TABLE LIST Table 1.1: Parameters calculated entirely outdoors Table 2.1: Indoor air calculation parameters Table 2.2: Outdoor air calculation parameters Table 2.3: Technical specifica
OVERVIEW
The urgency of the subject
The invention of air conditioning has drastically transformed human life, allowing for unprecedented comfort in warm climates While our ancestors mastered the art of warmth through fire, finding effective cooling solutions has been more complex Historical accounts reveal that Roman Emperor Elagabalus once had snow transported by slaves to cool his gardens, demonstrating early attempts to harness natural elements for indoor comfort.
- The air conditioning is very great; the sweltering weather of the previous few days has vanished, and the desert has been replaced with crisp, clean air
The gradual warming of the planet is significantly impacting human health and daily life, primarily due to the increasing reliance on air conditioning to cope with extreme weather conditions Air conditioning technology serves the essential purpose of allowing individuals to adjust indoor temperatures to their personal preferences, providing comfort in a changing environment.
- Here, air conditioning functions include dust filtering, temperature control, humidity, air movement, and components that are unhealthy for humans
In today's industrialized and modernized environment, air conditioning has become essential for homes, businesses, and industries, particularly in industrial parks The rise in air conditioning installations in public spaces is notable, as heat and harmful vapor emissions are prevalent in residential buildings, pharmaceutical and textile manufacturers, chemical factories, and printing enterprises Air conditioners play a crucial role in regulating harmful chemical emissions, ensuring they remain below safe levels for social use and protecting the health of the workforce.
Comfortable temperatures are essential for enhancing human comfort, which significantly influences learning, working, and productivity This is particularly evident in educational settings and future workplaces, highlighting the importance of an effective air conditioning system, such as the one implemented in the Khai Vy apartment building.
The goal of the subject
Calculation and design of the Air Conditioning and Ventilation system for Khai
Vy apartment building are determined with the following purposes:
+ Optimize the environment inside the building, ensuring that air conditioning and ventilation are designed so that residents and employees inside feel most comfortable and at ease
+ Through careful calculation and design, it will help the air conditioning and ventilation system operate efficiently from an energy perspective, helping to reduce electricity consumption and related costs
+ Ensure the system fully meets safety and environmental standards, and complies with design regulations and national technical standards on air conditioning and ventilation
To effectively address the specific needs of the project and the investor, it is essential to engage with the appropriate users This ensures that the air conditioning and ventilation system is tailored to meet the unique challenges of Ho Chi Minh City's crowded and dusty environment.
Our objective in selecting this topic is to develop an efficient, safe, and energy-saving air conditioning and ventilation system that ensures a comfortable and pleasant environment for all residents, staff, and tourists at Khai Vy Apartment.
Scope and objects of research
Research object: Khai Vy Apartment Building in Ho Chi Minh city.
Overview of air conditioning
Air conditioning (abbreviated as AC) is to create and maintain microclimate conditions in a room according to a predetermined formula that does not depend on external conditions
Air conditioners are diverse in design and meet many different spaces, so they are classified based on many criteria This system is classified as follows
Level 1 air conditioning system: This system is capable of maintaining internal microclimate parameters with all ranges of outdoor parameters, even at the most extreme times of the year in summer and winter
Level 2 air conditioning system: This system is capable of maintaining internal microclimate parameters with an error of less than 200 hours a year, equivalent to about 8 days a year That means that in a year, on the most extreme days in summer and winter, the system may have a certain error, but the number of those days is only approximately 4 days in a season
Level 3 air conditioning system: This system is capable of maintaining indoor calculation parameters with an error of less than 400 hours in 1 year
Air conditioning systems are chosen based on the specific needs and purposes of each construction project, with level 3 systems being the most commonly used due to their superior suitability for various applications.
Classified according to cooling medium characteristics
The system includes: central air conditioning, heat exchangers, condenser cooling equipment, water pumps
Central air conditioners generate cold water, which is delivered to heat exchangers in air-conditioned areas This water facilitates heat exchange, effectively regulating temperature and humidity levels within the space.
At the end of this step, the water circulates back to the central air conditioner and continues a new cycle
This system effectively meets the air conditioning needs of various spaces, accommodating different temperature and humidity levels Each individual area can be customized to achieve the desired temperature and humidity, depending on the specific control method employed for that space.
The installation space requirements for this system are not very high The distance between the false ceiling and the bottom of the beam is about 100 - 200 mm
This system includes: central air conditioning, air channels and cold air distribution, condenser cooling equipment
Central air conditioners generate cool air and distribute it to indoor spaces via ductwork, facilitating the exchange of heat and humidity This process effectively regulates the temperature and comfort levels in enclosed areas Once the cool air has circulated, it is returned to the central air conditioning unit through a separate duct, allowing the system to initiate a new cooling cycle.
Classified according to moist heat treatment method
Dry air conditioning system: Air is processed through surface exchangers while not increasing the moisture content of the air
Wet air conditioning system: Air is processed through mixed-type heat exchangers that increase, decrease, or maintain the air's moisture content constant
Historical records indicate that ancient Egyptians pioneered basic forms of "air conditioning" by sprinkling water on window sills and hanging reeds nearby This method utilized hot winds from outside to distribute the dew, effectively lowering indoor temperatures Additionally, the dew droplets helped mitigate the dry desert heat, creating a cooler environment indoors.
To maintain a comfortable indoor temperature during hot, dry days, the Romans implemented plumbing systems around their buildings, while the Persians developed wind towers and water tanks to effectively cool the air.
In the 17th century, significant scientific advancements enabled humanity to manipulate ambient temperatures, transitioning from summer to winter Cornelis Drebble (1572–1633) pioneered a model for cooling air through water evaporation and salt addition Despite presenting this innovation to King James I of England, it was not taken seriously at the time.
In 1758, Cambridge University chemistry professor John Hadley and Pennsylvania governor Benjamin Franklin pioneered the concept of heat reduction through evaporation, a groundbreaking discovery that laid the foundation for modern air conditioning systems.
British chemist Michael Faraday (1791–1867) successfully compressed and liquefied ammonia gas by 1820, discovering that liquid ammonia could evaporate and cool the surrounding air, a concept that later contributed to the development of air conditioning.
Air compression was also utilized in 1842 by physician John Gorrie (1803–
1855) to make cooling bandages for patients in Florida, USA Engineer James Harrison (1816–1893) created the first commercial ice maker in 1851, and it was initially used in Australia
In the past, it was customary to store ice during winter for summer use, and most scientists had adopted traditional methods for ice production Notably, the concept of generating cool air by circulating air over ice was not introduced by the US Navy until shortly before the assassination of President James Garfield in 1881 Following Garfield's assassination, he required hospitalization, during which millions of kilograms of ice were utilized to keep him cool.
In 1911, Carrier introduced a refrigeration formula with an optimal moisture ratio to the American Society of Mechanical Engineers, a cooling method that remains relevant today By 1914, the first home in Minneapolis was equipped with a Carrier air conditioning system, marking a significant milestone in residential cooling technology.
From 1917 to 1930, cinemas began utilizing air conditioning to provide cold air, significantly enhancing the viewing experience In 1922, Carrier replaced the hazardous refrigerant ammonia with the safer substance dielene, marking a crucial advancement in air conditioning technology Subsequent generations of air conditioners have become increasingly smaller and more portable, making them prevalent in various settings such as department stores and ships Additionally, air cooling equipment was widely adopted in many US federal buildings between 1924 and 1930.
Freon gas, developed in 1928 by American engineer Thomas Midgley, revolutionized refrigeration technology as a key condenser This innovative gas was widely used in air conditioning systems until 1994 In 1931, Schultz and Sherman introduced a compact air conditioner designed to cool a room by placing it on a window sill, further advancing the convenience of indoor climate control.
Introduction to computational work
Figure 1.1: Panorama of Khai Vy Grand Nest project
Khai Vy Grand Nest apartment project is located at 04, Dao Tri, Phu Nhuan Ward, District 7, Ho Chi Minh City
The apartment has a land area of: 77,354m 2
Number of floors: 34 floors - 1 basement (1st floor: Services, commercial and office 2nd & 3rd floor: Parking lot 4th floor: Utilities and services 5th-34th floor: Apartments & office-tel)
Figure 1.2: Distribution of floors of the apartment building
Choose a design option
Through analyzing the architecture and characteristics of Khai Vy Apartment, the group will choose a design plan for a level 2 air conditioning system and select a Multi system for the project
Because the project has an outside balcony design, it is easy to install the outdoor unit for convenient maintenance and relocation when not in use
Outdoor design parameters for Ho Chi Minh city area
Table 1.1: Parameters calculated entirely outdoors
Air conditioning level Hot season
1.6.2 Design parameters in the room
Indoor design parameters are designed for summer Because our country is a tropical country, winter parameters will not be considered
Calculate the representative room M2-02 on the 19th floor with the room area 86,69 [m 2 ]
COOLING LOAD CALCULATION
Calculate the cooling load for the 19th floor apartment
According to Vietnam's climate statistics from the Department of Statistics, we have:
According to outdoor calculation parameters for location in Ho Chi Minh City Temperature: t6,8℃
And we have the calculation parameters for the air inside the conditioned space as follows:
The Grand Nest Khai Vy project boasts a prime location along the Saigon River, offering stunning views and refreshing air from its three riverfront sides While providing a tranquil environment, it remains conveniently close to the city center The development features four luxury apartment blocks, designated as block U, block M, block S, and block V, with floors ranging from the 5th to the 34th The first floor is dedicated to service, commercial, and office spaces, while the second and third floors are designated for parking For our analysis, we focus on the important configuration load for the 19th floor, with plans to quickly calculate the remaining floors and utilize Daikin’s Heatload software for recalibration and comparison of results.
