DESIGN OF A SINGLE-ZONE HVAC SYSTEM FOR A PUBLIC LIBRARY IN DALLAS, TEXAS

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D

ESIGN OF A

S

INGLE-

Z

ONE

H

VAC

S

YSTEM FOR A

P

UBLIC

L

IBRARY

IN

D

ALLAS,

T

EXAS

ME 260 - Heating, Ventilation and Air Conditioning George Washington University

Spring Semester 1999

Submitted to: Professor A. M. Kiper Submitted by: John F. Woodall

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TABLE OF CONTENTS

Section Page Number

REFERENCES ... 1

INTRODUCTION ... 2

DESIGN CONDITIONS ... 2

BUILDING DESCRIPTION ... 3

CALCULATION OF SPACE HEATING LOAD ... 4

CALCULATION OF SPACE COOLING LOAD ... 9

AIR-CONDITIONING EQUIPMENT SELECTION ... 15

AIR-HANDLING/DISTRIBUTION SYSTEM DESIGN ... 17

SUMMARY ... 25

APPENDICES

APPENDIX A ... Circular of Requirements APPENDIX B ... Drawing Sheets APPENDIX C ... Calculations and Process Diagrams APPENDIX D ... Selected Component Vendor Data

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REFERENCES

1) "Heating, Ventilating, and Air Conditioning - Analysis and Design," Faye C. McQuiston and Jerald D. Parker, John Wiley & Sons, Inc., New York,  1994.

2) "1993 ASHRAE Handbook - FUNDAMENTALS," American Society of Heating, Refrigerating, and Air Conditioning, Inc., Atlanta, GA, Chapter 27 - Fenestration.

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INTRODUCTION:

This report documents the preliminary design of a single-zone air conditioning system for a public library in Dallas, Texas. The circular-of-requirements (COR) for the HVAC system is provided in Appendix A. When information, beyond that provided in Appendix A, was necessary for design calculations, assumptions were necessary. These assumptions are stated throughout this report, where applicable.

A significant portion of the formulas, tables, and design methodology needed for the design of the subject HVAC system were taken from Reference (1). However, the data used in Reference (1) is based on current ASHRAE standards, and the ASHRAE database. Therefore, this report adheres to current ASHRAE practice and standards.

Appendix B is a set of drawing sheets, which include various library arrangements, drawings, and system schematics. These drawings are referenced throughout the report where applicable. Appendix C contains system design calculations and process diagrams, and Appendix D contains vendor provided data for several selected HVAC system components.

DESIGN CONDITIONS:

The design indoor and outdoor conditions are prescribed in Appendix A, and are as follows:

Summer

Indoor: Outdoor: 75

o_{F dry bulb, 50% Relative Humidity}

2.5% Design Conditions

Winter

Indoor: Outdoor: 72

o_{F dry bulb, 30% Relative Humidity}

97.5% Design Conditions

For Dallas, Texas from Ref. (1), Table C-1, the Summer 2.5% design conditions are 100oF dry bulb and 75oF wet bulb, and the Winter 97.5% design conditions are 22oF and an assumed 0% Relative Humidity. Therefore, for Dallas, Texas, the library air-conditioning system design conditions become:

Summer

Indoor: Outdoor:

75oF dry bulb, 50% Relative Humidity 100 oF dry bulb, 75 oF wet bulb

Winter

Indoor: Outdoor:

72oF dry bulb, 30% Relative Humidity 22oF, assumed 0% Relative Humidity

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BUILDING DESCRIPTION:

The library, as described in Appendix A, is a one story, slab-on-grade floor, having a total floor area of 5,000 sq. ft., with internal dimensions of 100-ft by 50-ft, and 10-ft high ceilings. Windows, as described in Appendix A, occupy 50% of the north and south side wall areas (i.e. 0.50(10-ft x 100-ft) = 500 sq. ft. of windows on both the north and south walls

A nominal window size of 4-ft by 6-ft was selected from Ref. (2), Ch. 27, Table 5., which provides information on standard fenestration products. This window size best suits a library application, since they are tall - allowing sunlight in to aid in reading. A detail of the selected window is provided in Appendix B, and they have a face area of 24 square feet. Therefore, to best meet the COR window area requirements, twenty-one windows are installed on the north wall and twenty windows are installed on the south wall. This is a best fit to the COR window requirements, without resorting to using non-standard window sizes. In addition to the windows, doors are specified for the south and east walls. These doors are 6-ft by 7.5-ft solid core swinging metal doors, without a storm door.

A ten-foot by fifteen-foot Utility Room was chosen and placed on the West wall of the library. This Utility Room is illustrated in Appendix B. The Utility Room is not air-conditioned, and therefore, for design purposes the temperature in the Utility Room is chosen to be the average of the inside and outside temperatures.

Throughout the design process, surface areas of the building, exposed to the outside environment, are used in calculating space heating and cooling loads. Therefore, these physical dimensions of the various exposed surfaces of the library are tabulated in Table 1 and are based on the library sketches provided as Sheets 1 and 2 of Appendix B.

Table 1. Surface Areas Used for Calculations

North Wall

Window Area: 21 x (4-ft x 6-ft) windows = 504 ft2 Wall Area: 100-ft x 10-ft wall - (window area) = 496 ft2

South Wall

Window Area: 20 x (4-ft x 6-ft) windows = 480 ft2

Door Area: 1 x (6-ft x 7.5-ft) door = 45 ft2

Wall Area: 100-ft x 10-ft wall - (window & door area) = 475 ft2

East Wall

Door Area: 1 x (6-ft x 7.5-ft) door = 45 ft2 Wall Area: 50-ft x 10-ft wall - (door area) = 455 ft2 West Wall

Wall Area (to outside): 2 x (17.5-ft x 10-ft wall = 350 ft2 Wall Area (to Utility Rm): 15-ft x 10-ft wall = 150 ft2

Roof

Roof Area: 50-ft x 100-ft = 5000 ft2

Floor

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CALCULATION OF SPACE HEATING LOAD:

As described in Ch. 5 of Ref. (1), the space heating load is an estimate of the maximum probable heat loss of each space or room to be heated. This does not include the heating requirements due to the outside air entering the library from the HVAC system (to meet fresh air requirements). Fresh air is considered when determining the heating system characteristics, but is not a component of the space heating load.

The space heating load is composed of two major components. First, is the heating load required to handle transmission losses due to heat transfer from the air inside the library to outside, through the walls, floor, roof, and various windows and doors. Second, is the heating load required to account for outside air infiltration through doors, windows, etc.

