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# HVAC Systems ME 414

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This 43 page Class Notes was uploaded by Roman Jaskolski on Friday October 23, 2015. The Class Notes belongs to ME 414 at University of Idaho taught by Staff in Fall. Since its upload, it has received 10 views. For similar materials see /class/227906/me-414-university-of-idaho in Mechanical Engineering at University of Idaho.

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Date Created: 10/23/15

February 16 2008 Student I had several problems reconciling and applying the information from Topic 06 and chapter 6 from the text book Here is a summary of the topics that confused me and my attempt at resolutions Based on this information my homework for PS 3 is attached First computing hour angle In Topic06 lecture notes page 5 it states morning is positive 15 for each hour for example 1100am is m 15 However in the quotSolar Geometry Angle De nitionquot table it state East is negative The formula provided in the book Eq39n 66 produces negative values for morning values of the solar hour angle 03 On the intemet both approaches are used Compare the difference in the following w p 12 no 12 htt solardatuore onedu SolarRadiationBasicshtml HRA 15 LST12 htt wwwudeledui ert vcdrom SUNLIGHT SOLARTHTM Both approaches are correct What39s important is that whichever method is chosen the remaining formulas are structured based on this initial approach I chose to stay with the equation in the book meaning East is negative and a negative w value would mean it is morning Second r nmhnfinn the solar avimnfh I The new formula provided in the lecture notes is DsarccossinBsin7 sin5cosBcos7 The problem is this formula does not yield any negative values ie it cycles from 180 at 1201am to 0 at noon to 180 at 1159pm When compared to Fig 66a and the table of angle de nitions the morning or east positions should be negative values I corrected the formula by adjusting the sign of the result based on the sign of the solar hour angle If the solar hour angle is negative then the solar azimuth should be negative CORRECTED IssignmarccossinBsin7 sin5cosBcos7 This correction is only noticeable on problem 3b the southeast facing window which is how I identified the problem Without a negative solar azimuth the southeast window was under shadow incident angle greater than 90 degrees and a southwest facing window is in the sunlight With the correction the values behave correctly Third probleml definition of solar zenith Just a minor correction but an important one The solar zenith angle 65 is the measure from the earthlocation normal vector to sun position The quotSolar Geometry Angle De nitionsquot table states the angle is measured from the pole and this is incorrect Since it is the compiment of solar altitude which is measured from the horizon the compiment must be measured from the ground39s normal vector RB Here are some comments 1 Hour angle I agree with your discussion and have used the same equation as you However we ended up with different results I looked at your solution and it appears that the only difference is that you took the given local time to be dayIight saving time The problem statement gives the local time as EST Eastern Standard Time So please use standard time even though the month given is July 2 I see your reasoning here I was using the correction that Dr Steciak had developed but it apparently has a sign mistake I will need to check this further before I announce it to the class Did the Internet sites that you tried corroborate your corrected equation 3 I totally agree and wil correct the handout and announce it to the class Thanks for your thoughtful approach to PS3 Please work PS3 for EST and then re submit it to me April 28 2008 Student I39m working on PS 9 right now and have some questions on Problem 134 1 CON RB H N Since we don39t have a number for the length of the corridor problem statement says it is large compared with height and width how can we calculate the area of the work plane How to nd the re ectivity of the work plane Since the length of the corridor is large can we assume that the room ratio is just 1 Take the area of the work plan to be 8ft x L Here L is a variable that represents the length of the corridor that is being illuminated You will see that this L will become the answer that is requested in the part a For a corridor the work plane is the oor So use the given oor re ectivity Yes when you apply equation 134 and then take the limit as L goes to infinity you will nd a room ratio of 10 Topreo41AQ doe ME 414514 HVAC System37 Tupi 4 Indnnr Air Quality mdoor envrronrnent of a buxldmg may be eornfortable but unnea1tny chk Buxldmg Syndrome Both aspeets ofLAQ must be eonsrderedrn the desrgn HVAC system envrronrnents ASHRAE Standard 62 ASHRAE Standard 55 ASHRAE Handbook ofFundarnentals C nntaminatinn C antral Souree ehrnrnatronrnodr eauon Lhe rnost effeetrve soluuon Outdoor an 7 one of the most eornrnon methods OutdnmAu R emm Aw Exhaust Aw E Hem on Space E Cuuhng CUM nte Suppty Aw ow of eontarnrnants ventrlauon Tople04 IAQ oloe Gas removal by chemlcal means adsorption etc Pamculate removal by mechanlcal means lters Cnntaminztinn Health Cnncerns 1 Carbon dloxldeCO2 2 Carbon monomole co 3 Sulfur omoles 502 so 4 Nm gen oxldesNONON20 5 Radon o Volatlle organl eompounols VOCs 7 Pam lates Control 1 Source ellmlnanonmodlflcanon 2 Venulatlon outdoor an 3 Alemsmbuuon 4 Alrcleanlng OuldumAlv R elum A EXHEUS AW 03 l mu exhaust ex ltratim Healng cell Space Es e euneemrauun u space eumaminams N e eumaminam generaliun Cuuhng CD rate in space in llraliun WWW use rcupcenlrallun ul cunlamlnanls m Entering air 0 SupplyAu is esupplys mm mm male eluma rate ex llral nrale mum exhaust rate The text provldes an example ealeulauon uslng the contamlnate balance m Example 4 16 Top1204IAQ doe For an example othe erreetrve use of outdoor an refer to Frgure 4722 on p 170 m the text where Era effecuveness of outdoor an use V ate at