Various methods exist for calculating heat-humidity balance to assess cooling capacity Our team has opted for the Carrier method, which innovatively determines productivity (Q0) by calculating the total amount of sensible heat, distinguishing it from traditional approaches.
Qh and latent heat Qa of all heat sources emitted and entering the space to be conditioned
Khai Vy - Grand Nest District 7 is a mixed-use residential and commercial area in Ho Chi Minh City, characterized by its warm climate year-round, eliminating the need for heating systems With no significant seasonal variations between summer and winter, this vibrant district offers a unique living experience in a bustling urban environment.
Heat calculation formula according to the Carrier method:
- Heat emitted from lighting equipment and machinery Q3
- Sensible and latent heat emitted by humans Q4
- Sensible heat and latent heat brought into QN by fresh wind
Select indoor air calculation parameters for the space needing air conditioning as follows:
Table 2.1: Indoor air calculation parameters
Parameter T In (℃) 𝝋 𝑰𝒏 (%) I 2 (kJ/kg) d 2 (g/kg.kkk) T s2 (℃)
Select outdoor air calculation parameters for air-conditioned spaces according to [Document 3 – Appendix]:
Table 2.2: Outdoor air calculation parameters
(kJ/kg) d 1 (g/kg.kkk) T s1 (℃) m(h/year)
Heat flow through glass door Q1
The sun rises in the East and sets in the West, causing solar radiation on surfaces to vary throughout the day East-facing windows capture the most sunlight between 8 a.m and noon, while west-facing windows experience peak radiation from 4 p.m onwards.
5 pm, and if they are horizontal windows on the roof, they will receive maximum radiation at 12 noon
Modern architecture frequently incorporates expansive windows to enhance natural light and elevate the building's aesthetics The Khai Vy - Grand Nest mixed-use project also employs extensive glass features; however, this design choice not only offers visual benefits but also contributes to significant solar radiant heat According to [Document 1 – Equation 4.1 – Page 123], the amount of solar radiant heat transmitted through glass can be quantified.
Q11’: is the amount of time the application takes to pass through the glass into the room, W
Q11’= Fk× Rt× ε c × ε ds × ε mm × ε kh × ε m × ε r (W) [2.2.1]
Because the project uses glass different from basic glass and has curtains:
Q11’= Fk× Rk× ε c × ε ds × ε mm × ε kh × ε m × ε r (W) [2.2.2]
With: ε r =1 and Rt is replaced by heat radiated into the room other than the basic glass
Fk: is the surface area of a window glass with a metal frame (m 2 )
Rk: is the radiant heat of the sun through the glass window into the room (W/m 2 ) ε c : is the influence coefficient of altitude above sea level ε c = 1 + H
H: is the height of the room above sea level ε kh = 1,17 : frame influence coefficient ε mm = 1 : influence coefficient of haze ε m : Glass coefficient, depends on color and type of glass ε ds : is the influence coefficient of the observed air dew point temperature ts compared to the dew point temperature of air above sea level is 20℃ ε ds = 1 − (t s −20)
The instantaneous impact coefficient (nt) represents the average areal density of the entire coating structure, including walls, ceilings, and floors This is calculated using the formula gs = G' + 0.5G'', where gs denotes the average areal density in kg/m².
G’: The mass of the wall whose outer surface is exposed to solar radiation and of the floor relative to the ground (kg)
G’’: The mass of the wall whose outer surface is not exposed to solar radiation and of the floor which is not above ground level (kg)
From the outdoor calculation parameters for the Ho Chi Minh City area as follows:
Temperature calculation: tN=ttb max6,8℃
Looking at the t-d graph, we can find the dew point temperature ts0,9℃
To calculate the thermal transmittance, we start with the formula 1000 × 0.023 The coefficient εc, which accounts for the building's height above sea level, is considered insignificant, thus we set εc = 1 For optimal results, measurements should be taken on a clear day, as noted in Document 1, Page 124 The coefficient εkh for a door with a metal frame is 1.17, and for clear, flat glass that is 6mm thick, the value of εm is 0.94, as detailed in Table 4.3 of Document 1, Page 131.
Radiation properties and glass coefficients of different types of glass ε r – according to [Document 1- table 4.3] We get the parameters:
Table 2.3: Technical specifications of glass walls and windows
Window glass (clear, flat glass 6mm thick) 0,15 0,77 0,08 0,94
Radiation characteristics of curtains and sunshine coefficient ɛr – According to [Document 1- table 4.4] - for average color curtains we get the parameters
Table 2.4: Technical specifications of curtains
Ho Chi Minh City, situated at approximately 0° North latitude, experiences its highest average temperatures in April According to Document 1, Table 4.2, the solar radiation that penetrates a 6mm thick flat glass window into a room is measured in W/m².
Table 2.5: Hot sunlight radiates through the glass door into the living room of apartment M2-02 on the 19th floor
The glass door is facing East, we have it: RTb0
The glass door is facing North, we have it: RN9,7, Rk!,3(W/m 2 )
The glass door is facing South, we have it: RNT,5, Rk),3(W/m 2 )
The glass door is facing West, we have it: RN p4,55=, Rk75,6(W/m 2 )
Radiant heat through the glass of the apartment living room, 19th floor East direction: Fk=6,27 m 2
Q11’= Fk× Rk× ε c × ε ds × ε mm × ε kh × ε m × ε r = 6,27×375,6 × 1 × 1,17× 0,94 ×
1 = 2590 (W) Looking up according to table 4.11 [Document 1 - Page 143] we have:
- Volume of 1m 2 wall (broken concrete bricks) (0.25m thick): 1800× 0,25 450 kg/m 2
- Volume of 1m 2 wall (broken concrete bricks) (0.15m thick): 1800× 0,15 270 kg/m 2
- Volume of 1m 2 wall (broken concrete bricks) (0.14m thick): 1800× 0,14 252 kg/m 2
- Volume of 1m 2 concrete floor (reinforced concrete) (0.125m thick): 2400×0.125 = 300 kg/m 2
Table 2.6: Density of walls and floors
The concrete floor is 125mm thick
Because 4 sides of the building are covered with glass, there is no direct exposure to solar radiation
- Volume of walls exposed to solar radiation:
G’=0 Walls not exposed to solar radiation are 140 mm walls
- Wall mass not exposed to solar radiation and floor not higher than ground level:
- Caculate gs for living room, apartment M2-02, 19th floor: g s =0+0,5×753240
Get value gs = 500 kg/m 2 floor According to [Document 1 - Table 4.6 -Page
134] Instantaneous impact coefficient of solar radiation through glass with curtains inside
Table 2.7: Corresponding instantaneous impact coefficient
Therefore, the heat radiates throughout the living room of apartment M2-02 on the 19th floor In the East direction:
2.1.2.2 Heat is transmitted through the roof by radiation Q 21
Flat roofs for air-conditioned rooms come in 3 forms:
Type 1: Air-conditioned room located between floors in a building with air conditioning 𝛥t = 0, Q21 = 0
Type 2: Above the air-conditioned room is considered a non-air-conditioned room
Type 3: In case the roof has solar radiation (roof), the amount of heat transmitted into the room includes two components: due to solar radiation and due to the temperature difference between indoor and outdoor air
Floors from floors 5 to 34 will use case 1: 𝛥t = 0, Q21 = 0
Therefore, we only calculate heat transfer through the roof of the 34th floor
K: is the heat transfer coefficient
∆t td : is the equivalent temperature difference
In which; tN: is the outside air temperature tN6,8℃ tT: is the air temperature in the air-conditioned room tT&℃
𝜀 𝑠 : is the solar radiation absorption coefficient The ceiling and floor are filled with reinforced concrete, the surface is plastered with light cement mortar, so 𝜀 𝑠 =0,47 [Document 1 – Table 4.10 –Page 141] We have:
RT:is the heat radiated from the sun through the glass door into the room
RN: is the heat dissipation coefficient to the air 𝛼 𝑁 = 20 k: is the heat transfer coefficient through the roof [Document 1 – Table 4.9 –
Firgure 2.1: Distribution of background sequences
For living room, apartment M2-02, 19th floor
2.1.2.3 Heat flows through the wall Q 22
Heat transfer through walls Q22 also includes 2 parts:
- Due to the temperature difference between outdoors and indoors ∆t = (t N − t T )
- Because solar radiation shines on the wall, for example East, West, , however, this amount of heat is considered zero when calculating
Formula for calculating heat transfer through walls [Document 1 -Page 142]
Q2i: Heat transfer through walls, windows (glass) ki: Corresponding heat transfer coefficients of walls, doors, and glass [W/m 2 K]
Fi: Approximate area of walls, doors, and glass [m 2 ]
According to document 1- Page 142, we have: k = 1
The heat dissipation coefficient outside a wall, denoted as α N, varies depending on exposure; it is measured in watts per square meter per Kelvin (W/m²·K) for both indirect exposure in buffer spaces like corridors or toilets and direct contact with outdoor air Additionally, the indoor heat dissipation coefficient, α T, also measured in W/m²·K, plays a crucial role in understanding thermal dynamics The wall's material thickness, represented as δ i in meters, and its thermal conductivity coefficient, λ i in watts per meter per Kelvin (W/m·K), are essential factors influencing heat transfer efficiency.
