432781 𝐾/𝑊
2
·
168.53 𝑚 (298.15 𝐾− 285.15 𝐾) 24
1000
= 16. 46726435 𝑘𝑊ℎ/𝑑𝑎𝑦
Total Heat Load in Ante Room = 16.46726435 kWh/day
219
G. Floor of Cold Storage:
𝑅 = 3. 960970584 𝐾/𝑊
𝐴 = 60 𝑚
2
𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 283. 15 𝐾
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 255. 15 𝐾
𝑄 =
1
3.960970584 𝐾/𝑊
2
60 𝑚 (283.15 𝐾− 257.15 𝐾) 24
1000
·
= 10. 94365162 𝑘𝑊ℎ/𝑑𝑎𝑦
Total Heat Load in Cold Storage = 10.94365162 kWh/day
H. Floor of Chiller Room:
𝑅 = 3. 960970584 𝐾/𝑊
𝐴 = 80 𝑚
2
𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 285. 15 𝐾
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 273. 15 𝐾
𝑄 =
1
3.960970584 𝐾/𝑊
2
·
80 𝑚 (285.15 𝐾− 275.15 𝐾) 24
1000
= 6. 25351521 𝑘𝑊ℎ/𝑑𝑎𝑦
Total Heat Load in Chiller Room = 6.25351521 kWh/day
I. Floor of Ante Room:
𝑅 = 3. 960970584 𝐾/𝑊
𝐴 = 168. 53 𝑚
2
𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 298. 15 𝐾
220
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑇𝑒𝑚𝑝. − 283. 15 𝐾
𝑄 =
1
3.960970584 𝐾/𝑊
2
·
168.53 𝑚 (298.15 𝐾− 283.15285.15 𝐾) 24
1000
= 15. 3171549 𝑘𝑊ℎ/𝑑𝑎𝑦
Total Heat Load in Ante Room = 15.3171549 kWh/day
II.
Product Load
𝑄 =
𝑚𝐶𝑝 (𝑇𝑒𝑚𝑝. 𝐸𝑛𝑡𝑒𝑟 − 𝑇𝑒𝑚𝑝. 𝑆𝑡𝑜𝑟𝑎𝑔𝑒)
3600
𝑊ℎ𝑒𝑟𝑒:
𝑄 = ℎ𝑒𝑎𝑡 𝑙𝑜𝑎𝑑 (𝑘𝑊ℎ/𝑑𝑎𝑦)
𝑚 − 𝑡ℎ𝑒 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑎𝑑𝑑𝑒𝑑 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 (𝑘𝑔)
𝑐𝑝 − 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑒𝑎𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 (𝑘𝐽/𝑘𝑔 − 𝐾)
