hvac design for library-dallas-texas me 260 1999

8/14/2019 HVAC Design for Library-Dallas-Texas ME 260 1999

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DESIGN OF A SINGLE-ZONE HVACSYSTEM FOR A PUBLIC LIBRARY

IN DALLAS, TEXAS

ME 260 - Heating, Ventilation and Air ConditioningGeorge Washington University

Spring Semester 1999

Submitted to: Professor A. M. Kiper

Submitted by: John F. Woodall

May 4, 1999


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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 ofHeating, 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 airconditioning system for a public library in Dallas, Texas. The circular-of-

requirements (COR) for the HVAC system is provided in Appendix A. Wheninformation, beyond that provided in Appendix A, was necessary for designcalculations, assumptions were necessary. These assumptions are statedthroughout this report, where applicable.

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

Appendix B is a set of drawing sheets, which include various libraryarrangements, drawings, and system schematics. These drawings are

referenced throughout the report where applicable. Appendix C contains systemdesign calculations and process diagrams, and Appendix D contains vendorprovided 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 Humidity2.5% Design ConditionsWinter

Indoor:Outdoor:

72oF dry bulb, 30% Relative Humidity97.5% Design Conditions

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

SummerIndoor:Outdoor:

75oF dry bulb, 50% Relative Humidity100 oF dry bulb, 75 oF wet bulb

WinterIndoor:

Outdoor:72oF dry bulb, 30% Relative Humidity22oF, assumed 0% Relative Humidity


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BUILDING DESCRIPTION:

The library, as described in Appendix A, is a one story, slab-on-gradefloor, having a total floor area of 5,000 sq. ft., with internal dimensions of 100-ftby 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. ofwindows 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. Thiswindow size best suits a library application, since they are tall - allowing sunlightin 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 CORwindow area requirements, twenty-one windows are installed on the north walland twenty windows are installed on the south wall. This is a best fit to the CORwindow requirements, without resorting to using non-standard window sizes. Inaddition 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 Westwall of the library. This Utility Room is illustrated in Appendix B. The UtilityRoom is not air-conditioned, and therefore, for design purposes the temperaturein the Utility Room is chosen to be the average of the inside and outsidetemperatures.

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

Table 1. Surface Areas Used for Calculations

North WallWindow Area: 21 x (4-ft x 6-ft) windows = 504 ft

2

Wall Area: 100-ft x 10-ft wall - (window area) = 496 ft2

South WallWindow Area: 20 x (4-ft x 6-ft) windows = 480 ft

2

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

2

Wall Area: 50-ft x 10-ft wall - (door area) = 455 ft2

West WallWall 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

RoofRoof Area: 50-ft x 100-ft = 5000 ft

2

FloorFloor Area: 50-ft x 100-ft = 5000 ft

2


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CALCULATION OF SPACE HEATING LOAD:

As described in Ch. 5 of Ref. (1), the space heating load is an estimate ofthe maximum probable heat loss of each space or room to be heated. This doesnot include the heating requirements due to the outside air entering the library

from the HVAC system (to meet fresh air requirements). Fresh air is consideredwhen determining the heating system characteristics, but is not a component ofthe space heating load.

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

The heat gain due to the sun, as well as the heat gain from internalelectrical and mechanical equipment is not considered when calculating thespace-heating load. This is a good assumption, because the heating system

must be able to maintain the design indoor temperature on cloudy/overcast dayswhen minimal equipment is being used. Also, infiltration air is not consideredwhen determining the amount of fresh air required. Although infiltration air isoutdoor air entering the library, the full 20 cfm per person ventilation airrequirement must be met by the system, not by infiltration.

Transmission Load:Transmission losses are calculated individually for each building

component from the general equation:

)( oi ttUAq =&

where 'U' is the Overall Heat Transfer Coefficient for the component underinvestigation, A is the surface area of the component, 'ti' is the inside airtemperature, and 'to' is the outside air temperature.