R i : Thermal conductivity of the ith material layer of the wall structure, m 2 K/W
∆t: Temperature difference between indoor and outdoor air∆t =(36,8-26)
,8℃ tN: Temperature outside Ho Chi Minh city tT: The temperature is in the air-conditioned space k: heat transfer coefficient through the wall, calculated according to [Document
Table 2.8: Glass parameters in direct contact with outside air (15 mm)
Thermal coefficient of glass in direct contact with outside air:
Table 2.9: Wall parameters in indirect contact with outside air (140 mm)
Parameter Cement grout Hollow brick Cement grount
Heat dissipation coefficient outside the wall when in indirect contact with outside air:
For the living room of apartment M2-02 on the 19th floor (Because 3 walls of the living room are in contact with the air-conditioned space, Q22t and Q22c = 0)
2.1.2.4 Heat flows through the substrate Q 23
The heat transmitted through the background is calculated using the expression: [Document 1 – Page 145]
F: Base area Q23 k: Heat transfer coefficient through the soil, [W/m 2 K]
∆t: difference in outdoor and indoor air temperature
There are also 3 similar cases:
Case 1: Floors placed on the ground: take k concrete floors 300mm thick, t= tN- tT
Case 2: Floor located in the basement or room without air conditioning, take
∆t = 0,5 × (t N − t T ), i.e the basement or non-air condition room has a temperature equal to the average temperature between outside and inside
We have the coefficient k=2,15 W/m 2 K Check in table 4.15 [1, page 145]
2.1.2.5 Heat is emitted from the device Q 3
The heat emitted by the lamp is calculated according to the formula [Document
Q: is the total heat of the lamp (W)
The project mainly uses LED lights, the difference is insignificant compared to fluorescent lights, so Q is calculated according to the formula:
The instantaneous effect coefficient (nt) of a lamp, which measures the impact of light and human heat, is calculated to be 0.87 for a surface area of 500 kg/m² over 8 hours of operation Additionally, the simultaneous effect coefficient (nd), applicable to residential and large air conditioning projects, is established at a value of 1.
N: Total wattage recorded on the light bulb
The heat load due to office lighting is 12 W/m 2 floor according to [Document 1 page 146] Living room of apartment M2- 02, 19th floor Area is: 30,2m 2
2.1.2.6 Heat emitted by the machine Q 32
Heat emitted from machinery and electrical equipment such as televisions, computers, phones, fax machines, etc is calculated according to the formula [Document 1 - Page 146]
With Ni: electrical capacity recorded on the device [W]
Heat is emitted from machinery Q32 living room, apartment
2.1.2.7 Sensible heat and latent heat radiate from the body Q 4
Heat radiated into the room by people is mainly due to convection and radiation and is calculated according to the formula [Document 1 - Equation 4.9 -
In an air-conditioned room, the number of occupants (n) and the non-simultaneous effect coefficient (nd), which ranges from 0.75 to 0.9 for high-rise buildings, play a crucial role in determining thermal comfort For this analysis, we select nd = 0.9 The heat radiated from each person (qh) can be calculated, with the room temperature set at 26°C, leading to the determination of the current heat emitted by an adult (qhh) in watts per person, as referenced in [Document 1 - Table 4.18].
Calculate the current heat emitted by humans for the Living Room, apartment M2-02 floor has density 10m 2 person according to [Document 1 - Appendix F - Page
88] fresh air standards for air-conditioned rooms and F= 30,2 (m 2 ) there are people
2.1.2.8 Latent heat radiates from the body Q 4a
Latent heat emitted by humans is calculated according to the formula [Document 1 – Equation 4.20 – Page 148]
In an air-conditioned room maintained at 26℃, the latent heat emitted by each adult can be quantified as qa W/person, where n represents the total number of individuals present This calculation is based on the data provided in Document 1, specifically Table 4.18, which outlines the heat output per person in such environments.
We calculate the latent heat emitted from people in the living room of apartment M2-02 on the 19th floor
Total latent heat generated by people in other rooms of apartment M2-02 on the 19th floor
2.1.2.9 Sensible heat and latent heat are supplied by fresh air Q N
Air-conditioned rooms must ensure a supply of fresh air to maintain adequate oxygen levels for occupants Fresh air, characterized by higher enthalpy, temperature, and humidity compared to indoor air, introduces sensible heat (QhN) and latent heat (QaN) into the environment when brought indoors This process is essential for optimal indoor air quality and comfort.
Calculate the cooling load for the 1th floor reception hall
According to Vietnam's climate statistics from the Department of Statistics, we have according to outdoor calculation parameters for location in Ho Chi Minh City
And we have the calculation parameters for the air inside the conditioned space as follows:
2.2.2 Calculate the load for the reception lobby
Heat calculation formula according to the Carrier method:
- Heat emitted from lighting equipment and machinery Q3
- Sensible and latent heat emitted by humans Q4
- Sensible heat and latent heat brought into QN by fresh wind
Select parameters to calculate the air in the lobby for the space needing air conditioning as follows:
Table 2.11: Air calculation parameters in the reception hall
Parameter T In (℃) 𝝋 𝑰𝒏 (%) I 2 (kJ/kg) d 2 (g/kg.kkk) T s2 (℃)
Select outdoor air calculation parameters for air-conditioned spaces according to [Document 3 – Appendix]:
Table 2.12: Outdoor air calculation parameters
(kJ/kg) d 1 (g/kg.kkk) T s1 (℃) m(h/year)
Solar radiant heat through glass is determined according to [Document 1 – Equation 4.1 – Page 123]:
Q11’: is the amount of time it takes the application to pass through the glass into the room, W
Q11’= Fk× Rt× ε c × ε ds × ε mm × ε kh × ε m × ε r (W) [2.2.1]
With: ε r =1 and Rt is replaced by heat radiated into the room other than the basic glass
Fk: is the surface area of a window glass with a metal frame (m 2 )
Rk: is the radiant heat of the sun through the glass window into the room (W/m 2 ) ε c : is the influence coefficient of altitude above sea level ε c = 1 + H
H: is the height of the room above sea level ε kh = 1,17 : frame influence coefficient ε mm = 1 : influence coefficient of haze ε m : Glass coefficient, depends on color and type of glass ε ds : is the influence coefficient of the observed air dew point temperature difference ts compared to the dew point temperature of air above sea level is 20℃ ε ds = 1 − (t s −20)
The instantaneous impact coefficient (nt) represents the average areal density of the entire coating structure, encompassing walls, ceilings, and floors This average areal density (gs) is calculated using the formula gs = G′ + 0.5G′′, where G′ and G′′ denote specific density components of the structure.
G’: The mass of the wall whose outer surface is exposed to solar radiation and of the floor relative to the ground (kg)
G’’: The mass of the wall whose outer surface is not exposed to solar radiation and of the floor which is not above ground level (kg)
From the outdoor calculation parameters for the Ho Chi Minh City area as follows:
Temperature calculation: tN=ttb max6,8℃
Looking at the t-d graph, we can find the dew point temperature ts0,9℃
To calculate the influence of building height above sea level, we use a coefficient of ε c = 1, as its impact is negligible For optimal results, measurements should be taken on a clear day, avoiding cloudy conditions The metal frame door has a coefficient of ε kh = 1.17, while clear, flat glass with a thickness of 6mm is assigned a coefficient of ε m = 0.94, as detailed in Table 4.3 of Document 1.
