BUILDING ENVIRONMENTAL SYSTEMS
BUILDING ENVIRONMENTAL SYSTEMS ARE 346N
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This 137 page Class Notes was uploaded by Rowena Barton DVM on Sunday September 6, 2015. The Class Notes belongs to ARE 346N at University of Texas at Austin taught by Jeffrey Siegel in Fall. Since its upload, it has received 82 views. For similar materials see /class/181537/are-346n-university-of-texas-at-austin in Architectural Engineering at University of Texas at Austin.
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Date Created: 09/06/15
Objectives Ch 11 Ch I Bldg design loads gt Dclunniuc design Describe room distribution basics H 0 Select diffusers gjgjgjj Supply and return g mve gt duct sizing sum sync We airflow required Sun iSl dlxd Al39 or each lcrmlnui clcvicc Luv mu wquotpr Sec 184 and rclurn duels Duct srhcnmm m mm mm swmuumnu Sue UH hmnch mm Figure 181 Oullinc of forcedair distribution system design pruccr Liler and sections in Chapter 18 bar cover lhc various slaps Designing Room Air ow 0 Very complex problem 0 Pumped ow bouyant ow or mixed ow 0 What nondimensional parameters govern each regime Archimedes number Ar g LA T V7 L characteristic length In R g acceleration due to gravity Ins2 rninz T absolute temperature K R B 11quot UK 1 0R V air Velocity Ins rnin Colorful Fluigc Dynamics LO ms 200 Wmin J Supply A yI1InulylIHIVIIVHIIIIIva v rm rr39v Figure 1813 Velocity vcclur l39icld simulated using highReynoldsr number turbulence mndel from Laundcr and Spnlding 5 50 fllmin contour contour c Entrainment Coanda effect and buoyancy Figure 185 Examples of air jet behavior msidc Ounidc Hunting su mmm A mnlmlr Ell l39hcrmtvbhll I i Stlpmm 1 lump i Temp 39nullng 5n lmin r m39 Slrulificuliuu 5 a Q u r niennmuu rSuLpuiIIl K 4 L q temp i 39lcnux Figure 186 Airflow and roomlcmpcrnture distributions for a flourpcri niclcr diffuser Coo 39l hrnw 5 Wmquot ling 4 A a j x soumm V 4 6 in Tlicrmosiin u I 1 39 I 4613970quot J icmp Temp Heating 39rhmw 50 l39lmin i quot 4 A a D L sunmm a Siranimation 3 Thnrmoslal SCIpnlnl lamp Tc mp Figure 187 Airflow and roomtemperature dislrihutions for a coil ing diffuser Diffuser Selection Procedure V Qtot Qserr pAh pAt V maximum volumetric ow rate m3s 3min QM total design load W BTUhr Qsen sensible design load W BTUhr p air density kgm3 lbm 3 At temperature difference between supply and return air C F Ah enthalpy difference between supply and return air Jkg BTUlbm Select and locate diffusers diVide air ow among st diffusers Find Characteristic Length TABLE 182 Characteristic Length for Various Diffuser Types Diffuser fype Characteristic Length High sidewall grille Distance to opposing wall Circular ceiling diffuser Distance to closest wall or midway to nearest ceiling i user Sill grille Length of room in direction of airflow Ceiling slot diffuser Distance to nearest wall or midway to nearest diffuser Light trotfer diffuser Onehalt distance to nearest diffuser plus distance from ceiling to top of occupied zone Perforated ceiling diffusers Distance to nearest wall or midway to nearest diffuser Diversion ADPI air distribution performance index 0 Fraction of locations that meet criteria 3 OF ltEDTlt2 OF 0 Where EDT effective draft temperature 0 Function of Vand At Eqn 181 Ideal and Reasonable Throws TABLE 18 Diffuser Seleciion Guidelines for MixedFlow Rooms Room Loud Terminal TEEL Ta 25L for Maximum Range of Wm Biuhr H2 Maximum ADP ADPI ADP gt ISOL rmL High 250 80 l 68 sidewall 190 60 18 72 70 l 5 22 grilles 125 40 L6 78 70 39l 2 23 65 20 L5 85 80 39l 0 19 Circular 250 80 08 76 70 0 7 13 ceiling 190 0 0 8 83 80 0 7 12 diffusers 125 40 0 8 85 80 D 5 15 65 20 08 93 90 0 7 13 Sill Grille 250 80 L7 61 60 l 5 17 siroighi 190 60 17 72 70 l 4 17 125 40 1 3 86 80 l 2 18 20 09 95 90 U 813 r50 7 throw in SDlrlmln veloclfv rm ihrow lo 025mls velociiy Select Register Pick throw volumetric flow from register catalog Check noise pressure drop WILE 5 Typical Circular Ceiling Dmuser Caialog Dara n 1 Size Neck Velocily ffmin 400 500 600 700 500 900 1000 1200 l 8quot Total press in WG 033 052 075 101 130 166 205 292 Flow role fPmln 140 175 210 245 280 315 350 420 Throw 1391 4 4 5 6 7 8 9 11 NC 15 21 26 31 34 37 44 1039 10101 press in WG 027 043 062 084 108 138 170 243 Flow rule Wmln 220 270 330 380 435 490 545 65 Throw 11 4 5 b 7 8 9 10 12 NC 11 17 21 26 30 33 39 12quot 10101 press in We 026 042 060 081 105 134 166 236 Flow 1019 Illamin 315 390 470 50 63 705 785 940 h 5 6 7 8 10 1 l 12 14 1 1 17 22 26 30 33 39 14 10ch press in WG 038 061 087 118 152 194 240 342 Flow rme l Pmin 425 530 635 745 850 955 1060 1270 Throw 11 6 B 9 12 14 15 18 ll 18 23 28 32 36 40 46 QM 384 kBTUhr Ah 95 BTUlbma id Circular ccrllng ammo l Example 183 Qsen VQ tsz39 pAh CppAt Note omission in text RCIIIIII Ilr pnllc 31 ll whiff w Figure 189 Room with circular ceiling ilil39lusers for F39 183 Pressures 0 Static pressure pressure not due to velocity 0 Velocity pressure pressure due to velocity Total pressure sum of above Relationship Between Static and Total Pressure Q 8czvgt 1 quot MMUSPHERIC waissuRE I I c gt K mm m Essunz I um I 1 I sun I p mm In Almosanzw t I Kenn It mum DJ 7 I I I l I I l l I i l i r l I I I 1 I I I a p is I g l nmoswzm PREssuRE Fig 7 Prmm L lmugvs During Mm in Dum Duct Design Total and static pressure drops are proportional to square of velocity LVz LV2 1 7 l 7 AP f D28 pg pf D2 v Plot of pressure drop vs volumetric ow rate or velocity is called system characteristic System Characteristic net 5 slum clmmclcrisn n V v Figure 18I5 Fundamenlal duclv 39m characteristic showing x39cla lionship between lolal pressurt ms and duct mean velocity and volumetric flow rate Electrical Resistance Analogy AP bc Series AZ o A P Vr O N O quoth u b c AP AP Al AP Same ow mu Lquot m H gt i x a Series law Al all 2 Pamqu AZ u uh 4 I a b l l A 7 l l a 39 1 Same prasx ure Llrop 1 n 39 V V ulh u2h ab 1 Parallel flow Figure 1816 Series and parallel system characteristic curves monoquot LOSE Palm Frictional Losses Fin 9 AIR auArmrv Us 039 mm Noncircular Ducts L 77 l r 7 I 1 K lt I Am r MWW A if i u I 2171 A my b Figure 1819 Equivalent diameter ul rcclangulur and Ilal oval ducts Parallel concept to wetted perimeter Dynamic losses 0 Losses associated with 0 Changes in velocity Obstructions Bends Fittings and transitions 0 Two methods 0 Equivalent length and loss coefficients CO L ff11n C1 APFCUPV O oss C06 06 ts Vii m 23 300 380 45 690 15H R 0 l0 008 007 006 005 003 r 41 Damn 75 150 C0 016 012 CD342 Elbow 3 Gore an Degree rD D 075 100 15a 200 42 A 0 E053 Tet Drlt ur 10 in Cull39el 39 I r p mum QtQr LHA Abby 04 05 M u 01 059 u 93 mm 3 001 0 7x I In 04 Ill x 0 x7 05 m9 159 l 02 017 105 1 v 230 1 7 024 l 0quot 0 7 JIJ SI 4 UN 1 4 us 411 Ix M4 L1 Hu 0 10 N 0 I I quot4 10 743 x 47 HA gt3 0 405 mm 10x 1 m 01 um um L07 1 1 11 M W Z LU I 7 7 Example 187 0 Detennine total pressure drop from 0 to 4 um I nus 15 m 753 35 cm ID 5m 2 5 z m 4 15 m ID mm 1 Figure 1820 Schematic of round return ducl used in Ex 187 Conversion Between Methods TABLE 1817 Friclion Fodors for Galvanized Sheet Metal Ducts Duct Diameter eq 2 V2 g f 2 2 in cm Friction Factor D g g 4 10 0035 6 15 0023 B 20 0023 L H3 25 0022 e f 13 32 3 3 16 40 0016 50 Summary Select diffuser based on ADPI Given a layout of duct system 0 Calculate total pressure drop 0 Identify ttings contributing most to pressure drop 0 Deal with noncircular ducts Use fitting loss coefficient or equal length method Next class fans duct layout other equipment Objectives Describe vapor compression cycle basics Draw cycle on T s diagrams Compare real cycles to ideal cycles Basis for discussion of individual components Very thermoheavy for next few classes Air supply Refrigerant duck to moms lled Whine Condenser coil 39 mace or airhandllng unit quot LRetain air duct Inside air 3 4 f 4 M 4 Expansion A valve Compressor 4 J 2 1 k I 39 V I 7 Outside air Ef ciency 0 First Law Coefficient of performance COP COP useful refrigerating effectnet energy supplied COP qw 0 Second law Refrigerating efficiency 17R 17R COPCOP rev net 0 Comparison to ideal reversible cycle Carnot Cycle No cycle can have a higher COP All reversible cycles operating at the same temperatures T 0 T R will have the same COP For constant temp processes dq T ds COP TRToi TR Figure 31 Processes of the Carnot Specific emmpy s refrigeration cycle Get Real Assume no heat transfer or potential or kinetic energy transfer in expansion valve 39 COP hs39h2h439h3 Figure 33 The theoretical singlestagc vaporcompression Cycle Area Analysis of Work and Efficiency Figure 35 Areas