In order to determine the heating load due to transmission losses, OverallHeat Transfer Coefficients must be found for each building component exposedto the outside environment. Sketches of various building components exposed tothe outside are provided in Appendix B, Sheet 3. The calculation of the OverallHeat 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-componentsof the walls each have their own heat transfer coefficient, and the heat transfercoefficients for the convection losses on the outside and inside wall surfacesmust also be included. Using the values found in Ref. (1), Table 8-21, the overallheat transfer coefficient for the library walls is determined. A listing of each wallsub-component, it's classification (per Table 8-21), U-value, as well as a summedoverall heat transfer coefficient for the library walls is provided in Table 2. Itshould 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 inthe Utility Room. Since the Utility Room is not air-conditioned, the temperatureduring the winter is chosen as the average of the outside and space designtemperature.

Table 2. Wall Overall Heat Transfer Coefficient

Description Code NumberFrom Ref. (1), Table 8-21

Thermal Resistance, R(ft

2-hr-

oF/Btu)

Outside surface resistance A0 0.33

4" face brick A2 0.43Air space resistance B1 0.91

6" concrete block *(C3+C8)/2 0.91Inside 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 typeinsulating glass, with a nominal 1/4" thickness. From Ref. (1), Table 5-8a, theOverall Heat Transfer Coefficient for "Double Glass, 1/4 inch air space, aluminumframe, without thermal break" was selected as the best representative descriptionof the specified windows. From Table 5-8a, the Window Overall Heat TransferCoefficient 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 tobest represent the specified doors. However, Table 5-9 does not provide theOverall Heat Transfer Coefficient for a 35-mm thick door. The value available isfor 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 solidurethane foam core, without thermal break, is extrapolated as follows:

"75.1

40.0

4.25

"135

=

mmmm

U

where '0.40' is the Overall Heat Transfer Coefficient for a 1-3/4" door, per Table5-9. From this extrapolation, the door coefficient 'U' is 0.31 Btu/ft2-hr-oF.

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 theirthermal resistance's for a specific roof example. This example does not matchwith 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 ofTable 5-7b. The specified roof components for the library, their thermalresistances, and the roof Overall Heat Transfer Coefficient are provided in Table3.

Table 3. Roof Overall Heat Transfer Coefficient

Description Thermal Resistance, R(ft

2-hr-

oF/Btu)

Outside surface (15 mph wind) 0.17Built-up roofing, 0.375 in. 0.33

Concrete slab, lightweight aggregate, 2 in. 2.22

Inside surface (still air) 0.61Total 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 calculationfor a slab-on-grade floor uses a slightly different form of the general heat transferequation than previously stated. The appropriate equation for a slab-on-grade-floor is:

( )oi ttPUq =&

where, U' is the overall heat transfer coefficient expressed as heat transfer perunit length of building perimeter, per degree temperature difference. Theperimeter of the floor is (100-ft + 100-ft + 50-ft + 50-ft) 300 ft, assuming the UtiltyRoom 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 beendetermined, the total transmission heat loss from the building to the outside canbe calculated. Table 4 tabulates the overall heat transfer coefficients, and showsthe calculations used to determine the total transmission heat loss. As shown inTable 4, the total transmission heat loss is 146,411 Btu/hr.


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Table 4. Total Transmission Heat Loss

Uo

(Btu/hr.ft

2.oF)

U'

(Btu/hr.ft

.oF)

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 7,053West 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 theair-change method. The crack method determines the infiltration rate based onpressure differences between the outside air and inside (due to wind), along withthe characteristics of the windows, doors, etc. (i.e. their "cracks"). The air-change method assumes a value of air-changes per hour, and with thisassumption determines the infiltration rate for a known volume. For thispreliminary 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 bea 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, isas follows:

( ) 60/VACHQ =&

where Q is the infiltration rate in cfm, ACH is the assumed number of air changesper hour, and V is the gross space volume in cubic feet. The gross spacevolume of the library is (100-ft x 50-ft x 10-ft) 50,000 ft3. Therefore the infiltrationrate 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. Thesensible heating component due to infiltration is the sensible heat required toincrease the infiltrated air to the design indoor temperature. The latentcomponent is the latent heat required to humidify the infiltrated air to the designindoor relative humidity. Equation 7-8b of Ref. (1) is used to calculate thesensible infiltration heat load, and is as follows:


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( )

o

oip

sv

ttcQq

=

&

&

where Q is the infiltration volumetric flow rate (ft3/hr), cp is the constant-pressurespecific heat of air (taken as 0.24 Btu/lbm-oF), ti and to are the inside and outsideair 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 equationand our calculated infiltration air rate of 833 cfm (50,000 ft3 /hr), the sensibleheating 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 heatload, and is as follows:

( ) fgoio

liWW

v

Qq =

&

&

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 ofvaporization. From the Psychometric Diagram, the inside humidity ratio Wi is0.005 lbmv/lbma. The outside humidity ratio is assumed to be zero. Thisassumption is valid since outisde air during the heating season is very dry (lowrelative humidity) and therefore a conservative aproach would be to assumethere is no water vapor in the outside air. The enthaply of vaporization, fromTable A-1a of Ref. (1), at the indoor design temperature of 72 oF, is 1052.8Btu/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 tothe infiltration of outside air through windows, doors, etc. This is the total spaceheating load, and is tabulated in Table 5.

Table 5. Total Space Heating Load

Sensible Heat Load

( )hsq&

Latent Heat Load

( )hlq&

Transmission( )hrBtu/ 146,411 n/a

Infiltration( )hrBtu/ 49,423 21,680

Total

( )hrBtu/ 195,834 21,680


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Supply Air for Space HeatingIn order to determine whether the heating season or cooling season is the

driving season for the selection of the air-distribution system, the supply airrequirements for space heating must be calculated. This will be compared tosimiliar calculations for space cooling to determine a design airflow rate.

Appendix C contains a process diagram of the heating system on aPsychometric Chart, as well as detailed heating system calculations. Also,Appendix B, Sheet 4 is a detailed schematic diagram of the heating system. Apre-heater was considered necessary, to bring the temperature of the fresh airabove the dewpoint temperature of the air inside the utility room. This ensuresthat condensation will not form on the ducting leading into the furnace. As statedpreviously, the utility room temperature is assumed to be the average of theoutside design temperature and the desired indoor temperature, namely 47oF.

A summary of the results of the detailed calculations is provided in Table6. It is important to note that these calculations do not take into consideration theeffects 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 winterconditions, are based on steady-state heat transfer, and this basis is adequatefor selecting heating and humidification equipment. For the cooling load,

however, steady-state heat transfer would not adequately represent the heatgain, and would lead to oversized equipment. This is because the solarcontributions to the space cooling load have a considerable lag effect, before theheat is actually absorbed by the inside air.

There are several methods that can be used to calculate the spacecooling load. These include the "Transfer Function Method," and the "CoolingLoad 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 theTransfer 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 maximumcooling load occurs at 15:00 solar time, during the month of August. This

statement greatly simplifies the process of determining the design space coolingload. If the time that the maximum cooling load occurs was not provided, thespace cooling load would have to be determined for several times, in an iterativefashion, in order to select the maximum. Since the time of the library maximumspace cooling load is known, only one iteration is necessary. Each contributingcomponent of the space cooling load is calculated separately as follows, and isthen summed to provide the total sensible and latent space cooling loads. Theoverall heat transfer coefficients used for conductive heat transfer are identical tothose used when calculating the space heating load.

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

given as roof #9, dark color outside surface, without a suspended ceiling belowthe roof deck. The overall heat transfer coefficient, calculated previously is 0.30.For determining the CLTD value, Table 8-18 of Ref. (1) is used. The librarymaximum cooling load occurs during the month of August, however Table 8-18 isfor July 21st. A footnote in Table 8-18 states that "for design purposes, the datawill suffice for about two weeks from the 21st day of the given month (in this caseJuly)." Therefore it is assumed that Table 8-18, and all other CLTD tables forJuly 21st with similar footnotes, are sufficient for the design of the library coolingsystem maximum load during August.

