HVAC Design Project

UNIVERSITY OF NEW ORLEANS Department of Mechanical Engineering

Dr. Martin Guillot

ENME 4777 Design of Thermal Fluid Systems Fall 2016

Project 2

Cooling System Design for a Facility in

New Orleans, LA

Group I Team Leader: Luis Molina

Jason Gleason Michael Hebert

Benjamin Leblanc

Jeremy Peacock

Table of Contents Page #

I. SUMMARY ............................................................................................................................................................................... 1

II. PROBLEM ............................................................................................................................................................................... 2

III. ZONING.................................................................................................................................................................................... 4

III. DUCT DESIGN AND HVAC COMPONENTS ............................................................................................................... 20

IV. AHU/COOLING COIL ......................................................................................................................................................... 30

V. HYDRONICS .......................................................................................................................................................................... 46

VI. CHILLER PLANT ................................................................................................................................................................ 57

VII. COOLING TOWER ............................................................................................................................................................ 69

VIII. AIR COOLED CONDENSER ........................................................................................................................................... 73

X. CONCLUSIONS ...................................................................................................................................................................... 77

XI. REFERENCES ....................................................................................................................................................................... 78

XII: APPENDIX .......................................................................................................................................................................... 79

APPENDIX II CHILLER BROCHURE ............................................................................................................................................... 80 APPENDIX:III REVIT FILES................................................................................................................................................................ 87

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I. Summary The purpose of a HVAC system is to provide a healthy and comfortable environment by

means of controlling humidity and temperature. Also, HVAC systems are responsible for

distributing this conditioned air to multiple locations at efficient rates. The majority of the

conditioned air can be recirculated into the desired locations, but a portion of the supply

air must be mixed with outside air to provide an ample supply of oxygen in this indoor

environment.

The main determinant of a HVAC system requirement is the total load for a given space.

These loads are not constant, but change with the amount of occupants, and with the

weather. Since temperature and humidity differ with seasonal changes, the HVAC system

must also be capable to adapt.

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II. Problem

A new building is being added to a campus in New Orleans, LA. The building layout

information is given in the drawings provided, and the building spaces are shown in the

figures below. The set point temperature of each zone is 74 °F, 50% relative humidity, with

the assumption that required ventilation air for each zone is 25% of the total for that zone.

Even though it is not possible to maintain humidity at exactly 50% within each zone, it

should be in the comfort range.

The given loads in Table 1 are based on 95oF DB and 82oF WB conditions. The central

chiller plant must be capable of providing a maximum of 1000 tons of refrigeration, but will

be designed to be able to handle loads between 25% and 100% of maximum loads with

redundancy built in.

Table 1: Space Cooling Loads Space Sensible Load (Btu/hr) Latent Load (Btu/hr)

Classroom 1 16940 9648 Classroom 2 16940 9648 Classroom 3 16940 9648 Classroom 4 17497 9814 Classroom 5 17497 9814 Classroom 6 17497 9814 Classroom 7 19662 10786

Lobby 39177 19979 Lavatory L 6082 1915 Lavatory M 6082 1915

Main Hallway 10851 3902 Mechanical Hallway 2515 683

Mechanical Room Not Conditioned Not Conditioned Office 1 10984 1324 Office 2 9030 1045 Plenum Not Conditioned Not Conditioned

Conference A 6346 1940 Conference B 6346 1940

Figure 1 on the following page shows the floor plan layout of the spaces in Table 1 above.

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MechanicalRoom

8

MechanicalHallway

17

Classroom3

Classroom2

Classroom1

Lavatory M10

Lavatory L9

Classroom7

Classroom6

Classroom5

Classroom4

Conference B12

Conference A11

Lobby15

Office 214

Office 113

Main Hallway16

20.00' 10.99' 30.00' 30.00' 30.00' 15.00' 15.00' 29.00' 29.00'

30.99' 30.00' 30.00' 30.00' 15.00' 15.00' 58.00'

69.50' 10.00'

208.99'

27.15'

42.35'

29.75'

29.75'

150.99'

922.0 ft2 892.5 ft2 892.5 ft2 892.5 ft2 446.3 ft2 446.3 ft2

2456.3 ft2

446.3 ft2 446.3 ft2892.5 ft2892.5 ft2892.5 ft2327.0 ft2595.0 ft2787.4 ft2 787.4 ft2

1509.9 ft2

(Not Conditioned)

Total Heated/Cooled Space: 14525 ft2

Figure1: Floor Plan of Spaces to be Conditioned

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III. Zoning

For our design, we have completed the zoning and the thermodynamic analysis of the air

entering the AHU unit and going through the zones. We have decided to split up the

building into 13 different zones, as shown in the next pages. We have performed

calculations for these zones already, using an excel spreadsheet, as well as psychometric

analysis using psychometric tables. We have calculated mass flow rates necessary for the

next part of the project, which is to size, and layout the ducts. Additionally, we found the

total mass flow rate of the system, which is necessary to select an AHU. When splitting the

classrooms into each individual zone with its own VAV unit, we took into consideration

their sensible and latent loads. We decided to maintain every single classroom as its own

Zone, with its own VAV unit to ensure that the rooms were maintained at the right

temperature when occupied. Since classrooms are sometimes occupied at different times, it

didn’t make sense to group them in sets. Although adding an independent VAV unit for each

classroom makes the design and layout of the duct work more expensive, it benefits the

overall design by ensuring that one room is not freezing while the other is really hot. The

conference rooms were designed to be a single zone, since there are rarely used. Each office

was made into an independent zone, as they might be occupied at different times. Lavatory

M and Lavatory L are a single zone. The mechanical hallway, and the main hallway were

set to be a single zone.

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Table 2: Specifications for each Zone

Zones Area Sensible

Load (Btu/hr)

Latent Load

(Btu/hr)

Total Load (Btu/hr)

SHR Mass Flow

Rate (lbm/hr)

Supply Volumetric Flow Rate (ft³/min)

Exhaust Volumetric Flow Rate (ft³/min)

Return Air Volumetric Flow Rate (Ft³/min)

Zone 1 Classroom 1 16940.0 9648.0 26588.0 0.6371 3529.2 784.3 196.1 588.2

Zone 2 Classroom 2 16940.0 9648.0 26588.0 0.6371 3529.2 784.3 196.1 588.2

Zone 3 Classroom 3 16940.0 9648.0 26588.0 0.6371 3529.2 784.3 196.1 588.2

Zone 4 Classroom 4 17497.0 9814.0 27311.0 0.6407 3645.2 810.0 202.5 607.5

Zone 5 Classroom 5 17497.0 9814.0 27311.0 0.6407 3645.2 810.0 202.5 607.5

Zone 6 Classroom 6 17497.0 9814.0 27311.0 0.6407 3645.2 810.0 202.5 607.5

Zone 7 Classroom 7 19662.0 10786.0 30448.0 0.6458 4096.3 910.3 227.6 682.7

Zone 8 Lobby 39177.0 19979.0 59156.0 0.6623 8161.9 1813.8 453.4 1360.3

Zone 9 Lavatory L 6082.0 1915.0 7997.0 0.7605 1267.1 281.6 70.4 211.2

Lavatory M 6082.0 1915.0 7997.0 0.7605 1267.1 281.6 70.4 211.2

Zone 10

Main Hallway

10851.0 3902.0 14753.0 0.7355 2260.6 502.4 125.6 376.8

Mechanical Hallway

2515.0 683.0 3198.0 0.7864 523.958 116.4 29.1 87.3

Zone 11 Office 1 10984.0 1324.0 12308.0 0.8924 2288.3 508.5 127.1 381.4

Zone 12 Office 2 9030.0 1045.0 10075.0 0.8963 1881.3 418.1 104.5 313.5

Zone 13

Conference A

6346.0 1940.0 8286.0 0.7659 1322.1 293.8 73.4 220.3

Conference B

6346.0 1940.0 8286.0 0.7659 1322.1 293.8 73.4 220.3

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Figure2: Floor Plan of different zones

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Psychrometrics for each Space The following section includes psychometric analysis for each space using the psychometric chart.

Figure 3: Psychrometric Chart for Classrooms 1-3

Classrooms 1-3 all have the same sensible load and latent load; therefore, the figure above represents all of them.

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Figure4: Psychrometric Chart for Classrooms 4-6

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Figure 5: Psychrometric Chart for Classrooms 7

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Figure 6: Psychrometric chart for Conference Rooms

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Figure 7: Psychrometric chart for Lavatories

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Figure 8: Psychrometric chart for Lobby

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Figure 9: Psychrometric chart for Main Hallway

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Figure 10: Psychrometric chart for Mech Hallway

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Figure 11: Psychrometric chart for office 1

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Figure 12: Psychrometric chart for office 2

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Table 3: Psychrometrics Analysis for each Area

Psychrometric Analysis

Area Tdb Twb Relative Humidity (%) Humidity Ratio Enthalpy (Btu/lb) Classroom 1 74.0 61.8 50.0 0.01120 30.01 Classroom 2 74.0 61.8 50.0 0.01120 30.01 Classroom 3 74.0 61.8 50.0 0.01120 30.01 Classroom 4 74.0 61.8 50.0 0.01120 30.01 Classroom 5 74.0 61.8 50.0 0.01120 30.01 Classroom 6 74.0 61.8 50.0 0.01120 30.01 Classroom 7 74.0 61.8 50.0 0.01120 30.01

Lobby 74.0 61.8 50.0 0.01099 29.78 Lavatory L 74.0 61.8 50.0 0.01057 29.32 Lavatory M 74.0 61.8 50.0 0.01057 29.32

Main Hallway 74.0 61.8 50.0 0.01043 29.17 Mechanical Hallway 74.0 61.8 50.0 0.00996 28.66

Office 1 74.0 61.8 50.0 0.00930 27.93 Office 2 74.0 61.8 50.0 0.00930 27.93

Conference A 74.0 61.8 50.0 0.00943 28.07 Conference B 74.0 61.8 50.0 0.00943 28.07

2 73.8 65.0 60.7 0.01080 29.57 3 73.8 65.0 60.7 0.01080 29.57 4 95.0 82.0 58.0 0.02070 45.63 5 79.1 70.0 62.4 0.01328 33.59

1&6 55.0 54.0 94.0 0.00863 22.57

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Psychrometrics

The following psychrometrics treat the whole building as a single zone

Table 4: Psychrometrics for the whole building

Figure 13: Psychrometric Analysis Diagram

For the psychrometrics we proceeded by getting all the values for state 0 (Supply air). For

state 0, the values were obtained by using 95 F and 82 F and the psych app. State 3/4/5 all

have the same temperature as no thermodynamic process occurs between these states.

Additionally, the problem statement tells us that there is 50% Relative humidity after going

through each zone. State 1/0/5 undergoes a mixing process. We proceeded, by finding the

enthalpy, and relative humidity at state 1.

STATE: T(db) T(wb)

Relative humidity i W v

0 95 82 58 45.63 0.0207 14.44

1 79.1 67.7 55.9 32.05 0.01187 13.83

2 55.5 51.2 74.6 20.88 0.00696 13.12

3,4,5 74 61.8 50 27.52 0.00893 13.64

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For state 1

For calculating Enthalpy at State 1

𝑖1 = .75𝑖5 + .25𝑖0

𝑖1 = .75(27.52) + .25(45.63)

𝑖1 = 32.05 𝑏𝑡𝑢/𝑙𝑏

For calculating Humidity Ratio at state 1

𝑤1 = .75𝑤5 + .25𝑤0

𝑤1 = .75(0.00893) + .25(0.0201)

𝑤1 = 0.01187

For state 2

𝐶𝑝 = 𝐶𝑝𝑎 + 𝑊𝐶𝑝𝑣

𝐶𝑝 = 0.24 + (0.00893) (0.444)

𝐶𝑝 = 0.244

Finding the heat transfer rate from state 2 to state 3 allows us to find the

temperature

𝑞𝑠 = 𝑚 ̇𝑎𝐶𝑝(𝑇3 − 𝑇2)

220386 = 487916 𝑙𝑏𝑚/ℎ𝑟 + 0.244(𝑇4 − 𝑇2)

𝑇2 = 55 ℉

𝑞𝑡𝑜𝑡𝑎𝑙 = �̇�𝑎(𝑖3 − 𝑖2)

324201 = 487916

𝑖2 = 20.58 𝑏𝑡𝑢/𝑙𝑏 Using Enthalpy of state 2 and Tdb= 55F, gives the rest of the states' properties.

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III. Duct Design and HVAC Components

The design for this entire system is based on how much energy is required for each room

or zone to be maintained at a specific temperature. Various environmental factors

influence this energy requirement such as the amount of people, building materials, and

outside weather conditions. The environmental factors add or remove heat to the system,

and therefore the HVAC system must compensate for these ever-changing conditions.

Ideally, each room would have its own unit and be able to maintain a specific desired

condition. Realistically, this scenario is not feasible, so the next alternative must be

considered. This consideration is a single air handler unit which supplied all the rooms

with conditioned air. Each room has a device to control the flow rate of air and effectively

is able to maintain specific conditions.

The total load seen below in Table 1 was found to be 324,000 Btu/hr, which equates to a

capacity of 27 tons. This load is based on key factors such as room size and the amount of

people in each room.

