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Tiêu đề Calculating, Checking And Developing To Use The Revit Software For The Air Conditioning System Of The Son Nam Restaurant Hotel Center
Tác giả Le Duy Khanh, Nguyen Chi Nguyen, Nguyen Vinh Khanh
Người hướng dẫn Assoc. Prof. Dr. Dang Thanh Trung
Trường học Ho Chi Minh City University of Technology and Education
Chuyên ngành Thermal Engineering Technology
Thể loại graduation project
Năm xuất bản 2023
Thành phố Ho Chi Minh City
Định dạng
Số trang 90
Dung lượng 3,55 MB

Cấu trúc

  • Chapter 1 Overall summary of air conditioning (0)
    • 1.1. The importance of air conditioning for people (21)
      • 1.1.1 Its importance for people (21)
      • 1.1.2. Its importance of manufacturing (21)
      • 1.1.3 Introduce some typical of air conditioners (21)
        • 1.1.3.1. Local air conditioning system (22)
        • 1.1.3.2. Central air conditioning system (23)
    • 1.2 The importance of the topic (26)
    • 1.3. Calculation and test objective of the restaurant’s air conditioning system (27)
      • 1.3.1 Introduce the construction (27)
      • 1.3.2. Basic of calculation and design of air conditioner (28)
        • 1.3.2.1. Requirements (28)
        • 1.3.2.2 Calculation Parameter for the project (28)
  • Chapter 2 Calculation (29)
    • 2.1 Choose calculation parameters (29)
    • 2.2. Heat load calculation method by the Carrier method (31)
    • 2.3. Calculate sensible heat and latent heat (32)
      • 2.3.1. Heat through the glass (32)
      • 2.3.2. Heat through the roof top (36)
      • 2.3.3. Heat through the baffle (37)
      • 2.3.4. Heat through the floor (41)
      • 2.3.5. Heat from the lamp (42)
      • 2.3.6. Heat through the machine (42)
      • 2.3.7. Heat transfer from people (0)
        • 2.3.7.1. Sensible heat by people Q 4s (43)
        • 2.3.7.2 Latent heat by people Q 4l (43)
      • 2.3.8 Heat from fresh air (44)
      • 2.3.9. Heat from leakage air Q 5 (45)
      • 2.3.10. Checking dew point temperature (46)
  • Chapter 3: Establish and calculate the air conditioning diagram (47)
    • 3.1. Select the air conditioning diagram (47)
    • 3.2. Determine factors (49)
      • 3.2.1. Bypass factor ε bf (49)
      • 3.2.2. Effective Sensible Heat Factor (ESHF) ε sef (50)
      • 3.2.3. Grand Sensible Heat Factor (GSHF) ε st (50)
      • 3.2.4. Room sensible heat factor (RSHF) ε sf (50)
    • 3.3. Determine parameters on the diagram (52)
    • 3.4. Cooling capacity (54)
    • 3.5. Calculate heat load by Trace 700 Software (55)
      • 3.5.1. Introduce Trace 700 Software (55)
      • 3.5.2. Calculation operations on software (55)
    • 3.6. Machine and equipment overview (56)
      • 3.6.1 Overview (56)
      • 3.6.2. Select indoor unit (57)
      • 3.6.3. Select outdoor unit (58)
    • 3.7. Compare capacity Trace 700 Software and company design (59)
  • Chapter 4: Calculate ventilation system for the project (60)
    • 4.1. Fresh air supply system (60)
      • 4.1.1. The purpose of fresh air supply (60)
      • 4.1.2. Determine the speed of air inside duct (60)
      • 4.1.3. Calculate fresh air flow supply (60)
    • 4.2. Determine dimension of duct (61)
      • 4.2.1. The dimension of mental duct dimension (61)
      • 4.2.3. The dimension of flexible duct size (63)
      • 4.2.4. Calculate pressure loss (63)
        • 4.2.4.1. Friction loss (64)
        • 4.2.4.2. Calculate local loss (64)
      • 4.3.2. Calculate exhaust flow rate (67)
      • 4.3.3. Calculate pressure loss (68)
        • 4.3.3.1. Calculate friction loss (68)
        • 4.3.3.2. Calculate local loss (69)
      • 4.3.2. Select fan (69)
    • 4.4. Exhaust smoke system (70)
      • 4.4.1. Calculate exhaust smoke for room (70)
      • 4.4.2. Calculate exhaust smoke in halls and corridors (71)
        • 4.4.2.1. Calculate total flow rate (71)
        • 4.4.2.2 Calculate for the gain (72)
    • 4.5. Calculation of pressurization (72)
      • 4.5.1. The purpose of creating pressurization (72)
      • 4.5.2. Working mechanism of pressurization system (72)
      • 4.5.3. Building parameters (73)
      • 4.5.4. Calculate pressurization (73)
        • 4.5.4.1. Calculate flow rate (73)
        • 4.5.4.2 Calculate gain box (74)
      • 4.5.5. Select air supply fan (74)
    • 4.6. Comparison between Software and Company design consultancy (76)
  • Chapter 5: Design Air-conditioning and duct by Autodesk Revit Software (77)
    • 5.1. Definition of BIM – Revit (77)
    • 5.2. Reasons to use Revit (77)
    • 5.3. Modeling air conditioning and ventilation systems for the Restaurant Center (77)
      • 5.3.1. Introduce Revit (77)
      • 5.3.1. Insert architecture Revit file into the project (81)
    • 5.4. Modeling (83)
    • 5.6. Mass Dissection (83)
  • Chapter 6: Conclusion and Requisition (85)
    • 6.1. Conclusion (85)
    • 6.2. Requirements (85)
  • Appendix 2.1: Heat transfer through the glass (87)
  • Appendix 2.2. Heat transfer through the rooftop (87)
  • Appendix 2.4. Heat transfer from the lamp (88)
  • Appendix 2.5. Heat transfer from the device (88)
  • Appendix 2.6. Heat from people (0)
  • Appendix 2.7. Heat from fresh air (0)
  • Appendix 2.8. Heat from leakage air (89)