on the T s diagram representing the refrigerating effect and the work supplied for the theoretical singlestage cycle Comparison Between SingleStage and Carnot Cycles 0 Figure 36 T 1 A2 qcamot n R Ml wcarnat 7 Example RZZ condensing temp of 30 OC and evaporating temp of 0 OC Determine 1U a qcamar woumar b Diminished IR and TN f 3 excess w for real cycle Al 39 caused by throttling and superheat horn 0 7R 300 N 00 Tempuralure K 100 02 l 04 06 08 Entropy kJkg K 11 L2 Figure 37 Temperatureenlropy diagram for Refrigerant 22 show ing area deviations of theoretical singlestage cycle with respect to the Carnot cycle Subcooling and Superheating Refrigerant may be subcooled in condenser or in liquid line Temperature goes below saturation temperature Refrigerant may be superheated in evaporator or in vapor suction line Temperature goes above saturation temperature 0 Neither of this results in as useful heat transfer Loss of ef ciency Vapor Compression Cycle Energy input Cold uid Warm uid Coo ng Warm fIUId Liquid refrigerant Rcfrigemm 22 C ascadc condenser 1 Figure 317 Schematic diagrams for cascade cycle composed of two singlwslage cycles Multistage Compression Cycles Combine multiple cycles to improve efficiency Prevents excessive compressor discharge temperature Allows low evaporating temperatures cryogenics b 4 Expansion valve E Figure 311 Schematic diagrams for cycle with twostage compression water intercooier and ash inlercoolcr General strategies for cooling and heating loads Heating Loads 1 Calculate Z UA including slab edge and infiltrationventilation of items on first page to get an answer in BTUhr 0F or WOC 2 Multiply by design temperature difference 3 The result is your raw uncorrected heating load W Cooling Loads all steps should be in BTUhr 01 W 1 90899 9 Calculate cooling load similar to raw uncorrected heating load but using the cooling design temperature difference Calculate conduction through opaque surfaces walls and roofs which is typically done with a modified temperature difference Add solar gain through windows Add sensible internal gains The result is your raw sensible cooling load Calculate latent internal gains Calculate latent gains due to ventilation and infiltration The sum of 6 and 7 is your raw latent cooling load Heating Sensible only Formula Strategies ASHRAE1 Otherz Textbooks Walls Heating only q UAA T Calculate U Values of 252 7 25 12 Table 23A Sllbtl39act WindOWdOOI39 assembly from wall 25Table 4 areas components Look up assembly Some examples Table 214 last values in chapter 25 column is UValue below grade in BTUhrOF ftz walls 2514 2515 Windows Skylights q UAA T Look up U Values in 3 lTable 4 NFRC Celti ed Examples in 1156 rough Opening area NFRC guide or default Products Guide Chapter 2 frame type very important tables Roof Heating only q UAA T Calculate U Values of 25Table 4 Table 23A subtract skylight areas assembly from wall components Look up assembly Some examples Table 212 values in chapter 25 Doors q UAA T Look up U Values 3 lTable 6 NFRC Certi ed Examples in use rough Opening area Products Guide for Chapter 2 doors with glazing Slab Foundation edge q FPA T Look up F Value and 29Table 18 WA State Energy Table 2102ll Heating only multiply by perimeter Code data in P References section In ltration q MCAT M pQ Chapter 27 Tables 28 Ventilation q MCAT C 024 BTUlb OF Chapter 27 ASHRAE Standards Table 29A or 1007 Jkg K 62 622 Weather Data AT Use comfort zone for Chapter 28 on Tables 22A2 elevation latitutde appropriate indoor CD 2B22C conditions and 99 drybulb for outdoor tem erature ASHRAE 2005 Fundamentals Chapterpage number or ChapterTable number 2NFRC Cern edPraducts Guide Super Goad Cents HeaILass Referencde Manual by Ecotope Inc on class website d 3Tao and Janis 2008 Mechanical and Electrical Systems in Buildings 4 E Cooling Load sensible and latent 7 Everything on the previous page except for walls roofs etc which are replaced with the calculations below and slab edgebelow grade calculations which are omitted and the following additions Formula Strategies ASHRAE1 Other2 Textbook3 Conduction through q UA TE TD Use greater of TETD or Tables 2 13 opaque surfaces ie AT roofs walls doors q UACLTD Use Tables to determine 28Tables l and 2 etc Use any ofthe Cooling Load following formulae Temperature Difference q UATe Tn Calculate SolAir 29152916 Solar gain through q A XSHGFXSC SC Z SHGC SHGF 3lTable l3 SHGC SHGF Table 2 15A Windows Skylights 1997 Fundamentals NFRC SC 216 watch for shading dirty Table 29 3951 Certi ed windows shades and Products blinds etc Guide Sensible Internal Gains Include people lights 30Tables l23510 Table 217 appliances Latent Internal Gains Include people 30Tables l235 10 Table 217 appliances moisture sources Latent Ventilation q thgA W hfg hig from ASHRAE Chapter 29 30 ASHRAE Table 29A Infiltration 6Table 3 Standards 621 622 Weather Data AT AW will need Use comfort zone for Chapter 28 on CD Tables 22A2lB2 elevation latitutde to use psychrometric chart to get indoor and outdoor humidity ratios indoor conditions and 1 drybulb and coincident wetbulb for outdoor conditions 1ASHRAE 2005 Fundamentals Chapterpage number or ChapterTable number 2NFRC Certi ed Products Guide 3Tao and Janis 2005 Mechanical and Electrical Systems in Buildings 3quoti Ed 2C Cooling Systems 0 Text 41 45 Parts of46 Describe refrigerant conditions and components in different cycles ngh pressure high bo ingcondensmg pomt DB 77 F WE 64 F 110 F Rem eram 56 F Warm return air N200 psig compressor Sig giigag Vapor refrigerant Condenser Warmar A Condensing uid air orwarer P Liquid refrigerant capillary tube Low pressure low boiling point The purpose of an expansion valve in a vapor compression cycle is to A Lower refrigerant pressure before entering evaporator B Lower refrigerant pressure after leaving evaporator C V Raise refrigerant pressure before entering evaporator D Raise refrigerant pressure after leaving evaporator V Why is it desirable to have lower pressure in the evaporator Lowers the boiling point of refrigerant Raises the boiling point of refrigerant Lowers the pressure drop Raises the density of the refrigerant cap What is the phase of refrigerant when it enters the evaporator A Solid B Liquid C Gas Absorption Cycle Water is typical refrigerant Strong vacuum N01 16 psi or 00008 MPa 0 Boiling point of water Differences from vapor compression 0 No compressor Chemical eg LiBr hygroscopic 7 Absorbs water vapor turns it into a liquid 7 Regenerated by removing water gtgt Requires heat Hot va or NP Cv Dilute 00mg sulution water out Steam or hot water in Concentrated solution Chilled water Out In Cooling in orator recirculating pump Solution Pump Evaporative Cooling 0 Wet mediawater spray direct Raise absolute humidity lowers temperature of conditioned air 0 Cooling tower indirect Cools through a heat exchanger 0 Lowers temperature around condenser outside 0 Does not affect indoor absolute humidity 0 Can directly cool air in cold weather Why Chilled Water Big centralized equipment 0 Low noise 0 Access to cooling towers Easier maintenance Less chance of leak Easier to make changesadd buildings Small pipes for distribution Disadvantages of Chilled Water Water freezing Water treatment No condensing pump Bigger first cost Components of a Chilled Water System Chiller Can use either refrigerant cycle Heat rejection 0 Air cooling 0 Cooling tower Type Location 7 Legionella Pipe Sizing Example 1 inch pipe ID Water 45 F at 25 fps 0 Return water at 60 OF qMgtltCgtltAT qpXQXCXAT 2 m 1 15 F 0173 5 hr lb 3600s 1 2 112 BTU q 25 n 2 2 144111 1b F q43kBTUhr Heating 375 kBTUhr What about phase change 0 Common problem sizing an evaporator coil 0 How cold should an evaporator coil be 0 Too cold and water Will freeze out of air 0 Too warm and it won t dehumidify Relevant equation qprxhfg CXAT Often have to solve for both air and for refrigerant Often have to do mass balance on waterwater vapor Objectives Finish reviewing cooling systems 0 Size evaporator coils Review absorption cycle Introduce zoning and air systems Cooling Systems 41 45 parts of 46 Air Systems Chapter 3 Ch 6 Administrative HW4 is practice midterm Can be graded or ungraded If graded due one week before midterm 33 solutions posted on 33 graded like other HW out of 30 0 If ungraded solutions posted immediately Pipe Sizing Example 1 inch pipe ID Water 45 F at 25 fps 0 Return water at 60 OF qMgtltCgtltAT qpXQXCXAT q 15 F 1b 3600s 1 2 112 BTU 225 n m 21 a 0173 5 hr 2 144111 le q 43 kBTUhr Heating 375 kBTUhr What about phase change Common problem sizing an evaporator coil How cold should an