Table 5: Zone Loads

Area Sensible Load

(Btu/hr) Latent Load

(Btu/hr) Total Load

(Btu/hr) SHR

Classroom 1 16940.0 9648.0 26588.0 0.6371 Classroom 2 16940.0 9648.0 26588.0 0.6371

Classroom 3 16940.0 9648.0 26588.0 0.6371 Classroom 4 17497.0 9814.0 27311.0 0.6407 Classroom 5 17497.0 9814.0 27311.0 0.6407 Classroom 6 17497.0 9814.0 27311.0 0.6407 Classroom 7 19662.0 10786.0 30448.0 0.6458

Lobby 39177.0 19979.0 59156.0 0.6623 Lavatory L 6082.0 1915.0 7997.0 0.7605 Lavatory M 6082.0 1915.0 7997.0 0.7605

Main Hallway 10851.0 3902.0 14753.0 0.7355 Mechanical

Hallway 2515.0 683.0 3198.0 0.7864

Office 1 10984.0 1324.0 12308.0 0.8924 Office 2 9030.0 1045.0 10075.0 0.8963

Conference A 6346.0 1940.0 8286.0 0.7659 Conference B 6346.0 1940.0 8286.0 0.7659

Total 220386.0 103815.0 324201.0

The mass flow rates for each room was calculated by dividing the sensible load by the

specific heat capacity for air, which is 0.24 Btu/lb°F, times a temperature difference of 20

°F. This 20°F temperature variant is the different in the inside and outside temperature.

Next we were able to calculate the volumetric flow rates by dividing the mass flow rate by

the density of air at 70 °F multiplied by 60.

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Now that the volumetric flow rate is known, we can start designing the ductwork using

Revit. The key component to consider in the design of this duct system is noise from high

velocity flow rates. Large volumetric rates through small area ducts create this high

velocity noise-producing condition. Table 2 shows these calculated flow rates for each

room in the building.

Table 6: Zoning Flow Rates

Area Mass Flow Rate

(lbm/hr)

Supply Volumetric Flow Rate (ft³/min)

Exhaust/Fresh Volumetric Flow Rate (ft³/min)

Return Volumetric Flow Rate (Ft³/min)

Classroom 1 3529.2 784.3 196.1 588.2 Classroom 2 3529.2 784.3 196.1 588.2 Classroom 3 3529.2 784.3 196.1 588.2 Classroom 4 3645.2 810.0 202.5 607.5 Classroom 5 3645.2 810.0 202.5 607.5 Classroom 6 3645.2 810.0 202.5 607.5 Classroom 7 4096.3 910.3 227.6 682.7

Lobby 8161.9 1813.8 453.4 1360.3 Lavatory L 1267.1 281.6 70.4 211.2 Lavatory M 1267.1 281.6 70.4 211.2

Main Hallway 2260.6 502.4 125.6 376.8 Mechanical Hallway 524.0 116.4 29.1 87.3

Office 1 2288.3 508.5 127.1 381.4 Office 2 1881.3 418.1 104.5 313.5

Conference A 1322.1 293.8 73.4 220.3 Conference B 1322.1 293.8 73.4 220.3

Total 45913.8 10203.1 2550.8 7652.3

The supply flow rates are what actually feed each zone, and the air handler must produce

this total amount of airflow, which is 10,203 cubic feet per minute. The unit must also be

able to carry a load of 27 tons or 324,000 Btu/hr. In addition, the air hander must

incorporate water for a transfer medium that is supplied by a chiller. This water circulates

in order to remove heat from the system.

The return volumetric flow rates are what the system recirculates or pulls from each zone.

This rate of flow is determined by the amount of fresh air to be applied to the system. The

fresh air to be added to the system is a percentage of the supply volumetric flow rate, and

in this case was 25% of that amount. The return flow rate is then calculated by subtracting

the fresh from the supply. Higher flow rates in the plenum above the hall area are

acceptable because the noise produced will be less noticeable due to movement in the hall.

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With this in mind, flow velocities were kept below 1100 feet per minute in this main supply

line. For this desired quality to be met, the ducting was stepped down to smaller cross

sectional areas in order to keep velocities up as the supply line progressed to the opposite

end of the building. At the far end of the main supply line, the duct size was 28”x28” with a

flow velocity of 1100 feet per minute.

Variable air velocity equipment was used for each room in order to be able to maintain a

specific temperature, and offer efficiency. These pieces of equipment control the rate of

conditioned air entering each space by a thermostat controlling the fan speed inside of the

unit. The variable air velocity units in the hallways further reduce noise inside of each

conditioned zone. Flow velocity was also taken into consideration for the supply ducts

leaving the variable air velocity units, with the velocity goal to be under 650 feet per

minute. Table 3 shows gives the area verses velocity.

Table 7 shows each area’s ductwork with its specific size. As well as the velocity and area.

Table 7 Area and Velocity for Each Room Supply Line Trunk Line Diffuser Line

Area Size Velocity (ft/min)

Size Velocity (ft/min)

Size Velocity (ft/min)

Total Pressure Loss (in wg)

Classroom 1 14"x14" 576 12" Φ 499 8" Φ 562 0.11 Classroom 2 14"x14" 576 12" Φ 499 8" Φ 562 0.11 Classroom 3 14"x14" 576 12" Φ 499 8" Φ 562 0.11 Classroom 4 14"x14" 600 12" Φ 519 8" Φ 462 0.09 Classroom 5 14"x14" 600 12" Φ 519 8" Φ 462 0.13 Classroom 6 14"x14" 600 12" Φ 519 8" Φ 462 0.11 Classroom 7 15"x15" 584 12" Φ 581 9" Φ 516 0.12

Lobby 20"x20" 670 19" Φ 614 10" Φ 554 0.19 Lavatories 14"x14" 564 12" Φ 517 8" Φ 404 0.03 Hallways 14" Φ 579 12" Φ 579 8" Φ 646 0.28 Office 1 14"x14" 605 12" Φ 466 8" Φ 364 0.02 Office 2 14"x14" 605 12" Φ 385 8" Φ 301 0.02

Conference Rooms 14"x14" 588 12" Φ 539 8" Φ 421 0.03

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Table 8 shows each area with the diffuser flow velocity for the supply and the return.

Additionally, it specifies the number of diffusers for each area

Table 8: Duct Size per Ductwork Line

Supply Line Trunk Line Diffuser Line Area

Size Area (in²)

Velocity (ft/min)

Size Area (in²)

Velocity (ft/min)

Size Area (in²)

Velocity (ft/min)

Classroom1 14"x14" 196 576 12" Φ 113 499 8" Φ 50 562 Classroom2 14"x14" 196 576 12" Φ 113 499 8" Φ 50 562 Classroom 3 14"x14" 196 576 12" Φ 113 499 8" Φ 50 562 Classroom 4 14"x14" 196 600 12" Φ 113 519 8" Φ 64 462 Classroom 5 14"x14" 196 600 12" Φ 113 519 8" Φ 64 462 Classroom 6 14"x14" 196 600 12" Φ 113 519 8" Φ 64 462 Classroom 7 15"x15" 225 584 12" Φ 113 581 9" Φ 64 516

Lobby 20"x20" 400 670 19" Φ 284 614 10" Φ 79 554 Lavatories 14"x14" 144 564 12" Φ 79 517 8" Φ 50 404 Hallways 14" Φ 154 579 12" Φ 154 579 8" Φ 64 646 Office 1 14"x14" 121 605 12" Φ 79 466 8" Φ 50 364 Office 2 14"x14" 100 605 12" Φ 79 385 8" Φ 50 301

Conf. Rooms 14"x14" 144 588 12" Φ 79 539 8" Φ 50 421

Table 9 includes the face sizes and connections sizes for both the supply diffusers and

return diffusers.

Table 9: Duct Size per Ductwork Line

Diffuser Flow Velocity (ft/min)

Area Supply Return Number of Diffusers

Classroom 1 562 420 4 supply 2 return







Lobby 554 370 6 supply 3 return

Lavatories 404 390 1 supply 1 return (each)

Hallways 462 330 4 supply 1 return

Office 1 364 360 4 supply 2 return

Office 2 301 340 4 supply 2 return

Conference Rooms 421 400 1 supply 1 return (each)

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Table 10: Face Size and Connection Size for Diffusers

All of the dimensions were found through the use of the software “Revit”. Data for the

ductwork is included in the Appendix section of this report.

The following pages contains figures designed in Revit. The blue ducts are the supply lines,

and the pink ducts are the return lines. The main supply line running down the middle was

32”x36” at a maximum flow velocity of 1300 feet per minute just as the air leaves the air

handler.

Supply Diffuser

Return Diffuser

Face Size 12"x12" 24"x24" Connection Size

8" Φ 12"x12"

Lobby Uses 10" Φ Connect for Supply

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Figure 14 : Duct/HVAC Equipment Arrangment Front View

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Figure 15 : Duct/HVAC Equipment Arrangment Rear View

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Figure 16: Diffuser Arrangment

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Figure 17: Duct/HVAC Equipment Arrangment Top View

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Some other special pieces of HAVAC equipment can be seen in the above figures. The

bottom right corner of Figure 15 shows a mixing unit connected to the system. The mixing

unit is responsible for allowing fresh air into the system, while removing the same amount

of return air from the system. This aspect of the HVAC is critical for healthy indoor air

because with a source of fresh air, air quality would become toxic. This scenario is not seen

in our homes because of small indoor space in conjunction with the opening and closing of

doors on a daily basis.

Another interesting aspect for the duct arrangement is what to do with the return air from

the lavatories. Obviously, bathroom air recirculated throughout the building is

unacceptable for health and comfort reasons. This issue is addressed by making the

returns for both lavatories exhaust outlets, which means the pressure created inside the

lavatory from the supply line is forced to the outside. This exhaust flow rate for the

laborites is small enough to dismiss for the fresh/exhaust calculations.

Return lines also had to be taken into consideration for sizing because of noise concerns

just like the supply lines. The diffusers for the return side were 24”x24” with 12”x12”

connections and are twice as large as the supply side diffusers. This obviously allowed for

the return side to use half of the amount of diffusers as compared to the supply side. The

flow rate for the return side was kept under 400 feet per minute by using 14”x14” duct

coming off the diffusers. The main return line is 30”x30” at the end, and 12”x12” at the

farthest run.

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IV. AHU/Cooling Coil Air Handling Unit

Cooling Coil

The design of the AHU (Air Handling Unit) cooling coil was the first step in selecting a unit

for this project. In order to select the unit, it was essential that the cooling coil’s dimensions

were known so that minimum physical requirements were set. There are two main cooling

coil configurations that are typically seen in the industry: direct expansion and chilled

water. The main difference between the two is that for a direct expansion configuration,

refrigerant is used in the cooling coil tubes as compared to water in the chilled water

system. For our project, it is specified that chilled water is supplied to the AHU from an

outside source, therefore, a chilled water cooling coil configuration is used. Figure 18,

below, shows a typical setup for a counter- flow cooling coil. It is common for coil systems

of three or more rows of tubes for counter-flow to be assumed (Page 505, “Enter book

reference).

Figure 18: Counter-flow Cooling Coil Setup

Before the calculation process could be carried out, a few design parameters had to be

chosen based upon typical industry standards. The table to follow shows the parameters

that were assumed at the beginning of the coil design process.

Table 11: Cooling Coil Design Parameters

Property: Value: Units:

Air Face Velocity 500 fpm

Water Velocity 4 fps

Water Pressure

Drop 17 ft

Tube OD 0.625 in

Tube ID 0.545 in

Tube Thickness 0.04 in

Xa 1.25 in

Xb 1.083 in

fin thickness (y) 0.006 in

fins/inch 10 in^-1

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Air face velocities are usually within the range of 200-550 feet per minute (fpm). Velocities

over 500 can cause water droplets to leave outer edge of fin causing moisture carry over

which is when water droplets leave the outer edges of the fins. 500 fpm was chosen for this

design as this is the most commonly seen in chilled water systems [5].

The water velocity was chosen as 4 feet per second (fps) as these values are typically seen

in between 3-6 fps but do not exceed 8 fps. [5]

The max water pressure drop in the coil is about 25’ according to [5]. Typically, for chilled

water coils the WPD is between 15-20’. 17’ was arbitrarily chosen for the WPD in this

design.

Type L Copper was the material chosen in this design as it is the most common type of pipe

seen in chilled water systems. The size of this pipe is ½ inch. This size was not only just

common within most cooling coils but Figure 14-12 in textbook provides useful j-factor

and fin information that could not be located in other sources.

Using https://sizes.com/materials/pipeCopper.htms, the actual diameter for ½ type L

copper tubing is 0.625” with an inner diameter of 0.545” which leads to a thickness of

0.04”. The rest of the coil properties that were pre-determined in Table 11 were found

using the Figure below.

Figure 19: Fin and Heat Transfer Data for Cooling Coil

As it can be seen from the previous figure, the dimensions Xa and Xb, which represent the

vertical and horizontal distance between coil pipes, respectively, are provided. The

remaining fin and heat transfer can be found according to which fin pitch was chosen and

can be found in Table 12, below. Typical values for this range from 8 to 16 fins per inch. A

value of 10 was arbitrarily chosen and its corresponding values were interpolated

appropriately. A fin thickness of 0.006” is also found from Figure 19.

https://sizes.com/materials/pipeCopper.htms

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Table 12: Properties of Cooling Coil

Property Value: Units:

σ 0.555 N/A

Dh 0.0133 ft

α 170.624 ft^-1

Af/A 0.9193 N/A

With all of the appropriate design assumptions made, the cooling coil calculations could

now be carried out. The first step in this process is to calculate the Reynolds number for the

water flowing through the coils. The equation to determine the Reynolds number is as

follows:

𝑅𝑒𝐷 =𝜌�̅�𝐷

𝜇 (1)

In equation 1, 𝜌 is the water density at 45 ᵒF, �̅� is the average water velocity, D is the inner

diameter, and 𝜇 is the dynamic viscosity of the water. With the Reynolds number

determined, the next property to calculate is the Prandtl number, which can be calculated

using equation 2.