Nội dung

COP: Coefficient of Performance SAF: Supply Air Fun EAF: Exhaust Air Fan ACH: Air Exchange per Hour RSHF: Room Sensible Heat Factor GSHF: Grand Sensible Heat Factor ESHF: Effective Sensi

Overall summary of air conditioning

The importance of air conditioning for people

In our life and production and human activities, temperature plays an important factor

Temperature conditions affect the surrounding environment and production needs

Temperature is a factor closely related to and affects the surrounding environment

If changes in ambient temperature, both increasing or decreasing, directly affect children people, daily activities and production (for example, if the ambient temperature is high, then Air humidity also increases, more water vapor makes people feel Heat and discomfort affect the spirit and ability to work as well as live active) With the growing social life, people are more strict with surrounding conditions Convenience and comfort come first,require a more comfortable and fresh temperature working and living space Air conditioning solutions with modern technologies are extremely necessary

In the field of refrigeration for people at work or in daily life, the use of Air conditioning makes it possible to balance environmental factors that are related to each other according to needs will help people to be comfortable and improve the quality of life Factors such as temperature, humidity, O2, CO2, air circulation rate can filter the combined dirt filtration and ion generation for the conditioned medium must be controlled at a balanced level The end Combining the above factors and applying advanced technology to air conditioners will help save money save energy, achieve the necessary requirements

With increasingly stronger development conditions, in addition to the need to serve people production demand plays a role in the development structure of the economy Therefore, not to mention the role that air conditioners play The air conditioning industry has great strides appear more and more in other professions such as medicine, pharmaceutical industry, seafood, seafood processing, and the introduction of air conditioners in other industries to ensure the quality of products, chemicals, components and equipment will have stable temperature and humidity requirements In addition, air conditioning is available at all locations All buildings, commercial centers, supermarkets, schools, offices

1.1.3 Introduce some typical of air conditioners

After a long process of formation and development, today's air conditioning system is more and more developed and perfected in terms of functions as well as diversity in terms of structure and design Today's air conditioning systems are no longer only for cooling purposes but also play roles such as increasing humidity, filtering dust, treating air, heating, etc

According to the concentration, we can divide the air conditioning system into 2 types:

Local air conditioners are single air conditioners installed in small spaces such as small offices, small restaurants, as well as cafes and apartments Some air conditioners are shown in Figure 1.1 – 1.3 Local air conditioners include two main types: type 2 window conditioner or multi-system split air conditioner

Figure 1.2 Multi-split air conditioner

Figure 1.3 Free blow packaged air conditioner Advan tages:

- The machines operate independently of each other, so it is easy to operate as well as repair and maintain

- The machine works with lightly active, not long expectancy

- Can't get fresh air, so it is necessary to add a fan to get fresh air

- Usually only applied to small projects that do not have too strict requirements on temperature and humidity

- Affects the architecture and aesthetics of the building due to the installation of many outdoor and indoor units on the wall.

 VRV System (Variable Refrigerant Volume)

VRV system is indicated in Figure 1.4 It consists of an outdoor unit and indoor units

- Because it is equipped with an inverter, it is possible to adjust the cooling capacityQ

- VRV air conditioners can solve the compressor oil recovery problem Periodically the VRV air conditioner will switch to oil recovery mode, during this time the thermal expansion valve will open and the compressor will run at high pressure to return oil from wherever oil accumulates in the system Therefore, the condenser of the VRV system can be placed more than 50m higher than the indoor unit as well as each indoor unit can be 15m apart, the length of the refrigerant pipeline between the two units is more than 100m long

- The system has high reliability, ease, flexibility in adjustment and maintenance through the control system displayed on the computer

- The outdoor unit is cooled by wind, so the working efficiency is not high, so it will depend on the weather The number of indoor units is limited, so it is only suitable for medium capacity systems If it is large systems, people will prioritize using Water Chiller or central air conditioning systems

- In terms of cost, in the past, VRV systems were often the highest among air conditioning systems However, nowadays the system has trended down and is cheaper than water cooled system

Figure 1.6 Water Chiller System center Water Chiller System Center (Figure 1.6) mainly includes:

- Machine which is cooled water (Water Chiller) or cooled water machine usually from 12˚C degree to 7˚C

- Heat source for heating used for humidity control and winter heating, usually provided by hot water boilers or resistors in FCUs

- FCU (Fan Coil Unit) or AHU (Air Handing Unit)

- Fresh air system, return air, air transport and distribution

- Sound absorption and reduction system

- System of dust filtration, sterilization and disinfection for the air

Water-cooled chillers usually don’t need replacement as often as air-cooled chillers do They aren’t exposed to outdoor elements such as rain, snow, ice, and heat, which makes them less vulnerable

The film coefficient is 10 to 100 times better in water-cooled chillers versus air- cooled chillers This means that water-cooled chillers transfer heat more efficiently The result to businesses is a savings on energy costs

Air-cooled chillers need to stay outdoors in an open space with plenty of fresh air to operate Water-cooled chillers stay inside buildings, which makes them ideal for companies that don’t have access to enough outdoor space

Water-cooled chillers use water as a refrigerant instead of toxic chemicals This makes them safer for people who have contact with them

Air-cooled chillers are cheaper than water-cooled chillers because they don’t require parts like cooling towers and condenser water pumps Many companies feel that the longer lifespan and savings on energy costs make water-cooled chillers worth the initial high investment, however

The extra parts in water-cooled chillers also make installation more of a hassle This can mean higher labor costs for installation of water-cooled chillers as opposed to air- cooled chillers

Businesses need to have a mechanical room to house a water-cooled chiller This is to ensure that the chiller will function properly with its cooling tower and extra components.