evaporator coil be 0 Too cold and water will freeze out of air 0 Too warm and it won t dehumidify Relevant equation qpXQXhfg CXAT Often have to solve for both air and for refrigerant Often have to do mass balance on waterwater vapor Example Size an evaporator coil that will cool 1000 CFM of air from 85 0F and 50 RH to 60 OF Coil temperature is 40 OF everywhere in coil Refrigerant is RZZ and is just boiling everywhere in coil What is sensible capacity of coil What is latent capacity of coil What is total capacity of coil How much water is removed by the AC system What is refrigerant mass flow rate What is volumetric ow rate into coil What is volumetric ow rate out of coil What is the sensible heat ratio of the coil Variants on the problem Condenser rather than evaporator Easier why Phase change and temperature difference ie R22 enters the evaporator at 40 OF immediately boils and leaves the evaporator at 55 OF What C do you use Provide amount of condensate that is produced 0 How do you determine AW and or latent capacity Absorption cycle components 0 See problem on practice midterm Summary Need to account for phase change 0 Sometimes also temperature rise 0 Check when it occurs ie before or after phase change Total refrigerant energy change is same as total air energy change lf condensation occurs may need to calculate water volumemass Objectives Summarize cooling systems Zone buildings Select air systems Reading Assignment 31 32 34 Zone Criteria Coolingheating load requirement Ventilation requirement Zone use Client needs Building layout If none of above criteria are different than should likely be the same zone for simplicity Heat LossGain in Buildings Walls roofs etc lD conduction Finish Thursday s lecture and example Infiltration and ventilation convection Ground contact 3 D conduction Solar gains radiation Building individual pieces 0 We will bring them together In ltration Convection Air carries sensible energy 0 qMX C X AT BTUhr W M mass ow rate p X Q lbhr kgs 1 air density 0076 1b 3 12 kgm3 STP Q volumetric ow rate CFM m3s C speci c heat of air 024 BTU1b 0F 1007 kJkg K STP For similar indoor and outdoor conditions p and C are often combined q 108BTU minft3 thI x Q XAT 39 NOTE THAT THIS FORMULA IS UNIT SPECIFIC Potential Confusion Most of the world uses Q for both heat transfer ie BTUhr and for volumetric flow ie CFM Your textbook uses Q and CFM Unsatisfying from a units perspective So I will use q for heat transfer and Q for volumetric flow 0 Look at context in case of confusion Latent In ltration and Ventilation Can either track enthalpy and temperature and separate latent and sensible later 0 q M X AH BTUhr W Or track humidity ratio for just latent portion q M x hfg x AW hfg 1076 BTU1b 25 kJkg Mp gtlt Q lbhr kgs Where do you get information about amount of ventilation required ASHRAE Standard 6212007 Table 61 Table 29A is shorter form in text Available on website from library Hotly debated many addenda and changes TABLE 51 MINIMUM VENTILATION RATES IN BREATHING ZONE Ground Contact Receives less attention 0 3D conduction problem 0 Ground temperature is often much closer to indoor air temperature Only use for heating loads Use F value from simulations BTUhr F ft 0 Note different units from UValue Multiply by slab edge length Add to ZUA Still need to include basement wall area WA State Energy Code heat loss tables 0 In references section of website L NHE ATED SLABONGRADE FV A LUES INSULATION STRATEGY OVERALL EVALUE Uninsuluicd Slub 073 RAS Horizonlal Insulation r ZFI no Lb 0 70 RIU Horimnwl Insulation IZFH no In 070 R715 Horizontal Insulnnon ZR no lb 0 69 R6 IIorIznnml Insulllllun 4Fl no h 0 67 RVIO Horizontal Insulation I4FI no In 064 R715 I39Im izonml Insulunon 39JFU no b 0153 R6 Vertical Insulation IZFI 055 R710 VL rIlL uI Insulzlnon IZFL 054 RIS Verncul Insulation GIT I 52 INTEEJPlL quot 39C INSULATION g 39 k E k a mama 1m r i ll 1m g G a lle amp g a 39 nlE g l a l ml mi E F H lltlzllll ll Type Interior Insululiun Only Willi Thermal Break Interior Insulzzlinn Slab Dcpxh Below Grade 20 35 70 UN U P U P U F 1211 Ban 0095 0065 0586 0061 0546 0053 0 Si R19 Ban 0060 0 045 0 591 0042 0553 0037 0300 R21 Ban 0050 0042 0592 0040 0556 0035 050 RZS Butt 0 050 0 038 0593 0036 0557 0032 0505 What is the heat loss for a 20 gtlt 20 ft slab 35 ft below grade With R 19 insulation on the walls and a thermal break at a 50 oF temp difference gt 5053 q 80 ft x 0553 BTUhr F ft x 70 F 50 F 885 BTUhr q 400 ft2 X 0042 BTUhr F re x 50 F 842 BTUhr q 80 ft x 0042 BTUhr F 02 x 50 F 168 BTUhr q 80 ft x 0553 BTUhr F ft x 50 F 2212 BTUhr q 400 ft2 X 0553 BTUhr F ft x 70 F 50 F 4264 BTUhr Solar Gain increased conduction because outside surfaces got ot Use q UAAT Replace AT with TETD Tables 212 7 214 in TeXt 4 pm for a dark colored surface Take highest of TETD or AT very important 39 a A i ll CLTDr f VLFRLF Tamas l and 2 39 rwl39ASHH lE Permian9852th Glazing q UAATA gtltSCgtltSHGF Calculate conduction normally q UAAT Use UValues from NFRC Certi ed Products Directory 39 ALREADY INCLUDES AIRFILMS h n u u I r Use the UValue for the actual window that you are going to use Only use default values if absolutely necessary Tables 4 and 6 Chapter 31 ASHRAE Fundamentals USE ROUGH OPENING AREA NOT GLASS AREA r nd nfrr 39 aqn Glazing q UAATA gtltSCgtltSHGF Calculate conduction normally q UAAT Use UValues from NFRC Certi ed Products Directory 39 ALREADY INCLUDES AIRFILMS htt c dnfrco ubsearch sMainas Use the UValue for the actual window that you are going to use Only use default values if absolutely necessary Tables 4 and 6 Chapter 31 ASHRAE Fundamentals USE ROUGH OPENING AREA NOT GLASS AREA Solar Gain Through Windows 0 Add to conduction AgtltSHGFgtltSC orAXSHGFXSHGC SHGF solar heat gain factor Measure of how much energy comes through an average perfect window Depends on r Latitude r Orientation 7 Time of Day 7 Time of Year Tabulated inASHRAE Fundamentals 1997 Chapter 29 Table 15 Tao and Janis Table 215A for 40 latitude July 21 8 am Shading Coef cient or SHGC Ratio of how much sunlight passes through relative to a clean 18 thick piece of glass Depends on Window coatings Actually a spectral property Frame shading dirt etc Use the SHGC value from NFRC for a particular window Lower it further for dirt blinds awnings shading Only use default values if necessary ASHRAE Ch 31 Table 13 Text Table 216 thpCdnfrcogse archsearchdefaultaspx What is the total cooling load in Austin associated with the following window with a 40 F AT 10 ftz south facing ATI Windows Dimension 4 Horizontal Slider ALUA12OOO3O U 034 BTUhr F ft2 SHGC 032 typically divide by 2 for dirt etc 016 SHGF 215 BTUhr ft2 Oct 20th at noon 6 UAATA gtlt SHGC gtltSHGF q from solar gain More about Windows Spectral coatings lowe Allows visible energy to pass but limits infrared radiation 39 Particularly shortwave 39 Can see itwith amatchlighter in older Windows Tints Polyester films Gas fills All improve lower the Uvalue sHGco l Lowa coatings uanue 125 m sum m f m M VAaIwIrhnl a mummn galquot nansml e 7w gain vr 115 a m m unsinlleed mgL751 mm H quotarm I llnhl ummmeu l ARE 346N Midterm I Page 1 of 6 Name October 23 2007 ARE 346N Building Environmental Systems Midterm I Solutions October 23 2007 Closed book closed notes Important numbers Formulae 3413 BTUhr 1 W q MCAT atmospheric pressure 147 psi q M hfg 1 3 75 gallons R lk density of water 624 lbft3 E mc2 density of air assume constant 0075 lbft3 q UAAT SCSHGFA speci c heat of air assume constant 024 Btulb 0F p lv hfg enthalpy of vapor 7 enthalpy of liquid q M hfg AW A 7Ir2 M pQ 1 ton 12000 BTUhr 1 11 2 6 3 3 4 9 5 5 6 16 Total 43 NOTE THAT THERE ARE 50 POSSIBLE POINTS Write your name on every page as well as this one Ifyou use any additional sheets make sure that your work is easy to follow Do not write on the backs of any sheets of paper If you use any of the assumed data ie you can t get part of a question make sure that you make it clear that you are doing so NOTE THAT ASSUMED DATA MAY OR MAY NOT BE CORRECT SHOW ALL OF YOUR WORK AND CLEARLY REFERENCE WHAT TABLESFIGURES YOU ARE USING ARE 346N Midterm I Page 2 of 6 Name October 23 2007 1 Fill in the appropriate units from Column B for each of the quantities in Column A You will not need all of the entries listed in Column B 11 pts total 1 pt each speci c volume l power ton absolute humidity lb H20 lb air energy m thermal conductivity BTU 1 hr OF ft mass ow rate M volumetric ow rate m latent heat of vaporization BTU lb R value 10F f tZ hr BTU temperature difference E speci c heat BTUg lb OF 2 Air at 90 OF drybulb and 65 OF wetbulb ows across