whreh outsrde ans taken m VMZrate at whreh unused outsrde an is exhausted the fraetroh R reerreulauoh factor the fraetroh ofreturh an that is reerreulated Filter Siling L L hut t t resrstahee Hrgh resrstahee values rherease the fan srze thus rhereasrhg operatrhg eost RV t e Rtv R emm Aw Exhaust Aw VuCu OutuuuvAu Heatmg Cm Space 39 E Cuuhng Cm uu u a a a t t t Example 4717 shows how these equauorrs are usedm mter se1eetror Suppty Aw TopicO4 IAQdoc Filters Particulate Removal Ef ciency 7 a measure of the ability of the media to remove particulate matter from the air stream Air ow resistance 7 the pressure drop across the lter Dust holding capacity 7 the amount of particulate that can be held by the lter at a speci ed ow rate before the ef ciency drops dramatically Figure 423 p 171 shows typical performance data for air lters Thermal Comfort The human body controls its temperature by 0 Metabolism 7 the rate at which chemical energy is converted to thermal energy Table 42 p 148 assuming an average body surface area of 196 ftz Blood circulation 7 controls the rate at which thermal energy is carried to the skin for dissipation Respiration 7 air and C02 is expelled at the body temperature and in a saturated state Sweating 7 a method to increase energy transfer from the body when blood circulation is insuf cient Engineers like to quantify things Activity level 7 Table 42 p 148 1 met 184 BTUhrf tz Clothing 7 insulating value 1 clo 088 F f t2 hrBTU Dry bulb temperature 7 easily measured Humidity 7 can be quanti ed using a psychrometric chart Air velocity 7 can be de ned using uid mechanics Thermal radiation 7 mean radiation temperature globe temperatures operative temperature wet bulb globe temperature humid operative temperature Thermal radiation Cold and warm surfaces in the occupied space may cause a person to feel uncomfortable even through the conditioned air may be in the comfort range The reason for the discomfort is due to thermal radiation between the body and these surfaces The mean radiant temperature Tmn represents the average temperature of the surfaces in the space This temperature can be determined using a globe thermometer a dry bulb thermometer and an anemometer to determine the average air velocity in the room TopicO4 IAQdoc The mean radiant temperature in R or K can be estimated from T5 T CxVTg T Ta dry bulb temperature R or K TIg globe temperature R or K C mystery constant 0103 x 109 IP units or 0247 x 109 SI units 17 average air velocity ftmin or ms The book discussed radiation exchange between surfaces at different temperatures Equation 439 T ZEHTn n Where F is the view factor between a person and the nth surface at temperature Tquot The operative temperature is T0 hmnTa hmd Tr p hm hmd Where ham is the convective heat transfer coefficient and Ta is the dry bulb temperature There are problems with this derivation The effective radiative heat transfer coefficient hmd is not defined This arises from linearization of the StefanBoltzman Law for thermal radiation heat transfer You ll find a description of the methodology in your undergraduate heat transfer book However this detail does not seem to matter see next paragraph The Operative Temperature Top is the average of the mean radiant temperature and the ambient temperature weighted by their respective heat transfer coefficients For most applications the text says that the Operative Temperature can be simply taken as the average value without weighting The Operative Temperature is required to determine comfort conditions from the comfort chart Figure 416 p 154 Adjustments for conditions other than specified in the comfort chart can be determined using Figures 417 through 420 and Equation 445 T13 Heatngqurp HP doe rpump whreh 15 of prrmary mterest m a heat pump In some ehmates heat pumps ear be run m reverse for eoohhg More aboutthat m Topre 14 Coolmg Equrpmeht um r heat txansferto the evaporator and powerto the compressons also 1ndmated Basn eompohehts ofa meehamea1 por eompressroh heat pump Trausruoh 1 2 ersehtha1pre ex ausroh Trausruoh 2 3 ersotherma1 evaporatroh Trausruoh 3 4 e rsehtropre eompressr Trausruoh 4 1 ersobane eohdehsatroh one measure ofperformance of the heat pump 15 the Coemereht of Performance CO The cm2 15 the desrred output dmded by the requrred mput CW mqu 217m g quot Mvgy mm W T13 HeatlngEqulp HP doc losses Q Q W Typrcally the coP ofelectrrcally dnyen heat pumps ls between 2 to o dependlng on the temperatures of the applrcatron neural a l lutan m ap tact heattornsrdearr These arethemostcommon e s However outdoorarrrs cold when heat ls needed for the burldrng whrch results In a low coP Defrostrng eyaporator corls becomes necessary In humld clrmates Waterrsuurce heat pumps can be used to heat bulldlng zones Water ls crrculated to h The water temperature ls malntalned by an auxlllary heat source heat source A as the heat source and may locate a heat erchanger m the water table Surraoe vtmter unrts use a lake or stream as a source Each system has merrts and drawbacks Tn n ml 11Aer ralrumts t u malt dun t tum Lu utu t all Table 9 8 on p 466 ln the terthas a prorcon llst Heatpump performance depends on source temperature For alr source systems peak be supplled h L ltselfto a temperature bin analysis as shown ln Example 9 4 ln the tart s am ent temperature drops the buddmghemoad heatpumpcapmy heat load ofthe bulldlng rncreas heat pump bulld ng balance Warm balm empe me below thrs auxlllary heat must be hereasrng outdoor temperature mgt supplied perhaps by elecmcal reslstance heaters ME4145 l4 T15 PC only 2006mcd ME414514 HVAC Systems Topic 15 Controls Example of proportional control Determine the steadystate error when proportional control is used for the same system we analyzed under twopoint control The cooling load for the zone is a sensible load of 100000 BTUhr The maximum supply airflow rate is 20000 cfm 88900 lbhr and is supplied at 55 F The throttling range is between 65F and 75F and the value of the offset is the midrange setting of 05 at the setpoint of 