1. Aircon Design OT

AirConditioningSystemDesign 1

CHAPTER I

Introduction

HVAC (heating, ventilation, and air conditioning) is the technology of indoor and

automotive environmental comfort. HVAC system design is a major sub discipline

of mechanical engineering, based on the principles of thermodynamics, fluid mechanics,

and heat transfer. Refrigeration is sometimes added to the field's abbreviation as

HVAC&R or HVACR, or ventilating is dropped as in HACR (such as the designation of

HACR-rated circuit breakers).

HVAC is important in the design of medium to large industrial and office buildings

such as skyscrapers and in marine environments such as aquariums, where safe

and healthy building conditions are regulated with respect to temperature and humidity,

using fresh air from outdoors (Wikipedia).

In our daily life situations, air-conditioning system place an important role. It serves

as a sense of relaxation, gives comfort to human bodies, regulates the temperature in

working places and many others. Generally speaking, air-conditioning system is the way

of conditioning the air insid e of a system to provide the necessary quality of air. This

research aims to provide the way of designing an air-conditioning system by the use of

duct system and the calculation on proper derivation of sizes in each duct to be able to

balance the flow of air.

Air conditioning generally is understood to mean the simultaneous control of

temperature, relative humidity, air motion, air distribution, and ventilation within an

enclosure. Air-conditioning system are used in theaters, churches, auditoriums, schools,

restaurants, offices, homes, etc., to produce of effect comfort for occupants by

maintaining a temperature and relative humidity which will lie in the so-called comfort

zone (Kent, 1895).


CHAPTER II

CALL CENTER AIR CONDITIONING SITE

Isometric View Layout

The Figure shows the plan of call center site where an air conditioning system is to be installed, the air conditioning site dimensions are needed for the computation of cooling load of the system.

Figure 2.1: Layout for Air Conditioning Area

AirConditioningSystemDesign 3Top View Layout

Figure 2.2 shows the layout of the call center site in top view. The measurements in meters will be used for the surface area that is essential for calculating the heat loads in the walls.

Measurements in meters

Figure 2.2: Floor Layout of the Air Conditioning Site

AirConditioningSystemDesign 4Glass Door

The maindoor is made up of a double glass door with patch and fittings, opposite slide opening, it is made of tempered glass 9 millimeters thick to acquire higher heating resistance.

Figure 3.3: Glass Door Dimensions


CHAPTER III

COOLING LOAD CALCULATION

This chapter focuses on the cooling load computations that are present in the following area for application of air conditioning system. It includes heat transmissions on walls, windows and doors. Also the heat gain from the lightings, people, solar heat, appliances and many others. These values are needed for acquiring the appropriate capacity of an air conditioning unit.

I. Sensible Heat Loads

a. Thermal Transmission

These are the heat transferred through the structure due to temperature difference from the environment, from high temperature to low temperature for air conditioning system.

Exterior Walls

Table 3.1: Specification of Exterior Walls

Material Description , /

,/ Cement plaster 16 mm 1.39 Light weight aggregate

200 mm 0.38

Internal conductance

Surface emissivity of 0.9

0.120

Outside conductance

Heating season, 6.7 m/s

0.029

source: Refrigeration and Air Conditioning by: Stoecker pg.68


Figure 3.1: Wall insulation materials

Thermal Coefficient

10.02 0.12 1.390.0162 0.38

.

Ceiling Insulations

Table 3.2: Specification of Ceiling Insulations


Material Description /, /

,/ Gypsum board 16mm 1.39 Concrete (sand and gravel)

200 mm 0.18

Air Space 0.170 Internal conductance

Surface emissivity of 0.9

0.120

Outside conductance

Heating season, 6.7 m/s

0.029


Figure 3.2: Ceiling insulation materials

Thermal Coefficient

...... .

Window Glass

Table 3.3: Specification of Window Glass

Material Description U, Single glass 6 mm, heat absorbing 5.9


Thermal Coefficient

. Doors

Table 3.4: Specification of Glass door

Material Description U, Tempered glass door 9 mm 5.7

source: http://www.aisglass.com/flat_tempered.asp#5


Figure 3.3: Glass door materials

Thermal Coefficient

.

Summary of Thermal Coefficients The summary of all the necessary thermal heat conductivity is tabulate in Table 3.5 for the computation of heat load.

