Cooling Load Calculation Procedure Engineering Essay

cool down Load Calculation Procedure Engineering EssayThe total arrive of fondness energy that must be removed from a system by a temperature reduction mechanism in a unit sequence, fitting to the rate at which ignite is generated by people, machinery, and processes, plus the net flow of hop up into the system not associated with the change machinery.1 The sensible and latent heat conveying between the space snap and the surroundings can be classified as follows1. put heat gain qe, in Btu/h (W), represents the rate at which heat enters a conditi unityd spacefrom an external source or is released to the space from an interior source during a given time interval.2. Space alter ladle, often simply called the cooling send Qrc, Btu /h (W), is the rate at whichheat must be removed from a conditioned space so as to maintain a constant temperature and acceptable relative humidity. The sensible cooling load is equal to the sum of the convective heat communicate from the surfa ces of the construct envelope, furnishings, occupants, appliances, and equipment.4. Space heat extraction rate Qex, Btu /h (W), is the rate at which heat is actually removed fromthe conditioned space by the give vent system. The sensible heat extraction rate is equal to the sensiblecooling load moreover when the space oxygenate temperature remains constant.5. Coil load Qc, Btu /h (W), is the rate of heat bump off at the genus Helix. The cooling coil load Qcc,Btu/h (W), is the rate at which heat is removed by the chilled water flowing done the coil or isabsorbed by the refrigerant in align the coil.Cooling load usually can be classified into two categoriesexternal and internal.External Cooling Loads1These lashings argon formed because of heat gains in the conditioned spacefrom external sources through and through the make envelope or building shell and the division walls.Sources of external loads include the following cooling loads1. Heat gain entering from the exterior wal ls and roofs2. Solar heat gain transmitted through the fenestrations3. Conductive heat gain coming through the fenestrations4. Heat gain entering from the crock upition walls and interior doors5. Infiltration of outdoor(a) air into the conditioned spaceInternal Cooling Loads1These loads argon formed by the release of sensible and latent heat from the heat sources inside the conditioned space. These sources contribute internal cooling loads1. People2. Electric lights3. Equipment and appliancesFor 1-1 1 see the section of referencesCHAPTER 2COOLING LOAD CALCULATION PROCEDURE2The estimation of cooling load for a space involves calculating a surface by surface conductive, convective, and radiative heat balance for for each one populate surface and a convective heat balance for the room air. Based on the same underlying principles, the following methods have been developed for calculating the cooling load.Cooling Load by Transfer Function order (TFM).Total Equivalent Temperature D ifference (TETD) method.Cooling Load Temperature Difference (CLTD) method.Transfer Function Method (TFM)1The transfer aim method or weighting agentive role method is a simplification of the laborious heat balance method. The wide application of the TFM is due to the user-friendliness of the inputs and outputs of the TFM software and the saving of computing time. In the transfer take to the woods method, interior surface temperatures and the space cooling load were first calculated by the exact heat balance method for many representative constructions. The transfer function co streamlineds (weighting factors) were then calculated which convert the heat gains to cooling loads. Sometimes, transfer function coefficients were also developed through test and experiments.Calculation ProcedureThe counting of space cooling load victimisation the transfer function method consists of twosteps. number 1, heat gains or heat loss from exterior walls, roofs, and floors is calculated using re sponse factors or conduction transfer function coefficients and the solar and internal heat gains are calculated directly for the scheduled hour. Second, room transfer function coefficients or room weighting factors are used to convert the heat gains to cooling loads, or the heat losses to heating loads. As described in Sec. 6.2, the sensible infiltration heat gain is the instantaneous sensible cooling load. All latent heat gains are instantaneous latent cooling loads.The TFM is expressage because the cooling loads thus calculated depend on the value of transferfunction coefficients as well as the characteristics of the space and how they are change from those used to generate the transfer function coefficients. In addition, TFM assumed that the total cooling load