70 F The zone thermal capacitance is known The temperature response ofthe zone is given by the solution of a transient energy balance assuming a lumpedcapacitance model F R d CZETZ QS 7 mCpTZ 7 TS m pV Here the zone temperature is higher than the supply air temperature CZ 2000m thermal capacitance of the zone F BTU BTU Q 200000 new load suddenly QS 100000 Initial ss cooling load new hr introduced 3 mass flow rate of dr air 1b mmax 88900 y pajI 0075 3 ft ft3 Vol 20000 volumetric flow rate BTU max C 7 024 mm P lbF TS 55F supply air db temperature Tset 70F zone setpoint temperature M i 1039F throttling range 65 F to 75 F Avmax 05 offset set to midrange at the setpoint Equation for the gain H 1 7 0 Position of damper is open 1 or closed 0 KP AT throttling range 1 Kp 01 F Volsupply vo1maxKpTZ 7 Tset Avmax control law forthe flow rate At steady state there is no net cooling or warming of the zone Qs VOlsupply39paiI39Cp39Tz 7 Ts ME414514 T15 PC only 2006mcd Substitute the control law for the flow rate into the steady state cooling load and solve forthe steady state zone temperature first guess TZSS Tset Given Qs VOlmax39Go39EKp39Tzss 7 Tset T AVmax39pair39cp39Tzss 7 Ts TZSS FindTZss TZSS 65077 F The steady state error or offset eSS T255 7 Tset eSS 74923 F negative error means the zone temperature is lower than the setpoint The volumetric flow rate 3 valsupply VOImaX39Kp39TZSS T Tset T Avmax Vol 153148 mm The negative error also shows that the load ot 100000 BTUh is lowerthan the misrange value The controller has been set to meet a higher load with a zero error The volumetric flow rate for zero error is the midrange value of 10000 cfm The cooling load at midrange supply Vol max QSS 2 60pa cpTseti TS QSS 972gtlt 10 6 BTU hr The zone temperature response can also be found When the ventilation air is supplied at the supply setpoint temperature the rate of change of the zone temperature is with the control equation forflow inserted 1T Q5 7 KPT2 7 Tset T AVmax39IllmaXCpTZ TS dt Z Z The solution ofthis 1st order nonlinear ODE must be done numerically A conditionally stable explicit scheme is Euler forward At lsec i 0 1200 T0 TZSS t iAt 1 ME414514 T15 PC only 2006mcd Qnew 7 KP117 Tset AvmaxmmaXCpTi 7 TS T T At 11 1 C2 T1 F What if the range were decreased to SF with the same setpoint Note that the zone steadystate temperature changes a so 1 Given KP5 5F Qs VOlmax396039Kp539Tzss 7 Tset AV aXJ39pa 39Cp39Tzss 7 Ts T Find T T 67531 F 2555 255 2555 T50 TZSS5 Qnew T KpS39 T5 7 Tset AVmax 39m39max39cp39 T5 7 T5 1 1 T5 At CZ MEMMM T15 PC army was man Zane Tmpmmre F I 1 1 mm Tmemn lEIFrange SFrznge Mxniyulnlzd variable Supply 1 aw mlz zone Ihcn39nosm mmmm Loud mange Schemauc uf mam feedback cuntrm uup Topic05 U valuesdoc ME 414514 HVAC Systems Topic 5 Heat Transfer in Buildings Determining heatingcooling loads requires heat transfer calculations Q UAAt where U overall heat transfer coefficient A surface area exposed to the At At design At U is generally made up of conduction convection and radiative components U R T RT unit thermal resistance of the structure hrft2 F BTU or mzKW Uhas units of BTUhrftzJ F or WmzK The thermal resistance can be based per unit area Thermal resistance R IVA unit resistancearea hr FBTU or KW The Rfactor or Rvalues of various building materials are the unit resistances For example Rl9 insulation R l9 hrft2 FBTU Hence U 119 005263 BTUhrftzJ F If At 50 F A 100 ftZ Q UAAt 005263 BTUhrftzJ F x 100 ft2 x 50 F 2632 BTUhr Using R U 1AR The conductance C lR is tabulated in handbooks for specific materials Conduction 0 Properties and thermal conductivity of materials 7 Table 23 p 34 Note that R AXk and C lR where k is the thermal conductivity Topic05 U valuesdoc Convection R lh Where h is the convective heat transfer coefficient For vertical walls 10 S h S 60 BTUhrft2 F Where the lower value is for still air natural convection and the higher value is for a 15 mph wind forced convection Radiation Thermal radiation is difficult to quantify because of the 43911 power temperature relation For two surfaces that see each other and are separated by a nonabsorbing media 0 T14 T Q 1 8 1 1 8 1 21 Algl AIFTZ A282 Where 8 surface admittance N absorptance gray body behavior For a surface with high 8 high absorptance for a surface with low 8 high re ectance Figure 222 p 68 shows an instance where radiation may be important Table 211 p 59 shows representative values of s for different building materials The resistances of air spaces is a function of position heat ow direction air temperature and the space effective emittance E 181 182 1 E is derived whenl A2 andF12 1 1 8 1 1 8 1 8 1 11 1 12 2 2 81 1 821 81 ll82 l8182 l The effective emittance gives the effect of the surfaces on each side of the air space The resistances can be treated like and electrical circuit to compute the equivalent total resistance beware of parallelseries resistances Series R ZR Topic05 U valuesdoc Parallel L i R I R We will do a composite wall example where area weighting is used to combine series and parallel heat transfer paths We will do several examples of U calculations for different building components Vapor retardants should be installed near the warmest surface to inhibit moisture from entering the insulation Moisture migrates according to Fick s Law rhw DA dx where D mass diffusivity A surface area dCdx concentration gradient Moisture moves from areas of high concentration to areas of low concentration Points of clarification 1 Table 24 p 36 7 average unit heat loss values for basement walls 0 This table gives QLAT values for normal construction 0 More importantly 7 the unit heat loss values are adjusted to take into account the ground temperature swing L QsdZZIMLAT1 tg where tg d temperature swing ta 7 ATg See Figure 26 p 37 for ATg the amplitude ofthe ground 2 Table 28 p 41 7 uninsulated basement oors more than 2 below grade 0 Depth of the oor is not significant the manner of the wall insulation is Note U Uni 0025 BTUhrft2 F Insulation of the wall U132 0036 BTUhrft2 F