Radiation properties and glass coefficients of different types of glass ε r – according to [Document 1- table 4.3] We get the parameters:
Table 2.13: Technical specifications of glass walls and windows
Window glass (clear, flat glass 10mm thick) 0,2 0,8 0,08 0,94
Table 2.14: The hot sun radiates through the glass door into the 1th floor reception hall
The glass door is facing north, we have it: RT5
The glass door is facing East, we have it: RNp4,1, RkE6(W/m 2 )
The glass door is facing South, we have it: RN@, Rk& (W/m 2 )
The glass door is facing West, we have it: RN p4,1=, Rk= 456 (W/m 2 )
- Heat radiates through the reception lobby glass North: Fk1 m 2
Q11’= Fk× Rk× ε c × ε ds × ε mm × ε kh × ε m × ε r = 31×26 × 1 × 1,17× 0,94 × 1 887 (W) Looking up according to table 4.11 [Document 1 - Page 143], we have:
- Volume of 1m 2 wall (broken concrete bricks) (0.25m thick): 1800× 0,25 450 (kg/m 2 )
- Volume of 1m 2 wall (broken concrete bricks) (0.15m thick): 1800× 0,15 270 (kg/m 2 )
- Volume of 1m 2 wall (broken concrete bricks) (0.14m thick): 1800× 0,14 252 (kg/m 2 )
- Volume of 1m 2 concrete floor (reinforced concrete) (0.125m thick):
Table 2.15: Density of walls and floors
The concrete floor is 125mm thick
Because 4 sides of the building are covered with glass, there is no direct exposure to solar radiation
- Volume of walls exposed to solar radiation:
G’=0 Walls that are not exposed to solar radiation are walls 140 mm
- Wall mass not exposed to solar radiation and floor not higher than ground level:
- Calculate gs for ground floor reception room g s =0+0,5×3236688
1440 = 1123 (kg /m 2 ) Get value gs = 500 kg/m 2 floor According to [Document 1 - Table 4.6 -Page
134] Instantaneous impact coefficient of solar radiation through inside glass
Table 2.16: Corresponding instantaneous impact coefficient
Therefore, the heat radiates throughout the reception room facing north on the ground floor:
2.2.2.2 Heat flows through the wall Q 22
Heat transfer through walls Q22 also includes 2 parts:
- Due to the temperature difference between outdoors and indoors ∆t = (t N − t T )
- Because solar radiation shines on the wall, for example East, West, , however, this amount of heat is considered zero when calculating
Formula for calculating heat transfer through walls [Document 1 -Page 142]
Q2i: Heat transfer through walls, windows (glass) ki: Corresponding heat transfer coefficients of walls, doors, and glass [W/m 2 K]
Fi: Approximate area of walls, doors, and glass [m 2 ]
According to document 1- Page 142, we have: k = 1
The heat dissipation coefficient outside a wall, denoted as α N, varies depending on exposure to outdoor air, either indirectly, such as in buffer spaces like corridors or toilets, or directly In this context, α N is measured in watts per square meter per Kelvin (W/m²·K) Additionally, the indoor heat dissipation coefficient is represented as α T, also in W/m²·K Key factors influencing heat transfer include the wall material thickness (δ i), measured in meters, and the thermal conductivity coefficient of the wall material (λ i), expressed in watts per meter per Kelvin (W/m·K).
R i : Thermal conductivity of the ith material layer of the wall structure, m 2 K/W
∆t: Temperature difference between indoor and outdoor air ∆t =(36,8-25)
,8℃ tN: Temperature outside Ho Chi Minh city tT: The temperature is in the air-conditioned space k: heat transfer coefficient through the wall, calculated according to [Document
Table 2.17: Glass parameters in direct contact with outside air (15 mm)
Thermal coefficient of glass in direct contact with outside air:
Table 2.18: Wall parameter in indirect contact with outside air (140 mm)
Parameter Cement grout Hollow brick Cement grount
Heat dissipation coefficient outside the wall when in indirect contact with outside air:
For the reception hall (Because the two walls of the living room are in contact with the air-conditioned space Q22t and Q22c = 0)
2.2.2.3 Heat flows through the substrate Q 23
The heat transmitted through the background is calculated using the expression: [Document 1 – Page 145]
F: Base area Q23 k: Heat transfer coefficient through soil, [W/m 2 K]
∆t: difference in air temperature outside and in the lobby
There are also 3 similar cases:
Case 1: Floor placed on the ground: take k concrete floors 300mm thick, t= tN- tT
Case 2: Floor located in the basement or room without air conditioning, take
∆t = 0,5 × (t N − t T ), i.e the basement or non-air condition room has a temperature equal to the average temperature between outside and inside
We have the coefficient k=2,15 W/m 2 K Check in table 4.15 [1, page 145]
2.2.2.4 Heat is emitted from the device Q 3
The heat emitted by the lamp is calculated according to the formula [Document
Q: is the total heat of the lamp (W)
The project mainly uses LED lights, the difference is insignificant compared to fluorescent lights, so Q is calculated according to the formula:
The instantaneous effect coefficient (nt) of a lamp, which measures the impact of light heat and human heat, is calculated as nt=0.87 for a scenario with 500 kg/m² and 8 hours of operation Additionally, the simultaneous effect coefficient (nd), applicable only for residential and large air conditioning projects, is defined as nd=1.
N: Total wattage recorded on the light bulb
The heat load due to office lighting is 16 W/m 2 floor according to [Document 1 page 146] Heat calculation for ground floor reception room There is an area of 172 m 2
2.2.2.5 Heat emitted by the machine Q 32
Heat emitted from machinery and electrical equipment such as televisions, computers, phones, fax machines, etc is calculated according to the formula [Document 1 - Page 146]
With Ni: electrical capacity recorded on the device [W]
Heat is emitted from machinery Q32 living room, apartment
2.2.2.6 Sensible heat and latent heat radiate from the body Q 4
Heat radiated into the room by people is mainly due to convection and radiation and is calculated according to the formula [Document 1 - Equation 4.9 - Page 148]
In an air-conditioned room, the number of occupants (n) and the non-simultaneous effect coefficient (nd) play crucial roles in determining thermal comfort For high-rise buildings, the recommended non-simultaneous effect coefficient is between 0.75 and 0.9, with a selection of nd=0.9 for optimal calculations The current heat radiated by an individual, denoted as qh, is expressed in watts per person Based on data from [Document 1 - Table 4.18], the ideal room temperature for air conditioning is set at 26°C, allowing us to compute the heat output of an adult as qhh W/person.
Calculate the current heat emitted by humans to the reception hall:
2.2.2.7 Latent heat radiates from the body Q 4a
Latent heat emitted by humans is calculated according to the formula [Document 1 – Equation 4.20 – Page 148]
In an air-conditioned room maintained at a temperature of 26℃, the latent heat emitted by each individual, denoted as qa in watts per person, can be calculated based on the number of occupants present.
We calculate the latent heat emitted from people in the 1st floor reception hall
Q4a= 35×92= 3220 (W) Total latent heat generated by people in the 1st floor reception hall
2.2.2.8 Sensible heat and latent heat are supplied by fresh air Q N
Air-conditioned rooms must supply fresh air to maintain adequate oxygen levels for occupants Fresh air, characterized by higher enthalpy, temperature, and humidity than indoor air, introduces sensible heat (QhN) and latent heat (QN) when it enters the space This process is essential for ensuring comfort and air quality in air-conditioned environments, as outlined in the relevant formulas from Document 1.
In an air-conditioned room, the number of occupants (n) significantly influences the required irrigation air (l) needed per person per second The humidity levels (dN for indoor and dT for outdoor) are measured in kg/kg, while the respective air temperatures (tN for indoor and tT for outdoor) are recorded in degrees Celsius (℃) Understanding these parameters is essential for optimizing indoor air quality and comfort.
According to [Document 3- Appendix F], there are fresh air standards for offices l= 30 (m 3 /h) = 8 (l/s), Other rooms available l = 35 (m 3 /h) = 10 (l /s),
Calculated for the reception room
Heat appears that fresh air brings in QhN
Latent heat brought in by fresh air:
Heat is supplied by fresh wind QN:
2.2.2.9 Sensible heat and latent heat due to wind Q 5h and Q 5a
To optimize energy efficiency in air-conditioned spaces, it is essential to seal the environment to regulate the intake of fresh air However, air leakage through doors and windows remains a challenge, particularly during significant temperature differences between indoor and outdoor settings Cold air often escapes beneath doors, while warmer outdoor air infiltrates from above, leading to increased energy loss Understanding the dynamics of visible and latent heat due to wind is crucial for effective climate control.