Using Roof #9 and 1500 solar time (given time of maximum cooling load),the roof CLTD is selected from Table 8-18 as 46. This value, however, must beadjusted to reflect the differences in the library design temperatures and thosetemperatures used to derive the Table. The adjustment is as follows:

( ) ( )8578. ++= mr ttCLTDCLTDCorr

where, tr is the inside temperature, and tm is the maximum outdoor temperatureminus the (daily range/2). From Ref. (1), Table C-1, the daily range for Dallas,Texas is 20oF. And with the design indoor and outdoor temperatures listed in the"Design Conditions" section of this report, the roof CLTD is adjusted as follows:

( ) 54)8590(757846. =++=CLTDCorr

To find the sensible heat gain to the library due to transmission throughthe roof, the following equation is used:

( )CLTDUAqc =&

Therefore, with U=0.30, A = 5,000 ft2, and CLTD = 54, the library space coolingload 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 resistancewas 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 materialand 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 theLibrary walls' actual secondary material is face brick, and principle material is 6"heavyweight concrete block, a wall # of 7 was interpolated. Therefore using awall # 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 CLTDadjustment done for the roof. Table 7 contains the calculations for each wall,using the surface areas and overall heat transfer coefficients determinedpreviously for the space heating load. The effects of the Utility Room on theWest wall are considered to be negligible.

Table 7. Wall Space Cooling Load

Wall CLTD Corr. CLTD U A ( )CLTDUAqc =&

North 16 24 0.31 496 3,690South 24 32 0.31 475 4,712East 32 40 0.31 455 5,642West 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 gainthrough 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 1500hrs. These values, and their corrected values (corrected similar to the correctionto roof CLTD) are provided in Table 8, along with calculations to determine thespace cooling load due to windows and doors. The surface areas and OverallHeat Transfer Coefficients were determined previously for the space heatingload.

Table 8. Window & Doors Conduction Space Cooling Load

Item CLTD Corr. CLTD U A ( )CLTDUAqc =&

North Windows 14 22 0.65 504 7,207

South Windows 14 22 0.65 480 6,864South Doors 14 22 0.31 45 307East Doors 14 22 0.31 45 307Total Space Cooling Load due to Glass-Door Conduction = 14,685 Btu/hr

Glass (Solar) - To determine the Solar contribution to the space cooling loadfrom 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, forsingle-story buildings, using 4 walls, and carpet floor covering, the Zone Type isselected as 'A' for Glass Solar. Then from Ref. (1), Table 8-24, the SCL factorsfor the North and South windows are selected with a solar time of 1500 hrs. Asstated 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-typeinsulating glass, a Shading Coefficient of 0.39 is selected. The contribution tothe space cooling load from glass solar effects is calculated as shown in Table 9.The heat transfer areas were calculated previously when determining the spaceheating load.

Table 9. Windows (Solar) Space Cooling Load

Item A SC SCL ( )SCLSCAqc =& North windows 504 0.39 36 7,076South 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 heattransfer to the floor slab is negligible.

Internal Lights - To determine the contribution to the space cooling load fromthe internal lights, a Cooling Load Factor (CLF) must be determined andmultiplied with the instantaneous heat gain of the lighting. The instantaneousheat gain can be determined from the following formula:

( ) suFFWq 41.3=&

where, W is the total installed wattage, Fu is the usage factor, and Fs is anallowance factor. As stated in the COR (Appendix A) the product of W and Fu is4 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 istaken to be 1.2. Therefore with a floor area of 5,000 ft2, the instantaneous heatgain 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 theCOR, 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 ofhours the lights are on up until the maximum cooling load time is 7 hours. Usingthese 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 foundto be (0.965)(81,840) = 78,976 Btu/hr.

Equipment - To determine the contribution to the space cooling load from theequipment, a Cooling Load Factor (CLF) must be determined and multiplied with


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the instantaneous heat gain of the equipment. The instantaneous heat gain fromthe equipment is determined in a similar fashion to the calculation for internallighting. As stated in the COR, the Library equipment emits 0.5 Watts/ft2 of floorarea. 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 LibraryZone 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 thecontribution to the space cooling load from equipment is found to be3.41(0.5)(5,000)(0.93) = 7,928 Btu/hr.