increases the oor slab Uf ins 0045 BTUhrft2 F heat transfer This occurs because insulating the wall changes the heat ow path in the slab thus changing the thermal resistance Topic05 U valuesdoc Windows Flip over to Section 65 p 264 for a discussion of window U values The center ofa single pane window has convective resistances on both inside and outside surfaces in addition to the conduction resistance of the glass Add a second pane and you must include the radiation resistance between the two panes The edge of the glass has a different U value because of the in uence of the frame Table 64 p 267 and because of aluminum spaces used to separate double or triple pane windows Space between panes may be filled with Ar why Inner surfaces may have lows coatings or the glass itself may have additives to achieve lows characteristics Table 66 p 274 The best source of U value information comes from vendors Manufacturers such as Marvin Anderson etc have measured U values and supply this data with their window specifications Degree Days The balancepoint temperature and degree days are discussed in Section 811 Here is a table with additional data on degree days for selected North American cities DD 65 F 40 N 24 N is the number of hours for which the average outdoor temperature is below 65 F Hence the Degree Day units are Fday is a representation of heating needed Tll Air Diffusersdoc ME 414514 HVAC Systems Topic 11 Space Air Diffusion To appreciate how control of an air jet by diffuser design impacts mixing of supply air into the conditioned space study Fig 1111 on p 578 The velocity of conditioned air supplied to a space is much higher than acceptable the supply air temperature can be quite different than the space temperature and the supply jet may need to counteract natural convection air ows in the space Diffusers are used to counteract these problems The middle zone of a jet the region where the centerline jet velocity varies linearly with distance from the jet entrance is the most important in regards to mixing The relation between the centerline velocity and the initial velocity for a free jet is x x x M 9cm centerline velocity at any axial position x Fl V0 initial velocity at outlet Ag cross sectional jet area at initial velocity x axial distance form outlet 0 volumetric ow rate at outlet K proportionality constant varies from 6 for a free jet to l for ceiling diffusers By using the proper K and Ag these equations can be used to find the throw for any outlet Throw the distance from the outlet to where the maximum jet velocity has decreased to a specified values 50 100 or 150 ftmin The throw is an important consideration in diffuser selection The jet entrains quiescent air The induction ratio is 2 C V0 Sum C entrainment coefficient 2 for a round free jet Near surfaces the jet attaches to the surface because of a low pressure region between the jet and the surface Because entrainment cannot occur on the surface side the throw is increased If the jet is not isothermal buoyant forces become involved and the jet may drop cooling or cling heating T11 Air Diffusersdoc General air jet characteristics 1 P39er Surface effect increases the throw and decreases the drop compared to free space conditions Increased surface effect may be obtained by moving the outlet away from the surface somewhat so that the jet spreads over the surface after impact Increased surface effect may be obtained by spreading the jet when it is discharged Spreading the air stream reduces the throw and drop Drop primarily depends on the quantity of air and only partially on the outlet size or velocity Thus the use of more outlets with less air per outlet reduces drop Effect draft temperature EDT The effective draft temperature EDT is a good metric concerning people comfort The EDT is defined as EDT tr t MV V x r where The subscript X refers to local air stream dry bulb temperature or velocity The subscript r refers to the room dry bulb temperature or velocity 30 ftmin or 015 ms M is a constant 007 Fminft or 70 Csm Research shows that people will be comfortable when 3 S EDT S 2 and Vx S 70 fpm Basic Flow Patterns Air ow patterns from outlets have been classified p 366 The air ow patterns from these 5 outlet groups are drawn in Figures 114 through118 T1 1 Air Diffusersdoc Primary air Total and room air lsovel near ceiling Section XX Section XX Profile of Jet Section X X X 1 Cooling 1 Heating Plan outline High sidewall outlet gtA Supply lsovel outlet near ceiling 9 Section AA Section AA Section AA I Profile of Jet Plan outline Cooling Heating Ceiling outlet Figure 116 Air motion characteristics of Group A outlets Reprinted by permission from ASHRAE Handbook Fundamentals Volume 1997 Insulation optical 3 Model TBO only Ceiing module l H Ceiling module Model A H B C D E Figure 119 A typical Tbar 27 12 4 12 diffuser assembly Courtesy of 28 12 12 Environmental Corporation Dallas TX T1 1 Air Diffusersdoc Primary air Total and room air vvvvvw lsovel Stagnant l Profile Outline Heating Cooling Outlet in or near floor nonspreading vertical jet Figure 115 Air motion characteristics of Group B outlets Reprinted by permission from ASHRAE Handbook Fundamentals Volume 1997 Primary air ll I W lsovel Profile 39 outline 39 4 42 42 40 Total and room air Figure 114 Air motion 39 characteristics of Group C outlets Reprinted by permission from C quotquotg Heat ASHRAE Handbook Outlet in or near floor spreading vertical jet Fundamentals Volume 1997 T1 1 Air Diffusersdoc Primary air Profile Plan outline near floor Total and room air Figure 117 Air motion characteristics of Group D outlets Reprinted by 39 39 39 quot permission fromASHRAE Cooling Heating Handbook Fundamentals Outlet near floor horizontal discharge low Volume 1997 Total and room air Ceiling Stagnant 139 j Ceiling Figure 118 Air motion characteristics of Group E outlets Reprinted by permission From ASHRAE quot 39 Handbook Fundamentals Cooling F100 Heating Floor Volume 1997 Air distribution system design We will look at the selection and location of