an evaporator coil at the temperatures listed below For each case state whether the absolute and the relative humidity increase or decrease and whether and why this would be a good choice of coil temperature for a building air conditioning system 6 pts total 2 pts each a Coil temperature 35 OF Absolute Humidity decreasesi RH increasesi Good Choice Yes removes humiditv without freezing 39 b Coil temperature 55 OF Absolute Humidity Stays the same RH increases Good Choice No doesn t remove any moisture c Coil temperature 15 OF Absolute Humidity decreases RH increases Good Choice No will freeze condensate ARE 346N Midterm I Page 3 of 6 Name October 23 2007 3 A mechanical engineer specifies a multizone VAV system for a surgical facility There is no space or budget for reheat coils and explicit ventilation requirements Is this an appropriate decision 7 why or why not If not what system would you recommend instead and why 3 pts This is not an appropriate decision 1 pt VAV systems still need reheat coils a dual duct system would be a better choice 1 pt VAV also doesn t necessary supply enough air for ventilation a CAV system would be a better choice 1 pt The advantage of VAV is that it is more energy efficient Therefore I would recommend a dualduct CAV system 4 You are designing a building and want to know the total internal gains from a direct evaporative cooler ie a swamp cooler The cooler uses 1 gallon of water per hour and lowers the air temperature by 30 OF The air ow through the cooler is 1000 CFM powered by a 300 W fan All of the fan energy goes into the air stream as heat 9 pts total a What are the sensible gains or losses associated with this cooling system including the fan 4 pts b When calculating a cooling load does this evaporative cooler represent a sensible gain or a sensible loss Justify your answer 2 pts Sensible loss because more cooling than heat added by fan c What are the latent gains associated with the cooler 2 pts d What are the total gains for this cooling system 1 pts 1000 CFM air ow 300W fan 1 gph water Sensible Fan 300 W 3413 BTUhr 1024 BTUhr adds heat Cooler M C AT 1000 ft3min 60 mir hr 0075 lbft3 024 BTUlb OF 30 OF 32400 BTUhr removes heat Sensible gain fan 7 cooler 31376 BTUhr negative because system is removing heat Latent thg 1 galhr 1 1amp3 75 gal 624 lb 1amp3 1076 BTU1b 8952 BTUhr adds energy Total sensible latent 22424 negative because the cooler removes energy from the space ARE 346N Midterm I Page 4 of6 Name October 23 2007 5 Calculate these quantities 5 pts total You can assume that operative temperature dry bulb temperature in your calculations a Cooling load design temperature difference for Dallas 1 pt AT 98 F77 F 21 F b Heating load design temperature difference for Austin 1 pt AT 72 F 30 F 42 F c Ifthe TETD for a 1000 112 roofin Dallas is 10 F and the Uvalue is 01 BTUhr OF 112 What is the total cooling load associated With the roof 15 pts q UAAT 01 BTUhr F ftz X 1000 ft2 X 21 F 2100 BTUhr d What is the total cooling load associated With a 100 z south facing Window for July 21st at 8 am in Dallas With a measured Solar Heat Gain Coef cient of05 and a Uvalue of05 BTUhr OF 112 15 pts q UAAT SHGC or SC x SHGF xA 100 112 x 05 BTUhr F 112 x 21 F 05 x 28 BTUhr 112 1450 BTUhr 111 511112 11 11011 L111 111110 11 11111 11 11 11 11 11 M11111quot 11 1111111 11111111 11111 1 DEW Palm TEMPERATURE P 011111 1111 111111 1111 111111 1111 111111 1 a 111 21 111 11 11 DPERAYIVE TEMPmnunE quotF Numer Rama I1 mm 11 m moo b dry 1111 346N Midterm I Page 5 of6 Name October 23 2007 6 Consider the LiBr absorption cycle depicted in the gure below 201 lbmin of high pressure 24726 psia steam is the heat source at the concentrator and the building requires 100 tons of cooling at the evapomtor The pumps and other equipment required to run the cycle consume 193 kW of electricity 16 pm total 00an solution water Concentrated solution expansion Chilled water out Cooling In water in Evaporator recirculating pump pump a Add and label the expansion valve on the diagram 1 pt b The refrigerant pressure in the evaporator is 0122 psia the refrigerant enters as a liquid boils immediately and leaves as a superheated vapor 15 degrees Warmer than When it entered What is the mass ow rate of the refrigerant 4 pts If you can t get b assume 10001bhr q 100 tons 1200000 BTUhr MCAThg M 1200000 BTUhr 0452 BTUlb F15 0F 1065 BTUlb 11201bhr c What is the mass ow mte of the refrigerant leaving the concentrator and going to the condenser Is it solid liquid or gas Does it contain any ofthe LiBr salt 2 pts ARE 346N Midterm I Page 6 of 6 Name October 23 2007 The mass ow rate has to be the same as calculated in b ie 1120 lbhr It is a vapor and does not contain any of the salt d The pressure in the absorber is 0363 psia The solution leaving the absorber has a ow rate of 1400 lbhr and is 20 LiBr salt by mass How much energy was rejected to the cooling water in the absorber 3 pts q thg 11201bhr 1093 38 BTUhr 1180000 BTUhr e If the velocity in the pipe leaving the absorber must stay above 2 ftmin what is the largest pipe diameter that can be used Assume that pipe comes in 12 diameter increments and the density of the solution is 100 lbft3 2 pts Q vAmax Mp Amax Mpv 14001bhr1001bft3 2ftmingtlt1 hr60 min 07 ft2 A 71 124 dmax V4An 46 in Use 45 inch pipe e If the heat source steam returns to the boiler as a liquid at 400 OF what is the heat energy input to the refrigerant in the concentrator 2 pts If you can t get e assume 900 kBTUhr q Mhg 201 lbmin 60 minhr 1202 375 BTUlb 1000000 BTUhr fAssuming that the energy input to the cycle is the steam heat source part e and the pump electricity and the useful cooling output is at the evaporator what is the efficiency of the system Is this a reasonable value for an absorption cycle 2 pts energy input 1000000 BTUhr 3413 BTUkW 193 kW 1065 kBTUhr energy output 1200 kBTUhr Ef ciency 112 this is a little high but reasonable for an absorption cycle Objectives Select air systems Control of Air Systems Zoning Vary temperature constant ow Constant temperature vary ow Vary temperature vary ow May need reheat Temperature F Onoff systems 0 Residential and small commercial Vary volume by turning system on and off Control Operating differential 695 Time Constant Air Volume 0 Single zone constant air volume Fan always runs Vary temperature of air in response to space thermostat 0 Single zone constant air volume with reheat Often used for dehumidi cation Multizone constant air volume with reheat Good humidity control exible Not very ef cient Multizone constant air volume with reheat m RA Airdischarge 55 F l 39 39 Constantvolume l l l l l l l l r I l l L 7 n A Requires less cooling Variable Air Volume 0 Single zone VAV Varies air ow based on room thermostat Multizone VAV Central chilled air supply 0 Zone thermostats control ow to each zone 0 Also use reheat Why 0 Much smaller energy penalty than CAV sunumumwhunns numbslimmmdm gvmmmmu Dnva mum Mum umwnw mm m msunmpm sumunnwuml mailm Turqwmu mm mm mu mama by xemmmm m mm mm wm nmlu Minna mm a m mamway M will mm mm meoan nnmvnvum nv mm mm cy mm mm Em namuummw mama mmnm Unmanmuamn PAmumI k or nd nwmmumam w mm lin Wuva mm whama may mun Multizone VAV Vary the flow DualDuct Systems 0 Can be VAV or CAV 0 Two plenums with chilled air and heated air 0 Zone thermostats control ratio Separate duct for each zone Which is the best air delivery system for a health care facility in which proper ventilation is more important than operating cost A Constant temperature variable volume B Constant volume with reheat coils C VAV with reheat coils D VAV with dual duct Which of the following systems is the least energy efficient A Dual duct VAV B Dual duct CAV C Constant temperature VAV D Multizone CAV with reheat Equipment 0 Fans Ducts and plenums Cooling and heating coils Combustion heat exchangers Filtration Dampers controls mixing valves Humidi cation steam spray Dehumidi cation w desiccant Noise control Other Equipment Fan terminal unit Fan coil Two pipe or four pipe 39 PTAC Rooftop through wall 0 All components in one location 0 Heat pump electric coils gas red Summary Choice between air systems determined by Possibilitydesirability of reheat Energy ef ciency 0 Importance of humidity control 0 Availability of waste heat 0 Building size Objectives Design HVAC systems 0 Prepare for ArE465Review Space for Environmental Systems Roof for rooftop units TABlE 11 Range DfME syslems cosls af bulldlngs Pierrequot of Toiai Buiiding Casi Medium iii n was OfUKCllDaI39IW Lomnulercemeis EU 45 ED Depaninenx mes m 25 30 Hos Ialsneseamhr 30 Au so