𝑃𝑟 =𝜇𝑐𝑝

𝑘 (2)

Using both the Reynolds number and Prandtl number, the convective heat transfer

coefficient can be determined. This is achieved using the following equation:

ℎ̅𝑖 = 0.023𝑘

𝐷(𝑅𝑒𝐷)0.8(𝑃𝑟)0.3 (3)

where ℎ̅𝑖is the convective heat transfer coefficient. The thermal conductivity (k) is based

off the same value used in the previous equations at 45 ᵒF.

�̇�𝑎 = 𝐺𝑐𝐴𝑐 (4)

Where Gc can be calculated by:

𝐺𝑐 =𝐺𝑓𝑟

𝜎 (5)

And, Gfr is found by the following equation,

𝐺𝑓𝑟 = 𝜌𝑓𝑟�̅�𝑓𝑟 (6)

Note: the subscript “fr” refers to face of the coil and “c” refers to the coil itself. In the previous equations, �̇�𝑎 is the mass flow rate of the air, 𝐺𝑐 is the mass velocity of the

minimum flow area, and 𝜎 is the ratio of minimum flow area to frontal area. As previously

mentioned, this value was obtained from Figure 19, where it was interpolated according to

the fin pitch assumed.

With the inner tube diameter accounted for, the outside tube diameter is focused on which

will allow working towards calculated the air side heat transfer coefficient.

G r o u p I | 33

The first step in this process is to find the Reynolds number for the incoming airflow. The

equation is as follows:

𝑅𝑒𝐷 =𝐺𝑐𝐷

𝜇 (7)

In the previous equation, the diameter used is the OD and the dynamic viscosity value is that of air at a temperature of 45 ᵒF.

Equation 8, calculates the ratio of the total heat transfer area to the area of the bare, finless tubes. 𝐴

𝐴𝑡=

4

𝜋

𝑥𝑏

𝐷ℎ

𝑥𝑎

𝐷𝜎 (8)

Where Xa and Xb are the row height and width spacing found in Figure 19, respectively, and Dh

is the hydraulic diameter which can be found in Table 12. Remember, the hydraulic diameter

was determined using Figure 19. The value for this variable was determined according to the fin

pitch that was assumed.

In calculating the outside air heat transfer coefficient, the j-factor must be calculated. The j-factor

is read from Figure 20, but first the JP parameter must be calculated. The JP parameter is found

by:

𝐽𝑃 = 𝑅𝑒𝐷−0.4 (

𝐴

𝐴𝑡)

−0.15

(9)

Once the JP parameter was calculated, the j-factor could be evaluated using the figure below.

Using this value and fin pitch the j-factor determined was 0.007 (dimensionless).

Figure 20 Heat Transfer correlation for smooth plate-fin-tube coils

With the j-factor now found, the heat transfer coefficient of the air can now be solved for using

the following equation:

G r o u p I | 34

ℎ̅𝑜 = 𝑗𝐺𝑐𝑐𝑝 (𝜇𝑐𝑝

𝑘)

−2

3 (10)

where Cp, is the specific heat of air and ℎ̅𝑜 is the air heat transfer coefficient.

The next objective is to calculate the overall heat transfer coefficient. The first step in reaching

this value is to calculate the two dimensions: L and M. The equations for these two values

involve referring to the previously obtained Xa and Xb dimensions. L and M can be computed by:

𝐷𝑖𝑚𝐿 =𝑥𝑎

2 (11)

𝐷𝑖𝑚𝑀 =[(

𝑥𝑎2

)2

+𝑥𝑏2]

1/2

2 (12)

Using the previously obtained dimensions L and M, the 𝜓 and 𝛽 parameters can be found. The

equations for these two parameters are as followed:

𝜓 = 𝑀

𝑟 (13)

𝛽 =𝐿

𝑀 (14)

The variable “r” is the outer radius of the copper tubes.

Using the two previous found values, we can now calculate the ratio of the equivalent fin radius

to outer tube radius by the following equation: 𝑅𝑒

𝑟= 1.27𝜓(𝛽 − 0.3)

1

2 (15)

where 𝑅𝑒 is the equivalent fin radius.

The next parameter to be calculated is ∅, which is used in the calculation of fin efficiency. The

parameter can be calculated by:

∅ = (𝑅𝑒

𝑟− 1) [1 + 0.35ln (

𝑅𝑒𝑟⁄ )] (16)

Following this calculation, the final parameter, m, for the fin efficiency calculation can be found

using the following equation:

𝑚 = (2ℎ̅𝑜

𝑘𝑦)

1/2

(17)

where y is the fin thickness which is found from Figure 19, and k is the thermal conductivity of

aluminum.

With the previous parameters calculated, the fin efficiency can now be found by:

ɳ =tanh[𝑚𝑟∅]

𝑚𝑟∅ (18)

G r o u p I | 35

The surface effectiveness can be calculated using the fin efficiency and the ratio, 𝐴𝑓

𝐴, which can

be found in the graph. It is dependent on the fin pitch chosen in Figure 19. The equation for

surface effectiveness is as follows:

ɳ𝑠𝑜 = 1 −𝐴𝑓

𝐴(1 − ɳ) (19)

Before the overall heat transfer coefficient can be calculated, the ratio of the water-side to air-

side heat-transfer areas must be computed, which can be achieved by the following equation: 𝐴𝑖

𝐴𝑜=

𝜋𝐷𝑖

𝑥𝑎𝑥𝑏𝛼 (20)

where Di is the inner tube diameter and 𝛼 is a parameter found using Figure 19.

With all the necessary information needed to determine the overall heat transfer coefficient, the

following equation can now be completed: 1

𝑈𝑜=

1

ℎ̅𝑜ɳ𝑠𝑜+

1

ℎ̅𝑖(𝐴𝑖 𝐴𝑜⁄ ) (21)

Obtaining the overall heat transfer coefficient concludes the water to air cooling coil continuous

plate-fin-tube. The next step in the design process is to determine the geometric configuration of

the coil which requires the computation of the NTU and fluid capacity rates of the cooling coil.

The first step in the process of designing the geometric configuration of the coil is to determine

the mass flow rate of the air which can be determined by the equation to follow:

�̇�𝑎 = 𝜌�̇� (22)

where 𝜌 is the density of air and �̇� is the volumetric flow rate of the incoming air. The next

equation calculates the corrected, total heat transfer rate of the AHU. The equation for this is:

�̇�𝑡𝑜𝑡 = �̇�𝑡𝑜𝑡(𝑖5 − 𝑖6) (23)

where �̇�𝑡𝑜𝑡 is the total mass flowrate of all zones and i represents the enthalpy at states 5 and 6.

Using the corrected total load, both the heat capacity rate of air and water can be calculated using

the following equations:

�̇�𝑡𝑜𝑡 = 𝐶𝑎𝑖𝑟(𝑇6 − 𝑇5) (24)

�̇�𝑡𝑜𝑡 = 𝐶𝑤(𝑇𝑖𝑛 − 𝑇𝑜𝑢𝑡) (25)

where T represents the temperatures at state 5, state 6, entering the coil, and leaving the coil, Cair

is the heat capacity rate of the air and is also equivalent to Cmin, and Cw is the heat capacity rate

of the water which is equal to Cmax.

G r o u p I | 36

The NTU can now be determined by using the ratio of the previously found heat transfer rates

and the effectiveness of the heat exchanger which is to be calculated. This involves using Figure

21, which provides the proper information needed to determine the NTU.

The effectiveness of the heat exchanger is the ratio of the actual heat transfer rate over the

maximum possible heat transfer rate. The equation for this is as follows:

𝜀 =𝑇6−𝑇5

𝑇𝑖𝑛−𝑇𝑜𝑢𝑡 (26)

Using this effectiveness value, the NTU can now be obtained using the following figure.

Figure 21: Effectiveness of Cross-flow Exchanger with Fluids Unmixed

Using the previously calculated values, the NTU was determined to be 0.48. This value is then

used to determine the overall heat transfer area. This is done by the following equation:

𝑁𝑇𝑈 =𝑈𝑜𝐴𝑜

𝐶𝑚𝑖𝑛 (27)

Once the area is found, the volume can be evaluated using:

𝑉 =𝐴𝑜

𝛼 (28)

As mentioned in the beginning of this section, the air face velocity was assumed previous to the

design calculation process. Using this value, the area of the face of the coil can be solved for

using:

𝐴𝑓𝑟 =�̇�

�̅�𝑓𝑟 (29)

G r o u p I | 37

With this value calculated the next geometric property to be determined is the depth of the coil

which is computed using the following equation:

𝐿 =𝑉

𝐴𝑓𝑟 (30)

The number of rows in the system can be determined as well using the length previously

determined and the horizontal distance between the tubes. The equation is as follows:

𝑁𝑅 = 𝐿

𝑋𝐵 (31)

The value of NR must be an integer and a multiple of two.

This concludes the design of the geometric configuration of the cooling coil. The next important

design parameter is to calculate the pressure loss that will occur for the air flow through the coil.

The first step is to calculate the following ratio: 𝐴

𝐴𝑐=

𝛼𝑉

𝜎𝐴𝑓𝑟 (32)

The mean density is calculated next,

𝜌𝑚 =𝑃

2𝑅(

1

𝑇5+

1

𝑇6) (33)

where P is the atmospheric pressure and R is the heat exchanger parameter constant.

In order to calculate the pressure drop, the friction factor must be found. To find the friction

factor the parameter FP must be calculated. This can be done by the following equation:

𝐹𝑃 = 𝑅𝑒𝐷−0.25 (

𝐷

𝐷∗)0.25

[𝑥𝑎−𝐷

4(𝑠−𝑦)]

−0.4

[𝑥𝑎

𝐷∗ − 1]−0.5

(34)

Using this value for FP along with the amount of fins per inch which was chosen before the

design calculations, the friction factor can be obtained from the following figure.

G r o u p I | 38

Figure 22: Heat Transfer Correlation for Smooth Plate-Fin-Tube Coils

From reading the graph, the friction factor for this design was chosen to be 0.025.

With the friction factor now known, the pressure drop can now be calculated. The equation for

the pressure drop is as follows:

∆𝑃𝑜 =𝐺𝑐

2

2𝑔𝑐𝜌1[(1 + 𝜎2) (

𝜌1

𝜌2− 1) + 𝑓 (

𝐴

𝐴𝑐) (

𝜌1

𝜌𝑚)] (36)

where 𝑔𝑐 is the gravitational constant, 𝜌1 is the density of the air at the inlet, and 𝜌2 is the air

density at the exit of the AHU.

This now completes the air side pressure loss calculations for the air handling unit. The next

series of calculations will determine the pressure loss in the tube side of the system. The first step

in this process is to determine the flow cross-sectional area of the water.

𝐴 =𝐶𝑚𝑎𝑥

�̅�𝜌𝑐𝑝 (37)

The number of tubes can now be calculated for this AHU by using the following equation:

𝑁𝑡 =4𝐴

𝜋𝐷𝑖2 (38)

The value of this variable must be an integer. After the calculations were performed, this value

was determined to be 35.2 but rounded to 40. This effects the next equation which determines the

height of the cooling coil.

𝐻 = (𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑢𝑏𝑒𝑠)𝑥𝑎 = 40𝑥𝑎 (39)

The width of the coil is the next geometric property to be determined. This is done by the

following:

𝑊 =𝐴𝑓𝑟

𝐻 (40)

G r o u p I | 39

The length of the coil can now be determined,

𝐿𝑤 = 𝐾𝑊 (41)

where K is the loss coefficient associated with each bend and was predetermined before design

calculations to be 2.

In order to determine the lost head within the tubes, the Moody friction factor must be

determined. This can be done by using the friction factor and Reynolds number and reading the

following figure.

Figure 23: Moody Diagram for Pipe Flow Friction Factor

Once the friction factor was determined from the previous figure, the tube side pressure loss can

be determined:

𝑙𝑓𝑤 = 𝑓𝐿𝑤

𝐷𝑖

�̅�2

2𝑔𝑐+ 𝐾𝑁𝑏

�̅�2

2𝑔𝑐 (42)

where 𝑁𝑏 is the number of return bends in each circuit.

The final stage of the design calculations involves calculating the water flow rate and rate of heat

transfer of water flowing through the coils.

