PDHonline Course M199 (3 PDH)HVAC Calculations and Duct SizingInstructor: Gary D. Beckfeld, P.E.2012PDH Online PDH Center5272 Meadow Estates DriveFairfax, VA 22030-6658Phone & Fax: 703-988-0088www.PDHonline.orgwww.PDHcenter.comAn Approved Continuing Education Provider

www.PDHcenter.comPDH Course M199www.PDHonline.orgHVAC Calculations and Duct SizingGary D. Beckfeld, M.S.E., P.E.COURSE CONTENT1. Heat Conduction and Thermal ResistanceFor steady state conditions and one dimensional heat transfer, the heat qconducted through a plane wall is given by:q kA(t1 - t2)LBtuhr(Eq.1)Where:L the thickness of the wall in inchesA the area of the wall in square feet(t1 - t2 ) temperature difference across the wall in degrees Fahrenheitk thermal conductivity of the wall materialBtu-inhr-ft2-FEquation 1 can be put in terms of a unit thermal resistance, R L/k, or anoverall heat transfer coefficient, U 1/R, to giveq A(t1 - t2)R UA(t1 - t2 )(Eq.2)Note that the R in equation 2 is the factor often found on blanket insulationand other building products. Equations 1 and 2 are for a single material sothe resistance R must be modified for building walls of several materials.2. Building WallsWalls of buildings are constructed of several layers of different thickness,material, and area. Figure No. 1 shows a typical 2x4 framed house wall anda concrete wall with polystyrene insulation on both the interior and exteriorsurfaces. For the concrete wall, with the vinyl siding and drywall, there are Gary D. BeckfeldPage 2 of 21

www.PDHcenter.comPDH Course M199www.PDHonline.orgfive thermal resistance layers. In addition there are thermal resistances onthe inside and outside surfaces of a building wall due to convective aircurrents and radiation. These resistances are accounted for with filmcoefficients, f, given byfi 6.0 Btu/ (hr-ft2 -F) 1/Ri inside surface with still air(Eq.3)fo 1.63 Btu/ (hr-ft2 -F) 1/Ro outside surface with moving air (Eq.4)Then for the whole concrete wall, the thermal resistance to be used in theconduction equation 2 becomesR Ro R1 R2 R3 R4 R5 Ri(Eq.5)For the framed wall, similar thermal resistance equations would be writtenfor the heat path through the studs and through the insulation path betweenthe studs as indicated in Figure ED WALLCONCRETE WALLFIGURE 1In addition to the heat conducted through the walls, given by equation 2, abuilding can have heat gains or losses from the attic and basement. Gary D. BeckfeldPage 3 of 21

www.PDHcenter.comPDH Course M199www.PDHonline.org3. Building Attic and BasementAt equilibrium conditions, the heat loss or gain from the attic or basement,as indicated in Figure 2, is equal to the heat loss or gain to the buildingthrough the ceiling or floor. For an attic, equation 2 givesUr Ar (to - ta ) Uc Ac (ta - ti )(Eq.6)Where r roof properties, c ceiling properties, a attic propertiesa attic temperature, o outside temperature, and i inside temperature.Solving equation 6 for the attic temperature givesta (Ur Ar to Uc Ac ti ) / (Ur Ar Uc Ac )(Eq.7)After the attic temperature is found from equation 7, the heat conducted tothe building through the ceiling can be found from equation 6. Thisprocedure can also be used to estimate a basement (or attached garage)temperature and heat loss or gain to a building through the floor or wall.qqFIGURE 2 Gary D. BeckfeldPage 4 of 21

www.PDHcenter.comPDH Course M199www.PDHonline.org4. Building Heat LoadsThe total cooling or heating load of a building consists of two parts, thesensible heat, Qs , and the latent heat, Ql . The sensible heat load comesfrom the following sources:1. Heat conducted through the building (walls, ceiling, floor, windows).2. Internal heat from lights, computers, ovens, and other appliances.3. Infiltration of outside air through cracks around windows and doors.4. People in the building.5. Sun radiation through windows.The latent heat load to a building comes from the following sources: in the building.Infiltration through cracks, chimneys.Gas appliances, range, ovens.Dishwasher, other appliances.The sensible heat load results in an air temperature rise in the building. Tomaintain temperature requirements, the air in the building is circulated overa cooling coil at a certain rate determined from the equationQs 1.1 (w) (ts -ti )Where(Eq.8)w cubic foot per minute of air circulation flowi denotes inside air temperatures denotes supply temperature of the airThe latent heat load determines the amount of moisture that is added to theair in the building and must be removed from the air by the cooling coil tomaintain humidity requirements. This is found from the equationQl 4840 (w) (Gs - Gi )Where(Eq.9)Gi pounds of moisture per pound of air in conditioned spaceGs pounds of moisture per pound of a supply air Gary D. BeckfeldPage 5 of 21