can be calculated by simply adding the item-by-item components-the superposition principle. However, this assumption can cause some errors.Total Equivalent Temperature Difference (TETD) Method1In the total equivalent tem perature difference (TETD)/time-averaging (TA) method, heat gains of a trope of representative exterior wall and roof assemblies are calculated. The internal heat gains and conductive heat gain are calculated in the same look as in the TFM.The radiant fraction of each of the sensible heat gains is then allocated to a period including the current and successive hours, a total of 1 to 3 h for light construction and 6 to 8 h for heavy construction. The TETD/ TA method is also a member of the TFM family and is developed chiefly for manual calculation. TETD/TA is simpler in the conversion of heat gains to cooling loads. However, the time-averaging calculation procedure is subjective-it is more an art than a rigorous scientific method. Also the TETD/TA method inherits the limitations that a TFM possesses if the TFM is used to calculate the TETD.Cooling Load Temperature Difference (CLTD) Method 2CLTD is the method we used to calculate the cooling load of the project we were assigned. Th e CLTD method accounts for the thermal response in the heat transfer through the wall or roof, as well as the response due to radiation of part of the energy from the interior surface of the wall to objects and surfaces within the space. The CLTD method makes use of (a) the temperature difference in the case of walls and the roofs and (b) the cooling load factors (CLF) in the case of solar heat gain through windows and internal heat sources, i.e ,Q = U x A x CLTDCWhereQ is the net room conduction heat gain through roof, wall or blur (W)A is the area of the roof , wall or glass (m2)U is the overall heat transfer co efficient (kW/m2.K)CLTDC is the cooling load temperature difference (oC)For 1 2 see the section of referencesCHAPTER 3CALCULATING COOLING LOADS USING CLTDOutdoor Design Conditions2ASHRAE Table A 2, F1980, provides the outdoor material body conditions for various locations in many countries including India, Malaysia and Singapore. The summer radiation diagram column lis ts hourly temperature which is exceeded by 1%, 2.5 % and 5% of all the hours in the year.Selection of interior Conditions2In private homes, the indoor conditions may be chosen by the owner. But in public premisis, various codes and regulations and ordinances dictate the limits of the specific indoor design conditions. For some critical occupancy, such as, hospitals, nursing homes, computer rooms, clean rooms, etc. specific indoor design conditions will usually be established by the regulating authorities or the owners.Transmission gains2Heat transfer through the different components of the building envelop occurs primarily the process of conduction and convection and is generally referred to as transmission load. Transmission heat transfer is given by the following equationQ = = (U) (A) (TD)Where,Q is the heat transfer rate (W)Rt is the sum of the individual thermal resistances (m2.oC/W)A is the surface area perpendicular to heat flux (m2)TD is the design temperature difference bet ween indoors and outdoorsU = 1/Rt is the overall heat transfer co efficient (W/m2.oC)With, Rt = R1 + R2 + + Rm for resistance in series.The values of thermal resistances are provided for specific thickness for typical building materials usually designated by U. For materials that vary in thickness according to the application, specific conductivity k is listed in terms of unit thickness. The relation between the two isR =Wherek is the coefficient of thermal conductivity (W/m.K)L is the length of the conduction path (m).CLTD/CLF calculation2To account for the temperature and the solar variations, the concept of cooling load temperature difference (CLTD) is introduced. The CLTD is a steady state representation of the Byzantine heat transfer involving actual temperature difference between indoors and outdoors, mass and solar radiation by the building materials, and the time of day.The following relation makes corrections in the CLTDs for walls and roofs for deviations in design and solar conditions are as followsCLTDc = (CLTD + LM)k + (25.5 Tr) + (To 29.4) fWhereCLTDc is the corrected value of CLTD.LM is the colour adjustment for light coloured roof.Tr is the design room temeperatureTo is average outdoor temperature, computed as the design temperature less half the chance(a) range.f is attic strike out factorSolar heat gain2When solar rays impinge on a glass surface, some of the radiation is reflected back outside before penetrating the glass. Of that radiation which is not reflected, some is transmitted through the glass and some is absorbed by