The heat gain due to the sun, as well as the heat gain from internal electrical and mechanical equipment is not considered when calculating the space-heating load. This is a good assumption, because the heating system must be able to maintain the design indoor temperature on cloudy/overcast days when minimal equipment is being used. Also, infiltration air is not considered when determining the amount of fresh air required. Although infiltration air is outdoor air entering the library, the full 20 cfm per person ventilation air requirement must be met by the system, not by infiltration.

Transmission Load:

Transmission losses are calculated individually for each building component from the general equation:

) (t_i t_o UA q&= −

where 'U' is the Overall Heat Transfer Coefficient for the component under investigation, A is the surface area of the component, 'ti' is the inside air

temperature, and 'to' is the outside air temperature.

In order to determine the heating load due to transmission losses, Overall Heat Transfer Coefficients must be found for each building component exposed to the outside environment. Sketches of various building components exposed to the outside are provided in Appendix B, Sheet 3. The calculation of the Overall Heat Transfer Coefficients for various outside surfaces is as follows:

Walls - The walls, as described in Appendix A, are 4-in. face brick and 6-in

concrete block with a 1-in. air gap between them. The various sub-components of the walls each have their own heat transfer coefficient, and the heat transfer coefficients for the convection losses on the outside and inside wall surfaces must also be included. Using the values found in Ref. (1), Table 8-21, the overall heat transfer coefficient for the library walls is determined. A listing of each wall sub-component, it's classification (per Table 8-21), U-value, as well as a summed overall heat transfer coefficient for the library walls is provided in Table 2. It should be noted that the transmission heat loss for the portion of the West wall

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exposed to the Utility Room is calculated by using an assumed temperature in the Utility Room. Since the Utility Room is not air-conditioned, the temperature during the winter is chosen as the average of the outside and space design temperature.

Table 2. Wall Overall Heat Transfer Coefficient

Description Code Number

From Ref. (1), Table 8-21 Thermal Resistance, R (ft2-hr-oF/Btu)

Outside surface resistance A0 0.33

4" face brick A2 0.43

Air space resistance B1 0.91

6" concrete block *(C3+C8)/2 0.91 Inside surface resistance E0 0.69

Total Resistance 3.27

Wall Overall Heat Transfer Coefficient (U = 1/R), Btu/ft2-hr-oF 0.31

* Note: 6" concrete block was not available in Ref. (1), Table 8-21. Therefore, the thermal resistance values for 4" concrete block (C3) and 8" concrete block (C8) were averaged for use as a 6" concrete block thermal resistance.

Windows - The windows as specified are heat absorbing out, clear in type

insulating glass, with a nominal 1/4" thickness. From Ref. (1), Table 5-8a, the Overall Heat Transfer Coefficient for "Double Glass, 1/4 inch air space, aluminum frame, without thermal break" was selected as the best representative description of the specified windows. From Table 5-8a, the Window Overall Heat Transfer Coefficient is taken as 0.65 Btu/ft2-hr-oF.

Doors - The doors, as specified, are 35-mm nominal thickness, 6-ft by 7.5-ft,

solid core, swinging metal doors without storm doors. From Ref. (1) Table 5-9, the "steel door, solid urethane foam core, without thermal break" was selected to best represent the specified doors. However, Table 5-9 does not provide the Overall Heat Transfer Coefficient for a 35-mm thick door. The value available is for a 1-3/4" door. In the Table 5-9 notes, "moderate extrapolation" is allowed. Therefore, the Overall Heat Transfer Coefficient for a 35-mm steel door with solid urethane foam core, without thermal break, is extrapolated as follows:

" 75 . 1 40 . 0 4 . 25 " 1 35 =       mm mm U

where '0.40' is the Overall Heat Transfer Coefficient for a 1-3/4" door, per Table 5-9. From this extrapolation, the door coefficient 'U' is 0.31 Btu/ft2_-hr-o_F.

Roof - As specified, the roof is flat built-up roofing, roof construction code #9,

dark color outside surface, without suspended ceiling below the roof deck. Ref. (1), Table 5-7b provides a breakdown of individual roof components and their thermal resistance's for a specific roof example. This example does not match with the roof specified for the library; however, the components of the specified

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roof can be parsed from the individual roof components in the example roof of Table 5-7b. The specified roof components for the library, their thermal resistances, and the roof Overall Heat Transfer Coefficient are provided in Table 3.

Table 3. Roof Overall Heat Transfer Coefficient

Description Thermal Resistance, R

(ft2_-hr-o_F/Btu) Outside surface (15 mph wind) 0.17

Built-up roofing, 0.375 in. 0.33 Concrete slab, lightweight aggregate, 2 in. 2.22 Inside surface (still air) 0.61

Total Resistance 3.33

Roof Overall Heat Transfer Coefficient (U = 1/R), Btu/ft2-hr-oF 0.30

Floor - As stated on page 177 of Ref. (1), the transmission heat loss calculation

for a slab-on-grade floor uses a slightly different form of the general heat transfer equation than previously stated. The appropriate equation for a slab-on-grade-floor is:

(

ti to

)

P U q_&= ′ −

where, U' is the overall heat transfer coefficient expressed as heat transfer per unit length of building perimeter, per degree temperature difference. The perimeter of the floor is (100-ft + 100-ft + 50-ft + 50-ft) 300 ft, assuming the Utilty Room does not have a slab-on-grade floor.

Also, 'degree days' input for Dallas, Texas is needed to determine U'. From Table C-2 of Ref. (1), the yearly degree days total is 2363 for Dallas, Texas. Therefore, from Table 5-12 of Ref. (1), the U' for a 8" slab-on-grade floor, brick face, uninsulated, using the 2950 degrees day column, is 0.62 Btu/ft-hr-oF.

Now that all the individual overall heat transfer coefficients have been determined, the total transmission heat loss from the building to the outside can be calculated. Table 4 tabulates the overall heat transfer coefficients, and shows the calculations used to determine the total transmission heat loss. As shown in Table 4, the total transmission heat loss is 146,411 Btu/hr.

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Table 4. Total Transmission Heat Loss

Uo (Btu/hr._ft2.o_F) U' (Btu/hr._ft.o_F) Sur. Area (ft2₎ Perimeter (ft) Ti To delta-T qh North Wall 0.31 496 72 22 50 7,688 South Wall 0.31 475 72 22 50 7,363 East Wall 0.31 455 72 22 50 7,053

West Wall (to outside) 0.31 350 72 22 50 5,425

West Wall (to Utility Rm) 0.31 150 47 22 25 1,163

Windows 0.65 984 72 22 50 31,980

Doors 0.32 90 72 22 50 1,440

Roof 0.30 5000 72 22 50 75,000

Floor 0.62 300 72 22 50 9,300

Total Transmission Heat Loss = 146,411

Infiltration Air:

The amount of air that infiltrates the building during the winter conditions can be determined by several methods. These include the crack method and the air-change method. The crack method determines the infiltration rate based on pressure differences between the outside air and inside (due to wind), along with the characteristics of the windows, doors, etc. (i.e. their "cracks"). The air-change method assumes a value of air-air-changes per hour, and with this assumption determines the infiltration rate for a known volume. For this preliminary design, the air-change method is considered to be acceptable. Also, since the library is only a one story building, stack effects are not considered.