The following equation will calculate the mass flow rate of water through the tubes:

�̇�𝑤 = 𝜌𝑤𝐴�̅� (43)

With this value now calculated, the rate of heat transfer of the tube water can now be found:

G r o u p I | 40

�̇�𝑤 = �̇�𝑤∆𝑇 (44)

To convert this value into GPM involves finding a conversion factor which is accomplished by:

𝜌𝑤𝑐𝑝=500 (45)

This conversion factor is now used in converting the heat transfer rate into GPM by the

following conversion:

𝐺𝑃𝑀 =𝐵𝑡𝑢

ℎ𝑟⁄

500∗(𝑇𝑖𝑛−𝑇𝑜𝑢𝑡) (46)

This now concludes the series of design calculations used for this AHU design. A summary of

the calculated properties is to follow:

The following table presents the temperatures used through the calculation process:

Table 13: AHU Design Temperatures

Property Value Unit

State 1 79.1 F

State 2 55.5 F

Inlet H2O 45 F

Exit H2O 55 F

Delta T H20 10 F

The following table presents the geometric dimensions calculated during the design process:

Table 14: AHU/CC Dimensions

Property Value Unit

Volume of HT 7.11 Ft^3

Face Area 20.41 Ft^2

Depth of HT 4.18 in

Number of Rows 3.86 N/A

Actual Number of Rows 4.00 N/A

Number of Tubes 35.23

*USE

40

Number of Bends 9.00 N/A

Tubes per Row 10 N/A

H 50.00 in

H 4.17 ft

W 58.77 in

W 4.90 ft

Tube Length (Flow

Length) 39.18 ft

G r o u p I | 41

The following table presents the pressure losses calculated during the design process:

Table 15: Calculated Pressure Losses

Property Value Unit

Air Side Pressure Loss 2.602767252 lbf/ft^2

Air Side Pressure Loss 0.500532164 in*wg

Tube Side Pressure

Loss 9.144435757 ft

This last table presents the flowrates determined for the AHU/CC system:

Table 16: Determined System Flow Rates

Property Value Unit

Air Flow Rate 10203.1 cfm

Total Supply Mass

Flowrate 45913.8 lbm/hr

Total Load 512857.146 Btu/hr

water flowrate 51306.26176 lbm/hr

GPM 102.6125235 GPM

Air Handling Unit

After all design calculations were performed for the AHU and cooling coil, the data was

analyzed and an air handling unit was chosen accordingly. The unit itself was chosen based

upon the entering and exiting air mass flowrates. For this design, the incoming volumetric

flow rate was calculated to be 10203.1 cubic feet per minute, and the exiting air volumetric

flowrate was found to be 7652.3 cubic feet per minute. Using the flowrates, an appropriate

air handling unit was chosen from Dunham-Bush, model RT 26 This specific model is sized

for volumetric flow rates within the range of 7000-15600 cfm, and has a nominal face area

of 26 𝑓𝑡2. The unit’s shell and inside design resembles the following two figures.

G r o u p I | 42

Figure 24: Model of RT 26 Dunham-Bush Rooftop AHU

Figure 25: Interior Design of RT 26 Dunham-Bush Rooftop AHU

Dunham-Bush prides themselves on the excellent standard at which the outer casing of the

AHU is constructed. The AHU is designed to be protected from any water entry due to

weather. Standing flanges raise the unit 1” from the ground which eliminate the possibility

of any water entry from the ground level into the unit. The top panels are pitched to the

cabinet sides with no pockets to collect rain, and an overhang is installed on the unit so no

rainwater will drip onto the door tops. All exposed panels are constructed of 18-gauge steel

and are finished with climate proof coatings. The unit is ensured to be sealed with absolute

air tightness through the use of a gasket, which is closed cell foam. The cell foam method is

far superior to caulking and further ensures no heat loss will be experienced by the unit.

The modular interior design of the AHU, as seen in Figure 25, minimizes the static

pressure resistance due to its straight thru design that allows low velocity airflow through

the unit. This design is very versatile as there are a total of 22 possible modular sections

that can be selected.

G r o u p I | 43

The water cooled coils come with the following options: (The bolded options represent the

values calculated for this design)

Material: Type 5 Copper

O.D.: 0.625”

Rows: 3, 4, 5, 6, 8, 10

Fins Per Inch: 6, 8, 10, 12, 14

A picture of the cooling coil for the chosen AHU is in the following figure.

Figure 26: Cooling Coil System for RT 26 AHU

With unit size chosen, the coil face area is given as 25.5 𝑓𝑡2. This is a good size unit because

the calculated area needed for the designed coil from calculations was 20.4 𝑓𝑡2.

The next component that was chosen based off of the previous design calculations was the

supply fan. The supply fan was designed to handle an incoming air flow rate of 10203.1

cubic feet per minute with a total static pressure loss of 0.5 inches of water. The fan was

chosen from the following table:

G r o u p I | 44

Table 17: Supply Fan Table Provided by Dunham-Bush

As seen in the above table, the chosen fan for this design is a 4.22 BHP, 575 RPM size 500

forward curved fan. The forward curved fans come with an adjustable fan/motor base. The

motor and drive are all built into a fan and motor sub-assembly. Only 1 supply fan is used

in the AHU which give the unit a very quiet performance with low energy consumption.

The figure below depicts a forward curved fan:

Figure 27: Model of a Forward Curved Fan

The last component chosen for the calculated design parameters is the return fan. The fan

was designed for a flow rate of 7652.3 cubic feet per minute and the same pressure loss.

The return fan was picked according to the smallest possible fan that could be found. The

fan was chosen from the table to follow.

G r o u p I | 45

Table 18: Return Fan Table Provided by Dunham-Bush

As seen above, the chosen fan has an rpm of 595 and brake horsepower of 3.04. The fan

type is a size 450 forward curved fan.

Overall, the components picked for the AHU closely match the values received during the

design calculation process. When assembled, the AHU system should provide good

performance and quality, cooled airflow to the school building.

G r o u p I | 46

V. Hydronics

The hydronic system is responsible for transporting chilled water from the condenser to

the air handling unit (AHU), returning heated water from the AHU back to the condenser,

and transporting heated water from the evaporator to the cooling tower, as well as

returning the cooled water from the tower back to the evaporator.

The piping chosen for the hydronic system is schedule 40 commercial steel pipe. The

calculated values for each of the piping systems is presented in Tables 1 through 3 below.

The tables present the flow rates, pipe size, and losses for pipe lengths, fittings and valves.

The resources used in the calculations came from McQuiston’s Heating, Ventilating and Air

Conditioning, 6th edition, and follow the tables.

Table 19: Chilled Water Primary Loop

Chilled Water Primary

Loop

Volumetric Flow Rate (gpm) 2400 Velocity (ft/s) 10 Diameter (in.) 10

Friction Factor ft

Head Loss/ 100 feet 2.6

Small Pipe Diameter (in.) 3 0.017

Pipe Length (ft.) 71

Large Pipe Diameter (in.) 10 0.013

Total Pipe Head Loss 1.988 Chiller Diameter (in.) 19 0.012

Valves Qty. K

K

(coeff.)

Equivalent

Length (ft.)

Total Head Loss

(ft.)

Ball 9 3 0.014 4 1.01

Check 2 2 0.014 3 0.17

Globe 2 340 0.014 520 14.56

Fittings

Elbow (90°) 2 30 0.014 32 1.79

Tee (Branch) 7 60 0.014 60 11.76 Head Loss through straight pipe 1.99

Head Loss through chiller 28.00

Grand Total Head Loss (ft.) 59.28

G r o u p I | 47

Table 20 : Chiller to AHU

Chiller to AHU

Volumetric Flow Rate (gpm) 103 Velocity (ft/s) 4 Diameter (in.) 3

Friction Factor ft





Total Pipe Head Loss 46.33 Chiller Diameter (in.) 0.012

Valves Qty. K

K

(coeff.)

Equivalent

Length (ft.)

Total Head Loss

(ft.)

Ball 7 3 0.018 10.5 0.26

Check 2 2 0.018 1 0.05

Fittings

Elbow (90°) 17 30 0.018 8 3.40

Tee (Branch) 2 60 0.018 15 0.75 Head Loss through straight pipe 46.33 Head Loss through vertical pipe 14.00

Head Loss through AHU 10.00


G r o u p I | 48

Table 21 : Chiller to Cooling Tower

Chiller to Cooling Tower

Volumetric Flow Rate (gpm) 2900 Velocity (ft/s) 8.5 Diameter (in.) 12

Friction Factor ft





Total Pipe Head Loss 2.475 Chiller Diameter (in.) 0.012

Valves Qty. K (ft.)

K

(coeff.)

Equivalent

Length

Total Head Loss

(ft.)

Ball 2 3 0.013 10 0.15

Check 1 2 0.013 4.5 0.07

Globe 2 340 0.013 350 10.50

Fittings

Elbow (90°) 6 30 0.39 210 3.15

Tee (Branch) 1 60 0.78 70 1.05 Head Loss through straight pipe 2.48 Head Loss through vertical pipe 17.00

Head Loss through chiller 28.00


Figure 28: Formulas, Definition of terms and Values of ft.

G r o u p I | 49

Figure 29: Resistance Coefficients K for Various Valves and Fittings.

Figure 30: Equivalent Lengths L and L/D and Resistance Coefficient K.

G r o u p I | 50

Figure 31: Friction Loss of Water in Commercial Steel Pipe (Schedule 40).

The chilled water supply traveling from the chiller plant to the AHU consists of two loops,

primary and secondary. The primary pumping system services the return water from the

AHU to the chillers. The secondary pumping system services the AHU of the new building

with the chilled water and returns it back to the chilled water return header. The primary

and secondary pumping systems will each have two pumps piped in such a fashion as to

allow either redundancy, maintenance or demand.

Table 22 : Chilled Water Loop Pumps

Pump Flowrate, Q (gpm) Head, h (Ft. of H2O) Motor Speed, w (rpm) NPSHR

(ft.) NPSHA

(ft.)

CHWP-1 2400 60 1760 14 42

CHWP-2 2400 60 1760 14 42

CHWS-1 103 75 1760 13 41

CHWS-2 103 75 1760 13 41

The primary pumps chosen were TACO Model SFI8013-1760-100 for 60 feet of head. The

pump curve is presented below.

G r o u p I | 51

Figure 32: Primary Pump Curve

The secondary pumps chosen were TACO Model SKS3011-1760-7.5 for 75 ft. of head.

Figure 33: Secondary Pump Curve

G r o u p I | 52

The chiller to cooling tower system consists of two redundant pumps. Each pump will have

the capacity to handle the full load.

Table 23 : Chiller to Cooling Tower Pumps

Pump Flowrate, Q (gpm) Head, h (Ft. of H2O) Motor Speed, w (rpm) NPSHR

(ft.) NPSHA

(ft.)

CHWR-1 2900 63 1760 14 40

CHWR-2 2900 63 1760 14 40

The pumps chosen for the chiller to cooling tower were TACO Model SKS1213B-1760-100.

Figure 34: Cooling Tower Pump Curve

G r o u p I | 53

The piping arrangements were modeled using the Applied Flow Technology (AFT) Fathom

software to make a comparison with the hand calculations. The following figures present

the model for the primary and secondary loop. They include the evaporator side of the

chillers and air handling unit (AHU), followed by a screenshot of the output after running

the software.

Figure 35: Fathom Model for Primary and Secondary to AHU Loop.

G r o u p I | 54

Figure 36: Fathom Output for Primary and Secondary to AHU Loop.

G r o u p I | 55

The following figures present the model for the chiller to cooling tower loop. They include

the condenser side of the chiller and the cooling tower, followed by a screenshot of the

output after running Fathom.

Figure 37: Fathom Model for Chiller to Cooling Tower Loop.

G r o u p I | 56

Figure 38: Fathom Output for Chiller to Cooling tower Loop.

G r o u p I | 57

VI. Chiller Plant

The chiller system works as refrigeration cycle. Figure 1 is a diagram of the refrigeration

cycle for fluid 134-A. From Process 1-2 Isentropic compression takes place. The fluid leaves

the evaporator at state 1 in a low pressure state, and flows through the compressor and

becomes super-heated fluid. During Process 2-4, the refrigerant flows at constant pressure

from the compressor into the condenser where heat is rejected from the system. The

refrigerant fluid then exits the condenser during state 3, and enters the throttling valve at

state 4. During this process, Process 3-4, the refrigerant ends in a two-phase liquid vapor

mixture. Process 4-1 is the final process of the cycle where the refrigerant flows at

constant pressure through the evaporator.

For this project we proceeded to select a Chiller that would meet the requirements of 1000

Tonnage specified by the problem statement, and provide the 45 oF required for the AHU

unit. Additionally, we designed a chiller system that would meet the requirements. For the

design we ignored the losses caused by entropy. The chiller system was analyzed as an

ideal-vapor compression cycle.

Figure 39: T-S Diagram for Refrigerant 134-A

G r o u p I | 58

Figure 40: Direct Expansion Chiller Schematic

The figure above, shows the layout of the chiller system used in this project. The Primary

supply loop refers to the water going to the hydronic system.

Hand calculations were performed to determine the required refrigerant temperatures, the

mass flow rate of water, and the mass flow rate of the refrigerant, needed to meet the

requirements set by the AHU to satisfy the sensible and latent loads of each room in the

building. The 1000 Tonnage requirement set by the problem statement was also

considered. The process of finding the specified values is shown below:

1000 𝑇𝑜𝑛𝑠 ∗ 12,000 = 12,000,000 𝐵𝑡𝑢/ℎ𝑟

We assumed the temperature of the refrigerant to be 30 oF.

For state 1:

From R-134a Tables

𝑃1 = 40. 788 𝐿𝑏𝑓/𝑖𝑛2

ℎ1 = ℎ𝑔 = 106.01 𝑏𝑡𝑢/𝑙𝑏

𝑆1 = 𝑆𝑔 = 0.2196

For state 2:

We assumed isentropic process, so sg1=sg2

G r o u p I | 59

Since, The Temperature has to be above 95 oF, from the Table A-12E from Moran’s

Thermodynamics book, we selected pressure table p=140 lbf/in2 , Tsat= 100.56, S2=0.2161.