Transmissions , Exterior walls 0.47 Ceiling insulation 1.91 Window glass 5.90 Glass door 5.70

Heat Gain on the Exterior

In the computation of heat load in the exterior sides, the following values are needed namely, thermal conductivity, ambient temperature, desired temperature and cross sectional area.

AirConditioningSystemDesign 9Temperature Standards

Table 3.6: Temperatures Standards for Design

Classification Temperature ( C ) (a) Ambient temperature 35 (b) Desired temperature 25

source: (a) http://mb.com.ph/node/357745/heat-wave-not-likely (b) Refrigeration and Air Conditioning by Stoecker and Jones pg.65

Transmission Dimensions

Figure 3.4 shows the diagram of the transmission blocks involve in thermal heat gain of the exterior sides. The orientation are presented base on its sides for computation purposes.

Figure 3.4: Transmission Diagram

Table 3.7: Dimension of Transmission Materials

Transmission Block Length, m Width, m Area ( m2 ) Wall A and C 30 6 180 Wall B and D 80 6 480 Ceiling 80 30 2400 Windows 3 2 6 Doors 3 1.5 4.5


Transmission Areas

Nomenclature: A ~ Area

Exterior Walls:

A exterior wall = A wall A + A wall B + A wall C + A wall D

A Wall A = 180 4(A window)

A Wall A = 156 m2

A Wall B = 480 4(A door)

A Wall B = 462 m2

A Wall C = 180 4(A window)

A Wall C = 156 m2

A Wall D = 480 8(A window) 2(A door)

A Wall D = 423 m2

A exterior wall = 156 m2 + 462 m2 + 156 m2 + 432 m2

A exterior wall = 1197 m2

Ceiling:

AC = 80 x 30

AC = 2400 m2

Windows:

A window = A window + no. of glasses

A window = 6 m2 x 16

A window = 96 m2

Doors:

Adoor = 2 x Adoor

Adoor = 6 x 4.5 m2

Adoor = 27 m2


Table 3.8: Summary of Transmission Areas

Transmissions Total Area ( m2 ) Exterior wall 1197 Ceiling 2400 Window glass 96 Glass Door 27

Heat Gain

For the computations of heat gain through external walls the following formula for Q will be used. The total value of heat load in the system will be used to acquire the appropriate capacity of an air conditioning unit.

Nomenclature: Q ~ Heat Gain, W

U ~ Overall thermal coefficient; W/m2 0C A ~ Area of the wall, ground floor, floor, or roof; m2 T ~ Thermal difference, 0C

Exterior Walls

1.7715 119735 25 21204.86 Ceiling

1.91 240035 25 45,840

Windows Glass

5.9 9635 25 5664

Glass Door

5.7 2735 25 1539

AirConditioningSystemDesign 12Tabulated Exterior Heat Gain

Table 3.9: Heat Gain in the ExteriorTransmissions Heat Gain ( W )

Exterior Walls 21204.86 Ceiling 45840 Window Glass 5664 Glass Door 1539 Total 74247.86 Watts

b-1. Solar Load through Transparent Surface

In a transparent surface of a glass the heat load produce came from the solar radiance that passes through the transparency quality of the glass. The figure 3.5 shows the area of the sun light expose for a certain amount time and respective direction.

Figure 3.5: Solar Radiance in the Glass


Nomenclature: Q ~ Solar heat gain SHGF ~ Solar heat gain factor SC ~ Shading coefficient CLF ~ Cooling load factor A ~ Sunlit area ~ Wall azimuth angle ~ Solar altitude angle ~ Solar azimuth angle ~ Angle of aertical plane normal to the wall makes with south d ~ Depth recess of the window glass y ~ Depth of shadow cast horizontal projection above window x ~ Width of the shadow cast by vertical projection depth

Design Parameters

The design parameter shown in Table 3.10 provides the following design conditions for the computation of solar heat gain through the glass.