The importance of the topic

It is not difficult to realize that any new construction project today also needs an air conditioning system Especially in areas where the environmental temperature changes greatly and is not suitable for people's needs to live, study and work, the need for air conditioning is even greater Air conditioners have become familiar with the production process and life, present in all buildings, commercial centers, supermarkets, schools, offices, Not only that, air conditioners also plays an extremely important and indispensable role in specific working environments such as hospitals, factories, chemical research facilities and many other specific environments require a that strict atmosphere control system

Son Nam Restaurant is an integrated resort, hotel restaurant which is located at Nam Dinh Province The main role of this construction project is a restaurant for visitors who travel to Nam Dinh Province With the main purpose of being a restaurant, the need for a stable air conditioning system to create a green and clean space is extremely important

With the high demand of society, so my group have decided to choose this topic for our thesis called “Calculating, checking and illustrating by Cad and Revit software for air conditioning system of Son Nam Restaurant in Nam Dinh Province” by

AutoCAD 2D and Revit software” for this resort project In which the calculation is carried out by the Carrier method, and the LG calculation software is then compared with the parameters of the proposed item, calculating and selecting equipment for the entire system, and simulating the work with Revit software Through this project, the team has the opportunity to combine the learned knowledge to synthesize and serve the project, and at the same time, it is a better opportunity to help the team gain more practical experience in the implementation of the equipment design an air-conditioning system for a large building from which it is a good springboard for future projects Thereby, helping us be more confident in our future work and love our major more.

Calculation and test objective of the restaurant’s air conditioning system

The aim of this thesis is to calculate the restaurant’s data by two methods: Carrier and LG head load calculation, then compare it with the required parameter Arrange and install equipment for the project appropriately, to ensure cooling capacity for rooms and areas requiring air conditioning in the restaurant as well as other issues such as the cleanliness of the incoming air, quietness, air flow, Applying the technology to save energy, help optimize the performance of equipment, bring economic efficiency to businesses and investors, and strictly comply with state regulations and standards and regulations

Son Nam Restaurant (Figure 1.7) with the investor Son Nam Joint Stock Company (SON NAM JSC) The project is located on a beautiful campus, with the purpose of making a commercial space, an office and a wedding venue

By bringing the restaurant with modern, luxurious and environmentally friendly architecture and design, thereby creating the best space for visitors and changing the face of the city, becoming a highlight outstanding for the tourist area in SON NAM The project was built to fully meet the criteria of tourism and the needs of diners when choosing a place to eat

Figure 1.7 Son Nam Restaurant Center

1.3.2 Basic of calculation and design of air conditioner

 The air-conditioning system ensures the requirements for heat and humidity, standards and regulations, and ensures the existing architectural beauty of the building, especially without disrupting the architectural landscape of the building

 According to the requirements of the project

 Easy independent control for each separate area, high reliability, low operating and maintenance costs

 Compactly arranged into a system, convenient and easy to exploit and use, ensure aesthetics, and have development reserve

 It is necessary to calculate between investment costs and operating costs in order to provide the most effective solution

1.3.2.2 Calculation Parameter for the project

Ambient air speed has an effect on the intensity of heat exchange and metabolism (perspiration) between the human body and the surrounding environment When the speed is large, the moisture heat exchange rate increases

So when standing in front of the wind we feel cool and often the skin is drier than in a quiet place under the same conditions of humidity and temperature

Table 1.1.Parameters of temperature, humidity, air speed in the room

Table 1.2 Parameters outside and inside the building

Calculation

Choose calculation parameters

 Choose initial parameters for the construction

To have the initial parameters of the project, we need to consider the location of the place where the project is located, specifically here is the Son Nam Restaurant Center There are many ways and methods to calculate the cold load for the project, our team selected the method of calculating the cooling load according to the Carrier This method we need to calculate the cooling capacity Q0 by separately calculating the sum of the excess present heat Qht and the excess latent heat Qat of all heat sources that radiate osmosis and impact on the air-conditioned space.

To calculate for the air conditioning system, we need to determine the indoor comfort parameters as well as the outdoor conditions:

Noise level is considered an important factor causing environmental pollution, so it should be controlled, especially with comfortable air conditioning

The Ministry of Construction of Vietnam has issued TCXDVN 175 – 2005: Maximum permissible noise level in public buildings – Design standards Based on Table 1.3 &TCXDVN 175 – 2005: Maximum permissible noise level in public buildings – Design standards, we can choose the appropriate temperature and relative humidity according to Table 2.1 below:

Table 2.1 Indoor calculation parameters of air conditioning Son Nam Restaurant

 Fresh air standards and number of air changes

Fresh air standard is the amount of outside air needed to supply an air- conditioned room to ensure the amount of oxygen for humans to function normally, with the unit of measurement being m 3 /h/person or m 3 /h/m 2 of floor The amount of air supplied depends on the type of building, the function of the room and the number of people in the room

Table 2.2 Standard of outside air (fresh wind)

STT Construction Room name Area m 2 /person Fresh air standard m 3 /person m 3 /h/m 2

Living room Corridor area Hall

 Choose external parameters of the Restaurant

According to the standard, depending on the importance of the construction, the air conditioning system is divided into 3 levels:

Level 1: The air conditioning system must maintain indoor parameters in all ranges of outdoor humidity in both winter and summer Level 1 systems are used for particularly important construction

Level 2: The system must be able to maintain indoor parameters within a deviation of 200 hours a year Level 2 system is used for relatively important works

Level 3: The system must maintain indoor parameters within a deviation of no more than 400 hours a year Level 3 system is used in common buildings such as offices and houses

In fact, for most of the projects such as hotel, office, supermarket, library air

With the analyses above, based on the requirements of the investors as well as the characteristics of the construction so it is chosen to calculate is level 1 with the guarantee coefficient and the parameters of temperature and humidity are (table 2.3.)