Hosnllalskllnlml 25 3 35 Holelsuesldenmb 10 I 35 Hot I t manila 15 35 40 DII KEN eneval 20 25 35 o sthighrleth 25 35 5 310 abmarovle 30 4 50 gt Resldenllalslngleroricunanry w 15 20 Residential h nglSB 5 ID 15 lei ndlvlduatsmr 10 20 25 R IN parlmenl m 15 an SChOD S eteirenlarv 15 ID 30 smouis secundalv 1 15 35 Unlvenlnes ma cullcgcs39 m m w nulldmgs omev man muse used rm dassmums oHow me spine YWUIIEI I 6m miaimwe hunmnns sum as iamuamvies nmpulev enters lesidentes ck Ref Tao and Jams 2001 First cost VS Operating cost First cost Size of equipment Design parameters Operating cost Builtin equipment Operational parameters Energy analyses for optimum balance 1995 Commcid Bu lcing EndUse Splits HVAC 45 WE Mn Qemllon 3 L4 Spaniel ullrg 31 5 War Phatirg 7 Source DOE Energy Bill for Residential Buildings Healing and Other Cooling 45 Lighting 7 Water Heater TV VCR Clothes 11 DVD Washer amp R f i 2 Compuier amp DI h h e rlgera or Monitor We 5 was er 6 10 httpvawenergystargov Questions Who is your client How big do HVAC components need to be Where does it go What type of equipment will you need Who is Your Client Budget incl budgeting for design How do they de ne comfort Determine needs 0 Climate Ventilation Steam 0 Building type 0 Aesthetic a 9 393 0 DEW POINT TEMPERATURE F w 9506 L quot3939 3939 10 60 65 70 75 80 OPERATIVE TEMPERATURE F 0 90 HUMIDIT I RATIO lb water vapor per 1000 lb dry Ii Climatelocation Determines size of heating and cooling system Strongly influences ventilation costs Affects heating and cooling loads How big does it need to be Heating and cooling loads 0 Figure out most important components of loads Sensible and latent for cooling loads Anticipate future needs 0 Do not oversizeundersize without a good reason and without guring out impacts of doing so Ventilation Loads ASHRAE Standard 621 Table 2 TABLE 51 MINIMUM VENTILATION RATES IN BREATHING ZONE mi lIIqunI Rn i nm In Systems Heating Humidification Cooling Dehumidification Distribution Zoningair system Diffusersradiatorsconvectors Ventilation Heating Boiler vs furnace Heat transfer medium 0 Fuel source and storage 0 Dependence on electricity Auxiliary equipment Jt39t xssaa392iffu39iif quot Sizing Maintenance Simultaneous heating and cooling Humidi cation 0 Steam injection 0 Water injection air washer evaporative cooler Depends on Whole building or single zone Maintenance Cooling Absorption or vapor compression cycle Cooling transfer medium Split system PTAC chiller Heat rejection equipment and medium Centralized or decentralized Dehumidification Sizing 33132333332 Q Maintenance T Dehumidi cation Sizing Maintenance Cooling coil dedicated dehumidifier desiccant system Outdoor air Special needs for Whole building or single ZODC air from building Cooling coil to building Distribution Transfer media air or water Two pipe vs four pipe water system Location mechanical chase Noiseacoustics Leakage Firesmoke dampers Sizing Maintenance ZoningAir System 3 a What determines a zone CAV VAV Dual duct Reheat coils Controls Sizing Maintenance DiffusersRadiatorsConvectors Noise Comfort Aesthetics Uniformity of conditioning Sizing Maintenance DiffusersRadiatorsConvectors Noise Comfort Aesthetics Uniformity of conditioning Sizing Maintenance Ventilation Integration Efficiencyenergy use Heat exchanger Sizing Maintenance Distribution Responsible for pipe sizing usually velocity constraint Not responsible for duct sizing Where are you going to put it Does it fit within the existing space Does it conflict with another system 0 Especially structural or roof system Can you access it for maintenance Integrative Design Mechanical system design should occur simultaneously with 0 Space planning 0 Structural design 0 Plumbing design 0 Electrical design 0 Architectural design Commissioning and maintenance plan Objectives 0 Finish psychrometric processes ch 8 Skipping Chapter 9 0 Example from last class 0 Use protractor to calculate SHR 0 Describe direct airwater contact ch 10 0 Less focus on calculation details 0 Describe refrigerant cycles and realideal difference 0 Ch 31 35 I will only touch on material in later sections Air WashersEvaporative Coolers Heat and mass transfer is mutually compensating l 0 Can evaluate based on temperature drop 3 7 humidification or e 7 4 e 1 comparison to other A gf 1 energy exchangers Ham39s 1i quot mlmun line for r m39ngtllcr Why do we have Direct Contact Processes Humidification and dehumidification see 104 Heat rejection 0 Water has better heat transfer properties than air Non dimensional parameter Lewis number Le orD hchDcP Ratio of heat transfer to mass transfer Assume Le l for evaporative coolers he convection heat transfer coefficient h D mass transfer coefficient cp specific heat a thermal diffusivity D mass diffusivity Air Washer Sprays liquid water into air stream Typically air leaves system at lower temperature and higher humidity than it enters Elevation Figure 101 Air washer with single bank of spray nozzles Reprinted by permission from ASHRAE Heating thiluring Air Cnnditianing Guide Vol 28 1950 p 710 Schematic Elimiumor plums A gt 1 Wm Makeup mater 4gt Figure 02 Schematic diagram of air washer using directly recirr culated spray water Cooling Tower Similar to an evaporative cooler but the purpose is often to cool water Widely used for heat rejection in HVAC systems Also used to reject industrial process heat Drifl eliminatnrs gtgtgtgtgtgtgtgtgtgtgtgtgt m lt s H quot rtype llpa Emma 39 0 l Wautle A A Dismbu onimpac region b Cnld water gt lupump AAMM Figure 104 Illustration of a counterflow induceddraft cooling tower and types of ll Reprinted by permission from ASHRAE Syxlems and Equipmml 1996 IP amp SI pp 362 363 HIM NH 1 Air Water lit 1 W511 I l T1 1 MW W 5 in w I w Ix v i i r quot 1quot IL W 111w Ii w 2 111 AI r quot391quot I I I51W HI 2 quotMg Figure 105 Schematic diagram nl counlcr ow cooling lowcr i IV 1quot Figure 107 The Stevens diagram Reprinlcd from W 17 Carey and G J Williamson Gas Cooling and Humidificulion Design of Packed Towers from Small Scale 39l cslsquot Pmruelling oj39lw Inxli me QfMeL hmiicul Engineers 163 1950 49 Solution Requires f factor Can get from Stevens diagram page 272 Can also be used to determine Minimum water temperature 0 Volume of tower required Can be evaluated as a heat exchanger by conducting NTU analysis Real World Concerns We need to know mass transfer coefficients 0 They are not typically known for a speci c direct contact device Vary widely depending on packing material tower design mass ow rates of water and air etc 0 In reality experiments are typically done for a particular application 0 Some correlations are in Section 105 in your book Use with caution Summary Heat rejection is often accomplished with deVices that have direct contact between air and water Evaporative cooling Can construct analysis of these deVices Requires parameters which need to be measured for a speci c system Objectives 0 Describe vapor compression cycle basics 0 Draw cycle on T s diagrams 0 Compare real cycles to ideal cycles 0 Basis for discussion of individual components 0 Very thermoheavy for next few classes Vapor Compression Cycle Energy input Suction gas Hm gas Warm uid Heat rejection I Cool fluid Liquid refrigerant Ef ciency 0 First Law Coefficient of performance COP COP useful refrigerating effectnet energy supplied COP qw 0 Second law Refrigerating efficiency 17R 17R COPCOP rev net 0 Comparison to ideal reversible cycle Carnot Cycle No cycle can have a higher COP All reversible cycles operating at the same temperatures T 0 T R will have the same COP For constant temp processes dq T ds COP TRToi TR Figure 31 Processes of the Carnot Specific emmpy s refrigeration cycle Carnot VaporCompression Cycle 1 Y Isentropic lwb expander lgt T0 4 11 Condenser b l TR sentropic 3W compressor ltb O c f Figure 32 Carnot vapor compression cycle Get Real Assume no heat transfer or potential or kinetic energy transfer in expansion valve 39 COP hs39h2h439h3 Compressor displacement 151123 Figure 33 The theoretical single stage vaporcompression cycle Area Analysis of Work and Ef ciency IAT 39 Figure 35 Areas on the T r diagram representing the refrigerating effect and the work supplied for the theoretical singlcslagc cycle Comparison Between SingleStage and Carnot Cycles 1 A2 1 qcarnat 1A1A2l g wcarnot VIR Figure 36 Area deviations on the Ts diagram for the theoretical singlestage U 4 r cycle wilh respect re the Camel cycle Example R22 condensing temp of 30 OC and evaporating temp of 0 OC Determine a qcamot wcamot b Diminished JR and excess w for real cycle caused by throttling