diffusers and return grilles for HVAC systems Noise Diffuser noise caused by air ow can be annoying to occupants Some manufacturers rate their air diffusing equipment on the NC noise criteria scale Acceptable NC values for various occupancies are presented in Table 111 below In general a quiet diffuser has a value of NC lt 35 The diffuser is considered noisy when NC gt 55 T1 1 Air Diffusersdoc Table 111 Acceptable HVAC Noise Levels in Unoccupied Rooms Occupancy Noise Criteria Private residences NC 25 30 Apartments NC 30 35 HotelsMotels Individual moms or suites NC 30 35 Meetingbanquet rooms NC 30 35 Halls corridors lobbies NC 35 40 Servicesupport areas NC 40 45 Of ces 39 Executive NC 25 30 Conference rooms v NC 2530 Private NC 30 35 Openplan areas NC 35 40 Business machinescomputers NC 40 45 Public circulation NC 40 45 Hospitals and clinics Private rooms NC 25 30 Wards NC 30 35 Operating rooms 39 NC 25 30 Laboratories NC 3540 Corridors NC 30 35 Public areas NC 35 40 Churches NC 30 35 Schools Lecture and classrooms NC 25 30 Openplan classrooms NC 35 40 Libraries NC 35 40 Courtrooms NC 35 40 Legitimate theaters NC 20 35 Movie theaters NC 30 35 Restaurants NC 40 45 Concert and recital halls NC 15 20 Recording studios NC 15 20 TV studios 39 NC 20 25 Source Reprinted by permission from ASHRAE Hand book HVAC Applications Volume 1991 T11 Air Diffusersdoc Characteristic Room Length for Several Diffusers Table 11 2 Cl 39 quot Room Length for Several Diffusers Diffuser Type Cl 39 quot Length L High sidewall grille Distance to wall perpendicularto jet Circular ceiling diffuser Distance to closest wall or intersecting jet Sill grie Length of room in direction ofjet fow Distance to wall or midplane between Ceiling slot diffuser outlets Distance to midplane between outlet planes plus distance from ceiling to top of Light troffer diffusers occupied zone Perforated louvered Distance to wall or midplane between ceiling diffusers outlets Tll Air Diffusersdoc Air Distribution Performance Index ADPI ADPI 7 the percentage of locations in the occupied zone where 3 S EDT S 2 The objective of diffuser selection is to keep the ADPI as close to 100 as possible Table 113 Air Diffusion Performance Index ADPI oom For Load x5oIL for ADPI Terminal BTUh Maximum Maximum greater Range Device ftz ADPI ADPI than of x5olL High sidewall grilles 80 18 68 60 18 72 70 1522 40 16 78 70 12 23 20 15 85 80 1019 Circular ceiling diffusers 80 08 76 70 07 13 60 08 83 80 07 13 40 08 88 80 05 15 20 08 93 90 07 13 Sill grille Straight vanes 80 17 61 60 15 17 60 17 72 70 1417 40 13 86 80 1218 20 09 95 90 0813 Sill grille Spread vanes 80 07 94 90 06 15 60 07 94 80 0617 40 07 94 20 07 94 Ceiling slot diffusers for xwoL 80 03 85 80 03 07 60 03 88 80 0308 40 03 91 80 03 11 20 03 92 80 03 15 Light troffer diffusers 60 25 86 80 lt 38 40 10 92 90 lt 30 20 10 95 90 lt 45 Perforated and louvered ceiling diffusers 11 51 20 96 90 14 27 80 10 34 Tll Air Diffusersdoc Round diffusers High sidewall diffusers Tbar diffusers Return General procedure for selection of diffusers and grilles l 2 3 Use the ADPI selection guide Table 113 to find the recommended throwtolength 4 V39 Determine room size and air ow requirements Select the type of diffuser to use see the diffuser characteristics as a guide Figures 114 through 118 Determine the room characteristic length Table 112 ratio Calculate the throw Select a diffuser or grille from vendor catalogs Make sure NC requirements are met Table 111 and note the pressure drop TomeS Coohhg Loads do ME 414514 HVAC Systems 7 Tupi s Cnnlingands Hez nglnzds e a steady state aha1ysrs the worst ease used system may be shghuy dyersrzed systerh Heat gain ethe rate at whreh heatrs txansferredto or generated wrthrh a spaee rharhtarh rhdddr desrgh condmons raddant energy andre1ease heattothe surrduhdhgs equrprheht The heat extraetror rate ddffers from the eddhhg load beeause of thermostatre edhtrol Reean The govemmg pamal ddfferenual equatror for trahsreht onerddmenslonal heat transfer rh thrs wall mu nllisule Inside is 3 2 12 3x2 2849 1 rs temperature 5x5 trrhe and 4115 the therma1 ddffusxvxty The BCs Atx0 hA At x L M0 71 Wham TheIc at50zt ASHRAE Topic08 Cooling Loadsdoc The Transfer Function Method TFM 7 pages 2 to 4 The CLTDSCLCLF Method 7 pages 4 to 6 The Heat Balance Method 7 pages 6 to 13 The Radiant Time Series Method 7 pages 13 to 16 bP N The first two methods are no longer being actively supported by ASHRAE Because they have only very recently been replaced you may still meet engineers who use these methods usually via a computer program for TFM and hand calculations for CLDTSCLCLF In ME414514 we will learn the CLDTSCLCLF method and introduce the Radiant Time Series Method Cooling Design Conditions Outdoor design conditions report the ASHRAE recommended outdoor temperature at 1 and 25 value The percent value represents the percent of time the listed temperatures were met or exceeded during the cooling season June 7 September In Boise Idaho a typical year will have 88 hours 1 of 8760 hours in one year at 96 F dry bulb or higher and 219 hours at 94 F or higher The summer wind speed is 75 mph The hourly outdoor temperature oscillates between the outdoor design temperature and a minimum temperature calculated by subtracting the daily range from the outdoor design temperature Kreider limits the analysis to an average outdoor temperature t0 td 7 DR X t0 design outside dry bulb temperature DR the daily range from the design condition table X percentage of the daily range see table on p 4 For indoor design conditions ASHRAE suggests 75 F with I 50 Comfort conditions must be satis ed Chapter 4 Figure 416 p 164 Transfer Function Method TFM The transfer function method is based on two important concepts 1 Conduction Transfer Functions CTF 2 Weighting Factors WF or Room Transfer Functions RTF The WF or RTF relate the hourly cooling load to the heat gain and its past history and past values ofthe cooling load qc ngx V1q1 7A V2 9107 wlqc 7A W2 9507 Topic08 Cooling Loadsdoc The first