𝑇𝑒𝑚𝑝. 𝐸𝑛𝑡𝑒𝑟 − 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑛𝑡𝑒𝑟𝑖𝑛𝑔 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 (𝐾)
𝑇𝑒𝑚𝑝. 𝑆𝑡𝑜𝑟𝑎𝑔𝑒 − 𝑡ℎ𝑒 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑖𝑛𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝑠𝑡𝑜𝑟𝑎𝑔𝑒 (𝐾)
3600 − 𝑐𝑜𝑛𝑣𝑒𝑟𝑠𝑖𝑜𝑛 𝑓𝑟𝑜𝑚 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 (𝑠) 𝑡𝑜 ℎ𝑜𝑢𝑟𝑠 (ℎ)
A. Product Load in Cold Storage:
𝑚 − 17000 𝑘𝑔
Above Freezing
𝑐𝑝 − 3. 4 𝑘𝐽/𝑘𝑔 − 𝐾
𝑇𝑒𝑚𝑝. 𝐸𝑛𝑡𝑒𝑟 − 0°𝐶
𝑇𝑒𝑚𝑝. 𝑆𝑡𝑜𝑟𝑎𝑔𝑒 − (− 18°𝐶)
221
𝑄1 =
17000 𝑘𝑔(3.4 𝑘𝐽/𝑘𝑔−𝐾) (0°𝐶 −(−18°𝐶))
3600
= 27. 29444444 𝑘𝑊ℎ/𝑑𝑎𝑦
Latent Heat of Fusion
𝐿𝑎𝑡𝑒𝑛𝑡 𝐻𝑒𝑎𝑡 𝐹𝑢𝑠𝑖𝑜𝑛 − 233 𝑘𝐽/𝑘𝑔
𝑄2 =
17000 𝑘𝑔(233 𝑘𝐽/𝑘𝑔)
3600
= 1100. 277778 𝑘𝑊ℎ/𝑑𝑎𝑦
Below Freezing
𝑐𝑝 − 1. 67 𝑘𝐽/𝑘𝑔 − 𝐾
𝐹𝑟𝑒𝑒𝑧𝑖𝑛𝑔 𝑃𝑜𝑖𝑛𝑡 𝑜𝑓 𝐵𝑒𝑒𝑓 𝐶𝑎𝑟𝑐𝑎𝑠𝑠 − (− 1. 7°𝐶)
𝑇𝑒𝑚𝑝. 𝑆𝑡𝑜𝑟𝑎𝑔𝑒 − (− 18°𝐶)
𝑄3 =
17000 𝑘𝑔(1.67 𝑘𝐽/𝑘𝑔−𝐾) (−1.7°𝐶 −(−18°𝐶))
3600
= 128. 5436111 𝑘𝑊ℎ/𝑑𝑎𝑦
Total Heat Load in Cold Storage = 𝑄1+𝑄2+𝑄3
Total Heat Load in Cold Storage = 1256.115833 kWh/day
B. Product Load in Chiller:
𝑐𝑝 − 3. 4 𝑘𝐽/𝑘𝑔 − 𝐾
𝑇𝑒𝑚𝑝. 𝐸𝑛𝑡𝑒𝑟 − 283. 15 𝐾
𝑇𝑒𝑚𝑝. 𝑆𝑡𝑜𝑟𝑎𝑔𝑒 − 273. 15 𝐾
222
𝑄 =
17000 𝑘𝑔(3.4 𝑘𝐽/𝑘𝑔−𝐾) (283.15 𝐾 −273.15 𝐾)
3600
= 160. 5555556 𝑘𝑊ℎ/𝑑𝑎𝑦
Total Heat Load in Chiller Room = 160.5555556 kWh/day
C. Product Load in Ante Room:
𝑐𝑝 − 3. 4 𝑘𝐽/𝑘𝑔 − 𝐾
𝑇𝑒𝑚𝑝. 𝐸𝑛𝑡𝑒𝑟 − 298. 15 𝐾
𝑇𝑒𝑚𝑝. 𝑆𝑡𝑜𝑟𝑎𝑔𝑒 − 283. 15 𝐾
𝑄 =
17000 𝑘𝑔(3.4 𝑘𝐽/𝑘𝑔−𝐾) (298.15 𝐾 −283.15 𝐾)
3600
= 4. 816666667 𝑘𝑊ℎ/𝑑𝑎𝑦
Total Heat Load in Ante Room = 4.816666667 kWh/day
III.
Internal Load
A. Occupant Load
𝑄 =
(𝑂𝑐𝑐𝑢𝑝𝑎𝑛𝑡𝑠)(𝑇𝑖𝑚𝑒)(𝐻𝑒𝑎𝑡)
1000
𝑊ℎ𝑒𝑟𝑒:
𝑄 = ℎ𝑒𝑎𝑡 𝑙𝑜𝑎𝑑 (𝑘𝑊ℎ/𝑑𝑎𝑦)
𝑂𝑐𝑐𝑢𝑝𝑎𝑛𝑡𝑠 − 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑃𝑒𝑜𝑝𝑙𝑒 𝐼𝑛𝑠𝑖𝑑𝑒
𝑇𝑖𝑚𝑒 − 𝑊𝑜𝑟𝑘𝑖𝑛𝑔 ℎ𝑜𝑢𝑟𝑠 (ℎ𝑟)
𝐻𝑒𝑎𝑡 − ℎ𝑒𝑎𝑡 𝑙𝑜𝑠𝑠 𝑝𝑒𝑟 𝑝𝑒𝑟𝑠𝑜𝑛 𝑝𝑒𝑟 ℎ𝑜𝑢𝑟 (𝑊)
223
B. Occupant Load in Cold Storage:
𝑄 = ℎ𝑒𝑎𝑡 𝑙𝑜𝑎𝑑 (𝑘𝑊ℎ/𝑑𝑎𝑦)
𝑂𝑐𝑐𝑢𝑝𝑎𝑛𝑡𝑠 − 10
𝑇𝑖𝑚𝑒 − 2 ℎ𝑟𝑠
𝐻𝑒𝑎𝑡 − 565 𝑊
𝑄 =
(10)(2 ℎ𝑟𝑠)(565 𝑊)
1000
= 11. 3 𝑘𝑊ℎ/𝑑𝑎𝑦
Total Heat Load in Cold Storage = 11.3 kWh/day
C. Occupant Load in Chiller Room:
𝑄 = ℎ𝑒𝑎𝑡 𝑙𝑜𝑎𝑑 (𝑘𝑊ℎ/𝑑𝑎𝑦)
𝑂𝑐𝑐𝑢𝑝𝑎𝑛𝑡𝑠 − 10
𝑇𝑖𝑚𝑒 − 2 ℎ𝑟𝑠
𝐻𝑒𝑎𝑡 − 565 𝑊
𝑄 =
(10)(2 ℎ𝑟𝑠)(565 𝑊)
1000
= 11. 3 𝑘𝑊ℎ/𝑑𝑎𝑦
Total Heat Load in Chiller Room = 11.3 kWh/day
D. Occupant Load in Ante Room:
𝑄 = ℎ𝑒𝑎𝑡 𝑙𝑜𝑎𝑑 (𝑘𝑊ℎ/𝑑𝑎𝑦)
𝑂𝑐𝑐𝑢𝑝𝑎𝑛𝑡𝑠 − 4
𝑇𝑖𝑚𝑒 − 3 ℎ𝑟𝑠
𝐻𝑒𝑎𝑡 − 565 𝑊
224
(4)(3 ℎ𝑟𝑠)(565 𝑊)
1000
𝑄 =
= 6. 78 𝑘𝑊ℎ/𝑑𝑎𝑦
Total Heat Load in Ante Room = 6.78 kWh/day
E. Lighting Load
𝑄 =
(𝐿𝑎𝑚𝑝𝑠)(𝑇𝑖𝑚𝑒)(𝑊𝑎𝑡𝑡𝑎𝑔𝑒)
1000
𝑊ℎ𝑒𝑟𝑒:
𝑄 = ℎ𝑒𝑎𝑡 𝑙𝑜𝑎𝑑 (𝑘𝑊ℎ/𝑑𝑎𝑦)
𝐿𝑎𝑚𝑝𝑠 − 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑙𝑎𝑚𝑝𝑠 𝑖𝑛𝑠𝑡𝑎𝑙𝑙𝑒𝑑
𝑇𝑖𝑚𝑒 − 𝑊𝑜𝑟𝑘𝑖𝑛𝑔 ℎ𝑜𝑢𝑟𝑠 (ℎ𝑟)
𝑊𝑎𝑡𝑡𝑎𝑔𝑒 − 𝑝𝑜𝑤𝑒𝑟 𝑟𝑎𝑡𝑖𝑛𝑔 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑎𝑚𝑝
F. Lighting Load in Cold Storage:
𝐿𝑎𝑚𝑝𝑠 − 5
𝑇𝑖𝑚𝑒 − 24 ℎ𝑟
𝑊𝑎𝑡𝑡𝑎𝑔𝑒 − 50 𝑊
𝑄 =
(5)(24 ℎ𝑟𝑠)(50 𝑊)
1000
= 6 𝑘𝑊ℎ/𝑑𝑎𝑦
Total Heat Load in Cold Storage = 6 kWh/day
G. Lighting Load in Chiller Room:
𝐿𝑎𝑚𝑝𝑠 − 5
𝑇𝑖𝑚𝑒 − 24 ℎ𝑟
𝑊𝑎𝑡𝑡𝑎𝑔𝑒 − 50 𝑊
225
𝑄 =
(5)(24 ℎ𝑟𝑠)(50 𝑊)
1000
= 6 𝑘𝑊ℎ/𝑑𝑎𝑦
Total Heat Load in Chiller Room = 6 kWh/day
H. Lighting Load in Ante Room:
𝐿𝑎𝑚𝑝𝑠 − 5
𝑇𝑖𝑚𝑒 − 24 ℎ𝑟
𝑊𝑎𝑡𝑡𝑎𝑔𝑒 − 50 𝑊
𝑄 =
(10)(24 ℎ𝑟𝑠)(50 𝑊)
1000
= 12 𝑘𝑊ℎ/𝑑𝑎𝑦
Total Heat Load in Ante Room = 12 kWh/day
IV.