As stated on page 227 of Ref. (1), the number of air changes per hour (ACH) can vary from 0.5 (very low) to 2.0 (very high). The ACH is assumed to be a moderate to low value. Therefore for the Library design, an ACH of 1.0 is selected. Equation 7-10 of Ref. (1), which is used to determine the infiltration, is as follows:

( )

V /60 ACH

Q& =

where Q is the infiltration rate in cfm, ACH is the assumed number of air changes per hour, and V is the gross space volume in cubic feet. The gross space volume of the library is (100-ft x 50-ft x 10-ft) 50,000 ft3. Therefore the infiltration rate of outside air is 1.0(50,000)/60 = 833 cfm.

As stated on page 226 of Ref. (1), the infiltration air is considered a heating load. This load has a sensible component and a latent component. The sensible heating component due to infiltration is the sensible heat required to increase the infiltrated air to the design indoor temperature. The latent component is the latent heat required to humidify the infiltrated air to the design indoor relative humidity. Equation 7-8b of Ref. (1) is used to calculate the sensible infiltration heat load, and is as follows:

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(

)

o o i p s v t t c Q q_& = & −

where Q is the infiltration volumetric flow rate (ft3/hr), cp is the constant-pressure

specific heat of air (taken as 0.24 Btu/lbm-oF), ti and to are the inside and outside

air temperatures respectively, and vo is the outside air specific volume

(interpolated from the Ref. (1), Table A-2a as 12.14 ft3/lbm). From this equation and our calculated infiltration air rate of 833 cfm (50,000 ft3_{/hr), the sensible}

heating load due to infiltration is (50,000)(0.24)(72-22)/12.14 = 49,423 Btu/hr. Equation 7-9b of Ref. (1) is used to calculate the latent infiltration heat load, and is as follows:

(

i o

)

fg o l W W i v Q q& = & −

where Wi and Wo are the inside and outside humidity ratios respectively

(lbmv/lbma), and ifg is the enthalpy of vaporization or, the latent heat of

vaporization. From the Psychometric Diagram, the inside humidity ratio Wi is

0.005 lbmv/lbma. The outside humidity ratio is assumed to be zero. This assumption is valid since outisde air during the heating season is very dry (low relative humidity) and therefore a conservative aproach would be to assume there is no water vapor in the outside air. The enthaply of vaporization, from Table A-1a of Ref. (1), at the indoor design temperature of 72 oF, is 1052.8 Btu/lbmv. From this equation and our calculated infiltration air rate of 833 cfm (50,000 ft3/hr), the latent heating load due to infiltration is calculated as (50,000/12.14)(0.005)(1052.8) = 21,680 Btu/hr.

The space heating load due to transmission through the building roof, walls, floor, windows, and doors is combined with the space heating load due to the infiltration of outside air through windows, doors, etc. This is the total space heating load, and is tabulated in Table 5.

Table 5. Total Space Heating Load Sensible Heat Load

( )

q& s _h

Latent Heat Load

( )

q& l _h Transmission

(

Btu /hr

)

146,411 n/a Infiltration

(

Btu /hr

)

49,423 21,680 Total

(

Btu /hr

)

195,834 21,680

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Supply Air for Space Heating

In order to determine whether the heating season or cooling season is the driving season for the selection of the air-distribution system, the supply air requirements for space heating must be calculated. This will be compared to similiar calculations for space cooling to determine a design airflow rate.

Appendix C contains a process diagram of the heating system on a Psychometric Chart, as well as detailed heating system calculations. Also, Appendix B, Sheet 4 is a detailed schematic diagram of the heating system. A pre-heater was considered necessary, to bring the temperature of the fresh air above the dewpoint temperature of the air inside the utility room. This ensures that condensation will not form on the ducting leading into the furnace. As stated previously, the utility room temperature is assumed to be the average of the outside design temperature and the desired indoor temperature, namely 47oF.

A summary of the results of the detailed calculations is provided in Table 6. It is important to note that these calculations do not take into consideration the effects of fan heat to the system.

Table 6. Heating System Characteristics

Required Quantity of Supply Air to Space (at 120oF) 4,113 cfm

Required Furnace Capacity 308,775 Btu/hr

Required water supply to humidifier (at 47oF) 1.0 lbm/min

Required Pre-heater Capacity 40,896 Btu/hr

CALCULATION OF SPACE COOLING LOAD:

As stated on page 247 of Ref. (1), the heat loss calculations, during winter conditions, are based on steady-state heat transfer, and this basis is adequate for selecting heating and humidification equipment. For the cooling load, however, steady-state heat transfer would not adequately represent the heat gain, and would lead to oversized equipment. This is because the solar contributions to the space cooling load have a considerable lag effect, before the heat is actually absorbed by the inside air.

There are several methods that can be used to calculate the space cooling load. These include the "Transfer Function Method," and the "Cooling Load Temperature Difference, Solar Cooling Load, and Cooling Load Factor (CLTD/SCL/CLF) Method." The CLTD/SCL/CLF Method is a hand calculation

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method that uses extensive tables and charts that were derived using the Transfer Function Method. To determine the space cooling load for the Library, the CLTD/SCL/CLF Method is employed.

The Circular of Requirements (Appendix A) states that the maximum cooling load occurs at 15:00 solar time, during the month of August. This statement greatly simplifies the process of determining the design space cooling load. If the time that the maximum cooling load occurs was not provided, the space cooling load would have to be determined for several times, in an iterative fashion, in order to select the maximum. Since the time of the library maximum space cooling load is known, only one iteration is necessary. Each contributing component of the space cooling load is calculated separately as follows, and is then summed to provide the total sensible and latent space cooling loads. The overall heat transfer coefficients used for conductive heat transfer are identical to those used when calculating the space heating load.

Roof - For the roof, A CLTD value is required. As stated in the COR, the roof is

UA q_&_c =

Therefore, with U=0.30, A = 5,000 ft2, and CLTD = 54, the library space cooling load due to the roof is calculated to be 81,000 Btu/hr.