three terms are the instantaneous heat gains The last two are corrections to prevent double counting The V and w values are found in tables of room weighting factors once the zone is determined Tables 78 p 343 and 79 p 344 have tables for w and V q39l is the instantaneous heat gain at hour 0 note the A to indicate the prior hours and q39c iA is the cooling load or reduction of the heat gain from the prior hour De nition Comments 12 3 40 space gypsum Roof Construction Types Type Description 1 Outside surface resistance 12 slag or stone 3 8 felt membrane 1 insulation steel siding inside surface resistance A0 E2 E3 B4 A3 E0 2 Outside surface resistance 12 slag or stone 3 8 felt membrane 6 LW concrete inside surface resistance A0 E2 E3 C15 E0 3 Outside surface resistance 12 slag or stone 38 felt membrane 2 insulation steel siding ceiling air space acoustic tile inside surface resistance A0 E2 E3 B6 A3 E4 E5 E0 4 Outside surface resistance 12 slag or stone 3 8 felt membrane 8 LW concrete ceiling air space acoustic tile inside surface resistance A0 E2 E3 C16 E4 E5 E0 The above are ways to get the weighting factors form extensive tables Not done anymore Topic08 Cooling Loadsdoc Further categorization tables were published for determining zone types for single story perimeter room interior rooms or single zones Conduction Transfer Functions CTF are used to compute the instantaneous heat gain to the space The heat gain is computed as 410 AZ bn Imam Z dnq1 rnA At1 ch quot0 n1 quot0 Here b d and c are the conduction transfer function coefficients in Table 77 p 340 for roofs and walls In this table the codes refer to construction layers the text CD has all the definitions The sol air temperature te in the CTF equation includes convective and radiative gain effects and is defined as I A t2 t0 0 i h 0 0 t0 is the outside temperature tdesign 7 DR X 100 DR is the daily range and X is the percentage of the daily range at each hour of the day Percentage of the Daily Range X Time Tim Time Time hr Percent hr Percent hr Percent hr Percent 1 87 7 93 13 11 19 34 2 92 8 84 14 3 20 47 3 96 9 71 15 0 21 58 4 99 10 56 16 3 22 68 5 100 11 39 17 10 23 76 6 98 12 23 18 21 24 82 0c is the surface absorptivity h0 is the combined convective and radiative heat transfer coefficient I is the global irradiance on the surface Aqir is the correction for radiation exchange between the surface and the environment when the sky temperature is colder than the air temperature A qirh0 varies from 0 for vertical surfaces to 7 0F for horizontal surfaces CLT DSCLCLF Method This is a simplified method base on the Transfer Function Method Cooling loads are directly computed for any hour of the day using expressions for cooling loads due to Topic08 Cooling Loadsdoc 1 Walls and roofs Q UACLTD2 2 Fenestration heat gains Q A SC SCL 3 Internal heat gains Q Q1 CLE CLTD Cooling Load Temperature Difference SCL Solar Cooling Load factor CLF Cooling Load Factor CLTD SCL and CLF values are tabulated for every hour of the day Therefore hourly estimates of the load can be determined However this method is often used to get a quick estimate of the cooling equipment size Using this method the cooling system can be sized based on the time when the peak load of the major component of the cooling load occurs This peak is evident in the CLTD SCL and CLF tables CLTD 7 Walls and Roofs CLTD values can be found for at roofs and sunlit walls in the text CD These tables have been constructed based on standard indoor and outdoor conditions Standard Outdoor Conditions July 21 at 36 deg N latitude No exterior shading Ground re ectance of 02 Clear sky Outside surface of roofs and walls with a ratio of the absorptance to the film coefficient ccho of 030 Outside air maximum dry bulb temperature of 95 F 35 C with a daily range of 21 F 12 C Standard Indoor Conditions 0 Room dry bulb temperature constant at 78 F 26 C 0 Inside film coefficient for still air If the design conditions are different than these standard conditions the CLTD must be corrected CLTDcm 7 CLTDtab 78 7 ti to 7 85 ti actual indoor design temperature t0m t0 7 DIU2 t0 outside design db temperature DR daily range 85 the mystery constant 95 7 212 Topic08 Cooling Loadsdoc To use the tables in the CD you need to know the type of roof 13 types or wall 7 types that is the closest match to the structure you are analyzing We will do examples in class Fenestration The cooling load due to fenestration heat gains are made up of two pa1ts and must be calculated separately Conductive load Use the CLTD values for glass on the CD Radiative load Use the CLF and SHGF values for glass on the CD Shading is accommodated by either using the CLF values for interior shading for glass the glass has drapes on the inside or the SC from Topic 6 Recall that the SC values depend on the type of glass etc Internal Sensible Heat Gains The CD has tables for people appliances and lights We will do examples in class Then we will compare our hand calculations with the HCB program on the text CD For small buildings one central cooling system constant ow of air one thermostat including single family residences a simpli ed CLTD method is this following 1 Exterior walls 7 use daily average CLTD from 769 Roofs 7 average CLTD for sunlit hours from 767 Glass 7 average SCL for sunlit hours from 770 Peoplelights equipment 7 as usual with peak at 4 pm 59 Using either the TFM or CLTDSCLCLF size the cooling load for the peak load on the warmest day of the year Heat Balance Method The HBM requires a load calculation program and this method is not discussed in our text Different pieces of the cooling load problem are looked at in isolation A program puts all these pieces together for us to perform a complete cooling load analysis for a structure Hence engineers working in HVAC are in practice forced to rely on commercial software packages for a cooling load analysis Basically the HBM consists of heat balances on the interior and exterior surfaces of a single wall or roof element Each of the heat uxes 