V: Room volume [m 3 ] ξ: Experience coefficient Look up table 4.20 Experimental coefficients ξ 0,7 − 0,35 according to [Document 1] - Table 4.20 - Page 151 dN, dT: humidity of indoor and outdoor air [kg/kgkkk]
44 | P a g e tN, tT: Indoor and outdoor air temperature [℃]
Calculating the reception room, we can store the room V = 172 × 3 = 516
Table 2.19: Total load for reception lobby and equipment selection
SETUP AND CALCULATE AIR CONDITIONING DIAGRAM
Calculating air conditioning diagrams according to the t-d diagram is very commonly applied in Western capitalist countries In essence, defining diagrams according to t-d graphs is similar to I-d graphs
The air conditioning diagram is designed to illustrate the process of air state changes on the I-d graph It is developed using initial climate conditions, facility specifications, and heat balance calculations to assess productivity and identify tasks that must be addressed before introducing air into the room.
Air conditioning systems can be represented through various diagrams, including straight diagrams, single-stage diagrams, and two-stage diagrams Each type offers distinct advantages, making it essential to select the most suitable option based on the specific project requirements and the significance of the air conditioning system.
A straight diagram, or non-circulating diagram, refers to a system where treated outdoor air is supplied directly to an air-conditioned space and subsequently expelled outside This method emphasizes the direct flow of air without recirculation, ensuring a consistent supply of fresh air while effectively managing heat and moisture levels.
The air outside, characterized by state N(tN, φN), enters the humid heat treatment chamber through an adjustable valve Inside the chamber, the air is processed to achieve a specific state O and is then transported by a fan through a pipeline The treated air flows into the room via vents, where it absorbs additional heat and moisture, transforming into state V This transformation follows the process beam εT = QT/WT, resulting in the air reaching state V.T (tT, φT) Finally, the air is expelled outside through the exhaust ports.
2.3.1.2 Single-stage air circulation diagram
One-level diagrams are extensively utilized across various fields due to their significant advantages, including the maintenance of hygienic conditions, simplicity in operations, and cost-effectiveness Consequently, these diagrams play a crucial role in driving economic development.
Figure 2.3: Single-level air diagram
Working principle: The air outside has a state N(tN,ϕN) with traffic LN through the air intake with adjusting valve (1), it is brought into the mixing chamber (3) to mix
46 | P a g e with the return air in the correct state T(tT,ϕT) with traffic LT from the air return vents
The mixture from state C will be directed to processing equipment, where it undergoes a pre-set program to achieve state O Subsequently, it is propelled by a fan through an air channel into room 8 The air exiting the blower mouth reaches a specified state.
When V enters the room, the conditions shift from V to T (tT, ϕT) due to excess heat (QT) and moisture (WT) Subsequently, some air is expelled while the return fan (11) draws in a portion of the air through the suction vents (9) This air is then reintroduced into the room via the return line (10) for proper mixing.
2.3.1.3 Two-level air circulation diagram
This diagram enhances economic efficiency and energy conservation in industrial workshops The two-level cyclic scheme effectively addresses the limitations of the single-level cyclic scheme, although it requires a higher investment cost.
The system operates by drawing outdoor air at state N through the air inlet (1) into the mixing chamber (3), where it combines with return air at state T from the air return vents (2) This mixture achieves a new state C before being processed by the air handling device (4) to reach state O The processed air is then blended with additional air from the first mixing chamber (C1) in a second mixing chamber (6) and is subsequently propelled by the fan (7) through the wind channels (8) into the room.
After exiting the blower mouth, the air at state V absorbs excess heat and moisture, transitioning to state T Some of this air is expelled through the intake vents to the outside, while a significant portion is drawn back into channel (12) to mix with fresh circulating air.
When calculating on an I-d graph, points N, T, H are determined more easily thanks to the calculated quantities Point O is determined by the cutoff point of φ 95% and process rays ε t = Q
There are no process rays on the wetware ε t Determining point O must use many different quantities such as:
- Current heat coefficient includes 3 types: room current heat coefficient, total heat coefficient and effective heat coefficient
- Dew point on the device
Using a hygrometer with the above quantities achieves higher accuracy Because the O point depends on the type of indoor unit it can reach 𝜑 = 90% đến 100%
2.3.2.1 Origin point and current temperature coefficient SHF (Sensible Heat
- The heat coefficient scale is now located on the right side of the humidity chart SHF (ε hf )
2.3.2.2 Current temperature coefficient of the room RSHF (Room Sensible Heat Factor) 𝛆 𝐡𝐟
- The room temperature coefficient is calculated according to the formula: ε hf = Q hf
Qhf: Total sensible heat of the room (without sensible heat of fresh air) [W];
Qaf: total latent heat of the room (without the latent heat component of fresh air) [W]
To establish ε hf scratch G-ε hf, draw a line from point T parallel to road G-ε hf, intersecting at φ = 100% The location of point V will be on the CT segment, with 𝜑 values ranging from 90% to 100%, influenced by the heat and moisture exchange efficiency of the indoor unit.
Example calculation for the living room of room M2-02 on the 19th floor
Total heat of the room (no sensible heat brought in by fresh air:
Total latent heat of the room (no latent heat brought in by fresh air:
2.3.2.3 Total heat coefficient GSHF (Grand Sensible Heat Factor) 𝛆 𝐡𝐭 ε ht = Q h
Qh The available heat component includes heat from fresh air and incoming air,
Qt: The total amount of current heat and pressing heat includes heat brought in by fresh air and wind, or total residual heat: 𝑄𝑡 = 𝑄0, W
The total heat coefficient is the inclination of the process jet from the mixing point
To determine the dew point of the device, start by calculating the output and marking the current temperature coefficient on the scale Next, connect the ray G - ε ht and draw a line from point H parallel to G - ε ht, which intersects at S, where 𝜑 = 100% The inflow point V is then identified as the intersection of lines HS and CT.
* Calculation example for the living room of apartment M2-02 on the 19th floor
- The existing heat component includes heat brought in by fresh baskets and drafts: Qa= 2337 (W)
Total latent heat and excess heat, including heat brought in by fresh air
The total heat GSHF coefficient is ε ht = Q h
Qh: sensible heat component (including sensible heat brought in by fresh wind
QhN has an external state N);
Qa: Latent heat component (including latent heat brought into QaN by fresh air in outdoor condition N);
Qt: Total excess heat is used to calculate cooling capacity Q0= Qt [W]
The total heat coefficient measures the angle of the process jet from the mixing point to the inlet point, playing a crucial role in the cooling and dehumidification of air within the condenser This process involves the integration of fresh air with recirculated air to optimize efficiency.
After establishing the ε ht on the current heat coefficient scale, we connect points G with ε ht From the blending point H, we draw a line parallel to G-ε ht, which intersects φ 100% at point S, representing the device's dew point The blowing point, V, must lie on both lines HS and CT, making V the intersection of these two lines.
The bypass coefficient (ε BF) represents the ratio of air that flows through a cooling system without engaging in heat and moisture exchange to the total air volume circulating within the system It is mathematically expressed as ε BF = G H.
In addition, it is also possible to calculate the detour coefficient once the points have been determined O≡V, H, S ε BF = t o − t s t H − t s = I o − I s
GH: Air flow passes through the cooler but does not exchange heat and moisture with the system [kg/s], so there is still the state of mixing point H
Go: Air flow passing through the condenser with heat and moisture exchange with the condenser [kg/s] and reaching the state O
G: total air flow through the system [kg/s]
Use the software to calculate the load for the project
In Vietnam, famous brands such as Daikin, Toshiba, LG, things have cold load software We can know as Trane’s Trace 700 Software, Daikin’s Heatload
Each software has its own advantages and disadvantages Our group chooses Daikin Heatload Software to calculate the project of Khai Vy
2.3.1 The steps to enter the load parameters for the project
Figure 2.10 is the symbol of Daikin Heatload Software on the Window screen
Figure 2.10: Software interface displayed on the computer
- The first we open the software Daikin Heatload
- Then click on “Project Outline”:
+ Project name: Khai Vy-Aparterment
+ City/ Country (City/ Country): Vietnam/ Ho Chi Minh
Figure 2.11: Image selects Outline of the software
- We choose the location of the apartment
Figure 2.12: Select the position of the software
- At “Design Data” => Temp & Humid => OK
We will enter the name of the room, floor, area, equipment in the ETC room, in the boxes in the figure 2.13
Figure 2.13: Enter the parameters to calculate the cold load
We enter the measurements: Fresh air intake, Lighting, Window type, etc In the figure 2.14
Figure 2.14: Enter the relevant parameters in detail
After entering the data for calculation, click “OK” => Main Menu
Then, Sum/Print to calculate the cold load for the rooms to be calculated
Figure 2.15: Pictures of all rooms that have entered data to calculate the load
We will see the calculation results in “Table of Room Heatload”
Figure 2.16: Show the results of the software loading of the software
To see the graph of the refrigerated load, you need to provide every day, click
Figure 2.17: The graph of the Cold Load provides every day
2.3.2 Compare the results of Dakin Heatload Software and Design Calculation results
Table 2.23: The table compares the results calculated by Dakin Heatload software and design calculation results
Error between Manual and Heatload (%)
Error between Design and Daikin Heatload (%)
CALCULATION OF VENTILATION SYSTEM
Ventilation system for the 1st floor reception hall
According to VN-5687 standards, according to Appendix G, we have air exchange coefficient in the hall AC/H = 3,2 times/h
We proceed to calculate the wind flow and the size of the wind pipe for the reception hall
The volume of the front floor hall
3.1.2 Determine the size of the pipe
Pipeline size in the reception lobby of the 1st floor
For corrugated iron tubes located in the main tube (open cone):
- Suitable maximum wind speed in the main duct ωmax = 2,4 (m/s) [Document 1 - Page 295]
- Therefore we can calculate the pipe section is:
2,4 = 0,22 (m 2 ) Based on Table 7.3 equivalent diameter dtđ (mm) of the rectangular cross section a×b According to [Document 1- page 299; 300]
We choose the main pipe size close to the most standard size: 1500x150
For soft tubes of the soft pipe section
- The wind through the soft tube of the lobby:
- Soft pipe area (round tube):
We choose the size of the AB branches close to the most standard size
- Wind flow out of the wind mouth:
3.1.3 Calculate the pressure loss on the duct
The pressure loss on the duct is determined ΔP=ΔPms +ΔPcb +∆Pmg (Pa) [3.3]
In there: ΔPms: Friction impedance on pipes, Pa ΔPcb: Local impedance on piping accessories (numbness, elbows, ) ΔPmg: Losses in the wind gate
According to the friction method, the pressure loss in a meter of tube is estimated to be between 0.8 and 1 Pa/m To determine the fan head, we calculate the longest impedance in the pipe, measuring from the fan to the furthest wind door, which experiences the greatest length and pressure loss.