www.PDHcenter.comPDH Course M199www.PDHonline.orgOnce the sensible heat, Qs , and latent heat, Ql , are known, a sensible heatratio, SHR, can be found fromSHR Qs / (Qs Ql )(Eq.10)The sensible heat ratio is useful in finding G, the moisture content of air, atdifferent conditions using a psychrometric chart.5. Psychrometric ChartA psychrometric chart is a graphical representation of the properties of moistair and is a useful tool in air conditioning calculations. This chart can beviewed online and a copy can be printed out from the following orengineers/psychrometric chart.pdf .Figure 3 is a sketch of an air conditioning cycle plotted on a psychrometricchart. The numbered points on the diagram correspond to the numberedpoints in Figure 4, a sketch of a building air conditioning system. Thesepoints are determined in the air conditioning 1SHRYLPG MOISTUREATHEN50% RH.010295TEMPERATURE DBFIGURE 3 Gary D. BeckfeldPage 6 of 21

www.PDHcenter.comPDH Course M1991www.PDHonline.orgREHEATERBUILDING35 X 73580DB, 50%RH4COILCOMPRESSOR215% MAKEUP395DB,75DP15% EXHAUSTFIGURE 46. Air Conditioning CalculationsExampleA building, 35 feet wide and 73 feet long, is constructed with the type ofconcrete wall indicated in Figure 1. The concrete is 4 inches thick and thepolystyrene insulation is 2 inches thick on each side. The east and westwalls each have two windows. The north wall has 6 windows and the southwall has 9 windows. All windows are 2.5 feet by 4 feet. The roof andceiling are frame construction. The conditions inside are to be maintained at80 F db (dry bulb) temperature and 50% relative humidity. The outsideconditions are 95 F db and 75 F dp (dew point) temperature. The coolingload in tons is be found for selecting cooling and air handling units. Forventilation, it is assumed the air handler will take 15% of the required flowfrom the outside conditions. Supply air in the building is to be 65 F db.Some thermal conductivities, k, and thermal resistances, R, of buildingmaterials for this problem are shown in Table 1. Where the overall heattransfer coefficient, U, is given, it is assumed to include film coefficients.More extensive lists of these values are found in References 1 and 2. Gary D. BeckfeldPage 7 of 21

www.PDHcenter.comPDH Course M199www.PDHonline.orgTABLE 1MaterialSidingPolystyreneConcreteDrywallPine 2x4InsulationSheathing .5 inGlassFramed CeilingFramed Roofk Btu-in/(hr-ft2 -F)R (hr-ft2 –F)/Btu U Btu/(hr-ft2 –F)1.0.1710. Equation 5, the thermal resistance of the 4 inch concrete wall with 2inch insulation is12421R 6 1.0 .17 10 .17 .45 1.63 26 (hr-ft2 –F)/BtuEquation 2 gives the heat conducted through the wall area minus the windowarea assuming 8 foot high walls.East wallWest wallNorth wallSouth wallq ((8x35) -2(2.5x4)) (95 -80) / 26 150 Btu/hrq ((8x35) -2(2.5x4)) (95 -80) / 26 150 Btu/hrq ((8x73) -6(2.5x4)) (95 -80) / 26 302 Btu/hrq ((8x73) -9(2.5x4)) (95 -80) / 26 285 Btu/hrThere are 19 windows with 10 ft2 each for a total glass area of 190 ft2 .The heat conducted through the glass is given by equation 2 asq 1.13 (190) (95 -80) 3220 Btu/hrAlso for single sheet of unshaded glass, the radiation heat gain can besignificant. This heat gain can be found from the equationq A (SC) (SCL) Btu/hr Gary D. Beckfeld(Eq.11)Page 8 of 21

www.PDHcenter.comWherePDH Course M199www.PDHonline.orgA glass area in ft2SCL solar cooling load in Btu/ (hr-ft2 )SC shading coefficient assumed 1.0 for no shadingReference 1 indicates that for south facing windows, depending on thelatitude, the solar cooling load can be nearly 100 Btu per hr per sq ft andnearly twice that for glass facing east or west. Then the radiation heat gainsfor this building are, from equation 11East facing glassWest facing glassSouth facing glassq 20 (200) 4000 Btu/hrq 20 (200) 4000 Btu/hrq 90 (100) 9000 Btu/hrFor heat conducted through the ceiling from the attic, the attic temperature isfound from equation 7. Assume the roof pitch is 30 degrees and using datafrom Table 1 givesta .23(2555)(80) .22(2950)(95) 88 F.23(2555) .22(2950)Then from equation 6 the heat from the attic isq .23 (2555) (88 – 80) 4701 Btu/hrTo account for infiltration of outside air through cracks around windows anddoors, a leakage rate of 20 cubic foot per minute assumed. The resultingsensible heat gain inside is given by equation 8 asq 1.1 (20) (95-80) 330 Btu/hrFor the heat gain from electrical appliances and lights, Reference 2 gives theequationq 3.42 (Wattage)(Eq.12)For the medium size building considered here, Reference 1 gives an averageelectrical usage of 1.0 watts/ sq ft. This gives a heat gain ofq 3.42 (1.0) (2555) 8738 Btu/hr Gary D. BeckfeldPage 9 of 21