7 solar convection and radiation into an exterior surface vary throughout the day as the sun changes position in the sky Topre08 Coolrng Loads doe Fm tlu r t trnrnt heat conduction problem Solutron methods rnelude lumped parameter 7 small preees wth unlform temperaturertlme frequency response epenodre BCs Founer sene numerreal rntegratron efrnrte dlfferences or flnlte elements Zrtransforms rthe theory usedrn the HBM wa A Zrtransforms use erther response faetors or eonduetron transfer funetrons to t wall eonyeeuon eonyeeuon 501a eondueuon solar 9 eondueuon radrauon Edam CTFs replaee temperature hrstory wrth heat ux hrstory program for that task Basreally the heat balanees expressedm terms of CTFs must be solved for eaeh hour of the day The solutron proeedure may be rteratrye rnrtral guess of the surface mamx rnversron A eheek on the solutron yalrdrty ls obtarned by solyrng a steadyrstate problem The result should agree wrth the transmrssron equatron from Ch 5 q Unv lv EXTERIOR SURFACE IEEAT FLUX Snlar radiatinn e hrgh temperature souree UV or short wavelength EMIR insular 1 Gt Topic08 Cooling Loadsdoc Where 0c surface absorptivity of short wavelength EMR Gt total solar irradiation from Chapter 6 calculated on the 12 hour Exterior convection 7 Newton s Law of Cooling q convection he to tos where hC exterior convective heat transfer coefficient t0 outside air temperature tOS outside surface temperature In Topic 5 we used tabulated values for he when we were calculating overall U values for walls roofs etc Here a correlation is given for using a computer h qulSf enff where C turbulent natural convection constant At difference between exterior surface temperature and exterior air temperature a b V0 wind speed in mph or ms Exterior radiation 7low temperature source IR or long wavelength EMR The surface exchanges thermal radiation in the IR spectra with the ground and the sky qumdmtmn 80Fs7g T T0 F57sky T52 To The view factors F57g and Fssky are from Topic 6 with 0c the surface tilt angle 5 3 F l cosa ilcosltagt1 The 4 power terms are a nuisance numer1cally so the rad1atlon ux 1s lmearized hhg go F5i T34 To g t05 h T 402 Wk tsky to qumdzatzun hug tg tn hasky tsW I The sky temperature is approximated as To 7 108 R or To 7 6K The correction for a surface tilted at angle at is Topic08 Cooling Loadsdoc All of the exterior heat uxes can be combined to get a simplerlooking relationship FENESTRATIONS Direct diffuse and re ected solar gain through windows and skylights is calculated using the methods of Topic 6 INTERIOR HEAT GAINS People Lights Equipment People People contribute sensible and latent heat Tables are available for different occupancies and occupant activities qSJaeople N Fu 1 qlatJaeople N Fu qlat N number of people Fu use factor 10 for fully occupied room Sensible heat gain split 30 convective 70 radiative Latent heat gain split 100 convective 0 radiative The split between radiative and convective fractions is needed because convective heat is part of the instantaneous heat load whereas radiative heat contributes to the delayed load Lights q lights 341 W Fu F W total installed wattage Fu use factor 10 if all installed lights are on F5 fudge factor to compensate for less heat from uorescent bulbs Heat gain split Fluorescent lamps 41 convective 59 radiative Incandescent lamps 20 convective 80 radiative Heat lost to the return air plenum depending on whether or not the xture is ventilated or unventilated Not all of the heat from the light enters the conditioned space Equipment Topic08 Cooling Loadsdoc qimotor C P Em 11 Fu qimmor heat equivalent of equipment operation BTUhr or W P motor shaft power rating hp or W Em motor ef ciency as decimal not F1 fraction of rated load delivered Fu motor use factor 10 when the equipment is running C unit conversion constant 2545 Btuhhp or 1 WW Motor outside conditioned space driven equipment inside qimotor C Fl Fu Motor inside conditioned space driven out qimotor C 1 39 Em Em 11 Fu Equipment heat gain split 30 convective 70 radiative Electronic equipment heat gain split Printer89 convective ll radiative Copier 86 convective 14 radiative PCs 745 convective 255 radiative average Appliances hooded steam amp electrical qapplime 05 X 032 qi Where 032 is the fraction of radiative heat convective heat is removed from the occupancy by hoods q is the input rating Appliances hooded fuel red qappliance qi Where q is the input rating INTERIOR OPAQUE SURFACES Convection q convection he ti ti Topic08 Cooling Loadsdoc where hC exterior convective heat transfer coefficient tis inside surface temperature different for each surface and varies throughout the day ti inside air temperature design condition Radiation exchange with objects in the enclosure Modeling the details of all the radiation exchanges between all of the surfaces 7 walls ceiling floor 7 and objects furniture is theoretically possible but impractical because of the detail involved The text describes the mean radiant temperature balance method This model uses a fictitious surface to represent the view that one surface has of all other surfaces Recall from your undergraduate heat transfer class 0 The sum of the views from one surface to each surface in an enclosure is l 2F 1 o A planar surface cannot see itself F11 0 o The net radiation exchange in an enclosure is zero Hence for one surface in an enclosure the fictitious surface area is the sum of the areas of all the other surfaces the fictitious surface emissivity is an areaweighted average the fictitious surface temperature is an areaweighted average the view factor of the fictitious surface is calculated using the fictitious emissivity and area a radiation coefficient is calculated using the fictitious view factor and temperatures The thermal radiation calculated from all this fictitiousness is not correct and needs to be corrected balanced so that the net exchange is zero All of these calculations are amenable to a spreadsheet approach see Example 85 Radiation gains from people ghts and equipment We saw