People (sensible) - The sensible portion of the instantaneous heat gain due tothe Library occupants requires the determination of a Cooling Load Factor. First,from Ref. (1), Table 8-11, the instantaneous sensible heat gain for people seateddoing 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, atthe point of maximum cooling load (1500 hrs) the people will have been in thespace 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 portionof 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 immediatelyabsorbed by the space air, and does not require a CLF factor for adding it to thespace cooling load. From Ref. (1), Table 8-11, for people seated doing very lightwork, the latent heat gain is 155 Btu/hr per person. For 75 people, the latentportion 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 theAir-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 innature, and immediately becomes cooling load."

As stated in the "Design Conditions" section of this report, the outside airis 100oF dry bulb, and 75oF wet bulb. From the Psychometric Chart inconjunction 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 istaken as 0.24 Btu/lbm-oF.

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

( )

o

iop

sv

ttcQq

=

&

&

Therefore, the sensible portion of the space cooling load due to infiltration air isfound 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 alsodetermined in the same fashion as was done for the space heating load using thefollowing formula:

( ) fgioo

liWW

v

Qq =

&

&

From Ref. (1), Table A-1a, the latent heat of vaporization (i fg) 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 dueto infiltration air is found as (50,000)(0.0130 - 0.0092)(1037)/14.419 = 13,665Btu/hr.

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

Table 10. Total Space Cooling Load

Sensible Cooling Load

( )csq&

Latent Cooling Load

( )clq&

Roof Transmission

( )hrBtu/ 81,000 n/a

Walls Transmission( )hrBtu/ 18,694 n/a

Glass-Doors Conduction

( )hrBtu/ 14,685 n/aGlass Solar

( )hrBtu/ 16,810 n/a

Internal Lights( )hrBtu/ 78,976 n/a

Equipment( )hrBtu/ 7,928 n/a

People

( )hrBtu/ 17,273 11,625

Infiltration

( )hrBtu/ 20,806 13,665Totals

( )hrBtu/ 256,172 25,290

Therefore the total space cooling load is 256,172 + 25,290 = 281,462 Btu/hr.


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Supply Air for Space CoolingAppendix 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. Asummary 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 effectsof 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 theLibrary is a Freon (R-22) cycle operating at a 30oF evaporator temperature andwith a 10oF superheat before the refirgerant enters the compressor, and a 110oFcondensing temperature with 10oF of subcooling before the expansion valve.

Appendix B, Sheet 6, contains a detailed diagram of this system employed toprovided cooling and dehumidification to the Library. As shown on Sheet 6, acooling tower is utilized to transfer the heat from the condenser cooling water tothe ambient air, and the chilled water system provides chilled water to theair/water heat exchanger which is the cooling/dehumidification coil for the Library.

Appendix C contains detailed cooling system calculations for the LibraryAir-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 CPressure-Enthaply diagram correspond to the state points as shown on the Air-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 52o

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 Condensersize in terms of GPM of water flow and capacity. It is assumed, for thecalculations, 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 beequated to the Water Chiller (R-22 Evaporator) capacity.

From this known Water Chiller capacity, along with the known refrigerationcycle state points, the mass flow rate of Freon in the theoretical refrigerationcycle was determined. Then, using the calculated mass flow rate of Freon, the

Theoretical BHP of the Compressor was determined. And, using an energybalance on the refrigeration cycle (Qcondenser = Qevaporator + Wcompressor, see Sheet6, 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 acooling capacity of 33.8 tons. This Chiller meets the Library coolingrequirements with a 10% margin. Details of this Chiller Unit are provided inAppendix D. The selection methodology is provided in the Appendix Ccalculations, and any assumptions not stated here are stated in Appendix C.Table 12 summarizes the characteristics of the Air-Conditioning Systemcomponents.