Equipment Load
A. Fan Motors
𝑄 =
(𝐹𝑎𝑛𝑠)(𝑇𝑖𝑚𝑒)(𝑊𝑎𝑡𝑡𝑎𝑔𝑒)
1000
𝑊ℎ𝑒𝑟𝑒:
𝑄 = ℎ𝑒𝑎𝑡 𝑙𝑜𝑎𝑑 (𝑘𝑊ℎ/𝑑𝑎𝑦)
𝐹𝑎𝑛𝑠 − 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑎𝑛𝑠 𝑖𝑛𝑠𝑡𝑎𝑙𝑙𝑒𝑑
𝑇𝑖𝑚𝑒 − 𝑓𝑎𝑛 𝑑𝑎𝑖𝑙𝑦 𝑟𝑢𝑛 ℎ𝑜𝑢𝑟𝑠 (ℎ𝑟𝑠)
𝑊𝑎𝑡𝑡𝑎𝑔𝑒 − 𝑝𝑜𝑤𝑒𝑟 𝑟𝑎𝑡𝑖𝑛𝑔 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑎𝑛 𝑚𝑜𝑡𝑜𝑟𝑠 (𝑊𝑎𝑡𝑡)
1000 = 𝑐𝑜𝑛𝑣𝑒𝑟𝑡 𝑓𝑟𝑜𝑚 𝑤𝑎𝑡𝑡𝑠 𝑡𝑜 𝑘𝑊
226
𝐾𝑛𝑜𝑤𝑛:
𝐹𝑎𝑛𝑠 − 4
𝑇𝑖𝑚𝑒 − 24 ℎ𝑟𝑠
𝑊𝑎𝑡𝑡𝑎𝑔𝑒 − 75 𝑊
𝑄 =
4(24 ℎ𝑟𝑠)(75 𝑊)
1000
= 7. 2 𝑘𝑊ℎ/𝑑𝑎𝑦
Total Heat Load = 7.2 kWh/day
B. Meat Trolley Load
𝑄 = 𝑚𝐶𝑝𝑛∆𝑇
𝑊ℎ𝑒𝑟𝑒:
𝑄 = ℎ𝑒𝑎𝑡 𝑙𝑜𝑎𝑑 (𝑘𝑊ℎ/𝑑𝑎𝑦)
𝑚 − 𝑡ℎ𝑒 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑜𝑙𝑙𝑒𝑦 𝑤𝑖𝑡ℎ 𝑠𝑙𝑖𝑑𝑖𝑛𝑔 ℎ𝑜𝑜𝑘𝑠(𝑘𝑔)
𝐶𝑝 − 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑒𝑎𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑜𝑙𝑙𝑒𝑦 𝑤𝑖𝑡ℎ 𝑠𝑙𝑖𝑑𝑖𝑛𝑔 ℎ𝑜𝑜𝑘𝑠 (𝑘𝐽/𝑘𝑔 − 𝐾)
𝑛 − 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑟𝑜𝑙𝑙𝑒𝑦 𝑤𝑖𝑡ℎ 𝑠𝑙𝑖𝑑𝑖𝑛𝑔 ℎ𝑜𝑜𝑘𝑠
∆𝑇 − 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 (𝐾)
C. Meat Trolley Load in Cold Storage
𝑚 − 200 𝑘𝑔
𝐶𝑝 − 8. 75 𝑘𝐽/𝑘𝑔 − 𝐾
𝑛− 3
∆𝑇 − 273. 15 𝐾 − 255. 15 𝐾
𝑄 = 200 𝑘𝑔 (8. 75 (𝑘𝐽/𝑘𝑔 − 𝐾)(3)(273. 15 𝐾 − 255. 15 𝐾) = 26. 25 𝑘𝑊ℎ/𝑑𝑎𝑦
Total Heat Load in Cold Storage = 26.25 kWh/day
227
D. Meat Trolley Load in Chiller
𝑚 − 200 𝑘𝑔
𝐶𝑝 − 8. 75 𝑘𝐽/𝑘𝑔 − 𝐾
𝑛− 3
∆𝑇 − 283. 15 𝐾 − 273. 15 𝐾
𝑄 = 200 𝑘𝑔 (8. 75 (𝑘𝐽/𝑘𝑔 − 𝐾)(3)(283. 15 𝐾 − 273. 15 𝐾) = 14. 58333333 𝑘𝑊ℎ/𝑑𝑎𝑦
Total Heat Load in Chiller Room = 14.58333333 kWh/day
V.