earlier that people lights and equipment include a radiative load These contributions are summed for each hour and treated as uniform uxes over each interior surface All of the interior heat uxes can be combined to get a simplerlooking relationship Topic08 Cooling Loadsdoc INTERIOR TRANSPARENT SURFACES Fenestrations The calculation of transmitted and absorbed irradiation follows the process that we learned in Topic 6 We can re ne the calculation by considering absorptance in each pane of a multipane window The calculation procedure is identical that T 6 What is new are the coefficients to separately calculate absorptance in the outer pane and inner pane of a double pane window ZONE AIR Surface convection sum over all surfaces Qconv A q aconvection A he ti ti Internal heat gain convection 7 add up the contribution from people equipment and lights Infiltration V qsinr0 r an to t1 V Cp t0 t1 V 95mm ma W0 W hfg W1 hfg System heat transfer qsystem a b ti IMPLEMENTATION 1 determine all zone parameters surface areas thermal properties geometry 2 find all imposed uxes transmitted and incident solar radiation internal loads infiltration rates 3 calculate exterior and interior surface temperatures for each hour of the day 4 calculate the required system heat transfer by using the zone air heat balance stmm ZAjh lj hm t1 75me t0 t1 Qint 0nv J The difficulty as we have seen in this deceptively simple looking equation is finding the interior surface temperature I All of the exterior heat loads solar convection transmission and interior heat loads solar through fenestrations radiation from people equipment and lights must be included and modeled Topic08 Cooling Loadsdoc Radiant Time Series Method To apply this simplified approach to calculating the cooling load the analyst calculates the solair temperature as a method of combining convection radiation to the ground and sky and solar gain by opaque surfaces The steps for the RTS method 1 Calculate solar irradiation for each hour a Solar irradiation Topic 6 b Absorbed and transmitted SHG for semitransparent surfaces c Solair temperature 2 Calculate other heat gains for each hour a Conduction or transmission Topics 5 and 6 b Lighting equipment and people contribution c In ltration Topic 7 3 Divide the heat gains into convection instant cooling load and radiation delayed load portions Calculate the cooling load due to the radiation portion 5 Sum the convective loads and the radiative load 4 1 Calculate solar irradiation for each hour We know how to calculate irradiation to opaque surfaces and SHG for fenestrations 7 from all our hard work in Topic 6 The solair temperature is new Sol air Temperature The sol air temperature is de ned as T05 2T0 a h h 0 0 where ho is the combined exterior convective and radiative heat transfer coefficient I is the global solar irradiance on the surface Aqir is the difference between the radiation incident on the surface from actual surroundings and the radiation emitted by a blackbody at outdoor air temperature The thermal radiation correction term Aqiho is approximately 7 F 39 C for horizontal surfaces and 0 F 0 C for vertical surfaces Topic08 Cooling Loadsdoc 2 Calculate other heat gains for each hour 2a Conduction Heat Gains 23 q0ndmj A ZYPn IL int Ira quot0 where Y 12quot n1h response factor BTUhr ft2 F or Wm2 K re solair temperature for surface j 71 hours ago F or C In room drybulb temperature F or C assumed constant Aj surface area of surface j ft2 or In2 You need a program to find the Y s periodic response functions Conduction heat gain through windows can be calculated without PRFs since glass has a small thermal mass when compared to the building masonry qmndmwmd0w medowAwmdow ta0 Ira 2b Internal Heat Gains People Lights Equipment People People contribute sensible and latent heat Use tabulated data in for different occupancies and occupant activities queople N Fu 1 qlatJaeople N Fu qlat N number of people Fu use factor 10 for fully occupied room Sensible heat gain split 30 convective 70 radiative Latent heat gain split 100 convective 0 radiative The split between radiative and convective fractions is needed because convective heat is part of the instantaneous heat load whereas radiative heat contributes to the delayed load Lights qiligms 341 W Fu FS W total installed wattage Fu use factor 10 if all installed lights are on F5 fudge factor to compensate for less heat from uorescent bulbs 14 Topic08 Cooling Loadsdoc Heat gain split Fluorescent lamps 41 convective 59 radiative Incandescent lamps 20 convective 80 radiative Heat lost to the return air plenum depends on whether or not the xture is ventilated or unventilated Not all of the heat from the light enters the conditioned space Equipment qimotor C P Em F1 Fu qimmor heat equivalent of equipment operation BTUhr or W P motor shaft power rating hp or W Em motor efficiency as decimal not F1 fraction of rated load delivered Fu motor use factor 10 when the equipment is running C unit conversion constant 2545 Btuhhp or 1 WW Motor outside conditioned space driven equipment inside qimotor C Fl Fu Motor inside conditioned space driven out qimotor C 1 39 Em Em Fl Fu Equipment heat gain split 30 convective 70 radiative Electronic equipment heat gain split Printer89 convective ll radiative Copier 86 convective 14 radiative PCs 745 convective 255 radiative average Appliances hooded steam amp electrical qappliance 05 X 032 q Where 032 is the fraction of radiative heat convective heat is removed from the occupancy by hoods q is the input rating Appliances hooded fuel red qappliance qi TopicOS Cooling Loadsdoc Where q is the input rating 2c In ltration V game r an t0 tr V Cp t0 Ira 0 x 0 x 95mm ma W W hfg W hfg 3 Splitting Heat Gains Into Convective amp Radiative Portions ASHRAE tables are available for this 4 Determine Radiative Loads 23 gem Z rnq rmi 710 where rquot n1h radiant time factor and the q in the summation is the cooling load heat gain 71 hours ago 5 Sum the Loads

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