The frictional impedance of the pipe is determined by the formula [Document 1- Page 298]: ΔPms = l ΔPi (Pa) [3.3.1]
In there: l: Wind pipe length, m ΔPi: Friction pressure loss over 1 meter, Pa/m
Calculate the loss of pipe pressure in the reception hall
We have a tube length: L= 1300 mm= 1,3 m
According to [Document 1 - Page 308] We have ΔPi = 0,8 1 Pa/m, (select ΔPi
=1pa/m) So the total loss due to friction on the exhaust pipe in the toilet is:
We have the formula for calculating local losses
𝜌: density of air For air within air conditioning ρ ≈ 1,2 kg/m 3 ω: Detailed wind speed passes through the tube m/s
Losses in the wind mouth: ∆Pcb= 0,73 Pa
Calculation of FCU losses connecting the tube:
The frictional impedance of the pipe is determined by the formula [Document 1- Page 298]: ΔPms = l ΔPi (Pa) [3.3.1]
We have a tube length: L= 7196 mm= 7,2 m
According to [Document 1 - Page 308] We have ΔPi = 0,8 -1 Pa/m, (choose ΔPi
=1pa/m) So the total loss due to friction on the exhaust shaft in the toilet is:
We can calculate the pipe section:
Based on Table 7.3 equivalent diameter dtđ (mm) of the rectangular cross section a×b According to [Document 1 - page 299; 300]
We choose the main pipe size close to the most standard size: 750x300
For losses in cone we have:
2 = 3 (N/m 2 ) The heel loss leads to the main tube using Ashrae Ducting Fitting software to calculate: ∆Pcb=8 Pa
Air exhaust system for toilets
3.2.1 Purpose of exhaust air absorption
Vacuuming in the toilet is essential for minimizing odors and reducing the concentration of harmful substances It effectively limits moisture and prevents the buildup of bacteria on clothing and equipment in the restroom area, contributing to a cleaner and healthier environment.
To minimize odors and harmful substances in the toilet area, it is essential to install an exhaust fan Selecting the right type of exhaust fan that meets both technical and economic standards is crucial Following this, it is important to calculate the airflow, speed, and pressure within the air duct system to ensure optimal performance.
3.2.2.1 Calculation of office toilet air flow
According to the standard VN-5687 according to Appendix G we have air exchange coefficient in the toilet AC/H = 12 times/h
There are 30 apartments from the 5th to 34th floor and each apartment has from
Apartments typically feature one to two bathrooms, each equipped with a common exhaust fan and a shared pipeline leading to the external exhaust hole To ensure optimal ventilation, we calculate the wind flow and determine the appropriate size for the exhaust pipe for each bathroom.
Calculate the ventilation of the toilet of apartments M2-02 floor 19:
The volume of apartment toilet M2-02 floor 19
3.2.2.2 Determine the size of the pipe
Pipeline size in apartment toilets M2-02 floor19
For corrugated iron pipes in the pipe AB:
- The maximum wind speed is suitable for toilets ωmax = 4 (m/s) [Document 1 - Page 295]
- Therefore we can calculate the pipe section is:
Based on Table 7.3 equivalent diameter dtđ (mm) of the rectangular cross section a×b According to [Document 1 - page 299; 300]
We choose the size of the AB branch tube close to the most standard size: 100x100
- We recalculate the wind speed of the branch tube AB:
0,00675= 4 (m/s) For soft pipe sections AB
- The wind through the soft tube of the toilet M2-02 19th floor
- The area of cross -section of toilet water drainage pipes on the 19th floor is
Based on Table 7.3 equivalent diameter dtđ (mm) of the rectangular cross section a×b According to [Document 1 - page 299; 300]
We choose the size of the AB branches close to the most standard size:
- We recalculate the wind speed of the soft pipe AB:
Similarly, we can calculate the ventilation in the toilet of other apartments on the 19th floor (typical floor) with the area and height of the toilet almost unchanged
3.2.2.3 Calculate the pressure loss on the duct
The pressure loss on the duct is determined ΔP=ΔPms +ΔPcb +∆Pmg (Pa) [3.3]
In there: ΔPms: Friction impedance on pipes, Pa ΔPcb: Local impedance on piping accessories (numbness, elbows, ) ΔPmg: Losses in the wind gate
To easily illustrate the formula, our group will calculate the theory of pressure loss on the exhaust of the toilet on the 19th floor
Here, according to the friction method, it should be lost on a meter of tube (according to [document 1 - page 307] we choose Δpi = 0,8 - 1pa/m) We proceed with
76 | P a g e the longest impedance on the pipeline from the fan to the furthest wind door with the largest length and the largest pressure loss to determine the fan head
The friction impedance of the pipe is determined by the formula [Document 1- Page 298]: ΔPms = l ΔPi (Pa) [3.3.1]
In there: l: Wind pipe length, m ΔPi: Friction pressure loss over 1 meter, Pa/m
Calculate the loss of pipeline pressure in the toilet on the 19th floor
We have a tube length: L= 11381 mm= 11,381 m
According to [Document 1 - Page 308] We have ΔPi = 0,8 1 Pa/m, (chọn ΔPi
=1pa/m) So the total loss due to friction on the exhaust shaft in the toilet is:
Calculate the loss through cone 90 0
For local losses through nodes, we consider it ∆Pcb = ∆Pms Then the formula for calculating local loss is calculated as follows:
Ltd is the equivalent length of the accessory
Ltd = a×d a is the ratio between ltd and the size d of quail is determined by the table 7.4 [Document 1- 301] and Table 7.5 [Document 1 - Page 302]
On the 19th floor, we see that the wind pipe system is designed with 2 elbows
90 o The rectangle has no wing R = 1,25d; We have a wind mouth 100x100 with w/d 1 and a = ltd/d = 7 and a wind mouth 100x100 with w/d = 1,5 and a = ltd/d = 7,25
So, ltd1 = a×d = 7 x 100 = 700 mm = 0,7 (m) ltd2 = a×d =7,25×100= 725 mm = 0,725 (m)
∆Pcb1 = (ltd1 + ltd2 )x ∆pi = (0,7+0,725)×1= 1,425 (Pa) Calculation of local and branched resistance, branches:
∆p: Local impedance failure, N/m 2 n: Dynamic column coefficient, check table 7.7 to 7.10 [Document 1] pd: Table, look up Table 7.6 [Document 1]
We choose 𝜔 1 find table 7.1 [Document 1 page 295]
At B there is a bell -shaped column
Look up Table 7.7 [Document 1] We have NB = 1,02
With ω1= 4 (m/s), we find table 7.6 với: Pd = 9,6 Pa
At D there is 1 heel leading to the main tube:
We use Ashrae wind pipe installation software to calculate:
Calculate the loss in the intake, exhaust and soft pipes
Looking at the Reetech exhaust catalog and Ashrae Ductfitting software, we have:
Damage at the entrance: Pmg = 3,6 Pa (Choose single-layer ventilation doors) Damage at the exit: Pmg= 12 Pa (Choose ventilation doors)
Loss of branch pipe: Pfd = 2,5 Pa (Use ASLI software ASHRAE Air Duct Installation)
So the total loss will be: ∆Pcb3 = 3,6 + 12 + 2,5 = 18,1 Pa
So the total local impedance loss of the air intake ventilation system of the toilet on the 5th floor will be: ΔP=ΔPms +ΔPcb +∆Pmg ,381+1,425+9,792+8+3,6+12F,2 (Pa)
Based on calculated pressure loss We proceed to calculate and choose a fan for the toilet area
Pressure loss for fan selection calculations: ΔP = ΔPq× m = 46,2 ×1,1 50,82(Pa)
So with Q = 27 l/s, ΔP = 50,82 Pa, Look up by device LiOA, Then the fan has the corresponding parameters, we have the following results table
Table 3.1 Toilet exhaust fan parameters in the building
Region Brand Model Flow (m 3 /h) Velocity
Corridor Smoke System
3.3.1 The purpose is to smoke the hallway
The smoke extraction system is designed to effectively remove toxic smoke during emergencies, facilitating safe evacuation through elevators and emergency exits, ultimately safeguarding human lives.