Infiltration Load
𝑄 =
𝑣𝑜𝑙𝑢𝑚𝑒 𝑥 𝑒𝑛𝑒𝑟𝑔𝑦 𝑥 𝑐ℎ𝑎𝑛𝑔𝑒 𝑥 ∆𝑇
3600
𝑊ℎ𝑒𝑟𝑒:
𝑄 = ℎ𝑒𝑎𝑡 𝑙𝑜𝑎𝑑 (𝑘𝑊ℎ/𝑑𝑎𝑦)
( 3)
𝑉𝑜𝑙𝑢𝑚𝑒 − 𝑡ℎ𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑙𝑑 𝑠𝑡𝑜𝑟𝑎𝑔𝑒 𝑚
𝐸𝑛𝑒𝑟𝑔𝑦 = 𝐸𝑛𝑒𝑟𝑔𝑦 𝑝𝑒𝑟 𝑐𝑢𝑏𝑖𝑐 𝑚𝑒𝑡𝑒𝑟 (𝑘𝐽/𝑚³ ℃)
𝐶ℎ𝑎𝑛𝑔𝑒 − 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑣𝑜𝑙𝑢𝑚𝑒 𝑐ℎ𝑎𝑛𝑔𝑒𝑠 𝑝𝑒𝑟 𝑑𝑎𝑦
Δ𝑇 − 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑎𝑖𝑟 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒
𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑎𝑛𝑑 𝑎𝑖𝑟 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑖𝑛𝑠𝑖𝑑𝑒 (°𝐶)
3600 − 𝑐𝑜𝑛𝑣𝑒𝑟𝑠𝑖𝑜𝑛 𝑜𝑓 𝑘𝐽 𝑡𝑜 𝑘𝑊ℎ
A. Infiltration Load in Cold Storage
𝑉𝑜𝑙𝑢𝑚𝑒 − 285. 6 𝑚
3
𝐸𝑛𝑒𝑟𝑔𝑦 = 2 𝑘𝐽/𝑚³ − 𝐾
228
𝐶ℎ𝑎𝑛𝑔𝑒 − 4
Δ𝑇 − (− 8 𝐾)
3
285.6 𝑚 (2 𝑘𝐽/𝑚³ −𝐾 )(4) (−8 𝐾)
3600
𝑄 =
=− 5. 077333333 𝑘𝑊ℎ/𝑑𝑎𝑦
Total Heat Load in Cold Storage = -5.077333333 kWh/day
B. Infiltration Load in Chiller Room
𝑉𝑜𝑙𝑢𝑚𝑒 − 368 𝑚
3
𝐸𝑛𝑒𝑟𝑔𝑦 = 2 𝑘𝐽/𝑚³ − 𝐾
𝐶ℎ𝑎𝑛𝑔𝑒 − 4
Δ𝑇 − 10 𝐾
3
𝑄 =
368 𝑚 (2 𝑘𝐽/𝑚³ −𝐾 )(4) (10 𝐾)
3600
= 8. 177777778 𝑘𝑊ℎ/𝑑𝑎𝑦
Total Heat Load in Chiller Room = 8.177777778 kWh/day
A. Infiltration Load in Ante Room
𝑉𝑜𝑙𝑢𝑚𝑒 − 775. 238 𝑚
3
𝐸𝑛𝑒𝑟𝑔𝑦 = 2 𝑘𝐽/𝑚³ − 𝐾
𝐶ℎ𝑎𝑛𝑔𝑒 − 4
Δ𝑇 − 15 𝐾
3
𝑄 =
775.238 𝑚 (2 𝑘𝐽/𝑚³ −𝐾 )(4) (15 𝐾)
3600
= 25. 84126667 𝑘𝑊ℎ/𝑑𝑎𝑦
Total Heat Load in Ante Room = 25.84126667 kWh/day
229
Appendix D
(Project Documentation)
230
Project Documentation
231
232
233
234
235
Appendix E
(Curriculum Vitae)
236
JESTER CEDRICK A. ABREU
Address: Muzon, San Luis, Batangas
Email: [email protected]
Phone: 09086765109
PERSONAL DATA:
Date of Birth:
September 04, 2001
Place of Birth:
Biñan,San Pedro, Laguna
Age:
21
Gender:
Male
Civil Status:
Single
Religion:
Roman Catholic
Citizenship
Filipino
EDUCATIONAL BACKGROUND:
Tertiary:
Batangas State University TNEU - Alangilan
Alangilan, Batangas City
2020 - Present
Secondary:
Our Lady of Caysasay Academy
Taal, Batangas
2018 - 2020
San Luis Academy
Calumpang West, San Luis, Batangas
2014-2018
Primary:
Our Lady of Mount Carmel Academy
#77, Cupang, Muntinlupa City
2008 - 2014
237
LYN MARIZ T. BROSOTO
Address: Pacifico, Santa Teresita, Batangas
Email: [email protected]
Phone: 09667025282
PERSONAL DATA:
Date of Birth:
September 25, 2002
Place of Birth:
Taal, Batangas
Age:
20
Gender:
Female
Civil Status:
Single
Religion:
Roman Catholic
Citizenship