Table 2.3 The outdoor calculation parameters for air conditioner

Local k level h I kJ/kg tN ᵒC ϕN

Heat load calculation method by the Carrier method

Based on the calculated parameters inside and outside the building and factors such as the covering structure, the number of people in the air-conditioned space, and the geographical location of the building, we can calculate the capacity cold for each room, each floor and the whole construction by Carrier method

In the Carrier method, the cold load Q0 is calculated based on the total excess heat

Qht and the total excess latent heat Qat of all heat sources (see Figure 2.1) and impact on the air-conditioned room

Heat loss due to radiation , enveloping , and radiant heat only show sensible heat Particularly, the heat radiated by people , fresh air and leak air consists of two components, sensible and latent

Figure 2.1 The diagram of sensible and latent heat sources by Carrier Method

Calculate sensible heat and latent heat

We have the formula which is approximately exact to the heat through the glass:

Q # / The amount of heat radiated instantaneously through the class into the room

F / The surface area of glass window with steel frames (0 )

R 2 / Solar radiant heat through the glass door into the room Because our air conditioning system operate during sunny hours so we choose % & =% &,34

' /Effect coefficient of altitude above sea level. đ) / The dew point temperature.

,, / the influence coefficient of clouds

*+ / frame influence coefficient Wood frame *+ = 1, steel frame *+ = 1,17 , / glass coefficient

R C / Solar radiation arrives outside the glass

R 8 / Solar radiation through glass into conditioned space α 8 , α ? /The absorption coefficient of the glass and of the curtain τ 8 , τ ? / Coefficient through the glass and curtain ρ 8 , ρ ? / Reflectance coefficient of glass and screen

The project use basic glass and medium color (according to table 4.3 and

Table 2.4 Characteristics of the curtain and sun coefficient -

Table 2.5 Characteristics of the color and sun coefficient -

 Effect coefficient of altitude above sea level ( ' )

H /The height of the ground floor above to sea level (12m)

So base on the formula we calculate ' =1.000276 (we choose ' =1)

 The dew point temperature > HIA

10 " 0.13 >2.4A t L / dew point temperature of outside air , ℃

With paraments of outside air in Nam Dinh Province, J M = 32.8℃ , M = 53.4%

Figure 2.3 The result of dew point temperature

So base on the formula we have HI = 0.8336

 The influence coefficient of clouds ( ,, )

When sky have no clouds : ,, = 1, when sky have clouds : ,, = 0.85

We choose sky have no clouds : ,, = 1

Depend on the color and type of glass different from the base glass Checking the table 4.3 [1/page 133]

Because the glass that they use for the project is basic glass So we choose , = 1

Including the effect of base glass when there is a curtain inside the glass, when there is no curtain inside - = 1 We use normal curtain in this project so - = 0.65 Checking table 4.4 [1/page 134]

The construction uses steel frame so we take F

Because the air conditioning system works continuously, Nam Dinh city is located in the northern hemisphere, at latitude 20° Checking the table 4.2 [1/page 134]

Table 2.6 The maximum amount of radiation X penetrates through the glass window

Considering the coefficient: g L / Average area density of the entire covering texture of walls, ceillings, floors, kg/0

G’ / volume of walls whose outer faces are exposed to solar radiation and of floors above ground level

G’’ / volume of walls whose outer faces are not exposed to solar radiation and of floors that are not above the ground

The mass of 10 wall (0.2 m): 1800 " 0,2 = 360 kg/

The mass of 10 floor (0.3 m): 2400 " 0,3 = 720 kg/0

So we have Y ) ≥ 700 search the table 4.6 [1/page 136] we have nt suitable with four different directions at 7.pm (table 2.5)

Table 2.7 Factor instantaneous effect factor (with curtain)

Direction East West South North

For example: Calculate heat through the glass Q11 through the glass for the VIP room 1

2.3.2 Heat through the roof top

Heat through the roof top have three types of forms:

In case 1, the room to be calculated is located between floors in an air-conditioned building So ∆J = 0 e!H = 0

In case 2, if the room above the air-conditioned room needs to be calculated as a non- air-conditioned room, then:

In case 3, if the roof has solar radiation, the amount of heat entering the room includes the following two components: solar radiation and the temperature difference between the indoor and outdoor air

For the calculated construction is the Restaurant Center:

Ground floor, first floor and second is in case 1 => = 0

And the third floor is in case 3, so we only need to calculate heat transfer through roof top in the third floor and k we have in the table 4.9 [1/page 142]

Q21/ Heat flows into the conditioned space through absorption by the roof and the difference in air temperature between outside and inside

∆J td / The equivalent temperature difference is determined by the following expression:

∆J đ = (tN - tT) + h i "j k l k (2.9) ) / Solar radiation absorption coefficient (check table 4.10 [1/143]) tN– Temperature outside the air-conditioning area tN = 32,8℃ tT – Temperature inside the air-conditioning area tT = 25℃ n M / Absorption coefficient n M = 20 k / heat transfer coefficient through roof

Table 2.8 Heat transfer coefficient through the top roof k, W/m 2 K

Describe Thickness of the air

300mm thick concrete ceiling, 25mm thick cement-sand mortar layer with bitumen layer, 797 kg/m 2

Figure 2.4 Energy balance on roof surface

For example: Calculate heat transfer through the rooftop Q21 for the VIP room 1 in the second floor

Because the air-conditioning room is located between floors in the Restaurant so:

Other areas will be shown at the appendix 2.2 (appendix table 2.2)