Hallway smoke exhaust pipes play a crucial role in preventing the spread of smoke to vital areas, such as engine rooms and locations housing sensitive equipment By effectively managing smoke dispersion, these systems help minimize damage and deterioration of valuable assets.
According to standards [SS553:2009 Singapore page 31] with corridor smoke absorption coefficient AC/H = 10 times/h
Calculation of smoke extraction in the 19th floor hallway
- Volume of corridor area on the 19th floor:
F: corridor floor area (m 2 ) h: hallway height (m)
- The volume of smoke that needs to be removed from the unburned corridor of one floor is:
- The volume of smoke that needs to be removed from the fire corridor of one floor is:
Q1: The amount of smoke is sucked out into the 1-story hallway when there is no fire (m 3 /h)
Q2: The amount of smoke sucked into the 1-story hallway when burning (m 3 /h)
The air-conditioned hallway features a smoke extraction system that activates solely during fire or explosion incidents, with the exhaust fan designed to ventilate the first-floor hallway only in such emergencies.
- According to [Document 1- Page 294], the recommended wind speed is suitable for the main air duct: ωmax = 4 (m/s) Therefore, we can calculate the pipe cross section as follows:
We choose the main shaft tube size as follows: 850x650
There is no smoke door on section AB 𝑄𝐴𝐵 = 2,145×0,63 = 1,35 (𝑚 2 )
Maximum wind speed suitable for branch duct: ωmax = 5 (m/s) [Document 1 – Page 294]
- Therefore, we can calculate the pipe cross section as:
We choose the size of the AB branch pipe: 900x300
Table 3.2: Corrugated exhaust pipe specifications
Pipe dimensions from the design company, mmxmm
Table 3.3: Compare mechanical smoke extraction flow
Floor Total calculated flow (l/s) Design flow (l/s) Error (%)
Pressure loss on the air duct will consist of 3 components:
∆𝑝𝑚𝑠: Frictional impedance on pipes, Pa
∆𝑝𝑚𝑔: Impedance at air vents and equipment, Pa
3.3.4.1 Pressure loss due to friction
Using the uniform friction method, we determine that the loss per meter of pipe remains consistent, with a pressure drop of ∆pi = 0.8 - 1 Pa/m (as referenced in Document 1, Page 307) To assess the fan head, we calculate the impedance along the longest duct, which runs from the terrace fan to the mezzanine exhaust, where the greatest length and pressure loss occur The frictional resistance of the duct is then calculated using the appropriate formula.
At: ΔPms: Total friction loss throughout the duct, Pa; l: Air duct length calculates loss, m;
∆pi : Friction loss per meter of pipe, Pa/m;
We have the pipe length: l = 3500,30+18300+1650+7045+3940= 135935 (mm)
According to [Document 1 - Page 300] we looked up ∆pi = 0,8 - 1 Pa/m, (choose ∆pi = 1 pa/m)
So the total friction loss on the chimney shaft for the corridor is:
Calculate losses through circular and rectangular elbows
For local losses across nodes we consider ∆𝑝𝑐𝑏 = ∆𝑝𝑚𝑠 Then the formula for calculating local loss is calculated as follows:
In there: ltd is the equivalent length of the accessories ltd = a×d
83 | P a g e a is the ratio between ltd and size d of the air vent
According to the design of the smoke exhaust pipe there is 90 0 Elbow sized
850x650 There are no flow guides R = 1,25d Therefore, it is necessary to check with a rectangular elbow 90 0 with: w d = 1.31 => a = l td d = 7.16
According to [Text 1 – Page 308] we have ∆𝑝𝑖 = 0,8 - 1 Pa/m, (usually take
So the loss through the elbow on the corridor chimney shaft is:
Local impedance of numbness, branching
- Local loss of numbness, branching, collection, opening, etc calculated according to the formula:
At: n: Dynamic water column coefficient, see table 7.7 to 7.10 [Document 1] pđ : dynamic pressure column, (Pa), Look up table 7.6 [Document 1]
- Spindle: has 1 tray 90 0 nA = 1,75 (because ω1/ω2 =1) pdA = 58,3 Pa (We interpolate with 𝜔 = 9,87)
- At B: there is 1 entrance 90 0 nA = 0,25 (Coefficient when entering current) pdA = 28,5 Pa (We interpolate with 𝜔 = 6,875)
Branch DE: have 1 cone nA = 1,02 (for pressure loss when shrinking) pdA = 29,5 Pa (𝜔 = 7 m/s)
So the local pressure loss from the fan to the farthest floor:
Calculate local losses at intake ports, exhaust ports, flexible ducts, and sound absorbers
Look up the ASLI air door catalog and the experimental calculation table using it
ASHRAE Duct Installation Software, we have it:
- The loss at the intake air outlet is 3.8 Pa/1piece (size Wxd= 650x450)
- Loss when dissipating sound absorption 55 Pa
So the local loss is:
So pressure loss ∆P on the air duct:
∆P = ∆pms + ∆pcb1 + ∆ pcb2 + ∆pmg = 135,935 + 4,654+ 139,24 + 397= 676,8 (Pa)
Corridor smoke extraction system is available: Q = Q1= 1930 (l/s) Δp = 676,8 (Pa) (compared to design 700 Pa, error 3,31 %)
We look at the catalog and Fantech fan selection software and can choose an axial fan with parameters like Appendix 3b
Because this building has 2 identical areas So we also calculate 1 area and then take it for the remaining area
Table 3.4 Corridor exhaust fan parameters
Location Brand Model Speed rpm
Stair pressure system
3.4.1 The purpose of pressurizing the stairs
The purpose of pressurizing stairs is to keep smoke and toxic gases away from emergency exits, thereby helping everyone in the building escape safely.:
- The system ensures human safety
- Limit fire spread: So that fire prevention and fighting can be carried out
To maintain effectiveness during a fire, elevator shafts and stairwells must be kept at differential pressure, preventing smoke from the affected floor from infiltrating these areas This is crucial for fire fighting efforts, regardless of whether an air conditioning system is in operation.
Protect assets: Limit smoke spread into areas with valuable equipment that is particularly sensitive to smoke
3.4.2 Stair buffer chamber pressurization system
For each staircase we must install a pressure boosting system for that staircase Some requirements of the pressure generation system according to ISO 5687:2010 and QCVN 06 -2020 standards are as follows:
To ensure optimal conditions in the elevator chamber, maintain a pressure of at least 20 Pa when the door is open and no more than 50 Pa when closed Additionally, control the wind speed through the door to a maximum of 1.3 m/s to prevent unwanted air from entering the elevator chamber.
The elevator's pressure is regulated by a pressure reducing valve (PRD) to ensure it does not exceed 50 Pa If the pressure rises above this threshold, the PRD automatically activates to lower the pressure, maintaining safety and efficiency.
- Small door opening force: Must ensure that elderly and weak people can open the door (no more than 110N)
Staircase doors are designed as fireproof barriers, capable of withstanding heat and flames for a minimum of 1-2 hours Equipped with hydraulic hinges, these doors ensure automatic closure, while the strong pressure within the stairwell contributes to their ability to close continuously, enhancing safety in emergency situations.
- Booster fans are given priority power and use fireproof cables
- Operation of the entire system will be controlled directly from the automatic fire alarm panel
The Khai Vy Block Mercury building features four staircases and eight elevators, including a direct elevator connecting the basement to the rooftop To ensure optimal performance, it is essential to calculate the pressure increase for both the elevators and the buffer rooms, leading to the division of the system into two distinct parts.