Filipino
EDUCATIONAL BACKGROUND:
Tertiary:
Batangas State University TNEU - Alangilan
Alangilan, Batangas City
2020 - Present
Secondary:
Sta Teresa College
Bauan, Batangas
2018 - 2020
Saint Mary’s Educational Institute
Lemery, Batangas
2014-2018
Primary:
Sampa-Pacifico Elementary School
Pacifico, Santa Teresita, BAtangas
2008 - 2014
238
VOIE JERRSON M. DE CLARO
Address: Bagong Tubig, San Luis, Batangas
Email: [email protected]
Phone: 09206247667
PERSONAL DATA:
Date of Birth:
October 06, 2002
Place of Birth:
Bagong Tubig, San Luis, Batangas
Age:
20
Gender:
Male
Civil Status:
Single
Religion:
Roman Catholic
Citizenship
Filipino
EDUCATIONAL BACKGROUND:
Tertiary:
Batangas State University TNEU - Alangilan
Alangilan, Batangas City
2020 - Present
Secondary:
Lemery Senior High School
Lemery, Batangas
2018 - 2020
Saint Mary’s Educational Institute
Lemery, Batangas
2014-2018
Primary:
San Luis Central School
San Luis, Batangas
2008 - 2014
239
ANGELYN R. GONZALES
Address: Ilat South, San Pascual, Batangas
Email: [email protected]
Phone: 09707522068
PERSONAL DATA:
Date of Birth:
December 11, 2001
Place of Birth:
Batangas CIty
Age:
21
Gender:
Female
Civil Status:
Single
Religion:
Born Again Christian
Citizenship
Filipino
EDUCATIONAL BACKGROUND:
Tertiary:
Batangas State University TNEU - Alangilan
Alangilan, Batangas City
2020 - Present
Secondary:
University of Batangas
Hilltop Rd, Batangas, 4200 Batangas
2018 - 2020
Bauan Technical Integrated High School
Pedro Munoz Street, Poblacion, Bauan, Batangas
2014-2018
Primary:
Ilat Elementary School
Ilat South, San Pascual, Batangas
2008 - 2014
240
MARC REY R. MALATA
Address: Alupay, Rosario, Batangas
Email: [email protected]
Phone: 09354713754
PERSONAL DATA:
Date of Birth:
September 18, 2001
Place of Birth:
Rosario, Batangas
Age:
21
Gender:
Male
Civil Status:
Single
Religion:
Roman Catholic
Citizenship
Filipino
EDUCATIONAL BACKGROUND:
Tertiary:
Batangas State University TNEU - Alangilan
Alangilan, Batangas City
2020 - Present
Secondary:
Sto. Niño Formation and Science School
San Roque, Rosario, Batangas
2018 - 2020
Sto. Niño Formation and Science School
San Roque, Rosario, Batangas
2014-2018
Primary:
Alupay Elementary School
Alupay, Rosario, Batangas
2008 - 2014
241
242