To calculate the total heat transfer through the wall, we need to calculate the heat transfer through the following structures including:

Q r - heat tranfer through the wall, W

Q ] - heat tranfer through the door, W

Q s - heat transfer through the glass, W k u - Corresponding heat transfer coefficient of wall, door, windows, (W/m K) ∆J - The different temperature between indoor air and outside air (℃)

F u - The area of wall, door, windows, glass (m 2 )

 Heat tranfer through the wall q q = g q "$ q " ∆J (2.11)

If the wall contact directly to the wall: ∆J q = J M / J &

If the wall contact indirectly to the wall: ∆J q = >J M / J & A.0,7

Heat tranfer through the wall by the formula: g = 1 n1 M + Σ xy p p + 1n &

>W/0 KA n M = 20 W/m 2 K, the heat transfer coefficient outside when contact directly with the outside air n M = 10 W/m 2 K, the heat transfer coefficient outside when contact indirectly with the outside air n & = 10 W/m 2 K , the heat transfer coefficient inside x p / Thickness of the material layer of the wall structure y p / Heat transfer coefficient of i th material layer of the wall structure

Table 2.9 Material parameter of the wall project

Parameter Cement mortar Ordinary brick Cement mortar y (W/mK) 0,93 0,81 0,93 x>0A 0,015 0,22 0,015

The wall of the construction consists of 1 layer of bricks inside and outside covered with 2 layers of cement The thickness of cement mortar x ' = 0,015m, the layer of the brick inside: x z = 0,22m

- The temperature difference between the inside and outside when the wall is in indirect contact with the air is:

+ The heat transfer coefficient outside when contact directly with the wall 220mm the outside air: g = 1 n1 M + Σ xy p p + 1n &

+ The heat transfer coefficient outside when contact indirectly with the wall

+ The heat transfer coefficient outside when contact indirectly with the wall

+ The glass wall inside the Restaurant:

Table 2.10 Parameters of the glass

Type of glass Thickness (mm) Coefficient k (W/m 2 K)

The result of Q22w will be showed at the appendix 2,3 (Appendix table 2,3)

For example: Calculate the heat transfer through the wall for VIP room 1

Area of the wall 200mm contact indirectly to the wall: F = 15,87 m 2

Area of the wall 100mm contact indirectly to the wall: F = 10 m 2

Temperature contact indirectly to the wall:

Heat transfer through the wall in the VIP room 1 is: q = 15,87.5,46.1,98 + 10.5,46.2,5 = 308 (W)

 Heat tranfer through the door (

∆J / Temperature difference inside and outside the house, tn- tT, K k / Coefficient of heat transfer through doors, W/m 2 K

Table 2.11 Coefficient of heat transfer through wooden doors for the building

Material Thickness (mm) Heat transfer coefficient

The result of Q22d will be showed at the appendix 2,3 (Appendix table 2,4)

 Heat transfer through the glass,window }

∆J - Temperature difference inside and outside the house, tN- tT, K g } - Coefficient of heat transfer through glass, W/m 2 K

To determine the heat transfer coefficient through glass we need to consider the structure of glass layer, while we need to consider how many layer and distance is so we search the table 4.13 [1/page 147] to find kg

Table 2.12 Coefficient of heat transfer through glass door and window

Because the heat transfer through the glass in summer is higher than winter so we only need to calculate for summer to figure out cooled load for this project

The result of Q22w will be showed at the appendix 2.3 (Appendix table 2.3)

F / Area of the floor, (m 2 ) k / Heat transfer coefficient through the floor (W/m 2 K) Check table 4.15[1/page

∆J/ The different temperature between indoor air and outside air (℃)

Table 2.13 Coefficient of heat transfer through floor for the building

Structure of the floor or roof Characteristic of the surface

(mm) Season Lined with paper and carpeted The thickness of the concrete floor is 150mm with cement 25mm 175

Figure 2.6 The structure of the floor Consider three cases to calculate:

+ In case 1, the floor is on the ground, we choose heat transfer coefficient k of concrete floor with a thickness of 175 mm, ΔJ = J M / J &

+ In case 2, the floor is in the basement or in the non-conditioned room, ΔJ = 0.5>J M / J & A + In case 3, the floor is between two conditioned rooms then = 0

Fluorescent light : Q = Σ1,25 q ( F , P (2.15) qd is the standard of shine per m 2 of floor select qd = 12 according to [1/146]

If the total capacity of the light is unknown , the standard direction value is 10 ÷

The formula of the heat transfer of the light:

Q / the total heat of light ni/ the instantaneous effect factor of the light , check table 4.8 [1/page 138]

Because the lights work continuously, so ni = 1 ns / the simultaneous effect factor , only used for the buildings and the large air conditioning work sites

Because the construction is tall building, hotel, center so ns = 0,4

For example: Calculate heat from lamp Q for the VIP room 1

So the heat from lamp Q for the VIP room 1 is:

The heat transfer by machines and electrical tools such as televisions, radios, computers, hair dryers, irons, etc., are types that do not use electric motors, so they can be counted as radiant heat sources of lights

• p / the power capacity of the machine, W

For example: In the VIP room 1 in the second floor can be equipped devices such as:

Q4s - the sensible heat by people, W

Q4l - the latent heat by people, W

Q4s = nt" n " qs n – The quantity of people in the air-conditioning room nt – Human instantaneous effect factor According to the table 4.17 [1/150] for the Restaurant we choose nt = 1,3

For example: Calculate sensible heat by people for the VIP room 1 in the second floor:

So we have: n = 24 people qh = 70 [W/people] according to Table 4,18 [1/151], table 2,11

Q4l = n "‚ ƒ n – The quantity of people in the air conditioning room

‚ ƒ / The latent heat by one person, W/people

For example: Calculate latent heat by people for the VIP room 1 in the second floor: n = 24 people ql = 60 W/people according to the table 2.11