120℉ − 𝑇2

120 − 100.56℉=

0.2254 − 0.2196

0.2254 − 0.2161

𝑇2 = 107.87℉

120.25 − ℎ2

120.25 − 114.95=

0.2254 − 0.2196

0.2254 − 0.2161

ℎ2 = 116.94 𝑏𝑡𝑢/𝑙𝑏

ℎ𝑉𝑎𝑝𝑠𝑎𝑡 = 114.95 𝑏𝑡𝑢/𝑙𝑏

For State 3

𝑇3 = 100.56℉

𝑃3 = 140 𝑙𝑏/𝑖𝑛2

(Same Pressure Line)

105℉ − 100.56℉

105℉ − 100℉=

0.0930 − 𝑠𝑓3

0.0930 − 0.0898

𝑠3 = 0.0901584 𝑏𝑡𝑢/𝑙𝑏 𝑅

105℉ − 100.56℉

105℉ − 100℉=

46.01 − ℎ𝑓3

46.01 − 44.23

ℎ3 = 44.42936 𝐵𝑡𝑢/𝑙𝑏𝑚

For State 4

For throttling process h4=h3

𝑇4 = 30℉

𝑃4 = 40.788℉

ℎ4 = 44.429℉

G r o u p I | 60

Flow Rate of Refrigerant

𝑄�̇� = �̇�𝑅134𝑎(ℎ1 − ℎ4)

�̇�𝑅134𝑎 =12,000,000 𝑏𝑡𝑢/ℎ𝑟

(106.01 − 44.42936)

Work of Compressor

𝑊𝑐̇ = �̇�𝑅134𝑎(ℎ2 − ℎ1) (47)

𝑊𝑐̇ = 19,504,936(116.944 − 106.01)

𝑊𝑐̇ = 2,132,114.363 𝑏𝑡𝑢/ℎ𝑟

Heat Transfer of Condenser

𝑄�̇� = �̇�𝑅134𝑎(ℎ2 − ℎ1) (48)

𝑄�̇� = 195,069.9326(116.944 − 106.01)

𝑄�̇� = 14,145,425.94 𝑏𝑡𝑢/ℎ𝑟

Flow Rate of water for Condenser

𝑄�̇� = �̇�𝑐𝑤𝑐𝑝𝑐𝑤∆𝑇 (49)

�̇�𝑐𝑤 =14,145,425.94

(0.999)(10)

�̇�𝑐𝑤 = 1,415,958.522 lbm/hr

Flow Rate of water for Evaporator

�̇�𝑐𝑤 =12,000,000

(1)(10)

�̇�𝑐𝑤 = 1,200,000 𝑙𝑏𝑚/ℎ𝑟

G r o u p I | 61

The chiller we have selected is from Trane can provide the 1000 Tonnage needed to fulfill

the system requirements. We have decided to go with two units of CVHF-080L-142L/IECU,

so that there is redundancy in case one of them shuts off the HVAC system won’t stop

working. This specific water chiller we picked meets the requirements for the AHU water

flow rate, with a minimum flow rate of 1135 GPM Additionally, it has the required

temperatures towards the AHU unit of 45 F, and it meets the requirements for the cooling

tower. For purpose of redundancy we have decided to use to Chiller Plants working at

34.58% percent each. The selected Chillers work with a tonnage range of 1000-1300.

Figure 41: Chiller System CVHF-080L-142L/IECU

G r o u p I | 62

Table 24: Selected Chiller Specifications

Chiller Specs Model Series CVHF Tonnage Range for Model 350-2000 Ton Model Number CVHF-080L-142L/IECU Evaporator Size 080L Condenser Size 142L Tonnage 1300 Capacity Running For Each 38.46 % Evaporator Pipe Connection Size 10-inch Condenser Pipe Connection Size 10-inch Number of Passes 1 Pass Evaporator Maximum Flow Rate 8394 GPM Evaporator Minimum Flow Rate 1135 GPM Condenser Maximum Flow Rate 6407 GPM Condenser Minimum Flow Rate 1747 GPM Length 180.3 in. Height 121.6 in. Width 121.7 in. Total Volume Occupied 2,668,209.216 in3

Evaporator Temperature Water Outlet 45oF Evaporator Temperature Water Inlet 55oF Condenser Temperature Water Inlet 95oF Condenser Temperature Water Exit 85oF

Additionally, from selecting a chiller, we have decided to design one. For the design chiller

we used a similar approach to the first project.

The first calculation is the tube clearance. This is the clearance between adjacent tubes. The

equation to calculate this value is as followed: 𝐶 = 𝑃𝑡 − 𝑂𝐷𝑡 (50)

Where Pt is the tube pitch and ODt is the outer diameter of the tube. The tube pitch for our

design was chosen as a square pitch. This means that if you were to look at a cross sectional

view of the tubes, they would be arranged in square patterns.

The next parameter to calculate is the baffle spacing, which can be calculated as followed:

𝐵 = 𝐿

𝑁𝐵+1 (51)

Where L is the length of the tubes and NB is the number of baffles within the heat exchanger

which is user defined. The more baffles within a heat exchanger increases the effectiveness

because it improves heat transfer. The improved heat transfer is due to the more resistance

the incoming water encounters as it passes through the heat exchanger. This gives the

water a more directed path and allows it to better transfer heat between the tubes.

With the baffle spacing and clearance calculated, the flow area within the shell can be

calculated as:

G r o u p I | 63

𝐴𝑠 =𝐷𝑠𝐶𝐵

𝑃𝑡 (52)

Where Ds is the user defined shell diameter, C is the clearance previously calculated, B is the

baffle spacing also previously calculated, and Pt is the user defined tube pitch.

The tube flow area can be calculated as followed:

𝐴𝑡 =𝑁𝑡𝜋(𝐼𝐷𝑡

2)

4𝑁𝑝 (53)

Nt is the number of tubes within the exchanger which is reliant upon the shell diameter and the

number of passes. IDt is the inner tube diameter which is user defined along with Np which is

the number of passes.

With the flow areas calculated, the shell and tube velocities can now be found:

𝑉𝑠 =�̇�𝑠

𝜌𝐴𝑠 (54)

Where ms is the water flow rate into the shell, 𝜌 is the density of the water, As is the flow area of

the water into the shell.

The flow velocity can also be calculated for the tube fluid as followed:

𝑉𝑡 =�̇�𝑡

𝜌𝐴𝑡 (55)

Where mt is the mass flow rate of the refrigerant passing through the tubes, 𝜌 is the density of the

vapor within the two phase flow of refrigerant, and At, as previously defined, is the tube flow

area.

The next value to calculate is the equivalent diameter. The calculation of this value depends on

the pitch chosen, triangular or square. For our design, as previously mentioned, a square pitch

was chosen. Therefore, the calculation is as followed:

𝐷𝑒 =4𝑃𝑡2−𝑝𝑖(𝑂𝐷𝑡2)

𝑝𝑖(𝑂𝐷𝑡) (56)

Where Pt is the tube pitch and ODt is the outer diameter of the tubes.

The next value to calculate are the Reynolds number for the shell and tube. These are calculated

using the following formulas:

𝑅𝑒𝑡 =𝑉𝑡𝐼𝐷𝑡𝜌

𝜇 (57)

And,

𝑅𝑒𝑠 =𝑉𝑠𝐷𝑒𝜌

𝜇 (58)

Where Ret is the Reynolds number for the tube and Res is the Reynolds number for the shell. For

the tube, Vt is the tube fluid velocity, IDt is the inner tube diameter, 𝜌 is the density of the

refrigerant vapor, and 𝜇 is the dynamic viscosity of the vapor. For the shell Reynolds number, Vs

is the fluid in the shell, De is the equivalent diameter, 𝜌 is the density of the water, and 𝜇 is the

dynamic viscosity of the water within the shell.

The Nusselt number is then calculated after the Reynolds number. The shell side is the easier

calculation and is found using the following equation:

𝑁𝑢𝑠 =ℎ𝑜𝐷𝑒

𝑘𝑓= 0.36𝑅𝑒𝑠

0.55𝑃𝑟1

3 (59)

Where ho is the heat transfer coefficient in the tube, De is the equivalent diameter, and kf is the

thermal conductivity.

G r o u p I | 64

𝑁𝑢𝑡 = 𝐶1 [(𝐺𝑣𝐼𝐷𝑡

𝜇𝑙)

2

(𝐽∆𝑥𝑖𝑓𝑔𝑔𝑐

𝐿𝑔)]

𝑛

(60)

Where:

J = Joule equivalent

Δx = change in quality of R134a

ifg = enthalpy of vaporization

n = constant = 0.4 (0.3 if the fluid was being cooled)

𝜇 = dynamic viscosity of R-134a (Liquid)

𝐺𝑣 =mass flux

C1 = 8.2*10-3

L = length of tube

g = gravity

The Nusselt Number for the condenser is calculated differently than for the evaporator. This

equation depends on the value for the Reynolds number. The parameters for choosing the

equation can be seen in the problem statement. Since our Reynolds number was greater than

10,000, the equation used was:

𝑁𝑢𝑡 =ℎ𝑖𝐷𝑖

𝑘𝑙= 0.1 (

𝑐𝑝,𝑙𝜇𝑙

𝑘𝑙) (𝑃𝑟𝑙)

1

3 (𝑖𝑓𝑔

𝑐𝑝,𝑙∆𝑡)

1

6(

𝐺𝑣𝐼𝐷𝑡

𝜇𝑙(

𝜌𝑙

𝜌𝑣)

1

2)

2

3

(61)

Where:

ifg = enthalpy of vaporization

𝑐𝑝 = specific heat of R-134a

𝛥𝑇 = difference in the refrigerant temperature and the average water temperature.

𝑃𝑟 = Pandlt number

𝐺𝑣 = mass flux

𝜇 = dynamic viscosity of R-134a (Liquid)

𝜌𝑣 = density of vapor

𝜌𝑙 = density of liquid

𝑘𝑙 = thermal conductivity in liquid phase

The heat transfer coefficients can be calculated by the following three equations:

ℎ𝑖 =𝑁𝑢𝑡𝑘𝑡

𝐼𝐷𝑡 (62)

ℎ𝑡 =ℎ𝑖𝐼𝐷𝑡

𝑂𝐷𝑡 (63)

ℎ𝑜 =𝑁𝑢𝑠𝑘𝑠

𝐷𝑒 (64)

Where hi is the heat transfer coefficient for the tube, ht is the heat transfer coefficient for the

evaporator, and ho is the heat transfer coefficient for the condenser.

G r o u p I | 65

Nu is the Nusselt number for the appropriate component, tube or shell. K is the thermal

conductivity of either the fluid in the tube or shell, De is the equivalent diameter, ODt is the

outside tube diameter, and IDt is the inside tube diameter.

With the individual heat transfer coefficients solved for, the overall heat transfer coefficient can

now be calculated.

𝑈𝑜 =1

1

ℎ𝑡+

1

ℎ𝑜

(65)

Where ht and ho are the individual coefficients for the tube and shell.

The heat transfer area is the next variable to calculate and is done by the following equation:

𝐴𝑜 = 𝑁𝑡𝜋𝑂𝐷𝑡𝐿 (66)

Where Nt is the number of tubes, ODt is the outer tube diameter, and L is the length of the pipes.

The NTU (number of transfer units) can now be solved for. This involves the minimum heat

capacitance rate, Cmin, the heat transfer area, Ao, and the overall heat transfer coefficient, Uo.

𝑁𝑇𝑈 = 𝑈0𝐴0

𝐶𝑚𝑖𝑛 (67)

Now, we start to calculate our goal design values: effectiveness, outlet temperature, and pressure

drop.

The effectiveness can be calculated using the following equation:

𝜀 = 2 {1 + 𝐶 +1+𝑒

(−𝑁𝑇𝑈(1+𝐶2)1/2

)

1−𝑒(−𝑁𝑇𝑈(1+𝐶2)

1/2)

(1 + 𝐶2)1/2}

−1

(68)

Where C is a constant that is set to zero because of the presence of two phase flow.

The exit temperature of the water leaving both the evaporator and condenser can be calculated by:

𝑡𝑤,𝑜𝑢𝑡 = 𝜀(𝑡𝑅134𝑎 − 𝑡𝑤,𝑖𝑛) + 𝑡𝑤,𝑖𝑛 (69)

Where tw is the temperature of the water. TR134a is the temperature of the refrigerant, and e

is the effectiveness.

Lastly, the pressure drop can be calculated. First the friction factor must be calculated

which can be done using the chen equation:

𝑓 = ( 1

−2.0log (𝑒

3.7056𝐷−

5.0452

𝑅𝑒∗log(

1

2.8257∗(

𝑒

𝐷)

1.1098+

5.8506

𝑅𝑒0.8981)))

2

(70)

G r o u p I | 66

Where e in this case is the roughness of the material. Since we specified that the tube

material was carbon steel, the roughness was selected to be 0.00015. This allowed the

relative roughness (e/D) to be calculated using the specified absolute roughness value and

the value of the shell diameter. By using this method, the use of the Moody diagram is

eliminated and a more accurate friction factor can be used. The only other variable in this

equation is Re which is the Reynolds number in the tube.

With the friction factor calculated, the pressure drop within the tube can now be found

using the following equation:

∆𝑃𝑡 =𝜌(𝑉𝑡2)

2𝑔(

𝑓∗𝐿∗𝑁𝑝

𝐼𝐷𝑡+ 4𝑁𝑝) (71)

Where 𝜌 is the vapor refrigerant density, Vt is the tube fluid velocity, f is the friction factor,

L is the length of the tubes, Np is the number of passes, and IDt is the inside diameter of the

tubes

Table 25: Chiller Designed Specifications for Condenser

Shell Diameter 15.25 in. Inner Tube Diameter 0.56 in. Outer Tube Diameter 0.75 in. # of Tubes 137 Length of tubes 9’ Number of Passes 1 Number of Baffles 5 Pitch 1 Water Exit Temperature 95.251 oF Effectiveness 0.6588

Table 25. Shows the dimensions calculated using the spreadsheet from project 1. This

condenser can fulfill the requirements.