and superheat horn 9 quotit 300 Temperature K 100 l l l l v t 0 02 04 06 08 10 12 Eulropy kIkg K Figure 37 Temperature entropy diagram for Refrigerant 22 show ing area deviations of theoretical singlestage cycle with raspcct m the Carnot cycle Subcooling and Superheating Refrigerant may be subcooled in condenser or in liquid line 0 Temperature goes below saturation temperature Refrigerant may be superheated in evaporator or in vapor suction line 0 Temperature goes above saturation temperature 0 Neither of this results in efficient heat transfer 0 Loss ofef ciency Homework Comment Objectives Heat exchangers Fluid mechanics ducts and pitot tubes Introduction to psychrornetrics Any QuestionsAdministrative Issues Syllabus Website Homework Lecture notesVideos Feedback Heat Exchangers n Counter ow Cross ow gt Cross New Cross Haw 7397m lam Cocurrent ow lquotlvlllH I I2 lmwlluw luml uwlmnguva alanrd whh lmlh fluids unmixed lb Un nde wall nm llnitl mixtnl um lln ullnr unmiwll u I FILI In I l l Cmnvmrir llllJ39 lle al exrllangera a Parallel lm b Commrth Ref Incropera amp Dewitt 2002 ShellandTube Heat Exchanger Tulle outlet Shell lanl Balflcs Shell T ube outlet Il llCl IquotIl39lll l I39l SINll uml luln lll39ill Ixrlmnglr with Ulll slitll pass and one tnlw pass Prnss vuuIllerllnw llllNll nl39 optruliunl Ref Incropera amp Dewitt 2002 Heat Exchanger Analysis lvu lt Hm ummhmpy 1 It gt Cum 139iuidn39vipvl I 1h A Hm llmd viiquot rm 1th lncatlon I Lucnliun A Ir I Hm uid lcmpcmmrc m 1 A 1 7 I It Cold uid tcmpcramru Figure 213 Schematic temperature changes in a counlcr ow heal exchangcrv Heat Exchanger Analysis 0 We want to know the energy exchange between two uids 0 But the temperature difference is not constant Q UA Q heat transferred W BTUhr m UA area X U value K BtuhrF At mean temperature difference F C 0 The trick is coming up with appropriate expression for A tm Heat Exchanger Analysis um lt Hm ummhmpy A 1 gt CHM139llidn39viW gt II Hm Immmhmw lt m Locath I Lucilliurl A Counter ow Parallel tho tciHhi tca 13th At a y m m In ho tcj 111 Ehj to a Atb 7 9 Fluid Flow in Pipes Analogy to steady ow energy equation 0 Consider incompressible isothermal ow 0 What is friction loss 2 g f1 D2g Pitot Tubes Pilot Tube Research enter r Density V Velocity p Pressure Tclal pressure P Static pressure Pressure Transducer Measure ddference in total and static pressure Bernoulli s Equation static pressure dynamic pressure lolal pressure pr Solve for Velocity V2 202 D5 r Summary 0 Use relationships in text to solve conduction convection radiation phase change and mixedmode heat transfer problems 0 Calculate components of pressure for ow in pipes and ducts Any questions about review material Where are we going 0 Heating and cooling loads Psychrometrics Psychrometric terms Using tables for moist air Application of IGL Using psychrometric charts 71 7 75 77 Thursday Heating and Cooling Loads Need to know how big to size systems All terms will have units of power BTU hr or W Two types of terms 0 Terms that are power solar gain internal gains 0 Terms that need to be multiplied by an indoor outdoor temperature difference UA for conduction MC for air ows Contrast With Energy Usage Goal of heatingcoolingventilation load is to design systems that 0 Can handle extreme conditions through conservatism not oversizing Operate ef ciently at most conditions Overall Strategy Be conservative not overthetop Use 1 values Include all components 0 Focus on the components that are important Be prepared to read about current practices Reference Table Sources unless otherwise indicated are ASHRAE Handbook of Fundamentals 2005 Trost textbook and other sources Your textbook is a fine source but may not agree entirely with other sources Re nements heating load Treat below grade heat loss as a 2D conduction problem Section 143 Use solair temperature Section 153 for outside temperature when calculating conduction for above grade walls and roofs Calculate TETD explicitly Admin Website was down sorry now back up Lecture notes and Videos posted Makeup classes 0 Monday Wednesday Friday mornings and afternoons Do you have a con ict EVERY week Objectives Solve thermodynamic problems and use propert1es in equations Calculate heat transfer by all three modes including phase change Apply Bernoulli equation to flow in a duct and use a pitot tube Differentiate heat exchangers Review heating and cooling loads next week Units Pound mass and pound force lbm lbf on Earth for all practical purposes Acceleration due to gravity 0 g 9807 ms2 3217 fts2 Pressure section 25 for unit conversions Temperature section 26 for unit conversions Thermodynamic Properties p density mass V01ume Both functions oft P v spe01fic volume l p specific weight weight per unit volume refers to force not to mass specific gravity ratio of weight of volume of liquid to same volume of water at std conditions usually 60 F or 20 C and 1 atm Heat Units Heat energy transferred because of a temperature difference Btu energy required to raise l lbm of water 1 OF kJ Specific heat heat per unit mass BtulmeF kJkgOC For gasses two relevant quantities CV and CF Basic equation 210 Q cht Qhea transfer Btu kl c speci c heat At temperature difference Sensible vs latent heat Sensible heat Q cht Latent heat is associate with change of phase at constant temperature Latent heat of vaporization hfg Latent heat of fusion h hfg for water 100 OC 1 atm 1220 Btulbm h for ice 0 OC 1 atm 144 Btulbm Work Energy and Power Work is energy transferred from system to surroundings when a force acts through a distance ftlbf or Nm note units of energy Power is the time rate of work performance Btu hr or W Unit conversions in Section 27 1 ton 12000 Btuhr HVAC specific Where does 1 ton come from 1 ton 2000 lbm Energy released when 2000 lbm of ice melts 2000 lbm X 144 BTUlbm 288 kBTU Process is assumed to take 1 day 24 hours 1 ton of air conditioning 12 kBTUhr Note that it is a unit of power energytime How big is one ton of ice p 574 1bmft3 Thermodynamic Laws First law Second law Implications for HVAC 0 Need a refrigeration machine and external energy to make energy ow from cold to hot Internal Energy and Enthalpy lSt law says energy is neither created or destroyed So we must be able to store energy in a uid 0 Internal energy u is all energy stored Molecular Vibration rotation etc Formal de nition in statistical thermodynamics Enthalpy Composite energy sensible latent We always track this term in HVAC analysis 11 u PV h enthalpy Jkg Btulbm P Pressure Pa ps1 v specific volume mKkg ftKlbm Entropy 0 Not directly measurable Mathematical construct S entropy JK BTU R Revers1ble process dS dQ Qheattmnsfer LBTU dT T absolute temperature K R Note difference between s and S 0 Entropy can be used as a condition for equilibrium 0 What do enthalpy internal energy and entropy all have in common T s diagrams 0 db T ds vdP general property equation 0 Area under T s curve is change in speci c enthalpy under What condition Figure 25 Schematic heal removal processes on T coordinates Ideal gas law Pv RT orPVnRT R is a constant for a given uid 1545 lbf 8314 U R 777 M 1bm R M kggtK 39 For perfect gasses M molecular Weight gmol lbmmol Au CV1 P pressure Pa psi V v lu ftK CPAI v specific volume kag ftKlbm T absolute temperature K R CF CV R t temperature C F u intemal energy Jkg Btu lbm h enthalpy Jkg Btulbm n number of moles mol Mixtures of Perfect Gasses 39 mmx my PxVmeXT VVx Vy PyVmyRyT TTx Ty PPx Py Assume air is an ideal gas 0 70 0C to 80 0C 100 0F to 180 0F m mass g lbm P pressure Pa psi V volume m3 113 R material speci c gas constant T absolute temperature K R MassWeighted Averages o Quality x is mgmf mg Vapor mass fraction 0 C v or h or s in expressions below 0 x C 4 lfg s entropy JKkg BTU Rlbm Cx Q4ng mmasslbm h enthalpy Jkg Btulbm 7 V speci c volume m3kg 7 1 x C 4 Qg Subscnpts f and grefer to saturated liquid and vapor states and fg is the difference between the two Properties of water 0 Water water vapor steam ice 0 Properties of water and steam pg 675 685 0 Alternative ASHRAE Fundamentals ch 6 Nprrmt mm m quotl IHllmm m 1 mm kJkc K Ammu I mxun mm x lunml m mm m Luyunl 5m uynn nu 10mm nt Vn39m r t t h h V 3 M1994 013an 0020 VODIDIX 0 021215 00mm 7 an 250M 2612 n2 u 02st 00mm s 723 25427 1614 as n 0293 oan s 453 253 2515 07 a 02503 anm a m 27253 2517 to c 02517 oumn2n s 943 27639 2519 52 a 02735 00mm 7m 23mm 262123 o quot2859 atonmu s m 23529 2622 95 0 02955 rumnm s 252 2325 2674 M a 03m mama 5042 29170 2525 34 v 03255 000102 A 3 297m 1628 us 0 03399 a mum I 652 302 m 2529 7a Psychrometrics 39 What is relative