The room temperature of all air-conditioned spaces is specified as 24℃, check table 2.15 [1/page 151] to find sensible and latent heat of the areas to be calculated

When calculating, we choose the heat emitted by an adult person (Table 2.11) to get the best result when calculating the load

Table 2.14 Radiation heat from an adult

Activity level Place Heat from an man Average heat Temperature need to be conditioned

Office active Hotel 140 130 qs ql

We have formula for the sensible heat:

Qsh = 1.2 " ! " „ " (tN-tT) Latent heat:

Qlh = 3 " n " „ " (dN-dT) n / The number of people in the air conditioning room l / The air flow required for 1 people per second, l/s

Total heat that fresh air bring to:

QH = Qsh + Qlh dN, dT/ Specific humanity, g/kg;

For example: if the temperature in the room is 25˚C, relative humidity 60% so we have specific humidity is 11,9 kg/kg

Figure 2.7 The result of specific humanity For example: Calculate heat from fresh air for the VIP room 1 in the second floor n = 24 people l = 15 l/s according to table 4,19 [1/152] dN = 16,4 g/kg dT = 12 g/kg

So we have: QH = Qsh + Qlh = 3369 + 5140 = 8509 (W)

The air-conditioned space is sealed to limit heat loss, and at the same time controls the amount of air supplied to the room, but there is still the phenomenon of air leakage through windows and doors The above factors are to ensure that the air conditioning system will work properly with the calculation, and also provide enough cold for the building or construction

The sensible heat from leak air:

Q5S = 0.39"… " V " ( tN-tT ) The latent heat from leak air:

Q5L = 0.84"… " V " ( dN-dT ) The total heat:

V / The volume of the room, m 3

…/ The experience factor Check the table 4.20 [1/page 153], depends on the room volume (V/m 3 ) to find …

For example: Calculate the Heat from leakage air Q5 for the VIP room 1 in the second floor n = 24, … = 0,7 because the V = 232 m 2 < 500

Dew condensation occurs when the wall temperature is less than the dew point temperature of the air Therefore, to avoid dew condensation for rooms, we need to check for dew condensation on the walls of the rooms Because the temperature and humidity in the room is the same for all rooms, we check for dew condensation on the wall common to all

To prevent dew condensation, the heat transfer coefficient kt of the wall must be less than the maximum heat transfer coefficient kmax

Dew condensation on the wall of the enclosure occurs:

+ At the inner surface of the wall (the surface in contact with the air inside the air conditioner) in the cold season

+ At the outer surface of the wall (the surface exposed to the outdoor air) in the hot season The condition for condensation to occur is that the heat transfer coefficient k of the wall is equal to the maximum value of the heat transfer coefficient kmax, kt = kmax kmax is determined by: k ?†‡ = α C t C / t LC t C / t 2 >W/m KA α C – Coefficient outside, α C = 20 >W/m KA when contact directly to the air and α C = 10 >W/m KA when contact indirectly tN, tT – The temperature outside and inside the house tsN – The dew point temperature, with the outside parameters t = 32,8 so we have tsN = 22,57℃

Establish and calculate the air conditioning diagram

Select the air conditioning diagram

The air conditioning diagram is established based on the calculation factors of the balance of heat and humanity, and at the same time it is necessary to satisfy the conditions of comfort for people and the weather conditions of the considered building

The one-stage diagram is the most widely used because the system is relatively simple, ensuring hygienic requirements, uncomplicated operation and high economy, due to the heat utilization of the recirculating air Therefore, the cooling capacity and drying capacity are reduced compared to the straight diagram This diagram is used both in the field of comfort conditioning and air conditioning of workshops producing electronic, optical, computer components

The two-stage diagram is often used in comfortable air conditioners when the blowing temperature is too low, not meeting hygiene standards, so a second mixing chamber and a system of air extraction to this mixing chamber are required Investment and operating costs increase It is also widely used in production workshops when it is necessary to adjust both temperature and humidity at the same time, such as textile

One - stage diagram is used popularly because it is considerably simple

1 Air intake 2 Return louver 3 Mixed air chamber

4 Humid equipment 5 Blast 6 Duct liner 7 Diffuser 8 Room air conditioners

9 Recovery diffuser 10 Absorbed air duct 11 Recovery fan 12 Air outlet

Figure 3.2 One-stage diagram on I-d diagram

Principle operation: The outside air has a •>J M , M ) state with GN flow through air inlet with adjustable valve (1), it is taken into mixed air chamber (3) to mix with recovery air has Ž>J M , M ) state with GT flow from return louver (2) Mixed mixture have H state will transfer to humid equipment (4), where it is processed according to a program O state and the blast transfer by duct liner to room air conditioner The air after going from diffuser have V state to excess heat room QT and excess moisture WT and excess moisture WT

Advantages and disadvantages of the diagram:

Advantage: Due to the utilization of heat from the recirculating air, the cooling capacity and drying capacity are reduced compared to the diagram It is the most widely used because the system is relatively simple, ensuring hygienic requirements, uncomplicated operation and high economy.