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 tonsystem cooling capacity)

Theoretical Capacity of Condenser 415,012 Btu/hr(for 28.5 tonsystem cooling capacity)

Selected Chiller Capacity(at 90

oF Cond. Lvg wtr temp, 42

oF Cooler lvg wtr temp)

33.8 tons

Compressor Power for Selected Chiller 28.0 kW

Chilled Water Circulation Rate 81.0 gpm

Condenser Cooling Water Flow Rate 101.2 gpm


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AIR-HANDLING/DISTRIBUTION SYSTEM DESIGN:

For the Library, a low-velocity air distribution system was selected. Ahigh-velocity system would likely not meet the noise requirements for a libraryapplication, and are normally selected for more industrial-type applications. As

stated in the COR (Appendix A), the recommended air velocities for the library airdistribution system are:

Duct Sections (max) 1,000 fpmAir Supply Diffusers 400 fpmReturn Air Registers 200 fpm

Cooling Coil 500 fpm

Also, the combined supply and return air pressure drop should not exceed4.0 in w.g. This requirement is likely based on an independent cost analysiswhich compared costs due to duct size (non-recurring costs) and costs due to fan

power consumption (recurring costs). From this analysis, a maximum systempressure 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, thesupply air required was previously determined to be 4,113 cfm. And, in coolingmode, the supply air required was previously determined to be 11,026 cfm.Obviously, the cooling season supply air requirements will drive the airdistribution system design, since the required volumetric flow rate is much larger.

The process followed in developing the air distribution system is asfollows. First, the design airflow is selected. Based on the cooling seasonrequired 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 pressurelosses are determined. Then the layout of the supply and return air-ductingnetwork is accomplished. With the duct layout decided upon, the ducts are sizedto ensure that the system behaves as desired, and the supply and return systempressure losses are determined. Finally, with the total system losses tabulated, afan is selected. More detailed discussion for each of these steps in provided inthe following sections.

Diffuser SelectionAfter several iterations of diffuser number and size, 10 diffusers were

selected to supply air to the Library. The diffuser type was selected as CircularCeiling Diffusers. Sheet 7 of Appendix B shows the location of these 10diffusers. They are each located in the Library overhead, in the center of a 20 ftby 25 ft section of the building. Since there are 10 diffusers, the cfm per eachdiffuser 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. Thecharacteristic 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, sinceeach diffuser is serving a 20 ft by 25 ft section, the characteristic room length istaken to be between 10 -12.5 ft. From Ref. (1), Table 11-8, knowing the Roomload, 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 diffuserselected for an application, ensures that the occupants of the building do not feeldiscomfort. A point where occupants feel discomfort can be defined by aneffective draft temperature. And, per Ref. (1), page 439, the ADPI is defined asthe percentage of measurements taken at many locations in the occupied zoneof a space that meet the effective draft criteria. By selecting a diffuser so that thex50/L value falls in the defined range for the known Room load, it can be ensuredthat the occupants do not feel uncomfortable.

Since maximum ADPI occurs at a value of x50/L = 0.8, x50 = 0.8(L), the x50value is 0.8(12.5 ft) = 10. Also, a recommeded diffuser velocity of 400 ft/min wasstated 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 rateof 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 rangeprovided in Table 11-8, and is acceptable. Therefore, 24" Circular CeilingDiffiusers are selected for the Library. Also, the Noise Criteria Index (NC) islower 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 SelectionAs 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, withcharacteristics: 1080 cfm, NC =20, and 300 ft/min. The return grill velocity isabove the recommended velocity in the COR, however this is considered to beacceptable 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 thevelocity pressure is 0.006. Therefore the Po value determined for this grill isfound to be -0.033 + 0.006 = - 0.027" w.g. This value is stated as negativebecause it is on the suction side of the supply fan, and therefore the pressure isbelow 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 toeach 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 equivalentto the pressure drop from the start of the particular branch to the end of the mainduct 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 toeach diffuser.

To determine the pressure drops through each section, the EquivalentLength Method was employed. This method determined the pressure dropthrough selected duct sections, by calculating equivalent lengths for components

other than straight duct, to add to the length of straight duct. This allows a singleequivalent length (Le) value to be multiplied by Po per 100 ft of duct value todetermine the branch pressure loss.