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Walls - For the walls, a CLTD value must be determined. The wall resistance

was previously determined as R = 3.23. The wall # is determined from Ref. (1), Table 8-22b. With R = 3.23, and with face brick as the secondary wall material and C3 (4" heavyweight concrete block) as the principle material, the wall # is 5. And, with R = 3.23, and with face brick as the secondary wall material and C8 (8" heavyweight concrete block) as the principle material, the wall # is 10. Since the Library walls' actual secondary material is face brick, and principle material is 6" heavyweight concrete block, a wall # of 7 was interpolated. Therefore using a wall # of 7, and a solar time of 1500, the CLTD values for the North, South, East, and West walls were selected. They are also adjusted similar to the CLTD adjustment done for the roof. Table 7 contains the calculations for each wall, using the surface areas and overall heat transfer coefficients determined previously for the space heating load. The effects of the Utility Room on the West wall are considered to be negligible.

Table 7. Wall Space Cooling Load

Wall CLTD Corr. CLTD U A q&c=UA(CLTD)

North 16 24 0.31 496 3,690

South 24 32 0.31 475 4,712

East 32 40 0.31 455 5,642

West 22 30 0.31 500 4,650

Total Space Cooling Load due to Walls = 18,694 Btu/hr Glass & Doors (Conduction) - For the conduction portion of the heat gain

through glass windows and doors, a CLTD value must be determined. These CLTD values are determined from Ref. (1), Table 8-23, for a solar time of 1500 hrs. These values, and their corrected values (corrected similar to the correction to roof CLTD) are provided in Table 8, along with calculations to determine the space cooling load due to windows and doors. The surface areas and Overall Heat Transfer Coefficients were determined previously for the space heating load.

Table 8. Window & Doors Conduction Space Cooling Load

Item CLTD Corr. CLTD U A q&c=UA(CLTD)

North Windows 14 22 0.65 504 7,207

South Windows 14 22 0.65 480 6,864

South Doors 14 22 0.31 45 307

East Doors 14 22 0.31 45 307

Total Space Cooling Load due to Glass-Door Conduction = 14,685 Btu/hr Glass (Solar) - To determine the Solar contribution to the space cooling load

from windows, a Shading Coefficient (SC) and a Solar Cooling Load Factor (SCL) must be found. As stated in the COR in Appendix A, the windows are

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equipped with medium venetian blinds. Also, from Ref. (1), Table 8-25a, for single-story buildings, using 4 walls, and carpet floor covering, the Zone Type is selected as 'A' for Glass Solar. Then from Ref. (1), Table 8-24, the SCL factors for the North and South windows are selected with a solar time of 1500 hrs. As stated in the notes for Table 8-24, the data is applicable to the month of August. And, from Table 6-5, for windows that are heat absorbing out, clear in-type insulating glass, a Shading Coefficient of 0.39 is selected. The contribution to the space cooling load from glass solar effects is calculated as shown in Table 9. The heat transfer areas were calculated previously when determining the space heating load.

Table 9. Windows (Solar) Space Cooling Load

Item A SC SCL q_&_c = A

( )

SC SCL

North windows 504 0.39 36 7,076

South Windows 480 0.39 52 9,734

Total Space Cooling Load Due to Glass (Solar) = 16,810 Btu/hr Floor - As stated on page 177 of Ref. (1), during summer conditions the heat

transfer to the floor slab is negligible.

Internal Lights - To determine the contribution to the space cooling load from

the internal lights, a Cooling Load Factor (CLF) must be determined and multiplied with the instantaneous heat gain of the lighting. The instantaneous heat gain can be determined from the following formula:

( )

W FuFs

q& =3.41

where, W is the total installed wattage, Fu is the usage factor, and Fs is an

allowance factor. As stated in the COR (Appendix A) the product of W and Fu is

4 Watts/ft2 of floor area. The lights as stated in the COR are fluorescent type, and from Ref. (1), page 280, for general applications of fluorescent lights, Fs is

taken to be 1.2. Therefore with a floor area of 5,000 ft2_{, the instantaneous heat}

gain is calculated as 3.41(4)(5000)(1.2) = 81,840 Btu/hr.

From Table 8-25a, the Library is Zone Type 'B' for lights. As stated in the COR, the lights are on at 8AM, and off at 11 PM. Therefore they are on for 15 hrs each day. The maximum load occurs at 1500 hrs, therefore, the number of hours the lights are on up until the maximum cooling load time is 7 hours. Using these values and Zone Type 'A', from Table 8-27, the CLF is found to be 0.965. Therefore the contribution to the space cooling load from internal lights is found to be (0.965)(81,840) = 78,976 Btu/hr.

Equipment - To determine the contribution to the space cooling load from the

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the instantaneous heat gain of the equipment. The instantaneous heat gain from the equipment is determined in a similar fashion to the calculation for internal lighting. As stated in the COR, the Library equipment emits 0.5 Watts/ft2 of floor area. Also stated in the COR is that the equipment is on from 8AM to 5PM. Therefore, the equipment is on for a total of 9 hrs, and is on for 7 hours at the maximum cooling load time of 1500 hrs. From Ref. (1), Table 8-25a, the Library Zone Type is 'B' for equipment and people. With these values, using Ref. (1), Table 8-26, the CLF for equipment is found to be 0.93. Therefore the contribution to the space cooling load from equipment is found to be 3.41(0.5)(5,000)(0.93) = 7,928 Btu/hr.

People (sensible) - The sensible portion of the instantaneous heat gain due to

the Library occupants requires the determination of a Cooling Load Factor. First, from Ref. (1), Table 8-11, the instantaneous sensible heat gain for people seated doing very light work is 245 Btu/hr per person. And from Ref. (1), Table 8-25a, the Library Zone Type is 'B' for equipment and people. As stated in the COR, the people enter the space at 8AM and leave at 9PM (13 hours in space). And, at the point of maximum cooling load (1500 hrs) the people will have been in the space for 7 hours. With this data, a CLF of 0.94 is determined from Ref. (1), Table 8-26. Therefore, for 75 people called out in the COR, the sensible portion of the space cooling load due to people is (75)(245)(0.94) = 17,273 Btu/hr.

People (latent) - The latent portion of the instantaneous heat gain is immediately

absorbed by the space air, and does not require a CLF factor for adding it to the space cooling load. From Ref. (1), Table 8-11, for people seated doing very light work, the latent heat gain is 155 Btu/hr per person. For 75 people, the latent portion of the space cooling load due to people is (75)(155) = 11,625 Btu/hr.

Infiltration Air - As calculated previously for the space heating load, using the

Air-Change Method, the infiltration air rate is 833 cfm. As stated in Ref. (1), page 331, "No CLF factors are necessary, since infiltration is convective in nature, and immediately becomes cooling load."