Disadvantages: The scheme has air recirculation, so the investment cost increases compared to the straight diagram The system requires a level 2 dryer to heat the air when hygiene conditions are not satisfied

Table 3.1 The parameters at N and T point from calculation parameter

Start point G is determined on humidity chart at t = 24˚C and = 50% Sensible heat coefficient scale ε h is put in

Determine factors

3.2.1 Bypass factor ε bf εbf (Bypass Factor) is the ratio between the amount of air that pass through the indoor unit but does not exchange humidity heat with the total amount of air blowing through the indoor unit εbf = T T •

GH – The air flow through the indoor unit but does not exchange humid heat with the unit, so there is still the state of the H mixing point

GO – Air flow through indoor unit with humid heat exchanger and reaches O state

G – The total air flow through conditioner, (kg/s) This factor is chosen from table 4,22[1/162], εbf = 0,1

Figure 3.3 Bypass Factor is showed in t-d diagram

3.2.2 Effective Sensible Heat Factor (ESHF) ε sef εsef = ‘ i’“

)•– = )– + ε —W " )+ ƒ•– – Effective room latent heat ƒ•– = ƒ– + ε —W " ƒ+ ε —W –Bypass Factor

Qsh – Sensible heat factor from fresh air, W

Qlh – Latent heat factor from fresh air, W εsef = ‘ i“ V ˜ ™š "‘ i›

3.2.3 Grand Sensible Heat Factor (GSHF) ε st εht = ‘ i“

Qs – Component of sensible heat (included Q5s state), W.

Ql – Component of latent heat (included Q5l state), W.

3.2.4 Room sensible heat factor (RSHF) ε sf

It is the ratio between the visible and latent heat components of the room, excluding the visible and latent heat components carried by fresh and intrusive air into the conditioned space. εsf = ‘ i“

Qsf/ Total sensible heat of the room (no sensible heat from fresh air), W.

Qlf/ Total latent heat from the room (no latent heat from fresh air), W.

Figure 3.4 Room sensible heat factor line (RSHF)

Figure 3.5 t-d diagram is illustrated relationship between RSHF, GSHF, ESHF

For example: Calculate factors for the VIP room 1 in the second floor:

Effective Sensible Heat Factor (ESHF): εsef = ‘ i“ V ˜ ™š "‘ i›

>‘ i“ V˜ ™š "‘ i› AV>‘ ”“ V˜ ™š "‘ ”› A= > V , AV> Œ V , V , A= 0,75 Grand Sensible Heat Factor (GSHF): εht = ‘ i“

‘ i“ V‘ ”“ = ‘ i“ V‘ i› œ‘ i“ V‘ i• •V>‘ ”“ V‘ ”• A= > V AV> Œ V V A= 0,60 Room Sensible Heat Factor (RSHF): εsf = ‘ i“

Determine parameters on the diagram

To determine the cooling capacity, the air flow to the indoor unit and the temperature to blow in, we must have the initial calculation parameters One-stage cyclic diagram with points N, T, H, O, V, S with current heat coefficients, detour coefficients are introduced in Figure 2.11, calculation of one-stage diagram is followed by:

- Determine total sensible heat and latent heat of the conditioned space brought in by the fresh air.

- Determine total latent and lack heat of required air-conditioning

- Draw TS parallel to line G- εhef cut = 100% at S

- Through S draw a line parallel with G-εht cut line NT at H, we determine the point mixed point H

- Through T draw a line parallel with G - εhf cut line SH at O After neglecting heat loss from fan and liner duct we have V ≡ O is blown point

Figure 3.7 Psychometric chart The processes take place on the t - d graph:

T – H: is the process of returning air from the room

H – S: It is the process of exchanging humidity heat with the cold coil of the FCU

S – T: Is the process by which the air exchanges heat with the air in the room to obtain a state T

 Determine air flow through indoor unit

After calculate and determine parameters above, we need to check the negative room temperature and blowing temperature

If ∆J ¡& < 10 K then hygiene standard and conduct to calculate air flow

If ∆J ¡& > 10 K then not hygiene standard then we need to use some alternative solutions to decrease negative blowing temperature because blowing temperature is too low it will influence to people health After checking then conduct to calculate Check sanitary conditions:

∆tVT = tT – tV = 16 – 12 = 4 < 10 => Hygiene standard

Cooling capacity

The cooling capacity of the air conditioning system can be checked using the formula:

G - Mass flow rate of air passing through the indoor unit, (kg/s)

G =¦.L (kg/s) ¦ – The density of air, (¦ = 1,2 gY/0 A

To determine air flow through indoor unit we use formula: § = , > ‘ i’“ ¨ ‰ i A.> ‰h ™š A= , > ‰ , A.> ‰ , A • = 723 (l/s) = 0,723 m 3 /s

)•– – Sensible efficient factor, W tT, tS – Dew temperature and indoor temperature, ℃

Calculate heat load by Trace 700 Software

Air conditioner companies entering the Vietnamese market today are very diverse such as Daikin, Toshiba, Trane, LG, Most of the companies have their own thermal load calculation software Daikin has DACCS – HKG (Daikin HeatLoad), Trane has Trace 700,…

Step 1: Open the Trace 700 Software

When we click to Open, it will open an existing Project, New to create a new Project

Figure 3.10 Add parameters for rooms

After completing parameters, we click to Calculation and View Results

Figure 3.11 Calculate and View Results

Machine and equipment overview

Air-conditioning system of Son Nam Restaurant is using Trane ducted floor- mounted air conditioner with many its advantages

- Screw compressors are available from 5 to 20 tons with excellent reliability and high efficiency

- All units are run 100 percent test before leaving the production line

- Trane split systems offer single and dual compressors enabling the right equipment for the job application and saving on operating costs

Choosing indoor unit is based on cooling capacity in each area Must choose a machine with sufficient cooling capacity required in the correct working mode calculated If due to the requirements of the investor or the important nature of the project, sometimes it is necessary to have a reserve cold capacity The total cooling capacity of the selector (Q0N) must be greater than the capacity refrigerant (Q0) is in working mode because the actual cooling capacity is not as fixed as the value stated on each machine

Figure 3.12 Ceiling concealed duct from Trane

Figure 3.13 Trane indoor unit parameters

Based on parameters that we calculate as well as compare to the company design, we choose air indoor unit:

Table 3.3 Indoor unit parameters for rooms

Floor/area name Model name Cooling capacity (BTU)