Table 3.10: Design Parameters for Solar Heat Gain

Variable Description Critical Date(a) April 20Location(b) 15N Latitude Critical Time(c) 3:00 pm Solar Altitude Angle, (c) 46 Solar Azimuth Angle, (c) 271 Wall Azimuth Angle, 32 Depth, d 97 mm

source: (a) http://newsinfo.inquirer.net (b) http://www.mapsofworld.com/lat_long/philippines-lat-long.html (c) Carrier Handbook of Air Conditioning System Design

AirConditioningSystemDesign 14Window Glass Specification

Table 3.11: Specification of Window Glass

Variable Description Type Heat absorbingShading Translucent, Light

Venetian BlindsThickness 6 mm Dimension(L x W) 3m x 2m

source: Refrigeration and Air Conditioning by Stoecker and Jones pg.76

Shading from Side Reveal

tan 97tan 32 60.61

Shading from Top Reveal

tan cos

97 tan 46cos 32 118.44

Sunlit Area

2 0.118443 0.0.06061 .

AirConditioningSystemDesign 15Solar Heat Gain Factor

Table 3.12: Tabulation of Solar Heat Gain Factors

source: Air Conditioning Principles and Systems by: Edward G. Pita

228

.

Shading Coefficient (Table 4-11, Stoecker, p.76)

. Cooling Load Factor (Table 4-12, Stoecker, p.77)

.

718.24 0.530.725.532 .

AirConditioningSystemDesign 16b-2. Solar Load on Opaque Surfaces

The solar heat gain for an opaque wall is described as a portion of solar energy that is reflected and the remainder is absorbed. In the energy absorbed some is converted and some radiated to the outside. The remainder of the absorbed solar energy is transmitted to the inside by conduction and temporarily stored.

Solar Heat Gain through an Opaque Wall

Nomenclature: Qow ~ Solar heat gain through the opaque wall Uw ~ Heat transfer coefficient of the wall CLTD ~ Cooling load temperature difference A ~ Surface area t1 ~ Inside temperature t2 ~ Outside temperature

Table 3.13: Tabulation of Heat Gain through the Opaque Walls

Direction U ( )

CLTD Area (m2)

TemperatureInside (C)

Temperature outside (C)

Heat Gain (W)

A South 1.77 16 156 25 35 6074.64 B East 1.77 20 462 25 35 21261.24 C North 1.77 8 156 25 35 3865.68 D West 1.77 11 423 25 35 12728.07

Total 43929.63 source: Refrigeration and Air Conditioning by Stoecker and Jones pg.82

Total Solar Heat Gain through Opaque Walls

.


c. Heat Gain through Infiltration

The air infiltration is the unwanted entry of the outside air directly inside the building, resulting from natural forces, such as wind and buoyancy due to the temperature difference of the environment the in system.

. Nomenclature: ~ Heat gain through infiltration, kW ~ Rate of infiltration air through opening doors

~ Outside temperature ~ Inside temperature

3600

680303600 1

4 1.23 4 35 25 49.2

d. Heat Emission from Occupants

The human has its heat and this heat is necessarily included for the computation of heat gain by the system. The more people a system possesses the more heat is being gain.

AirConditioningSystemDesign 18Nomenclature: ~ Heat gain per person ~ number of occupants ~ Cooling load factor per person

Table 3.14a: Design Conditions

Variable Description Type of Space Office Activity Office work Working time 12 hours Occupancy 10 m2/occupant

source: Refrigeration and Air Conditioning by Stoecker and Jones pg.73-74

Table 3.14b: Heat Emission from the Occupants

Heat Gain per Person, W

No. of Occupants

Cooling Load Factor

Heat Emission (Watts)

150 384 0.92 52992 source: Refrigeration and Air Conditioning by Stoecker and Jones pg.73

Total Heat Emission from the Occupant

e. Heat Gain from Electric Lights Lighting produces heat that is also calculated for the cooling load. A light level of

500-750 lux is usually sufficient, depending on the difficulty of the visual tasks done in the factory (Graham, 1984).

Nomenclature: ~ Lighting Capacity, Watts ~ Ballast Factor ~ Cooling Load Factor for Lighting ~ Number of lightings

Table 3.15: Lighting Material Specifications


Variable Description Lighting capacity 40 W Light level 750 Lux, Lumen/m Led light lumen 3050 lm Area 30m x 80m = 2400 m Ballast factor (a) 1.25 Cooling load factor (b) 1.0

Source: (a) Air conditioning principles and systems by Pita p. 137;

(b)http://www.amazon.com/dp/B002CZ15FK/ref=asc_df_B002CZ15FK2173156?smid=A2E4DH1S65