As stated in the "Design Conditions" section of this report, the outside air is 100o_{F dry bulb, and 75}o_{F wet bulb. From the Psychometric Chart in}

conjunction with Ref. (1), Table A-2a, the outside air conditions are: 100oF, 32% relative humidity, and a specific volume of 14.419 ft3/lbm. The specific heat is taken as 0.24 Btu/lbm-o_F.

The sensible portion of the space cooling load due to infiltration air is determined the same way as determined previously for the space heating load with the following formula:

(

)

o i o p s v t t c Q q_& = & −

Therefore, the sensible portion of the space cooling load due to infiltration air is found as (50,000)(0.24)(100-75)/14.419 = 20,806 Btu/hr.

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The latent portion of the space cooling load due to infiltration air is also determined in the same fashion as was done for the space heating load using the following formula:

(

o i

)

fg o l W W i v Q q_& = & −

From Ref. (1), Table A-1a, the latent heat of vaporization (ifg) at 100oF is 1037

Btu/lbm. And from the Psychometric Chart, Wo = 0.0130 lbmv/lbma, and Wi =

0.0092 lbmv/lbma. With this data the latent portion of the space cooling load due to infiltration air is found as (50,000)(0.0130 - 0.0092)(1037)/14.419 = 13,665

Btu/hr.

A summary of all calculated space cooling loads, and the tabulated resulting total space cooling load are provided in Table 10.

Table 10. Total Space Cooling Load Sensible Cooling Load

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Supply Air for Space Cooling

Appendix C contains a process diagram of the cooling system on a Psychometric Chart, as well as detailed cooling system calculations. Also, Appendix B, Sheet 5 is a detailed schematic diagram of the cooling system. A summary of the results of the detailed calculations is provided in Table 11. It is important to note that these calculations do not take into consideration the effects of fan heat to the system.

Table 11. Cooling System Characteristics

Required Quantity of Supply Air to Space (at 55oF) 11,026 cfm Required Cooling/Dehumidification Unit Capacity 341,775 Btu/hr

Coil Sensible Heat Factor (SHF) 0.84

AIR-CONDITIONING EQUIPMENT SELECTION:

As stated in the COR (Appendix A), the refrigeration cycle chosen for the Library is a Freon (R-22) cycle operating at a 30oF evaporator temperature and with a 10oF superheat before the refirgerant enters the compressor, and a 110oF condensing temperature with 10oF of subcooling before the expansion valve. Appendix B, Sheet 6, contains a detailed diagram of this system employed to provided cooling and dehumidification to the Library. As shown on Sheet 6, a cooling tower is utilized to transfer the heat from the condenser cooling water to the ambient air, and the chilled water system provides chilled water to the air/water heat exchanger which is the cooling/dehumidification coil for the Library.

Appendix C contains detailed cooling system calculations for the Library Air-Conditioning system as well as a diagram showing the process on a Freon (R-22) Pressure-Enthalpy diagram. The state points shown on the Appendix C Pressure-Enthaply diagram correspond to the state points as shown on the Conditioning System Schematic (Sheet 6, Appendix B). Also, on the Air-Conditioning System Schematic, it is shown that the Chilled Water enters the Cooling/Dehumdification coil at 42o_{F, and leaves it at 52}o_{F. And, as shown,}

cooling water enters the condenser at 82oF and leaves it at 92oF.

The detailed calculations in Appendix C show the process of calculating the Theoretical BHP of the Chiller Compressor, Water Chiller (R-22 Evaporator) size in terms of chilled water circulation rate, and the Water-Cooled Condenser size in terms of GPM of water flow and capacity. It is assumed, for the calculations, that the heat transfer in the chilled water piping is negligible. With

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this assumption, the design cooling/dehumidification unit capacity can be equated to the Water Chiller (R-22 Evaporator) capacity.

From this known Water Chiller capacity, along with the known refrigeration cycle state points, the mass flow rate of Freon in the theoretical refrigeration cycle was determined. Then, using the calculated mass flow rate of Freon, the Theoretical BHP of the Compressor was determined. And, using an energy balance on the refrigeration cycle (Qcondenser = Qevaporator + Wcompressor, see Sheet

6, Appendix B) the Theoretical capacity of the Condenser was determined.

To meet the cooling and dehumidification needs of the Library, a York, Model LCHWC 35, Reciprocating, Packaged Liquid Chiller was selected with a cooling capacity of 33.8 tons. This Chiller meets the Library cooling requirements with a 10% margin. Details of this Chiller Unit are provided in Appendix D. The selection methodology is provided in the Appendix C calculations, and any assumptions not stated here are stated in Appendix C. Table 12 summarizes the characteristics of the Air-Conditioning System components.

Table 12. Air-Conditioning System Characteristics

Cooling/Dehumidification Capacity 28.5 tons

(341,775 Btu/hr) Cooling/Dehumidification Capacity

(With 10% margin) 31.4 tons

Theoretical BHP of Compressor 21.5 kW

(for 28.5 ton system cooling capacity)

Theoretical Capacity of Condenser 415,012 Btu/hr

(for 28.5 ton system cooling capacity)

Selected Chiller Capacity

(at 90o_{F Cond. Lvg wtr temp, 42}o_{F Cooler lvg wtr temp)} 33.8 tons

Compressor Power for Selected Chiller 28.0 kW

Chilled Water Circulation Rate 81.0 gpm

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AIR-HANDLING/DISTRIBUTION SYSTEM DESIGN:

For the Library, a low-velocity air distribution system was selected. A high-velocity system would likely not meet the noise requirements for a library application, and are normally selected for more industrial-type applications. As stated in the COR (Appendix A), the recommended air velocities for the library air distribution system are:

Duct Sections (max) 1,000 fpm Air Supply Diffusers 400 fpm Return Air Registers 200 fpm Cooling Coil 500 fpm

Also, the combined supply and return air pressure drop should not exceed 4.0 in w.g. This requirement is likely based on an independent cost analysis which compared costs due to duct size (non-recurring costs) and costs due to fan power consumption (recurring costs). From this analysis, a maximum system pressure drop of 4.0 in w.g. may have been selected.

The air distribution system should be designed for the maximum volumetric flow rate that the system will operate with. In heating mode, the supply air required was previously determined to be 4,113 cfm. And, in cooling mode, the supply air required was previously determined to be 11,026 cfm. Obviously, the cooling season supply air requirements will drive the air distribution system design, since the required volumetric flow rate is much larger.

The process followed in developing the air distribution system is as follows. First, the design airflow is selected. Based on the cooling season required cfm, a design airflow rate of 11,000 cfm is selected. The velocity limitations on the system were provided in the COR, and are listed above. Next, the diffuser and return grills are selected, and their corresponding pressure losses are determined. Then the layout of the supply and return air-ducting network is accomplished. With the duct layout decided upon, the ducts are sized to ensure that the system behaves as desired, and the supply and return system pressure losses are determined. Finally, with the total system losses tabulated, a fan is selected. More detailed discussion for each of these steps in provided in the following sections.