Small feast room TWE120BD 120,000 380-415 R22/R4

Big feast room TWE240BD 240,000 380-415 R22/

From catalogue we can choose standard outdoor unit, each outdoor unit can integrate with many indoor units

We choose outdoor unit for first, second and third floor: According to cooling capacity calculation, we choose 6 outdoor unit with 240kW cooling capacity and 1

Compare capacity Trace 700 Software and company design

Table 3.4 Comparison of manual cooling load

Tolerance between Carrier and design company

Tolerance between Trace 700 and Carrier

Calculate ventilation system for the project

Fresh air supply system

4.1.1 The purpose of fresh air supply

The fresh air intake vent is used to bring fresh air into the building The fresh air intake vent is an important part of the HVAC system because it helps to provide fresh air for the occupants of the building and also helps to keep the air inside the building clean

4.1.2 Determine the speed of air inside duct

Airspeed is the quantity of primary interest because it directly affects the system When the air speed is high, the fan power is large, the noise level will be large, but the advantage of the duct size is small and vice versa Therefore, we need to consider to determine the appropriate wind speed in the pipeline for the system to operate stably, but also to achieve the appropriate noise level along with economic efficiency

4.1.3 Calculate fresh air flow supply

The fresh air supply system for the public area is provided by PAU located on the mezzanine floor Exhaust air from the air-conditioned spaces is returned to the heat recovery wheel to exchange heat with the fresh air outside the house, saving operating energy

The fresh air valves use the motor-mounted type to control the fresh air supply to the air-conditioned spaces to save energy

Flow rate is determined by the formula: unit m 3 /h

N – The number of people in air conditioning area (person)

„ M – Fresh air flow rate need to provide for 1 person in 1 hour (m 3 /h person) According to appendix F

For example: Calculate fresh air flow rate for The Rental Area at the first floor

Follow appendix F – TCVN 5678: 2010 With density 1 person is 3 m 2 /person then the number approximate is • = = 80 and fresh air flow rate for 1 person is 30 m 3 /h

We calculate fresh air flow rate is provided by formula 3.1 is:

 Comparing parameters between Calculation and company design

Table 4.1 Fresh air flow rate parameters

Determine dimension of duct

4.2.1 The dimension of mental duct dimension

 We use Duct Size Software to calculate size for duct

Figure 4.1 Duct Checker Pro interface

- Flow rate is flow air rate through liner duct, (L/s)

- Heat loss is loss pressure through duct, (Pa/m)

- Velocity is the speed go through duct, (m/s)

- Equivalent diameter is diameter of duct, (mm)

- Duct size is dimensions length and width of air duct (mm)

For example: Calculate the dimension of duct for The Rental Area 4 in the first floor Based on Cad drawing, we determine fresh air duct dimension of the area

Figure 4.2 The dimension of duct

Figure 4.3 The parameter dimension from Duct Checker Pro Software

Table 4.2.The dimension of duct for building

Area/Duct section Air flow rate

4.2.3 The dimension of flexible duct size

The dimension is determined by: d = ơ đ.¯ - , (3.2)

Q – Flow air rate through duct, (m 3 /s) v – Speed velocity, (m/s) Choose v = 3 ÷ 3,5 m/s, we choose v = 3,2

For example: Calculate dimension of flexible duct size for the restroom in first floor With Q = 63,8 L/s = 0,0638 m 3 /s we have d = °4Q π v = °4.0,0638π 3,2 = 0,15 0

So d = 0,15 m we select flexible size diameter is ∅150mm

Table 4.3.Dimension of the flexible duct

Big feast and lobby area

Air duct pressure loss includes two component:

∑ ´ ,) – Friction loss on pipe (Pa)

∑ ´ 'z – Pressure loss (Pa) Pressure loss is calculated by ASHRAE Duct Fitting Database

Friction loss of air ducts in fresh air supply system will create impedance and be calculated by appendix [2]

∑ ´ ,) – Loss due to friction in duct (Pa)

∆´ – Loss due to friction is calculated on 1 meter (Pa/m), ∆´ = 1 Pa/m

For example: Calculate friction loss on air supply duct in the second floor ( we only consider the highest loss of the duct) So we have:

The total length of fresh air duct is 26,5m, so the friction loss is: ả ´ ,) = „ ∆´ = 26,5.1 = 26,5 àe Table 4.4 Friction loss in fresh air supply

Floor Section Length Friction loss

Friction loss in the duct ∑ ã áạ

Local loss of the circle and rectangular elbow is:

∆´ 'z : The local loss through circle and rectangular elbow, Pa; l a] :The equivalent length of the duct, Rectangular duct: l a] = e " H ( , is determined by table 7.5 [1/page304], with d a] is selected from table 7.3 [1/303]

Local loss for tee, wye, receiver:

∆p = ! " ´>ẵA n: the coefficient pressure, detemined from table 7.7 to table 7.10 [1/307] p: velocity pressure, Pa

+ To make it easier during the calculation process, we will calculating the local loss on ASHRAE Duct Fitting Database

Figure 4.3 ASHRAE Duct Fitting Database interface

To local loss to connector we assume ∆´ 'z = ∆´ ,) [1/page 374]

So the formula to calculate local loss is determined by:

∆p \ằ = l a] " ∆p u l a] - The equivalent height (table 7.4 and 7.5 [1/375]) l a] = e H a - the ratio of l a] and the dimension d of the connector

Table 4.6.Local loss through connector 90°

Figure 4.4 Fresh air second floor duct liner second floor in CAD drawing

 Calculating local loss due to the wye

 Calculating the local loss due to the receiver:

The local loss through the receiver designed on the supply air duct above the fresh air supply system is calculated according to Table 7.7 [1/page305]

Most of the fresh air duct have angle a

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