GTFP&tag=nextagusmp0403791-20&linkCode=asn&creative=395105&creativeASIN=B002CZ15FK

Table No. shows the specification of the lighting materials that is used with an

additional factor for lightings. Also the selected design of the light level is 750 lux. The lux (symbol: lx) is the SI derived unit of illuminance or illumination. It is equal to one lumen per square metre. Lux is the symbol for light level which is the basis of the design of the lighting load of the cold storage. ( Brillianz Company UK, 2006)

Total lumen = light level x area = 750 lumen/m x 2400 m = 1,800,000 lumen =

,,/ 590.16 590lamps

401.251.0590 f. Heat Gain from Appliances

The appliance also emits heat as it consumes the electricity to produce power. This heat gain is also essential for achieving the total heat gain in the system. The common appliance present in the system is the computers which has a common wattage of 300W as stated in ASHRAE 2011 and there is a computer in every cubicle.

. ) Power Consumption = 300 Watts (ASHRAE 2011)

No. of Appliances = (12)(2)(3)(2) = 144 cubicle + 4 units = 148 Computers

300148 2. Latent Heat Loads

AirConditioningSystemDesign 20a. Latent Heat from Infiltration Air

Nomenclature: ~ Rate of infiltration air, m3/s ~ Humidity ratio outdoor air ~ Humidity ratio indoor air

Table 3.16: Humidity Ratio Conditions Type Humidity Ratio (grains/lb)

Indoor 106.40 Outdoor 131.60

Source: Psychometric Tables and Charts by: Stoecker

0.68 0.68 35.315

0.68 6.45 35.315

60/ 131.6 106.40 234196/ b. Latent Heat from Occupants

Table 3.14b: Heat Emission from the Occupants

Heat Gain per Person, W

No. of Occupants

Cooling Load Factor

Heat Emission (Watts)

150 384 1 57600 source: Refrigeration and Air Conditioning by Stoecker and Jones pg.73

Total Latent Heat from Occupants

Total Cooling Load


74247.86 1515.66 43929.63 49200 52992 29500 44400 68632 57600

422017.15

422017.15 3516.7

CHAPTER IV


AIR DISTRIBUTION SYSTEM

This chapter is mainly about the methods of supplying appropriate air conditioning system for the building site, including the various computations for volume flow, mass flow and duct diameters. These values are relevant in providing an efficient distribution of air in the scope of the system.

Figure 4.1: 3D Ducting Layout

A. Duct Sizing

In duct sizing there are many options or methods for the design of duct. The most accurate in the two methods is the equal friction method where the friction in the main duct follows all throughout the latter part of the duct system which is the same.

Table 4.1: Design Conditions for Duct Sizing

Variable Description Main duct velocity (a) 4 7 m/s Main branch velocity 3 6 m/s Density of Air (b) 1.18425 kg/m3Sensible heat gain 295.78515 kW Temperature of room 25C Temperature of supply air 35C

source: (a) http://www.engineeringtoolbox.com/equal-friction-method-d_1028.html (b) Refrigeration and Air Conditioning by: Stoecker and Jones

Determination of Volume Flow Rates:


.

Nomenclature: ~ Volume flow rate, m3/s ~ Total sensible heat gain, kW ~ Density of air, kg/m3

295.785151.184251.006235 25

. Main Duct Dimensions

The main ducts are the ventilation mechanism that holds the total air volume flow of the whole system. It is the key in finding the appropriate machine capacity of the fans to be able to supply the whole air conditioned area.

Nomenclature: A ~ Cross sectional area of duct Q ~ Volume flow rate of air ~ Average velocity, (5.5 m/s for average speed)

Using two main ducts for the system with equal performance we arrive

24.822

12.41

12.415.5

2.26


4

42.26

. Equivalent square duct dimensions

1.61 . .

For the orientation of diffusers, one main duct is composed of 20 diffusers that are placed proportionally with each other. Considering two main ducts are to be used in the design for a total number of 40 diffusers to ventilate the system.

12.4120

0.6205/ 2 12.41 20.6205 . /

The formula for computing the next duct dimension is the same as the calculations above, the rest of the values follows. In addition, to acquire the various volumes of the duct assume the same friction losses in each resized duct and use the friction loss chart for their values. (Carrier Air Conditioning System Design p.190)

AirConditioningSystemDesign 25Duct Sizing Dimensions

In the sizing of the ducts the tabulated values in Table 4.2 are essential for providing the right volume flow of air to maintain comfort cooling from the occupants.