Diffuser Selection

After several iterations of diffuser number and size, 10 diffusers were selected to supply air to the Library. The diffuser type was selected as Circular Ceiling Diffusers. Sheet 7 of Appendix B shows the location of these 10 diffusers. They are each located in the Library overhead, in the center of a 20 ft by 25 ft section of the building. Since there are 10 diffusers, the cfm per each diffuser is 11,000 cfm / 10 diffusers = 1,100 cfm per diffuser.

The Room load is the system cooling load divided by the room floor area. This is determined to be (341,775 Btu/hr) / (100 ft X 50 ft) = 68.4 Btu/hr-ft2_{. The} characteristic room length (L), per Ref. (1), Table 11-7, for Circular Ceiling

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Diffusers is the distance to the closest wall or intersecting jet. Therefore, since each diffuser is serving a 20 ft by 25 ft section, the characteristic room length is taken to be between 10 -12.5 ft. From Ref. (1), Table 11-8, knowing the Room load, the x50/L = 0.8 for maximum ADPI, and the x50/L can range from 0.7 to 1.3.

The x50 value is the length from the diffuser where the air velocity is 50

ft/min. And, the ADPI value is a value which when met by a particular diffuser selected for an application, ensures that the occupants of the building do not feel discomfort. A point where occupants feel discomfort can be defined by an effective draft temperature. And, per Ref. (1), page 439, the ADPI is defined as the percentage of measurements taken at many locations in the occupied zone of a space that meet the effective draft criteria. By selecting a diffuser so that the x50/L value falls in the defined range for the known Room load, it can be ensured

that the occupants do not feel uncomfortable.

Since maximum ADPI occurs at a value of x50/L = 0.8, x50 = 0.8(L), the x50

value is 0.8(12.5 ft) = 10. Also, a recommeded diffuser velocity of 400 ft/min was stated in the COR. Therefore, from Ref. (1), Table 11-3, a 24" diffuser was selected, with 1260 cfm @ 400 ft/min. This airflow rate is close to the design rate of 1100 cfm. From Table 11-3, the Radius of Diffusion for 50 ft/min is 15 ft. Therefore our actual x50/L = 15 ft / 12.5 ft = 1.2. This x50/L value falls in the range

provided in Table 11-8, and is acceptable. Therefore, 24" Circular Ceiling Diffiusers are selected for the Library. Also, the Noise Criteria Index (NC) is lower than the recommended noise level in a Library of NC = 35-40, as stated in Ref. (1), Table 11-1. Per Table 11-3, the ∆Po for these diffusers is 0.024" w.g.

Return Grill Selection

As stated in the COR, the recommended return air register velocity is 200 ft/min. After several iterations, ten return grills, each sized for 1,100 cfm were chosen. From Table 11-6, each is chosen to be a 24" by 24" grill, with characteristics: 1080 cfm, NC =20, and 300 ft/min. The return grill velocity is above the recommended velocity in the COR, however this is considered to be acceptable since the Noise Criteria value (NC = 20) is well below the levels specified for Library applications. The ∆Ps for this grill is -0.033" w.g., and the

velocity pressure is 0.006. Therefore the ∆Po value determined for this grill is

found to be -0.033 + 0.006 = - 0.027" w.g. This value is stated as negative because it is on the suction side of the supply fan, and therefore the pressure is below atmospheric pressure.

Supply Air Ducting Layout/Sizing

Appendix B, Sheets 7 & 8, show the layout of the supply air duct system. To determine the sizes of these ducts, in order to properly deliver 1,100 cfm to each of the ten diffusers, the Balance Method was employed. For this method, first the pressure drop through the longest run is determined (main duct run). Then, the additional branches are sized so that their pressure drop is equivalent to the pressure drop from the start of the particular branch to the end of the main duct run. This ensures that the pressure drop is equal for every path the air can

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travel through the system, which ensures the proper amount of air is delivered to each diffuser.

To determine the pressure drops through each section, the Equivalent Length Method was employed. This method determined the pressure drop through selected duct sections, by calculating equivalent lengths for components other than straight duct, to add to the length of straight duct. This allows a single equivalent length (Le) value to be multiplied by ∆Po per 100 ft of duct value to

determine the branch pressure loss.

Table 13 shows the Equivalent Length calculations for the supply system braches. Equivalent lengths for components other than straight duct are determined using Ref. (1), Figures 12-25 and 12-26. It is not necessary to include the diffuser pressure losses in each branch for sizing purposes, since each branch has a diffuser of the same size and airflow rate. The section designations in Table 13 correspond to the duct section labels as shown in Appendix B, Sheet 8.

Table 13. Supply System Equivalent Length Calculations

Section Individual Lengths Le

Main Run

A-B 35' (plenum) + 5' (duct) + 10' (90oelbow) + 15' (duct) + 10' (90oelbow)

+ 3' (duct) + 5' (Main Branch) 83'

B-C 3' (duct) + 5' (Main Branch) 8'

C-D 17' (duct) + 5' (Main Branch) 23'

D-E 3' (duct) + 5' (Main Branch) 8'

E-F 17' (duct) + 5' (Main Branch) 23'

F-G 3' (duct) + 5' (Main Branch) 8'

G-H 17' (duct) + 5' (Main Branch) 23'

H-I 3' (duct) + 5' (Main Branch) 8'

I-J 17' (duct) + 5' (Main Branch) 23' J-T 3' (duct) + 5' (45oelbow) + 3' (duct) + 5' (45oelbow) + 10' (duct) 26'

Branches

B-K 35' (Take-off branch) + 3' (duct) + 5' (45oelbow) + 10' (duct) 53' C-P 35' (Take-off branch) + 3' (duct) + 5' (45oelbow) + 10' (duct) 53' D-L 35' (Take-off branch) + 3' (duct) + 5' (45oelbow) + 10' (duct) 53' E-Q 35' (Take-off branch) + 3' (duct) + 5' (45oelbow) + 10' (duct) 53' F-M 35' (Take-off branch) + 3' (duct) + 5' (45oelbow) + 10' (duct) 53' G-R 35' (Take-off branch) + 3' (duct) + 5' (45oelbow) + 10' (duct) 53' H-N 35' (Take-off branch) + 3' (duct) + 5' (45oelbow) + 10' (duct) 53' I-S 35' (Take-off branch) + 3' (duct) + 5' (45oelbow) + 10' (duct) 53' J-O 35' (Take-off branch) + 3' (duct) + 5' (45oelbow) + 10' (duct) 53'