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

Table 13. Supply System Equivalent Length Calculations

Section Individual Lengths LeMain Run

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

oelbow)

+ 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' (45

oelbow) + 10' (duct) 26'

BranchesB-K 35' (Take-off branch) + 3' (duct) + 5' (45

oelbow) + 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 todetermine the Pressure Drop per 100 ft, for each section. Iteration wasnecessary to ensure that the duct velocities remain below the recommendedvelocities. As stated in the COR, the recommeded maximum duct velocity is1,000 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 arecharacteristic of a low-velocity duct system. However, since the recommendedmaximum duct velocity was considered to be important to avoid any noiseproblems in the Library, it was adhered to. This resulted in low pressure dropsthroughout the supply system. Table 14 shows the resulting supply system duct

sizes for the main run, and Table 15 shows the resulting supply system ductsizes for the corresponding branches. These values were determined after muchiteration, and as shown in Tables 14 & 15, the system is balanced and the ductvelocities remain under the prescribed limit. Also, these Tables show the supplysystem 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 ftSection

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 0.0153Total 0.0804

Table 15. Supply System Branch Sizing

Branch Ducts

Section

Loss

Equivalent

to Section:

Required

Delta-PoLe (ft)

Delta-

Po/100 ftCFM 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/SizingThe 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 thelongest 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." Ascan be seen on Sheet 9, sections A-O and A-J both have branches (i.e. the shortducts leading to the return grills). Therefore in addition to the braches beingsized to match the pressure drop from their branch point to the end of the longestrun, both "main runs," A-O and A-J, must have equal pressure drop. This is truebecause both sections originate from the same plenum.

The pressure drop through section A-O was calculated first, then sectionA-J was sized to match the pressure drop through A-O. Once this wasaccomplished, the branches to the other return grills were sized using thebalance method, for their corresponding "main run." Table 16 provides theequivalent lengths of all runs in the return ducting, used in sizing calculations,

determined in a similar fashion as the supply ducting. The return grill pressuredrops are not included in the equivalent length calculations because each runhas an identical grill and airflow rate, and therefore, an identical pressure loss.The section designations in Table 16 correspond to the duct section labels asshown in Appendix B, Sheet 9.

Table 16. Return System Equivalent Length Calculations

Section Individual Lengths LeMain Run (A-O)

A-P 35' (plenum) + 50' (duct) +10' (90oelbow) +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' (90

oelbow) +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' (90

oelbow) +1' (duct) 31'

BranchK-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 dropthrough each path that the return air can travel, with the equivalent lenghtsprovided in Table 17, the pressure drop through section A-O was calculated afteriteration, and is provided in Table 17. Also Table 17 provides the pressure dropcalculations 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 bothconnected to the same plenum, on the suction side of the fan, equal quantities ofair should travel through them, namely 5,500 cfm in each run. As depicted inTable 17, slightly smaller duct sizes were used in section A-J, to provideadditional pressure drop to equalize with section A-O (section A-O has anadditional sizty feet of equivalent duct length).

Table 17. Return System Main Duct Pressure Drop

Balance Method (Main Duct Run A-O)

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

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

Section

Delta-PoA-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 eachreturn grill. As shown in Table 18, the branches from section A-O, and thebranches from section A-J are sized so that their pressure drop matches thepressure drop of the duct run that begins at the respective branch point to theend 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

SectionRequired

Delta-PoLe (ft)

Delta-


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

SectionRequired

Delta-PoLe (ft)

Delta-


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 SelectionTo 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 theindividual pressure drops calculated previously. Also, assumed pressure dropsfor the cooling coil and filters are neccessary. As stated in previous sections andTables, the supply system pressure drop is 0.0804" w.g., the diffuser pressuredrop is 0.0240" w.g., the return system pressure drop is -0.0876" w.g., and thereturn grill pressure drop is -0.0270" w.g. The cooling coil pressure drop isassumed 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 fanare included in the supply and return system pressure drop calculations. In orderto ensure the fan is capable of delivering air through the duct system, a margin of15% 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 selectionpurposes. As can be seen in Table 19, the total pressure drop does not exceed


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

Table 19. System Total Pressure Drop

Item PoDiffuser 0.0240" w.g.