Figure 4.2: Friction Loss Chart

The Friction Loss Chart shown in Figure 4.2 is a useful Chart in getting the velocity in the equal friction method. It is a three way process of acquiring variables. Specifically air volume flow, velocity and diameter of duct


Table 4.2: Tabulated Duct Dimensions

Duct Air Volume Flow m3/s

Velocity m/s

Duct Diameter, m

Rectangular Length, m

Rectangular height, m

A 12.410 5.691 1.666 1.5 1.454 B 11.169 5.640 1.588 1.5 1.320 C 9.928 5.488 1.518 1.5 1.206 D 8.687 5.335 1.440 1.5 1.085 E 7.446 5.030 1.373 1.5 0.987 F 6.205 4.929 1.266 1 1.259 G 4.964 4.573 1.176 1 1.085 H 3.773 4.319 1.055 1 0.874 I 2.482 3.963 0.893 1 0.626 J 1.241 3.252 0.697 1 0.382

Main Branch Dimension

For the computation in sizing of ducts for the main branches consider same air volume flow in each diffuser. Tabulated values of the dimension in each branch are shown in Table 4.3.

Nomenclature: ~ Air volume flow for the branch ~ Velocity of the branch

1.2414.5

0.275

40.275

.

AirConditioningSystemDesign 27Equivalent Rectangular Dimensions

0.275 .

Table 4.4: Main Branch Dimensions

Branch Air Volume Flow

Diameter

Rectangular Length

Rectangular height

A1 1.241 0.592 0.524 0.524 A2 1.241 0.592 0.524 0.524 B1 1.241 0.592 0.524 0.524 B2 1.241 0.592 0.524 0.524 C1 1.241 0.592 0.524 0.524 C2 1.241 0.592 0.524 0.524 D1 1.241 0.592 0.524 0.524 D2 1.241 0.592 0.524 0.524 E1 1.241 0.592 0.524 0.524 E2 1.241 0.592 0.524 0.524 F1 1.241 0.592 0.524 0.524 F2 1.241 0.592 0.524 0.524 G1 1.241 0.592 0.524 0.524 G2 1.241 0.592 0.524 0.524 H1 1.241 0.592 0.524 0.524 H2 1.241 0.592 0.524 0.524 I1 1.241 0.592 0.524 0.524 I2 1.241 0.592 0.524 0.524 J1 1.241 0.592 0.524 0.524 J2 1.241 0.592 0.524 0.524

B. Pressure Losses

The pressure losses are the opposing force that is cause by friction. Friction is a mechanism that resists or opposes the direction of the force, these values are needed to acquire the total pressure loss in the system which is an essential variable in computing the machine capacity of an air handling unit.

AirConditioningSystemDesign 281. Converging Duct System

Converging air duct has a gradual decrease in size or dimensions.

Nomenclature: ~ Pressure loss, pa ~ Velocity of air at point 2, m/s ~ Cross sectional area at point 1, m2 ~ Cross sectional area at point 2, m2 ~ Density of Air, kg/m3

1.1845.640

2 2.1801.810

1

. The rest of the system follows and the values are tabulated in Table 4.5.

Table 4.5: Tabulation of pressure losses in duct

Duct Density (kg/m3)

Velocity (m/s)

Diameter (m)

Area A(m)

Area B(m)

Pressure loss(Pa)

AB 1.184 5.691 1.666 2.180 1.981 18.835 BC 1.184 5.640 1.588 1.981 1.810 17.834 CD 1.184 5.488 1.518 1.810 1.629 16.853 DE 1.184 5.335 1.440 1.629 1.481 14.981 EF 1.184 5.030 1.373 1.481 1.259 14.386 FG 1.184 4.929 1.266 1.259 1.086 12.383 GH 1.184 4.573 1.176 1.086 0.874 11.045 HI 1.184 4.319 1.055 0.874 0.626 9.300 IJ 1.184 3.963 0.893 0.626 0.275 6.653

Total 115.617

AirConditioningSystemDesign 292. Pressure Drop for Sudden Contraction

When the duct size is abruptly reduced in the direction of flow in a duct section a sudden contraction occurs. The flow of patterns consists of a separation of the fluid from the wall upon entering the reduced sectional area.