The ducts were sized using these equivalent lengths for each branch. Using a Figure similiar to Ref. (1), Figure 12-29, an iterative process was used to determine the Pressure Drop per 100 ft, for each section. Iteration was necessary to ensure that the duct velocities remain below the recommended velocities. As stated in the COR, the recommeded maximum duct velocity is 1,000 ft/min. For the large ducts needed for the 11,000 cfm supply system, the

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1,000 ft/min velocity limit kept the duct velocities well below velocities that are characteristic of a low-velocity duct system. However, since the recommended maximum duct velocity was considered to be important to avoid any noise problems in the Library, it was adhered to. This resulted in low pressure drops throughout the supply system. Table 14 shows the resulting supply system duct sizes for the main run, and Table 15 shows the resulting supply system duct sizes for the corresponding branches. These values were determined after much iteration, and as shown in Tables 14 & 15, the system is balanced and the duct velocities remain under the prescribed limit. Also, these Tables show the supply system pressure drop calculated as 0.0804" w.g.

Table 14. Supply System Main Branch Pressure Drop Balance Method (Main Duct Run A-T)

Section CFM Le (ft) De (in) V (ft/min) Delta-Po/100 ft

Section Delta-Po A-B 11,000 83 45 996 0.025 0.0208 B-C 9,900 8 43 982 0.027 0.0022 C-D 8,800 23 41 960 0.029 0.0067 D-E 7,700 8 39 928 0.027 0.0022 E-F 6,600 23 36 934 0.030 0.0069 F-G 5,500 8 33 926 0.032 0.0026 G-H 4,400 23 30 896 0.035 0.0081 H-I 3,300 8 26 895 0.045 0.0036 I-J 2,200 23 22 833 0.053 0.0122 J-T 1,100 26 16 788 0.059 0.0153 Total 0.0804

Table 15. Supply System Branch Sizing Branch Ducts Section Loss Equivalent to Section: Required Delta-Po Le (ft) Delta-Po/100 ft CFM De (in) V (ft/min) B-K B-T 0.0596 53 0.113 1,100 14 1,029 C-P C-T 0.0575 53 0.108 1,100 14 1,029 D-L D-T 0.0508 53 0.096 1,100 14 1,029 E-Q E-T 0.0486 53 0.092 1,100 14 1,029 F-M F-T 0.0417 53 0.079 1,100 15 896 G-R G-T 0.0392 53 0.074 1,100 15 896 H-N H-T 0.0311 53 0.059 1,100 16 788 I-S I-T 0.0275 53 0.052 1,100 16 788 J-O J-T 0.0153 53 0.029 1,100 18 622

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Return Air Ducting Layout/Sizing

The layout of the return air system is shown in Appendix B, Sheets 7 & 9. The balance method was employed to determine the pressure drop through the longest run (Section A-O, per Appendix B, Sheet 9), in similar fashion as the supply air ducting. However, in this case, there is more than one "main run." As can be seen on Sheet 9, sections A-O and A-J both have branches (i.e. the short ducts leading to the return grills). Therefore in addition to the braches being sized to match the pressure drop from their branch point to the end of the longest run, both "main runs," A-O and A-J, must have equal pressure drop. This is true because both sections originate from the same plenum.

The pressure drop through section A-O was calculated first, then section A-J was sized to match the pressure drop through A-O. Once this was accomplished, the branches to the other return grills were sized using the balance method, for their corresponding "main run." Table 16 provides the equivalent lengths of all runs in the return ducting, used in sizing calculations, determined in a similar fashion as the supply ducting. The return grill pressure drops are not included in the equivalent length calculations because each run has an identical grill and airflow rate, and therefore, an identical pressure loss. The section designations in Table 16 correspond to the duct section labels as shown in Appendix B, Sheet 9.

Table 16. Return System Equivalent Length Calculations

Section Individual Lengths Le

Main Run (A-O)

A-P 35' (plenum) + 50' (duct) +10' (90o_{elbow) +8' (duct) + 5' (Branch Main)} _108'

P-Q 15' (duct) + 5' (Branch Main) 20' Q-R 20' (duct) + 5' (Branch Main) 25' R-S 20' (duct) + 5' (Branch Main) 25' S-O 20' (duct) +10' (90o_{elbow) +1' (duct)} _31'

Main Run (A-J)

A-B 35' (plenum) + 8' (duct) + 5' (Branch Main) 48' B-C 15' (duct) + 5' (Branch Main) 20' C-D 20' (duct) + 5' (Branch Main) 25' D-E 20' (duct) + 5' (Branch Main) 25' E-J 20' (duct) +10' (90o_{elbow) +1' (duct)} _31'

Branch

K-P 50' (Branch Main) + 1' (duct) 51' L-Q 50' (Branch Main) + 1' (duct) 51' M-R 50' (Branch Main) + 1' (duct) 51' N-S 50' (Branch Main) + 1' (duct) 51' B-F 50' (Branch Main) + 1' (duct) 51' C-G 50' (Branch Main) + 1' (duct) 51' D-H 50' (Branch Main) + 1' (duct) 51' E-I 50' (Branch Main) + 1' (duct) 51'

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Using the Equivalent Length Method to calculate the pressure drop through each path that the return air can travel, with the equivalent lenghts provided in Table 17, the pressure drop through section A-O was calculated after iteration, and is provided in Table 17. Also Table 17 provides the pressure drop calculations for the other "main run," section A-J. And, as can be seen, the pressure drop through both runs is the same. Therefore, since they are both connected to the same plenum, on the suction side of the fan, equal quantities of air should travel through them, namely 5,500 cfm in each run. As depicted in Table 17, slightly smaller duct sizes were used in section A-J, to provide additional pressure drop to equalize with section A-O (section A-O has an additional sizty feet of equivalent duct length).

Table 17. Return System Main Duct Pressure Drop

Balance Method (Main Duct Run A-O)

Section Delta-P_o A-P 5,500 108 32 985 0.035 0.0378 P-Q 4,400 20 30 896 0.035 0.0070 Q-R 3,300 25 26 895 0.045 0.0113 R-S 2,200 25 22 833 0.053 0.0133 S-O 1,100 31 16 788 0.059 0.0183 Total 0.0876

Balance Method (Main Duct Run A-J)

Section Delta-Po A-B 5,500 48 31 1,049 0.037 0.0178 B-C 4,400 20 28 1,029 0.044 0.0088 C-D 3,300 25 24 1,050 0.054 0.0135 D-E 2,200 25 20 1,008 0.065 0.0163 E-J 1,100 31 14 1,029 0.100 0.0310 Total _0.0873

Once the two "main runs" were sized, the respective braches were sized, using the Balance Method, to ensure that approximately 1,100 cfm enters each return grill. As shown in Table 18, the branches from section A-O, and the branches from section A-J are sized so that their pressure drop matches the pressure drop of the duct run that begins at the respective branch point to the end to the respective longest run. The duct sizes, selected for each branch, provide the necessary pressure drop, so that the return system is balanced. Based on the calculations in Tables 17 and 18, after much iteration, the return

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system is considered balanced. The pressure drop, through the return system, not including the pressure drop through the grill, was calculated as - 0.0876" w.g.