Supply Ducting 0.0804" w.g.Filters 0.1500" w.g.

Cooling Coil 0.2500" w.g.Return Ducting 0.0876" w.g.

Return Grill 0.0270" w.gTotal 0.619" w.g.

Total (with 15% margin) 0.7119" w.g.

Because the fan vendor data provided includes selection tables in terms ofsystem static pressure loss, the total system pressure drop must be convertedinto static pressure. To do this the velocity pressures before and after the fanmust be determined. Figure 1 depicts where the velocity pressures must becalculated. The static pressure delta is detemined as follows:

( ) ( )1122 vvo PPPPP =

( ) ( )1212 vvo PPPPP +=

( )12 vvos PPPP=

=

2

2

2

2

40054005

VVPPos

1 2

Figure 1. Fan Velocity Pressure

A SWB, Model SWB-24, Centrifugal Fan is selected as the fan for theLibrary HVAC system. As shown in the vendor data in Appendix D, the fan inletarea is 5.582 ft2 , and the fan outlet area is 3.3 ft2. With the design airflow rate of11,000 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 dropcan be translated into a static pressure delta as follows:

=

22

4005

971,1

4005

333,37119.0sP

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

SUMMARY:

Using the methods outlined in the ME 260 HVAC course, Spring Semester1999, a preliminary HVAC system design was performed for a Library to belocated in Dallas, Texas. The heating and cooling systems were designed, andthe air-distribution system, based on the cooling system airflow requirements wasalso designed. Appropriate equipment components, such as a pre-heater,furnace, humidifier, cooling and dehumidification unit, reciprocating compressorchiller 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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P

Lo

Pre

Tit

100 ft

PLAN VIEW

10 ft

15 ft

35 ftUtility Room


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Lo

Pre

Tit

SOUTH WALL ELEVATION

NORTH WALL ELEVATION

EAST WALL ELEVATION

WEST WALL ELEVATION


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Lo

Pre

TitFLOOR CROSS-SECTION

WINDOW DETAIL

4 ft

6 ft

WALL CROSS-SEC

1

Outside

4 face brick

ROOF CROSS-SE

Outside

Inside

3.75 built-uproofing

8 block wall, brick face,uninsulated

slab-on-grade floor


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P

Lo

Pre

TitH

Space

Furnace

Humidifier

qh

mw

0

1

2

4

Qo = 1500 cfm

to = 22o

Fo = 0%

0

Pre-heater

t2 = 120oF

qph


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Lo

Pre

TitCo

Space

Cooling andDehumidification

Unit

qc

0

1

2

3

4 5

Qo = 1500 cfm

to = 100oF db

75oF wb

t2 = 55oF


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Lo

Pre

Tit

Evaporator

Condenser

ExpansionValve Freon-22 Refrigeration Cycle

Supply Air

CooliTow

1

2

4

3

Qevaptchilled water, in = 52

otchilled water, out = 42oF

tcooling water, out = 92tcooling water, in = 82oF

Qcond


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Lo

Pre

Tit

PLAN VIEW

F

Return Ducting

Supply Ducting

24 x 24 Return Grill

24 Circular Ceiling Diffuser

1500 cfm exhaust fan required

Space Division Lines for Diffuser Selection


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P

Lo

Pre

Tit

C

D

E

F

G

K M

P Q R

L

B

5

15

3

3 3 3

3 3

10

10

10

10

10

10

17 17 17

A

Plenum


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Lo

Pre

Tit

AB C D E

P RQ S

15 20 20

50

Plenum

15 20 20

FHG I

K ML N

The ducts connecting the main runsand the return grills are 1 in length

Note:

8

8


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APPENDIX C

Calculations and Process Diagrams


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APPENDIX D

Selected Component Vendor Data


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hvac design for library-dallas-texas me 260 1999

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