- System 1: Pressurize the elevator chamber in zone 1
- System 2: Pressurize the elevator chamber in zone 2
Because Systems 1 and 2 are the same So we will only calculate system 1 and it will also account for system 2
Table 3.5 Door and staircase parameters of the building
System Door style Size (mm) Acreage (m 2 )
3.4.4 Calculate the flow rate of the stair cage pressurization
Determine the air flow supplied to the stairwell to create a pressure difference to prevent smoke from entering the stairwell as:
- Traffic with 3 doors open at the same time
According to BS 5588 standard page 53 we have:
- Air flow to the outside due to air leaking through doors when all doors are closed
AE: Clearance area in pressurized space (m 2 )
P: Pressure level in pressurized space
Based on BS 5588 standard page 66, we get the gap area in the pressurized space as follows:
Table 3.6 Leakage area through door Door type Leakage area (m 2 )
Single door for pressurized space 0.01
Single door in the outer door frame 0.02
- The air flow blowing into the stairwell of system 1 is:
The suitable wind speed for the inlet is: ω = 13 (m/s) according to ISO 5687 standard
Therefore, we can calculate the pipe cross section as follows:
Based on table 7.3 - Document 1, choose the tube size closest to the standard size: We choose the main shaft tube 1000 mm x 800 mm = 0,8 (m 2 )
𝑆 = (m/s) Airflow for 40 mouths From there, the flow rate of each air outlet can be determined as follows:
We have a 1-door airflow, we look up the ASLI catalog, we get it:
Table 3.7 Specifications of single-layer air vents
System Acreage m 2 Size mm Speed m/s Pressure loss Pa
System Q(l/s) Calculate pipe size (mm)
Compare the design fan flow with the results we just calculated:
Table 3.8 Comparison table of stair pressurization flow
Pressure loss on the air duct will consist of 3 components:
∆P𝑚𝑠 : Frictional impedance on pipes, (Pa)
∆P𝑚𝑔 : Impedance at air vents and equipment, (Pa)
3.4.6.1 Pressure loss due to friction
We have traffic Q = 10,472 m 3 /s = 10472 l/s, diameter equivalent to the tube 1000x800 mm have dtd = 976 mm
According to [Text 1- page 300], we look up chart 7.24 and the loss per 1m of pipeline is 2,3 (Pa/m) System 2 is similar
The frictional resistance along the gas pipeline is determined according to the formula:
∆pms = l×∆pi (Pa) At: ΔPms: Total loss due to friction throughout the pipe (Pa) l: Air duct length calculates loss (m) Δpi : Friction loss per meter of pipe (Pa/m)
So the total loss due to friction is: ∆pms = 119 × 2,3 = 273,7 (Pa)
Local loss through air vents 90 0
For local losses through nodes, we consider ∆𝑝𝑐𝑏 = ∆𝑝𝑚𝑠 Then the formula for calculating local loss is calculated as follows:
∆p𝑐𝑏 = 𝑙𝑡𝑑× 𝑝𝑖 (𝑁/𝑚 2 ) At: ltd is the equivalent length of the accessory (m)
𝑙𝑡𝑑 = 𝑎× 𝑑 a is the ratio between ltd and size d of the air door
According to the design, the supply pipe has 2 elbows 90 0 There are no flow guides R=1,25d; We have 2 air vents w = 1000 mm and d = 800 mm ; w/d = 1,25 So we look up table 7.5 [Document 1]:
𝑑 = 7.125 ltd = a×d = 7.125×0,8 = 5,7 m According to [Document 1], we have pi = 2,4 Pa/m
So the elbow loss on the main shaft of the stair pressurization system is:
Loss of local numbness, branching, heels, shoes
The system uses pipes to spread sand on the surface 1000 x 800 mm So the local loss of numbness, branching, collection, and hole opening will be zero
=> Loss in air vents and sound absorption:
Looking at ASLI and ASHRAE Ductfitting's vent catalogs, we have: 35 single- layer vents to pressurize stairs: 22 Pa/1 piece
Low loss of 86 pa/1 piece
Total local loss: ∆p = ∆pms + ∆pcb1 + ∆pcb2 + ∆pmg = 856 + 9,74,2 + 616 1491,5 (Pa)
To address the challenge of a large head size, we opted to split System 1 into two fans, selecting smaller diameter fans for easier installation and maintenance Utilizing the flow and pressure loss parameters provided, we referenced the Fantech fan selection catalog and software to choose an appropriate centrifugal fan for the system (see Appendix 3).
Because zone 1 and zone 2 of this project have the same characteristics So we can have the same fan properties from system 1 to system 2
Table 3.9 Technical specifications of booster fan
IMPLEMENTATION OF AIR CONDITIONING SYSTEM USING REVIT
INTRODUCTION TO REVIT
Revit is a widely recognized software utilized by engineers and architects for project completion, offering robust support for Building Information Modeling (BIM) This program allows users to efficiently generate and visualize design concepts, streamlining the modeling process.
Professionals interested in Building Information Modeling (BIM) often turn to Revit software due to its remarkable efficiency in design representation Revit allows users to visualize designs as parameterized objects, enabling the storage of comprehensive data within a single model This data can then be utilized to generate countless perspectives, enhancing the overall design process.
Specifically, when there are any changes in the project, the corresponding relationships will also change automatically and synchronously (statistical tables, sections, views, plans, )
Revit software is also divided into three main parts:
❖ Revit Architecture for architectural design
❖ Revit MEP: design of mechanical and electrical systems, fire protection
SIMULATION OF AIR CONDITIONING SYSTEM BY REVIT
Use Revit software instead of CAD-2D software to clearly understand the air conditioning system of the building in Figure 4.2
Figure 4.2: Arrange equipment on the 1st floor
Figure 4.3: Arrange equipment of apartment on the 19th floor
Volume extraction in Revit is essential for managing input materials effectively It's crucial to base orders on actual conditions rather than solely relying on software data The initial overview provided by Revit, as illustrated in Figure 4.6, showcases three tables detailing the volumes of pipes, ducts, and fittings Due to the extensive length of these tables, only representative samples are presented.
Figure 4.4: Table of mass from revit
CONCLUSIONS AND RECOMMENDATIONS
CONCLUSION
The team conducted a thorough analysis of the cold load for a typical 19th-floor apartment (M2-02) by employing both manual calculation methods and Daikin's Heatload calculation software The manual load calculation yielded a result of 10.32 W, while the Heatload software provided a slightly lower result of 9.9 W.
W, the result of the calculated result Manual compared with the load calculation by
The Heatload software exhibits a maximum error margin of 4.07% For the reception hall's first floor, manual calculations yield a result of 39,365W, while the Daikin Heatload software reports 36,279W, resulting in a discrepancy of 7.84% between the two calculations.
In addition, we also use many other software to support calculations such as
The Fantech and ASHRAE teams have selected machines tailored to meet the project's specific needs Following thorough calculations, they determined that the error results fall within an acceptable range when compared to the project's design requirements.
This error may be due to the following reasons:
- Because in the process of measuring the area data, the area is not completely accurate
- Differences in the use of company standards versus group standards
Material and pipe sizes can vary significantly, leading to different calculations across companies Experienced designers often apply a redundancy factor to ensure the load is appropriate for the equipment selected for the project.
RECOMMENDATIONS
The project is currently under construction, and the fortunate team had the opportunity to visit the site We hope to gather more project information and compare it with actual data from the construction site, allowing us to gain valuable insights into calculations and audits This knowledge will be essential for advancing our work in the future.
Once again, we would like to express our sincere thanks to the professors of the
Department of Heat and Refrigeration Technology of Ho Chi Minh City University of
Technology and Education In particular, my team would like to express our sincere thanks to PhD Doan Minh Hung He has always accompanied us during the past time
[1] Nguyen Duc Loi (2009), “Gas Design Textbook of Air Conditioning Systems” Educational Publishing House, 339 pages
[2] Vo Chi Chinh “Textbook of Air Conditioning”, Hanoi Science and Technology Publishing House, 492 pages
[3] Standard VN 5687-2010 Ventilation and air conditioning
[5] Catalog of air outlets of REETECH
[7] Standard Singapore Code of Practice 13
[8] HTS GL05- Table of calculations based on experience by software ASHRAE Duct Fitting
[9] Vietnamese Standards on Fire Prevention 06 – 2020
APPENDIX Appendix 1a Khai Vy project
Appendix 2a Floor area to be cooled
Appendix 2b Table of state of living room
Appendix 2c Table of state of bedroom
Appendix 2d The status parameters of the moisture points
Appendix 2e Use Heatload’s Daikin software
Error between Manual and Heatload (%)
Error between Design and Daikin Heatload (%)
Appendix 2f Device used for aparterment floor 19th
Apendix 2g FCU used for floor 1st
Appendix 3a Using performance curves work
Region Brand Model Flow (m 3 /h) Velocity
Appendix 3d Technical specifications of booster fan
Appendix 3e Properties of fan AP0804KP9/23