Table 18. Return System Remaining Branch Duct Sizing

Branches From Section A-O

Section Required Delta-P_o Le (ft) Delta-P_o/100 ft CFM De (in) V (ft/min) K-P 0.0498 51 0.098 1,100 15 896 L-Q 0.0428 51 0.084 1,100 15 896 M-R 0.0315 51 0.062 1,100 16 788 N-S 0.0183 51 0.036 1,100 18 622

Branches from Section A-J

Section Required Delta-P_o Le (ft) Delta-P_o/100 ft CFM De (in) V (ft/min) B-F 0.0696 51 0.136 1,100 14 1,029 C-G 0.0608 51 0.119 1,100 14 1,029 D-H 0.0473 51 0.093 1,100 15 896 E-I 0.0310 51 0.061 1,100 16 788 Fan Selection

To select a fan for the Library HVAC system, the total system pressure drop must be determined. This is done by summing absolute value of all the individual pressure drops calculated previously. Also, assumed pressure drops for the cooling coil and filters are neccessary. As stated in previous sections and Tables, the supply system pressure drop is 0.0804" w.g., the diffuser pressure drop is 0.0240" w.g., the return system pressure drop is -0.0876" w.g., and the return grill pressure drop is -0.0270" w.g. The cooling coil pressure drop is assumed to be 0.2500" w.g., and the pressure drop due to various filters in the system is assumed to be 0.1500" w.g. Plenum losses before and after the fan are included in the supply and return system pressure drop calculations. In order to ensure the fan is capable of delivering air through the duct system, a margin of 15% in pressure drop is added to account for any additional losses overlooked, as well as degradation of the duct system over time from fouling. Table 19 shows a tabulation of system pressure losses, and a total ∆Po for fan selection

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the 4.0" w.g. maximum per the COR, and the relatively low total pressure drop is assumed to be due to the relatively stringent duct velocity restrictions.

Table 19. System Total Pressure Drop

o s P P P P =∆ − − ∆               −       − ∆ = ∆ 2 2 2 2 4005 4005 V V P P_s _o 1 2

Figure 1. Fan Velocity Pressure

A SWB, Model SWB-24, Centrifugal Fan is selected as the fan for the Library HVAC system. As shown in the vendor data in Appendix D, the fan inlet area is 5.582 ft2 , and the fan outlet area is 3.3 ft2. With the design airflow rate of 11,000 cfm, the inlet and outlet velocities are determined as 11,000/5.582 =

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1,971 ft/min, and 11,000/3.3 = 3,333 ft/min. Therefore the total pressure drop can be translated into a static pressure delta as follows:

              −       − = ∆ 2 2 4005 971 , 1 4005 333 , 3 7119 . 0 s P

Therefore, the static pressure delta is 0.2615" w.g. As can be seen in the Appendix D vendor data, A SWB-24 model fan operating at 1,281 rpm, with a max BHP of 5.011, provides 10,813 cfm at a static pressure of 0.2500" w.g. This centrifugal fan is ideally suited for this library application. It should also be noted that an exhaust fan rated at 1,500 cfm is required to vent a portion of the return air to the outside, in order to allow the air distribution system to intake 1,500 cfm of fresh air to meet the Library fresh air requirements.

SUMMARY:

Using the methods outlined in the ME 260 HVAC course, Spring Semester 1999, a preliminary HVAC system design was performed for a Library to be located in Dallas, Texas. The heating and cooling systems were designed, and the air-distribution system, based on the cooling system airflow requirements was also designed. Appropriate equipment components, such as a pre-heater, furnace, humidifier, cooling and dehumidification unit, reciprocating compressor chiller unit, and fan were sized and selected. No cost analysis was performed, however this report can be used as a basis for future cost analyses if necessary.

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APPENDIX A

Circular of Requirements

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APPENDIX B

Drawing Sheets

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ME 260 HVAC

Sheet Public Library A/C System Design

Location: Dallas, Texas

1 Prepared By: JFW 5/4/99 Title: Scale = None

N

S

E

Scale = None

Heating & Humidification Schematic

Space

Furnace

Humidifier

q

s

= 195,834 Btu/hr

q

_l

= 21,680 Btu/hr

q

_h

m

_w 0’ 1 2 3 4 5 Q_o= 1500 cfm t_o= 22o_F φ_o= 0% 0

Pre-heater

t₂= 120o_F t₃= 72o_F φ3= 30%

q

_ph

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ME 260 HVAC

Sheet 5 Public Library A/C System Design

Title:

Scale = None

Cooling & Dehumidification Schematic

Space

Cooling and

Dehumidification

Unit

q

_s

= 256,172 Btu/hr

q

_l

= 25,290 Btu/hr

q

_c 0 1 2 3 4 5 Q_o= 1500 cfm t_o= 100o_{F db} 75o_{F wb} t₃= 75o_F φ3= 50% t₂= 55o_F

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ME 260 HVAC

6

Title:

Scale = None

A/C System Schematic

Evaporator

Condenser

Compressor

Expansion

Valve

Freon-22 Refrigeration Cycle

Supply Air

Cooling

Tower

1

2

4

3 W

_compressor

Q

_F

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ME 260 HVAC

7

Title:

Scale = None

Air Distribution Schematic

PLAN VIEW F

Return Ducting Supply Ducting

24” x 24” Return Grill

24” Circular Ceiling Diffuser 1500 cfm exhaust fan required

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ME 260 HVAC

8

Title:

Scale = None

Supply Ducting Details

C D E F G I K M P Q R H J O L N B T S 5’ 15’ 3’ 3’ 3’ 3’ 3’ 3’ 3’ _3’ _3’ _3’ 10’ 10’ 10’ 10’ 10’ 10’ 10’ 10’ 10’ 10’ 17’ 17’ 17’ 17’ A Plenum

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ME 260 HVAC

9

Title:

Scale = None

Return Ducting Details

A B C D E P Q R S 15’ 20’ 20’ 20’ 50’ Plenum 15’ 20’ 20’ 20’ F _G _H _I _J K L M N O

The ducts connecting the main runs and the return grills are 1’ in length

Note:

8’

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APPENDIX C

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APPENDIX D

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