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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. 2 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 3 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 4 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: 5 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 6 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 7 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 8 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. 9 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. 10 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. 11 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 12 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. 13 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. 14 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 15 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 16 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 17 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. 18 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 19 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 20 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. 21 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. 22 CONCEPTUAL PARADIGM OF THE STUDY Figure 2. Conceptual Paradigm of the Study 23 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. 24 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. 25 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. 26 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 27 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. 28 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 29 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 30 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 31 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 32 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. 33 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. 34 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 35 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. 36 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. 37 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 38 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 39 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. 40 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 41 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). 42 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. 44 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 45 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. 46 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. 47 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 48 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 49 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. 50 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 51 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 52 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 53 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