humidity RH 39 What is humidity ratio w What is deWpoint temperature Id 39 What is the Wet bulb temperature 1 39 How do you use a psychrometric chart 39 How do you calculate RH hint partial pressure 39 Why is w used in calculations 39 How do you calculate the mixed conditions for two volumes or streams of air E my Ma mmmmmmmmmy 2 Thermodynamic Properties of Refrigerants What is a refrigerant Usually interested in phase change What is a definition of saturation Enthalpy of liquid is the same as saturated liquid at same temperature ASHRAE Fundamentals ch 20 has additional refrigerants Homework Assignment 1 Review material from chapter 2 Mostly plug and chug Depends on your memory of thermodynamics and heat transfer You should be able to do any of problems in Chapter 2 after Videos Problems 22 26 212 214 220 222 216a Due on Tuesday 23 15 weeks Objectives Differentiate compressors and expansion valves Ch 4 Introduce heat exchangers ch 1 l 0 Next two weeks 0 Real strength of textbook Compressor Workhorse of the system Several types all compress gas with varying degrees of efficiency 0 Far from isentropic our assumption earlier CD mv for ideal conditions Compressor piston displacement Wshaft work done by shaft W elec electric power requirements gt gt Reciprocating Compressor d L39 Presaurev P a 1 1 I 1 1 1 1 1 I Vquot cnnst i 1 I Vquot Icnnst Figure 44 Schematic pressurevolume diagram for a reciprocating compressor nyxncm vulume V Reciprocating Piston compressing volume PVquot constant C For all stages if we assume no heat transfer Can measure n but dependent on many factors Often use isentropic n in absence of better values 0 R12 n 107 0 R22 n 112 0 R717 n 129 1 Z Condensng pressure 4 u Horsepowcr per lun Tons capacity and horsepower w 40 0 20 40 Saturation suction temperature F Figure 47 Capacity and horsepower characteristics for a typical Refrigerant12 condensing unit equipped with a constant displace ment compressor Rotary Compressors Higher ef ciency lower noise and Vibration 0 Cylinder rotating eccentrically in side housing 39I IJI Em V1quot l Doubleacting E E b Figure 48 Rullingpistonlype rotary compressor Reprinted by permission from ASPIRAE Syrlemx 11ml Equipnlenl1996lP amp SI p 349 Rotary Compressor CD szcomp QzBz g 773x comp All 5 Figure its Rolling pistonlype rotary compressor Reprinted by pernussron from ASHRAE Syxzemx and Equipmen1996 IP amp SI Scroll Compressors 0 One scroll is fixed 0 The other scroll wobbles inside compressing refrigerant 39 FixedWquot Often requires heat transfer from refrigerant to cool scrolls Orbiling scroll Figure 412 Sketch of the two scrolls showing various stages of compression Scroll Compressors Constant displacement Higher ef ciency but harder to manufacture Close tolerance between scrolls Ugly to analyze see text for details Screw Compressors Rotating meshed screws One or two screws l l s39 4 Discharge J I HighAprcssure cusp Figure 415 Twinscrew compressor Reprinted by permission from ASHRIE Syrians and Equipment 19 IP amp SI 139 3417 Inlel pun Figure 414 Schematic of a singlescrew Compressor Reprint permission from ASHRAE Equipnmil 1988 p 1216 scharg rm I Cnmpresslun a 1 Figure 416 Compression process Reprinted b permission from ASHRAE Syslwnx um Eruipnmzl 1996 IP 84 SI p 3417 Summary Many compressors available ASHRAE Handbook is good source of more detailed information 0 Very large industry Where are we going 0 Expansion valves Coils evaporators and compressors Expansion Valves Throttles the refrigerant from condenser temperature to evaporator temperature Connected to evaporator superheat Increased compressor power consumption Decreased pumping capacity 0 Increased discharge temperature Can do it with a fixed orifice pressure reducing device but does not guarantee evaporator pressure Thermostatic Expansion Valve TXV 0 Variable refrigerant ow to maintain desired superheat Diaphmgm Bulb prim Spring pubss Evap coil Ncudle and scat chmlc z bulb gt Adjusungsmw Figure 418 Thermostatic expansion valve TEV AEV Maintains constant evaporator pressure by increasing ow as load decreases Spnng Dmphngm Swing Evap Needer and seal press Figure 420 Automatic expansion Valve Summary Expansion valves make a big difference in refrigeration system performance Tradeoffs Cost refrigerant amount Complexitymoving parts Objectives 0 Introduce coils Analyze cross ow heat exchangers Calculate heat exchanger efficiency Analyze heat exchangers Derive n analysis Combined resistances Kays and London 1955 Moist surfaces Fins that are cooling and dehumidifying Nondesign conditions Other considerations Coils AKA Extended Surfaces AKA Compact Heat Exchangers Fins added to refrigerant tubes 0 Important parameters for heat exchange Figure 111 Schematic illustrations of finned tubing Some HX Heat Exchanger truths All of the energy that leavesenters the refrigerant entersleaves the heat transfer medium 0 If a HX surface is not below the dew point of the air you will not get any dehumidification Water takes time to drain off of the coil 0 Heat exchanger efficiency varies greatly What about compact heat exchangers 0 Analysis is very complex 0 Assume at circularplate fin Figure 1113 Approximation method for treating a rcclangularplalc tin of uniform thickness in terms 0139 a flat circularplate fin of equal area Overall Heat Transfer Q UOAOATm pipc wall iliickncss x I air lcmperalurc 1 n n hasc ltmpcralurc Pipe inside surface area Arr mean snr39 rea 11 m 15 pipc unlur suriuce not outside surface area Anquot lemlr emluu n pipe inner surface Fin surfacr area AF Icmpcrnmrc If hul uid lcmpcraluru In mean Fm lcinpcralure Hm fluid I Figure 1114 Schematic illustration of a nnedlube heat exchanger Heat Exchangers Parallel ow a Counter ow L Cross ow m I VILIHH I I2 lll39nsnrlluw luml uwlnmgmm alanrd whh lumh fluids unmixed HI Un nde mlh nm llnitl mlxml um lln ullnr unmixed FILI In I l l Cmnvnlrir lulJc lira exrllangers a Parallel lm b COllllll39l DW Ref Incropera amp Dewitt 2002 Heat Exchanger Analysis 3911quot lt O uid ri l3939 lt I 11 39gt CulLl uid Ill cm gt I hm lt H0 fluid quot 1quot I39M I Localinn B Lucullun A 0 01 uid rcmpcrulurc 1n h A 1 1V Told uid mm permurc Figure 213 Schematic lempcrature changes in a counlcr ow heal exchangerl Heat Exchanger Analysis Loculmn D gt Counter ow Parallel Atm he tci 39 th tc0J Atm Ata Atb ml Inga hi tco Atb Counter ow Heat Exchangers l h i th 0 R Atm I39Qh o t6quot Qh l tag Qua tci 1n ha tcj Q t Eh tc 0 P 00 ci QM tci Hm 1 mm quotlv39 51 h Cold X fluxd Wu Figure 112 Schematic counterflow heat exchanger Counter ow Heat Exchanger leap tci IR 1M Atm What about cross ow heat exchangers Atm F Atm f Cun39ecriou mer F Finer 5 39urreetiun factor Iquot AimAM for u singlerpuss en sv39 1w heal exchanger nne I luid 1 mixed ulher I39luiLI ll unmixed Reprinted from R A liuwmzm A 39 Mueller and M Nngle quotMean 39l39empermure Difference In Designquot ASMIf 39I rm 62 1041 289 with permissinn of lhe puhlisher Amer Soc Meeh Engrs Heat exchanger performance 113 0 NTU 7 absolute sizing of transfer units 0 s 7 relative sizing effectiveness Criteria pm lt quotll0M Mc pm llcm NTU 114 U04 0 39hhcm s P RP Cr quot2 quotquotWm bomb We 5 TABLE quot2 Effectiveness Expressions for Heat Exchangers of Various How Con guroiions exp U UU 9 Counter ow a 39 7 c7egt in NW 7 cm Parallel flow 5 L gpL N uir l i c i Crossflow s cTil 7 exp 7011 exp NTU cmquot uid is unmixed r cmx uid is mixed Cross ow r 5 l 7 exp 7 I U 7 exp 17NTUc r rmm fluid is mixed 0mm uid is unmixed C jec ves Bldg design load gt nommim Differentiate fans 51112 0 Select fan appropriate for rjjgjyquotj Liii3ii 2li quot 1 539 e 39 Luczninn system REM grille we gt mumi dlxd Al39 or each lcrmlnul clcvicc 0 Calculate system Luv um supply Sec 1amp4 curve for multl um tennina duct System Duclsrhcmnlit m design memo am nvnn and hue 1M Imm duals Figure 181 Outline of forcedair distribution system design pruccr 1 q Liler and sections in Chapter 18 lhai cover lhc various s up Fans 0 Driving force to move air in buildings 0 Raise pressure and produce ow 0 Two main types Centrifugal Axial Relevant Fan Parameters Total pressure rise Static pressure rise 5 Eggm a Power requirement x gas air density P total pressure E iclency WM total power WM sha power rhAP Wm Wm 7 quottot W p sha o Note 27m does not account for motor e iciency Balance Point System characterer Balance point Tum pressure Voiumcrric ow rate Figure 1824 Balance point between a fan and duct system Pressure in wg Fan Performance I Z 4 6 10 Capacny x 104 cllm 12 l4 16 Figure 1822 Conventional curves for backwardcurved blade fan Reprinted by