G r o u p I | 67

Table 26: Thermodynamic Properties for Condenser

Table 27: Chiller Designed Specifications for Evaporator

Shell Diameter 23.25 in. Inner Tube Diameter 0.56 in. Outer Tube Diameter 0.75 in. # of Tubes 341 Length of tubes 9’ Number of Passes 1 Number of Baffles 5 Pitch 1 Water Exit Temperature 45.782 oF Effectiveness 0.307

Condenser Design

Quantity Magnitude Units

Condenser Heat Transfer Rate 7072712.97 Btu/hr

Refrigerant Flow Rate 97534.9663 lbm/hr

Condenser Water Flow Rate 70868.7964 lbm/hr

Temperature of R134 Condenser 100.56 oF

Inlet Temperature of Water 85 oF

Outlet Temperature of Water 95 oF

Average Temperature of Water 90 oF

Water Properties

Specific Heat 1 Btu/lbm*R

Density 62.12 lbm/ft ^3

Dynamic Viscosity 0.0005117 lbm/ft*s

Kinematic Viscosity 0.000008027 ft^2/s

Thermal Diffusivity 0.0059599 ft^2/hr

Prandtl Number 4.844 -

Refrigerant Properties

Liquid Specific Heat 0.3552424 BTU/lbm*R

Liquid Thermal Conductivity 0.0443048 BTU/h*ft*R

Liquid Dynamic Viscosity 0.000114446 lbm/ft*s

Vapor Dynamic Viscosity 9.25515E-06 lbm/ft*s

Liquid Density 72.08712 lbm/ft^3

Vapor Density 2.963 lbm/ft^3

G r o u p I | 68

Table 27 shows the specifications of an evaporator that can meet the requirements. It was

found through the process of using the excel spreadsheet from project 1.

Table 28: Thermodynamic Properties for Evaporator

Evaporator Design Quantity Magnitude Units

Evaporator Heat Transfer Rate 6000000 Btu/hr

Evaporator Water Flow Rate 600000 lbm/hr

Temperature of R134 Evaporator 30 F

Inlet Temperature of Water 55 F

Outlet Temperature of Water 45 F

Average Temperature of Water 50 F

Water Properties

Specific Heat 0.97663 Btu/lbm*R

Density 62.403 lbm/ft ^3

Dynamic Viscosity 0.0000273 lbm/ft*s

Kinematic Viscosity 0.00001407 ft^2/s

Thermal Diffusivity 0.0055772 ft^2/hr

Prandtl Number 8.8027 -

Refrigerant Properties

Liquid Specific Heat 0.3185 BTU/lbm*R

Liquid Thermal Conductivity 0.05565 BTU/h*ft*R

Liquid Dynamic Viscosity 0.0001883 lbm/ft*s

Vapor Dynamic Viscosity 0.00000496 lbm/ft*s

Liquid Density 81.8 lbm/ft^3

Vapor Density 0.866 lbm/ft^3

G r o u p I | 69

VII. Cooling Tower

Cooling Tower Calculations

Temperature Hot H20= 95 oF

Temperature Cold H20= 85 oF

Wet Bulb Temperature of Air =82oF

Heat Load =Qhcond=14,145,425.94 Btu/hr

Mass Flow Rate of H20= 1,415,958.552 lbm/hr

Mass Flow Rate of H20= 2827.82 gpm

Rate of Evaporation

𝑅𝑎𝑡𝑒 𝑜𝑓 𝐸𝑣𝑎𝑝𝑜𝑟𝑎𝑡𝑖𝑜𝑛 = �̇�ℎ20(∆𝑇) ∗ 0.0008 (72)

2827.82 𝑔𝑝𝑚 ∗ 10 ∗ 0.0008= 22.62256 gpm

Rate of Drift Loss

𝑅𝑎𝑡𝑒 𝑜𝑓 𝐷𝑟𝑖𝑓𝑡 𝐿𝑜𝑠𝑠 = �̇�ℎ20 ∗ 0.0002 (73)

𝑅𝑎𝑡𝑒 𝑜𝑓 𝐷𝑟𝑖𝑓𝑡 𝐿𝑜𝑠𝑠 = 2827.82 ∗ 0.0002

𝑅𝑎𝑡𝑒 𝑜𝑓 𝐷𝑟𝑖𝑓𝑡 𝐿𝑜𝑠𝑠 = 0.565564 𝑔𝑝𝑚

Rate of Blowdown

We determined the contaminate concentration to be 4

𝑅𝑎𝑡𝑒 𝑜𝑓 𝑏𝑙𝑜𝑤𝑑𝑜𝑤𝑛 = 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑒𝑣𝑎𝑝𝑜𝑟𝑎𝑡𝑖𝑜𝑛−(𝐶𝑜𝑛𝑡𝑎𝑚𝑖𝑛𝑎𝑡𝑒 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛−1)∗𝐷𝑟𝑖𝑓𝑡 𝑙𝑜𝑠𝑠

(𝐶𝑜𝑛𝑡𝑎𝑚𝑖𝑛𝑎𝑡𝑒 𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛−1) (74)

𝑅𝑎𝑡𝑒 𝑜𝑓 𝑏𝑙𝑜𝑤𝑑𝑜𝑤𝑛 = 22.623 − [(4 − 1) ∗ 0.565564]

(4 − 1)

𝑅𝑎𝑡𝑒 𝑜𝑓 𝑏𝑙𝑜𝑤𝑑𝑜𝑤𝑛 = 6.9754

G r o u p I | 70

Cooling tower Efficiency

𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦% = (𝑇ℎ𝑜𝑡 𝑤𝑎𝑡𝑒𝑟−𝑇𝑐𝑜𝑙𝑑𝑤𝑎𝑡𝑒𝑟

)∗100

𝑇ℎ𝑜𝑡 𝑤𝑎𝑡𝑒𝑟−𝑇𝑤𝑒𝑡𝑏𝑢𝑙𝑏

(76)

𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦% = (95℉ − 85℉) ∗ 100

95℉ − 82℉

𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦% = 76.923

Make up water Requirement

𝑀𝑎𝑘𝑒 𝑢𝑝 𝑤𝑎𝑡𝑒𝑟 𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑚𝑒𝑛𝑡 = ∑ 𝐿𝑜𝑠𝑠𝑒𝑠 (77)

𝑀𝑎𝑘𝑒 𝑢𝑝 𝑤𝑎𝑡𝑒𝑟 𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑚𝑒𝑛𝑡 = 22.62256 + 0.565564 + 6.9754

For the cooling tower we used the mass flow rate of determined from the thermodynamic

analysis as previously shown in the report. By finding our mass flow rate and using the

temperatures determined by the chiller we selected. With the specifications of 85 F and 95

F and a mass flow rate of 1,415,958.522 lbm/hr. We were able to plug in those values into

Baltimore’s Air Coil Company’s software to select a cooling tower.

G r o u p I | 71

Figure 41: Cooling tower input data

The data plugged in gave us the following cooling tower with its specs displayed below.

Table 29: Cooling Tower Properties

Cooling tower Specifications Design Conditions

Model Information XESE-122-10o Flow Rate 2,827.82 USGPM

Number of Units 2 Hot water Temp 95

Fan Motor 30.00 HP/Unit Cold Water Temp 85

Unit Length 11'0.975" Wet Bulb Temp 82

Unit Width 21' 0.650 Tower Pumping Head 6.71 psi

Unit Height 16'05.00" Reserve Capability 0.45%

Air Flow 165,430 Operating Weight 33,370 lbs

G r o u p I | 72

Figure 42: Single Unit cooling tower

G r o u p I | 73

VIII. Air Cooled Condenser The air cooled condenser will serve to reject the heat gained from the building into the

atmosphere. It is an alternative to using a regular water cooled chiller with a cooling tower.

Using an air cooled condenser will eliminate the need for a cooling tower entirely, therefore

saving space and water consumption. However, the lower heat transfer capabilities of air in

comparison to water mean that the system is less efficient than a cooling tower setup. Therefore,

the air cooled condenser setup will require more power than the cooling tower setup would to

remove the same amount of heat from the refrigerant.

The air cooled condenser operates by having the refrigerant’s temperature raised through the

compressor, similarly to the water cooled setup. It then enters the condenser at 125°F where it

runs through the condenser tubes and is cooled by a fan moving air over the tubes at 325000 cfm.

The refrigerant then exits the condenser at 115°F where it then moves through a throttling valve

to lower its temperature back to the required 45°F for the air handling unit.

The condenser uses ¾ in Type K copper tubing with copper fin attachments for increased heat

transfer area. The design values for the air cooled condenser are shown below in tables 30-32.

Table 30: Air Cooled Condenser Temperature, Heat Transfer, and Refrigerant Flow Rate

Air

Temperature

inlet, Tai [°F]

Air

Temperature

Outlet, Tao

[°F]

Refrigerant

Temperature

Inlet, Tri [°F]

Refrigerant

Temperature

Outlet, Tro

[°F]

Condenser

Heat

Transfer, Q

[BTU/hr]

Refrigerant

Mass Flow

Rate, �̇�

[lbm/hr]

95 105 125 115 512860 160270

Table 31: Design Dimensions for Air Cooled Condenser

Tube

Length,

L [in]

Number

of

Tubes,

Nt

Number

of Tube

Passes,

Np

Tube

Outer

Diameter,

Do [in]

Tube

Inner

Diameter,

Di [in]

Surface

Area,

As [ft2]

Tube

Pitch,

Pt [in]

Fin

Pitch,

Nf

[fin/in]

168 61 1 0.75 0.652 590.9 1 4

Table 32: Calculated Values for Air Cooled Condenser

Tube

Fluid

Velocity,

Vt [ft/s]

Air

Volumetric

flow rate,

𝑉�̇� [cfm]

Overall

Heat

Transfer

Coefficient,

U [BTU/hr-

ft2-°F]

Pressure

Drop, p

[ft H2O]

4.7 325000 103 11.74

The calculation procedure for the design of the air cooled condenser is shown below.

To begin, the inlet and outlet temperatures for both fluids were chosen. The air was chosen at

95°F inlet as that is the temperature for a typical summer day in southern Louisiana. The

refrigerant was chosen to be 125°F at the inlet so as to always be above the air temperature.

G r o u p I | 74

The temperature difference for both fluids was chosen to be 10°F as this is a typical design value

for a cross flow heat exchanger.

The required heat transfer for the condenser was already known before the design, so the mass

flow rate for the refrigerant could be calculated using equation 78 below.

�̇�𝑟 = �̇�/𝐶𝑝𝑟∆𝑇𝑟 (78)

Where �̇�𝑟is the mass flow rate of the refrigerant, �̇� is the heat transfer of the condenser, 𝐶𝑝𝑟 is

the specific heat of the refrigerant, and ∆𝑇𝑟 is the temperature difference of the refrigerant

between he inlet and outlet of the condenser.

Next the log mean temperature difference is calculated using equation 79 below.

𝐿𝑀𝑇𝐷 =∆𝑇𝑖−∆𝑇𝑜

ln (∆𝑇𝑖−∆𝑇𝑜) (79)

After this a value for U, the overall heat transfer coefficient, is assumed in order to calculate the

total heat transfer area that is required. The value of U will be iterated to produce a valid result.

The calculation for the heat transfer area is shown below in equation 80.

𝐴𝑠 =�̇�

𝐿𝑀𝑇𝐷∗𝑈𝑎𝑠𝑠𝑢𝑚𝑒𝑑 (80)

Next the tube dimensions are chosen in order to calculate the single tube and fin surface area,

which is used to calculate the required number of tubes. This is all shown below in equations 81

and 82.

𝐴𝑠𝑡 = (𝜋𝐷𝑜𝐿) + (2𝐻𝐿) (81)

Where 𝐴𝑠𝑡 is the surface area of a single tube, 𝐷𝑜 is the tube outer diameter, 𝐿 is the tube length,

and 𝐻 is the fin height.

𝑁𝑡 = 𝐴𝑠/𝐴𝑠𝑡 (82)

Next, the Tube velocity must be found as it needs to be within the typical design value of 2-5

ft/s. This was done by first choosing the inner tube diameter so as to find the tube’s flow area,

which is then used to find the velocity. This is shown below in equations 83 and 84.

𝐴𝑐 = 𝑁𝑡𝐷𝑖2 𝜋

4 (83)

Where 𝐴𝑐 is the flow area of the tubes and 𝐷𝑖 is the inner tube diameter.

𝑉𝑡 = 𝑚𝑟/̇𝑁𝑡

𝑁𝑝/

𝜋

4𝜌𝑟𝐷𝑖

2 (84)

Where 𝑁𝑝 is the number of tube passes and 𝜌𝑟 is the density of the refrigerant.

Next the mass flow rate of the air must be calculated. This was done using equation 85 below.

𝑚𝑎̇ =�̇�

𝜌𝑎 (85)

Where 𝑚𝑎 is the mass flow rate of the air, �̇� is the volumetric flow rate of the air, and 𝜌𝑎 is the

density of the air.

Next the flow area for the air must be calculated.

𝐴𝑓 = 𝑁𝑡(𝐷𝑓(𝐿 − 𝐿𝑁𝑓𝑡𝑏) + 2(𝑁𝑓 − 1)𝐻 (𝑡𝑏

2+

𝑡𝑡

2)) (86)

Where 𝐷𝑓 is the fin diameter, 𝑁𝑓 is the number of fins per inch, 𝑡𝑏 is the fin width at the base,

and 𝑡𝑡 is the fin width at the tip.