the glass. The remaining radiation is refracted slightly and goes on to heat the contents of the room. If there is external shading, such as with blinds or drapes or shades, a portion of t radiation entering the room is confined to the area immediately adjacent to the window and has a diminished ready on the conditioning of the room.All of these effects are accounted for to some degree by the following relation fo r calculation of cooling loads due to solar radiationQSHG = A(SC)(SHGF)(CLF)WhereQSHG is the solar radiation cooling load (W)A is the open glass area (m2)SC is the shading co efficient for various types of furnish and shadingsSHGF is the maximum solar heat gain factor for specific orientation of surface, latitude and calendar month (W/m2)CLF is the cooling load factor, dimensionless.Internal Loads2Lighting is often is the study space cooling load component. The rate of heat gain at any instant, however, is not the same as the heat equivalent of power supplied directly to these lights. Only part of the energy from lights is transferred to the room air by convection, and thus becomes the cooling load. The remaining portion is the radiant heat that affects the conditioned space only after having been absorbed by walls, floor, furniture, upholstery, etc. and released after a time lag. The cooling load imposed by these sources is given byQ appliances= P(CLF)WhereQ appliances is the co oling load due to equipment of appliances (W)P is the input operating power rating of the appliance or equipment (W)CLF is the cooling load factor (dimensionless) depending on operating hours, room construction, and air circulation.Occupancy2The people who occupy the building give off thermal energy continuously, the rate of which depend on the level and type of activity in which they are engaged. For the sensible portion of the heat released, a cooling load factor similar the one applied to lights and appliances has been developed to account for the lag in time between occupancy and the observed cooling load. The sensible cooling load due to people is therefore,QS = (N)(GS)(CLFS)WhereQS is the sensible cooling load due to occupants (W)N is the digit of occupantsGS is the sensible heat gain depending on activity and time for entry (W)CLFS is the cooling load factor (dimensionless) for people.The latent heat gain from occupants is found byQl = (N)(Gl)WhereQl is the latent heat gainN is the number of occupantsGl is the latent heat gains from occupants depending on activity and time from entry airing/Infiltration2Heat gain from breathing and infiltration needs to be considered in the cooling load calculations.General Design Guidelines2The general procedure compulsory to calculate the space cooling load is as followsconstruction configuration an characteristics Determine the building location, orientation and external shading, building materials, external surface colour and shape. These details are usually obtained from building plans and specifications.Outdoor design conditions Obtain the outdoor weather data for the building location and select the outdoor design conditions.Indoor design conditions Specify temperature, humidity, air velocity, etc.Operating schedules obtain a schedule of lighting, occupancy, internal equipment, appliances and processes generating heat load.Date and time Select the time of the day and month to estimate the cooling load. Several different times of the day and several different months need to be analyzed to determine the peak load time. The particular day and month are often dictated by peak solar conditions.For 2 see the section of referencesCHAPTER 4DATA FOR THE MAIN MOSQUEMosque 1st FloorFigure 4.1 First floorFigure 4.2 Window type 1 (WT1)Figure 4.3 Window type 2 (WT2)Figure 4.4 Doors 1 (D1)4.2 Mosque end FloorFigure 4.5 Ground floorFigure 4.6 Window Type 3 (WT3)Figure 4.7 Window Type 4 (WT4)Figure 4.8 Door 2 (D2)General Information analog 32, Longitude 72 7Main Mosque, College of Electrical and Mechanical Engineering, Rawalpindi, Pakistan argues33cm brick, 1.5 cm cement on both sidesRoof15 cm concreteConstruction skim color paint on both sidesGlass0.5cm black shadedLighting 176 tubes each 18W, 8hrs per dayLighting 24 tubes each 40W, 8hrs per dayOccupancy300 people moderately activeCeiling fan18 ceiling fans each 75 W, 8hrs per dayBracket fan9 bracket fans, each 40W, operating 8hrs per dayVentilation7.5 x 300 = 2200 liters/secNote Suspended ceiling was broken therefore we considered it as the refrigerating space.U Value CalculationRoof contributionsL/km/(W/m.K)RW/m2.KReference TableTable title extracurricular air0.044Table A6Surface conductance (W/m2.oC) and resistance (m2.oC/W) for airConcrete 10cm0.15/0.510.2943Inner air0.160Table A6Surface conductance (W/m2.oC) and resistance (m2.oC/W) for airTotal0.498U = 1/R = 1/0.498 = 2.01WallComponentL/km/(W/m.K)RW/m2.KReference TableTable titleOuter air0.044Table A6Surface conductance (W/m2.oC) and resistance (m2.oC/W) for airBrick0.33/0.321.0315Table A7Typical thermal properties of prevalent building and insulating materialsCement20.015/0.720.0417Table A7Typical thermal properties of common building and insulating materialsInside air film0.120Table A6Surface conductance (W/m2.oC) and resistance (m2.oC/W) for airTotal1.237U = 1/R = 1/1.237 = 0.808For 3 see referencesGlassComponentL/km/(W/m.K)RW/m2.KReference TableTable titleGlass materia l0.005/0.050.1Table A7Typical thermal properties of common building and insulating materialsOuter air0.044Table A6Surface conductance (W/m2.oC) and resistance (m2.oC/W) for airInner air0.12Table A6Surface conductance (W/m2.oC) and resistance (m2.oC/W) for airTotal0.264U =1/R = 1/0.264 = 3.79Description of appliancesItemsGround floor (Qty)First Floor (Qty)Ceiling fans126Tube lights76 (small)4 (large)Wall fan9Area CalculationsEastern Wall AreasWall area108.11514 m2Door area26.3 m2Windows area12.76 m2 veritable wall area69.055 m2 true(a) glass area30.4 m2Note The area for aluminum in the windows is not accounted for.4.6.2 Western Wall AreasWall area88.4816 m2Windows area6.583 m2 positive wall area81.899 m2Actual glass6.003 m2Note The area for aluminum in the windows is not accounted for.4.6.3 Northern Wall AreasWall area52.45 m2Windows area12.61 m2Actual wall area39.84 m2Actual glass area11.62 m2Note The area for aluminum in the windows is not accounted for.4.6.4 confederationern Wall AreasWall area52.45 m2Windows area6.58 m2Door area1.86 m2Actual wall area44.01 m2Actual glass area7.58 m2Note The area for aluminum in the windows is not accounted for.4.6.5 Roof AreasTotal roof aream24.7 CLTD field CalculationTo = 47 16/2 = 39To is the average outside temperature on design day equal to our design temperature minus half of daily temperature range.ExposureCLTDLM4K525.5TrTo29.4F6CLTDcNorth70.50.525.525.53929.4113.35East1500.525.525.53929.4117.1South11-2.20.525.525.53929.4114West1100.525.525.53929.4115.1Roof361.10.525.525.53929.4127.05For 4, 5 6 see references.CHAPTER 5COOLING LOAD CALCULATIONS FOR THE MAIN MOSQUEBuildingMain mosqueLocationEME College, RWPMonthJune Day 22 Time 0200hrsPsychrometric analysisItemTdbTwbRHEnthalpy (h)Sp. Humidity (w)Outside474170%183.870.0528Inside25.517.8250%57.350.012Difference21.523.1820%126.520.041Daily range = 16Transmission LoadItemDescriptionArea(m2)U factorCLTDc (oC)Qtransmission(W)WallsNorth39.840.80813.35429.75South44.010.8081 4497.8East69.0050.80817.1953.4West81.8990.80815.1999.2Roof194.152.0127.0510556.03GlassNorth11.623.7920.1885.2South7.583.7920.1577.44East30.43.7920.12315.84West6.0083.7920.1457.3Total transmission cooling load (W)17671.89Solar light beamDescriptionArea(m2)SCSHGFCLFQSHGNorth0South7.580.941890.36484.8East0West6.0030.946950.552228.6Total solar radiation gain cooling (W)2713.4Internal loadsItemInput (W)CLFQapplianceLights15280.08122.2Appliances171011710Total internal gain cooling (W)1832.24OccupancyNumberSHG/LHGCLFQlQsSensible300750.4911025Latent30055116500Total Occupancy gains, Qoccupancy (W)27525Ventilation/Infiltrationm3/sCFMT/wQlQsSensible2.21.23T=21.558.18Latent2.23010w=.0417180Total Infiltration / Ventilation load (W)7238.2Grand total cooling loadsQlQsTotal latent load (W)23680Total sensible load (W)33300.61Total load (kW)56.98Tons of infrigidation16.3CHAPTER 6RESULTS AND RECOMMENDATIONSResultsFollowing the CLTD method we calculated cooling load to be 16.3 Tons. In which main con tribution was from people present in the mosque (almost half the contribution) and heat conduction through walls and windows glass. The contribution from each mode is shown in fig 6.1.Fig 6.1 Contribution from each mode of heat transfer in cooling loadRecommendationsUse 6 ACs each of 2.5 Ton and one of 1.5 ton, we will need all the ACs thumbed ON during Jumma prayers only. On normal days we will switch ON 3 or 4 ACs depending on the number of occupants. As the number of occupants decreases the required cooling load also decreases. For different values of occupants required cooling load has been calculated and shown in the fig 6.2.Fig 6.2 Relation between number of occupants and cooling loadNormally 30 people are present in the mosque at prayer times so we need only 9.2 tons of refrigeration. We will switch ON three ACs of 2.5 tons and one of 1.5 tonsWe can minimize the Cooling load byIncreasing the glass thicknessBy using opaque sheets on the outer side of the windows and doorsBy u sing reflecting and insulating material on the roof, reflective material will reduce solar radiation and insulating material will minimize conduction ontogeny trees on southern side