permission from ASIIRAE Journal 141 1972 Centrifugal fans suPnom nm24mmcmgtr1gt2m 2052 v29sz mgtnxltltgtmux ncmltmn 1022me ncmltmn momsimo ammo ncmltmo was gtzmor TYPE uiuim immamai gam g v d d n w u o N s Axial F ans NLET CONE OR GUIDE VANE INLET BELL WHEEL ROTOR MOTOR 4W1 W54 Housmc CASING INNER CYLINDER SWEPT AREA SWEPI AREA RAND OUTLET NOSE COVER PLATE SPINNER NDER AREA OF NNER CVL AREA F FAN Mole The swepl area raho m axraw tans s equwewem m me blast alea vaha n eemmugar vans Fig1 mm 139 quotL39nmlmnum AXIAL FANS m preucrency Simp z crreuxer nng mace wake arvemun39 m Unmed m lawrpressure apphcatmns Op mum design rs c use m made mpsend lomls UsuaHy nw Dust mpeHers have me pr mare News at smumh amen mm wheel g smg a mrckness attached 1p re anve y smaH hub g Pmmary energy vansler by ve omiy pressure a 4 Somewhat more ef uent and capab e of deve upmg CyHndr rz lube wrm c use c earanoe p made ups g mere usem 503th pressure men propeHer can usuaH hasd 1p 3 b adss mm emu Drs rvg er w mrckness crass sectmn D A 5 Hub rs usuaUy 955 men nernne an up dwamaker seed meae desrgn g39 es mediumem manpressure CyHndH39ca lube wrm dose clearance a made ns 2 cepepmcy 3 ad aquot lean Gmde vanes upstream Dr dmmslream fmm rmpeuer Mes ef uent pr mese lens have emu mades I ncrease pressure pepepmy and ef mency 3 B ades may have xed adjustable er eemreuepre I A 3 wequot I l gt Run rs usueuy greeeermen new an hp dwameter PROPELLEN A m m 3 VANEAXKAL 39I39uhli 2 Fun Laws mlrprndonl Eu39iuhhs baby NIN3 In11 N r DMDEI IN 7 PI P mlDz In1 1 plpl rwlnHmmlm Wm In1 Inp315 pgpl u x I13I 1quot 91sz DZ1 QIIQIF p411 Dz1mquot GI99 mm performance data Be very careful L Suburipl I dunmcuhc mrluhlc lur mu hm undurumudumliun Suhxnm admum am my 41 and H uim m mung mm at mung2 l D blade diameter Q volumetric ow rate p gas air density p total pressure W rotation speed Controlling Air Flow m cm V39r umemr 11va late a V3 0 mm Vane Fun spied Tutal p1 essuze Volumemc ow at b Vrmabh m s esd Figure 1823 E ca of variable inlet vanes m ian spccd on total prcs bum characlvns c Curve Effect on Fan Curve m gt 4 m1 l39m I igurc I827 Flo ummv In munn n In A pr mmrul n 17 tunwad umm X mimm um Flow 20 40 60 Damper pauiion D degrees ope Figure 1825 Performance of pal allelrblade dampers Used in air the IWO air streams mixing applications Where the erccnlage of Dampers Pcrcem nf mustaucc needs to be conLrolled without changing the tmal air volume u w m Iknupm pmmmv u mum plum A l 112 Tum I r0si1c i W39 l VAV Operation x iglual 12m cum Nux b3 lame 3939 39 mszom ruz39vc CUIVC Initial 39Ualzmcc prim Volumetric flnw rate Figure 1828 Illustration ofquot AV operation Question How does fan power consumption vary with volumetric air flow rate 0 Different answer for most residential smaller and commercial larger systems 39 D blade diameter mot tot mot m mass ow rate ofair quotSha p gas air denslty P total pressure Wm total power WWIB sha power Duct Layout Overview Layout and select diffusers Use maximum air ow rates Locate air handling equipment and do all connections 0 Space requirements Aesthetics Acoustics Assign fittings Duct Layout cont Find pressure drop data for all components and duct fittings 0 Complete rough sizing using either Equal friction Static regain method 0 Complete second sizing Check sound levels Accessibility for maintenance Equal Friction method 0 Small to medium size buildings Low pressure and low velocity Shaded area on friction loss diagram Figure 1829 Pressure distribution in system sized with cqualAfrlc tron method before balancing branches 2 and r Example 189 Modificationsz Cooling coil at Pt 1 real diffusers N Clt35 Pam mmd uct mnsnion U L 20 so some CD371 an l 2 4 5 6 OF isn rzn i 2011 V H 9 7 Fan thle39g 6m x I n mm u i 4 u ZHO 39r m m suo it3mm 2 b Figure 18330 Duct lem sclismutic for Ex 189 20 in me H Ms W Al u lZ39 Ak is w M lE39 AR 1211 lE Ali LEI UTE ll Ha nl balanle hand I Ilcnmmzndld m tablet GB system huh Am in some in um Pramquot Ln ln lnchzs 01mm 5 d an 1 WE lnnm aluminum R2 Ill n wdnsi ASHME 3542 Incl 39nm39mn uhquot Whalle me an all upnmd um mulllul Wm M Round Dilluser Page 59 G lFADZiS l Tl as GlUAOZ SH G lFAOZ ASlUTl G l UA02S l7 GlFADEDSlJm lUA03DS M cm W GFAUJESWT Giquaes 11 emu swm mum 4517 G l FAD3SS2 l m GlUAOZGSZl Overall Procedure A Make initial decision about straight duct sizes use gray zone on chart 2 Calculate pressure drops for all ducts and fittings 3 Calculate total pressure drops for each run 4 Adjust duct sizes as necessary to get equal total pressure drops in each run 5 Find a fan that meets your total pressure requirements Summary Duct design and layout is very ugly iterative process 0 Can use T method see ASHRAE Fundamentals or reference in Text Practice experience are key Software can help Static Regain Method Large buildings high velocity high pressure high noise Use terminal boxes and exible ducts Design system so that total pressure drop and velocity pressure drop are the same 0 Reduce velocity as you move towards terminal boxes Other Issues Other equipment Duct leakage Noise Material costs IAQ impacts Cleaning and maintenance Objectives Terminology Types of controllers 7 Differences Controls in the real world 7 Problems 7 Response time vs stability Motivation Maintain environmental quality 7 Indoor air quality 7 Occupant comfort 7 Material protection 39 Conserve energy Protect equipment History 0 Process controls Selfpowered controls Pneumatic and electromechanical controls 0 Electronic controls Direct digital control DDC Terminology Sensor Measures quantity of interest 0 Controller Interprets sensor data 0 Controlled device Changes based on controller output Controller a m 3 3 I I Control 7 Temperature valve m5 sensor 0 395 Air Heating flow Figure 213 Temperature Controller sensor a Controller g g E T E E I I Control 7 Temperature Control valve W sensor valve V o g I J Air ung Radiant ow I floor 0 U Dlrect Indlrect Closed Loop or Feedback Open Loop or Feedforward Set Point Desired sensor value Control Point Current sensor value Error or Offset Difference between control point and set point TwoPosition Control Systems Used in small relatively simple systems Controlled device is on or off It is a switch not a valve Good for devices that change slowly Temperature F Control Operating differential Time Anticipator can be used to shorten response time Control differential is also called deadband Humble Honeywell T87 50 years old Innards are deceptively complicated Elegant design 0 Signi cant patent issues NotSoHumble Chronotherm III 0 DDC 7 measures temperature many times every second R 0 Incorrectly accounted magi LE for wall temperature 0 Wide swings in air temperature m I r lFl l39r HFlII carat1 Modulating Control Systems 0 Used in larger systems 0 Output can be anywhere in operating range 0 Three main types 7 Proportional P 7 PID Proportional Controllers 0AeKP O is controller output A is controller output with no error KP is proportional gain constant e iaror o t e K P 0AeKP Very big gain leads to big changes in output and instability Goal is to pick biggest possible gain and still have have a stable system SET POINT CONTROL POINT SET POINT OFFSET gt VARIABLE VARIABLE CONTROL POINT CONTROLLED CONTROLLED TIME gt TIME gt Unstable system Stable system Issues with P Controllers 0 Always have an offset 0 But require less tuning than other controllers 0 Very appropriate for things that change slowly ie building internal temperature Proportional Integral PI OAeKPKi edt Ki is integral gain gt I l SET POINT If controller is tuned properly offset is CONTROL POINT reduced to zero CONTROLLED VARIABLE TIME gt Figure 2 18a SET POINT CONTROL POINT CONTROLLED VARIABLE TIME gt Issues with P1 Controllers 0 Scheduling issues Require more tuning than for P 0 But no offset Proportional Integral Derivative PID 0AeKPKifedtKd Improvement over PI because of faster response and less deviation from offset Increases rate Of CI39I39OI39 COI39I39CCthl l as CI39I39OI39S get 121139ng 0 But HVAC controlled devices are too slow responding Requires setting three different gains Input iqmad Change Time i Offset Proportional T Time PI Time PID Time Ref Kreider and Rabl Figure 12 5 The Real World 50 of US buildings have control problems 7 90 tuning and optimization 7 10 faults 25 energy saVings from correcting control problems Commissioning is critically important Practical Details Measure what you want to control Verify that sensors are working Integrate control system components Tune systems Measure performance Commission control systems
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