Next, the Reynolds number for the air side must be calculated. This is shown in equation 87

G r o u p I | 75

𝑅𝑒𝑑𝑜 = 𝐷𝑜𝑚𝑎̇

𝜇𝑎𝐴𝑓 (87)

Where 𝑅𝑒𝑑𝑜 is the Reynold’s number for the flow of air outside the tubes, and 𝜇𝑎 is the viscosity

of the air.

The Reynolds number is then used in conjunction with a tabulated Prandtl number to calculate

the Nusselt number for the flow outside the tubes. This is shown in equation 88.

𝑁𝑢𝑜 = 0.027𝑅𝑒𝑑𝑜0.805𝑃𝑟𝑎

1/3 (88)

Where 𝑁𝑢𝑜 is the Nusselt number for the flow of air outside the tubes and 𝑃𝑟𝑎 is the Prandtl

number of the air.

The definition of the Nusselt number shown below in equation 89 is then used to calculate the

outside heat transfer coefficient.

𝑁𝑢 =ℎ𝐷

𝑘 (89)

Where h is the heat transfer coefficient and k is the thermal conductivity.

Next, the inside heat transfer coefficient was found through an empirical relation shown below in

equation 90.

𝑁𝑢𝑖 = 0.1(𝐶𝑝𝑟𝜇𝑟

𝑘𝑟)(𝑃𝑟𝑟

1

3)(𝑖𝑓𝑔𝑟

𝐶𝑝𝑟(𝑇𝑟𝑖−𝑇𝑟𝑜))

1

6(𝐷𝑖�̇�

𝐴𝑐𝑁𝑡∗

𝜌𝑙

𝜌𝑣

1

2) 2

3 (90)

Where 𝑁𝑢𝑖 is the Nusselt number for the flow inside of the tubes, 𝜇𝑟 is the viscosity of the

refrigerant, 𝑘𝑟 is the thermal conductivity of the refrigerant, 𝑃𝑟𝑟 is the Prandtl number for the

refrigerant, 𝑖𝑓𝑔𝑟 is the enthalpy of vaporization for the refrigerant, 𝐶𝑝𝑟 is the specific heat of the

refrigerant, 𝑇𝑟𝑖 − 𝑇𝑟𝑜 is the temperature difference between the inlet and outlet for the

refrigerant, 𝜌𝑙 is the liquid density of the refrigerant, and 𝜌𝑣 is the vapor density of the

refrigerant.

This Nusselt number is then used to calculate the inside heat transfer coefficient

Next we need to calculate the fin efficiency using equation 91 below

𝜂 =

√ℎ𝑜𝑃𝑘𝑓𝐴𝑠𝑓(𝑇𝑤−𝑇𝑎)tanh (√ℎ𝑜𝑃

𝑘𝑓𝐴𝑐𝑓)

ℎ𝑜𝐴𝑠𝑓(𝑇𝑤−𝑇𝑎) (91)

Where 𝑃 is the perimeter of the fin, 𝑘𝑓 is the thermal conductivity of the fin, 𝐴𝑠𝑓 is the fin

surface area, 𝑇𝑤 is the wall temperature of the tube, 𝑇𝑎 is the temperature of the air, and 𝐴𝑐𝑓 is

the cross sectional area of the fin.

Lastly, the overall heat transfer coefficient is calculated using equation 92 below.

𝑈 =1

1

ℎ𝑜𝜂+

1

ℎ𝑖𝜂+

ln (𝐷𝑜𝐷𝑖

)𝐷𝑖𝜋𝐿

2𝑘𝑓𝐿

(92)

Where 𝑈 is the overall heat transfer coefficient.

The pressure drop for the tubes was also calculated using equation 93 below

𝑝 = (𝜌𝑟𝑉𝑡

2

2∗32.2) (

𝐿𝑓𝑁𝑝

𝐷𝑖) + 4𝑁𝑝 (93)

Where 𝜌𝑟 is the refrigerant density and 𝑓 is the friction factor for the tubing obtained from a

moody chart.

G r o u p I | 76

The design values in the previous tables were found by iterating the dimensions for the fins and

tubes, the assumed overall heat transfer coefficient which is used to determine a surface area, as

well as the flow rate for the air.

In order to meet the building’s required load, the following air cooled condenser was selected. It

is a CAUJC60G20002 Trane 60-ton Air Cooled Condenser. A picture of it is shown below in

Figure 1.

Figure 43: Trane CAUJC60G20002 Air cooled Condenser.

It is able to provide the 512860 BTU/hr needed to keep the refrigerant temperature where it

needs to be. It also uses less air and is about 3 feet less in length than the design specifications. It

is also capable of operating within any ambient temperature ranges that would present

themselves within southern Louisiana, with ambient entering air temperatures up to 115 °F. It

also has similar head losses to the design specifications, with head losses of 5 to 10.9 ft of head

depending on the flow rate. It also comes with variable air volume controls as an option, for

when it doesn’t need to operate at full capacity.

G r o u p I | 77

X. Conclusions

The overall process of designing the HVAC System started with the zoning and psychometrics of each room. The psychometrics were performed using the psych app, an excel spreadsheet and the psychometric charts. The sensible, and latent loads were provided for each space by the problem statement. The next step from the psychometrics was to determine the velocities, and flow areas in order to create the ductwork. For the ductwork, we used the software Revit. The Revit outputted the necessary diameters for the return and supply ductwork. Thermodynamic analysis was for performed the whole building to determine the required temperatures and specifications of the AHU. Before the calculation process was carried out, a few design parameters had to be chosen based upon typical industry standards. After we determined the required temperatures needed for the Air Handling Unit, we were able to determine the required mass flowrates needed from the chiller plant. After that we were able to design hydronic system that connected the chiller plant with the Air handling unit. For the hydronic system we ended up selecting 2 pumps that will make the water flow from the chiller to the AHU. The primary pump will cycle water through the chiller, while the secondary pump will supply the AHU. There are two of each pump to allow redundancy in case of failure or service. The primary pumps have a flow rate of 2400gpm and the secondary has a 103 gpm flowrate. They both run with a motor speed of 1760 rpm. The chiller to cooling tower loop also has redundancy with both pumps being able to handle the full load independent of the other. The flow rate of the cooling tower loop pumps is 2900 gpm, with a motor speed of 1760 rpm.. For the chiller plant, we first started with a thermodynamic analysis to determine the specifications for the chiller plant. Once we calculated, the mass flow rate of the refrigerant 134-A, along with the temperatures, we selected 2 chiller plant from TRANE. Each chiller has a capacity of 1300 Tons. However, we will be running each one at 500 tons to meet the 1000 tonnage max requirement. Overall, each chiller plant will be running at 38.5% of there full capacity. Alternatively, we designed a chiller plant using the spreadsheet from After that, with the specifications of 85 °F and 95°F and a mass flow rate of 1,415,958.522 lbm/hr., we inputted those values into the Baltimore Air Coil Company software to determine the specifications of the cooling tower. The efficiency of the cooling tower was calculated to be 76.923%. As an alternative we had selected to use air cooled condenser. The unit would rest on the rooftop of the building and supply the 512860 BTU/hr needed to keep the refrigerant at the

required temperature

G r o u p I | 78

XI. References

1. McQuiston, Faye C.; Parker, Jerald D.; Spitler, Jeffery D. Heating, Ventilation, and Air

Conditioning. 6th Edition. New York: John Wiley & Sons, 2005

2. Cengel, Yunus. Heat and Mass Transfer: Fundamentals and Applications. 5th Edition.

New York: McGraw-Hill, 2015

3. Ingadóttir, Sigríður Bára. "Air Cooled Condensers." A-to-Z Guide to Thermodynamics,

Heat and Mass Transfer, and Fluids Engineering (2014): Web.

4. http://www.tacohvac.com/products/variable_speed_products/selfsensing_series/i

ndex.html

5. Posted October 25, 2012 by Jim Shiminski. "Custom Air Handling Units | Coil Selection

Guidelines." DAC SALES. N.p., n.d. Web. 07 Dec. 2016

6. Mitrovic, Jovan. Heat Exchangers: Basics Design Applications. Rijeka: InTech, 2012. Print

G r o u p I | 79

XII: APPENDIX

G r o u p I | 80

APPENDIX II CHILLER BROCHURE

G r o u p I | 81

G r o u p I | 82

G r o u p I | 83

G r o u p I | 84

G r o u p I | 85

G r o u p I | 86

G r o u p I | 87

APPENDIX:III REVIT FILES

Return Line Total Pressure Loss Calculations by Sections

Section Element Flow Size Velocity Velocity Pressure

Length Loss

Coefficient Friction

Total Pressure Loss

Section Pressure

Loss

1

Fittings 683 CFM - 0 FPM 0.00 in-wg - 0.50538 - 0.00 in-wg 0.06 in-wg

Air Terminal 683 CFM - - - - - - 0.06 in-wg

2

Duct 683 CFM - 490 FPM - 6' - 0 19/32" - 0.02 in-

wg/100ft 0.00 in-wg

0.01 in-wg Fittings 683 CFM - 490 FPM 0.01 in-wg - 0.687508 - 0.01 in-wg

3 Fittings 683 CFM - 0 FPM 0.00 in-wg - -29.315397 - -0.01 in-wg -0.01 in-wg

4

Duct 4309 CFM 36"x36" 480 FPM - 41' - 4 23/32"

- 0.01 in-

wg/100ft 0.00 in-wg

0.02 in-wg

Fittings 4309 CFM - 480 FPM 0.01 in-wg - 1.176644 - 0.02 in-wg

5 Fittings 4309 CFM - 0 FPM 0.02 in-wg - 4.094168 - 0.07 in-wg 0.07 in-wg


7

Duct 7654 CFM 24"x43" 1070 FPM - 6' - 2 3/4" - 0.04 in-

wg/100ft 0.00 in-wg

0.01 in-wg


8

Duct 7654 CFM 43"x26" 990 FPM - 0' - 7 3/8" - 0.04 in-

wg/100ft 0.00 in-wg

0.08 in-wg


9

Duct 10722 CFM 43"x26" 1380 FPM - 4' - 11 5/8" - 0.07 in-

wg/100ft 0.00 in-wg


10

Fittings 10722 CFM - 0 FPM 0.24 in-wg - 0 - 0.00 in-wg 0.00 in-wg

Equipment 10722 CFM - - - - - - 0.00 in-wg

11



12

Duct 220 CFM - 400 FPM - 9' - 4 3/32" - 0.03 in-

wg/100ft 0.00 in-wg

0.01 in-wg



14

Duct 1802 CFM 24"x24" 450 FPM - 15' - 1 25/32"

- 0.01 in-

wg/100ft 0.00 in-wg


G r o u p I | 88


16

Duct 2410 CFM 28"x28" 440 FPM - 27' - 3 19/32"

- 0.01 in-

wg/100ft 0.00 in-wg

0.00 in-wg



18

Duct 3018 CFM 30"x30" 480 FPM - 27' - 0 19/32"

- 0.01 in-

wg/100ft 0.00 in-wg



20

Duct 3626 CFM 32"x32" 510 FPM - 28' - 8 7/8" - 0.01 in-

wg/100ft 0.00 in-wg

0.01 in-wg



22



23

Duct 220 CFM - 400 FPM - 9' - 7 7/8" - 0.03 in-

wg/100ft 0.00 in-wg

0.01 in-wg




26

Duct 1582 CFM 23"x23" 430 FPM - 20' - 8 19/32"

- 0.01 in-

wg/100ft 0.00 in-wg

0.00 in-wg



28



29

Duct 588 CFM - 420 FPM - 7' - 0 3/32" - 0.02 in-

wg/100ft 0.00 in-wg

0.01 in-wg


30 Fittings 588 CFM - 0 FPM 0.00 in-wg - -7.236559 - 0.00 in-wg 0.00 in-wg

31

Duct 2882 CFM 32"x32" 410 FPM - 18' - 1 1/4" - 0.01 in-

wg/100ft 0.00 in-wg

0.00 in-wg


32

Duct 3346 CFM 32"x32" 470 FPM - 31' - 7 1/2" - 0.01 in-

wg/100ft 0.00 in-wg

0.02 in-wg



34



G r o u p I | 89

35

Duct 588 CFM - 420 FPM - 7' - 7 1/32" - 0.02 in-

wg/100ft 0.00 in-wg



37

Duct 2293 CFM 28"x28" 420 FPM - 26' - 11 5/32"

- 0.01 in-

wg/100ft 0.00 in-wg

0.00 in-wg



39



40

Duct 588 CFM - 420 FPM - 8' - 1 31/32" - 0.02 in-

wg/100ft 0.00 in-wg

0.01 in-wg



42

Duct 1705 CFM 24"x24" 430 FPM - 27' - 1 25/32"

- 0.01 in-

wg/100ft 0.00 in-wg

0.00 in-wg



44



45

Duct 314 CFM - 340 FPM - 8' - 10 9/16" - 0.02 in-

wg/100ft 0.00 in-wg

0.00 in-wg



47

Duct 695 CFM 16"x16" 390 FPM - 17' - 0 7/32" - 0.02 in-

wg/100ft 0.00 in-wg



49

Duct 906 CFM 18"x18" 400 FPM - 20' - 6 19/32"