Nomenclature: ~ Contraction Coefficient

1.1845.640

2 0.010

. The rest of the values follow and the tabulation for pressure drop for sudden contraction is shown in Table 4.6.

Table 4.6: Tabulation of Pressure Drop For Sudden Contraction

Duct Velocity m/s

Density kG/m3

Area 1 m2

Area 2 m2

Cc Pressure Loss (Pa)

A 5.691 1.184 2.180 1.981 0.909 0.193 B 5.640 1.184 1.981 1.810 0.914 0.168 C 5.488 1.184 1.810 1.629 0.900 0.220 D 5.335 1.184 1.629 1.481 0.909 0.168 E 5.030 1.184 1.481 1.259 0.850 0.466 F 4.929 1.184 1.259 1.086 0.863 0.365 G 4.573 1.184 1.086 0.874 0.805 0.728 H 4.319 1.184 0.874 0.626 0.716 1.733 I 3.963 1.184 0.626 0.382 0.610 3.793 J 3.252 1.184 0.382 0.275 0.720 0.948

Total 8.782 Pa

3. Turns or Elbows

Most common elbows used in duct system are 90 degree turn which accumulates pressure losses as the air pass through.


For Geometric Factor refer to (Refrigeration and Air Conditioning by: Stoecker and Jones

. .

.

. .

4. Friction Loss

As the air travels through the duct system there is a corresponding pressure drop opposing force in a unit of length which is needed to get the total pressure drop.

Nomenclature: ~ Length of duct, m ~ Pressure drop per meter, Pa/m ~ 1.0 Pa/m (Carrier Handbook of Air Conditioning)

121/

Table 4.7: Tabulation of Friction Loss in a Length of Duct

Duct Length (m)

Pressure Drop (Pa)

A 12.723 12.723 B 4.232 4.232 C 5.000 5.000 D 5.000 5.000 E 10.000 10.000 F 13.000 13.000 G 5.000 5.000 H 5.000 5.000 I 5.000 5.000 J 3.986 3.986

Total 68.941 Pa

AirConditioningSystemDesign 31Total Pressure Drop

115.617Pa 8.782Pa 79.12 68.941Pa

. A total of 272.46 Pa of pressure drops in one main duct. Consider two main ducts in the system with equal specification and measurements.


CHAPTER V

AIR CONDITIONING SYSTEM

This chapter provides the necessary specification mechanism that is appropriate to the air conditioning system to ba applied. The machineries must be compatible and reliable to make the system a comfort air conditioning area.

A. CHILLER

A chiller is a machine also known as a heat exchanger which removes the heat from a liquid by vapor compression or absorption refrigeration cycle to main the comfort temperature of air desired. In the Air Conditioning System it is appropriate to choose two chillers with same specification to provide the corresponding cooling load that the system requires. So when the other chiller is malfunctioning. The other chiller is still operating.

Table 5.1: Water Cooled Chiller Source source: http://www.aquaair.net/HighCapacityChillerSystems.pdf

AirConditioningSystemDesign 33Chiller unit specification OM60-4VIHD Cooling Capacity

60 tons [ 720,000 BTU/H ] [ 180,000 KCAL/H ] at 45/ F ( 7.2/ C ) leaving water temperature and 55/ F ( 12.8/ C ) returning water temperature. Chiller unit flow rate will be approximately 180 gpm. Condenser flow rate ( each ) is to be approximately 60 gpm entering at a maximum temperature of 90/ F ( 32/ C ). All ratings are at a fouling factor of 0.0005. Heating Capacity `54 Kw [ 184,410 BTU/H ] [ 46,103 KCAL/H ] of total heating capacity at 120/ F ( 48.9/ C) leaving water temperature and 100/ F ( 37.8/ C ) returning water temperature. Construction & Ratings

The chiller unit shall be constructed in accordance with ARI Standard 590-86 and shall comply with all applicable NEC and ASME codes for water cooled chillers. Compressors

The chiller unit will have four, 15 ton Bitzer semi-hermetic compressors. Each compressor will be equipped with suction and discharge valves. Input voltage to the compressor motor will be 208-3-60. Power consumption of each compressor is approximately 14.1 kW each. Refrigerant to be used is R-22 . Capacity Control