www.PDHcenter.comPDH Course M199www.PDHonline.orgHuman occupants of the building contribute sensible heat according to theiractivity. From Reference 1, 200 Btu/hr per person is estimated for lightactivity. For two people in the building, this gives a heat gain ofq 2 (200) 400 Btu/hrAdding all the heat gains from conduction and internal sources givesTotal sensible load Qs 35274 Btu/hrHuman occupants also contribute to the latent heat gain in the building. Forlight activity people produce a latent gain of about 180 Btu/hr per person sothis heat isq 2 (180) 360 Btu/hrThe assumed infiltration of 20 cubic foot per minute of outside air bringslatent heat into the building which is given by equation 9 asq 4840 (20) (G3 - G2 )where the quantities G3 and G2 are the moisture content of the outside andinside air respectively. Referring to a copy of a psychrometric chart or toFigure 3, locate point 2 representing the inside design conditions of 80F dband 50% relative humidity. From point 2, project a horizontal line to theright to the moister content scale and read and read G2 .011 lb moistureper lb dry air. Similarly locate point 3 for the outside conditions of 95 F dband 75F dew point. From the chart obtain G3 .019 lb moisture per lb dryair. Then the latent heat from the infiltration isq 4840 (20) (.019 - .011) 774 Btu/hrAppliances such gas ovens, ranges, and dishwashers add to the latent heatload in a building. These contributions are estimated from the data ofReference 2 as follows:DishwasherGas ovenGas range Gary D. Beckfeldq 420 Btu/hrq 1200 Btu/hrq 5600 Btu/hrPage 10 of 21

www.PDHcenter.comPDH Course M199www.PDHonline.orgAdding the latent heat gains givesTotal latent load Ql 8354 Btu/hrWith the sensible and latent heat loads determined, equations 8 and 9become35274 1.1 (w) (t2 -t1 )8354 4840 (w) (G2 -G1 )The first of these equations determines the flow rate, w, required of the airhandler. The temperature t2 is the inside design condition 80F and t1 is thedesired supply of 65F air at the outlet of the air handler. The flow rate isw 35274 / ((1.1) (80 -65)) 2137 ft3 /minThe second equation determines G1 , the moisture content of the air at theair handler outlet. G2 .011, the moisture content of the 80F and 50% RHair in the space. The moister content isG1 .011 – 8354/ (4840x2137) .0102 lb/lb dry airWith this value, the point 1 shown in Figure 4 can be located on thepsychrometric chart. From .0102 on the moisture scale extend a horizontalline to intersect with a vertical line at 65F. This point 1 can also be locatedor checked using the sensible heat ratio given by Equation 10SHR 35274/ (35274 8354) .81From .81 on the SHR scale draw a line through point 2 and extend it tointersect the 65F line to again define point 1. Extending the horizontal linefrom this point to the 100% relative humidity curve locates point 5 shown inFigures 3 and 4. This gives the dew point temperature of 59F. This is therequired cooling coil temperature of the air conditioning unit.Point 4 shown in Figures 3 and 4 is the condition of the air entering thecooling coil. This point can be found from the following equation: Gary D. BeckfeldPage 11 of 21

www.PDHcenter.comPDH Course M199www.PDHonline.orgG4 G2 f (G3 - G2 )(Eq.13)Where the G’s are the moisture contents at the various points and f is thefraction of the air flow taken from the outside. Here f .15 so at point 4G4 .011 (.15) (.019 - .011) .012 lb/lb dry airFirst draw a line between points 2 and 3 on the psychrometric chart. Thenintersect a horizontal line draw from .o12 on the moisture scale to definepoint 4.The points of the air conditioning cycle on the psychrometric chart are nowused to read the enthalpy of the air at points 1, 4, and 5. These areh1 27.2 Btu/lbh4 33.2 Btu/lbh5 25.8 Btu/lbThese values are used in calculating the refrigeration load and reheater load.The refrigeration load is given byRe w (Den) (h4 - h5 )WhereBtu/hr(Eq.14)w (2137 ft3 /min) (60 min/hr) 128,220 ft3 /hr, air flowDen .075 lb/ft3 , air densityThenRe 128,220 (.075) (33.2 – 25.8) 71162 Btu/hrOrTons of refrigeration 71162 / 12000