- 0.01 in-

wg/100ft 0.00 in-wg

0.00 in-wg



51

Duct 1117 CFM 20"x20" 400 FPM - 16' - 6 5/16" - 0.01 in-

wg/100ft 0.00 in-wg



53



54 Duct 381 CFM - 360 FPM - 11' - 10 - 0.02 in- 0.00 in-wg 0.00 in-wg

G r o u p I | 90

27/32" wg/100ft


55

Duct 381 CFM 12"x12" 380 FPM - 23' - 7 5/32" - 0.02 in-

wg/100ft 0.00 in-wg

0.01 in-wg



57



58

Duct 211 CFM - 390 FPM - 9' - 8 23/32" - 0.03 in-

wg/100ft 0.00 in-wg

0.01 in-wg



60



61

Duct 211 CFM - 390 FPM - 10' - 0 5/32" - 0.03 in-

wg/100ft 0.00 in-wg

0.01 in-wg



63



64

Duct 608 CFM - 440 FPM - 6' - 9 1/16" - 0.02 in-

wg/100ft 0.00 in-wg

0.01 in-wg



66



67

Duct 608 CFM - 440 FPM - 7' - 0 3/4" - 0.02 in-

wg/100ft 0.00 in-wg

0.01 in-wg



69



70

Duct 608 CFM - 440 FPM - 7' - 4 15/32" - 0.02 in-

wg/100ft 0.00 in-wg

0.01 in-wg



72



G r o u p I | 91

73

Duct 454 CFM - 370 FPM - 14' - 7 19/32"

- 0.02 in-

wg/100ft 0.00 in-wg



75

Duct 1362 CFM 22"x22" 410 FPM - 21' - 1 3/8" - 0.01 in-

wg/100ft 0.00 in-wg

0.01 in-wg


76



77

Duct 454 CFM - 370 FPM - 15' - 2 1/2" - 0.02 in-

wg/100ft 0.00 in-wg

0.00 in-wg



79

Duct 908 CFM 18"x18" 400 FPM - 14' - 9 5/32" - 0.01 in-

wg/100ft 0.00 in-wg



81



82

Duct 454 CFM - 370 FPM - 8' - 0 3/4" - 0.02 in-

wg/100ft 0.00 in-wg


83

Duct 454 CFM 13"x13" 390 FPM - 36' - 5 7/32" - 0.02 in-

wg/100ft 0.01 in-wg

0.02 in-wg



85



86

Duct 464 CFM - 330 FPM - 7' - 0 13/16" - 0.01 in-

wg/100ft 0.00 in-wg

0.01 in-wg



88

Duct 3068 CFM 24"x26" 710 FPM - 10' - 10 31/32"

- 0.03 in-

wg/100ft 0.00 in-wg



Critical Path : 88-9-10 ; Total Pressure Loss : 0.63 in-wg

G r o u p I | 92

Main Supply Line Total Pressure Loss Calculations by Sections

Section Element Flow Size Velocity Velocity Pressure Length

Loss Coefficient Friction

Total Pressure Loss

Section Pressure

Loss

1

Duct 823 CFM 16"ø 590 FPM - 0' - 1 13/32" - 0.03 in-

wg/100ft 0.00 in-wg

0.10 in-wg



2

Duct 9114 CFM 32"x36" 1140 FPM - 28' - 9 1/32" - 0.04 in-

wg/100ft 0.01 in-wg


3

Duct 10072 CFM 32"x36" 1260 FPM - 9' - 7 3/8" - 0.05 in-

wg/100ft 0.01 in-wg


4

Duct 10723 CFM 32"x36" 1340 FPM - 18' - 8 1/32" - 0.06 in-

wg/100ft 0.01 in-wg

2.14 in-wg



5

Duct 823 CFM 16"ø 590 FPM - 0' - 1 13/32" - 0.03 in-

wg/100ft 0.00 in-wg

0.08 in-wg



6

Duct 7438 CFM 32"x36" 930 FPM - 26' - 7 11/32" -

0.03 in-wg/100ft 0.01 in-wg


7

Duct 8291 CFM 32"x36" 1040 FPM - 3' - 4 21/32" - 0.04 in-

wg/100ft 0.00 in-wg


8

Duct 823 CFM 16"ø 590 FPM - 0' - 4 11/32" - 0.03 in-

wg/100ft 0.00 in-wg

0.11 in-wg



9

Duct 5762 CFM 28"x28" 1060 FPM - 22' - 6 9/32" - 0.05 in-

wg/100ft 0.01 in-wg


10

Duct 5762 CFM 32"x36" 720 FPM - 3' - 2 1/32" - 0.02 in-

wg/100ft 0.00 in-wg


G r o u p I | 93

11

Duct 6615 CFM 32"x36" 830 FPM - 3' - 4 21/32" - 0.02 in-

wg/100ft 0.00 in-wg


12

Duct 592 CFM 16"ø 420 FPM - 0' - 4 11/32" - 0.02 in-

wg/100ft 0.00 in-wg

0.06 in-wg



13

Duct 4087 CFM 28"x28" 750 FPM - 27' - 0 23/32" -

0.03 in-wg/100ft 0.01 in-wg


14

Duct 4939 CFM 28"x28" 910 FPM - 3' - 6 1/32" - 0.04 in-

wg/100ft 0.00 in-wg


15

Duct 441 CFM 16"ø 320 FPM - 0' - 1 1/2" - 0.01 in-

wg/100ft 0.00 in-wg

0.02 in-wg



16

Duct 974 CFM 18"x20" 390 FPM - 19' - 5 11/32" -

0.01 in-wg/100ft 0.00 in-wg


17

Duct 974 CFM 20"x20" 350 FPM - 0' - 0 3/4" - 0.01 in-

wg/100ft 0.00 in-wg




20

Duct 2877 CFM 28"x28" 530 FPM - 3' - 8 9/16" - 0.01 in-

wg/100ft 0.00 in-wg


21

Duct 3494 CFM 28"x28" 640 FPM - 2' - 11 9/32" - 0.02 in-

wg/100ft 0.00 in-wg


22

Duct 533 CFM 16"ø 380 FPM - 0' - 2 9/16" - 0.01 in-

wg/100ft 0.00 in-wg

0.01 in-wg



23

Duct 533 CFM 16"x16" 300 FPM - 21' - 1" - 0.01 in-

wg/100ft 0.00 in-wg


24 Duct 533 CFM 18"x20" 210 FPM - 5' - 3 21/32" - 0.00 in- 0.00 in-wg 0.00 in-wg

G r o u p I | 94

wg/100ft


25

Duct 958 CFM 16"ø 690 FPM - 0' - 1 1/32" - 0.04 in-

wg/100ft 0.00 in-wg

0.13 in-wg



26

Duct 853 CFM 16"ø 610 FPM - 0' - 1 1/32" - 0.03 in-

wg/100ft 0.00 in-wg

0.09 in-wg



27

Duct 853 CFM 16"ø 610 FPM - 0' - 1 1/32" - 0.03 in-

wg/100ft 0.00 in-wg

0.08 in-wg



28

Duct 853 CFM 16"ø 610 FPM - 0' - 3 1/32" - 0.03 in-

wg/100ft 0.00 in-wg

0.09 in-wg



29

Duct 617 CFM 16"ø 440 FPM - 0' - 3 1/32" - 0.02 in-

wg/100ft 0.00 in-wg

0.05 in-wg



30

Duct 1903 CFM 20"ø 870 FPM - 2' - 0 1/4" - 0.05 in-

wg/100ft 0.00 in-wg

0.01 in-wg




32

Duct 651 CFM 16"ø 470 FPM - 1' - 4 3/4" - 0.02 in-

wg/100ft 0.00 in-wg

0.10 in-wg



Critical Path : 4-3-2-7-6-11-10-9-8 ; Total Pressure Loss : 2.43 in-wg

G r o u p I | 95

Classroom 1 Total Pressure Loss Calculations by Sections



Total Pressure Loss

Section Pressure Loss

4

Duct 392 CFM 12"ø 500 FPM - 5' - 7 1/4" - 0.03 in-wg/100ft 0.00 in-wg


5 Fittings 784 CFM - 0 FPM 0.03 in-wg - 0 - 0.00 in-wg 0.00 in-wg

6

Duct 784 CFM 14"x14" 580 FPM - 14' - 0 3/32" - 0.04 in-wg/100ft 0.01 in-wg


7

Fittings 784 CFM - 0 FPM 0.03 in-wg - 0 - 0.00 in-wg

0.00 in-wg Equipment 784 CFM - - - - - - 0.00 in-wg


11



12


0.01 in-wg



15


0.01 in-wg



16


0.01 in-wg



18


0.00 in-wg Fittings 0 CFM - 0 FPM 0.00 in-wg - 0 - 0.00 in-wg

19


0.01 in-wg



Critical Path : 7-6-5-11-16 ; Total Pressure Loss : 0.05 in-wg

G r o u p I | 96




Total Pressure Loss


3



4




6



7



10



11



12


0.01 in-wg



15


0.01 in-wg



18


0.01 in-wg



19


0.01 in-wg




G r o u p I | 97




Total Pressure Loss



4




6

Duct 784 CFM 14"x14" 580 FPM - 13' - 11 5/32" -

0.04 in-wg/100ft 0.01 in-wg


7




11



12


0.01 in-wg



15


0.01 in-wg



18


0.01 in-wg



19


0.01 in-wg




G r o u p I | 98




Total Pressure Loss



7


0.01 in-wg




9


0.01 in-wg



15


0.01 in-wg



16






23



24



25

Duct 812 CFM 14"x14" 600 FPM - 13' - 5 17/32" -

0.04 in-wg/100ft 0.01 in-wg


26


0.01 in-wg



Critical Path : 24-25-4-21-23-26 ; Total Pressure Loss : 0.04 in-wg

G r o u p I | 99




Total Pressure Loss


3




9


0.02 in-wg



13



14


0.02 in-wg




16


0.01 in-wg




25



26



27

Duct 406 CFM 12"ø 520 FPM - 6' - 9" - 0.04 in-wg/100ft 0.00 in-wg


28



29



G r o u p I | 100






Total Pressure Loss




8



10


0.01 in-wg



12


0.01 in-wg



14



15


0.01 in-wg



16



19



20




G r o u p I | 101

22





25


0.01 in-wg







Total Pressure Loss


3




6


0.00 in-wg



8

Duct 0 CFM 12"ø 0 FPM - 0' - 11 13/32" -

0.00 in-wg/100ft 0.00 in-wg


9


0.00 in-wg



11


0.00 in-wg



18



22 Duct 456 CFM 12"ø 580 FPM - 7' - 2 15/32" - 0.05 in- 0.00 in-wg 0.00 in-wg

G r o u p I | 102

wg/100ft


26



29

Duct 912 CFM 15"x15" 580 FPM - 13' - 3 15/32" -

0.03 in-wg/100ft 0.00 in-wg




36



37



38



39



40


0.00 in-wg



Critical Path : 26-29-4-30-22-37-11 ; Total Pressure Loss : 0.07 in-wg

G r o u p I | 103

Lavatories Total Pressure Loss Calculations by Sections



Total Pressure Loss



4




6

Duct 564 CFM 14"x14" 410 FPM - 13' - 11 5/32" -

0.02 in-wg/100ft 0.00 in-wg


7




11



12


0.00 in-wg



15


0.00 in-wg



18


0.00 in-wg



19


0.00 in-wg




G r o u p I | 104

Office 1

Total Pressure Loss Calculations by Sections



Total Pressure Loss


4




6



7



10



11



12


0.00 in-wg



15


0.00 in-wg



16


0.00 in-wg



18



19


0.00 in-wg



G r o u p I | 105


Office 2 Total Pressure Loss Calculations by Sections



Total Pressure Loss


3



4




6

Duct 420 CFM 14"x14" 310 FPM - 10' - 6 31/32" -

0.01 in-wg/100ft 0.00 in-wg


7



10



11



12


0.00 in-wg



15


0.00 in-wg



18 Duct 105 CFM 8"ø 300 FPM - 6' - 1 1/2" - 0.02 in-wg/100ft 0.00 in-wg 0.00 in-wg

G r o u p I | 106



19


0.00 in-wg




Conference Rooms Total Pressure Loss Calculations by Sections



Total Pressure Loss


3




7


0.00 in-wg




9


0.00 in-wg



13


0.00 in-wg




17



G r o u p I | 107



24



25

Duct 588 CFM 14"x14" 430 FPM - 13' - 5 17/32" -

0.02 in-wg/100ft 0.00 in-wg


26


0.00 in-wg




Lobby Total Pressure Loss Calculations by Sections



Total Pressure Loss


3




7



10



13


0.00 in-wg Air Terminal 302 CFM - - - - - - 0.00 in-wg

14



15



16



G r o u p I | 108

17



21



23



25



28






38


0.04 in-wg



40



41



45



47




Hallways Total Pressure Loss Calculations by Sections

G r o u p I | 109



Total Pressure Loss


1



2



3

Duct 155 CFM 12"x12" 160 FPM - 0' - 8" - 0.00 in-wg/100ft 0.00 in-wg

0.06 in-wg




5


0.06 in-wg




7

Duct 465 CFM 14"ø 430 FPM - 33' - 0 23/32" -

0.02 in-wg/100ft 0.01 in-wg


8


0.06 in-wg




10



11



12

Duct 155 CFM 14"ø 140 FPM - 30' - 10 29/32" -

0.00 in-wg/100ft 0.00 in-wg


Critical Path : 1-2-4-3 ; Total Pressure Loss : 0.08 in-wg

G r o u p I | 110

Documents

HVAC Design Project