Chiller unit capacity control will be achieved through the use of four variable frequency drive ( VFD ) units, one for each compressor. The VFD will vary the compressor motor speed from a maximum of 100% of capacity to a minimum of 70%. The VFD requires an input power supply of 208-3-60. The maximum output power will be 208-3-60 to the compressor motor. The VFD output will be regulated by a 4-20ma signal to the VFD from the PLC. The VFD voltage/frequency output will be varied based upon chilled water outlet temperature. The VFD will also control the compressor motor so that there is no current inrush, during starting, above the motor's standard running amperage. Cooler

The unit is equipped with four plate style heat exchangers, each of 15 tons capacity. Each plate heat exchanger has a single water and refrigerant circuit. Construction of the unit is of #316 stainless steel. The material used to braze the plates together is copper. Maximum test pressure for both circuits is 635 psig. Each plate will be individually insulated with 1/2" thick closed cell insulation. Condenser The unit is equipped with four shell and tube marine condensers. The shell is constructed of ASME spec SA-53 steel pipe. Shells are shot blasted and cleaned before assembly. Tubes are high performance enhanced surface seamless 90/10 Cupro-Nickel tubes to ASME spec SB-359. Tubes are roller expanded into double grooved tubesheets

AirConditioningSystemDesign 34to assure tight joints. Tubesheets are 90/10 Cupro-Nickel to ASME spec SB-171 Alloy 706. Tube supports are quality steel plug welded to the shell. Heads are cast bronze with integral pass partitions, ASME spec SB-62. Gaskets are die-cut providing effective sealing between tubesheets and machined heads. The refrigerant side is constructed and tested in accordance with Section VIII, Division 1 of ASME Code for unfired pressure vessels. Shell side design pressure ( refrigerant side ) is 350 psig at 250/ F. Tube side ( water side ) is 150 psig at 150/ F. Every condenser is tested per ASME Code prior to shipment. Seawater connections are 2" Class 150 PVC schedule 80 flanges. Water flow to the condenser will be regulated by a compressor discharge pressure actuated water regulating valve. A pressure relief valve ( set for 350 psig ) on the shell is standard. Immersion Heater Elements

The unit is equipped with a three stage, 18 element, 54 Kw 5" flange style immersion heating element. The heater elements are rated at full wattage on 208-3-60 power input. The elements are constructed of copper with a maximum watt density of 50 watts per square inch. The element heater tank will be constructed of steel pipe to ASME specifications. All welds will be by MIG welding procedure. The tank will be equipped with a 5" 150lb ANSI raised face welding neck flange to accept the 5" flange style immersion heater. The tank design rating pressure is 150 psig at 200/ Fahrenheit. The tank will be equipped with a ASME water pressure relief valve. Refrigerant Circuit

Each of the four refrigerant circuits shall include a discharge line check valve, liquid line ball valve, replaceable core liquid line filter drier with access fitting for refrigerant charging, combination moisture indicator and sight glass, liquid line solenoid and thermal expansion valve. All suction lines will be covered with a minimum of 1/2" closed cell insulation.

Control Panel / Electrical Box

The unit will have a NEMA 12 type enclosure for all of the electrical components. The chiller unit will be controlled by a programmable logic controller ( PLC ). The user interface for this PLC will consist of a touch screen mounted on the front of the electrical box. B. HVAC FANS

AirConditioningSystemDesign 35 The capacity For an HVAC is determine by the volume flow rate of air which is 12.4m3/s and the total pressure Drop it can handle. By choosing the right specification for ducts fan will help provide the appropriate air flow in the system until to the outermost part.

Figure 5.2: Centrifugal Ventilation Fan, HVAC

source: http://www.tradezz.com/buy_10365340_ChaoYue2-26-Low.htm

Table 5.1: Specification of Centrifugal Ventilation Fan

Model Capacity Wattage Speed Static Pressure ChaoYue2-134

825~62205 m3/h

2.2~410 kW

960~2900 rpm

557~1570 Pa

source: http://www.tradezz.com/buy_10365340_ChaoYue2-26-Low.htm

Other specification: Low noise High pressure Large air flow capacity

High Efficiency Specifically designed for

supplying air The product brand ChaoYue2-134 can be an option for its specifications meets the standards for the air conditioning system design of the project this values will be useful to maintain comfort air condition inside the building.

Documents

1. Aircon Design OT